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PITMAN PUBLISHING INC 1020 Plain Street, Marshfield, Massachusetts Associated Companies Pitman Publishing Pty Ltd, Melbourne Pitman Publishing New Zealand Ltd, Wellington Copp Clark Pitman, Toronto
©JMH1II, 1982 AMS Subject Classifications: (main) 34-XX, 35-XX (subsidiary) 20-XX, 22-XX
British Library Cataloguing in Publication Data
Hill, J. M. Solution of differential equations by means of one-parameter groups. 1. Differential equations I. Title 515.3'S 0A371 ISBN 0-273-08506-9
Library of Congress Cataloging in Publication Data
Hill, J. M. Solution of differential equations by means of one-parameter groups (Research notes in mathematics; 63) Bibliography; p. 1. Differential equations—Numerical solutions. 2. Groups, Theory of. I. T. II. Series. 0A371.H56.
515.3'5 82-621 ISBN 0-273-08506-9 AACR2.
Australian Cataloguing in Publication Data
Hill, J. M. (James M.) Solution of differential equations by means of one-parameter groups. Includes bibliographical references.
ISBN 0858968932. 1. Differential equations. 2. Groups, Theory of. I. Title. (Series: Research notes in mathematics; 63).
515.3'S
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Reproduced and printed by photolithography in Great Britain by Biddies Ltd, Guildford
JMHi11 University of Wollongong
Solution of differential equations by means of one -parameter groups
Pitman Advanced Publishing Program BOSTON LONDON. MELBOURNE
Preface
The trouble with solving differential equations is that whenever we are
successful we seldom stop to ask why.
The concept of one—parameter
transformation groups which leave the differential equation invariant provides the only unified understanding of all known special solution techniques.
In these notes I have attempted to present a fairly concise and
self—contained account of the use of one—parameter groups to solve differential equations.
The presentation is formal and is intended to appeal
to Applied Mathematicians and Engineers whose principal concern is obtaining solutions of differential equations.
I have included only the essentials of
the subject, sufficient to etiable the reader to attempt the group approach
when solving differential equations.
I have purposely not included all known
results since this would inevitably lead to unnecessarily reproducing large portions of existing accounts.
For example, for ordinary differential
equations, the account of the subject by L.E. Dickson, "Differential equations from the group standpoint", is still extremely readable and is recommended to the reader interested in pursuing the subject further.
For partial
differential equations the books by G.W. Bluman and J.D. Cole, "Similarity methods for differential equations", and by L.V. Ovsjannikov "Group properties of differential equations", contain several applications and examples which I have not reproduced here.
The first two chapters are introductory.
Chapter 1 gives a general
introduction with simple examples involving both ordinary and partial differential equations.
In Chapter 2 the concepts of one—parameter groups
and Lie series are introduced.
Just as ordinary methods of solving
differential equations often require a certain ingenuity so does the group approach.
In order to establish some familiarity with the group method I
have attempted to exploit our experience with linear equations. are aware that linear differential equations for the transformation
x1 =
f(x),
y1 =
g(x)y
devoted to implications of this result.
y(x)
Most of us
remain linear under
and Chapter 3 of these notes is
In Chapters 4 and 5 I have tried to
relate the usual theory for the group method with the results obtained in the third chapter.
In this respect these notes differ from most accounts of the
subject and I believe that a number of results given, especially in Chapter 3 are new.
The remaining two chapters are devoted to partial differential equations. For the most part the theory is illustrated with reference to diffusion related partial differential equations.
The theory for linear partial
differential equations is introduced in Chapter 6 for the classical diffusion or heat conduction equation and the Fokker—Planck equation. equations are treated in Chapter 7.
Non—linear
For partial differential equations the
group approach is less satisfactory since for boundary value problems both the equation and boundary conditions must remain invariant.
In these notes
we principally consider only the invariance of the equation and view the group method as a means of systematically deducing solution types of a given partial differential equation.
Although these notes appear as a research monograph they actually represent advanced teaching material and in fact form the basis of a post—graduate course given at the University of Wollongong for the past six years. therefore included numerous examples and exercises.
I have
In addition to the
exercises I have used the problems at the end of each chapter to conveniently
locate standard results for differential equations.
On occasions I have also
used these problems to include summaries of theory which is already adequately described in the literature. The existing theory of the solution of differential equations by means of one—parameter groups is by no means complete. the subject are highlighted in the text.
Many of the inadequacies of
When it does work it is very easy
and it is therefore an area of knowledge which every Applied Mathematician ought to be aware of.
Whatever the limitations of the group method may be,
it will always represent a profoundly interesting idea towards solving differential equations.
I hope these notes prove to be useful and complement
the existing literature.
James M. Hill, The University of Wollongong, Australia.
Acknowledgement
The author wishes to thank all of his students over the
past six years who have discovered various errors, spelling mistakes and omissions in a preliminary draft of these notes.
He is also grateful to Mrs.
Kerrie Gamble for her careful and thoroughly professional typing of the manuscript.
For
Desley,
and
Ruth.
Contents
1.
Introduction 1.1 1.2
2.
3.
First 4.1 4.2
4.3 4.4 4.5 4.6
5.
5.4
of standard linear equation8
First order equation y' + p(x)y = q(x) Second order homogeneous equation y" + p(x)y = 0 Third order equation y"+ p(x)y' + q(x)y = 0 Fourth order self—adjoint equation y" + [p(x)y']' + q(x)y = 0 Problems
order differential equation8 Infinitesimal versions of y' and y' = F(x,y) and the fundamental problem Integrating factors and canonical coordinates for y' = F(x,y) The alternative problem The fundamental problem and singular solutions of y' = F(x,y) Invariance of the associated first order partial differential equation Lie's problem and area preserving groups Problems
Second 5.1 5.2 5.3
groups and Lie series
One—parameter transformation groups Lie series and the commutation theorem Problems
Invariance 3.1 3.2 3.3 3.4
4.
Ordinary differential equations Partial differential equations Problems
One-parameter 2.1 2.2
1
and higher order differential equation8
Infinitesimal versions of y" and y" = F(x,y,y') Examples of the determination of and Determination of the most general differential equation invariant under a given one—parameter group Applications Problems
1
6 9
12 12 18
21
26 26 30 33 36 39
48
49 51
55 58 59 62 66
77 77 79
83 87 92
6.
Linear partial differential 6.1 6.2
equations
Formulae for partial derivatives
Classical groups for the diffusion equation 6.3 Simple examples for the diffusion equation
6.4 Moving boundary problems 6.5 Fokker-Planck equation 6.6 Examples for the Fokker—Planck equation 6.7 Non—classical groups for the diffusion equation Problems
7.
7.4
99
ioi 103 105 109 116 120 123
partial differential equations
135
Formulae for partial derivatives Classical groups for non—linear diffusion
136 140
Non-linear
7.1 7.2 7.3
97
Non—classical groups for non—linear diffusion Transformations of the non—linear diffusion equation Problems
References
146 148 150
159
1 Introduction
Although
a good deal of research over the past two centuries has been devoted
to differential equations our present understanding of them is far from complete.
These notes are concerned with obtaining solutions of differential
equations by means of one—parameter transformation groups which leave the equation invariant.
This subject was initiated by Sophus Lie [1] over a
hundred years ago.
Such an approach is not always successful in deriving
solutions.
However it does provide a framework in which existing special
methods of solution can be properly understood and also it is applicable to linear and non—linear equations alike.
In formulating differential equations
the Applied Mathematician inevitably makes certain assumptions.
Using group
theory these assumptions can be seen to hold the key to obtaining solutions of their equations.
The purpose of this chapter is to present a simple introduction to the subject for both ordinary and partial differential equations by means of simple familiar examples.
For ordinary differential equations comprehensive
accounts of the subject are given by Cohen [2], Dickson [3], Page [4] and more recently Bluman and Cole [5] and Chester [6].
For partial differential
equations the reader may consult Bluman and Cole [5] and Ovsjannikov [7] where additional references may also be found.
1.1
ORDINARY DIFFERENTIAL EQUATIONS
In order to illustrate some of the ideas developed in these notes we consider a simple example.
It is well known
differential equation
that
the 'homogeneous' first order
2
2
dx
(11)
xy
can be made separable by the substitution
u(x,y) =
y/x
and the resulting
solution is given by
=C, where
(1.2)
denotes an arbitrary constant.
C
We might well ask the following
questions:
Why does the substitution
Question I
equation for Question 2
u(x,y) =
y/x
lead to a separable
u ?
How do we interpret the degree of freedom embodied in the
arbitrary constant
C
in the solution?
Answers to these questions can be provided within the framework of transformations which leave the differential equation unaltered.
Consider
the following transformation,
x1ex,
y1ey,
C
where
C
is
C
(1.3)
an arbitrary constant.
We notice that (1.1) remains invariant
under (1.3) in the sense that the differential equation in the new variables x1
and
is identical to the original equation, namely
y1
2
2
—= x1 +y1 dy1
x1y1
dx1
(1.4)
.
Moreover we see that (1.3) satisfies the following: gives the identity transformation
(1)
c = 0
(ii)
—c
characterizes the inverse transformation
(iii)
if
x2 =
e6x1,
y2 = e6y1
x1 =
x,
y1 =
y, y =
x =
eCy1,
then the product transformation is also a
member of the set of transformations (1.3) and moreover is characterized by the parameter 2
c+5,
that
is
x2 = e
c+5
x, y2 = e
c+cS
y.
A transformation satisfying these three properties is said to be a one—
parameter
group
of transformatione.
We
observe that the usual associativity
law for groups follows from the property (iii).
With this terminology
established we might answer the above questions as follows: Answer 1 u
because
u(x1,y1) =
The substitution
y/x
leads to a separable equation for
is an invariant of (1.3) in the sense that
u(x,y)
u(x,y)
since, yl
u(x1,y1) =
u(x,y) =
= u(x,y)
—=
(1.5)
,
and it is this property which results in a simplification of (1.1).
In
general we shall see that if a differential equation is invariant under a one—parameter group of transformations then use of an invariant of the group results in a simplification of the differential equation.
If the
differential equation is of first order then it becomes separable while if the equation is of higher order then use of an invariant of the group permits a reduction in the order of the equation by one. Answer 2
From (1.2) and (1.3) we see that we have 2
y1
— =C xl
log x1 —
+ c
(1.6)
,
so that the degree of freedom in the solution (1.2) resulting from the arbitrary constant
C
is related to the invariance of the differential
equation (1.1) under the group of transformations (1.3) which is characterized That is, the transformation (1.3) permutes
by the arbitrary parameter
c.
the solution curves (1.2).
In general we shall see that for every one—
parameter group in two variables there are functions
u(x,y)
and
v(x,y)
such that the group becomes u(x1,y1) =
u(x,y)
,
v(x1,y1)
=
v(x,y)
+ c
.
(1.7) 3
Moreover if a first order differential equation is invariant under this group then in terms of these new variables =
4(u)
u
and
v
it takes the form,
(1.8)
,
and consequently has a solution of the form
v +
=
C
(1.9)
,
for appropriate functions
and
In order to give the reader some indication of the usefulness of the above we consider the following non—trivial equation, (1.10)
This is an Abel equation of the second kind (Murphy [8], page 25) which we see is not readily amenable to any of the standard devices.
However the
equation is clearly invariant under the group
x1ex, C
y1=e—Cy,
and therefore we choose
(1.11)
u(x,y) = xy
as the new dependent variable and the
differential equation (1.10) becomes, (1.12)
which can be readily integrated.
It is worthwhile emphasizing that not all
equations can be solved in such a simple manner. —
Consider for example,
(1.13)
,
which arises in finite elasticity (see Hill [9]).
This equation is again an
Abel equation of the second kind but in this case there is apparently no simple group such as (1. 11) which leaves the equation invariant.
In this general introduction it may be appropriate to mention here possible research areas for which group theory has not yet been applied.
4
The reader
might well like to bear these problems in mind with a view to developing results in these areas. Research area 1
Differential—difference equations.
It is well known that formal solutions of linear differential—difference equations, for example
-y(x—x0)
=
where
(1.14)
,
is a constant, can be expressed as
x0
r
y(x)
C.e
=
—W4X
(1.15)
,
j
are arbitrary constants and
where
denote the roots of
w =
If the equation is non—linear then there are no such general methods of Consider for example Hutchinson's equation which can be written as
solution.
4x)
=
y(x)[l
—
y(x—x0)]
(1.16)
.
This equation arises in theory of populations (see Hutchinson [10]).
What are
the implications of group theory, if any, for equations of this type?
(See
problems 19 and 20 of Chapter 4). Research area 2
Differential equations invariant under transformations which
cannot be characterized as one—parameter groups. A differential equation occurring in fluid dynamics is Tuck's equation (see Tuck [11]),
— 2
dt2
dx + dt
(5+3x)
+ 3x(1—x)
4x(1+x)(dtJ
It can be verified that if the usual way we let y
3x(1-.x)
dx
(1+x)
y =
+ 2y
x(t)
dx/dt
(117)
(1+x)
is a solution then so is
x(t)'.
If in
then (1.17) becomes
+ (5+3x) 4x(1+x)
2
'
which is again an Abel equation of the second kind.
(1 18)
From the invariance
5
property of (1.17) we can deduce that (1.18) remains invariant under the transfo rmat ion
(1.19) which clearly cannot be characterized as a one—parameter group.
Can we use
such invariance properties to determine solutions of differential equations? Research area 3
Abel equation of the second kind;
As we have already Indicated one of the most frequently occurring differential equations which is not always amenable to standard devices is the Abel equation of the second kind.
The general equation can be expressed
in the form (see Murphy [8], page 26) y
Equation
dx
=
a(x)
(1.20)
+ b(x)y .
(1.20) with arbitrary functions
a(x)
and
b(x)
would appear to be
a problem worthwhile studying.
1.2
PARTIAL DIFFERENTIAL EQUATIONS
Unlike ordinary differential equations the success of the group approach for partial differential equations depends to a considerable extent on the accompanying boundary conditions.
That is, the group approach is only
effective in the solution of boundary value problems if both the equation and boundary conditions are left unchanged by the one—parameter group.
For the
most part we confine our attention to specific differential equations rather than boundary value problems.
For any particular boundary value problem we
should always first look for any simple invariance properties.
These may be
more apparent from the physical hypothesis of the problem rather than its mathematical formulation.
If no such invariance
can be found and if the
problem merits a numerical solution then the group approach might still be
6
relevant as a means of checking the numerical technique with artificially imposed boundary conditions which permit an exact analytic solution. As an illustration we consider a boundary value problem for which both the partial differential equation and the boundary conditions are invariant under a simple one—parameter group.
Consider the problem of determining the source
solution for the one—dimensional diffusion or heat conduction equation for c(x,t), namely 2 3c
=
(t
—°°
>
x
<
< 00)
(1.21)
.
3x
The source solution of (1.21) is a solution which vanishes at infinity for all times and initially satisfies c(x,O) = where
c0ô(x)
(1.22)
,
is a constant specifying the strength of the source and
c0
ô(x)
is
We observe that both of (1.21) and (1.22)
the usual Dirac delta function.
are left unchanged by the transformation c
x1 = e x where
£
t1
,
denotes
2c
= e
t
c1 = e
,
—c
c
(1.23)
,
an arbitrary constant and we have made use of the
elementary property of delta functions, cS(Ax)
=
,
for any non—zero constant
(1.24)
.
Thus if
A.
and (1.22) then we have also
c1 =
is the solution of (1.21)
c =
Clearly this is the case if
has the functional form
4,(x,t)
for
=
t
_1
some function
(1.25)
,
ij
of the argument indicated.
Upon substituting (1.25)
into (1.21) we obtain the ordinary differential equation
+
+
= 0
,
(1.26) 7
where
E
and primes indicate differentiation with respect to
denotes
Equation (1 .26) can be reduced to the confluent hypergeometric equation (see
Murphy [8], page 321).
However the solution vanishing at infinity can
be readily verified to be simply, = Ae
A
where
(1.27)
,
denotes an arbitrary constant.
This constant is determined from
(1.22), namely
c(x,t)dx = c0
(1.28)
.
(1.27)
From this equation, (1.25) and
we find that the required solution of
the boundary value problem (1.21) and (1.22) becomes
c(x,t)
=
c0
e
—x2/4t
(4irt)
This
½
(t
>
solution is of course well known.
—°° < x
< co)
.
(1.29)
For our purposes it firstly serves as
a specific non—trivial boundary value problem for which the differential equation and boundary conditions are both invariant under a one—parameter group.
Secondly it serves to illustrate that knowledge of a one—parameter
group leaving the equation invariant enables, at least in the case of two independent variables, the partial differential equation to be reduced to an ordinary differential equation.
For more independent variables knowledge of
a group leaving the equation unchanged reduces the number of independent variables by one.
In these notes we give the general procedure for determining the group such as (1.23) whIch leaves a specific equation invariant.
We also give the
general technique for establishing the functional form of the solution such as that
8
given
by (1.25).
PROBLEMS
1.
Determine in each case the constants
a
and
8
such that the one—
parameter group ac x1 = e x
y1 = e
,
y
leaves the following differential equations invariant.
Use an invariant
of the group to integrate the equation.
(a) E =
3,12
+ By3
and
(A
B
are constants)
(b)
+ By = 0
(c)
x(A + xytl)
2.
Verify that,
(A, B
and
n
are constants)
y1=e —2c y,
x1=x+C,
is a one—parameter group of transformations and hence integrate the
differential equation (1 — 2x —
3.
logy)
2y = 0
Integrate the differential equation (x—y)2
=
(A
is a constant)
by observing that the equation admits the group
x1x+C, 4.
Given that
p(x)
y1=y+C. is a solution of the linear differential—difference
equation (1.14) show that p(x—x0) y(x)
p(x)
=
is a solution of the non—linear differential—difference equation dy(x)
=
y(x)[y(x)
—
y(x—x0)] 9
5.
Show that the transformation e
y(x)=
x
f(eXX0)
reduces equation (1.16) to the differential equation
f(t)
df(t) —
f(At)
dt
where 6.
t=e x-xO and X=e -xO
Show that with
w =
3(1-x)
=
dx
(1+x)
y/x
the differential equation (1.18) becomes
+
+
(1-x) (1+x)
Show further that the substitution 2
4
s =
(1—x)/(1+x)
2
(s —1)
w
2
dw SW — = 35 + 2w + — ds 4
and observe that the transformation (1.19) becomes Si
7.
yields
w1 =
—w
and
= —S.
Observe that the partial differential equation (1.21) remains invariant under the transformation c
x1 = e x
t1
,
= e
2€ t
,
c1 = c
so that the equation admits solutions of the form
Deduce the ordinary differential equation for
c(x,t) where 8.
A
and
Continuation.
4
c(x,t) =
and hence show that
2
"4dy + B
= A,j'
B
denote arbitrary constants.
For the non—linear diffusion equation
D(c)j—) where the diffusivity
D
is a function of
c
only, use the one—
parameter group and functional f6rm of the solution in the previous problem to deduce the ordinary differential equation 10
+ dD(4)
9.
Continuation.
+
For the case
= 0
D(c) = c
show that
the
ordinary
differential equation of the previous problem remains invariant under the group
41e 2c
c
=
and that with
the equation reduces to the Abel equation of
the second kind
+ p2 + p =
p = 10.
+
+
and
log
y =
Show that the singular solution
E.
corresponds to the solution
Continuation.
4(e)
constant.
Show that the transformation
p =
reduces the Abel
equation in the previous problem to q
+ fJq +
+
= 0
which has the same form as equation (1.20). 11.
By calculating the quantity, ac1
2 ac1
at1
show directly, using the chain rule for partial derivatives, that the classical diffusion equation (1.21) remains invariant under the following transformations, (i)
x1 = x + et,
t1 = t,
c1c
exp[_ —
2
1
(ii)
x1 = (1—ct)
=
(1—ct)
—
4(1—ct)
11
2 One-parameter groups and Lie series
In
this chapter we introduce the concepts of one—parameter groups and Lie For one—parameter groups there are two important results.
series.
the method of obtaining the global
Firstly
of the group from the infiniteBimal
Secondly the existence of canonical
coordinateB
for the group.
For
Lie series the important and remarkable result is the so—called Coninutation
These concepts are discussed below.
theorem.
2. 1
ONE—PARAMETER TRANSFORMATION GROUPS
In the (x,y) plane we say that the transformation =
is
f(x,y,c)
group of
a one-parameter
(i)
(identity) the value x =
(ii)
f(x,y,O)
(2.1)
,
tran8forn?ation8
y
=
the following properties hold:
characterizes the identity transformation,
c = 0
,
if
g(x,y,O)
(inverse) the parameter —c characterizes the inverse transformation, x =
(iii)
g(x,y,c)
y1 =
,
f(x1,y1,—c) x2 =
(closure) if the
,
y = g(x1,y1,—c)
f(x1,y1,cS)
two transformations
,
y2 = g(x1,y1,cS)
then the product of
is also a member of the set of transformations
(2.1) and moreover is characterized by the parameter x2 =
f(x,y,c+5)
,
c+tS
,
that
is
y2 =
Again we remark that the usual associativity law for groups follows from the closure property.
12
(a)
x1 = x ,
(b)
x1 = eCx
Some simple examples of one—parameter groups are: y1 = y + c ,
y1 = eCy
(translation group), (stretching group),
y sin c
x1 = x cos c
(c)
,
y1 = x sin c + y cos c
x1 = x
In order to show (c) forms a group we have inmiediately
when
=0
c
and
y = y1 cos c — x1 sin c
characterizes
the inverse and (ii) is satisfied.
x2 = x1 cos
— y1 sin d
(x
x2 =
and
cos c — y sin
=
x cos(c+5) — y
=
(x cos c — y sin c)sin
=
x sin(c+5) + y cos(c+tS)
y1 = y
so that —c
For (iii) we see that if
y2 = x1 sin d + y1 cos d
c)cos
and
group).
