Elliptic Differential Equations: Theory and Numerical Treatment (Springer Series in Computational Mathematics, 18)

The bound erably.

Remark 4.4.8'.

0<

<

I':

—

(4.4.7)

This estimate via 0(1 log hj) reflects the logarithmic singularity of the — = singularity function

ExercIse 4.4.9. For the proof of inequality (7) define := 1/4

log(Ix

—

—

+ 2h2)/logl6,

Uh(X)

sh(x,O)

and carry out the following steps: for all XE Thh, (a) sh(x, htih(O) = h2, (b) 0 for all x 1R2 (longer calculation !), (c) —Lthuh(x) 0 for fixed (d) fIn,

0

sh(x,e).

(e)

Let

be defined as in (2.6b). The solution of the discrete potential

equation

htth =0 in flu,

is given by Uh := through defined on the aid of reads

(if one

on Ph = continues the grid function, at first only lLh

on Ph; cf. Remark 4.2.5). The representation with

uh(x)=h2

XEQh.

Since Sph(x) vanishes at all points x far from the boundary, it suffices to extend

the sum over the points near the boundary. If one sums over the neighbouring boundary points of or (1,1)}

not acorner point

instead of over the points near the boundary, then the definition of

results

in the representation uh(x) = —h

XE

S?h,

(4.4.8)

where 8; is the backward difference with respect to the direction of the normal

57

4.4 Properties of the Matrix Lh

=

— hn,x)J

— hn ranges over all

C f's,). The variable

= 0 for E (note that points next to the boundary.

Remark 4.4.10. Equation (8) corresponds to the representation in Theorem approximates the integral f'... 3.2.10. The summation

Finally we want to take a closer look at the estimate for the solution IIqhik0/8 (cf. (ic)). According to through contains the right-hand side fh(x) = 1(x) of the discrete (2.6b), fh + The x E 1h hidden in Poisson equation and the boundary values foLlowing theorem gives a bound in which these components are separated.

Theorem 4.4.11. According to (2.6b), let = f and The discrete solution

be constructed from = fh + of the Poisson boundary value

problem can be bounded by


+ max

max

<

8 XEOh

xEf.,

(4.4.9)

PROOF. Set

:= and := The estimate for the first results in the first surnmand in (9). To bound summand in Uh = U'h + use the inequality (4a) with uZ, ço1 = and =0, = 0. The corresponding inequality (4.4.10)

+

for the solution of the boundary value problem (2.2a,b) has not been mentioned until now, but will be proved in a more general context in Section 5.1.3. The maximum norm fl. in (9) can be replaced by the Euclidean norms

Jh2 V

Here, h2

and fe,, h

/h x€QA

lco(x)P.

V

and 5,, correspond to each other.

Theorem 4.4.12. Under the assumptions of Theorem 11 there holds

tIL'

—'

+

1

1

S

+

1

(4.4.11)

PROOF. (1) It suffices to consider the case of the potential equation (i.e., f 0). Let the restriction of on f'h result in the grid function

Equation

4 Difference Methods for the

58

= Ltcoh be given by the rectangular Let the mapping matrix A: uh = Açoh. According to Equation (8) the entries of A read for x E

= Since

— hn,x) =

=

A 0 (cf. Remark 8), one obtains the row sum norm IIAIk0 for the choice of ph(x) = 1 in all x fl',. as = IIAwhIko = 111U00

The column sums of A are s(e) :=

The grid function Vh boundary has the values — hn),

8(e)

The

= 1 (why?), so it follows that

then reads

solution vh =

(2) for

E fl.

XE

— hn),

h2L'l

1.

= at the points

— —

hn)

hn near the

n normal direction,

E

be a point of the left or right as is implied by Remark 7. Let e = E = 0 or 1). As mentioned in the proof of (ic), Lhwh boundary (i.e., 1 and x(l Since 0, then so is wh — z)/2. holds for wh(z,y) := hence vh < h2wh. In particular we have the following estimate

=

—

hn)

—

hn)

= h2h(1

—

h)/2

at the point near the boundary

e

or from the lower boundary one obtains the same estimate if are the row sums one uses wh(x, y) := y(l — y)/2. Since the column sums of AT, we have proved

=

E FL)

h—t/2.

(3) We have IIATAH2 = p(ATA) IIAT(IOOIIA,I Exercise 4.3.20a, (3.lOd), (3.lOa)) so that the solution Uh = satisfies the following estimate

=

u2(x) =

h2 XEOk

<

(cf.

=

h2

xEflA

= h2


¶' Since

the assertion follows.

2

S

4.5 Convergence

59

4.5 Convergence The discrete solution Let U,, be the vector space of the grid functions on u C°(S1) cannot be compared directly because of the different domains of definition. In difference methods it is customary to compare both functions on the grid Th,,. To this end one must map

the solution u by means of a "restriction" u

(u : continuous solution)

U,,

(4.5.1)

to U,,. In the following we choose R,, as restriction on

(R,,u)(x) = u(x) for all x

(4.5.2)

The limit h

Ois made precise as follows. Let H C be a subset with accumulation point zero: 0 11. For example, the step sizes considered so far form the set H {1/n:n For each h H let (4 be equipped with the norm 11

DefinItion 4.5.1. (Convergence) The discrete solutionsE (with respect to the of norms ft. h E H) to u if —

R,,uII,, —' 0.

converge (4.5.3a)

We have convergence of order k if —

R,,uI(i,

= O(hk).

(4.5.3b)

The proof of convergence is usually carried out with the aid of the concepts of "stability" and "consistency". The discretisation {Lh: h H} is said

lobe stable

if

sup

hEll

<00.

(4.5.4)

For the discretisatjon defined in Section 4.2, the stability has been proved in (4.lc) with respect to the row awn norm, and in (4.ld) with respect to the spectral norm. The grid function in = is the restriction

fh = R,,f

(4.5.5a)

of f, where in (2.6b) k,, was chosen as the restriction to 12k,:

(!4f)(x) = 1(x) for all x

i?h.

(4.5.5b)

The formulation of consistency becomes simpler if instead of the matrix L,, one considers the difference operator —La,, on which it is based (cf. Remark 4.2.5). Let I),, be a general difference operator, and let —Dhu,, = .1,, discretise = I. The discretisation described by the triple (1),,, R,,, !4,) is said to be

4 Difference Methods for the Poimon Equation

80

consistent of order k with respect to fi operato.), if

(consistent with the Laplace

-

K?

—

for all u E

(4.5.6)

II

Here K is independent of h and u.

Remark 4.5.2. Let R,, and formula

be given by (2) and (Sb). The five-point isconBistent of order 2: estimate (6) holds with k =2 and K = 1/6.

PROOF. The expansion (1.4c) can be applied In the x and y directions and yields

=

+ 14) with

+

IR,I (4.5.7)

I

Theorem 4.5.3. Let the discretisation (Dh,

be consistent of onier k.

Let the matrix Lh associated to the difference opelntor D, (cf. Remark 4.2.5) be stable wsth respect to Jj Then the method is of onier k q UE

-

PROOF. wh :=

satisfies the difference equations

—Dhwh = —Dhuh + DhRAU =

fh ÷

iflah. Because of the —

boundary condition wh =0 on we have that Ra4*i) (cf. Remark 4.2.1). (4) and (6) imply (3b).

I

If one substitutes the concrete values obtains

1/8 and K = 1/6 one

Corollary 4.5.4. Let the continuous solution of the boundary value problem (2.2a,b) belong to C4(Th) Let u4 be the discrete solution defined in (2.4a,b). Then the convergence of ub to u is of second order (4.5.8)

—

The

u

can be weakened to

Corollary 4.5.5. Under the condition u

we

also have

—

PROOF. The remamder term R4 in (1.5b) may also be written as

(4.5.8')

4.5 Convergence

=

h'

—

61

,i"(x)J(x ± h —

The Lip6chitz estimate — u"(x)I — so that in (7), (6), and (8) the norm

implies R4

1

by

If, however, one further weakens u E C3.1(Th) to u one obtains a weaker order of convergence.

Corollary 4.5.6. Under the condition u

C'(Th), 2 <

2 < a

s <4,

converges

4,

of order a —2: (4.5.8")

IIt4h —

where K, := 1/[2s(s — 1)1 for 2

a 3 and K, :=

3<8<4.

1/(2s(s — 1)(a

—

2)J for

The proof results from

Exercise 4.5.7.

Show

8Kah'211ti11c.(m),

—

2< 5

4.

Even though the proofs of convergence are simple, the results remain unsatisfactory. As can be seen from Example 2.1.3, the continuous solution of the boundary value problem (2.2a,b) generally does not even satisfy u E C2(Th), although one needs at least u E C'(i7) with a > 2 in Corollary 6 for convergence. Stronger results can be obtained an analysis which will be discussed in Section 9.2. That errors of the order of magnitude of 0(h2) occur even under weaker is shown in

Example 4.5.8. If one

solves

the difference equation (2.4a, b) for -4u =

1

the values ua(4, 4), which are shown in the first column of Table 1. The exact solution u(4, 4) = 0.0736713... results from a representation that one can find in Example 8.1.11. The quotients of the errors = u(4, 4) — u,(4, 4) C2(Th.

has furthermore the asymptotic expansion ua(4,4) u(4,4)+ h2e(4,4) + 0(h4). The error term e(4, 4) independent of h, is eliminated by using the Richardson extrapolation

fll\

:=

if

ull\

Ill

The extrapolated values are already very accurate for

column of Table 1).

h=

1/16 (cf. the last

4 Difference Methods for the Pouson Equation

62

Thble 4.5.1 Difference solutions for Example 8 Quo-

Quotient

h

1/8 1/16 1/32 1/64

0.0727826 0.07344576 0.0736147373 0.0736571855

8.89x 0.0736668 0.07367106 0.07367133

0.250

5.66x105

tient

4.5x106

06

0.249

4.6 Diacretisations of Higher Order The five-point formula (2.11) is of second order. Even if the solution u belongs than 0(h2) would to C(17) wIth 8>4, no better bound for result. An obvious method for constructing difference methods of higher order

-

is the following. As an anaatz for the discretisation of the one-dimensional equation u" = choose

/

(DhU,l)(x) =

+ vh).

The Thylor expansion provides

ap =

(DhRhu)(x) = Ea,.h"_2Q'(x) +

The2kj-Iequataonsao=ai=as=a4=•=a2a=Oanda2=Iforma linear syateni for the 2k + 1 unknown coefficients c,. For k = 1 one obtains the usual difference formula for k =2 a difference of fourth order results:

= for u" to the x andy coordinates, one obtains

If one applies this

for -4 the difference star 1

h2 — 12

—16 1

—16

60

—16

1

(4.6.1)

—16

I

(cf. (2.12)). The difference scheme (1) is of fourth order, but presents difficulties at points near the boundary. To set up the difference formula at

4.6 Discretisations of Higher Order

63

(h, h) E ap,, for example, one needs the values uh(—h, h) and uh(h, —h) out(cf. Figure 1). One poesibility would be to use scheme (1) only at points side

far from the boundary and to use the five-point formula (2.11) at points near the boundary. The above complications do not occur if one limits oneself to compact

nine-point formulae; by thin one means difference methods (2.12) which 1" (cf. Fig. 1). An 0 only for —1 afi are characterised by "cap 1, leads, however, to a, —1 ansatz with the rune free parameters a negative result: there is no compact nine-point formula with (Dhu)(x) + 0(h3). In this sense the five-point formula is indeed optimal. Neverthelesa. nine-point procedures of fourth order can be obtained if one also selects the right side f,, of the system of equations (2.6a,b) in a suitable manner. N

*

Difference Scheme (1) .

•: Compact nine-point formula

N.,.

N

FIgure 4.0.1. Scheme (1) and the compact nine-point formula

If one applies the compact nine-point scheme

h2

:= —'--i--

—1

—4

—1

—4

20

—4

—1-4-1

(4.6.2)

to u E C8(Th, the Thylor expansion results in —Dhu =

(4.6.3)

For the special choice ci the restriction fh

-

Rhf :=

1

via

1/2 1/2

4

1/2

1/2 f,

(4.6.4)

i.e.,

fh(X,y) =(khf)(x,y) [f(z — h, y)

+ f(x + h, y)+

f(z, y — h) + f(x, y + h) + 8f(x,

(4.6.4')

Equation

4 Difference Methods for the

64

one obtains the expansion fb(X, y) =

f(x, y) +

f = —4ti, agrees with (3) up to 0(h4). The matrix Lb of the system of equations which results after the elimihas the entries nation of the boundary values uh(x) = zE (20/6 if x —

= (Eh, 0) or (0, ±h),

otherwise.

The right side of the system of equations Lau4 = Qh 'lb

:=

fh := R,,f according to (4'),

The dlscretisation (Dh, Rh,

(4.6.7)

:= —

with D, from (2), Rh from (5.2),

from

(4') is called the mehrstellen method (cf. Collatz lii). Exercise 4.6.1. Let D, and Lb be defined respectively by (2) and (6). Prove

that (a) Lb Is an M-matrix; 1/8 (stability), (b) (c)

20h2/3; ifuE

(consistency).

Theorem 4.8.2.

Let Sib be defined as in Section 4.2. Let Ub be the solution pivvided by the mehratelien method Lhuh = with Lb from (6), and from (7). Let the solution is of the boundary value problem (2.2a,b) belong to C6(fl) or C6"(37). Then the following error estimates hold in the respective

huh

11h4

11h4 —

<

—

(4.6.8a)

In the aise of the potential equation, i.e. / =0, we even have Huh

s/u E

—

Kh611u11(,.(Th,

huh — RhuhIoo

(4.6.8b)

reap. U

PROOF. As in Theorem 4.5.3, (8a) results from Exercise (Ib,c). In the case thatf=OtheO(h4)-termalsovanishesin(3). U

65

of the Neumann l3otmdaiy Value Problem

4.7 The

4.7 The Discretisation of the Neumann Boundary Value Problem directly in the differThe Dinchlet boundary conditkrn u(x) = p(x) was not necessary. A different situation arises was ence method; a diacretisation for the Neumann boundary value problem in

s'l=(O,l)x(O,l), Ou/ôn=ço

on

1.

(4.7.1)

The normal derivative, which reads explicitly ths

forx==(x,1)Ef',

for x=(x,O)€f, t9u

for x=(0,y)Ef,

forxr(1,V)Ef',

Like the Laplace operator, must be replaced by a difference. We will investigate three different discretisations.

4.7.1 One-sided Difference for au/en The Poisson equation leads to the (n —

1)2

= (1/h —

1)2

equations

= 1(x) for all x E Os, which require the values of

:=

U

for all x E

(4.7.3)

where

:= I'h\{(O,O), (0,1), (1,0), (1, 1)}.

To obtain a further 4(n—1) equations for {uh(x):XE we replace, at all x the normal derivative eu/en = by the backward difference

Ouh(X) :=

L. I

— tLh(X — lin)J =

for

xE

(4.7.4)

• Discretisation of the Neumann condition

.

Discretisation of the Poseson equation

FIgure 4.7.1.

If one inserts the corresponding normal directions n for the four sides of the square one obtains

66

4 DIfference Methods for the Pofeson Equation h

h

J

—u,(h,y)I =

—

1 for

—

J

h 2h

1— h

(4.7.4') equations for as many unknowns.

Equations (3) and (4) yield (n + 1)2

Exercise 4.7.1. After a rescaling of equations (4) to h18;uh(x) = these equations, together with (3), form a system Lhuh x that Lh is symmetric and satisfies (3.la).

Show

can be eliminated As in the Dirichiet problem the variables uh(X), x E with the aid of (4) in (3). At the point (h, y) near the boundary, for example, Equation (3) becomes h2f3tLh(h, v)

—

tih(h, v — h) — tih(h, y + h) — uh(2h, i')J

= f(h, v) -t-h

v).

—1

The star h2 —1

4

h2

—1

thus becomes

h2

3

3

—1

—1

0

—1

respectively, the left, right, upper and lowe boundaries. At the corner points one even has to replace two boundary values, so that, for example, at near,

—1

(h, h)

Q,,

the star reads W2

0

2

—1

.

Except for the special case

0 h = 1/2, one obtains for the (n — equations

= with

= h2

1)2

values u,(x), x

the system of (4.7.5a)

(4 lfxES?hisfarfromtheboundary, 2

if x = (h, h), (h, I —

3

otherwise

h), (1

—

h, h), (1

—

h, 1

—

(4.7.5b)

= f —h2 and

= fh + 4°h,

If

Thh are neighbours,

fh(x) = (f4,f)(x) := 1(x), (Ph(X) =

(4.7.5c)

4.7 The D

67

tisatlon of the Neumann Boundary Value Problem

Remark 4.7.2. (a) L,. is symmetric and satisfies the sign condition (3.Ia). points of (b) With lexicographical arrangement of the grid form

T-I

L,, has the

1 T-I

-I

Lh=h2

..

3

—1

—1

4

-I

T

—I

-I T—I

—1

T= 4

—1

—1

3

—1

like the continThe matrix Lh is singular because the system L1ti, = uous boundary value problem (1), is, in general, not solvable. The analogue of Theorem 3.4.1 reads as follows.

Theorem 4.7.3. The system of equations (5a) is aauable if and only if

f(x) =

—h2

(4.7.6)

h

XEQh

Any two solutions of(5a) con only differ by a constant:tjL—4 = ci, CE

PROOF. Evidently, L,1 = 0 holds, i.e., 1 E kernel(Lh). Furthermore, Theorem 4 will then imply dim(kernel (Lh)) =1. This proves kernel(Lk) = {cl: C E

It)

(4.7.7)

and thus the second part of the aseertiori. (ãa) is solvable if and only if the Because scaler product vanishes for all v E kernel (LI) = range of LI = Lh and (7) (1,

=

0

or

qh(x)

=0

(4.7.6')

XEOk

is sufficient and

According to Definition (5c), (6) and (6') agree. •

Let conditIon (6) be satisfied. System (5a) can be solved as follows. Select and normalise the solution Uk (determined except for

an arbitrary xo E one constant) by

(4.7.8)

Let ilk be the vector without the component Let L,, be the submatrix of L,, in which the row and column with index x0 have been left out. Let be constructed likewise. Then

Methods for the Poisson Equation

4

68

(4.7.9) is a system with (n — 1)2

— 1

equations and unknowns.

Theorem 4.7.4. The system of equations (9) is solvable; in particular, LA L;14a, supplemented by

is a symmetric M-matrix. Under condition (6), (8), yields the solution

of system (5a).

is symmetric. In flh\{XO} PROOF. (a.) As a principal submatrix of any two grid points can be connected by a disin of neighbouring points so

that LA is irreducible. For all x E üh\{XO} there holds (3.4b); at neighbouring points of x0 we even have (3.4a), so that LA is irreducibly diagonally dominant. According to Criterion 4.3.10, LA is an M-matrix, thus nonsingular. is the solution of (5a), one can assume (8) without loss of generality (b) If restricted to also solves Equation (9) and has to agree so that with the unique solution

As a corollary of Theorem 4 one obtains that rank(LA) (n

—

j)2

— 1,

i.e., dim(kernel (LA)) =

1

=

and henos (7) holds.

Another possibility for the solution of Equation (5a) is to pass to an extended system of equations (4.7.lOa)

= with where

LII,

11

—

=

{tihl

—

fQhl

j,

(4.7.1Db)

a can be prescribed arbitrarily.

Theorem 4.7.5. Equation (iDa) is always solvable. If for the last corn.. ponent of the solution we have A = 0, then condition (6) is satisfied and uA represents the solution of system (5a) which is normahsed by a. However, sf A 0 hoLd, one am interpret Uh as Uh(X) = solution of L,uh = stI&ere = — Al belongs to 1(x) =1(x)— A andf and

satisfy condition (6).

PROOF. I is linearly independent of the columns of LA so that rank ILA, 11 = 1 = (n — 1)2. Likewise, (11,0) is linearly independent of the rows of ILA, 1) so that rank (Li) = (n — + 1, i.e., is nonsingular. The other rank (LA) ÷

statements can be read from (lob).

Attention. One should either use Equation (lOa,b) or Equation (9), after

— (ITqh/1Tl)I. As a justification of this recomfirst replacing by mendation note that the condition for solvability of the continuous problem is

ity condition (6). For einooth functions I and Equation (6) can be shown to hold up to a remainder of order 0(h). Thus it is generally unavoidable to

4.7 The Discretisation of the Neumann Bound&y Value Problem

69

= In the case of Equareplace f,, and qh by fh —Al and q,, tion (lOa,b) this correction is carried out implicitly. If, however, (9) is used without any correction, the resulting solution can be interpreted as a solution with = qh(x) for x Xo and qh(xo) of Lhuh = qh(x). — That is, here too, an implicit correction of is carried out, with the difference that the correction is not distributed over all components as before, but is concentrated on If Equation (6) is satisfied up to order 0(h), then qh(xo) and qh(Xo) differ by 0(h1). Therefore the solution of Equation (9) contains a singularity at the point Xo. Theorem 4.7.6. (Convergence) Let u E

be the solution of the

Neumann problem (1). Let

be the solution of Equation (lOa). = Then there exists a c E IR and cvnstants C, C' independent of u and h such

that huh —

S

Cfhf

—

+ (4.7.11)

PROOF. (a) We have A=

+

Because fa I dx + f1-. error

—

1)_2.

di = 0, the first factor consists only of the quadrature

(b) According to Theorem 5, is the solution of Lhuh = corresponds to the difference equations lflI?h, The difference wh := uh —LlhWh =

=

8ntIh=W

LIhRhU + fh —

Al (4.7.12a)

Ifl(ih,

=

on

The errors of consistency are Ilchfloo

(of. Remark 4.5.2),

+ Since the solution

Al. This

satisfies

—

+ LI hR -

—

(cf. Lemma 4.1.1).

exists, condition (6) is satisfied with ch and

(4.7.12b)

70

4 Difference Methods for the Poisson Equation

a,.

ci with c = lTwh/1TI is the solution of (12s,b) normalised Then wh Wh = 0. The application of the following Theorem 4.7.7 to Equation by U (12a,b) provides the inequality (11). Theorem 4.7.7. (Stability) Let condition (6) be soLüfie4. Let the solution of(3), (4) be norinatised by iTuh = 0. Then there exist constants C1,C2 independent of u and 4 such that C1

Dt&hlloo

max If(x)I + C2 max

(4.7.13)

xE!,,

xEOj,

The proof of this theorem, which corresponds to Theorem 4.4.11, will be supplied in Sect. 4.7.4.

47.2 Symmetric Difference for 8u/On As can be seen from Theorem 6, the one-sided difference 8; causes the error term 0(4). It seema obvious to replace 8; by a symmetric difference. Th this end the five-point diacretisation is set up at afi points x 47h U Ph (cf. (2.lc)): (4.7.14a) Rhf in Thh, = 1k where Rh is the restriction to Thh. The symmetric difference is defined by + hn) — uh(x — lrn)J

=

in x E 1). (4.7.14b)

we assign two normal directions to the corner points, so that for each of the corner points two equations of the form (14b) can be set up. In the corner x = (0,0) one has, for example, the normals Here,

(o\

(—1

The corresponding equations (14b) also contain different (!) values 0) = For x E F4 the difference forand ço(0,0+) = mula (14a) needs the values in points x + hii outside of These can be

eliminated with the aid of (14b) so that a system of equations Lhuh = remains for the (n + 1)2 components uh(X), x E Thh.

ExercIse 4.7.8. (a) If the grid points of i74

are arranged in lexicographical

order, L,, has the form

4-2

T-21 —I

L4=h2

T

.-•

—I

—1

.. —I

,

T

—I

-21T

T=

4

—1

.. —1

4

—1

-24

4.7 The Discretl8ation of the Neumann Boundary Value Problem (b) Lh is not symmetric, but DhLh with Dh = diag {d(x)d(y)}, d(O) 1/2, d(.) 1 otherwise, is symmetric.

71

d(1) =

The analogue of Theorem 3 reads:

Theorem 4.7.9. Equation (14a,b) is solvable sf and only if = 0 for D,, in Exercise 8b. Any two solutions may differ by only a constant. The formulation of —h2

= 0 with the aid off and

reads:

d(x)d(y)f(x, y) = 2h >2 d(z)d(y)p(z, y),

>

(4.7.15)

(x,v)Erh

where the sumrnand for the corner points occurs twice in the second sum, and both the different bmits for are taken into account.

Remark 4.7.10. The sums in (15) represent summed trapezoidal formulae. Thus, from 4od.l' 0 follows Equation (15), except for a remainder

/

Theorems 4 and 5 can be transferred without difficulty. The convergence theorem, Theorem 8, becomes

Theorem 4.7.11. LetuE the solution of

be a solution of(1). Let li,,

=

With

)A)

huh —

We

=

gence of second

}.

be

have conver-

(4.7.16)

—

The proof is essentially the same as for Theorem 6. An additional technimust also be of treating the difference equation and the boundary diacretisation separately as in (12a,b) one should directly analyse the equations Lhuh = q,., from which the values uh(x + hit) outside of fl have already been eliminated. cal difficulty is the fact that the consetency error Cletermined in x E although uls only defined in

4.7.3 SymmetrIc Difference for ôu/On on en Offset Grid If we offset the above grid by h/2 in the x and v directions, we obtain the grid

in Figure 2. The points near the boundary of V. We set

are

at a distance h/2 from

72

4 Difference

Methods for the Poleson Equation

€Zor

:=

1,

(ci. Figure 2). To each point near the boundary x—hn/2 (x an outlying neighbour x + hn/2.

"h) corresponds

0000

r * — S— —0. —.

OS

°!'

•

S

S

S

0: Points in

'

Figure 4.7.2.

.:PointsmFh

Offaet grid

The discretisation of the Poleson equation (1) is

= 1(x)

in x

(lh,

(4.7.17a)

(4.7.1Th) in x E "h• h1[tzh(x + hn/2) — uh(x — hn/2)j = The difference (iTh) is symmetric with respect to the boundary point x and neverthekes agrees with the backward difference 8; at the grid point x-i-hn/2.

Remark 4.7.12. After eliminating the values u4(x + hn/2), x E Ph, we where Remark 2 is also valid for this obtain a system of equations L,uh = matrix L,,. In contrast to Section 4.7.1, Lh is of size n2 x n2.

The Theorems 3,4,5, and 7 hold analogously. Theorem 6 hoLds with the inequality (16) instead of (11).

4.7.4 Proof of the Stability Theorem 7 In the case of Dirieblet boundary values the stability statement of Theorem 4.4.11 foLlows immediately from the maximum principle and the bound for The corresponding statement of Theorem 7 for Neumann boundary valnsa, however, cannot be proved that easily. In the literature one can only find

weaker estimates which on the right-hand side of Equation (13) contain an However, this factor is not to be avoided if one uses additional factor I without condition (6) being satisfied. Equation (9), =

Let the discrete Green function gh(x,e), XE QhUPh,

E

be

defined by

=

(4.7.18a)

forx=el fh'/(4_4h) ifxisneartheboundazy %.O

otherwlsei

t.O

1

if x is far from the boundaryj

for XE

and 8; act on x.

where

73

of the Neumann Boundary Value Problem

4.7 The

exists

(4.7.18b)

1)

=

EXEOk

0

proves the

solvability condition (6').

Lemmi 4.7.13. For arbitrary qp, with *Lh(X)

=

0

(cf. (6')), (4.7.19)

h2

represents a solution of Lhuh =

=

PROOF. At points far from the boundary x E S?h, (Lhua)(x) =

qh(x). At points near the boundary x E ilh, from 8uh = 0 and (6'), one has

the Identity

=

=

= qh(x).

— 4—4h

U

Equation (18a,b) determine 9h up to a constant.

Theorem 4.7.14. The Green function C(i + lIog(Ix

om be so selected that

for

-t-

(4.7.20)

E

This inequality corresponds to the bound (4.7) in the Dirichiet case. Be-

foreTheoreml4isproved,wewanttoshowthatTheorem7followsfromit. K1 and logously K2 we obtain h2

xEaa

From (19), (20), and

(cL (5c)) thus follows the estimate

= fh +

Ki max $f(x)( + 2K2 max

+ — If Üh := =0, one can see that

XEflA

is the solution of

2inf{Ijua so

—

E

Jt}

that the inequality (11) and Theorem 7 are proved. It remains to prove Theorem 14.

=

normalised by

74

Equ&tIon

4 DIfference Methods for the

11h LemmA 4.7.15. Let e be one oJ the unit vectors (1,0) or (0,1). Let also. For all these a and let there exist a function be such that + he E with the properties Gh(X) =

(1

(.4.7.21a)

—1

)

otherwise

8;C,1=0

(4.7.21b) (4.7.21c)

< hC'/(Ix — + h),

Then Theorem 14 holds, andthus also

where C' does not depend oneand Theorem 7.

PROOF. Fbr

let

E

be defined as

(

1

on f's, For

= E+he,

or (0,1). Srnce

'I

—1 L

0

agrees with Gh m Lemma

15.

otherwise) For

one

(21c) implies the estimate

=

If one considers first the case x1 =

1/k coost

a solution of

(1 + log(k2h)

=

—

C"[1 +

x2

and one applies

log(k1h)), one obtains —

EJ + h)) + I

—

+ h)fl.

In the general case, this bound also comes out, but one must choose the connection

such that Ix —

min{lx —

Since we know that

x—

for SllXEflb, 4 —4h

1= (4-4h)/h the equations (18a,b) satisfies the inequality (20). Because are also satisfied, i.e., is the Green function. We shall construct the function basic building block is the discrete

needed In Lemma 15 explicitly. The

singularity function

on the

grid in 1R2

{(x, v) E LAmmi 4.7.16. and

x/h, y/h

7}.

The singularity function defined by sh(x,e) := Oh(x

-

4.7 The Diecretisatlon of the Neumann Boundary Value Problem

ah(x)

1

:= m;:i — —

for all

II ('

1

—1

dr1i

+ + x2,12)/(2h)) +

(W

d

= h-2

E Qh has the property

75

d

=

and

0 otherwise.

—

and

Note that

:= exp(i(xith +

PROOF. Let

fT

1)

= 4h2e(x, ,,)tsm2øii/2) +

fT

I4w2

forx=O

0

0, 02 0 one defines the For multi-indices a = (a1,a2) E Z2 with partial difference operators of order Ial = aj +02 by =

(4.7.22)

Starting with the representation in Lemma 16, Thomée (11 proves

Lemma 4.7.17. I(ô°ah)(x)I

+

for all x E Qh.

For the construction of the function Gh In Lemma 15 we select, without

lcesofgenerality, e=(1,0)andkeepeEQh withe+heEQh.The function := hOuh(x —

=

—

— 0h(X — — he)

(4.7.23)

satisfies (21a) but not the boundary condition (21b). Let the symmetrisation operators and 8, be defined by

y) :=

S,uh(z,y) :=

y)+Uh(h—x,

h—y)j

for (x,y) E Qh. The function cill......AC'

symmetric with respect to the axes y = h/2 and z = h/2. Thus we have (21b) on the left and lower boundary: Is

= Furthermore, satisfies conditIon (2 La) just sa we define the operator by

=

0

(4.7.25)

does. For each 13

Equation

4 DIfference Methods for the

76

(Ppuh)(x,y) = g32L)J,

where L = 2— 2h, and set (4.7.26)

—

2 there holds

Lemma 4.7.18. For /3= (/3i,132) wsth

2, x E flh.

for all

Here, Küind

(4.7.27)

heETh,.

ndeni of the choice of the

PROOF. According to the definition we have 0) = 0.

(4.7.28a)

implies preserves the symmetry, i.e., Uh = The operator and consequently = = Thus we have that

= 0.

=

for different with

Is the linear combination of According to Lemma 17, xE

hCK'91313

= 2,x E

for alt

2. Let Ial=

so that (27) follows for

(4.7.28b)

1.

(4.7.28c)

DaGh(x) can be written in the

form

+ with x0 = x,

= x,

flxk

—

-

2hCK"I/313

= for

h.

Each summand has the form = 1, x E

(4.7.28d)

Likewise one infere from (28a) and (28d) the inequality (27) for Iciij =0, which

proves Lemma 18.

U

Since E,E12\{o} (4.7.29)

Gh(X)

pEZ'

= 0 in for all (0,0), Gh aho satisfies Equation (21a). As already mentioned in the proof of (28b), = = S,G5

exists. Since

4.7 The DiscreUsatlon of the Neumann Boundary Value Problem

77

holds so that G,, also satisfies Equation (25): 8Gh =0 at the left and lower boundary. The proof of 8;Gh =0 at the lower boundaries requires

Lemma 4.7.19. Gh is L-periodic: Ch(x,y) = Gh(x+ L,y)

L).

be abbreviated to

PROOF. Let Definition (29) says that

(4.7.30a)

= lim

k-.oo

Now

can be written as a sum over the differences

(x-f-131L,

+ vh,/32L) from wh = 0 from ph = 0 to ph = x, and over 2 the distance between the arguments to vh = y. For - 1)L. Each summand is thus, by Lemma 17, bounded he is always by O(h2/((piI — 1)2) = O(h2/13?). As a sum of such terms then

estimated by

2.

for

can be (4.7.30b)

We want now to show that the following equation (30c) also holds:

E

= Inn

=

(4.7.30c)

urn 1—k
The

The sums (30a) and (30c) differ by Dh := values

value of the sum Dh is then bounded, from (30b), by (2k + 1)O(h/k2) = O(h/k), so that the limits of the doubLe sums in (30a) and (30c) are the same. Changing the variable th to th -1 then transforms (30c) Into

-k$i (4.7.30d)

The first part of the identity 132L) — GZ(/31L

—

÷ kL, thL) —

=

+ L,

+ (Yh'(kL + L,

is elementary; the second results from the symmetry

(4.7.30e)

As above, = the summands of the last sum can be bounded by O(h/k2) so that (30e)

vanishes for k

oo.

Together with (30d) one obtains

4 DIfference Methods for the Poleson Equation

78

Gb(X, y) = lim

+ ,91L +

= lim

L,y +

—

'yØ(x+L,v)=Gk(x+L,y).

k.-.oo

S

The proof of Gh(X, v) = Gh(x, y + L) is analogous.

Since L =2- 2h, the symmetry

= S,Gh and the periodicity yield

Likewise we show 8,1 G,(1,y) = 0 on the upper boundary (l,y) E that 1) =0 on the right boundary. Thus (21b) is proved. It remains to show (21c): The +h) in for S llflhIcio <1 arelinearcombinationsofh8;orh(x—E) forE 0h, i.e.,

whith are generated by S,, and P,,. Fbr all E there holds so that from Lemma 17 foLlows

hC'/(Ix —

+ h) for

UJ3IIOO 1,

k-El1

XE Qh.

On the other hand, Lemma 18 shows that

K/l$13 = hK'

h

I

—

+ h).

IIPlk.2 The assumptions of Lemma 15 are hence satisfied: Thus theorems 14 and 7 are proved.

