COMPUTATIONAL METHODS IN ENGINEERING BOUNDARY VALUE PROBLEMS
This is Volume 145 in MATHEMATICS IN SCIENCE AND ENGINEERING A Series of Monographs and Textbooks Edited by RICHARD BELLMAN, University of Southern California The complete listing of books in this series is available from the Publisher upon request.
COMPUTATIONAL METHODS IN ENGINEERING BOUNDARY VALUE PROBLEMS T.Y.Na Department of Mechanical Engineering University of Michigan-Dearborn Dearborn, Michigan
@
1979
ACADEMIC PRESS
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United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. 24/28 Oval Road. London NWI 7DX
Library of Congress Cataloging in Publication Data Na, Tsung Yen. Computational methods in engineering boundary value problems. (Mathematics in science and engineering; v. Includes bibliographies and index. 1. Boundary value problems--Numerical solutions. I. Title. II. Series. TA347.B69N3 515'.35 79-51682 ISBN 0-12-512650-6
PRINTED IN THE UNITED STATES OF AMERICA
79 80 81 82
987654321
CONTENTS
Preface
CHAPTER
IX
1
Introduction
1.1
Introduction Methods of Solution Numerical Integration of Initial Value Problems Concluding Remarks References
1.2 1.3
1.4
CHAPTER
2
Method of Superposition
2.1 2.2
Introduction Reduction of Linear Boundary Value Problems to Initial Value Problems Reduction of Third-Order Boundary Value Problems to Initial Value Problems Concluding Remarks Problems References
2.3 2.4
CHAPTER
3
Method of Chasing
3.1 3.2
Introduction Derivation of Equations of Chasing By JonesSecond-Order Differential Equations Application of the Method Third-Order Differential Equations
3.3 3.4
I
3 7 II II
13 13
21 25 27
28
30 31 32 42 v
vi
CHAPTER
CHAPTER
CHAPTER
Contents
3.5
Concluding Remarks Problems References
4
The Adjoint Operator Method
4.1 4.2 4.3 4.4
Introduction Second-Order Differential Equations Third-Order Differential Equations Concluding Remarks Problems References
5
Iterative Methods-The Shooting Methods
5.1 5.2 5.3 5.4 5.5
Introduction Newton's Method Parallel Shooting Quasi Linearization Concluding Remarks Problems References
6
Iterative MethodsThe Finite-Difference Method
6.1 6.2 6.3
Introduction Finite Differences Solution of Boundary Value Problems by Finite Difference Second-Order Differential Equations Third-Order Differential Equations First-Order System and Newton's Method Concluding Remarks Problems References
6.4 6.5 6.6 6.7
CHAPTER
7
Method of TransformatlonDirect Transformation
7.1 7.2
Introduction Transformation for a Given Group of Transformations
48 49 51
52 54 62 65 66 69
70 71 76 84 90 91 92
93 93 96 98 107 126 132 134 135
137 143
Contents
vII
7.3 7.4
CHAPTER
8
Method of TransformatlonReduced Physical Parameters
8.1 8.2 8.3
Introduction Reduced Physical Parameters Application to Simultaneous Differential Equations Application to an Eigenvalue Problem Concluding Remarks Problems References
8.4 8.5
CHAPTER
9
Method of TransformatlonInvarlance of Physical Parameters
9.1 9.2
Introduction Boundary Value Problem with Two or More Parameters Systematic Search of Multiple Solutions Thin Struts with Large Elastic Displacement Problems References
9.3 9.4
CHAPTER
Extension of the Transformation Method for a Given Group of Transformations Uniqueness of the Solution Problems References
10
Method of Parameter Differentiation
10.1 10.2 10.3
Introduction Nonlinear Algebraic Equations Parameter Differentiation Applied to Differential Equations Application to Simultaneous Equations The General Parameter Mapping (GPM) of Kubicek and Hlavecek Method of Continuation of Roberts and Shipman Concluding Remarks Problems References
10.4 10.5 10.6 10.7
155 165 173 174
177 177 192 197 202 204 206
208 209 221 221 231 232
233 234 246 252 260 264 267 267 270
Contents
vIII CHAPTER
CHAPTER
Index
11
Method of Invariant Imbedding
11.1 11.2 11.3 11.4 11.5 11.6
Introduction Concept of Invariant Imbedding Isothermal Packed-Bed Chemical Reactor Radiation Fins Solution of Falkner-Skan Equation Concluding Remarks Problems References
12
Integral Equation Method
12.1 12.2 12.3 12.4
Introduction Linear Boundary Value Problems Nonlinear Boundary Value Problems Concluding Remarks Problems References
272
273 275 279 280 286 286 288
289 290 298 303 304
305 307
PREFACE
Two-point boundary value problems occur in all branches of engineering and science. In these problems the boundary conditions are specified at two points. To complicate the matter, the governing differential equations for a majority of such problems are nonlinear; since analytic solutions do not in general exist, solutions have to be obtained by numerical methods. Methods for the numerical solution of such problems can be separated into two groups, the iterative and the noniterative methods. For linear boundary value problems, solutions can always be obtained noniteratively. For nonlinear problems, iteration is usually needed. It should be emphasized however that there are many methods by which iteration of the solution can be eliminated, thus resulting in considerable savings in computation time. Three chapters in this book are devoted to iterative methods, including the shooting method, the finite-difference method, and the integral equation method. A total of six noniterative methods will be given. Following the order of presentation, these are the methods of superposition, chasing, adjoint operators, transformation, parameter differentiation, and invariant imbedding. This book is written for engineers and scientists who are interested in obtaining numerical solutions of boundary value problems in their particular fields. Emphasis is therefore placed on the computational instead of the mathematical aspects of the methods. All are presented in sufficient detail that the reader can follow through the examples and duplicate the numerical results, which are all tabulated. The author has therefore deviated from the axiomatic approach adopted in some other textbooks written by mathematicians. The intuitive approach presented here should be of greater help to those engineers and scientists whose principal need is the application of these methods, and not the mathematical theory. Learning the logical sequence of steps and trying them on the computer are the only ways to acquire a numerical technique. From the point of view of an Ix
x
Preface
educator the material covered in this book should occupy as important a place in a modern engineering curriculum as did the method of separation of variables a few years ago. It is difficult to acknowledge all the help given to the author in the preparation of this book. Above all, I thank Professor D. S. Jones, of the University of Dundee, who reviewed the manuscript twice, made many valuable suggestions, and was instrumental in the final publication of this book. Dr. Tuncer Cebeci of Douglas Aircraft Company provided constant encouragement and frequent help during the course of the work. This book was influenced by many useful discussions over the years with Dr. A. G. Hansen, President of Purdue University, who, in the successive roles of teacher, colleague, and friend, has initiated and greatly supported my interest in the general area of applied mathematics. I appreciate greatly the detailed comments of a reviewer for Academic Press. Thanks are also due to Professors I. S. Habib and G. M. Kurajian, both of the University of Michigan-Dearborn, for their considerable advice and administrative assistance. Last, but certainly not least, I wish to thank my wife, HwaSung, and our children, Arthur, Helen, and Patricia, for their understanding and encouragement and for the long hours they had to endure when, during the course of this work, I had to be away from them. TSUNG-YEN
NA
CHAPTER
1
INTRODUCTION
1.1
INTRODUCTION
The aim of this book is to describe in detail various methods for the solution of linear and nonlinear boundary value problems. A boundary value problem differs from an initial value problem in that the boundary conditions are specified at more than one point and in that solutions of the differential equation over an interval, satisfying the boundary conditions at the end points, are required. Consider, for example, the following boundary value problem: d
2T
dx 2
+
P dT X
dx
dT(O)
+ f( T) = 0,
d;-=O,
T(l)=l
(1.1)
which results from an analysis of the heat conduction through a solid with heat generation. The function f(T) represents the heat generation within the solid; this, in general, is a function of the local temperature T. The constant p is equal to 0, 1, or 2 depending on whether the solid is a plate, a cylinder, or a sphere. We shall now discuss the solutions of Eq. (1.1) for different values of p and f(T). A simple case occurs when the solid is a flat plate with constant heat generation, i.e.,
p=O
and
f(T)=q/k
where k is the heat conductivity and q is the heat generation per unit volume. Equation (1.1) then becomes dT(O)
d;- = 0, T(l) = 1
(1.2)
which is a linear differential equation with constant coefficients. The solution of this problem is simple and is reduced to the determination of 1
1. Introducllon
2
the two integration constants by the two algebraic equations which result from the two boundary conditions. The analytic solution of Eq. (1.2) is
T= -(qj2k)x 2+ C\x+ C2 The two algebraic equations for the determination of C\ and C2 are obtained from the two boundary conditions, which are
-(qj2k) + C 1 + C2 = I
and
The temperature distribution is therefore
T = I + (qj2k)(1 - x 2) The two-point nature of the problem enters at the point where the integration constants are determined by a set of algebraic equations. Consider next the case in which the solid is a cylindrical rod and the heat generation is linearly proportional to the temperature T. For this case,
p=l
and
and Eq. (1.1) becomes
2T d +.!. dT + {32T = 0 dx 2 x dx '
dT(O) = 0 dx '
T(l) = I
(1.3)
which is a linear differential equation with variable coefficients, the solution of which becomes complicated. One common method is the power series method. As is well known, the solution of Eq. (1.3) can be written in terms of the Bessel functions as
T= C 1JO({3x) + C2YO({3x) The two boundary conditions give C2
=0
C\J o({3) + C2YO({3) = I
and
which can be solved for C\ and C2 • The final form of the solution is therefore
T(x)
=
J o({3x)jJo({3)
While the series method of solution is standard, there are problems where the application of this method becomes either complicated or impossible. As an example, it is impossible to obtain the solution by the series method if the coefficients are solved from another differential equation (see, e.g., Section 2.2.2). For such cases, recourse must be made to numerical methods.
3
1.2 Methods of Solution
As a third example, let us consider the heat conduction through a sphere with heat generation proportional to the exponential of the temperature [l]. For this case, p = 2 and the heat generation is written as
f(T) = a.eT where a. is a constant. Eq. (1.1) then becomes dT(O) --=0 dx '
T(l) = I
(1.4)
The nonlineality of the heat generation term changes the nature of the problem to a nonlinear boundary value problem. For such problems, numerical methods are almost the only choice.
1.2 METHODS OF SOLUTION
We shall be concerned in this book with various methods of solving those linear and nonlinear boundary value problems whose solutions cannot be obtained analytically. Such methods can be classified into two categories, depending on whether or not the method involves an iterative process. They are Iterative methods: 1. Shooting methods (Newton's method, parallel shooting method, and quasi linearization) 2. Finite-difference methods 3. Integral equation method II. Noniterative methods: 1. Method of superposition 2. Method of chasing 3. Method of adjoint operators 4. Method of transformation 5. Parameter differentiation 6. Invariant imbedding I.
The shooting method is sometimes referred to as "the garden hose method." The principle recalls the situation of a gardner with the nozzle of a garden hose in his hand trying to water a distant plant. It usually takes a few adjustments before the jet of water finally hits the target. For a given boundary value problem, the missing initial condition is first assumed, and the resulting initial value problem can then be solved by one of the standard forward integration techniques. For a complete treatment of
4
1.
Introduction
initial value problems, the reader is referred to texts on this subject (e.g., that by Lambert [2]). The accuracy of the assumed initial condition is then checked by the boundary condition at the second point. If this boundary condition is not satisfied, another value may be assumed and the process is repeated again. This process is continued until satisfactory accuracy is achieved. For this kind of iterative method, the natural question is whether or not there is a systematic method by which new values of the missing boundary condition can be chosen so that the solution will rapidly converge to the final solution. Such methods do exist in the literature, and three of them will be presented in Chapter 5, namely Newton's method, the parallel shooting method, and the method of quasi linearization. The finite-difference method for the solution of boundary value problems is perhaps one of the most widely used. This discrete method consists of converting the set of ordinary differential equations into a finite set of algebraic equations by replacing the derivatives of the dependent variables by appropriate finite differences and subsequently solving the resulting algebraic equation to get approximations of the solution at nodal points. For linear boundary value problems, the algebraic equations are linear and the solution can be obtained in one step. For nonlinear boundary value problems, an iterative process is required, since now the problem has to be linearized. One commonly used method is to linearize the differential equation by using such methods as quasi linearization before derivatives are replaced by finite differences. Another approach is to replace derivatives by finite differences; the nonlinear algebraic equations are then linearized by Newton's method. Both techniques will be treated in detail in Chapter 6. In the integral equation method, the boundary value problem is replaced by an integral equation, which in turn is solved by numerical quadrature formulas. The derivation of the equivalent integral equation of a boundary value problem in general involves the determination of the Green function. For linear boundary value problems, the solution can be obtained without iteration. For nonlinear boundary value problems, however, an iteration process is required. This may be achieved by linearizing either the original boundary value problem or the nonlinear integral equation. The method will be treated in Chapter 12. The method of transforming linear boundary value problems to initial value problems by the method of superposition is well known and can be found in such standard texts as Collatz [3]. It is based on the principle of superposition, following which the solution of the boundary value problem is replaced by the solution of two or more initial value problems. Combining these solutions then yields the solution of the original equation. Chapter 2 will briefly summarize this method.
1.2 Methods of Solution
5
Not so well known for the transformation of boundary value problems to initial value problems is the method given in Chapter 3 of chasing, which was developed by Gel'fand and Lokutsiyevskii [4] in the Steklow Mathematics Institute of the Academy of Science U.S.S.R. In this method, the missing boundary condition at the second point is chased by creating a system of new differential equations governing the parameters in the boundary condition and integrating these equations from the first point. Once the missing boundary condition at the second point is found, the required solution can then be chased backward from the second point. The method of the adjoint operator for the solution of linear boundary value problems offers another simple alternative [5]. It is based on the idea that every set of ordinary differential equations is associated with a companion set of ordinary differential equations, known as the adjoint equations, defined as the set of linear ordinary differential equations whose coefficient matrix is the negative transpose of the coefficient matrix of the original set of ordinary differential equations. These adjoint equations provide the link between the initial and the terminal boundary conditions of the original differential equations. Details of this method will be outlined in Chapter 4. The method of transformation bases its concept on an interesting idea, introduced by Toepfer in 1912, of solving Balsius's equation [6] in the boundary layer theory. The successful application of this type of transformation makes one wonder whether it can be applied to other equations. No progress was made for 50 years, until 1962, when Klamkin [7], following the same reasoning, extended this method to many similar types of equations. Major extensions were made possible when the method was reexamined from the point of view of transformation groups by Na [8, 9], in whose work the steps followed were systematized and the missing initial condition was identified as the parameter of the transformation group. From this point of view, Toepfer's [6] and K1amkin's [7] transformations belong to the so-called "linear group" of transformations. By introducing other groups of transformations, the method was seen to be applicable to other types of equations. In addition, equations with boundary conditions specified at finite intervals can also be treated by this method. Details of the above will be treated in Chapter 7. For a complete treatment of the theories of transformation groups, the reader is referred to [10-13]. Although the concept introduced in Chapter 7 is useful, the class of equations which can be treated in this manner was still limited. Many attempts were made to extend the method. The first extension was made for those nonlinear two-point ordinary differential equations where a physical parameter appears either in the differential equation or in the boundary conditions and where solutions for a range of the physical
6
1.
Introduction
parameter are sought. By replacing the physical parameter by a "transformed" parameter, the method developed in Chapter 7 can then be followed. This extended method will be treated in Chapter 8. The extension is of significance since, in most physical problems in which a physical parameter appears in the formulation, it is always of interest to obtain the solutions for the complete range of the physical parameter, instead of a single solution for a particular value of the parameter. Although the extended method given in Chapter 8 greatly extended the range of application of the method, it can treat only problems involving one physical parameter. For problems with more than one parameter, introduction of transformed parameters cannot lead to useful solutions. Apparently, something of a fundamental nature is involved. This point was resolved recently (see, e.g., [14]) by investigating the invariance of the physical parameters under the transformation. As a result, problems with any number of parameters can be solved by this method. Details of this will be given in Chaper 9. Also given in Chapter 9 will be a systematical way of searching for multiple solutions. The method of parameter differentiation in reducing a boundary value problem to an initial value problem has been developed only for a few years, even though the idea was used in other contexts. Basically, the method involves the solution of a differential equation where a physical parameter appears either in the differential equation or in the boundary conditions. Starting with the known solution for a certain value of the parameter, the solution of the differential equation for other values of the parameter may be obtained by integrating the rate of change of the solution with respect to the parameter. Each step in the calculation involves only a small perturbation in the parameter. In this way, the equations solved are linear differential equations which can be solved noniteratively by the methods given in Chapters 2-4. The resulting solution can then be perturbed again, and in such a way the solution for a wide range of the parameter can be constructed without iteration. Chapter 10 gives details of this method. The method of invariant imbedding has a longer history than the methods of transformation and parameter differentiation for the transformation of boundary value to initial value problems. In basic concept this approach differs from the classical ones in that the study of a particular solution of a differential equation is carried out by studying a family of solutions. This may, at first sight, appear to complicate rather than simplify the problem; its justification lies in the fact that a bridge spanning the particular problem and other members of the family is constructed from which the characteristics of the particular member of the family can be obtained by studying the relation between neighboring
1.3 Numerical Integration of Initial Value Problems
7
solutions. Due to the thorough coverage of this method in the literature [15, 16], only a brief treatment will be given in Chapter 11.
1.3
NUMERICAL INTEGRATION OF INITIAL VALUE PROBLEMS
With the exception of Chapters 6, 11, and 12, all the chapters in this book rely heavily on the approximate numerical integration of initial value problems. The reader is therefore assumed to have a certain basic knowledge of this subject. An initial value problem is a differential equation whose boundary conditions are specified at a single point. In a numerical approach the value of the dependent variable and its derivatives are calculated at discrete values of the independent variable. By approximating derivatives by discrete expressions, the solution of an initial value problem can be obtained by "marching" out from the specified initial conditions. Modern computers can give numerical solutions of systems of large numbers of ordinary differential equations, given the complete set of initial conditions, with accuracy and speed. The numerical procedure for the numerical integration of initial value problems may be classified into two groups, namely the one-step and the multistep methods. Let us consider the nth-order ordinary differential equation in) = f(x,y,y',y", ... ,y(n-I»)
(3.1)
subject to the initial conditions
(where in) = d"y/dx n) and divide the interval [xo,xf ] over which the independent variable x is divided into I subintervals. The mesh or step size is given by (3.2)
from which Xi + 1
= Xi + h
(3.3)
To solve an initial value problem means to find approximate values of the dependent variable and its derivatives at the mesh points of the interval [x o' xf] on which the solution is sought. A method is called a one-step method if y/~\ can be calculated with only the knowledge of Yi(v),
i
V
)
8
1. Introduction
The method is therefore self-starting, meaning that only the boundary conditions at the initial point are needed for the evaluation of yfv), y1v),
Y3(v) , .... Two such methods, namely, Taylor's method and the Runge-Kutta method, will be mentioned here. The numerical algorithm of Taylor's method can be written as
(3.4) where
y(v) y(v+ I) ' Tk(V) = Tk(x. ,' I 'I
••• ,
=y~V+I)+.f!-y~V+I)+ 2!
I
I
y(n-I») I
+ •
•
•
n v 2 (n-I) P - I)! Yi
h - -
(n -
(3.5)
It is known that Taylor's method is conceptually very simple. Its difficulty lies in the necessity of taking the partial derivatives of the function f. The numerical algorithm of the Runge-Kutta method is perhaps the most widely used scheme, because of its low truncation error. The most frequently used fourth-order formulas for the numerical integration of Eq. (3.1) can be obtained as follows [3]: Let us assume that the solution of y
1. Let us define
v = _h y(v) v p! V
v
V - _h y(v)
v,i - p!
i
(3.6) (3.7)
and use the notations
(3.8) tJa)
= vv,i + a(p
t 1 )VV+I,i + a 2(p t 2)VV+2'i
+ ... +an-v-1(n-l)v . P n-I,l
(3.9)
1.3 Numerical Integration of Initial Value Problems
9
2. Next, we calculate the k's by (3.10) hn
k2 = ~
t2 (
i; -
k4 =
(
Xi
(I)
(I)
k, + '2h ,to '2 + k, 2n ,t) '2 + 2n - I
(n)I '
t)+ 2~~2 G),·· . t)+ :1 (n ~ ,tn _
I(
t)+ ~2 (n ~
l(
2)'
(3.12)
I ))
~;
~ 2)' tn-I(I) + k 3(n ~
I))
(3.13)
3. Next, we calculate k(v) _
(n + 2) (n + 1)(In + 2) JI
X
{(n - JI)2k l + 2(n - JI)(k 2 + k 3) + [2 - (n - JI)]k4 }
(3.14)
4. We then get
(3.15) 5. Finally, we get (v)
Yi+ I
_
-
JI!
y;; vv• i + I
(3.16)
By putting n equal to the order of the differential equation under consideration, the above sequence of equations gives the formulas necessary for the computing scheme. As an example, let us consider the case of a second-order differential equation. For this case, n = 2, and Eqs.
1. Introduction
10
(3.6)-(3.16) give 1. VI
Vo=Y, VO,i
= Yi'
= hy'
VI,i
= hy;
2.
3.
= t (k I + k 2 + k 3 ) k' = t(k l + 2k 2 + 2k3 + k 4 ) k
4. VO,i+1
= VO,i + VI,i + k,
VI,i+1
= VI,i
+ k'
5. Yi+1
= VO,i+I'
Y;+I
= (l/h)vI,i+1
Similar expressions can be obtained by setting n equal to the order of differential equation under consideration. Collatz [3] tabulated the sequence of equations for first-, second-, third-, and fourth-order differential equations. All the initial value problems in the examples to be presented in this book were solved by using the Runge-Kutta algorithm. One feature of one-step methods is that the step size can be changed easily at different x. For problems of the boundary layer type (see Section 7.1.1), where the range of independent variable is from zero to infinity, a variable-grid integration saves considerable computational time. There are, however, disadvantages of one-step methods. Compared with multistep methods, all one-step algorithms have a larger number of function evaluations. One-step methods need a small step size for desired
References
11
accuracy because of higher truncation errors than multistep methods which mean more computing time. The most widely used multistep method is the predictor-corrector procedure of Adam and Multon [17, 18]. Two steps are needed in this scheme. The normal procedure is to predict with an "open" formula and then correct with a "closed" formula. A detailed treatment of this and other methods for the solution of initial value problems and their relative merits is beyond the scope of this book. Interested readers should consult standard texts on this subject, such as [2, 17-19]. 1.4 CONCLUDING REMARKS
In this book, the various methods of analysis will be applied to boundary value problems from many branches of engineering and presented in great detail, including tabulated numerical results and figures. Emphasis will be placed on the solution techniques, instead of the mathematical theories of the methods. For most problems encountered in day-to-day engineering, the methods work well. For those problems where difficulties occur, the reader should refer to either the references listed in the chapter where the method is presented or to mathematical texts on this subject [20-23]. This book is based on work reported in the literature and in other texts on this subject [15-23], but no single text covers all the methods treated here; noniterative methods are particularly often overlooked. In view of the frequent occurrence of nonlinear boundary value problems in engineering and the sciences and the need for a text outlining the computational details of the various techniques, this book will serve a useful purpose. REFERENCES 1. Na, T. Y., and Tang, S. C; A method for the solution of conduction heat transfer with
nonlinear heat generation, Z. Angew Math. Mech 49, 45-52 (1969). 2. Lambert, J. D., "Computational Methods in Ordinary Differential Equations," Wiley, New York, 1973. 3. Collatz, L., "The Numerical Treatment of Differential Equations," pp. 184-186, Springer-Verlag, New York, 1966. 4. Berezin, I. S., and Zhidkov, N. P., "Method of Chasing in Computing Method" (0. M. Blum and A. D. Booth, trans.), Vol. II, Pergamon, Oxford, 1965. 5. Goodman, T., and Lance, G., The numerical integration of two-point boundary value problems, Math. Comput. 10, 82-86 (1956). 6. Toepfer, K., Bemerkung zu dem aufsatz von H. Blasius "Grenzschicht" in Flussigkeiten mit Kleiner Reibung, Z. Math. Phys. 60, 397-398 (1912). 7. Klamkin, M. S., On the transformation of a class of boundary value problems into initial value problems for ordinary differential equations, SIAM Rev. 4,43-47 (1962).
12
1.
Introduction
8. Na, T. Y., Transforming boundary conditions to initial conditions for ordinary differential equations, SIAM Rev. 9, 204-210 (1967). 9. Na, T. Y., Further extension on transformation from boundary value to initial value problems, SIAM Rev. 10, 85-87 (1968). 10. Hansen, A. G., "Similarity Analyses of Boundary Value Problems in Engineering," Prentice-Hall, Englewood Cliffs, New Jersey, 1964. II. Na, T. Y., and Hansen, A. G., "General Group-Theoretic Transformations from Boundary Value to Initial Value Problems," Univ. of Michigan, Dearborn Campus, Rep. 07457-18-T, 1968. 12. Na, T. Y., and Hansen, A. G., Similarity analysis of differential equations by lie group, J. Franklin Inst. 292,471-489 (December 1971). 13. Ames, W. F., "Nonlinear Partial Differential Equations in Engineering," Vol. II, Academic Press, New York, 1972. 14. Scott, T. c., Rinschler, G. L., and Na, T. Y., Further extension of an initial value method applied to certain nonlinear equations in fluid mechanics, J. Basic Eng., Trans. ASME 94, 250-251 (1972). 15. Meyer, G. H., "Initial Value Methods for Boundary Value Problems, Theory and Application of Invariant Imbedding," Academic Press, New York, 1973. 16. Lee, E. S., "Quasilinearization and Invariant Imbedding," Academic Press, New York, 1968. 17. Henrici, P., "Discrete Variable Methods in Ordinary Differential Equations," Wiley, New York, 1962. 18. Gear, C. W., "Numerical Initial Value Problems in Ordinary Differential Equations," Prentice-Hall, Englewood Cliffs, New Jersey, 1971. 19. Ceschino, F., and Kuntzmann, J., "Numerical Solution of Initial Value Problems," Prentice-Hall, Englewood Cliffs, New Jersey, 1966. 20. Fox, L., "The Numerical Solution of Two-point Boundary Value Problems," Oxford Univ. Press, Oxford, 1957. 21. Keller, H. B., "Numerical Methods for Two-Point Boundary Value Problems," GinnBlaisdell, Waltham, Massachusetts, 1968. 22. Robert, S. M., and Shipman, J. S., "Two-Point Boundary Value Problems: Shooting Methods," Elsevier, New York, 1972. 23. Bailey, P. B., Shampine, L. P., and Waltman, P. E., "Nonlinear Two-Point Boundary Value Problems," Academic Press, New York, 1968.
CHAPTER
2
METHOD OF SUPERPOSITION
2.1
INTRODUCTION
The transformation of linear ordinary differential equations from boundary value to initial value problems by the method of superposition is well known and can be found in most books on numerical solutions of differential equations (e.g., see [I)). In Section 2.2, the application of this method will be illustrated by applying it to a second-order differential equation. Application to problems in engineering and sciences will be made in two problems, namely the analysis of an isothermal chemical reactor [2] and the analytical interpretation of electrostatic probe measurements in rocket exhausts [3]. To further demonstrate the technique and illustrate the proper form of the superposition functions, the method will then be applied to the third-order differential equation, in Section 2.3.
2.2
REDUCTION OF LINEAR BOUNDARY VALUE PROBLEMS TO INITIAL VALUE PROBLEMS
For linear ordinary differential equations, it is in general possible to reduce the boundary value problem to two or more initial value problems. Using one of the shooting methods, such as ths Runge-Kutta method, the initial value problems can be integrated. Combining these solutions then gives the solution of the original boundary value problem. Iteration is eliminated as a result of this transformation. Consider, for example, a second-order linear ordinary differential equation, dy dx
-2
+ flex)
dy -d x
+ f2(X)Y = rex)
(2.1) 13
2.
14
Method of SuperposItion
subject to the boundary conditions
y(a)
= Ya'
y(b) = Yb
(2.2)
which is a boundary value problem. To transform Eq. (2.1) into an initial value problem, let us assume (2.3)
where IL is a constant to be determined. Substitutingy defined in Eq. (2.3) into Eq. (2.1), we get
dYI [ dxz
] [d YZ dYz ] +fl(X) dYI dx +fZ(X)Yl - r(x) + IL dx z +fl(x) dx + fz(x)Yz = 0 (2.4)
from which two equations are obtained:
dYI -z dx
dYI x
+fl(x) -d +fz(x)YI = r(x)
(2.5)
and
dyz dYz dxz + fl(X) dx + fZ<x)Yz
=0
(2.6)
The first boundary condition in Eq. (2.2) is next transformed to
YI(a)
+ ILYz(a) = Ya
from which
Yz(a) = 0
(2.7a, b)
Differentiating Eq. (2.3) and setting x equal to a, we get
dy(a)
dYI(a)
dYz(a)
~ = ~ +IL~
(2.8)
If the two unknown boundary conditions are set equal to
(2.9a, b)
dYl(a)/ dx = 0, then Eq. (2.8) gives
dy(a)/dx
= IL
(2.10)
Thus the unknown constant IL is identified as the missing initial slope. As a final step, the boundary condition at the second point is transformed to
15
2.2 Reduction to Initial Value Problem-Linear Equations
from which (2.11) Thus the solution of Eq. (2.1) consists of the following steps: 1. Integrate Eq. (2.5) with the boundary conditions given by Eqs. (2.7a) and (2.9a) from x = a to x = b. The value of Yl(b) is obtained. 2. Integrate Eq. (2.6) with the boundary conditions given by Eqs. (2.7b) and (2.9b) from x = a to x = b. The value of Y2(b) is obtained. 3. From Eq. (2.11) calculate Il, which, according to Eq. (2.10), is the missing initial slope. 4. The solution of the original differential equation can be calculated from Eq. (2.3). 2.2.1 Isothermal Packed-Bed Tubular Reactors
The tubular reactor is a conduit through which a mixture of chemically reacting fluid flows. Such reactors are of fundamental importance in the design of chemical plants. Consider an isothermal first-order reaction A ~ B, the rate of which, according to the theory of chemical kinetics, is given by kC A' where k and CA are the rate of reaction and the concentration of species A, respectively. In the analysis of such systems, it is customarily assumed that the axial dispersion can be described by an effective diffusion coefficient Eo giving a diffusion flux of (2.12) A balance of the fluxes of species A in and out of an infinitesimal volume (see Fig. 2.1) will give total flux into left-side surface = vC A + N total flux out of right-side surface = [vC A + d(vC A ) ] rate of disappearance of species A = kC A dx
_x
I
I
N _ _ N+dN
Fig. 2.1 Fluxes of species A.
+ (N + dN)
2. Method of Superposition
16
where v is the axial velocity of species A. Both k and v are assumed to be constants. Conservation of species A requires that ( vCA+ N) = {[ vCA+ d ( vCA)]
+ (N + dN)} + kCAdx
Upon simplification and applying Eq. (2.12), we get d 2C A dC A - v - -kC =0 E -a dx 2 dx A
(2.13)
If we define y = CAICAO'
= xl L,
Z
N pe
= vLI Ea,
R
= kLlv
Eq. (2.13) then becomes dy
-2 -
dz
dy N pe -d - N peRy = 0 z
(2.14)
If CAO is the concentration of the fluid entering the tube, the flux at the entrance of the tube is vCAO' from whence it is transported downstream by convection and diffusion, i.e., vC AO= vCA(O)
+ N(O)
or vC AO= vCA(O) - Ea
dCA(O) dx
or, in dimensionless form, 1 dy(O) 1 =y(O)- - N pe dz
(2.15)
If the length of the tube is denoted by L, then at x = L, the boundary condition is
or, in dimensionless form, dy(I)ldz = 0
(2.16)
which means physically that the length of the tube is long enough for A to react to form B. Based on Eq. (2.12), dC AI dx = 0 if the flux of A is zero. Eq. (2.14), subject to the boundary conditions (2.15) and (2.16) represents a boundary value problem. To solve this equation noniteratively, let us shift the variables z and y by
z = 1 - s,
y=l-f
2.2 Reduction to Initial Value Problem-Linear Equations
17
Eq. (2.14) and its boundary conditions then become d~ ds
-2
= RNpeJ -
dJ N pe -d - RNpe s
(2.17)
subject to the boundary conditions
s = 0: s
= I:
dJ(O) ds
(2.18)
--=0
J(I) + _I dJ(I) = 0 N pe ds
(2.19)
To apply the method of superposition outlined in Section 2.2, we write
J(s) = O(s) + f3cf>(s)
(2.20)
where f3 is a constant to be determined. The two initial value problems can then be written as
dO (0)
~
= 0, 0(0) = 0 (2.21)
and
d 2cf> ds
-2
= RNpecf> -
dcf> s
N pe -d '
d~~O)
= 0, cf>(0) = I
(2.22)
The constant f3 in Eq, (2.20) can be determined by the boundary conditions at s = I, which give NpeO(I) + (dO(I)/ ds)
f3
= - N pecf>( I) + (dcf>( I)/ ds)
(2.23)
To illustrate the steps followed in the method, consider the case with Npe = 1.0 and R = 1.0. Integration of Eqs. (2.21) and (2.22) from s = 0 to s = I gives solutions 0 and cf> as functions of s. These solutions are shown in Fig. 2.2. At s = 1, we get
0(1) = -0.3973,
cf>( I) = 1.3972
which can be substituted into Eq, (2.23) to give
f3 = 0.5324 The solution of Eq. (2.17) can therefore be calculated by Eq, (2.20), which is also plotted in Fig. 2.2. No iteration is needed.
18
2.
for
Method of Superposition
e or ¢
1.5
1.0
0.5
0.0
-0.5
Fig. 2.2
2.2.2
Solutions of
(J,
>, and
f
(N pe
= 1.0, R = 1.0).
Electrostatic Probe Measurements In Rocket Exhausts
The example given in the preceding section is a second-order differential equation with constant coefficients. The same method can be applied to differential equations with variable coefficients, or even to equations with coefficients whose values are calculated from another equation. We shall now present such a problem, which results from the analysis of flow around an electrostatic probe [3]. In the analysis of the performance of solid-propellant rockets, it was found necessary to measure accurately the distribution of local electron and ion densities in the exhaust plume. The analytical procedure for interpreting probe measurements leads to the following ordinary differential equation for the solution of concentration of charged species:
I+
I
E
+ f3
n.; R
d
2N
d1j2
+j
dN = 0 d1j
(2.24)
The boundary conditions are 1j
= 0: N = 0;
1j=00:
N=I
The independent variable 1j is defined by 1j = y( u 00/ VX)I/2, where U 00 and v are, respectively, the mainstream velocity and viscosity. Also, N is the normalized density of charged species, n / n 00' with nand n 00 representing the number density at any point and at infinity, respectively. The function j, which relates to the flow around the probe (stagnation point flow in the viscous flow theory), is given by the solution of another
2.2 Reduction to Initial Value Problem-Llnear Equations
19
nonlinear ordinary differential equation (2.25)
subject to the boundary conditions
1j
= 0: f(O) = df(O)/ d1j = 0;
Since the solutions of Eq. (2.25) are known [4], the function f appearing in Eq. (2.24) can be calculated as a function of 1j. Eq. (2.24) is therefore a linear second-order ordinary differential equation with variable coefficients. We shall now apply the method of superposition for the solution of this equation. Let us separate N as (2.26)
Substituting Eq. (2.26) into Eq. (2.24) and separating the resulting equations into two groups of terms in the same way as in Eq. (2.4), we get two initial value problems:
dO (0) d1j
0(0) = 0,
--=1
(2.27)
and
>(0) = 0,
d>(O) d1j
--=-1
(2.28)
which gives
dN(O)/ d1j
= 1-
s
(2.29)
The parameter s in Eq. (2.26) can be evaluated by using the boundary condition at 1j = 00, which gives
N ( 00) = 0( 00) + s>( 00 ) = 1 or
s = [1
- 0( 00 ) ] / >( 00 )
(2.30)
In problems of this type, where the second boundary condition is given at infinity, no problem is experienced in determining these quantities at infinity, such as 0(00) and >(00). In general, the solution approaches such limits at a finite distance.
20
2.
Method of Superposition
The solution of Eg. (2.24) therefore consists of the following steps: 1.
From [4], the second-order derivative of the function f at 1/ = 0 is
d'f(O)/ d1/2
= 0.927680
which, together with the two boundary conditions at 1/ = 0 given with Eg. (2.25),
f(O) = df(O)/ d1/ = 0 enables the solution f of Eg. (2.25) to be obtained by forward integration. 2. Integration of the two initial value problems, Egs. (2.27) and (2.28), gives 8(1/) and cj>(1/). In particular, we obtain 8(00) and cj>( (0). 3. Substituting 8( 00) and cj>( (0) into Eg. (2.30), the parameter s can be calculated. 4. The solution of N(1/) can be found by combining the solutions of the two initial value problems, according to Eg. (2.26). The concentration gradient on the surface of the probe, I, is given by
1= dN(O)/d1/ Using Eg. (2.29), we get
1= dN(O)/ dn = 1 - s
(2.31 )
Knowing s from step 3, I can be calculated. Figure 2.3 shows I plotted as a function of (l + €)Nre/(l + fJ)R. This curve agrees with that given in [3]. 0.7
0.5
0.3
0.1
o
5
10
(l+f)Nre/(l+iJ)R
Fig. 2.3 Surface concentration gradient of the probe.
21
2.3 Reduction to Initial Value Problem-Thlrd-Order Equations
2.3
REDUCTION OF THIRD-ORDER BOUNDARY VALUE PROBLEMS TO INITIAL VALUE PROBLEMS
Consider the third-order ordinary differential equation d 3y dy dy dx 3 + fl(x) dx 2 + f2(x) dx + f3(x)y = rex) subject to the boundary conditions:
yeO) = 0,
dy(O)/ dx = 0,
y(l) =
(3.1)
°
(3.2)
The solution can be obtained by assuming
y(x) = y,(x) + 1tY2(x)
(3.3)
As in the solution of second-order differential equations one begins with the integration of two initial value problems, namely d 3y, dy, dy, dx 3 + fl(x) dx 2 + fix) dx + f3(x)YI = rex) (3.4)
dy,(O) --=0 dx '
YI(O) = 0,
(3.5)
and
Y2(0) = 0,
(3.7)
until x = 1. The constant It can then be calculated by using the boundary condition at x = 1, which gives (3.8)
Finally, the solution of Eq. (3.1) can be calculated by using the Equation (3.3) since It is now a known constant. As an illustration, consider the differential equation
dy ely dx 3 - 7 dx + 6y = 6
(3.9)
subject to the boundary conditions:
yeO) = 0,
dy(O) - - =0 dx '
y(l) =0
Comparison with Eq. (3.1) shows that
fl(x) = 0,
f2(x) = -7,
f3(x) = 6,
rex)
= 6
22
2.
Integration of Eqs. (3.4) and (3.6) from x =
YI(I)
= 1.3509
and
Method
0' Superposition
°to x = 1 gives
h(l)
= 0.8007
which can be substituted into Eq. (3.8) to give /l =
- Yl(I)lh(l) = -1.6871
The solution of the given differential equation can therefore be calculated by using Eq. (3.3) since now YI(X), h(x), and the constant J-t are known. Because of its simplicity, no details will be given. 2.3.1
Three-Point Third-Order Differential Equations
It should be noted that the form of Eqs. (2.3) and (3.3) is not the only possible form. As a general rule, the number of unknown constants should be equal to the number of missing initial conditions. For example, if the boundary conditions specified in Eq. (3.1) are changed to a three-point form,
dyeD)
~=o,
y(b)
= 0,
dy(c) dx
--=0
(3.10)
then a solution involving two constants must be defined, i.e.,
y(x) = Yl(X) + J-tYix) + AYJ(X)
(3.11)
since there are two missing initial conditions at x = 0. To get the solution, we first integrate three (instead of two) initial value problems, namely
dYI dYI dYI dx 3 + fl(x) dx 2 + f2(x) dx + f3(x)YI = rex) dYl(O) = dx 2
°
dY2 dY2 dh dx3 + fl(x) dx 2 + f2(x) dx + f3(x)h =
°
YI(O)
= 0,
yiO) = 1,
°
dYI(O) -- = dx '
dh(O)
~ =0,
dY2(0) dx 2 =0
dY3 dY3 dY3 dx3 + fl(x) dx 2 + f2(x) dx + f3(x)Y3 YJ(O) = 0,
°
dY3(0) -- = dx '
=
°
dY3(0) =1 dx 2
(3.12)
(3.13)
(3.14)
2.3 Reduction to Initial Value Problem-Thlrd-Order Equations
23
and the constants JL and A. are found by using the two boundary conditions at x = b and x = c, respectively. We get y\(b)
+ JLY2(b) + A..Y3(b) = 0
dy\( c) dY2( c) dY3( c) - - +JL-- +'11.-- =0 dx dx dx
(3.15)
Since y\(b), Y2(b), Yib), dy\(c)/ dx, dY2(c)/ dx, and dY3(C)/ dx are known from the solution of Eqs. (3.12)-(3.14), the solutions of Eq. (3.15) give JL and A.. With these two constants known, the solution of the original equation can be found by Eq. (3.11). 2.3.2 Sandwich Beam Analysis
Beams formed by a few lamina of different materials are known as sandwich beams. In an analysis of such beams subject to uniformly distributed load along the entire length, Krajcinvic [5] found that the distribution of shear deformation I/; is governed by the linear ordinary differential equation d 31/; 2 dl/; - - k -+a=O (3.16) 3 dx dx where k 2 and a are physical constants which depend on the elastic properties of the lamina. For the free ends, the condition of zero shear bimoment at both ends leads to the boundary conditions dl/;(O)/dx = dl/;(I)/ dx
=0
(3.17)
From symmetry considerations,
1/;0)=0
(3.18)
Eq. (3.16), subject to the boundary conditions (3.17) and (3.18), constitutes a three-point boundary value problem. While the formulation of the problem is simply an application of the principle of minimum potential energy, details of the derivation are too involved to be included here. We shall apply the method of superposition to solve the problem. First, the solution is separated in accordance with the following equation: (3.19)
24
2. Method of Superposition
The three initial value problems, Eqs. (3.12)-(3.14), are d J, 1. _'1'_1
dx J
-kz
d,l, _'1'_1
dx
+a = 0 (3.20)
d\[J\(O) --=0 dx '
(3.21)
(3.22)
To find J.L and A., Eq. (3.15) is followed. We then get
\[JI( t ) + J.L\[Jz( t) + A.\[JJ( t)
=0
d\[JI(l) d\[Jz(l) d\[JJ(l) ~+J.L~+A.~=O
(3.23a) (3.23b)
from which
(d\[JJ(l)j dX)\[Jl( t) - (d\[J\(l)j dx)tf;J( t)
= (dtf;z(l)jdx)tf;J(t) _ (dtf;J(I)jdx)tf;z(t)
(3.24)
A. = (dtf;\(I)j dx)tf;z( t) - (dtf;z(l)j dx)tf;\( t) (dtf;z(l) j dx)tf;J( t ) - (dtf;J( 1) j dx)tf;z( t )
(3.25)
J.L
Solutions based on this method are obtained for a = 1 and k They are shown in Table 2.1. Calculations are also made based on the exact solution, tf;
= 5 and 10.
= aJ (sin t k - sinh k~) + az (~ - t) + aJ tanh t k ( cosh k~ - cosh t k) k
k
k
(3.26) The values of tf; calculated based on Eq. (3.26) are identical to those obtained by the method of superposition.
2.4
25
Concluding Remarks TABLE 2.1
Sample Solutions of Eq. (3.16) (for a = 1.(0) ~
",m
5.00
0.0 0.2 0.4 0.6 0.8 1.0
- 0.0121 - 0.0092 - 0.0033 0.0033 0.0092 0.0121
10.00
0.0 0.2 0.4 0.6 0.8 1.0
- 0.0040 - 0.0029 - 0.0010 0.0010 0.0029 0.0040
k
2.4 CONCLUDING REMARKS
The method of superposition described in this chapter is found in the mathematical literature as the method of complementary functions [6] and, in a slightly different form, as the method of particular solutions [7-10]. A few remarks about the latter method are in order at this point. Consider a third-order differential equation, written in terms of a system of first-order equations, dy dx = u,
du dx
=
1),
subject to the boundary conditions y(O) = Yi'
u(O)
= Ui '
y(l)
=
Yj
(4.2)
Miele and co-workers [7-10] assumed the solution to be y(x)
= b1i l ) + b 2y(2)
u(x)
= b1u(1) + b2u(2)
v(x)
=
b1v(l)
(4.3)
+ b2v(2)
where (y(i),u(i),v(i» satisfy the differential equation (4.1) and the boundary conditions (4.2). Substituting the solution from (4.3) into Eq.
26
2. Method of Superposition
(4.1), we get b
1(
d (I) -U(I) ) + b ~
( d (2) ~ -U(2) ) =0 2dx
dx
du( b ( -I)- v 1 dx
(I») +b ( -dU(2) - - v (2») =0 2dx
(4.4) (4.5)
bl ( d~~l) +1I V(I) + 12U(l) + 13y(l») + b2( d~~2) +1I V ( 2) + 12U(2) + 13/ 2») = rex)
(4.6)
Since (/i),u(i),v(i) are solutions of Eq. (4.1), Eqs. (4.4) and (4.5) are satisfied identically and Eq. (4.6) gives the following conditions between b, and b2 : b l + b2=1 (4.7) We will now show the equivalence of the method of complementary functions and the method of particular solutions. With the condition (4.7), the solution (4.3) can be written as y(x)
= y(l)(x) + b2[ /
u(x)
= U(I)(X) + b2[ U(2)(X) -
u(l)(x)]
vex)
= v(l)(x) + b2[ V(2)(X) -
v(l)(x)]
2)( X) - y(l)(x)]
(4.8)
or, since (/i) ,u(i) ,v(i) are unknown functions, Eq. (4.8) can be written in another form as:
= y(l)(x) + b2/ 3)(X) u(x) = u(\)(x) + b2u(3 )( x )
y(x)
vex)
where /3)
(4.9)
= V(I)(X) + b2v(3 )(x )
= /2) - y(l), etc. Similarly, Eqs. (4.4)-(4.6) can be written as d (I) -U(I) ) ~ ( dx
d (3) +b ( ~ 2
dx
-u(3) ) =0
(4.10) (4.11)
(4.12)
27
Problems
Separating terms with and without b2 , we then get two sets of differential equations, namely l d i ) _ u(l)
dx dV(I)
~
=0
dU(I) - V(I)
'
dx
=0 (4.13)
+ flv(l) + f2 U(I) + f 3 i
l)
= rex)
and dV(3) _:J'_ _
dx
dv(3) dx
u(3) = 0
'
d (3) _u_ dx
V(3)
=0
+fI V(3) + f 2 u(3) + t.3 y(3) = 0
(4.14)
which are the equations to be used in the method of complementary functions. The above shows that these two methods are in essence identical. PROBLEMS
1. A viscoelastic fluid is a fluid which possesses memory. In an analysis of the flow about an infinite horizontal plate in such a fluid (Na and Sidhom [11]), the following boundary value problem was obtained: p
d
2F
dTJ2
_ (>-'0/32 + /3)F= 0,
F(O) = Uo, F(oo) = 0
Solve the equation by the method of superposition and compare with the exact solution F
=
°
~) ( V------;--
U exp -
TJ
Note: Introduce the following transformation ~
= TJ
W
o/3 2+ /3 p
,
F
g=Uo
before applying the numerical method.
2. In an analysis of the mass transfer on a rotating disk in a nonNewtonian fluid (Greif and Anderson [12]), the concentration of the diffusing species was found to be governed by d
2C
ds 2
+1.9 [( 7+5n) +.§.] dC =0 2 + 2n s ds '
C(O) = 0, C(oo) = Crr;;
28
2.
Method 01 Superposition
Solve the equation by the method of superposition for n = 0.2, 1.0, and 1.5 and compare the result with the exact solution 7 + 5n ) / 6 + 6n
I s c/ Coo = y( "3'"3
( 1) r"3
where r is the gamma function and y is the incomplete gamma function. 3. In an analysis of the heat transfer in the radial flow between parallel circular disks [13], the following equation was obtained:
d~
+7) d7)2
2
df
d7)
3exrif= 0
,
f(O) = 1, f(oo) = 0
Solve the equation by the method of superposition for ex = 0, 5, and 10. Answer: The missing initial slopes are 1'(0) = - 0.77633, - 1.83778, and - 2.29040, respectively. 4. An electric circuit contains the following elements in series: an inductance L(l H), a resistance R(1000 ohm), and a capacitance C(6.25 X 10- 6 F). The initial charge in the circuit is zero and a constant emf of 24 V is applied at t = O. If the current at t = 0.001 sec is 0,031 A, it is desired to find the current at t = O. The boundary value problem for the solution of this problem is:
1!Q + 1000~ + 2 dt
dt
6.25
Q
X
10- 6
=24 '
Q(O) = 0
. dQ(O.OOl) 1(0.001) = dt = 0.031 Answer: i(O) = dQ(O)/ dt
=
5 A.
REFERENCES l.
2. 3. 4. 5. 6. 7.
Collatz, L., "The Numerical Treatment of Differential Equations," pp. 184-186, Springer-Verlag, New York, 1966. Lee, E. S., Quasilinearization, nonlinear boundary value problems and optimization, Chern. Eng. Sci., 21, 183-194 (1966). Maise, G., and A. J. Sabadell, Electrostatic probe measurements in solid-propellant rocket exhausts, AIAA J. 8, 895-901 (1970). Moore, F. K., "Theory of Laminar Flow," p. 127, Princeton Univ. Press, Princeton, New Jersey, 1964. Krajcinvic, D., Sandwich beam analysis, J. Appl. Mech. 39, 773-778 (1972). Robert, S. M., and Shipman, J. S., "Two-Point Boundary Value Problems: Shooting Methods," Chapter 4, Elsevier, New York, 1972. Heideman, J. C., Use of the method of particular solutions in nonlinear two-point boundary value problems, Part I, uncontrolled systems, Aero-Astronaut. Rep. No. 50, Rice Univ. (1968).
References
29
8. Heideman, J. C., Use of the method of particular solutions in nonlinear two-point boundary value problems, Part II, controlled systems, Aero-Astronaut. Rep. No. 51, Rice Univ. (1968). 9. Miele, A., Method of particular solutions for linear two-point boundary value problems, Part I, preliminary examples: Aero-Astronaut. Rep. No. 48, Rice Univ. (1968). 10. Miele, A., Method of particular solutions for linear two-point boundary value problems, Part II, general theory, Aero-Astronaut. Rep. No. 49, Rice Univ. (1968). II. Na, T. Y., and Sidhom, M. M., J. Appl. Mech. 34,1040-1042 (1967). 12. Grief, R., and Anderson, J. A., Phys. Fluids 16, 1816-1817 (1973). 13. Na, T. Y., and Chambers, R. C., ASME papers 67-WA/HT-16 (1967).
CHAPTER
3
METHOD OF CHASING
3.1 INTRODUCTION
The method of chasing was developed by Gel'fand and Lokutsiyevskii in the Steklov Mathematics Institute of the Academy of Science, U.S.S.R. It first appeared in the English literature in the book "Computing Methods" by I. S. Berzin and N. F. Zhidkov [1]. Briefly, the method starts by creating an ordinary differential equation, based on the form of the boundary condition at the initial point, which is one order less than the order of the given differental equation and the coefficients of which involve unknown functions. The number of such unknown functions is, as a rule, equal to the order of the given differential equation. This point will be made evident as we progress with the introduction of the method. Differentiating the created equation will make it the same order of differentiation as the given differential equation. Equating the coefficients of these two differential equations will lead to a system of first-order differential equations which can be integrated to give the solutions of the unknown coefficients. In particular, the solutions at the end point, together with the boundary conditions at this point, enable the complete set of boundary conditions at the end point to be evaluated. This process is called "forward chasing." With the complete set of boundary conditions at the end point now known, the original differential equation can be integrated backward as an initial value problem from the end point back to the initial point. This process is therefore called "backward chasing." Iteration can therefore be avoided. In this chapter, we will present the method for solution of second- and third-order linear differential equations. Examples with details of the analysis will also be given. 30
3.2 Derivation of Equations of Chasing
31
3.2 DERIVATION OF EQUATIONS OF CHASING BY JONEStSECOND-ORDER DIFFERENTIAL EQUATIONS
To illustrate the method, consider the linear second-order differential equation
d~
= p(x)y + q(x)
dx
(2.1)
where p(x) and q(x) are continuous functions. The boundary conditions are dy(a) (2.2) ~ = aooy(a) + a lO dy(b)
~
= /3ooy(b) + /310
(2.3)
where a oo, a lO , /300' and /310 are constants. We now consider a linear first-order differential equation, dy dx = ao(x)y + a)(x)
(2.4)
and choose ao(x) and a)(x) so that y still satisfies Eq. (2.1). Differentiating Eq. (2.4) with respect to x, we then get dy dx2
dao
dy
da,
= dx Y + dx + ao dx
(2.5)
Replacing dy/ dx by the right-hand side of Eq. (2.4), we get
;~
=(
~:o + a5)y + ( ::)
+ aOa) )
(2.6)
From comparison with Eq. (2.1), it is seen that the following equations must be satisfied, dao(X) ~ da)(x)
~
+ [ao(x)]
2
= p(x)
(2.7)
+ao(x)a)(x) = q(x)
(2.8)
As a first step, Eq. (2.7) and (2.8) are integrated, as an initial value t The proof of this section was given by Professor D. S. Jones of the University of Dundee (private communication).
32
3. Method 01 Chasing
problem, with the initial conditions given by
and the range of x from a to b. The two quantities (XoCb) and (X1(b) are obtained. From Eq. (2.4), we write (2.9)
On the other hand, the boundary condition at x = b, Eq. (2.3), gives dy(b)
~
= f3ooy(b) + 1310
(2.10)
Since (Xo(b) and (X1(b) are now known quantities, Eqs. (2.9) and (2.10) can be solved for y(b) and dy(b)j dx: y(b)
= 13 10 - (X](b) (Xo( b) - 1300
dyeb)
f3oo(Xl( b) - f3lO(XO( b)
dx
1300 - (Xo( b)
(2.11 ) (2.12)
As a result, Eq. (2.1) is tranformed into an initial value problem, since now Eq. (2.1) can be integrated backward from x = b by using the initial condition given by Eqs. (2.11) and (2.12). Another approach is to integrate Eq. (2.4) using (2.11) as the initial condition. 3.3 APPLICATION OF THE METHOD
The method described in Section 3.2 will now be applied to three examples. To demonstrate its accuracy, solutions calculated by the method of chasing are compared with the exact solutions. The method starts by solving Eqs. (2.7) and (2.8) for (Xo(x) and (XI(X) from x = a to x = b; the values of these functions at the end points, (Xo(b) and (X](b), are then obtained. This process is called forward chasing. From Eqs. (2.11) and (2.12), the boundary conditions at the second point, x = b, can be calculated. With y(b) and dy(b)j dx known, we can integrate Eq. (2.1) backward from x = b to x = a as an initial value problem. Alternatively, Eq. (2.4) can be integrated from x= b to x = a usingy(b) as initial condition. This is the backward chasing process. Details of the numerical results will be presented in the examples.
33
3.3 Application of the Method
3.3.1 A Simple Boundary Value Problem
Consider the solution of the boundary value problem
dy
-
dx 2
= -y + xcosx
(3.1)
subject to the boundary conditions
dy(O)
dY(1T/2) = -5Y(1T/2) + 2 dx
d;- = 3y(0) + 2,
(3.2)
For this problem, the exact solution is available and can be written as
y = -0.73cosx - 0.441 sinx + Hx 2sinx + xcosx)
(3.3)
At x = 1T /2, we get
Y(1T/2) =0.175
dY~:2) = 1.122
and
(3.4)
This problem will now be solved by the method of chasing and the same two quantities will be sought. From comparison with Eq. (2.1), it is seen that
p(x) = -1,
q(x) = xcosx
(3.5)
Eq. (2.7) and (2.8) therefore become
dao(x)
dX =
da,(x)
dX = xcosx - ao(x)a,(x)
-1 - aJ(x),
(3.6)
The boundary conditions are
ao(O) = 3, Eq. (3.6) can be integrated from x in Figs. 3.1 and 3.2. Now, since
a,(O) = 2
= 0 to x = 1T /2. The
results are shown
(3.7) we get
dy( 1T /2) = ( :!!.-) (:!!.-) ( :!!.- ) dx ao 2 y 2 + a, 2
(3.8)
Also, from the boundary condition at the second point,
dY~:2)
=
-5Y(1T/2) + 2
(3.9)
34
3. Method
0' Chasing
2
o 0.5
1.51
x
Fig. 3.1 uo(x) versus x from the solution uo(w/2) = -0.333. 2
o 0.5
1.0
1.5
x
Flg.3.2 ul(x) versus x from the solution u,(w/2) = 1.184.
Solving Eqs. (3.8) and (3.9), we get
7T )
2" =
2-aJ(7T/2) 5+a
o(7T/2) = 0.176
(3.10)
dY(7T /2) dx = -5Y(7T/2) + 2 = 1.122
(3.11 )
y(
which agrees with the exact solutions. 3.3.2 Heat Conduction In an Inllnlte Plate with Heat Generation
As another example, a classical problem from heat conduction will now be solved by this method. Consider an infinite flat plate which separates a
3.3 Application of the Method
35
I I
I
I I I I
Too
dx
x
Too
I
rFig. 3.3
I
2Q--i-----I
Schematic diagram of the plate.
fluid at a temperature of T OCJ' as shown in Fig. 3.3. Heat is generated within the plate at a constant rate, qs' It is required to find the temperature distribution in the plate. To formulate the problem, the first law of thermodynamics can be written as (rate of conduction into control volume) + (rate of heat generation in the control volume) = (rate of conduction out of control volume) which, referring to Fig. 3.3, becomes
{Aqx} + {qsAdx} = {A(qx + dqx)} where A is the area. Simplifying the above equation and introducing the Fourier law of conduction, qx = -kdT/dx the first law of thermodynamics becomes
d
2T
dx?
+ qs = 0 k
(3.12)
where total differentiation is used since the temperature is a function of x only. Eq. (3.12) will be solved with the following boundary conditions: because of symmetry, only half of the plate is considered. The boundary condition at x = 0 is based on the same consideration, since at the plane midway between the two surfaces of the plate, the temperature gradient must be zero, i.e., x =0: dT(O)/ dx = 0
3. Method of Chasing
36
The second boundary condition is written in the usual way when the solid surface is in contact with a fluid at a different temperature with negligible radiation: x
= I:
- k dT (l) / dx
= h[ T (l) -
Too]
where k is the heat conductivity, h is the convective heat transfer coefficient, Too is the temperature of the fluid away from the surface of the plate, and I is half the thickness of the plate. Introducing the dimensionless quantities
_ x x= I '
0=
T- T 00
qJ2/k
Eq. (3.12) becomes d
20
dX2
+ 1= 0
(3.13)
subject to the boundary conditions
x = 0:
dO(O)/ dx = 0
x = 1:
- dO(I)/ dx
= NbiO(I),
where N bi is the Biot number. The exact solution of Eq. (3.13) is
o= t (1 The boundary conditions at x
=I
O(l)=l/Nbi,
x2 ) + 1/ N bi
(3.14)
are therefore
dO(l)/dx = -I
(3.15)
We will now use the method of chasing to find the same two boundary conditions. Comparing Eq. (3.13) and its boundary conditions with Eqs. (2.1)-(2.3), we identify that p(x) = 0, lXoo
= 0,
lX lO
= 0,
q(x) = -I (300
= - Nb j ,
(310
=0
(3.16)
Eqs. (2.7) and (2.8) then become
dlXO(X)/ dX = - [ lXO(X)]2 dlX1(X)/ dx = -I - lXO(X)lXJ(X)
(3.17) (3.18)
subject to the boundary conditions (3.19)
3.3
37
Application of the Method
Eqs. (3.17) and (3.18) can now be integrated from which we get a o(1) and a l (1). Now, since
x = 0 to x = 1, from
dO(x)/ dx = ao(x)O(x) + a(x)
(3.20)
we can set x = 1 and get dO (1)/ dx = ao(1)O (1)
+ a( 1)
Also, from the boundary conditions at the second point, dO(l)/dX The boundary conditions at
x=
0(1) = _
= -NbiO(l)
(3.21)
1 can now be solved. They are a(l)
(3.22)
N b i + ao(1)
and dO(I)
Nb ia l(l)
dx
N b i + a o(1 )
--=
(3.23)
As illustrations, numerical results for three values of N bi are tabulated in Table 3.1, along with those calculated from exact solutions, Eq. (3.14). The agreement is excellent. TABLE 3.1
Comparison of Solutions with Exact Solutions By present method N bi
0(1)
dO(I)/ dx
0.5 1.5 3.0
2.ססOO
0.6667 0.3333
i.oooo i.oooo
t.oooo
Exact solutions 0(1) 2.ססOO
0.6667 0.3333
dO(l)/riX
t.oooo
i.oooo r.oooo
The above represents forward chasing for the purpose of obtaining the complete set of boundary conditions at the second point, as shown in Table 3.1. With the boundary conditions at the second point known, the second step is called backward chasing where integration is carried out backward from the second boundary point to the initial point. In the present example, it means integration from x = 1 to x = O. Equation (3.13) can now be integrated backward from x = 1 to x = O. The results are shown in Fig. 3.4. Physically, the curves in Fig. 3.4 represent the dimensionless temperature distribution for three values of N bi '
38
3. 2.5
Method of Chasing
r--=:::::::::==---------,
2.0
S ee " "-§ 0
1.5
.0
'J:
.E
Nbi = 1.5
"0
2
B ~
~
1.0
E
"
I-<
0.5
0.0 0.0
0.4
0.8
Fig. 3.4 Temperature distributions.