On inverting we obtain
so (i) is satisfied.
x = x1 cos c + y1 sin c
(rotation
then we have
5 —
(x
d +
(x sin c + y cos c)cos d
sin c + y cos c) sin d
sin(c+tS)
and
and therefore (iii) is satisfied. The functions
=0
c
and
g(x,y,c)
are referred to as the global form
If for small values of the parameter
of the group. since
f(x,y,c)
we expand (2.1) then
c
gives the identity we have dx
x1 = x + c
+ 0(c2)
y1 = y + c
,
+ 0(c2)
where
If
indicates
0(c2)
(2.2)
,
c=O
c=O
terms involving only and
we introduce functions
and higher powers of
c2
c.
by
dy
dx c
=
= fl(x,y)
,
(2.3)
,
c=O
c=O
then we obtain x1 = x +
+ 0(c2) ,
y1 = y + cfl(x,y) + 0(c2) ,
and (2.4) is referred to as the infinitesimal form of the group. property
form of
of one-parameter tran8formatwn groups is the
group we can deduce
that
(2.4) The crucial
given the infinitesimal
the global form by integrating the following
13
of differential equations,
autonomous dx1
dy1 =
subject
=
,
(2.5)
,
to the initial conditions,
x1 = x
y1 = y
,
e = 0
when
(2.6)
.
A proof of this result can be found in Dickson [3].
Here we merely indicate
its validity by means of a simple example.
For the rotation group (c) we have dx1
dy1 ,
e = 0
and therefore on setting F(x,y) =
—y
=
,
we have from (2.3)
x
Thus in this case we need to integrate dy1
dx1 ,
subject
Introducing the complex variables
to the initial conditions (2.6).
z=x+iy and z1=x1+iy1 weobtain dz1
= iz1 and thus log z1 =
ic
+ log z
where we have used the initial conditions (2.6). z1 = e1Cz
Imaginary parts of rotation group (c). r =
(x
then we have
2
2½
+ y ) z =
On equating real and
we can readily deduce the global form of the
If we introduce polar coordinates
,
re 10
0 =
tan
and from
—1
(y/x)
z1 = e
ic
z
we see that the global form of
the rotation group (c) can be written alternatively as 01 = 0 + c. 14
That is, in terms of
defined by
(r, 0)
(r,0)
r1 =
r
and
coordinates the rotation group has
the appearance of the translation group.
This is a general property of one-'
parameter transformation groups. For
any given one-parameter transformation group (2.1) there extBtB
and v(x,y)
functwn8 u(x, y)
u(x1,y1) The function
u(x,y)
=
auch that the global form of the group becomes
v(x1,y1)
,
=
v(x,y)
(2.7)
+ c .
is said to be an invariant of the group while together
u(x,y)
are referred to as the canonical coordinateB of the group.
(u,v)
Methods for finding
—=
From (2.5) we obtain
dx1 dy1
(2.8)
n(x1,y1)
which we suppose integrates to yield
u(x1,y1) —
constant
so that from the
Initial conditions (2.6) we deduce the first equation of (2.7) and
Alternatively
known.
eliminating
c
u(x,y)
from
and (2.1)2.
We note that if
and suppose that from
u(x1,y1) =
u(x,y), namely a =
u(x,y)
we can deduce the explicit relation
Now for the purposes of integration in (2.5),
y1 =
is an
u(x,y)
In the integration of (2.8) let
v(x,y).
is
may be deduced directly from (2.1) simply by
invariant then so also Is any function of Method for finding
u(x,y)
a
is a
constant and from (2.5)i we have dx
(2.9)
.
=
If for some constant fX
x0
we define
by
dt
(2.10)
•
=
0
then from (2.6) and (2.9) we can deduce (2.7)2 where and hence
v(x,y)
v(x,y) =
Is known.
15
Example 1
Consider,
X1
(1+ex)
(1+cx)2y
=
'
(2.11)
.
The reader should verify that (i), (ii) and (iii) of the definition of a one— parameter are Indeed satisfied. x1 = x — cx2 + 0(c2) ,
so that from (2.4)
For small values of
and
fl(x,y) =
differentiatIng (2. 11) with respect to
-
(1+ex)
2
2xy.
Alternatively on
we have
c
dy1
2
= —x1
we have
y1 = y + 2cxy + 0(c2)
= —x2
dx1
c
-i-— = 2(1+cx)xy
,
and (2.5) confirms the given expressions for
=
2x1y1
E(x,y)
and
(2.12)
.
fl(x,y).
From
(2.12) we have X1
dx1 dy1
u(x,y) = x2y
which on integrating gives v(x,y) = x
we see that
-1
as an invariant while from (2.12)i
satisfies (2.7).
could be deduced directly from (2.11) by eliminating
Example
2
and
by the relations,
v
Show that
r
2
Alternatively the invariant
x y
C.
are related to the canonical coordinates
and
—
u
213
'
—
where the Jacobian is given by
On differentIating (2.7) with respect to dx1
r 16
dy1
+
dx1
=0
,
C
we obtain dy1
+
=1 -s----
,
(2.
14)
where
and
u1
denote
v1
and
u(x1,y1)
(2.5) and (2.14) we have on replacing
n=0
+
(x1,y1)
by
respectively.
From
(x,y),
Ti = 1
÷
,
v(x1,y1)
and (2.13) can be deduced immediately from these relations. A transformation in the
Example 3
(x,y)
plane is area preserving if
a(x1,y1) =
1
(2.15)
.
Show that (2.1) is area preserving if and only if —
u
only.
From (2.4) and (2.15) we can deduce on equating terms of order
c,
(2.16)
ax
From (2.13) and (2.16) we can deduce
1 a(x,y)
=o
and the required condition follows.
lit may be of Interest to note that since the set of area preserving transformations forms a group, the infinitesimal condition (2.16) is precisely the same as the global condition. respect to
c
That is, if we differentiate (2.15) with
we have dy1
dx1
y1
=0
and on multiplying this equation by
dx1
,y1,
we obtain
dy1
÷ ,
' —0
17
so that we have a
dx1
dy1
a
=0
+
(2.
.
17)
On using (2.5) we see that (2.17) is the same condition as (2.16)fl
2.2
LIE SERIES AND THE COMMUTATION THEOREM
Suppose we have the group (2.1) with infinitesimal version (2.4). the differential operator
We define
by
L
(2.18)
.
L =
Now for any function
$(x1,y1)
which does not depend explicitly on
c
we
have
dx
d41
de
dy1
de
ax1
denotes
where
L1
(x1,y1).
From (2.5) and (2.19) we obtain
(2.20)
,
denotes the differential operator
with
(x,y)
replaced by
d3
= L1 (L1($1))
,
—i-
= L1 [L1
(2. 21)
.
then by Maclaurin's expansion we have
•(c) =
=
4(O) +
c=O
de
and thus from (2.20) and (2.21) with
dc
c=O
c = 0
That is, we have
+
+
c=O
we obtain 3
2
=
18
L
Similarly we have
d2
If we let
(2.19)
de
4(x1,y1).
= L1(41) where
ay1
+
+ ...
(2.22)
,
= n=O
and we refer to such a series as a Lie series.
We notice that we can write
(2.22) as =
(2.23)
,
provided we interpret the differential operator
as the series operator,
)
in particular if we take
to be
4(x,y) e
x
cL
y
x
and
y
then from (2.23) we obtain,
(2.24)
.
On combining (2.23) and (2.24) we have the remarkable result, CL
x, e
CL
y) = e
CL
(2.25)
,
which is called the Commutation
theorem of Lie series (see
and Knapp
t121, page 17).
To illustrate (2.24) consider the rotation group (c). differential operator L = —y so that
a
L(x) =
In this case the
is given by
L
a
+ x
r
—y
and
L(y) = x
and the global form of the group can be
deduced from (2.24) using the expansions,
k2k
cos
£
= k=O
(—1) C (2k)!
k2k+l
(—1) C
sin c = k=O
(2k-i-1)i
It is worthwhile noting that using Lie series we can give a formal
Example 4
solution of any autonomous system of differential equations given initial values.
That is, consider =
F(X,Y)
,
= G(X,Y)
,
19
and
X
Y =
a,
at
t =
0.
The formal solution of this initial value
problem is x = etMa
y
,
F(a,8)
(2.26)
,
is defined by
M
where the operator N =
=
+
Consider two simple examples.
Firstly, the single differential equation,
dX
dt In this case we have M = -a2 —a-,
and
-a2,
M(a) =
M2(a) = 2a3
and in general
N(a)=(—1)n n!an+1 Hence
x =
etha =
and thus for
(a) = a
n0
lati <
n=O 1
we obtain the solution,
a 1-Fat
Secondly, consider (2.27)
A and
where
M= so that
20
B
are constants.
In this case we have
M(a) = (Aa + N2(cz) =
K2cz
M(8) = (Ba
,
,
N3(a) = K2(Aa
—
A8)
= K28
+ B8)
= K2(Bcz — A8)
,
M4(8)=K48, M5(cz)
= K4(Acz + B8)
M5(8) = K4(Ba — A8)
,
K = (A2 + B2)½. From these results and (2.26) we can
and so on, where
deduce the solutions, 2KX = [Ka + (Mi + B8) ]eKt + [lca - (Aa + B8) ]eKt 2KY = [K8 + (Ba —
A8)]eKt +
(Ba —
—
which of course could be established by more elementary methods (for example, differentiate (2.27)i with respect to [The
and make use of (2.27)2).
t
following problems which arise in continuum mechanics are useful
exercises in manipulating Lie series.] PROBLEMS
1.
tn cylindrical polar coordinates is
(r,O)
a transformation in the plane
area preserving if a(r1,01)
r r1
a(r,o)
Following the note at the end of Example 3, differentiate this equation with respect to
dr1 +
Hence
4
which
1
dr1 +
= 0
deduce there exists a function 4(r1,01,c) dc
tf
and show that,
c
r1 ao1
'
dc
r1 ar1
does not depend explicitly on r1 =
r,
01 = 0
when
such that
c=
0
C
is
the solution of this system for a one—parameter group with
4
as 21
an invariant, that is =
2.
4(r,0)
In the above problem obtain the one—parameter group given that,
where
Ar20 + BO
(a)
=
(b)
4(r,O) = Ar20 +
(c)
4(r,0) =
A
and
Br2log
sin
(0 +
r
0 cos 0)
denote arbitrary constants.
B
[Answers:
3.
(a)
r1 =
(b)
r1 =
(c)
r1 =
+ +
01 =
,
[r2
=
— Ac cos20]½,
r + Bc
01 = 0
Consider,
log(1 + cx)
x1 =
dx1
cx)y
dy1
and express in terms of
,
Calculate
y1 = (1 +
,
(x1,y1).
Hence or
otherwise deduce that this is not a one—parameter transformation group. 4.
Consider the one—parameter group, x1 =
f(x,y,c)
=
y1 =
g(x,y,c)
= y
x +
+
O(c)
+ cn(x,y) + 0c )
and introduce the operators,
P=L+w, Q=L-w, where
w
is defined by,
3x
Let
4(x,y)
notation
22
and
denote arbitrary functions and agree to use the for the Jacobian, that is
We consider the following Lie series,
e(x,y) = n=O
= n=O
= n=O
=
n0
Show that =
(f)
!
=
,
+
=
(ii)
[Hint: for (ii) start by considering 5.
If we suppose that
Continuation.
(e
Co
eL 4,
e
cL
r
=
fl
c
n=O show that,
(i)
k
= k=O
(ii)
;
=
Hence deduce,
= = n=O
[Hint: for (ii) start by considering P(;,1)
and use (i) and (ii) of
previous problem.] 6.
Continuation.
Observe that in particular,
(x1,y1) = (e
cL
x, e
eL
y) = e
1
and therefore, 3.
2
(x1,y1) =
1
+ ew +
P(w) +
P2(w) + ...
Verify 23
2
= Lw +
L(w) +
L (w) +
2
(,c1,y1)1 =
(ii)
7.
— Lw —
1
Q(w)
In cylindrical polar coordinates
—
(r,8)
r1 =
f(r,0,c)
=
+
+ 0(c2)
=
g(r,0,e)
= 0 +
+ 0(c2)
r
Q
(w) +
consider the one—parameter group,
and introduce operators,
P1=L+w1 , where
w1
w 1
and
w2
ar
ao'
are defined by w
2
r
Suppose that
£. L"(r)
r1 = =
A
r
=
=
n0
= n=O
a(r,0) =
=
Verify,
L(f1)
=
(i)
(ii)
A = 8.
eEl'2l
ContInuation.
Observe from Problem 6,
p=e cP1 1. Suppose that, =
24
r
a(r,0)
r
a(r,0)
P2=L+w2,
Verify,
(j)
(ii)
where
XklJk
n
o
k=O =
P3
! is given by
P3 = L +
+
Hence conclude, r1 a(r1,01) = r
=
[Hint: for (ii) start by considering
3 Invariance of standard linear equations
It
is well known
that
linear differential equations for
y(x)
remain linear
under transformations of the form,
x1 =
f(x,c)
,
y1 =
g(x,c)y
(3.1)
.
Throughout this chapter we consider only transformations (3.1) which we suppose form a one—parameter group such that infinitesimally we have x1 = x +
+ 0(62) ,
(3.2)
y1 = y + cfl(x)y + 0(62) .
We look for groups (3.1) which leave standard linear equations Invariant and deduce the form of the differential equation in terms of canonical coordinates (u,v) (see (2.7)).
initially the reader may well consider this
approach clumsy and irrelevant for such equations.
The object of the
exercise being to demonstrate the group approach in familiar situations with a view to the reader obtaining some insight into the relation between solutions and groups leaving the equation invariant.
Moreover even linear
equations are not always readily solved and the results obtained in sections 3.3 and 3.4 by this method are non—trivial and have not been given previously.
3.1
FIRST ORDER EQUATION
y' + p(x)y =
g(x)
For first order differential equations our primary objective is to introduce new variables such that the equation becomes separable.
For equations
invariant under a one—parameter group of transformations the appropriate new variables are the canonical coordinates (u,v) of the group. this procedure with the standard first
+ p(x)y = 26
q(x)
.
We illustrate
equation,
(3.3)
For convenience we Introduce the function
p (t)
by
s(x)
dt
s(x) = e where
(3.4)
Is some constant.
x0
With this definition the solution of (3.3) Is
known to be given by rx
s(x)y
—
s(t)q(t)dt
= C
(3.5)
,
J
"Co
where
Is a constant.
C
We now deduce (3.5) by finding a group (3.1) which
Invariant,
that Is
dy1
+ p(x1)y1 = q(x1)
Prom this equation and (3. 1) we deduce
+
+
f'(x)P(f)}
which becomes (3.3) provided p(x) =
+
g(x) q(f)
Y =
and
f(x)
f'(x)p(f)
are such that
g(x)
q(x)
=
,
,
g(x)
=
q(f)
From these equations and f(x) = x +
p(f) = we
cE(x) ÷ 0(c2)
+
p(x) +
0(c2)
obtain on equating terms of order
+
+
= 0
+
,
1
q(f)
,
(3.6)
+ cn(x) + 0(c2) , =
q(x) +
+
0(c2),
c
= 0
—
Hence we have n
where
+ C1
=
,
is a constant.
+
+
= C1
,
(3.7)
Thus
27
s(t)q(t)dt
= q(x)s(x) .IX
+
0
(3.8)
C1 —
where
is a further constant.
C2
[Of course in obtaining these results WE Now the
have had to solve an equation of the type (3.3), namely
global form of the group (3.1) is obtained by integrating (see (2.5) and (2.6) dx1
dy1 =
= n(x1)y1
,
subject to the initial conditions
(3.9)
,
x1 =
x,
y1 = y
when
c =
0.
From (3.8)2
and (3.9) we have dx1
dy1 1
= C1 —
—a-
and therefore, by integrating this equation we obtain, y1s(x1) = e
where
C1c
ys(x)
(3.10)
,
is defined by (3.4).
s(x)
From (3.8)i and
we have
q(x1)s(x1)dx1 = dc
(3.11)
,
+
and there are two cases to be considered. Firstly if
*.
C1 # 0
then (3.11) gives
1s(t)q(t)dt +
s(t)q(t)dt
+
and from this equation and (3. 10) we can deduce that our canonical
coordinates (u,v) (see (2.7)) are given by
28
+c
8(x) y
u(x,y) =
+
{c1
v(x,y)
log
=
+
In these coordinates it can be readily verified that the differential equation (3.3) becomes du —= dv
Cu 1
—
1
which is separable and integrates to give
log(1
—
— C1u) =
—
log
C3
is,
that
(1—C1u)e Civ where the
v
C3
C3
is a constant.
arbitrary
constant
if
Secondly
C1 = 0
C
This equation can be reconciled with (3.5) in (3.5) is
(C2 —
then from (3.10)
where
C3)/C1.
and (3.11) we have canonical
coordinates
u(x,y) — s(x)y ,
v(x,y)
—
J—
2jx0
s(t)q(t)dt
and in these coordinates (3.3) becomes du dv
C
Again our equation in canonical coordinates to
separable and can be integrated
give u — C2v =
C
which can also be reconciled with (3.5) where the constant both equations.
C
is the same in
We remark that (3.3) is invariant under other groups, in
addition to those considered here. 29
3.2
= 0
SECOND ORDER HOMOGENEOUS EQUATION y" +
For second order linear homogeneous equations we can without loss of generality (see problem 8) consider the normal form of the equation, namely 2
+ p(x)y = 0
(3.12)
.
We shall assume
and
are two linearly independent solutions of
(3.12) and for convenience we suppose their Wronskian is unity, that is —A
A
—
and
Our obj ective here is to relate
to a one—parameter group (3. 1)
For higher order linear differential equations
which leaves (3.12) invariant.
the use of canonical coordinates (u,v) simplifies the equation to one with constant coefficients. From (3.1) we have
+
= -.g-
f'dx
dx1
and
+
=
dx1
f'
dx
f'
+
-
14
X
f'
Clearly if (3.12) is to remain invariant there can be no term involving On equating the coefficient of
y'
to zero in (3.14) we obtain
y'.
f'(x)/g(x)2
constant, which must be unity if (3.1) is a one—parameter group and therefore we have f'(x) =
g(x)2
.
From (3.14) and (3.15) we find that the differential equation 2
d y1
+ p(x1)y1 = 0 dx1
becomes
30
(3.15)
invariant provided
and thus (3.12) remains
p(f)g4
+
2
—
=
p(x)
(3.16)
.
Equation (3.15) and (3.16) constitute two equations for the determination of From (3.6), (3.15) and (3.16) we find on equating terms of
the group (3.1). order
c, 'I,
=
2
+ p'E =
+2pE
The equation (3.17)2 for
E(x)
0
(3.17)
.
is a formally self—adjoint third order
differential equation (sometimes called anti self—adjoint, see Murphy [8], page 199) with first integral —
which
can
+ pE2 = constant ,
be verified by differentiation.
problem 20
or Murphy [8],
(3.18) It is well known
(see either
page 200) that the general solution of (3.17)2 is
given by = A$12
where is
A, B
+ 2B+1$2 + Cd,22
and
C
(3.19)
,
denote arbitrary constants.
the general solution consider for example =
+
'
=
+
[In order to see that (3.19) =
then we have
+
Observe that from the original differential equation we have =
and
—
substitution
gives zero.]
of this expression and those
for E'
and
into (3.17)2
Further from (3.12) and (3.19) we can deduce
31
+
=
+
+
=
+
+
,
— 2pE
and on substitution into (3.18) we find on using (3.13) that (3.18) becomes
(2EE"
—
= (AC — B2)
+
(3.20)
.
Now the global form of the one—parameter group (3.1) is obtained by solving the differential equations
dy1
dx1
E'(x1) 2
subject to the initial conditions
x1 =
x,
y1 = y
and
e =
0.
We find that
suitable canonical coordinates (u,v) are given by =
v(x,y)
u(x,y) =
E(x) ½ where
is some constant.
x0
f
In terms of (u,v) we find that the differential
equation (3.12) becomes
+
-
= 0
+ But from (3.20) we see that the differential equation finally becomes 2
+ (AC — B2)u = 0
(3.21)
.
dv
Thus for example, if
(AC — B2)
is positive the general solution of (3.12)
is given by
y(x) = E(x)½{C1 cos{K
where
C1, C2
(3.22)
+ C2 sin[K
are arbitrary constants and
K = (AC — B2)
.
We
have therefore
established the relationship between the general solution of (3.12) and the infinitesimal version of the one—parameter group of transformations leaving 32
In a sense, (3.22) is the 'inverse' of (3.19).
(3.12) invariant.
From
(3.19) we see that if we know a group leaving (3.12) unaltered then essentially we know a quadratic relation between the linearly independent solutions of (3.12).
We remark that solutions of (3.12) in the form of (3.22)
have been given previously although the function
appearing in (3.22)
has not been identified with the one—parameter group leaving the differential equation invariant (see for example Coppel [13], page 19). As an illustration of the above, consider the simple Euler equation, 2
i-i + dx
2
4x
2
= 0
In this case the equation is clearly invariant under the group and
y1 = y = x½
so that and
=
x.
we see from (3.19)
u = C1v + C2.
A = 1, B =
C
= 0
and
This expression confirms our
linearly independent solutions since in this case
3.3
= eCx
Since we have linearly independent solutions
q2(x) = x½ log x
therefore from (3.21) we have
x1
u =
and
v =
log
x.
THIRD ORDER HOMOGENEOUS EQUATION y" + p(x)y' + q(x)y =0
We see from problem 12 that for third order linear homogeneous differential equations we can without loss of generality consider the equation 3
+ p(x)
dx
X
+ q(x)y = 0
In this section we suppose that
(3.23)
.
41(x)
and
are linearly independent
solutions of the second order equation
+
y = 0
(3.24)
,
such that their Wronskian is unity.
We are concerned with finding one—
parameter groups of the form (3. 1) which leave (3.23) invariant.