4.8

with (21a-c) exists. U

Discretisation in an Arbitrary Domain

4.8.1 Shortley-Weller Approximation Let the boundary value problem

—4u=f

in

47,

cuP

(4.8.1)

be given for an arbitrary domain 47. If one places a square grid of step size h over 47 one obtains

:= {(x,y)

E 47:

x/hE Z and y/h E Z}

(4.8.2)

as the set of grid points. By contrast, the set of boundary points must be defined differently from the case of the square. The left neighbour point of (x, y) 47h reads (x — h, y). If the connecting segment {(x — y): t9 (0, l]} does not lie completely in 47, there exists a boundary point (x —

P with (x,y) E

E (0, 1J,(x —t9h,y) €47 for all t9 E [0,s), (4.8.3a)

4.8 Dlseretisation in an Arbitrary DomaIn

79

which now, instead of (x — h, y), is called the left neighbour point of (x, y) (cf. Figure 1). Likewise the right, and upper neighbour points can be boundary points of the following form:

(x+8h,y)E I' (z,y—sh)Ef' with(x,y)Eflh,sE(O,1),(x,y—6h)EQforall

(4.8.3b)

for all '9€ jO,s).

E

(4.8.3d)

We set

:= {boundary points which satisfy (3a), (3b), (3c), or (3d)}.

(4.8.4)

grid point (x,y) poseeseing a neighbour from 1'h, is said to be near the boundary. All other points of are said to be far from the

A

boundary. As can be seen in Figure 1, a point may be near the boundary although (x ± h, y) and (z, y ± Ii) belong to Oh (namely if not all connecting segments lie completely in .0).

•

near the boundary .E •EQhfarfromboundary

U

ciE

boundary

U. •a FIgure 4.8.1.

auid "h

ExercIse 4.8.1. For a convex domain .0 show that

Fh={(x,y)EF: x/heZory/hEZ}. If one wishes to approximate the second derivative u"(x) with the aid of the values of u at x' <x <x", one can use Newton's divided differences: I

I

I

u (x) = 2[ —

U

,

—

JJ(x" — x') + Rem.

(4.8.5)

ExercIse 4.8.2. Show that (a) the Taylor expansion

IRemI

1 (x" — x)2 + (x

max{z" — z,x

—

(4.8.6) —

3

br the remainder in Equation (5)11 u

C3((z',x"J).

4 Difference Methods for the Poleson Equation

80

(b) In Equation (6) one can replace the norm of C3((x', x"I) by

that of

C'2"Ux',x"D.

(c)Ifx" = x-t-h andx' = x—h thedifferencein(5) agrees wlththe u8ual second difference 884n(x). To set up the difference equation for —4u = / at (z,y) four neighbouring points (x

—

(x + Srh,V), (x,y

(x,y + s0h)

—

U

Qh we use the

1,

1',, 0 <

defined above. The subscripts to the a are suggest respectively left, right, under and over. Fbr points far from the boundary = 1 holds; for points near the boundary (x,y) at least one neighbour lies on 1h and the corresponding as

distance a.hmaybesmalierthanh.Equation(&)withx'=x-a,handx"= x + a,.h provides an approximation for

Analogously one can replace

by a divided difference. One obtains the difference scheme of Shortley

and Weller (lj:

—+— —

i4z,y)—

2

+

, ,

Sj)

2

301.50 +

)*4(x)Y —

(4.8.7)

Remark 4.8.3. The discrete boundary value problem aesumes the form on = := with (14f)(x) = 1(x) for x ah, tih=fP OflFh.

(4.8.8a) (4.8.8b)

Theflvecoelflcientson thirightaideofEquation(7)deflnethematrixelements for = x and for the four neighbours of x. Otherwise we set =0. The right side of the system of equations

=

= fli +401,,

(ph(X)

Again 40h(X) =0 holds for points x

(4.8.9a)

qh,

:=

Qa

—

E

far froin the boundary.

(4.8.9b)

4.8 D

etisation in an Arbltr&y Domsan

81

Theorem 4.8.4. Let 17 be bounded and contained ma strip (xo,xo+d)xlR or belonging to the Shortley.. Weller

lRx (im, yo + d) of width d. For the matrix

discretisation the following holds: is generally not symmetric; (a) is M-matviz with (b) (4.8.10)

d2/8.

PROOF. (a) Let x = bour x' = (x + h,y) = —2h—2/(1 + a,)

be near the boundary, but its neighbe far from the boundary. Then it holds that < 1. Other than in Exercise if —h2

(x, y) E

diagonal) becomes 4.7.1, in general no scaling can be found so that DhLh symmetric. need not neceesarily be irreducible and hence irreducIbly diagonally (b) dominant. But the weaker condition in Exercise 4.3.8 is satisfied and proven the M-matrix property. (c) For the proof of (10) we use Theorem 4.3.16. If the domain 17 lies in the w := (x — z0)(x0 + d — x)/2. strip (x0, zo + d) x R we select The reinwnder in Equation (5) contains only third derivatives that vanish for = -1: w. Thus Dhwh agrees with

The corresponding system of equations reads

:= ía -I- wa with = 1 and Theorem 4.3.16 proves

Ia = I and qj 0. Thus we have

flw,i$co (d/2)2/2 = d2/8. Exercise 4.8.5. Prove the analogue of the derivative (4.4.9):

If(x)I

HuaNco

With (10) stabIlity haS been proved. The order of comistency however is only 1, for at points near the boundary := DhRhu is

—

of the order of magnitude 0(h'). Here, R4 is the restriction to U has been defined in (8a). We want to show that nevertheless there is

convergence of order 2. The difference solution uh = and the solution ii

:=

-

between the discrete of (1) satisfies

= —

= 0 on

in 17k,

(4.8.lIa)

(4.8.llb)

Equation

4 Difference Methods for the

82

of the x difference (rasp. y difference), in turn,

is the error of is split into

:=

'},

Analogously one defines

c'1 + The errors

:=

are described by

+

:=

(5.7):

<

With K :=

define

v,=K1 For

XE

(4.8.12a)

vh=O OflPh,

from the boundary èh(x) =

far

ë,,:=Lhvh.

0; for XE

near

the boundary;

however, we have

near boundary).

(x E

= (x

for example, the case x = (z, y) Ph. According to Exercise

- sjh,y)

11h

near the boundary with

the x difference has the

error

= The analogous estimate for gives IcL(x)I =0 at the x E far from the boundary one has i.e., 0 it follows that —vh wI

Because 4(x) = Since

K= Equations (12a,b) together with

(4.8.12b)

+

=

=

- R,,u prove

Theorem 4.8.6. Let (1 the asswnptwn of Theorem 4. The for the Shortkij- Weller method is second onier huh

+

—

1

if ti

E

(4.8.13)

+

4.8 Dlseretlsatlon In an Arbitrary Domaãn

83

Exercise 4.8.7'. Show that if one uses the Shortley-Weller discretisation for all grid points near the boundary, but the mehrstellen method from Section 4.6 for all points far from the boundary, we obtain a method with convergence 0(h3). of third order: —

4.8.2 InterpolatIon at Points near the Boundary near the Instead of discretising the Poiseon equation at grid points x E boundary, one could also try to determine uh(x) by interpolation from the is near the boundary, neighbouring points. If, for example, x = (x, y) E flh U I),, then linear interpolation yields (z — s1h,y) fh and (x +

tLh(X,y) = Iszuh(X + 8rh,y) + srUh(X

+ Si).

—

Thus, at the point x we set up the equation

+ Sr)Uh(X,V) — siuh(x + srh,y) — SrUh(X

—

s1h,y)

=

(4.8.14a)

0.

= (x — sgh, y) should be a boundary point, one can replace by is a boundary point, we choose If, however, (x,y + s0h) or (z,y — interpolation in the y direction: Since

÷ At any x

y + s0h) — Souh(z, y —

So)Ua(X, y) —

=0.

(4.8. 14b)

0h far from the boundary the five-point formula (2.3) is used:

= fa(x) =

1(x)

(x far from boundary). (4.8.14c)

Theorem 4.8.8. Let (2 satisfy the assumptions of Theorem 4. Let the di.cretisatlon be given by (14a-c) with the choies between (14a) and (14b) being made in such a way that always (at least) one boundary point is used for the inteipolation. Let the system of expiation. resulting after the elimination of the boundary values Lh is an (in general unE f's, be Lhu,, = = syinmefric) M-matrix tuhsch satisfies the estimate (10). The discrete solutions with the order 2, if u E tih

+

flua —

(4.8.15)

PROOF. The M-matrix property and (10) are proved as in Theorem 4.2. Let

xE

be a point near the boundary at which (ida) is used. The error of

interpolation is

y) := (sz + Ich(x,V)I

y) — 8lRhU(x 1

+

j5r31(8i + Sr)h

y) — 2

—

sth, y),

84

Equation

4 Difference Methods for the

With this cL and K of Theorem 6.

:= h2flu11C1,I(Th),

one can essentially just repeat the proof U

By rescaling and adding the equations (14a,b), one obtains

+ 8, h2 ((iiSgSr —

÷

So +

Uh(x,V)

/

1

1

8,.

So

—

!Uh(Z — sjh,y) Ii

—Uh(X + s,.h, y) — —uh(x, y + s0h)

= 0.

—

(4.8.16)

Using this device one can obtain a symmetric matrix Lh, even at arbitrary .12.

ExercIse 4.8.9.

Show that the diacretisatiori (14c), (16) leads to asymmetric M-matrix. The estimate of convergence reads —

+

Rg,t&floo

In (14a,b) linear interpolation was chonen because the values at the neighbouring points were sufficient for it. Constant interpolation by

u(x,y) = u(z — a1h,y) =

a1h,y),

if(x,y)

Ila,(x — s1h,y) E f's,

is less desirable since it only provides first-order convergence: = 0(h). By contrast, interpolation of higher order is very applicable indeed (cf. Pereyra-Proekurowakl-Widlund (1)). However, it is desaibed by an equation wblch also contains points at an distance of 2h. Higher boundary appraiu-

mations are in particular then necessary If one wants to apply extrapolation methods (cf. Marthuk-Shaidurov [1, p. 162 ff.J).

5 General Boundary Value Problems

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations 5.1.1 Posing the Problem In Section 1.2 we have already formulated the general linear differential equation of second order: (5.1.la) Lu=f in I? with

We

(5.1.lb)

+ a(x).

+

L=

mentioned that, without lose of generality, one can assume

(5.l.lc)

o,,(x) so that the matrix A(x) :=

(5.1.ld)

•,,

is symmetric. A differential operator apparently more general than (ib) is I ,

82

L =>2 N 8x0x +

8

i— +

82

liii

j

"

+>2

I

8

+

8 j

+0.

(5.1.2)

But since, for example, the operator = + (2) can be described in the form (ib), provided the coefficients are sufficiently often differentiable. According to Definition 1.2.3, Equation (1) is elliptic in S? if all elgenvalues of A(x) have the same sign. One can assume without lose of generality that all elgenvalues are positive so that A(x) is positive definite (cf. Exercise 4.3.22a). is elliptic m (1 if

forallxEO, $,3=1

(5.1.3a)

5 Ceneral Boundary Value

86

: = 1) and it must be For any x E 1? there exists c(x) := positive (c(x) is the smallest eigenvalue of A(x)!). Hence one can also write in the form (3a'):

E

c(x) > 0 for all x E 1?,

E

(5.1.3a')

1

Definition 5.1.1. The equation (is), or the operator L, is defined to be uniformly elliptic in 1? if Q} > 0 (c(x) from (3s')).

Inf{c(x): x

(5.i.3b)

On P =81? we impose the following Dirichiet boundary value condition:

onl'.

(5.1.4)

5.1.2 Maxhnum Principle In general, the maximum principle does not hold for the equation Lu =1, nor is the solution of the boundary value problem (is), (4) uniquely determined.

Example 5.1.2. Let 1? = (O,w) x (O,w), = 0, f = 0, Lu = + 2u. = sin(x)sin(y) are solutions of the boundary Then both u = 0 and value problem. The second solution assumes its maximum at the interior point

(ir/2,ir/2) EQ. In the above example the coefficient 0(x) = sign. As soon asaO, we have

2

(ci. (ib)) has the wrong

Theorem 5.1.3. (Maximum-minimum principle) Let a(x)

0 in the

domain 1?. Assume the coefficients of the elliptic opemtor (ib) are continuous

mO. LetuEC2(Q) .atisfyLu=fandbenonconstant. Then we have: in 4?, no negative minimum of u exists in 1?; (a) if f (b) if f 0 in 1?, no positive maximum exists in 0.

PROOF. The proof is based on Hopf's lemma, which can be studied, for example, in Heliwig 11,111-1.11. Here we give a shorter proof, which however, the stronger condition 0(X) <0 instead of a(x) 0. We assume

that u has a minimum at x' E 1? although f 0. Consequently we have = 0, i. e.,

f(x) = (Lu)(x) =

(5.1.5a)

"4

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations

and the Hesse matrix B :=

must be positive semidefinite

(here, a matrix B is 8aad to be positive semide finite if it is synunetric and satisfies 0 for all E From Exercise 4 follows

tr(A(x)B) =

0.

(5.1.5b)

ij= 1

If the minimum u(x') were negative, one would obtain from (5a,b) the result f(x') a(x)ti(x') > 0 in contradiction to the assumption I 0. Part (b) is proved analogously.

ExercIse 5.1.4. The trace is defined by tr(A) := (a) 0 and tr(A) 0 if A is positive semideflnite; (b) tr(AB) = tr(BA) = (c) tr(A.B) 0, if A and B are positive semidefinite.

Prove that:

Hint for (c): B112AB112 is positive semidefinite; Exercise 4.3.22d.

As in the case of the potential equation, from Theorem 3 follows the

Corollary 5.1.5. Let I? be bounded (not necessarily a domain); furthennore, as*

that the conditions of Theorem3hold JffOin (JJ and

sf there exists a negative minimum [positive maxzmurnj of u in on the boundary 8fl.

it must lie

Remark 5.1.6. The continuity of the coefficients and a of L in Theorem 3 and COrOllary 5 can be replaced by the assumption a <0 in 1'? or by the stipulation: In every compact set K C (1 let a and be uniformly elliptic.

5.1.3 UnIqueness of the Solution and Continuous Dependence Lemma 5.1.7. Let I? be bounded, let the coefficient, of L be continuous and

let(3a) be valid anda
ma,

oni' (i=1,2).

(5.1.6)

IIftf2rnaandwl_<4O2onr,thenalaoui
— satisfies the equations Lv = 12—11 0 in a and Oon I'. FromCorollary5and the fact thatv>Oonfone

infers that no negative minimum of v exists. Thus v

0, i. e.,

u1.

U

88

5 General Boundary Value Problems

Theorem 5.1.8. (Uniqueness) Under the conditions of Lemma 7 the solution of the boundary value problem Lu =f infi andu—ço on!' is uniquely determinetL

12=1,

PROOF. Let u1, u2 be two solutions. Lemma 7 with fi

=

Thus, Ul=u2.

= U

The next theorem states that the solution depends L be unsformly elliptic in Si. Under the conditions of Lemma 7 the foil owing holds: IIui

—

—

'P2IIoo

+ Mflf1 — f21 100

(5.1.7)

andu2 of(S). In this inequality the vnimberM depends only on for K := sup{Iaij(x)I, a(x)I :x E I1}, on the dhpticity constant defined by m := inf{c(x):x E Si) >0 (cf. (3b)) and on the diameter of the domain

0. PROOF. aelect a 0 so that

ma2—K(a-f-1) 1 and we define

w(x) :=

1100 +

—

jf'

—

—

/2(100.

on F. From the thoice of KR(z) we have

w(x)v(x)

forxEf'.

Furthermore, the selection made of a ensures:

(Lw)(x) = —

<0 ai(x) a+

— f2H00

1 —{ma2—K(a+I)}((fi —1111 — 121100 fi(x) —

=v,

Ifoneuses —

u2(x)

v(x)

/2(x) = (Lv)(x).

=w, theresultisvwin 0,1 .e.,

w(X) I(wlioo

— 402

lIoo + Mflfj

f2Iloo,

5.1 DirichIet Boundary Value Problems for Linear Differential EquatIons

89

v so that (7) follows.

U

Analogously one proves —w

where M :=

Exercise 5.1.10. (a) Let Si be bounded; let the coefficients of L be continuNow show that L is uniformly elliptic in Si if and only if L is elliptic ous in

=/2 without the

(b) Theorem 9 holds for the special case

a strip in It's be described by

where x

E

Ut" is a boundary point of

E

Ut" with l'il =1 is a unit vector,

if

and S is the strip width. Show that inequality (7) holds with M := Si C S and a are as in the proof of Theorem 9. (d) Under the conditions of Theorem 9 show that

(5.1.8)

+

Exercise 5.1.11. Let the coordinate transformation (' x E Th E be continuously differentiable. To the operator and the inverse L (in the x coordinates) let Ii correspond in the coordinates. Show that if L satisfies the conditions of Lemma 7, or of Theorem 9, then so does L'.

The right side f and the boundary values ço are not the only parameters on which the solution u depends. We next investigate how the solution depends on the coefficients and a of the differential operator L.

Theorem 5.1.12. Let the coefficients of V and L" Let

and

be

resp.

be solutions of

LV=L"u"=f in Si, u1=u11 =p on Let

L' saiofpthe conditions of Theorem9andtet U" betongtoC2(Th). With

M from (7) we then have lu'

—

M{

114 —

(

lItLllc2(Th (5.1.9)

+ + ha' —

I/al, =

the conditson u11 e

then just u" E C°(Th)

C1(i7)flC2(fl) u sufficient; i/also al =

C2(ii) will do.

90

5 General Boundary Value Problems

can be bounded PROOF. Set f' := L1ts11. Then = by the right side of (9) without the factor M. Theorem 9 applied to = 1' and Vu' = implies His' — u'%, — filoo.

f

5.1.4 Difference Second

Methods for the General Differential Equation of

Order

For notational reasons we limit ourselves to the two-dimensional case n =2. General domams 17 E E2 require special discretisations at the boundary as explained in Section 4.8. Here we only want to discuse the difference formulas at points far from the boundary. Therefore it will suffice to base our comments on the unit square

17=(0,1)

x(O,1).

LetLbegnby(Ib).Ifonewihestoobnadifferencemethodof consistency order 2, the following choice suggests itself, which will later be modified for reasons of stability.

a11(z, y)08; + 2a12(z, + +

+ afl(x, + a(x,y)

(5.1.10)

a12(x,y)/2

= h2

alj(x,y)

all(x,y)

—2[aii(x,y) +

—012(x,y)/2 1

0 a2(x,v) 0 —al(x,y) 0 —02(x,y)

0

0

a,(x,y) + 0 0

0

0

0

a(x,y)

0

0

0

Remark 5.1.13. The difference method (10) is a nine-point scheme of consistency order 2. Let Lh be the mMth of the system of difference equations In 17,, associated to (10) (cf. Remark 4.2.5):

Lg,u,,=q,,.

(5.1.11)

The solvability of Equation (11), for arbitrary q,,, Is equivalent to uniqueness. In the continuous case, uniqueness was essentially given by the condition a 0

arid the ellipticity condition a,ian >42 (cf. Theorem 8). These conditions are in general not sufficient to guarantee the solvability of Equation (11). We thus replace (10) with another discretisation.

Theorem 5.1.14. For

the coefficients of L let (3a,b) hold, and asswne that

1a12(x, y)j

min{aii (x, y),

p)}

(5.1.12)

5.1 Dizichlet Boundary Value Problems for Linear Differential Equations

Then there exists for L a seven-point difference method

0 in a.

and a(x, y)

91

of consistency oirder 1 such that the associated mattiz Lh is an M-matrix the resulting Equation (11) is solvable. except for the sign. In Note that Condition (12) foLlows from (3a) only ii

>0.

=

PROOF. At the grid point (x, y) E sia (ef. (4.2.la)) we abbreviate the onOj, and a. The principal part and a(x,v) to efficients is discretised by the following difference (x1 = z,x2 = y) stars:

82._21 (JXj

1

—2

82..2

,

1

1

0 —2

0

0

1

a —1

—i 2

i —1

,

if

a 1 1

0

1

0, we thus obtain

012 2

82

E

8z1

:h2

1

0

—0j3

2(012 —011

—ar) 011—012 0

:= min{a12,O} then both for If one introduces := max{a12,0} end 0 and for 012 <0 we get the seven-point star

4.2

—aj2 011 — 10121

4

— 10121

2(10121 — 011 — afl)

— 10121

.

(5.1.13')

— 10121

ExercIse 5.1.15. Let Lh be the matrix belonging to the difference formula (13') (note that the coefficients in (13') depend on the spatial coordinates). Assume (3) and (12) hold. Prove that —Lh satisfies the conditions of Exercise 4.3.8a.

The first derivative terms ward differences, 2

8

are replaced by the forward and backif 03

:h_1

—aj

0: 0

a0 0

element is negative, the others nonnegative. If one adds these terms to (13'),

5 General Boundary Value Problems

92

the resulting matrix —Lh satisfies the conditions of Exercise 4.3.8a. Therefore —Lh is an M-matrix.

it is easy to see that the difference formula (13') is of second order of consistency; (14), however, contains the one-sided differences so that the total discretisation is only of first order.

The condition (12) can be avoided if one allows larger difference stars, which also contain the values u(x1 ± vh, £2 ± ph) for fixed v, p E Z (but > 1) (cf. Bramble-Hubbard 11)). Layton—Morley 111 point in general out that with weaker conditions than (12) one may still obtain a matrix Lh, which, though not an M-matrix, does have a positive inverse. To obtain a method of consistency order 2, one must discretise E as in (10). The following corollary shows that —Lh is also an M-matrix when is sufficiently small.

Corollary 5.1.16. In addition following hold:

the assumptions of Theorem 14 let the

to

h

(s=1,2).

(5.1.15)

fvvm (10) together with (13') leads to a Then the discretisation of seven-point difference method of second order of consistency such that -L,, is an M-matriz.

PROOF. -Lh satisfies (4.3.la) and is irredudbly diagonally-dominant. ExercIse 5.1.17. The condition su 1a121 + 8ufflcient. Construct a counterexample with a11 = variable, and lad =6 so that L,, is singular.

•

instead of (15) is not 1, 0, h = 1/3,

=

Considerably weaker conditions for the nonsangulanty of L,, than in Theorem 14 and Corollary 16 are needed in Section 9.2 (cf. Exercise 9.2.6, Corollary 11.3.5).

is not a symmetric matrix. Symmetry of L,, is to be exin general, pected only if L is also symmetric: L = Ii. Here the formally adjoint differential operator L' which is associated to L in (2), is defined by

L=

1111

[aj

82

+

0

8

+

82

1

1

_E [(li

8

+

8

+a. (5.1.16)

It is easy to see that a 8ymmetric operator can always be written in the form

8 vxj

(5.1.17)

A difference method for this is given for the case n=2 and a12 =0 by the

five-point star

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations

93

0

o —

a22(x,y

—an(x,y+

—

0

o

o

a(x,y)

0 0

o

0

0

+0

o

(5.1.18)

.

Theorem 5.1.18.

The difference method (18) is Let a22 E The associated matrix L,, is symmetric. If a.,, > 0 consistent of order (ellipticity) and a <0, then —L,, is a positive definite M-rnatrix.

PROOF. (a) For the proof of consistency, expand

v(x + h/2) :=

a11(x + h/2, y)Lt4x + h, y) — u(x, y)j

(y fixed)

around x + h/2, and then expand v(x + h/2) — v(x — h/2) around x. (b) The symmetry results from for x, E = (cf. Exercise 4 .2.2c). (c) L,, is irreducibly diagonally dominant so that Criteria 4.3.10 and 4.3.24 are applicable. U

The mixed term — a12(xB) 2a12(XA)

2

0

can be discretised by

+

0 2a12(Xc)

a12(xC) + a12(XB) —21a12(xA) + a12(xC))

a12(XA)

+ 012(XD)

—a12(xc) — a12(xD)

(5.1.19)

with xA = (x —

(x+

—

XB = (x

—

+

h), 'cc = (x +

and

x0 =

h).

Another way of writing (19) is:

(5.1.19')

where b :=

a12.

Exercise 5.1.19.

(a) For the coefficients of L in (17) let the following hold: > 0, = a21 0, a11 + a12 > 0, and also + h/2) + a12(x — h/2, y + h) + a12(x -f h/2, y) > 0, and a 0. Show that the difference scheme which is described by (18) and (19) has consistency order 2 and that the ansociated matrix —L,, is a symmetric, irreducibly diagonally-dominant and positive definite M-matrix. (b) What is the suitable discretisatton for the case a12 0?

a11

>

0,

Exercise 5.1.20. L reads

The difference formula from Theorem 14 for the operator

+

a when

5 General Boundary Value Problem8

94

0, 0. Let the associated matrix be Lh. Then prove that the 0, transposed matrix LZ describes a difference method for the adjoint operator V and also possesses consistency order 1.

012

In general it is possible to show for regular difference methods that a discretisation of L'. The role of regularity is demonstrated in

is

Example 5.1.21. Let Lu := u" + au' in 11 = (—1, 1) with a(x) 0 for 0 for x > 0. According to (14) au' is discretised for 0 and a(z) z and for x > 0 by a(x)8u(x). Let the associated matrix z 0 by + Lh,1, where Lh,2 and Lh,1 correspond to the terms u" and be Lh = Lh,2 should be a discretisation of au' respectively. According to the above, —(av)'. But the differences at x = 0 and x = h are h'ta(—h)vh(—h) — a(O)vh(O)

h'[a(O)vh(O) + a(h)Vh(h)

—

.-(av)' — h10(O)Vh(0) + 0(h),

—

o(2h)vh(2h)1 = —(av)'

+ h10(0)Vh(0) + 0(h);

is a possible discretisation of thus they are not consistent. Nevertheless, has order of L'v = v" — (av)', for it can be shown that the error — magnitude 0(h). To prove stability one has to show const. Obviously it is sufficient to prove this inequality for sufficiently small h. In the proof of Theorem 9 we used the fact that

—Lwl mO,

on!'

for w(x) := exp(2Ra) — — + R)). Let Dhuh(x) be the difference equations from which Lhuh results after elimination of the boundary values. We set we,, := 2Rhw; i. e., wh(x) 2w(x) for x The followrng holds:

- RhL)W - 2RhLW 2 -

-Dhwh =

Each consistent difference method satisfies fl(DhRh small ho we thus have

- RhL)w. -

-' 0.

For

sufficiently

—Dawh(x)l (XE 0h, h
x

£7,, far from the boundary with —Lhw,,(X)

to the boundary, —Dhwh(x) also contains the sum which is not contained in —L,,wh(x). For the discretisations from Theorem 14, Corollary 16, Theorem 18, and Exercise 19, however, 1. For

Qh close

—Lhw,,>—D,,w,,1 holds for all grid points. Theorem 4.3.16

5.2 General Boundary Conditions

95

Theorem 5.1.22. The ducretssations from Theorem 14, Coroflary 16, Theorem 18, and Exercise 19 are stable under the conditions posed there, i.e., const for alt h H = {1/n:n E IN}. According to Theorem 4.5.3 the methods converge. The order of convergence order of consistency.

agrees

with the corresponding

5.1.5 Green's Function The idea of representing the solution by the Green function can be repeated for the general differential equation (Ia,b). The Green function (of the first x) is singular at = x and satisfies kind)

=

=0 foe

0,

x

forxEIoreEf'. Here, L' is the adjoint differential operator (16). If L L', then g is no longer x). Under suitable conditions the solution of (la-c), symmetric: g(x, (4) can be represented as

u(x) = —

x)

L

where B = Bt =

is a boundary differential operator (n, are the components of the normal vector n = F). Only when the e principal part of L agrees with is B the normal derivative. In the discrete case the inverse L' again corresponds to the Green function g(., .).

5.2

General Boundary Conditions

62.1 FormulatIng the Boundary Value Problem Let the differential equation be given by (1.la,b). The Dirichlet boundary condition (1.4) can be written in the form Bu

on I'

(5.2.la)

where B is the identity (to be precise: the trace on F). In more general settings B can be an operator — a so-called boundary differential operator — of order I: B

Ebs(x)o/ax. +bo(x),

If one introduces the vector b(x) = (bj(x), .. the form

XE F.

(5.2.lb)

By can be written in

5 General Boundary Value Problems

96

or B = bTV + b0.

Bti = (b(x), Vu(x)) +

(5.2.1W)

0 there results what is known as on F, also known as the boundary

Example 5.2.1. (a) From b = 0, b0(x)

the Dirichiet condition u = condition of the first kind.

(b) The choice b = n, ho = 0 characterises the Neumann condition, also

called the boundary condition of the second kind. (c) Equation (Ia) with (b,n) 0, ho 0, is known as the mixed, or the boundary condition of the third kind. Occasionally one just means by a mixed boundary condition: B = unTV ÷ ho = o'O/On + ho with or(x) 0. Remark 5.2.2. The case (b,n) = 0 is excluded in general. For (b,n) = 0, bTV is a tangential derivative. The boundary condition Bu = w is then very similar to a Dirichiet condition. The condition "u = on F" implies "Bu = := Bço on F" (Why is Bço defined?).

—x)/2

(a)

FIgure 5.2.1. (a) Boundary value problem with changing boundary condition type (b) Dirichiet problem in a disk with a cut

The normal derivative B = 8/8n (i.e. b = n) is important in connection with L = —4. For the general operator L in (l.lb) the so-called conormal derivative B with

b = An (A = (°ij)jJ1,...,n, aq as in (I.lb)) is of greater importance, as we will see in Section 7.4. Statements about existence and uniqueness of the solution always depend

on L and B. We have seen already that for L = —4, B = I (Dirichlet condition) uniqueness is guaranteed (cf. Theorem 3.1.2), while the problem associated to L = —4, B nTV = 8/t9n is, in general, not solvable (cf. Theorem 3.4.1).

5.2 GeneraL Boundary Conditions

97

The coefficients of B depend on position. Of course, B(x) 0, i.e., b(x) = o and bo(x) = 0, must not occur for any x E I'. But it is poesible that b(x) = 0 (and bo 0) in C I' and b(x) 0 in t'\'y. Then there is a on the piece -y and a boundary condition of fIrst Dirichiet condition u

order on the remaining boundary piece f'\-y. At the points of contact between

and I'\'y the solution generally is not smooth (it has singularities in the derivatives).

0 with y radius 1. Let the differential equation and boundary conditions be given as in Figure la. The boundary condition changes its order at x = y = 0. The (cf. (2.1.3)). Check that solution in polar coordinates reads: u = r 1/2 = O(r112) and =

Example 5.2.3. Let .0 be the upper semicircle around x =

The same singularity as in Example 3 occurs in the problem described This Dirichlet problem and in Figure ib; the solution is also r112 Example 3 are closely connected with each other.

Figure 5.2.2. A domain symmetric with respect to

Example 5.2.4. Let 5? =

U Q2 U be as in Figure 2: let the reflection of S?j in -y result in S?2. If one seeks a solution of Lu = in (1, Bu on

f

OS?, and if together with u the function II reflected in 'y is also a solution, one expects u = This solution then satisfies Lu = in Ru = on

Ou/On=Oon'y.

f

While the boundary condition Bti = on OS? may be of physical origin, Example 4 shows that a Neumann condition may also have a geometric basis. Another geometrically justified boundary condition is the following. Let 5? be given as in Figure 3: and 'Y2 are parts of F =85? with

98

5 General Boundary Value Problems

= {(x,,y):yi y Y2},

Then, in addition to Bu =

on

boundary condition u(x1,y)

t4z2,y),

f'\(i'i

i =

U 72), we

uz(Zj,y) = Uz(X2,y)

1,2.

can require the periodic for

Yl y Y2,

(5.2.2)

The solution is periodically continuable in the x-direction (with and period x2 — x1). The origin of periodic boundary conditions is discussed in Example 5, on

Example 5.2.5.

(a) Let if be an annutus which is described by the p0-

lar coordinates r E (ri, r2), w E [0, 2ir). fransformation of the differential equation to polar coordinates gives as the image domain the rectangle 11 = (i-i, r2) x (0, 27r). The original boundary conditions on if become bound-

ary conditions at the upper and lower boundaries while Equation (2) describes the periodicity of the angular variable x E (0, 2ir).

(b) Instead of on (1' C Ui?, one can also define a boundary value problem on a part of the 2-dimensional surface of a 3-dimensional body. If if lies, for example, on the surface of the cylinder E e IR}, the +
5.2.2 Difference Methods for General Boundary Conditions The boundary conditions posing the least difficulties are the periodic boundary conditions. Define as the set of grid points in .0 and on (but not on

ci. Figure 3).

Z2Z Figure 5.2.3. Discretisation for periodic boundary conditions

5.2 General Boundary Conditions

The difference equation at a grid point (x1, y) E

99

has a left neighbour

Replace uh(zj — h,y) in the difference equation by (x1 — h,y) outside uh(x2 — h, y). By doing so we have transferred the periodicity onto the differ-

ence solution without explicitly discretising Equation (2). Of course, the step size must be chosen so that X2 — x1 is a multiple of h.

Exercise 5.2.6. For periodic boundary conditions the structure of the matrix in the square (1 = (0, 1) x (0, 1). Assume Dirichlet Lb changes. Let L = conditions for the upper and lower boundaries, and for the lateral boundaries the periodicity condition (2). In analogy to (4.2.8) exhibit the form of the matrix Lb for a lexicographical arrangement of the grid points. In Section 4.7 we described the discretisation of the boundary condition nTV = a/on for the case that .11 is a rectangle and I coincides with the grid. In the same situation one can discretise the general boundary condition (la,b) as follows. B

outer

inner Line

(z,y)

s

Figure 5.2.4. Discretisation of Bu =

fl 1' be a grid point on I. The point of intersection Note that + b1 0, since bTV is not to be a tangential derivative. Fbr u E the directional derivative bTVu = (b, Vu) in is approximated by Let

+ h,

E

1h =

drawn in Figure 4 is given by

—

+

+

On the other hand, one obtains interpolation up to order 0(h2):

+ 0(h).

—

under the same conditions by

+ h,

h

(5.2.3a)

—

(5.2.3b)

100

Boundary Value Problems

5

The combination of (3a) and (3b) leads to the difference formula h)}

—

—

:=

I/h2 + +

(5.2.3c)

uh(X,y)

= with

=

Lemma 5.2.7. Let U E C2(TJ) one has

be the restriction to the grid poinis

O(h11uflc3(m)),

—

—

For a function

i.e., the discretüation of the boundary condition has consistencij order 1.

Equation (3c) has the form =

—

—

h,V+ h).

(5.2.3c')

fh(x) (before We assume that the difference equations satisfy the inequalities the elimination of the values E (5.2.4a)

S 0 correspond to the sign condition (4.3.la). The > 0 and second inequality in (4a) agrees with (4.3.4b). The corresponding inequalities for (3c') read: Here

ct?O, C20,

(5.2.4b)

and the boundary equations (3c') 0, (4a,b) implies be combined into Ahuh =

Let the difference equations in x E given for

E

Therefore, Ah is an M-matrix if A,, is irreducible and (4.3.4a) holds for at least one x E siderations explain why the boundary discretisation should The above satisfy the conditions (4b). (4b) holds for the case b0 =0 if and only if b2/b1 E [—1,0J, (4b) can be satisfied if one interpolates between and (0, 11. If b2/b1 i.e., if the tangential and + h). If, however, — h (instead of > and component is greater, one could interpolate between k. Fbr the case b2/b1 1, however, it is more practical kh), where 1b2/bd + h) between to interpolate at the point + h) and + h,V + h). Generally one should choose as interpolation point the point of intersection of the line with the dotted straight line in Figure 4. In the preceding discussion we started with the case fl = (0, 1) x (0, 1). Iii? is a general region, two discretisation techniques offer themselves (cf. Sections 4.8.1 and 4.8.2).