3.3.3 Heat Transfer In a Fin
Consider the fin shown in Fig. 3.5, which receives heat through the left end at the rate of Qo Btuy'hr, the ambient temperature being Too' In the design of fins, it is a good approximation to assume that the temperature over every cross section perpendicular to the axis of the fin is uniform. The temperature at the base (x = 0) of the fin is known as T s ' If we consider an infinitesimal length dx of the fin, taken as the control volume, conservation of energy requires that (Conduction into control volume from left-side surface) = (conduction out of control volume from right-side) + (convection out of the side surfaces)
(3.24)
Referring to the infinitesimal control volume shown in Fig. 3.5, Eq. (3.24) can be wriHen as (3.25)
where A and P are the cross-sectional area and the perimeter, respectively, and ii is the convective heat transfer coefficient.
3.3 Application of the Method
39
Fig. 3.5 Schematic diagram of the fin.
Simplifying Eq. (3.25) and introducing the Fourier law of heating,
qx = - kdT/dx we get m = hP/kA
(3.26)
The boundary conditions are written based on the following physical observations: at x = 0, the temperature is known to be equal to T s ' i.e., x
= 0:
T= T,
On the other surface, x = L, the plate is exposed to air and the only mode of heat transfer is by convection. Equating the conduction out of the solid to the convection into the fluid, we get
x= L:
-kdT(L)/dx = ii[T(L) -
Too]
If the following dimensionless quantities are introduced,
x = x/ L, Eq. (3.26) and its boundary conditions become d 20/ dx 2 = f30
(3.27)
subject to the boundary conditions
x = 0:
0(0) = 1
(3.28)
x = l:
- dO(l)/ dX = NbiO(l)
(3.29)
40
3.
Method of Chasing
where
f3 = m2L 2 ,
N bi = fiLl k
Eq. (3.27) appears to be simple to solve. However, if we compare Eqs. (3.27)-(3.29) with Eqs. (2.1)-(2.3), it becomes evident that the equations developed in Section 3.2 cannot be used directly. More specifically, the boundary condition at x = 0, Eq. (3.28), cannot be reduced to the form of Eq. (2.2). A simple way to overcome the difficulty is by rewriting Eqs. (2.2) and (2.4) as dy(a) (3.30) Yoo -;IX = yea) + YIO and
dy
Yo(x) dx
= Y + y((x)
(3.31)
where (3.32) and
Yo(x) = I/o:o(x),
YI = o:((x)/o:o(x)
(3.33)
Eqs. (2.7) and (2.8) for forward chasing become
dyo(x)
~
dYI(x)
~
=I
(3.34)
- yJ(x)p(x)
= yo(x)q(x) - yo(x)y((x)p(x)
(3.35)
subject to the boundary conditions
yo(a) = Yoo,
YI(a) = YIO
(3.36)
To get the boundary conditions at x = b, Eq. (3.33) is again introduced and Eqs. (2.11) and (2.12) become
O(b) dO( b) dx
-
=
YI(b)
(3.37)
1+ yo(b)Nbi
=-
Nbiy(( b)
(3.38)
1+ yo(b)Nbi
Referring to the present example, where
p(x)=f3,
q(x)=O, f3 0 = - I INbi'
Yo=O,
f3( =
°
y(=1
3.3 Application of the Method
41
Eqs. (3.34) and (3.35) become
dyoCx)
dx
dYI(x) dX
= 1 - /3yUx)
(3.39)
= - PYo(x)Yl(x)
(3.40)
subject to the boundary conditions
=1 (3.41) Eqs. (3.39) and (3.40) are integrated from x = to x = 1, and Yo(1) and YI(1) are then known. To get the boundary conditions at x = 1, Eqs. (3.37) Yo(O)
= 0,
YI(O)
°
and (3.38), with b replaced by 1, can be used. The results are tabulated in Table 3.2 for a few values of N bi and p. Also included in Table 3.2 are the values of 0(1) from the exact solution. It is known that the exact solution of Eq. (3.27), subject to the boundary conditions, Eqs. (3.28) and (3.29), is
{jj cosh{jj (1 - x) + Nbisinh{jj (1 - x) o= ----....:..---=-----.::..:........----=-----....:... {jj cosh {jj + N bi sinh {jj
(3.42)
At x = 1, we then get (3.43)
The process of forward chasing is now complete. To get the temperature distribution, the backward chasing is used which is simply integration of Eq. (3.27) with the boundary condition given by the values of 0(1) and dfJ(1)/dX listed in Table 3.2. Details are left to the reader as an exercise. TABLE 3.2 Boundary Conditions at x = I
By chasing
Exact solutions
f3
N bi
9(1)
d(9XI)jdX
9(1)
2.0
0.5 1.5 3.0
0.3494 0.2364 0.1592
0.1747 0.3546 0.4775
0.3494 0.2364 0.1592
1.0
0.5 1.5 3.0
0.4693 0.3025 0.1973
0.2347 0.4537 0.5919
0.4693 0.3025 0.1973
42
3. Method
0' Chasing
3.4 THIRD-ORDER DIFFERENTIAL EQUATIONS
The method developed in the previous sections for the second-order differential equations can now be extended to third-order differential equations. In Section 3.4.1, we will treat those third-order differential equations with one boundary conditions given at the initial point and two at the end point. This is a simple extension of the second-order case. In a more complex case, given in Section 3.4.2, two of the three boundary conditions are given at the initial point. It is necessary for this case to introduce two systems of equations instead of one. Application of the method given in Section 3.4.2 to the solution of a third-order differential equation resulting from an analysis of a sandwich beam will be given in Section 3.4.3. 3.4.1
Third-Order Differential Equations with One Boundary Condition at the Initial Point
Consider the simple case where the differential equation is given by d 3y dx
-3
dy x
= P(X) -d + Q(X)Y + R(x)
(4.1)
subject to the boundary conditions dj>(O) --2-
dx
dy(O)
= a oo -d- + aIOY(O) + a 20 x
(4.2) (4.3)
dj>(l)
dy(l)
----;;;z = Yoo d;- + YlOy(l) + Y20
(4.4)
Since Eq. (4.1) is third order, we therefore consider a second-order differential equation, dj> dy dx 2 = ao(x) dx
+ a(x)y + a2(x)
(4.5)
Differentiating Eq. (4.5) and eliminating dj>/ dx 2 by using Eq. (4.5), we get
Proceeding as in the second-order case, we choose ao(x), a(x), and az(x) such that y still satisfies Eq. (4.1). It is seen that the following
3.4 Third-Order DI"erentlal Equations
43
equations must be satisfied: dao dx
z
+ a o + a l = P(x)
da t dx + aoa l da z dx
(4.7a)
= Q(x)
(4.7b)
+ aoaz = R(x)
(4.7c)
As a first step, Eqs. (4.7) are integrated as an initial value problem, with the initial conditions given by az(O) = azo
and the range of integration from x = 0 to x following quantities will be obtained:
ao(l),
(4.8)
= I. In particular, the
a1(1),
From Eq. (4.5), we write dy(l)
----;J;l = ao(l)
dy(l)
~
+at(l)y(l) + az(l)
On the other hand, the boundary conditions at x give dy(l)
dy(l)
dy(l)
dy(l)
(4.9)
= I, Eqs. (4.3) and (4.4),
----;J;l = f300 ~ + f3lOy(l) +
f3zo
(4.10)
----;J;l = Yoo ~ + YlOy(l) + Yzo
(4.11)
The complete set of boundary conditions at x = l.namely y(l),dy(l)/ dx, and dy(l)/ dx', can now be solved from Eqs. (4.9)-(4.11). Using these conditions, Eqs. (4.1)-(4.4) can be integrated from x = I to x = 0 (the backward chasing) as an initial value problem. Due to the similarity between this case and the case of second-order differential equations treated in Section 3.3, no example will be given here. 3.4.2
Third-Order Differential Equations with Boundary Conditions Given at Three Points
Consider the third-order differential equation given by d 3y dy = P(x) -d dx x
-3
+ Q(x)y + R(x)
(4.12)
44
3.
Method of Chasing
dY(O) , + aIOY~O) + a20 dx
(4.13)
Subject to the boundary conditions
dy(O) x
- d - = aOO
--2-
dY(b) dy(b) y(b) = (300""dF + (310 ~ + (320 dy(c)
dy(c)
~ = Y00""dF
+ YlOy(c) + Y20
(4.14) (4.15)
where the boundary conditions are specified at three points. For this case, we have to introduce two equations, instead of one, as in the previous section. First, let us introduce a second-order differential equation:
dy -d x
=
dy ao(x) - 2 + a](x)y + ai x) dx
(4.16)
Differentiating Eq. (4.16) with respect to x and eliminating the term dy / dx2 by using Eq. (4.16), we get
3y d I { [dao(X) ] dy dx3 = [a 1- d;- -aO(x)al(x) dx o(x)]2 dao(x)
dal(x) ]
+ [ aleX) d;- -alex) - ao(x) d;- y (4.17)
Comparing Eq. (4.17) with Eq. (4.12) shows that, if y in Eq. (4.16) still satisfied Eq. (4.12), the coefficients must be equal, which gives
dao(x)
d;- = I - ao(x)a](x) - P(x)[ao(x)] da](x)
d;-
= -
2
(4.18)
2 [al(x)] - P(x)ao(x)a](x) - ao(x)Q(x)
da2(x)
d;- = - al(x)ai x) - P(x)aO(x)a2(x) - ao(x)R(x)
(4.19) (4.20)
Eqs. (4.18)-(4.20) can now be integrated as an initial value problem since the initial conditions are given by
ao(O)
= aoo,
a 1(0)
= a lO ,
a2(0)
= a20
(4.21)
45
3.4 Third-Order Differential Equations
Next, another second-order differential equation, corresponding to the boundary condition at x = a, is introduced:
dy dy Y = f3o(x) dx2 + f31(X) dx + f3 2(x)
(4.22)
Differentiating Eq. (4.22) with respect to x and eliminating dy / dx2 by using Eq. (4.22), we then get:
f30 - f3 o( df3./ dx) + f31( df3o/ dx) + f3~ dy
+
(df3o/ dx) + f3l
f3J dx f32( df3o/ dx) + f3l f3 2 - f3o( df32/ dx) 2
f3 0
f3J
y (4.23)
We again choose f3's such that y in Eq. (4.22) still satisfies Eq. (4.12). By comparing the coefficients of Eqs. (4.12) and (4.23), this condition requires that the corresponding coefficients must be equal. Equating these pairs of coefficients and rearranging the terms, we then have
df30 2 dx = - f3. - f3 oQ(x)
(4.24)
df3l dx
=I
(4.25)
df32 dx
= - f30f32Q(X) - f3o R(x)
- f30 f3. Q(x) - f3 oP(x)
(4.26)
The boundary conditions are (4.27)
Since the end point is at x = c, integration of Eqs. (4.18)-(4.20) and (4.24)-(4.26) is to end at this point. In particular, we get the following quantities:
ao(c), f3 o( c), At x
= c, Eqs. (4.16) and dy(c)
-d-
x
(4.22) give
dy(c)
= ao(c) - - 2 - +al(c)y(c) + a2(c) dx
dy(c) dy(c) y(c) = f3o(c) ~ +f3I(C) dX +f32(C)
(4.28) (4.29)
46
3. Method of Chasing
Also, from the boundary condition at x = c,
dy( c)
d:v( c)
~ = Yoo ~ +YIOY(c) + Y20
(4.30)
Equations (4.28)-(4.30) can be solved for the complete set of boundary conditions at y = c. This completes the forward chasing. Equations (4.28)-(4.30) can then be solved for y(c), dy(c)/ dx, and d:V(c)/ dx': By use of these three boundary conditions at x = c as the initial conditions, Eq. (4.12) can be solved by forward integration. This completes the backward chasing. 3.4.3
Sandwich Beam Analysis
The method introduced in Section 3.4.2 will now be applied to the solution of the sandwich beam equation, which was treated in Section 2.3.2. Starting from the differential equation
d 31fJ 2 dlfJ -k -+a=O 3 dx dx
(4.31)
and its boundary conditions
dlfJ(O) dx
dlfJ(l) dx
(4.32)
--=--=0
'
we can identify that
P(x)=k 2, aoo = a lO = a20 = 0,
Q(x)=O,
R(x)=-a
fJoo = fJlO = fJ 20 = 0,
b=L
Yoo
= YIO = Y20 =
°
c=l
The two systems of initial value problems, Eqs. (4.18)-(4.20) and Eqs. (4.24)-(4.26), become
dao(x)
---;tX = I da)(x)
- ao(x)a)(x) - k 2[ ao(x)]
2
2
---;tX = - [a1(x)] - k 2a o(x)a)(x) da2(x)
---;tX = - a)(x)a 2(x) - k 2a o(x)a 2(x) + aao(x)
(4.33)
47
3.4 Third-Order Differential Equations
subject to the boundary conditions
0: 0(0) = 0: 1(0)
= O:z(O) = 0
and d(30(x) dx
= _ (3 (x)
d(3l(x) dx
= 1 - e(3
1
(x)
(4.34)
0
d(3z(x) _ (3 ( ) dx -a 0 x
subject to the boundary conditions (300 )
= (310 ) = (3z0) = 0
As an example, let us consider the case
k = 5, a=1 Integration of Eqs. (4.33) and (4.34) as initial value problems gives the solutions of the o:'s and the (3's, as shown in Table 3.3. TABLE 3.3
Solution of the a's and f3's (k x
ao
al
a2
0.0 0.2 0.4 0.6 0.8 1.0
0.ססOO
0.00 0.00 0.00 0.00 0.00 0.00
0.ססOO
0.1523 0.1928 0.1990 0.1999 0.2000
= 5, a = I)
130
0.0141 0.0294 0.0360 0.0385 0.0395
-
13.
132
0.ססOO
0.ססOO
0.0051 0.0217 0.0541 0.1105 0.2053
0.1042 0.2350 0.4259 0.7254 1.2100
0.ססOO
-
0.0002 0.0014 0.0050 0.0130 0.0284
Next, the complete set of boundary conditions at x = 1 will be sought. For this problem, Eqs. (4.28)-(4.30) become d1[;(l) --=0 dx 0: 0
d~(1)
( 1) --z- + 0:(1)1[;(1) + O:z(l) = 0 dx
1[;(1) = (30(1)
d~~l) dx
+ (3z(1)
48
3. Method of Chasing
which can be reduced to
ao(1 ) f3 2( I) - 130(1 )a2( I) IjJ ( I) = a 1(1) 130 (1 ) + a o(1) dljJ(1) dx
--=0
(4.35)
+ a 2( I) a 1(I )130(1 ) + ao(1)
d~(1)
a 1(I) f32( I)
Since all the a's and f3's on the right-hand side of Eq. (4.35) are known from Table 3.3, we therefore have 1jJ(1)
= 0.0121,
dljJ(l)
d;- =0.0,
d~(1) = -0.1973 dx?
which can bs used as the boundary conditions for the integration of Eq. = I to x = O. The solutions from this final step ljJ(x) were found to be identical to those tabulated in Table
(4.31) in the backward chasing from x 2.1.
3.5
CONCLUDING REMARKS
The method described in this chapter offers a very interesting alternative for the conversion from a boundary value problem to an initial value problem. One other advantage of this method which has not been discussed so far is the property that under certain conditions the solutions of the equations of forward and backward chasing, i.e., ao(x) and a1(x), increase slowly with x even if the solution of the original equation y(x) does, thus making it possible to use larger intervals of the independent variable for the same accuracy. This can be explained by investigating the behavior of the solution of the homogeneous part of Eq. (2.1),
z" = p(x)z
(5.1 )
and the boundary condition at x = a, which is z'(a) = aooz(a)
(5.2)
where primes represent differentiation with respect to x. Let us first differentiate the product zz'; we get d(zz') ~ =(zy+zz"
(5.3)
49
Problems
Replacing z" in Eq. (5.3) by the right-hand side of Eq. (5.1), we then get
d(zz')
~
2.r cr[
= p(X)z2(x) + [z'(x)J
2
Eq. (5.4) is now integrated from a to x twice, and we get
Z2 =
(5.4)
1
p( 1J)Z2(1J) + Z,2(1J) Jd1J }d~ + [2lX OO(X - a) + ]Z2(a) (5.5)
where the boundary condition (5.2) is used in the evaluation of the integration constant. We now consider the case in which both p(x) and lXoo are greater than zero. For this case, the integrand is greater than zero (so is the constant term) and therefore z(x) will increase as x is increased. Consequently, z(x) can become very large if the lower bound of p(x) is large. The consequence is that for sufficient accuracy in the solution, the step size has to be very small. Next, let us consider the equation of chasing. We notice that by using the transformation
lXo(x) = z'(x)j z(x)
(5.6)
Eq. (5.1) becomes Eq. (2.7) and the boundary condition (5.2) is transformed to lXo(a) = lX OO ' Thus, if z(x) increases rapidly, lXo(x) will show a sharp decrease accordingly. Eq. (2.8) is linear in lX), and its solution is X
lX)(X) = exp( - i lXO(X)dX){ lX lO + i
X
exp(i~lXO(1J)d1J)q(~)d~}
(5.7)
Due to the factor on the right-hand side of Eq. (5.7) involving a negative exponential the solution will decrease rapidly with x. Due to this property, the solutions of lXO(X) and lX)(X) can be obtained by using larger step sizes, especially when x is large. The same is true for the inverse chasing using Eq. (2.4) and the initial condition (2.11). For nonlinear boundary value problems, the differential equation has to be linearized before this method can be applied. The method is suitable for boundary value problems where the interval of integration is finite.
PROBLEMS
1. Finish the solutions of heat transfer through a fin discussed in Section 3.3.3. Calculate solutions of Eqs. (3.27)-(3.29) for {3 = 0.1, 5.0, 10.0 and Nbi = 0.5, 1.5, 3.0, respectively and compare with the exact solution, Eq. (3.42).
50
3.
Method of Chasing
2. In the solution of the problem of heat transfer through the fin, Eqs. (3.27)-(3.29), the boundary conditions at x = 0, Eq. (3.28), are not given in the form of Eq. (2.2), and it was therefore found necessary to modify the equations derived in Section 3.2. This was done in a simple way by following Eqs. (3.31)-(3.38). The same equations can be derived, in a more
basic manner, by replacing Eq. (2.4) with Y
= yo(x)dy /
dx
+ y(x)
and following the same steps as in Section 3.2. Repeat the derivation and get Eqs. (3.34) and (3.35). 3. The motion of a heavy particle sliding without friction along a massless straight rod which rotates about its points in a vertical plane with constant angular velocity w can be represented by the following boundary value problem: d 2r - 2 - w2r dt
=
.
-gsmwt '
where rand t are the distance of the particle from the center of rotation and time, respectively. The constant g is the gravitational acceleration. The subscripts 0 and 1 represent conditions at t = 0 and t = t (, respectively. Suppose the angular speed is w = 12 rpm and the particle, which is at r = 5 ft initially, reaches r = 15 ft at t = 0.05 sec. Find the initial velocity of the particle by the method of chasing. Answer: 225.3 ft/sec. 4. In the analytical prediction of the temperature distribution in an incompressible Newtonian fluid held between two coaxial cylinders, the outer one is rotating at a steady angular velocity. Bird et al. [2] found that the dimensionless temperature 0 is given by the differential equation
dO) + 4N ~41 = 0 I1 d~d (ed~ subject to the boundary conditions:
O( ,,) = 0,
0(1)= 1
Solve the equation by the method of chasing and compare the results with the exact solution:
[
In~ eN] - [(N+ 1)- -N] -In"
0= (N+ 1)- -
,,2
The values of " and N are 0.9 and 0.4, respectively.
References
51
5. Derive the necessary equations for the solution of the fourth-order boundary value problem by the method of chasing. Solve problem 4 of Chapter 2 by this method.
REFERENCES Berzin, I. S., and Zhidkov, N. P., Method of chasing, in "Computing Methods" (0. M. Blum and A. D. Booth, trans.), Vol. II, Pergamon, Oxford, 1965. 2. Bird, R. B., Stewart, W. E., and Lightfoot, E. N., "Transport Phenomena," p. 326, Wiley, New York, 1960. I.
CHAPTER
4
THE ADJOINT OPERATOR METHOD
4.1
INTRODUCTION
The adjoint operator method was developed by Goodman and Lance [1]. As in the two methods treated in Chapters 2 and 3, the missing initial conditions of a linear boundary value problem can be found by a onesweep process. The general concept can be introduced as follows. Consider a general system of linear, ordinary differential equations, which can be written in matrix form as
Y = A(t)Y + F(t)
(1.1 )
where the dot denotes differentiation with respect to the independent variable. The coefficient matrix is a square matrix defined by
A=
(1.2)
The two column matrices Y and F(t) are
Y=
(1.3)
Yn and
(1.4) respectively. 52
53
4.1 Introduction
The boundary conditions are given at two points, namely
Yi(O) = (Xi Yi(T) = f3i
for for
i
= 1, ... , r
i = r
+ 1, ...
,n
(1.5)
To convert the above boundary value problem to an initial value problem, the so-called adjoint system of equations to Eq. (1.1) will be introduced. The definition of the adjoint equations of Eq. (1.1), as given by Bliss [2], Wright [3], and Goodman and Lance [1], can be written as
-x = AT(t)X
(1.6)
where the superscript T on A denotes the transposed matrix and the column matrix X is defined by
We next differentiate the product XTy with respect to the independent variable, which gives
; (XTy) = XTy + XTy Substituting
Y and XT from Eqs. (1.1) and
(1.7)
(1.6) into Eq. (1.7), we get
; (XTy) = XT(AY + F) - (ATX)Ty = XTAY + XTF - XTAY
(1.8)
where the property that AIT = A is applied. The first and the third terms on the right-hand side of Eq. (1.8) are equal in magnitude and opposite in sign and can therefore be canceled. This property is the key to the success of this method, since now the right-hand side does not involve Y. Next, Eq. (1.8) is integrated over the interval (0,T), and we get (1.9)
Eq. (1.9) can now be used to determine all the missing initial conditions of Eq. (1.1). We recall from Eq. (1.5) that (n - r) initial conditions are missing. Also, in the definition of the adjoint system of equations, Eq. (1.6), no boundary conditions are specified. Since (n - r) initial conditions are needed, Eq. (1.6) is integrated (n - r) times from t = T to t = 0, each time employing a different set of initial conditions at t = T. These (n - r)
54
4. The AdJoint Operator Method
sets of solutions will then be combined with Eq. (1.9) to solve for the (n - r) missing boundary conditions. Details of this last step are quite
abstract and can only be fully understood by actually working through a few examples. 4.2
SECOND-ORDER DIFFERENTIAL EQUATIONS
We will now introduce the details of the method by solving two second-order differential equations, namely the magnetohydrodynamic Couette flow and the bending of a beam. Numerical solutions obtained by this method are compared with exact solutions and are found to be very accurate. 4.2.1
Magnetohydrodynamlc CoueUe Flow
Magnetohydrodynamics involves the study of the motion of electrically conducting fluids subject to the influence of electric fields, magnetic fields, or both. To analyze such problems, use must be made of the hydrodynamic laws and concepts, together with the basic relations of the electromagnetic theory. In order to understand the interaction of electric, magnetic, and hydrodynamic forces in the fluid, let us consider a simple flow problem, known as the Couette flow, in which an electrically conducting fluid, subject to a uniform magnetic field in the y direction, flows between two parallel flat plates, one of which at rest and the other moving in its own plane with a velocity U, as shown in Fig. 4.1. y
y=L ====t===:;=====r===== -
y=O
U
=======:!!5::======:--_X Fig. 4.1 The Couette flow.
The governing differential equations, which include the equations of continuity and momentum, can be written as Continuity:
~=o
ax
(2.1)
4.2
55
Second-Order Differential Equations
Momentum:
pu -au
ax
= -
2 ap + J-t (a - u + -a2u ) ax ax 2 ay 2
-
aB 2u 0
ap
(2.2) (2.3)
0=- -
ay
where u and p are the velocity and the pressure, respectively. The physical properties p, J-t, a, and Bo are the density, the viscosity, the electrical conductivity and the imposed magnetic field, respectively. Physically, the extra term in the x direction momentum aBJu, represents the electric body force. Other than this term, the basic equations are identical to those for the Couette flow of a Newtonian fluid. Equation (2.1) means that the velocity u is independent of x. Similarly, Eq. (2.3) shows that p is independent of y. This reduces Eq. (2.2) to
d 2u aBJ I dp ----u=-dy 2 pv pv dx
(2.4)
The boundary conditions are y
= 0:
u
= 0;
y
= L: u =
U
In terms of dimensionless quantities
u = u/ U,
Y/=y/L
Eq. (2.4) becomes
(2.5) subject to the boundary conditions
u(O) = 0;
y/=l:
u(l)=l
(2.6)
where the two dimensionless quantities M and P are defined as a
)1/2
M
= ( pv
P
= 1-( dp ) L U
dx
BoL 2
ov
(Hartmann number) (Pressure number)
Equation (2.5), subject to the boundary conditions (2.6), will now be solved by the adjoint method.
56
4. The Adjoint Operator Method
First, Eq. (2.5) is written as a system of first-order differential equations, dU 1
--;J;j = U z = (0)u 1 + (l)u z + (0)
(2.7)
duz
--;J;j = MZu 1 + P = (Mz)u 1 + (O)u z + (P)
(2.8)
subject to the boundary conditions Tj = 0:
u1(0) = 0;
where u 1 = U. In matrix notations, Eqs. (2.7) and (2.8) can be written as
u = A(Tj)u + F(Tj) where the dot represents differentiation with respect to n; matrix of the unknowns,
(2.9) U
is the column
A(Tj) is the square matrix A=
[~z ~]
and the forcing function F(Tj) is another column matrix,
We next introduce the adjoint system,
-x = ATX
(2.10)
where the column matrix is defined as
and AT is the transpose of A. For this example, the transpose of A is
AT=[~ ~Z] The adjoint differential equations are therefore dx.] dTj
= - MZxz
(2.11 )
dx z/ dn
= - XI
(2.12)
57
4.2 Second-Order DIfferential Equations
The steps outlined in Eqs. (1.7) and (1.8) can be carried out as follows. Multiplying Eqs. (2.7), (2.8), (2.11) and (2.12) by xl> x 2 ' u., and u2 , respectively, and summing up these four expressions, we then get d
dTJ (x)u) + X2U2) = PX2
(2.13)
Integrating Eq. (2.13) from 0 to I, we get
[x.(I)u)(I) + x 2(I)u2( 1) ] - [x)(O)u)(O) + xiO)u2(0)] =
L
1 pX2dTJ (2.14)
Since one boundary condition at TJ = 0 is missing, Eq. (2.14) and the adjoint equations (2.11) and (2.12) will be applied once. First, let us substitute the known boundary conditions into Eq. (2.14); we get
[x.(l) + x 2(1)u2(1)] - [X2(0)U2(0)] = L·PX2dTJ
(2.15)
Now, u2 (0) is the missing initial condition of Eqs. (2.7) and (2.8), which can be solved from Eq. (2.15):
u2(0) = x)O) { [x.(I) + x 2(I)u2( 1) ] - L1pX2dTJ}
(2.16)
In Eq. (2.16), uil) is an unknown quantity, whereas the two quantities x.(l) and x 2(l) are the initial conditions for the integration of Eqs. (2.11) and (2.12) and can be assigned arbitrarily. The natural choice of xiI) is zero, since this will set the second term on the right-hand side of Eq. (2.16) to zero no matter what the value of uil) may be. The simplest choice of x)(l) is, of course, unity. Equation (2.16) therefore becomes
u2(0) =
X2~0)
(I - L PX2dTJ )
(2.17)
1
To summarize, the solution of Eqs. (2.7) and (2.8) takes the following steps. 1. Equations (2.11) and (2.12) are integrated backward from TJ
TJ
= 0, using the boundary conditions x 1( 1) = I and x 2(l) = o.
= I to
2. With the solutions of x. and x 2 known from step 1, u2(0) can be found by Eq. (2.17). 3. Since both u 1(0) (given) and uiO) (found in step 2) are known, Eqs. (2.7) and (2.8) become an initial value problem and can be integrated to get the final solution. As an example, consider the solution for M
= 2 and
P
= 4. Solutions of
58 XI
4.
and
X2
The Adjoint Operator Method
from step 1 are tabulated in Table 4.1, from which we find x 2 (0) = 1.8129
Substituting the solutions from Table 4.1 into Eq. (2.17), we find (1 - 0.6904) 1.8129 = -0.9717
u2 (0) =
Finally, using the two boundary conditions u2 (0)
= - 0.9717
Eqs. (2.7) and (2.8) can be integrated as an initial value problem. The results are shown in Table 4.2. Also shown in Table 4.2 are numerical TABLE 4.1
Solutions of XI and x 2 from Step 1 (for M = 2 and P = 4)
1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00
1.0000 1.0200 1.0810 1.1853 1.3372 1.5428 1.8103 2.1505 2.5769 3.1067 3.7612
0.0000 0.1007 0.2054 0.3183 0.4440 0.5875 0.7546 0.9520 1.1875 1.4707 1.8129
TABLE 4.2
Solutions of Eqs. (2.7) and (2.8) (for M = 2 and P = 4) Exact solution
By adjoint u1
'1/
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
-
0.0000 0.0778 0.1185 0.1238 0.0940 0.0279 0.0773 0.2257 0.4233 0.6781 1.0000
u2 - 0.9717 - 0.5885 - 0.2289 0.1214 0.4767 0.8511 1.2596 1.7186 2.2466 2.8648 3.5979
u. -
0.0000 0.0777 0.1185 0.1239 0.0941 0.0279 0.0772 0.2256 0.4232 0.6779 1.0000
59
4.2 Second-Order Differential Equations
values of
U1
U
based on the exact solution of Eqs. (2.7) and (2.8), I
=
P [2 sinh t MYJ sinh t M (YJ - I) ] sinh MYJ + -2 sinhM M coshtM
The agreement is seen to be excellent.
4.2.2 Bending of a Beam
A beam of length L is simply supported at both ends, as shown in the figure below. It is desired to find the deflection y of the beam as a function of x if the beam has a constant load W per unit length.
r-
x w
Fig. 4.2
Schematic diagram of the beam.
The total weight of the beam is WL. Each end therefore supports half of the weight, i.e., t WL. To find the bending moment M at any location x, let us consider the force to the left of x. The force t WL at A has moment - t(WL)x. The force Wx due to the weight of beam to the left of x has moment (Wx)(1 x). The total bending moment M T at x is therefore the summation of the above, which can be substituted into the equation of motion dy £ 1 -2 = M T dx to give £1 dy = dx 2
1 2
Wx 2
-
1 2
WLx
(2.18)
The boundary conditions are x = 0:
y = 0;
x = tL:
dy(tL)/ dx = 0
The first boundary condition means that the deflection of the beam is zero at the end. The second boundary condition is due to symmetry of the deflection curve.
60
4.
The AdjoInt Operator Method
If we introduce
x=
xl L,
J=YIL
then Eq. (2.18) becomes dy I dx 2 = (3(x 2 - x)
(2.19)
subject to the boundary conditions x= 0: J = 0;
x=t:
dyldx=O
where We shall now solve Eq. (2.19) by the present method. First, the equation has to be written as a system of first-order differential equations as.
= h = (O)y. + (I)h + (0) dhl ds = (3(S2 - s) = (O)Yl + (O)h + (3(S2 -
(2.20)
dYll ds
s)
(2.21)
where the boundary conditions are
s
= 0: Yl(O) = 0;
s
=
t:
h(t ) = 0
The coefficient matrix is
A=[~ ~] Its transpose is therefore
The transposed system of equations is therefore
or
dz.
-=0 ds
dZ 2
d'S =
(2.22) (2.23)
-z.
Next, Eqs. (2.20)-(2.23) are multiplied by z I' the summation of the resulting equations is
Z2'
Y., and h, respectively; (2.24)
61
4.2 Second-Order Differential Equations
Integrating Eq. (2.24) from
°to 1-, we get
[Yl( 1- )Zl( 1-) + Yz( 1- )Z2( 1-)] (1/2
= f3 J
o
- [Yl(O)ZI(O) + Yz(0)Z2(0)]
Z2(S2 - s)ds
(2.25)
Substituting into Eq. (2.25) the known boundary conditions, Yl(O)
YzO) = 0, we get
Yl( 2"I) Zl(I) 2"
-
Yz(0)Z2(0) -_ f3 J(1/2Z2(S 2 - s) ds o
°
= and (2.26)
We therefore choose the boundary conditions for the integration of Eqs. (2.22) and (2.23) to be (2.27) Eq. (2.26) then gives
f3
yiO) = - Z2(0)
(1/2
2
Jo Z2(S -
s)ds
(2.28)
The solution of Eqs. (2.20) and (2.21) can therefore be summarized as follows. First, Eqs. (2.22) and (2.23) are integrated from s = 1- to s = 0, using the boundary conditions in (2.27); next, the missing boundary condition, Yz(O), is calculated from Eq. (2.28); finally, the solutions of Eqs. (2.20) and (2.21) can be found by integrating these two equations subject to the boundary conditions Yl(O) = and YiO) just found. Results for the missing boundary condition Yz(O) are tabulated in Table 4.3 for four values of f3 using this method. With these values of the missing boundary condition, Eqs. (2.20) and (2.21) are integrated, and the results are tabulated in Table 4.4. Physically, these give half the deflection curve of the beam. For beams of the same cross section and made of the same material, a larger value of f3 means either a larger load W or a longer beam, or both.
°
TABLE 4.3
Y2(0) for a Few Values of {3's
0.1 0.5 1.0 5.0
0.00833 0.04167 0.08333 0.41667
4. The AdJoint Operator Method
62 TABLE 4.4
Deflection of Beams Deflection curve, Yl(S)
4.3
S
f3 = 0.1
f3 = 0.5
f3 =
0.0
0.ססOO
0.ססOO
0.ססOO
0.ססOO
0.1 0.2 0.3 0.4 0.5
0.0008 0.0015 0.0021 0.0025 0.0026
0.0041 0.0077 0.0106 0.0124 0.0130
0.0082 0.0155 0.0212 0.0248 0.0260
0.0409 0.0773 0.1059 0.1240 0.1302
1.0
f3 = 5.0
THIRD-ORDER DIFFERENTIAL EQUATIONS
The method developed in the previous section will now be extended to the solution of third-order differential equations. Even though the idea remains the same, there are certain points which need detailed explanation. Accordingly, we shall give full details of the solution of a third-order differential equation. The example chosen is the sandwich beam analysis treated earlier. Sandwich Beam Analysis
The third-order differential equation resulting from an analysis of the sandwich beams can be written as dYI 2 dYI - -3 - k -+a=O dx dx
(3.1)
subject to the boundary conditions dYl(O)
dYl(1)
~ = ~ =0,
Yl(!)=O
(3.2)
The physical explanation of the problem and its formulation was discussed in Section 2.3.2. As in the preceding section, Eq. (3.1) is first written as a system of first-order differential equations: dYI
dt
= Yz,
dYz
dt
= Y3'
dYJ
dt
= kY2 -
a
(3.3a-c)
In matrix notations, Eq. (3.3) can be written as
y = Ay + f
(3.4)
63
4.3 Third-Order Dlnerentlal Equations
where
YI Y= Y2 , [ h
f
~
=[
I
o
e
-a
!]
The adjoint differential equation is therefore, i = -ATx
(3.5)
where AT is the transpose of A, i.e.,
o o I
Eq. (3.5) can be written as
dx, dt
= 0,
dX2 dt
=
dX3 dt
2
-XI - k X3'
=
-X 2
(3.6a-c)
Multiplying Eqs. (3.3a)-(3.3c) and (3.6a)-(3.6c) by XI> X2 ' x 3'YI>Y2, and h, respectively, and adding these six equations, we then get d dt (xly\
+ x 2Y2 + x 3h) =
(3.7)
-ax 3
Integrating Eq. (3.7) from t = 0 to t = I and substituting the given boundary conditions into the resulting equation, we then get [X\(l)y\(I)
+
x 3(I)h(l)] - [xl(O)y\(O)
+
x 3(0)h(0)]
=
-afo\X 3dt
(3.8) Inspection of the first three terms on the left side of Eq. (3.8) shows that if the adjoint equations (3.6) are integrated using the boundary conditions x\(I)
= 0,
(3.9)
Eq. (3.8) becomes
(3.10) This is an algebraic equation in y\(O) and Y3(0), since integration of Eq. (3.6), subject to the boundary conditions (3.9), will yield xl(t), x 2(t), and x 3(t) for the range of t = I to t = O. In particular, xl(O) and x 3(0) are known from these solutions. For the solution of two missing initial conditions, y\(O) and h(O), another algebraic equation relating them will be needed. This equation can
64
°
be obtained by integrating Eq. (3.7) from to condition is specified at t = !. We then get [x 20
4. The Adjoint Operator Method
!
(since the third boundary
)yi!) + x 3( !)Y30)] - [XI(O)YI(O) + X3(0)h (0) ] = - a
L
I/ 2 x 3dt
(3.11) where the given boundary conditions have been substituted into the equation. Inspection of the first three terms of Eq. (3.11) shows that if Eq. (3.6) is integrated backward from t = ! to t = 0, using the boundary conditions
XIO) = I,
X2
( !) = 0,
(3.12)
then Eq. (3.11) becomes (1/ 2
X1(0)Yl(0) + x 3(0)h (0) = a J o
x 3 dt
(3.13)
Since xl(t), xit), and xiI) are now known solutions, Eq. (3.13) gives the second algebraic equation for the solution of YI(O) and h(O). Once YI(O) and Y3(0) are solved, the complete set of boundary conditions at t = is known, since yiO) is given. Equation (3.3) can then be integrated as an initial value problem. As an example, consider the solution for
°
k = 1.0 and a = 1.0 Integration of Eq. (3.6) from t = I to t = 0, using the boundary conditions specified in Eq. (3.9), gives the first set of solutions, which is tabulated in Table 4.5. TABLE 4.5 First Set of Solutions of Eq. (3.6)
1.00 0.80 0.60
0.40 0.20 0.00
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
1.0000 1.0200 1.0809 1.1852 1.3370 1.5425
-
0.0000 0.2013 0.4107 0.6365 0.8879 1.1749
Carrying out the integration in Eq. (3.10), we get
- 1.1749h(0)
=-
0.54292
(3.14)
Next, Eq. (3.6) is integrated from t =! to t = 0, using the boundary conditions (3.12). The results are summarized in Table 4.6.
65
4.4 Concluding Remarks TABLE 4.6
Second Set of Solutions of Eq. (3.6) x2
XI
0.5 0.4 0.3 0.2 0.1 0.0
i.oooo i.oooo i.oooo i.oooo i.oooo i.oooo
-
x3
0.ססOO
0.ססOO
0.1002 0.2013 0.3045 0.4107 0.5211
0.0050 0.0201 0.0453 0.0811 0.1276
Carrying out the integration in Eq. (3.13), we get y\(O) + 0.127613(0)
= 0.0211
(3.15)
The missing boundary conditions can now be solved from Eqs. (3.14) and (3.15), and the results are YI(O) = -0.0379,
Y3(0) = 0.4621
(3.16)
As a final step, solutions of Eq. (3.3) can be obtained by forward integration, since now it is an initial value problem. Table 4.7 shows the results. Agreement with the data calculated by the two methods discussed in the preceding chapters is excellent. TABLE 4.7 Solutions of Eq. (3.3)
0.0 0.2 0.4 0.6 0.8 1.0
YI
Y2
- 0.0379 - 0.0299 - 0.0112 0.0112 0.0299 0.0379
0.0730 0.1087 0.1087 0.0730
0.ססOO
0.ססOO
Y3 0.4621 0.2701 0.0888 - 0.0888 - 0.2700 - 0.4621
4.4 CONCLUDING REMARKS
The adjoint operator method outlined in this chapter is a very simple numerical scheme for the solution of linear boundary value problems. The examples given were presented in such a way that the reasoning which led to the choice of the initial conditions of the adjoint equations was made clear. In general, the number of backward integrations required is equal to the number of the unknown boundary conditions at the starting point. Therefore, if a third-order differential equation has one boundary condition given at the starting point and two boundary conditions at the end
66
4. The Adjoint Operator Method
point, we should interchange the roles of the two points, i.e., treat the end point as the starting point and vice versa. In this way, the number of backward integrations required is reduced by one. For a discussion of the convergence properties and the error estimate of the method of adjoint, the reader is referred to Roberts and Shipman [4] (see Section 6.7). The methods of superposition, chasing, and adjoint operator treated in the last three chapters provide three alternatives for the solution of linear boundary value problems. Basically, all three methods give a systematic way of finding the set of missing initial conditions by integrating a set of equations in a process of one or several passes. Solutions are obtained without iteration. There is very little difference among the three methods with regard to ease of programming, computational time, accuracy, and stability, with the exception that the equation itself may posses certain special properties that make one of the' methods ineffective. For example, Roberts and Shipman [4] cited the case in which an equation is stable on integrating forward but is unstable on integrating backward; in this example the adjoint method would not be a good choice. As another example, the method of superposition is a clear choice if the interval of integration is from zero to infinity. One particular problem that might sometimes occur in the solution of certain linear boundary value problems is that the columns of the matrix of coefficients of the linear equations developed by any of the above methods may be linearly dependent. Should this occur, it will be very difficult to compute the solution with sufficient accuracy. One effective way to overcome this difficulty is the GramSchmidt orthonormalization scheme [5]. For nonlinear equations, it is necessary first to linearize the nonlinear differential equations. Among the most commonly used methods are quasilinearization and Newton's variation method. Details of these linearization procedures are given in detail in the next two chapters. PROBLEMS
1. It is desired to find the deflection of a simply supported beam loaded by a concentrated load, as shown in Fig. 4.3. The differential equation for y
t
p
a
Ii
I
Cb----tx D.
Fig. 4.3
Diagram of problem I.
I
67
Problems
the solution of the deflection curve is given by £1 dYl dx?
dYz
£1 dx z
= =
Pb 1
X
Pb - I x - P(x - a)
for
x
for
x
>a
The boundary conditions are
x = 0: Yl = 0 x = a: YI = Yz,
x = I:
dYl dYz dx = dx
Yz = 0
Solve the problem by the adjoint operator method and compare the results with the exact solutions [6].
Hint: First, introduce the following dimensionless variables,
x = xl I, The differential equations will be in dimensionless form with two dimensionless parameters, namely, a II and b I I. Solutions can therefore be sought for a few pairs of values of these two parameters. 2. In an analysis of the hydrodynamic extrusion with controlled follower block clearance [7], the equation for the solution of the velocity w is given by
subject to the boundary conditions w(a) = 0,
w(b) = 0
Find the velocity distribution by the adjoint operator method. The following data are given: Q (flow rate) = 0.24 in. 3/sec, a (outer radius) = 0.513 in., c (clearance) = 0.00025 in., b (inner radius) = a-c. Answer: The missing slope at r = a is dw(a)1 dr = - 2322. 3. Gas A diffuses into a liquid and an irreversible, first-order, homogeneous reaction occurs, A + B = AB. The concentration of A can be
68
4.
The Adjoint Operator Method
solved [8] from the differential equation d 2c A - D - - +kc A =0 dz 2
subject to the boundary conditions
cA(O) = cAo'
dcA(L)1 dz = 0
Solve the problem by the adjoint operator method for kL 2 I D = 10- 2 and 10- 3 • Hint: Introducing the dimensionless quantities CA =
z= zlL
cAl CAo ,
the differential equation becomes d 2cAI dz 2 - (kL 2I D
)cA = 0
subject to the boundary conditions
The exact solution is cosh';kL21 D (1 - z)
cosh';kL 2 I D 4. A Newtonian fluid flows through the space between two concentric cylinders [9], both of which are rotating at different but steady speeds. The velocity distribution of the fluid can be solved from d
2u
dr 2
+ .!!.... ( !!. ) = 0 dr
r
subject to the boundary conditions u(r J ) = rJw l ,
u(r2) = r 2w2
Solve the problem by the adjoint operator method. The constants are specified as follows: r J = 0.75 in., r2 = 1.00 in., WI = 3 rpm, and W 2 = 10 rpm. Answer:
2W2r~ - wl(r~
+ rD
r~ - r~
5.
Solve problems 3 and 4 of Chapter 3 by the adjoint operator method.
References
69
REFERENCES I.
2. 3. 4. 5. 6. 7. 8. 9.
Goodman, T. and Lance, G., The numerical integration of two-point boundary value problems, Math. Comput., 10, 82-86, 1956. Bliss, G. A., "Mathematics for Exterior Ballistics," pp. 63-71, Wiley, New York, 1944. Wright, F. B., Jr., "The Adjoint Method in Analog Computation," Advisory Board on Simulation, Tech. Note 48, Univ. of Chicago, Chicago, Illinois, 1954. Roberts, S. M. and Shipman, J. S., "Two-Point Boundary Value Problems: Shooting Methods," Elsevier, New York, 1972. Todd, J., "Survey of Numerical Analysis," p. 347, McGraw-Hill, New York, 1962. Timoshenko, S., "Strength of Materials," pp. 168-169, Van Nostrand-Reinhold, Princeton, New Jersey, 1956. Kukarni, K. M., and Schey, J. A., ASME paper 74-Lubs-12 (1974). Bird, R. B., Stewart, W. E., and Lightfoot, E. N., "Transport Phenomena," p. 532, Wiley, New York, 1960. Schlichting, H., "Boundary Layer Theory," p. 80, McGraw-Hill, New York, 1968.
CHAPTER
5
ITERATIVE METHODSTHE SHOOTING METHODS
5.1 INTRODUCTION
In Chapters 5 and 6, the solution of boundary value problems (both linear and nonlinear) by iterative methods will be treated. These methods can be classified into two groups, namely shooting methods (Chapter 5) and finite-difference methods (Chapter 6). Even though the treatment in these two chapters is brief, the methods that are included are presented in sufficient detail for the reader to make meaningful application of these techniques to problems arising in sciences and engineering. In a shooting method, the missing (unspecified) initial condition at the initial point of the interval is assumed, and the differential equation is then integrated numerically as an initial value problem to the terminal point. The accuracy of the assumed missing initial condition is then checked by comparing the calculated value of the dependent variable at the terminal point with its given value there. If a difference exists, another value of the missing initial condition must be assumed and the process is repeated. This process is continued until the agreement between the calculated and the given condition at the terminal point is within the specified degree of accuracy. For this type of iterative approach, one naturally inquires whether or not there is a systematic way of finding each succeeding (assumed) value of the missing initial condition. Three such methods will be covered in this chapter: Newton's method (Section 5.2), the parallel shooting method (Section 5.3), and the method of quasi linearization (Section 5.4). All these methods are essentially based on the same principle, Newton's method of solving nonlinear algebraic equations and all therefore possess the same two important properties of monotone convergence and quadratic convergence. For most problems, Newton's method and the quasi-linearization method are equally efficient. However, due to the fact that Newton's method adjusts only the missing 70
5.2 Newton's Method
71
initial condition whereas in the method of quasi linearization the function at every point within the interval of integration is adjusted systematically, the latter is expected to exhibit better convergence. The method of parallel shooting is usually applied to problems where the solution is extremely sensitive to the assumed initial slope. 5.2 NEWTON'S METHOD
In this method, the differential equation is kept in its nonlinear form and the missing slope is found systematically by Newton's method. This method provides quadratic convergence of the iteration and is far better than the usual "cut-and-try" methods. Consider the second-order differential equation
dyjdx 2 = f(x,y,dyjdx)
(2.1)
subject to the boundary conditions
y(O) = 0,
y(L)
=A
(2.2a, b)
First, Eq. (2.1) is written in terms of a system of two first-order differential equations:
dyjdx = u,
duj dx = f(x, y, u)
(2.3)
We denote the missing initial slope by
dy(O)j dx = s
or
u(O)
=
s
(2.4a, b)
The problem is to find s such that the solution of Eq. (2.3), subject to the initial conditions (2.2a) and (2.4), satisfies the boundary condition at the second point, (2.2b). In other words, if the solutions of the initial value problem are denoted by y(x,s) and u(x,s), one searches for the value of s such that
y(L,s) - A = >(s) = 0
(2.5)
For Newton's method, the iteration formula for s is given by
or
s(n+ 1) = s(n) _
y(L,sn) - A ay(L,sn)jas
(2.6)
72
5.
Iterative Methods-The Shooting Methods
To find the derivative of y with respect to s, Eqs. (2.3), (2.2a), and (2.4) are differentiated with respect to s, and we get
.
dU dx
and
= _at
YeO) = 0,
where
Y=
aylas,
ay
at U
Y+ _
au
U(O) = 1 U=
(2.7) (2.8)
aulas
(2.9)
The solution of Eq. (2.3), subject to the boundary conditions (2.2a, b), can therefore be obtained by the following steps. 1. Assume a value of s for the missing initial slope, (2.4). Let us denote this approximate value of s by s( I) • 2. Integrate Eq. (2.3), subject to the boundary conditions (2.2a) and (2.4), as an initial value problem from x = to x = L. 3. Integrate Eq. (2.7), subject to the boundary conditions (2.8), as an initial value problem, from x = to x = L. 4. Substituting the values of y(L,s(l» and Y(L,S(I» into Eq. (2.6), we get S(2) = s(l) - [y(L,S(I») - A ]IY(L,s(I»)
°
°
the next approximation of the missing initial slope S(2) is obtained. 5. Repeat steps 2-4 until the value of s is within the specified degree of accuracy. An example will be given in the next section. Unsteady Flow 01 a Gas through a Porous Medium
Consider the unsteady flow of a gas through a semi-infinite porous medium. The medium is assumed to be filled with gas at a uniform pressure Po initially. At time t = 0, the pressure at the surface is suddenly reduced from Po to PI and is thereafter maintained at this pressure. The governing differential equation for the unsteady isothermal flow of a gas can be derived as follows. Consider the infinitesimal volume shown in Fig. 5.1. Conservation of mass requires that (mass flux into the volume)
(mass flux out of the volume)
(rate of accumulation of mass in the volume)
(2.10)
or, in mathematical form [ (puA)] -
a(pu)]} {[pu + ~ dx A
a = at
[p(A dx ep) ]
(2.11 )
73
5.2 Newton's Method
where A and u are the area and the velocity of gas, respectively, and ep is the porosity of the medium, defined as the volume of the pore spaces per unit volume of the medium. Eq. (2.11) can be simplified to give ap
apu
ep at = - ax
(2.12)
x dx
Fig. 5.1 Semi-infinite porous membrane.
For flow in a porous medium, the velocity is related to the pressure gradient through Darcy's law, k ap u= - - (2.13) fL
ax
where the constants k and fL are the permeability of the medium and viscosity of the fluid. In addition, for an isothermal gas the density is related to the pressure through the equation of state, (2.14)
p=pRT
where R is the gas constant. Substitution of Eqs. (2.13) and (2.14) into Eq. (2.12) yields
a(
ax
p
ap)
ax
=
epfL ap
k at
subject to the boundary conditions
p(x,O) = Po,
p(O,t) = PI>
p(oo,t)=po
Kidder [1] introduced the similarity transformation x
z=-
It
and
"" )1/2 't"IL ( -4pok-
(2.15)
74
S. Iterative Methods-The Shooting Methods
where a
= 1 - (PUP5)
Equation (2.15) and its boundary condition become
d\v + 2z dw = 0 dz 2 (1 - aw)I/2 dz and
w(O) = 1,
(2.16)
w(oo) = 0
In order to compare the results with those from [2], a transformation of variables TJ = z/(I - a)I/4 and f(TJ)=w-I is introduced, and Eq. (2.16) and its boundary condition become
d~ +
dTJ2
2TJ df (l-[a/(I-a)]j}1/2 dTJ
=0
(2.17)
and
f(O) = 0,
f( 00) = -1
To apply Newton's method, Eq. (2.17) is written first-order differential equations,
df -=u dTJ '
(2.l8a, b) ill
terms of two
du 2TJu dTJ = - {l - [ a/ (1 - a)] f} 1/2
(2.19)
subject to the boundary conditions
f(O) = 0,
f(oo)
= -1
(2.20a, b)
Let us denote the missing slope by
df(O)/ dTJ = s
or
u(O) = s
(2.21a, b)
Equations (2.19), (2.20a), and (2.2Ib) can be differentiated with respect to s to give
dF = U dTJ dU
2TJ ( auF dTJ = - {l - [ a / (1 - a)] j} 1/2 2( 1 - a){ 1 - [ a / (1 - a)] j} (2.22)
F(O) = 0,
U(O) = 1
(2.23)
75
5.2 Newton's Method
where F=
u = au/as
of/as,
(2.24)
As an example, consider the solution of Eq. (2.19) for a = 0.154. As a first approximation, let us assume the missing initial slope to be
df(O)
~
= u(O) = s(l) = -1.00
(2.25)
where the superscript (1) indicates the number of iteration. Equation (2.19), subject to the boundary conditions (2.20a) and (2.25), can therefore be integrated as an initial value problem. A summary is given in Table 5.1. TABLE 5.1 First Approximation of the Solution"
r»
11
0.0 0.4 0.8 1.2 1.6 2.0
-
u(l)
0.0000 0.3814 0.6655 0.8234 0.8892 0.9096
° Superscripts give
-
j<1)
11
1.0000 0.8582 0.5436 0.2559 0.0894 0.0231
2.4 2.8 3.2 3.6 4.0
-
- 0.0044
0.9143 0.9151 0.9152 0.9153 0.9153
- 0.0006 - 0.0001 - 0.0000 - 0.0000
the number of iterations.
With f and u known, the initial value problem defined by Eqs. (2.22) and (2.23) can be integrated. The results are summarized in Table 5.2. TABLE 5.2 Solutions of Eqs. (2.22) and (2.23) 11
F(I)
U(l)
11
F(I)
U(I)
0.0 0.4 0.8
0.0000 0.3817 0.6688 0.8334 0.9056 0.9297
1.0000 0.8611 0.5567 0.2740 0.1023 0.0287
2.4 2.8 3.2 3.6 4.0
0.9358 0.9369 0.9371 0.9371 0.9371
0.0060 0.0009 0.0001 0.0000 0.0000
1.2 1.6 2.0
From Tables 5.1 and 5.2, we get
f(4.0) = -0.9153
and
F(4.0) = 0.9371
The second approximation of S(2) can be calculated from Eq. (2.6): S(2)
= s(l)
_
f(4.0) + 1 = -1.00 _ 1 - 0.9153 = -10904 F(4.0) 0.9371'
76
5.
Iterative Methods-The Shooting Methods
Using this second approximation of s, Eq. (2.19) can again be integrated and the preceding processes repeated. This process is continued until the change in value of s is within a certain prescribed accuracy. It should be noted, however, that the convergence of the method is so fast that it takes only a few iterations. For the present numerical example, the missing initial slope s reaches four-digit accuracy of s = - 1.0903 in the second iteration. In fact the first iteration gives a value of s = - 1.0904, which is only 0.01% larger. The final solutions (result of the second iteration) are summarized in Table 5.3. TABLE 5.3 Results of the Second Iteration
0.0 0.4 0.8 1.2 1.6 2.0
u(3)
f(3)
7j
-
0.0000 0.4158 0.7259 0.8988 0.9711 0.9937
-
1.0903 0.9360 0.5940 0.2808 0.0988 0.0257
r»
7j
2.4 2.8 3.2 3.6 4.0
-
0.9990 0.9999 1.0000 1.0000 1.0000
u(3)
-
0.0049 0.0007 0.0001 0.0000 0.0000
5.3 PARALLEL SHOOTING
For certain problems, the solution is quite sensitive to the initial guessing and the method of parallel shooting has been found to be quite suitable. Based on this method, the interval is divided into a number of subintervals and each succeeding approximation is adjusted simultaneously to satisfy the boundary conditions and appropriate continuity conditions at the points dividing the subintervals. Consider again the system of differential equations given by Eqs. (2.3) and (2.2),
dyf dx
= u,
du f dx
= f(x,y,u)
(3.1 )
subject to the boundary conditions
yeO) = 0,
y(L) = A
(3.2)
As an illustration of the method, we arbitrarily divide the interval into two subintervals, [0, t L] and [t L, L]. Following the steps outlined in Section 5.2, the following initial value problems will be integrated in each interval: dil) / dx
= u(l),
(3.3)
5.3 Parallel Shooting
77
with the boundary conditions y(I>C0) = 0,
(3.4a, b)
and dy<") / dx =
(3.5)
U(II) ,
with the boundary conditions y<")(tL)
= a,
(3.6)
where, again, the constant s is the missing slope to be found. The constants a and b in (3.6) are the unknown initial conditions whose values will be
found in the solution process. To find the three unknown constants s, a, and b, the following conditions are imposed: From the continuity condition,
y<1)( t L) = y<")( t L)
and
which give and or cf>1(s,a) = y<1)(tL,s) - a = 0
(3.7)
cf>2(s,b) = u(l)(tL,s) - b = 0
(3.8)
and Also, the boundary condition at x = L gives cf>3(a,b)
= y<")(L,a,b) -
A
=0
(3.9)
To start the solution, a set of values of s, a, and b is assigned. Let us denote these first approximations by s(\), a(I), and b(\). Naturally, cf>1(S(I)
,a(\» =1= 0
cf>is(\), b( I» =1= 0 cf>3(a(l) ,b(I» =1= 0
Let us next denote ths second approximation by S(2), a (2) , and b(2) and require that cf>1(s(2),a(2»
=0
cf>2(S(2) ,b(2»
=0
cf>la(2), b(2»
=0
78
5. Iterative Methods-The Shooting Methods
(3.10)
By keeping only the first-order terms, Eqs. (3.10) give acp
( asl
(I) )
acp
(S(2) - s(l») + (
acp
aal
acp
(I)
(I) )
(a(2) - a(l») = - CPI( s(l), a(l»)
(I)
2) 2) ( as (S(2) - s(l») + ( ab (b(2) - b(l») =
(a:) acp
(1)
(a(2) - a(l») +
(a: acp
)(1)
- CP2(S(l) , b(l»)
(3.11 )
(b(2) - b(l») = -CP3(a(l) ,b(l»)
Substituting the cp's from Eqs. (3.7)-(3.9) into Eq. (3.11), we get
(
a (I) )(1) ~s
=
(a~:I)
(S(2) - s(l») - (a(2) - a(l»)
a(l) - y(l)(tL,s(l») f)(S(2) - s(l»)
= b(l) - u(l)(tL,s(l»)
( =A
(3.12)
a (II) )(1) ~a
- y
(a(2) - a(l») +
(
a (II) )(1) ~b
(b(2) - b( I»)
79
5.3 Parallel Shooting
or, in matrix form, - I
[ (aY'" las)'"
(au(I) jasf)
=
0
(ay
0 a( I)
-
y(l)O L, s(l»)
b( I)
-
u(l)(t L, s(l»)
A -
0 - I
(ay
so, ]
s<" ]
a(2) - a(l) b(2) - b(l)
(3.13)
y
from which -I
S(2)
s(l)
(aY(l) jas)(I)
- I
0
a(2)
a(l)
+ (au(I) jas)(I)
0
- I
b(2)
b(l)
x
0 a(l) -
Y(l)(t L,s(l»)
b(l) -
u(l)(t L, s(l»)
(ay
(ay(II) jab )(1)
(3.14 )
A - y<1I)(L,a(l) ,b(I»)
Therefore, the second approximations S(2), a(2), and b(2) are solved from Eq. (3.14). For higher-order approximations, the same process can be repeated until a certain prescribed accuracy is reached. An example will be given in the next section to illustrate the computational details. Diffusion In a Chemical Catalytic Converter
One model for analysis of a catalytic converter is a flat plate catalytic bed separated from the bulk fluid motion by a stagnant film of small thickness 6. Gas A diffuses from the bulk flow to the catalyst surface at z = 6 where it undergoes an instantaneous heterogeneous reaction A ~ t A2 • The system is shown in Fig. 5.2. We are interested in the variation of the mole fraction of A, x A' with the distance z. A mass balance of gas referred to an infinitesimal layer of the fluid film gives (3.15) dNAjdz = 0 where N A is molar flux of gas A.