From a further differentiation of (3.14) we obtain 33
= ___g__
dx13
-
+
f'3 dx3
f'3
-
ft
ft
+ 3gf"2
-
dx
f'4
3g'f"2 -
+
+
'f"
+
y
(3.25)
.
f'
ft
If (3.23) is to remain invariant we require the coefficient of
y"
in (3.25)
From this condition we deduce
to be zero.
f'(x) =
g(x)
(3.26)
,
and (3.25) becomes d3y
__! dx1
=
3
L
3
-
+
2
gdx
3
g
3
4dx g
2.&_. -
+ g
5
3
g
4
g
Using this equation and 3 dy1
dy1
i—
+ dx1
+ q(x1)y1
= 0
I
we obtain on multiplying by —
g2
the equation,
+
+
—
4gtg" + p(f)gg' + q(f)g3]y
=
o.
+
+
For invariance this equation must be identical with (3.23) and therefore f(x)
and
g(x)
as well as satisfying (3.26) must also satisfy
-
+ p(f)g2
=
p(x) (3.27)
+
-
+ p(f)ggt
From (3.6) and (3.26) we obtain 2E"t +
+
q(f)g3 =
=
q(x)
while from (3.27) we have
+ ptE = 0 (3.28)
F" + pE" +
+ q'E = 0
and these equations are only consistent if
34
=D
—
where then
(3.29)
,
Clearly If (3.23) is self—adjolnt (see problem 14)
is a constant.
D
p' = 2q
and (3.29) is trivially satisfied with the constant
However if (3.23) is not self—adjoint then as well as satisfying (3.29). general solution of
zero.
must be both a solution of
In terms of solutions of (3.24) the
is given by
+
=
Cx)
D
+
(3.30)
,
and we have —
where
A, B
= 4(AC — B2)
+ and
denote arbitrary constants and we have used the fact that
C
the Wronskian of for a given q(x) =
where
E(x)
I
is unity.
and
p(x)
(3.31)
,
we need to assume
+
If (3.23) is not self—adjoint then is given by
q(x)
D
(3.32)
,
X
is given by (3.30).
From the equations dx __.i
— r(
dy )
— r'( x1 )
we find that suitable canonical coordinates (u,v) are given by
v(x,y)
,
u(x,y) =
=
for some constant
x0.
f
In these coordinates the differential equation (3.23)
can be shown to become 3
+
—
+ V
+
+
+
— 0
which on using (3.28)i, (3.29) and (3.31) finally becomes
35
3
(3.33)
V
dv
Thus for third order differential equations (3.23) which are not self— adjoint, we can for a given function
p(x)
obtain a one—parameter group
(3. 1) which leaves the equation invariant provided the function
the form (3.32) for suitable constants
A, B, C
and
D.
q(x)
has
If this is the case
then the solution of (3.23) reduces to solving the third order linear equation (3.33)
the
with
If (3.23) happens to be seif—adjoint then
constant coefficients.
general solution can be obtained in the usual way for such equations from
the solutions of (3.24) (Murphy [8], page 200). For a simple illustration based on the example at the end of the previous section, suppose that
q(x)
from (3.32)
q(x)
=
x3{[A
for some constants
p(x) =
x2
then
41(x) = x½,
= x½ log x
and
must be given by + 2B log x + C(log x)2] 3D A, B, C
and
D.
ii
If this is the case (3.33) can be
solved in a straightforward manner and hence the solutions of (3.23) can be deduced.
3.4
FOURTH ORDER SELF-ADJOINT EQUATION y" +
[p(x)y']'
+ q(x)y = 0
From problems 16 and 17 we deduce that the general fourth order seif—adjoint equation
can be taken as +
In
+ q(x)y =
this section we suppose
(3.34)
0 .
41(x)
and
42(x)
are linearly independent
solutions of (3.35)
such that their Wronskian is unity.
36
On a further differentiation of (3.25) we find that the coefficient of y" Is zero provided, 2
g(x)3
f'(x) =
(3.36)
,
in which case we obtain, 4
(
1
g
dx14
g
3g2dx2
g2
(3.37)
From the equation 4 dy1
d
+
dy1
p(x1)
i—
+ q(x1)y1
= 0
and (3.37) we deduce that if the resulting equation is to be identical with (3.34) then we require
and
f(x)
to satisfy
g(x)
p(x) =
p(f)
gg'
[gig" -
(3.38)
+
+
g
g' g"'
3
g
2
-
32 g
+
g '2g" 3
g
3
g
From (3.6), (3.36) and (3.38) we have,
+
+
= 0 (3.39)
+ pE:']' +
and
=
+ 2q'E = 0
The two equations (3.39) are consistent only if
D
where
D
is a constant.
Thus in general for a given
(3.40)
p(x)
we need to 37
q(x)
assume
q(x) =
is given by
9p(x)2
D
(3.41)
,
dx
where
is a solution of
and has the general form (3.30) where
are linearly independent solutions of (3.35).
and
constants
A, B
and
are as in (3.30) then we have
C
+
—
Moreover if the
= (AC — B2)
(3.42)
.
The global form of the one—parameter group (3.1) can be deduced from dy1
dx1
= E(x1)
subject
=
,
to the initial conditions
x1 =
x,
y1 = y
when
c =
0.
Suitable
canonical coordinates (u,v) are given by =
v(x,y)
u(x,y) =
X0
where
x0
Making use of the result given in problem 18
is some constant.
3
(replacing
a(x)
and
with
B'(x)
and
E(x)1
respectively) we find
that in terms of (u,v) the differential equation (3.34) eventually becomes 4
2
+ 10(AC — B2) dv
+ [D + 9(AC — B2)2]u = 0
where we have made use of
(3.40) and (3.42).
is given by (3.41) then (3.34) constant coefficients. independent solutions
(3.43)
,
dv
Thus provided
q(x)
can be reduced to a linear equation with
Evidently the approach presupposes that the linearly and
+2
of the associated equation (3. 35) can be
readily obtained.
As an illustration suppose that solutions of (3.35) are
38
+1(x) =
1
p(x)
and
Is zero.
+2(x) =
x.
The linearly independent Hence
E(x)
is of the
form,
= A + 2Bx + Cx2 and
the
above approach is effective provided
q(x)
is given by
D
q(x)=
24'
(A + 2Bx + Cx )
for
some constants
A, B, C
D.
Consider for example
A =
0,
B
—½
C=D=1. Wehave
and
= x(x—1)
and
and
(3.43)
=
[x(1—x)f4
has the general solution
= (C1 + C2v)e
where
q(x)
,
C1, C2, C3
2
and
+ (C3 + C4v)e C4
are constants.
The solution of the original
equation can now be readily deduced.
[In the following problems
s(x)
is assumed to be defined by (3.4).
Also for
problems 4,5,6 and 7 a further arbitrary constant could be introduced into the condition restricting the coefficients.
In these problems we have assumed
that this constant has been absorbed into the constant defines
x0
in (3.4) which
s(x).]
PROBLEMS 1.
For Bernoulli's equation, + p(x)y =
q(x)y"
(n # 1)
show that =
q(x)s(x)
1-n
{(1_n)Ci
L0
+
C2}
= C1 —
39
2.
Continuation.
If the constant
C1
is non—zero deduce that suitable
canonical coordinates (u,v) are s(x)y
u(x,y)
=
+
v(x,y)
(1-n)C1
=
+ C2}
log
and therefore the differential equation becomes du
n-i
—C1).
Integrate this as a separable equation to obtain = C3
(1 —
and 3.
hence
deduce the solution of the original equation. If the constant
Continuation,
u(x,y)
s(x)y
=
,
v(x,y)
C1
is zero show that
= 1
2jx0 and that the differential equation becomes du dv
xi
2
Integrate to obtain u
and
with
40
1-n
—
C2(i—n)v =
C4
show that the same solution is obtained as in the previous problem C4 = (C2 — C3)/C1.
4.
Show that the generalised Riccati equation
E + p(x)y = q(x)
+ r(x)y2
remains invariant under (3.1) provided
r(x) =
q(x)s(x)2.
If this is the case show that —p(x)
1
—
ThX1
q(x)s(x)
q(x)s(i)
and that suitable canonical coordinates are, rx
u(x,y)
=
s(x)y
v(x,y)
,
s(t)q(t)dt
= 0
Hence show that the differential equation becomes du
2
and therefore the solution of the original equation is
y(x)
5.
=
+
(x)
c]
Show that the Abel equation of the first kind,
+
p(x)y
=
q(x)
+ r(x)y3
is invariant under the group of the previous problem provided r(x) =
q(x)s(x)3.
—1+u du dv
6.
Show that the differential equation becomes
3
Show that
+ p(x)y =
q(x)
+ r(x)log y
is invariant under (3.1) provided
q(x) =
r(x)log
s(x).
Show that, =
/
1
r(x)s(x)
= '
/
r(x)s(x)
'
41
u(x,y) = s(x)y
,
s(t)r(t)dt
v(x,y) =
and that the differential du
7.
=
log
0
equation becomes
u
Verify that the differential equation + p(x)y = q(x)ym +
admits
r(x)y"
the group (3.1) provided
r(x) =
q(x)s(x) n—rn
.
case deduce that, =
1
q(x)s(x)
fl(x)
1—rn
—p(x)
=
q(x)s(x) rx
u(x,y)
=
s(x)y
,
v(x,y)
I
=
I
s(t) 1-rnq(t)dt
JxO
and that the differential equation becomes du ——U dv
8.
rn
+U n
Show that the linear honxgeneous second order equation
can
be reduced to normal form either by,
(i) changing the dependent variable to rx -½1
y* where
a(t)dt
y=e
in which case we have
+ {b(x) or
42
(ii)
a(x)2
-
changing the independent variable to
= 0
x* where
If
this is the
(5 —I
fx
x*=j
a(t)dt
ds,
u
e
JxO
in
which case we have b(x)
dx*
9.
For the equation,
Continuation.
g
2
dx
0
dx* dx
3x 2
2
(1—x
(1—x )
2 )
show that the reductions to normal form given in the previous problem give rise to the following equations,
+
+
3
= 0
(')
dx
(ii) 10.
(1—x )
+
dx*
n(n+2)
(1—x
)
= 0
(1+x* )
With the notation of section 3.2 consider the non—homogeneous equation 2
+ p(x)y
q(x)
dx
If this equation is to remain invariant under the same group which leaves the homogeneous equation unaltered then show that
q(x)
must be
given by 3 2
q(x) = q0
where
q0
is a constant.
Hence show that the equation corresponding
to (3.21) becomes 2 du
2
+ (AC—B )u = q0
dv
43
11.
satisfies
If
,
where
K
satisfies
w(x) =
is a constant, show that
the non-
linear second order equation d2w
+ p(x)w
dx 12.
K2 =
The following three operations leave a linear differential equation linear,
y =
where
a(x)y*
changing the dependent variable to
(ii)
changing the independent variable to
(iii)
multiplication of the equation by a non—zero function
Show that by choosing
B(t) 3A(t)
=e
a(x)8'(x)
the
x*
where
x* = y(x)
such that
dt
0
,
y(x)
3
=
A(x)] —1
general linear third order equation + B(x)
A(x)
can
y
and
a,
fX
+ C(x)
+ D(x)y = 0
(*)
,
be reduced to an equation of the form, +
13.
y*
(i)
+ b(x)y =
a(x)
Continuation.
.
(**)
A second order linear equation is self—adjoint if it is
of the form
t[P(x) Show
+ Q(x)y =
that any second—order
self—adjoint by any previous problem.
44
0
linear differential equation can be made
one of the operations (i), (ii) and (iii) of the
14.
Continuation.
A third order equation is formally seif—adjoint (or
anti seif—adjoint) if it has the form (Murphy [8], page 199)
+
t{P(x)
+ Q(x)
+
=0
Show that the general equation (*) is self—adjoint if and only if
3dA
B(x)
ldf
D(x)
,
1dB
Make use of this result and the reduced equation (**) to show that no
combination of the operations (1), (ii) and (iii) of problem 12 can make a third order equation seif—adjoint unless it is originally self—adjoint. 15.
Continuation.
Show that if a third order equation is self—adjoint then
it remains seif—adjoint under (i), (ii) and (iii) of problem 12 provided
16.
is a constant multiple of
Cx(x)
Continuation.
y(x)8'(x)
Show using the operations of problem 12, that the general
fourth order equation
A(x) —f + B(x) can
+ E(x)y = 0
,
(+)
be reduced to one of the form, + a(x)
+ b(x)
dx
X
dx
by choosing
8
and
y fX
2
cz(x) 8'(x) 17.
+ D(x)
+ C(x)
Continuation.
3
= e
+
c(x)y
= 0
,
(++)
to be such that
B(t)
0
d ,
y(x) = [cz(x)8'(x)4A(x)] —1
A fourth order equation is self—adjoint if it has the
form
-44P(x)
+
t[Q(x)
+ R(x)y
= 0
Show that (+) is self—adjoint if and only if
45
B(x) = 2 18.
12 can
+
make (++) self—adjoint unless it is seif—adjoint
If (++)
originally.
—
Show that no combination of (i), (ii) and (iii) of
ContinuatIon.
problem
D(x) =
,
is self—adjoint show that these operations give
to another self—adjoint equation provided a(x)
rise
is a constant
multiple of y(x)13'(x).
[For the second part, if (++) the
is self—adjoint we have
b(x) =
a'(x)
and
equation becomes
+
dx*
(+4+)
+ R*y* = 0 ,
dx*
where, p*
= cx28'3
Q* =
cx28"+ 2cia'8" +
(ci" +
R* =
where
cz, 8,
a
and
acx"
c
+
(4cta"
Continuation.
2cx'2 + acx2)8'
a'ci'+ ccx)
are all functions of
differentiation with respect to
19.
—
x
x, primes denote
and we have taken
If in the previous problem the functions
are such that
iai
a
show that (4++) admits the factorization L2[A3L2y*] = 0
where
L2
is the second order operator defined by
L2y*
where
46
+ A2y*
dx*k A2
and
A3
are given by
y = a(x)
and
c(x)
Acx28' 20.
,
Verify by differentiation that the third order self—adjoint equation of problem 14 admits the first integral P(2yy" If
y'2)
+ P'yy' +
and
•1(x)
=
constant
are linearly independent solutions of the second
order equation + 2P'
and if
is given by
y(x)
A, B
where
and
P(2yy" — where
denote arbitrary constants then deduce that
C
y'2)
P'yy'
+
=
4(AC—B2)Pw2
+ Qy2
is the Wronskian of
w(x) U)
+ Qy = 0
d
namely
and
cp1
—
Hence conclude that this expression for
y
gives the general solution
to the seif—adjoint equation of problem 14. 21.
In the notation of section 3.3 deduce from (3.26) and (3.27) the equation,
dP(x))
[q(f) —! dP(f))g(x)3 = —
f
Hence deduce the condition (3.29). 22.
In the notation of section 3.4 deduce from (3.36) and (3.38) the equation,
-
—
{q(f)
—
.1.
=
— 9p(x)2 —
3
Hence deduce the condition (3.40).
47
4 First order differential equations
In
this chapter we discuss Lie'B fundamental problem (see Lie [1]) of finding
a one—parameter group which leaves a first order ordinary differential equation unaltered.
That is, for a given
F(x,y)
we wish to determine a one—
parameter group, x1 = x +
+ 0(c2)
(4.1)
y1 = y + cfl(x,y) + 0(c2)
,
such that the differential equation, =
F(x,y)
remains invariant.
(4.2)
,
This problem is by no means solved.
Much of the
literature is concerned with the alternative problem of finding differential equations which are left invariant by a given one—parameter group.
For this
aspect the reader should consult the standard tables of differential equations and their associated groups (see for example either Dickson [3], page 324 or Bluman and Cole [5], page 99).
We shall also consider the alternative
problem but with a view to situations not previously discussed.
For the
fundamental problem we highlight the role of singular and special solutions of (4.2) and we refer the reader to the discussion given by Page [4](page 113). Integral curves of (4.2)
z(x,y) =
constant,
evidently satisfy the first
order partial differential equation az
r + F(x,y)
= 0
.
(4.3)
We consider the invariance of (4.3) under a one—parameter group in the three variables
(x,y,z)
which we relate to integrating factors of (4.2).
sense this result provides a generalization of Lie's famous result for
48
In a
integrating factors (see problem 1).
This section deals briefly with the
group approach to partial differential equations and therefore the reader is perhaps best advised to avoid it until familiar with the material on partial differential equations described In
subsequent chapters. In the final
section of this chapter we attempt the solution of Lie's fundamental problem. Since the two functions
are not completely determined
and
by the single constraint (4.6) Lie's problem is rather to propose a second independent constraint on the group (4.1) which is in some sense compatible wIth (4.6) so as to simplify the subsequent analysis.
Here we propose that
the assumption that (4.1) Is area preserving may be such a constraint. Although the results obtained are by no means conclusive, different forms of Lie's problem are generated which at least convey some insight into the fundamental difficulties associated with the problem.
4.1
INFINITESIMAL VERSIONS OF PROBLEM
y'
AND
y' =
We calculate the infinitesimal version of
y'
F(x,y)
AND THE FUNDAMENTAL
as follows.
From (4.1) we
have,
dy 1
dx1
+0(c2) —
dx +
and on dividing through by d 1
—
dx1 —
dx 1
dy)
dx + dx
ay dx)
+
+
we obtain,
+
2
ay dx
Hence on using the binomial theorem for the denominator we have dy =
where
+ c'rr(x,y,y') + 0(c2)
rr(x,y,y')
,
(4.4)
is given by
49
dx
and this
is the infinitesimal version of
y'.
If (4.1) leaves (4.2) invariant then from (4.4), dy1 =
and
F(x1,y1) =
F(x,y) +
+
+ 0(c2)
we obtain
+ crr(x,y,y') =
F(x,y) +
cfr
and therefore from the terms of order
9x
+ 0(c2)
+ ii
c
we have
(4.6)
,
9y
where we have used (4.1) and (4.5).
differential equations determine two functions The functions
is
for
that E(x3y)
and
(4.6) is satisfied and
Lie 's fundamental problem for first order
a given
and n(x,y)
n(x,y)
can
F(x,y)
how can we systematically
such that
(4.6) is
satisfied.
be completely arbitrary provided
Equation (4.6) always admits the solution
#
However, this solution does not serve our purpose since in this case
=
when we come to deduce the global form of (4. 1) we need to solve dx1
dy1 =
=
,
and thus we are led back to our original problem (4.2). in the following section, it is also evident that
=
Further from (4.8)
is not an
acceptable solution of (4.6). If
and
fl(x,y)
are known functions then we show in the following
section that the condition (4.6) reduces to the existence of an integrating factor for the differential equation (4.2). 50
Moreover for given
E(x,y)
and
we may view (4.6) as a first order partial differential equation for the determination of
F(x,y).
Thus we may determine classes of differential
equations invariant under a known one—parameter group and this is the alternative problem which is discussed in the section thereafter.
4.2
INTEGRATING FACTORS AND CANONICAL COORDINATES FOR
If we introduce
by
A(x,y)
A =
y' =
F(x,y)
then (4.6) simplifies considerably
—
and we obtain
÷ F
afx
2
—O
4
a good deal simpler than (4.6), the interesting
aspect of (4.6) has been removed since (4.7) does not involve either If we introduce
directly. p(x,y)
by
p(x,y)
p =
A1
or
then we have
I
n(x,y)
=
F(x,y)E(x,y)
—
(4.8)
'
and (4.7) can be shown to become, (Fp) = 0
(4.9)
.
Hence if we write the original differential equation (4.2) as dy —
F(x,y)dx
= 0
(4.10)
,
then from (4.9) we see that
p(x,y)
is an integrating factor for (4.10).
This result is due originally to Lie [1] (see problem 1) and is generally given some prominence in the literature.
However, from the point of view of
actually solving differential equations the use of canonical coordinates is preferable.
Moreover as we have seen in the previous chapter canonical
coordinates can be used with higher order equations and therefore we will emphasize their use here.
From (4.8) and (4.9) we see that there exists a function
9z_ —
—F
9z '
z(x,y)
1
9y — (ri—FE)
.
such that
(4.11) 51
But we have
az
dx
dz =
dy—Fdx
dy
=0
=
Thus
where we have used (4. 10) and (4.11).
z(x,y) =
C
where
is a
C
constant represents the integral of (4.2) and we have using (4.11)
1
ay
ax
(4.12)
.
Thus if we introduce the operator a
L =
+
L
by
a ri
L(z) =
then (4.12) gives
and from the Commutation theorem (see (2.25)) we
1
have n
z(x1,y1) = e
r
z(x,y) =
C
n
L (z)
L
n0 and therefore z(x1,y1) = From
z(x,y)
+ c
(4.13)
.
thie equation and (2.7) we eee that if a fir8t order differential
equation iB invariant under a one-parameter group then the required integral
hae the form
z(x,y) (u, v)
where
=
v(x,y)
+
(4.14)
,
are the canonical coordinatee of the group and
i8
eome
function of u only. In
order to obtain (4.14) more directly we suppose that in terms of
canonical coordinates the differential equation (4.2) becomes dv
=
But clearly if this equation is invariant under then
4s
must be independent of
from the equation
52
v
u1 = u
and
v1 =
v + c
and the result (4.14) follows immediately
dv
=
$(u) Solve the differential equation,
Example 1
Y
dx
(x+x2+y2)
In this case we have — ax
—(1+2x)y
=
222' (x+x +y )
ay
(x+x2—y2)
222'
(x+x +y )
and from (4.6) we need to find =
(x+x2—y2)fl —
Unfortunately =
Try
1
and
(that is,
(x+x2+y2)
n
+y
and
n(x,y)
such that ,,2
—
(x+x2+y2)2
—
+
must now be determined by trial and
n(x,y)
constant)
then
F
error.
must satisfy
— (x+x2—y2)
=
Unfortunately even at this stage we cannot systematically solve this equation since the solution by Lagrange's method involves solving the original differential equation (see problem 3).
arrive
at the solution
=
x/y.