5.2 General Boundary Conditions

101

F

.--

Line

—

S — h)

•: Points in U: Points in Th

FIgure 5.2.5. FIrst boundary discretisation

First option for discretisation: and lj, be chosen as in Section 4.8.1. At near-boundary points the Let differential equation is approximated by a difference scheme which in the case

of L =

corresponds to the Shortley-Weller method. For this one needs As in Figure 5 let the values of Uh at the boundary points E a boundary point. Again we can set up an equation — sjh, be E 1h has the same position as + h, in analogous to (3a—c). In Figure 5 Figure 4. In general one has to use the point of intersection of the line which is also presented by + tb (t IR), and the dotted straight line in Figure —LI

5.

Second option for discretisation: Let consist of all points of be the grid aa used above. Now let that are far from the boundary. For all (x, y) Qh difference equations (with equidistant step size) are declared which involve For each point a boundary discretisation must be found (cf. 1h := Figure 6). Equation (3a) can be set up with and instead of + and The arguments of the coefficients b1 and b2 are + h, This point results implicitly from

—1) =

—

€1'.

(5.2.5)

Remark 6.2.8. (a) At the grid points adjacent to the boundary, the first discretisation requires a difference scheme for Lu = f with nonequidistant

stepsizes. The second discretisation requires an approximation of the nonlinear problem (5). From a programming viewpoint both procedures are undesirable because of the case distinctions. (b) If the vector b from B approaches the tangential direction, both methods fail since the straight line no longer intersects the dotted straight line from Figures 5 and 6. Here, the second discretisation fails earlier than the first one.

102

5

General Boundary Value Problems TI

Line

(1+ h,

• Points in

I Points in FIgure 5.2.6.

Second

boundary dlecretisation

Another option for avoiding the difficulties described above consists in using variational difference equations, at least near the boundary (cf. Remark 8.6.1).

Occasionally it is also poesible to simplify the boundary conditions by using coordinate transformations. Let Bu = be prescribed on C P. If one finds a transformation (x, y) 0 ,) audi that the equations = and are satisfied on one obtains for the transformed problem Bu oOu/On + bou = on a vertical boundary piece (8/On = The discretisation may be earned out as described in Section 4.7. The same reasoning as in the preceding sections proves stability and

convergence: Remark 5.2.9. Let the difference method

= ía satisfy the conditions := K1 + of Theorem 5.1.22 so that 1 holds for (K> o is constant; wh is as in the proof of Theorem 5.1.22). Let the boundary discretisation (3c') satisfy

For sufficiently large K, one then obtains (Bhti,h)(x) 1 for x E Hence it follows that 1, where is the M-matrix defined immediately after Equation (4b). Since conat (cf. Theorem 4.3.16), stability has been proved. The order of convergence is the minimum of 1 (order of consistency of ; cf. Lemma 7) and the order of consistency of = Ia. In general it is not poesible to approximate bTV by symmetric differences.

To construct a discretisation of Bu = w with consistency order 2 despite this fact, set: :

—

=

E

5.3 Boundary Problems of Higher Order

103

If, in order to remove the first two are three points in = in terms in the Taylor expansion, one uses the differential equation and the tangential derivative of Bu = one obtains a discretisation of Bramble-Hubbard [2) order 2. For the special case L = 4, B = 0/On + may be chosen such that the inequalities proved that the three points (xe, hold and guarantee stability. However, in general, the + C2 ÷ 4 0, c0 where

f

f'h. Nonetheless, the do not lie in the direct neighbourhood of construction of the discretisation appears too complicated to be recommended for practical purposes. (xe,

5.3 Boundary Problems of Higher Order 5.3.1 The Biharnionic Differential Equation In elastomechanics the free vibration of rods Leads to (ordinary) differential equations of second order if longitudinal vibrations (= compression waves) or torsional vibrations are involved. By contrast, transversal vibrations ( bending waves) result in an equation of fourth order. Correspondingly, the bending vibration of a plate leads to a partial differential equation of fourth

order. This is the biharmonic equation (plate equation)

42u=f infl

(5.3.1)

where 42 = 04/8x4 + 284/8z20y2 + 04/8y4. Here u describes the deflection of the plate perpendicular to the surface. If the plate is firmly clamped at the edge, one obtains the boundary conditions

u=

and Ou/On =

on F

(5.3.2)

with = 0. A btharmonic probLem (1), (2) also results from the transformation of the Stokes equations in 11 C in.2 (cf. Remark 12.1.5). The differential equation (1) may be combined with other boundary values than (2). An example is: is

=

and

4u =

on F

(5.3.3)

(simply supported plate). Other examples can be found in (7.4.12b,e).

Exercise 5.3.1. Show that if one solves the Poisson equations 4v = f in

£7,

solution of the boundary value problem (1), (3). Why can Problem (1), (2) not be handled likewise?

Remark 5.3.2.

The solutions of L12u =

minimum pnnciple (counterexamnple: u

0

do not satisfy a maximum-

x2 + y2 in £7 = KR(0)).

5 General Boundary Value Problems

104

5.3.2 General Linear Differential Equations of Order 2m multi-index) of order IaI = The partial derivative D° (a + is defined in (3.2.5). A differential operator of order 2m has the form

(x

L=

Q)

+ (5.3.4a)

and defines the differential equation of order 2m:

Ltt

f

in (2.

(5.3.4b)

Ellipticity has been explained thus far oely for equations of second order (cf. Definition 1.2.3).

Definition 5.3.3. The differential operator L (with real-valued coefficients

aa)iasaidtobe elliptic (oforder2rn) atxE Qif (5.3.5a)

for

Here,

We set o for all

of degree is an abbreviation for the polynomial Evidently (5a) is equivalent to P(x,e) = 1. For reasons of continuity either E IR' with

P(x, <0 must bold. Without loss of generality, we may assume that P(x,e) > 0; otherwise we scale with the factor —1 (changing from Lu(x) = 1) is compact it follows f(x) to —Lu(x) —1(x)). Since the set E min{P(x, = 1} > 0 and this justifies the formulation of that c(x) (Sa) as

c(x) >0.

(5.3.5b)

frzl.2m

Definition 5.3.4. The differential operator L is said to be uniformly elliptic in I? if inf{c(x):x E (1} >0 for c(x) from (5b). ExercIse 5.3.5. (a) L from (1.lb) into the notation of (4a). What axe the coefficients Ga for L 112? (b) Prove that the btharmonic operator 112 is uniformly elliptic. (c) Let aa be real-valued. Why are there no elliptic operators L of odd order? (d) If the coefficients Ga are sufficiently smooth one can write L from (4a) in the form L = (cf. (1.lb) and (1.2)).

For in = 1 (equation of order 2) we have used one boundary condition; for the biharmonic equation (in = 2) two boundary conditions occur. In general one needs m boundary conditions

of Higher Order

5.3 Boundary

b,0DaIL=

B,u

(j = 1,2,. ..,rn) on P

105

(5.3.6)

lcxl<m,

with boundary differential operators B1 of order 0 m3 <2m. arbitrarily, Remark 5.3.6. The boundary operators B2 cannot be but must be independent of each other (they must form a so-called "normal system"; cf. Lions-Magenes [1, p. 113] and Wioka Definition 14.1]). In particular the orders in1 must differ pairwise.

We have a Dirichiet boundary condition if B, = forj 1,•• (ci. (2)).

=

(O/On)-'1

The representation of the solution using the Green function will not be discussed at this point. However, it is remarkable that the Green function (and in particular the singularity function) for L is continuous whenever 2m > n. For L = for example, the singularity function is (cf. Wloka [1, Exercise 21.91). yj Ix —

5.3.3 Dlscretisation of the Biharmonic Differential Equation The simplest difference formula for L

is the 13-point difference

method

1

h4

2

—8

2

—8

20

—8

2

—8

2

,

(5.3.7)

which can be represented as the square of the five-point star Let the grid C £7 and the boundaries 1h c r, c in Figure 1. The differenee equation

from (4.2.3). be defined as

1

1

I

Dhuh=fh inQh requires the values of uh in boundary conditions (cf. (2)) Uh(X)=401(X)

£7,, U P,, U

for

7h• These are determined by the

x€f,

:= ju,,(x + hn) — Uh(X — hn)J/(2h) = for

(5.3.8a)

(5.3.8b) cP2(x)

(5.3.8c)

______

106

5 General Boundary Value Problems

0000000

000000000 on.... •00

Qh:. -yb:

00 Sf0 on....• so a a

a • • •

000000000 0000000

a 0

=

Ui U

Figure 8.3.1. Grid 8r the biharmonic equation In 11= (0,1) x (0,1)

As in Equation (4.7.14b) we assign to the corner points two normal directions each. Correspondingly, to each corner point belong two equations (8c). Note that two (m =2) boundary layers occur, fh and 'yb, and that two boundary conditions are given.

Remark 5.3.7. If with the aid of (8b,c) one eliminates in Equation (8a) the unknowns (uh(e):C e rh U'yh}, one obtains a system of equations Lhuh = Li, is not an M-matrix since the sign condition (4.3.la) for E also is in general not 0 (cf. (4.3.lb)). is violated. But Thus the methods of the proof in Section 4 cannot be used here. Instead we will prove the analogue of Theorem 4.4.12.

Theorem 5.3.8. Let ui, be the discrete solution of the biharmonic equation in the square a = (0,1) x (0,1) defined by (8a—c). The matrix Lh described m Remai* 7 is symmetric and positive definite. It satisfies

64h4, If

= 0 on "i, (cf (8b)), then

1/256.

(5.3.9a)

be estimated by 1

—

(5.3.9b)

+

where the norms are defined as follows: Iui,(x)t21"2,

:= Ih

Let the sum ample, the value and once as

in

(5.3.9c) XEA

xEOk

contain the corner points twice, where, for ex-

at x = (0,0) should be intempreted once as (of. comment on (8c)).

5.3 Boundary Problems of Higher Order

107

PROOF. We limit ourselves to the most important part, the inequality (9b). {(x,y) E 1R2:x/h,y/h Z} we set In the infinite grid Qa Uh(X)

Uh(X)

tih(x) =0 for x

for x E 0h,

Qh\Qh.

After partial summation one obtains

= >

= xEQh

= — E(8v)(x)(8u)(x)

(First consider the identities

in the one-dimensional case.) From the definition of iih it = 0, and U f'h). On 17h one has uh = 0 on

follows that

Uh(X+hfl)0,

tih(Xhfl)=Uh(x—hD), uh(X + hn)

= uh(x — h(n) +

for x E f'i,. If one sets 0uh(x) at

xE

— hn)/h — Uh(X — hn)1/h = —uh(x

:= [Uh(X)

then

= —h'8uh(x), for

= —2h'8uh(x) +

x E Ph, implies that

(up1, Laup1)Qh

where (.,

(us,

—2h 1(a;uh,

1-

and (.,

are

the scalar products belonging to the norms (9c).

According to ab one

a2 + b2/4

(a, bE JR)

(5.3.10)

estimates IOn

÷

:S

Ico2lr'h/4.

Thus one obtains

+

(un, fh)f',,

+ 2h1(a;up1, W2)I',,

+ Theorem 4.4.12 shows that (11) implies

16211fhHph.

j

In the general case we

0

therefore

IILlhUhIIflh

(5.3.12b)

write Up, as t4 +uj,t,

+

=

(5.3.12a)

0, (11) implies the inequality IIup1IIs).,,

f/' = 0.

+

IIUhlfQh II4huaIInb/16. Thus for so that

Iltiafla,. For

(5.3.11)

where

=0,

=0,

= fp, and U

together with (12a,b) implies (9b).

5 General Boundary Value Problems

108

Theorem 5.3.9. (Convergence) Let the solution of the bihannonic boundThen the ary value problem (1), (2) in 17 = (0,1) x (0,1) satisfyu E discrete solution of equations (8a—c) is convergent (at least) of order

+

huh — RhulIab

where Rh is the restriction to

(5.3.13)

U 1h•

PROOF. The Thylor expansion shows that

4uhRi,u =

+

+

+ h4R,

IRI

flh. (5.3.14)

Outside 17h one can specify Rhu arbitrarily since these values do not appear in (13). In l'h we set

Rhu(x-fhn) := u(x — hn) +

h4

+

—

+

2h5 8

x E Ph.

The choice is made so that (14) also holds for x E Th. Applying

to

yields

=

DhR,1u

=

+

(kh Is the restriction to flh). The difference Wh := satisfies

in 17h Uk — R,,u

(defined in Thh)

Dhwh = Dhuh — DhRhu

+

= Rh(f —

=: 9h,

(5.3.15a)

wh(x) = uh(x) — (Rhu)(x)

= = = =:

— u(x)

= 0 for x E

(5.3.15b)

— —

in

X E 1h.

If one sets

+ (5.3.15c)

and 'I'h = instead of fh and = in Theorem 8 (note that çoj = 0!) then (9b) implies assertion (13).

I

By using more complicated methods one can replace the factor h1/2 in (e > 0 arbitrary) so that the order of convergence is

(13) by i.e., almost 2.

Remark 5.3.10. Inequality (9a) shows conda(Lh) = O(h4). In general a difference method for a differential equation of order 2m leads to the condition

cond(La) =

(5.3.16)

5.3 Boundary Problems of Higher Order

This indicates greater sensitivity to

109

errors at higher orders 2m

(cf. §4.4 in Stoerili and Bulirsch—Stoer[1J). Difference methods for boundary value problems of fourth order with vailable coefficients and general domains fl are discus9ed in an paper by Zlámal

[1). There too convergence of oider 0(h3/2) is shown. By contrast, 0(h2)convergence was proved by Bramble (1J for the 13-point star (7) in a general domain if the boundary conditions are suitably discretised.

6 Tools from Functional Analysis

To enable readers without previous knowledge of functional analysis to follow

the next chapters we summarise here all definitions and results that will be needed later on.

6.1 Banach Spaces and Hubert Spaces 6.1.1 Normed Spaces Let X be a linear space (alternative term: vector space) over K where K = IR or K = C. In the following the normal case K = IR is always intended. K = C occurs only in connection with Fourier transforms. The notion of a norm . fl: X —. oo) is explained in Definition 4.3.12. The linear space X equipped with a norm is called a nor med space and is denoted by the pair (X, fl. Whenever it is clear which norm belongs to X, this norm is called Ii lix and one writes X instead of (X, fl. lix).

Example 6.1.1. (a) The Eudlidean norm H from (4.3.13) and the maximum norm (4.3.3) are norms on lit". (b) The continuous functions form the (infinite-dimensional) space If 1? is bounded, all ii E C°(fl) are bounded so that the supremum norm (2.4.1) is defined and satisfies the norm axioms. If ii is unbounded, the bounded, continuous functions form a proper subset 11C°(& of C°(fl). Instead of jJ . we use the traditional notation fl. (c) The Holder-continuous functions introduced in Definition 3.2.8 form the normed space (C'(J7), ii The norm defines a topology on X: A C X is open if for all x E A there exists an >Osothatthe "ball" in A. We write

x or x = urn

Example 6.1.2. Let I,,, I to

if

—

0.

f

The limit procees f, —' (with respect denotes the unifonn convergence known from analysis.

6.1 Banach Spaces and Hilbert Spaces

Exercise 6.1.3. The norm fl. reversed triangle inequality holds —

IlvIlI

X —* [0, oo)

fix — till

111

is continuous; in particular the

for

x,y E X.

(6.1.1)

As Example la shows, several norms can be defined on X. Two norms on X fi and ifi o < C
are said to be equivalent if there exists a number

(jzfi

HlxlD
for all x E X.

(6.1.2)

Exercise 6.1.4. Equivalent norms lead to the same topology on X.

6.1.2 Operators Let X and Y be normed spaces with the norms fi. fix rasp. ff fly. A linear mapping T: X —. Y is called an operator. If the operator norm .

:= sup{IlTxffy/(IzlIx: 0

xE X)

(6.1.3)

is finite, T is said to be bounded (ci. (4.3.9)).

Exercise 6.1.5. (a) T is bounded if and only if it is continuous. (b)

If

is a norm.

With the addition (Tj + T2)x = T1x + T2x and the multiplication by scalars, the bounded operators form a linear space which is denoted by L(X, Y). (L(X, Y), II 11Y4-x) is a normed space. With the additional multiplication (T1T2)x = T1 (T2x), L(X, X) even forms an algebra with unit I, since I always denotes the identity: Ix = z.

Exercise 6.1.6. Prove that fiTzffy HTHyi_xIlxflx If T1 E L(Y,Z) and T2

for all x

write T for the operator norm).

(6.1.4a)

L(X,Y), then T1T2 E L(X,Z) and fiT1

We

X, T L(X, Y).

E

L(X,Y) if

(6.1.4b) 0 (convergence in

112

6 Tools from Functional Analysis

6.1.3 Banach Spaces E X:n 1} is said to be Cauchy convergent, or is called A sequence k} —'0 for k —' 00. A space a Cauchy sequence, if SUp{IIXn — X is said to be complete if any Cauchy-convergent sequence converges to an x E X. A Banach space is a complete norrned space. is not is complete, but Example 6.1.7. (a) is a Banach space. (b) Let D c (C°(D), ((.

rational numbers).

(c) IN and the Holder spaces (C'(D), It Hc.(D)) II lIc'(D)) for k 8 > 0 as well as Ii Ilck.1(D)) are complete, thus Banach spaces.

PRooF of (b). If {u,1} is Cauchy convergent then there exists a limit u(x) := for all xE D. Since converges uniformly to u, t& must be continuous, i.e., u C°(D). U Denote by L°°(D) the set of functions that are bounded and locally iiflegrable on D. Here we do not distinguish between functions which agree almost everywhere. In this case the supremuin norm is defined by IHILOQ(D)

inf{sup{Iti(x)1 X

E

D\A} A set of measure zero}}

(L°°(D), H IIL'.'(D)) is a Banach space.

Exercise 6.1.8. Let X be a normed space and Y a Banach space. Show that L(X,Y) is a Banach space.

Exercise 6.1.9. Let X be a Banach space and Z C X a closed subepace. The quotient space X/Z has as elements the clasass I = {x + z: z Z}. Show that X/Z with the norm 1(111 Inf{IIx + zIIx: z Z} is a Banach space.

A set A is said to be dense in (X, II fix) if A C X and A = X, i.e., every x X is the limit of a sequence E A. If (X, fi lix) is a normed, but not complete, space, one calls (X, iI• 112) the completion of X, if X is dense in X and iIxII.t = llxlIx for all x E X. The completion is uniquely determined up to isomorphism, and can be constructed in the same way as one constructs as the completion of Lemma 6.1.10. Let be a nonned Linear space and subspace of a Banach space (Y, fi. lly) with IlxlIy CtlxiIx for all z X C Y. Let each Cauchy sequence in X with -. 0 aLso satisfy -.0. Then

there ezütsacompletionX ofXinY:XCXcY.

Frequently it is not easy to describe the image Tx of an operator T E L(X, Y) for all elements x of the Banach space X. The following theorem permits a considerable simplification: It suffices to investigate T on a dense set X0 C X.

6.1 Banach Spaces and Hilbert Spaces

113

Theorem 6.1.11. Let Xo be a dense subspace (or just a dense subset) of the normed space X. Let Y be a Banach space. (a) An operator To E L(X0,Y) defined on the subspace Xo with x E Xo} has a unique continuation T L(X,Y), sup{IIToxIIy/flxllx:O i.e., Tx = Toz for all x E Xo. x (b) For x,, E X0, x X) holds Tx = (c) = IIToIIr4_xo. PROOF. For x E Xo, T is defined by Tx = Tox, while for x X\Xo there = E X0 and Tx = —, X, exists a sequence exists and is independent of the choice of It remains to show that urn XmJIX, the the sequence E X0. Since — Tozm(Jy sequence T0x1,, is Cauchy convergent. Because of the completeness of Y there Xo with exists y Y with y. Similarly, for a second sequence y'. x, there exists for the same reason a y' E Y such that and Since y' — y = lim(T04 — — Toxn(Iy — —, 0, Tx is well defined by (b). Part (c) follows from — for E XO. IITxHy/IIxIIx =

Exercise 6.1.12. Let X0 be dense in (X, fi. (() Let Il( be a second norm on Xo, that is equivalent with fl . II on Xo. Show that the completion of X0 with respect to on X.

results in (X,((f.((() and

and (((•ffl are alsoequivalent

A corollary of the "open mapping theorem" (cf. Yoeida ti, §H.5J) is the following, important result:

Theorem 6.1.13. Let X, Y be Banach spaces. Let T and surjective. Then

L(X, Y) be

E L(Y, X) also holds.

Exercise 6.1.14. Let X, Y be Banach spaces, and T E L(X, Y) be injective. Show that: (a) Y0 := range(T) := {Tx:x E X} with as norm is a Banach subspace of Y. (b) T1 exists on Y0 and is bounded: E L(Yo,X).

Let X and Y be Banach spaces with X C Y. Obviously the inclusion denoted by I : z X x E Y is a linear mapping. If it is bounded, i.e., I E L(X,Y)

that is, ((xIly S CIIxI)x

for all x E X,

(6.1.5)

then X is said to be continuously embedded in Y. If furthermore X is dense in (Y,($ then X is said to be densely and continuously embedded in Y.

6 Tools from Functional Analysis

114

Exercise 6.1.15. Let X c V C Z be Banach spaces. Show that if X is [densely audi continuously embedded in Y and Y is (densely audi continuously embedded in Z, then X is (densely andj continuously embedded in Z.

6.1.4 Hilbert Spaces A mapping (., .): Xx X

K (K =

IR or

K = C) is called a scalar product

on X if

(z,z)>O t7..x+y,z) = A(x,z)+(y,z) for all A€ K,x,y,z

forallz,yEX, where

X,

(6.1.6a) (6.1.6b) (6.I.6c)

denotes the complex conjugate of A.

Exercise 6.1.16. Show that (a) lizil

(x,x)1/2 is a norm on X. For the

proof use

(b) the Schwarz inequality: flxfl Ifvll for sit z,y

(6.1.7)

X.

Hint: consider Uz — Ayfl 0 for A = flxfl2/(x,y), if (x,y)

0.

(c)(.,.):X xX-9Kiscontinuous. A Banach space X is called a Hubert space if there exists a scalar product ')x on X such that x X are said to be orthogonal (x I v) if (x,v)x = 0. X

If A C X is a subset of the Hubert space X then the orthogonal space A' :={XEX:(x,a)x =Oforall aEA}definesaclosedsubspaceofX. Lemma 6.1.17. Let U be a dosed subspace of a filbert space X. Then X mn be decomposed into the direct sum X = U U', se., x

a

u

u

v E U'.

= flull3c +

Exercise 6.1.18. Let A be a subset of the Hubert space X. Prove the equivalence of the following statements: (a) A is dense in X; (b) A' = {O}; there exists an a

A with (a,z)X

0.

___________

6.2 Sobolev Spaces

115

6.2 Sobolev Spaces in the following .12 is always an open subset of

6.2.1 L2([2) L2(12) consists of all Lebesgue-measurable functions whose squares on 12 are Lebesgue-integrable. Two functions u, V E L2(fl) are considered to be equal (u = v) if u(s) = v(x) for almost all z E 0, i.e., for all x E S?\A, where the = 0. exceptional set A has Lebesgue measure

Theorem 6.2.1. L2(12) forms a Hubert space with the scalar product (6.2.1)

:= (u,V)L2(f1) := and the norm luk := 11U11L2(fl)

:=

v'L

(6.2.2)

Iu(x)I2dx.

Lemma 6.2.2. The spaces Coo(S?)flL2(12) and

are dense in L2(Q).

Here

:= {u E C°°(i2): supp(u) compact, supp(u) CC 12}.

supp(u) := {x E Q:u(x)

(6.2.3)

0) denotes the support of u. The double zndusion fl

if cc $2 indicates that 777 lies in the interior of Si (i.e., 12' C Si and OS? = 0).

Let D° be the partial derivative operator (3.2.5b). In the following we need so-called weak derivatives which are defined in a nonclassical way.

DefinItion 6.2.3. for the latter v

u E L2(0) has a (weak) derivative v := D°u L2(fl) holds:

(w,v)o

for all WE

L2(S?) if

(6.2.4)

Exercise 6.2.4. Show the following: (a) Let u have a weak derivative D"u E L2(Q). If the classical derivative Dcv. exists in (1' C $2, it coincides there (almost everywhere) with the weak one. (b) If u has the weak derivative Va = LPu E L2(S?), and if has the weak derivative = E L2(S?) then is also the weak D(r+&derivative of u; i.e., = (c) Let f? c be bounded and 0 E 12. u(x) := (a E IR) has weak fIrst derivatives in L2(12) if a =0 or 2a + ii >2.

6 Tools from Functional Analysis

116

C°°(Q) let L2(fl) norm. Then v = D0u. (d) Fbr

—4 u E

L2(Q) and

v

L2(f1) in the

For further applications we consider

Example 6.2.5.

37k, where the bounded subdomains Let 37 = are disjoint and have piecewise smooth boundaries. Let Ic IN. A function (1 i N) has a (weak) kth u E Ck_1(37) with restrictions u derivative v0 = D°u L2(I?), Ic, which coincides with the classical one

PROOF. The (Ic — 1)st derivatives exist as classical derivatives so that the assertion need only be shown for k = 1. Let D0 = 0/Ox,, w E Cr(s?). The assertion is obtained via integration by parts:

= _Jwx,udX

—(D0w,u)o =

=

-

= since every x E normal direction

=

J J was, dx =

>

di]

{L.

J

—

wv0 dx

also belongs to

=

f

for some k

dx = (w, va)o, I with opposite

The inequality I(u, v)oI IuIoIvIo (cf. (17)) reads explicitly

L For v =

1

(6.2.5a)

the result is

u(x) dxj For a

(u(x)I2dxj

Iu(x)12

(j&(fl): measure of .0).

(6.2.5b)

L°°(fl),u,v E L2(D) one now has I

f

a(x)u(x)v(x) dxl

(6.2.5c)

IIUIIL2(fl)11V11L2(O).

6.2.2 H"(Jl) and Let Ic E IN U {O}. Let Hk(Q) C L2(Q) be the set of all functions having weak derivatives U-'u L2(J?) for

:= {u E L2(J7): EY'u

L2(S2)

for IaI

k}.

6.2 Sobolev Spaces

117

The Sobolev space denoted here by Hlc(a) is denoted by

in

some other places. Usually, there Hk(i1) is a space which only coincides for sufficiently smoothly bounded S? with

Theorem 6.2.6. Hlc(a) forms

a Hilbert space with the sailor product

(6.2.6)

(u,v)Jfh(D) :=

(u,v)k and the (Sobolev) norm lulk

lies densely in HIc(rl).

fl

Lemma 7, whose proof can be found, for example, in Wioka 3.5J permits a second definition of the Sobolev space

Theorem

Remark 6.2.8. Let Xo := {u

C°°((1): jujk
Definition 6.2.9. The completion of

in L2(S?) with respect to the

norm (7) is denoted by

Theorem 6.2.10. The Hilbert space is a subspace of Hk(ü) with the same sailer product (6) and some norm (7). For is dense in k = 0 there holds

(6.2.8)

PROOF. (a) Cr(Q) C Xo (cf. Remark 8) implies (b) According to the definition CT(.Q) is dense in norms (2) and (7) coincide. Lemma 2 proves (k arbitrary) (8) follows. C Hk(a) C 6.2.11. For £7 bounded,

C Hk(S2) C L2(fl). (c) For k 0 the = L2(.Q). Because of

and 1/2

(6.2.9)

ftLlk,o

lat=k

are equivalent norms in

PROOF. (a) Evidently,

I

Ik,O

<

IIHM(n).

6 Tools from Functional Analysis

118

(b) Let exists an

= k —1 set v := D"u E C000(.Q). For an with thus KR(O). Foreach XE

There

Rwith Qc -=

2

f

+ H)]

-R

2RJ —R

4RiuI o, = k—I yields follows likewise for all I j < k, and

I? yields is k-fold derivative. Summation over all o with

(cf. (Sb)). integration over x Now

...

thus all u E

Since

=

for

5

the statement follows from Exercise 6.1.12.

U

Exercise 6.2.12. Show that (a) for bounded (2 and k 1, Hk((2) and are different. Hint: Consider the constant function u(x) = 1 and use Lemma 11. (b) Lemma 11 holds even if (2 is bounded in one direction, i.e., if (2 lies on a strip {x E IR": xk1 < R} (k E {1,. .

Theorem 6.2.13. Letm 1. There exist constants C = C(m) and 17(e) = i)(C, in) snch that for aU 0

Ic

(6.2.lOa)

5 m, u E

+ 77(€))uIo for all e > 0,0 k < m,u E

PROOF. (a) Partial integration for

= Since also u by with

—

= I shows that

= it follows that (n + 1, one obtains in the same way

IuI?
(b) Let

(6.2.lOb)

0 be fixed. Set

wj. (bc) leads to 2z1 —

If one replaces

(1 <1 < m; u E

(6.2.lOc)

:= log IuII,wj := IlVm..+ (in —l)vo)/nz and z1 — z:+j

:= log C, where zo =

z = (z1,. , one obtains Az all. Here A is the M-matrix (4.1.9b) for h.= 1.A' implies = exp(vj) + Wi), i.e., (lOs) with C := (c) Elementary calculation shows that for each e > exists an Oo) such that

a9b'° ca + 17(e)b Formula (lOb) is a corollary of (lOa,d).

0, 0

for all a, b 0.

0 Oo < 1 there (6.2.lOd)

6.2 Sobolev Spaces

119

By similar means, together with (5.3.10), one proves

+

m,

> 0,

Remark 6.2.14. The set

E

k m, u

(6.2.lOe)

uE

(6.2.lOf)

C°°(47):supp(u) compact, IuIk
in Hk(.Q).

PROOF. Let u E Hk(fl), e >0. According to Lemma 7 there exists a function with a(x) = 1 E C°°(Q) with Iu — u4k <e/2. There exists a E 2. For sufficiently large R, one also has for lxi 1, a(x) = 0 for lxi <e/2. Thus there exists v(x) = a(x/R)ue(x) E

S

withlu—vlke. Since "supp(u) compact" already implies "supp(u) cc

Corollary 6.2.15.

=

we obtain

for all k 0.

The Leibniz rule for derivatives of products proves

Theorem 6.2.16. uEHk(fl).

for all a E

IlatLHHk(a)

Theorem 16 together with the substitution rule for volume integrals shows

Theorem 6.2.1T. (Transformation theorem). Let T: (1—. if be a oneto-one mappsng onto if with T and ldetdT/dxl 6>0 in f?. We write v = u o T for v(x) = n(1'(x)). Then u Hk(if) [u Hk(1l) [€ Hok(S2)j and

also impises u o T flu o

6.2.3 Fourier For tt E

(6.2.11)

TIIHb(a)

and one defines the Fourier-transformed function €i by dx.

(6.2.12)

Note that ü is described by a proper integral since the support of u is bounded.

120

6 Tools from Functional Analysis

ForR—'oo

Lemma 6.2.18. IR(u; y) :=

J

dx]

[JR'

uniformly to u(y) on supp(u).

PROOF. It suffices to discuss the case n = tion with respect to results in IR(u; y) = (1/7r)

J(x

(by Fubim's theorem). Integra-

sin(R(x — y))u(x) dx.

—

=

Then IR(1;y) =

1

1

for all R >0. Since u

C°°(LR"),

then also w(x, y) := [u(x)—n(y))/(x—y) E The estimates w(x, y) = and Wz(x,y) = O(1/x2) hold uniformly for y E supp(u). Partial integration yields —

u(y);y) = IRU

—

y)w(.,y);y) =

sin(R(x — y))w(x,y)dx

= .-(1/ir) J cos(R(x — y))Wx(X,y) dx/R

O(1/R).

The statement follows froni

IR(u;y) = u(v)IR(1;y) + IR(u() — u(y);y) = u(y) +O(1/R). Lemma 6.2.19. (s E L2(JR') and

PROOF. Lemma

18

•

= lulo for all u

shows

J [J

[J =

J

IR(u;y)u(v)dy —.

J

U

Iu(y)I2dy.

Lemma 6.2.20. The inverse Fourier transformation

=

u is defined

by(13):

J lim

J

(6.2.13)

6.2 Sobolev Spaces

PROOF. We have from Lemma

18

=

J

121

dy

J

J

IL"

= u(x) + O(1/R).

Theorem 6.2.21.

with E = = 1, i.e., 7 is an isometric mapping of L2(IR') onto itself The scalar prodtict satisfies (u,v)o = (ü,ii)o for all u,v E

PROOF. Since Cr(1R') is dense in L2(JR") (cf. Lemma 2), 7 can be contin(cf. Theorem 6.1.11). The norm estimate follows —+ ued to 7: are interchangeable (cf. (12) and from Lemma 19. The roles of 7 and (13)); thus E L(L2(JRYI), L2(IR")) also holds. The second statement results from

=

—

—

=

Exercise 6.2.22. Prove that: (a) With

+

=

—

—

.. .

=

U

there holds

for u E

(6.2.14)

(b) There exists C = C(k) such that E IR".

Lemma 6.2.23. (a) Juk = I%JEcrI
equivalent to 1u1

holds for all u E 1k is

:= 1(1 +

(6.2.15)

PROOF. (a) It suffices to show the statement for is E 15, Theorem 6.1.11):

-

-

= laIk

(b) The statement follows from Exercise 22b.

(cf. Corollary

= V

U

Lemma 6.2.24. Let8a,, be the difference opelntorOhju(x) := where e, is jth unit vector. If u and l49hjUjk C for all h > 0, 1 j n, then is E Hk+1(lRn) holds. Conversely, I8hjUIk Nk+1 holds for all u E H'+1 (lit"), h > 0.

6 Tools from Functional Analysis

122

follows +Se,))(e) = PROOF. (a) From for h Hence, since 4h2s1n2((,h/2) = 1/h. Summation over j = 1,.. for I

= follows ,

n and

then gives

integration over

(1u1+i)2 =

J(1 + J

...

+

+

.

(1

+

+

+

J

J

..

<1 J111/h 1/h vanishes for h The integral over (b) For the converse use (%ju)(x) =

6.2.4

H'(a) for Real

a

0

so that the statement follows. S (x + the,) cit.

0

Let 8 0. Fur a = It's one can define the following scalar product (16a) and the Sobolev norm (16b) for all u E Cr(JR.Th):

(u,v) :=

J(l +

(6.2.16a)

(6.2.16b)

Iut := 11(1 +

also defines the Sobolev space H(IR') for noninteger order s. On the basis of Lemma 23 and Exercise 6.1.12 the newly The completion in

defined 11(r) for a

U {0} agree with the Sobolev spaces used until this

poht.Letl7cR'.ThenumbersOcanbedeconipoeedaas=k+Awith 1. Wedeflne

:= IaIk

+

{f —

Jf

IuI. := IIUIIH.(sl) := The norm

—

dxdY]7 (6.2.17a)

(a = k + A, 0< 6 < 1).

(6.2.17b)

is called the Sobolev-SlobodeddT norm. One can define the

The properties of these spaces are summarised in the following theorem (cf. Adams 111, Wloka [11):

6.2 Sobolev Spaces

123

the norms (16b) and (17b) Theorem 6.2.25. Let a 0. (a) For i? = are equivaient, i.e., both norms define the same space
if a, u wheret=3EINU{O} ort>s. (d) aD°(bu) E

H(.0), a

for a t. C be reaL by 1ITUc.(m), with

(e) H'(.O) C Ht(S?), (f) In (lOa,b) k and m (g) In (11) replace

t > k if k

b

E C(i)),

t4.

the norm (16b), as well as (17b), that the Exercise 6.2.26. Check, u(x) = 0 for (xj > I belongs to characteristic function u(x) = I in

IP(IR) if and only ifO 8<1/2.