80
5. Iterative Methods-The Shooting Methods z=O
----1---------------
z=8
""
catalyst surface
Fig. 5.2 The model.
From Fick's law, the molar flux N A is related to the mole fraction gradient dx AI dz through the relation N A= -CD(dxAldz)+XA(NA+NA2)
(3.16)
On the surface of the catalyst, the chemical reaction A~tA2
takes place where, for each mole of A, there will be 0.5 mole of A 2 produced. We can therefore write
tNA
N A2= -
Eq. (3.16) therefore becomes
(1 - tXA)N A = - CD (dxAI dz)
(3.17)
Substituting N A from Eq. (3.17) into Eq. (3.15), we get
2x
d A dz 2
1
+ 2(1 - tXA)
( dXA
)2
=0
(3.18)
xA=O
(3.19)
dz
The boundary conditions are z=~:
The physical meaning of the second boundary condition is simply that gas A is completely reacted as it reaches the surface. By introducing ~ = z I~, the equation becomes dimensionless, and we finally get the following second-order boundary value problem:
2x
d A de ~=O:
+
1 2(1 -
tx
XA=X AO;
A
( dXA )
d~
~=1:
)2 =
0
xA=O
(3.20) (3.21a,b)
To solve Eq. (3.20) by the method of double shooting, we observe the following steps:
5.3 Parallel Shooting
81
1. Assuming the first approximation of the constant s, a, and b. Let us choose S(I)
= -
a(l)
=
b(l)
= - x Ao/ (2 . IN)
x
x Ao/
(2 · IN)
/2
Ao
where IN is the number of grid points between ~ = 0 and ~ = t. 2. We integrate the following two initial value problems: d
(I)
2= d~
x~l(O) =
( U(l»)2
du(l)
--=d~ 2 - x~)
(I) u,
u(l)(O)
x A O'
=
(3.22)
S
and d (II) _ xA (II) ~-u , X(II)( 1. A 2)
( U(II»)2
du(lI)
--=d~
= a,
U(II)(
t)
2-
(3.23)
X~I)
= b
where, for the calculation of the next iterates, we use S = s( I), a = a( I), and b(I). The solutions of x~) and U(I) from ~ = 0 to ~ = t are then obtained, as well as the solutions of X~I) and U(II) from ~ = to ~ = 1. The results are shown in Table 5.4.
b=
t
TABLE 5.4
Solution of Eqs. (3.22) and (3.23) (xAo = 0.25) u(l)
~
x5!)
0.0 0.1 0.2 0.3 0.4 0.5
0.2500 0.2497 0.2495 0.2492 0.2490 0.2487
-
0.0025 0.0025 0.0025 0.0025 0.0025 0.0025
~
x5!l)
0.5 0.6 0.7 0.8 0.9 1.0
0.1250 0.1247 0.1245 0.1242 0.1240 0.1237
U(II)
-
0.0025 0.0025 0.0025 0.0025 0.0025 0.0025
3. Let us define Ys =
ax~)
b=
a
au(l)
Us =
ax~')
----as ' Y = -aa- Y ----as ' u, = ----a;- , u, = au(lI)
ax~1)
ab au(lI)
----ab
(3.24)
5. Iterative Methods-The Shooting Methods
82
Differentiating Eq, (3.22) with respect to s, we get
ar,
au,
2U(I)
(U(I»2
d~ = x)!) _ 2 Us - (2 _ X)!»)2 r,
d~ = Us,
(3.25)
subject to the boundary conditions
Ys(O) = 0,
U.(O) = I
Differentiating Eq. (3.23) with respect to a, we get
au,
ar,
d~ =
2U(II)
d~ = X)!I) _
Ua'
2
u, -
(U(II»2 (2 _ X)!I»)2
r,
(3.26)
r,
(3.27)
subject to the boundary conditions
YaU) = 1, Differentiating Eq. (3.23) with respect to b, we get
ar,
au,
d~ = Ub'
(U(II»2 X)!I) _ 2 U; - (2 _ X)!I»)2 2U(II)
d~ =
subject to the boundary conditions The three initial value problems can now be integrated. For Ys and Us' the range of integration is from ~ = 0 to ~ = t. For (Ya, Ua) and (Yb, Ub)' the range of integration is from ~ = to ~ = l. The results are shown in Table 5.5.
t
TABLE 5.5 Solutions of Eqs. (3.25}-(3.27)(x A O = 0.25) ~
0.0 0.1 0.2 0.3 0.4 0.5
r,
Us
0.ססOO
i.oooo
1.1203 1.2551 1.4061 1.5752 1.7647
0.1053 0.2233 0.3554 0.5035 0.6694
~
r,
0.5 0.6 0.7 0.8 0.9 1.0
i.oooo i.oooo i.oooo i.eooo t.oooo i.oooo
u,
r,
o,
0.ססOO
0.ססOO
i.oooo
0.ססOO
0.1049 0.2216 0.3514 0.4956 0.6560
1.1119 1.2364 1.3747 1.5285 1.6995
0.ססOO
0.ססOO 0.ססOO 0.ססOO
83
5.3 Parallel ShootIng
4. From steps 2 and 3, we get the following quantities: x~)(
t ) = 0.2487
x~I)(I)
Ys ( t)
= 0.1237
u(l)(t) = - 0.0025
= 0.6694
Us(t) = 1.7647
Ya(l) = 1.0000
Ua(l) = 0.0000
Yb(l) = 0.6560
Ub(l) = 1.6995
These quantities can now be substituted into Eq. (3.14) to give
from which S(2)
= s(l)
-
±{x~I)(I) + [U(I)( t) -
+ [x).I)( t) a(2)
= a(l)
-
±{[
= b(l)
-
±{[
a(l)]
Yb(l)
Ya(l)}
u(l)(t) -
- [x).I)( t) b(2)
a(l)]
b(l)]
b(l)]
Ys(t) Yb(l) + x).II)(l) Ys(t)
Use t) Yb(I)}
x).I)(t) -
a(l)]
(3.28)
UsC!)YaCI)
where (3.29) Equation (3.28) then shows the second iteration of s, a, and b to be
= -0.1380 a(2) = 0.1581
S(2)
b(2)
=
-0.2416
5. Using the second iterates S(2), a(2), and b(2), the second iteration of the solutions of Eqs. (3.22) and (3.23) can be obtained. This is shown in Table 5.6.
84
5.
Iterative Methods-The Shooting Methods
TABLE 5.6 Second Iteration of the Solutions of Eqs. (3.22) and (3.23) ~
x~)
0.0 0.1 0.2 0.3 0.4 0.5
0.2500 0.2362 0.2222 0.2081 0.1940 0.1797
u(l)
-
0.1380 0.1391 0.1402 0.1413 0.1424 0.1435
~
x~[)
0.5 0.6 0.7 0.8 0.9 1.0
0.1581 0.1338 0.1092 0.0842 0.0589 0.0333
(XAO
= 0.25) u(II)
-
0.2416 0.2447 0.2480 0.2512 0.2546 0.2579
6. Repeat steps 2-5 until desired accuracy is reached. Table 5.7 gives the results of iterations in terms of s(i) , a(i) , and b(i) • TABLE 5.7 Results of Iterations of s(i), a(i), and b(i) (x AO = 0.25)
Number of iterations
s(i)
I" 2 3 4 5 6 7 8 9
-
0.0025 0.1380 0.2037 0.2255 0.2317 0.2333 0.2337 0.2338 0.2338
a(i)
0.1250 0.1581 0.1358 0.1307 0.1295 0.1293 0.1292 0.1292 0.1292
b(i)
-
0.0025 0.2416 0.2584 0.2538 0.2512 0.2503 0.2501 0.2500 0.2500
"First approximation.
°
With s, a, and b thus found, Eqs. (3.22) and (3.23) can be integrated one more time to get x A and u for the range of s from to 1. The physical significance of the solutions is that x A is the mole fraction of species A as a function of the distance from the edge of the diffusion layer. 5.4
QUASI LINEARIZATION
Consider the nonlinear second-order differential equation
y"
= !(x,y,y')
(4.1)
subject to the boundary conditions
y(o) =0,
y(L)=A
(4.2)
where the letters y' and y" represent, respectively, dy/ dx and dy / dx', This notation is needed in this section.
5.4 Quasi Linearization
85
Let us rewrite Eq. (4.1) as
cf>(x,y,y',y") = Y" - f(x,y,y') = 0
(4.3)
To derive the recurrence equation, let us denote the nth and the (n + l)th iterations by Yn and Yn+ l ' respectively and require that, for both iterations, cf> = O. For the nth iteration, this gives
Y; - f(x, Yn' Y~) = 0 For the (n
(4.4)
+ l)th iteration, we get
cf>(x,Yn+l'Y~+I'Y;+l)=cf>(X,Yn'Y~,Y;)+ (:; )n(Yn+l- Yn) + ( :;,
t(Y~+l
-
Y~) + ( a;'1
t
or,
- (
) n(Y;+I - Y;) + ...
=
0 (4.5)
~~) n(Yn+l - Yn) - ( a~ (Y~+I - Y~) + (Y;+I - Y;) = 0
(4.6)
Substituting j f from Eq. (4.4) into Eq. (4.6), we get
" I - (af) Yn+ ay' nYn+ 1 I
(
af) nYn+ 1 ay
af) /n - f(x,Yn,Yn) - (af ay ) /n - ( ay ' _
I
(4.7)
I
The boundary conditions are
Yn+I(O) = 0,
Yn+I(L)
=
A
(4.8)
Equation (4.7), subject to boundary conditions (4.8), is a linear boundary value problem, the solution of which can be obtained by one of the methods treated in Chapters 2-4. 5.4.1 Radiation Fin of Trapezoidal Profllet
Consider the radiation fin shown in Fig. 5.3. The fin is assumed to liberate heat to its surrounding only through radiation. By using the one-dimensional form of the energy equation, the following differential equation is obtained for the solution of the temperature distribution: d
2U
dR 2
+ [1
4
_ 0 t a n a ] dU /3U R+p - (I-R)tana+O dR - (l-R)tana+O-
(4.9) t
See Keller and Holdrege [3].
86
5. Iterative Methods-The Shooting Methods
annular fin of trapezoidal profile
Fig. 5.3 Annular fin of trapezoidal profile. (Reprinted from [3] with permission of the American Society of Mechanical Engineers.)
subject to the boundary conditions
U(O) = 1,
- 0 ( dU) dR R - I -
The dimensionless quantities in the above equation are related to the physical variables through the following definitions: T
U = TB
'
=
f3
(rT
-
rB)mTJ
k cos ex
R= (4.10)
where T B , €, (J, and k are, respectively, the temperature at rB , the emissivity of the fin, Planck's constant, and the heat conductivity. The radius of the base and tip are labeled r B and rT , respectively, and ex is the angle of inclination of the top surface. The linearized equation of Eq. (4.9), corresponding to Eq. (4.7), is then d2U ( dR 2 -
)(i+I)
[1
tan ex
+ R + P - (1 - R )tan ex + ()
l
4f3( u3 ) [ (1- R)tanex + () i
]
.
U(·+I) = -
J( dU)(i+l) dR
3f3(U 4 ) ( i ) (1- R)tanex + ()
(411)
.
87
5.4 Quasi Linearization
The boundary conditions become R
= 0:
= 1;
U(i+I)
R
= 1: ar-:» /dR = 0
To solve Eq. (4.11), the method of superposition, treated in Chapter 2, will now be used. Let us write = V+ sW
U(i+I)
(4.12)
and set s
=
au": 1)(0)/ dR
Equation (4.11) can be separated into two initial value problems: 2V
d + [1 dR 2 R +P
tan a ] dV (1 - R )tana + 0 dR 3{3( u 4f ) V = - (1 - R )tan a + 0
4{3( U 3f ) - [ (1 - R )tan a + 0
]
V(O) = 1,
dV(O)/ dR
(4.13)
=0
and 2W
d dR 2
+ [1
tan a ] dW (I-R)tana+O dR
R+p
-
4{3( u 3 f ) [ (1 - R )tana + 0
W(O)
= 0,
]
(4.14)
W-O -
dW(O)/ dR
=1
As a numerical example, the solutions of Eq. (4.9) are worked out for the following values of the parameters: p = 0.5,
0=0.05,
{3
= 0.1
By assuming an initial approximation of U(I)
=1
higher-order iterates can be obtained by integrating the two initial value problems defined by Eqs. (4.13) and (4.14) from R = 0 to R = 1. The value of s can be computed from the boundary condition at R = 1, i.e., s
= - V'(I)/ W'(I)
Combination of the two solutions, V and W, then gives the second iterate U(2). This process is continued until the desired accuracy is reached. A summary of the solutions is given in Table 5.8.
5. Iterative Methods-The Shooting Methods
88
TABLE 5.8 Sample Solutions of Eq. (4.9) (a = 6°, p = 0.5, () = 0.05, f3 = 0.1)
U U) a
R
i= 1
0.0 0.2 0.4 0.6 0.8 1.0
i.eooo i.oooo t.ocoo t.oooo i.oooo
ai
i.oooo
i= 2
i=3
i=4
i.oooo
i.oooo 0.9049 0.8410 0.7971 0.7692 0.7584
i.oooo
i.oooo
0.9044 0.8400 0.7956 0.7673 0.7564
0.9044 0.8400 0.7956 0.7673 0.7564
0.9185 0.8665 0.8328 0.8127 0.8052
i=5
is the number of iteration.
5.4.2 A Nonlinear Dynamics Problem
We shall now consider another nonlinear boundary value problem which is slightly different from the one treated in the last section. Consider a mass m acted upon by a nonlinear force which equals, say, - xe - X, where x is its position on the x axis. Initially, the mass is located at the origin x = O. We are to find the initial velocity and the duration of motion such that when the particle reaches the point x = x o, the velocity of the mass is exactly zero. From Newton's second law of motion, the problem can be represented by the following boundary value problem: d 2x/dt2= -xe- x
(4.15)
subject to the boundary conditions t=O:
x=O
t= T:
x
= x o,
(4.l6a) dx/ dt
=0
(4.16b)
The unknowns are therefore dx(O)/ dt and T. The quasi-linear form of Eq. (4.15) is . . d 2X (i+ I) + [(1 _ x(i»e-x('l]x(i+I) = _(x 2)(I)e- x(i )
dt?
(4.17)
subject to the boundary conditions
=0
t = 0:
X(i+I)
t = T:
x(i+ I) = x o, dx(i+ I) / dt = 0
Because of the fact that the duration of time T is also unknown, Eq. (4.17) will now be solved by the method of superposition with slight
5.4
Quasi Linearization
89
modification. Let us first separate
x(i+ I)
X(i+I)
into two solutions by
= Y + sZ
(4.18)
By identifying s = dx':" 1)(0)/ dt
Eq. (4.17) can be separated into two initial value problems, as 2
d y
dt 2
+ [(1 _ y(o) d
2Z
x(i»e-x(i)] Y
= 0,
+ [(1 -
dt 2 Z(O) = 0,
=
_(x 2)(i)e- x(I)
dY(O)/ dt
=
° °
x(i»e-x(I)]Z =
(4.19)
(4.20)
dZ(O)/ dt = 1
= T gives Y(T) + sZ(T) =
The boundary conditions at t
X
(4.21)
o
and dY(T) dZ(T) dt +s dt
=
°
(4.22)
Eliminating s from Eqs. (4.21) and (4.22), we get Y(T)
dZ(T) dY(T) dt - Z(T) dt
= Xo
dZ(T) dt
(4.23)
Equation (4.23) and one of (4.21) and (4.22) are the equations needed for the evaluation of the unknown duration T and the parameter s. The solution of Eq. (4.15) therefore consists of the following steps. l.
An initial approximation is assumed. Let us choose x
= xo(l
- e- t )
which gives X~I), for n = 1, ... , N. 2. The initial value problems defined by Eqs. (4.19) and (4.20) are integrated from t = with the increment of t equal to h. During the integration process, the condition (4.23) is checked for each t step. Integration will stop when Eq. (4.23) is satisfied. We have now the unknown time duration T as well as Y(T) and Z(T). 3. The value of s is then formed from Eq. (4.21). 4. Combination of the two solutions through the relation (4.18) then gives X~2), n = 1, ... , N. 5. Steps 2-4 are repeated, until the desired accuracy is reached.
°
90
5. Iterative Methods-The Shooting Methods
A summary of the numerical solutions thus obtained is given in Table
5.9.
TABLE 5.9
Sample Solution of Eq. (4.15) (xQ= 0.5)a First approximation
X (i)
X(I)
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5
0.0000 0.1967 0.3161 0.3884 0.4323 0.4590 0.4751 0.4849 0.4908 0.4945
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45
;=2
i = 3
;=4
0.0000 0.0589 0.1 175 0.1755 0.2328 0.2890 0.3440 0.3977 0.4497 0.5000
0.0000 0.0586 0.1168 0.1746 0.2316 0.2877 0.3427 0.3965 0.4490 0.5000
0.0000 0.0586 0.1168 0.1746 0.2316 0.2877 0.3427 0.3965 0.4490 0.5000
a After the second iteration, the unknown duration T is found to be T= 0.45.
5.5 CONCLUDING REMARKS
In this chapter we have described two shooting methods, the simple shooting method (Newton's method) and the double-shooting method. The simple shooting method replaces the usual "cut-and-try" searching technique and provides a systematic way of obtaining the missing initial conditions. It generally leads to quadratic convergence of the iteration and decreases computing time. Of course, the rate of convergence depends on how close the initial approximation is to the correct value. In general, there is no universal procedure guiding the initial approximation. In practical engineering problems, however, there are ways to estimate the initial approximation. One such example is the solution of a boundary value problem which contains a physical parameter either in the differential equation or in the boundary condition. If the solution for a certain value of the parameter is known, solutions corresponding to a neighboring value of the solution can be obtained by choosing this solution as the first approximation. One difficulty which arises for certain problems in the simple shooting method is the sensitivity of the solution to the initial guess. This phenomenon of instability arises because the rapidly growing solutions of the initial value problems magnify the truncation and round-off errors. One way to overcome this difficulty is to use parallel shooting. In this method, the interval of integration is divided into a number of subinter-
Problems
91
vals, with the appropriate initial value problems integrated over each subinterval. Adjustment of the initial guess is then made simultaneously to satisfy the boundary conditions and the continuity conditions at the subdivision points. A detailed treatment of the various mathematical aspects of the two shooting methods is given in Chapter 2 of Keller [4]. The quasi-linearization technique outlined in Section 5.4 not only linearizes the nonlinear boundary value problem but also provides a sequence of functions which in general converges quadratically to the solution of the original equation. Starting from a rough first approximation for the dependent variable, the solution of the original nonlinear boundary value problem can be obtained through a sequence of successive iterations of the dependent variable. The idea of quasi linearization is basically a generalization of the well-known Newton-Raphson method. It inherits the two important properties of that method, namely quadratic convergence and monotone convergence. The mathematical problem of existence and convergence for the quasi-linearization technique is beyond the scope of this book. Interested readers should consult the various texts specifically on the method of quasi linearization [5-7]. PROBLEMS
1. In an n analysis of the flow through a uniformly porous channel, Terril [8] found that the dimensionless velocity can be found from the solution of the following boundary value problem:
subject to the boundary conditions
f(O) = 1,
df(O)
-dyt - = 0 ',
f(l) = 0,
where K is an eigenvalue. Find the solution of f for R = 15.014 by (a) Newton's method, (b) parallel shooting, and (c) quasi linearization. In part (c), the linearized equation is solved by all three methods, namely the method of superposition, the method of chasing, and the method of adjoint operator. Note: The problem can be solved by differentiating the differential equation to make it of fourth order. This problem includes all the basic concepts learned in chapters 2-5. Answer: d'1(l)/ dyt2 = -10.162
92
5. Iterative Methods-The Shooting Methods
2. In an analysis of the problem of phase change of solids with temperature dependent thermal conductivity, Cho and Sunderland [9] found that the temperature distribution can be found by solving the following boundary value problem:
.if... dTJ
{(l + {30) dTJdO } + 2TJ dTJdO
=
0(0) = 0, 0(00) = 1
0,
Solve 0 for {3 = 4.00 and 1.441 by (a) Newton's method and (b) the method of quasi linearization. Answer: dO (O)j dTJ = 1.751 and 1.388 for {3 = 4.00 and 1.441, respectively. 3. The stress in the spherical membrane of a spherical cap is found to be governed by the following boundary value problem [10]: d 2F
e
>-..2
de
32 F
8
- + - - 2= ~
= 0:
F= 0;
dF
2d~-(l+/I)F=0
Solve the equation by the method of parallel shooting for /I = 0.3 and = 0.991. Answer: F(l) = 0.2932. 4. Solve the problem treated in Section 4.1 by Newton's method. 5. Solve the problem treated in Section 4.1 by quasi linearization, with the linearized equations solved by the method of chasing. >-..
REFERENCES I.
2. 3. 4. 5. 6. 7. 8. 9. 10.
Kidder, R. E., Unsteady flow of gas through a semi-infinite porous medium, J. Appl. Mech. 79, 329-334 (1957). Na, T. Y., An initial value method for the solution of a class of nonlinear equations in fluid mechanics, J. Basic Eng., Trans. ASME, 503-509 (1970). Keller, H. H., and Holdrege, E. S., Radiation heat transfer for annular fins of trapezoidal profile, J. Heat Transfer, Trans. ASME 92, 113-116 (1970). Keller, H. B., "Numerical Methods for Two-Point Boundary Value Problems," Chapter 2, Ginn-Blaisdell, Waltham, Massachusetts, 1968. Bellman, R. E., and Kalaba, R. E., "Quasilinearization and Nonlinear Boundary Value Problems," Elsevier, New York, 1965. Radbill, J. R., and McCue, G. A., "Quasilinearization and Nonlinear Problems in Fluid and Orbital Mechanics," Elsevier, New York, 1970. Lee, E. S., "Quasi Linearization and Invariant Imbedding," Academic Press, New York, 1968. Terril, R. M., Aeronat. Q. 15,299-310 (August 1964). Cho, S. H., and Sunderland, J. E., J. Heat Transfer, Trans. ASME 96, 214-217 (1974). Perrone, N., and Kao, R., J. Appl. Mech. 38,371-376 (1971).
CHAPTER
6
ITERATIVE METHODS-THE FINITE-DIFFERENCE METHOD
6.1
INTRODUCTION
In this chapter the method of finite differences for the solution of boundary value problems will be treated. In this method, the derivatives in the differential equations are replaced by appropriate finite differences and the differential equation is therefore reduced to a system of algebraic equations. The solution of the algebraic equations then gives the dependent variable at discrete values of the independent variable. The method is preferred over the shooting method for solving numerically sensitive twopoint boundary value problems. For problems which necessitate finding two or more missing initial conditions, this method is more efficient, since the computation in the shooting method is often quite laborious. For the solution of the system of algebraic equations of the finitedifference representation of the differential equation, a very efficient method of factorization will be given in detail. This method is especially suitable for the solution of boundary value problems where the coefficient matrix of the algebraic equations is tridiagonal. Furthermore, the size of the coefficient matrix in actual engineering analyses is in general very large, which necessitates too many computer memory locations and involves too many unnecessary computations if standard methods of matrix inversion, such as Gauss-Seidal's elimination, are used. Details of the derivation of the necessary equations are given. By following through the steps twice in Sections 6.4.1 and 6.5.1, the reader will be able to make meaningful use of the final equations and, if the need arises, derive the corresponding equations for those differential equations not covered in this chapter. 6.2 FINITE DIFFERENCES
To obtain numerical solutions for a differential equation, the continuous variables should be replaced by discrete variables. This is accomplished by 93
94
6. Iterative Methods-The Flnlte·Dlnerence Method
replacing derivatives by finite-difference ratios. The result is a system of algebraic equations, which are then solved on a digital computer. To introduce the concept, consider a continuous function u(x), which is plotted in Fig. 6.1. Let us divide the x axis in finite intervals at a distance ax apart. For three arbitrary adjacent points, namely, xn-I> x n' and x n+ I' the corresponding values of U are un-I> Un' and un+)' respectively. u(x)
I-...._..L-_....L._---I.../\
2
x=O
,.I...-_...l-_....L._---I.._
3
Fig. 6.1
n-l
n
x
n+1
Finite-difference grids.
In terms of un and ax, the values of un+) and un-) can be written in series form as U = U + ( du ) ax n+ 1 n dx n
2U) (ax)2 dx2 n 2!
+( d
+ ...
(2.1)
2U) (ax )2 _ . . . dx 2 n 2!
(2.2)
and U n-)
= U _ ( du ) ax + ( d n
dx n
From Eqs. (2.1) and (2.2), the finite-difference form of the derivatives can be obtained. From Eq. (2.1), we get
(~~ t
=
Un+~: Un
- (
~~~
t ~~ -...
or
(2.3)
95
6.2 Finite Dlnerences
Similarly, from Eq. (2.2), we get 2 du ) = un - un- 1 + ( d U) ~x _ ... ( dx n ~x dx? n 2!
or (
~~ ) n~
un
~=n-l
(2.4)
Equations (2.3) and (2.4) are known as the foward difference and the backward difference, respectively. In both forms, the first term which is truncated is (dluj dx2)n(~xj2!). Since this term contains ~x to its first power, the truncation error is first order. To get the second-order correct finite-difference form of the first-order derivative, we subtract Eq. (2.2) from Eq. (2.1). Upon rearranging, we get du ) un+ 1- un- 1 ( dx n= 2(~x)
_ ( d 3U) (~X)2 + ... dx 3 n
6
or
(2.5) which is called the central difference. Since the first term truncated involves (~X)2, the truncation error is second order. To get the finite difference analog of the second-order derivative, Eq. (2.2) is added to Eq. (2.1) and the resulting equation is solved for the second-order derivative:
or
(2.6) which is again correct to second order. Similarly, the third-order derivative is given by
(2.7) A simple example will be given in the next section to outline the idea of solving boundary value problems by finite-difference methods.
96
6. Iterative Methods-The Finite-Difference Method
6.3 SOLUTION OF BOUNDARY VALUE PROBLEMS BY FINITE DIFFERENCE
In this section, the idea of solving boundary value problems by the finite-difference method will be introduced through a simple example. Consider again the heat transfer through a fin, as shown in Fig. 3.5, Section 3.3.3, with the exception that the surface at x = L is insulated. An energy balance then gives the differential equation for the solution of the dimensionless temperature distribution,
d2() / dX2
= A8
(3.1 )
subject to the boundary conditions
x = 0:
0(0)
= 1;
We will now divide the range of pivot points defined by
x= I:
x
dO(I)/dx=O
(3.2)
(from 0 to 1) into N intervals with
Xo = 0,
n
= 1, ... , N
°
To replace Eq. (3.1) by finite differences, we replace at xn by On and its second-order derivative by the finite-difference representation Eq. (2.6). Equation (3.1) then becomes
(On+l - 20n + 0n_l)/(t:.x)2= AOn or
0n-l For n
[2 + A(t:.X)2]On + 0n+l = 0
= I, Eq. (3.3) becomes 00 -
[2 + A(t:.X)2]01 + 02 = 0
From the first boundary condition, 00
= N,
(3.4)
= 1, Eq. (3.4) therefore gives
- [2 + A(t:.X)2]01 + O2 = -I For n
(3.3)
(3.5)
Eq. (3.3) gives
0N-l -
[2 + A(t:.X)2]ON + 0N+l = 0
(3.6)
The second boundary condition in Eq. (3.2) indicates that a good approximation is
0N-l = 0N+l Equation (3.6) therefore becomes
20N- 1-
[2 + A(t:.X)2]ON = 0
(3.7)
6.3 Solution of Boundary Value Problems by Finite Dlnerence
97
°°
Equations (3.5), (3.7) and Eq. (3.3) for n = 2,3, ... , N - 1 are N equations for the solutions of N unknowns I' 2 , ••. , ON' In matrix-vector form, this system of equations can be written as
[A]8 = s
(3.8)
where
8=
a 1
01
SI
O2
S2
03
s=
s3
(3.9 )
ON-I
SN_I
ON
SN
a 1
a
[A] =
(3.10) a 2
a
with for
n = 2,3, ... , N
(3.11)
and (3.12) The coefficient matrix [A] is known as a tridiagonal matrix due to the fact that all elements of [A] are zero except those three along the diagonal. To find the vector 8, the problem involved is the inversion of the coefficient matrix [A], i.e., 8 =[Arls
Two commonly known methods of matrix inversion are Gauss-Seidel iteration and the relaxation method. While such methods are standard for those problems where the number of intervals N is small, they require a large memory space in the digital computer and too many unnecessary computations when the number N of intervals is large. Such is thecase for most problems in engineering and science where the independent variable
98
6. Iterative Methods-The Finite-Difference Method
has to be divided into a large number of intervals in order to obtain accurate solutions. For this reason, these two methods will not be treated here. Instead, a very effective factorization scheme will be given which makes use of the tridiagonal nature of the coefficient matrix [A]. It requires the minimum number of computer memory spaces and no unnecessary computations. The method will be given in the next two sections with complete details of the derivation included so that the reader can use the final equations meaningfully. It will also enable the reader to derive the necessary sequence of equations for new problems. The method is due to Thomas [1] and Keller [2,3]. 6.4 SECOND·ORDER DIFFERENTIAL EQUATIONS 6.4.1 Linear Differential Equations
Consider the second-order linear differential equation
d~ dx
+A(x) ddY +B(x)y = C(x) x
(4.1)
subject to the boundary conditions
y(O) =
y(L) = 8
lX,
Equation (4.1) will now be written in finite-difference form with the grid points defined by
n = 1,2, ... , N where N is the total number of intervals and X N = L. The variable y and its derivatives at x n of the net are given by y= Yn dy dx
=
Yn+1 - Yn-I 2h
Yn+1 - 2Yn + Yn-I -dy2 = -'--------'----'-2 dx
h
Equation (4.1) and its boundary conditions therefore become 1
h2 (Yn+1 - 2Yn + Yn-I) +
A(~
----v;- (Yn+1 -
Yn-I) + B(X)Yn = C(x)
or
anYn-1 + bnYn + cnYn+ 1= 'n
(4.2)
99
6.4 Second-Order Differential Equations
where an
cn
= I - thA(xn ) ,
b; = h 2B(xn ) -1,
= I + t hA (xn ) ,
rn
= h 2C( x n )
The boundary conditions are written as Yo= a,
YN
=8
(4.3)
In matrix-vector form, Eqs. (4.2) and (4.3) can be written as Ay =
(4.4)
$
where
y=
YI
SI
Y2
S2
$=
rJ -
r N- J -
SN-J
YN-J
aa J r2
8cN -
J
and bJ
C1
az
b2
c2
.A= a N- 2
bN -
2
CN- 2
aN-J
bN _ 1
The matrix A whose elements are nonzero only on the diagonal of the two adjacent codiagonals is known as a tridiagonal matrix. A very efficient factorization procedure can be used to solve Eq. (4.4). We assume that A is nonsingular and can be factored into A=LU
(4.5)
where
1 a N- 2
f3N-2 a N- J
f3N-J
100
6.
Iterative Methods-The Finite-Difference Method
and
U=
(4.7) YN-2
I
The unknowns (f3n'Y n)' n = I, ... , N - I, are therefore found from Eq. (4.5) to be related through the equations
= b, f3 1YI = c 1 f31
n = 2,3, ... , N - I n
= 2,3,
(4.8)
... , N - 2
In terms of Land U, Eq. (4.4) becomes
LUy= s
(4.9)
Uy= z
(4.10)
Lz= s
(4.11)
Let Equation (4.9) can be written as or
(4.12)
The unknown elements of z are therefore found n=2,3, ... ,N-I
(4.13)
Once z is known, Eq. (4.10) can be solved for y since U is already
101
6.4 Second-Order Dlnerentlal Equations
known. Writing Eq. (4.10) in expanded form,
(4.14) YN-2
YN-2
1
YN-l
we then get
n = N - 2,N - 3, ... ,3,2,1
Yn = zn - YnYn+l'
(4.15)
which are the solutions of Eq. (4.4). As a summary, the solution of Eq. (4.1) involves the following steps.
1. Reduce the given differential equation to its corresponding finitedifference form. 2. Compare with Eq. (4.2) to identify an' bn, en' and Tn' 3. Calculate fin and Yn (for n = 1,2, ... , N - 1) from Eq. (4.8). 4. Calculate Zn (for n = 1,2, ... , N - 1) from Eq. (4.13). 5. Calculate Yn (for n = N - 1 N - 2, ... , 3,2,1) from Eq. (4.15), which are the required solutions. 6.4.2 Conduction through Fins
Consider the fin discussed in Section 6.3. The energy equation and the boundary conditions are (4.16)
and
x=O:
0(0)= 1;
x = 1: dO(l)/dx = 0
In terms of Eq. (4.1), we find
A(x) = C(x) = 0
B(x) = -"A
and
Also, from the boundary conditions, a
= 1,
L
=1
and or
102
6.
Iterative Methods-The Flnlte-DIUerence Method
we therefore have
(2'-; n .-; N - 1) (1 .-; n
«N
- 2)
(2'-; n .-; N - 1) With these quantities identified, the steps discussed earlier can be followed to find the finite-difference solutions of Eq. (4.16) for different values of A's. The results are tabulated in Table 6.1. TABLE 6.1 Solution of Eq. (4.16) for Two A's A 2.0
X
(J
0.0 0.2 0.4 0.6 0.8 1.0
1.0000 0.7175 0.4924 0.3068 0.1457 0.0037
A
x
(J
6.0
0.0 0.2 0.4 0.6 0.8 1.0
1.0000 0.6050 0.3579 0.1984 0.0874 0.0022
6.4.3 Nonlinear DIfferentIal EquatIons
For nonlinear differential equations, one way to obtain the solution is to linearize the differential equation before it is written in finite-difference form. An effective method is the method of quasi linearization outlined in Section 5.4. Starting from a first approximation, higher-order iterates can be obtained by solving the linearized equation. Two examples will be given in the next two sections to illustrate the details of the method. 6.4.4 The Pendulum
Consider the pendulum, shown in Fig. 6.2, which consists of a concentrated mass m suspended by a weightless cord of length I. The force of gravity has two components. The radial component is balanced by the tension in the cord, whereas the tangential component serves as the driving force for the vibrational motion. From Newton's second law of motion, we write (4.17)
6.4 Second-Order Differential Equations
103
I0 I I
I I
I I
I I
I
I s I-- - ...; "\ I
\0
I
I I I Fig. 6.2
mg
Schematic diagram of the pendulum.
Replacing s by I(), Eq. (4.17) becomes:
-d
2()
dt 2
g. + -sm()=O
(4.18)
I
Suppose the positions of the pendulum are known at two instants of time. Consider, for example, the case in which these two positions are t = 0:
() = 0;
t = tt=
() =
()f
The problem is to find () as a function of time t. Let us first introduce the dimensionless time
T=t/~g/I Equation (4.18) and its boundary conditions then become
. () = 0 -dZO2 + SIn dT
(4.19)
subject to the boundary conditions T
= 0: () = 0;
(4.20)
Based on the method of quasi linearization, the iteration equation relating successive iterates can be obtained as follows. Let
f= f(T,(),()")
=()" + sin() = 0
(4.21)
6. Iterative Methods-The Flnlte-DIUerence Method
104
where the notation 0" represents d 20jdr 2 • Differentiating Eq. (4.21) according to
where the superscript i represents the number of iteration, we get (O"f+ I)+ (cos 0 (i»)0 (i+I)
= O(i) cosO(i) - sinO(i)
(4.22)
The boundary conditions are r
= 0: 0(i+1) = 0;
Comparing this with Eq. (4.1) and its boundary conditions, we find
C(r) = O(i) cosO(i) - sinO(i)
B ( r) = cos 0 (i),
A(r) = 0,
IX
(4.23)
= 0,
The solution of Eq. (4.22) therefore follows the same steps as outlined in Section 6.4.1. To start the solution, a first approximation of the solution of Eq. (4.19) is assumed. One natural choice is 0(1) = 1 The solution of the second iteration, 0(2), can be found from Eq. (4.22). For the next iteration, the solutions 0(2) are substituted in Eq. (4.23) for a new set of the coefficients, and the solutions of 0 (3) are then obtained, again from Eq. (4.22). The process can be continued until the desired accuracy is reached. Numerical solutions of Eq. (4.19) are obtained for the case of If = 0.5 sec and Of = 15° and are tabulated in Table 6.2. TABLE 6.2
Angular Displacement of the Pendulum, 9 9 (i) T
i= I
i=2
i=3
i= 4
0.0 0.1 0.2 0.3 0.4 0.5
1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
0.0000 0.0596 0.1159 0.1686 0.2173 0.2618
0.0000 0.0545 0.1085 0.1614 0.2127 0.2618
0.0000 0.0545 0.1085 0.1614 0.2127 0.2618
105
6.4 Second-Order Differential Equations
6.4.5
Natural Convection Flow of Powell-Eyring Fluids between Vertical Flat Plates*
The natural convection behavior of pseudoplastic fluids is of interest to engineers working on the fluid mechanics of non-Newtonian fluids. A pseudoplastic fluid which obeys the Powell-Eyring model flows between two vertical parallel plates as a result of the bouyancy force, as shown in Fig. 6.3.
I
I I
I yl
~x
x=-/
I
x=/
Fig. 6.3 The flow system.
The governing differential equations, which consist of the equations of momentum and energy, can be written as Momentum:
ds / dx
+ peg(T -
Energy:
T m) = 0
(4.24) (4.25)
where p, e, g, T, and T m are, respectively, the density of the fluid, the coefficient of volume expansion, the gravitational acceleration, the temperature of the fluid, and the mean temperature of the fluid. The shearing stress 7' is related to the velocity gradient through the equation 7'
- I -1 dv = fL -dv + -l'nh Sl dx
B
cdx
(4.26)
where fL, B, and c are constants. Equation (4.25) gives T = Tm * See Bruce and Na [4).
-
t( T 2 - T 1) ( x/I)
(4.27)
6. Iterative Methods-The Finite-Difference Method
106
Combining Eqs. (4.24), (4.26), and (4.27) and introducing the dimensionless variables
x* = ~
v* =
/ '
v
aNgrN pr/ /
we finally get
x*
(4.28)
where N gr t:
= aNgrN pr/ el 2 ,
= gp2e/ 3( T 2 ~
= p,Bc,
a
T m)/ p,2
= k/pc p
The boundary conditions are:
°
x* = - I: v* = 0; x* = 0: v* = which refer to the velocities at the points x = -/ and x = 0, respectively. Introducing s = x* + I, Eq. (4.28) and its boundary conditions become
(s - I)
1+[I/~~(t:dv*/ds)2+ with the boundary conditions
v*(O) = 0,
v*(I) =
(4.29)
I]
°
Linearizing Eq. (4.29), we get
t:2~(S - 1)(dv*(I)Ids) ]( dv* cj>(~cj> + 1)2 ds
d2v*(i+I) _ [
ds2
= ~(s - I)cj> _ t:2~(S - I) ( dv*(i) )2 (I + ~cj»
cj>(~cj>
+ 1)2
ds
)(i+I)
(4.30)
subject to the boundary conditions V*(i+
1)(0) = 0,
where
dv*(i»)2 ]1/2 cj>= [( t : - - +1 ds
(4.31)
6.5
107
Third-Order Differential Equations
Comparing Eq. (4.30) with Eq. (4.1), we can identify
A(s) = -
[
E:2~(S - 1) dv *(i) cj>(~cj> + 1)2 ~
]
B(s) = 0 C(s) =
(4.32)
~(s - l)cj> _ E:2~(S - I) ( (I + ~cj» cj>(~cj> + 1)2
dV*(i)
)2
ds
The steps that should be followed in solving Eq. (4.30) are therefore the same as those used in Section 6.4.3. As a numerical example, consider the solution of Eq. (4.30) for E: = 0.01, ~ = 10. By assuming a first approximation, v(l)
= sin ex
the solution, accurate to the fourth digit, is obtained in three iterations. The results are shown in Table 6.3. TABLE 6.3 Solutions of Eq. (4.29) for e = om,
s 0.0 0.2 0.4 0.6 0.8 1.0
i
=I
0.0000 0.5878 0.9511 0.9511 0.5878 0.0000
g = 10
i = 2
i=3
0.0000 0.0436 0.0582 0.0509 0.0291 0.0000
0.0000 0.0437 0.0582 0.0509 0.0291 0.0000
6.5 THIRD-ORDER DIFFERENTIAL EQUATIONS 6.5.1
Linear Differential Equations
Consider the third-order linear differential equation
dy
dx3
dy
dy
+ A (x) dx2 + B(x) dx + C(x)y = D(x)
(5.1)
subject to the boundary conditions
yeO)
= LX,
dy(O)/ dx = E:,
dy(L)/dx
=A
(5.2)
108
6. Iterallve Methods-The Finite-Difference Method
Equation (5.1) will now be written in finite-difference form with the grid points defined by n
Xo= 0,
= 1,2, ... , N
where N is the total number of intervals and X n = L. The variable Y and its derivatives at x n of the net are given by
Y = Yn dy Yn+1 - Yn-l 2h dx ~ dy Yn+1 - 2Yn + Yn-I -=------dx 2 h2
(5.3)
d) Yn+2 - 2Yn+ 1 + 2Yn-1 - Yn-2 2h3 dx 3 ~ Equation (5.1) and its boundary conditions therefore become
+ 2Yn-1 - Yn_2)/2h 3 + A(X)(Yn+1 - 2Yn + Yn_I)/h 2 +B(X)(Yn+1 - Yn_I)/2h + C(x)Yn = D(x)
(Yn+2 - 2Yn+1 or
Yn+2
+ [2M (x) + h2B(x) -
2 ]Yn+ I
+ 2h[ h2C(x) - 2A (x) ]Yn
+ [2 + 2hA(x) - h2B(x) ]Yn-I - Yn-2 = D(x)
(5.4)
The boundary conditions become
Yo
= £X,
(YI - Y _1)/2h
= e,
(YN+ 1 - YN_I)/2h
=A
(5.5)
By putting n = 0, I, ... , N in Eq. (5.4), we get N + I algebraic equations for the solutions of N + I unknowns Yn (n = 0, I, ... , N). However, the finite-difference form of the third-order derivative d) / dx 3 involves five adjacent values of Yn' For n = 0, two "unwanted" values of Yn' namely, Y - I and Y -2' will appear in the equation. Similarly, for n = N, another two "unwanted" values of Yn' namely YN+I and YN+2' will appear in the equation. For the determination of these four additional unknowns, only three equations can be written based on the three boundary conditions. Numerical oscillation will occur if a fourth equation is not properly chosen. Fox [5] suggested that the difficulty can be overcome by replacing Eq. (5.1) by two equations, dy d 2p dp dx = p, dx2 = - A (x) dx - B(x)p - C(x)y + D(x) (5.6)
6.5 Third-Order Differential Equations
109
subject to the boundary conditions
y(O) = a,
p(L) ="A
p(O) = e,
(5.7)
Replacing these equations by the finite difference form, we get
(Yn - Yn-I)/h = t(Pn + Pn-I) (Pn+1 - 2pn + Pn_I)/h 2 = -A(X)(Pn+1 - Pn_I)/2h - B(x)Pn - C(x)Yn + D(x) or
Yn-I + thpn-I - Yn + thpn =
°
a"yn + bnPn-I + cnPn + dnPn+1 = h
(5.8)
2D(x
n)
(5.9)
where
an = h2C(xn),
t hA (x n),
b; = 1 -
cn = h2B(xn) - 2, d; = 1 + thA(xn) Here, the first equation is replaced by the center-difference equation around the point X n - I/ 2• It will be seen later that this will reduce the "unwanted" values to three, instead of four, so that the number of unknowns is equal to the number of equations available. The boundary conditions can be written as Yo=a, For n
=
Po=€,
PN="A
(5.10)
0, Eqs. (5.8) and (5.9) become
Y - I + t hp - I
-
Yo + t hpo = 0
aoYo + bop_I + coPo + doPI = h
2D(x
o)
or, since Yo = a, and Po = e, we get
Y_I + thP_1 = a - th€ bop-I + doPI = -aoa Similarly, for n
(5.11) Co€
+ h2D(xo)
(5.12)
= 1, Eqs. (5.8) and (5.9) give - YI + t hPI = - a alYI
t he
+ ClPI + d l P 2 = - b l € + h 2D (x l )
(5.13) (5.14)
For 2 ,.; n ,.; N - 1, Eqs. (5.8) and (5.9) can be used without any change.
6. Iterative Methods-The Finite-Difference Method
110
For n = N, Eqs. (5.8) and (5.9) give (5.15) (5.16) where the boundary condition PN = A has been taken into consideration. The above steps eliminate Yo' Po' and PN as variables. We therefore have 2N + 2 equations for the solution of 2N + 2 variables, namely n = 1,2,3, ... , N - 1
P-I'
Y-I'
In matrix-vector form, these 2N + 2 equations can be written as (5.17)
A~=t
where
-I]
[Y P-I
[;: ] [;~ ]
~=
[ YN-I ] PN-I
[~:I ]
t=
[;~ ] [:: ] [:~ ] [ rN- 1 ] SN_I
[;: ]
with ro = LX
-
th€,
(5.18) for 2".; n ".; N - 1 (5.19) for
and A is given by Eq. (5.20) on page Ill.
2".; n ".; N - 1
111
6.5 Third-Order Differential Equations
.-------. 0
-s I
00 '-----'
,-------,
-e
-IN
-e
-IN
I
.----------,
..s:::
I
'
..r::; 0
0""; 00 '-----'
.-------. O~ 00 '-----'
..s:::
-IN
0
..r::; 0
..s:::
-IN
C' r;j
,....-----, 00
00
'----------J
II
J '-----'
.----------,
I
J
r;j'
-IN
o~
..r::;'
'----------J
6. Iterative Methods-The Flnlte-Dlnerence Method
112
The above suggests that the vectors written as
l)=
l)o l)1
l) and I and the matrix A can be
to II
t=
(5.21)
IN
l)N
and [ B o]
[Co]
[ AI]
[B I]
[CI ]
[ A 2]
[ B 2]
[ C2]
A= [AN-I]
[BN-
I ] [CN-I] [AN] [BN]
(5.22)
where
tb h ]' o
iNl [Al]=[~ ~l
[Bn] = [ ~nI
tcnh ]'
I
[Cn] = [~
1l
O
[An] = [ ~
tb h ]'
2
n
so =[Y-I] P_I'
s, = [ ~: l
1
l)N=[~:ll
In =
0< n
c
[ ::],
(5.23) - 1,
N
In this form, Eq. (5.17) will be in exactly the same form as Eq. (4.4) in
113
6.5 Third-Order Differential Equations
Section 6.4. Again, the matrix A is factorized into (5.24)
A=LU
where
(5.25)
L=
[AN-d
[,8N-I]
[AN]
[,8N]
and
(5.26)
U=
where
[,8n]=[(,8n)ll (,8n )21 [Yn] = [(Yn)ll (Yn)21
(,8n)12]
(,8n )22 ' (Yn)12] (Yn)22'
O
O
The unknown matrices [,8n] and [Yn]' n = 0, ... , N, are therefore found from Eq. (4.8) to be related through the equations
[ ,80] = [ Bo]
(5.28)
[ ,80][Yo] = [Co]
(5.29)
[Bn] - [An][ yn-d, n = 1,2, ... , N [,8n][ Yn ] = [Cn], n = 1,2, ... , N - 1
(5.30)
[,8n]
=
(5.31)
6. Iterative Methods-The Finite-Difference Method
114
From Eq. (5.28), we get
({30)\I [ ( {30)21 which gives
({30)\I = 1, ({30)21 = 0,
({30)12 = th ({30)22 = bo
(5.32)
From Eq. (5.29),
( {30)\I [ ( {30)21 from which
(5.33)
From Eq. (5.30), for n
= 1,
we get
({3I)\I = - 1, ({31)21 = al> From Eq. (5.31), for n
({3I)\I [ ({31)21
({3I)12=t h ({31)22 = c i
(5.34)
= 1,
({31)12 ] [ (YI)\I ({31)22 (YI)21
from which
For 2
(YI)\I = 0,
(YI)12 = hdl/(a1h + 2c l)
(YI)21 = 0,
(r1)22 = d l / ( talh + c 1)
1,
th] _ [ 1 cn
°
(5.35)
6.5 Thlrd·Order Dlfterentlal Equations
115
from which, we get
(f3n)11 = -1 - [(Yn-I)ll + th(Yn-l)21] (f3n)12 = th - [(Yn-l)12 + t h(Yn-I)22] (f3n)21 = an - [bn(Yn- I)21] (f3n)22 = en - [bn(Yn- I)22] From Eq. (5.31), for 2
~
n
~
(5.36)
(2 c n ~ N - I)
N - 1,
from which we get
(Yn)11 = 0, (Yn)21 = 0,
(Yn)12 = -dn( f3n) 12 / [( f3n )I1 ( f3n )22- (f3n)21(f3n)l2] (Yn)22 = dn(f3n)I1/[(f3n)I1(f3n)22 - (f3n)21(f3n)l2] (for 2
For n
= N,
~
n
~
(5.37)
N - 1)
Eq. (5.30) gives
which gives
(f3N)11 = -I - [(YN-l)11 + t h(YN-l)21] (f3N)12 = - [(YN-l)12 + t h (YN- l)22] (f3N )21 = aN - bN(YN- I)21
(5.38)
(f3N )22 = dN - bN(YN- l)22 Equations (5.32)-{5.38) are those necessary for the evaluation of the elements of the matrices [f3n] and [Yn]. In a manner similar to that of Section 6.4, we can write Eq. (5.17) as
=t
(5.39)
U~ =Z
(5.40)
LU~
If we now define
116
6. Iterative Methods-The Finite-Difference Method
where
Z=
(5.41)
Eq. (5.39) becomes
Lz or
=I
(5.42)
[AN-I] [f3N-I] [AN]
[f3N]
ZN_I
IN-I
ZN
IN
which gives
[ f3o]Zo = 10
(5.43)
[f3n]Zn = In - [An]Zn-1
(5.44)
From Eq. (5.43),
o '0 1 [ oI tbOh ] [ wV o ] = [ So from which (5.45) From Eq. (5.44), for n
= 1,
( f3 \)11 [ ( f3\)21 from which
(f3I)lIVI + (f31)12 wl ='1 (f3\)21 Vl + (f31)22 WI = Sl
6.5 Third-Order Differential Equations
117
we then get V1
=
For 2
r l ( f3')22 - SI( f3')'2
,
(f3')1I(f31)22 - (f3,b(f3')'2
< n < N,
we write
from which
+ (f3n)12 Wn = rn - Vn-, - thwn_, (f3n)2,Vn + (f3n)22 Wn = Sn - bnwn- I
(f3n)lI Vn
The solution of these equations is
1 =1
vn =
wn
n n
~ Wn_,) -
{(f3n)22(rn - vn- , -
(f3n)12(sn - bnWn_,)}
{(f3n)lI(Sn - bnwn_,) - (f3n)2,(rn - vn- , -
~ wn-,)}
(5.47)
s, = (f3n )11( f3n )22 -
(f3n )21( f3 n )12 Once Z is found from Eq. (5.42), we can return to Eq. (5.40) to find the final solution. Based on Eq. (5.40),
(5.48)
U«'j=Z
or
«'jo «'j,
[IJ [YoJ [IJ
[Y' J
ZO ZI
(5.49)
[1J
[YN-d
[IJ
«'jN-' «'jN
ZN-' ZN
from which «'jN
=
ZN
(5.50)
«'jn
=
z, - [ynJ«'jn+l
(5.51)
118
6. Iterative Methods-The Flnlte·Dlfference Method
From Eq. (5.50),
we therefore get (5.52)
From Eq. (5.51), Yn ] [ r,
= [ Vn ] wn
_ [ ('Yn)ll (Yn)21
from which we get Yn = vn - (Yn)llYn+l - (Yn)12Pn+l P« = wn - (Yn)2 1Yn+l - (Yn)22Pn+1 To illustrate, let us consider the linear boundary value problem
dy
-
dx
3
dy
+4-
dx
2
dy - 6y = I dx
+ -
(5.53)
(5.54)
subject to the boundary conditions dy(O) dx For this problem, we can identify y(O) = 0,
A(x) = 4,
--=0
'
B(x) = I,
dy(l) dx
--=1
C(x) = -6,
D(x) = I,
ex = e = 0, A= I, L= I The quantities in Eq. (5.17) can therefore be written as
-6h 2 ,
an
=
Cn
= h2 - 2,
bn
= 1- 2h,
d; = I
+ 2h (for 0 "
n " N)
and rn
= 0,
s; = h
2
(for 0 " n " N - I)
For n = N, we have
rN=-th,
sN=2
Once these coefficients are identified, the steps for the solution of Eq. (5.54) are the following.
1. Calculate ({3n)ll' ({3n)12' ({3nhl' ({3nb, (Yn)ll' (Yn)12' (Ynhl' and (ynb by using Eqs. (5.32)-(5.38), for 0 " n " N.
&.5 Thlrd·Order Differential Equations
119
2. Calculate V n and W n by using Eqs. (5.45)-(5.47), for 0 ..;; n ..;; N. 3. Calculate Yn and Pn by using Eqs. (5.52) and (5.53) for 0 ..;; n c N. Step 3 gives the solution of Eq. (5.54) at the pivot points. Details of the numerical calculations are left to the student as an exercise at the end of the chapter. 6.5.2 Nonlinear Differential Equations
As in Section 6.4.3, the equation is linearized before it is written in finite-difference form. An effective method is the method of quasi linearization outlined in Section 5.4. Starting from a first approximation, higherorder iterations can be obtained by solving the linearized equation. The method will now be illustrated by applying it to two examples in engineering. 6.5.3 Hlemenz Magnetic Flow
Flows in which the velocity of the incoming fluid is perpendicular to a plane surface is known as Hiemenz flow [7]. If, in addition, the fluid is electrically conducting, the flow is then called Hiemenz magnetic flow. The solution of this problem is of interest since it is one of the few exact solutions of Navier-Stokes equations in magnetohydrodynamics. The Navier-Stokes equations for such flows are [6] Continuity (5.55) Momentum
au + v -a au Y
u -a x
2
aZu
2
= a x + V - 2 + paB (ax - u)
ay
(5.56)
subject to the boundary conditions y =0: y = 00:
u = 0, u = ax
v=0
where u, v, v, p, and a are, respectively, the x component of the velocity, the y component of the velocity, the viscosity, the density, and a constant characteristic of the incoming flow. Also, a and B are, respectively, the electrical conductivity and the magnetic induction. The first two boundary conditions are based on the physical observations that there is neither slip nor mass transfer on the surface, whereas the boundary condition at infinity means that the velocity of the fluid approaches a linear relation with x.
6. Iterative Methods-The Finite-Difference Method
120
Hiemenz [7] separated the variables by assuming dF u=axdq
and
(5.57)
where
T/=,fa/vy
(5.58)
It can be shown that Eq. (5.55) will be satisfied identically and Eq. (5.56) becomes 3F 2F d +F d + dF + M(I _ dF) = (5.59) dT/3 dT/2 dn dq
1-(
0
)2
subject to the boundary conditions F = 0, dF/ dn = 0 dF/ dT/ = I where the dimensionless constant M is defined by T/
T/
= 0: = 00:
M = aB 2/ap
For M equal to zero, the problem is reduced to the case of flow of Newtonian fluids. The solution of this case is known as Hiemenz flow. To solve Eq. (5.59) by the finite-difference method, we first linearize it by the method of quasi linearization. Thus let us denote Eq. (5.59) by G, i.e., 3F d 2F G= d +F +1_(dF)2+ M dF)=O dT/ 3 dT/2 dT/ dT/
(I_
For convenience, we write G = F''' + FF"
+ I - (F')2+ M(I - F') = 0
(5.60)
where the primes represent differentiation with respect to T/. By following the linearization scheme
( o~~"
ri)[
(F"')U+I) - (F"f)]
+(
00::',
r)[
(F"t+ 1) - (F,,)(i)]
+( ;:' r)[(F')U+I)- (F')U)] +(~~ r)[(F)U+l)- (F)(i)] =0 (5.61)
(5.62)
6.5 Third-Order Differential Equations
121
where the superscripts in parentheses are the numbers of iteration, subject to the boundary conditions
arv:» /d'l'/ = 0
'1'/
= 0:
F(i+I)
'1'/
=
dF(i+I) /d'l'/ = 1
00:
= 0,
Equation (5.62) can now be written in the form of Eq. (5.6), as
('+ I)
dFd'l'/I d7(i+l) ---=--d'l'/2
+ (F(i»
dp(i+I) - - - (2P
(5.63) ( dp(i) )
+ M)p(i+I) + _
d'l'/
F(i+I)
. dP
= F(l)
(5.64)
In finite-difference form, Eqs. (5.63) and (5.64) become Fn(~il)
+ (th)p~~+II) -
F;i+l)
[~h(p~21 - P~~I) ]Fy+I)
- [(2p~i)
+ (th)p~i+l) = 0
+ [1 -
(5.65)
thFY)]p~~+II)
+ M)h 2 + 2 ]p~i+ I) + [1 + ~hFY)]p~~+/)
= IhF(i)(p(i) 2 n n+1
2(p(i»)2 2 2M _p(i») n-I _ h n _ h - h
(5.66)
From this equation on, the steps outlined in Section 6.5.1 between Eqs. (5.8) and (5.53) can be followed. By taking the first approximation to be the solution of Eq. (5.59) for M = 0, solutions of Eq. (5.59) for other values of M are obtained by this method. The results are summarized in Table 6.4, where the missing initial slopes are tabulated. It is seen clearly that it takes only a few iterations before an accuracy of up to the 4th digit is reached. TABLE 6.4 Selected Solutions of Eq. (5.59) for Different Values of M Number of iterations
I
Q
2
3 4
5 6 Q
Missing intital slope, d 2F(0)j drj
M=0.05
M = 0.1
M = 0.5
M = 1.0
M=2.0
1.2326 1.1688 1.1743 1.1715 1.1717 1.1717
1.2326 1.1916 1.1973 1.1947 1.1949 1.1948
1.2326 1.3593 1.3637 1.3618 1.3621 1.3620
1.2326 1.5393 1.5407 1.5392 1.5394 1.5394
1.2326 1.8385 1.8339 1.8328 1.8330 1.8330
Initial approximation.
6. Iterative Methods-The Finite-Difference Method
122
6.5.4
Hamel's Problem: Flow In a Wedge-Shaped Region
Hamel's problem [8] is one of the very few exact solutions of the Navier-Stokes equations in fluid dynamics. Consider an incompressible fluid that flows radially outward in a trough between two nonparallel plane walls, as shown in Fig. 6.4. It is desired to find the velocity distribution at any radial location r. For such flows, it is expected that the velocity components in both 0 and Z directions, VIJ and vz' are zero. The Navier-Stokes equations then become
Fig. 6.4 Hamel flow.
Continuity (5.67) Momentum PVr
OVr op a;:- = - a;:- +
0= _
p. 02v r 00 2
1. op + r
(5.68)
r2
00
2p. r2
oV
r
00
(5 69) .