However, with some persistance we can
Thus the global form of the one—parameter
group is obtained by solving
dx1x1 dc
y1'
dy1 1
dc
subject to the initial conditions
x1 =
x,
y1 = y
when
£
= 0.
We obtain
y1=y+C, and the reader should verify that the given differential equation is indeed invariant under this group.
,
and
the
v=y
Canonical coordinates (u,v) are given by
,
differential equation becomes
53
dv
—
1
1
— (1+u2) dyj
1x
Thus the solution
s
v + tan 1u =
C
or
y+tan
x
Alternatively if we write the differential equation as (4.10), namely
ydx
dy—
=0,
(x+x+y) then (4.8) gives the integrating factor
as
ji(x,y)
— 2
—
2
(x +y )
and we obtain dy
+ (x dy —ydx)
22
= 0
(x +y )
This integrates to give the previously obtained result. Example 2
ObtaIn a function
a(x)
or class of functions such that the
differential equation 2
remains invariant under a one—parameter group.
From (4.6) we have
_*(a2+2ay2+y4) This condition simplifies if
and
=
+
+ 2n(x)y2 =
=
—
and we obtain + y2]
From this equation we deduce on equating coefficients of powers of = Ax
provided 54
a(x)
+
B
,
fl(x)
=
takes the form,
—A
y,
(Ax+B)
where A, B
and
dx1 —a-
2' are
C
all constants.
From
dy1
= -Ay1
(Ax1 + B) ,
we deduce that suitable canonical coordinates u = (Ax+B)y and
v
,
(u,v)
are given by
log(Ax+B)
=
from the original differential equation we obtain
dv_
1
du
2
(u
+ Au + C)
This equation is separable and can be integrated for given values of the constants
4.3
A
and
C.
THE ALTERNATIVE PROBLEM
For a given
and
fl(x,y)
such that (4.6) is satisfied? differential equation in
F
can we obtain the most general
F(x,y)
We solve (4.6) as a first order partial (see problem 3).
The characteristic equations
are
= n(x,y)
,
,
(4.15)
,
(4.16)
and
dT
3x
(3y
3X)
ay
and in order to obtain the most general
independent
F(x,y)
integrals of (4.15) and (4.16).
we need to deduce two
In general (4.16) is a Riccati
equation which we solve using the known solution of (4.6), namely
=
Making the substitution (see problem 4) (4.17)
we obtain 55
di
418
ay'
ax
lay
which is linear and can be solved in the usual way. Obtain the most general first order differential equation
Example 3
invariant under a one—parameter group of the form,
y1=g(x)y
x1f(x) ,
Infinitesimally we have
+ 0(c2)
x1 = x +
+ 0(c2)
y1 = y +
,
and therefore the characteristic equations (4.15) and (4.16) become =
= n(x)y
,
(4.19)
,
and
=
n'(x)y
+ (r1(x)
(4.20)
.
From (4. 19) we have
dx
and therefore y s(x) = A where
A
(4.21)
,
is a constant and
s(x)
is defined by
s(x) = e
(4.22)
for some constant
4! +
From (4.19)i, (4.20) and (4.21) we obtain
x0.
—
dx
which integrates to give =
where
B
+
is a constant.
equation is obtained from
56
B
(4.23)
,
Hence our most general first order differential B =
that
is
•[s(x) y]
=
dx
In this case we can
verify
that suitable canonical coordinates
(u,v)
are
given by 1X
u(x,y) =
s(x)y
v(x,y)
,
=
dt
0
and that the differential equation becomes du
[Notice
=
this example generalizes problems 4, 5, 6 and 7 of Chapter 3.] Obtain the most general first order differential equation
Example 4
Invariant under the one-parameter group, =
where
k
ky
fl(x,y)
,
=
ky
is a constant.
In this case we have from (4.15)
dx and
therefore (4.24)
where
A
is a constant.
Making use of the result given in problem 6 we have
on performing the integration
(4.25)
where
B
is a constant and
s(x)
is defined by (4.22).
Since w =
we have from (4.17)
w =
and hence
—
n(x))
the required differential equation is
57
s(t)k )
—
dt +
= s(x)k
denotes an arbitrary function.
where
dt]
—
Suitable canonical coordinates are
—
dt
u(x,y) = y
v(x,y)
,
dt =
and on using du
du_ dx
-
1
dv
dv
1
dx
du
we see that the differential equation becomes ku du — e dv 4(u)
The remaining sections of this chapter are devoted to various aspects associated with Lie's fundamental problem and the condition (4.6).
4.4
THE FUNDAMENTAL PROBLEM AND SINGULAR SOLUTIONS OF z(x,y) =
Suppose the integral
C
F(x,y)
y' =
of (4.2) is solvable for
y
so that we
have y =
S(x,C)
(4.26)
.
But from (4.13) we see that if (4.2) is invariant under the one—parameter group (4.1) then y1 = and
S(x1,
therefore —
C-I-c)
on equating terms of order
C
we have
lasi +
where the partial derivatives in brackets refer to two arguments =
58
x
and —
C.
y
as a function of the
From this equation and (4.2) we deduce
F(x,y)E(x,y)
.
(4.27)
Now
A =
satisfies (4.7) and using (4.27) we see that (4.7) could be
—
deduced alternatively in the following two ways. Firstly, (4.7) follows from differentiating (4.2) partIally with respect to
In the bracket notation for the partial derivatives (4.2) becomes
C.
= F(x,y)
(4.28)
,
on partially differentiating with respect to
and
C
we obtain,
laxlaCJJ But we have, +
=
from which (4.7) can be deduced.
Secondly, (4.7) follows from the compatibil-
ity of the two equations (4.27) and (4.28) which the reader can
readily
verify.
We see from (4.27) that if and
E(x,y)
n(x,y)
fl(x,y0) =
y
y0(x)
Is a singular solution of (4.2) then
must be such that
F(x,y0)E(x,y0)
(4.29)
.
Hence, if as is often the case a singular solution of (4.2) is known, then (4.29) might well suggest the general nature of
and
n(x,y).
These
considerations indicate that singular solutions of first order differential equations perhaps play a more vital role than has been previously considered.
4.5
INVARIANCE OF THE ASSOCIATED FIRST ORDER PARTIAL DIFFERENTIAL EQUATION
In this section we consider the invariance of the associated first order partial differential equation (4.3).
We use the group approach for partial
differential equations which is described in detail in subsequent chapters. We look for a one—parameter group of transformations in three variables (x,y,z)
which leaves (4.3) invariant.
We use the convention that subscripts
59
denote partial differentiation with
x, y
and
as three independent
z
variables.
Suppose that the one—parameter group x1
+ 0c2
=x+
y1 = y + CT)(x,y,z) + 0(c2) z1
=z+
(4.30)
,
÷ 0(c2
— 3z1
leaves (4.3) unaltered. 3z
3z
3x
ax1
3y
c[c
3x1
—
+
=
and
as follows,
! .iL
÷
= 3x1
We calculate — and
+
therefore —
=
p+
c{c
+
+ 0c2
—
—
(4.31)
Similarly, az
=
3y1
_L
+
ax 3y1
__! ay
3y1
+
=
+
c +
+ s
+ 0(c2)
÷
—
and hence —
=
÷
+
—
13)2
+ —
(4.32) If
60
z =
is a solution of (4.3) then by invariance we have
=
therefore
z =
+ n(x,y,z)
=
and
also satisfies
C(x,y,z)
(4.33)
Now from,
+ F(x1,y1)
i—
=0
and (4.30), (4.31) and (4.32) we can deduce
+ FC
where
0
and we have used
=
—
+ F(ri-FF)
=
=
(4.34)
,
-F0.
From (4.3) and (4.33) we have
C =
and therefore (4.34) gives + —p--
(FO) = 0
(4.35)
.
Hence
0, that is
if
z
satisfies an equation of the type (4.33) as well as (4.3) then (4.35)
can
be deduced immediately since from (4.3) and (4.33) we have
is an integrating factor for (4.10).
-FC —
Clearly
C '
—
and (4.35) follows from the compatibility of these equations. In the above we have used the so—called non—classical approach for partial differential equations described in subsequent chapters.
We have shown that
the first order condition for invariance of (4.3) under (4.30) is equivalent to the existence of an integrating factor for (4.10).
Moreover this condition
conveys no more information than the condition for the compatibility of (4.3) and (4.33).
If we apply the claasical approach for partial differential
equations then on equating coefficients of 00 and deduce that
C =
$(z)
and that
derivatives as given in (4.7).
A =
—
0
to zero in (4.34) we
satisfies (4.7), with partial
Thus the classical approach. gives rise to the
61
well known result that if where
4.6
p
is an integrating factor then so also is
is the integral of the differential equation.
z
LIE'S PROBLEM AND AREA PRESERVING GROUPS
In this section for a given differential equation (4.2) we attempt to solve (4.6) assuming that the one—parameter group (4.1) is area preserving.
That
is, we assume there exists a sufficiently continuous and differentiable function
such that
G(x,y)
,
=
=
(4.36)
.
—
From (4.6) and (4.36) we obtain the second order partial differential equation for
G(x,y) 2
= ,(G,F) + F2 aG
+ 2F
(4 37)
2
which we require to solve for a prescribed function
F(x,y).
we can solve this equation by introducing two functions
In principle
A(G,F)
and
B(G,F)
such that
p = A(G,F)
= B(G,F)
,
The compatibility condition for
(4.38)
-
G(x,y)
together with (4.37) yields two
equations for the determination of the first order partial derivatives of F(x,y)
and the compatibility condition for this function gives the final
equation for
A(G,F)
and
Although the equation obtained is no more
B(G,F).
tractable than (4.37) the analysis does merit some simplifying features which would seem worthwhile reporting.
The following analysis should be contrasted
with other possible restrictions concerning the nature of the one—parameter group.
For example if
and
are assumed to be given as the
gradient of some function then this assumption appears to compound the subsequent analysis rather than simplify it.
62
The simplifying features
associated with (4.36) may not be due to the fact that the group happens to be area preserving but rather to the fact that (4.36) is embodied in the general expressions for G(x,y)
E(x,y)
(see (2.13)).
and
More precisely
is an invariant of the group and (4.36) results from (2.13) in the
case when the Jacobian In (2.13) Is a function of
u
only.
In order to solve (4.37) be means of (4.38) we need to assume that the Jacob Ian
—
a(G,F)
(4 39)
is non—zero and finite.
We also need the following elementary relations B
J
Writing the compatibility equation for
+ a(G,F)
G(x,y)
in the form
=0
we obtain
A + FB
F2
we see from (4.7) that (4.37) can be written as — 0 —
which on simplification yields, (4.42)
We note that it is in the derivation of (4.42) that the assumption (4.36) appears to significantly simplify the analysis. On solving (4.41) and (4.42) for
aF/ax
and
we obtain
63
aF 3x
aC
aC
I
Fac
IF3F
(4.43) 3F
where
3C
is given by
H(G,F)
H(G,F)
(444)
—
=
that
We note from (4. 38) and (4.43)
the given differential equation (4.2)
becomes
(4.45)
From the above equations we find after a long calculation that the
compatibility condition for
F(x,y)
becomes
(4.46) which can be written
as
(4.47) On comparing this equation with (4.7) we see that
(4.47)
is the statement
that (4.45) remains Invariant under the one—parameter group with infinitesimals
and =
—
B)
fl*(G,F)
given by
fl*(G,F) = —c
,
(4.48)
.
Thus an integrating factor for dF
is therefore
—
so that
64
(4.49)
— H(G,F)dG = 0 ,
the
.—
n*Y'.
n* = c2 3(A:B)
Now we can verify that —
compatibility condition for
(4.50)
F(x,y)
reduces to the statement
that the differential form
dG +
dF
— C
(.
2a(A,B)
)
a(G,F)
Problems 13 and 14 illustrate the above analysis
is an exact differential.
with two simple solutions of (4.47). expression for the Jacobian
For specific examples we need an
defined by (4.39).
J
From (4.38), (4.39),
(4.43) and (4.50) we find that —
j
=
C = A + FB
Using
(4.52)
.
dA +
(
we can simplify (4.51) to give — C dB —
— 0
c2 a(A,B) a(G,F)
Thus with
C1
=
dA +
the condition (4.47) is equivalent to the statement that dB = 0
AB
(4.53)
,
That is the compatibility condition for
is an exact differential.
F(x,y)
becomes
laB'
t'aA)
j
a(A,B)
a(A,B)
where the functions —
'
a(A,B)
are defined by
and —
1
(A+FB) A
4)
a(A,B)
—
F
455
- (A+FB)
particular method of solution of (4.54) is outlined in problems 15, 16, 17
and 18.
It is worthwhile noting that the differential forms (4.51) and (4.53) are consistent with that obtained from the requirement that
(A+FB)1
integrating factor for (4.9), provided we make use of the for
SF/ax
and
aF/ay.
Since from (4.40) and using
C = A + FB
must be an (4.43)
we have 65
dy—Fdx_ (A+FB)
C dF
ax JC
—
+
F
and (4.43), (4.44) and (4.52) yields precisely (4.51). PROBLEMS 1.
If the differential equation M(x,y)dx + N(x,y)dy = 0 is invariant under (4.1) show that
the
infinitesimal condition is
equivalent to the existence of an integrating factor
ji(x,y)
where
1
=
2.
is an integrating factor for both of the differential
If
equations,
M(x,y)dx ÷ e =
show that
3x
= 0
tan'(M/N) +
v20 =
3.
N(x,y)dy
and
N(x,y)dx —
M(x,y)dy
= 0
satisfies
= 0 ay
For the quasi—linear first order partial differential equation a(x,y,z)
÷ b(x,y,z)
= c(x,y,z)
(*)
,
show that the general solution is given by p =
where
is
•
an arbitrary function and
p(x,y,z)
and
o(x,y,z)
two independent integrals of the system of differential equations dx
[From
=
a(x,y,z)
p =
constant
dT 66
x
dT and
=
a =
b(x,y,z)
dz ,
= c(x,y,z)
constant we have
+bay +cOz =0,
are any
where subscripts denote partial differentiation with x, y three independent variables.
and
z
as
These two equations together with (*)
constitute three homogeneous equations for
a, b
and
c.
For non—trivial
solutions the determinant vanishes and this condition can be shown to become = o
from which the required condition fOllows.
derivatives are with = p
x
4.
If
+ p
as the independent variables, that is
y
=a+ a
,
etc.]
,
y
is a known
y0(x)
and
x
In the Jacobian partial
solution
q(x)
+ p(x)y =
of the Riccati equation
+ r(x)y2
show that the substitution
y = y0
+w1 gives rise to the linear
equation
+ 5.
—
p(x)]w
=
—r(x)
Show that the most general first order differential equation which admits the group =
is
=
,
-
-
=
where
s(x)
n(x)y +
-
s(t)dtj
is given by 1X
dt
s(x) = and
e
is an arbitrary function of the argument indicated.
67
6.
Using (4.15)2 and
(4.18),
make the substitution w =
and deduce the
equation, 1 —
7.
Show that the most general first order differential equation invariant under the group n-i
E(x,y) =
=
,
ri(x)y n
is -
dt + = s(x)
5.
is as defined in problem
Using canonical coordinates rx
u(x,y) =
s(x)y
,
v(x,y)
1
=
i
n1 s(x) n—i y
s(t) n—i dt
show that the differential equation becomes du dv 8.
u
n
For the Riccati equation given in problem 4, show that i y(x)=— r(x)z(x)
the
substitution
dz —, dx
gives rise to the linear equation,
+
+ —
q(x)r(x)z
= o
Deduce the normal form of this differential equation. 9.
Continuation.
E(x,y) =
Show that Riccati equation of problem 4, admits the group
E(x)
,
=
+
,
+
2rt
provided, (re)
68
and
-
+
=
satisfies
From these equations deduce that 1
+ 4{qr where
section
1
=
Can you reconcile this result with that of
a constant.
is
C
1
the equation,
3.2.
[The following three problems sununarize the three criteria given by
Dickson [3] (page 313) for the invariance of a differential equation under a one—parameter group.] 10.
Show that a
first order ordinary differential equation is invariant
under the group x1 = x +
if
+ 0c2
,
y1 = y + cfl(x,y) + 0(c2)
(**)
and only if Lz =
where
a
L =
and 11.
is the integral of the equation,
z(x,y)
+
is the operator
a
denotes an arbitrary function.
I
Continuation.
The differential operator associated with
(+)
N(x,y)dx + N(x,y)dy = 0 ,
is
L
given by P = N
The
ax
- M ay (LP)
is defined by
(LP) = LP — PL
Show that, (i)
(LP) =
—
.
69
the differential equation (+) is invariant under (**) if and only
(ii)
if the commutator (LP) is a constant multiple of the operator 12.
The first extension of the operator
Continuation.
L
is
L'
P.
where
1+71_L 3y' and where
is
ii
-
+
=
ii
given by
-
y'2
Show that the first order differential equation F(x,y,y') = 0
remains invariant under (**) if and only if L'F = 0
[The following six problems relate to section 4.6.] 13.
Assuming that
where
A =
f(G)
,
f
and
g
B =
g(G)F'
are functions of
C
only, show that equation (4.47)
simplifies to yield f"(l+f/g) +
f'(f/g)'
= 0
where primes denote differentiation with respect to equation and show that
J — ag
where
C.
Integrate this
is the integration constant.
a
Hence from the relations (4.40) deduce that fF
1
Co
where
and
C0
are further integration constants.
From these
results show that the original differential equation (4.2) in this case has the form —
for 70
(ay+B)h[ctx + log(ay+8)]
some arbitrary function
h
,
of the argument indicated.
Observe that
this equation can be solved by the substitution p = ax + log(ay+8) 14.
Assuming that
is identically zero show from (4.44) and (4.47)
H(G,F)
that
{Z'(F)[Z(F)G+m(F)]
B =
—
[L(F)m'(F)—m(F)&'(F)]}
C
where
+ n(F)
]½
2R(F)
[Z(F)G+m(F)]½
=
£, m
and
n
denote arbitrary functions of
denote differentiation with respect to
F.
F
Show that
and here primes J =
£(F)/2
and
from the relations (4.40) deduce
x =
+
£(F)
for some functions
and
p(F)
p(F),
'
+ q(F)
= £(F)
With s(F) = q(F) — Fp(F)
q(F).
show
that in this case the original differential equation (4.2) is the well known Clairaut's equation (Murphy [8], page 65) =
dx
+ (dxJ'
which has general solution 15.
y =
yx + s(y)
Assuming there exists some function
w(A,B)
for some constant such that
4s
and
y. iji
as
defined by (4.55) are given by
show that equation (4.54) becomes 2
a(A,B)
where the Laplacian
V2 =
V
2
a(A,B)
(**)
is given by
+
Show that (**) can be written alternatively as 71
________
v2 a(w,G) =
—
16.
(***)
a(A,B)
a(A,B)
a(A,B)
ContinuatIon.. From (4.55) and (*) conclude that
w(A,B)
satisfies the
first order partial differential equation aw
aw
BaA
1,
tan'(B/A)
+
AaB and
hence w =
denotes an arbitrary function of the argument indicated.
f
Introducing polar coordinates R = (A2+B2)½
tan'(B/A)
e =
,
show that F = where to 17.
[B—Ag(R)]/[A+Bg(R)]
g(R) =
Rf'(R)
and the prime denotes differentiation with respect
R
Continuation.
coordinates observe that (***) of problem 15
(R,e)
In
becomes 1 a(w,G) = 1 fa(w,V2G) — a(V2w,G)t, a(R,e) R a(R,e) R a(R,e) J
where
V
2
is given by
v2
R
R
On using w — e
+
f(R)
R aR
where 18.
72
g(R)
Continuation.
R
ae2
show that (****) simplifies to give
R2 ae2
RJaeJ
aR
is as defined in problem 16.
With
G = R2
and
h(G)
g(G½)
show that
G½[F+hGJ
G½[1_Fh(G)] A
B
,
{[1÷F2][1÷h(c)2])½
Hence show that
J =
(1+F2)/2
{[1+F2][1+h(G)2]}½
and that the relations (4.40) yield,
apart from arbitrary additive constants 2G½[F+h(G)]
2G½[ 1—Fh(G)]
X
{[1+F
2
2½' ][1+h(C) ]}
2
{[1+F
][1+h(G) ]}
Hence conclude that the original differential equation (4.2) in this case is
fy
=
-
xg[(x2+y2)½/2]j
+ yg[(x2+y2)½/2]J
dx
which Is solved using polar coordinates (see Murphy [8], page 67). 19.
Given the one—parameter group x1 = x +
y1 = y + cfl(x,y) + 0(C2)
+ 0(c2) ,
show that y1(x1—x0) = y(x—x0) +
Y(x_x0))]
—
(x—x0)
+ fl(x—x0, y(x_x0))} + 0(c2)
[This result can be verified by two distinct methods. (i)
Suppose that
y =
S(x,C)
y(x—x0) = S(x—x0,C) = S(x1—x0,C+c)
=S
and
and
,
,c-I-c) + 0(c2)
and
+
y(x-x0) +
where the partial derivative of x—x0
C
then
y1(x1—x0)
= S(x—x0,C) +
=
y1 = S(x1,C+c)
(x_xO) + S
+
+ 0(€2)
with respect to
C
has arguments
and is found from (4.27) to be given by 73
y(x—x0)) —
=
(x-x0)
y(x—x0))
from which the required result follows. Alternatively we have
(ii)
d
y(x—x0) = e
y(x)
and we require to find d
-x0
y1(x1—x0) = e
1 y1(x1)
x = x1
+ u(c )
From
A2
—
we have
+0ic2)
dx1{1
dx
and therefore
dx
dx1
Hence if we define the differential operators D
D
dx'
1
D1
and
D2
by
2
then we require to evaluate y1(x1—x0) = and type.
Knapp [12]
x = x
74
—
x0 ,
y = y(x—x0)
e
that
D2y(x+t) = In
(page 40) give formulae for operators of this
Observe that, e
and
+ cn(x,y)] +
-
(x,y)
(x+t)
order to calculate the order of
c
term arising from
we use the integral given in [12] (page 40).