6.2.5 Trace and Extension Theorems The nature of boundary value problems requires that one can form boundary values (restriction of u to I' = 8$?, or trace of u on F = 8$?) in a meaningful way. As can be seen easily, a Holder-continuous function u E has a restriction "I E if only F is sufficiently smooth. But, from o H(1') does not follow E H(fl. Since the equality u = v on only means that u(x) = v(x) almont everywhere in .0, and F is a set of measure zero, u(x) u(x) may hold everywhere on F. Also, the boundary value 0(X) (x F) cannot be defined by a onntinuous extension either since, for example, o H'(.O) need not be continuous (cf. Exercise 4c).

r

The inverse problem for the definition of is extension: does there exist, for a given boundary value on 1', a function 0 E If'($?) such that ço ulr? If the answer is negative there exists no solution u H(.0) to the Dirichiet boundary value problem. First we study these problems on the half-space

Functions u restricted to F.

will first be continued to Il E

Theorem 6.2.27.

,

(6.2.18) and then zi

0. There exists an extension operator such that for aU u the extension

Let a

E

andu coincide on

PROOF. For s = =I

0,

= 11 = u

on

set = u on IR!. and = 0 otherwise. Since the mapping defined by := 11 is bounded:

1.

Ti(xi,

For $ .

1

define

=

—xe)

u(x1,.. .

by

for x,, > 0

124

6 Tools from Functionai Analysi8 1 one obtains, for example, For larger s one uses higher interpolation p. 101]). U (ef. Exercise 9.1.13, cf. Wloka

(i.e., by reflection on F). For s = i.e.,

.. , —xc)

formulae for

In the following the restriction is defined only on

is

written in the form -yu. At first, 'y

(-yu)(x) := u(x)

C

for all (x) E

1'.

(6.2.19)

x' =

We write x = H4(P) =

(x1, .

f' is identified with

. .

Theorem 6.2.28. Let a> 1/2. Then 'y from (19) can be continued to Thus we have in particular l'vuIa_i,2 In the mae n = 1, i.e., -yu = u(0), we have I'yul C,, lul..

for u

PROOF. It suffices to show h'u1_i,2 for tz (cf. Theorem 6.1.11 and Theorem 25a). Let the Fourier transforms of u E and tu := -yu = and i1 = be Fourier transform). can be written as product where acts on x,. Therefore and on x' E has the properties = and tui = W(.,0). According to Lemma 20, = W(e',O) = has the representation

=

for all.e'

:= (1 + Since u E Lemma 23b). Inequality (5a) yields

(6.2.20) lies in L2(IR") (cf.

+

=

J

The first factor may be computed as I(.(1 + with K4 = + dx < oo, since 8 > 1/2. The second is := IIU(e', (Fubin.i's theorem). Together because IIUIIL2RØ_I = we have:

(1 +

for

Integration over

J(i

+

E

results in =

I

=

i-

E

6.2 Sobolev Spaces

125

(cf. (16b)). In the = is proved with and the integration over is = u(0), represents already U

Thus jwI,_j/2 case n = 1, not required.

= 0. Evidently, Theorem 28 describes the restriction 14(., 0) = -yu to E 1R, with the for any other similarly we have is continuous tresp. Holderu(., same constant C8. The mapping continuous] in the following sense.

Theorem 6.2.29. For s> 1/2 the following statements hold: IR,u E

= 0 for all

—

(6.2.21a)

<

—

for all u E

0

(6.2.21b)

—

A < 1, A s — 1/2 (and for A =

1

ifs > 3/2).

PROOF. (a) Let

u E H8(IR") and The function ço,, is continuous in JR and set := since converges uniformly to — u(,x)I,_112 — t4,, for all x JR. Thus (21a) follows. = + e) — has the Fourier transform (b)

so that = = — := in the proof for Theorem 28 set first integral in the eethnate of now reads

f(i +

+

e/2)

=

(1

+

As uui

W(.,0). The

l/2—* J(i + t2)' sin2(i7t)

The decomposition of the last integral into subinte+ with ,, = grals over 1/u shows and fri The remainder of the argument follows the same lines as in the proof of Theorem 28. U Up to this point we have obtained H8(Wt") by completion of in

The next theorem shows that for suffithentiy large $ one can also complete in fl L2(IR") so that contains only claseical functions (i.e. continuous, HOlder-continuous, [HOlder-J continuously differentiable functions).

Theorem 6.2.30. (Sobolev's lemma) IN

Li {0}, s > k + n/2 and H'(IR") C

PROOF. (a) Let a

c for 0 < t

holds fork E IN, a

t + n/2.

t + n/2, 0
IRTh we want to show iu(x) — u(y)I

C independent of x,y. The coordinates of the 1R" can be rotated so that x = (x1, 0,. . 0), y = .

(yr, 0,..

,

,

0). An (n— 1)-fold application of Theorem 28 to tzQ, u(., 0), u(., 0,0)

126

6 Tools from Functional Analysis

etc., results in u(.,O,0, ,0) desired estnnate with C = u and fl . 11c(R") Cfl

Theorem 29 provides the . . .

Then u

.

thus C°(IR') follows from

C

(b) Let a

t + n/2, 1 < t < 2, u E H(IR"). Part (a) is applicable to for 1: Dau E (cf. Theorem 25d,e) with

1. Thus u

etc.

We return to the statements of Theorems 27 and 28. For all u E = u(.,0) agrees with Completion in the restriction

ylekis

This proves the following corollary:

Corollary 6.2.31. Let a >

1/2. For the restriction -iu := u(.,0) we have

'y E

With the restriction to = 0 one evidently loses half an order of differentiability. Conversely one gains half an order if one continues w E suitably in

Theorem 6.2.32. Let a> 1/2, w (t& u such that 1t4,

There eztsts a function and = w, i.e.,

w= PROOF. Let = and zZ' = equivalent to (20). For

=

+

:=

K. :=

be the Fourier transforms.

=

w is

+ WI2 +

J(i ÷

one checks that (20) and 1u1 = to IR!L proves the parenthetical addition.

hold. Restriction of u U

If one replaces with ageneral domain 1? C x then {0} = becomes F 8!?, and the necessity arises of defining the Sobolev

space JP(fl. We begin with the

Definition 6.2.33. Let 0< t IRU{oo} Iresp. k e t4U{0}J. We write (1 Ct (Ii Ck.1j if for every x F := OS? there exists a neighbourhood U C such that there exists a bijective mapping —, K1(O) = < 1} E with

6.2 Sobolev Spaces

127

çoE Ct(TJ),

ço1 E C'(K,(O)) Ck.t(U), E

fl F) = ço(U

(6.2.22a)

= 0),

K,(O):

n Si)

(6.2.22b) (6.2.22c)

O},

K,(0):

çp(U fl (IRTh\Q)) =

<0}.

(6.2.22d)

K,(0) is a circle (ball) if I is the Euclidean norm. For the maximum norm Ki(O) is a square (cube). Likewise, K,(0) can be replaced by another circle KR(z) or any rectangle x x Here

I

Figure 6.2.1.

Covering neighbourhoods of I' and I?

Example 6.2.34. Let 11 be the circle K,(0) C A neighbourhood off = (1,0) is U' from Fig. 1. The mapping XE U' i-s := E (—1,1) x (—1, 1) with (1 — Z2 = (1 — is bijective and satisfies (22a-d) with t = oo. The same holds for any x E I'. Thus (1 C°°.

ExercIse 6.2.35.

(a) The rectangle Si = (xi, x',') x and the L-domain from Example 2.1.4 are domains Si E C°". (b) The cut circle in Figure 5.2.lb does not belong to C°".

Lemma 6.2.36. Let 1? E Ct (5? e Ck.1)

(J

C

U° CC Si,

i SN),

(6.2.23a)

(JUi =F

(6.2.23b)

bijectivcfori = 1,.. ,N,

(6.2.23c)

(1

E Ct(aj(Ut flU3)) (resp. i

a bounded domain. Then there

N

Ut open, bounded (0 i N),

On W (1

be

(1 i
exsstsNElN,

orj' E

N) are defined mappings

flU,))I. (6.2.23d)

with the properties (22a-d).

6 Tools from Functional Analysis

128

according to PROOF. For every x E F there exist U = U(x) and cc = (i €N) be to U(x)flf'. Let be the restriction Definition 33. Let V' be an the open sets {x (1: dint (x, F) >1/i) Cc fl Let

open covering of the compact set Th. Therefore there exists a finite covering N) and at most one V' which is denoted by i (1 through U' = the statements follow from U0. If one sets U1 := U' n F, (22a-d).

N} fulfilling the conditions (23b.d) i A set of pairs {(U,,aj): 1 (resp. coordinate system for F.

18

called a

In Example 34 one has N = 4. The map E U1; and likewise sin((3 + el)7r/2) e U4, where in each case —1 < one obtains = + 1.

is given by = = cos((3 + ej)ir/2), < 1. On n

Lemma 6.2.37. (Partition of unity) Let

I N} satisfy (23a).

There exist ftinctions

E

C U',

for alt x E

=1

(6.2.24)

The general construction of the can be found, for example, in Wioka [1, §1.2). In the special case of Figure 1 one may proceed as follows. Let o-(t) :=O and for ti := exp(1/(12 — 1)) for t E (—.1,1). Then E 1 and supp(a) = [—1,11. In U' from Figure 1 one defines, for example,

:=

for x =

u(2r —

These cj(x) satisfy (24). A function u on F can be written in the form is parametrisable over C This in turn makes poasible

Definition 6.2.38. Let .0 E

Each summand —. IR. :a4(Ug) CC

and

Assume

fE

etc.

(23b-d) and (24). Let a< t E IN [a k-fl) or 8< t Sobolev space H(F) is defined as the set of all functions U: F (u,u) (1 i N). E

satisfy

N, t > 1. The IR

such that

Theorem 6.2.39. (a) H3(f') is a Hubert space with the sci'4ar product (u,

:= (u,v)ff.(f) :=

o

1 N} is another C'coordinate system off and {&,) another partition of unity, then the space 11(F) defined &y this is equal (b) If

to H(F) as a set. The norms of H'(F) and

are equivalent.

6.2 Sobolev Spaces

PROOF. For (b): 'fransformation Theorem 17 Eresp. 25g) ; for I?

129

C°", cf.

Wloka [1, Lemma 4.5).

The trace and extension theorems (Corollary 31 and Theorem 32) can be extended to any domain with a sufficiently smooth boundary. ')' now denotes the restriction to P 811: 7U = U Ii'.

Theorem6.2.40. LetQECwithl/2<8<_tEINOfl/2<8
L(H8(Q), y belongs to (a) The restrtction -yu of u E t/2( H8 — F)). there extsts an extension u e H'(11) with vi = (b) For each vi E

(c) For each w

H'(Jl) there exists a continuation Ew E H8(IR') with

E E L(H8(J7), and czj be as in PROOF. The proofs follow the same pattern. Let has its [resp. w8 = Lemma 36. The suminand onto [resp. 114 and can be mapped via 40j (reap. lresp. support in There Corollary 31 and Theorem 32 hold. The x {O} = can be mapped back continuation to or restriction to again. The first statement of the theorem is proved in detail as follows. o On the grounds of Theorems 17, 25g and ü4 := Let := Lies Thus the restriction 'y÷üj := the function üj belongs to := in H8 /2(IR"') (cf. Corollary 31) and has c4(U1) as support. Set on =0 on f'\U1. According to Definition 38, vi := on (i one finds belongs to H8_h/2(fl. Since represents the restriction (Ct(i?) is dense in H'(Q)): for all u E

uE

w(x) = E

P. Since all the partial mappings are bounded, one finds that

I.7u1,l/2 = 1w15_i,2 C3JuJ8.

I

Remark 6.2.41. (a) Under the conditions of Theorem 40 and the additional condition s > IaI + 1/2 there exists a restriction -1D°u E of the derivative of u E H8(Q). (b) For each uE with

s
Theorem 6.2.42. For .0 E C°" holds HJ($1) = {u

PROOF. (a) u 0. Since H

H1(11): ufr = 0}.

is the limit of v,, all of which satisfy uf1, with respect to H'/2(fl, it follows that u E

Analysis

6 Tools from

130

can = 0. The proof that u (b) Conversely, let u H'(a) and be divided up as follows: (ba) By using the partition of unity over {Ut: 0 i N} (cf. Lemma 37), the

statement reduces to the case I? = (bb) Without lose of generality one can amuine n = 1: Si = > 0 there exists a (be) According to Remark 14, for each E 0 there exists with finite support such that iii — ü41 q/2. For each

forz,7, 3/ti, := The function supp(u€) finite, and lu

—

—

(1

1.

0

satisfies

—

E C00(11t÷),

Thus X := {v

=

0, 0,

supp(v) finite} is dense in {u E 111(ffLF):u(O) = O}. were also dense in X with (bd) The statement would be proved if Let v E X. Evidently, respect to := holds for all E 0 for , —' 0. Since > 0. Set 0(17) := ilv'I1L3(O,,7) and note that 0(77) v(z) = for x it remains to estimate the variables liv — VnhiLa(Oe) + (p,, — 1)v', one obtains and liv' — t4iIL2(o,,,). Because of v, — Vt =

Ii

— v'lf

llL0..(O,n)ilt)11L2(O,,7)

+

+ 0(77). Smce

0 x

v(x) =

xlv

and hence Iv — v,,l? liv —

When Si

(5b) nnphes the estimate iv(x)i for all Then the statement follows from U + liv — C0(i1) 0.

C', the normal direction n exists at all boundary points.

Analogously to Theorem 42 one proves

Corollary 6.2.43. For (1 E C' and k

fI holds

H0k(a)={uE

=0 foro <


1).

6.3 Dual Spaces 6.3.1 Dual Space of a Normed Space Let X be a normed, linear space over JR. As a dual space, X' denotes the space of all bounded, linear mappings of X onto IR: L(X,IR).

6.3 Duai Spaces

131

According to Exercise 6.1.8, X' is a Banach space with the norm (dual norm) IIx'IIx'

The elements x'

zE X}.

=

:

(6.3.1)

X' are called linear functiouals on X. Instead of z'(x)

(application of x' to x) one also writes (x, X')XXX', or (x', X)X'xX, and calls

the dual form on X x X':

(•,

(X,X')XXX'

= (x',x)x'xx

x'(x).

Lemma 6.3.1. Let the Bariach spaceX be embedded densely and continuously in the Banach space Y. Then Y' is continuously embedded in X'.

PROOF. (a) Let y' E Y'. Because X C Y, y' is defined on X. (b) Since X is a dense subapace of Y, according to Theorem 6.1.11 and (1.5),

for each y' E Y' holds: Ity'IIy' = sup Iy'(x)I/(jxfly

sup Iy'(x)I/IIzfIx

i.e., Y' is embedded continuously in X'.

To the transposed matrix in the finite-dimensional case corresponds the

dual mapping (or the dual operator). Lemma 6.3.2.

Let X and V be normed and let T E L(X,Y). For each

E Y' (X,X')XXXI

for all z E X

(6.3.2)

defines a unique x' X'. The linear mapping y' x' deft nes the dual operator T': Y' X' with T'y' = z'. The following holds: T' E L(Y', X') and

= flTfly'...x.

(6.3.3)

PROOF. If one writes (2) in the form y'(Tx) = x'(x), one can see that = y' o T. (3) follows from the definitions of the norms: = =

sup

sup

sup

I Example 6.3.3. Let SI = (0,1), X = (c°(.??),fl and x SI. The mapping '-U' u(x) E IR is a functional: E (the soE called delta function). The Laplace operator LI to L(C2(Th), .

6 Tools from Functional Analysis

132

is char-

is applicable to

The dual mapping LV = acterised by

u E C2(fl).

ExercIse 6.3.4. Let S

L(X, Y), T

Exercise 6.3.5. Show that if T

L(Y, Z). Show that (TS)' = S'T'.

L(X, Y) is surjective, then T' is injective.

6.3.2 Adjoint Operators Let X be a Hubert space (over IR). Every YEX defined by (x, Y)x

is a linear functional

X'

wIth

IIxi

ilylix. The converse also holds

(cf. Yosida

Theorem 6.3.6. (Riesz representation theorem) Let X be a Hubert space andfEX'. There existsauniqne yjEXsuchthat

f(s) = (x,yj)x for ails E X and ilfIix'

liv,lix.

Corollary 6.3.7'. Let X be a Hilbert space. (a) There exists a one-to-one correspondence (the Riesz isomorphism) that preserves the norm, i.e., E L(X,X') with JXJJ = =

liJxilx'-x

IlJx 11x4-x' 1. (b) X" is a Hilbert space with the scahir product (x',y')x' = , ,1/2 The dual non's Iix lix' from (1) agrees with the norm induced by (x ,x (c) One always identifies X with X" because x(x') := x'(x). Prom this follows

=

Jx =

T" =

TforT€ L(X,Y) ifY= Y" is also aHiibert

space.

(d) One can identify X and X': X = X', .Jx = I. Let X, Y be Hilbert spaces and T L(X, Y). The mapping defined by E L(Y, X) is called the operator adjoint toT and satisfies

= (x,Ty)x for allxE X, yE Y;

= IITUxi-y. (6.3.4)

The adjoint and the dual operator only coincide (i.e., T' = T') if X' is identified with X and Y' with Y.

T€ L(X,X) issaidtobe selfadjoint (or symmetric) ifT=T. T E L(X,X) is called a projection if T2 = T. It is an orthogonal projection if furthermore T is selfadjoint.

Remark 6.3.8. Let X0 be a closed subepace of the Hubert space X. An orthogonal projection is given by Tx := y

X0 with

6.3 Dual

liz — vilx = inf{iiz —

133

Spaces

(6.3.5).

Xo}.

If conversely T L(X, X) is an orthogonal projection with the range Xo := {Tx: z E X} one has (5) for y Tx. An orthogonal projection always has the norm IIT1IT_x 1.

PROOF. (a) x can be decomposed uniquely into x = y+z (yE X0, z Xd) X0 implies y (cf. Lemma 6.1.17). y is the unique solution of (5). x thus T2 = T. The analogous decomposition x' y' + z' shows (x,T*XI) = (Tx,x') = (y,x') = (y,y'+z') = (y,y') = (y+z,y') = (x,Tx'), henceT= T'. (b) Let T be an orthogonal projection with range Xo. Let x = y + z be split = as above. T2 T shows Ty = y. For each y' X0 holds (Tz, y') = (z, (z,Ty') = (z,y') = 0, thus Tz E Xi-. Together with Tz E Xo follows Tz 0

sothatTx=Ty+Tz—y. (c) Tx

y and ilxti3c =

+

iivlft

prove IlTflx4_x

U

1.

6.3.3 Scales of Hilbert Spaces We assume:

VCU

are Hubert spaces with a continuous and dense embedding. (6.3.6)

Lemma 6.3.9. densely

Under assumption (6) U' is embedded continuously and

in V'.

PROOF. The continuity of the embedding U' C V' was established in Lemma 1. For proof that U' is dense in V', we use Exercise 6.1.18d (A := U',X := V'). Let 0 v' V' be arbitrary and u := E V C U. According to the definition, u' := JUU U' C V' is characterised by u'(x) (x, U)U for all

x€U.

follows

(v',u')V' = According to Corollary 7d

obtains the Gelfand triple

= ,

U

= u'(u)

>0.

U

and U' can be identified. By this one

V c U C V' (V c U continuously and densely embedded).

(6.3.7)

Corollary 6.3.10. In a Gelfand triple (7) V and U are also continuously arid densely embedded in V'.

PROOF. For U C V' see Lemma 9, for V C V' see Exercise 6.2.25.

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6 Tools from Functional Analysis

Attention. Likewise one could identify V with V' and one would obtain U' c V' = V c U. But it is not possible to identify U with W and V with V' simultaneously. In the first case one interprets x(y) = (y, x)uxrj' for x, y E U as x)u (in particular for x, y E V U), in the second case as (y, z)v.

Exercise 6.3.11. Let (7) be true. Set W := {Jç'u:u

U} and define (x, y)w := (Jvx, Jvy)u as the scalar product on W. Show that (a) W is a Hubert space; (b) W C V is a continuous and dense embedding; (c) (v, w)v = (v, Jvvi)u for all v E V, w W; (d) l(z,y)vI HxIIuIlyIIw for all x,y E W. Because U = U' the scalar product (z, y)u can also be written in the form

y(x) = (x,y)uxu'. If z E V, then y(x) = (x,y)Vxv' also holds. That means that (x,y)u = (X,y)vxv' for all x V,71 E U C V'. Likewise one obtains (x, Y)u = (X,y)v'xv for all x U and y E V. The dense and continuous embedding U C V' proves

Remark 6.3.12. Let V C U C V' be a Gelfand triple. The continuous extension of the scalar product (., .)u to V x V' (V' x VJ results in the dual XVI. Therefore the following notation is practical form (., •)v v'

for x€V,y€V', (x,y)v'xv = (z,y)u for x E V',y E V. In connection with Sobolev spaces one always chooses U := L2((1) so that the embeddings read as follows: H0($2) c L2(Q)

c

H8($2) c L2(a) c (118(fl))'

(a 0),

(6.3.8a)

(a 0).

(6.3.8b)

ExercIse 6.3.13. Show that (8a) and (8b) are Gelfand triples. The dual space of

is also denoted by

or

:= H3(J?) :=

(a 0).

The norm of H(.Q) according to (1) reads: := sup{I(u,v)L2(fl)I/1v18:0

where (u,v)L2(a) is the dual form on

Remark 6.3.14. (a) Let (1 = :=

for a 0

v

x H'(Q) (cf. Remark 12).

The norm dual to v

€

6.4 Compact Operators

135

equivalent to and has the representation (2.16b) with —s instead of s. (b) The Fourier transform shows for all 8 IR. E (c) au E H8(Q), if u H8(fl), a C'(i7), where = Isi E IN U {O} or t > (s(. i8

6.4 Compact Operators Definition 6.4.1. A subset K of a Banach space is said to be precompact [compact] if each sequence E K (i E IN) contains a convergent subsequence

fand

E

Kl.

Another definition of compactness reads: Each open covering of K already contains a finite covering of K. Both definitions are equivalent in metric spaces (cf. Dieudooné (1, (3.16.1)]). For the terms "relatively compact" and "precompact" see also Dieudonné (1, (3.17.5)].

Remark 6.4.2. (a) K C

is precompact [compactJ if and only if K is bounded (and complete]. (b) Let X be a Banach space. The unit sphere {x X: flx(f 1} is compact

if and only if dim(X)
Exercise 6.4.4. When is the identity I L(X, X) compact? Lemma 6.4.5. Let X, Y, and Z be Banach spaces. (a) Let one of the mappings T1

L(X, Y), T2

L(Y, Z) be compact. Then T2T1

L(X, Z) is also

compact.

(b) T e L(X, Y) is compact

if and only if T' e L(Y', X') is compact.

PROOF. (a) Let K1 := {x

X: jIz(Ix 1}. If T1 is compact, i.e., T1(K1) is precompact, then T2(T1(K1)) is also precompact and thus T2T1 is compact. If, however, T2 is compact, one proves the assertion as follows. Since scaling does not change compactness, 1 can be assumed without loss of generality. Hence T1 (K1) is a subset of the unit sphere in Y and therefore T2(T1(K1)) is precornpact. (b) cf. Yosida §XJ.

I

A special type of compact mapping is a compact embedding:

Definition 6.4.6. Let X C Y be a continuous embedding. X is said to be compactly embedded in Y if the inclusion I L(X,Y), Ix = x is compact.

136

6 Tools from Functional Analysis

Together with Definitions I and 3 one obtains: X C Y is compactly 1 contains a subsequence E X with

embedded if every sequence convergent in Y.

Example 6.4.7. Let I? be bounded. for s > 0.

is a compact embedding

C

<1 are equi-continuous and with PROOF. Functions E uniformly bounded. The assertion is based on the theorem of Arzelà-Ascoli (cf. Yosida [1, §111.3)).

Analogous results can be obtained for Sobolev spaces (ci Adams (1, p. 144), Wloka 11, §7)):

Theorem 6.4.8. Let 5? C IRTh be open and bo'ande.i (a) The embeddings C

s > t) are compact. (s, t E C°'1. The embeddings Hk(S?) C

(b) Fhrther, leti?

(k,l E NU TO),

k> 1) are compact.

(c) Let 0 t,r> 1) or 5? E the embedding 11(fl) C

(k + 1 > t). Then

ie compact.

Remark 6.4.9. In Theorem 8b one can replace .0 E C°' by the "uniform cone property" (cf. Wioka [1, §2.1)). To ensure Si E it is su.fficient that the boundary OS? is piecewise smooth and the inside angles of possible corners are smaller than Indented corners (cf. Figure 2.1.1) are thus permitted while a cut domain (cf. Figure 5.2.lb) is excluded. In Section 6.5 the following situation wdl arise:

V C U C V' Gelfand triple,

T E L(V', V).

(6.4.1)

Because of the continuous embeddings, T also belongs to L(V', V'), L(U, U), L(V, V), and L(U, V).

Theorem 6.4.10. Let (I) hold. Let V C U be a compact embedding. Then T L(V',V'), T L(U,U), T L(V,V), T e L(V',U), andT L(U,V) ore compact.

PROOF. As an example, let us do T

L(U, V). Since the inclusion I L(V, U) is compact we see I L(U, V') is also compact (cf. Lemma 5b). T L(U, V), as the product of the compact mapping I L(U, V') with T E L(V', V), is compact (cf. Lemma 5a). U

Exercise 6.4.11. Let dirnX
L(X,Y) is

6.5 Bilinear Forms

137

The significance of compact operators T L(X, X) lies in the fact that the equation Tx — Ax = y (with z,y E X, y given, x sought) has properties analogous to the finite-dimensional case.

Theorem 6.4.12. (Riesz-Schauder theory) Let X be a Banach space; let T E L(X,X) be compact. (a) For each A

(C\{O} one of the following alternatives holds:

(i) (T —

AI)'

L(X,X) or (ii) A is an eiger&value.

y has a unique solution x E X for all In case (i) the equation Tx — Ax T) := y E X. In case (ii) there exists a finite-dimensional eigenspace the eigenvalue problem E(A,T)\{O} solve {O}. All x kernel (T — Al)

Tx =

0. Ax, x and, if not (b) The spectrum o'(T) of T consists by definition of all There exist at most countably many eigenvalues which L(X, X), A = 0. e

can only accumulate at zero. A E a(T) if and only if A E o(T'). dim(E(A,T)) = dim(E(5,T'))
Lemma 6.4.13.

Let X C Y C Z be continuously embedded Banach spaces and let X C V be compactly embedded. Then for every e > 0 there exists a such that (6.4.2) flxIIy dUxIIx + C€IIxIIz for all z X.

X PROOF. Let e > 0 be fixed. The negation of (2) reads: There exists := X we thus 00. For with (llxiJIy — eIlx4Ix)/IJxdIz 0 and flysfIy > 1 have — ')/IIv4Iz —. oo. From this one infers UydIz for sufficiently large i. Since 1/c and X C Y is a compact embedding, a subsequence ye,, converges to y' V. Now, J)y1JIy > 1 implies Jjy*Jjy 1, i.e., y 0. On the other hand, also converges in Z to y since Y C Z is continuously embedded. —+ 0 gives the contradiction sought: y = 0.

6.5 Bilinear Forms In the following let us assume that V is a Hubert space. The mapping a(•, .): V x V —* Ut is called a bilinear form if a(x, y + Az) = a(x, y) + Aa(x, z),

a(x + Ay, z) = a(x, z) + Aa(y, z)

138

6 Tools from Functional Analysis

for (x, y, z a(x, Ày) =

V,

A E JR) (in the complex case one speaks of sesquilinear forms:

a(.,.) is said to be continuous (or bounded) if there

exists a C3 such that ja(x,y)I

CsIIxJIvI(yIIv

for all x,y

(6.5.1)

E V.

Lemma 6.5.1. (a) operator A

To a continuous bilinear form one can assign a unique L(V, V') such that

a(x,y) =

for all x,y E V,

C5.

(6.5.2)

(b) Let V, and V2 be dense in V. Let a(',.) be defined on V1 x V2 and satisfy (1) with "x E Vj,y E V2" instead of "x,y V'. Then can be extended uniquely to V x V 80 that (1) holds with the same Cg for all x,y E V.

PROOF. (a) Let x E V be fixed. := a(x,y) defines a functional c°x E V' with IIcPxIIv' Cs(Ixflv. One sets Ax := çox for x V. hAzily' Cshlxilv proves l(AiIv.._y Cs. The definitions show (AX,y)VIXV = (coz,y)v'xv = = a(x,y). Conversely, for each A L(V,V'), a(x,y) := (Ax, y)v'xy hIAxIlvslIyhiy (b) According to Theorem 6.1.11, A is also uniquely determined if a(.,

is only

given on V1 x V2. Then (Ax, y)v' v represents the extension. The proof shows

ilAilv'_v =sup{la(x,y)i:x,yE V,Uxilv = iiyhiv = 1}.

(6.5.3)

A is called the operator that is associated to a(.,.). Thebilinear form a(.,.) adjoint to a(.,.) is given by a*(x, y) := a(y,x) (x, y V). The bilinear form is said to be symmetric if a(.,•) = a*(., .). Exercise 6.5.2. Show that (a) If A belongs to a(.,.), then A' belongs to a(.,.). (b) If a(.,.) is symmetric, then A = A'.

Lemma 6.5.3. Let A E L(V, V') be the operator associated to a continuous bilinear form a(.,.). Then the following statements (i), (ii), (ui) are equivalent:

(I) A' E L(V',V) (ii) e, e' >

0

exist such that

inf{sup{ia(x,y)l:yE E

= 1}:xE V,hlxiiv = 1}=e >0,

(6.5.4a)

= l}:y

(6.5.4b)

liyilv

1}

=

e'

>0;

(iii) the inequalities (4a) and (4c) hold:

sup{la(x,y)i:xe V,lJxhlv = If one of the statements (i)—(iii) holds, then

1}

>0.

(6.5.4c)

139

6.5 Bilinear Forms

(6.5.4d)

(e,e' from (4a,b)).

1/11A'IIv.._vi

C

Prom (4a) follows

l',IIxIIv =

inf{sup{Ia(x,y)I:y E V,flyIlv = 1}:x

1}

e >0.

(6.5.4e)

Conversely, (4a) follows from (4e) wsth a possibly larger e > 0. (4e) and (4c) conditions. (4e) is equivalent to are also called the

for all x

V, lIyItv = 1) cHxIIv

sup{Ia(x, y)I: y

V,

(6.5.4e')

because (4e) is equal to (4e') for all x V, IIxIIv = 1. The scaling condition tIxIIv = 1 can emdently be dropped. The left-hand side in (4e') agrees with the definition of the dual norm of Ax so that (4e) and (4e') ore also equivalent to (4e"): (6.5.4e") for all x V. hAzily'

PROOF. (a) "(i) mf{

}

Ia(x,y)I

= inf aEV

Uxllviiyiiv

.

— mf sup —

L(V',V) exist. Then (4a) follows from

(ii)": Let

i(Ax,y)l

_.

iixilvhiyhlv

I(AA'x',y)i

p€V IIA'ziivliviIv

=

— —

c

II A1 lu—i sup i(x',y)I iv' i,€v

= lIE sup 11A'x'ilv/hlx'flv'l

CV

=

C.

Because A'' =

In the same way one shows (4b) with e' =

(A')', (3.3), and V" (b) "(ii)

hIyHv

V, it follows that e = (iii)": (4c) is a weakening of (4b).

(i)": e > 0 in (4a) proves that A is injective. We wish to show (c) "(iii) with that the image W := {Ax: x V} C V' is dosed. For a sequence the From (4a) one infers W there exists zI, E V with = w1,. that via (4e) and (4e") (with x := — flwg, — WpiIv'/f. — There is Cauchy convergent, this property carries over to Since —. in V. The continuity of A E L(V, V') proves exists an x' E V with = Ax,, —+ Ax' so that w' = Ax' E W. According to Lemma 6.1.17 one Wt. If A were not surjective (thus W V'), there can decompose V' into would exist a w W1 with w 0. Then y Jv'w = E V would satisfy 0 (cf. Theorem 6.3.6, Corollary 6.3.7). Since a(z,y) = (Ax,y)v')
Iw—

f

6 Tools from Functional Analysis

140

It will be shown that for interesting cases conditions (4a) and (4b) are equivalent (cf. Lemma 17). A particularly simple case is in

Exercise 6.5.4. Show that if dim V < 00, then (4a) implies the statement (4b) with e' = e and conversely. Definition 6.5.5. A bilinear form is said to be V-elliptic if it is continuous on V x V and there is a constant CE such that

a(x,x)

for all x

(6.5.5)

V.

Exercise 6.5.6. Show that: (a) If W C V is a Hubert subspace with the norm equal (or equivalent) to V, then a V-elliptic bilinear form is also W-elliptic. IR be continuous. If (b) Let a(., .): V x V holds for all x E V0 where Vo is dense in V, then (5) follows with the same CE. (c) Let a(•, .): V x V IR be continuous, symmetric, nonnegative (a(v, v) 0) and satisfy (4a,c). Then a(.,.) is V-elliptic with e2/Cg (e from (4a,b), Cs from (1)). Hint: First prove that Ia(u,v)1 (a(u,u)a(v,v)11/2 (cf. Exercise 6.1.16).

Lemma 6.5.7. V-ellipticity (5) implies (I) and (4a,b) with e =

e'

CE.

PROOF. Lets E V, IIxIIv = 1. V, IIyIIv = 1} Ia(x,x)I CE proves (4a) with e CE. (4b) follows analogously. U

The combination of Lemmata 1, 3, 7 together with IIA''tIv,_vi IIA1Hv_v' (cf. Lemma 6.3.2) proves

Theorem 6.5.8. Let the bilinear form be V-elliptic [or satisj*j (1),

(4a,c)J.

Then the corresponding operator A 8atIsfies the conditions L(V, V')1

A

A1 with

L(V', V),

IIA'Iiv"_v = IIAIIv'÷_v Cs,

(6.5.6)

IIA''llv._v' = 11A'IIv._vi C'

from (1) and C' = 1/CE fre.sp. C' = 1/c = 1/c']. With the help of the bilinear form a(.,.) and a functional I

V' one can

formulate the following problem:

Find x E V with

a(x,y) = 1(y) for all y

V.

(6.5.7)

According to Lemma 1 one can write (7) in the form (Ax—f, y)V'xv = 0

if and only if A' E L(V', V). Hence one obtains Theorem 6.5.9. Let the bilinear form be continuous (cf. (1)) and satisfy the stability condition (4a,c) [it is sufficient that a(.,.) is V-ellipti4 Then Problem

6.5 Bilinear Forms

(7) has exactly one solution x

141

with

A'f. This satisfies

C= 1/e= ife' [resp. C= 1/CEI. Corollary 6.5.10. Under the assumptions of Theorem 9 the analogous statement holds with the same estimate for the adjoint problem

Find xEV with a'(x,y)=f(y) broil yEV. PROOF.

(cf.

(6.5.8)

U

Exercise 2a, Theorem 8).