Equation (5.67) means that the product ro, is independent of r. For two-dimensional flows, this condition implies that ro, = vF( 0)
(5.70)
where the right hand side is multiplied by the viscosity v so that F(O) is dimensionless. Substituting this form of o, into the momentum equations, we get (5.71) (5.72) The pressure p can be eliminated by differentiating Eq. (5.71) with respect to 0 and Eq. (5.72) with respect to r and then taking their
123
6.5 Third-Order Differential Equations
difference. We then get
3F d + 2F dF + 4 dF dO 3 dO dO
=0
(5.73)
The boundary conditions are (no slip on wall) (symmetry)
0= w: 0=0: and from Eq. (5.68)
ap + -J.L -av = 0 a, ,2 ao 2 2
o= w:
- -
r
In terms of F(O), the boundary conditions become
0= w: 0=0:
F(w) = 0, dF(O)/ dO
(5.74)
=0
We now introduce the change of variable ep = which becomes 3F d + (2F + 4) dF = 0 dep3 dep
w -
0 into Eq. (5.73), (5.75)
subject to the boundary conditions
ep = 0: ep = w:
d 2F(0)/ dep2
F(O) = 0, dF(w)/ dep
=-
K
=0
Since Eq. (5.75) is nonlinear, it has to be linearized before the factorization scheme outlined in Section 6.5.1 can be applied. Let us introduce the iterates dF(i) / dep, etc. For higher-order iterates, we have
r»,
(
d 3F(i) ) dep3
d3F(i+ I) dep3
.
- - - - - - + (2F(I) + 4)
( dF(i+ I) dF(i) ) - -dep dep
+ ( 2 d~epi ) ) (F(i+l) - F(i» = 0 or
d 3F (i+I) + (2F(i) + 4) dF(i+ I) + (dF(i») .:.:......::._2 - - F(i+ I) dep3
dep
dep
= _ ( 2F(i) dF i ) + 4 dF (i) ) + (2F(i) + 4) dF (i) + ( 2 dF (i) ) r» dep
dep
dep
dep
6.
124
Iterative Methods-The Finite-DIfference Method
or + (2F(i)
+ 4)
+ ( 2 dF (i)
(i+ I)
dF dep
dep
)
F(i+ I)
= 2F(i)
(i)
dF dep
(5.76) Equation (5.76) can now be written in terms of the following system of differential equations:
r"
dF(i+I) _-,----_ = dep
(5.77)
I)
Replacing by finite differences as before, we get F(i+I) - F(i+I) - lh(p(i+l) +p(i+I») n
n-I
2
n
n-I
=0
h- 2(p(i+l) _ 2p(i+l) +p(i+I») = -(2F(i) n+ 1
n
n
n-I
+ 4)p(i+I) n
or F(i+I) n-I
a F(i+ I) n n
+ lhn(i+I) 2 Tn-I
F(i+I) n
+ lhn(i+I) =0 2 Tn
+ bnp(i+ I) + C p(i+ I) + d p(i+ I) n-l n n n n+ 1
(5.79)
= 2h 2F(i)n(i) n r n
(5.80)
where an
=
For n
= 2h 2(F(i) n + 2)
- 2,
d; = 1 (5.81)
= 0, Eqs. (5.79) and (5.80) give rv: I) + 1 hn(i+ I) _ F.(i+ I) + 1 hn(i+ I) = 0 -I 2 T -I 0 2 TO
(5.82)
-2h 2p(i) n'
rs: 0
ao
bn- -1'
I)
cn
+ b0 p(i+ I) + C p(i+ I) + d. p(i+ I) = 2h 2F.(i)n(i) -I 0 0 0 1 0 rO
From the boundary conditions at ep FJi+l)
(5.83)
= 0,
= 0,
Equations (5.82) and (5.83) therefore become F(i+I) -I
( bo + c0 )p(i+l) -I
2K + hn(i+I) = lh T -I 2
+ d.0 p(i+l) 1 -
c0 Kh
(5.84) (5.85)
125
&.5 Third-Order Differential Equations
Similarly, for n = 1, -Fl i+ l) aIF?+I)
+ hp(i+l) = -
-!hzK
(5.86)
+ ib, + CI)p(i+l) + dIP!i+l) = 2h zFl i>P(i) -
blhK
(5.87)
For 2 " n " N - 1, Eqs. (5.79) and (5.80) can be applied without any change. For n = N, Eqs. (5.79) and (5.80) give F(i+I) N-I
a F(i+ I) N
N
+ Ihip(i+I) 2 N-I
=-
F(i+I) N
+ bN p(i+ I) + d p(i+ I) N-I N N+I -
Ih
(5.88)
2
2h ZF(i) N
cN
(5.89)
where the boundary condition at cp = w, namely PN = 0, has been taken into consideration. as variables • We The above steps eliminate £,(i+I) o 'p(i+l) 0 ' and p(i+l) N therefore have 2N + 2 equations for the solution of 2n + 2 variables, namely I») , n = 1,2,3, ... , N - 1 ( F n(i + I)r rp(i+ n F(i+ I)
(i+
(i + l )
I)
P -I , PN+I' We have again reduced the solution of the present problem to the same linear form as given in Eq. (5.17). The steps between Eqs. (5.17) and (5.53) can now be followed. To start the solution of Eq. (5.75) for a given pair of values of wand K, we first assume a first approximation of the solution. Without any information related to the solution, we arbitrarily select -I
,
F = cp, P=0 (5.90) With Eq. (5.90) as the first approximation, higher-order iterates can be obtained by the steps outlined in this section. The results are summarized in Table 6.5. TABLE &.5 Solutions of Eq. (5.75) for w = 0.087 rad and K
= 0.10
F(i)
n
>
(rad)
0.000 0.017 0.035 0.052 0,070 0.087
i = I
i=2
i=3
i=4
0.0000 0.0175 0.0349 0.0524 0.0698 0.0873
0.0000 O.oI74 0.0352 0.0529 0.0705 0.0881
0.0000 0.0174 0.0352 0.0529 0.0705 0.0880
0.0000 0.0174 0.0352 0.0529 0.0705 0.0880
126
6. Iterative Methods-The Finite-Difference Method
6.6 FIRST-ORDER SYSTEM AND NEWTON'S METHOD
The method presented in Section 6.5, that of replacing a third-order differential equation by a first-order equation and a second-order equation [see Eq. (5.6)], will now be improved one step further. Consider again Eq. (5.1), which will now be replaced by a system of first-order differential equations as follows: dy dx =p,
dp dx
-=q
~;
'
+A(x)q + B(x)p + C(x)y = D(x) (6.1)
One advantage of this is that the finite-difference representation of Eq. (6.1) no longer has "unwanted" values of Yn [such as y _ 1' P_ l' and PN + 1 in Eqs. (5.6)-(5.16)]. To outline the method, an example will be given in the
next section. Linearization of the differential equation in this example will be achieved by using another method, namely that of Newton. This process has been applied successfully to a wide class of viscous flow problems (See [9-16] and the references cited in those works.) 6.6.1
Natural Convection Boundary Layer Flow
Fluid flows due entirely to the density gradient created by temperature differences are usually called natural convections. Such flows occur both in the vicinity of external surfaces and within channels in which the fluid flows. The governing differential equations for such flows consist of the equations of continuity, momentum, and energy, as follows:
(6.2) (6.3)
Vx
ao se a20 ax + vy ay = IX ay 2
(6.4)
subject to the boundary conditions
y=o: y
where 0 = (T - T
= 00:
Vx
= 0,
vy
= 0,
Vx
= 0,
0=
°
0=1
.u«.- Too)' By introducing the stream function, de-
6.6
First-Order System and Newton's Method
127
fined by alj; v = --
ax
y
and the transformation 1]
2
1/4
= y(gef3/ 4v x) ,
Eqs. (6.2)-{6.4) become
d~ -2 ( -df -dy3 +3fdTJ dTJ 2 dTJ d 2g
-2
dTJ
)2+g=O
(6.5)
dg
+ 3Npr f -d = 0
(6.6)
TJ
subject to the boundary conditions df(O)
--=0
TJ = 0:
f(O) = 0,
TJ=OO:
--=0
dTJ
df(oo) dTJ
g(O) = 1
'
g(oo) = 0
'
We now replace Eqs. (6.5) and (6.6) by a system of first-order equations, namely df dTJ = u,
du dTJ = v,
dg dTJ = t
dv 2 dTJ +3fV-2U +g=O
dt dTJ
(6.7)
+ 3Nprft = 0
The boundary conditions are TJ = 0: 1]
= 00:
f(O) = 0,
U(O) = 0,
U( (0) = 0,
g( (0) = 0
g(O) = 1
Next, the derivatives are approximated by centered-difference gradients and averages centered at the midpoints of the net, defined by TJo = 0;
'IJ.i
= 'IJ.i - 1 + hj ,
j = l , 2, ... , J;
TJJ
= TJ oo
(6.8)
128
6. Iterative Methods-The Finite-Difference Method
Equation (6.7) then becomes
(fj - fj_l)hj- 1= Uj-(1/2)
(gj - ~_l)hj-l = 0-(\/2)
(6.9)
1 (vj - vj_l)hj- + 3fj-(1/2)Vj-(1/2) - 2(Uj-(1/2l+ ~-(1/2) = 0 (0 - 0-1)/f
l
+ 3Nprfj-(1/2)0-(1/2) = 0
where Uj-(1/2) = t(uj + ~-l)' etc. Equations (6.9) are nonlinear algebraic equations and therefore have to be linearized before the factorization scheme can be used. Let us write the Newton iterates as follows: For the (i + l)th iterates, we write (i+ I) = (i) Jj
J)
+ ~(i) 'lJ'
etc.
(6.10)
for all dependent variables. By substituting these expressions into Eq. (6.9) and dropping the quadratic and higher-order terms in ~!?), etc., a linear tridiagonal system of equations will be obtained, as follows: ~fj - ~fj-l - th)(~uj
(r l)j-(l/2)
(6.11)
+ ~Vj_l) = (r 2)j-(1/2)
(6.12)
t!y( ~0 + ~0-1) = (r 3)j-(l/2)
(6.13)
~Uj - ~Uj-l - t!y(~Vj
~~
-
~~-l
-
+ ~Uj_l)
=
Ul)}Vj + (2)}Vj- 1+ (3)}fj + (4)}fj-l + Us)}uj + (6)}Uj- 1 + (~7)}~ + US)}gj-l = (r4)j-(1/2)
(6.14)
(f31)}0 + (f32)}0-1 + (f33)}fj + (f34)}fj-l + (f3s)/% + (f36)}~-1 (6.15) subject to the boundary conditions ~fo
= 0,
~Uo
= 0,
129
6.6 First-Order System and Newton's Method
where
al)j = 1 + ~fy(jj + fj-I)'
aZ)j = (KI)j - 2, (K3)j = (K4)j = ~hjvj_(I/Z)
(Ks)j
(K7)j
=
a6)j
=
-2hjuj_(l/Z)'
(f3I)j = 1 + ~ Nprfyfj-(I/Z) ,
= as)j = t hj
(f3Z)j = (f3I)j - 2
(f33)j = (f34)j = ~ N prhj~_ (I/Z)' (f3 S)j = (f36)j = 0, (r l)j_(1/Z)
= fj_1
(r 3)j-(I/Z) = ~-I
- fj -
+ hjuj_(I/Z)'
(rz)j_(I/Z) = uj_ 1 - uj + fyvj_(I/Z)
gj + hj~_(l/z)
(r 4)j_(1/Z) = vj_1 - vj - hjgj_(I/Z) - 3fyfj_(l/z)vj-(I/Z) + 2hiUj-(l/z/ (rs)j_(I/Z) = ~-I - ~ - 3Nprfyfj-(I/Z)~-(I/Z) Equations (6.11)-{6.l6) correspond to Eqs. (5.8)-{5.1O). The difference is that we now have five equations instead of two. In vector-matrix form, Eqs. (5.8)-{5.10) can be written as AlJ = r
(6.17)
where
[ AI] [ Bz]
[ CI ]
[ A z]
[ Cz ]
A=
[BJ - I ] [AJ-d [ CJ - I ] [ BJ ]
[ 81 ]
[r l ]
[ 8z ]
[ rz]
lJ=
(6.18)
r=
[ 8J - d [8J ]
[A J]
[rJ -I] [rJ]
6.
130
Iterative Methods-The Flnlte-DIUerence Method
where in Eq. (6.18) the elements are defined by
o [A1J=
o o
0
-a l
o o
( f3Z)1
(3)1 ( (33)1
o
(~Z)I
o
0
- 1
o o
(~6)j
(~g)j
(~3)j
( f3g)j
( (36)j
( (33)j
o
0 0 0 0 0
o o o
o
o o
o
o o
-aj
(4)j ( (34)j
aZ)j
o
2<'j<,J
o
o o
-aj 1
o (~5)j
( (37)j
=
1
- 1
0 0 0 0 100 (7)j 0 0 (f35)j 0 0
~vo
~Uj_1
&0
~&-I
~fl
,
[~J
=
0 0 0 0 0
~fj
~VI
~v)
&1
~~
('I)j-(I/Z) ('z»)-(l/Z)
['jJ =
('3»)-(I/Z) , ('4»)-(l/Z) ('5»)-(I/Z)
o o
o
- 1
[Aj ] =
[~d
1
o o
-aj
[c;]=
o
-al
l<'j<,J
l<'j<,J-l
2 <, j <, J,
131
6.6 First-Order System and Newton's Method
Now, we let
A=Lu
(6.19)
where
r-.
[ a2 ]
[ B2 ]
L= [BJ ]
[f l ] [1]
[ 1]
[aJ ]
[f2 ]
u=
where [1] is the unit matrix and [a;), and [f;] are 5 X 5 matrices whose elements are determined by the following equations:
[al]=[Ad [AI][f l ] =[C 1 ]
raJ = [A
j ] -
[BJ[fj -
I] ,
)
= 2,3, ... , J
(6.20)
[aj][fJ=[Cj], )=2,3, ... ,J-l Details of this step are similar to those between Eq. (5.28) and Eq. (5.38). Equation (6.19) can now be substituted into Eq. (6.17), and we get
LUc5 = r
(6.21)
Uc5 = W
(6.22)
LW=r
(6.23)
If we define
Eq. (6.21) becomes
132
6.
Iterative Methods-The Finite-Difference Method
where
[ WI] [ W2 ]
w = [W3 ] [ W4 ]
[ Ws] and the [Jfjl are 5 X I column matrices. The elements W can be solved from Eq. (6.23):
[£XI][WI] =[r)] [ £xj ] [
Jfj] = ['j] - [ BJ [ Jfj- I]
(6.24)
Details of this step are similar to those between Eq. (5.43) and Eq. (5.47). Once the elements of Ware found, Eq. (6.22) then gives the solution (j, the elements of which are obtained by the following relations:
[8J ] = [WJ ] [ 8j ] = [ Jfj] - [fj ] [ 8j + I ]
(6.25)
Details of this step are similar to those between Eq. (5.50) and Eq. (5.53). Once the elements of (j are found, Eq. (6.10) can be used to find the (i + l)th iteration. A typical numerical solution for N pr = 0.72 is given in Table 6.6. TABLE 6.6 Solutions of Eqs. (6.5) and (6.6) (Npr = 0.72) Numbers of iterations
Initial slopes
p
d2j(O)/ drj2
- dg(O)/ drj
1
0.6800 0.6741 0.6742 0.6742
0.5000 0.5050 0.5048 0.5048
2 3 4
6.7 CONCLUDING REMARKS
The finite-difference method of solving two-point boundary value problems converts the set of differential equations into a finite set of algebraic
6.7 Concluding Remarks
133
or transcendental equations. The solution of the set of algebraic or transcendental equations yields approximations to the solution of the original differential equations at discrete points. If the original ordinary differential equations are linear, the finite difference equations will be linear algebraic equations. If the ordinary differential equations are nonlinear, the resulting finite difference equations will be nonlinear algebraic or transcendental equations. There are two ways to linearize the problem. One is by linearizing the differential equations before they are written in finite-difference form by methods such as quasi linearization. Another is first to write the differential equation in finite difference form and then to linearize the resulting nonlinear algebraic or transcendental equations. Both approaches were considered in this chapter. Finite-difference methods have been proved to be a very useful technique for solving numerically sensitive two-point boundary value problems. This is due to the fact that the finite difference equations incorporate both specified initial and terminal conditions in the final set of equations, and thus the resulting solutions of these equations are constrained to satisfy these boundary conditions This differs from a shooting method, where the terminal boundary condition never enters into the solution during the forward integration process. Another difference is that the solution is produced simultaneously at all points, whereas in a shooting method the solutions at different points are generated in sequence. We may therefore expect the solutions near the terminal point to be less accurate than those obtained near the initial point because of the propagation of round-off errors. The discussion of the finite-difference representation of differential equations is brief. There are many finite-difference formulas in which a differential equation can be approximated. The choice is made based on a compromise among such factors as simplicity, size of the truncation error, and stability. For a detailed study of the relative merits of various formulas, the reader is referred to [17-22]. The linearized algebraic equations exhibit banded matrix structure. The factorization scheme developed by Keller, as outlined in this chapter, is the most efficient way to solve such equations. Its value extends far beyond the solution of boundary value problems of ordinary differential equations. It has been demonstrated [9-16] how this method can be applied to partial differential equations of the parabolic type. Almost identical steps can be followed for hyperbolic and elliptical partial differential equations. For a discussion of the stability, uniqueness and error analyses, the reader is again referred to texts specifically on the finite difference method and, in particular, Chapter 3 of the classic text of Keller [2].
134
6. Iterative Methods-The Finite-Difference Method
PROBLEMS
1. Finish the problem given at the end of Section 6.5.1: dy dx 3
-
d2y
dy
+ 4 - 2 + - - 6y = I dx dx '
y(O)
= 0,
dy(O) dx
--=0
'
dy(l) dx
--=1
Answer: d2y(O)/dx 2 = 2.8858. 2. In an analysis of heat transfer over thin needles, Narain and Uberoi [23] found that the temperature is given in terms of the following boundary value problem: 2ep 2ep dep(a) !!... (z d ) + 1. ep d 2 = 0 ep(a) = 0, - - = 0 dz ' dz dz? 2 dz ' Solve ep by the method of finite differences for a = 0.1. Answer: d 2ep(a)/ dz 2 = 1.28883. 3. The effect of radiation on the external flow over flat surfaces has been studied recently by Viskanta and Grosh [24]. The temperature profile near the wall can be obtained by solving the energy equation, 3
40 d [( dT/ 1+ 3N
)
dO ] dO dT/ + Npr! dT/
where the function
f
is the solution of the following initial value problem: f(O)
= 0,
= 0, df(O) dT/
0(0)= Ow, O(oo)=OOCJ
--=0
'
dz.!(O) dT/2
= 0.46960
Solve the problem by the method of finite differences for N pr = I, N = 10, and Ow = 0.1. Answer: dO (0)/ dT/ = 1.212. 4. Solve problem I, Chapter 5, by the method of factorization. Note: The student should derive the equations by following the steps shown in Sections 6.4.1 and 6.5.1 since this is a fourth-order differential equation. 5. Solve the problem of the sandwich beam treated in Section 4.3 by the finite-difference method. Note: This is a three-point boundary value problem. Some modifications are needed before applying the equations in Section 6.5.1. 6. The deflection of a cantilever beam under a concentrated load can be found by solving the following boundary value problem [25]:
-dz.!2 + Ascosf= 0 ds
'
dfd~)
= 0, f(l) = 0
References
135
Solve the problem by the method of finite differences for A = 8. Answer: df(l)/ds = -3.194. 7. (a) Write Eq. (5.1) as a system of three first-order differential equations,
dy dx
= u,
du dx
= v,
dv dx = - A (x)v - B(x)u - C(x)y
+ D(x)
Replace the derivatives and the variables by
=Yn-Yn-I dY ) ( dx n _ 1/2 h '
Yn -
1/2
~
Yn+Yn-1 2
etc.
and derive the series of equations corresponding to the derivation in Sections 6.5.1 and 6.6.1. (b) Re-solve the problem of Hiemenz flow using the equations derived in (a).
REFERENCES 1. Thomas, L. H., Elliptic problems in linear difference equations over a network, Watson Sci. Comput. Lab. Rep., Columbia Univ., New York, 1949. 2. Keller, H. B., "Numerical Methods for Two-Point Boundary Value Problems," Chapter 3, Ginn-Blaisdell, Waltham, Massachusetts, 1968. 3. Keller, H. B., A new difference scheme for parabolic problems, in "Numerical Solutions of Partial Differential Equations" (B. Hubbard, ed.) Vol. 2, pp. 327-350, Academic Press, New York, 1971. 4. Bruce, R. W., and Na, T. Y., Natural convection flow of Powell-Eyring fluids between two vertical flat plates, ASME paper 67 WA/HT-25, presented at the 1967 ASME Winter Annu. Meet., Pittsburgh, Pennsylvania, 12-17 November 1967. 5. Fox, L., "The Numerical Solution of Two-Point Boundary Value Problems," pp. 148-150, Oxford Univ. Press, Oxford, 1957. 6. SjuKic, OJ. S., Hiemenz magnetic flow of power-law fluids, J. Appl. Mech. 822-823 (1974). 7. Hiemenz, K., Die Grenzschicht an einem in den gleichforrnigen Flussigkeitsstrom eingetauchten geraden Kreiszylinder, Dingl. Polytech. J. 326, 321 (1911). 8. Schlichting, H., "Boundary Layer Theory," pp. 99-100, McGraw-Hill, New York, 1968. 9. Cebeci, T., Berkant, N., Silivri, I., and Keller, H. B., Turbulent boundary layers with assigned wall shear, Comput. Fluids 3, 37-49 (1975). 10. Cebeci, T., and Keller, H. B., Separating boundary layer flow calculations, J. Comput. Phys. (in press). 11. Keller, H. B., Accurate difference methods for nonlinear two-point boundary value problems, SIAM J. Numer. Anal. 11, 305-320 (1974). 12. Keller, H. B. Some computational problems in boundary layer flows, Proc. 4th Int. Con! Numer. Methods Fluid Mech. 1-21, Boulder, Colorado, 1974. 13. Keller, H. B., and Cebeci, T., Accurate numerical methods for boundary layer flows. I: Two dimensional laminar flows, Proc. 2nd Int. Con! Numer. Methods Fluid Dynamics, Berkeley, California, 1970; "Lecture Notes in Physics," Vol. 8, pp. 92-100, Springer, Berlin, 1970.
6. Iterative Methods-The Finite-Difference Method
136
14. Keller: H. B.; and Cebeci, T., An inverse problem in boundary layer flows: Numerical determination of pressure gradient for a given wall shear, J. Comput. Phys. 10, 151-161 (1972). 15. Keller, H. B., and Cebeci, T., Accurate numerical methods for boundary layer flows. II: Two dimensional turbulent flows, AIAA J. 10, 1193-1199 (1972). 16. Keller, G. B., Numerical methods in boundary layer theory, Annu. Rev. Fluid Mech. 10, 417-433 (1978). 17. Collatz, L., "The Numerical Treatment of Differential Equations," pp. 184-186, Springer-Verlag, New York, 1966. 18. Henrici, P., "Discrete Methods in Ordinary Differential Equations," Wiley, New York, 1962. 19. Henrici, P., "Error Propagation for Difference Methods," Wiley, New York, 1963.
20. Hildebrand, F. B., "Methods of Applied Mathematics," Prentice-Hall, Englewood Cliffs, New Jersey, 1952. 21. Varga, R. S., "Matrix Iterative Analysis," Prentice-Hall, Englewood Cliffs, New Jersey, 1962. 22.
Issacson, E., and Keller, H. B., "Analysis of Numerical Methods," Wiley, New York,
1966. 23. Narain, J. P., and Uberoi, M. S., J. Heat Transfer, Trans. ASME 94,240-242 (1972). 24. Viskanta, R., and Groth, R. J., Int. J. Heat Mass Transfer 5, 795-806 (1962). 25. Bisshopp, K. E., and Drucker, D. C., Q. Appl. Math. 3, 272-275 (1945).
CHAPTER
7
METHOD OF TRANSFORMATIONDIRECT TRANSFORMATION
7.1
INTRODUCTION
In this chapter, a method of transformation from nonlinear boundary value problems to initial value problems will be presented. The method was first introduced by Toepfer [I] in 1912, in his attempt to solve Blasius's equation in the boundary layer theory by a series expansion method. No progress was made for half a century, until 1962 when Klamkin [2], following the same reasoning, extended the method to a wider class of problems. Major extensions were made possible only when the concept was reexamined and interpreted by Na [3, 4] in terms of transformation groups. Chapters 7-9 will be devoted to the various applications of this concept. Even though the method is sometimes referred to as the method of transformation groups, perhaps the only group concept used in the method at this stage is that the "operation" of a transformation is a group. Attempts have been made [II] to derive groups of transformations which can reduce boundary value problems to initial value problems by using the invariant properties of a differential equation under a so-called infinitesimal transformation group. The theory is, however, incomplete at this time and considerable work is needed before the method can be established. Until that happens, it is much simpler to view the method as a straightforward transformation. We will, however, keep the phraseology of a group of transformations instead of a family of transformations, even though the latter may be more familiar to the reader. For a complete review of the various aspects of the group-theoretic method, the reader is referred to Ames [5], Na and Hansen [6], Bluman and Cole [7], Ovsjannikov [8], and Na [9, 10]. Basically, the method starts by defining a group of transformations. The particular transformation within this group of transformations which can reduce the boundary value problem to an initial value problem is identified 137
138
7.
Direct Transformallon
by the requirement of independence of the given differential equation of the parameter of the transformation and by the identification of the parameter of transformation as the missing boundary condition. The method is simple and its application is straightforward. Blasius Solution In the Boundary Layer Theory
To introduce the background which led to the method discussed in this chapter, let us consider the flow of an incompressible fluid over a semiinfinite flat plate, as shown in Fig. 7.1. Such a flow is usually called the boundary layer flow since the viscous effects are limited to a thin layer near the surface, across which the velocity changes from zero at the wall to the main-stream velocity U00 at the edge of this layer. The analysis of such a flow requires the application of the Navier-Stokes equations, which are derived from the laws of conservation of mass and of momentum. By introducing the boundary layer assumptions (i.e., the existence of a very thin layer and that u» v), the Navier-Stokes equations become
~ + av =0
ax
u~ + v au ax
(1.1)
ay
ay
= v
2u
a
ay2
(1.2)
subject to the boundary conditions
y=o:
u=v=o
y
u
= 00:
=
U oo
where u and v are the velocity components in the x and y direction, respectively, and v is the viscosity of the fluid. The boundary conditions at the wall (y = 0) are based on the physical condition that neither slip nor mass transfer occurs on the surface. The boundary condition at the edge of
--
y
Uoo
-----
~~======~========--x
Uoo
Fig. 7.1
Boundary layer over a plate.
u
7.1
Introduction
139
the boundary layer simply means that, mathematically, the velocity component u approaches the mainstream velocity U 00 asymptotically. Equations (1.1) and (1.2) were solved by Blasius [11] in 1908. As the next step, a stream function tf; is introduced such that u = atf;jay, v = -atf;jax
(1.3)
Besides the physical reason of introducing this function (i.e., constant tf; lines are streamlines), the mathematical significance is that the equation of continuity, Eq. (1.1), is satisfied identically and the momentum equation becomes atf; a~ ay ax ay
atf; a~ ax ay 2
a3tf; ay 3
------=p-
(1.4)
subject to the boundary conditions atf; atf; ax ay atf; -=U ay 00
y=O: y
-=-=0
= 00:
In other words, the solution of two equations with two variables (u, v) is reduced to the solution of one equation with one variable (tf;). Blasius introduced the following transformation: (1.5) and
f('q) = tf; j ~pxU 00
(1.6)
which transforms Eq. (1.4) to d:J d q3
1
d~
+ '2 f d q2 = 0
(1.7)
subject to the boundary conditions
f(O) = df(O)j dq = 0,
df(oo)jdq = 1
The class of transformations in which a nonlinear partial differential equation is transformed to a nonlinear ordinary differential equation is known as the class of similarity transformations. The foundation of the method is contained in the general theories of continuous transformation groups that were introduced and treated extensively by Lie in the latter part of the last century [12]. We will not treat this subject in this book. Interested readers should refer to the works of Birkhoff [13], Morgan [14], Hansen [15], and Na and Hansen [6] for details.
140
7.
Direct Transformation
Equation (1.7) is a boundary value problem. In order to follow the reasoning which led Toepfer [1] to introduce a transformation of variables and transform the problem to an initial value problem, solution of Eq. (1.7) by series expansion should be given in detail. Let us assume the solution of Eq. (1.7) to be
f(1j) =
00
2:
n=O
Cn1jn
(1.8)
Before substituting the solution into the differential equation, some of the coefficients can be evaluated from the boundary conditions. From the two boundary conditions on the surface,
1j
= 0:
f(O) = 0, df(O)/ d1j = 0
we get
CO=C(=O
(1.9)
Next, let us introduce the notation d~(O)/d1j2
=A
(1.10)
where, of course, the constant A is an unknown quantity. In terms of A, the coefficient C2 is given by
C2 = A/2!
(1.11)
Equation (1.8) can now be substituted into Eq. (1.7), and we get
n~3 n(n -
l)(n - 2)Cn1jn-3+
t L~o
Cn1jn)[ n~2 n(n - I)C n1jn-2] = 0 (1.12)
or, in expanded form,
[ (3)(2)( I) C3 + ! Co(2)( I) C2 ]
+ [(4)(3)(2)C 4 + !Co(3)(2)C3 + H2)(I)C2Cd1j
+ [(5)(4)(3)Cs + ! Co(4)(3)C4 + H2)(I)C2C2 + ! C(3)(2)C3]1j2 + [(6)(5)(4)C 6 + ! Co(5)(4)Cs + H2)(I)C2C3 +! C(4)(3)C4 + t Ci3)(2)C 3]1j3 + ... = 0 (1.13) By setting the coefficients of Eq. (1.13) equal to zero and employing Eqs. (1.9) and (1.11), we get C3 = C4 = 0,
Cs = -A 2/5!(2)
c, = IIA 3 / 8! (2)2
(1.14)
7.1
141
Introduction
The solution, Equation (1.8), can therefore be written as
!(TJ) =
ATJ2
A~5
2! -
S! (2)
11 A3TJ 8 37SA4.ryll 8! (2)2 - II! (2)3
+
+...
(US)
which still involves the unknown constant A. In principle, the boundary condition at the second point should be applied to determine the constant A. However, this cannot be done directly here due to the fact that the boundary condition at the second point is given at TJ = 00. Blasius [11] used a matching method, in which the solution given by Eq. (US) applied to the region close to the wall is matched to an asymptotic expansion for large TJ. The process is tedious and is of no interest here. By rearranging the variables, Eq. (US) can be written as
37S(A 1/3
t
TJ ---+ ... II! (2)3
(1.16) which suggests a transformaion of the form F
= A-1 / '1,
(1.17)
Equation (1.16) then becomes
e
e
lIe
2!
S! (2)
8! (2)2
37Sg 11 II! (2)3
F=----+---
+...
(1.18)
The significance of the transformation of variables will become obvious if the differential equation, Eq. (1.7), and its boundary conditions are transformed. From Eq. (1.7), we get 3F
d de
+ 1. F d 2
2F
de
= 0
(1.19)
The boundary conditions at the wall are transformed to
g= 0:
F(O) = 0, dF(O)j dg = 0
(1.20)
Using this transformation, Eq. (1.10) gives d 2F(0)j de = I
(1.21)
The transformed equation, Eq. (1.19), subject to the boundary conditions (1.20) and (1.21), is an initial value problem. The solution of Eq. (1.19) can be obtained by a forward integration scheme. The relation between (TJ,j) and (g,F) is defined in Eq. (1.17). If the constant A can be found, the solution of (TJ, j) can be calculated from the
7. Direct Transformation
142
solution of (g, F). To this end, the boundary condition at infinity df( (0)/ dTj
=1
is transformed according to Eq. (1.17) and we get A-2/3[dF(00)/dg] = 1 or
]3/2
1
A = [ dF( 00/ dg)
(1.22)
Thus, the solution of Eq. (1.7) consists of three steps. 1. Solve Eq. (1.19), subject to the boundary conditions (1.20) and (1.21), as an initial value problem by forward integration. In particular, we get the value of dF( (0)/ d g. 2. Compute A by Eq. (1.22). 3. Compute f(Tj) by using Eq. (1.17). The above demonstrates an important idea of solving boundary value problems, that of solving a complementary initial value problem. In this way the solution becomes noniterative. Discussion of the computational details and the physical interpretations of the results are of interest but will be delayed until Section 7.2.1. While the value of such a transformation was obvious, no further work was done to extend the idea to other types of problems until 1962, when Klamkin [2] published an extension of this method by following the same reasoning as given in Toepfer's method. For example, it was found that for a general second-order ordinary differential equation of the form ~ A; (d2r _J ;=1 dTj 2 N
)m
i (
-dlf dTj
)n fT'Tjs,= 0 i
(1.23)
subject to the boundary conditions f(O) = 0, df( (0)/ dTj
=k
a transformation (1.24) enables the solution of Eq. (1.23) to be obtained by first solving the initial value problem
(1.25)
143
7.2 Transformation for a Given Group of Transformations
subject to the boundary conditions F(O) = 0,
dF(O)/d~
=I
and then calculating the constant A from _ / A- k
dF(oo)
(1.26)
d~
Finally, the solution of Eq. (1.23) can be calculated by using Eq. (1.24). The constant a in the transformation (1.24) is determined by requiring that, upon transformation, Eq. (1.25) should be independent of A. Although Klamkin's method represents an important extension of the technique, the method is still quite limited for the following three reasons. First, only the types of transformations similar to those given above are known to be applicable. Second, the boundary conditions at the second point have to be given at infinity. Third, the boundary conditions at the initial point must be homogeneous. To remove some of the limitations, the present method has to be interpreted in terms of the concepts of transformation groups. This will be the content of the next section.
7.2 7.2.1
TRANSFORMATION FOR A GIVEN GROUP OF TRANSFORMATIONS The Blasius Equation In Boundary Layer Flow
The Blasius equation, Eq. (1.7), which Toepfer treated, will now be reconsidered from the point of view of transformation groups. The equation is dy I d~ -+-f-=O d'l3 2 d'l2
(2.1)
subject to the boundary conditions f(O)
= df(O)/ d'l = 0,
Let the given group of transformations be the linear group defined by
(2.2) where A is the parameter of transformation and a\ and a 2 are constants to be determined. Under this group of transformations, Eq. (2.1) becomes d 3'f
-
d'l3
I
+-
2
d 2(
f-J
d'l2
= A "2- 3"1
dy-
-
dri 3
I
+-
2
A 2" 2-
2r
dJ "'1 _ 2
2
dri
=0
(2.3)
144
7.
Direct Transformation
We first require that the transformed differential equation be independent of the parameter of transformation A. This requires that the powers of A in the two terms be equal, i.e., 0: 2
30:1 = 20: 2
-
-
20: 1
(2.4)
Equation (2.3) gives
(2.5) From Eq. (2.5), we get
d3J
dij3
1 -
+ 2" f
d2J
dij2
=0
(2.6)
In the second step, the missing boundary condition is set equal to the parameter of transformation, i.e.,
(2.7) This step is somewhat arbitrary, but it does not place any limitation on the method since A is an unknown constant to begin with. Under the linear group of transformations, this condition becomes (2.8) and we further require that the transformed boundary condition be independent of A. This is possible if 0: 2 - 20: 1 = 1
(2.9)
Equation (2.8) then gives
d2J (0)/ dij2 = 1
(2.10)
Equations (2.4) and (2.9) give (2.11) Finally, the parameter A can be evaluated from the boundary condition at infinity, which gives (2.12)
or, using Eq. (2.11), A
-
= [ df ( 00) /
dij ]
-3/2
(2.13)
145
7.2 Transformation for a Given Group of Transformations
The other boundary conditions are transformed simply to
J(O) = dJ(O)/dij = 0
(2.14)
Equation (2.6), together with the boundary conditions (2.10) and (2.14), constitutes an initial value problem, the numerical solution of which is shown in Fig. 7.2. df(OO) = 2.0852 2
-- - -- - - -
dTj
-
- --....:-:.-;;;..--------...........
df dTj
2
4 Tj
Fig. 7.2
Solution of Eq. (2.6).
It is seen that as 'ij approaches infinity dJ/ dij approaches the value of 2.0852, which is taken as dj(oo)/dij. Substituting into Eq. (2.13), we get
A = [ dJ(00)/ dij] -3/2 = 0.332068 which is the missing boundary condition at Tj = 0, as defined by Eq. (2.7). Since a\, a 2 , and A are all known, the solution of the original equation, Eq. (2.1), can be completed by using Eq. (2.2). It will be of interest to give a physical interpretation of Fig. 7.3, which shows df/ dTj as a function of Tj. From Eqs. (1.3) and (1.6), we get
u=
~;
=
= frlJ: 00
a~ Uvxu f) oo
df dTj
aTj
ay
146
7. Direct Transformation
1.0
df dry
-
-
-
-
-
-
-
-
-
-
-=..;;-;:;..=00..-------
0.5
0.0
L-..._ _--'-
...J...-_ _- - - - ' ' - -_ _- - ' -
o
3
2
-'
4
ry
Flg.7.3 Solution of Eq. (2.1).
or (2.15) u/ U 00 = df/ d'1 Thus, the curve given in Fig. 7.3 is a plot of the velocity distribution as a function of 1/. Referring to Fig. 7.1, if the fluid and the mainstream velocity Uoo are given and the velocity distribution at a distance x = X o is needed, Eq. (1.5) then gives the relation between 1/ and y, namely (2.16)
Since U00' P, and X o are known, Eq. (2.16) gives the corresponding value of 1j for any y. The curve in Fig. 7.3 can then be used to read u/ U 00 at this value of y. The method can now be summarized as consisting of the following steps. 1. A group of transformations is defined. From the point of view of transformation groups, it is immediately recognized that the class of transformations treated by Toepfer (I] and Klamkin [2] is the familiar linear group of transformations. The immediate conclusion is that there must be other groups. Questions as to whether there is a deductive way to derive the particular group of transformations may also arise (16]. At the present time, these theories are still incomplete and we have to proceed with the method based on a given group of transformations. 2. The differential equation is required to be independent of the parameter of transformation, which should lead to only one relation between U 1 and u2 • This step offers the first test as to whether or not the method is applicable. For example, the Falkner-Skan equation d:t I d~ [ 1- ( -df -+-f-+/3 d1/3
2
d1j2
d1/
)2] =0
7.2 Transformation for a Given Group of Transformations
147
is independent of the parameter of transformation in the linear group defined in Eq. (2.2) only for a l = a 2 = O. The method thus fails to apply. 3. The unknown boundary condition is set equal to the parameter of transformation. Another relation connecting a. and a 2 can be written from which a l and a 2 can be solved, leaving A as the only unknown in the transformation. 4. The parameter of transformation is determined from the boundary condition at the second point, such as in Eq. (2.13). One interesting case will be when the boundary condition at infinity is homogeneous, i.e., when the right-hand side of Eq. (2.12) is zero. One example is found in the flow of a laminar non-Newtonian jet [17]. Another case is when the boundary condition at the second point is given at a finite distance instead of at infinity. Both cases will be discussed later. 5. As a last step, the boundary conditions at the initial point are transformed. Here, it is required that they be homogeneous. Otherwise, the method cannot be applied. For example, if f(O) = k, then upon transformation A a>j(o) = k which is independent of A only if a 2 is equal to zero. We will discuss in the next two chapters how problems with nonhomogeneous boundary conditions at the initial point can be treated. At this point, a recent work by Klamkin [18] should be cited. In this work, Klamkin tried to remove the requirement that the boundary condition at the initial point be homogeneous. To illustrate his reasoning, let us consider one example given in Klamkin [18]. d~ 1 df 1 -+--=a+{3dT/2 T/ dT/ We will, however, write the boundary conditions as
F
df(O)j dT/
= af(O) + b,
f( 00) = 1
After a linear transformation, Eq. (2.2), the differential equation becomes
A a2-2a,( d2j + drj2
~
d])
T/ drj
= a
+ {3A - 2a2~
I'
We will consider three cases. Case 3 is that treated by Klamkin. Case 1 If neither a nor {3 is zero, invariance of the differential equation gives two equations, namely 2a 2 = 0 a 2 - 2a. = 0, which means the problem cannot be transformed to an initial value problem since now both a. and a 2 are zero.
7. Direct Transformation
148
Case 2 If £x equals to zero (i.e., the differential equation does not contain constant term), then one equation results from the invariance of the differential equation, namely £x 2 - 2£x 1 = -2£x 2
and the method presented above can be applied. Since we now have one equation for the solution of £XI and £x2' only one equation relating £XI and £X 2 should come from the boundary conditions. This is the reason why the boundary condition at the initial point should be homogeneous, since the boundary condition at the second point always gives one such relation. Case 3 If both £x and f3 are zero (i.e., for differential equations whose transformed form, like the two terms on the left-side of the transformed equation above, has exactly the same powers of A, which can then be factored out for all values of £XI and £X2 without the necessity of imposing any relation between £XI and £x2), then no relation between £Xl and £X2 will emerge from the differential equation. This leaves the two boundary conditions to provide the two equations necessary for the solution of £XI and £X 2• We now set
f(O) = A
The first boundary condition gives By putting
J(O) = I
and
dJ(O)j dij
= I
we get two equations £X 2
=I
A "'-"'aA
+b
which, together with the transformed boundary condition at the second point, A ",
= I j J(00 )
are the equations required for the solution of
£XI' £X2'
and A.
The above cases show clearly that by placing limitations on the form of the differential equation, some of the restrictions imposed on the method can be removed. 7.2.2 The Brachlstochrone Problem In Optimal Control
The brachistochrone problem is well known in the field of optimal control and orbital determination. It involves the determination of an optimal path between two points in a gravitational field.
149
7.2 Transformation for a Given Group of Transformations
Consider two points A and B in a space containing a constant gravitational force field, as shown in Fig. 7.4. Point A is the initial point at (0, Yo), which is a fixed point. Point B lies anywhere on the terminal line x = X T' with the restraint that dy/ dx = 0 at this point. It is the purpose of the analysis to find a frictionless path from A to B along which a particle will slide in minimum time. ,---------,---- x
Yo
A
B Y
Fig. 7.4 The optimal path.
Since gravitational force is the only force acting on the mass, the time of descent T is given by T
~ rx,
V
Jo
2gy
fix
(2.17)
or, if the integrand is defined by (2.18)
we get
T = LXTFdX
(2.19)
To minimize the time of descent T, Euler's equation is applied to the integrand F, i.e., aF
d ( ay' aF) -_0
ay - dx
(2.20)
Performing the operation, we get 2y dy + dx 2
[I + ( dxdy )2] = 0
(2.21 )
The boundary conditions are y(O)=Yo,
dy(xT)/dx=O
(2.22)
Equation (2.21), subject to the boundary conditions (2.22), is a two-point nonlinear boundary value problem. Before this problem can be trans-
150
7. Direct Transformation
formed into an initial value problem, a simple transformation is introduced: (2.23) The reason for introducing this transformation will become evident later. Equation (2.21) now becomes
[1 + (d-dqf )2] =0
d~ + 2fdT/2
(2.24)
subject to the boundary conditions
df(O) / dT/ = 0,
f(l)
=
S
(2.25a, b)
where S = yo/xT • To reduce the equation to an initial value problem, let us introduce the linear group of transformation (2.26) and require that Eq. (2.24) be independent of the parameter of transformation A; the following relation between a\ and a 2 is obtained. (2.27) Equation (2.24) is transformed to
2J d2j + d1j2
[1 + ( d1jdJ )2] 0 =
(2.28)
The first boundary condition, (2.25a), becomes
dJ(O)/ d1j = 0
(2.29)
Next, the missing boundary condition is chosen to be
f(O) = A
(2.30)
which, upon transformation, becomes
VVe therefore get (2.31) and
J(O) = 1
(2.32)
This is the step which made the transformation (2.23) necessary. In its original form, Eq. (2.21), the missing boundary condition is dy(O)/dx. It is
7.2
Transformation for a Given Group of Transformations
151
therefore necessary to set
dy(O)/dx = A which, under the linear transformation x
= A a 'x,
y
= Aa,y
becomes
A a,-a'dy(O)/ dX = A Because of the fact that 0:\ = 0: 2 , from the differential equation, this boundary condition cannot be reduced to a form independent of A. By introducing the transformation (2.23), this difficulty is bypassed. Now, Eqs. (2.27) and (2.31) give 0:\
=
0: 2
=1
(2.33)
Finally, the boundary condition at the terminal point gives
1= s/ A
at
1i = 1/A
(2.34)
Eliminating A from Eq. (2.34), we get
s = 1/1i The noniterative solution of Eq. (2.24) for a given value of s therefore proceeds as follows: Eq. (2.28) is solved for J(1j) as an initial value problem using the boundary conditions (2.29) and (2.32), until the ratio 1/1i equals the given value of s. The values of lor 1i at the point can be substituted into Eq. (2.34) to get A. Once A is known, the solution of (2.24) can be calculated from the J(1i) solution through the definition of the transformation, Eq. (2.26). Consider for example the case in which s = 3. Integration of Eq. (2.28) is carried out until the ratio 1/1i is equal to 3, from which we get
1i = 0.324,
1 = 0.9721
at this point. This is shown in Table 7.1. The parameter of transformation can be calculated from Eq. (2.34) by A = 3/1= 3/0.9721 = 3.0864
or A = 1/ 1i = 1/0.324 = 3.0864
which are the same result. Finally, the solution of the original equation, Eq. (2.24), can be calculated through the definition of the transformation, Eq. (2.26). These solutions are also included in Table 7.1. We have therefore found the optimal path which satisfies all the restraints.
7.
152
Direct Transformation
TABLE 7.1
Sample Solutions
7.2.3
fj
J
'1/
f
0.000 0.030 0.060 0.090 0.120 0.150 0.180 0.210 0.240 0.270 0.300 0.324
1.0000 0.9998 0.9991 0.9980 0.9964 0.9944 0.9919 0.9889 0.9855 0.9816 0.9773 0.9721
0.000 0.092 0.185 0.277 0.369 0.462 0.554 0.646 0.738 0.831 0.923 1.000
3.0864 3.0763 3.0742 3.0707 3.0658 3.0596 3.0519 3.0429 3.0324 3.0205 3.0071 3.0000
Longitudinal Impact of Nonlinear Vlscoplastlc Rods
As another example, the problem of longitudinal impact of nonlinear viscoplastic rods will be considered [19]. Figure 7.5 shows the rod under consideration. The coordinate axis will be chosen with the origin at the end of the bar and the Ox positive axis directed along the bar. We will suppose that for t < 0 the bar is at rest, while for t = 0 the end of the bar is subjected to a constant velocity impact, so that for t > 0 the particles of the bar are no longer at rest. The bar will be considered thin in the sense that lateral inertia-that is, inertia in directions transverse to the generatrices of the bar-can be neglected. Assuming the compressive stresses to be positive and the displacement of particles parallel to the axis of the bar to be equal in a given transverse plane for the whole cross section, the equations for longitudinal motion can now be derived.
o
(E Fig. 7.5
x
)
())
-l f-
dx
Schematic diagram of the rod.
If one considers that the element of the rod for which t = 0 is limited by the cross-sectional planes at x and x + dx, at time t these planes will possess the coordinates x* and x* + dx*. Denoting the displacement by u, we have x* = x + u. Neglecting the higher-order derivatives of u, this gives dx* = (l + au/ax) dx. If the cross-sectional area is A and the density is p, both at time t = 0, the equation of motion for the element of length dx* in
7.2 Transformation for a Given Group of Transformations
153
the Lagrangian system of coordinates is
a2u aF af2--a:x
A
p
.
(2 35)
where F(x, t) is the force acting on a constant cross section of coordinate x at time t. In terms of particle velocity v(x, t) and stress O'(x, t), Eq. (2.35) can be written as
av
aO' ax
Pat = -
(2.36)
The constitutive equation which establishes a relation between the strain e and the stress 0' and which describes the mechanical properties of the material considered, i.e., viscoplastic material, must be combined with Eq. (2.36).
ae
at
=
D(.!!.-. - l)P
(2.37)
0'0
This type of constitutive relationship represents plastic deformation phenomena on account of high-speed impact. The constants D and p depend on the type of materials under consideration. For small deformations, the strain displacement relationship can be written as
-ae = - -av
at
(2.38)
ax
The boundary and initial conditions for the problem under consideration are
v(O,t)=vo v(oo,t) = 0, v(x,O) =
°
vo > 0, t
>
t>o
°
(2.39)
where vo, a constant, is the velocity of impact. Eliminating 0' and e from Eqs. (2.36)-(2.38), yq(
::~ )( - ~~
r
=
~~
(2.40)
where q = 1/P and y = 0'0/ pD q. By introducing the similarity transformation (2.41)
where 'IJ
= kx f t'",
m = l/(q + 1)
k =
1 l)vg
(yq)l/(q+
g-
O/(q+ I)
154
7.
r:
Direct Transformation
Eq. (2.40) and auxiliary conditions (2.39) can now be written as
d:J _
dTj2
m'rl(·r
df
dTj
=0
(2.42)
subject to the boundary conditions
f(O)
=
f(oo) = 0
1,
(2.43)
The closed-form solution of Eq. (2.42) can be obtained by integrating Eqs. (2.42) and (2.43), giving (2.44a) where
13 = -H (p C
- 1) j (p + 1)],
= { 413 [
0 = P j (p - 1)
]2}1/(1-2/J)
f(O)
(2.44c)
f(O - t)
'TT
(2.44b)
and I'( ) is the gamma function. From Eq. (2.44a), the missing initial slope 1'(0) is found to be 1'(0)= -ljc/J= -ljcp/(p-I)
or
'0 = _{
f()
413 [ 'TT
f(O) f( 0 - t)
j2}P/(P+I)
(2.44d)
It is seen to be a function of p. From the physical point of view, solution (2.44a) gives the particle velocity as a function of x and t. Substituting solution (2.44a) into Eq. (2.38) and integrating over t, an expression of the nominal compressive strain will be obtained. We will now attempt a conversion of the boundary value problem into an initial value problem. Equations (2.42) and (2.43) can be written as
d
2g
-+mTj dTj2
where g = 1 -
( dg dTj
)2-
q
=0
f. Also, the boundary conditions can g(O) = 0,
g(oo) = 1
(2.45) be written as (2.46)
7.3
155
Extension of the Transformation Method
Invariance of Eq. (2.45) under the linear group of transformations 1j = A ""'ii,
gives a relation between
(Xl
and
(2.47)
(X2:
(X21 (X I = (q + 1)1(q - I) Further, let dg(O) I d1j
(2.48)
= A. This leads to another relationship, (X2 -
(XI
= I
(2.49)
From Eqs. (2.48) and (2.49), (XI
=
t(q - I),
(X2 =
t(q
+ I)
(2.50)
To find A, the boundary condition (2.46) at infinity is utilized, i.e., giving A
= [g(oo)r 2 / (Q+ I)
(2.51)
The value of 1'(0) obtained by this technique is compared with the value of 1'(0) from Eq. (2.44d). Table 7.2 lists the values obtained by the two methods [19]. The agreement is good. TABLE 7.2 Comparison of Solutions"
p
- 1'(0) using Eq. (2.51)
- 1'(0), exact
2.0 3.0 4.0 5.0 6.0 7.0
5.718 4.730 4.550 4.500 4.489 4.488
5.734 4.748 4.348 4.436 4.386 4.354
a Reprinted from [191 with permission of the Industrial Mathematics Society.
7.3
7.3.1
EXTENSION OF THE TRANSFORMATION METHOD FOR A GIVEN GROUP OF TRANSFORMATIONS Spiral Group of Transformations
It was pointed out that the class of transformations in the works of both Toepfer [1] and Klamkin [2] belong to the linear group of transformations. The question naturally arising is whether or not there are other groups of
156
7.
Direct Transformation
transformations which will lead to similar results when applied to other types of equations. As an example, consider the equation of heat conduction with exponential heat generation by Na and Tang [20], as follows 2T
d dr2
+ P + 1 dT + f3e T = 0 r
(3.1)
dr
subject to the boundary conditions
= 0,
dT(O)j dr
T(I)
=0
where p equal to - 1, 0, or 1 corresponds to plane, cylindrical, or spherical coordinates, respectively. For this equation, a linear group of transformations, such as those defined in Eq. (2.2), will lead to the conclusion that (Xl and (X2 must both be zero for Eq. (3.1) to be independent of the parameter of transformation. In other words, the method is not applicable under the linear group of transformations. As a second trial, suppose we introduce the spiral group of transformations, defined by
(3.2) Equation (3.1) becomes 2T
d &2
+ P + 1 dT + f3e T = r &
e- 2<X1a( d
2T
~2
+ P + 1 ddTr_ ) + f3 e f
=0
<X2
ae T (3.3)
Equation (3.3) will be independent of the parameter of transformation a if the powers of e are the same, i.e.,
(3.4) Equation (3.3) becomes d 2T p+1 - 2+ - - dT
df
f
df
+f3e T = 0
(3.5)
The boundary condition at r = 0 can be transformed to dT(O)j di = 0
(3.6)
Following the same steps as before, the unknown boundary condition is set equal to the parameter of transformation, i.e., T(O) = a
which, upon transformation, gives
7.3 Extension of the Transformation Method
157
This boundary condition is seen to be independent of the parameter of transformation if (3.7) Thus we get
T(O) = 0
(3.8)
Equations (3.4) and (3.7) then give al
= -
L
a2 = 1
(3.9)
To find the parameter of transformation, the boundary condition at the second point is used, which gives (3.10) or
T= -
a
at
r = e a/ 2
(3.11)
The evaluation of the parameter a from eq. (3.10) or (3.11) is not as easy as in Eq. (2.13), since now the second boundary condition is given at finite distance instead of at infinity. To get the parameter a, it is eliminated from Eq. (3.11), which gives (3.12) Thus, the solution of Eq. (3.1) takes the following steps. Equation (3.5) is solved as an initial value problem using the boundary conditions given in Eqs. (3.6) and (3.8). One T-r curve is obtained. Another curve is given by Eq. (3.12). The intercepts of these two curves give values of T and r which can be used to determine the parameter a from Eq. (3.10) or (3.11). Solutions of Eq. (3.5) with the boundary conditions given in Eqs. (3.6) and (3.8) are plotted in Fig. 7.6 for five values of /3. Also plotted in this figure is the T-r curve from Eq. (3.12). For a given value of /3, the intercepts of these two curves give a pair of values of T and i, For /3 > 1.7, there are two intercepts, and thus two solutions are possible. For /3 = 1.7, the curves are tangent to each other and there is only one possible solution. For values of /3 smaller than this critical value, no solution exists. The duality of solutions in the first case (/3 > 1.7) is typical of many nonlinear ordinary differential equations. For most physical problems, however, only one solution is physically meaningful. Figure 7.7 is a plot of such solutions. Physically, they are the temperature distributions. The stability criteria for the determination of the physically meaningful solution will be exploited in detail in Section 7.4.
7. Direct Transformation
158 0
-2
4
T
-6
-8
-10 0
40
20
60
r Fig. 7.6 Graphic determination of a. (Reprinted from [20] with permission of AcademieVerlag, Berlin.)
0.25 .------.-----,,..----,-----,....----,
0.20
0.15 T
0.10
0.05
0.2
0.4
0.6
0.8
Fig. 7.7 Temperature distribution. (Reprinted from [20] with permission of AcademieVerlag, Berlin.)
7.3 Extension of the Transformation Method
159
7.3.2 The Eigenvalue Problem of Ames
At this point, an interesting alternative approach for the solution of the problem of heat conduction with exponential heat generation will be outlined. This method is due to Ames [21]. Consider again Eq. (3.1) and its boundary conditions, (3.13) dT(O) --=0 dr '
T(l)
=0
(3.14a, b)
Ames introduced the idea by thinking of f3 as an eigenvalue, say f3 = w2 • Then f3 is eliminated from Eq. (3.13) by setting R = er, which gives d
2T
dR 2
+
P
+1
R
dT +e T
dR
=0
(3.15)
However, w will now appear in the boundary condition. dT(O)/ dR
= 0,
T(w) = 0
(3.16a, b)
We next introduce a spiral group of transformation similar to Eq. (3.2) with the exception that the parameter of transformation is now identified as w, i.e., (3.17) Like Eq. (3.3), Eq. (3.15) is transformed to d 2T p + 1 dT -+- - +eT=O dIP R dR
(3.18)
and the relation between a, and a 2 is found to be -2a,
= a2
(3.19)
The boundary condition at R = 0, Eq. (3.16a), is transformed to dT(O)/ dR = 0
(3.20)
Next, we deviate from Ames [21] slightly by setting the unknown boundary condition at R = 0 equal to the parameter of transformation plus an arbitrary constant y, i.e., T(O) = w + y
which, upon transformation, gives
T (0) + a2w = w + Y
160
7.
Direct Transformation
We then get (3.21) and
T(O) = y
(3.22)
Equations (3.19) and (3.21) give (3.23) Finally, to find the eigenvalue w, the boundary condition at R = w, Eq. (3.16b), is transformed, which gives
T+w=O
at
(3.24)
Eliminating w in Eqs. (3.24), we get
Re T/ 2 = - T
(3.25)
Thus, for a given value of y, Eq. (3.18), subject to the boundary conditions (3.20) and (3.22), is integrated as an initial value problem. As the integration proceeds, Eq. (3.25) is checked. At the point where Eq. (3.25) is satisfied, the outer edge of integration is reached and the eigenvalue w can be calculated from either expression in Eq. (3.24). Once w is known, the corresponding value of p can be found by the definition p = w 2 • In other words, for a given value of the missing initial condition T(O) (now, written as w + y), there is a corresponding value of p. The difference between this approach and that of Section 7.3.1 is that in the latter a value of p is given and the missing initial condition T(O) is sought, whereas in Ames's approach [21], the missing initial condition T(O) is given and the problem becomes one of determining p. It should be noted that the case discussed by Ames [21] is one of the special cases of the above treatment, namely for y = 1. Details of the above are left to the reader as an exercise. (See problem 6 at the end of this chapter.) 7.3.3
Transformations from Homogeneous to Nonhomogeneous Boundary Conditions
The method developed up to this point depends on one important condition. The boundary condition at the second point must be nonhomogeneous if the second point is at infinity. If this boundary condition is homogeneous, the method will fail in the last step, namely the calculation of the parameter of transformation. Referring to the example discussed in Section 7.2.1, it is evident that, if the boundary condition at the second point is df( 00)/ dYj = 0 instead of df( (0) / doq = 1, then the parameter
7.3
161
Extension of the Transformation Method
flow
--I--F\-----FI----- x
Fig. 7.8
Schematic diagram of the jet.
A can not be calculated from its transformed form, Eq. (2.12). For such
problems, one more transformation is necessary. This method will be illustrated by an example. The outflow of an incompressible fluid from a narrow two-dimensional slit into a mass of the same fluid (see Fig. 7.8) will be considered. For such flows, the boundary layer equations can be applied, which are
~+
ax
av ay
=0
(3.26)
au uau - + vau ax ay- = pay-2 2
(3.27)
with the boundary conditions
y=O:
v =0,
y=
u=O
00:
au = 0 ay
Since the pressure in the surrounding fluid is constant, the momentum flux J remains a constant, i.e., J
oo = PJ_+oo u 2 dy= const
By introducing the following dimensionless quantities,
x=x/L, Y=~Nre y/L, u= u/Uo, v=~Nre «i u; Nre=pUoL//-L and introducing the stream function 1/1, defined by
u = 31/1/3y,
v = -31/1/3x Eq. (3.26) is satisfied identically and Eq. (3.27) becomes 31/1 3~ a1/l a~ -----
ay ax ay
3x ay2
331/1
ay3
(3.28)
162
7.