=
y(x-x0)
We have
O[Dy(x+T)]*
+ U
dT + 0(c2)
where the 'star' in the integrand denotes that (x,y) in the square bracket becomes (x—x0—T, y(x—x0.-T)).
If in the integral we make the
substitution p
then
x0
x
T
we have
=
y(x—x0)
(x-x0)dp + 0c2
(p,y(p))
+
and thus = y(x—x0) ÷ c
(x-x0)
E(x.-x0, Y(x.-x0))]
+ 0c2 The result now follows since =
crl(x-x0,
y(x.-x0)) ÷
0(c2
Note that we can check the validity of this second method by using the integral given in [12] to evaluate x1 — x0 =
+ cE(x,y)] + 0(c2)
Proceeding as above we obtain x1
x0 = x — x0 + cE(x,y) + 0(c2)
which of course is the desired result.] 20.
Continuation. x1 = x +
Obtain a one—parameter group
+ 0c2
,
y1 = y + cTl(x)y + 0c2
which leaves the following differential—difference equations invariant, (a)
dx
(x) = —y(x—x0)
75
(b)
(x) =
y(x)[1
(e)
(x) =
y(x)[y(x)
y(x—x0)] —
y(x—x0)]
Can we use these groups to simplify or integrate the equation? [Answers:
(a)
x1=x+ac,
y1=e8cy,
(b)
x1=x+C,
y1=y,
(c)
x1e
where
a
and
-1
$
(1—c
—aE:
),
cic
are arbitrary constants. d
y(x—x0) = e
y1=e y,
dx
y(x) =
r L
n=0
(—x ) 0 n!
Notice that since,
n y
(x)
differential—difference equations are really 'infinite' order differential equations.]
76
5 Second and higher order differential
equations First
order differential equations can be Invariant under an infinite number
of one—parameter groups.
Second and higher order equations differ in that
they are invariant under at most a finite number of groups. equations are invariant under at most 8 while for differential equations are invariant under at most Dickson [3], page 353).
n >
2,
Second order nth order
n + 4
groups (see
Higher order equations also differ from first order
ones in that if there exists a one—parameter group leaving the equation invariant then this group can be systematically determined.
Much of the
literature is concerned with obtaining the most general second order differential equation invariant under a given group and again we advise the reader to consult standard tables of such differential equations (see for In the first section of this chapter we
example Dickson [3], page 349).
deduce the condition (5.8) for a second order differential equation to be invariant under a one—parameter group.
In the next section we give four
examples making use of this condition.
In the section thereafter we give
examples of the determination of the most general second and higher order differential equations invariant under a given one—parameter group.
In the
final section we give three applications from the nuclear industry due to Axford [14], [15], [16] and [17].
5.1
We
INFINITESIMAL VERSIONS OF
y"
AND
y"
=
consider the general second order differential equation 2
t I
dx
2
y, dxJ
'51 77
and look for a one—parameter group, =
x +
+ 0c2
,
which leaves (5.1) invariant.
y1 = y + cfl(x,y) + 0(c2) ,
(5.2)
Throughout this chapter we shall use the
notation (5.3)
80
that
from results given in the previous chapter we have
(5.4)
= z + cvr(x,y,z) + 0(c2)
where
ir(x,y,z)
is given by
ay
ax
In
order to calculate the infinitesimal version of
y"
we proceed as
follows, 2
d y1 2
dx1
d
dx
dy1
dx1
dx dx1
and thus we have
ax
dx2
ay
az dx2
lax
(5.6)
ay
)dx2J
If (5.1) is left invariant by (5.2) then on using F(x1,y1,z1) =
F(x,y,z)
+
+
+
+ 0(c2)
and (5.1) and (5.6) we deduce the condition that (5.2) leaves (5.1) invariant, namely
tax
78
ay
az
tax
ay
ax
ay
az)
(57)
If
involves powers of
F
then generally we can determine
z
from (5.7) by equating coefficients of the powers of
and
On using (5.5) we
z.
find that (5.7) becomes
(ay
axJ
(5.8)
axjazj
ay
ax2 aXaY
ax az
ax
ay azj
=0. In the following section we illustrate with examples how solutions and
of (5.8) can be deduced.
5.2
EXAMPLES OF THE DETERMINATION OF
Example 1 n(x,y)
Show that if
AND
ri(x,y)
is independent of
F(x,y,y')
y'
and
then
take the following forms,
p(x)y
E(x,y) = If
((x,y)
+
,
n(x,y)
is independent of
F(x,y,z)
z
=
p'(x)y2
+
+
.
(5.9)
then (5.8) becomes
(5.10)
ax2 + ILI1
-
2 -
2
3
ay2
lay2
=0, and from the coefficients of
axay
'
and
we have
ay2 79
We notice that from the coefficients of
and (5.9) follows inmiediately.
we also have
z°
and
z
3p"(x)y +
3p(x)F
—
[p'(x)y2
[p(x)y + =
Hence either
p(x)
(5.11)
+
[p"(x)y2 + rf'(x)y + t"(x)J
+
—
+
is identically zero and
and
=
F(x,y)
is
a solution of the partial differential equation
+ or
p(x)
t(x)J
+
=
—
is non—zero in which case
F(x,y)
+
+
t"(x)J
must be a linear function of
y
(see problems 1 and 2). Example 2
y" = 0
Show that
is invariant under precisely 8 one—parameter
groups.
From (5.8) with
identically zero we have
F
2
2
2
(5.12)
2
and as in the previous example we deduce that of the form (5.9).
From the coefficients of
must be
and z
and
z°
in (5.12) and (5.9)
we deduce =
and
2fl'(x)
,
p"(x)
=
rf'(x)
=
t"(x)
hence
p(x) =
C1x
+ C2 ,
T)(x) = C3x + C6 , where
C1, C2, ..., C8
= C3x2
t(x)
+ C4x + C5
= C7x + C8
denote arbitrary constants.
gives rise to a one—parameter group leaving example the group generated by
80
= 0
y" = 0
C3, that is take
Each of these constants invariant.
C3 =
1
Consider for
and assume all the
other constants are zero.
dx1
From
dy1
2
X1)'1
dc
x1 = x
and
y1 = y
,
c = 0
when
we find that the global form of the group
is
x We
x (1—cx)
= 1
y 1 = (1—cx)
have +
= (1—ex)
so that clearly y" = Show
Example 3
=
is
cy
0
,
=
remains
(1-cx)3
invariant
under this group.
that the differential equation
Xf +
,
not invariant under any one—parameter group. With
F = xy + e
+
z
we see that (5.8) becomes
3
z2 -
- 2
tay2
+ ez
From the term involving
,
so that C3
ax
ay
= C1x + C2
we deduce
ax' and
denote arbitrary constants.
involving
eZ
n(x,y) = C1(y—x) + C3 From the coefficient of
where z°
C1, C2
and
in the term not
we have
—C1xy = y(C1x + C2) + x[C1(y—x) + C3] 81
which is clearly only satisfied if
C1 = C2 = C3 = 0
.
Hence there is no one-
parameter group leaving the given differential equation invariant. Obtain the most general invariant one—parameter group for the
Example 4
second order differential equation 2
+ p(x)y = 0
(5.13)
.
dx
As in example 1 with
and
y)
F(x,y) = —p(x)y
are given by (5.9) and from (5.11)
y)
we deduce
p"(x) + p(x)p(x) = 0 =
p(x)C(x)
C"(x) +
fl"(x) +
,
= 0
+
(5.14)
,
= 0
and hence the given differential equation is invariant under 8 distinct one— parameter groups.
We notice that the group arising from
that considered in section 3.2.
The group arising from
the invariance of (5.13) under the addition to We consider
p(x)
in more detail.
y
and
ri(x)
merely reflects
of any solution of (5.13).
For this group we find that suitable
canonical coordinates are
v(x,y)
,
u(x,y) =
=
and we have du d(uv)
=
dx —
Hence on using (5.13) and (5.14)1 we obtain
= p(x)
-
p"(x)y
= 0
,
dx2
and therefore we have u = C1uv + C2
where
82
C1
and
C2
are constants.
is
From this equation we readily deduce
1X
y = C1p(x)
dt
+ C2p(x)
p(t) 2
I
which is a well known result for the general solution of (5.13). It is also worthwhile noting that embodied in (5.14)2 =
i)
(ax) = 0,
is the group C
=x,
y1 = e y
which reflects the invariance of all linear homogeneous equations under stretchings of
In this case suitable canonical coordinates are
y.
v(x,y)
u(x,y) = x , and
with
w(u)
w(u)
the
=
=
log
y
defined by
dv
differential equation (5. 13) becomes the first order Riccati equation
(Murphy [8], page 15), dw 2 1-+w +p(u)0.
5.3
DETERMINATION OF THE MOST GENERAL DIFFERENTIAL EQUATION INVARIANT UNDER A GIVEN GROUP
In the notation of section 5.1 suppose have
deduced two independent
invariants
and
for given A(x,y)
and
B(x,y,z)
n(x,y)
we
(say) of the
characteristic equations dx
where
=
dT
Tr(x,y,z)
=
dz ,
is defined by (5.5).
= ir(x,y,z)
,
(5.15)
The basic result for the determination
of the most general second order equation invariant under this group is that this equation is given by =
where
•(A,B)
,
is an arbitrary function of the arguments indicated.
(5.16) Clearly (5.16) 83
is of second order and is invariant under the given group.
In order to see
that there can be no more general equation than (5.16) we refer the reader to the comment following problem 5.
We remark also that the most general third order differential equation invariant under the given group is given by
B,
=
where
'V
(5.17)
,
denotes an arbitrary function. Obtain the most general second order differential equation
Example 5
invariant under the group x1
y1 =
f(x)
g(x)y
and hence deduce the most general linear invariant second order equation. From example 3 of the previous chapter we have already obtained
s(x)y
A =
where
B =
—
-
—=
(5.18)
,
are as previously defined.
and
s(x),
fl(x)y]
2ri(x))
Now
-
X
,
dA
(5.19)
and on incorporating the denominator of (5.19) into the arbitrary function of (5.16) we deduce the required second order equation
+ fE'(x)
fl(x)2
— 2
(5 20)
E(x)
dx2
where
=
—
denotes an arbitrary function and
A
and
B
are given by (5.18).
The most general linear homogeneous second order differential equation invariant under the given group is obtained by taking
to be given
by = aA
84
+
,
(5.21)
where
cx
and
are
B
We find from (5.18), (5.20) and (5.21) that
constants.
this equation becomes + b(x)y = 0
+ a(x)
where a(x) =
2
—a +
{fl(x)2 —
and on eliminating
between these two later equations we obtain —
+
1
da —
a(x)2)
=
a
— B2
Hence this result is consistent with that obtained in section 3.2 (see also problem 8 part (i) of Chapter 3).
Find the most general second order differential equation
Example 6
invariant under the group =
where
k
ky
ky
=
,
is a constant.
From example 4 of Chapter 4 we can deduce
A—y+logs(x) B =
-
n(x))
with the usual definition for dB = dA
-
s(x).
ok
dt}
(5.22)
,
On differentiating we can show
+
-
k
n'(x)1
—
ldxJ
—
J
and thus the required differential equation is
+
k
=
+
— (5.23)
85
denotes an arbitrary function and
where
A
and
are given by (5.22).
B
Hence a differential equation of the form (5.23) can be reduced to the first order equation
+ kB +
— 0
Obtain the most general linear third order equation of the form
Example 7 3
+
p(x)
dx
+ q(x)y = 0
X
(5.24)
,
which is invariant under the group given in example 5. In the notation of example 5 we have from (5.18) and (5.19) =
B
s(x)
+
—
+
(x))y
—
On differentiating this equation with respect to
result by
+ (dB)2}
+
+
=
+
+ 3n(x)2 —
-
-
+
the
and multiplying the
we obtain
B
B{B
Hence
A
—
31i(x)E'(x)
—
-
most general third order equation invariant under the given group
is obtained by equating the expression on the right—hand side of this equation to
1P1(A,B,dB/dA) = ciA
B,
where fl(x) =
86
a,
and —
y/3
y
+
.
The most general linear equation arises from + yB
are constants. and we find
In order to obtain (5.24) we require
—+
/
(x) =
3
-
q(x) =
+
+
+
+
and thus
which agrees with the result obtained in section 3.3.
APPLICATIONS
5.4
In this section we consider three specific second order differential equations which arise from various problems in the nuclear industry.
These applications
are due to Axford [14], [15], [16] and [17] and we refer the reader to these papers for the full motivation and detailed analysis.
For additional
applications the reader should consult Bluman and Cole [5] (page 116).
For
our purposes the problems considered illustrate the scope and limitations of the group approach for problems arising from a practical context. Example 8.
Reactor core optimization
In Axford [14] for the problem of
determining the appropriate fuel distribution which minimizes the ratio of the critical mass of the core to the reactor power when the power output is prescribed, the following differential equation is obtained
- y'2
+ where
y(x)
—0
+
denotes the non—dimensional thermal flux, primes denote
differentiation with respect to two cases considered are geometry while For
cx =
(5.25)
,
0
a =
1
ci
x
and
ci
and
are known
constants.
The
= 0 which corresponds to assuming a slab
corresponds to a cylindrical geometry.
equation (5.25) becomes 87
(5.26)
and it is instructive to solve this equation first by standard devices and then by the group approach. we let
y'
z =
Since (5.26) does not depend explicitly on
x
and we obtain in the usual way the first order differential
equation
y
dy
(5.27)
z
We recognise this as an equation of the Bernoulli type and therefore set w = z2
and deduce w = C1y2 - 28y3
where
C1
denotes an integration constant. dy
½
On integrating
= dx
we obtain in a straightforward manner C'
(5.28)
y(x) = + C2)]
B[1 + where
C2
constant
denotes a further integration constant and we are assuming the C1
is positive.
Alternatively we may deduce the general solution
(5.28) from the group approach in the following way.
We observe that (5.26)
remains invariant under the two one—parameter groups,
x,=x-i-c,
y,=y,
(5.29)
x,ex,
y,e —2cy.
(5.30)
and
The group (5.29) is merely the formal statement that (5.26) does not depend explicitly on
x
again set
y'
88
z =
and
therefore since
and obtain (5.27).
y'
is an invariant of this group we
However the group (5.30) means that
(5.27) is invariant under
y1=e —2c y,
z1=e —3c z,
and therefore we select
u* = zy
3
2
as the new dependent variable and (5.27)
becomes the separable first order differential equation du* —
(u*2÷213)
—
yu*
dy —
This equation readily integrates to yield
C1/y
=
- 2B
from which the solution obtained previously can be deduced. For
cx
non—zero equation (5.25) is still invariant under (5.30) and we
therefore select
u = yx2
substitution, t =
log
as the new dependent variable. and
x
p =
du/dt
With this
equation (5.25) becomes the Abel
equation of the second kind (Murphy [8], page 25) p
=
2(cx-1)u
- Bu2 -
(cx-1)p +
This equation can be reduced to standard from (see equation (1.20)) by the p = qu
substitution q
which for for
= 2(cx—1)
cx =
1
find
We
.
—
— (cz—1)
(5.31)
,
can be integrated to finally obtain the following solution
y(x), namely 2C 1
y(x) =
+ where again
C1
other values of
and cx
(5.32)
+ C2
denote integration constants.
We note however for
the solution of (5.31) is by no means apparent.
The next two examples discussed in detail in [15], [16] and [17] arise from the steady state heat conduction equation with non—linear thermal
89
conductivity
and non—linear source terms
k(T)
divlk(T)grad T] + S(T) = 0 where
T
S(T), that is
(5.33)
,
denotes the temperature.
These problems arise in the context of
thermal instability phenomena in rods and plates in the sense that if the rate of energy produced by the heat source exceeds the rate at which energy can be transferred out across the boundary then a steady state temperature distribution cannot exist (see Axford [15]).
The particular thermal
conductivity and source terms considered are, k(T) =
k0(T/T0)1
(5.34)
,
S(T) = S0(TIT0)6 , where
y, 6, k0,
S(T)
and
= S0 exp(T/T0)
(5.35)
,
denote constants.
T0
Power law conductivity and source term
Example 9.
In Axford [15] and [16]
the following differential equation in non—dimensional variables is deduced from (5.33), (5.34) and y1 {y" +
where
a
and
+ yy '
a, y
a =
1
and
(5.
,
are constants and again
slab geometry while three constants
= 0
+
a = 0
corresponds to the plate or
corresponds to a rod or cylindrical geometry. 6
36)
The
encompass a wide variety of physical behaviour
and the full analysis of (5.36) involves consideration of a number of special cases which are detailed in [15] and [16].
Here we restrict our attention to
results which can be deduced rapidly and simply from the group approach.
For
any practical problem we first examine simple groups leaving the equation invariant before using the theory given in the first section of this chapter and equation (5.8).
For the examples of this section, this aspect is
summarized in problems 9, 10 and 11.
90
If we look for a simple stretching group
y1=eacy,
c x1=ex,
(5.37)
leaving (5.36) invariant then we fin.d 2
(5.38)
a =
provided take
S #
1
u =
+ y.
Assuming for the time being that this is the case we
as the new dependent variable so that equation (5.36) becomes
+
+ (cx+2a)xu' + a(cx+a—1)u} +
and with the substitutions
t =
log
and
x
p =
du/dt
this equation becomes
+ (a+2a+2ay—1)pu + a(a+a+ya—l)u2 + yp2 +
up
= 0
= 0
In general this is again an Abel equation of the second kind. cases give rise to standard equations. is homogeneous while if
a
1
and
cS =
For example if
S =
(1+y)(a+3)/(cx—1)
(5.39)
.
However special
y + 3
the equation
the equation is of
the Bernoulli type. If
cS =
1
+ y
then (5.36) becomes (5.40)
which is invariant under the one—parameter group x1 = x
C
y1 = e y
,
Thus we take
w =
y'/y
(5.41)
.
as the new dependent variable (see example 4) and
(5.40) becomes
+
w + (1+y)w2 +
= 0
(5.42)
,
which is a Ricatti equation (Murphy [8], page 15). are special cases which can be readily solved.
Clearly
a = 0
or
y =
—1
For the final example we
consider the equation arising from (5.33) for the case of constant thermal
91
conductivity and exponential source term. Example 10.
Constant coliductivity and
exponential
source term
With non—
dimensional variables Axford [15] and [171 obtains the equation
y" + For
a
=
(5.43)
.
zero this equation can be readily integrated by means of the
standard substitution
z =
dy/dx.
a
For
non—zero we look for a one—
parameter group leaving (5.43) invariant of the form £
x1 = e x We find
a =
—2
y1 = y + ac
,
(5.44)
.
and therefore on eliminating
as an invariant of the group.
u =
With
c
from (5.44) we obtain
u
as the dependent variable
(5.43) becomes
x2(uu"—u'2) + axuu' + 2(1—a)u2 =
and the usual substitutions = p2 + (1—a)up —
up
For
a =
1
t
=
log
and
x
2(1—a)u2 —
p =
du/dt
yield,
(5.45)
.
equation (5.45) is the same as equation (5.27) and therefore the
solution can be deduced from (5.28).
However for
a #
1
(5.45) must be
solved as an Abel equation of the second kind.
The examples of this section illustrate how simple groups leaving the equation invariant may be utilized to reduce the order of the differential equation.
The resulting differential
standard type
equations may or may
not
be of a
with a simple solution.
PROBLEMS 1.
In the notation of example 1 of section 5.2 show that if zero then F(x,y) =
92
G(x)y
+ H(x)
p(x)
is non-
where
p"(x) CC) X — —
and deduce that
-
for constants
C1
and
Continuation.
If
p(x)
that
F(x,y)
— —
—
3p(x) must be such that
and
p(x),
+
2.
H( X )
p(x)
P(X)
dt
= C1
PX
+
p(t)
C2.
is identically zero and
=
E'(x)
is given by 1X
+
F(x,y) = ½
Jx0
E(x) where 3.
If
4
p(x)
show
E(t)
denotes an arbitrary function of the argument Indicated. is non—zero but arbitrary show that the differential equation
2
+ p(x)y2 = 0 dx
is invariant under at most 6 one—parameter groups.
Show that
2
+
=0
dx
can be reduced to a first order Abel equation of the second kind (Murphy [8], page 25). 4.
Find the most general second order differential equations which are invariant under the one—parameter groups given in problems 5 and 7 of Chapter 4.
5.
Show that the second order differential equation F(x,y,y',y") = 0 is invariant under the one-parameter group x1 = x + cE(x,y) + 0(c2)
,
y1 = y +
+ 0(c2
,
(*)
93
if and only if L" F = 0
where
L"
is the second extension of the operator
where
a
a
L
and
ir
a
a
are the infinitesimal versions of
a
and
y'
and
y"
(Note that if
respectively and are defined by (5.5) and (5.6). B(x,y,y')
L, namely
A(x,y),
are three independent integrals of the
C(x,y,y',y")
equations =
1(x,y)
=
,
= 7T(x,y,y')
,
=
,
then the most general second order equation invariant under (*) takes = 0
the form
or
C =
[For a detailed discussion of the following problems the reader should consult Dickson [3], page 358.] 6.
Given two one—parameter groups with operators L1 =
11(x,y)
a
+
a ri1(x,y) -b-—
L9 =
show that the first extension of the commutator
E2(x,y)
a
+ ii2(x,y)
(L1L2)
a
is identical to
the commutator of their respective first extensions, that is
(L'1L'2).
(See problems 11 and 12 of Chapter 4.) 7.
Continuation. y" =
F(x,y,y')
operators
L1
Show that if the second order differential equation, is invariant under two one—parameter groups with and
group with operator 8.
Continuation.
If
then it is also invariant under the one—parameter
L2
(L1L2). L1
and
L2
that there exists an operator invariant and is such that (L1L3) = aL1 + bL3 94
leave L3
y" =
F(x,y,y')
invariant show
which also leaves the equation
for some constants 9.
a
and
b.