Lit be continuous. Let V0 be dense in V. Exercise 6.5.11. Let a(., .): V xV Show that the solution x E V of problem (7) is already uniquely determined by "a(x, y) = f(y) for all y E V0". The same holds for (8).

Problem (7) may be equivalent to a variational problem:

Theorem 6.5.12. Let a(.,.) be V-elliptic and symmetric; furthennore, let

f E V'. Then

J(x) := a(x, x) — 2f(x)

(x

(6.5.9)

V)

assumes its unique minimum for the solution x of Equation (7).

PROOF. Let x be the solution of (7). For arbitrary z E V set y :=

z —

x.

From

J(z) = J(z + y) = a(x + y, x + y) —2f(x + y) a(x, x) + a(x, y) + a(y, x) + a(y, y) — 2/(x) = J(x) + a(y, y) + 2ja(x, y) — 1(y))

=

J(x) + a(y, y) .1(x) + one can read J(z) > J(x) for all z

—

2f(y)

.1(x) + CEIIZ

U

x.

The term "V-elliptic" seems to indicate that to elliptic boundary value problems correspond V-elliptic bilinear forms. In general this is not the case. Rather, V-coercive forms will be assigned to the elliptic boundary value problems. Their definition necessitates the introduction of a Gelfand triple (cf.

(6.3.7)):

V C U C V'

(U = U', V C U continuous and densely embedded).

Definition 6.5.13. Let V C U C V' be a Gelfand triple. A bilinear form is said to be V-coercive if it is continuous and if there exists oK and CE > 0 such that

a(x,x)

for

all xE V with Ck >0.

lR

(6.5.10)

142

6 Tools from Functional Analysis

a(x,y) +CK(x,y)u with CK from (10). Let

Exercise 6.5.14. Set a(x,y)

V' be the inclusion. Show that (a) the coercivity condition (10) is I: V equivalent to the V-ellipticity of a.

A + CkI to a(.,.).

L(V, V') is associated to a(.,.), then so is A Why does A L(V, V') hold?

(b) If A

The results of Riesz-Schauder theory (Theorem 6.4.12) transfer to A as soon

as the embedding V C U is not only continuous but also compact.

Theorem 6.5.15. Let V C U C V' be a Gel/and tr*ple with compact embedding V C U. Let the bilinear form a(.,.) be V-coercive with correspond:ng V' be the inclusion. operator A. Let I: V (a) For each A E C one of the foil owing alternatives holds: (i)

(A—Xf)1 €L(V',V)

(ii)

A is an eigenvo.lue.

and (A'—Xiy' EL(V',V),

f

f

in case (i) Ax — Ax = and A'x' — = are uniquely solvable for all I E V' (i.e., a(x, y) — A(x, Y)u = 1(y) and a(x',y) v)u = f(y) for all E(A) := y V). in case (ii) the,e exist finite-dimensional eigenspaces {0} such that E'(A) kernel (A' — kernel (A — Al) and {0} Ax = Ax

for x

E(A),

a(z,y) = A(x,y)u

i.e.,

for all y

V,

(6.5. lie.)

for x*EEI(A),

a*(x*,y)=X(x*,y)u forall y€V.

i.e.,

(6.5.1 ib) (b) The spectrnm o'(A) of A consists of at most countably many eigenvalues

a(A) if and only if A

which cannot accumulate in C. A

<00. dimE(A) = (c) For A E r(A), Ax — Ax = only f I E'(A), i.e.,

f E V'

o'(A'). Fhrthermore

has at least one solution x E V if and

(f,x)v = 0/or all x' E E'(A).

PROOF. With V C U, V C V' is also a compact embedding, i.e., the inclusion I: V —+ V' is compact. A + CKI with from (10) satisfies A + CKI E L(V, V'), (A + CKI)1 e L(V', V) (cf. Exercise 14 a). Lemma 6.5.4 shows that K := (A + V —+ V is compact. Hence the RieszSchauder theory is applicable to K K—

id =

—j41

—

= =

—

jsl. Since

+ CxIY' {A +

—

—Al)

with A = 1//h — CK, the statements of Theorem 6.4.12 transfer via K —

I

6.5 Bilinear Forms

143

has measure zero so that under the conditions of Theorem 15 the solvability of Ax — Ax = is guaranteed for almost all A. ProbLem (7) is solvable if not "accidentally" 0 E c(A).

Remark 6.5.18. The spectrum

f

Lemma 6.5.17. Under the conditions of Theorem 15 the inequalities (4a,b) are equivalent.

PROOF. (4a) proves that A is injective, i.e., 0 (1(A). Theorem iSa shows L(V', V) so that (4b) follows from Lemma 3.

I

A'

Gen-

Evidently a(.,.) remains V-coercive if one adds a multiple of (., erally, there holds

Lemma 6.5.18. Let a(.,.) be V-coercive where V C U C V'. Then a(.,.) + b(.,.) is also V-coercive if the bilinearformb(.,.) satisfies one of the following conditions: (a) for every e > 0 exists C( such that

for all x E V.

jb(x, x)f

(6.5.12a)

(b) Let the embeddings V c X and V C Y be continuous, with at Least one of them compact. Let the following hold: for all x E V.

Ib(x,x)I

(6.5.12b)

(c) Let the embeddings V C X, V C Y be continuous. Let (12b) hold. For such that or Uv assume that for every e >0 there exists a or

÷

iS

for x

eIlxIIv +

V.

(6.5. 12c)

PROOF. (a) Select e = CE/2 with CE from (10). Then a(•, .)+b(.,.) satisfies the V-coercivity condition with CE/2 > 0 and CK + instead of CE and CK. (b) Lemma 6.4.13 proves (12c).

(c) Let the first inequality from (12c) hold, for example. Since the embedding V C Y is continuous, Cy exists with CyflxlIv. Choose C' = in (12c):

+

5

K

Since

I

7 Variational Formulation

'7.1 Historical Remarks In the preceding chapters it was not possible to establish even for the Dirichiet problem of the potential equation (2.1.la,b) whether, or under what conditions, a classical solution u C2(Q) n C°(fl) exists. Green took the view that his Green's function, described in 1828, always exists and that It provides the solution explicitly. This is not the case. Lebesgue proved in 1913 that for certain domains the Green function does not exist. Thomson (1847), Kelvin (1847), and Dirichiet offered a different line of

reasoning. The Dirichiet integral

J

(7.1.1)

=J

describes the energy in physics. With boundary values u = on I' given, one seeks to minImise 1(u). This variational problem is equivalent to

I(u,v) :=

=0 for all v with v =

0

on

(7.1.2)

The proof of the equivalence results from 1(u+v) = I(u)+21(u,v)+I(v) and 1(v) 0 for all v (cf. Theorem 6.5.12). Green's formula (2.2.5*) provides I(u, v)

= f0 v/lu dx =0 for all v with v =0 on F such that /lu =0 follows.

Thus, like (2), the variational problem 1(u) = miii is equivalent to the Dirichiet =0 in a, u = on I'. problem The so-called Dirichiet principle states that 1(u), since it is bounded from below by 1(u) 0, must take a minimum for sonie u. According to the above considerations this would ensure the existence of a solution of the Dirithlet problem. In 1870, WeierstraB argued against this line of reasoning, stating that while there may exist an inflinum of 1(u) over {u E fl = on r} it need not necessarily be in this set. Fur example, the Integral J(u) := fu2(x)dx in {u C°U0,11):u(0) = 0,u(1) = 1} never takes the

value inf J(u) = 0. Further, the following example due to Hadainard shows that no finite infirnum of the Dirichiet integral need exist. Let r and be the polar coordinates In the circle Q K1(0). The function = in .0 but the integral 1(u) does not exist.

7.2 Equations with Homogeneous Dirichiet Boundary Conditions

145

The above difficulties disappear if one seeks the solutions in the more suitable Sobolev spaces instead of in C2(Il) fl C°(17).

7.2 Equations with Homogeneous Dirichiet Boundary Conditions In the following we investigate the elliptic equation (7.2.la)

in Si,

Lt.i = g

L=

(7.2.lb) PIm

of order 2m (ef. Section 5.3; Exercise 5.3.5d). The principal part of L is

=

(7.2.2)

According to Definition 5.3.4, L is uniformly elliptic in Si if there exists

0

such that for all x E Si,

Attention. In the case that only

E IR".

(7.2.3)

is assumed, one needs to

replace "for all XE Si" by "for almost all x

Si".

We assume the homogeneous Dirichiet boundary conditions (7.2.4)

which are only meaningful if F =

is

sufficiently smooth. Note that in the

standard case m = 1 (an equation of second order) condition (4) becomes U =0. Since with u = 0 on F the tangential derivatives also vanish, not only the kth normal derivatives (k m — 1) but also all the derivatives of order $ m —1 are equal to zero:

inxEf forlal<m—1.

(7.2.4')

Condition (4') no longer requires the existence of a normal direction. According to Corollary 6.2.43, (4') can also be fonnulated as (7.2.4")

Let it E be a classical solution of(la) and (4). To derive the variational formulation we take an arbitrary v and consider

146

7 VariatIonal fbrmulation

(Lu, v)0 =

the integrand vanishes in the proximity of I' so that one can integrate by parts: E C8°(.O),

Dau)dx

without boundary terms occurring. Thus we have found the variational formulation

IaI,I$Im

f

dx

= Ja

g(x)v(x) dx

(7.2.5)

for all v

since Lu = g. If conversely a with boundary conditions (4) satisfies condition (5), then the partial integration can be reversed and f0(g — Lu)vdx = 0 for all v E Cr(a) proves Lu = g. This means that a classical solution of the variational problem (5) with boundary condition (4) is also a solution of the original boundary value problem. Hence the differential equation (la,b) and the variational formulation (5) are eqwvalent with respect to classical solutions. We introduce the bilinear form a(u, v) :=

J

(7.2.6)

J ag(x)v(x)dx.

(7.2.7)

and the functional

/(v)

As remarked above, the boundary condition (4) for classical solutions u means that a E Thus the "variational formulation" or "weak formulation" of the boundary value problem (1), (4) reads as follows: find

a

with a(u,v)

1(v)

for all v E

(7.2.8)

A solution of problem (8) which, according to the definition, lies in but not necessarily in is called a weak solution.

ExercIse 7.2.1. (a) Let 0 be bounded. Show that any claasical solution aE fl is also a weak solution. (b) With the aid of Example 2.4.2 show that this statement becomes false for unbounded domains.

Theorem 7.2.2.

Let

bOunded on

x

E

The bilinear form defined by (6) se

7.2 Equations with Homogeneous Dirichiet Boundary Conditions

PROOF. Let u,v E

147

The inequality (6.2.Sc) yields

const Since

to Hr(.O) x

is dense in (ci. Theorem 6.2.10), a(.,.) has an extension and is bounded by the same constant (cf. Lemma 6.5.lb). U

The function f(v) is also defined and bounded for v E if, for example, 9 E L2 ((1). According to Exercise 6.5.11, the variational formulation (8) is equivalent to the following one:

with a(u,v) = f(v)

find u

One can regain the form Lu =

for all v E

(7.2.9)

/

by applying Lemma 6.5.1. Let L E H—m(a)) and f E Hm(12) = be defined by a(u,v) = and 1(v) (f,v)ff-.,.(a)XH.,,(a) for all v E Equation (9) states that Lu—f. (7.2.9') While (la) represents an equation Lu g in C°(I2) (i.e., for a classical solution), (9') is an equation in Hm(fl). Theorem 6.5.9 guarantees unique solvability of Equation (9) if a(.,) is We first investigate the standard case m = 1 (equations of order 2tn = 2).

Theorem 7.2.3. Let 12 be bounded, m (uniform

+

)i31

a(u, u) PROOF.

1, aafi

L°°(12). Let L

(3)

elhpticity) and be equal to the principal part L0, i.e., aap = 1. Then the form a(.,.)

Since

=

=

1

e' > 0.

0 for

(7.2.10)

= one can identify a and /3 according to {1, . .• ,n}. Fbr fixed x 12 use (3)

= 8/Ox, with indices i,j with

=

=

E

Integration over $2 yields a(u, u)

= 1Vu12 dx.

Since

(cf. Lemma 6.2.11), (10) followscC0. with e' =

Corollary 7.2.4.

dx

U

The condi*ion "12 bounded" may be dropped s/for

Example 7.2.5. The Helmholtz equation the bilinear form

+u=

f in

a=

1? leads

to

148

7 Variational Fbrmulatlon

a(u, v)

liE

(x) + u(x)v(x) j dx =

ff(Vu, Vv) + uvj dx.

a(ts,v) is the scalar product tn Ha(S2) (and H'(S?)). The fact that a(u, u) = proves the (S?)-ellipticity. Exercise 7.2.6. Let the assumptions of Theorem 3 or Corollary 4 be satisfied, except for the fact that the coefficients aao and ao$ = 1) of the = first derivatives are arbitrary constants. Show that inequality (10) holds unchanged.

Theorem 3 cannot easily be extended to the case m> 1. Theorem 7.2.7. Let the coefficients of the principal part be constant8: 0afl = m. assume that constforlaj= =Oforo < 2m —1, aoo 0 for a = f3 =0. Let L be uniformly elliptic (cf. (3)). Further let either £2 be bounded or aoo >0. Then a(.,.) is

PROOF. We continue u through u =0 onto Exercise 6.2.22, and inequality (3) show that

a(u,u)-Jaoou2dx= a

Theorem 6.2.21,

>2 JaafiD0'uD6udx (aI,(PI=m

=

=

=

>2

E

>2

11

>2

ef Let aoo > 0. There exists an e' > 0, so that for all E From this follows

e' E1cz1<m

fri2 and

—

(cf. Lemma 6.2.23). IfS? is bounded, use Lemma 6.2.11.

Having shown the of the form a(., •), to apply the general proofs after Theorem 6.5.9.

we

I

are now able

Theorem 7.2.8. (Existence and uniqueness of weak solution8)If a(.,.) is Hr(fl)-eflipgic then there exists a solution u E (9) which satisfies

iUim

(CR

(6.5.5)).

of Problem

(7.2.11)

7.2 EquatIons with Homogeneous Dirichiet Boundary Conditions

Hm(.0) and u =

(11) holds for all I is equivalent to

Since

149

(cf (9')), inequaLity (11)


(7.2.11')

1/CE.

The term "variational problem" for (9) goes back to the following statement, inferred from Theorem 6.5.12:

and symmetric bilinear Theorem 7.2.9. Let a(.,.) be an form. Then (9) is equivalent to the variational problem

find u E

such that

J(u) J(v)

for all v E (7.2.12a)

where

J(v) :=

Attention. If a(.,.) is either not (9) remains meaningful although the

(7.2.12b)

—1(v).

or not symmetric, Problem does not minimize the functional

J(u).

Example 7.2.10. The Poisson equation —i.lu = f in S7, u 0 on F, leads to a(u,v) = f0(Vu(x), Vv(x))dx. For a bounded domain .0, a(.,.) is HJ(Q)-elliptic (cf. Theorem 3), so that for any / E

there exists exactly one (weak) solution u E H01(Q) of the Poisson equation. This is also the solution of the variational problem f0 fVul2dx — f(u) = miii. A weaker condition than

a(u,u)

is the

—

Theorem 7.2.11. Letm = 1, and let the coefficients condition (3) of uniform ellipticity. Then a(.,.) is

satisfy

PROOF. We write L as L L1 + L11, with Lj satisfying the conditions of Theorem 3, resp. Corollary 4 ifS? is not bounded, and L,1 containing only derivatives of order

1. Then we can apply the following lemma.

U

Lemma 7.2.12. Let a(,.) is

with aap E

is also

a'(., + a"(.,.) be decompose4 such that a'(.,.) or perhaps only

contains only derivatives of order

2m — 1.

Then a(.,.)

150

7 Variational Formulation

Y= PROOF. (6.5.12c) follows from (6.2.lOb) with X = V = and U = L2(f1), so that Lemma 6.5.18c proves the aseertion. • The generalisation of Theorem 11 to arbitrary m 1 requires stronger conditions on the coefficients of the principal part.

Theorem 7.2.13. (Carding's theorem) Let L be uniformly elliptic (cf. (3)) and assume let the coefficients a,,p with E L°°(a). a(.,.) is = m be uniformly continuous in fl. Then tat If conversely aa$ E C(s?) holds for tat = 1131 = m and E L°°(12) otherwise,

then from the

follows uniform etlipticity (3).

Details of the proof can be found in Wioka 11, Theorem 19.21. The proof given there also holds for unbounded S?, since the coefficients are uniformly continuous. For the first part of the theorem one uses a partition of unity. The significance of coercivity lies in the following statement.

Theorem 7.2.14. Let (1 be bounded and a(.,.) be

Then one of the following alternatives holds: (i) Problem (9) has exactly one (weak) solution u E (ii) The kernels E = kernel(L) and E = kemel(L') are k-dimensional for a k IN, i.e.,

a(e,v)=G,

a(v,e*)=O

Further, the eigenualue problem

a(e,v) =

A(e,v)L2(O)

for all v E

(7.2.13)

has countably many eigenvalties which do not accumulate in C.

PROOF. Since for bounded I? the embedding V := C U := L2(Q) is compact (cf. Theorem 6.4.8b), Theorem 6.5.15 is applicable. U

7.3 Inhomogeneous Dirichiet Boundary Conditions Next, we consider the boundary value problem

Lu = g in fl,

u = w on 1,

where L is a differential operator of second order (i.e., m = formulation of the boundary value problem reads: find u E

H'(fl)

with u

(7.3.1) 1).

on f' such that

a(u,v) = f(v) for all v E

The variational (7.3.2a) (7.3.2b)

7.3 Inhomogeneous Dirichiet Boundary Conditions

151

But according to Section 6.2.5 the restriction uk of is IP (i1) on I' is well defined as a function in H1/2(F). Thus "is = coon F" must be understood as the equality = in H1/2(I'). In contrast to the preceding section one uses a(.,.) in (2b) as a bilinear form on H1(fl) x Hd(S?). It is easy to see that a(.,.) is well-defined and bounded on this product.

Remark 7.3.1.

For the solvability of Problem (2a,b) there exists a iso

H'(i7) with

it is necessary that: (7.3.3)

isoll-

If a function iso with Property (3) is known, a second characterisation of the weak solution results: Let

iso

satisfy (3);

find w

that

a(iu,v) = f'(v) := 1(v) — a(uo,v) for all v

(7.3.4a)

(7.3.4b)

Remark 73.2. The variational problems (2a,b) and (4a,b) are equivalent. If iso and w are the solutions of (4a,b), then is = iso + w is a solution of Problem (2a,b). is is a solution of (2a,b), then, for example, iso = u and w = 0 satisfy Problem (4a,b).

Exercise 7.3.3. Show that f'

H'(fl) for f' from (4b) and

Ill—i Ill—i +

(7.3.5)

with C5 from ja(u,v)l CsIuljIv(t (cf. (6.5.1)).

Remark 7.3.4. The problem (1) and the variational fonnulation (2a,b) have the same classical solutions if such exist. PROOF. It suffices to assume v in (2b). Integration by parts can be carried out as in Section 7.2 and proves the assertion. U Theorem 7.3.5. (Existence and uniqueness). Let problem (2.9) (with homogeneous boundary values) be uniquely solvable for all f H1(Q). Then Condition (3) is sufficient, and necessary, for the unique solvability of Problem

PROOF. If there exists a solution is E H'(.Q) of (2a,b) then (3) is satisfied. However, if (3) holds, one obtains via (4a,b) a unique solution since (4b) agrees with (2.9). U

Remark 7.3.6. Under the condition 0 C°", (3) is equivalent to E II 112(f).

(7.3.6)

152

7

Variational

PROOF. If vo satisfies Condition (3) then Theorem 6.2.40 shows that E H'/2(F). If conversely 'p H'/2(f), then the same theorem guarantees an extension E H1(Q) to a with uoIj' = 'p and luoli 5

(7.3.7)

U

Let inequality (2.11') hold in the case of homogeneous boundary values. Equation (5) shows Iuui S lt'oli + Iwli S Itioli + (Ill—i + C'IuoIj)/€ for the solution of Problem (4a,b). The estimate (7) proves

Theorem 7.3.7. Let a

C°". Let the bilinear form be restricted to H'(a)x and let it satisfy (2.11'). Then for every f H1(Q) and 'p H'12(f) there exists exactly one solution u H'(a) of Problem (2a,b) with

+

<

(7.3.8)

ExercIse 7.3.8.

Let a(.,.) be symmetric and HJ(Q)-elliptic. Show that Problem (2a,b) with / = 0 is equivalent to the variational problem: Find ti H'(S?) with tip = 'p such that a(is,u) becomes minimal (cf. (1.1)).

7.4 Natural Boundary Conditions The bilinear form a(.,•) defined in (2.6) is also well-defined on Hm(S?) x Hm(Q). In analogy to Theorem 7.2.2 there holds

Theorem 7.4.1. Assume

E

The bilinear form defined by (2.6)

is bounded on Hm(Q) x Hm(Q): ia(u,v)I S Eap 113c*$Ik.o(O)ItImIVIm 101 all u,v E Hm(O). Now let / be a functional from (Hm(.O))'. Equation (2.7) with g for example, describes such a functional; but (2.7) is only a special case of the functional f subsequently defined in (la), which we want to use as a foundation in the following.

Exercise 7.4.2. Let r be sufficiently smooth and let g

L2(a),

L2(r)

hold. Show that

1(v) :=

J

g(x)v(x) dx + j 'p(x)v(x) dl',

v

lf'(Q),

(7.4.la)

defines a functional in (H'(Q))' with ill Ii(H'(ci))' S L2(D) + UcoIiL2(r)). This implies f E (Hm(a))' for all m> 1. More precisely, the following also holds:

7.4 Natural Boundary Conditions (II))' + IkoItH—1/2(Ii).

153

(7.4. ib)

Frequently, variational problems with a physical background have the form:

such that

find u€Hm(A2),

a(u,v) = 1(v) for all v

(7.4.2a)

Hm(fl).

(7.4.2b)

from Section 7.2, u Hm(47) In contrast to the condition u E contains no boundary condition. Nevertheless, Problem (2a,b) has a unique solution if a(•,•) is Hm(.O)-elliptic. This condition is easy to satisfy. 7.2.4 Theorem 7.4.3. Under the conditions of Theorem 7.2.3 or a(-,) is H'(fl)-elliptsc: a(u,u) for all u E H1(i2). Problem (2a,b)

(with m =

1)

has exactly one solution which satisfies the estimate (3): (7.4.3)

It&li

PROOF. The

same

as for Theorem 7.2.3 or Corollary 7.2.4, and Theorem

7.2.8.

U

Corollary 7.4.4. (a) A unique solution which satisfies the estimate (3), also exists if instead of H1(12)-ellspticity one assumes: a(.,.) is H'(i2)-coercive, O C°" is bounded, A = 0 is not an eigenvaltie (i.e., a(u, v) 0 for all

yE H'(Q) imphesu=O). (b) The combination of inequalities (3) and (ib) results in luli

+

for the solution of (2a,b), if f is defined by (is) with

H'/2(f).

(7.4.3')

IIWIIff—1/2(r)J

g

(H'(Q))',

PROOF. (a). According to Theorem 6.4.8b, H'(O) is compactly embedded in L2(37) so that the statement of Theorem 7.2.14 can be transferred. If A = 0

is not an eigenvalue then L1 6.4.12).

L((H1(a))', H'(Q)) holds (cf. Theorem U

To find out which classical boundary value problem corresponds to the variational formuiation (2a,b), we assume that (2a,b) has a classical solution u E Hm(Q) 11 Further, v E can be assumed also (cf. Lemma 6.5.ib). For reasons of simplicity we limit ourselves to the case m 1. Under the assumption E C1 (1 the following general Green formula is applicable:

154

7

a(u, v)

VariatIonal Formulation

=

+

+

J

{

=

ij=1

+ aoouvl

4=1

4=1

—

÷

+ aoou} v

[

+

+f

(17.4.4)

Here, the are the components of the normal direction n = n(z), x E F. We define the boundary differential operator

B :=

(7.4.5)

when L is described by (2.lb). Equation (4) becomes a(u,v) = f0vLtidx + vBu di. By the formulation of the problem, a(u. v) agrees with 1(v) from (is). If we first choose v E C H'(f?) the boundary integrals drop out and we obtain Lu = g as in Section 7.2. By this the identity a(u,v) = f(v) = f1pvdi for all v E H'(i7). According to Theorem reduces to runs over the set H'/2(I') if v runs over H'(O), so that we have 6.2.40b,

dl'

fr %b(Bu —

0 for all

H'/2(fl; thus Bu =

This proves

Theorem 7.4.5. Let F be sufficiently smooth. A classiosJ solution of the problem (2a,b) with / frvm (is) is also the claaaieal 8olution of the boundary value problem

Lu=g

mu, Bu=,o

on 1',

(7.4.6)

and conversely.

The condition Bit = is called the natural boundary condition. This results from the fact that in (2b) (as distinct from (2.9)) the function v may assume arbitrary boundary values. Note that the bilinear form determines L as well as B. Exercise 7.4.6. Show that the bilinear form from Example 7.2.10 for ghas as the natural boundary condition the Neumann condition

=

=0.

00,1 be bounded and a(.,.) be Let Then the statements of Theorem 7.2.14 holds with Hm(fl) instead of

Theorem 7.4.7.

Example 7.4.8. Let (1 a(u,v) =

0°" be a bounded domain. The bilinear form associated to the Helmholtz equation

—%iu+cu=f in!? (c>0),

9u/On=O onf,

is H'(fl)-elliptic since a(u, u) inin(1, C)IUI1. For c =0, however, a(.,.) is only H'(f?)-coercive. As is known from Theorem 3.4.1, the Neumann boundary

7.4 Natural Boundary Conditions

155

value problem for the Poisson equation (i.e., for c = 0) is not uniquely solvable. According to alternative (ii) in Theorem 7, there exists a nontrivial elgenspace E = kernel(L). u E E satisfies a(u, u) = 0, thus Vu = 0. Since (1 is connected, it follows that u(x) = coust and therefore dimE = 1. Since a(.,.) is symmetric, E :== kernel(L') coincides with E. According to Theorem 7, the Neumann boundary value problem a(u, v) = 1(v) (v II'(fl)) is solvable if and only if f I E, i.e., f(1) = 0. If f(v) := f0g(x)v(x)dx, 1(1) = 0 reads as gdx = 0. If, however, / is given by (Ia), the integrability condition reads f(1) = g dx + f1. çodl' = 0 (this is Equation (3.4.2), in which I should be

replaced by —g).

Remark 7.4.9. While the classical formulation of a boundary condition such as ôu/ôn = 0 requires conditions on the boundary 1', the problem (2a,b) can be formulated for arbitrary measurable 0. In the following, we proceed in the opposite direction: does there exist, for a classically formulated boundary value problem Lu = g in .0, Bu = on f', with given L and B, a bilinear form a(., .) such that (2a, b) is the corresponding variational formulation? This would mean that the freely prescribed boundary operator B represents the natural boundary condition. For m = 1 the general form of the boundary operator reads

B=

+ bo(x)

(x

1).

(7.4.7)

With bT = (b1, .. . , (ci. (&2.lb,b')). Here one can also write B = bTV is not allowed to be a tangential derivative (cf. Remark 5.2.2):

(b(x), n(x))

0

for all x E 1.

(7.4.8)

(ac) p to the equivalent scaled equation (cf. (5.1.ld)). By passing from Bu oBu = with o(x) = (n(x), A(x)n(x))/(n(x), b(x)), one can ensure that (n, ob) = (n, An). Thus in the following it is always assumed that b already satisfies (n, b) = (n, An).

Remark 7.4.10. Let m = I. Let (8) hold. Let A(x) be the matrix A

PROOF. Because of (8) a is well defined. For Bu = and oBu = a 0 is required. This is guaranteed by the uniform ellipticity: (n, An) eInl2 = Theorem 7.4.11. (Construction of the bilinear form). Let m = 1. Let L and B be given by (2.lb) and (7), with b satisfljing condition (8). Then there exists a bilinear form a(.,.) on H1(s?) x H1(0) such that to the variational problem (2a.,b) corresponds the classical formulation Lu g in 0,

Bu=p onf.

7 VariatIonal

156

Fonnulation

PROOF. The bilinear form we seek is not uniquely determined. We will give two possibilities for its construction. First we discuss the absolute term in (7). On the basis of Remark 10 we assume (n,b) = (n,An). (a) Let the vector function 13(x) := (13i(x),. . be arbitrary E The differential operator

:=

+

—

+

maps every u C'(S2) into zero: Lju = 0. Thus the operator L can be replaced by L + L1 without changing the boundary value problem. Let a(.,.) be constructed according to (2.6) from the coefficients of L + Lt. Equation (5) shows that the boundary operator associated with o(.,•) has the absolute term (7.4.9)

If = 0, the choice of = is successful. Otherwise, two other options are available. (aa) Select such that on I' the following holds: /3j(x) = Since ml = 1, the term (9) then agrees with bo(x). The practical difficulty in this method consists in the need to construct a smooth continuation on £1 of the boundary values x E F. (ab) Set /3, = -ao' and add a suitable boundary integral: a(ti, v) :=

f

+ [000

—

+

dx

(7.4.10)

The integration by parts described above shows that the boundary operator associated to (10) reads

b=

+ b0.

(7.4.11)

(b) The operator (11) can be written in the form B = bTV+b0 with b An. Since (ii, b) = (ii, An) = (n, b) has been assumed already, d := b — b is

orthogonal to n. To change b to b there are again two options.

(ba) Define the n x n-matrix A' on F by A'

dziT —

i.e., = njdj. This A'(x) is skew-symmetric: A'T = —A'. Continue A'(x), which is at first only defined on F, to a skew-symmetric matrix A' E C'(i7). [Here we have the same practical difficulty as in step (aa).J The entries of A' —

define

L2 :=

E

8

8

8

7.4 Natural Boundary Conditions

157

Again, L,2u = 0 holds for all u E C2(Q), since —

(a,

+

—

=

+

—

= 0.

L2 without changing the boundary value problem. The coefficients belonging to L + L2 result in a boundary operator B whose derivative terms read + = t(A + A')n]TV. By (dnT — ndT)n d = b — b , since the construction of A' we have A'n = (n, n) = 1 and (d, n) 0. Since we also have An = b, the derivative term L + L2 does not change the in B gives bTV as desired. The transition L absolute term in B, so that from (11) follows B = bTV + b0. (bb) Let B = bTV + b0 be given (cf. (11)) such that d = b b is orthogonal to n. From this the boundary operator T := dTV is the derivative in a tangential direction if n = 2 [resp. in the tangent hyperplane if n 3J. if I' of v is sufficiently smooth then the restriction H'/2(f'). By Remark 6.3.14a, one can show that T E L(H"2(r'), H-'/2(f')). is well-defined for H'/2(f), in Since T(ulp) Hh1'2(f), particular, for = with v H (fl). Thus

Thus L can be replaced by L ÷

b(u,ts) := is

j(vtr)T(ulr)dI'

a bilinear form bounded on H'(fl) x H1(a). We add b(.,.) to a(.,.) in

(10). Integration by parts yields the boundary operator B + T = bTV + bo+ U

Remark 7.4.12. Let the bilinear form (2.6) be H'(.Q)-coercive (cf. Theorem 3). Let its coefficients, as well as the boundary 1', be sufficiently smooth

(€ C'). Then the constructions of the preceding proof again result in an B7jH1(Q)-eoercive form.

PROOF. We go through the steps. (a) In step (aa) only terms of lower order are added so that Lemma 7.2.12 is applicable. As for step (ab), see step (bb). (ba) In step (ba) one adds b(u,v) Here + f0 Lemma 7.2.12 is also applicable, since the skew symmetry of A' results in 6(u,u) (bb) In the construction in step (bb), Lemma 7.2.12 is applicable analogously. This is easiest to understand in the case n = 2. Let I' be described by {(z,(s),x2(8)): 0 < s 1}. If Xl,X2 E C'flo, 1J) and =

if we have d E C'(f') for the d in T = dTV, then b(.,.) has the representation b(u,v) = where = u(s) = v(x,(s), x2(8)), r E C1((0, 1)). Thanks to periodicity, integration by parts yields in b(u, v) = — 10' (rV)'lZ ds without boundary terms, so that and

b(u,u)

ds

—

=

This implies

158

7 Vsriational Fbrmulation Ib(u, u)I

CIITIICI(Io,l])

for all s C (1/2,1) (cf. Theorem 6.2.40a). Since apply Lemma 6.5.lSc.

EIUII —

one can

U

The case m 2 has been excluded in this section (except for Theorem 1). Boundary value problens of order 2m require m boundary conditions = on 1' (j = 1,... ,m) (cf. Section 5.3). For m 2 the proof of Hm(fl)coercivity becomes more complicated. In order to carry over Theorem 7.2.13, one needs in addition the so-called condition of Agmon (cf. Wloka 1, Theorem 19.3J, Lions-Magenes [1, p. 2101). The resulting complications can be seen with the aid of the biharmon Ic

equation. Example 7.4.13. (a) To the variational problem: find a(u, v) :=

C

ff2(fl) with

dx = 1(v)

J

(7.4.12a)

for all v E H2(Q), corresponds the

=

g

in Q,

=

formulation

and

=

on F.

(7.4.12b)

But the bilinear form a(.,.) Is not H2(.Q)-coercive. (b) To the variational problem: find u C H2(i1) fl HJ(Q) with

for alive (7.4.12c)

(a(.,.) as in (12a)) corresponds the claseical formulation

in!?,

u=O

(7.4.12d)

The bilinear form is fl (c) The boundary conditions in

= g in

S?,

u =0 and

=

p

on F

(7.4.12e)

are admis8ible.

this boundary value problem cannot be written in the present form as a variational problem. A variational formulation for (12e) reads:

Find uEH2(fl)flH01(f1) suththat

forallv€H2(!?)with (7,4.12f)

7.4 Natural Boundary Conditions

159

(a(.,.) as in (12a)). But this does not agree with the present concept since u and v belong to different spaces.

PROOF. The equivalence of the variational and the classical formulation can be shown via integration by parts: a(u, v)

=

j

2

u dx + j

f

01)

-

dl'.

The noncoercivity in part (a) results as follows. Let (1 C IRTh be bounded. For all a E IR., ... = sin(axj)exp(ax2) lies in H2(i7) and satisfies = 0, hence also a(ua,ua) = 0. If a(.,.) were coercive, there would exist a C with 0 a(ua,uo) for all a, i.e., The contradiction results from 102ua/OxiIo = O2IUaIO for sufficiently large a. Natural and Dirichiet conditions can occur together. In Example 13b, u = is a Dirichiet condition and = ço a natural boundary condition. Even in the case m = 1 both sorts of boundary conditions can occur. 0

Example 7.4.14. Let 'y be a nonempty, proper subset of I'. The boundary value problem

=

g

in 0,

= 0 on 'y,

Ou/On =

çø

in the variational formulation reads as follows: find u

a(u,v) :=

on

(7.4.13)

114(0) such that

= 1(v) :=

for ally E where H4 (.0) :={uE H1(l)): ur=Oon-y}. Equation (13) is occasionally termed a Robin problem.

Exercise 7.4.15. Let a(u,v) := f0[(Vu, Vv) +cuvldx with c >0 on V x V with V := {u E H'(0):u constant on f'} be defined. Show that (a) a(.,.) is V-elliptic. (b) The weak formulation: u V1 a(u,v) = for all v V corresponds to the problem —4ui- cu = g in (1,

u constant on 1',

f

fcodr.