Direct Transformation
subject to the boundary conditions
a~
y=o:
ay2
y=
atl'
00:
ay
=0
'
atl'
-=0
ax
= 0
Equation (3.28) can be transformed from a partial differential equation to an ordinary equation by introducing the following similarity transformations: '1'/ =
y/X 2 / 3,
Equation (3.28) becomes
dy d'l'/3
+ .1 f d~ + .1 ( d'l'/2
3
3
df dI]
)2 = 0
(3.29)
with the boundary conditions '1'/ = 0:
f(O)
'1'/=00:
--=0
= 0,
df(oo) d'l'/
The equation of momentum flux is transformed to J
L J+ (d-df )2d'l'/ !N '1'/
= pUJ - V1 V re
(3.30)
00
-00
Equation (3.29) is a nonlinear boundary value problem with the boundary condition at the second point being homogeneous. As a result, the method cannot be applied directly before this boundary condition is transformed to nonhomogeneous form. To overcome this difficulty, it is necessary to introduce a transformation 'I'/=B'ij,
f=CJ
It can be shown by simple substitution that Eq. (3.29) will be independent of the parameter of transformation B if BC= I
Equation (3.29) then becomes 3J d d'ij3
+.1 - d2J +.1 ( dJ )2 = 3 f d'ij2 3 d'ij
0
(3.31)
163
7.3 Extension of the Transformation Method
and its boundary conditions at '1'/ = 0 become
J(O) = 0, d2j(O)j drj2
'ij=0:
= 0
(3.32)
Inspection of the boundary conditions of Eq. (3.29) shows that dfj d'l'/ starts from certain constant values at '1'/ = 0 and decreases to zero as '1'/ approaches infinity. Therefore, if dfj d'l'/ is integrated from zero to infinity, a finite constant will be obtained, i.e., area
=
i X
d'l'/
00
= const
Under the linear transformation, we then get
i
area = C
oo
~
drj = C[J(oo) - J(O)]
If we now identify
C
= area
and also, from the boundary condition (3.32), ](0) = 0, we get '1'/
= 00:
J(oo)
= 1
(3.33)
From Eq. (3.30), we get
U~L 3J+00( -dJ )2 drj
J=p--C 'N VlY re
-00
drj
(3.34)
Since J is a given quantity, Eq. (3.34) will be used to determine the constant C. Thus, the solution of Eq. (3.29) is reduced to the solution of Eq. (3.31) subject to the boundary conditions (3.32) and (3.33). The boundary condition at the second point has now been transformed to a nonhomogeneous form. Without repeating the same process of deriving the transformation from boundary value to initial value problem, the following is a summary of the solution of Eq. (3.31). First the initial value problem
'1'/*
= 0:
f*(0) = 0,
df*(O)
d:;;*
=
1,
7.
164
Direct Transformation
is solved (tabulated in Table 7.3) and, in particular, we know f*( 00), which can be used to calculate the parameter of transformation, A =
[f*(oo)r 2 = 0.1667 TABLE 7.3
Solution of the Initial Value Problem
TJ·
r
dj*/dTJ·
TJ·
0.00 1.25 2.50 3.75 5.00 6.25 7.50
0.0000 1.1517 1.8864 2.2304 2.3681 2.4198 2.4386
i.oooo
r
0.7789 0.4069 0.1708 0.0652 0.0240 0.0087
8.75 10.00 11.25 12.50 13.75 15.00
2.4454 2.4479 2.4487 2.4490 2.4491 2.4491
dj*/dr,· 0.0031 0.0011 0.0004 0.0001 0.0001 0.ססOO
Finally, the solution of Eq. (3.31) can be obtained by ij
= A -1/21]*,
The result is shown in Table 7.4. TABLE 7.4
Solution of Eq. (3.31)
r;
J
dJ/ d1j
r;
J
dJ/ d1j
0.000 3.062 6.123 9.185 12.246 15.308 18.369
0.ססOO
i.oooo
0.4702 0.7702 0.9106 0.9669 0.9880 0.9957
0.1298 0.0678 0.0285 0.0109 0.0040 0.0015
21.431 24.492 27.554 30.616 33.677 36.739
0.9984 0.9995 0.9998 0.9999 0.9999 0.9999
0.0005 0.0002 0.0001 0.ססOO 0.ססOO 0.ססOO
The physical quantity of interest in such an analysis is the velocity distribution. In terms of and ij, the following expression can be derived:
J
u
-
1
[J2/3j(pX)I/3J = 22 / 3
[
- 2
2
-
~ (dfj) dfj dfj 00
df
df
]
(3.35)
and (3.36)
165
7.4 Uniqueness of the Solution
In a given two-dimensional jet, the total moment J is always known. Given the fluid, the viscosity can be found from a table of properties of fluids. The integrand in Eqs. (3.35) and (3.36) is found to be 0.112. Equations (3.35) and (3.36) therefore become
u/ {J2/3 /(VX)1/3} = 0.01960dJ/ d1j
and
'II = 0.1400(J/ V2)1/3Y /
x2/ 3
which together with Table 7.3 gives the velocity at any point (x, y). 7.4 UNIQUENESS OF THE SOLUTION
It was found in the previous section that Eq. (3.1) may have two, one, or no solutions depending on whether the value of f3 is, respectively, smaller than, equal to or greater than the critical value (f3 = 1.7). For the case of dual solutions (f3 < f3 cr) ' the question is raised as to which of the solutions is physically meaningful. This duality of solutions is also found in the solution of other equations in engineering problems. Extensive discussion on the stability of the equations governing the flow in chemical reactors can be found in [22-25]. Aris [24] and Luss [25] gave a stability criterion for a class of more general nonlinear ordinary differential equations. Due to the frequent occurrence of this problem in the solution of such equations, a detailed description of the theory is presented by using the solution of Eq. (3.1) as an example. A direct way of answering the question is to formulate a corresponding transient problem with a homogeneous initial condition. Since it is a well-posed mixed initial and boundary value problem, its solution is unique [26]. The bounded solution, which is approached by the transient as the time becomes infinity [27], is called the steady solution. Thus the transient equation must be solved, and this involves one more independent variable, the time coordinate. To avoid the added complexity of solving a nonlinear partial differential equation, a simple method will be presented in this section which enables the determination of the physically meaningful solution without solving the transient equation completely. Using a steady solution as the initial condition, the stability of the subsequent temperature can be investigated. From the stability criterion the unique solution can then be determined. The corresponding transient equation for the stability of the solutions of Eq. (3.1) is
oT = r -.! a;or (r 0orT ) + f3e T
(4.1 )
7. Direct Transformation
166
with the initial and boundary conditions
T(r, 0) = To(r)
(4.2)
°
aT(O,r)/ar = 0, T(I,r) = (4.3) in which T is the dimensionless temperature, rand r are dimensionless radial and time coordinates, respectively, and f3 is a parameter. The relationship between the actual values (T', r', t, and a) and the dimensionless values of T, r, r, and f3 is
= '{T',
kt/pcR 2 ,
f3 = '{R 2a/ k where R, k, p, and c are the radius, the heat conductivity, the density, and the specific heat, respectively, and a and '{ are physical constants in the definition of the function of heat generation T
r = r' / R,
r =
[heat generation per unit volume (Btu/ft? sec)]
= aeYT'
Thus, the dimensions of a and '( are British thermal units per cubic foot per second and inverse degrees Rankine, respectively. Assuming To(r) to be a solution of the steady equation Eq. (3.1), i.e.,
1. i£ (r r dr
o) + f3e
dT dr
To
=
°
(4.4)
with the boundary conditions T~(O) =
(4.5)
0,
then To(r) is also a solution of the transient problem defined by Eqs. (4.1}-(4.3). Incorporating an often used definition for stability (infinitesimal stability), let To(r) and To(r) + 8T(r, r), satisfying the boundary conditions in Eqs. (4.3), be two solutions of Eq. (4.1) for which the initial conditions at r = are To(r) and To(r) + 8To(r), respectively. The expression To(r) + 8T(r, r) is inserted in the transient problem defined by Eqs. (4.1}-(4.3) and powers of 8T are neglected. When e lJT is expanded into a series (8T)2 1+ 8T+ - - + ...
°
2!
we obtain a linear variational partial differential equation with initial and boundary conditions for 8t:
a(8T) _ I a [ a(8T)] ---- r--
ar
r
ar
ar
8T(r, 0) = 8To(r) = E:(r) a(8T)
a;:-
I
r=O
= 0,
(4.6)
(say)
(4.7)
°
(4.8)
8Tlr=1=
7.4 Uniqueness of the Solution
167
If all solutions of ~T(r, 'T) of the above system are bounded, then To(r) is said to be stable; otherwise it is unstable. Since To(r) is the specified initial condition and is a known function of r, Eq. (4.6) is linear and the method of separation of variables is applicable. Assuming a solution of the form ~T(r,'T) = X('T)Y(r) (4.9)
and substituting in Eq. (4.6), we have
[1r (rY')' + [JeToy]-Y1 = -a
-x = X
(4.10)
in which the dot denotes d/ dr, the primes d/ dr, and a is a parameter to be determined. Equation (4.10) represents two ordinary differential equations in X('T) and Y(r), respectively. The equation for x('T) is
X +cxX =
0
(4.11)
which has a solution of the form X
= Ae-
(4.12)
aT
where A is an integration constant. For a bounded solution of Eq. (4.6), the real part of a in Eq. (4.12) cannot be negative. The equation for Y(r) is
1r (rY')' + [JeToy+ aY=O
(4.13)
and the boundary conditions are obtained by substituting Eq. (4.9) in Eq. (4.8): Y'(O)=O,
(4.14)
Y(l)=O
Eq. (4.13) with the boundary conditions expressed by Eq. (4.14) defines a Sturm-Liouville problem [28], which states that there is a discrete set of real distinct values a!' a 2 , ••• of the parameter a for which the system in Eqs. (4.13) and (4.14) has nontrivial solutions. These real values a!' a 2 , ••• are known as the eigenvalues, and corresponding to each eigenvalue a j there exists a nontrivial solution Yj(r) known as the eigenfunction of the system. Since a;'s are real and distinct, they may be arranged as follows: al
< a 2 < a3 < . . .
(4.15)
The solution of the variational equation, Eq. (4.6), can be written in the form of a series: ~T(r,'T) =
00
L: Aje-aiTj(r)
(4.16)
j=l
The solution
~T(r,'T)
must satisfy the initial condition in Eq. (4.7) or 00
fer) =
L: A j Yj(r)
j=!
(4.17)
168
7.
Direct Transformation
The integration constants A j can be determined by Eq. (4.17) and the orthogonality property of the eigenfunctions Yj(r). Since the variation c:(r) of the initial condition is arbitrary, all the eigenfunctions Yj(r) may participate in Eq. (4.17). Therefore, a bounded solution l5T in Eq. (4.16) can exist only if the smallest eigenvalue a j is not less than zero. From Eq. (4.15), this condition is satisfied if a l ;> O. In other words, the solution To(r) to the transient problem defined by Eqs. (4.1}-(4.3) is stable if the smallest eigenvalue of the associated system defined by Eqs. (4.13) and (4.14) is equal to or greater than zero. This is the stability criterion. The stability criterion can be applied to determine the physically meaningful solution of the steady problem defined by Eqs. (4.4) and (4.5). If To(r) is the physically meaningful solution of the steady problem, it must be a stable solution of the transient problem defined by Eqs. (4.1}-(4.3). Using To(r) as the initial condition from the definition of stability, we see that the subsequent solution to the transient problem will not change; therefore To(r) satisfies the requirement of the steady solution. In checking stability, the sign of the smallest eigenvalue in the system of Eqs. (4.13) and (4.14) must be found. The method of Galerkin [29] will now be used to locate the eigenvalues approximately, since exact solution of the system is difficult. This approximate solution is, however, accurate enough to find the sign of the smallest eigenvalue. In this method, the nontrivial solution of the system is assumed to have the form M
Y(r) = ~ bjcl>j(r)
(4.18)
1
where M is a definite positive integer, and cl>j is a set of orthonormal functions with respect to the weight function r satisfying the boundary conditions in Eqs. (4.14). The function cl>j may be chosen to be cl>j(r)
= JoCA./)/ NjI/2
(4.19)
in which J o is the Bessel function of order zero, >-y is the jth root of = 0, and
J 0(A.)
(1
2
Nj = J rJo(A./) dr
o By Galerkin's method, Y(r) can be used in Eq. (4.18) to reduce the Sturm-Liouville system in Eqs. (4.13) and (4.14) to a matrix eigenvalue problem:
(I Cij] - «I l5ij])(b) = 0
where l5ij is the Kronecker delta, Cij = A.;l5ij - f3
.r
r2e T0cl>j(r)cI>i r) dr
(4.20)
7.4
169
Uniqueness of the Solution
and [eij] is a symmetric matrix of order M. A standard subroutine in a digital computer may therefore be used to locate its smallest eigenvalue, say a). This a) is equivalent to the smallest eigenvalue for the system in Eqs. (4.13) and (4.14). As an example, consider the case in which the parameter f3 = 0.7 in Eq. (4.4). From [20], we realize 0.7 < f3er' so there are two possible steady solutions of Eqs. (4.4) and (4.5). These two solutions are denoted by T)(r) and T2(r). Table 7.5 shows T)(r) and Tir) in terms of the radial coordinate r, TABLE 7.5
Table of Steady Solutions r
TI(r)
T 2(r)
r
TI(r)
T 2(r)
0.0 0.1 0.2 0.3 0.4 0.5
0.204 0.202 0.195 0.185 0.170 0.151
4.696 4.526 4.072 3.482 2.868 2.281
0.6 0.7 0.8 0.9 1.0
0.128 0.101 0.071 0.037 0.000
1.739 1.244 0.792 0.379 0.000
With To(r) = T)(r) in Eq. (4.13), the smallest eigenvalue a) is found by Galerkin's method to be equal to 4.96, which is greater than zero. It is seen that T)(r) is a stable solution of the transient problem; therefore T)(r) is the physically meaningful steady solution. With To(r) = Tir) in Eq. (4.13), the smallest eigenvalue a l is found to be - 25.09 (which is less than zero); therefore Tir) is the unstable solution and it is not meaningful. The uniqueness question is thus removed. Even though the discussion above is adequate for a stability analysis. of the multiple solutions to a given differential equation, a numerical solution of the transient partial differential equation, Eq. (4.1), will now be given to verify the conclusions. Since the equation is nonlinear, the finite-difference method [30] of numerical solution is used. Laying a grid system on the r-1" plane with uniform mesh size I1r and 111" in Fig. 7.9, the difference approximation to Eq. (4.1) is T!+) - T! 111"
T!:/ - 2T!+1 + T!~ll
1
(I1r)2
il1r
----:.....:..-=-------.:.-----..:...~+-
T!:l) -
itt;
211r
+f3expTf+I,
< M, i > 0 (4.21) where TI = T(i I1r,j 111"). At r = 0, i = 0, the second term on the right in the 0< i
above equation is indeterminate since i I1r = 0 and T!:)) = T!~/. from the
170
7.
Direct Transformation
! tu
t
o
t:1J.
11------
I-
M t:1J-= I
------1
r
Fig. 7.9 The grid system.
boundary condition aTlar = 0 at r = O. Applying L'Hopital's rule, Eq. (4.1) may be written in the form aT =2
aT
2
aT2 ar
+f3e T
r=O
(4.22)
'
The difference approximation is T/+
I
-
T/
_----,--__ = 2
Tj+1 1+1
~T
zrr ' + Tj+1 I
+ f3 exp T/+ I
I-I
(~rf
i
= 0, j > 0 (4.23)
The initial and boundary conditions at r = 1 are
1';0 = To{i ~r)
(specified)
(4.24)
Tit = 0
(4.25)
Solving the system in Eqs. (4.21) and (4.23}-{4.25) gives T at any grid point. Since the system involves a nonlinear term f3 exp T/+ I direct solution of these simultaneous algebraic equations is impossible. We rearrange Eqs. (4.21) and (4.23) and use Eq. (4.25): (a - 2b)T6+ 1 + 4bT{+1 (b
+ c)Tj+ 1+ aTj+ 1+ (b I-I I (b
=
T6/~T
- c)Tj+ I = Tj/~T 1+1 I
+ f3exp T6+ 1
+ fJexp tt: I 1
+ c)Ti+ I + .at: I = rtM-I I~T + f3exp ttr I M-2 M-I M-I
in which
a = I/~T + 2/(~r)2, b = - 1/(~r)2, c
= 1/(2r~r)
(4.26)
7.4 Uniqueness of the SolutIon
171
For the U + l)th time step, T/ (i = 0, 1, ... , M) is known either from the initial condition or through the computation of the jth time step; therefore, the right-hand side of Eq. (4.26) is known except for the term f3exp Tf+ I. For the first iteration we replace this term by f3exp T/; therefore, the right-hand side of Eq. (4.26) is approximately determined and Tf+1 (i = 0, 1, ... , M - 1) can easily be solved from the linear equations of which the coefficient matrix is tridiagonal. The second iteration is computed by using T/+ 1 from the first iteration in the term f3 exp T/+ I, and the iteration goes on until the difference between subsequent values of Tf+ 1 is insignificant. The tabulated values of T1(r) and Tz{r) given in Table 7.1 are now used as the initial conditions to find the transient solutions. The results are summarized in the following examples. As a first example, take initial condition to be To(r) = T1(r) and intervals t::.r = 0.025 and t::.T = 0.Ql. The solution stays the same as the initial condition T1(r), even after 50 time steps. This demonstrates the stability of the transient solution T1(r). Next, the initial condition To(r) = Tz{r) may be chosen, with intervals t::.r = 0.05 and t::.T = 0.001. The transient solution before T = 0.09 is plotted in Figure 7.10. The temperature increases as does the dimensionless time T, 8.0
r-----,------.-----.-------r----...,
6.0
4.0
2.0
0.2
0.4
0.6
0.8
Flg.7.10 Transient solution for T.(r) = T 2(r), fJ = 0.7.
1.0
172
7.
T -+
0.2
Direct Transformation
00
r -__ ----'
0.1
0.2
Flg.7.11
0.4
0.6
Transient solution with To(r)
0.8
1.0
= 0, f1 = 0.7.
and overflow occurs in the computer due to such a large T during computation of e T at 7' = 0.09. This demonstrates the instability of the transient solution Tir). To demonstrate further that T(r) shown in Table 7.4 is the physically meaningful steady solution, the present numerical scheme is used with initial condition To(r) = 0 and intervals Ll7' = 0.001 and Llr = 0.05 and the transient solution plotted in Figure 7.11. This approaches the steady solution as time increases. In comparison with the temperature shown in Table 7.4, the transient solution computed in this case is very close to T(r).
The above analysis of the transient partial differential equation further demonstrates the validity of the stability criterion. Thus, without solving the transient problem entirely, it is possible to determine the physically meaningful solution among all possible solutions to the steady equation by investigating stability. The stability criterion depends upon the sign of the
Problems
173
smallest eigenvalue of the Sturm-Liouville system in Eqs. (4.13) and (4.14). This eigenvalue can be easily located, by Galerkin's method, for example. PROBLEMS
1. For equations of the type where y is absent, i.e., F[
2: a.( i
I
dy dx?
)m/(
dy dx
i
)n xPi] = 0
where F is an arbitrary function, introduction of the spirial group of transformations, x = xea,a, y = Y +a 2a will always lead to the same equation a l = 0 in order that the differential equation be independent of the parameter of transformation a. As an example, consider the boundary value problem dy dx 2
+ 2(
dy dx
)2 + 1 = 0'
O
dyd(x ) = 0, y(l) = 0
Solve the problem as an initial value problem by following the same steps as those discussed in Section 7.3.1 and compare the results with the exact solution, y = 0.5(ln cos If x - In cos If). Hint: a\ = 0 and a 2 = 1; the parameter of transformation is given by a = - y(l). 2. Consider the boundary value problem, y3 dy dx?
+ sin(y
dy ) dx
+ 1= 0
'
dy(O)
~
=0, y(I)=O
Reduce it to an initial value problem. Hint: Even though a sine function is present in the differential equation, no difficulty will be encountered since one relation between the a's, namely, 2a 2 - al = 0 will be able to reduce the differential equation to a form independent' of the parameter of transformation. Answer: yeO) = 0.90168. 3. In an analysis of the unsteady flow of power-law fluids, Habib and Na [31] obtained the following boundary value problem: d 2F
-2 d1]
dF )2-N _ + 21] ( -d - 0, 1]
F(O) = 0, F(oo) = I
Transform the problem to an initial value problem and find the solutions for N = 0.25, 0.75, 1.25, and 2.0. Answer: The missing slope dF(O)/ dn is 0.6574, 1.0200, 1.1992, and 1. 3090, respectively.
174
7.
Direct Transformation
4. In a note, Pifke and Goldberg [32] applied the Foeffl-Hencky large deflection membrane equations to a uniformly loaded annular membrane, fixed at the outer boundary and supported at the inner boundary, and reduced the governing equations to the following nonlinear boundary value problem:
2 d~~) - (1 + v)f(l) = 0, f(A 2) =
°
Transform the problem into an initial value problem and compare the solutions with the series solution f(~)
= taicJ>(l - cJ> - tcJ>2 - ticJ>3 - HcJ>4 - #cJ>5 - -'f2fcJ>6 + ... )
where cJ>=(~-A2)/ai
and a l is given by the following table: 0.0
0.01
0.1
0.2
0.3
0.4
1.7245
1.724
1.713
1.678
1.625
1.553
Hint: First, let cJ> = 1.. 2 -~. We then get (Xl = 3 and (X2 = 4. 5. Consider again the problem of gas diffusion given in problem 3 of Chapter 4. Transform the boundary value problem 2 d A -2--
C
dZ
2
_ (kL -D ) CA-O,
to an initial value problem and compare the solutions with the exact solution cosh ~kL 2 / D (1 - z) cosh ~kL2/ D for kL / D = 10- and Hint: The independent variable z has to be transformed to s = 1 - z before the method of transformation is applied. 6. Referring to Section 7.3.2, find the value of f3 for y = - 0.6 and - 0.5. 2
2
10- 3 •
REFERENCES I.
Toepfer, K., Grenzschichten in Flussigkeiten mit kleiner Reibung, Z. Math. Phys 60, 397-398 (1912).
References
175
2. Klarnkin, M. S., On the transformation of a class of boundary value problems into initial value problems for ordinary differential equations, SIAM Rev. 4,43-47 (1962). 3. Na, T. Y., Transforming boundary conditions to initial conditions for ordinary differential equations. SIAM Rev. 9, 204-210 (1967). 4. Na, T, Y., Further extension on transforming from boundary value to initial value problems, SIAM Rev. 10, 85-87 (1968). 5. Ames, W. F., "Nonlinear Differential Equations in Engineering," Vol. 2, Chapter 2, Academic Press, New York, 1972. 6. Na, T. Y., and Hansen, A. G., Similarity analysis of differential equation by Lie group, J. Franklin Inst. 292471-489 (1971). 7. Bluman, G. W., and Cole, J. D., "Similarity Methods for Differential Equations," Springer-Verlag, New York, 1974. 8. Ovsjannikov, L. V., "Group Properties of Differential Equations" (G. Bluman, trans.), California Institute of Technology, 1967. 9. Na, T. Y., Group-theoretic analysis of unsteady one-dimensional gas dynamics equations. Part I: Similarity solutions by Lie group, Symp. Group-Theoretic Methods in Mech., Calgary, Canada (17-19 August 1974). 10. Na, T. Y., Group-theoretic analysis of unsteady one-dimensional gas dynamics equations. Part II: Non-similar solutions by Lie series, Symp. Group-Theoretic Methods in Mech., Calgary, Canada (17-19 August 1974). II. Goldstein, S., "Modem Developments in Fluid Dynamics, " Oxford Univ. Press, London, 1957. 12. Lie, S., and Engel, F., "Theorie der Transformationsgruppen," Vol. 1-3, Taubner, Leipzig, 1890. 13. Birkhoff, G., "Hydrodynamics," Princeton Univ. Press, Princeton, New Jersey, 1950. 14. Morgan, A. J. A., Reduction by one of the number of independent variables in some systems of partial differential equations, Q. Appl. Math. 3, 250-259 (1952). 15. Hansen, A. G., "Similarity Analyses of Boundary Value Problems in Engineering," Prentice-Hall, Englewood Cliffs, New Jersey, 1964. 16. Na, T. Y., and Hansen, A. G., General group-theoretic transformations from boundary value to initial value problems, NASA Rep. CR-61218, 1968. 17. Lemieux, P. F., and Unng, T. E., The laminar two-dimensional free jet of an incompressible pseudoplastic fluid, J. Appl. Mech. 35, 810-811 (1968). 18. Klarnkin, M. S., Transformation of boundary value problems into initial value problems, J. Math. Anal. Appl., 32, 308-330 (1970). 19. Seshadri, R., and Na, T. Y., Invariant solution for nonlinear viscoplastic impact, Ind. Math. 25, 37-44 (1975). 20. Na, T. Y., and Tang, S. C., A method for the solution of heat conduction with nonlinear heat generation, Z. Angew. Math. Mech. 49!, 45-52 (1969). 21. Ames, W. F., and Adams, E., Exact shooting and eigenparameter problems, Nonlinear Anal. Theory, Methods Appl. 1, 75-82 (1976). 22. Raymond, L. R., and Admundson, N. R., Some observations on tubular reactor stability, Can. J. Chem. Eng. 42, 173-177 (1964). 23. Amundson, N. R., Some further observations on tubular stability, Can. J. Chem Eng. 43, 49-55 (1965). 24. Aris, R., On stability criterion of chemical reaction engineering, Chem Eng. Sci., 24, 149-169 (1969). 25. Luss, D., Sufficient conditions for uniqueness of the steady state solutions in distributed parameter systems, Chern. Eng. Sci., 23, 1249-1255 (1968). 26. Garabedian, P. R., "Partial Differential Equations," Wiley, New York, 1964.
176
7.
Direct Transformation
27. Copple, C., Hartree, D. R., Porter, A., and Tyson, H., The evaluation of transient temperature distribution in a dielectric in an alternate field, J. Inst. Electr. Eng., 85, 56-66 (1939). 28. Churchill, R. V., "Fourier Series and Boundary Value Problems," McGraw-Hili, New York, 1963. 29. Crandall, S. H., "Engineering Analysis," McGraw-Hili, New York, 1956.
30. Bruce, G. H., Peaceman, D. W., Rachford, H. H., Jr., and Rice, J. D., Calculation of unsteady-state gas flow through porous media, Am. Inst. Mining Metal. Eng. Trans. 198, 79-91 (1953). 31. Habib, I. S., and Na, T. Y., AIAA J. 11,238-239 (1973). 32. Pifke, A. B., and Goldberg, M. A., AIAA J. 2, 1340-1342 (1964).
CHAPTER
8
METHOD OF TRANSFORMATIONREDUCED PHYSICAL PARAMETERS
8.1 INTRODUCTION
In Chapter 7 we showed how Toepfer's ideas can be considerably clarified and generalized in a deductive way by first introducing a group of transformations and then imposing the condition that the transformed differential equation be independent of the parameter of transformation, thereby obtaining the particular transformation that reduces the boundary value problem to an initial value problem. Many important extensions were made possible. It is the purpose of this chapter to demonstrate still one more extension of the method for a particular type of problem. This extended method can be applied to the class of equations for which certain physical parameters appear either in the boundary conditions or in the differential equation and, most important, numerical solutions for a range of these parameters are required. This type of equation can be found in most engineering problems, where interest is usually in the effect of changes in certain physical parameters, and not just in that solution for a particular set of values of the parameters. In such cases, solutions to the equation for a range of the parameters are needed. The method will be presented through a few examples. 8.2 REDUCED PHYSICAL PARAMETERS 8.2.1 Boundary Layer Control
For flow over external surfaces, such as airplane wings or turbine blades, there exists a very thin viscous layer near the surface of the body, known as the boundary layer. The solution of the conservation equations governing the flow in this layer permits the determination of the viscous shearing force by the fluid on the surface. It is well known that the boundary layer separates from the surface before the trailing edge is reached if the 177
178
8.
---
Reduced Physical Parameters
y
-
c~u v(x.O)=-~ 2 x Fig. 8.1 Schematic diagram of the flow. C < 0, suction; C > 0, blowing.
pressure increases in the flow direction, causing a considerable increase in drag. It is therefore desirable to prevent separation by adopting suitable boundary layer control measures, one of which is removal of the fluid from the boundary layer before it is given a chance to cause separation. Figure 8.1 shows a schematic diagram of boundary layer flow with suction. From the mathematical point of view, the only difference between the present problem and the boundary layer analysis in Section 7.2 lies in the fact that the boundary condition of the velocity component v in the y direction is not zero after the introduction of surface mass transfer. It can be shown [1] that if the velocity of suction is of the form (2.1)
the boundary layer equations, Eqs. (l.l) and (1.2) of Chapter 7, are similar and can be transformed by Eqs. (1.5) and (1.6) of Chapter 7 into the following equation:
d'1 1 d~ -+-f-=O d1/3 2 d1/2
(2.2)
subject to the boundary conditions 1/ = 0:
f(O)
= C,
df(O)/d1/
= 0;
1/ = 00:
df(oo)/d1/ = 1
Since the only difference between this problem and Eq. (2.1) of Chapter 7 is the boundary condition of f( 1/) at 1/ = 0, all steps between Eqs. (2.1) and (2.13) of Chapter 7 remain valid. However, the boundary conditions at 1/ = 0 will now be transformed to ij
= 0:
J = A -a 2 C, dJ(O)/dij = 0
which can be independent of A only if (X2 = O. From Eq. (2.4) of Chapter 7, also equals zero. This means the method cannot be applied. The difficulty can be overcome by modifying the original method if solutions of Eq. (2.2) for a range of values of C, instead of for a particular
(X\
8.2
179
Reduced Physical Parameters
value,are sought. The key step of this method lies in the introduction of a E, defined by
E = A - e = A -1/3e
(2.3)
a2
The boundary conditions at 'ij = 0 now become 'ij=0:
J(O)
= E,
dJ(O)/ dij
=0
(2.4)
From Eq. (2.1), the constant e is a dimensionless constant, its value being directly proportional to the velocity of suction. (If e is negative, we speak of blowing.) In other words, when e is taken to be zero, the velocity of suction is zero. As e increases, the velocity of suction increases linearly with e, even though the suction velocity still follows the same relation, according to which v(x,O) is inversely proportional to x. Therefore, if a family of solutions of Eq. (2.2) is sought for a range of values e, the effect of the strength of blowing on the flow can be investigated. The constant E, based on Eq. (2.3) and Eq. (2.7) of Chapter 7, is the product of d~(O)/ d'l2 raised to the - t power and C. Inspection of Eq. (2.2) and its boundary conditions shows that for a given value of C, there is a corresponding value of d~(O)/ d1J2. Equation (2.3) therefore gives the relation between E and C. Since we are interested in the solutions of Eq. (2.2) for a range of values of C, these solutions correspond to a definite range of values of E. To start the solution, values of E, instead of C, are assigned and Eq. (2.6) of Chapter 7 is integrated with the boundary conditions given by Eq. (2.10) of Chapter 7 and Eq. (2.4). Again, this is an initial value problem. From the solution, dJ( (0) / dij can be found, and the value of A is then computed from Eq. (2.13) of Chapter 7. Finally, the value of C can be determined from Eq. (2.3). By changing values of E, solutions for other values of C can be obtained. Numerical solutions of Eq. (2.2) are shown in Table 8.1 for C from -7 to 7 with increments of I. Consider, for example, the solution corresponding to C = 4. Solutions of Eq. (2.6) of Chapter 7 with boundary conditions (2.10) of Chapter 7 and (2.4) are plotted in Fig. 8.2. It is seen that dJ/ dij approaches a constant value of 0.4864 at 'ij = 4.2. Application of Eq. (2.13) of Chapter 7 then gives A = 2.9478. The value of C corresponding to this solution (C = 4) can then he computed from Eq. (2.3), which gives C = 5.7353. The solution of the original equation, Eq. (2.2) can be obtained by using the transformation defined in Eq. (2.2) of Chapter 7 since now ai' a 2 , and A are known constants, namely - t, t, and 2.9478, respectively. Other data in Table 8.1 are obtained in the same manner, without the need of a trial-and-error process.
8. Reduced Physical Parameters
180
TABLE 8.1 Numerical Results of Eq. (2.2)Q
E
dJ( 00)/ rEf
A[= 1"(0)]
-7 -6 -5
34.7939 26.4395 19.3835 13.6107 9.0884 5.7530 2.0852 0.8682 0.4864 0.3304 0.2841
0.004872 0.007356 0.011718 0.019915 0.036498 0.072471 0.332068 1.236130 2.947783 5.265859 6.603934
-4 -3 -2
o 2 4
6 7
c -
1.18671 1.16688 1.135673 1.084224 0.995123 0.833843 0.0 2.146437 5.735332 10.438571 13.133052
Reprinted from [2] with permission of the American Society of Mechanical Engineers. Q
0.6
.-----...,-----,---r--.,.------. 0.4864
0.4 1.,-
~
I~
C =4
0.2
B=O
0'--_-'-_---1_ _. . L . - _ - ' - _ - - - J
o
2
4 7)
Fig. 8.2 Initial value solution of Eq. (2.6) of Chapter 7. (Reprinted from [2] with permission of the American Society of Mechanical Engineers.)
Figure 8.3 shows the variation of the initial slope d~(O)/ d1J2 with C (not
C). If the problem is to find solutions of Eq. (2.2) for a range of C from - 1 to 10, then the solutions presented in Table 8.1 are sufficient. The only point to be noted is that the assigned values of C in the solution cover the required range of C.
181
8.2 Reduced Physical Parameters 8.-------r------,----r------,-----r------,
6
2
OL..-......=.....L..._ _----' -2
0
2
-'--_ _---I.
4
-'--_ _---'
6
8
10
c Fig. 8.3 Initial slope 1"(0) of Eq. (2.2). (Reprinted from [2] with permission of the American Society of Mechanical Engineers.)
Physically, the local skin friction is given by
Tw(X) = /-t(ilujily)y=o= /-tUoo~Uoojpx 1"(0) Data from Fig. 8.3 are therefore needed in order to calculate the viscous drag. 8.2.2 Design of Chemical Reactors
Consider again the design of a tubular reactor except that a nth order chemical reaction occurs. It is desired to determine the rate of conversion of the reactant as a function of three dimensionless physical parameters, namely the axial Pec1et number N pe' the reaction rate group R, and the order of reaction. The formulation of the problem is identical to that developed in Section 2.2.1, except now that the rate of disappearance of species A is kC'),.dx. The governing differential equation is therefore dy dy (2.5) - 2 - N pe -d - NpeRyn = 0 dz z subject to the boundary conditions
z = 0:
1 = y(O) __1_ dy(O) N pe dz
z=l:
dy(l) = 0 dz
(2.6)
182
8. Reduced Physical Parameters
To transform this equation to an initial value problem [3], let us first shift the independent variable z by
z=I- s
(2.7)
Equation (2.5) and its boundary conditions become dy
-2
ds
dy
+ N pe -d - NpeRyn = 0 s
(2.8)
and
s = 0: s = I:
dy(O) ds
=0
+
1= y(l)
(2.9a) _1_ dy(l) N pe ds
(2.9b)
We next define a linear group of transformations
s = A a,s,
y=Aa~
(2.10)
where III and 112 are constants to be determined and A is the parameter of transformation. Equation (2.8) is then transformed to
If we now set
(2.12) and
if =
RA(n-l)a2+ a,
(2.13)
Eg. (2.11) becomes I dy N pe ds2
--n + -dY -Ry =0
ds
(2.14)
and the boundary condition (2.9a) is transformed to
s=O,
dY(O)
--=0 ciS
(2.15)
Next, the parameter of transformation A is set equal to the missing boundary condition at s = 0, i.e., yeO) = A
(2.16)
183
8.2 Reduced Physical Parameters
Upon transformation, we get
A aJi(O) = A and
yeO) = I
(2.17a, b)
As a last step, the boundary condition at the second boundary point, Eq. (2.9b), is transformed to I = A [HI)
s = I:
+
_1_ N pe
dY~l) ds
]
which can be solved for A as A
= [Y(l) + _1_ N pe
dJ~l) ds
]-1
(2.18)
The transformation from a boundary value to an initial value problem is thus complete. To illustrate the solution procedure, consider the solution of Eq. (2.8) for n = 2 and N pe = 1.0 and a series of values of R in the range for R = 0.0 to 10.0. It is desired to find the variation of the missing boundary condition yeO) within this range of R. The results might be either a y(O)-R plot or a tabulation. For such purposes, the interval t::.R is of no importance as long as enough points are calculated within the specified interval of R. In fact, the interval t::.R does not even have to be a constant. This freedom in the choice of t::.R makes the present noniterative method possible. The following procedure is used. 1. Assign arbitrarily a value of R, say R = 0.1. 2. Solve Eq. (2.14) with boundary conditions (2.15) and (2.17b) numerically as an initial value problem from s = 0 to s = 1. The values of HI) and dY(l)/ ds are found to be
HI) = 1.0374
and
dy(I)/ ds = 0.06533
respectively. 3. The value of A is then calculated by using Eq. (2.18), which gives A
= (1.0374 + 0.06533)-1 = 0.9069
From Eq. (2.16), it is seen that A is the required missing boundary condition, i.e., yeO) = A = 0.9069. 4. The value of R is calculated by Eq. (2.13) as R
= RA (1- n) = (0.1)(0.9069)- 1 = 0.11026
5. Assign other values of R and repeat steps 2-4.
184
8.
Reduced Physical Parameters
6. Repeat steps 2-5, increasing R at intervals dR = 0.05 until R reaches the maximum value desired, such as R = 10. The results are summarized in Table 8.2. TABLE 8.2
Family of Initial Slopes for n = 2 N pe
Ii 0 0.1 0.3 0.5 0.7 0.9 l.l 1.2 1.3 1.4 1.5 1.7 1.8 1.9 2.1
2
5
10
R
y(O)
R
y(O)
R
y(O)
R
y(O)
0 0.1103 0.3976 0.7867 1.2953 1.9442 2.7583 3.2362 3.7667 4.3545 5.0043 6.5123 7.3827 8.3400 10.546
1.0 0.9069 0.7546 0.6356 0.5404 0.4629 0.3988 0.3708 0.3451 0.3215 0.2997 0.2610 0.2438 0.2278 0.1991
0 0.1104 0.4030 0.8150 1.3833 2.1575 3.2046 3.8578 4.6161 5.4961 6.5181 9.0873 10.696
1.0 0.9055 0.7445 0.6135 0.5061 0.4172 0.3433 0.3111 0.2816 0.2547 0.2301 0.1870 0.1683
0 0.1107 0.4122 0.8685 1.5722 2.6866 4.5149 5.8650 7.6580 10.078
1.0 0.9032 0.7279 0.5757 0.4452 0.3350 0.2436 0.2046 0.1698 0.1389
0 0.1109 0.4184 0.9113 1.7552 3.3390 6.7334 10.021
1.0 0.9019 0.7170 0.5487 0.3988 0.2695 0.1634 0.1198
Note that each column terminates when R ;;. 10.0.
It is to be noted that each set of values in the table is calculated noniteratively. The only difference between this method and other iterative methods is that values of R instead of R are assigned. As a result, the intervals between the calculated values of R are not constant. However, it is a family of solutions that is needed, and not a single solution for a particular value of R. If a y(O)-R curve is to be plotted, the data given in Table 8.2 are sufficient. To plot this curve, the values of R need not be spaced equally. If equally spaced values of R are desired, a simple interpolation of the data given in Table 802 can be made. With al> a 2 , and A known, A
= y(O) (from Table 8.2)
the variation of concentration with axial distance from the entrance of the reactor can be determined by using Eq. (2.10). 8.2.3
Nonlinear Diffusion Equation In Biology
In this section the method of transformation will be applied to the solution of a nonlinear boundary value problem in biology. We will
185
8.2 Reduced Physical Parameters
consider the problem in which the enzyme E with concentration Eo interacts with the substrate S with concentration C to form a complex E· S which may either dissociate into its components or decompose to regenerate E and liberate product P [4]:
The local rate of consumption of substrate can be expressed in terms of the local concentration of substrate by the Michaelis-Menten equation v = K 2EoKC/(1
where
K is defined
+ KC)
as K= K1/(K_ 1 + K2 )
The concentration of substrate, C, can be found by solving the steadystate one-dimensional diffusion, d 2C p + 1 dC K E KC - -2 + - - - = - -2 -o - dr r dr D(1 + KC)
(2.19)
subject to the boundary conditions r = 0:
dC I dr = 0
r= ro:
D(dCldr)r=ro= H(C o- Cj )
where D and H are the diffusivity and the mass transfer coefficients, respectively. The constant p equals - 1, 0, or 1, depending on whether the geometry is a plate, a cylinder, or a sphere. Introducing the following dimensionless radius and concentration,
r = r Ir o'
C= CICo
(2.20)
Eq. (2.19) becomes
d 2C + P + 1 dC dr 2
r
di
=
fiC
1 + k*C
subject to the boundary conditions i
= 0:
r = 1:
dC =0 dr
( ~; ),=1 = N(1
-
C'=I)
where N
= Hrol D
(2.21)
186
8. Reduced Physical Parameters
To transform Eq. (2.21) to an initial value problem, a linear group of transformation is introduced: C = A a2C* (2.22) Under this transformation, Eq. (2.21) becomes d 2C* + P + 1 dC* = A 2al /3C* dr*2 r* dr* 1 + k* A a2C*
(2.23)
If we simultaneously set
= k*A
k'
a2
(2.24)
and (2.25)
Eq. (2.23) then becomes d 2C*
p + 1 dC* /3C* -- + - - - = --=-...,....-d;2 r di 1 + k' C* The boundary condition at r = 0 is transformed to r = 0: dC* / di = 0 To get the missing boundary condition at r = 0, let us put
r=O: Upon transformation, this becomes
(2.26)
(2.27)
C=A A a2C*=A
r=O: which can be independent of A if a2
=1
(2.28)
We then get
r = 0:
C* = 1
(2.29)
The parameter of transformation A can be found by the boundary condition at i = 1, from A (dC* / dr)r=l = N(l - AC;:*=I)
or (2.30)
For a given pair of values of /3 and N, the solution of Eq. (2.21) for a range of values of k* consists of the following steps. First, a value of k' is assigned. Equation (2.26) is then integrated with boundary conditions at r = 0 given by Eqs. (2.27) and (2.29), from which A can be computed by using Eq. (2.30). With A known, the solution of the original equation, Eq.
187
8.2 Reduced Physical Parameters
(2.21), can be computed from Eq. (2.22). The value of k* corresponding to this particular solution can be computed by Eq. (2.24). Other values of k' are assigned and the steps are repeated until the required range of k* is covered. As an illustration, consider the concentration profile of a cylinder (p = 0) for f3 = 12, N = 1, and a range of k* from 5 to 20. Let us consider the solution for k' = 1.0. Equation (2.26), subject to the boundary conditions (2.27) and (2.29), is an initial value problem. The solutions are shown in Fig. 8.4 and 8.5. 4.-----r---~---..,..._--.,...--____.
3
2.7216 ___
c' 2
0'--_ _....1...-_ _- - ' 0.0
0.2
Fig. 8.4
0.4
- ' - -_ _- ' -_ _- - - '
0.6
r
Solution of C·.
f3 =
0.8
1.0
12; k' = 0.5.
8.----..,..._---r-----r------r----,
6
dC' dr
4
2
o "-===-l-__...l.-_ _-L_ _--l..._ _---l 0.4 o 0.2 0.6 0.8 1.0 Fig. 8.5
Solution of dC· / dr.
f3 =
12; k· = 0.5.
188
At ,
8. Reduced PhysIcal Parameters =
I, we get C*(l)
= 2.7216,
dC* / di
= 3.8078
The value of A can then be computed as A
= N/ [
dC;,(I) +NC*(I)]
= 0.1532
from which the value of k* is found to be: k*
= k' / A
= 6.5274
The value of the concentration outside the cylinder for k = 2000 liter/mole is Co = k* /
K = 0.00326 mole/liter
In the final step, the solution of the original equation, Eq. (2.21) can be computed from Eq. (2.22). The results are shown in Fig. 8.6. Physically, Fig. 8.6 shows that the concentration of substrate is 0.1532 Co at the center of the cylinder, increases to 0.4168Co at the surface immediately inside the cyclinder, and reaches Co across the cylinder surface. By assigning other values of k', solutions for other values of k* can be generated. For k' in the range of 1.0-10, the corresponding range of k* is found to be from 6.529 to 19.304. Since we are looking for k* from 5 to 20,
0.4168-
0.4
0.3
0.2
' " 0.1532 0.1
0.0 L -_ _.l..-_ _...l.-_ _--L_ _--L_ _----l 0.0
Fig. 8.6
0.2
0.4
0.6
0.8
1.0
Solution of Eq. (2.21). f3 = 12; k* = 6.5274; N = I.
8.2
189
Reduced Physical Parameters
the above solutions cover this range. Table 8.3 summarizes the results. For solutions corresponding to other ranges of k*, [4] should be consulted. TABLE 8.3
Summary of Numerical Solutions (f3
8.2.4
= 12, N = I)
k'
k*
E(O)
E(l)
1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0
6.529 8.609 10.146 11.477 12.704 13.870 14.997 16.098 17.179 18.247
0.1532 0.2323 0.2957 0.3485 0.3936 0.4326 0.4668 0.4970 0.5239 0.5480
0.4168 0.4797 0.5255 0.5625 0.5935 0.6200 0.6432 0.6636 0.6817 0.6979
Deflection of a Rotationally Symmetric Membrane
Consider the analysis of the large deflection of homogeneous, isotropic, shallow elastic membranes of revolution, symmetrically loaded, as shown in Fig. 8.7. The governing differential equations for the solution of the dimensionless radial stress S, and the dimensionless vertical displacement W can be writtenas [5]
1.p -E' .. dp
[PS
p -E' . . (W _ A. 2 r dp 2
)]
= -
1
o Fig. 8.7 Schematic diagram of the membrane.
(2.31 )
190
8. Reduced Physical Parameters
and P -d
do
[1-p
-d
dp
(p 2] S) r
- AP -dW dp
1(
+ -2 -dW dp
)2 = 0
(2.32)
where Eq. (2.31) originates from vertical equilibrium considerations and Eq. (2.32) is obtained from a compatibility relation. In Eqs. (2.31) and (2.32), the dimensionless radius p and the loadgeometry parameter A are defined as p= rib
where b, E, t, a, and p are the outer radius of the spherical cap in the projected plane, the modulus of elasticity, the thickness of the membrane, the radius of the sphere, and the uniform pressure intensity, respectively. The dimensionless stresses S, and So are given by Nrl ap): and Nol aps, respectively. The dimensionless radial and vertical displacements are and respectively. The nonvanishing field equations associated with Eqs. (2.31) and (2.32) are the meridional equilibrium equation
d
dp (pSr) - So = 0
(2.33)
and the stress-displacement relations S r
-s = 0
dU _ AP dW dp dp
+ .1 ( dW 2
dp
)2
(2.34)
(2.35) So - vSr = Ulp where 11 is Poisson's ratio. The boundary conditions are given as follows. At the pole, the symmetry condition gives p = 0:
dWI dp = 0
and the regularity condition gives p
= 0:
S,
is finite
The clamped condition at the outer edge gives p=l:
U=W=O
By integrating Eq. (2.31), solving for d[ W -
(l Ap2)]/ dp from
the result-
8.2 Reduced Physical Parameters
191
ing equation, and substituting this relation into Eq. (2.32), we get ~2
d2F
)0..2
8
de + 32F 2 =
(2.36)
where ~
= p2,
(2.37a, b)
The boundary conditions are F=O
~=O: ~
2 dF/ d ~ - (1 + p)F = 0
= 1:
(2.38)
Once the solution of F is obtained from Eq. (2.36), Eq. (2.37b) can be used to calculate Sr' Integrating Eq. (2.31) twice, we then get )0.. W= - -(l-~)+ 2
L 4F~ I
~
-d~
(2.39)
Solution of Eq. (2.36) was given by Goldberg [5] by an iterative method. We will now show how this problem can be solved noniteratively [6]. We introduce the linear group of transformations, F= A lX 2F
(2.40)
Equation (2.36) is independent of A, i.e., reducible to d 2F P )0..* -+--=-
dP
32P
8
(2.41 )
if, simultaneously, we put a2
2a l
-
= 2a l
-
2a 2
(2.42)
and (2.43) The boundary condition at
~
= 0 becomes
~=O:
By choosing
th~
F(O) = 0
(2.44)
missing boundary conditions to be dF(O) ~=A
(2.45)
and requiring it to be independent of A, we get a2 -
al
=1
(2.46)
8. Reduced PhysIcal Parameters
192
and dF(O) --=1 d~
(2.47)
From Eq. (2.42) and (2.46), we get a l = 3,
a2
=
4
(2.48)
The boundary condition at the second point then gives, upon transformation, 2~ dF I d~
= (1 + p)F
and
- 1/3
A = (1/g)
(2.49)
We now seek solutions of Eq. (2.36) for the range of values of ;\ from 0 to, say, 1.5. Since a family of solutions, covering a range of values of the parameter, is sought, the method developed in this chaper can be applied. The solution procedure can be summarized as follows. By assigning a value of ;\*, Eq. (2.41) can be integrated as an initial value problem since its boundary conditions, Eqs. (2.44) and (2.47) are given at the same point. The parameter of transformation A can then be calculated from Eq. (2.49). The solution of the original equation, F(g), can be obtained from the definition of the transformation, Eq. (2.40). Finally, the corresponding value of ;\ is calculated from Eq. (2.43), which gives
;\ = ;\* I /2 I A
(2.50)
Selected results are shown in Table 8.4 and agree with those presented in Goldberg's work. The third column in that table is Sr(l). We note that at ~ = 1 Eq. (2.37b) shows that Sr(l) = F(l). TABLE 8.4
Selected Results of Solutions of Eq. (2.36)
A*
A
Sr(l)
A*
A
Sr(l)
0.01 0.05 0.10 0.13
0.233 0.545 0.825 0.991
0.3233 0.3149 0.3063 0.2932
0.16 0.19 0.20 0.22
1.171 \.388 1.474 1.694
0.2755 0.2626 0.2579 0.2418
8.3 APPLICATION TO SIMULTANEOUS DIFFERENTIAL EQUATIONS
The method developed in the preceding sections can be applied to the solution of simultaneous differential equations in which two missing boundary conditions need to be found. An example is found in the
193
8.3 Application to Simultaneous DIUerential Equations
boundary layer flow of a viscous, electrically conducting incompressible fluid past a semi-infinite flat plate in the presence of a magnetic field, which was treated originally by Greenspan and Carrier [7] and Glauert [8] and cited by Klamkin [9] and Ames [10] as an example to which Toepfer's method cannot be applied. Following Glauert [8], the governing differential equations for the steady incompressible magnetohydrodynamic flow are Momentum equation: (q·V)q= -(llp)Vp+pV 2q+(lllp)JxH
(3.1)
Maxwell equations:
(3.2)
Jla=E+llqxH
(3.3)
J = curlH
divq = divH = divE = curlE
(3.4)
where p, q, J, H, and E are the pressure, velocity, current, and magnetic and electric intensities, respectively. The boundary conditions are
y=O:
q=O,
Hy=O
y=oo:
P=Po'
q=Uoi,
H=Hoi
where the electric field may be taken as zero everywhere. Using Eq. (3.3), Eqs. (3.1) and (3.2) may be written as (q. V)q = -
(11 p) V(p + 11lH2) + P V2q + (Ill p)(H' V)H q X H = (I I all) curl H
(3.5) (3.6)
From Eq. (3.4), we write q = curl 1[; (x, y)k
(3.7)
H = curl A (x, y)k
where k is a unit vector in the z direction. By substituting Eq. (3.7) into Eqs. (3.5) and (3.6) and introducing the usual boundary layer assumptions, we get 31[;
01[; o~ 01[; o~ 0 Il (OA 02A oA 02A) oy oxoy oy2 oy oxoy - ox oy2 =p oy3 +
01[; oA oy
ax -
P
01[; oA 02A ox oy = all oy2
ax
(3.8) (3.9)
194
8. Reduced Physical Parameters
By introducing the similarity transformation, 1/J
= (UOVX)1/2f('q)
A = Ho(vx/ UO) I/ 2g (,q) 1/ =
H UO/VX)1/2y
Eqs. (3.8) and (3.9) become
d'1 d~ d 2g + f - -f3g- =0 d1/3 d1/2 d1/2 2g d d1/2
(3.10)
+ €(f dg _ g df ) = 0 dn
(3.11)
d1/
subject to the boundary conditions
g(O) = 0, dg( 00)/ d1/
f(O) = df(O)/ d1/ = 0, df( 00 ) / d1/ = 2,
=2
To reduce the equations to an initial value problem [11], a linear group of transformations is again introduced
f= Bf,
1/ = ATj,
g= cg
(3.12)
where A, B, and C are constants to be determined. Under this group of transformations, Eqs. (3.10) and (3.11) become
-d1 -f3 (AC2)_d~ -dl +ABfg - =0 ~3
B
~2
2
fig -d g +€AB ( f_~2
~
~2
df -g~
)
=0
(3.13)
(3.14)
As before, we require that the differential equation to be independent of A, B, and C. However, only three conditions are needed for the determination of the three constants A, B, and C in the transformation, and two equations will be obtained from the boundary conditions at infinity. It is
therefore necessary to keep the number of conditions from Eqs. (3.13) and (3.14) at one. In view of the fact that solutions of Eqs. (3.10) and (3.11) for a range of f3 are needed, Eqs. (3.13) and (3.14) can be made invariant, i.e., independent of the constants if, simultaneously,
AB = 1
(3.15)
and
(3.16)
8.3 Application to Simultaneous Differential Equations
195
Equations (3.13) and (3.14) then become
d 3J
_ d2j
__ d 2g
-dij3 +f-f3g=0 dij2 dij2
(3.17)
(3.18) It should be emphasized that although 13 is a function of A and C, we will reverse the process by assigning values of 13 and solving the equations as an initial value problem based on the assigned value of 13. The value of f3 which corresponds to this solution will then be computed from Eq. (3.16). This is possible since the solutions of Eqs. (3.10) and (3.11) for a range of f3 and E: are sought. In the second step, the missing boundary conditions are normalized by putting
and
(3.19)
and thus
dg(O)/ dij = 1
and
(3.20)
These two boundary conditions, together with the other three at 1j = 0, namely
g(O)
J(O) = dJ(O)/ dij = 0,
=
0
(3.21)
enable us to solve Eqs. (3.17) and (3.18) as an initial value problem. Finally, the remaining two conditions needed for the determination of the constants A, B, and C, can be obtained by using the boundary conditions at 1j = 00, which give
B dJ(oo)
C dg(oo)
---=2 A dij
and
---=2 A dij
1
and
C =2[ dg(oo) A dij
or
A =[
2
dJ(oo) ]1/2 dij
]-1
Equation (3.15) and the above two equations then give A = B-1 =
[1
-
df ( 00) ] 2 dij
1/2
(3.22)
196
8. Reduced Physical Parameters
and
_[ dj(oo)
C- 2 - -
err;
]1/2[ dg(oo) ]-1 err;
(3.23)
As an illustration, details of the solution of Eqs. (3.10) and (3.11) for e = 0.50
and
f3
=
0.2-0.8
will be given [11]. The solution takes the following steps 1. Assign a value of 13 (instead of f3), e.g., 13 = 0.20 2. Integrate Eqs. (3.17) and (3.18) numerically as an initial value problem with boundary conditions given by Eqs. (3.20) and (3.21). Figure 8.8 shows the variation of dj/ err; and dg / err; with 'ij. Both are seen to approach constants at large 'ij. These will be taken as]'( (0) and g'( 00). We then have
dj( (0)/ err; = 1.84798,
dg((0)/ err; = 2.0467
3. The parameters A, B, and C in the transformation can be computed. From Eq. (3.16), we get f3 =
13A -2C -2 = 13[
dg(00 ) / dj( (0) ]2=0.2439
err;
err;
3.0 . - - - - - . . . . - - - , - - - - - , - - , - - - - , - - - - - . - - , - - - - - - - ,
g;) (00) = 2.04067 f;) (00) =
1.84798
\ _ _ _
2.0
_::::.--....L------i
I",
100
is I",
1,-
1.0
0.0
L..-_-'-_--'-_---'_ _..L...-_---'--_--l._ _...l...------I 2 4 8 6
o
l)
Fig. 8.8
Solution of Eqs. (3.17) and (3.18, for differentiation.)
£
= 0.50, P= 0.20. (Subscripts represent
197
8.4 Application to an Eigenvalue Problem
4. The initial slopes can be calculated from Eq. (3.19) d~(O)/ dYj2
= B / A 2 = 1.1259,
dg(O)/ dn
= C/ A = 0.9801
5. Other values of 13 can be assigned and steps 1-5 repeated until the solution covers the required range of {3. (For this example, the required range of (3 is from 0.2 to 0.8.)
Selected solutions are presented in Table 8.5. TABLE 8.5 Selected Solutions of Eqs. (3.10) and (3.11) 1",,(0)
0.2 0.4 0.6 0.8
1.84798 2.12768 2.56630 3.33271
2.04067 2.27483 2.64417 3.28990
0.24388 0.45724 0.63697 0.77958
1.12590 0.91135 0.68799 0.46489
0.98007 0.87919 0.75638 0.60792
Physically, the velocity components can be written as u=
1
df
"2 Uo dYj
(3.24)
,
and the magnetic field components are (3.25)
Also, the total electric current in the boundary layer is given by /.: =
where
J = _ ifA = _
ay2
foooJ dy
(3.26)
"04 ( Uvx )1 /2 ddYj22g o
From Eqs. (3.24)-(3.26), it is seen that once the solutions of Eqs. (3.10) and (3.11) are obtained, the physical quantities u, v, B x ' By, and J to ta1 can be calculated. 8.4 APPLICATION TO AN EIGENVALUE PROBLEM
The initial value method will now be applied to the solution of an eigenvalue problem resulting from an analysis of the finite deflections of a
8. Reduced Physical Parameters
198
L
/
Fig. 8.9 Schematic diagram of the elastic bar. (Reprinted from [Ill with permission of the American Society of Mechanical Engineers.)
nonlinearly elastic bar [12] clamped at one end and axially loaded at the free end, as shown in Fig. 8.9. If we assume that plane cross sections remain planar during deformation, the local bending moment at a typical section can be expressed as M= -Pv
(4.1)
where P is the axial load and v is the transverse deflection of the bar. On the other hand, from the momentum-curvature relation for the bar, another expression for M is (4.2) where Do and D 1 are the characteristic flexual rigidities, CPo is a material constant, and cp = dO/cis
(4.3)
In Eq. (4.3) 0 is the slope of the bar and s is the arc length along the deformed axis. Equating the right-hand sides of Eqs. (4.1) and (4.2), noting that sinO = do] ds
(4.4)
and differentiating the resulting equation, we get _dcp = _ P sin ds
O/{D [1 __(_DI_CP-_p_V)_2 ]} 0
(Do - D 1)DoCP5
(4.5)
199
8.4 Application to an Eigenvalue Problem
If we introduce the dimensionless quantities, u=v/L,
z=s/L,
1[;=L
p=CPoL
e=D,/D o,
y=I/(I-e)p2
"A=P/Per>
(4.6)
where Per is the Euler buckling load, defined as
Per
= 'TT 2Do/ 4L 2
Eqs. (4.3)-(4.5) become
du . B -=s1o
(4.7)
dz
dB dz
= 1[; 2
![ (
-d1[; = - ( -'TT ) "A sin B 4
dz
I-
(4.8)
2 'TT "Au 4
+ e1[; )2
I p2(1 _ e)
subject to the boundary conditions
u(o) = 0,
HO) = 0,
B(O) = a,
B(l) =
°
(4.9)
(4.10)
A family of solutions of Eqs. (4.7)-(4.10) is now sought in which all parameters, namely e, a, and p, vary in a range of values. By solution of these equations is meant the value of "A which satisfies these equations and their boundary conditions. The method originally used by Oden and Childs [13] involves iteration for every point calculated. From Fig. 8.9, the angle a is seen to be the end slope. The problem under consideration is one of specifying the end slope and then determining the load required to maintain a given slope characterized by the unknown load parameter. Introducing a linear group of transformations as
B = A a3if, where a" a 2 , a 3 , and a 4 are constants to be determined and A is the parameter of transformation, Eqs. (4.7)-(4.9) become dU =sin(A a3B) A al - a2_ dE
A a3-a2
~~ = A a4{f
Aa4-a2
d~ = -('TT )"A Sin( A a3if ) /(1dz 4 'I
2
(4.11)
(4.12)
I p2(1 -
e)
['TT
2
4
Ma'u+e.Aa4{f]2j (4.13)
200
8.