For the second order differential equation
(+)
x dx
dx2
y
functions
that for all
y
the only one—parameter groups leaving (+) invariant
f(y)
take the form
g(x)
i,(x,y) =
n(x,y) =
where
denotes an arbitrary constant and
A
functions of
=
Continuation.
g(x)
and
h(x)
are
such that
x
+
10.
+
+ A}
-
If
—
2g'(x)}
f(y) —
nf'(y)
show that if
f(y) =
6
y +
1
the only one—
parameter group leaving (+) invariant is given by =
If
iS =
y + 1
x
n(x,y)
,
= (1+y—6)
show that the differential equation (+) Is invariant under
the group (++) where
and
g(x)
h(x)
satisfy the following differential
equations
g" +
=0
+ a(2—a) N—
h" + 11.
h' +
= 0
If
f(y) = Be3'
ContInuation.
and
y = 0
show that for
a
1
the
only one—parameter group leaving (+) invariant is given by (-H-) with g(x)
and
h(x)
given by
95
4A
2A
where g(x)
A
and
is
the arbitrary constant in problem 9.
12.
B
Show that
ac_
h(x) = 4A
log x + 2(2A-B)
is a further arbitrary constant.
the
classical diffusion equation 2 ac 2'
ax
admits travelling wave solutions of the form —
B(x)]
is a constant and
A(x)
c(x,t) =
where
w
A(x)sin[wt
A" = AB'2
where
and
B(x)
satisfy
AB" + 2A'B' + k2A = 0
,
k = (w/D)½.
Observe that
+ k2A2 — 0
remains invariant under the one—parameter group
x1=x,
C A1=eA,
and deduce the second order differential equation
w" + 6w' + 4w3 + 2k2(w' + w =
96
a =
are given by
h(x)
g(x) = —2Ax log x + Bx ,
where
For
A'/A.
1
show that
6 Linear partial differential equations
For partial differential equations the calculations involved in the determination of a one—parameter group leaving the equation invariant are generally fairly lengthy.
In order to keep these calculations to a minimum
we first consider a restricted class of one—parameter transformation groups applicable to linear partial differential equations. considered in the following chapter. variable
c
Non—linear equations are
Specifically for a single dependent
and two independent variables
x
and
t
we consider
transformations of the form x1 =
f(x,t,c)
= g(x,t,c) c1 =
h(x,t,c)c
where the functions
= x
=
t
+
+ 0(c2)
+ cn(x,t) + 0(c2)
= c +
f, g
(6.1)
,
+ 0(c2) and
h
do not depend explicitly on
c.
If the
transformation (6.1) leaves a given partial differential equation invariant and if
then from
c =
c1 =
4(x1,t1)
on equating terms of order
c
we have
+
fl(x,t)
For known functions
=
E(x,t), ri(x,t)
(6.2)
.
and
t(x,t), equation (6.2) when solved
as a first order partial differential equation, yields the functional form of the similarity solution in terms of an arbitrary function.
This arbitrary
function is determined by substitution of the functional form of the solution into the given partial differential equation.
In the case of two independent
variables the resulting equation is an ordinary differential equation.
For 97
more than two independent variables the procedure reduces the number of independent variables by one.
In the following section we give the formulae for the infinitesimal versions of the partial derivatives
ac/ax, Dc/Dt, D2c/Dx2, D2c/axat
and
Although we make no use of the last two partial derivatives, they
D2c/3t2.
are included for completeness.
For the remainder of the chapter we
principally consider groups of the form (6.1) and the corresponding solutions of diffusion type equations.
In particular we consider the classical
diffusion equation 2
(6.3) Dx
and
the
Fokker—Planck equation which we assume given in the form,
DC =
where
p(x)
D
p(x) and
DC1
q(x)
D
+ j—(q(x)c)
(6.4)
,
are functions of
x
only.
In the determination of
groups leaving an equation invariant there are two methods, termed and non—classical.
The classical approach equates the infinitesimal version
of the given partial differential equation to zero without making use of equation (6.2).
The non—classical procedure which is considerably more
complicated makes use of (6.2) and includes the classical groups as special cases.
For the most part we obtain results from the classical procedure.
However in the final section we discuss the non—classical approach with reference to equation (6.3).
The results given in this chapter for (6.3) are
due to Bluman and Cole [18] and Bluman [19J(see also Bluman and Cole 15], page 206) while the general equation (6.4) was first studied by Nariboli [20 1. [20] several special examples are analysed in detail. present some new results for equation (6.4). how the most general function 98
q(x)
In section 6.5 we
We show for arbitrary
p(x)
can be found such that (6.4) admits a
In
classical group of transformations leaving the equation invariant.
6.1
FORMULAE FOR PARTIAL DERiVATIVES
For the one—parameter group of transformations (6.1) we assume that the Jacobian, a(x1,t1)
a(X,t)
=
ax1
(6.5)
=
is non—zero and finite. =
1
+
+
From (6.1) and (6.5) we have
+
(6.6)
.
Now for the partial derivative ac1/ax1 a(c1,t1)
ac1
= 3(x1,t1)
we have
a(C1,t1) = J
(6.7)
a(X,t)
and on substituting (6.1)2, (6.1)3 and (6.6) into (6.7) we obtain
=
3x1
3x
+
+
—
a(x,t)
+
+ 0c2
which simplifies to give (6.8)
Similarly from a(c1,x1) 1
at1
= — a(x1,t1) = — J
(6.9)
a(x,t)
we obtain
L=ft÷
c{c*+
If we introduce
+0c2
(c
and
ff2
.
(6.10)
by
99
f
IL
z
ie
31
(89)
pUP
=
fxe
(o19) +
=
+
+
+
(zr9)
•
iapio
' '3ZC —
I
—
z
ç
Ix UIo1J
(99)
'(zr9) e
=
no
(ii
+
+
-
2ujsn 1(ir9)
XC
xe
xe
+
—
1
+
woij
——
13e
xe —
1OJ
001
+
+
—
—
ç
— —
(11''x)e
'
I
—
c
(E19) suaflr?nba
xe
xe
xe
xc
xe
xe
+
-
Z9
-
+
NOISfkIdIQ NOLLVflöa
U0T439S
UI
9) (01
I
dnoi2
a3flpap 9q4
(E9)
30 aq4 UOTST13TP
(ci
9)
+ xe
'Xe
xe
xe
+
•
asn 30
31
PUT3
9q4
l9pUfl
UOT
UOTSfl33TP
(r9)
9q4
SUOT43Un3
ipns
xe
z
+
xe
xc
+
(cr9)
'
(cr9)
ST
30 dnoi2
30
STU.L ST
1Cldurrs
30
10I
derivatives to zero.
we deduce that
From the coefficient of
the coefficient of =
11
(t)x
n(t) while
from
we have
ac/st
+ p(t)
n =
(6.16)
,
where the prime here denotes differentiation with respect to denotes an arbitrary function of
t.
t
and
p(t)
On equating the coefficient of
to zero and making use of (6.16) we deduce that
[flutX2
=
denotes
+ P'(t)x) + o(t)
(6.17)
,
where
a(t)
of
in equation (6.15) and using (6.17) we obtain
c
—
1
[n" (t)x2
a further arbitrary function of
+
phI(t)x)
+
a'
(t)
+
n"(t)
t.
From the coefficient
=
from which it is apparent that we require = 0
p"(t)
,
= 0
a'(t)
,
=
—
From these equations it is now a simple matter to deduce the classical group of the diffusion equation, namely = K + 5t + Bx + yxt
where
T1(x,t) =
a +
C(x,t) =
-y
a,
y,
x
2
5, A
Sx
ti
+
-
and
T+ K
A
denote six arbitrary constants and for
comparison purposes we have adopted the same notation used in Bluman and Cole [5] and [18].
Some of these constants give rise to standard or even
trivial solutions of (6.3).
However it is instructive for the reader to
deduce the global form of the one—parameter group and the resulting similarity solutions of the diffusion equation. 102
The constants
K,
a
and
A
represent
respectively the invariance of (6.3) under translations of stretching of
c
The constants
(see problem 1).
B,
y
x
and
and 6
t
and
are
considered in the examples of the following section.
6.3
SIMPLE EXAMPLES FOR THE DIFFUSiON EQUATION
The general classical similarity solution of (6.3) is obtained from (6.2) and (6.18) with all the constants in (6.18) non—zero.
For purposes of illustration
it is useful to consider the solutions arising from one non—zero constant with the others taken to be zero. Example
= 1,
B
I
a = y = 6 =
=
K
0.
=
In this case the global form
of the one—parameter group is obtained by solving
dt1
dx1
dc1
— x1
2t1
—
— 0
subject to the initial conditions x1 = x when
c =
t1
,
0.
=
t
c1 = c
,
(6.19)
,
In this case we find
x1=ex,
c1c,
t1=e2ct,
so that clearly the constant simultaneous stretchings of
B
x
reflects and
the invariance of (6.3) under From (6.2) we obtain the partial
t.
differential equation
+ 2t
x
=0
which on solving gives rise to the functional form previously considered in Chapter 1 (page 7). 6 = 1,
Example 2
a = B=
=A
= K =
0.
In order to deduce the global
form of this group we require to solve dx
dt
=
t1
dc
= 0
x
=
— T c1 103
with initial conditions (6.19). We find = x + ct,
t1 =
t
c1 = c exp[_
,
—
Further
from (6.2) the functional
t
x
=
—
form of
the solution is obtained by solving
C
which yields c(x,t) =
denotes
where
(6.20)
,
an arbitrary function of
t.
On substituting (6.20) into
(6.3) we readily deduce the ordinary differential equation
+
= 0
and therefore
— denotes
where
From this equation and (6.20) we
an arbitrary constant.
see that the constant
also
iS
gives rise to the well known source solution
(1.29).
y = 1,
Example 3 dx1
cx =
=A
8=
dt
=
x1t1
,
= K =
dc
=
,
0. x2
=
together with the initial conditions (6. 19).
t
+
c1
,
(6.21)
From (6.21)2 we have (6.22)
=
and
In this case we have
therefore
(6.21)i becomes
dx1
x1t
dc
(1—€t)
which on integration yields X1 104
= (ict)
(6.23)
Using (6.22) and (6.23) in (6.21)3 and integrating the resulting equation we find
2) c1 =
(6.24)
exp 4(l—ct)J
In order to determine the functional form of the corresponding similarity solution we have from (6.2)
(6.25)
.
On solving this equation we find that
c(x,t) =
e
—x2/4t
(6.26)
, —
t
where
r
denotes an arbitrary function of the argument Indicated.
On
substitution of (6.26) in (6.3) we find that we have simply =
so the constant
c(x,t) =
gives rise to the solution
y e
—x2/4t
+
'
x —
t
which again includes the source solution (1.29) as well as the solution of (6.3) which is the derivative of the source solution with respect to
x.
Thus although no new solutions are obtained by considering separately the constants in (6.18), these simple examples illustrate the basic procedure in simple terms.
In order to obtain non—trivial results we need to consider
the full group (6.18).
This is done in the following section with reference
to moving boundary problems (see also problems 6, 7, 8 and 9).
6.4
MOVING BOUNDARY PROBLEMS
Problems involving the classical diffusion equation (6.3) and an unknown moving boundary
x =
X(t)
occur in many areas of science, engineering and
105
industry (see Ockendon and Hodgkins [21] and Wilson, Solomon and Boggs [22]). The literature on these problems is scattered throughout many diverse disciplines and it is not possible here to consider the subject in detail. The purpose of this section is to identify the moving boundaries which remain invariant under the classical group (6.18).
x =
X(t)
These boundaries
relate to most of the exact analytic results which are available for such problems and therefore might provide a useful guide to the solution of other problems with unknown boundaries.
Typically a moving boundary problem takes the form
O<x<X(t) (6.27)
c(X(t),t) = 0 together with either prescribed on
(X(t),t)
,
—k(t)
(or a linear combination of these)
or
c
=
and prescribed initial data for
x = 0
c
and
X.
We note
that the dot denotes differentiation with respect to time and that for some problems the initial condition on
c
may
not
Such problems are
be present.
However it is instructive to observe the precise nature
clearly non—linear.
of this non—linearity for those problems which can be transformed to fixed Assuming that both
boundary value problems.
zero and that the prescribed data for explicitly involve
'r =
,
=
and with
t
c(x,t) =
=
then
(l,'r)
and
on
or
5C(t)
x = 0
are never does not
we can make the transformation.
X(t)
C(p,'r)
c
X(t)
(6.28)
,
the moving boundary problem (6.27) becomes
,
—
0 < p < 1
,
(6.29) C(1,T) = 0
106
,
(1,T)
=
Thus in principle we have transformed (6.27) to a fixed boundary value problem except that now the equation to be solved is non—linear, although not of the usual type of non—linearity with which we are familiar. for prescribed data
c(0,t) = c0
and
X(0) = a
where
c0
For example,
and
a
are
constants and no initial condition we supplement (6.29) with
T(0)
C(0,i) = c0 , In this
case
X(t)
C =
/2d0}
4(p).
We can readily deduce
(6.31)
,
is a root of
b =
b = c0e
and
(6.30)
.
a solution exists in the form
= c0
where
= a
—b/2
—hcJ2/2
tie
i(t)) is
(or
+ 2bt)2
X(t) = (a2
do1
(6.32)
,
given by
(6.33)
.
We observe that (6.33) actually includes the well known moving boundary X(t) =
as a special case (see Crank [231, page 99).
Moreover
Langford [24] has generated general solutions of the diffusion equation with a moving boundary of the form (6.33).
These solutions have been further
generalized by Bluman [19] and 'I'ait [25].
moving boundaries
page 235) which
X(t)
Here we simply consider the general
due to Bluman 119] (see also Bluman and Cole [51,
remain invariant under (6.18).
Applications of the resulting
similarity solutions involve complicated eigenfunction expansions and we refer the reader to the original papers [19], [24] and [25] for these details as well as for related references. When
deducing
the functional form of the solution
c(x,t)
from (6.2) we
require two independent integrals of the system of differential equations,
107
= E(x,t)
= ?(x,t)c
= n(x,t)
,
(6.34)
.
,
If we suppose
that
dx =
6 35
dt
admits the integral
w(x,t) =
constant then w
is the similarity variable
and from dc = c dt fl(x,t)
with
x
(6 36)
replaced by
x =
x(w,t)
we can deduce (on treating
w
as a
x
and
constant) that the solution takes the functional form, c(x,t) =
w
If we from
in both
and
being regarded as a function of
now consider the boundaries
x1 =
X(t1)
x =
X(t)
t.
left invariant by (6.1) then
we have
dX =
(6
dt
so that the similarity variable w
38)
defines the invariant boundaries from the
equation
w(X(t),t) = where
(6.39)
,
denotes an arbitrary constant.
In particular for the classical
group we can deduce from (6. 18)1, (6.18)2 and (6. 38)
df
K+ót
X
3. )
(6.40)
J
In the integration of (6.40) there are four cases which must be considered separately. Case (i)
ay.
In this case we find that the invariant boundary takes
the form X(t) = At + B + 108
,
(6.41)
where the constants A =
A
and
,
B
are defined by
B
K8—&Z
=
(6.42)
czy-8 We
observe that (6.41) contains
as a special case and that since it
contains six arbitrary constants
8,
y,
6, K
and
it may perhaps be
utilized as an approximate expression for an unknown moving boundary. 82 =
Case (ii)
ay,
Case (iii)
8
=y
X(t) = Kt +
=
0,
t
#
0.
6
(6.43)
.
In this case we find
62 +
=8=y
Case (iv)
w =
For this case we have
+ w0(t+B) —
X(t) =
simply
y # 0.
(6.44)
.
=
0,
6 #
0.
For this case the similarity variable is
and the invariant boundaries are therefore
t =
constant.
In Bluman [19] (and Bluman and Cole [5], page 235) the above moving boundaries are exploited for an inverse moving boundary problem in the sense that the heat input on the moving boundary is not prescribed but rather
determined so as to be consistent with the assumed special form of moving boundary.
The resulting solutions of (6.3) corresponding to the above four
cases are outlined in problems 6, 7, 8 and 9.
6.5
FOKKER-PLANCK EQUATION
In this section we consider classical groups of the form (6.1) which leave the Fokker—Planck equation (6.4) invariant.
Nariboli [20] considers in some
detail a number of special cases of (6.4) and also gives extensive references to physical and biological applications of (6.4).
Bluman [26] gives a
detailed analysis of a boundary value problem for the special case of (6.4) with
p(x)
equal to a constant (see also, Bluman and Cole [5], page 258).
109
Here we first obtain the general form of the group for arbitrary This analysis has not been given previously.
q(x).
results for particular functions introduce the functions = JX
dY
1(x)
and
1(x)
,
J(x)
and
p(x)
J(x)
p(x)
We then illustrate these It is convenient here to
q(x).
which we define by
+ g(x)
=
(6.45)
p(y)
in the interval under
p(x)
Further throughout this section primes denote differentiation
consideration.
with respect to the argument indicated.
From the formulae for the transformed partial derivatives (6.8), (6.10)
we find that the
and (6.13) and making use of (6.4) to eliminate condition for invariance of (6.4) becomes
- lulic. -
+
C
Ic.
+
—2
—
=
+
2
+
is a function of
t
matter to deduce that =
ri'(t)
+ çq'(x))
(6.46)
.
=
n(t),
that
it is a simple
From the coefficient of
only.
is given by
P(X)½ 1(x) + p(t)P(X)½
denotes an arbitrary function of
integral defined by (6.45)i.
110
c(Eq"(x)
we have immediately that
From the coefficient of
obtain
q'(x)c}
—
k
+ (p"(x) +
p(t)
+
2
2
+ (p'(x)
where
(p'(x)
(6.47)
t
and
From the coefficient of
1(x)
is the indefinite in (6.46) we
where
p'(t)
—
—
=
—
—
denotes a further arbitrary function of
o(t)
defined by (6.45)2.
+ a(t)
.acji). t
and
On substituting the above expressions for
into the equation obtained by equating the coefficient of
c
(6.48)
, J(x) E,
is
and
in (6.46) to
zero we obtain the equation,
n"(t) 1(x)2 + p"(t) 1(x) —
n'(t)
=
where the
+
function
o'(t)
—
p(x)½
+
,
(6.49)
is defined by
2p(X)½ J'(x) + J(x)2 -
=
n(t)
4q'(x)
.
(6.50)
In the analysis of (6.49) there are two distinct cases to consider. Case (1)
=
where p(t)
In this case
C11(x)2 + 2C21(x) + C3 ,
C1, C2 and
non—zero.
p(t)
denote arbitrary constants and the functions
C3
and
n(t),
are obtained by solving the following equations,
o(t)
4C1r1'(t)
—
(6.51)
= 0 3C
p"(t) —
o'(t) —
Case (ii)
C1p(t) —
n"(t) 4
p(t)
=
,
—
zero.
n'(t)
34
(6.52)
—
In this case we have
+
=
2C11(x)2 + C3
which upon integration gives
=
C11(x)
2
C4
+ C3 + 1(x)
2
(6.53)
111
where
C4
and
ri(t)
denotes a further arbitrary constant.
a(t) are
determined by solving the following equations,
0
n"(t) — 4C1T1'(t)
In
(6.54)
"I
——
For this case the functions
n
p(x), we obtain the most general
both cases f or given
q(x)
such that
(6.4) admits a classical one—parameter group of transformations leaving the equation invariant, by solving
4q'(x)
+ J(x)2 —
where for case (1)
f(I)
= C112
f(I)
=
f(I)
(6.55)
,
is given by
(6.56)
+ 2C21 + C3 ,
while for case (ii) we have C4
f(I)=CI 2
(6.57)
1
We solve (6.55) by taking defined by (6.45)1.
I
as the independent variable where
1(x)
is
From (6.55) we have
(6.58)
,
and on using J =
(log
+
equation (6.58) simplifies to yield the Ricatti equation f(I)
where
u
u =
If
112
for
(6.59)
,
is defined by
(log
the time
being
—
(6.60)
.
we assume that
u =i1(I)
is the solution of the
Ricattl equation (6.59) then from (6.60) we see that for a given function p(x)
the function
q(x), such that (6.4) admits a classical group, is given
by
q(x) =
=
p'(x)
-
,
1(x)
and we observe that in general Clearly,with
f(I)
(6.61)
,
J
2
contains four arbitrary constants.
given by either (6.56) or (6.57), equation (6.59) has
a number of simple solutions for special values of the constants and
C4.
C1, C2, C3
For solutions of the Ricatti equation we refer the reader to
Murphy [81 (page 15).
Here we give the general solution of (6.59) in terms
of confluent hypergeometric functions, that is solutions of the second order different ial equation
zw"(z) + (c—z)w'(z) — aw(z) = 0 where
a
and
c
(6.62)
,
are constants (see Murphy [8], page 331).
If
c
is non—
integer then (6.62) has linearly independent solutions,
1F1(a,c;
w1(z) =
z)
(6.63)
w2(z) = where
2—c; z)
is defined by
1F1(a,c; z)
k
(a)
1F1(a,c;
k
k=0
where the symbol (a)o =
(6.64)
,
a) =
Is defined by
(a)k
1
(6.65)
(a)k
=
a(a+1)(a+2)
We observe that if
a
(a+k—1)
(k
1)
is a negative integer then the series (6.64) reduces.
to a polynomial expression.
113
(6.59)
In order to reduce we make
the
to a second order linear differential equation
transformation
u(l)
(6.66)
and (6.59) becomes
(6.67)
We consider the two cases separately. Case (i)
non—zero and
p(t)
f(I)
given by (6.56).
In this case we let
2
C
,
v(I)
z)w'(z)
—
z =
=
(6.68)
,
+ and (6.67) becomes
zw"(z) +
—
where the constant
a =
Case (ii)
z =
1
= 0
(6.69)
,
is given by
a
—ci
fcc 1
+
aw(z)
(6.70)
3/2 8C1
zero and
p(t)
—i- i2
v(I)
,
given by (6.57).
f(I)
=
In this case we let
(6.71)
,
and (6.67) gives 2m —
zw"(z) + where the constants
+
a =
and
a
in
z]w'(z)
,
m =
in
aw(z) = 0
—
,
(6.72)
are given by
—
(1+C4)½)
.