(7.4.14)

which is also called an Adler problem. Finally we want to point out the difficulty of classically interpreting a weak solution. In the variational formulation (2a,b) the right-hand sides g and ço of the differential equation and the boundary condition are combined in the functional 1. In the variational formulation the components g and ço are

7

160

VarIational Fbrniulatlon

indistinguishable! u E H'(a) has fist derivatives in L2((J), whose restrictions to V do not have to make sense. That is why Bu cannot be defined in general;

cannot be viewed as an equality in the space H'/2(f) although

Ru =

H'/2(F)

(cf. Corollary 4b). But even if there Is a classical solution, the following paradox arises. Let u to (Ia) define be a classical solution of Lu =0 in 1?, Bu = on F.

One may also view u as a solution E (H1(fl))' by f,(v) := in 0, Bu = 0 on I. These equations may even be interpreted of Lu = classically in the following way: there exist E C°°(0) with 4 in = 0. Then (H1(i7))'. Let be the claseical solution of = converges in H'(O) to the above-mentioned classical solution u. Incorporating the boundary values Ru = in the differential equation corresponds to a modification of the discretised problem as used Lu = in Section 4. The difference equations Dhuh = fh in and the boundary on F,, resulted in the system of equations L,,u, = conditions uh (cf. (4.2.6b)). If one defines by in 0h, uh = 0 on = fh + then satisfies the equations D,,Ti,, = in iZ,, = 0 on Just as the functional f cannot be uniquely separated into g and and cannot be reconstructed from In contrast to the discrete case, the separation off into g and is possible, however, provided stronger conditions than g (H'(.Q))' are imposed on g (for example, g L2(fl)).

4

8 The Method of Finite Elements

In Chapter 7 the variational formulation was introduced only for the purpose of proving the existence of a (weak) solution. It will now turn out that the variational formulation is the foundation of a new method of discretisation.

8.1 The Ritz-Galerkin Method Suppose we have a boundary value problem in its variational formulation:

Find u€V, so that a(u,v)=f(v) foiallv€V,

(8.1.1)

and V = H'($?) (cf. where we are thiniciug, in particular, of V = Section 7.2, Section 7.4). Of course, it is assumed that a(.,.) is a bounded bilinear form defined on V x V, and that / V': Io(u,v)I

CsfItLUvIIVftv

for

'€

E

(8.1.2)

Difference methods arise through discretising the differential operators. Now we wish to leave the differential operator hidden in a(.,.) unchanged. The RitzGalerkin discretisation consists in replacing the infinite-dimensional space V with a finite-dimensional space VN:

VNCV,

dhnVN=N
(8.1.3)

is still a Banach space. Since VN C V, both VN equipped with the norm fl . a(u,v) and f(v) are defined for u,v E VN. Thus we may pose the problem (4):

Find uN E VN, so that a(uN,v) = 1(v)

for all v E VN.

(8.1.4)

The solution of (4), if it exists, is called the Ritz-Galerkin solution (belonging to VN) of the boundary value problem (1). To calculate a solution one needs a basis of VN. Let (b1, a basis, i.e.,

VN= span{bI,...,bN}. For each coefficient vector v = {vl,...,vN}T we define

.

..,

bN} be such

(8.1.5)

(8.1.6)

8 The Method of Finite Elements

162

Remark 8.1.1. P is an isomorphism between 1RN and VN. The inverse is thus well-defined on VN. P* VN Lemma 8.1.2. Assuming (5) the problem (4) is equivalent to

foralli=1,...,N.

Find tLNEVN, sothat

(8.1.7)

PROOF. Putting v = v=

E VN

in (4) gives (7). On the other hand, suppose that is arbitrary. Then (7) and the linearity of and f

give (4):

—1(v) =

—1(b1)) = 0.

We now seek u so that uN = Pu. The following theorem transforms the problem (4) [reap. (7)) into a system of linear equations.

Theorem 8.1.3. Assume (5). The NxN-matrixL f = (fi,.. IN)T are defined by .

(L11) and the N-vector

,

:= a(b,,b1) := f(b1) Then the probLems (4) and

(i,j = 1,.. .,N),

(8.1.8a)

(i = 1,..., N),

(8.1.8b)

Lu=f

(8.1.9)

are equivalent. If u is a solution of (9), then uN : Pu solve8 the problem (4). In the opposite direction, if is a solution of(4), u 1, a solution of(9) (cf. Remark 1).

PROOF. (4) is equivalent to (7). In (7) put

a(uN, b1)

=

uja(bj,

=

=

=

= Pu = Eu,bj; then and thus Lu = f.

I

In engineering applications where the boundary value problems arise from continuum mechanics (cf. the first paragraph of Section 5.3.1), one calls L the

stiffness matrix. The connections between L and a(.,.), on the one hand, and between f and f(.), on the other, are clear from

Remark 8.1.4. If(u, v) := u1v, is the usual scalar product, then a(u,v) = and f(v) = (f,v) with u = Pu,v = Pv. A trivial consequence of Theorem 3 is:

Corollary 8.1.5. The Ritz-Galerkin discretisation (4) has a unique solution for each f E V' exactly when the matnx L in (8a) is nonsingular.

8.1 The Rits-Galerkin Method

163

The Gelfand triple V C U C VI corresponds to the finite-dimensional where the spaces are the same as sets:

situation VN C UN C

VN = UN

VN.

However, the three spaces have different norms: VN := (VN,

UN

fly),

IIu),

(VN, U

V, :=

(VN, II.

and UN has fi flu as a norm, while the dual norm

VN has fi

sup{I(v,u)uJ/IJulJv:O

u E VN}

for v E V'

(in particular for v VN) is defined on V', but is only a norm on = 0 for each v orthogonal to VN. By Lemma 6.5.1 there is an operator associated to a(., .): VN x VN

since

JR

LN:VN

so that

a(u,v) = (LNU,v)u

for all u,v E Vpj.

(8.1.lOa)

for (., •)VJXVN (cf. Remark 6.3.12).

Here we are writing (.,

P is defined by

The map

(Pu,v)

for uE VI,vEJRN.

(8.1.IOb)

C U (orthogonal space ExercIse 8.1.6. (a) Show: The kernel of P is JRN is an isomorphism. relative to fi flu). The map P: Vk JRN exists, we can define V,, (b) Since P

Cp :=

:=

U E

JRN where fi .fl Is the Euclidean norm. Show: The matrix P'P:

VN}, •

has an

inverse with spectral norm

IRP'PY'II (c) Show: QN :=

P(P*P)tP*:U -, U

is the orthogonal projection (relative to fl• flu) onto UN = VN. In addition we have QN E L(V', V).

Lemma 8.1.7.

Let LN E L(VN,Vk) be the operator conespondin9 to IR, and L L(V,V') that fora(.,.):V x V JR. The foUowirig relationships hold between the operators L, and the stiffness x VN

mat,ix L:

164

8

The Method of Finite Elements

L = PLP =

PLP

LN =

P'1LP1:VN

LN

the restriction of QNL and QNLQN to VN.

vii,

PROOF. For all u, v E iPI' we have

(Lu,v) = a(Pu,Pv) = (LNPU,PV)U = (P*LNPU,V) = By Remark 6.3.12 we can write (f, v)u for 1(v). If v = P'v, it follows = (P'f, v). Usmg Remark 4 we then have from (lOb) that 1(v) = (f,

f=

Pf

(8.1.lOc)

for the f in the right side of Eq. (9). In the case of a continuous variational problem the V-ellipticity guarantees its unique The same condition is also suffictent in the discrete case:

Theorem 8.1.8. Auume (3). Suppose the bilinear form is V-elliptic: a(u,u) CEIIuIft, for ail u E V with CE > 0. Then the mat'*r L in (9) is nonsmgt4ar and the Ritz-Goier*in solution

E VN satisfies

(8.1.lla)

IIU?ullv

PROOF. Lisnonsingularsince

and

thus

(Lu, u) = a(Pu, Pu) CEIIPUU?, >0 and so, in particular, Lu 0. By &ercise 6.5.6a a(.,.) is also VN-eIIiptiC with the same constant CE. From Theorem 6.5.8 there holds (ha) then results from 1/CE, i.e., HU"flv

ExercIse 8.1.9. Show

II! fly' for any f

V'.

Exercise 8.1.10. Show: (a) If a(.,.) is symmetric then so is L.

(b) If a(.,.) is symmetric and V-elliptic then L is positive definite. Under the same assumptions the Ritz-Galerkin solution ut1 solves the following variational problem (cf. Theorem 6.5.12): J(uM) J(u) :=a(u,u)—2f(u)

for all u

VN.

Example 8.1.11. (Dirichiet Problem) The boundary value problem is

8.1 The Rltz-Galerkin Method

infl=(0,1)x(O,1),

.—zlu(x,y)=1

165

u=O

The weak formulation is given by (1) with V =

a(tt,v) :=

J (Vu,Vv)dxdy =

f

JvdxdY.

1(v)

The functions bj (x, y) = sin(irx) sln(wy), bs(x, y) = sin(wx) sm(3iry),

b2(x, y) = b4(x, y) = sin(3wx) sin(3,ry),

fulifi the boundary conditions and so belong to V = Hd(i7). They form a basis of V4 := span {b1,. ..,b4}. The matrix elements = can be worked

out to be L11 =

=

= 5w2/2

=

In addition the chosen basis is o(., .)-orthogonal:

=

L is diagonal. Furthermore one may calculate dxdy, getting

=

fi =4/ir2, Hence

u=

12

L'f has the components

Ui

=

u2 =

=

u4 = 8/(81ir4),

and the Ritz-Galerkin solution is then

=-.[sinwzsiniry + +

sin

+sinirzsin3iry)

sin 3iry].

The Ritz-Galerkin solution and the exact solution for x = v = 1/2 are

=

= 0.07219140...,

=

/[(i + 2v)(1 + 2p)((1 + 2p)2 + (1 +

= 0.0736713.

Example 8.1.12. (Natural boundary conditions) Let the boundary value problem be

8 The Method of Finite Elementa

166

Ou/ôn=Oon F.

= (0,1) x (0,1),

= ir2cosirx in

The solution is given by u = cos wx + const. The weak fonnulation is in terms of (1) with V = H'(s?),

a(u,v) :=

+

:=ir2Jv(x,y)cosirxdxdv.

1(v)

The boundary value problem has a unique solution in

W:={v€ V:Jvdxdv=0}. The basis functions

bi(x,y) = z are

—

b2(x,y) = (x —

1/2,

in W. The stiffness matrix L and the vector f are then L—

Ii

1/41

—2

1

9/80]'

1

so that

—L''f—1 — [—20+240/7r2 The solution is t/'(x, y) =

(3

—

60/w2)(x

—

1/2)

(20 — 240/ir2)(z

—

—

The Ritz-Galerkin solution satisfies the boundary condition öu/8n = the differential equation only approximately:

y)/On =

For x =

1/4

12

—

120/ir2

0

and

—0.16.

the approximation is v."(1/4,y) = —7/16 + 45/(4ir2) =

0.70236..., whereas u(1/4, y) = oosir/4 = 0.7071

...

is

the exact value.

In the following we shall consider the case in which o(.,.) is no longer V-elliptic, though it is V-coercive. That a(.,.) is V-coercive guarantees that either problem (1) is solvable or A = 0 is an elgenvalue. Even if one assumes V-coercivity and the solvability of the problem (1) one can not deduce the solvability of the discrete problem (4).

Example 8.1.13. a(u,v) := f(u'v' — lOuv)dx is 1)-coercive and a(u,v) = 1(v) := f01 gvdx (v E J.1o1(0, 1)) has a unique solution. Let VN spanned by bj(x) = x(1 — z) E V = Hd(0, 1) (i.e., N = problem (4) is not solvable since L = 0.

1).

Then the discrete

If one replaces the space V in Lemma 6.5.3 with VN, then there follows from Exercise 6.5.4

8.2 Error Estimates

Theorem 8.1.14.

167

The problem (4) is solvable for all f E V' and has a

unique solution, Ut", which satisfies the estimate S

(8.1.llb) EN

if and only if VN,UVIIV = 1}:u

VN,(IUIIV = 1} = CN >0. (8.1.12)

Since (4) is equivalent to the system of equations (9), one has the

Corollary 8.1.15. The matriz L is nonsingtdar if and only if (12) holds. Exercise 8.1.16. The requirement (12) is equivalent to

flullv

VN,HVUV

EN

1}

forallu€ VN.

(8.1.12')

and

= 1/EN

(LN as in (lOa)).

(8.1.12")

Warning: The condition (12) for VN does not follow from the analogous condition (6.4.5a) for V. See, however, Theorem 8.2.8.

The requirement (12") guarantees the existence of L', but it does not which is the deciding say anything about its condition, cond(L) = factor for the sensitivity of the system of equations Lu = f. For example, if (i = 1g.. .,N) then one one chooses for a(u,v) := u'v'dx the basis = iii (i+j —1). The conditioning obtains the very badly conditioned matrix of L is optimal if one chooses the basis to be a(•, )-orthogonal: a(bj,b,) = as is the case in Example 11, up to a scaling factor.

8.2 Error Estimates For difference methods the solution u and the grid function uh are defined on is, on the other hand, directly different sets. The Ritz-Galerkin solution,

comparable with is. One can measure the error due to discretisation With IIu — u?JUv or with

—

uNItu.

Let is be the solution of (1.1): a(u,v)

f(v) for v E V. Suppose — by

chance or because of a clever choice of V,1 — that u also belongs to VN; then

is also satisfies (1.4). That means: The discretisation error is zero if We shall now show: The "closer" that is is to VN the smaller is the discretisation error. is E

8 The Method of Finite Elements

168

Theorem 8.2.1. (Ceo) Assume (1.2), (1.3), and (1.12) hold. Letu E V be a solution of the problem (1.1), and let uN E VN be the Ritz-Oatev*in solution of (1.4). Then the following estimate holds: III'

— u"IIv

(1+ Cs/CN) iflf

(8.2.1)

— wIly

WEVN

with Cs from (1.2) and from (1.12). Note iflfwEyN IIu—wlIv is the distance of the function u from VN; it will be abbreviated in the following to

d(u,VN) := ml flu—wIly.

(8.2.2)

WEVN

PROOF. If u satisfies a(u,v) = /(v) for all v

V then it does in particular

for all V E VN. Since we also have a(uN, v) = f(v) for v E Vpj, it follow8 that

forall VEVN. For arbitrary v,w E VN with lIvIlv = —

= a((t/'

w, v)

1

(8.2.3)

we can therefore conclude

— tij + [u — w], v)

= a(u — w, v)

and

w,v)l 5 From (1.12') we then obtain —

— wIIvIfvIlv = CsIlu

—wily

—

why.

E VN,hIvhIv = 1} (Cs/CN)flu—wllv.

The triangle inequality then gives flu

flu

—

wily +

—

uNhiv (1+ Cs/EN)hIu —

SincewEVNisarbitrarywededucetbeaeeertion(I).

U

In Theorem I the unique solvability of the problem (1.1) was not assumed,

but only the existence of at least one solution. If the discretisation error is supposed to converge to zero one makes use of a sequence of subspaces V that converge to V In the following sense:

Theorem 8.2.2.

:=VN4 c V(iEt4) be a sequence of subspaces with lim

= 0 for all u

V.

Asumethat(1.12)holdswitheN4

(8.2.4a)

in addition assume (1.2) (continuity of a(.,.)). Then there exists a unique solutionu of the problem (1.1) and the Ritz-Galer*in solution ud := to

iIu—u'hIv--.O

for i—.oo.

169

8.2 Error Estimates

Sufficient to ensure (4a) is

UvidensemnV.

(8.24b)

PROOF. (a) Assume first the existence of a solution u. (aa) The estimate (1) implies the convergence: d(u,

ilu — u'IIv (1 +

—.0.

(ab) We now wish to show that (4a) follows from (4b). The inclusion ¾-i C Now d(u, will be a null sequence if for each implies d(u, V1) d(u,

e > 0 there exists an i such that d(u, is, for each u E V and e > 0, a w E U,

e. From the assumption (4b) there with e. Therefore we — wily have w€ for an i N. That d(u,V3) e then proves (4a). — why The convergence u hnplies the uniqueness of the solution u. (b) The next thing to show is that the image W := (Lv: v V} C V' of the operator L:V —. V1 associated to a(-, -) is closed. Foreachf E Wthereisau E V with Lu = f, so that part (a) of this proof suffices to show the convergence --. u. Since Itu'IIv it follows that hiutiv = urn 11f110'/€. Let in V' and f,, = Lug. The Cauchy E W be a sequence with —. convergence Ii!" — shows —0 so that ti/v the limit E V exists. The continuity of L L(V, V') shows that = limf,. = lirnLu,, = Lu, and thus that f E W. Hence W is closed. (c) In order to demonstrate the existence of a solution to the problem (1.1) we have to show the surjectivity of L: V —. V'. If L were not surjective (Le., W V'), there would be an I E W-1- with hIlly' = 1. Let Jv: V —. V' be the Rieaz isornorphism (ci. Corollary 6.3.7). Set v := E V. It follows that

f

f=

f(v)=

—(f,f)v' =

for all u E V. Let E satisfy v) = 0, i.e.,

0

1,

be the Ritz-Galerkin solutions. They must also a(u',v) — 1(v) =

We split up v into may

+

guarantee that

I

=

1.

where v' Vi. By (4a), —.0. This shows

with

v in

place

of u, one

/(vi)+o(ui,w)_f(wi) w1) — f(wi)

and I

=

v) — f(v)l [Cshtu'ljv + hhfik"j llw'(Iv.

Since Ilu'IIv IIfhIv'R is uniformly bounded and Hw'hIv —. (I this amounts to a contradiction. Therefore £ must be surjective, so that for each f E V'

there exists a solution u to the equation Lu =

I, i.e., to the problem (1.1). •

170

8 The Method of Finite Elements

>o from

Corollary 8.2.3. The requirement (1.12) with CN,. CR if a(.,.) is V-elliptic: a(u,u) is satisfied with

Theorem 2

is dense in V.

Exercise 8.2.4. Show that (4a) implies that

Let QN be the orthogonal projection onto VN (cf. Exercise 8.1.6c). The V V', QN:V' —' factors L:V V

(8.2.5a)

C V.

Exercise 8.2.5. Show there is also the representation SN =

PL'P'L.

(8.2.5&')

Lemma 8.2.6. SN is the projection onto VN and is c*zIled the Ritz projection. It sends the solution u of the problem (1.1) to the Ritz-Galerkin solution uN VN: uN = SNU. Assuming (1.2) and (1.12) we have (8.2.5b)

IISNIIv+-v CS/CN.

A definition of 5N equivalent to that in (5a) is SNU E VN and O(SNU,V) = a(u,v)

for all v

VN, u

(8.2.5c)

V.

PROOF. (a) Since

a(u,v) =

=

= (Lu,QNV)U = (QNLU,v)u

for all v VN, it follows that SNtI in (5c) is the Ritz-Galerkin solution for the QNLu. Conversely one may right-hand side f := QNLU E V', i.e., SNU = argue similarly, and so show the equivalence of the definitions (5a) and (5c).

projection. Inequality (5b) follows from IISNuIIv proof of (1.11)) and IILuII'v CsfIuIfv = from (1.2) (cf. Exercise 8.1.9).

(cf. the with C8

Remark 8.2.7. Let a(.,.) be V-elliptic and symmetric. IIIvIlIv := a(v,v)'/2 is a norm equivalent to f$. liv. The Ritz projection SN is, with respect to lily, an orthogonal projection onto Vp,. Thus, in particular, we have IlISp,iiiv..-v

1.

(8.2.5d)

PROOF. The scalar product associated to is a(., .), so that it is to be shown that a(SNv, w) = a(v, SNW). (5c) Implies a(SNv, Sqw) = a(v, SNW), since SNW E VN. The symmetry of a(.,.) and exchanging v and w give

8.3 Finite Elements

171

a(SNv,w) = a(w, SNV) = a(SNW, Spjv) = a(SNV, SNW), so that a(SNv, w) = a(v, SNW). (5d) results from Remark 6.3.8.

Remark 7 shows once again that the Ritz-Oalerkin solution = SNU is the best approximation to u in VN in the sense of the norm fly. This

is equivalent to the variational formulation J(uN) J(v) for all

v

VN (cf. Exercise 8,1.lOb).

The condition (1.12), with e,, e > 0, is difficult to prove, except for V-elliptic bilmear forms. However, the following theorem shows that this condition does hold for subepaces approximating well enough.

Theorem 8.2.8. Let the bilinear for,n a(.,.) be V-coercive, where V C U C V' is a continuous, dense, and compact embedding. Let Problem (1.1) be solvable for all f V'. Assume that (4a) holds for the subspaces C V. For large enough i the stability condition (1.12) is then satisfied with > 0. The proof of this will be postponed to a supplement to Lemma 11.2.7.

8.3 Finite Elements 8.3.1 Introduction: Linear Elements for 11 = (o,b) As soon as the dimension N = dim V?, becomes larger the eseential disadvantage of the general Ritz-Calerkin method becomes apparent. The matrix L is in general full, i.e., L11 0 for all i,j = 1, . ., N. Therefore one needs N2 integrations to obtain the values of = o(b,, = fe,. .., whether exactly or approximately. The final solution of the system of equations Lu = f requires 0(N3) operations. As soon as N is no longer small the general Ritz-Galerkin method therefore turns out to be unusable. A glance at the difference method shows that the matrices Lh which occur there are sparse. Thus it is natural to wonder if it is poesible to choose the basis {b1, . . ,bp,r} so that the stiffnees matrix = is also sparse. The best situation would be that the were orthogonal with respect to a(.,.): a(b,,bi) for i j. However, such a basis can be found only for special model problems such as the one in Example 8.1.11. Instead we shall base our further considerations on .

.

Remark 8.3.1. Let the bilinear form a(.,.) be given by (7.2.6). Let B be the interior of the support ri B, =

of the basis function i.e., := A sufficient condition that ensures = a(bj, b1) = 0 is

PROOF. The integration

•

= f0... can be restricted to B1 fl B,. In order to be able to apply Remark 1 the basis functions should have

as small supports as poesible. In constructing them in general one goes about

172

8 The Method of Finite Elements

it from the desired goal: one defines partitions of Q into small pieces, the socalled finite elements, from which the supports of b4 are pieced together. As an introduction let us investigate the one-dimensional boundary value problem

u(a)=u(b)=O.

—u"(x)=g(x) fora<x
(8.3.1)

Assume we have a partition of the interval [a, bI given by a = x0 <x1 <...< XN÷1 = b. Denote the interval pieces by 14= (x4_i,x4) (1 i N + 1). For let us choose the piecewise linear functions: the subspace VN c

C°([a,bI):the restriction of u to 13(1

VN ={u

i < N+ 1)

islinear;n(a)=u(b)=O} See

(832)

the figure below.

(a)

a'

a=zo

a'i

Z2

a'3

a'4

FIgure 8.3.1. (a) a piecewise linear function,, (b) a basis function

The continuity "u C°([a, bl)" is equivalent to continuity at the nodes x4 (1 i N) : u(x4 + 0) = u(x4 — 0).

Remark 8.3.2. u E VN is uniquely determined by its node values u(x1) —x)+U(x4+t)(x—s4)j/(x4+1 —x4 for zE

u(x) =

(8.3.3a)

u(ZN+1) = 0. From Theorem 6.2.42 we condude that VN C (a, b). The (weak) derivative u' L2(a, b) is piecewise constant:

where u(xO)

u'(x) = (u(x4+1)

—

u(xi)j/[z4÷i

—

x4 for a'

14+1.

(8.3.3b)

The basis functions can be defined (see Figure ib) by b4(x) =

I

(a'

— x4_1)

—

— 1.0

—

a'4)

for

for


otherwise.

1 I N (8.3.4)

8.3 Finite Elements

Remark 2 then has 88

173

consequence

a

The supports

Remark 8.3.3. One has the representation u =

= 1x1_i,xi+i1. The B1 from Remark 1 is now

of the functions b1 are

(xl_l,xl+1).

B1

f(v)

The weak forirnilation of the boundary value problem (1) is a(u, v)

with

j

a(u,v)

b

u'v'dx,

fgvdx

1(v)

for u,v E

H01(a,b).

By Remark 1 we have, for the matrix elements, L1, = 0 as soon as 2, since then B1 fl B1 = For Ii —21 1 one obtains

= L11

=

a(b1_i,b1)

a(b1, b1)

1

= / (x1

=J = 1/(xg

1

Xs_1 Zj —

dx +

—

—X4_1)

dx = —1/(xj —xs_1),

J

—

dx (8.3.5a)

+ t/(x1+i —x1),

= —1/(x1÷i —x1). The right-hand side f =

(ft,.. ,IN)T .

is given by (5b):

fZ4+j

gbgdx i

=

1

x1 —x1_1

j

g(x)(x—xg_i)dx+

1

xi+1 —x, I (8.3.5b)

Remark 8.3.4. The system of equations Lu = f, given in (5a,b), is tridiagonal, thus, in particular, sparse. Fbr an equidistant partitioning x1 a + ih with h := (b — a)/(N + 1) the coefficients are = —1/h, L1,1 = 2/h, hi = hf0' +th)+g(xg —th)1(1 —t)dt. If g€ C°(a,b), then by the intermediate value theorem f1 = hg(x1+&1h), with <1. Therefore the system of equations Lu = f is, up to a scaling, identical with the difference equation by := = g(x1 + &1h) Lhuh = fh in Section 4.1, if one defines instead of by fh(xl) = g(xj): then L = hLh,f = hi,,.

This example shows that it is possible to find basis functions with small supports so that L is relatively easy to calculate. In addition, the similarity is apparent between the resulting discretisation method and difference methods. Finally consider the numerical evaluation of the integral (5b),

Exercise 8.3.5. Show that the

quadrature formula for

with b1(x) as weight function and one support point is:

gb1 dx

174

8 The Method of Finite Elements (8.3.5c)

What is the formula for a partition of equal intervals? If one replaces the Dirichiet boundary conditions in (1) by the Neumann

condition then V = H'(a, b). The subapace VN C H'(a, b) results from (2) after removal of the condition u(a) = u(b) 0. In order that dim VN = N the numbering has to be changed: The partition of (a, b) is given by a =

<

X2<,..4(VNb. Exercise 8.3.6. Let Ou(a)/&z = ga,Du(b)/On = gb, with a(.,.) as before. Suppose the partition of (a,b) Is equidistant: x4 = a+(b — a)(i — 1)/(N — 1). What is the form of the equation Lu f? Show that Lu = f has at least one solution

b

8.3.2 LInear Elements for £7 C It2 We shall assume:

Qc

is a polygon.

(8.3.6)

As shown in Fig. 2 we divide Si into

t

number of

N := number of Inner

sodas (number of all sodas)

Definition 8.3.7. r := {Tj,.

. .

is called an admimible triangulation of

£1 if the following conditions are fulfilled:

(1 i

t) are open triangles ("finit, elements")

(8.3.7a)

(8.3.7b)

U for

= the

(8.3.7c)

iseither

8.3 FinIte Elements

empty, or a common side of the elements a common edge of the elements

1)

ii)

in)

175

and 1',, or and T3.

(8.3.7d)

Remark 8.3.8. (a) The conditions (7a) and (7c) imply the polygonal shape of S?, i.e., the assumption (6). (b) Figure 2 shows only admissible triangulations. An example of an inadmissible triangulation would be, e.g., a square partitioned as follows:

Let r be an admissible triangulation. The point x is called a node (of T) 1-. One distinguishes interior and boundary nodes, according to whether x £2 or x OS?. Let N be the total number of interior nodes. We define VN as the subapace of the piecewise linear functions:

if x is a vertex of one of the T1 E

with a linear function, i.e., u(x,y) = an +

on T1}. (8.3.8)

+

Remark 2 can be applied to the two-dimensional case under consideration: (5?). (b) Each function is uniquely Remark 8.3.9. (a) it is true that VN C determined by its node values u(Xj) at the interior nodes xi, 1 i $ N.

PROOF. (a) Example 6.2.5 shows that VN C H'(fl). Since u = 0 on 8.0, Theorem 6.2.42 proves that VN c HJ(S?). (b) Let x, x', x" be the three vertices of E r. The linear function u(x, y) is uniquely determined on T1 by the values u(x), u(x'), u(x").

a

The converse to Remark 9b is as follows:

Remark 8.3.10. Let x' (1
(8.3.9a)

If T E r is a triangle with the vertices x' = x' = (x', y'), x" = (s", then — —

On all T E

r

I

— '\i'"

(Xj —x')(y"

'\ —

7/)

that do not have an

/ —

1'tI

(xi, Vi) I

I

(vi _7/)(xII

as a vertex,

as in (9a)) and

—

x')

=

on T.

(8.3.Qb)

0.

PROOF. Obviously, using (9b) we get a linear function defined on all T E r, which satisfies (9a.). In addition, (9a) forces continuity at all nodes. If the

176

8 The Method of Finite Elements

= (x,,y,) andxk =(xk,yk) are directly connected by the side of a triangle, then (9b) provides the representation + s(xk —xJ)) = sbj(xk) + with a (0,11 for both triangles that have this side in common. (1 — C°(Q). Applying Thus is also continuous on the common edges, so that this consideration to two boundary nodes x',x" with = = 0 we

I

Remark 8.3.11. (a) The dimension of the subepace determined by (8) is the number of interior nodes of T. (b) The support of the basis function is

=

T E r has x' as a wrner}.

(c) Let B, be the interior of There holds fl B, = the nodes and are directly connected by a side.

0

if and only if

Corollary 8.3.12. a(u, v) = f0(Vu, Vv) dx is the bilinear form associated to the Poisson equation. The interala = a(bj, b4) = >k (yb,7 dx are to be taken over the following Tk:

ifi=j; (i)aZZTa withx' ifi1&j. (ii) all Tk withx4 andx1 (iii) We have Lgj=O ifx' andx1 are not directly connected by the side ofa triangle.

FIgure 8.3.3. Bask

for the node (z07y0)

We get an especially regular triangulation when we first divide into squares with sides of length h, and then divide these into two triangles (0). The first and second triangulations in Figure 2 are of this sort. We call them "square grid triangulations". The corresponding basis function is depicted in Figure 3. One therefore expects that the matrix L corresponds to a 7-point formula. For the Laplaes operator, however, one finds the well-known 5-point formula (4.2.11) from Section 4.2:

8.3 FinIte Elements

177

Exercise 8.3.13. Let r be a square grid triangulation. Further let a(u, v) = Vv) dx. The basis functions are described by Figure 3. Show that one has for the entries of the stiffness matrix L

=

= —1

4,

if x' x3 = (0, ±h)

or

(±h,0),

from (4.2.8).

L agrees with

h2Lh holds in the case of the Poisson equation

Although L

=

the finite-element discretisation and the difference method still do not coincide,

since h2fh has h2g(x') as components and so differs from f

dx.

V

'7

(0,1) x3

(1,0)

(0,0)

FIgure 8.3.4. R.eferenoe triangle T

The integration ft... dx over the triangle T1 E r seems at first difficult. However, for each i one can express fT. ... dx as an integral over the reference triangle T in Figure 4. The details are in

ExercIse 8.3.14. Let x' =(x',y') (i = 1,2,3) be the vertices oITE r, and let T be the unit triangle in Figure 4. Show: (a) (b)

,j)

+

x1

(x2 —

— x') + x1)(? _yL)

—

—

x1) maps T onto T. xl) for all

(y2 —y1)(x3 —

IR.

(c) The substitution rule gives

Jv(x,Y)dxdY= (8.3.10)

In general one evaluates the integral dq over the unit triangle numerically. Examples of integration formulae can be found in Schwarz (1, §2.4.3) and Ciarlet [1, §4.1). The necessary ordering of the quadrature formulae is discussed in the same place by Ciarlet (14) and by Wztsth [1).

178

8 The Method of Finite Elements

In contrast to the difference methods finite-element discretisation offers one the possibility of changing the size of the triangles locally. The third triangulation in Figure 2 contains triangles that become smaller as they are nearer to the intruding corner. This flexibility of the finite-element method is an essential advantage. On the other hand, one does obtain systems of equations Lu = f with more complex structures, since (a) u can no longer be stored in a two-dimensional array, (b) L cannot be characterised by a star as in (4.2.12).

Remark 8.3.15. If one replaces the Dirichiet condition u =

0 on 817 by natural boundary conditions, then the following changes take place: (a) N = dim VN s the number of all nodes (inner and boundary nodes). (b) VN c H'(Q) is given by (8) without the restriction "u = 0 on 811". (c) In Remarks 9—li it should read "nodes" instead of "inner nodes". calculated in Exercise 13 are valid only for inner (d) The matrix elements nodes For a Neumann boundary value problem in the square 17= (0,1) x (0,1), L coincides with h2DhLh from Exercise 4.7.8b. (e) In calculating = one has to take account of possible boundary integrals over 811 fl if 811 fl 0.

8.3.3 Bilinear Elements for 11 c

1R2

The difference procedures In a square grid are near to partitioning 1? into squares of side h (cf. Figure 5a). If, more generally, one replaces the squares by parallelograms one obtains partitions like thoee in Figure 5a,b. An admissible

partition by parallelograms is described by the conditions (7a—d), if in (7a) the expression "triangle" is replaced by "parallelogram".

(a)

(b)

FIgure 8.3.5. PartitIon of 0 Into parallelograms If one were to define the subspace VN by the condition that u must be a linear function in each parallelogram, then there would be only three of the four vertex values that could be arbitrarily assigned. In the case of the partition in Figure 5b one can see that the only piecewise linear function u with u = 0 on 81118 the null function. Thus in each parallelogram u must

8.3 Finite Elements

179

be a function that involves four free parameters. We next consider the case of

a rectangle parallel to the axes P = (z1,x2) x (yj,y4 arid define a bilinear function on P by (8.3.lla) + a2z)(aS + a4y). u(x,y) = Then u is linear in each direction parallel to the axes — thus, in particular, along the sides of the rectangle. For an arbitrary parallelogram P such as in Figure 6a, the restriction of the function (1 la) to the side of a parallelogram is in general a quadratic function. Therefore one generalises the definition as follows. Let i-+

x1 +

— x1) +

—

(8.3.llb)

x1)

be the mapping taking the unit square (0,1) x (0,1) onto the parallelogram P (cf. Figure 6a,b). A bilinear function is defined on P by u(z, y) :=

y)),

(a +

v(e,

+ 5i).

(8.3.1 ic)

it is not

to calculate v(4r'(x1 y)) explicitly, since all the integrations can be carried out over (0, 1) x (0, 1) as the reference parallelogram (cf. Exercise 14).

y

'1

£1

FIgure

(1,1)

(0,0)

(i,0)

x

8.3.6. (a) Parallelogram

partition into parallelograms one by

VN := {u E C°(Th): on all P a bilinear ftinction} VN :=

0.

(b) Unit square as reference parallelogram

If ir = is an defines VN C H'(a) [rasp. VN C

reap.

(0,1)

u coincides with

C°(Th: ii =0 on 0S?; on all P E with a bilinear function}

Here N = dixnVN is the number of nodes [rasp. inner nodasj.

(8.3.12a) u coincides

(8.3,12b)

8 The Method of FInite Elements

180

AbilinearfünctiononPEwislinearalongeac*lsldeofP. Continuityin the node points thus already implies continuity in $1.

Remark 8.3.16. The Remarks Sa, 9, 10, and 11 hold with appropriate thanges.