Reduced Physical Parameters
Following the general approach, let (4.14)
=X
(4.15)
p/A a 4= p
(4.16)
M
2a 2
and Equations (4.11 )-(4.13) then become .-dii = sm()
(4.17)
dZ
dO dZ
-
(4.18)
= tJ;
(4.19)
subject to the boundary conditions
z=O:
il(O) = 0, f(O) = 0, 0(0) = a
(4.20)
0=0
(4.21)
where, without loss of generality, the transformed parameter X is set to unity and, for convenience, we put {3
= l/p2
(4.22)
As an illustration, consider the family of solutions for the pair of values of the parameters e = 0.25 and a = 60° [12]. The solution is obtained by the following steps. First, a value of (3 of 50 is assigned. Equations (4.17)-(4.19) are integrated as an initial value problem, using the boundary conditions given by Eq. (4.20). The solutions of 0 is plotted in Figure 8.10, which shows that the solution of 0 intercepts the z axis at z = 0.5858. From the boundary condition (4.21), it may be seen that the value of A a2 = 1/0.5858. From Eqs. (4.22) and (4.15), p=
A=
pA -a2 = _1_
fi
--!., = 0.5858 A
2
,;so
= 0.08
X = _1_ = (0.5858)2 = 0.34 2a
A 2a 2
A
2
The solution for one value of {3 is now completed. Next, the value of {3 is increased by, say, t::.{3 = 25. Then, by repeating steps 2 and 3, another set of
201
8.4 Application to an Eigenvalue Problem
1.00
0.75
I",
0.50
0.25
o
o
0.2
0.1
0.3
0.4
0.5
0.6
z
FIg. 8.10 Solution of 9 as a function of t. (Reprinted from [I2l with permission of the American Society of Mechanical Engineers.)
solutions is obtained: {3
= 75,
p
= 0.07,
A = 0.33
By assigning a series of values of {3, a family of solutions is generated for
€
= 0.25 and a = 60°. For illustration, five of these solutions are given in
Table 8.6 (see the second block in the first column). TABLE 8.6
Results of Varying E and a O
a
a
p
A
a
p
A
0.25
20
0.5334 0.3136 0.2322 0.1893 0.1624
0.8534 0.6884 0.5932 0.5373 0.5009
0.75
20
0.5684 0.3192 0.2626 0.2278 0.2037
0.9691 0.9168 0.8967 0.8824 0.8716
0.25
60
0.3087 0.2172 0.1505 0.0800 0.0700
0.4764 0.4247 0.3852 0.3400 0.3300
0.75
60
0.5724 0.3223 0.2665 0.2321 0.2082
0.9829 0.9348 0.9232 0.9159 0.9107
Reprinted from [I2l with permission of the American Society of Mechanica:I Engineers.
8. Reduced Physical Parameters
202
8.5 CONCLUDING REMARKS
The examples discussed in the preceding sections demonstrate the various applications of the method when a family of solutions of a boundary value problem for a range of values of a physical parameter is required. From a strictly mathematical point of view, the method really is not "noniterative" if the solution of the boundary value problem for only one value of the physical parameter is sought. This is true except that in engineering applications, the solution corresponding to one value of the physical parameter is in general useless. What an engineer really needs is a family of solutions for the complete range of values of the physical parameter, which can then be tabulated or plotted for the designer to use. In this context, it does not make any difference which values of the parameter are chosen in the table or in plotting the curves as long as they are within the range of values of the physical parameter and are fairly evenly distributed within the range. One feature of this method is that it can treat those boundary value problems with nonhomogeneous boundary conditions. Another approach to the treatment of nonhomogeneous boundary conditions, due to Klamkin [14], was discussed in Section 7.2.1. Note, however, the limitation imposed on Klamkin's method, namely that the differential equation has to be in such a special form that it is invariant for all values of the parameter and the constants in the definition of the transformation. Unfortunately, very few boundary value problems in engineering satisfy such a strict limitiation. As another useful application, we will outline a method of obtaining general solutions for certain boundary value problems where the physical parameter appears only in the boundary condition at the second point. Consider the problem of the large deflection of an annular membrane under pressure [6]. The boundary value problem for this problem can be written as
f
2
d~ _
dT/2 - -
1 2 32 T/
(5.1 )
f=O
(5.2)
2 df - (1 + p)f = 0
(5.3)
dry
It should be noted that the physical parameter A appears only in the boundary condition at the second point, Eq. (5.3). Solutions of the above boundary value problem for a family of values of A are sought.
8.5 Concluding Remarks
203
Introducing the linear transformation 1]
= A a'ij,
(5.4)
Eq. (5.1) becomes
A 3a 2 -
2-
2a
'F dij2 d'f
= _ -.L A 2a ' ij2 32
(5.5)
For Eq. (5.5) to be invariant, the powers of A should be equal, i.e., 3a 2
= 2a l
2a]
-
(5.6)
It is to be noted here that the possibility of imposing a relation between a l and a 2 , Eq. (5.6), relaxes one of the conditions required in Klamkin's method [14), since for Klamkin's method to work the powers of A on the two sides of Eq. (5.5) must be the same (i.e., either 3a 2 - 2a l or 2a l on both sides). The transformed differential equation is therefore 2-
F d'f
= _
dij2
-.L ij2
(5.7)
32
The first boundary condition is transformed to
J(O) = 0
'ij=0:
(5.8)
We choose the missing initial condition as the parameter of transformation, i.e., df(O)/drj = A which, upon transformation, becomes
(dJ(O)/ dij) A a2- a,
'ij=0:
=
A
(5.9)
The boundary condition (5.9) is independent of A if a2 -
al
=1
(5.10)
which leads to 'ij=0:
dJ(O)/ dij = 1
(5.11)
From Eqs. (5.6) and (5.10), aI
=3
and
a2 = 4
Finally, to find the parameter of transformation A the boundary condition at the second point gives
2(dJ/dij)e - (1 + p)A1e = 0
(5.12)
8. Reduced Physical Parameters
204
Eliminating A in Eq. (5.12), we get ,\2= 1-
[
2Tje(d!/dii)] e (1 + nl:
(5.13)
To get the general solution for all A's, we integrate Eq. (5.7) as an initial value problem, using Eqs. (5.8) and (5.11) as the initial conditions. Starting from Tj = 0, values of ,\ are calculated by using Eq. (5.13) at every point. For small values of Tj, the right-hand side of Eq. (5.13) is negative and no real values of ,\ exist. As Tj is increased, the right-hand side of Eq. (5.13) continues to increase until at Tj = 12.4632 it reaches zero. For this case, ,\ = 0. From this point on, every step in the integration of the initial value problem will lead to a new positive value of '\. A general table can therefore be generated which represents the solution of all A's. For details, the reader is referred to [6]. In the next chapter, we will explore the invariant properties of the physical parameters for those boundary value problems involving more than one physical parameter. PROBLEMS
1. Extended surfaces transferring heat by radiation are of great importance in the design of space vehicles since this is the only means for disposing of waste heat. From the standpoint of mass reduction, it is generally agreed that tapered fins represent one of the best alternatives. In the mathematical analysis of such fins, Keller and Holdrege [15] found that the temperature distribution can be solved by the energy equation: 2U
d dR 2
+[
_1_ _ tana R +p (1 - R) tan a
4
] dU _ /3U + 8 dR (1 - R) tan ex
-0
+8 -
subject to the boundary conditions
U(O)
=
1.0,
where
R=
T
U=T' B
and a is the angle of taper. The radii r, rB , and rT are the radial position, the radius of fin base, and the radius of fin tip, respectively. Also, T, T B , and ZT are the temperature at any point in the fin, the temperature at the base, and the fin thickness at tip. Transform the problem into an initial value problem for /3 = 0.1, ex = 6. 0, 8 = 0.05, p = 10.0, and N from 0.1 to 1. [Hint: First, let s = I - R.
205
Problems
Upon transformation, it will become clear that a, = 0, a 2 = I, and 13 = /3A 3 • A is found by A = [U(l)r l . ] Answer: N = I, 1/ U(l) = 0.49507; N = 0.10014, 1/ U(l) = 0.78843. 2. The problem of the melting of a solid due to a sudden increase of surface temperature is inherently difficult, due to variable heat conductivity and moving boundaries. In an analysis of such a system, Cho and Sunderland [16] obtained the following nonlinear boundary value problem:
~ [(1+/30):~ ]+21):~
=0,
0(0)=0, 0(00)=1
Solve this problem as an initial value problem for /3 = 4.0 and 1.441 [17]. [Hint: = 0, 13 = A "'2/3.] Answer: dO (0)/ d1) = 1.751 and 1.388, respectively, for /3 = 4.00 and 1.441. 3. In an analysis of the deflection of a cantilever beam, Bissopp and Drucker [18] obtained the following nonlinear boundary value problem for the solution of the deflection angle of the beam axis:
a,
d 20/ds 2 +
/3 cosO = 0,
= 0, dO(I)/ds = 0 Reduce the problem to an initial value problem for /3 = O.oI to 0.5. [Hint: a 2 = 0 and 13 = /3A 2"',.] Answer: dO (0)/ ds = /3. 0(0)
4. Flow through a porous medium is an important branch in fluid mechanics. In an analysis of the unsteady flow of gases through a porous medium, the following boundary value problem is obtained [I]:
d2w + 2z 2 dz ~I - aw
dw = 0 dz '
I () 0 w (0) =,woo=
Solve the problem by the initial value method for 0.05 ..;;; a ..;;; 0.5. Hint: First, the equation is transformed by introducing the following variables: = z/.;I=a , f(1))=w-1 It will then follow that /3 = [a/(I - a)]A "'2
1)
Answer:
Selected numerical answers are given below:
df(O)/ d1) 0.04272 0.15406 0.27194 0.36450 0.49940
-
1.12055 1.09820 1.07092 1.04609 1.00241
206
8. Reduced Physical Parameters
5. Consider again the flow of Newtonian fluid between two coaxial cylinders as in problem 4 of Chapter 3. Transform the boundary value problem
Id(dB)
~ d~ ~ d~
+ 4N ?I
= 0,
B(k) = 0, B(l) = I
into an initial value problem for N = 0.1 to 1.0 and k = 0.9 and compare with the exact solution
B = [(N + I) - N Ie]
-
[(N + I) - N I k2](ln~/lnk)
6. Determine the eigenvalues of the harmonic equation d~
-
dx 2
+AU=O
'
U(O) = 0, u(l) =
°
by the method of Section 8.4 and compare with the exact solutions,
u(x) = sin nex where n = 1,2,3, . .. and A = n 27T2 [19].
REFERENCES I. Schlichting, H., "Boundary Layer Theory," Chapter 14, McGraw-Hill, New York, 1968. 2. Na, T. Y., An initial value method for the solution of a class of nonlinear equations in fluid mechanics, J. Basic Eng., Trans. ASME 92,503-509 (1970). 3. Lin, S., and Fan, L. T., "Examples of the Use of the Initial Value Method to solve Nonlinear Boundary Value Problems," AIChE J. 18, 654-656 (1972). 4. Na, H. S., and Na, T. Y., An initial value method for the solution of certain nonlinear diffusion equations in biology, J. Math. Biosci. 6,25-35 (1970). 5. Goldberg, M. A., An iterative solution for rotationally symmetric nonlinear membrane problems, Int. J. Nonlinear Mech. 1 169-178 (1966). 6. Na, T. Y., Kurajian, G. M., and Chiou, J. P., General solution of the problem of large deflection of an annular membrane under pressure, Aeronaut. Q. London 27, 195-200 (August 1976). 7. Greenspan, H. P., and Carrier, G. F., The magnetohydrodynamic flow past a flat plate, J. Fluid Mech. 6, 77-96 (1959). 8. Glauert, M. B., A study of the magnetohydrodynamic boundary layer on a flat plate, J. Fluid Mech. 10,276-288 (1961). 9. Klamkin, M. S., On the transformation of a class of boundary value problems into initial value problems for ordinary differential equations, SIAM Rev. 4,43-47 (1962). 10. Ames, W. A., "Nonlinear Ordinary Differetiential Equations in Transport Processes," Academic Press, New York, 1968. 11. Na, T. Y., An initial value method for the solution of MHD boundary layer equations, Aeronaut. Q. 21, 91-99 (1970) 12. Kurajian, G. M., and Na, T. Y., An initial value method for the solution of an eigenvalue problem in solid mechanics, J. Appl. Mech. 39, 854-855 (1972).
References
207
13. Oden, J. T., and Childs, S. B., Finite deflection of a nonlinear elastic bar, J. Appl. Mech. 37,48-52 (1970). 14. Klamkin, M. S., Transformation of boundary value problems into initial value problems, J. Math. Anal. Appl. 32, 308-330 (1970). 15. Keller, H. H., and Holdrege, E. S., J. Heat Transfer, Trans. ASME 92,113-116 (1970). 16. Cho, S. H., and Sunderland, J. E., J. Heat Transfer, Trans. ASME 96,214-217 (1974). 17. Chiou, J. P., and Na, T. Y., J. Ind. Math. 28, No.2 (1978). 18. Bisshopp, K. F., and Drucker, D. c., Q. Appl. Math 3, 272-275 (1945). 19. Bedford, G. G., SIAM J. Numer. Anal. 6,99-103 (1969).
CHAPTER
9
METHOD OF TRANSFORMATIONINVARIANCE OF PHYSICAL PARAMETERS
9.1
INTRODUCTION
In Chapter 8, the method developed in Chapter 7 was extended to the treatment of nonlinear boundary value problems where a physical parameter appears either in the differential equation or in its boundary conditions and solutions for a range of values of this parameter (instead of a single solution for a particular value of this parameter) are desired. This improved method has been successfully applied to the solution of many nonlinear boundary value problems, among which are Blasius's equation with suction or slip [1], the unsteady flow of a gas through a porous media [1], magnetohydrodynamic boundary layer flow [2], nonlinear diffusion in biology [3], the axial diffusion in tubular chemical reactors [4], shockinduced flow [5], and the laminar boundary layer of pseudoplastic fluids [4,6]. This extension is of significance since in most physical problems in which physical parameters appear in the formulation, it is always of interest to obtain the solutions for the complete range of all physical parameters, instead of a single solution for a particular set of values of these parameters. Although this improved method has been successful, it can treat only problems involving one physical parameter. For example, in solving Blasius's equation, the method applies if either suction or slip is present but not both. The class of problems to be treated in this chapter offers other examples in which this improved method cannot be applied. Further extensions of the method will be made in this chapter in several ways. First, the method is applied to problems involving more than one physical parameter. This is done by investigating the invariance of these parameters under transformation-a property not heretofore exploited. Second, a limitation imposed previously, namely that the differential equations be independent of the physical parameters, is relaxed. As a 208
9.2 Boundary Value Problems with Two or More Parameters
209
result, a much wider class of problems can be treated by this method without iteration. As illustrations, the method is applied to the solution of four nonlinear boundary value problems: boundary layer flow with mass transfer and slip (7], the onset and propagation of surface combustion [8], flow over a slender body of revolution [9], and the thin strut with large deflections [10-12]. 9.2 BOUNDARY VALUE PROBLEMS WITH TWO OR MORE PARAMETERS 9.2.1
Blasius's Equation with Mass Transfer and Slip
In Section 8.2, a brief discussion was given of one of the methods of preventing separation in a boundary layer, that is, suction of the fluid from the boundary layer. There is in the literature another method of suppressing separation, namely slip; i.e., the solid wall may be moved by some mechanical means relative to the body to which it is attached, thus reducing the difference between the velocity of the solid wall and that of the mainstream. Even though this idea is still of little practical value because of the complications involved in moving such surfaces, there is experimental evidence [13] on aerofoils to show that slip is vety effective in avoiding separation. From the mathematical point of view, the only difference between this problem and the one treated in Section 8.2.1 lies in one boundary condition: 1'(0) is no longer equal to zero. The differential equation to be solved is therefore (2.1)
subject to the boundary conditions
f(O) = C,
df(O)j dn = B,
where B equals the slip velocity divided by the mainstream velocity. Other quantities are the same as those in Section 8.2.1. To transform Eq. (2.1) to an initial value problem, a one-parameter linear transformation is again introduced: (2.2)
By following exactly the same steps as before, Eq. (2.1) can be trans-
9.
210
Invarlance of Physical Parameters
formed into the following initial value problem: 3j
d dij3
I - d2J dij2 = 0
(2.3)
+ 2" f
subject to the boundary conditions 1j = 0:
j=
C,
dj/ dij
= B,
d2J/ dij2 = I
(2.4)
where
C = CA -1/3,
B = BA
-2/3
(2.5)
and _ ] -3/2 A = [ df ( 00) / dij
(2.6)
Thus, to start the solution, values of Band C, instead of Band C, are assigned and Eq. (2.3) is integrated as an initial value problem with the initial conditions given by Eq. (2.4). From this solution, dj( 00)/ dij can be found and the value of A computed from Eq. (2.6). Knowing A, the values of Band C can be calculated from Eq. (2.5). By changing values of Band C, solutions for other values of Band C can be obtained. This noniterative method has proven to be extremely useful in problems for which the solution is desired over a given range of certain physical parameters. In the solution of Eq. (2.1), for example, one may be interested in the variation of skin friction d~(O)/ dTJ2 for a given range of surface mass transfer B and slip C. The results sought may be illustrated in the form of a series of curves of d~(O)/ dTJ2_C at a few constant values of B, or the results may be tabulated. Without the present method, each point on the curves must be solved by iteration. The above scheme was applied in Chapter 8 to problems involving only one parameter. For problems with more than one physical parameter, as in Eq. (2.1) with B =1= 0 and C =1= 0, it is difficult to plot the series of curves just mentioned. The difficulty lies in the fact that Band C are functions of both Band C so that lines of constant B do not correspond to lines of constant B. In other words, when the transformed initial value problem is solved, the solution is based on the assigned values of Band C (not Band C). There is no way of knowing which pair of values of Band C this particular solution corresponds to until the last step. Even if we keep B constant and vary C for the calculation of a family of solutions, the corresponding values of Band C will be such that neither B nor C for this family of solutions remains constant. For this reason, presentation of such data in tabulated form (which requires that either B or C be kept at a constant value) is also difficult.
9.2
211
Boundary Value Problems wIth Two or More Parameters
To overcome this difficulty, an important invariant property among the physical parameters under the transformation will be introduced which enables the method to be extended to the solution of equations with more than one physical parameter present [7]. From Eq. (2.5), it is seen that the following invariant property can be obtained by eliminating A:
B/C 2 = li/c 2 Let us define this ratio as fJ. The boundary conditions therefore become
11 = 0:
(2.7)
A table can now be constructed as follows. Select a value of fJ and solve the auxiliary equation, Eq. (2.3), subject to the boundary conditions (2.7), for a range of values of C. For each solution, the corresponding values of C and d~(O)/ d.,,2 are listed. After a family of solutions for one value of fJ is obtained, the process can be repeated by assigning another value of fJ, and a family of solutions corresponding to this second value of fJ can then be generated. This procedure is carried out for a few constant values of fJ, and the results are plotted as in Fig. 9.1. A few trial solutions will show what range of C to use to obtain the desired limits of C. In this problem, a separate set of solutions is necessary for C = 0, B =F 0 since this yields fJ = 00. Data for plotting the constant-fJ lines are also tabulated in Table 9.1. TABLE 9.1 Tabulation of d1(0)j m,2 a
C (j
- 0.4
-0.2
0
0.2
0.4
0.7
0.0 0.2 0.4 0.6 0.8 1.0 2.0 3.0 4.0
0.1956 0.2012 0.2053 0.2083 0.2106 0.2109 0.2016 0.1748 0.1340
0.2616 0.2624 0.2630 0.2637 0.2643 0.2646 0.2658 0.2653 0.2634
0.3321 0.3321 0.3321 0.3321 0.3321 0.3321 0.3321 0.3321 0.3321
0.4060 0.4053 0.4044 0.4036 0.4024 0.4014 0.3954 0.3875 0.3779
0.4832 0.4764 0.4688 0.4604 0.4514 0.4417 0.3886 0.3114 0.2271
0.6037 0.5663 0.5230 0.4744 0.4211 0.3633
a Reprinted from [7) with permission from the American Society of Mechanical Engineers.
Table 9.1 may be used in many ways. For example, suppose a plot of = 0.7. To do this
d~(O)/ d.,,2_B is desired for a constant value of C, say, C
212
9. Invarlance of Physical Parameters ~=o
1.0
0.5
0.2
0.1
0.05 L..-.l..-_....I..-_....l...._---l._----J'---I_..I...-L-....L...----JL.I.-_....LL_----IL..----L..I...-_....L.._--L. 2.0 -0.5 o 1.0
c Fig. 9.1 Solution of Eq. (2.1) in terms of f3 and C. (Reprinted from [7] with permission of the American Society of Mechanical Engineers.)
one simply reads values of d~(O)/ d1J2 from the table under the column for C = 0.7. The corresponding values of B follow from B
= C 2f3 = 0.81f3
Another interesting plot might be B-C for constant d~(O)/ d1J2. Such a plot is obtained by interpolating along each row of constant f3 to find the C corresponding to the desired value of d~(O)/ d1J2. On the other hand, with the tables, any future solution of Eq. (2.1) for a given Band C may be obtained by double interpolation to find the value of the missing initial condition. If more accuracy is desired, the table may be improved in the neighborhood of the solution of interest by feeding back the results of the first integration. Whatever the application, the important point is that without this method, each value on any table or graph must be calculated by iteration.
213
9.2 Boundary Value Problems with Two or More Parameters
9.2.2 Onset and Propagation of Combustion
Following a series of accidental fires in space capsules in the mid-1960s, a study was made [8] of the pertinent parameters governing the onset and propagation of various modes of combustion that may take place in the interior of a spacecraft. The model used is a Rayleigh-type fluid model in which the surface temperature experiences a sudden increase. The fluid motion is created by the emergence of the mass flux leaving the surface as a result of a simple one-step irreversible reaction. The conservation laws can be formulated to give the following governing differential equations: Continuity equations of the mixture:
ap at
a(pv) ay
- + - - =0
(2.8)
Species continuity equation:
ac;
ac;
a(
ac; )
(2.9)
Pat +pv ay = ay pD; ay where i = 1,2, ... , N (N is number of species). Energy equation:
ah
ah
a(
Pat + pv ay = ay
sr
ac; )
A ay + p ~ hiD; ay
(2.10)
where p, v, C;, h, and T are the density, velocity, mass fraction of species i, enthalpy, and temperature, respectively. The boundary conditions for the species C; can be written as
y=O: y=
00:
(2.11) C; = C;e'
aCjay = a2Cjay 2 = 0
(2.12a, b)
The boundary conditions for the enthalpy are
t
h(y,t)=h e , forally
t ;;;. 0:
h(O, t) = hw ' h( 00, t) = he
(2.13) (2.14a, b)
By introducing the surface reaction kinetics equation, all gaseous components are then related, leaving one of the species as the variable (here we choose Co, the mass fraction of oxygen). It is usually assumed that
p2D / p;De = const = y
214
9. Invarlance o' Physical Parameters
where the subscript e refers to the external stream conditions. Equation (2.8) then leads to the conclusion that v is independent of y, which can therefore be related to the surface reaction rate. This eliminates v as a variable. The number of dependent variables is now reduced to two, namely Co and h, which can be solved by the species continuity equation of oxygen and the energy equation. However, by introducing a new variable, (2.15)
these two equations will be combined into one equation, which is dM d1/
m,] dM =0 d1/ (2.16)
where a is a physical parameter relating the mass flux at the wall to the mass addition availability and H w is the enthalpy ratio at the wall. The boundary conditions then become 1/ = 0:
M(O) = 0;
M(oo)
1/=00:
= 1
(2.l7a,b)
The problem has now been reduced to the solution of M as a function of the two parameters, namely, a and H w' A family of solutions, covering the range of values of the parameter, is sought. To illustrate, let us assume that solutions for the following ranges of values of a and H ware required: a:
0.1-0.75,
Due to the fact that a family of solutions is needed, instead of a single solution for a particular pair of values of a and H w the noniterative method developed in the present chapter is especially useful. Introducing the linear group of transformations, (2.18) Eq. (2.16) is transformed to
2M + 2A2at[ii _ .!. aA a2-2a,( dM
d d1j2
2
d1j
)
;;=0
(2.19)
9.2
Boundary Value Problems with Two or More Parameters
215
which is seen to be independent of A if we put
a= GA,
(2.20)
The boundary condition at T/ = 0 is transformed to
'ij=0:
M=O
(2.21)
Let us now assign the missing boundary condition to be
dM(O)/ dT/
=A
(2.22)
or, in transformed form, A "2-"1 dM (0)/ dij = A
which will be independent of A if we put a2
-
a1= I
(2.23)
We therefore get
dM (0)/ dij = I
(2.24)
Solving Eqs. (2.20) and (2.23), we get The boundary condition at the second point then gives
(2.25) The invariant property of the physical constants will be invoked. From Eq. !2.20), it is seen that (since a 1 = 0)
aHw = aHw/(Hw - I) we can therefore define a parameter f3 as
f3 = aHw = aHw/(Hw - I)
(2.26)
Equation (2.19) finally becomes 2M
d + 2('ij _ !i + ('1/ a'ij dM dij) dM = 0 dij2 2 Jo aM - f3 dij dij
(2.27)
The solution takes the following steps. From the definition of f3, Eq. (2.26), and the required range of a and H w' the range of values of f3 is found to be approximately from 0.1 to 1.6. Let us calculate solutions for four values of f3: f3 = 0.35, 0.85, l.l, and 1.6. For each value of f3, Eq. (2.27) is integrated as an initial value problem for a range of values of a.
216
9.
Invarlance of Physical Parameters
For each solution, the corresponding values of ex and dM(O)/ dry are calculated. Following this procedure, the results for these four values of f3 are obtained. For illustration, the family of solutions with f3 = 0.35 is shown in Table 9.2. In general, a few trial solutions will show what range of ex to use to obtain the desired range of ex. TABLE 9.2
Solutions of Eq. (2.16) for f3 = 0.35 (j
a
M'(O)
(j
a
M'(O)
0.005 0.010 0.050 0.Q75 0.100 0.125 0.150
0.0045 0.0089 0.0465 0.0716 0.0983 0.1271 0.1586
1.1233 1.1183 1.0759 1.0475 1.0170 0.9836 0.9457
0.175 0.200 0.212 0.214 0.216 0.218 0.220
0.1944 0.2388 0.2684 0.2748 0.2823 0.2916 0.3063
0.9001 0.8377 0.7900 0.7787 0.7652 0.7477 0.7182
Using the Lagrangian method, Table 9.2 can be reduced to a form where the values of ex are at 0.005, om, 0.05, 0.10, ... , etc. The same will be done for three tables with f3 = 0.85, 1.10, and 1.6. The results are summarized in Table 9.3. All comments given in Section 9.2.1 related to the usefulness of such tables apply equally well here. TABLE 9.3
Tabulation of Solutions a
0.005 0.010 0.050 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.100 1.200
f3
= 0.35
1.1227 1.1171 1.0718 1.0150 0.8922 0.7208
= 1.6
f3 = 0.85
f3 = 1.1
f3
1.1239 1.1196 1.0853 1.0445 0.9687 0.8988 0.8323 0.7672 0.7014 0.6312
1.1091 1.1077 1.0873 1.0486 0.9776 0.9208 0.8647 0.8085 0.7523 0.6962 0.6422 0.5892
1.1243 1.1204 1.0894 1.0530 0.9865 0.9358 0.8855 0.8353 0.7851 0.7385 0.6990 0.6614 0.6257 0.5902 0.5556
9.2
Boundary Value Problems with Two or More Parameters
9.2.3
217
Transverse Curvature Effects of Flow past a Slender Parabola of Revolution
As another example, consider the flow of incompressible fluid over a slender parabola of revolution. When the velocity of the mainstream is large, a boundary layer will be formed and the boundary layer equations can be written as [9] Continuity equation:
a(ru)
a(rv)
--+--=0
(2.28)
u~ + vau = !:.. ~ (r au ) ax ay r ay ay
(2.29)
ax
ay
Momentum equation:
subject to the boundary conditions
y=O: y =
00:
u = 0, v = vw ( x) u = Uco
where u and v are the velocity components in the x and y direction, respectively, and /I is the viscosity. The distance r is defined as
r(x,y) = ro(x) + ycos(}
(2.30)
Figure 9.2 shows the coordinate system. By introducing the Levy-Lees transformation [9], l:.=
..
L x
0
ro2
-dx L2 '
1]=
fi LY -
co
/I~
0
r -dy L
---
Fig. 9.2
Schematic diagram of the body.
9. Invarlance of Physical Parameters
218
and
.s: _ U oo
df
= _ L r;c;-f (a1j df + oo
v
- d1j ,
r
VIIU
!;,
ax d1j
'5
.1)
L 2 2~
Eq. (2.28) is satisfied identically and Eq. (2.29) becomes
+ .1 f
d:J d1j3
2
d~ + t -.!£ ('n d~ ) = 0 d1j'/ d1j2
d1j2
(2.31)
where Land U 00 are the characteristic length and the mainstream velocity, respectively, and the transverse curvature parameter is defined as
= (2Lcos(}/'5)~/I~/Uoo
t
The boundary conditions are 71 = 0: 71=00:
f(O) = C, df(O)/ d1j df(00)/d1j=1
=
0
where C is a constant related to the dimensionless suction (or blowing) velocity and t is the transverse curvature parameter. To transform Eq. (2.31) to an initial value problem, a linear group of transformations, f= Aa>j is again selected. Under this transformation, Eq. (2.31) and the boundary conditions at 71 = 0 transform to (2.32) (2.33)
In addition, the missing initial condition is chosen to be 71
= 0:
and, in transformed form, becomes 'ij=0:
(2.34)
Under the requirement of the independence of the differential equation on A, Eqs. (2.32)-(2.34) produce the following relations: a 2 - 3a)
= 2a 2 - 2a) = a 2 - 2a),
Again, one is faced with
a)
=
a2
a2 = 0,
a 2 - 2a l = I
= 0 as the only solution.
9.2
Boundary Value Problems with Two or More Parameters
219
As in the first example, the answer is to define two auxiliary constants (2.35) Making these substitutions yields the four equations 0: 2 - 30: 1 = 20:2 - 20: 1 = 0: 3 + 0:2 - 20: 1 , Thus, the «'s are The initial value problem is now 3J d d2j t- d (_ d2j ) + -1 f- + - TJ-2 =0 3
crrf
2
dij
drj
drj
(2.36)
with the boundary conditions
J(o)
1j = 0:
c, dJ(O)/drj = 0, d2j(O)/ drj2 = 1
=
Another table may be created, just as in the preceding two sections. A new parameter is found by eliminating A from the relation
i = A -1/3t ,
C = A -1/3C
to yield
/3=t/c=i/c For each solution, A is found from the transformation of the remaining boundary condition A
= [ df- ( 00 ) /
] -3/2
(2.37)
drj
Figure 9.3 is a plot of 1"(0) for a range of /3. As before, a separate solution is required for the case where C = 0, t =F O. This solution is shown in Fig. 9.4. The numerical data from these curves are collected in Tables 9.4 and 9.5, respectively. All comments related to the possible applications of Table 9.3 remain valid here. TABLE 9.4 Selected Solutions of Eq. (2.31) for C = 0
0.0 0.5 1.0 1.5 2.0 3.0
0.33206 0.45863 0.56656 0.66459 0.75706 0.93445
5.0 7.0 9.0 11.0 13.0
1.27874 1.61625 1.94841 2.27428 2.60909
9. Invarlance of Physical Parameters
220
TABLE 9.5 Selected Solutions of Eq. (2.31), C*-O C
f3
0.0
2.0
4.0
6.0
8.0
15.0
0.0 1.0 3.0 5.0 7.0 9.0
0.3321 0.3321 0.3321 0.3321 0.3321 0.3321
1.1694 1.4426 2.0574 2.6943 3.3258 3.9512
2.1074 2.4970 3.6488 4.8555 6.0456 7.2146
3.0771 3.5500 5.2048 6.9476 8.6658 10.3544
4.0596 4.6033 6.7360 8.9945 11.2236 13.4147
7.5330 8.2812 11.9684 15.9416 19.8771 23.7535
f3
- 25.
-15.
-8.0
- 6.0
-4.0
- 2.0
6.0476 10.0227 13.9400 17.7961
1.3553 3.6219 5.8820 8.1002 10.2795
1.1211 2.8811 4.6269 6.3380 8.0177
0.0671 0.8756 2.1061 3.3192 4.5063 5.6701
0.1194 0.6199 1.2813 1.9293 2.5625 3.1848
C
-0.5 -2.0 -4.0 -6.0 -8.0 - 10.0
21.7830 27.9359
c Fig. 9.3
d1(O)/ ttr,2 versus C for a range of
f3.
9.4
Thin Struts with Large Elastic Displacement
221
2.0
1.0
o
o
4
Fig. 9.4
d2f(O)/ rJr,2 for C =
12
8
o.
9.3 SYSTEMATIC SEARCH OF MULTIPLE SOLUTIONS
In this section, an example will be given which includes an extension of the method in three ways. First, the invariant property among physical parameters is included, as in the previous two sections. Second, the limitation imposed previously on the differential equations, that they must be independent of the parameter of transformation, is relaxed. Third, a systematic way of searching for multiple solutions is introduced. The example further demonstrates the versatility of the transformation method. It provides a highly efficient procedure for searching for multiple solutions and analyzing the properties of these solutions. 9.4 THIN STRUTS WITH LARGE ELASTIC DISPLACEMENT
Consider an eccentrically loaded thin strut (see Fig. 9.5) experiencing large elastic deformations and where the end supports are simultaneously pinned and restrained by torsional bar springs of modulus k [10]. The differential equations resulting from the analysis are Euler:
dO ds
( a7T 2/ L 2) sin( TTS / L)
M
=
[I + (7T2a2/L2)cos2(7TS/L)f/2
£1 -
(4.1)
Geometry:
y
. du a7T ( TTS di = sin 0 - L cos L
)
(4.2)
222
9. Invarlance of Physical Parameters Q
--l
T
p
Q/2 Q/2 f----------L - - - - - - - - - - 1
Fig. 9.5 Schematic diagram of the strut. (Reprinted from [II) with permission of the American Society of Mechanical Engineers.)
Geometry: d~ =
(4.3)
cosO
ds subject to the boundary conditions
(4.4) where a, E, I, L, M, s, uy' 0, and ~ are the amplitude of initial curvature, modulus of elasticity, moment of inertia of cross section at any location, length of strut, bending moment, distance along deflected center line, y displacement, angular inclination of the centerline, the x location, respectively. Since the strut is symmetric, the boundary conditions can now be rewritten as
(4.5)
O(tL)=O,
For the column, the function M can be written as M
= - PUy - t Q~ - Pa sin(7Ts/ L) - N pe + kO(O)
Introducing the following dimensionless quantitites, t=
s/ L,
K= kL/ Elo,
a = a/ L,
€
=
e/ L,
~=~/L
(4.6)
223
9.4 Thin Struts with Large Elastic Displacement
Eq. (4.1) becomes dO = _
dt
2N 7T pr Y
_
~
~
7T2Npli.
2Npe 1110 -
1110 2N 7T prf.
KO(O)
1110
1110
- -- + -- -
1110 sm er
asin 7Tt [1+7T 2a2 cos27Tt]3 /2 7T2
(4.7)
For the special case in which the midspan lateral load is zero, i.e., Q = 0, and there is no initial curvature, i.e. a = 0, and considering the column whose moment of inertia I is given by 10 (a constant), Eq. (4.7) is simplified to
~ =
-7T
2N
pr Y
-
2N 7T prf.
+ KO(O)
(4.8)
Also, for this special case, Eqs. (4.2) and (4.3) become
dy I dt = sinO
(4.9)
dVdt = cosO
(4.10)
subject to the boundary conditions
(0) = 0,
yeO) = 0,
~(O) = 0
(4.11)
We will consider here only the simultaneous solution of Eqs. (4.8) and (4.9). Once the solution of 0 is known, the solution of ~ can be obtained by simply integrating Eq. (4.10), since it is independent of Eqs. (4.8) and (4.9). Huddleston [10] presented an iterative method for solving Eq. (4.8) and (4.9). The problem is different from other examples treated so far in that multiple solutions exist which are physically possible. Following the previous sections, the linear transformation is introduced as t
= A ali,
0 = A a20,
y
= A a3Y
(4.12)
where 0:\, 0: 2 , and 0:3 are constants to be determined and A is the parameter of transformation. Equations (4.8) and (4.9) become
dO di
=-
7T2N ~(al+a2)
PA a2
d-.2:. = A a,-a3 sin A a20 di
(y+
Af.( 3
)
+ KAaIO(O)
(4.13) (4.14)
224
9.
Invarlance of Physical Parameters
This method requires that the differential equations be independent of the parameter A. An inspection of Eqs. (4.13) and (4.14) shows that this cannot be satisfied unless a l = a2 = a 3 = 0, which, in turn, means the identity transformation. To overcome this difficulty, this restriction is relaxed and the following are introduced:
f* = f/ A a"
N*pr = N pr A 2a , '
Equations (4.13) and (4.14) then become -
d~ = dt d-
2
a:r
N* A (y
TT
+ f*) + K*O(O)
_
.2::. = sin A azO di
(4.16) (4.17)
Following the method developed in previous sections, the unknown boundary condition at t = 0 is set to equal to the parameter of transformation or its negative value, i.e., 0(0)
= ±A
(4.18)
which, upon transformation, gives and thus 0(0) = ± 1
and
(4.19)
Next, the invariant properties of the physical parameters will be invoked. In other words, combinations of the physical parameters are sought which are invariant under the transformation defined by Eq. (4.12). These are and
f*K*
= fK
(4.20)
This means that the above combination of parameters remains the same under the transformation. Thus we define
t/J = f*K* =
fK
(4.21)
and the three parameters N;r' f*, and K* are then replaced by N;r' a, and
t/J. Also, generality is not lost if the transformed parameter N;r is taken as unity. Equations (4.16) and (4.17) then become 2 dO TT C t/J _ -_ = - - (y + va ) ± dt A {ci
fly
-
- = sinAO
di
(4.22) (4.23)
9.4 Thin Struts with Large Elastic Displacement
225
subject to the boundary conditions
i = 0: As the last step,
0'1
J (0) = 0, 0 (0) = ± I
is found by using the boundary condition at t 0=0
(4.24)
= t; thus
at
or 0'1
= {In(0.5) - In[ i (where 0 = O)]} / In A
(4.25)
The procedures for the solution of Eqs. (4.22) and (4.23) for a given set of values of the parameters 0' and I[; are now described. The physical quantity of interest is the deflection at midspan, i.e., y( t). The following steps are taken: First, consider the case of 0(0) = + 1. Then, a value of A is assigned. Equations (4.22) and (4.23), subject to the boundary conditions (4.24), are integrated until 0 = O. The values of i and J at this point are obtained, which are substituted into Eq. (4.25) and the constant 0'1 calculated. The values of N pr and E can be calculated from Eq. (4.15) as N pr = A
-2a,
(4.26)
and E
= ~O'/ N pr
(4.27)
The midspan deflection yO) can be calculated from Eq. (4.12) and the value of J obtained earlier (i.e., value of J where 0 = 0): (where 0 = 0) Another value of A is then assigned and the above steps repeated, resulting in another set of values of N pr' E, and YO). This may be continued until the required range of N pr is covered. The deflection y( t) may now be either plotted or tabulated as a function of N pr for constant values of 0' and 1[;. Next, the same steps are repeated, except that now the boundary condition of 0(0) is taken as negative unity. Another set of yO )-Npr data is obtained for the same pair of values of 0' and 1[;. It is of interest to note that the integration curve of Eqs. (4.22) and (4.23) intercepts the 0 axis more than once. This particular property explains why published numerical solutions exhibit the fact that for a given set of physical constants there may be more than one solution. Details of these are now demonstrated by a numerical example. Consider the numerical solution for 0' = 0.05 and I[; = 0.5. First, the case
228
9. Invarlance of Physical Param......
of 0(0) = + I is considered. Equations (4.23)-(4.24) become
d~
= _
dt
ely
-
di
7(2
A
(y + vIa") +
-.:L
(4.28)
via"
-
(4.29)
= sinAl}
subject to the boundary conditions
i = 0:
y(O) = 0, 0(0) = I
(4.30)
Since a and If; are given, A is the only parameter in the solution of Eqs. (4.28)-(4.30). A family of solutions is now calculated by assigning a series of values of A. For example, if A = 0.65 is selected, then the initial value problem represented by Eqs. (4.28)-(4.30) can be integrated. The integration curve is shown in Fig. 9.6 for i ranging from 0 to approximately 5.6-a range sufficient to show the properties of the solution. 1.0 0.5 10:.
-t
0.0 6
·0.5 ·1.0 Fig. 9.6 Integration curve for {j versus i (0: = 0.05, 1/1 = 0.5, A = 0.65). (Reprinted from [111 with permission of the American Society of Mechanical Engineers.)
The integration curve shows that the first intercept with the i axis is at a point where i equals approximately 0.394. Substituting this value of i into Eq. (4.25), we get a\
= In(0.5/0.394)/ln(0.65) = -0.552
(4.31)
The corresponding value of N p r from Eq. (4.26) is N pr
= (0.65)2(0.552) = 0.6216
(4.32)
It is to be noted that a 3 = a\ = -0.552 from Eq. (4.15). Also noting that a2 = I [from Eq. (4.19)], we can employ the transformation, Eq. (4.12), to
9.4 Thin Struts with Large Elastic Displacement
227
calculate the midspan deflection y( t):
yO) = 0.1785 The above is the solution corresponding to the first intercept of the integral curve with the i axis. As the integration is continued the values of if change smoothly from plus to minus values, and then change back to positive. Each time the value of if changes sign, an intercept with the i axis is obtained, which then leads to a solution. Six such interceptions are shown in Fig. 9.6. The remaining five such points occur at i = 1.425, 2.456, 3.487, 4.51, and 5.549. By following the same steps as in the evaluation of the solutions corresponding to the first interception, the solution for these points can easily be obtained. Table 9.6 shows a summary of the results for all six intercepts. TABLE 9.6 Summary of Solutions for", = 0.5, a = 0.05, A = 0.65 Q
i(U = 0)
y(!)
0.394 1.425 2.456 3.487 4.518 5.549
- 0.552 2.431 3.695 4.509 5.110 5.587
0.6216 8.1245 24.1297 48.6374 81.6478 123.1604
0.1785 - 0.1030 0.0287 - 0.0421 0.0156 - 0.0264
Q Reprinted from [II) with permission from the American Society of Mechanical Engineers.
Of interest are the deflection curves y(t) for the above six possible solutions, as plotted in Fig. 9.7. They are calculated directly from the definition of the transformation, as discussed above. Due to the symmetry of the problem, only one-half of the strut is shown with t = a and t = t representing the left end and the midspan of the strut, respectively. To discuss the significance of Fig. 9.7, the original equations, Eqs. (4.8) and (4.9), are rewritten in terms of N p r , lX, and 1/;, as follows:
dO / dt = -
7T
N pr( y + ~lX / N pr
2
dy/ dt = sinO,
) + 1/;~N prl lX (0(0»
yeO) = 0, (0) = 0
(4.33) (4.34)
Since the missing boundary condition is defined as the parameter of transformation, i.e.,
0(0) = ±A
228
9. Invarlance
0' Physical Parameters
0.2
;>.,
0.1
-0.1
-0.2 ' - - - - - - - - - ' - - - -........- - - - - ' - - - - - - ' - - - - - - - - - ' 0.0 0.1 0.2 0.3 0.4 0.5
Fig. 9.7 Deflection curves (ex = 0.05, '" = 0.5, A = 0.65). (Reprinted from [II] with permission of the American Society of Mechanical Engineers.)
Eqs. (4.33) and (4.34) become
dO j dt = -
'1T
2
N pr( Y
+ ~aj N pr ) ± tf;~Npri a (A)
dyj dt = sinO
(4.35) (4.36)
subject to the boundary conditions
yeO) = 0,
0(0) = ±A
which is seen to be an initial value problem. For the example just described, the plus sign was taken in both Eqs. (4.35) and (4.36) and the boundary conditions for the parameters a
= 0.050,
tf;
= 0.5,
A
= 0.65
The six solutions just obtained (shown in Table 9.6) are the solutions corresponding to six values of N pr ' To get another solution, the value of A is changed from 0.05 to another value (e.g., 1.85). For each value of A, the above steps can be followed and six solutions, corresponding to six values of N pr ' are obtained. If the Npr - yO) data thus calculated are collected for the first interception and
0040
__ (1+)
(1.)
~
co ~
-.
-t
0.10
.......
/
(2·)
.......
/
.-
-
:::r 5'
(3·)
.......
......
fa
............ I
-. .......
/'
-
-_
I.-
.......
(<;+.
~
(5·)
~
......
!f
...... ........
:;
'im
(4·)
iii
(6·)
1::'.
!.
n
c,
e
+,'
iii
'U
§
om
3
CD
a
102
10
103
N pr
Fig. 9.8 Family of solutions yO) versus Np.(9(0) = + 1, '" = 0.5, a = 0.05). For curves labeled (1 + )-(6 +), ordinate is + yO); for (1 - )-(6 -), ordinate is - y(!). (Reprinted from [11] with permission of the American Society of Mechanical Engineers.)
N
~
230
9. Invarlance 01 Physical Parameters
plotted on yO)-Npr coordinates, a curve will be obtained. Since six interceptions are considered, six curves are obtained. This is shown in Fig. 9.8, where the curves are labeled (l +), (2 +), (3 +), (4 +), (5 +), and (6 + ). The plus sign after the number indicates that these are the solutions of Eqs. (4.22) and (4.23) for which 0(0) was taken to be + A. Next, consider the case in which 0(0) is taken to be - A, which means that the minus sign in Eq. (4.22) is selected. The same procedure can be followed and another six curves obtained. These are also included in Fig. 9.8, labeled as (1 - ), (2 - ), (3 - ), (4 - ), (5 - ), and (6 - ). It should be noted that, although more than six intercepts can be located, it is not necessary to proceed further since a definite pattern is clearly shown in Fig. 9.8. Before proceeding with more discussion of the curves shown in Fig. 9.8, it should be pointed out that each point in Fig. 9.8 is computed noniteratively. This is not the only feature characterizing this new extension of the method. Another feature of this method is that it provides a systematic way to locate all solutions for a given set of values of physical parameters. In order to compare the solutions obtained by the transformation method with those of Huddleston [10], consider the solution corresponding to the following values of the physical parameters:
K=5,
E:
=
0.1,
(4.37)
From Fig. 4 of Huddleston [10], there are three solutions:
yO) = 0.375, -0.17, -0.43
(4.38)
In terms of the notations of this section, the parameters in Eq. (4.37) become: ex
= 0.05,
1/; = 0.5,
(4.39)
For these values of the parameters, Fig. 9.8 gives the same three solutions. By drawing a vertical line at a distance of 5 on the N pr axis, it intercepts the y( t )-Npr curves at three points, resulting in three values of y(!-). The number of times the vertical (constant N pr ) line intercepts the y(!- )-Npr curves is the number of solutions for this given set of values of the physical parameters. Figure 9.8 then shows that there may be one, two, three, four, or five solutions for a given pair of values of ex and 1/; (in this example, ex = 0.05 and 1/; = 0.5). The above establishes the method as a highly efficient procedure for searching for multiple solutions and analyzing the properties thereof. It has been used to construct working tables for use by designers [13].
Problems
231
PROBLEMS
1. Using data in Table 9.5, plot d1(0)j dq2_ C curves at a few values of the transverse curvature parameter t. 2. Recent developments in hypersonic flight, missile reentry, rocket combustion chambers, power plants for interplanetary flight, and gas-cooled nuclear reactors have focused attention on thermal radiation as an important mode of heat transfer. It is therefore of fundamental importance to analyze the coupled effects of convection and radiation. For this reason, Viskanta and Grosh [14] presented an analysis on the combined convection and radiation in boundary layer flows. The temperature distribution is found to be governed by the energy equation 2 d (0 dTJ2
+
~) + 3N
N
pr
2
f
dO = 0 dTJ
subject to the boundary conditions
0(00) = 1 where the function f is given by Blasius's equation, Eq. (1.7) of chapter 7. Solve the problem by the initial value method described in this chapter. Hint:
N = NB
-3 P,
OW = OwB -P,
NO; = NO;
Also, use the following approximate solution for f:
f= tTl - tTJ3 Answer:
N
Ow
dO (O)j dTJ
0.1
1.5 2.0
- 0.04225 - 0.06971
1.0
1.5 2.0
- 0.00381 - 0.01507
10.0
1.5 2.0
- 0.00553 - 0.00658
3. Consider again the analytical prediction of the temperature distribution of a Newtonian fluid held between two coaxial cylinders, problem 4 of
9. Invarlance of Physical Parameters
232
Chapter 3; the resulting boundary value problem is
I1 dgd(dO) g dg + 4N g41
O(k) = 0, 0(1)=1
= 0,
Solve the problem by the method of transformation treated in this chapter and compare the results with the exact solutions,
o= [ (N + l) -
N / e]
- [ (N + 1) -
N / e](in Uln k)
for Note:
k = 0.1,0.2, ... , 0.9; Introduce the tranformation
N
= 0.1, 1.0, 10.
and put
N = NA- 2a"
k = kA -a,
REFERENCES 1. Na, T. Y., An initial value method for the solution of a class of nonlinear equations in fluid mechanics, J. Basic Eng., Trans. ASME 92, 503-509 (1970). 2. Na, T. Y., An initial value method for the solution of MHD boundary layer equations, Aeronaut. Q. 21, 91-99 (1970). 3. Na, H. S., and Na, T. Y., An initial value method for the solution of certain nonlinear diffusion equations in biology, J. Biosci. 6, 25-35 (1970). 4. Lin, S. H., and Fan, L. T., Examples of the use of the initial value method to solve nonlinear bounary value problems, AIChE J. 18, 654-656 (1972). 5. Vimala, C. S., and Nath, G., An initial value method for the study of shock-induced laminar compressible boundary layers, ASME paper 73-FE-4 (1973). 6. Nath, G., Initial value method for a pseudoplastic fluid in a laminar boundary layer over a porous flat plate with suction, J. Appl. Mech. 95, 1123-1124 (1973). 7. Scott, T. C., Rinschler, G. L., and Na, T. Y., Further extensions of an initial value method applied to certain nonlinear equations in fluid mechanics, J. Basic Eng., Trans. ASME 94,250-251 (1972). 8. Liu, S. W., Onset of surface combusion in still atmosphere, AIAA J. 7, 734-736 (1969). 9. Mark, R. M., Laminar boundary layers on slender bodies of revolution in axial flow, Guggenheim Aerosp. Lab., California Instit. of Technology, Hypersonic Wind Tunnel Memo. No. 21 (July 1954, unpublished). 10. Huddleston, J. V., A numerical technique for elastica problems, J. Eng. Mech. Proc. ASCE 94,1159-1165 (1968). II. Na, T. Y., Kurajian, G. M., and Holbert, D. L., Analysis and design of thin struts with large elastic displacement, Part I, J. Eng. Ind. 96,917-922 (1974). 12. Kurajian, G. M., Na, T. Y., and Holbert, D. L., Analysis and design of thin struts with large elastic displacements, Part II, J. Eng. Ind. 96, 923-930 (1974). 13. Favre, A., "Contribution a l'etude experimentale des mouvement hydrodynamiques a deux dimensions," Ph.D. thesis, pp. 1-192, Univ. of Paris, Paris, 1938. 14. Viskanta, R., and Grosh, R. J., Int. J. Heat Mass Transfer 5,795-806 (1962).
CHAPTER
10
METHOD OF PARAMETER DIFFERENTIATION
10.1
INTRODUCTION
The method of parameter differentation is known in the mathematical literature as the method of continuation. According to Lichtenstein [I], literature on this concept can be traced as far back as 1869 (Schwarz). Earlier work on this subject has been found in the field of conformal mapping. Application of this method to engineering problems, however, started in the early 1960s as a result of the availability of high-speed digital computers. Basically, the method involves the solution of a differential equation where a physical parameter appears either in the differential equation or in the boundary conditions. Starting from the known solution for a certain value of the parameter, the solution of the equation for other values of the parameter may be obtained by integrating the rate of change of the solution with respect to the parameter. Each step in the calculation involves only a small perturbation in the parameter. In this way, the equations are linearized and, as such, can be solved noniteratively by the methods given in Chapters 2-4. The resulting solution can then be perturbed again, and the solution corresponding to another increment of the parameter can be calculated. This process can be repeated until the solutions for the complete range of the parameter are obtained. Before we introduce the method for the solution of nonlinear differential equations, an interesting application of the concept to the solution of nonlinear algebraic equations will be given in Section 10.2. The general idea of parameter differentiation will be derived in Section 10.3. This will be followed by the application of the method to the solution of four nonlinear boundary value problems. The general parameter mapping (GPM) of Kubicek and Hlavacek [2] will then be outlined in Section 10.5 and the method of continuation of Roberts and Shipman [3] sketched in Section 10.6. 233
234
10. Method of Parameter Differentiation
10.2 NONLINEAR ALGEBRAIC EQUATIONS
This method of solving nonlinear algebraic equations was due independently to Kane [4] and Yakolev [5]. By its use, the solution of a system of nonlinear algebraic equations can be obtained by integrating a system of ordinary differential equations of the first order. To introduce the idea, let us consider the algebraic equation
f(x) = 0
(2.1)
whose solution is sought. As a first trial, let us arbitrarily choose
x=k
(2.2)
which, of course, does not satisfy the given algebraic equation. We will now apply the method of parameter differentiation to find the solution of Eq. (2.1) using this first approximation. First, we introduce an equation
f(x)
=
f(k)(l -
(2.3)
7")
where k is the first approximation given by Eq. (2.2). The notation f(k) means replacing x in Eq. (2.1) by k. The variable 7" is introduced as a new "independent" variable, which changes from 0 to 1. For 7" = 0, Eq. (2.3) becomes f(x) = f(k) which means that x is equal to its assumed value k, as given by Eq. (2.2). For 7" = 1, Eq. (2.3) becomes f(x) = 0 which means x now satisfies the given algebraic equation, Eq. (2.1). We therefore expect that as 7" changes from 0 to 1, the value of x will be changed from its (arbitrarily) assumed value of k to a value which satisfies the given algebraic equation. This points out the need for a differential equation giving dx / dr, Such an equation can be obtained by differentiating Eq. (2.3) with respect to 7", which gives
fxCx)dx/d7" = -f(k)
(2.4)
(where fx(x) = df/ dx). The boundary condition is 7"
= 0:
x
=k
(2.5)
As a simple example, consider the algebraic equation
f(x) = 3x 2 - 2x - 1 and an arbitrarily selected first approximation, x
(2.6)
= 2, which means
k
= 2.
10.2 Nonlinear Algebraic Equations
235
For Eq. (2.6), we have
f(k) = f(2) = 3(2)2- 2(2) - I = 7
fAx) = 6x - 2,
Equation (2.4) and its boundary condition, (2.5), therefore becomes
= -7 j(6x - 2)
dxj dt
(2.7)
subject to the boundary condition T
=0:
(2.8)
x=2
Equation (2.6) can then be solved by integrating Eq. (2.7) from T = 0 to T = I will be the solution. In general, the differential equation has to be integrated numerically, as will be done for Eq. (2.7). Table 10.1 gives a summary of the solution. At T = I, we get x = 1.000. The solution of Eq. (2.6) is therefore x = 1.000.
T
= 1. The value of x at
TABLE 10.1 Solution of Eq. (2.7) or
x
or
x
0.0 0.2 0.4
2.000 1.854 1.691
0.6 0.8 1.0
1.507 1.288 1.000
For Eq. (2.6), there are two independent solutions. If we choose k to be closer to the other solution (x = - t), the value of x will approach to the value of - t as T is increased from 0 to I. This point will be fully discussed later in the chapter. The above can now be generalized to a system of n simultaneous algebraic equations as follows. Consider the system of n simultaneous algebraic equations governing n unknowns, say, the elements of a n X I matrix x with (2.9) F(x) = 0 where F(x) is a n X I matrix. To find the matrix x satisfying Eq. (2.9), x is regarded as a function of a parameter T, i.e., x = X(T). The value of T is assumed to vary from zero to unity, and at T = 0,
x(O) = k
(2.10)
where k is an arbitrarily selected n X I matrix. Next, it is required that x( T) satisfy the equation F(x) = F(k)(1 - T)
(2.11)
236
10. Method of Parameter Differentiation
Differentiating Eq. (2.11) with respect to the parameter 'T, the result is (2.12)
FxCx)dxjd'T = -F(k) which can be solved for dx j dr as
dxj dt
= - FxCX)-IF(k)
(2.13)
The inversion of the matrix and the rapid and accurate integration of a system of ordinary differential equations are rather standard. Thus, using the boundary conditions specified in Eq. (2.10), Eq. (2.13) can be integrated from 'T = 0 to 'T = 1. The matrix x(l) will be a solution of Eq. (2.9) since at 'T = 1 Eq. (2.11) is reduced to Eq. (2.9). A simple example will be given to illustrate the steps. Consider the system of equations sin(x) - x 2)sinx\ + sin(x) - x\)sinx 2 = 0 cosx 1 - 2cosx 2 + 3cosx) = 0
(2.14)
sin x, - 2sinx2 + 3sinx) = 3 - 3 1/ 2 The set of equations corresponding to Eq. (2.12) is
. . dX1 [sm(x) - x 2)cosx\ - cosfx, - x\)smx 2 ] d'T .
.
dX2
+ [sm(x) - x\)cosx 2 - cost x, - x 2)smx\] d'T
dx)
+ [cos(x) - x 2)sinx\ + cos(x) - x l)sinx2] d'T
= - [sin(k) - k 2)sink l + sin(k) - k\)sink 2] . dx, . dX2 smx\ d'T - 2smx2 d'T
.
(2.15)
dx)
+ 3smx) d'T = coek, - 2cosk2 + 3cosk)
dx\ dX2 cos Xl d'T - 2 cosx 2 d'T
dx)
+ 3 cosx) dt
= -(sink\ - 2sink 2 + 3 sink) - 3 + 3\/2) Kane [4] arbitrarily selected the initial approximations
and integrated Eq. (2.15) from 'T = 0 to 'T = 1. In less than 3 seconds calculation time on a Bourroughs 5500 computer, the following solutions
10.2 Nonlinear Algebraic EquatIons
237
are obtained XI
= -0.68968,
X2
= 0.47982,
x 3 = 1.2300
The same solutions were obtained by assigning another set of initial conditions, A different solution, namely XI
x 2 = 2.66178,
= 3.83127,
X3
= 1.91159
resulted from using the initial conditions xl(O) = k, =
7T,
x 2 (0) = k 2 = 27T/3,
This is due to the property that a set of transcendential equations possesses infinitely many solutions. The method will now be applied to a few engineering problems. 10.2.1
Determination of the EqUilibrIum Composition of a Gas Mixture at High Temperature
The theoretical determination of equilibrium compositions in the flow of chemically reacting fluids through channels is of great importance in various branches of engineering. The calculation is complicated by the fact that it is necessary to obtain the simultaneous solutions of a system of nonlinear algebraic equations. Although special iterative techniques have been developed for the solution of such systems of equations, the parameter differentiation method is seen to be especially suitable for such calculations. Let us consider a simple example. Suppose one mole of CO2 at 10 atm is heated in a steady-flow constantpressure process. It is required to find the equilibrium compositions at a total pressure of 10 atm and a range of temperature from 1000 to 6000 K. The equilibrium composition is assumed to consist of CO 2 , CO, and 02' From the given conditions, we can write (2.16)
where the left-hand side represents the initial condition and the terms on the right-hand side are the chemical species at equilibrium. The unknown constants a, /3 and yare the number of moles of the species at equilibrium. From a balance of the element C, we get a+/3=1
Similarly, a balance of the element 2a
° gives
+ /3 + 2y
Both relations are based on Eq. (2.16).