(6.73)
+ We note that when
114
C4
is zero the soluticn for this case coincides with that
for case (1) when the constant
is zero.
C2
As a simple Illustration of the above, consider case (I) when the constants and
C1, C2
are such that
C3
a, as given by (6.70)
For case (I) we have from (6.66), — 1)
u(I) =
and when
a = —(1/2)
value —(1/2).
and (6.68)
(6.74)
,
the linearly Independent solutions of (6.69) obtained
from (6.63) are essentially (that Is, apart from an arbitrary multiplicative constant)
*
w1(z) = e
z
eY
I
— z
—i-j
dy
y
(6.75) =
From (6.74) and (6.75) we deduce that for case (I) with
a = —(1/2)
the
solution of the Ricatti equation (6.59) becomes -
(1_z){C5
u(I)
dY) - z½ezl
—
(6.76) z z
C5
denotes a further arbitrary constant and
z
as a function of
I
is defined by (6.68)i.
In the following section we consider a number of special cases of (6.4).
The first example is due to Bluman [261 who gives a similar analysis relating to the confluent hypergeometric functions but with reference only to the special case
p(x) =
1
and
q(x)
arbitrary.
The remaining examples are due
to Nariboli [201.
115
6.6
EXAMPLES FOR tHE FOKKER—PLANCK EQUATION
Example
1(x)
= x
J(x)
,
arbitrary. In this case we have from (6.45)
q(x)
p(x) = 1,
4
q(x)
=
while from (6.50) we obtain 2
4(x) =q(x)
—2q (x)
Further (6.49) becomes ri"(t)
X
8
2
+
p"(t) 2
X
—
—
ri'(t)
=
a
+
÷
q'(x)
As before there are two cases to consider. Case (1)
In this case
non—zero.
p(t)
q(x)
must satisfy the Ricatti
equation q(x)2
— 2q'(x)
p(t) and
and Case
(ii)
p(t)
are obtained from (6.52).
a(t)
zero.
In this case
+ 4,(x) = 2C1x2 + C3
$' (x) so
= C1x2 + 2C2x + C3
that
q(x) 2
Bluman
+ C3 +4
= C1x2
a(t) are obtained from (6.54).
and
and
[See
C
2q'(x)
—
[26]
or Bluman
and Cole [5] (page 258)
for a full discussion and
application of this example.] Example 5 from case (i)
p(x) = 1,
of the
C1=b2, 116
q(x)
=
bx
where
b
previous example we have
C2=0,
C3=—2b,
is a constant.
In
this
case
and therefore
p(t)
n(t),
4b2ri'(t)
—
b2p(t)
p"(t) —
are
a(t)
determined from the equations
= 0
= 0
' \/ +
——
4
/
and
b
2
From these equations we readily deduce n(t) =
a +
+
bt +Ke—bt
p(t) = óe
o(t)=ybe —2bt where
a, 8,
y,
+A,
6, K
and
denote six arbitrary constants.
A
Altogether
using (6.47) and (6.48) we obtain E(x,t) =
bx(8e2bt
ri(x,t)=a+8e C(x,t) = A ÷
2bt
+ Kebt)
+
—
+ye —2bt
bye—2bt
bóxe
—
bt
2 2
— b x
2bt
which is in agreement with the result given by Nariboli [20].
As a simple illustration consider the case constants zero.
8
=1
and the remaining
The global form of the one—parameter group is obtained by
solving dx1 =
dt1
2bt1 ,
=e
dc1
2bt1 ,
2 2 2bt1 = —b x1e c1
subject to the usual initial conditions (6.19). X1
x
=
(l—2cbe
I
1
—
log(l—2cbe
2bt )
)
222bt cbxe (1—2cbe
Further
=
2bt ½
We find
2b )
the partial differential equation (6.2) becomes
117
bx
ac
-b22 xc
ac
+
which has similarity variable
w = xe
—bt
and the functional form of the
solution is given by
c(x,t) = e On
—bx2/2
—bt )
substituting this functional form into (6.4) with
p(x) =
I
and
q(x) = bx
we obtain simply = 0
,
so that
+
c(x,t) = where
The more general solution
denote arbitrary constants.
and
types for this example are summarized in problem 12. Consider the equation,
Example 6 ac
2
a —i
=a
(xc) + b
a
(xc)
ax
where
a
and
denote arbitrary constants.
b
In this case we have the
following results:
q(x)=a+bx,
p(x)=ax, ½
J Cx) = b
,
1(x) = 2 2
b —x
½
+ 3[a) ½
3a
From these results we find from (6.49) that a(t)
are determined from —
b2ri'(t)
=—
Altogether 118
we find,
= 0
,
p(t)
is zero and
ri(t)
and
=
fl(x,t) =
—
where
bx[8e1)t
—
+ ye
b[Bebt
and
3, y
x,
bt
a + A
—
—bt
ye_bt)
—
—
denote four
.A
arbitrary constants.
The various
solution types for this example are summarized in problem 13. ConsIder the equation
Example 7
2 a2 = —i (x c)
+ b
(xc)
ax
where
b
p(x)=x2, 1(x) = and
(6.49)
In this case we have
is again an arbitrary constant.
log
q(x)=(b+2)x, x
J(x)
,
=
(b+3)
,
4(x)
=
(b+1)2
becomes,
(log x)2 +
p"(t)
log
—
n"(t)
a'(t)
n'(t) (b+1)2
Thus we obtain, =
x
+ yt2
+
p(t) = K + cSt
a(t) = where
a,
+
-
5, K
13, y,
-
and
A
+ A
denote six arbitrary constants.
From (6.47)
and (6.48) we find x + (K-h5t)x T)(x,t) =
=
a + —
—
+ (log x)2 —
(b-i-i)2
{S +
(a+2$t+yt2) —
log x
—
(b+3)(K+ót) +
A
and the various solution types are simunarized in problem 14. 119
Finally in this section we remark that Nariboli [20] also details the analysis for the following two equations., 2
ac =
a
r
i(1—x ) Ct
ax
ac
1
a
at
m
where
ma rm
1x
4ax For these equations we have
denotes an arbitrary constant.
respectively in the notation of (6.4) p(x) = (1—x
p(x) We
x
22 )
q(x)
,
= —4x(1—x
mi-I
mx
q(x)
,
2 )
m
refer the reader to [20] for the subsequent analysis of these equations.
6.7
NON-CLASSICAL GROUPS FOR THE DIFFUSION EQUATION
In this section for the diffusion equation (6.3) we make use of equation (6.2) so that the left—hand side of (6.15) depends on ac/ax.
We then equate the coefficients of
c
c
and
only through ac/ax
c
and
to zero and obtain
two equations for the determination of the non—classical groups of (6.3). For the one—parameter group (6.1) we introduce
A(x,t)
and
B(x,t)
defined
by
B(x,t)
,
A(x,t) =
(6.77)
,
=
so that (6.2) becomes
—Ac—B-—. ax at
(6.78)
On differentiating (6.78) partially with respect to
x
and making use of
(6.3) it is a simple matter to deduce 2
axat
r
(
ax
jax
(6.79)
On substituting (6.78) and (6.79) into (6.15) and then equating to zero the 120
______
coefficients of
and
c
then remarkably the resulting equations
ac/ax
simplify to give,
aAa2A ax
2A3B ax'
2
(6.80)
2 ax
ax
2
3x
These equations determine the non-classical group of (6.3).
Although equations
(6.80) are non—linear and clearly a good deal more complicated than the original problem, we observe that any special solution of (6.80) can be employed to reduce the diffusion equation to an ordinary differential equation. If we assume that
for some function
B =
integrating (6.80)2 with respect to of
I'(x,t)
then on
and neglecting an arbitrary function
x
t, we have a2
(6.81)
ax
On substituting these expressions for
A
and
namely
single equation for the determination of aci,
2
atax
ax
2
49x
into (6.80)i we obtain a
B
2
a3ci
8
ax
ax
2
ax
2
2
+
8
ax
2 ax)
+
ax axat
= 0
We simply note that the classical
This equation is clearly complicated.
group (6.18) arises from the case A(x,t) = 0 ,
where
h(x,t)
B(x,t)
=
—
(6.82)
.
lxxx = 0
and that if
C(x,t)
is zero then
(6.83)
,
denotes any solution of (6.3).
The resulting solution
c(x,t)
satisfies ac ax
h 121
Example 8
•(x,t) = ci log x
If
a = —(1/2)
a =
or
A(x,t) = 0
—(3/2).
a = —(1/2) then
If
B(x,t)
,
then from (6.82) we have either
—
=
so that (6.78) becomes ac at
which
19c_
has general solution
c(x,t)
2
=
j- +
=
On substituting this functional
t
form
into (6.3)
= 0 ,
and
therefore +
c(x,t) =
where If
denote arbitrary constants.
and
a = —(3/2) A(x,t) =
+
—
then
B(x,t)
,
=
—
I
and (6.78) becomes ac
3ac
3
x
2
c(x,t)
=
x4(w) ,
w =
j— +
3t
and from (6.3) we deduce c(x,t) =
where
122
and
+
3t)
+ q1x
denote arbitrary constants.
we find that
PROBLEMS 1.
Show for the diffusion equation that the one—parameter group arising from the constants
and
K
A
in (6.18) (that is, with
=
y
= 6 =
0)
becomes
t1t+ac,
X1=x+KC,
c1=ecc,
and that the functional form of the solution of (6.3) is c(x,t) = e
Ax/K
4(czx—Kt)
and relate this
Hence deduce the ordinary differential equation for result to problem 12 of Chapter 5. 2.
For the diffusion equation with the constants
and
6
cx =
y
= K =
A
= 0
and
8
non—zero in (6.18), show that the global form of the one—
parameter group becomes, + t
=
,
t1
c1 = c exp
=
+4
Show also that the functional form of the solution is
c(x,t)
tJ
48
and obtain the resulting ordinary differential equation for 3.
With
where
c(x,t) =
y =
(x/X(t))2
4.
deduce from the diffusion
equation (6.3) X(t)2
+ =
Hence conclude that if
X(t)
+ YX(t)X(t))}
takes the form
X(t) = (a+28t)½
where
a
and
8
are
arbitrary constants then (6.3) admits separable
solutions of the form 123
c(x,t) =
and
satisfies
4)(y)
y4"(y) +
—
denotes a further arbitrary constant.
a
Show by a simple change
of independent variable that this equation reduces to the confluent hypergeometric equation (see Langford [241). 4.
With
where
c(x,t) = dT =
X
=
1
X(t) =
(x+2Bt+yt2)1
(*)
and 1
=
2• xXt
exp
4X(t)
½
X(t)
verify that the diffusion equation (6.3) simplifies to give
aT
2
ap
2
where
5.
4
½ With
Continuation.
where
c(x,t) =
defined by equation (*) of the previous problem and
exp {-
=
show that the multi—dimensional diffusion equation, ac
—
a2c
2+
ax
(2v+1) ac x
ax'
becomes
ap
where
124
5
-
+ (2v+1)
=
p
ap
4
is as previously defined (see Taft [25]).
p,
T
and
X
are as
is given by
6.
if
82
82 > cxy show from (6.18) that the equation
and
corresponding to (6.37) becomes
T
I8+yt—(8
(a+28t+yt
I
2 ¼
A
—cry)
2
[frf-yt+(8 -cry')
)
½
(A2+yw2) +
times exp {_
where
2
is given by (6.42)1
and
w
and
(cx+28t+yt2)½} , p
are defined by
x—(At+B) L (cz+28t+yt )2
=
U
[62+A2(cry 82)])
21
1
2(8
where
B
is given by (6.42)2.
By substituting (**) into the diffusion
equation (6.3) deduce the following ordinary differential equation for
+
+
where the constants D = If
E
82 < cry
= 0
and
D
E
are given by
[A2(82—ory')—62] — A
=
show that in place of the square bracket in (**) we have
+
exp
[The solutions for
4(w)
]) tan1
+
can be expressed in terms of confluent
hypergeometric functions (see Bluman and Cole [5], page 215).] 7.
If
82 = cry
and
y # 0
deduce from (6.18) and (6.36) the following
functional form
c(x,t) =
where
w,
L
(t+8)½ and
exp
-
w2(t+e)} ,
+
(***)
-
M are given by
125
=
(4
I
1
+
L = ½(K—58)
Further
1
1
+ 12J (t+8)f (t+8) M
,
(6.3)
show from
¼(62+28+4A)
=
satisfies,
and (***) that = 0
—
[The solutions of this equation can be expressed in terms of Airy functions (see [5], page 217).] 8.
If
8
=y
and
= 0
a
show from (6.18) that (6.37) becomes
0
÷ At -
c(x,t) =
wt}
,
{-
where
w
is given by 2
(4
c5t
=x
—i--
—
Kt
Deduce from (****) and (6.3) that
=0
+ K4'(w) +
[This
satisfies
equation also has solutions expressible in terms of Airy functions
(see [5], page 218).] 9.
If
a =
8=y
= 0
and
CS #
0
show from (6.18) that the source solution
results, namely
I
c(x,t)
exp
(t+Ic)½
=
r
4(t+K)
denotes an arbitrary constant.
where 10.
(
i
Show that the equation
I can be transformed to the classical diffusion equation — 2
126
'
by means of the transformation, c(x,t) =
a(x)C(y,t)
if and only if
takes the form,
p(x)
p(x) = (C1x + C2) where 11.
and
C1
y =
,
4/3
denote arbitrary constants.
C2
Observe from example 5 that the similarity variable
w
for the classical
group of ac =
2 ac
a
+ b
(xc)
ax
Is
obtained by integrating,
dj
bt —bi (&e+Ke
x
)
dt —
y,
where the constants
and
K
are the same as those used In
By making the transformation
example 5.
p= e
&
2br
show that the IntegratIon of (+) can be effected by considering separately four distinct cases In a completely analogous manner to the corresponding integration for the classical diffusion equation.
Deduce
In each case the similarity variable. 12.
WIth reference to example 5 establish the following solution types for which
0
as
t
0
and
c(x, t)
is given by
c(x,t) = (I)
= —y = 1,
=
= XT + ó* + K* —
=
0.
6*(12_1)½
T(T2_1)A '2exp
bo*[w(T2_l)½
+
127
= —a = 1,
(ii)
y
0
(T2_1)½
+ K*
2 T—1
T
times
—i---
=
=
=0.
x+
=
bA*4(w)
—
•"(w) +
T —
exp {_
205*+2K*)WT(T2_1)cj}
=0
— bw'(w) —
(iii) B=—y=1, a=0. TX
+ tS* + K*T2 , (T4_1)1 T2_1I
T
—
{
=
=0
—
(iv)
B
=y =
—(x/2)
= 1.
+
(13=
(T2—1)
(T2—1)2 2 2
=
(T21)½ -
{ LT2_1) b {(1+2X*) + 4(tS*+K*)bw — bw2}
(v)
=
TX
03=
=
1,
a =
+ tS* + y1(T2—y2) 2
22½'
[(T —1)(T —ji )]
128
(tS*+K*)21fl
+
exp
= 0
*(w,t) =
2(1—i.i
2
T
k[(w2+12)12
exp {
1
22
times
b{(1+2A*)(1+V2)
—
2(2K*
2
)
—
q)"(w) —
—
V
2
X*
+ (1+112)6*)
—
22
'
= 0
—
(1+112)
=
2(1-jA
)
In each of the cases
Te
2 )
is defined by
bt
6*, K*
and the constants based on
I
1
'
2
and
K
6,
A
A*
and
are appropriately redefined constants
respectively and do not necessarily refer to
the same constant in each case. 13.
WIth reference to example 6 establish the following solution types where is given by
c(x,t)
c(x,t) = (1)
XT (A)
= T(T—1)
= (T—1)
—
+
(ii)
w =
A*—1
(t—
=
(i—i)
= 0
exp {—
+ (2a—bw)4'(w) — b(A*+1)4(w) = 0
bix 2
(iii)
=
,
= T
(121)
w =
XT
(11)2
exp
a(t —1)J
(T+1)
aw4"(w) + 2a4'(w) —
(iv)
(i—i)
ij,(w,t)
=0
2bA*
=
T
(T—1)
2
exp J—
bt2x a(T—1)
+
(T_1)1 J
129
+
(v)
2
U) = (r—1)(wr—1)
{-
(T-1)(UT-1)
=
ac$"(w) ÷ In
=0
+
+
each of the above cases
is
T
—
4
= 0
defined by
T=ebt and 14.
A*
denotes an arbitrary constant.
With reference to example 7 establish the following solution types where c(x,t)
is given by c(x,t) =
(i)
a=B=O, y=1. K
2t —
*(w,t) =
exp
{x +
(w)
K
—
(ii)
a—I = 0,
*(w,t)
—
—
130
t-
—
÷
(w)
+
—
2(b+3)t]}
=0
= (1/2).
tAexp
ir
+ (b+3)cS
4'(w) — A4(w) = 0
+ (b+1)
jt
½ —
(2c5+b+3))'
(iii)
I =
B = 0,
W =
log
ci =
1.
x — Kt —
(St2
+ (b+3)
= exp
(K-I-b+3)t2
-
+
(iv)
+
B = 0,
cx =
K +
(k?)]t (S2t3
(b+3) 4[A
+
K+
2
(b+3)2
+T14)(W)0.
y =1.
(S+Kt+logx =
P(t)A
—
exp
t2+u2)¼
—
(b+3_K)(t2-I.p2)½}
(22 (4
4"(w) +
15.
—
For the boundary value problem, ac
I
c(x,O)
c(x,t)
where
c0
(t>O,
a
c05(x—x0) ac ,
and
ox
x0
(x,t)
+ 0
as
x +
and
provided
the functions
satisfy = 0
Hence with
±00
denote arbitrary constants, show that the initial
condition remains invariant under (6.1) fl(x,t)
—°°<x<°°)
,
fl(x0,O)
10 E 1(x0)
= 0
,
1(x0,O)
=
show that the functions
(x
p(t)
and
a(t)
131
in (6.47) and (6.48) satisfy
p(O) = —
= 0 ,
ri"(O)
a'o'
16.
r2 + 0
8
10
p'(O) 2
—
0
2
For case (I) of section 6.5 show that one group leaving
Continuation.
the boundary value problem of the previous question invariant is, C(x,t) =
2p(x)1
82 = C1.
c(x,t)
where
,
fl(x,t)
= 0
+ (C2/B)(1—cosh(Bt)) — 81 cosh(Bt) — J sinh(8t),
=
where
sinh(Bt)
Hence show that the solution takes the form
4(t) (I)
exp
"½
=
812
1
—h--
coth(8t) +
denotes an arbitrary function of
28 sinh(8t)
3
From the partial
t.
differential equation deduce that =
c(x,t)
+
C)2
exp
[p(x)sinh(8t)]
B
+ + C2
c21
+
where
a = [c1c3
—
/
4C1
1
Blo
28 sinh(8t)
denotes an arbitrary constant.
and
Showthatas t÷O,
c(x,t)
[p(x)Bt]½
exp
I '.
t
and observe that for given 1(x) the constant the condition J
132
c(x,t)dx
— c0 .
is determined from
17.
For case (ii) of section 6.5 show that the functions
Continuation. n(t)
and
o(t)
are given by
.2 (st) sinh
fl(t) =
822 o(t) =
C
= C1.
where again
sinh
—
—
2
(st)
Deduce from (6.2) that the functional form of the
solution of the boundary value problem of question 15 becomes Yt
c(x,t) = exp {—
where $(w)
is the similarity variable,
w =
y = C3/4
and
satisfies the differential equation C4
÷
—
—i
4(w)
= 0
4w With
and
ci
defined by 4(w)
=
=
so that —
+
where
n = (1+C4)½/2.
correct behaviour as
where
0
,
Show further that the solution 0
t
previous question when
=
=0
[i(ci)
is given by
1(d))
is a constant and
function of the first kind.
giving the
and agreeing with the result of the
C2 = C4 = 0
+
$(w)
I
n
denotes the usual modified Bessel
Verify that as
t + 0
4t
133
18.
Obtain the solutions of the boundary value problem of
Continuation.
question 15 for the following three equations: 2
2
=a
(ii)
(iii)
where
(xc) + b
and
(xc)
(x2c) + b
=
a
(xc)
b
denote arbitrary constants.
Answers:
a'
(i)
c(x,t)
(ii)
c(x,t) =
where
b
C01
m(t)
m(t) =
1½
—bt ,
b(x—x0e
c(x,t) =
1
L
are defined by
y(t)
4Ort)'ixI
134
exp
)
2(l=e2bt)
=
b
a(1—e
(iii)
—bt2
exp
_2btJ
y(t)
and
1
Ut)
+ (b+1)t]2
1
—
4
t
7 Non-linear partial differential equations
For non—linear partial differential equations we need to consider more general transformations which leave the given equation invariant.
chapter for a single dependent variable x
and
In this
and for two independent variables
c
we consider one—parameter groups of the form
t
f(x,t,c,t)
=
x+
+
0(c2)
= g(x,t,c,c)
=
t + cn(x,t,c)
+
0(c2)
+
0(c2)
x1 =
c1 =
h(x,t,c,c)
= c
+
For known functions
fl(x,t,c)
and
(7.1)
,
?(x,t,c)
the similarity
variable and functional form of the solution are obtained by solving the first order partial differential equation,
+
fl(x,t,c)
= ?(x,t,c)
(7.2)
.
In the first section we give formulae for the infinitesimal versions of the first and second order partial derivatives of completeness we also give formulae for make no use of these results.
c(x,t). and
Again for
a2c/at2
although we
In the second section we deduce the classical
groups of the non—linear diffusion equation, namely ac =
where
D(c)
a
I
1D(c)
ad
rj
(7.3)
'
denotes an arbitrary function of
c.