Exercise 8.3.17.

Let the bilinear form aesociated to be a(u,v) = f0(Vu, Vv) dx. Aesume (1 = (0,1) x (0,1) is divided into squares of side h as in Figure 5a. What are the basis functions characterised by (9a)? Show that the matrix L coincides with the difference star

—1

—1

—1

-1

8

-1

—1

—1

—1

FIgure 8.3.7. A coinbinatlon of triangles and parallelograms

Remark 8.3.18. Thangle and parallelogram divisions can also be combined. A polygonal domain a can be divided up into both triangles and parallelograins (ci. Figure 7). In this case VN is defined as {n E

u linear on the triangles, bilinear on the parallelograma}.

8.3.4 Quadratic Elesnents for fl c Let 'r be an adinindble triangulation of a polygonal domain I?. We wish to

increase the dimension of the finite-element sulispace by allowing, instead of linear functions, quadratics

u(x, y) = afl +

+

+ ajtx2 +

+

on

Er

(8.3.13)

so that

VN={UEC°(fl):u=0000a; with a quadratic function).

u coincides (8.3.14)

8.3 FInite Elements

(a)

Figure 8.3.8.

181

(b)

Nodes for

(a) a quadratic ansatz on a triangle, (b) a quadratic a.nsatz of the serendipity clam on a parallelogram

(a) Let x1,x2,x3 be the vertices of a triangle T E r,

Lemma 8.3.19.

while x4, x5, x6 are the midpoints of the sides (cf. Figure 8a). Each function

quadratic ont is detennined by the values {u(x3):j=1,...,6}. (b) The restriction of the function (13) to a side of T E r gives a onedimensional qt&adratic function, which is uniquely deterinine4 by three of the nodes lying on this side (e.g., u(x'),u(x4),u(x2) in Figure 8a). (c) flu is quadratic on each E r and continuous at all nodes (i.e., vertices of triangles and mid-points of sides), then u is continuous on 11. PROOF. (a) The quadratic function can be obtained as the uniquely defined interpolating polynomial of the form (13). Part (b) is also elementary.

(c) Let T and T be neighbouring triangles in Figure 8a. Since UIT and coincide on x1, x4, x2, by part (b) they represent the same quadratic function

onthecocninonsideofTandT. We call all [inner] vertices of triangles and mid-poiiits of sides tinner] nodes. By Lemma 18a we can find (or each inner node a basis functIon By Lemma 18c

belongs to VN. This proves

Remark 8.3.20. The number of inner nodes is the dimension of Vjy Lfl 14. Each u E VN admits the expression u = function belonging to the node is tharacterised by (9a).

where the basis

In the sequel we wish to assume that, as in Figure 7, both triangles and parallelograms are used in the partition. On the triangles the function u E VN are quadratic. The functions on the parallelograms P must satisfy the following conditions:

(a) u(z,y)Ip must be uniquely determined by the values at the 4 vertices and 4 mid-points of the sides (cf. Figure 8b). (/3) The restriction to a side of P gIves a (one-dimensional) quadratic function (of the arc length).

By condition (a) the ansatz must contain exactly 8 coefficients. The quadratic (13) has only 6 while the biquadratic

8 The Method of Finite Elements

182

has one parameter too many. If one omits the term ansatz one obtains for the unit square the function

=

+

+ as?? +

+

in the biquadratic

+ aaq2 + are2,i +

(8.3.15)

the so-called quadratic ansatz of the serendipity class. The restric= 0, 1] give a quadratic function in tions to the sides = 0, 1 [resp. in (1 Ib) (cf. Figure 6a,b) yields the function defined on the parallelogram P, which still satisfies the conditions (a) and (f3). If one were to use the full biquadratic ansatz one would one additional node which one could choose as the barycentre of the parallelogram. Cubic ansatzes can be carried out in the same way in triangles and parallelograms n Iresp.

The mapping

tt(x,y) =

(cf. Schwarz f 1]).

8.3.5 Elements for Si C It3 In the three-dimensional case assume isapolyhedron.

(8.3.16)

The triangulation from Section 8.3.2 corresponds now to a division of .0 into tetrahedra. it is called admissible, if (7a—d) hold in the appropriate sense: (7a)

becomes "Z(1 5 iS t) areopen tetrahedra"; in(7d)itnowshouldread "For

j,

Tg ii T, is either empty, or a common vertex, side, or face of and T,". Each linear function al + + asy + a4z is uniquely determined by its values at the 4 vertices of the tetrahedron. As a basis for the space I

VN={tIEC°(Th): u=O on OS?; u islinear on each tetrahedron Tj (1

I 5 t)}

one chooses with the property (9a): = 1, = 0 (j 1). The support of consists in all tetrahedra that share x' as a vertex. The dimension N = dim VN is again the number of inner nodes (i.e., vertices of tetrahedra).

As in the two-dimensional case the linear ansatz may be replaced by a quadratic one. Instead of a tetrahedron one can use a parallelipiped or a triangular prism with corresponding ansatses for the functions (cf. Schwarz 111).

8.3.6 Handling of Side Conditions The space V which lies at the foundation of the whole matter may be a subspace of a simply discretisable space W V. A given function w W belongs to V if certain side conditions are satisfied. Before we describe this situation in full two examples will be provided as

8.3 Finite Elements

183

Example 8.3.21. If one wishes to make the Neumann boundary-value probon 1' uniquely solvable by the addition of = gin (1 and 8u/8n lem u(x) dx =0, one can choose the space V to be the side condition

V = {t&

H'(a): J u(x) dx = 0).

(8.3.17a)

For a bounded domain (1 the form a(u, v) := f0(Vtt, Vii) dx is V-elliptic. The weak formulation (1.1) with 1(v) := if9 and satisfy the integrability condition equation = g, Oufôn = /(1) = 0 (cf. (3.4.2)). But even when f(1) 0 there exists a weak solution of dx. the corrected equation —4u = := g(x) — [fag dx +

Example 8.3.22. The Adler problem of Exercise 7.4.15 uses

V = {u

H1(a): u constant on r}.

(8.3.17b)

In both cases W = H1(.Q) is a proper superset of V. Let r be a trianboundarynodes.Let WhcWbe the space of linear triangular elements (cf. Remark 15): Nh := dim Wn =

+ 4d.

(8.3.18a)

The basis functions (bi: 1 i will be described as usual by C = öjj (cf. (9a)). Astheflnita-elementsubspaceofV we define Vh := Whfl V, which isof smaller dimension: Vh:_—W4flV;

Mh:—Nh—dirnVh.

In the Examples 21 and 22 we have Mh = 1, reap. MA = difficulty in the numerical solution of the discrete problem (19),

Find i1

E V,

the

with a(,i',v) = 1(v) for all v

Vh,

for example, in Example 21 no

(8.3.18b)

- 1. The (8.3.19)

belongs to

V,andthuaalsononetoVh,amcefobidx>o. In principle, it is poesible in the case of (17a) to construct as new basis functions linear combinations := V, which have again localised (but a bit larger) supports. In the case of (17b) it is still relatively emiple to find a practical basis for Va: The basis functions b4 which belong to Inner nodes are also in since &j H01(Q) C V. As further elements one

uses bo(x) := E'bj(x),

where E' denotes the sum over all boundary nodes. In spite of this, even in this case, it would simply be easier if one could work with the standard basis of WA. In order to treat the problem (19) with the aid of E we reintroduce the notation = Pw (w a coefficient vector, of. (1.6)).

8 The Method of Finite Elements

184

Eachv€Vh,C Vh} = P'Vh. Thus problem (19) is equivalent to: Find u E V with a(Pu, Pv) = f(Pv) for all v E V.

-

From (1 8a,b) we have dim V = dim Vh = Nh Mh = dim V are described by Mh linear conditions spaces

(8.3.19')

- Mh.

The

N,,

=

0

(1 I Mh);

(8.3.20)

j=l

V = Ker C = {w E where C =

(CU)

= O},

is an Mh x Nh matrix. In the case of Example 21 we have

Mh=1,

ci,=Jb,(x)dx

For Example 22 let x0,. ..,xMh, with Mh = Then C can be defined as follows:

= -1,

1, be the boundary nodes.

CU =0 otherwise.

The variation of v over V in (19') can be IRNh

(8.3.21)

by the variation of w over

if one couples the conditions (20) by using Lagrange multipliers

(1

which can be put together to form a vector A =

(A1,... ,AM,,). The resulting formulation of the probleni is:

with Cu =0 and find u E and A a(Pu, Pw) + (A, Cw) = f(Pw) for all w where (A,

=E

(8.3.22a) (8.3.22b)

is the acalar product in

Theorem 8.3.23. The problems (19) and (22a,b)

are

equivalent in the fol-

lowing sense. IfuA is a solution pairfor(22a,b) then u is a solution of(19') and Pu is a solution of (19). Conversely, if uh = Pu isa solution of (19) u and A solve

then there exists precss ely one A E

Before we prove this theorem, we give a matrix formulation which is equivalent to (22a,b)

Remark 8.3.24. Let L and f be defined as in (1.8a,b). Further, that B := CT with C from (21). Then (22a,b) is equivalent to the system of equations

FL

Biful

IBT oJ [Aj

Ff1

[oJ

(8.3.22')

8.4 Error Estimates for Finite Element Methods

PROOF. Lu + BA = f is equivalent to (BA,w) = (f,w) = f(Pw) for all w E JR.

185

Pw) + (A, Cw) = (Lu, w) + = 0 is the same as

Note BTU

Cu=O. PROOF (of Theorem p3). (a) Assume a solution of (22a,b) is given in terns of u, A. Then Cu = 0 implies

:= Pu E Vh. Equation (22b) holds in particular for all w V, U E V and so that (19') follows since Cw 0. (b) (19') implies I — Lu E V1. Since V1 = (KerC)1 = (KerBT)'- = Im(B), there exists a A E IRMh with f — Lu = BA, so that (fl'), and thus (22a,b) are we have KerB = {0}, so satisfied. For the Nh x Mh matrix 13 with rank U that the solution is unique. Note

that in the case of Example 21 the system of equations (22') is

essentially identical to (4.7.lOa,b).

8.4 Error Estimates for Finite Element Methods 8.4.1

for Linear Elements

In this section we shall restrict ourselves to the consideration of the linear elements from Section 8.3.2 and therefore assume:

(8.4.la) (8.4.lb) (8.4.lc)

r is an admissible triangulation of 0 C VN is defined by (3.8), if V = VN is as in Remark 8.3.15b, if V = H'(i7).

By Theorem 8.2.1 one has to determine d(u, VN) = WE VN}. To do this one begins by looking at the reference triangle from Figure 8.3.4.

Lemma 8.4.1. Let T =

0, + , <1 }. For all u

H2(T) there

hoLds

lu(0,0)12 + Iu(0, i)J2 + Iu(1,0)12 +

J.

(8.4.2)

PROOF. (a) First it has to be shown that the bilinear form a(u, v) := u(0, O)v(O, 0) + u(0, 1)v(0, 1) + u(1,0)v(1, 0) +

D°v)L3(T)

continuous on H2(T) x H2(T) and is H2(T)-coercive. The continuity follows from the continuous embedding H2(T) C C°(T), which implies tu(x)l for all x T (cf. Theorem 6.2.30). Thus we have (1 + 3O)IuI2IvI,. is

186

8 The Method of Finite EUements

Since T E C°", one may apply Theorem 6.4.8b: 112(T) is oompactly embedded In H'(T). By Lemma 6.4.13, there exists a constant C,12 such that

=

Since

H2(T).

the estimate

—

a(u,u)

for all

+

+ Cii2IuIo)

—

—

shows that a(.,.) is H2(T)-coercive. (b) Since the embedding 112(T) C

Theorem 6.5.15. The operator A E

L2(T) is also compact one may apply 112(T)') which is associated L(112(T)', 112(T)) or has )'. = 0 as an

with a(.,.) either has an inverse A'

112(T). In the latter case e must, in =0 particular, satisfy the equation a(e,e) =0. From this follows that Furthermore, for all lal =2, so that e must be linear: e(x, y) = a + /3x + from a(e,e) = 0 one may conclude that e(O,O) = e(1,0) = e(0,1) = 0, so that e =0 in contradiction to what was just assumed. From Lemma 6.5.3 and with a(u,u) Exercfse 6.5.6c one may show the + 3O)J > 0 and C from part (a). Thus we CE := have assertion (2) with C := 1/CE.

eigenvalue with an eigenfunction 0

e

:= hT := {(x,y): xv 0, x + y h}. Fbr each Lemmi 8.4.2. Let uEH2(Th)therehoidswithlf3l2 [u2(0, 0) + u2(O, h) + u2(h, O)J

(8.4.3)

+ leI=2

with C independent of

6 JP(T). The derivatives with respect = each = (x,y)/h are related by :=

PROOF. Let

to (x,y) and to

lID

and h.

= JJ =

=

jf

+ v2(1,0) + v2(O, 1) +

Since v(0,0) = u(0,0), v(1,0) = u(h,0), v(0, 1) = t40,h), and one may conclude (3).

=

187

8.4 Error Estimates for Finite Element Methods

Lemma 8.4.3. Let T be an arbitvury truzngle with Interior angles ao >0. Side lengths 2, there holds and For each u

+

C(cio)

>

(8.4.4)

I(D°u11t3(f)

otT (8.4.5)

where C(ao) depends only on a.o and not on u,

PROOF. Let h

or h.

be one of the lengths of the sides of T. In a way

similar to that in Exercise 8.3.14a, let Th T be the linear transformation := belongs to H2(Th). Under that maps Th onto T. Then the conditon (4) we know that the determinant E is bounded both above and below. From Ip,'=I$l

(3), and

C2(oo)

111

IaI=2

there then follows (5).

U

From Lemma 3 follows the important result:

Theorem 8.4.4. Assume that conditions (Ia-c) hold for r, VN, and V. Let czo be thea whiiehüthe maximum length of the sides of all T1€r. Then IIU—vflHk(a) (f (ao)h2'11u11u2(a) fork = 0,1 and all u H2(a)nV. (8.4.6)

PROOF. For a u

u E H2(t?)

one chooses v :=

VN, i.e., v

implies the estimate T4Er

= JaI=2

fl

if V =

VN with v(x') = u(x1) at the

188

8 The Method of Finite Elements

Let

S

k now proves the inequality (6).

Summation over

be the longest side of T e r, then h := rnax{hp:T E r}

is

the longest side-length which occurs in -r. The parameter h is the essential one and will be used as an index in the sequel: r = rh. Denote by Pr the radius of the inscribed circle in T E r. The ratio hp/p1' tends to infinity exactly when the smallest interior angle tends to zero. If one has the bound max{hT/p.r: T E T} const for a family of triangulations, then one calls the family quasi-uniform. In this condition the constant should be the same for all -r in the family {-rh). A family is called uniform if it fulfils the stronger requirement h/ T E r} const. A family {rh} which is both unform and for which h —.0 is called regular by Ciarlet 11, §3.1J. In the construction of triangulations it should be noted that with increasing refinement —.0) the interior angle may not also be reduced. Strategies for the systematic construction of quasi-uniform sequences V4 can be found, for example, in Hackbusch (1, §3.8.2J. The case k = 1 of Theorem 4 may now be written as follows:

Theorem 8.4.5. Assume conditions (la—c) hold for a sequence of quasiunsform triangulations ri,. Then there ezuts a constant C, such that for all and Vh = there holds the following estimate: h=

inf lu — vii ChIui2

for all u E H2(1l) fl V.

yE V,,

(8.4.6')

The combination of this theorem with Theorem 8.2.1 gives

Theorem 8.4.6. Assume conditions (la—c) hold for a sequence of quasitriangulation, ri,. Let the bilinear form fulfil conditions (1.2) and (1.12). Suppose the constants CN := > 0 from (1.12) are bounded below > 0 (cf. Corollary 8.2.3). Let Problem (1.1) have the solution by q,, fl V. Let uE E V,, be the finite-element solution. Then there ezzsts a constant C, which does not depend on u, h = or v, such that — uhil (8.4.7) ChIuI2. The combination of the Theorems 5 and 8.2.2 can be written:

Theorem 8.4.7. Assume conditions (la-c) hold for a sequence of quasi-

-'

untform triangulation. with 0. Let the bilinear form fulfil (1.2) and 0. Then the problem (1.1) has a unique solution u E V, (1.12) with and the finite-element solution conveives to U: E

>

IlL —

PROOF. Let u there exists U(

0

(v

—+ oo).

(8.4.8)

V and e >0 be arbitrary. Since 112(Q) n V is dense in V, V with Iu — u4i e/2. From (6) and h, —. 0 we

H2(1?)

for Finite Element Methods

8.4 Error with

see there are i' and

<e/2; thus Its —

—

189

e. This

proves (2.4a).

Theorem 7 proves convergence without restrictive conditions on u. On the other hand the order of convergence 0(h) in the estimate (7) requires the assumption that the solution u V lies in H2(S1). As will later become clear, this assumption is hardly to be taken as fulfilled in every situation. A weaker assumption is u V fl H8(Q) with s E (1,2). The corresponding result is as follows.

Theorem 8.4.8. Assume that in the assumptions of Theorem 6 u E V fl

112(0)ssreplacedbyu€VflH(17)withl

LeU? be sufficiently

smooth. Then there holds —

u"ti

(8.4.9)

The proof uses a generalisation of Theorem 4:

Lemma 8.4.9. Under the assumptions of Theorem 4 and suitable conditions on 0, there holds

inf fu—vfi C"(ao)h''1u15 forsELl,2J,uEH9(.Q)flV.

(8.4.10)

PROOF. The proof is based on an interpolation argument that will not be further explained here. Equation (10) holds for a = 2 (cf. (7)) and for a = 1, since infju — vii S — Ok = luli. From this follows (10) for a E (1,2) with the norm of the interpolating space [H1(0) fl V, 112(0) fl (cf. Lions-Magenes [1)), which under suitable assumptions on 0 coincides with

H(0)nV.

U

For a =

1

the right-hand side of (10) becomes ccest.Ni. In fact the

estimate inf{Its—vIj:v E V,j is the best poesible. To prove this ,chooee (orthogonal with respect On the other aides = 2 is the maximal value for which the estimates (9) and (10) can hold. Even u E C°°(0) permits no better order of approximation than 0(h)! The Ritz projections SN : V —. VN introduced in (2.5a) will now be written Sh : V —, Vh. The inequality (9) becomes ju — Shuil/iul. and this proves

I

Corollary 8.4.10. Assume conditions (la-c) fori-, Vp, and V. Let the bilinear form fulfil (1.2) and (1.12). The Ritz pmjeetion S,, satisfies the estimate

Ch'.

UI —

(8.4.11)

Under the asumptions of Lemma 9 there holds —

Sp,i1Jj1(n),...jj.(0)flv

for all 1

$

2.

(8.4.11')

190

8TheMethodolFlniteElementa

8.4.2 L2 and H' Estimates for Linear Elements According to Theorem 6 0(h), is the optimal order of convergence. This result seems to contradict the 0(h2) convergence of the five-point formula (cf. Section 4.5), since the finite-element method with a particular triangulation is almcot identical to the five-point formula (cf. Exercise 8.3.13). The reason norm. The estimate for this is that the error u — tie' is measured in the I in Theorem 4 suggests the conjecture that the
also be imposed on the adjoint problem to (1.1)

Find u E V with o(u,v) = 1(v) for all v E V,

(8.4.12)

which usas the adjoint bilinear form a'(u,v) := a(v,u). For f L2(fl) C V', The regularity condition is: the value 1(v) becomes

ForeathfEL2(O) theproblem(12) has a solution is E H2(1l) fl V with

C,4f10.

8413

In Section 9 we shall see that this statement holds for sufficiently smooth domains fl. Theorem 8.4.11. (Aubin—Nitsche) Assume (13), (1.2), and (1.12) with

> 0, and It' — vfj
for all is

H2(i1)

V.

=

(8.4.14)

Let problem (1.1) have the solution is E V. Let E Vh C V be the finiteelement solution. Then) with a constant C1 independent of is and h, —

CihIuli.

(8.4.15a)

If the solution is also belongs to 1P(.t?) fl V, then these is a constant C2, independent of is and h, such that (8.4.15b)

—

&om Theorem 4, it is sufficent to ensure (14) thai Vh be the space of finite elements of an admissible and quasi-uniform Inangulation. PROOF (cf. Nitache [1J). For each e := is —

E L2((l) define w

H2(S?) fl V

as the solution of (12) for I = a(v,w) = (e,v)L2(O)

for all v E V.

(8.4.16a)

8.4 Error Estimates for Finite Element Methods

191

Corresponding to w there is, by (14), a w' E V4 with (8.4.16b)

— w"li CohlwI2 COCRhIeI0. Equation (2.3) is a(e, v) = 0 for all v

V4; therefore, in particular, we have

a(e,w?&)

= 0.

(8.4.16c)

From (16a—c) we obtain

=

=

(e, —

a(e, w)

= a(e, w — w4)

w"Ii CsJefiCoCahlelo,

and thus feb

(8.4.l&J)

CSCJOCRh feb1.

From (2.1) one deduces febi

—

'11v S (1 + Cg/e4) inf fu — vlv vEVh

(1 +

so that (15a) follows with C1 := CSCOCR(1 + If u E H2(fl), one may use the inequality (7), Icli Chlul2, and deduce (lSb) with C2 := CSCOCRC.

I

Corollary 8.4.12.

The ineqtwlitics (ISa) and (lSb) aie properties (17a) and (17b) of the Ritz projection S4:


JJI —

flI

to the (8.4.17a)

C2h2.

—

(8.4.17b)

The estimates (17a) and (17b) may also be proved directly. The definition

of S4 and the connection between S4 and S4 may be found in

Exercise 8.4.13. lb a(.,.) let the operator L: V —, V', the Ritz projection S4, and the matrix L be associated. Show: (a) To the adjoint bilinear form a(.,.) belong the stiffness matrix V and the Ritz projection P(LT)_1P*L. (8.4.18a) (b) There hokls

S4 =

(8.4.14b)

PROOF. Second proof of Theorem 11: Let (u E H2(a): u is a solution of (12) for an I

be the range of norm

Since

L2(J1)) C V

E L(L2(Q),H2($1)). Equip the space H2(f1) with the

L2(fl)' = L2(Q) we have that L—1 E L(L2(f1),

is

8 The Method of Finite Elements

192

equivalent to L' E L(H?(fl)',L2(Q)) (ci. Lemma 6.3.2),

so

that there is a

Ca with (8.4.19a)

Cci.

The assumptions (1.2) and (1.12) of Theorem 11 can be brought over without

change of the constants to the adjoint problem, so that the statement of For Theorem 8.2.1 can be written in the form flI — ShIIv,..Jp(a) the adjoint operator we have the estimate —

(8.4.19b)

&) UH2(0'4—V'

The basic assumption (1.2) may be written

C3.

IILIIv'.—v

(8.4.19c)

From Equation (18b) one may read the expression

=

I—

I

= L'(I

—

=

—

—

Sh)'L.

From (19a—c) one has —

—

(8.4.19d)

Therefore we have proved (17a) with C1 = CaCpC5. Since S5 is a projection,

sois I—S5, so that = II(' — III —

111 —

Ill

—

From (17a) and (11) we deduce (ITh).

U

The regularity condition (13) is weakened in the following theorem:

Theorem 8.4.14. Assume (1.2), and (1.12) with CN = (10) for all I a 2. In the place of(13) we assume for some t E 1). If V = H1(a) then assume E

s/V = then assume finite-element solution satisfies the inequality htIIuIJ,

flu — u5J(,

(0

t 1

> 0, and in-

fl V). Then the 2).

(8.4.20)

The Ritz projection satisfies —

(8.4.21)

PROOF. (For the case V = H'(Q)) In (19a-c) replace L2(I1) by Ht(S?), and := {u by with is the solution to! s replaced by 2—ti Ht(I?)'}. Because of (11'), (19b) becomes

8.5 Generalisations HI

IRI —

As in (19d) one shows that

—

193

ShIIv._H.2—.(n)

Combining this with

ShIIH.(O)+.V

Ch'1 (cf. (11')) there follows (21), and thus (20). I For nonlinear boundary value problems, estimates of the errors in terms are ofinterest. For this we refer to Ciarlet of other norms (e.g., L°°(a), and Schatz [li. El.

8.5 Generalisations 8.5.1 Error Estimates for Other Elements The estimates for the errors for linear functions on triangular elements proved in Section 8.4 are also true for bilinear functions on parallelograms and for combinations of both sorts of elements (cf. Section 8.3.7). The proofs are along analogous lines. Even for tetrahedral elements in a three-dimensional region S? C 1R3 the same results can be carried over. For the proof one notices that u J12(Q) has well-defined nodal values even for a C since 2> n/2 (n =3, dimension of fl; cf. Theorem 6.2.30). For quadratic elements (cf. Section 8.3.4) one expects a correspondingly improved order of convergence. In general one can show the following: If the ansatz function is, in each E r, a polynomial of degree k 1 (i.e., u(x) = on T,), then

focallu€

d(u,Vh)—in({Iu-.vIl:vE

(8.5.1)

(cf. Ciarlet El, Theorem 3.2.lJ). If one uses parallelograms as a basis and uses ansatz functions that are polynomials of degree at least k, then (I) also holds. For example, biquadratic elements and quadratic ansatz functions of the serendipity clase fulfil this requirement for k =2. The result corresponding to Theorem 8.4.6 follows from Theorem 8.2.1:

Theorem 8.5.1.

Let Vh fulfil (1), and let the bilinear form satisfy (1.2) and (1.12) with =: > 0. Assume problem (1.1) has a solution a E V n H16+l(fl). Then the finite-element solution a4 E satisfies the

inequality

Ia

— u411

The Ritz projection satisfies

Chklalk+l.

—

(8.5.2)

Chk.

The estimate (4.9) now holds for a [l,kJ if a H'(S7)nV. With suitable regularity conditions, there are, as in Theorem 8.4.11, the error estimates Ia — II*L

u"lo

— U3iIflb-L(fl),

(8.5.3a) (8.5.3b)

194

8 The Method of Finite Elements

'-regularity: For each f E For (3b) one needs, for instance, adjoint problem (4.12) has a solution u E

(.0) the

Remark 8.5.2. Under suitable assumptions the inequality (4.20) holds for should be understood 1 — k
8.5.2 F'inlte Elements for Equations of Higher Order 8.5.2.1 Introduction: The One-Dimensional Biharmonic Equation All the spaces of finite elements, Vp,, constructed so far are useless for equations of order 2m> 2, since Vp, Hm(Q). According to Example 6.2.5,

in order to have V,, C H2(.0) it is necessary that not only the function u change continuously between elements. The ansatz but also its derivatives functions must therefore be piecewise smooth and globally in C'(.0). introduce the one-dimensional biharmonic equaAs a model problem tion

u""(x)=g(x) forO <x <1, u(0)=u'(O)=u(l)=u'(l)=O which becomes in its weak formulation

a(n,v)=f(v) a(u, v)

where

= Jo/ u"v" dx,

1(v) =

j Jo

(8.5.4) gv

dx.

Divide the interval .0 = (0,1) into equal subintervals of length h. Piecewise linear functions (cf. Figure 8.3.1) can be viewed as linear spline functions, so

that it isnatural todefine Vp, aathespaceof cubic splines (withti=u'=O for x = 0 and x = 1) (cf. Stoer 11, §2.4) and Stoer-Bulirsob [1, §2.4J). We can take as basis functions B-splines, whose supports in general consist of four subintervals (ci Figure 1). Since cubic spline functions belong to C2(0, 1), 1) n they are not just in 1), but even in 1).

•

$

U

U

U

•

S

FIgure 8.5.1. B-spline

Simpler yet than working with spline functions is to use cubic Hermite

interpolation: C'(O, 1):u cubic on each subinterval, u(0) = til(0) = u(1) = u'(1) = 0).

Vp, := {u

(8.5.5)

195

8.5 Ceneraliaation8

To each of the inner nodes x3 = jh there are two basis functions and

.

I

S

S

= 0 for

= 1,bM(Z,) =

=

= = with Figure 2). i) (ci. k = 1,2 (j U

U

&

(a)

(b)

FIgure 8.5.2. Hermite basis functions

The support consist of only two subintervals. The expressionB are (h

b14(x)

—

Ix —

—

x,j + h)/h3 for Zj4 X

for z

and u21 (0 < I < 1/h) be the coefficients of the Exercise 8.5.3. Let Show that the coefficients E expression u(x) = + of the finite-element solution of the problem (4) are given by the equations + 24ti1g — —

where

6tLl,s+jJ +

=

+

= +

f2ü

(856)

+ = ih.

=

The system of equations (6) differs completely from the difference equations (cf. Sectioh 5.3.3), since in Equation (6) there appear values of the of the derivatives at the functions at the nodes together with the values nodes.

8.5.2.2 The Two-Dimensional Case The ansatz (5) can be carried over to (1 C JR2 if one starts with a partition into rectangular elements. The ansatz function is bicubic:

vj'=o on

each rectangle of the partition.).

At each inner node one may prescribe the four values u, quently there are four unknowns and four basis functions

Consey), .

. .

y)

belonging to each node. The latter are products (j, k = 1,2) of the one-dimensional basis functions described in Section 8.5.2.1 (cf. MeisMarcowitz

Figure 15.221, Schwarz (1, §2.6.1)).

If one starts from a triangulation r, a fifth-order ansatz gives what is wanted: u(x) = on Er. The number of degrees of freedom is

8 The Method of Finite Elements

196

the 21 (the number of v Z2 with v 0, lvi 5). For the nodes one points x1, . . ,x6 in Figure 8.3.8a. At the vertices xt,x2,x3 one prescribes 6 2}. The 3 remaining degrees of freedom result from givvalues {D"u(x2): at the midpoints of the sides x4, x5, x6. ing the normal derivatives Notice that if two neighbouring triangles (T and T in Figure 8.3.8*) have the same vertex values 2, j = 1, 2} and the same normal derivative at x4, then both u and Vu are continuous on the shared side, i.e., According to whether V = V = H2(Q) n or Vh C V = H2(ul) one has to set u, or the first derivatives, to be zero at the boundary nodes .

Remark 8.5.4. The finite-element space Vh C H2(Q) described here can of course be used for differential equations of the order 2m =2.

8.5.2.3 Estimating Errors Instead of (5.1) one obtains inf{Iu —

E

Chk_m+liuIk÷l for all

V,

(8.5.7)

where k m depends on the order of the polynomial ansatz (e.g., k = 3 for cubic splines, cubic (resp. bicubicj Hermite interpolation). Here m = 2 for the biharmonic equation. As in Theorem 8.5.1, there follows from (7) the error estimate —

t4hlm

for the finite-element solution one gets lu

—

(8.5.8*)

E Vh.

(2m

—

k

Under suitable regularity assumptions

— 1


m


(8.5.8b)

(cf. Remark 8.5.2). The maximal order of convergence 2(k — m) + 2 results for

solution of the adjoint problem (12) with I

Hk+l(a).

must belong to

8.6 Finite Elements for Non-Polygonal Regions Since the union of triangles and parallelograine only generates polygonal regions a polygonal shape was assumed in (3.6). The finite-element method is,

however, in no way restricted to just such regions. On the contrary finite elements can readily be adapted to curved boundaries.

8.6 Finite Elements for Non-Polygonal Regions

(c)

(b)

(a)

197

Figure 8.6.1. Curvilinear boundaries

Let V = H'(Q) and let .0 be arbitrary. The triangulation 'r can be so chosen that (3.7a,b,d) hold and the "outer" triangles have two vertices on 1' = OS? as in Figure 1. In the convex case of Figure la one extends the linear functions defined on T onto T := TU B. In the case that the boundary is concave (see Figure ib) one should replace T by T := T\B. One copes correspondingly in the situation of Figure ic. The nodes and the expressions for the basis functions remain undisturbed by these changes. is the union of the closures of all the inner triangles E r and all modified triangles T which lie on the boundary. All the properties and results of Sections 8.3—8.4 extend to the new situation. The only difficulty is of a practical sort: To calculate L and f one has to work out integrals over the triangle T which involves arca. Assume now V = H0' (0). The previous construction will not be a success, since the extensions of the linear functions to T will not vanish on the boundary piece r'flOT. Thus Vb C will not be satisfied. As long as the region is convex, only the situation shown in Figure la occurs, and

may be extended to B C T by uh = 0. in the case of Figure Ib, one must set the values at the inner nodes to zero, so that uh =0 on T = T\B, and in particular = 0 on f'flO(T\B). All in all, what results is that the support of any uh V,, is in a polygonal region inscribed in 0. One interpretation is the following: In the boundary value problem, .0 should be replaced by an approximating region 0h C 5? (cf. Theorem 2.4.6). The finite-element solution described agrees with that which would result from the smaller region. However, the error estimate in Theorem 8.4.4 only holds for the smaller region IIu — ChIIuIIjJ9(fl). Since v =0 on J?\S?h for any v V,, one should also estimate IltSIJgl(Q\ah). Now .Q\flh is contained in a = {x Oa:dist(x,P) 6} of width 5= max{dist(x,f'):x i2\S?h}. One can estimate as follows: —

= IIUIIHI(O\flh)

II1LIIHI(s1)

If Q\0, consists only of arc segments B as in Figure la (for instance, in the case of a convex region), and ill? E C", then S = 0(h2). From this follows the estimate iflfvEVh ((U — VHHI(n) Ch2lfr4IH2(fl), as for a polygonal

198

8 The Method of Finite Elements

region. If, however, as in Figure lb the whole triangle is part of Q\Qh, then 6 becomes 0(h) and the approximation worsens to infVEvh Vu — VIIHI(Q) Ch1/2fluHJp(l7) (cf. Strang-Fix

p. 192)).

In order to adapt the trianglar or parallelogram elements better to a curved boundary one can use the technique of isoparametric finite elements. Fiom Figure 8.3.4 we see that the reference triangle T may be mapped

into an arbitrary triangle T E r by an alfine transformation

'-+

x'

—x') . The linear fresp. in the case of §8.3.4, quadratic) of a lin= T can be expressed as the image, function u(x) on ear [resp. quadratic) function v on T. We now replace the affine transformation of triangles by a more general quadratic mapping

(

Oj +

+

+

+

+ açfl7 +

+ +

÷ T + 012,72)

C

The image T is a curvilinear triangle. The coefficients . . . ,012 are uniquely determined by the nodes x11... ,x6 of T which are the images of the vertices

and midpoints of the sides of the reference triangle T (cf. Figure 2). The triangulation r used so far can be replaced by a "triangulation" by triangles with curved sides, if neighbouring elements have the same arca as common boundaries and the midpoints also coincide. The ansatz functions on T E is linear Iresp quadratic] where f have the form u(x) = in and,7. The resulting finite-element space is the space of isoparametric linear (reap. quadratic] elements (cf. Strang-Fix [1), Ciarlet [1), Sdiwarz 11], Zienkiewicz[11).

x6

FIgure 8.6.2. MappIng of the reference tangle T to the curvilinear triangle T

In general, there is no reason to use curvilinear triangles in the interior of .0. As in Figure 3, one chooses ordinary triangles in the interior (i.e., the quadratic transformation is again taken to be linear). A boundary triangle, like T in Figure 3, is, on the other hand, defined as follows: x' and x2 are the vertices of T which lie on V. One chooses another boundary point x4 between x1 and x2 and requires that the boundary of the triangle cuts the boundary I' of the region at x1,x4, and x2.