= 2
(2.17) (2.18)
238
10. Method of Parameter Differentiation
Since there are three unknowns (a, 13, and y), one additional equation is required, which can be obtained from the equilibrium equation co 2:;;=co+ !02 From the law of mass action, we get
13ft
I
Ja + 13 + Y
a
or
(2.19)
fi = K (2.20)
where K is the equilibrium constant and p is the total pressure. In the present example, p = 10 atm. Table 10.2 gives the equilibrium constants at different temperatures for the reaction (2.19). TABLE 10.2 Equilibrium Constant of Reaction (2.19) as a Function of Temperature
T(K)
InK
T(K)
1000 2000 3000
- 23.535 -6.641 -1.117
4000 5000 6000
InK 1.593 3.191 4.239
In summary, three simultaneously algebraic equations are obtained for the solution of a, 13, and y at a given temperature, which are
11 = a + 13 - I, 12 = 2a + 13 + 2y 13 = f32 yp - K 2a2(a + 13 + y)
2,
(2.2Ia, b) (2.2Ic)
We will now apply the method of parameter differentiation as outlined in Eqs. (2.9)-(2.13). The set of equations corresponding to Eq. (2.12) is da dr da 2 dr
- K
2(3a2
+
df3 dr
+ 2af3 + 2ay)
= - { f3hop
+
df3 dr
=
-(ao + 130
-
I)
(2.22)
+2
dy dr = -(2ao + 130 + 2yo - 2)
da dr
+ (2f3yp - K 2(2) dr + (f31J - K 2(2) dr
df3
- K2a5(a o + 130 + Yo)}
(2.23) dy
(2.24)
where ao, 130' and Yo are the (arbitrarily) assumed first approximation. In other words, Eqs. (2.22)-(2.24) will be integrated from 'T = 0 to 'T = I subject to the boundary condition 'T
= 0:
a(O) = a o,
13(0) = 130,
yeO) = Yo
(2.25)
10.2
Nonlinear Algebraic Equations
239
The last step involves the inversion of the coefficient matrix in Eqs. (2.22)-(2.24) [refer to Eqs. (2.12) and (2.13)] so as to express da] dr, df3/ dr, and dy / dt explicitly in terms of a, f3, and y and the integration of the resulting system of linear ordinary differential equations. Both are standard numerical steps. As a numerical example, the solutions of Eqs. (2.22)-(2.24) at a temperature of 4000K and the set of first approximation
a(O) = ao = 0.538, f3(0) = f30 = 0.462, y(O) = Yo = 0.200 (2.26a) are listed in Table 10.3. By choosing another set of first approximations, namely, '1"
= 0:
a(O) = ao = 0.500, f3(0) = f30 = 0.500, y(O) = Yo = 0.500 (2.26b) the same solutions are obtained, as shown in Table 10.3. The values of a, f3, and y at '1" = 1 are the solutions of Eq. (2.21). They are a = a(l) = 0.453, f3 = f3(I) = 0.547, y = y(l) = 0.273 (2.27) '1"
= 0:
By following the same procedure, the equilibrium composition at other temperatures can be calculated. Table 10.4 is a summary of such calculations. TABLE 10.3 Solutions of Eqs. (2.22)-(2.24) a.
Using the first approximation (2.26a)
b.
Using the first approximation (2.26b)
'T
a
fJ
y
'T
a
fJ
y
0.0 0.2 0.4 0.6 0.8 1.0
0.53800 0.51928 0.50147 0.48450 0.46829 0.45281
0.46200 0.48071 0.49852 0.51550 0.53170 0.54719
0.20000 0.21556 0.23066 0.24535 0.25965 0.27359
0.0 0.2 0.4 0.6 0.8 1.0
0.50000 0.49290 0.48487 0.47573 0.46524 0.45308
0.50000 0.50710 0.51513 0.52427 0.53476 0.54691
0.50000 0.45355 0.40756 0.36213 0.31738 0.27345
TABLE 10.4 Equilibrium Composition at Different Temperatures T(K)
a(l)
fJ(l)
y(l)
1000 2000 3000 4000 5000 6000
0.981 0.981 0.739 0.251 0.068 0.025
0.019 0.019 0.261 0.749 0.932 0.975
0.010 0.010 0.130 0.374 0.466 0.487
240
10. Method of Parameter Differentiation
10.2.2 Analysis of Flow In Pipe Networks
The analysis of pipe network systems in which both the magnitude and direction of the flow in the pipes are to be predicted is of interest to civil, chemical, and mechanical engineers. Such flow problems are usually solved satisfactorily by assuming a constant friction factor and employing the widely used relation between the flow rate Q and head drop Hi - He' the Hazen and Williams formula (2.28) where Hi and He are the total head at the inlet and exit sections of the pipe and I and d are the length and the diameter, respectively, of the pipe. The flow rate is represented by Q. The constant Chw is an empirical constant known as Hazen-Williams coefficient. Equation (2.28) will now be used for the solution of pipe network problems. A simple example will be given below, one which was solved in [6] by the classical Hardy-Cross method. Consider the pipe network shown in Fig. 10.1 where all the pipes are 2000 ft long and have a Hazen-Williams coefficient of 100. It is required to calculate the magnitude and direction of the flow rate in the five pipes. B
7 ft3Jsec
o
A
-+---<
>---+---
7 ft3 Jsec
c Fig. 10.1 The pipe network.
If Eq. (2.28) is applied to all five pipes, then we have H A - H B-- k AB QI.85 AB'
HB
-
H C -- k BC QI.85 BC'
HC
-
(2.29)
H D -- k CD QI.85 CD
where the k's are calculated from their defintion, given in Eq. (2.28), i.e., (2.30) For the problem under consideration, Chw = 100 and the lengths and
241
10.2 Nonlinear Algebraic Equations
diameters of each pipe are given. The five k's are found to be k AB = 4.5492,
k AC = 1.8755,
k BD = 54.466,
k BC = 13.456
kCD = 4.5492
Next, the heads can be eliminated from Eq. (2.29) as follows. For the pipe loop ABCA, the net head drop is zero, i.e.,
(H A
-
H B)
+ (H B - He) - (H A - He) = 0
which gives (2.3 I) Similarly, for the pipe loop BDCD, the net head drop is zero, i.e.
(H B - H D )
-
(H c - H D )
-
(H B - He)
=0
which gives kBDQri~5 - kCDQ~~5 - kBCQri~5
=0
(2.32)
In addition to the above, conservation of mass requires that QAB + QAC
=7
ft3/ sec
(2.33)
QBD + QCD
= 7 ft 3 / sec
(2.34)
QAB
=
(2.35)
QBC + QBD
Equations (2.31)-(2.35) are those required to solve for the five flow rates, QAB' QAC' QBC' QBD' and QCD' The system of equations corresponding to Eq. (2.12) is therefore I 85 k .
-
-
I 85 k .
--
0.85 ABQAB -
BD -
dQAB
~
+
8 0.85 dQBC 0.85 dQAC 1. 5 kBCQBC ~ - 1.85k ACQAC ~
(k AB Q ABO 1.85 + k Q 1.85 k AC Q ACO 1.85) BC BCO QO.85 BD
(2.36)
dQBD - I 85 k QO.85 dQCD _ I 85 k QO.85 dQBC dt . CD CD dt . BC BC dt
(k BD Q BDO 1.85 - k CD Q CDO 1.85 - k BC Q BCO 1.85) dQAB
~
dQAC
+
~
+
~
-
~
dQBD
~
dQAB
~
-
dQBC
~
dQCD dQBD
(2.37)
=
-(QABO+ QACO-
7)
(2.38)
=
-(QBDO+ QCDO-
7)
(2.39)
=
-(QABO -
QBDO)
(2.40)
QBCO -
242
10.
Method of Parameter DI"erentlatlon
Equations (2.36)-(2.40) are integrated from boundary conditions T = 0: QAB(O) = QABO' QAc(0) = QBO(O) = QBOO'
T
=
°
to
QACO'
T
= I, subject to the
QBc(0) = QBCO
Qco(O) = Qcoo
(2.41)
which involves the inversion of the coefficient matrix and forward integration, both of which are standard numerical techniques. The results are summarized in Table 10.5. TABLE 10.5 Summary of Solutions of Eqs. (2.36)-(2.40) T
QAB
QAC
QBC
QBD
QCD
0.0 0.2 0.4 0.6 0.8 1.0
3.000 2.877 2.751 2.623 2.491 2.356
4.000 4.123 4.249 4.377 4.509 4.644
1.000 0.968 0.938 0.908 0.878 0.849
2.000 1.908 1.813 1.715 1.613 1.507
5.000 5.092 5.187 5.285 5.387 5.493
The solutions of the original equations, Eqs. (2.31)-(2.35), are therefore the values of the Q's at T = 1, i.e., QAB = QAB(I) = 2.356,
QAC
= QAC(l) = 4.644
QBC = QBC(l) = 0.849,
QBO = QBo(l) = 1.507
Qco = Qco( I) = 5.493
The accuracy of the method can be seen by substituting the above solutions into Eqs. (2.31 )-(2.35). 10.2.3 Vibration of a Cantilever SUbject to a Tensile Following Force
In this section, we will demonstrate how efficiently the method of parameter differentiation can be applied to a search for multiple solutions. Consider a uniform Bernoulli-Euler beam clamped at one end and located at the other by a tangential tensile force of constant magnitude P, as shown in Fig. 10.2. The governing differential equation for the amplitude of small transverse vibrations about the x axis is a4y
a2 y ax 2
a2 y at
E I -4 - P - + p =0 2
ax
(2.42)
where E, I and pare Young's modulus, the moment of inertia, and the linear density, respectively.
243
10.2 Nonlinear Algebraic Equations x
I
p
"rr?r,.,.,..,~---)'
Fig. 10.2 Schematic diagram of the beam.
Introducing the dimensionless variables
~=x//,
r=t(EI/p/4)1/2,
(2.43)
Eq. (2.42) becomes a4y a2y -k 2 _
ae
ae
a2y
+-
ar 2
=0
(2.44)
Following the usual procedure for solving such problems, we introduce the solution (2.45) y(~,r) = Y(~) sin cor and invoke the boundary conditions dY(O)
YeO) = ~ =
d 2Y(I)
de
=
d 3Y(l)
de
=0
The following frequency equation is obtained k 4 + 2w 2(l + COSh~2COS~I) - wk2sinh~2sin~1 = 0
(2.46)
where and
~2 = [(w 2 + -;i-k 4) 1/ 2 + t k 2]
1/2
The solutions of Eq. (2.46) will give the natural frequencies for various values of k 2 . (Note that for a given beam, the variables /, E, and I are constants, and k 2 is therefore directly proportional to the load P.) Anderson and King [7] solved for the first three natural frequencies for a range of values of k 2 from 0 to 100, the solution procedure requiring iteration. Their results are shown in Fig. 10.3.
10. Method of Parameter Differentiation
244 100
80
w
60 40
20
20
0
80
60
40
100
k2
Fig. 10.3 Dependence of the frequencies on the follower force. (Reprinted from [7] with permission of the American Institute of Aeronautics and Astronautics.)
Equation (2.46) will now be solved for w 2 by the method of parameter differentiation [8]. The differential equation in dw/ dr, corresponding to Eq. (2.4), is
[""'{I + COShA,COSA,) - k'sinhA,sinA,
+
w3
(
sinhA2cosA( _ coshA 2sinA( )
(w + tk4)1/2
A2
2
e
w2
2(w2 + tk4)1/2 = - [
where
( coshA
2sinA(
AI
+ sinhA2cosA I
A2
A(
k4+ 2wJ( 1 + cosh A20 cos A10)
wo is the first
-
)]
dw
dr
woe sinh A20 sin A10 ]
(2.47)
approximation, i.e.,
= 0:
w(O) = wo
tk2f/2,
A20 =
7"
and
AIO = [(w~ + tk4)1/2 -
(2.48)
[(w J + tk4)1/2 + tk2f/2
For a given value of k 2, Eq. (2.47) can be integrated, using the boundary condition (2.48), from 7" = 0 to 7" = 1; the value of w(7") at 7" = 1 will be the solution of Eq. (2.46). The solution, however, depends on the value of W o chosen. By choosing different values of wo, the integral curve w(7") will approach different natural frequencies at 7" = 1, depending on which one is closest to the
245
10.2 Nonlinear Algebraic Equations
assumed value. For the present problem, in which multiple solutions are sought, we can simply start from W o = 0 as the first solution and then systematically increase the value of W o by Llwo each time. The solutions will approach the first eigenvalue for the first few solutions, then the second eigenvalue, etc. This is demonstrated in Table 10.6. For the first three values of wo' namely 1, 5, and 10, the solutions all approach to the first eigenvalue of 1.9112. For the next three values of wo, the solutions all approach the second eigenvalue of 27.18. For Wo equal to 67, 80, and 100, all solutions approach the third eigenvalue, 68.71. This process can be continued if other eigenvalues are required. TABLE 10.6 Solutions of w(r) for Various Values of Wo (k 2 = 20)0 r
w(r)
r
w(r)
0.0 0.2 0.4 0.6 0.8 1.0
1.0000 1.2409 1.4405 1.6141 1.7695 1.9112
50
0.0 0.2 0.4 0.6 0.8 1.0
50.000 47.432 44.697 41.516 37.250 27.182
5
0.0 0.2 0.4 0.6 0.8 1.0
5.0000 4.5996 4.1398 3.5934 2.9047 1.9112
67
0.0 0.2 0.4 0.6 0.8 1.0
67.000 67.387 67.749 68.089 68.410 68.716
10
0.0 0.2 0.4 0.6 0.8 1.0
10.0000 9.2058 8.2707 7.1022 5.4592 1.9112
80
0.0 0.2 0.4 0.6 0.8 1.0
80.000 78.572 76.929 74.964 72.451 68.714
25
0.0 0.2 0.4 0.6 0.8 1.0
25.0000 25.5830 26.0660 26.4830 26.8530 27.1870
100
0.0 0.2 0.4 0.6 0.8 1.0
100.000 96.455 92.605 88.079 81.996 68.712
40
0.0 0.2 0.4 0.6 0.8 1.0
40.ססOO
38.6260 37.0300 35.0700 32.3920 27.1840
o Reprinted from [8] with permission from the American Institute of Aeronautics and Astronautics.
246
10. Method of Parameter Differentiation
10.3 PARAMETER DIFFERENTIATION APPLIED TO DIFFERENTIAL EQUATIONS
In this section, a general description of the method of parameter differentiation as applied to nonlinear differential equations will be presented. Consider the boundary value problem (3.1) dy/dx 2 = f(x,y,dy/dx,A) with the boundary conditions
f3oy(L) + f3(dy(L)/ dx) = f32 (3.2a, b) where A is a physical parameter. The problem to be solved can be stated as follows: Given the solution of Eq. (3.1) for one particular value of A, namely A = Ao, find a family of solutions of Eq. (3.1) corresponding to the range of values of A: AO
A new dependent variable cf>(x) can now be defined by differentiation of the original dependent variable y(x) with respect to A, i.e. cf>
= ayfaA
(3.3)
Assuming that the function f in Eq. (3.1) is continuously differentiable with respect to all variables, Eq. (3.1) can now be differentiated with respect to A to give (3.4)
The boundary conditions, upon differentiating with respect to A, become lXocf>(O)
+ lX1(dcf> (0)/ dx) = 0,
f3ocf>(L) + f31(dcf>(L)/dx) =
°
(3.5a, b)
Equation (3.4) is now linear since f and its derivatives are calculated with values at A = AD, the solution of which can be obtained noniteratively by one of the three methods treated in Chapters 2-4. Once cf>(x) is known, Eq. (3.3) can be integrated to give
y(X)IAo+/lA = y(X)IA o+ cf>IAoIlA
(3.6)
Thus, based on one set of solutions y(X)IA o the solution of y(x) at AD + IlA can be calculated by Eq. (3.6). Since the solutions of Eqs. (3.4) and (3.6) do not require iteration, the solutions are obtained noniteratively. The above process can be continued to get solutions for AD + 2M, AD + 31lA, etc. The method will now be applied to a few engineering problems.
247
10.3 Application to OIHerentlal Equations
10.3.1 Reactor Deslgnt
Consider again the isothermal packed-bed reactor in which a chemical reaction of the form
(3.7)
A+A~B
takes place. Assuming negligible influence of the packing on the reaction except for its contribution to the axial mixing, the nondimensional differential equation for the fraction y of A remaining is of the form (see Sections 2.2.1 and 8.2.2 for the formulation)
-
I
dly
N pe
-
ds 2
dy
+ - -Ryn=o ds
(3.8)
subject to the boundary conditions
s
= 0:
s
= 1:
dy(O)
=0
ds 1 = y(l)
+ _1_ N pe
dy(l)
(3.9a, b)
ds
The problem is to find solutions of Eq. (3.8), subject to the boundary conditions (3.9), corresponding to a given value of N pe and a given order of reaction n for a range of values of R. The parameter is therefore identified as R. Differentiating Eq. (3.8) and its boundary conditions (3.9) with respect to R, we get 2ep+N
d ds2
pe
dep N R n-I,/.,-N n ds - pe ny 'I' peY
(3.10)
and the boundary conditions
dep(O)
ep(l) + -
~=O,
1
N pe
dep( 1) = 0 ds
--
(3.11)
where ep is defined as
ep
= dy / dR
(3.12)
Equations (3.10) and (3.11) correspond to Eqs. (3.4) and (3.5), respectively. Equation (3.10) is now linear and can be separated into two initial value problems by the method of superposition (Chapter 2). Briefly, we set
ep = F + l\G t See Na and Habib [9).
(3.13)
248
10.
Method of Parameter Differentiation
and identify the unknown parameter A as A =
(3.14)
Equation (3.10) can then be separated into the following equations: 2F
d ds
-2
dF - N pe R ny n-1F = N peY, n + N pe -d s
2G
d ds
-2-
dG - N pe R ny n-1G =, 0 + N pe -d s
dF(O)
F(O) = 0,
----;]S =0 (3.15)
G(O) = I,
~
dG(O)
=0
(3.16)
The value of A can be evaluated by using the boundary condition at s = I, Eq. (3.11), which gives
NpeF(I) + dF(I)/ds - A = N peG(I) + dG(I)/ ds
(3.17)
The solution procedure can be illustrated by a numerical example. Consider the solutions of Eq. (3.8) for N pe = 5, n = 0.25 and a range of R from R = 0 to 0.6, with the solution of Eq. (3.8) for R = 0 given as y = 1. It should be noted that, in applying this method, the solution for one value of the parameter is needed as a starting solution. With this in mind and with the function y(s) in Eqs. (3.15) and (3.16) known, Eqs. (3.15) and (3.16) can now be integrated for R =!:iR to give F and G along with their derivatives for 0 ,;;; s ,;;; 1. These values can then be used to evaluate A from Eq. (3.17) and subsequently f from Eq. (3.13). As a final step, Eq. (3.12) can be integrated to yield the solution of Eq. (3.8) for R = !:iR as follows y(S)IR=AR = y(S)IR=O
+
This procedure can be repeated to calculate the solutions of Eq. (3.8) for R = 2 !:iR, 3 !:iR, ... , etc. [9]. A summary of the missing boundary condition y(O) is given in Table 10.7. Physically, it is the (dimensionless) concentration of A at the exit section of the chemical reactor. TABLE 10.7
Missing Boundary Condition y(O) of Eq. (3.8) (IlR = 0.02)
R
y(O)
R
y(O)
0.00 0.10 0.20 0.30
1.0000 0.9017 0.8068 0.7155
0.40 0.50 0.60
0.6280 0.5445 0.4653
10.3 Application to Differential Equations
249
10.3.2 Boundary Layer Flow over Wedges
We will now apply the method of parameter differentiation to the boundary layer equations governing the flow of fluids over surfaces other than a semi-infinite flat plate. Such surfaces include, among others, those commonly found in engineering problems, as shown in Fig. lOA.
-
,
~--
/r~Mm, -
Fig. 10.4
Examples of similar boundary layer flows.
The governing differential equations, derived from the Navier-Stokes equations, are identical to Eqs. (1.1) and (1.2) of chapter 7 except for an additional term, resulting from the fact that the mainstream velocity, i.e., the velocity at the edge of the boundary layer, is now a function of x. We therefore have
ax av ay = 0 «u, au au ) p(u - +v- =pU ax ay e dx ~ +
(3.18)
a +p.2u
ay2
subject to the boundary conditions
y=O: y
= 00:
u
= 0,
v
=0
u= Ue(x) = Uoo(x/Lt
If we now introduce the stream function
and
\fI such that
(3.19)
250
10. Method of Parameter Differentiation
Eq. (3.18) is satisfied identically and Eq. (3.19) becomes
al/; a"-lf; al/; a"-lf; au, a3l/; - - - - - -- = U - +vay axay
ax ay2
e
dx
ay3
(3.20)
subject to the boundary conditions
y = 0:
al/;/ax = 0, al/;/ay = 0
y = 00:
al/;/ay = Uoo(x/ L)m
Introducing the similarity transformation
1) =~Hm + I)Uoo/vL m (y/x(l-m)/2)
(3.21 )
and (3.22)
Eq. (3.20) can be transformed from a partial differential equation to an ordinary differential equation. We therefore have
d:t +fd~ -d1)3 d1)2
+[3 [ 1- ( -df )2] =0 d1)
(3.23)
subject to the boundary conditions 1)
= 0:
f= 0, df/d1) = 0
1)
=
df/ d1) = 1
00:
where [3 is related to m through the equation [3
= 2m/(m + 1). Equation
(3.23) is known as Falkner-Skan equation, the solution of which for
various values of [3 has attracted the attention of both applied mathematicians and aeronautical engineers. The only family of noniterative solutions of Eq. (3.23) available in the literature is that found by Rubbert and Landahl [10] using parameter differentiation. If the solutions of Eq. (3.23) for a family of values of [3 are required, the parameter is naturally taken to be [3 and a new dependent variable g( 1) is defined as (3.24)
where the differentiation is performed by holding 1) constant. Differentiating Eq. (3.23) and its boundary conditions with respect to [3, there results
251
10.3 Application to Differential Equations
and the boundary conditions
g(O) = 0,
dg(O)/ dn = 0,
dg(oo)/dYJ = 0
(3.26)
Integration of Eq. (3.24) can be started from the case where fJ equals zero, i.e., the Blasius equation, the solution of which was given in Section 7.2.1 by an initial value method. With this starting solution of f(YJ) known and substituted into Eq. (3.25), the function g(x) can be solved from the linear differential equation, Eq. (3.25). Even though the boundary conditions, Eq. (3.26), are given at two points, the method given in Chapter 2 can be applied as is summarized below. Let us write (3.27) where C is a constant to be determined. Substitution of g(YJ) from Eq. (3.27) into Eq. (3.25) then gives two differential equations as follows:
2F 3 d F +f d + d~ F _ 2fJ df dF = ( df dYJ3 dYJ2 dYJ2 dYJ dYJ dYJ F(O) = 0,
dF(O)
----chI = 0,
)2_ 1
d 2F(0) dYJ2 = 0
d2G d~ dlf dG d 3G - 3 + f - + - G-2fJ- - =0 dYJ dYJ2 dYJ2 dYJ dl) G(O) = 0,
dG(O)
~ = 0,
d 2G(0) dYJ2 = 1
(3.28) (3.29) (3.30)
(3.31)
The boundary condition at infinity, Eq. (3.26), then gives C
= _ dF(oo) / dYJ
dG(oo) dYJ
(3.32)
For the solution of f( YJ) at fJ = 6.fJ, the function f('1) and its derivatives appearing in Eqs. (3.28) and (3.30) are furnished by the solution of Eq. (3.23) corresponding to fJ = 0 (known from Blasius's solution, Section 7.2.1). Integration of Eqs. (3.28)-(3.31) then gives F( YJ), G('I), and their derivatives. In particular, we obtain dF( 00)/ dYJ and dG( 00)/ dYJ, which can be substituted into Eq. (3.32) to get C. With C known, the function g(YJ) can be found by Eq. (3.27), which is then substituted into Eq. (3.24) to get f('l) at fJ = 6.{3. This is accomplished by writing Eq. (3.24) as
g(YJ)
= (f(YJ)I,8=Il,8 - f(YJ)I,8=o)/ 6.fJ
from which (3.33)
252
10.
Method
0' Parameter Differentiation
This process can be repeated to calculate the solution of f( '1]) for [1 = 2/),.[1, 3/),.[1, ... , etc. Table 10.8 gives typical solutions by this method with the starting solution at [1 = O. Rubbert and Landahl [10] used this method to get the solution of Eq. (3.23) for the range of [1 from - 0.19884 to 0 and from 0 to 1. (See also the solution obtained by Smith [11].) TABLE 10.8 Selected Solutions of Eq. (3.23)
d'f(O)/ dr,2
f1
Present method
[12]
0.00 0.05 0.10 0.15 0.20 0.25
0.4696 0.5321 0.5886 0.6406 0.6886 0.7340
0.4696 0.5870 0.6867
Tan and DiBiano [12] solved Eq. (3.23) with a nonzero value of f('I]) at = 0 (physically, this represents mass transfer across the boundary). The boundary conditions are then
'I]
f(O) = -K,
df(O)j d'l] = 0,
df( oo)j d'l] = 1
(3.34)
The solutions for a range of values of K corresponding to a given value of [1 were sought. For this case, they chose to vary the parameter K while keeping [1 constant. The procedures followed by Tan and DiBiano [12] were identical to those used by Rubbert and Landahl [10]. 10.4 APPLICATION TO SIMULTANEOUS EQUATIONS 10.4.1 Finite Deflections of a Nonlinear Elastic Bar-Eigenvalue Problemt
This problem was treated in Section 8.4 (see Fig. 8.9), where the equations were solved noniteratively by the method of transformation. We will now solve the same equations by using the method of parameter differentiation. tSee Kurajian and Na [13].
10.4 Application to Simultaneous Equations
253
The governing differential equations can be written as
= sinO dO/dz = t/;
(4.2)
dA./dz = 0
(4.3)
(4.1)
duf dz
p2(1- lO)('IT 2/4)A.sinO
dt/;
dz = - p2(I-lO)-[('IT2A.u/4)+lOt/;t
(4.4)
subject to the boundary conditions
u(O) = 0,
t/;(O)
= 0,
0(0) =
lX,
0(1)
=0
(4.5)
where e and lX are physical constants. For a given pair of values of lO and lX, the eigenvalues A. for a range of values of p are sought. Thus the parameter is identified as p. Equations (4.1)-(4.5) are differentiated with respect to p, and we get
dU/ dz = (cos 0 ) C
(4.6)
dC/dz
P
(4.7)
dL/ dz = 0
(4.8)
dP/ dz = fo - fLL - fcC - fv U - fpP
(4.9)
=
subject to the boundary conditions
U(O) = 0,
P(O) = 0,
C(O) = 0,
C(l) = 0
(4.10)
U= du dp ,
C= dO dp ,
dt/; P=dp ,
L = dA. dp
(4.11)
where
and
fo = 2p(1 - lO)( '17 2/ 4)A.sinOG 2/ F 2 fL = p2(l - lO)( '17 2/4) sinO( F + 2A.'IT 2/ 4uG)/ F 2 fe = p2(l - lO)( '17 2/ 4)A. cos 0/ F fv = 2p 2( 1 - lO)('IT 4/16)A. 2sinOG/ F 2 fp = 2p2(1 - lO)('IT 2/4)A.sinOGlO/ F 2
and
F = p2(l - lO) - ('IT 2A.u/4 + lOt/;)2 G = 'IT 2A.u/4 + lOt/;
254
10.
Method of Parameter Differentiation
Equations (4.6)-(4.9) are now linear and can therefore be reduced to two initial value problems by separating the variables as follows:
U = U I + sU z,
C = CI
+ sC z,
P = PI + sP z,
L = L I + sl., (4.12)
where s
= L(O)
(4.13)
The two initial value problems are
= (cosB)Ci' dCII dz = Pi' dLII dz = 0 dPII dz = fo - fLLI - feCI - fUUI - fpP I UI(O) = PI(O) = C(O) = LI(O) = 0
dUll dz
(4.14)
(4.15) (4.16)
and
dUzl dz
= (cosB)C z,
dCzI dz
= Pz,
dLzl dz
dPzI dz = - fLLz - feCz - fuUz - fpP z Uz(O)
= Pz(O) = Cz(O) = 0, Lz(O) = I
=0
(4.17)
(4.18) (4.19) (4.20)
The value of s can be evaluated by using the boundary condition at
z
=I
(4.21) from which (4.22) As an example, consider the solutions of Eqs. (4.1)-(4.4) for IX = 40° and e = 0.25 and a range of p from 0.1349 to 0.45. The solution of these equations at the first value of p, namely 0.1349, is known as p = 0.1349, A. = 0.3822. Since the eigenvalue A. at p = 0.1349 is known, the dependent variables u, B, and 1f; for the range of z from 0 to 1 can be obtained by integrating Eqs. (4.1)-(4.4). Using these functions in Eqs. (4.14)-(4.21), these equations can now be integrated to give Ui, Ci ' Pi' and L i from z = 0 to z = I (i = 1,2) for p = 0.1349 + Llp. The constant s can now be evaluated by Eq. (4.22). With s and the solutions of Eqs. (4.14)-(4.21) both known, U, C, P, and L can be calculated from Eq. (4.12). As a final step, integration of Eq. (4.1I) gives the solutions of Eqs. (4.1)-(4.4) for p = 0.1349 + Llp
u( z) Ip=0.1349Hp = u( Z )lp=0.1349 + U( Z )IEq. (4.12) Llp
10.4
255
Application to Simultaneous Equations
and similar expressions for 0 (z) and t[;(z) at p = 0.1349 + Ap. It is to be noted that the eigenvalue A remains constant between z = 0 and z = 1 and is given by Eq. (4.13). This procedure can be repeated to calculate the solutions of Eqs. (4.1)-(4.4) for p = 0.1349 + lAp, 0.1349 + 3Ap, ... , etc. Selected results are shown in Table 10.9. Also shown are the same solutions calculated by using the method of transformation group (see Section 8.4). TABLE 10.9 Selected Results for the Eigenvalues"
0.50
0.75
a
p
A(fRi
A(PD)"
40°
0.5180 0.2772 0.2260 0.1951 0.1789
0.8051 0.6914 0.6641 0.6471 0.6381
0.8012 0.6900 0.6634 0.6469 0.6380
80°
0.5284 0.3321 0.2606 0.2068 0.1901
0.8377 0.7722 0.7469 0.7274 0.7212
0.8347 0.7710 0.7464 0.7271 0.7211
40°
0.5618 0.3145 0.2594 0.2256 0.2072
0.9467 0.8899 0.8751 0.8656 0.8603
0.9440 0.8892 0.8747 0.8655 0.8603
80°
0.4598 0.3393 0.2808 0.2448 0.2247
1.0570 1.0358 1.0251 1.0184 1.0146
1.0560 1.0353 1.0249 1.0183 1.0146
a Reprinted from (13) with permission from the American Institute of Aeronautics and Astronautics. b TR = Method of Transformation. C PD = Method of Parameter Differentiation.
10.4.2 Natural Convection over a Semi-Infinite Vertical Plate
Fluid motion resulting from the action of a gravitational field is known as natural convection. Consider, for example, the flow over a semi-infinite flat plate, as shown in Fig. 10.5. The fluid velocity at points far away from the plate is essentially zero. Near the plate, however, there will be fluid motion if the temperature of the plate is different than that of the fluid,
10.
256
Method 01 Parameter Differentiation
x u
"'-----
)'
Fig. 10.5 Schematic diagram of the plate.
which happens as a result of a change in density of the fluid near the plate. It is of interest to find the rate of heat transfer as a function of the temperature difference, the dimensions of the plate, and the physical properties of the fluid. The governing differential equations include the equations of continuity, momentum, and energy in differential form, which are as follows:
av =0 ax ay au au a2u u-+v-=p-+g ax ay ay 2 2 0 uao - + vao- = aaax ay a/ ~ +
(4.23)
Tw
-
Too
Too
0
(4.24) (4.25)
subject to the boundary conditions
y = 0: Y = 00:
u u
= 0,
= 0,
v
= 0, 0 = 1
0 =0
where u and v are the velocity components in the x and y directions, respectively; 0 is the dimensionless temperature defined by and g, Tw ' TOO' and a are, respectively, the gravitational acceleration, the temperature of the plate, the temperature of the fluid far away from the plate, and the thermal diffusivity.
10.4 Application to Simultaneous Equations
257
If we introduce the stream function l/; such that u = al/;jay,
and the similarity transformation
C=[g(Tw
-
T oo)j4v 2Too f /4 (4.26)
Eqs. (4.23)-(4.25) and their boundary conditions become 2F 3F d + 3F d _ dYJ 3 dYJ 2
2( dF )2 + 0 = 0 dYJ
20 d + 3N F dO dYJ2 pr dYJ
=0
(4.27) (4.28)
subject to the boundary conditions 1) = 0:
F(O) = 0, dF(O)j d1) = 0, 0(0) = 1
1)=00:
dF(oo)jd1) =0, 0(00)=0
where N pr is the Prandtl number. It is for different values of the parameter N pr that the solution is sought. The first step is therefore differentiation of the differential equations and their boundary conditions with respect to N pr• We obtain 2g 3g 2F d +3d g + 3F d _ 4 dF dg + 8 = 0 (4.29) dn3 d1)2 d1)2 d1) dn d~ ~ ~ d8 -2 +3F- +3Npr g - +3Npr F - =0 d d dYJ d1) YJ YJ
(4.30)
subject to the boundary conditions YJ = 0:
g(O) = 0, dg(O)j d1) = 0, 8(0) = 0
YJ = 00:
dg(oo)jdYJ = 0, 8(00) = 0
where g=aFjaNp r
and
8=aOjaNpr
(4.31)
Equations (4.29) and (4.30) are linear and can be separated into three initial value problems as follows. Let g = g\ + Ag2 + P.g3 8 = 8\ + A82 + p.83
(4.32)
258
10. Method of Parameter Differentiation
We get
S (0) = 1
SiO)
dS (0) =0 dry
_ I-
= 0,
S(O)= 3
dS (0) =0 dry
_3_
The boundary conditions in the three sets of initial value problems were based on the given boundary conditions at 1/ = 0 and are separated in such a way that and
A = dS(O)/ d1/
Finally, the boundary conditions at infinity yield dgkX) A d1/
dg3( (0) dg1( (0) +p. d1/ =d1/
AS2(OO) + P.S3(OO) = - S)(oo)
10.4
Application to Simultaneous Equations
259
from which A= p,=
Sl( 00)(dg3(00)/ dry) - S3( 00)( dg 1( 00)/ d1j) S3( 00)( dg2(00)/ d1j) - S2( 00)(dg3(00)/ d1j)
(4.33)
S2( 00)( dg 1( 00)/ d1j) - Sl( 00)(dg2(00)/ d1j)
(4.34)
S3( 00)(dg2(00)/ d1j) - S2( 00)(dg3(00)/ d1j)
The solution procedure involves exactly the same steps as discussed in Section 1004.1 and is therefore omitted here [14]. Solutions based on this method are shown in Table 10.10. For each set of the solutions shown, the initial slopes d 2F(0)/ d1j2 and dO (0)/ d1j are known for the starting value of N pro The other solutions are then obtained by parameter differentiation. The results agree closely with those of Ostrach [15]. TABLE 10.10 Selected Solutions of Eqs. (4.27) and (4.28)Q
Present method
»;
d b
2F(0)jdrl2
d8(0)jdrl
O.72 0.60 0.50 0.40 0.30 0.20 0.10 0.06 0.04 0.01
0.6760" 0.6946 0.7131 0.7354 0.7633 0.8009 0.8590 0.8961 0.9221 0.9887
-
0.5046 0.4725 0.4420 0.4066 0.3641 0.3101 0.2326 0.1864 0.1556 0.0817
3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00
0.5713 c 0.5312 0.5036 0.4827 0.4660 0.4522 0.4405 0.4304 0.4215
-
0.7165 0.8145 0.8898 0.9517 1.0047 1.0512 1.0930 1.1390 1.1658
zoo-
Ostrach [15] d8(0)jdrl
0.6760
-0.5046
0.9862
-0.0812
0.5713
- 0.7165
0.4192
- 1.1694
QReprinted from [14] with permission from the International Journal of Heat and Mass Transfer, copyright 1974, Pergamon Press, Ltd. b !:J.Npr = -0.005 in this set of solutions. c Given starting solutions from Ostrach [15]. Other solutions were obtained by parameter differentiation. d !:J.Npr = 0.05 in this set of solutions.
260
10.
Method 01 Parameter Differentiation
Physically, the rate of heat transfer is given by
qw= -k(aT/ay)y=o= -kCX- 1/ 4(dO(0)/dq) or
qw/kCX-I/4= -dO(O)/dTJ It may therefore be seen that the values of dO (0)/ dTJ obtained by this method for different values of N pr are the data needed for the calculation of the rate of heat transfer. 10.5 THE GENERAL PARAMETER MAPPING (GPM) OF KUBICEK AND HLAVECEK
In this section, the method of general parameter mapping (GPM) of Kubicek and Hlavacek [2] will be outlined [16-20]. Even though their method is similar to the one developed in the preceding sections of this chapter, it does offer an alternative approach to the solution of such problems. Consider a second-order differential equation
dy / dx? = f(x, y,dy/ dx,a)
(5.1)
subject to the boundary conditions
aoY(O) + f3ody (O)/ dx = Yo
(5.2)
a1y(b) + f31 dy(b)/ dx = Yl
(5.3)
where a is a physical parameter for different values of which the solution is sought. Let the unknown initial condition be denoted by A, i.e.
yeO) = A which can combine with the boundary condition (5.2) to give
yeO) = A,
f3 ody (O)/ dx = Yo - aoA
(5.4)
where A is a function of the parameter a. We will now derive the differential equation governing the function A (a). First, the differential equation, Eq. (5.1), and its boundary condition (5.4) are differentiated with respect to the parameter a, and we get
d 2q dx 2
af
=
ay q +
af dq af ay' dx + aa
(5.5)
(where, y' = dy I dx) subject to the boundary conditions
q(O) = 0,
dq(O)/ dx = 0
(5.6)
10.5 General Parameter Mapping
261
where
q = ay/aa
(5.7)
Next, Eq. (5.1) and its boundary conditions (5.2) and (5.4) are differentiated with respect to A, and we get
d]J af af dp dx2 = ay p + ay' dx
(5.8)
subject to the boundary conditions
x = 0:
p(O) = 1, dp(O)/ dx = - ao/ /30
where p = ay/aA. The boundary condition at x
(5.9)
= b, Eq. (5.3), can be written as (5.10)
since y is a function of both A and a. Along the curve A (a) the following equation holds
(aF/aA)dA + (aF/aa)da = 0 from which
aF/aF da = - a;; aA
dA
(5.11)
Substituting F from Eq. (5.10) into Eq. (5.11), we then get (5.12)
which is the differential equation governing the missing initial slope A as a function of the physical parameter a. The integration of Eq. (5.12) can be performed numerically, since the right-hand side of the equation can be calculated from the integration of the initial value problems defined by Eqs. (5.5), (5.6), (5.8), and (5.9). The computational details of the method will now be illustrated by an example. For applications of the method to other problems, e.g., the solution of simultaneous differential equations and solutions near a branching point, the reader should refer to the works of Kubicek and Hlavacek [2,16-20] and the references therein. The Falkner and Skan Equation
Consider again the boundary layer flow over wedges, treated in Section 10.3.2. The boundary layer equations, upon transformation of variables,
262
10. Method of Parameter Differentiation
are reduced to a nonlinear ordinary differential equation f (d - )2] dTf
-d:J3 +fd'i -2 + (3 [ 1 dTf
=0
(5.13)
f(O) = 0, df(O)/ dTf = 0
(5.14)
dTf
subject to the boundary conditions Tf = 0: Tf = 00:
df( 00)/ dTf = 1
(5.15)
We will now solve the same equation by using the method of GPM [16]. Let us denote the missing slope by (5.16)
For different values of {3, the values of A will also be different, which means that A is a function of {3. The method of GPM can now be used to find a differential equation for A ((3). First Eqs. (5.13), (5.14), and (5.16) are differentiated with respect to A; we get
with the boundary conditions 11
g(O) = 0, dg(O)/ dTf = 0, d 2g(0)/ dTf2 = I
= 0:
(5.18)
where
g=
af/ aA
(5.19)
Next, Eqs. (5.13), (5.14), and (5.16) are differentiated with respect to {3, and we get 3S
d dTf 3
2S d11 2
+ d'i S + f d dTf
2
+ [I _ ( df
dn
)2] _2{3 dTfdf
dS = 0 dl1
(5.20)
with respect to the boundary conditions Tf = 0:
S(O) = 0, dS(O)/ dl1 = 0, d 2S(0)/ dTf2 = 0
(5.21)
where
s=af/ a{3 \
(5.22)
Finally, the boundary condition at the second point, (5.15), is written as F(A, (3)
= df(oo)/dTf -
I
(5.23)
10.5 General Parameter Mapping
263
since dj/ dry is a function of A and f3. Along the curve A (f3), the following equation is always true: (aF/aA)dA + (aF/af3)df3 =
°
from which dA df3
= _ aF/aF af3 aA
(5.24)
Substituting F from Eq. (5.23) into Eq. (5.24), we then get dA = _ dS(oo)/dg(oo) df3 dry dry
(5.25)
which is the differential equation for A (f3). Starting from the given solution of Eq. (5.13) for f3 = 0, the solution of Eq. (3.13) for f3 = 11f3, 211f3, 311f3, ... , can be obtained as follows. First Eqs. (5.17), (5.18), (5.20), and (5.21) are integrated as two initial value problems. The value of Tj which can be considered to have reached Tj = 00 is determined by checking the ratio dS(Tj) /dg(Tj) dTj dTj When this ratio reaches a constant value, the corresponding value of Tj is considered to have reached Tj = 00. We then get dS( 00 ) / . dg( (0) dTj dn
(5.26)
During the integration process, the functions j, dj/ dn, and d~/ dTj2 in Eqs. (5.13) are determined from the previous values of f3. Second, the ratio just obtained, (5.26), is substituted into Eq. (5.25) to give dA/ df3, from which A(nl1f3)
= A[(n - 1)11f3] - dSd
oo
Tj
)!
dgd,OO) Tj
I
11f3 (5.27)
(n - I) 1l,B
Once A (n 11f3) is known, Eq. (5.13) becomes an initial value problem, since we can now integrate Eq. (S.13) subject to the boundary conditions (5.14) and (5.16). Typical solutions based on the above scheme are given in Table 10.11. Solutions for other values of f3 are tabulated in [21]. It is seen that the method of GPM is quite accurate as compared with the method of parameter differentiation treated in Section 10.3.2 and offers another approach to the solution of the equation.
264
10.
Method of Parameter Differentiation
TABLE 10.11 Selected Solutions of (5.13)
(3
GPM
Iterative [221
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
0.4696 0.5322 0.5887 0.6406 0.6889 0.7342 0.7770 0.8177 0.8566
0.4696 0.5870 0.6867
0.8544
10.6 METHOD OF CONTINUATION OF ROBERTS AND SHIPMANt
The concept of continuation can be traced, according to Lichtenstein [1], as far back as 1869 (Schwartz). Early applications of this method occurred in conformal mapping. Subsequent work on the subject for functional equations can be found in the writings of Bernstein [23], Leray and Schauder [24], Leray [25], Friedricks [26], Ficken [27], and the references therein. Roberts and Shipman [28-30] have contributed greatly in applying this method (in conjunction with other techniques) to various engineering problems. Basically, the concept is identical to that behind the methods discussed earlier, that is, the solution of the operator equation F(x,A) = 0 is sought for a certain value of Awhen the solution for a particular value of the parameter, say AD, is known. The basic difference between this method and the ones developed earlier lies in the fact that in the former integration is made over a smaller range of the independent variable than the given range. Once the solution for the smaller range is obtained, the range of integration can be extended progressively until finally the given range is reached. Due to the availability of a complete treatment of the subject with detailed numerical examples by Roberts and Shipman [3,28-30], we will illustrate the general application of the method through an example. Troesch's Problem
The boundary value problem posed by Weibel [31] and Troesch [32] was originally derived following an analysis of the confinement of a plasma column by radiation pressure. Due to the severe limitations imposed, there is little physical interest in the solution. The considerable interest in the t See, in-particular, [31.
265
10.6 Method of Continuation
literature in recent years [33-38] is mainly generated by numerical analysts, who consider the problem as an excellent example for testing different numerical schemes for solving unstable nonlinear boundary value problems. Troesch's boundary value problem, in terms of dimensionless variables, can be written as dy / dx?
= n sinhny
(6.1)
subject to the boundary conditions y(O) = 0,
y(l)
=I
(6.2)
To introduce the general procedure, the quasi-linearization process is first applied, following the reasoning outlined in Section 5.4. The linearized equation of Eq. (6.1) is therefore dy(P+
I) /
dx?
+ lpy(p+ 1) = ~p)
(6.3)
/P+1)(1)= 1.0
(6.4)
subject to the boundary conditions /P+ 1)(0)
= 0,
where the superscript v indicates the number of iterations and p lp) = - n 2coshn/ ) ~p)
= n sinh nip)
- ny(p) cosh ny(p)
To start the solution, we first put v = 0 in Eqs. (6.3) and (6.4). The first approximation of the dependent variable /0) must be assumed. One way to do this is to write Eq. (6.1) as dy(O) / dx"
and the boundary condition at x
= n sinh n/O)
(6.5)
=0
/0) =
0
(6.6)
The rate of convergence of the assumed first approximation /0) to the final solution depends on how close the assumed initial slope dy(O) / dx is to its final value. Consider the case where n = 0.1, the final value of the initial slope is known to be 0.9984. However, for the purpose of illustrating the method, we choose a considerably different value, namely d/°l(O)/ dx
=5
(6.7)
The method of continuation then offers a systematic way of approaching the final solution. The steps to be taken are the following.
266
10.
Method of Parameter Differentiation
1. Integrate Eq. (6.5), subject to the boundary conditions (6.6) and (6.7) as an initial value problem. From the boundary conditions, Eq. (6.2), the final solution y'<'{x) should change fromy = at x = to Y = I at x = 1. However, since the initial slope in the first approximation is approximately five times larger than the final solution,y<°)(x) is expected to reach I long before x = I is reached. Table 10.12 shows that this happens at x = 0.20. We will arbitrarily choose [0,0.2] as the range of integration.
°
°
TABLE 10.12 Solutions of Eqs. (6.1) and (6.2) by the Method of Continuation (n = 0.1)
x
p=o
p=1
p=3
p=5
p=7
p=9
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
0.0000 0.5000 1.0001
0.0000 0.5000 1.0000
0.0000 0.2499 0.4999 0.7499 1.0000
0.0000 0.1666 0.3332 0.4998 0.6644 0.8332 1.0000
0.0000 0.1249 0.2498 0.3447 0.4996 0.6246 0.7497 0.8748 1.0000
0.0000 0.0998 0.1997 0.2996 0.3994 0.4994 0.5994 0.6994 0.7995 0.8997 1.000
2. Usingy<°)(x) obtained in step I as the first approximation, solve Eq. (6.3) subject to the revised boundary conditions and by any method discussed earlier. It is to be noted that the same boundary conditions are specified at a shorter range, [0,0.2]. We will now get the first iteration yf1) and lj>f) in Eq. (6.3) for the interval [0,0.2]. For the extended range [0.2,0.3], we assume that y< I) (x) is constant and equal to y< I) (0.2). Equation (6.3) is then solved (by putting p = 1) with the boundary conditions and over the extended interval [0,0.3]. 4. Repeat step 3 for another extended range, say, [0,0.4]. 5. Continue step 4 until the range reaches the specified range [0, 1.0], at which point the boundary value problem defined by Eqs, (6.1) and (6.2) will be solved.
267
Problems
Numerical solutions based on the above procedure are shown in Table 10.12. For problems with larger values of n where the solution becomes unstable, the reader should refer to [33-38]. 10.7 CONCLUDING REMARKS
The various methods treated in this chapter, even though different in detail, are based on the same general idea. Basically, we seek the solution of a boundary value problem which depends on a parameter. The solution is sought for a certain physical parameter based on the solution of the problem for another value of this parameter. The general approach is to start from the given solution and find the solution for a nearby value of the parameter, using the given solution as a first approximation. This process can be continued with the parameter increased each time by a small increment until the required value of the parameter is reached. Due to the fact that the solution sought during the process corresponds to a continuous change in the value of the parameter, differentiation of a given boundary value problem with respect to the parameter can be performed. For nonlinear boundary value problems, this amounts to a linearization process in which a nonlinear boundary value problem is reduced to a linear boundary value problem and the iteration can thus be eliminated. The solution for each increment of parameter can then be solved in one sweep. For boundary value problems which are sensitive to the initial condition, this method "guides" the solution from one parameter to another parameter. The mathematical justification of the continuation method is treated quite extensively by Roberts and Shipman [3,29]. For a fairly complete summary of the mathematical literature on this method, the reader is referred to Wacker and Engl [39]. PROBLEMS
1. In an analysis of a stirred reaction cooling coil [40], the temperature of the fluid is found to be governed by the nonlinear equation
T- 2exp(2100/T) - 1.11
X
1011 = 0
Solve for the temperature by the parameter differentiation method. T = 551.774° R. 2. In determining the azeotropic point of a binary solution, the following nonlinear algebraic equation was obtained [41]:
Answer:
AB[ B(l -
X)2 -
Ax 2J/[x(A - B) + By+ 0.14845 = 0
268
10. Method of Parameter Differentiation
where A = 0.38969 and B = 0.55954. Solve the problem by the method of parameter differentiation. Answer: x = 0.691471. 3. The behavior of fluids which do not follow the Newtonian model is of interest to chemical engineers. In the analysis of the natural convection of such fluids between parallel plates, Bruce and Na [42] found that the velocity of the fluid is governed by the following boundary value problem: d 2v -dS 2
=
s- 1 -~--;:::=======-=
v(O) = 0, v(l) = 0
Find the velocity distribution for a = 0.01 and f3 from 0.1 to 1.0. The missing initial slope for a = O.oI and f3 = 0.1 is given as dv(O)/ds = 0.0299. Answer: The missing initial slope for a = 0.01 and f3 = 1.0 is 0.1642. 4. In an analysis of the heat and mass transfer and exothermic chemical reaction occurring simultaneously in a catalyst, Kubicek and Hlavacek [2] found that the concentration of the reactant is governed by the following boundary value problem: dy dx2
a dy dx
+~
= 8yexp
[
yf3(1 - y)
1+
f3 (1 -
y)
]
,
d~~O) = 0,
y(l)
=1
Find the missing boundary condition y(O) for 8 in the range of 0.0-0.14. Assume y, f3, and a equal to 20, 0.4, and 0, respectively. The starting solution is chosen as that with 8 = 0.00, with the missing initial condition given by y(O) = 1. Answer: y(O) = 0.8 for 8 = 0.13734; y(O) = 0.9 for 8 = 0.11429. 5. The analysis of the flow in a two-dimensional channel with porous walls through which fluid is uniformly injected or extracted leads to the following boundary value problem for the solution of the velocity distribution [43]:
f(l) = 1, df(l)/ dry = 0
f(O) = 0, d~(O)/ dry 2 = 0;
Solve the equation for R from 0.2 to 2.0 by the parameter differentiation method. The starting solution is chosen as R = 0.133121, where the two missing slopes are and
269
Problems
Answer: For R = 1.31066 and 2.15638, d~(l)jd'IJ2 = -3.13103 and - 3.2392, respectively, and
d~(1) =3-gR+~R2 d'IJ3
35
35,770
6. In an analysis of the stagnation point shock layer, it was found that the total enthalpy is governed by the following boundary value problem [441: 2H _1_ d +x dH -K(Hf= 0 H(O) = 0, H(I) = 1 N re dx' dx ' Solve H for K from 0.01 to 0.1 with N re = 10 and n = 5. The starting solution is chosen that K = 0.01, N re = 10, and n = 5. The missing slope for the starting solution is dH(O)j dx = 2.5145. Answer: For K = 0.1, N re = 10, and n = 5, the missing initial slope is dH (0) j dx = 2.41065. 7. The boundary layer equations for steady, axisymmetric flow of a compressible and electrically conducting fluid in the stagnation region of an insulating porous blunt body of revolution has been solved by Nath [45]. The boundary value problem is
d~ +fd~ +f3{g[I+M(I_gdf)J_(df)2)=0, ~2 ~ ~
~3
d 2g
-2
d'IJ
f(O) = K,
df(O) j dry
=
0,
g(O) = 0,
dg +Nprf- = 0 d'IJ
df(oo)jd'IJ = I,
g(oo) = 1
where f and g are dimensionless stream and enthalpy functions, respectively, M is the magnetic parameter, K is the mass transfer, N pr is the Prandtl number, and f3 is the pressure gradient parameter, which equals t for this problem. Find the solutions of the equations for K = 0.5 and N pr = 0.72 and for a range of values of M from 0 to 2. The missing initial slopes for M = 0 are d~(O)j d'IJ2 = 0.9278 and dg(O)jd'IJ = 0.7086. Answer: M = 2, d~(O)j d'IJ2 = 1.1881 and dg(0) j d'IJ = 0.7389. 8. The boundary layer equations for compressible fluids, after the application of a similarity transformation, can be written as
d~- +f3 [ s- ( -df )2] =0, -d~ + f d'IJ3 d'IJ2 d'IJ subject to the boundary conditions
f(O) = 0,
df(O) j d'IJ = 0,
df(oo)jd'IJ = 1,
S(O) = Sw'
S(oo) = I
10.
270
Method of Parameter Differentiation
Solve the problem by the method of parameter differentiation [46] for = 0.2 and f3 = 0-2.0. For f3 = 0, the missing slopes are d'i(O)/ dTJ2 = 0.4696 and dS(O)/ dTJ = 0.3757. Answer: Choose f3 as the parameter. For Sw = 0.2 and f3 = 2.0, the missing initial slopes are d'i(O)/ dTJ2 = 0.9489 and dS(O)/ dTJ = 0.4339. Sw
REFERENCES I. 2. 3. 4. 5. 6. 7. 8. 9. 10. II. 12. 13. 14. 15. 16. 17. 18. 19. 20.
Lichtenstein, L., Kontinuitatsmethods im Gebiete der konformen Abbildung, in "Encyklopadie der Mathematischen Wissenschaften, IIC3 (Potentialtheorie, konforme Abbildung)," section 46, pp. 346-352, Leipzig, (1970). Kubicek, M., and Hlavacek, V., Solution of nonlinear boundary value problems-Va: A novel method: General parameter mapping (GPM), Chern. Eng. Sci. 27, 743-750 (1972). Roberts, S. M., and Shipman, J. S., "Two-Point Boundary Value Problems: Shooting Methods," Chapter 7, Elsevier, New York, 1972. Kane, T. R., Real solutions of sets of nonlinear equations, AIAA J. 4, 1880-1881 (1966). Yakolev, M. N., Solution of systems of nonlinear equations by a method of differentiation with respect to a parameter, USSR Cornput. Math. 4, 198-203 (1964). Streeter, V. L., "Fluid Mechanics Handbook," McGraw-Hill, New York (1961). Anderson, J. M., and King, W. W., Vibration of a cantilever subject to a tensile follower force, AIAA J. 7, 741-742 (1969). Chiou, J. P., and Na, T. Y., Solution of lateral vibration of a cantilever subjected to a tensile force by parameter differentiation, AIAA J. 13, 1233-1234 (1975). Na, T. Y., and Habib, I. S., Noniterative solution of a boundary value problem in reactor design by parameter differentiation, Chern. Eng. Sci. 29, 1669-1670 (1974). Rubbert, P. E., and Landahl, M. T., Solution of nonlinear flow problems through parameter differentiation, Phys. Fluids 10,831-835 (1967). Smith, A. M. 0., Inst. Aeronaut. Sci. fund paper FF-IO (1954). \ Tan, C. W., and DiBiano, A study of Falkner-Skan problem with mass transfer, AIAA J. 10, 923-925 (1972). Kurajian, G. M., and Na, T. Y., Solution of an eigenvalue problem by parameter differentiation, AIAA J. 13, 220-222 (1975). Na, T. Y., and Habib, I. S., Solution of the natural convection problem by parameter differentiation, Int. J. Heat Mass Transfer 17,457-459 (1974). Ostrach, S., An analysis of laminar free-convection flow and heat transfer about a flat plate parallel to the direction of the generating body force, NACA Rep. No. IIII (1953). Kubicek, M., and Hlavacek, V., Solution of nonlinear boundary value problems-III. A noval method: Differentiation with respect to an actual parameter, Chern. Eng. Sci. 26, 705-709 (1971). Kubicek, M., and Hlavacek, V., Solution of nonlinear boundary value problems-VIII. Evaluation of branching points based on shooting method and GPM technique, Chern. Eng. Sci. 29, 1695-1699 (1974). Kubicek, M., and Hlavacek, V., Solution of nonlinear boundary value problems-IX. Evaluation of branching points based on the differentiation with respect to boundary conditions, Chern. Eng. Sci. 30, 1439-1440 (1975). Kubicek, M., Evaluation of branching points for nonlinear boundary value problems based on the GPM technique, Appl. Math. Cornput. 1, 341-352 (1975). Kubicek, M., and Hlavacek, V., Direct evaluation of branching points for equations arising in the theory of explosions of solid explosives, J. Cornput. Phys. 17,79-86 (1975).
References
271
21. Na, T. Y., General parameter mapping of the Fa1kner-Skan equations, paper No. II, presented at the 11th Ann. Meet. Soc. Eng. Sci., Duke Univ., North Carolina, 11-13 November, 1974. 22. Cebeci, T., and Keller, H. B., Shooting and parallel shooting methods for solving the Falkner-Skan boundary layer equation, J. Comput. Phys. 7,289-300 (1971). 23. Bernstein, S., Sur la generalisation du probleme de Dirichlet, Math. Ann. 82, 253-271 (1906); 89, 82-136 (1910). 24. Leray, J., and Schauder, J., Topologie et equations fonctionelles, Ann. Sci. Ecole Norm. Sup. 51, 45-78 (1934). 25. Leray, J., Les problemes de representation comforme d'Helmho1z; theorie des sillages at des prones, Commemt. Math. Helv. 8, 149-180,250-263 (1935/1936). 26. Friedrichs, K. 0., "Functional Analysis," Chapters 9 and 10, Inst. Math. Mech., New York Univ. Press, 1950. 27. Ficken, F. A., The continuation method for functional equations, Commun. Pure Appl. Math. 4,435-456 (1951). 28. Roberts, S. M., and Shipman, J. S., Continuation in shooting methods for two-point boundary value problems, J. Math. Anal. Appl. 18, 45-58 (1967). 29. Roberts, S. M., and Shipman, J. S., Justification for the continuation method in two-point boundary value problems, J. Math. Anal. Appl. 21, 23-30 (1968). 30. Roberts, S. M., Shipman, J. S., and Roth, C. V., Continuation in quasiiinearization, J. Optimization Theory Appl. 2, 164-178 (1968). 31. Weibel, E. S., Confinement of a plasma column by radiation pressure, in "The Plasma in a Magnetic Field" (R. K. M. Landshoff, ed.), Stanford Univ. Press, Stanford, California, 1958.\ 32. Troesch, B. A., Intrinsic difficulties in the numerical solution of a boundary value problem, Internal Rep. NN-142, TRW Inc., Redondo Beach, California, 1960. 33. Robert, S. M., and Shipman, J. S., Solution of Troesch's two-point boundary value problem by a combination of techniques, J. Comput. Phys. 10232-241 (1972). 34. Jones, D. J., Solutions of Troesch's and other two-point boundary value problems by shooting techniques, J. Comput. Phys. 12, 429-~34 (1973). \ \35. Kubicek, M., and Hlavacek, V., Solution of Troesch's two-point boundary value problem by shooting technique, J. Comput. Phys. 17,95-101 (1975). 36. Chiou, J. P., and Na, T. Y., On the solution of Troesch's nonlinear two-point boundary value problem using an initial value, method, J. Comput. Phys. 19,311-316 (1975). 37. Troesch, B. A., A simple approach to a sensitive two-point boundary value problem, J. Comput. Phys. 21, 279-290 (1976). 38. Roberts, S. M., and Shipman, J. S., On the closed form solution of Troesch's problem, J. Comput. Phys., 21, 291-304 (1976). 39. Wacker, H. J., "Continuation Methods," Academic Press, New York, 1978. 40. Bilous, 0., and Amundsen, N. R., Am. Inst. Chem. Eng. J. 1,513 (1955). 41. Shacham, M., and Kehat, E., Chem. Eng. Sci. 27,2099-2101 (1972). 42. Bruce, R. W., and Na, T. Y., ASME paper 67-WA/HT-25. 43. Terril, R. M., Aeronaut. Q. 15,299-310 (1964). 44. Nerem, R. M., AIAA J. 4,539-541 (1966). 45. Nath, G., AIAA J. 11, 1429-1432 (1974). 46. Na, T. Y., AIAA J. 11, 1790-1791 (1973).