Equation (7.3) is well
known in the literature (see for example, Knight and Philip [27], Munier et al [28] and Shampine [29]). D(c) = cm
In particular the power law diffusivities
have received a good deal of attention (see Babu and Genuchten
[30], Munier et al [28] and Crundy [31]).
The results of section 7.2 are due 135
to Ovsjannilcov [7] and Bluman and Cole [5] (page 295).
In section 7.3 we
briefly consider the non—classical approach for the non—linear diffusion equation (7.3)
Although we present no new results in this section the
governing equations are summarized for the reader interested in pursuing the matter further.
In section 7.4 we give two results for equation (7.3) due to
Munier et al [28].
Although not directly related to group methods these
results involve important transformations of non—linear diffusion equations. The first result shows that every equation of the form (7.3) can by a sequence of transformations be reduced to an equation of the form D(x)
ac
1
(7.4)
,
so that the power law diffusivity
D(c) =
c2
plays an important role.
The
second result due to Munier et al [28] is that the most general inhomogeneous and non—linear diffusion equation with diffusivity
D(x,c)
which can be
reduced by transformations to the classical linear diffusion equation (6.3) is given by equation (7.53).
7.1
FORMULAE FOR PARTIAL DERIVATIVES
In this section we deduce the infinitesimal versions of the partial derivatives
ac/ax,
ac/at,
a2c/ax2,
a2c/axat
and
a2c/3t2.
We again use
the convention that subscripts denote partial differentiation with c
as three independent variables.
ax
cax'
x
at
t
x, t
Thus for example we have,
catS
Now either directly from (7.1) or from the definition of a one—parameter group we have
136
x = x1 —
+ O(C2)
t = t1
+ 0(c2)
—
and
and therefore ax
=
up to order
we obtain ax
+
lx +
—
1
c
=
÷ 0(c2)
+
(7.5) at L
First
=—
cii +ri —
for
at = 1
+
—
+
+
we have
ac/ax
3c
x
1
at
and using (7.1) and (7.5) we obtain aC1
=
+ c1c
+
+
11
+
I
x
I.
+
+ s
ft)}
+ 0(c2)
÷
which simplifies to give
I
3C = —
+
3c
c
÷
3c ac} —
—
+ 0
—
2
)
(7.6) Similarly for
we have
ac/at
aC
a
at so that from (7.1) and (7.5) we obtain ac1 =
+ c
3t1
x
+
+
c
I.
+ 01c2)
which
becomes aC1
= 3c
÷ c{
ac +
ac
ac
—
c
÷
0c2 (7.7) 137
Again for convenience we introduce
+
=
-
and
-
-
—
—
112
such that
ac
—
(7.8)
ac
ac
+
112 =
'
so that we have simply
=
2
+ 0 c)
÷
(7.9)
.
We observe that (7.8) can be written alternatively as
axax (7.10) —
at
For the infinitesimal version of 2 a
c1 =
=
T
1
ax
+
1
which using (7.5) and (7.9) gives
Cl x
+
+
C(11
nx +n
+
c
ad1
On simplifying this result we obtain a2 c1
2
=
+
I
+
C
From the first equation of (7.8) we have
138
+ 0(c2)
.
(7.11)
= XT
XX
+ )
xe
3X
—
Z
xe
ie ulolJ (11'L)
z
=
b
(zr'L)
+
+
3X
xe
'Xe +
(
33
—
—
—
xe
Ixe) -
+
ei't
xe
-
33U
-
+
J
U
xo
XQ)
Ixeiie +
(crL)
ioj =
uioij e
(c'L)
z
=
pirn
Z(8.1)
z =
+
ei't
JLe)
+
—
+ 3
1
x
+ U
+
ezrnpep
2UTSn z
=
Uj
(Xe lexe
JLeJ
+ 3
—
+
J
+
+ U
(cri)
ptw (cl'L)
woia
i)
(8
eq
jo
L)
uz.toqs
('iT
eq eqi ewes 2uisn
10 L) (cT
eqi
z
=
+
'ie'xe
+
ix
+
— —
33
eJ
—
ie
-
+
U
X3
—
xc
-
-
xc
—
ie
-
—
+
J
lie xc z
-
UZ
+
+
J
(91L) U93
B
z J
+
lie
ie
ie
1
—
(ielxe
-
+
ie
-
-
+
+
{iexe1ie
(LIL) (LIL) puB
pue
ZL
'WDISSVID
aq
LUoiJ
(EVL)
iCq
x
U
MOISIkMIcI
(EL) z =
xc aiaqz.i Os?'
inoq2noiqi
[Xe)
(8rL)
'
qipt i3adsal oi
indicated.
Now using
D'(c1)
÷ EçD'(c) + 0(c2)
D(c)
D(c1) = =
D'(c)
+ C1D"(c) + 0(c2)
we find from (7.6), (7.7) and
(7. 13)
that (7. 18) remains invariant under (7.1)
provided
+
= D(c)
—
—
+
—
nxx
ac
ac
ac
-
+
+
x
Ô(c)
(7.19) where the last term,
involving the two curly brackets, arises from
eliminating
by means of equation (7.18).
On equating the
coefficients of the various partial derivatives in (7.19) to zero we obtain the following equations: 2
c
2
nx=o' cc
—
+ D'(c)fl
c
= 0
0
141
2
in
,
c
ac ac t
C
t
x
0
XC
+ D'(c)fl X = 0
—
xx
+2D(c)C xc +2D'(c)C =
—D(c)?
,
C
+ D(c)T)
—D(c)ri
c
=0,
D'(cfl
+
t
x
x
xx
xx
+
D'(c)
D(c)
=0
0,
(7.20) (7.21)
,
=0.
(7.22)
From the first seven of these equations we can readily deduce =
n = n(t)
,
+ D(c)
c
—
=
cc
=
0.
fD(c) 1
—
ax so that 2
t.
But from (7.21) and
(7.21)
From
n'(t)'
we obtain
,
(7.25)
)
either ax
=
(7.26)
,
or the diffusivity '
ID'(c)J
D(c)
is such that
,,
=0
is
D(c) = a(c+8)m
142
and
+ 4(x,t)
n'(t) — 2
and therefore =
x
deduce
(7.24) we can
where
(7.24)
,
denotes an arbitrary function of
where
that
(7.23)
,
a, B
and
,
m denote arbitrary constants.
(7.27) If (7.26) holds then from
(7.20) and (7.25) we can readily deduce
x + K
=
fl(x,t,c)
= 6 +
yt
,
A
denote arbitrary constants.
(7.28)
= 0
where
and
y, 6, K
(7.28) is applicable to all functions Alternatively if
—
D(c).
has the form (7.27) then from (7.25) we have
D(c)
(c+B)[2
=
We note that the group
r)'(t))
(7.29)
,
and on substituting this expression into (7.20) we find
(7.30)
,
_D(c)(3
while substitution of (7.29) into (7.22) and making use of (7.30) gives
(7.31)
.
=
ax We see that (7.30) and (7.31) give rise to two
=
while for
a2
we have,
fl"(t)
=
K + Ax
fl(x,t,c) = 6 +
in
= 0
we have
in
F(x,t,c) =
while for
namely for all constants
= n"(t) = 0
m = —4/3
Thus for all
cases,
yt
,
=
(c+8)(2X—y)
= —4/3
we have
(7.32)
143
E(x,t,c) = K + Ax + px2
where
rl(x,t,c) =
tS
C(x,t,c) =
-
+ yt
+ 2A
5, K, A
y,
(7.33)
,
denote
and
-
y) The resulting
arbitrary constants.
ordinary differential equations corresponding to (7.28), (7.32) and (7.33) are given in problems 1, 2 and 3.
Source solution of (7.3) with a power law diffusivity.
Example 1
(7.34)
,
such that
vanishes at infinity while initially satisfies
c(x,t)
c(x,O) = c0
where
c0
and
delta function.
Consider
tS(x)
(7.35)
,
denote arbitrary constants and
m
cS(x)
is the usual Dirac
Noting the elementary property (1.24) of delta functions we
see directly that (7.34) and (7.35) remain invariant under the one—parameter group
x1ex, c
that
t1=e (m+2)c t
,
c1=e —c c
is,
E(x,t,c) = x
,
ri(x,t,c)
=
(m+2)t
This equation corresponds to (7.32) with y = (m+2).
with
B
= —c
,
B =
=K
We note that the more general case with
=
0, D(c)
(7.36)
.
A =
1
and
given by (7.27)
non—zero appears not to admit a simple group leaving (7.35)
From (7.2) and (7.36) we see that the similarity variable and functional form
of the solution are obtained by solving x
r + (m+2)t
We find that
144
=
-c
where
i
,
c(x,t) =
(7.37)
,
On substituting (7.37) into (7.34) we obtain
n =
+
= 0
vanishes
and since
at infinity, the constant of integration is zero
and we obtain
+
= 0
A further integration gives
2)m
I
=
where
'
2(m+2),)'
—
(7.38)
denotes an arbitrary constant.
C
With
C
given by
2
C
(7.39)
= 2(m+2)
zero for
we take
c(x,0)dx
>
so that the condition
= c0
J
becomes
I m
1
2
w1 —
u
2i
= c0
j
1
Thus with
w =
(cos0)
I
J
where
sin
0
and using the formula
d0= 12
r(x)
constant
C
m
denotes the usual gamma function, we can readily deduce that the in (7.38) is given by
145
In
C
2
13
-
in
—
2(2-I-in)
0
(2-fin)
in
r[i
(7.40
.
+
Thus altogether from (7.37) arid (7.38) we have that the source solution of (7.34) is given by 1
1
c(x,t) =
1
t
c(x,t) = 0
where
n =
I
mx
—
2 I
- 2nt
2(m+2)t ,
(m+2)1,
wi
ii)
'
<
(7.41)
J
>
w =
C
,
Is given by (7.40) and
is defined
by equation (7.39).
7.3
NON—CLASSICAL GROUPS FOR NON-LINEAR DIFFIJSI ON
Although there are no known non—classical groups of (7.3) we derive here the governing equations in order to illustrate the non—classical approach in a non—linear context.
With
A(x,t,c) ,
A(x,t,c) =
and
B(x,t,c)
B(x,t,c)
defined by
(7.42) =
we have from (7.2)
(7.43)
On differentiating (7.43) partially with respect to (7.43) to eliminate
and
ac/st
x
and using (7.18) and
respectively it is a simple matter
to deduce
=
2
+
-s-fl-) +
(7.44)
— Bx
+
—
We remind the reader that the subscripts refer to partial derivatives of functions of three Independent variables
146
x, t
and
c.
Writing
(7.45)
and substituting (7.43) and (7.44) Into (7.19) we obtain the following cubic expression in
0, namely —
+
—
—
+
=
—
—
2D(c)
[A
+
—
+ 2D'(c)O
+
—
-
+ D"(cfl02
+
+
[B
(A—B0—D'(c)02)
+
(7.46)
.
On equating to zero the coefficients of following equations for the determination of D(c)Bcc —
o2}
— Bc)
-
—
—
+
[Ac — B
D'(c)B
0
00
and
A(x,t,c)
and
we obtain the B(x,t,c):
= 0
2BB
o2,
(7.47) 0
+ 2[D(c)A ]
B
,
xc
t
00
,
A
t
=
D(c)A
We observe that with
xx
—
— 2AB
D'(c) 2AB = D(c)B — 2BB + AB xx x c D(c)
x
+ A2 D'(c)
D(c) = 1,
D(c)
A =
a(x,t)c
and
B =
b(x,t)
(7.47) reduce precisely to (6.80) for the two functions
a(x,t)
equations and
b(x,t).
Equations (7.47) are recorded for purposes of illustration and we make no attempt here to obtain special solutions.
147
TRANSFORMATIONS OF THE NON-LINEAR DIFFUSION EQUATiON
7.4
The most widely known tranformation of a non—linear partial differential equation is for Burgers' equation (see for example [32], [33] and [34]) 2
3u
u
the classical diffusion equation, assuming that
(7.48)
D
is a
In this section we give two Important results for the non—linear
constant.
diffusion equation (7.3) due to Munier et al [28].
A number of related
transformations can be found in the literature (see for example, Knight and Philip [27] and Storm [35]).
The first result due to Munier et al [28] is that every non—linear
diffusion equation of the form (7.3) can be transformed to the following equation with a simpler non—linearity, namely
—i
v where
v(c,t)
(7.50)
,
is essentially the flux associated with equation (7.3).
see this we define
(7.51)
.
On multiplying (7.3) by
D(c)
partially with respect to
=
I1D(c)
We now introduce 3v
by
u(x,t)
u(x,t) = D(c)
j—j
v(c,t)
x
and differentiating the resulting equation we find
(7.52)
.
u(x,t)
so that (7.52) becomes
I '
148
To
which on usIng (7.3) and (7.51) simplifIes to give (7.50).
We note that the
equivalence of (7.4) and (7.50) is readily seen. The second result due to Munier et al [28] is that the most general inhomogeneous and non—linear diffusion equation with diffusivity
D(x,c)
which can be transformed to the classical diffusion equation (6.3) takes the form
at
A.
(7
ax
and
y
6
denote arbitrary constants.
53)
In order to see that
(7.53) can be reduced to the classical diffusion equation we can without loss of generality consider the equation, ac
a
,
at
ax
llcJ
Instead
ac ax
of working with (7.54) with independent variables
(x,t)
we consider
the same equation
for
aw
a
at
ac
(7.55)
acf
ltwJ
v(c,t). Making the transformation w(c,t)
=
(7.56)
v(c,t)
it is a simple matter to show that (7.55) becomes av
2av 2
=
v
(7.57)
.
ac
This
is clearly the same equation as (7.50) with
by introducing
x
D(c)
unity and therefore
such that
v(c,t) E u(x,t)
,
(7.58)
equation (7.57) is equivalent to the classical diffusion equation (6.3) for c(x,t).
149
PROBLEMS
1.
For the non—linear diffusion equation (7.3) show that the similarity variable and functional form of the solution corresponding to the group x + K,
E(x,t,c)
rl(x,t,c) =
2(t+5),
0
are respectively x+K (t+5)
Hence show that the resulting ordinary differential equation is = 0
D(4)4"(w) + D'(4)q'(w)2 2.
For the non—linear diffusion equation with m
D(c) =
show that the similarity variable and functional form of the solution corresponding to the group =
(1+A)x + =
K
,
rl(x,t,c)
=
2(t+6)
2A —
are given respectively by
=
Ix +
(t+6)
K ,
-
c =
2
Show that the resulting ordinary differential equation is
+
+
(i-i-A)
u4'(w) —
= 0
Show that this equation can be reduced to a first order ordinary differential equation by observing that the above equation remains invariant under the one—parameter group of transformations c
'so
2dm
3.
For the non—linear diffusion equation with D(c) =
the special case of the group
consider
jix2 ÷ (l+A)x ÷ K,
=
C(x,t,c) =
=
2(t+6),
(c+8)(2px+A)
—
for which the constants
fl(x,t,c)
K,
A,
and
ji
m satisfy
(A÷l)m = 4jiK
In this case show that the similarity variable and functional form of solution are =
c =
(t+6) -½
1
exp
-2
÷ (A÷I))
-8 ÷
exp
Show that the resulting ordinary differential equation is — —
÷ IL
= 0
which can be reduced to a first order ordinary differential equation by observing that the above equation remains invariant under the group
41e—3c/2
e
4.
Show that the non—classical approach applied to a
F(x)
Ii
gives rise to the following four equations for the one—parameter group (7.1), B
B
cc
C
151
A +
+ c2[F(x)—3JBB
= 0
(AB)] + F'(x)B2 + {2[1]
-
—
BA}
-
2
2
F(x) [A + AA I + t c
where
A(x,t,c)
and
F' (x)AB
—
AA
c
=
2 C
- 2AB
x
2 c
are defined by (7.42) and subscripts
B(x,t,c)
denote partial derivatives with respect to the three independent variables 5.
x, t
and
c.
The non—linear axially symmetric diffusion or heat conduction equation in cylindrical and spherical regions can be transformed into an equation of the form
kac
a2c
ac
(*)
ax
where
k =
and
1
k
regions respectively.
2
corresponds to cylindrical and spherical
By considering the classical invariance of (*)
under the one—parameter group
x + cE(x) + 0c2)
x1
t + cri(t) + 0c2 =c
+ CC(x,c) + 0(c2)
show that for both values of constant
k
a group exists if either
f(c)
or if
f(c) = where
a, B
situation 152
and
m
denote arbitrary constants. In the latter
show that for both values of
k,
is
= x, where
y, ó
n(t)
and
= (nry+2)t +
denote further
X
For the special case of
k =
?(x,c)
cS,
arbitrary
=
yc +
X
(**)
,
constants.
show that a more general group than
1
(**) exists provided f(c) =
and deduce for example that (*) admits groups of the form,
where again 6.
=
(log
y
and
Continuation.
x +
(Wy)]
x +
y)(c+$)
cS
n(t)
,
=
2t +
S
denote further arbitrary constants.
Deduce the similarity variables, functional forms of the
solutions and the resulting ordinary differential equations for the groups given in the previous problem. 7.
Observe that 2
ax
2
xax
remains invariant under the group, c
x1 = e x,
t1 = e
(2+mn)c
c1 = e
t,
ne
c
Use this group to deduce the source solutions of (***) for k = 2
n =
given that for these values of
k
the group with
k =
n =
—2
1
and and
respectively leaves the appropriate initial condition invariant
—3
as well.
[This is because with rectangular cartesian coordinates appropriate initial condition for
k =
1
(X,Y,Z)
the
is
c(X,Y,O) = c0S(X)ó(Y)
while for
k = 2
we have
c(X,Y,Z,O) = c0cS(X)S(Y)S(Z)
,
153
where as usual
denotes a constant specifying the strength of the
c0
source.]
8.
Show that 2 ôc
2 ac
13cac
at'
ax2
remains invariant under x1 = x +
+
0(c2)
y1 = y +
+
0(c2)
=
t+
+
0(c2)
2 y--1—-1--- +0(c),
provided
satisfy the Cauchy—Riemann equations,
and
E1(x,y)
namely
and 9.
x, (3, y
and
ô
ô = 0
For
Continuation. =
x
denote arbitrary constants. and
and
2
2
given by
= 2xy
—
deduce the following similarity variables and functional form of the solution,
22 (x+y)
I
yx
,
2
2
y
2
2
(3(x+y) Hence from (+) deduce the following partial differential equation for
aw)
154
aT
B
10.
For
Continuation.
= e
and
= 0
nx = e sin fly
nx
cos fly
given by
and
,
deduce the following similarity variable
and functional form of the
solution, = e
-nx
sin ny
,
T
exp ft
t
=
4(w,t) —
c(x,y,t) =
e
-nx cos
log(sin ny)
fly —
Hence from (-I-) deduce the following partial differential equation for 4(w,t)
a 2
n
3T1T 3TJ
2
(nw)
2
e
2
[Theabove six problems are due to Nariboli 11.
Show that the source solution for Burgers' equation, namely au
+ u 3u = D
2 3u
3x
u(x,0)
= u0 5(x)
x-'±°°,
as
where
u0
is a constant, remains invariant under the one—parameter
group x1
= e cx
t1
,
= e
2c
t
,
u1 = e
—c
u
Hence deduce that u(x,t) =
and that -2De [c
—w
2
/4D
+
]55
where w =
and
is a constant..
C
condition that the constant —
C
is given by
C
sinh(Du0)
—
12.
Deduce from the initial
Show that the classical groups of the non-linear wave equation, 2 ac
2
=
f(c) 2ac —i
where
f(c)
(-H-)
,
ax
at
is non—constant, are summarized by the following three cases:
f(c) arbitrary =
yx + 6
rI(x,t,c) =
yt: + K
= 0
(ii)
f(c) =
= yx
yt
=
C(x,t,c) (iii)
+ 6 + Xmx
+ K
=
f(c) = cx(c+8)2
F(x,t,c) =
yx +
T1(x,t,c) =
yt
t(x,t,c)
6
+ 2Xx + jix2
+ K
+
In each case deduce the similarity variables, functional forms of the solution and the resulting ordinary differential equations. 13.
Continuation.
The fundamental solution of the non—linear wave equation
(+1-) of the previous problem satisfies the initial data
(x,0)
c(x,0) = 0 ,
156
=
6(x)
.
(-H-I-)
Show that for all wave speeds
the fundamental solution remains
f(c)
invariant under the one—parameter group C
x1 = e x
t1 =
,
e
C
c1 = c
,
t
and hence takes the form c(x,t) =
4(xt1)
Deduce from (++) that
satisfies
4)
f(4))24)"(w)
w24)"(w) + 2w4)' (w) =
where
14.
the ordinary differential equation
xt'.
w=
For the linear case
Continuation.
deduce that
f(c) =
f0
where
f0
is a constant
is given by
4)'(w)
C 2
where
denotes an arbitrary constant.
C
Hence show that the
fundamental solution in this case is given by
x+f0t
c(x,t)
—
0
Can you 15.
ft 0
determine the constant
Continuation.
C
from the initial data (4-f-f)?
f(c) = c
For the case
show that the ordinary
differential equation of problem 13 remains invariant under the one—
group
parameter
w1=ew,
C
C
Hence with,
±
Tlogw,
deduce the Abel equation of the second kind, 2
P
'p
+
+ ' (1—'p
)
= 0
.
('—'p ) 157
___
16.
Obtain the classical groups and resulting solutions of the following partial differential equations: (1)
the telegrapher's equation, 2
a
where (ii)
2
a
and
D
where
5
and
D
2 ac
2' are constants.
the non—linear Burgers' equation, aU
where (iv)
are constants.
the diffusion equation with convection,
ax_D
(ill)
2'
U
2
+ u
D
au
= D
2 au
is a constant.
Barenblatt's equation (see Barenblatt et al [37]) for flow in fissured rocks, 2
3 2
ax
where
158
a
and
$
are constants.
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