CHAPTER
11
METHOD OF INVARIANT IMBEDDING
11.1 INTRODUCTION
The concept of invariance was first applied to the transformation of a boundary value problem to an initial value problem by Ambarzumian [1] in the study of atmospheric scattering problems. Chandrasekhar [2] extended the concept to problems of radiative transfer and gave it the name "principle of invariance." This concept has been extensively studied and generalized in recent years by Bellman, Kalaba, and co-workers [3-8] and the methodology given its present name of "the principle of invariant imbedding." This method has been found extremely useful in various fields of physics and engineering. Chief among them are applications in neutron transport theory [5,9], radiative transfer [2,5], random walk and scattering [6], wave propagation [6,7], rarefied gas dynamics [3], Hamilton's equation of motion [8], and the flow in chemical reactors [10]. A fairly complete bibliography prior to 1962 can be found in the books by Wing [9] and Bellman et al. [5]. For more recent works, the books by Lee [10], Meyer [11], and Scott [12] can be consulted. In its basic concept, the invariant imbedding approach differs from the classical one in that the study of a particular solution of a differential equation is carried out by studying a family of solutions. While this may appear to complicate rather than to simplify the problem, its justification lies in the fact that a bridge spanning the particular problem and other members of the family may be constructed and from it the characteristics of the particular member of the family can be obtained by studying the relation between neighboring processes. This subject will only be touched upon in this chapter because of its thorough coverage in the books just mentioned and in the large number of works reported in the literature. 272
11.2 Concept
0' Invariant Imbedding
273
11.2 CONCEPT OF INVARIANT IMBEDDING
The theories of invariant imbedding enable the transformation of boundary value problems into initial value problems by introducing new state variables and imbedding a specific problem in a family of similar problems. A brief introduction of this concept will be given in this section. Consider the second-order ordinary differential equation d 2x/dt 2 = g(x,dx/dt,t)
(2.1)
subject to the boundary conditions x(O)
= 1,
dx(l)/ dt
=0
The problem is to find the missing initial slope, dx(O)/ dt. First Ell. (2.1) is written in the form of a system of two first-order differential equations dx f dt
= y,
dy/dt
= g(x,y,t)
(2.2)
The boundary conditions are therefore x(O)
= 1,
y(l)
=0
(2.3)
The missing initial slope to be found is now y(O). The boundary value problem defined by Eqs. (2.2) and (2.3) is a member of the following family of boundary value problems dx f dt
= y,
dy/dt
= g(x,y,t)
(2.4)
subject to the boundary conditions
x(a) = c,
y(l)
=0
(2.5)
where a and c are real constants. By choosing different pair of values of a and c in the first boundary condition of (2.5), a family of boundary value problems can be generated. The boundary value problem defined by Eqs. (2.2) and (2.3) is seen to be one member of the family identified by a = 0 and c = 1. For the family of boundary value problems, the following are observed 1. All members of the family satisfy the same differential equation, Eq. (2.4). 2. All members of the family satisfy the same boundary condition at t = 1, namely
y(l)
=0
3. The only difference between members lies in the values of a and c.
274
11. Method of Invariant Imbedding
Based on observation 3, the missing initial slope y(a) is a function of both a and c, i.e, ,
y(a) = r(c,a)
(2.6)
By changing the values of c and a, we can make Eq, (2.6) represent the missing initial slopes for all the members of the family of solutions. In particular, we consider the relation between the initial slopes r(c,a) of two neighboring members of the family of solutions at the same value of c, say, cj' Consider now two members of the family, member I:
dx f dt = y,
dyf dt = g(x,y,t)
x(a;) = c)'
y(l)
=0
(2.7) (2.8a, b)
member 2:
dx/dt=y, x(a;
+~)
dy/dt=g(x,y,t)
= c)'
y(l)
=0
(2.9)
(2. lOa, b)
The missing slopes of these members, based on Eq, (2.6), are therefore member I: (2.11a, b)
member 2: y (a; + ~)
= r ( CJ' a; + ~)
From Eqs, (2.8a) and (2.10a), (2.11) becomes member I:
y(a;)
=
r[x(a;),a;]
member 2:
(2.12a, b)
y(a; +~) = r[x(a; + ~),a; +~] To find the relation between the two missing initial slopes, (2.12), both sides of Eq. (2.12b) will now be expanded in Taylor series. Expanding the left-hand side of Eq, (2.12b), we get
y(a; +~) = y(a;) +
dy(a;)
-cit ~ + O[ ~2]
Replacingy(a;) by r(cl,a;) and dy(a;)/ dt by
dy(a;)
-cit = g[ cl,r(cJ'a;),a;],
(2.13)
11.3 Isothermal Packed-Bed Chemical Reactor
275
based on (2.4), Eq. (2.13) becomes
y(ai +~) = r(cj,ai) +
g[ cj,r(Cj,ai),ai]~ +
O[~2]
(2.14)
To expand the right-hand side of Eq. (2.12b), we first expand
dx(a;) x(ai +~) = x(ai) + ----;]I ~ + O(~2) which can, in turn, be written as
x(ai +~) = cj + y(ai)~ + O(~2) = cj + r(ai'c)~ + O(~2) based on Eqs. (2.8a), (2.4), and (2.11a). The right-hand side of Eq. (2.12b) can therefore be written as
r[x(ai + ~),ai +~]
=
r[ cj + r(Cj,ai)~,ai + ~]
(2.15)
Substituting these equations, (2.14) and (2.15), into Eq. (2.12b), we finally get
r(cj,a;) +
g[ cj,r(cj,a;),a;]~ = r[ cj + r(cJ'ai)~,ai + ~]
(2.16)
Equation (2.16) is of fundamental importance in the noniterative solution of boundary value problems. Details of its application will be given in the next two sections. 11.3
ISOTHERMAL PACKED-BED CHEMICAL REACTOR
Consider the problem of the isothermal packed-bed chemical reactor treated in Section 8.2.2, but now with the concentration at the inlet section specified. The differential equation and its boundary condition can therefore be written as d 2x dx n_ -Npe -d -NpeRx - 0 dz z
-2
(3.1)
and
x(O) = 1,
dx(l)/dz
=0
(3.2a, b)
To apply the method of invariant imbedding to the problem, Eqs. (3.1) and (3.2) are first written as a system of first-order differential equations dx f dz
=
(3.3)
y,
with the boundary conditions
x(O) = 1,
y(l) = 0
(3.4)
11. Method of Invariant Imbedding
276
As in the preceding section, let us define a family of boundary value problems by (3.5) dx f dz = y, dyf dz = Npe(Rx n + y) subject to the boundary conditions
x(a) = c,
y(l)
=0
For this family of boundary value problems, Eqs. (2.16) can be written as
r(cj,ai) + {N pe[ r( cj,a;) + Rc}]}ll = r {[ cj + r( cj,ai)ll ],ai + ll} Solving for r(c.,ai ) , we get
r( cj' ai) = 1 + ~pe ll {r[ cj + r( cj' ai)ll,ai + ll] - NpeRc}ll} For sufficiently small ll, the following approximation can be made
r[ cj + r( CjA)Il, ai + ll] = r[ cj + r( ci' a, + ll)ll, a, + ll]
(3.6)
We then get
where c~=Cj+r(ci'ai+ll)1l
(3.8)
To discuss the significance of Eq. (3.7), let us refer to Fig. Il.l. Consider the point q; the values of C and a representing this point are cj and a.. Similarly, the values of C and a at the point pare cj and ai + ll. By the same reasoning, the values of C and a represented by the point ~ are cd, and ai + s. Let us consider the two points q and ~ in Fig. 11.1. The two members of the family of boundary value problems, Eq. (3.5), are, for point q,
dx/dz=y, and, for point
dx] dz = y,
dy/dz=Npe(Rxn+y),
x(ai)=cj, y(I)=O (3.9)
dy / dz = Npe(Rx n + y),
x(ai + ll) = Cd'J y(I)
~,
=0 (3.10)
Since the only differences between these two members of the family of boundary value problems are the starting boundary conditions, each point in Fig. 11.1 represents one member of the family.
277
11.3 Isothermal Packed-Bed Chemical Reactor C
I
I
:C1 __ I"
_ _ _ _ _ _ _ e1.
I I
/1
I
"1
d, I
_ ( Y - r.
ql"" ~" -------?!"---p / I
"1 I
/
I I
I
I
I
I
I
L.....
I
----l.
Fig. 11.1
'r OJ +A) ...
....l-
C-Q
0
plane.
Now, let us consider Eq. (3.9) and (3.10). The missing slopes for these two members of the family are, respectively,
r( cl ' aj )
for point q
r(c<1,a + i:l) j
for point ~
based on Eq, (2.6). With this in mind, Eq. (3.7) is seen to be the relation connecting these two missing slopes. Therefore, if the missing slope of the boundary value problem at point ~, r( Cd' aj + i:l), is known, the missing slope of the boundary value problem at p6int q, r(cl,a j ) , can be calculated by using Eq, (3.7). We now divide the a-c plane (see Fig. 11.2) into a finite number of grid
~
Flg.11.2
d3
»> .>
d2
~
do
d]
Grid points.
a
278
11. Method of Invariant Imbedding
points by dividing a and
C
into ai and cj with i, j = 1,2, ... , and
ai + 1 - a i =~,
cj + I
-
cj
=
~c
Based on the above, if the initial slopes of those members of the family with a equal to a, + ~ and C equal to the series of values 0'C I'C2 '
•••
are known, the initial slopes of the corresponding members of the family with a equal to a, and C equal to the same series of values, 0'C 1'C2 '
•••
may be computed by the following steps 1. Since the initial slopes r(cO,a i + ~),r(cl,ai + ~),r(c2,ai + ~), ... of the members of the family are known for all c/s, we can calculate Cd 1'Cd 2' ••• by Eq. (3.10) for j = 0, 1,2, ... , from which the points do, d., d 2 , . . • in Fig. 11.2 can be located. 2. Compute the missing initial slopes at points ~ by linear interpolation between r(cpai +~) and r(cj+1,ai + ~). 3. Compute the missing initial slopes of the members of the family of boundary value problems, r(cj,ai), at a = ~ by using Eq. (3.7) for all values of cj ' 4. Repeat steps 1-3 to get the missing initial slopes at ai - ~ for all 5. Repeat steps 1-4 until a = is reached. The missing initial slopes for those members of the family of boundary value problems with a = and for all c/s can be calculated.
«»
°
°
As a numerical example, consider the solution for the special case in which N pe
= 6,
R =2,
So = 0.1,
Co
= 0,
~C
= 0.1
The missing initial slopes r(cj,O) can be calculated by following steps 1-5 for all c/s. For this particular example, C = 1. Since the values of Co and ~c are taken to be 0.0 and 0.1, r(c lO , 0) is the required initial slope. We get
r(clO, O) = -1.42231 By using more accurate iteration methods, the initial slope is found to be - 1.4100. The error by using this noniterative method is seen to be less than I %. More accurate answers can be obtained by taking smaller ~a and
Sc.
Initial slopes corresponding to other values of N pe and R can be obtained in a similar manner. Such data are plotted in Fig. 11.3 [to].
279
11.4 Radiation Fins
0.6
1.0
Flg.11.3
0.2
o
a
Missing initial slopes.
11.4 RADIATION FINS
It is generally recognized that the rate of heat transfer from a hot body to a cooler liquid may be increased by extending the body surface through the addition of fins. In analyzing the heat transfer characteristics of a constant-thickness, circular fin due to radiation, Chambers and Somers [13] found that the temperature of the fin can be solved by the following boundary value problem:
d20 +
err 2
(p - 1) dO _ y04 = 0 (p - l)r + 1 err
0(0) = 1,
dO (1) --=0
err
(4.1)
(4.2)
where
0= TIT'
"
r = (r -
rj)/(ro - r;),
y (a dimensionless parameter)
= (r o - ri€GTj k{)
The variables T, T;, r., ro, e, G, k, and () are the temperature of the fin, the temperature of the base, the inner radius, the outer radius, the emmisivity, Planck's constant, the heat conductivity, and the thickness of the fin, respectively. The formulation of this problem is similar to that given in Section 3.3.3 except that cylindrical geometry is needed in this problem.
280
11.
Method 01 Invariant Imbedding
From the mathematical point of view, the problem to be solved is the determination of the missing initial slopes dO(O)/ di for different pairs of values of p and y. Once these are known, the rate of heat transfer can be calculated from the definition dT q =-kW dr
I r
r
r,
ta,
=-
(r o - r;)
dO(0) df
--
(4.3)
To solve Eqs. (4.1) and (4.2) by the method of invariant imbedding, the corresponding form of Eq. (2.16) for this problem can be written as
r(cj,a;) + (yc/ - {(p - 1)/[(p - l)a; + 1]}r(ci'a;»)L'1 (4.4)
=r{cj+r(ci'a;)L'1,a;+L'1}
Solving for r(cj,a;) and using again the approximation (3.6), Eq. (4.4) gives
r(cj,a;) =
r[ cj + r( cjA + L'1)L'1, a; + L'1 ] - yc/L'1 1 _ (p _ 1)L'1/[(p _ l)a; + 1]
(4.5)
Following the same steps as those outlined in Section 11.3, the missing initial slopes of Eqs. (4.1) and (4.2) are found for a range of values of p and y, as tabulated in Table 11.1. The results agree closely with those in [13]. TABLE 11.1 Missing Initial Slopes
11.5
p
y
J.5
0.4 0.8 J.2 1.6
dlJ(O)jdf -
0.3487 0.5587 0.7158 0.8445
p
y
3.0
0.4 0.8 1.2 1.6
dlJ(O)jdf -
0.5085 0.7782 0.9694 J.l212
SOLUTION OF FALKNER-SKAN EQUATION
The Falkner-Skan equation in the boundary layer theory [14,15] is one of the most important equations in fluid mechanics. It is a third-order nonlinear two-point ordinary differential equation with one of the boundary conditions specified at infinity. None of the methods of solution, except the method of invariant imbedding to be presented in this section, can reduce it to an initial value problem. In terms of the similarity variables, the Falkner-Skan equation can be written as
fm + ffH + f1 (1 - ff) = 0
(5.1)
11.5 Solution of Falkner-5kan Equation
281
subject to the boundary conditions ~=
(5.2)
f= f~=O
0:
h=l
~=oo:
(5.3)
where the subscripts in Eq. (5.1) represent differentiation with respect to the independent variable ~. The boundary condition at infinity, Eq. (5.3), is usually replaced by ~
= L,:
f~
= 1 and t« = f
(5.4)
where f is an infinitesimal quantity arbitrarily specified in accordance with the degree of accuracy required of the solution. The quantity ~oo is the distance from the plate where the boundary condition f ~ 00) = 1 can be considered satisfied. The introduction of a fourth boundary condition, namely f H = f, poses no problem since an unknown parameter ~oo is introduced at the same time. The missing boundary condition at ~ = 0, namely f~~O), is the quantity to be sought. Equation (5.1), in its present form, cannot be treated directly by the method of invariant imbedding. For convenience, a transformation of the independent variable is defined by
(5.5) Equation (5.1) then becomes
fTf'/'"
=
f."." + /3(1 - f,n
(5.6)
and the boundary conditions become TJ
= 0:
TJ
= ~oo:
f.,,(O) = - 1, f.".,,(O) = f(~oo)
= f."((,,,) =
°
f
(5.7) (5.8)
The purpose now is to find the missing boundary conditionfTf'/(~oo) and ~oo. Following the concept of invariant imbedding, a wider class of problems is considered, in which the boundary conditions are given by TJ
= 0: TJ
= a:
f.,,(O) = -1, f.".,,(O) = f(a) = c, f.,,(a) =
° f
(5.9) (5.10)
where the parameter ~oo is replaced by a for the sake of uniformity of notation. Thus by assuming different values of the starting point a, say, a = 0, A, 2A, ... , there will be a family of problems. The boundary conditions f(a) = c is added to enlarge the family further by including solutions with different starting positions f(a), where c can be taken to be 0,<5,2<5,
282
11. Method of Invariant Imbedding
38, ... , with the notation 8 representing the step length in e. All comments leading to Eq. (2.6) are valid. The enlarged family of solutions therefore includes solutions starting from a = 0, bo, 2bo, ... , with the starting positions at f(a) = e = 0,8, 28, .... Each member of the family differs, therefore, from other members of the family by its starting value a and its starting position e. The solution to Eq. (5.6), subject to the boundary conditions given by Eqs. (5.9) and (5.10), thus depend on both a and e. Since neighboring processes are related to each other, the missing boundary condition to Eq. (5.6), f1/1/(LJ, can be determined by examining the relationship between neighboring processes .. Let us now define the missing second-order derivative of f of the system represented by Eq. (5.6) and its boundary conditions, Eqs. (5.9) and (5.10), where the process begins at 'T/ = a, fT/T/(a, e), by
fT/T/(a,e)
=
(5.Il)
r(a,e)
We will now consider r(a,e) as the dependent variable and a and e as independent variables, and an expression for r(a, e) as a function of a and e will now be sought. By using a Taylor series, it can be shown that
r(a + bo,e)
=
r(a,e) + fT/T/1/(a,e)bo + O(bo 2)
(5.12)
From Eq. (5.6), fm(a, e) can be written as
fm(a, e) = cr(a,e) + p {I where
-
[s(a,e)f}
(5.13)
s(a,e) = f1/(a,e)
(5.14)
Equation (5.12) therefore becomes
r(a + bo,e) = (1 + ebo)r(a, e) + p {I - [s(a,e)]2}bo On the other hand, using the boundary condition f(a, e) that
= c, it
r( a + bo, e) = r[ a + s, f( a + bo, e)] = r[ a + bo, f( a) + I; (a, e)bo ]
(5.15) is found (5.16)
or
r(a + bo,e) = r[a + bo,e + s(a,e)bo]
(5.17)
Equating the right-hand sides of Eqs. (5.15) and (5.17), we get
r[a + ts,« + s(a,e)bo] = (1 + ebo)r(a, e) + P{1- [s(a,e)f}bo
(5.18)
By the same reasoning, an expression for the first-order derivative can be derived:
f1/[ a + bo, f(a,e) + f1/(a,e)bo] = f1/(a, e) + r(a,e)bo
283
11.5 Solution of Falkner-5kan Equation
or
s[ a + t::.,e + s(a,e)t::.]
=
s(a,e) + r(a,e)t::.
(5.19)
Equations (5.18) and (5.19) can be used directly to obtain the missing initial condition r( a, e) for the whole family of solutions. An alternate approach is to expand the left-hand sides of these equations to get 2
er(a,e)+P{1-[s(a,e)]}=
ar(a,e) ar(a,e) aa + ae s(a,e)
(5.20)
and
r(a,e) =
as(a,e) as(a,e) aa + ae s(a,e)
(5.21)
Equations (5.20) and (5.21) can then be used to solve r(a,e) and s(a,e). Since the original difference equations, Eqs. (5.18) and (5.19), preserve the physical characteristics of the process, they are used in this section. To illustrate the solution steps, let us consider the following values in Eq. (5.6):
P=
t::. = 0.025,
I,
e
= 0,8,28, ... ,
8
= 0.05
(5.22)
As discussed by Lee [10], there is very little sophistication involved in choosing the grid values. Experience, computer memory capacity, and accuracy requirements play major roles in the selection of these values. The solution of Eq. (5.6) takes the following steps. 1. To start the solution, the parameter a is set to zero, and the boundary conditions give the following numerical data:
f(O,e) = e,
fT/(O, e) = s(O,e) = -1,
fT/T/(O, e) = r(O, e) = e
and, from Eq. (5.6),
fm(O,e)t::. = er(O,e)t::. + P[1- S2(0, e) ] where e takes the values of 0,8,28, ... , and e is chosen to be an infinitesimal quantity, which is assigned to be 0.0001 in the numerical data given in this section. In Fig. 11.4, these are the points at (0,0), (0,8), (0,28), ... , on the a = line. 2. The equation
°
d(O, e) = e + s(O,e)t::. is used to compute the values of f at all d's in Fig. 11.4. Once these points are fixed, the corresponding values of r(a,e) and s(a,e) at these points can be obtained directly from Eqs. (5.18) and (5.19), respectively. Finally, the third-order derivative can be computed by using the differential equation, Eq. (5.6).
11. Method of Invariant Imbedding
284
(0,38)
(0,28) <(;
<;-
d(0,28)
.; i;::;;
(0,8)
(0,0) (0,0)
(~,
0)
(2~,0)
(3~,
0)
a
Flg.11.4
Grid points on the c-a plane.
3. Once the above are known at all d points, the values of these quantities at the points (il, 0),(il, 8), (il, 28), ... , can be determined by interpolation or extrapolation. This gives values of sand r of those members of the family starting at these points. 4. Steps 2 and 3 are repeated by adding il to the value of a. The values of sand r at the new starting position a for the same grid points in the c direction can be computed. In general, at the starting point of a = nil, with starting positions at c = 0,8,28, ... , the following quantities are obtained: s(nil, 0), s(nil,8), s(nil,28), r( nil, 0), r( nil, 8), r( nil, 28),
. .
(5.23)
The important point to be noted is that if the differential equation, Eq. (5.6), is solved by forward integration with an iteration scheme subject to the boundary conditions '1/ = 0:
f.,,(O) = -1 f(nil) = 0, f.,,(nil) = s(nil,m8)
(5.24)
where m = 0, 1,2, ... , then the missing initial slopes f.".,,(nil) or r(nil,m8) thus obtained should be the same as those given in Eq. (5.23). Owing to the fact that the values of r(nil, m8) are obtained without the need of iteration, the value of the invariant imbedding concept is now obvious.
11.5 Solution of Falkner-5kan EquatIon
285
1.0 ...----.,-------r_;::-----r------,
0-
~
0.5
oL-_ _...L_ _---l o 100
..L-_----l---l
200
a
Fig. 11.5 Variations of s(a, 0) with a;
°
f3 = I.
5. After values of sand r are computed at each new station of a, the condition f1J(a) = [i.e., f1J(~co) = 0] has to be tested to see if ~co has been reached. Since the other boundary condition at 11 = a is f(a) = 0, the starting position is at c = and the condition is therefore
°
f1J(a)
=
s(a,O)
=
°
(5.25)
Figure 11.5 gives the variation of s(a,O) for this example. If it equals zero, then
s(a,O) = s(~co'O) and the value of ~co can be determined. The value of r(a,O) at this point will be the missing initial slope h1J(~co)' For this problem, ~co is reached between 186Li and 187Li, as shown in Fig. 11.5. The value of s(a, 0) starts from - 1 at a = and reaches a value of zero at approximately 11 = 4.66. The corresponding value of r( a, 0) is therefore the missing initial slope for the solution of the Falkner-Skan equation for f3 = 1. We thus get
°
f1J1J(~co)
= r(a,O) = 1.242340
The result is within 1% of accuracy as compared with the exact value of 1.232587. By using the present method, however, iteration is eliminated. Other values of f3 can be assigned and steps 1-4 repeated, and the missing boundary condition for each value of f3 can be obtained in one sweep without the need of an iteration process. Numerical results for f3
286
11. Method of Invariant Imbedding
from 0 to 2 are collected in Table 11.2. Also included in the table are values of the missing boundary condition from published data in the literature [14,15), which were computed by iteration except at {3 equal to zero. The percentage errors are also included. In each case, the error is seen to be less than 1%. TABLE 11.2 Selected Examples of the Solution
1(/..0) Error
f3
By this method
From [13, 14]
(%)
0.00 0.05 0.10 0.20 0.60 0.80 1.00 1.60 2.00
0.4675750 0.53178230 0.5895354 0.6915536 1.0042250 1.1294110 1.2423400 1.5327980 1.6995770
0.4695999 0.5311298 0.5870348 0.6867079 0.9958359 1.1202660 1.2325870 1.5215120 1.6872170
- 0.431 0.123 0.426 0.706 0.842 0.816 0.791 0.742 0.733
11.6 CONCLUDING REMARKS
In this chapter the subject of invariant imbedding is treated very briefly because of the availability of many texts [3-5, 10-12) dealing specifically with the various aspects of this concept. When invariant imbedding was first introduced, it was considered as a concept and not as a technique. It reformulates the original boundary value problem by introducing new variables or parameters. Subsequent developments have established it as a new initial value method for the solution of boundary value problems. For a rigorous derivation of the theoretical aspects of the invariant imbedding method, the mathematically oriented readers should consult the book by Meyer [11). PROBLEMS
1. In an analysis of finite bending of a thin-walled tube [16), the following boundary value problem was obtained:
df(O)/ d~ = {3(0) = f( 'TT /2) = {3( 'TT /2) = 0
287
Problems
Solve the problem by the method of invariant imbedding for 3.0. Answer: The dimensionless momentum is given by m
= -40: 7T
0:
=
1.0 and
L"'/2 fcos(~ + f3)d~ 0
The answers are m = 0.8650 for 0: = land m = 0.7250 for 0: = 3.0. 2. Consider again the deflection of a cantilever beam under a concentrated load P treated in problem 6 of Chapter 6. The boundary value problem is given by d~/ ds 2 + A.sCOs f= 0,
df(O)/ ds = 0, f(l) = 0
Solve the problem by the method of invariant imbedding for A = 8. df(l)/ds = -3.194. 3. The analysis of heat and mass transfer in a porous catalyst is described by the following boundary value problem [17]: Answer:
dy dx 2
[
yf3(l - y)]
dy(O)
= 8yexp 1 + f3(1 _ y)'
~ = 0, y(l) = 1
Solve the problem by the method of invariant imbedding for y = 20, f3 = 0.4, and 8 = 0.1376. Answer: y(O) = 0.7928. 4. Consider the flow between two concentric rotating cylinders [18]. The pressure distribution can be obtained by solving the following boundary value problem:
~(rdP)I/2+.!!....(.!. dP)I/2=0 dr 2
dr
dr
r dr
Solve the problem by the method of invariant imbedding and compare with the exact solution,
P = !pa 2(r2 -
rD -
2paf31nr/r l
-
!pf32(I/r 2 -
l/rD
where The following physical quantities are given: p (density of the fluid) = 62 Ibm/ft3, WI (angular velocity of the inner cylinder) = 10 rpm, w2 (angular velocity of the outer cylinder) = 30 rpm, r l (radius of the inner cylinder) = 5 in., r2 (radius of the outer cylinder) = 6 in., PI (pressure at r l ) = 20 Ibr/in. 2
288
11.
Method of Invariant Imbedding
REFERENCES 1. Ambarzumian, V. A., "Theoretical Astrophysics," Pergamon, Oxford, 1958. 2. Chandrasekhar, S., "Radiative Transfer," Dover, New York, 1960.
3. Arosesty, J., Bellman, R., Kalaba, R., and Ueno, S., Invariant imbedding and rarefied gas dynamics, Proc. Natl. Acad. Sci. 50,222 (1963). 4. Bellman, R., Kalaba, R., and Prestrud, M. C., "Invariant Imbedding and Radiative Transfer in Slabs of Finite Thickness," American Elsevier, New York, 1963. 5. Bellman, R., Kagiwada, H. H., Kalaba, R. E., and Prestrud, M. c., "Invariant Imbedding and Time-Dependent Transport Processes," American Elsevier, New York, 1964. 6. Bellman, R., and Kalaba, R. E., Invariant imbedding, random walk, and scattering, J. Math. Mech. 9, 411 (1960). 7. Bellman, R., and Kalaba, R. E., Wave branching processes and invariant imbedding, Proc. Natl. A cad. Sci. 47, 1507 (1961). 8. Bellman, R., and Kalaba, R. E., A Note on Hamilton's equations and invariant imbedding, Q. Appl. Math. 21, 166 (1963). 9. Wing, G. M., "An Introduction to Transport Theory," Wiley, New York, 1962. 10. Lee, E. S., "Quasilinearization and Invariant Imbedding," Academic Press, New York, 1968.
II. 12.
13. 14.
15. 16. 17. 18.
Meyer, G. H., "Initial Value Methods for Boundary Value Problems," Academic Press, New York, 1973. Scott, M. R., "Invariant Imbedding and Its Applications to Ordinary Differential Equations: An Introduction," Addison-Wesley, Reading, Massachusetts, 1973. Chambers, R. L., and Somers, E. V., Radiation in efficiency for one-dimensional heat flow in a circular fin, J. Heat Transfer 81, 327-329 (November 1959). Elzy, E., and Sisson, R. M., Tables of similarity solutions to equations of momentum, heat and mass transfer in laminar boundary layer theory, Eng. Exp. Station, Bull. 40, Oregon State Univ., February 1967. Moore, F., "Theory of Laminar Flow," Princeton Univ. Press, Princeton, New Jersey, 1964, p. 127. Perrone, N., and Kao, R., J. Appl. Mech. 38, 371-376 (1971). Kubicek, M., and Hlavacek, V., Chern. Eng. Sci. 29, 1695-1699 (1974). Schlichting, H., "Boundary Layer Theory," pp. 80-81, McGraw-Hili, New York, 1968.
CHAPTER
12
INTEGRAL EQUATION METHOD
12.1
INTRODUCTION
In the integral equation method, the boundary value problem is replaced by an integral equation, which in turn is solved by numerical quadrature formulas. The derivation of an equivalent integral equation for a boundary value problem is in general complicated. The procedure followed involves the determination of the Green's function for a linear boundary value problem. To introduce the procedure, let us begin with a simple example. Consider the boundary value problem
dy/dx 2 + Y = 0 yeO) = 0,
(1.1 )
y(l) = 0
(1.2)
Integrating Eq. (1.1) from 0 to x, we get
dy = _ rxydx+ e dx Jo
(1.3)
Another integration from 0 to x then gives
y = - fax faxY dx + ex + d
(1.4)
Now, we recall from calculus that
Jo . . . LX f(x)dx·: . dx = (n -
1
x
0
(n times)
(n times)
I)!
LX(x 0
~r- If(~)d~
(1.5)
Equation (1.4) can then be written as
y(x) = - fa\x - ~)y(~)d~+ ex + d
(1.6) 289
290
12.
Integral Equation Method
To determine the two constants of integration c and d the two boundary conditions, Eq. (1.2), are used, which gives
d=O
and
Equation (1.6) then becomes
y(x)
= x L\l - ~)y(~)d~- LX(x - ~)y(~)d~
or
y(x)
= L\(1- ~)y(~)d~+.fX(1 - ~)y(~)d~- L\x - ~)y(~)d~
or
y(x)
= LX~(1- x)y(~)d~+
L 1
X(1 -
~)ymd~
(1.7)
If we now define a kernel as ~(l - x) K(x,~) = { x(l _~)
when when
<x ~ >x ~
Eq. (1.7) can be written as
y(x) = iIK(X,~)Y(~)d~
(1.8)
which is a Fredhom integral equation of the second kind. To solve Eq. (1.8) for y(x), numerical methods for approximating the solution of integral equations can be applied [1]. Thus the solution of a boundary value problem is reduced to the numerical solution of an integral equation. We will consider the solution of nonhomogeneous ordinary differential equations with variable coefficients in Section 12.2. It will be shown that the general solution can be expressed in terms of a so-called Green's function. Application of this method to nonlinear boundary value problems will be treated in Section 12.3. 12.2
LINEAR BOUNDARY VALUE PROBLEMS
Consider the differential equation Ly=>(x)
(2.1)
where L is the operator L
d [
d ]
= dx p(x) dx
d? dp d - q(x) = p(x) dx 2 + dx dx - q(x)
(2.2)
12.2 Linear Boundary Value Problems
with p(x)
> 0 and
291
q(x) ;;;. O. The homogeneous boundary conditions are x=a: x
= b:
The Green's function (2.1) is given by
lX1y(a)+l3ldy(a)/dx=0
(2.3)
+ 132dy(b)/dx = 0
(2.4)
lX 2y(b)
G(x,~)
is a function such that the solution of Eq.
(2.5) The Green's function is defined such that, for a given ~, it is given by G1(x,O when x < ~ and by G2(x,O when x > ~ and has the following properties. 1. The functions G 1 and G2 satisfy the equations LG 1 = 0
for
x
<~
(2.6)
for
x
>~
(2.7)
and
2. The function G1 satisfies the boundary condition prescribed at x = a, Eq. (2.3), and the function G2 satisfies the boundary condition at x = b, Eq. (2.4). 3. The function G is continuous at x = ~, i.e., (2.8) 4. The derivative of the function G has a discontinuity of magnitude - 1/ at the point x = ~, i.e.,
pm
dG2(X,~) dx
I x=€
dGI(x,~) dx
I
= -
x=€
_1_ p(~)
(2.9)
We will now show that if a Green's function exists which satisfies the above conditions, then the solution of the original differential equation, Eq. (2.1), can be written in the form given in Eq. (2.5) [1]. To start, let u(x) be the solution of Ly = 0 which satisfies the prescribed boundary condition at x = a, and let v(x) be the solution of Ly = 0 which satisfies the boundary conditon at x = b. The same is true if we multiply u(x) and v(x) each by a constant. The first two properties will be satisfied if we define the Green's function as when when
x x
<~ >~
(2.10)
The two constants can be found from the third and fourth properties. Substituting the Green's function from Eq. (2.10) into Eqs. (2.8) and (2.9),
292
12. Integral Equation Method
we get Czv(~)
-
CZV'm -
=0
(2.11)
CIU'(~) = - P~~)
(2.12)
CIU(~)
from which
= V(~)/p(~)[ V(~)U'(~) Cz = U(~)/p(~)[ VmU'(~) C\
V'(~)U(~)]
(2.13)
V'mU(~)]
(2.14)
where and Further manipulation is necessary to simplify Eqs. (2.13) and (2.14). Since u and v satisfy Ly = 0, we can write -d [ p(x) -du ] - q(x)u = dx dx
°
(2.15)
and - d [ P(x) -dv ] - q(x)v = 0 dx dx
(2.16)
Multiplying Eqs. (2.16) and (2.15) by u and v, respectively, taking their difference, and setting x = ~, we get { v(x)
:x [P(X) ~~ ]} x=~
-
{
u(x)
:x [P(X) ~~ ]} x=~ = 0
or
{ -!L dx
[P(X)(u dv -v du )]} =0 dx dx x=~
from which p(~) [u(~)v'(~) - v(~)u'(~)]
= const
The constants c\ and Cz in Eq. (2.13) and (2.14) can therefore be written as c\ = Av(~) and Cz = Au(~) where A is a constant. The Green's function in Eq. (2.10) can therefore be written as G x ~
( ,)
Av(~)u(x) when x < ~ = { Au(~)v(x) when x > ~
(2.17)
12.2 Linear Boundary Value Problems
293
Once the Green's function is found, we can show that the solution given in Eq. (2.5) satisfies the differential equation, Eq. (2.1). To this end, we write Eq. (2.5) as
y(x) = - faxG(x,~)(~)d~-
.r
G(x,~)(~)d~
= -A faxv(~)u(X)(~)d~-A ibu(~)V(X)(~)d~
(2.18)
Differentiating Eq. (2.18) twice to get dy/ dx and d) / dx2 and substituting these expressions into Eq. (2.1), we get
A {faX[ Lv( x)] u(~)(~)d~ + i +A {P(X)
AP~X) (X)} -
b[
LU(x)] v(~)(~)d~ }
(x) = 0
(2.19)
Since the first two terms of Eq. (2.19) are identically zero and the third and the fourth terms cancel each other, the solution given in Eq. (2.5) satisfies the differential equation (2.1) identically. Following the same reasoning, it can be proved that the solution, Eq. (2.5), satisfies the boundary conditions, Eqs. (2.3) and (2.4). A natural extension of the above theory is obvious. If the differential equation is in the form
Ly = r(x)y(x)
+ s(x)
(2.20)
the corresponding integral equation can be written as (2.21 )
The examples presented in the next two sections illustrate the computational details in the application of this method. 12.2.1
Heat Conduction through Cone-Shaped Fins
Consider a cone-shaped fin protruding from a wall at surface temperature Let the radius at the base and the length of the fin be Rand L, respectively. The fin is surrounded by fluid at a temperature of Too. The energy equation for the solution of the temperature distribution can be written as [2]
r..
294
12. Integral Equation Method
subject to the boundary conditions dT(O)j dx = 0
T(L) = T;
and
where hand k are the convective heat transfer coefficient and the heat conductivity of the fin, respectively. Next, we introduce the dimensionless variables _ x x= L'
T- T 0= T _; , s
00
Equation (2.22) then becomes 2
d 0 dx 2
+ ( ~) dO = ( "!..2 )(0 _ 1) X
dX
x
(2.23)
subject to the boundary conditions dO(O)jdx = 0
and
0(1)=0
To find the Green's function, we first write Eq. (2.23) in the form
:x { ~: x
2
}
= m 2x (0 - 1)
(2.24)
Comparison with Eq. (2.1) shows that p(x)
= x 2,
q(x)
=0
The integral equation can therefore be written as
(2.25) where the Green's function G(x, ~ can be derived as follows. Consider now the equations and or and The solutions are To find the constants, the boundary condition at x = 0 is first applied to 0 1, i.e.,
12.2 Linear Boundary Value Problems
295
from which c\ = O. Next, the boundary condition at x = I is applied to i.e.,
()2'
from which
c4 The functions
()l
and
-
()2 therefore
()\ = c2
c3 = 0 become ()2 =
and
c3( 1 - I/x)
Based on Eq. (2.17), the Green's function can be written as
= AC2C3( 1 - I/~), G2 ( x,~) = AC2C3( I - 1/ x), To find the constants, the jump condition at
dG2(_X,~) I
x=~
dx
<~ for x > ~ x = g, namely for x
G\(x,~)
_dGl(_X,~) I -_ - _1_ x=~
dx
p(~)
(2.26) (2.27)
(2.28)
can be used. Substituting the Green's function, (2.26) and (2.27), into Eq. (2.28), we get
from which AC 2C3
= -I
The final form of the Green's function is therefore _ {(I/~)-I G(x,~) = (I/x) - I
x<~ x>~
for for
(2.29) (2.30)
Once the Green's function is derived, the solution of Eq. (2.24) can be obtained by approximating f(x) by u/x), where u/x) is defined as the solution of the following system of algebraic equations:
u;= -
J
2: m G(x;,xj )xj (uj 2
I ).:lj
(i = 1,2, ... ,J)
(2.31)
)=1
or, in expanded form, (I
+ Cl:ll)U 1 + Cl:\2U2 + + (1 + Cl: 22)u2 +
Cl:2\U 1
+ Cl:iJuJ + Cl:2JuJ
(2.32)
296
12.
where
Integral Equation Method
J
ai) =m2~G(Xjl~)Doj'
f3i
= m2 2: ~G(Xi,~)Doj j=1
and, based on the trapezoidal rule and constant x grid, Do] =DoJ
= tDox;
Doj=DoX
for j=2,3, ... ,J-I
(2.33)
Numerical solutions of Eq. (2.32) based on this method are shown in Table 12.1 for m 2 = I and 10. The solutions agree with the exact solution [2]. For example, the values of 0(0) from exact solutions for m 2 = I and 10 are 0.3713 and 0.9621, respectively. The corresponding values of 0(0) from the present method with J = 100 are 0.3702 and 0.9600, as shown in Table 12.1. TABLE 12.1 Solutions of Eq. (2.23) (m 2 = I and 10)
12.2.2
m2
x
8
m2
x
8
1.0
0.0 0.2 0.4 0.6 0.8 1.0
0.3702 0.3085 0.2393 0.1655 0.0869 0.0031
10.0
0.0 0.2 0.4 0.6 0.8 1.0
0.9600 0.9089 0.8144 0.6556
OAOOO 0.0020
Sandwich Beam Analysis
We will now apply the integral equation method to the solution of the sandwich beam equation treated in Section 2.3.2. If we define
f= dtf;/dx
(2.34)
Eq. (3.16) can be written as
d':J/dx 2
-
k':J+ a = 0
(2.35)
subject to the boundary conditions
f(O) = f(l) Once the solution of f is obtained, the unknown function tf; can be solved by integrating Eq. (2.34) over x. We will consider first the solution of J. First, Eq. (2.35) can be written as
d':J/ dx 2 = k':J - a
(2.36)
Comparing with Eq. (2.20), it is seen that
p(x)=I,
q(x)=O,
r(x)=k 2 ,
s(x)=-a
12.2 Linear Boundary Value Problems
297
The integral equation can therefore be written as (2.37)
which is seen to be Fredhom integral equation of the second kind. To find the Green's function, let us consider the equations
=0
and
dz.tl/ dx? = 0
and
Lfl
Lfz
=0
or The solutions are
fl = C1X + Cz and fz = c3x + c4 To find the constants, the boundary condition at x = 0 is first applied to fl' i.e., fl(O) = 0 from which Cz = O. Next, the boundary condition at x = 1 is applied to fz, i.e.,
fz(1) =0 from which c3 + c4
=
O. The functions
fz
t. and fz therefore become = ci x -
1)
The Green's function can therefore be written as GI(X,~)
= ACIC3X(~ - 1),
for x
<~
Gz(x,~)
=
for x
>~
ACIC3~(X
- 1),
based on Eq. (2.17). To find the constants, the jump condition at x ACIC3~
-
ACIC3(~
= ~, Eq. (2.9) gives
- 1) = -1
from which
AC1C3 = -1 The final form of the Green's function is therefore
G(x,~) =
{ ~(1 -
X ( l - ~)
x)
for x for x
<~ >~
(2.38)
With the Green's function known, we can proceed to solve f(x) from the integral equation, Eq. (2.37). The function u/t) which is to approximate
298
12.
Integral Equation Method
!(X) is defined as the solution of the system of algebraic equations ]
1; = -
L:
)=1
G(x;,xj)(kt - a)Llj,
i
= 1,2, ... , J
(2.39)
or
(1 + (Xll)fl + (X12!2 + ... + (Xu!] (X2l!1 + (1 + (X22)!2 + . . . + (X2Jh
(2.40)
where ]
P; = L:
)=1
aG(x;,x)Llj
and Llj is defined in Eq. (2.33). Solutions of Eq. (2.40) for a = 1 and k = 5 and 10 are shown in Table 12.2. The solutions of If; in the third column are obtained by integrating Eq. (2.34). The solutions agree with those in Section 2.3.2. TABLE 12.2 Sample Solution of Eq. (2040) (a = I) k
x
'"
f
5.0
0.0 0.2 004 0.6 0.8 1.0
- 0.0120 - 0.0092 - 0.0033 0.0033 0.0092 0.0120
0.0000 0.0246 0.0326 0.0326 0.0246 0.0000
25.0
0.0 0.2 004 0.6 0.8 1.0
- 0.0040 - 0.0029 - 0.0010 0.0010 0.0029 0.0040
0.0000 0.0086 0.0098 0.0098 0.0086 0.0000
12.3 NONLINEAR BOUNDARY VALUE PROBLEMS
In this section, we will consider nonlinear boundary value problems and show how the method developed in the last section can be used to reduce nonlinear boundary value problems to integral equations [3].
299
12.3 Nonlinear Boundary Value Problems
Consider the nonlinear boundary value problem
= f(x,y,dy/dx) <x1y(a) + f31dy(a)/dx = 0 <xzy(b) + f3 zdy(b)/dx = 0
Ly x
= a:
x
= b:
(3.1) (3.2) (3.3)
where L is defined in Eq. (2.2). To solve Eq. (3.1) by the integral equation method, an iterative process is required. Let the number of iterations be denoted by v and write the iteration equation as Lid I)
= f(x, ipl, dip) / dx)
(3.4)
subject to the boundary conditions <x1iP+t)(a) <XZip+1)(b)
+ f3tdip+1)(a)/dx = 0 + f3 zdi
p+ 1)(b)/dx
=0
(3.5) (3.6)
Comparing Eq. (3.4) with Eq. (2.1), we see that the function j(x, i p ) , dip) / dx) can be identified as <j>(x) for the solution of the (v + I)th iteration, i + I). The solution of Eq. (3.4) can therefore be written as p
(3.7)
where v = 0, I, .... The example presented in Section 12.3.1 illustrates this method. Even though the above scheme converges for any initial approximation, the rate of convergence may be slow. The situation can be considerably improved if Newton's method is applied. The steps involved include replacing Yj in the nonlinear integral equation by Yj + dYj' expanding in powers of dYj' and dropping all higher-order terms. A linear integral equation in dYj will be obtained, which can be solved by replacing the integral equation by a system of algebraic equations as shown in Section 12.2. The example presented in Section 12.3.2 illustrates the details of the application of this iterative scheme. 12.3.1 Heat and Mass Transfer In a Porous Catalyst
In an analysis of the solutions of enthalpy and mass balance for a catalytic reaction within a porous catalyst particle in a chemical reaction,
12.
300
Integral Equation Method
Kubicek and Hlavacek [4] obtained the following nonlinear boundary value problem:
dy2 dt
yf3y
=a(l+y)ex p(-
)
(3.8)
y(l)=O
(3.9a, b)
1 - f3y
subject to the boundary conditions dy(O)/ dt = 0
and
For this example, the operator L is
(3.10)
L = d 2/dt 2 i.e.,
q(t) = 0
pet) = 1,
To find the Green's function, we consider the two equations
LYI = 0
Lh
and
=0
or and The solutions are
Yl = cit + c 2
h = c3t + c4
and
For YI' the boundary condition at t
= 0,
(3.9a), is applied, which gives
cl = 0
For h, the boundary condition at t = 1, (3.9b), is applied, which gives c4 = - c3 The functions Y I and h then become
YI=C 2
and
h=c3 ( t - I )
Based on Eq. (2.17), the Green's function can be written as
G(t
AYI(t)h(~) = AC2C3(~ - 1) ={ , AYI(~)h(t) = AC2C3(t - 1) ~)
for for
t t
<~
>~
(3.11 )
To find the constants, we substitute the Green's functions (3.11) into the jump condition at t = ~, Eq. (2.9), which gives
AC2C3 = -1 The final form of the Green's function is therefore given by (~- 1) G(t,~) = { (t - 1)
for for
t t
<~ >~
(3.12)
12.3
Nonlinear Boundary Value Problems
301
Once the Green's function is found, the solution of Eq. (3.8) can be written in terms of the integral equation as (3.13)
where By choosing as first approximations
/O}(t) = t 2 - I Eq. (3.13) can be integrated to give higher iterations v = 1,2 .... Table 12.3 gives a typical solution for IX = 0.1, y = 1.0, and f3 = 0.5. Four iterations are required for the solution to converge. TABLE 12.3 Solution of Eq. (3.8) (0: = 0.1, y
= 1.0, f3 = 0.5)
p=2
p=3
p=o 0.0 0.2 0.4 0.6 0.8 1.0
-
P
1.0000 0.9600 0.8400 0.6400 0.3600 0.0000
-
= I
0.0105 0.0260 0.0321 0.0289 0.0175 0.0000
-
0.0496 0.0473 0.0413 0.0314 0.0177 0.0000
-
0.0491 0.0471 0.0412 0.0314 0.0177 0.0000
p=4 -
0.0491 0.0471 0.0412 0.0314 0.0177 0.0000
12.3.2 The Explosion of Solid Explosives
The energy equation which governs the temperature distribution in the explosion of a solid explosive can be written in terms of dimensionless variables as [5] 20 d 2 dO [0] (3.15) dt2 + dt + IX exp I + (0/ y) = 0
t
subject to the boundary conditions
t = 0:
dO (0)/ dt = 0
t = 1:
N nuO(I) + dO(I)/dt = 0
where N nu is the Nusselt number. Equation (3.15) can be written in the form
~ (t 2 ~~ ) = - IXt 2exp [
1+
to/y) ]
(3.16)
302
12. Integral Equation Method
Comparing with Eq. (2.1), we see that p(t) = t 2 ,
q(t) = 0
For this equation, the Green's function can be found to be
( I/ ~ - I) + 1/ N nu G(t,~) = { (l/t - I) + 1/ N nu
for for
<~ t >~ t
(3.17)
The integral equation is therefore
O(t) = LIG(t,~)aeexp(
y
~ 0)d~
The functions U:i(t) which are to approximate solution of the system of equations ui
vu,J_ ) dj' = LJ «oi); ~)~2exp( __ )=1 y + uj
O(~)
i
(3.18) are defined as the
= 1,2,
... ,)
(3.19)
where, based on the trapezoidal rule and constant grids, d J=dJ = ! l l t ;
)=2,3, ... ,)-1.
~=Ilt,
Equation (3.19) is the system of nonlinear algebraic equations for the solution of the u;'s. Iteration is therefore required. To derive equations for the iteration process, we use
(3.20) where (v)
ri
=
J
2
LaG(ti'~)~exp
)=1
[
Yu~v) J
Y+
]
(v)
Uj
~
_
(v) Ui
(3.21)
Equation (3.20) can be written in the form
(I - fll)IlU J - f12llu2 - f2J ll u(
+ (I -
f22)IlU 2 -
- fIJlluJ - f2JIlUJ
(3.22) r(v) J
12.4 Concluding Remarks
303
where (3.23) By assuming a first approximation, e.g., U(l)
= t 2 - (N nu + 2)/ N nu
(3.24)
Eq. (3.22) can be used to solve for the next iterations U;(2), uP), ... , until a prescribed accuracy is reached. Numerical results for N nu = I, a = 1, and y = 0.1 are given in Table 12.4. TABLE 12.4 Numerical Solution of Eq. (3.15) (Nnu = I, a: = I, 'I Iterations of
0.00 0.04 0.08 0.12 0.16 0.20
= 0.1)
ul")
v= I
v=2
v=3
v=4
0.3000 0.2998 0.2994 0.2986 0.2974 0.2960
0.0233 0.0231 0.0222 0.0208 0.0188 0.0163
0.0223 0.0221 0.0213 0.0200 0.0181 0.0156
0.0223 0.0221 0.0213 0.0200 0.0181 0.0156
12.4 CONCLUDING REMARKS
The integral equation method outlined in this chapter has two basic steps, the conversion from a boundary value problem to an integral equation and the numerical solution of the resulting integral equation. For linear boundary value problems, the solutions can be obtained in one sweep, whereas for nonlinear boundary value problems iteration is required. Conversion of boundary value problems to integral equations using Green's function is not the only procedure available. As an illustration of other procedures, let us consider Blasius's equation in the boundary layer theory, Eq. (1.7),
/'" + t if" = 0
(4.1)
where primes represents differentiation with respect to '1'/. Dividing Eq. (4.1) by f" and integrating the resulting equation, we get (4.2)
304
12.
Integral Equation Method
Integrating Eq. (4.2) over 1/, we get f' =
lX
~1jexp[ -
i ~1jI(~)d~]
d1/+ /31
The boundary condition f'(0) = 0 gives /31 = O. To get condition f'( 00) = I gives
(4.3) lX,
the boundary
Integrating Eq. (4.3) over 1/ one more time and determining the integration constant by the boundary condition 1(0) = 0, we get
or
(4.5) which is the integral equation for the solution of 1(1/). This offers another procedure for the conversion of boundary value problems to integral equations. For the numerical solution of the resulting integral equation, only two methods are used in the examples of this chapter. For a complete treatment of the various approximate and numerical methods of solution of the integral equations, the readers are referred to Hildebrand P], Courant and Hilbert [6], Kantorovich and Krylov [7], and Green [8]. For those who are interested in the mathematical theory of the integral equation method, such as the convergence and uniqueness properties, a fairly complete treatment can be found in Keller [3]. PROBLEMS
1. Consider a straight fin the thickness of which increases linearly with x from 0 at x = 0 to 2 Y at x = L. The boundary value problem resulting from such an analysis (see problem 2.26 of [2]) can be written as 2T xd- + -dT - m (2T - T )=0 dx' dx 00'
dT(O) -d -x =0 '
T(L) = T
S
Solve this problem by using the integral equation method for L = 2 in., = 200°F, Too = 70°F, and m 2 = 1.37/ft. 2. The boundary value problem resulting from an analysis of the heat transfer in a transparent stagnation point shock layer was treated by Chiou T;
References
305
and Na [9] using the method of transformation. The differential equation is 2H
d +SR dH -LRH n = 0 ds 2 ds
subject to the boundary conditions
H(O) = 0,
H(l) = 1
where Rand L are physical constants and n is an integer. Solve this problem by the method of integral equation for n = 5, R = 10, and L = 0.01,0.1, 1.0 and 10.0. Hint: Make a transformation s*
=/R s
to reduce the equation to the form of Eq. (2.1). 3. In an analysis of the coalescence of two closely spaced drops at different potentials, Taylor [10] and Ackerberg [11] introduced the following boundary value problem:
dy
1 dy
dx
x dx
dy(O)
f3
- 2+ - - = -
y2'
d;- = 0, y (1) = 1
Solve the problem by the integral equation method for f3 = 0.1 and 0.3. 4. In an analysis of the stresses and deflections of an annular diaphragms, Pifke and Goldberg [12] gave the following boundary value problem:
~
= 1: 2
df
d~ - (1
+ v)f= 0
By introducing a transformation "7 = ~ - A2 , Na et al. [13] was able to get the general solution of this problem. Solve this problem by using the integral equation method (v = 0.3).
REFERENCES Hildebrand, F. B., "Methods of Applied Mathematics," Chapter 4, Prentice-Hall, Englewood Cliffs, New Jersey, 1954. 2. Myers, G. E., "Analytical Methods in Conduction Heat Transfer," Chapter 2, McGrawHill, New York, 1971. 3. Keller, H. B., "Numerical Methods for Two-Point Boundary Value Problems," Chapter 4, Ginn-Blaisdell, Boston, Massachusetts, 1968. 4. Kubicek, H., and Hlavacek, V., Solution of nonlinear boundary value problems-Part VIII, Chern. Eng. Sci. 29, 1695-1699 (1974). I.
306
12. Integral Equation Method
5. Hlavacek, V., Marek, M., and Kubicek, M., Modeling of chemical reactors-X. Multiple solutions of enthalpy and mass balances for a catalytic reaction within a porous catalyst particle, Chern. Eng. Sci. 23, 1083-1097 (1968). 6. Courant, R., and Hilbert, D., "Methods of Mathematical Physics," Vol. I, Wiley, Interscience, New York, 1953. 7. Kantorovich, L. V., and Krylov, V. I., "Approximate Methods of Higher Analysis," Wiley, Interscience, New York, 1958. 8. Green, C. D., "Integral Equation Methods," Barnes and Noble, New York, 1969. 9. Chiou, J. P., and Na, T. Y., Int. J. Math. Ed. Sci. Tech. 9,253-256 (1978). 10. Taylor, G. I., Proc. R. Soc. London Ser. A 306,423-434 (1968). II. Ackerberg, R. c., Proc. R. Soc. London Ser. A 312,129-140. 12. Pifke, A. B., and Goldberg, M. A., AIAA J. 2, 1340-1342 (1964). 13. Na, T. Y., Kurajion, G. M., and Chiou, J. P., Aeronaut. Q. 27, 195-200 (August 1976).
INDEX
A Adjoint differential equation, 53, 56, 63 Adjoint operator method, 3, 5, 52, 69 Atmospheric scattering, 272 Axial diffusion in reactors, 208 Azeotropic point, 267
B Bending of beam, 59 Blasius equation, 5, 137, 138,303 with mass transfer and slip, 209 with suction or slip, 208 Boundary layer control, 177 flow with mass transfer and slip, 209 of pseudoplastic fluids over wedges, 249 Brachistochrone problem, 148
c Catalytic converter, 79 Chasing backward, 30, 37, 41, 46 forward, 30, 37, 41, 46 method, 3, 5, 11,30,51 Chemical reactors, 181 Coalescence of two drops, 305 Complementary functions method, 25 Conduction through fins, 101
Conformal mapping, 233 Continuation method, 233, 264 Continuity method, 6 Continuous transformation groups, 139
o Deflection of cantilever beam, 134, 205, 287 Deflection of membrane, 189 Difference backward, 95 central, 95 foward,96
E Eigenvalue problem, 159, 197, 212 Electrostatic probe measurements, 18 Equilibrium composition, 237 Explosion of solid explosives, 301 Extended surfaces, 204
F Factorization, 98 Falkner-Skan equation, 146,250, 261, 280 Fin heat transfer, 38 Finite bending of tube, 286 Finite deflections of elastic bar, 207 Finite-difference method, 3, 4, 70, 93
307
308
Index
M
Flow in channel with porous walls, 268 in chemical reactor, 272 in pipe networks, 240 through porous channel, 72 through porous medium, 205 over slender body of revolution, 209 between two cylinders, 287 Foeffl-Hencky membrane equations, 174 Fredhom integral equation, 290
Magnetohydrodynamic couette flow, 54 Magentohydrodynamic flow, 193, 209 Mass transfer on a rotating disk, 27 Matrix inversion, 93 Melting of solid, 205 Motion of heavy particle, 50 Multistep method, 7
G
N
Galerkin method, 268 Gauss-Seidal's elimination, 93 General parameter mapping, 260 Gram-Schmidt orthonormalization, 66 Green's function, 291, 295, 297, 300, 302
Natural convection between parallel plates, 268 of Powell-Eyring fluids, 105 over vertical plate, 255 Neutron transport theory, 272 Newton's method, 3, 4, 71, 128 Newton's variation method, 66 Nonlinear equation algebraic, 4, 234 diffusion in biology, 184,209 dynamics problem, 88
H Hamel's problem, 122 Hamilton's equation, 272 Hardy-Cross method, 240 Heat and mass transfer in catalyst, 287, 299 Heat conduction with heat generation, I, 34 Heat transfer in radial flow, 28 over thin needles, 134 Hiemenz magnetic flow, 119 Hydrodynamic extrusion, 67
o One-step method, 7 Onset and propagation of combustion, 209, 213
p
Infinitesimal transformation, 137 Initial value problems, 7 Integral equation method, 3, 4, 289 Invariance of physical parameters, 208 Invariant imbedding method, 3, 6, 12, 272, 288 Isothermal packed-bed reactor, 15,275
L Laminar non-Newtonian jet, 175 Linear group of transformations, 5, 143, 155 Longitudinal impact of rod, 152
Parallel shooting method, 3, 70, 76 Parameter differentiation method, 3, 6, 233 Parameter of transformation, 143, 155, 157, 177 Particular solutions method, 25, 28, 29 Pendulum, 102 Phase change of solids, 92 Predictor-corrector procedure, II Principle of invariance, 272
Q Quasi-linearization method, 3, 12, 66, 70, 84, 92,120
309
Index
R
T
Radiation fins, 279 Radiation in boundary layer flows, 231 Radiative transfer, 272 Random walk and scattering, 272 Rarefield gas dynamics, 272 Relaxation method, 97 Runge-Kutta method, 8, 13
Taylor's method, 8 Thin struts with large deflections, 209 with large displacement, 221 Transformation groups,S, 137, 139, 177 method, 3, 5, 177,208 Transverse curvature effects, 217 parameter, 218 Tridiagonal matrix, 97 Troesch's problem, 264 Two-dimensional jet, 161
s Sandwich beam analysis, 23, 28, 46 Search of multiple solutions, 221 Shock-induced flow, 208 Shooting method, 3, 70 Similarity analyses, 12 transformation, 139 Simultaneous algebraic equations, 235 Slender parabola of revolution, 217 Spherical cap, 92 Spiral group of transformation, 155 Stability analysis of multiple solutions, 165 criterion, 165 Stagnation point shock layers, 269 Stirred reaction cooling coil, 267 Stresses of diaphragm, 306 Sturm-Liouville system, 167 Superposition method, 3, 4, 13
u Uniqueness of solution, 165 Unsteady flow through porous media, 208 of power-law fluids, 173
v Vibration of cantilever, 242 Viscoelastic fluid, 27 Viscoplastic rods, 152
w Wave propagation, 272
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