8.7 Additional Remarks

199

XX) X4

FIgure 8.8.3. Isoparainetrlc triangulai diseection

8.6.1. Section 5.2.2 shows that, in the case of other than Dinchlet boundary conditions, the construction of difference schemes is increasingly

complicated. Instead one may restrict the usual difference methods to the inner grid points, and near the boundary discretise by using (e.g., isoparametric) finite elements.

87 Additional Remarka it is not poasible to consider here many of the details about, and modifications of, finite-element methods. In the following, individual cases are just treated summarily.

8.'T.l Non-Conformal Elements Condition (1.3), VA c V, characterises conformal finite-element methods. V are called non-conformal. An example Discretisations for which VA of a non-conformal element is the "Wilson rectangle". For this method one and assumes x starts with a diseection into rectangles the function to be quadratic on Rj: VA := {u quadratic on

u continuous at vertices of the rectangle R4.

Notice that u VA is required to be continuous only at the nodes, and not in the interior of 11. A function which is not continuous along a side of cannot belong to H'(fl): VA V = H1(a). A possible basis for Vh is the following. For each node x3, let bj be the basis function belonging to the bilinear elements (cf. Section 8.3.3). The quadratic ansatz chosen here has two

200

8 The Method of Finite Elements

additional degrees of freedom. Because of this one chooses for each rectangle two further basis functions = (Xis, X25) x

y) := (x

—

—

x),

y) := (y —

—

so that continuity is ensured there if are extended by zero outside

which vanish at the vertices of

and

The first problem which arises for non-conformal elements is the definition as in (1.8a), = of the stiffness matrix L. A definition in terms of V! Therefore, in addition to discretising V to is not possible because bj, a Vh, one must replace a(.,.) by a bilinear form ah(., .): Vh x Vh —i ff1.. one defines ah(•,•) by partition of £? into rectangles

ah(u, v)

>2 >2fRs IaI,t$Im

dx,

Notice that a(u,v) = ah(u,v) for u,v V. Conformal finite-element methods always give consistent discretisations. For nonconformal methods one does not always have consistency. For this problem, in

if a(.,.) is given by (7.2.6).

particular for the so-called "patch test" see Strang-Fix [1, p. 174], Ciarlet §4.2], Sturnmel (2), Gladwell-Wait [1, p. 83—92], and Thoinasset [11.

That an ansatz which seems completely reasonable can possibly yield completely false discretisations is shown by

Exercise 8.7.1. Let r be an admissible triangulation. Assume that Vh {u constant on each T E r}. Show that (b) inf{Iu — yb: v E V,1} ChIuIj for all u E V = (Vu, Vv) dx and ah(u, v) := (c) Let a(u, v)

is the result for

J,(Vu, Vv) dx. What

=

Non-conformal elements are, in particular, of importance for equations of higher order, since they are lees complicated than conformal elements.

8.7.2 The ¶frefftz Method In Theorem 6.5.12 it was shown that for symmetric and V-elliptic bilinear forms the weak formulation (1.1) of the boundary value problem is equivalent to the variational problem

Find u E V with J(u) = min{J(v):v

V}

(8.7.1)

(cf. also Theorem 7.2.9, Exercise 7.3.8). Another variational problem

Find w E W with K(w) =

E W}

(8.7.2)

is called the dual or complementary variational problem to (1), if both have the same solutions u = w V fl W.

8.7 Additional Remarks

201

f E L2((1) in .0, u =0 on F, A special dual variational lVvI2dx for v E V := Hd(fl). Leads to J(v) = problem is due to Trefftz (1926): For example, the Poisson problem, —Llu

K(w) :=

H'(fl):-Ltw

fVwI2dx for WE W := {w

f}

(cf. Velte [1, p. 911). Notice that v E V satisfies the boundary condition v =0

on F, while for vi E W no boundary condition is required though it must satisfy the differential equation Lw = I.

8.7.3 Finite-Element Methods for Singular Solutions The error estimates provided so far, such as for instance (4.7), depend on the assumption u E H2((2) fl V, winch will be discussed more carefully in Section 9.1. That this assumption does not always hold is shown by

Example 8.7.2. (a) The Laplace equation of Example 2.1.4 has the solution u =

(b) The solution u =

r213

= —

0

in the L-shaped region This function only

in Example 5.2.3 only belongs to IP(.Q)

for a <3/2.

If one replaces, for example in the L-shaped region, the Laplace equation = I in .0, u = 0 on F, then one may still show that for f E L2(a) the solution u may be split up as u = u1 + — ir)/3) with u1 E H2(fl) fl and a E Jt by the more general Poisson equation,

(cf. Strang-Fix [1, p. 275ff.]). In this equation x(r) C°°(fl) is a specifiable function with = 1 in 0 r 1/4, x(r) = (I for r 1/2, such that is ensured. An ordinary discretisation by linear x(r)r2"3sin(...) triangular elements can only attain Iu — = 0(h'), a < 2/3. However, if one includes the function sin(...) or similar functions in Vh, then the approximation of u in V1, can be improved: d(u, Vh) = 0(h), so that — uhli = 0(h). For such matters refer to [1] and Blum-Dobrowolski [1). See also Gladwell-Wait [1, p. 119] and Hackbusch L41. A completely different approach using finite elements which does not need

explicit knowledge of the singularity Is the following. Let r be a triangulation whose triangles become finer near where the singularity is expected, for exampie, at a re-entrant corner (cf. Figure 2.1.1, 5.2.1). For analysis of the errors see Schatz-Wahlbin

8.7.4 Adaptive Thangulatlon The sort of triangulation mentioned in the last paragraph can be most easily obtained by adaptive methods. So far r = rh has been given depending on a given "step size" h. In an adaptive procedure one starts with a triangulation

202

8 The Method of Finite Elements

r0 and uses a-pceteriori error estimates. Those triangles of r0 that lead to the largest errors are then divided further so that a new triangulation r is formed. One repeats this procedure until the errors are satisfactory. For the literature on this we refer to, for example, Rhernboldt's work [11 and the paper and Rhemboldt referred to therein, as well as to Eriksson—Johnson of El'.

8.7.5 Hierarchical Bases Let us give the fundamental idea in the one-dimensional case. Let Vh C 1) be, as in (3.2), the space of piecewise linear functions for an equidis-

tant division of the interval f2=(O, 1) intopartsoflengthh. Forh= 1/2 we have dim(V112) = 1, and b1 as in Figure la is the only basis function. Clearly V114 C V1,2. For V114 one takes over b1 E V112 as a basis function, and adds

the functions b2, b3 as in Figure ib; {b1, b2, b3} is a basis for Vj,4. Similarly (cf. Figure ic) gives Thus b1, b2, b3 supplemented by b4,.. . , V1,8 C a basis — the so-called hierarchical basis — of V118. In contrast to the basis functions that have as yet been used, the supports of the are not necessarily of length 0(h); e.g., supp(bi) = [0,1]. From this it follows that the stiffness matrix has more entries than is otherwise usual. In spite of this, the above choice of basis does have important advantages: is a submatrix of the stiff(i) A matrix L = L(Th) already calculated for ness matrix L = for Vh, so that (a global or local) refinement of the triangulation requires only a minimum of additional effort. (ii) The condition cond(L) = IILII 11L'II of the stiffness matrix is better than in the standard case (for example, cond(L) = 0(logh) instead of

0(h2); cf. Section 8.8). (iii) There are suitable methods available for solving the system of equations Lu = f. Further details on this can be found in Yserentant [1], Bank-DupontYserentant [1,, and Hackbus& (9]. b1

(a)h= 1/2

(b)h= 1/4

(c)h=1/8

FIgure 8.7.1. A hierarchical basis

8.7.6 Superconvergence In difference methods the error is given by a mesh function Uh — RhU (uh: the difference solution, u: the exact solution), while for finite-element methods the

8.8 Properties of the Stiffness Matrix

203

u can be used is defined on the whole region 12, and therefore solution as the error. A grid function u.1 could in principle be extended to a function ii" defined (x': a grid point of 17h) on all 17, for example, by diooeing the values as the nodal values of the piecewise linear or bilinear finite elements, and thus with u5 := Pu5 (for P see (1.6)). In addition to the errors interpolating at the gnd points, there are now the interpolation errors u — PRhu as well: u5 —u = — = P(uh — R5u) — (u — PR5u). As soon as the — Rats, one has interpolation errors are of the same order of magnitude as for Ptih — u the same convergence assertions, otherwise Pu5 — u displays worse convergence than — Rhu. Conversely, for finite-element methods, one can ask if one has the same 0(h2), R5: restriction to the convergence properties (e.g., huh — R5uD nodes) for the values at the nodes u5 {u"(x4): nodes } or for suitable averaged values. If one has indeed better convergence than for u5 —u, one speaks of superconvergence. This notion can also be applied to difference quotients of node values, if these are more precise than (ha(r= 1). For this see Lesaint-Zlárnal [1], Bramble-Schatz [1], Thomée [2], and Louis [1).

Greater precision can be achieved with extrapolation techniques which, under suitable conditions, can be applied to finite-element problems (cf. BlurnRannacher 12]).

8.8 Properties of the Stiffness Matrix The properties of the matrix L5 have been carefully studied for difference procedures since the solvability of the difference equations and the analysis of convergence properties depend on them. In the case of finite-element discretisation we obtain the corresponding assertions in another way. The factors that control the solvability and convergence are the subapace V5 and the operator Vh —.

V51,

which was introduced as LN in (1.lOa).

The stiffness matrix L, for a given Vp,, depends on the basis {b1,. chosen. Let

N5 :=dim(V5).

The isomorphism P :

V5 defined in (1.6) maps the coefficient vector Fbr finite elements, in general, the coefficients coincide with the nodal values: = (1 i S Np,). The connection between the matrix L and the operator L5 : Vh —4 is, according to Lemma

v to Pv =

E

8.1.7,

L = P'LhP,

Lh =

P'1LP1.

(8.8.1)

The definition of in (1.lOb), (P*u, v) = (u, also depends on the choice of the scalar product (.,.) in Let us choose here the usual

204

8 The Method of Finite Elements Nh

(8.8.2a)

(u, v)

= In Exercise 8.1.10 we have already established the following properties of L: If a(.,.) is symmetric [and V-ellipticl then L is symmetric (and positive definitel. Notice that the difference methods do not always have these properties. For the Poisson problem there is a symmetric form, but nonetheless the matrix Lh in Section 4.8.1 is not symmetric (cf. Theorem 4.8.4). The condition cond(L) plays an important role in the solution of the equation Lu = f. We would like to show that, under standard conditions, we (cf. Remark 5.3.10). First we have to decide on the have cond(L) = ". Up to a scaler factor fundamental norm of 1/2

N,,

(n: dimension of 17 C

:=

(8.8.2b)

is the Eucidean norm coming from the scaler product (2a). The corresponding matrix norm (8.8.2c) sup{IfLvIIh/IIvIIh:O V E is the usual spectral norm of L. As an alternative to

let us introduce

the norm := tIPuIILa(rn for u E

(8.8.3)

In some important cases and are equivalent (uniformly in h). As consider the linear elements. an

Theorem 8.8.1. Let

h —40 be a sequence of quasiunifonn triangtdations.

Let Vh be the space of linear elements defined in (3.8) with the usual basi8 (see

(3.9a)). Then there is a sonstant Cp, which does not depend on h, such that

CplIuIIp.

(8.8.4)

The basis for the proof of this is

Lemma 8.8.2. Let T =

0, +

'i):

< 1) be the unit ti'iangle (see

If u is linear on T, then

+u2(1,0)-t-u2(0,1)J sJJ

T

(8.8.5)

1

..[ti2(O,O) + u2(1,0) + u2(0, 1)].

PROOF. Put ui 1= u(0,0), uj + (u9

—

+ (us —

is9

:= u(1,0),

it follows that

u(0, 1).

u(e,,i) =

8.8 Properties of the Stlthiees Matrix 2 1

U2

1

1

205

fui

i 2

us)

The eigenvalues of the symmetrIc 3 x 3 matrix are Therefore the right-hand side is and

=

A2 -=

I and A5 = 4.

U

4E

dx = ET.EI.,, IT. (Pu)2 dx. as To prove Theorem 1 write as in Exercise 8.3.14a. be the linear maps from T to : T —, Let u', u", u" be the values of u = Pu at the vertices of T,. The inequality 0 < C1h2 < Idet4
Let

+ uhI2

+ uIhI2) j IPuI2 dx = det çôJ

Jj

,1))12

(8.8.6)

By definition of a quasiuniforni triangulation all the angles of the triangles are 0. Let M > 2w/go. Bath node belongs to at least one triangle, and to at most M triangles. Since u', U", U" are the components of the vector u, the summation in Equation (6) over all T1 E Th gives

J (Pu)2dx

=

U

max(MC2/6, 24/C1)1/2.

so that inequality (4) holds with Cp

be derived from the square grid triangula8.3.13). Show that tion (cf. Exercise Hulla

Exercise 8.8.3. Let Vh C The matrix

M := P'P, i.e.,

:=

J0

dx

(8.8.7)

called the mass matrix when it occurs in engineering applications. Concrete bounds in the inequality (4) have been given by Wathen [1J for various specific elements. is

Remark 8.8.4. (a) Inequality (4) is equivalent to 11M111

IIMII

(b) We have fluflp Iih

IIuIIh

PROOF. (a) The calculation

= (Mu,u)

S

= f0(Pu)2 dx = (Pu, Pu)o = (P Pu, u)

UMII(u,u) =

(b). The next follows from

= is positive definite.

proves the first inequality in

=

u) = since

M

206

8 The Method of Finite Elements

(.3, using (b), we deduce the (b) From (.3 and inequality (4). Conversely (Mu,u)112 = IuPjp CpIIuIIa = C3, since M j8 positive definite (cf. Lemma 4.3.25). implies that U follows from II"IIh Similarly, Since Vh is finite-dimensional Ch := sup{IuIi/IuIo:0

is finite.

ti

One says that Vh satisfies the inverse estimate if C, = O(h'), that is,

HiCihT'Ho for a1lu€Vh where C1 does not depend on it as h —,

(8.8.8)

0.

Theorem 8.8.5. Let rh be a sequence of qwzsiunifonn triangulations. Then the space of finite elements inirvdueed in (3.8) sati.tfies the inverse estimate.

PROOF. As in Theorem 1, the proof follows by transforming the integrals dx for and 1 into integrals over T. Here we have fT in addition to transform D = into (ci. the proof of Lemma 8.4.2). U Theorem 8.8.6.

Suppose 0 C

Assume that (1.2), (8), and IIUIIp

CPIIUHh hold. Then we have

IILII

(8.8.9)

PROOF. Multiply the inequality (u, Lv)

by

a(Pu,Pv) CgIPuliJPvIi CgC7h2IPuloIPvIo

and set u := Lv:

= hTh (Lv, Lv) and thus liLvila

that is (9).

Theorem 8.8.7. Assume (1 C Ft'. Let (1.12) hold with fl.flh
0L'II PROOF. We know Section 8.1), so that

=

e > 0 and (8.8.10)

1/c (ci.

and

=

U

=

8.8 Properties of the

Matrix

207

Since h

=

=

=

sup

sup

I(P1v,f)I/IvIi

Vj,

sup I(v, f)I/IPvlj

= sup


=

we have that 11L'fUh

for all f; thus we conclude (10).

1

The combination of Theorems 6 and 7 leads to

Theorem 8.8.8. Assume (1.2), (1.12) with Ch

e

> 0, (4), m =

1,

and (8).

Then we have oond(L)

(8.8.11)

The ideas involved in the proof can, in principle, be carried over to the case 2m > 2, i.e., to boundary value problems of higher order. The inverse estimate becomes for all U E

(8.8.12)

In order that inequality (4) hold one must be careful in defining the norm

If all the components u are nodal values (as for instance is the case for the spline a atz for VhC then ji. llaombedefined as in (2b). However as soon as the components u involve derivatives (D'Pu)(x'), the in (2b) must be replaced by For example, when the Hermits functions of (5.5) are used then u has the components and (cf. Exercise 8.5.3), where = (Pu)(x4) and = The appropriate definition of fl. Ha is with n = I, + V .

since (0, 1) C

Exercise 8.8.9. Let V c U C 1" be a Celfand triple and assume V5, C V. Show that the inverse estimate llullv S CIhmIlulIu for all u

(8.8.13a)

implies Hulk,

CihmIIujlvi and holly

Cih_2mlluhlv, for all u E

(8.8.13b)

9 Regularity

9.1 Solutions of the Boundary Value Problem in H8(O), 8> m 9.1.1 The Regularity Problem The weak formulation of a boundary vaLue problem

Lu—g lull, Bu=p on!'

(9.1.1)

uEV, a(u,v)=f(v) forallvEV

(9.1.2)

as

was, in Section 7, the basis upon which we were able to answer the questions of existence and uniqueness of the solution. Here, by existence of a solution we understand the existence of a weak solution u V. The error estimates in Section 8.4 made it dear that the statement u E V is not enough. Under this assumption we can only show 0. The — uhllv more interesting quantitative estimate Iu — uhIi = 0(h) as in, for example, Theorem 8.4.6 for V = HJ(a) or V = H1(i7), requires the assumption u

112(Q) n V. The assertion u E 112(Q) or, more generally u E H'(s?), is a statement of regularity, i.e., a statement about the smoothness of the solution, which will be examined in greater detail in this section. The regularity proofs in the following sections are very technical. To make the proof ideas clearer, let us sketch the proof of inequality (4) below for the

Helmholtz equation = fin Sic u =OonI'. Step 1: 5? = Since the bilinear form for this situation a(u,v) =

Vv) dx+f0 uvdx is H'(1R2)-elliptic, (4) holds for s = m = 1. Weshall prove (4) by induction for 8 = To this end we take the derivative of the differential equation with respect to x, + = If I then H3(fl) and the equation by the induction as+v= sumption, has a unique solution v with IvL_i If one sets up this inequality for v = and likewise for v = the result is IuIa IuIe_i + IuxIe_i + Iui,Ia_i 3C,—11f(8_2. Thus, (4) has been shown for S. Step 2: 5? = = lRx(O, oo). As above, we can obtain the estimate since alsosatisfies = and the boundary condition

9.1 Solutions of the Boundary Value Problem in H'(Ii), s > m

=

0

implies

on 1. This is not the case for u1. But

209 E

E

This property, however, results from the differential u— — E equation, Step 3: Let 11 be arbitrary, but sufficiently smooth. As in Section 6.2.1, Ills decomposed into (overlapping) pieces il, which can be mapped into 1R2 or

show

f

is a partition Correspondingly one splits the solution u into Exsu of unity). Then the arguments from steps I and 2 prove inequalities for which together result in (4). Note that only a sketch of the proof was given. Some of the steps of the + proof are incomplete. For example, might not the equation = have a solution in 1R2, E L2(1R2), which does not belong to H' (1R2) and +v= hence does not coincide with the solution v H'(1R2) of In the following, always let s m. The boundary value problem (2) with of problem is sald to be H3-regular if each solution u V= and satisfies the estimate (2) with f H3_2m(Sl) belongs to

(9.1.3)

+

IuI.

If L is the operator associated with a(.,

then it is also said that L is

H'-regular. Remark 9.1.1. (a) H's-regularity always holds. with (b) Let the variational problem (2) have a unique solution u I-rn for all / E H-m(Il). If the boundary value problem is H3< C01f IUIm satisfies the inequality regular, then the weak solution of (2) with f luIs

(9.1.4)

Cjf13_2m.

(c) Let L be the operator associated with a(.,.). (4) is equivalent to H8(Q)) and (4'): IlL

—1


(9.1.4)/

PROOF. (a) Set Cm = 1. (b) In (3) estimate IUIm by C0 I/I-rn and use the fact that because of s in the embedding H3_2m(S?) C Hm(Q) is continuous = C,(1+C0C'). U such that C'IfI._2m. Thus we obtain (4) with The following remark shows that a perturbation of a(.,.) by a smooth term of order <2m does not change the H3-regularity.

Remark9.1.2. LetLbeH'-regularforallt an operator of order 2m —1: 5L

for all

The assumption on ÔL holds in particular if 5L belongs to the bilinear form a"(.,.) in Lemma 7.2.12 and its coefficients 0afl are sufficiently smooth: for example, €

9 Regularity

210

PROOF. (By induction) For 8 = m the statement follows from Remark Ia. Assume the assertion for 8 — 1. Let u be the weak solution of (L + ÔL)u = hence also the solution of Lu = I := — öLu. Under the

I

I

induction assumption the following already holds: It&L_i such

+JUIrn]

that Ifla—2m IfL—2m

II

+

+

ItLimi.

of the H-regularity of L, u belongs to H'(fl) and satisfies lu). U 1 I.-2m + IUIm). Together these inequalities result in statement (4).

Because

9.1.2

Regularity Theorems for

=

The domain (1

is distinguished by the fact that it has no boundary, and hence no boundary arnditions either.

Theorem 9.1.3. Let m

IN, (2= lIt". Let the bilinear form

f°

a(ti,v) := be

Hm(IR")-coercive. For some k E L°°((2)

for all a,

(9.1.5)

IN let the following hold:

with H max(O, k +

—

in).

(9.1.6)

Then everij weak solution u E Hm(IR") of the problem

a(u,v) = (f,v)o for all v with /

belongs to ItLIm+k

(9.1.7)

and satisfies the estimate


(9.1.8)

PROOF. (a) First we must investigate the start of the induction, k = 1. Let be the difference operator

8=

i.th unit vector ,1 i n.

:=

Let u be the solution of (7).

d(u,v) := where ap(u,v) :=

fm,.

v E H"'(lR") set + ap(Oh,gu,v),

(9.1.9a)

9.1 Soluttose of the Boundary Value Problem

:= _a+ u+]v dx = we see that

—h

211

=

ouôvdx = fm,.

Prom

Let

in H(fl), B > in

dx

av Du

f

d(u,v) =

P.

—

IaI.n

181="

J

+

—

—

P.

(9.1.9b)

Since IvI,÷i (ef. Lemma 6.2.24), one bounds the first sum in (9b) with CIUlmI8VIm_i < CIUI ml Vim. One obtains the same bound for the second sum:

for v

ld(u,v)i ClliimlVlm However, if a(.,.) is

(9.1.9c)

ap(.,.) is also (cf. Lemma 7.2.12). For

CE >0 and CK from (6.5.10) we thus obtain

op(8u,Ou) +

=

—a(u,a2u) + d(u,Ou) +

(9.1.lOa) The first summand can be transformed according to (7): a(u, 02u)o = (f, 02u)o and bounded by The inequality JOvl. lvl.+i for 8= rn—i and v = Ou yields

8u)l

(9.1.lOb)

ifI_m+lIOh,it4im.

By (9c), the bound for the second suxnmand in (lOa) reads:

(9.i.iOc)

Finally we have S lOh,iUlm_liOh,iUIm 5

(9.i.lOd)

(lOa.d) yields S LifI-m+1 + (Cd + CK)lUlml/CE for all h

>0, 1 5 i 5

Lemma 6.2.24 shows that u e Hm+l(JRn) and inequality (8) for k = 1. (b) Now let k = 2. In (9a) substitute EYT8v with ii'l = 1 for and integrate by parts:

9 Regularity

212

d(u, u)

=

a(u, D78v)

v)

—

E Jit',

dx

si —

+

EJ (Dau(x

J

—

4ei))

+ (Oaap)

+ +

+

+

As in the first part of the proof, from Id(u,v)I D78v)o(

(D°v) dx.

—

CdIUIm+lIvIm and

If12_mID7OVIm_2 1112-mi Vim

follows

[If I-m+2 + (Cd + — 1, h > 0, 1 I n. Thus, the differences formly bounded so that Lemma 6.2.24 proves u

for all hi CEll

are uniand IUim+2

+ IUIm+jJ. If one uses the estimate lUlrn+1 CEIfi_m+i + proved in (a), (8) follows for k — 2. +

tUImI

(c) An induction argument is carried out analogously for k 2.

Corollary 9.1.4. The aoercivity in Theorem 3 ean be replaced (a) by the sufficient conditions/rem Theorem 7.2.11; (b) by the assumption that for some A the bilinear form ap(u,v) + A(u,v)o satisfies the assumption (6.5.4a,b).

Corollary 9.1.5. If, in

a(.,.) is Htm(It")-elliptic or if a(.,•) satis-

fies condition (6.5.4a,b), then in Theorem 3 the estimate (8) should be replaced by

CklfI_m+k.

PROOF. ByRemark2.

(9.1.11)

U

Corollary 9.1.6. Let a(.,.) be Hm(IR')-coercwc. Let the conditions (6) and be satisfied fork f4 with k > s + n12 n/2. Then the weak jE solution u of (7) belongs to C(11t"). Hence for s 2m, the weak solution is (1180 a classsa11 solution.

PROOF. The statement results from Sobolev's lemma (Theorem 6.2.30). U

9.1 Solutiom of the Boundary Value Problem in IP(fl), 8> m

213

Let the conditions (6) and Ut Then the weak solution of problem be satssfied for all k E sotssfled in particular if f belongs The conditions (7) belongs to constant. and the coefficients to

Corollary 9.1.7. Let a(.,.) be

/

Let

Theorem 9.1.8. Let a(.,.) in (5) be

0<0<6<1, t:=k+i9.

s=k+0, let

For the coefficients (Q =

aap E Ct

(1R0) with

to u e

with k

The generalisation of u real s > 0 reads as follows:

Hm(S[j)

E L°°(fl) otherwise. (9.1.12)

m,

for k +

belongs to

Then each weak solution of (7) with f E and satssfle8 the estimate IUIm+a


(9.1.13)

PROOF. The beginning of the induction is given byk = 0, i.e., 0< s
Ru(x) := Raju(x) := where

=1, and

unit vector. Here

is the

= -.s(1

()

- a)(2 - 8).

..

(p

-1-

is the binomial coefficient.

ExercIse 9.1.9. Let 0 < 8 < 1. Show that (a) The operator adjoint to is

R'u(x)

WhJu(x) =

—ph.,).

(c) Fbr the Fburier transform we have

= Hint:

+

=

((1

—

= [(1

—

214

9 Regularity

IR, h > 0,

(d)

1 j n,u

fUk+.. Hint: Make use of the norms I k and I. similarly (1 + (6.2.15)), and show

(cf.

By analogy with (9) one obtains a(u,

d(u, v)

—

ap(RLu, v)

=

+ p=1

/

—

—

= — iihe,) + to ap(Ru, v). Since we have it follows that

since the first summand in

—

— phe3) belongs

—

—

—

—

—

oap(x)I

id(u, v)I CIUImI VIm [1

+ ht'

C'iUimIVIm,

= and same consideration as in Theorem 3 yields

for it is true that

C[(P

e_M

C'EIfi_m+.+itLimj

= O(ha_t). The

for all h >0, 1 j n

(cf. Exercise 9d). To obtain this estimate for jUIm+, we write

(i +

+

f

(i +

The second integral converges to zero for h -s 0. The first one can be estimated

by

plus the sum

c'J

(1

j=1

cuf

3

(1

+ 3

= C"

Of

+

+ (1 +

215

9.1 SolnUons of the Boundary Value Problem in H(Q), a> m

and (13) follow. As in Lemma 6.2.24, u E (2) As in the proof of Theorem 3, one carries out induction over k and inves= k. v), tigates d(u, v) := a(tL, RD1'v) —

I

Exercise 9.1.10. Generalise Corollary 6 with the aid of Theorem 8.

9.1.3 Regularity Theorems for $1 in (6.2.18) is characterised by x,, >0. As in Section 7, we The halfspace limit ourselves to the following two cases: either a Dirichiet problem is given for arbitrary m 1, or the natural boundary condition is posed for m =1.

Theorem 9.1.11. (Homogeneous Dirichiet problem). An analogue to Theorem 3 holds for the Dirichlet problem:

a(u,v) = (f,v)0 for all V E

UE

(9.1.14)

Theorem 8 can also be carried over f one eXdUdC8 the values 8=1/2,3/2,..., m — 1/2.

For the proof we need the following highly technical lemma.

The nonnl•I.

Lemma9.1.12. is equivalent to

:=

(9.1.15)

+ loI=m

PROOF. The relatively elementary case s m is left to the reader. For 0<8< m too, the proof would be considerably simpler if in (15) one were to by that of of replace the dual norm) . AccordStep 1. First we prove the statement for .0 = IR" instead of $1 = is equivalent to ing to Theorem 6.2.25a, for .1? IR" the norm :=

=

Since

0 < C0(1 ÷

I

m)2)h/2

÷

+ < 1+(

and

1e12)(' + +


are also equivalent.

the following extension Step 2. For the transition to .0 = H'(lR") must be investigated, where x = It":

9 Regularity

216

(4,u)(x):—u(x)

for

:=

(4,u)(x',

+ u(x',

<0.

for

We list the properties of 4, as

Exercise 9.1.13. Let the coefficients

of 4' be selected as the solution = (_l)k (0 k L — 1). Sho

a,..(vk +

the system of equations

that (a) U E (b) 4,E

yields 4,u E

for k=0,1,...,L.

(c) The operator adjoint to 4, reads

= u(x', xn)+> (d) (8/ôx,2)k(4,*u)(xI,0) = 0 for k = (e) for E 6.2.43.

for a =

(f) 4,E

for x',,>

_VXn)]

0,1,..., L —2 and U E k = 0,1, L — 1. Hint: cf. Corollar . . .

1

,

—L,2—L,...,0,l,...,L.

Step 3, (provab. L results from D° E via continuation arguments from Remark 6.3.14b) so that remair to be shown. Step 4. luL I4,uI. 1114'uIIL is true according to Step 1 of the proof. Tb

inequality llJ4,uJjL Cfr., which would finish the proof, reduces to for IcwI = m,

UE

= m. For

Let

f

L

S

i.

one verifies

[(—v) "u(x

=

,

+ (—i)

u(x', —xe/v)]

As in Exercise 131 one shows that

otherwise, E

over to real a [1 + m — L, L — m] except for the cases — $ E IN (i.e a = —1/2, —3/2,...) (cf. Liona-Magenes [1, p. 54 ffJ). Let L 2m + 1 an V

one infers from

E ffm—'(JRTh) Since

(Dau,

the estimate I(Da4,u, v)oI =

for all V E

C :=

It?Im_s,

I

and therefore (16) with

=

9.1 Solutions of the Boundary Value Problem in F1'(ø), 8> m

217

U

PROOF. (of Theorem 11) (a) First let k s = 1. The proof of Theorem ,n — 1) and implies 3 can be repeated for the differences U = the existence of the derivatives Ou/Ox, E j n. Thus one has = m except for a = (0,... ,0,m). for all (b)Weset . m, is the coefficient from the bilinear form (5). The remainder of the proof runs as follows. In part (c) we will show that

where

$Fa(v)I S CIvIm_i

m, V

for

(9.1.17)

= (w,D°v)o = (_l)rn(Daw,v)o, (17) means that E and IDaWIl_m C for (at = in. According to Lemma 12 it follows that w E The coercivity of a(.,.) implies uniform ellipticity i.e., (cf. Theorem of L = Hence it follows from 7.2.13). For = (0,... ,O, 1) one obtains aoa(x) Accordand wE E J'yJ = 1, that Dàu E ((a( m,a a) belong to ing to part (a) all other derivatives anyway so that u E has been proved. = 0, (c) Proof of (17). For each a a there exists a with ('y( = 1, o < y a (component-wise inequalities). Integration by parts yields Since

Fa(V) =

dx

+ a&&(D

J

—

and thus

for v E


with Ca :

CEIUIm

for all v

+

Here we used the fact that DTu E

aceording to part (a). Since a.

is dense in (17) follows for a There remains to investigate a a. We write

F6(v) = a(u, v) — a(u, v) with &(u, v) =

where E'

dx,

J

the summation over all pairs (o,f3) there exists with

represents

(&, a). For each

h'I=l, As above, one integrates each summand with

= in by parts:

9 Regularity

218

dx

= —

dx

J

—

dx

J C=

for v E gether one obtains Iâ(u,v)I

C(u). Alto-

CIVIm_l. Together with Ia(u,v)I =

IfI_rn+iHrn—i, (17) also follows for a = & (d) By induction for (k CIvlm_k, and from this u E

2,...) one proves in the same way IFa(V)I For real a >0, 8

one proves correspondingly 1F0(t')I

1/2,..., rn— 1/2,

CIVIm_a and hence u

The generalisation of Theorem 11 to inhomogeneous boundary values

reads as follows.

Theorem 9.1.14. Let the biliriearfortn a(.,.) from (5) be For an 8 > 0, a {1/2,. . . , m — 1/2} either let (6) hold if a = k E IN, or be the weak solution of the inhomogeneous (12), ifs N. Let u E Dirichiet problem

a(u,v) = (f,v)o 8'u/8n1=çoj

for all v E

(9.1.18a)

forl=0,1,...,m—1,

(9.1.18b)

where

f

(0 I m

(9.1.19)

and satisfies the inequality

Then it belongs to

IUIm+. C.

— 1).

[if i...m+a +

÷ IUtm].

(9.1.20)

PROOF. For m =

1 Theorem 6.2.32 guarantees the existence of ito which satisfies the boundary conditions (18b) (for m> 1 ci.

it — ito is the solution of the homogeneous Wioka [1, Theorem 8.8J). w problem a(w,v) = F(v) (f,v)o — a(uO,v) (cf. Remark 7.3.2). Theorem 11 may also be carried over to the right-hand side F(v) under discusaion here U and yields w E Hm+a(lRfl) (instead of (1'

By similar means one proves

Theorem 9.1.15.

(Natural boundary conditions) Let the bilinear

form a(.,•) from (5) be N, or (12) ifs 8 = k

N. Let it

Fbr 8 > 0 either let (6) hold if be the weak solution of the

problem

a(u,v)= 1(v) =1 9(x)v(x)dx+jp(x)v(x)df' for ally €

9.1 SolutIons of the Boundary Value Problem in H'(S?), s> m

219

where H'—1

for

( I-I

>

H' 1/2(f)

—

fors<1J' and satisfies the estimate

Then u belongs to

(9.1.21)

It4i+a C.ELt9IIH.—1 + 1p1...t/2 +

If a boundary value problem is in the form (1): Lu = g, Bu = with B = then according to Theorem 7.4.11 one e > 0 on I = bTV+bo, an associated variational formulation and apply Theorem 15. can find

9.1.4 Regularity Theorems for General 11 C JR" The following theorems show that the above regularity statements also hold for (1 C IR" if Q is bounded sufficiently smoothly.

Theorem 9.1.16. be

Let Let

a

8

for some t 0. Let the bilinear form (5) 0 satisjlj

ift€14;0s
for all a,13,'y with I-il max(O,t + (III — m), if t E IN,

E

(9.1.22)

for

e

with f

if t

IN.

of the problem

Then each weak solution u E a(u, v)

otherwise,

> m — 1,

f f(x)v(x) dx

for all v

Hm+'(a) belongs to

and satisfies the estimate

C.

(If I—m+. +

(9.1.23)

For inhomogeneous boundary conditions

=

with

the statement a and the estimate (20).

instead of u

•,m —1)

Hm(a) implies the statement a

PROOF. (a) Let {U':i = 0,1,...,N} with U' CS? be a covering of S? as with in Lemma 6.2.36. Let C U', be E the associated partition of unity from Lemma 6.2.37. There exist derivatives

which map U' (i 1) into such that a'(OU' n I) c By contrast, U° lies in the interior of 5? so that In 81)° =0. The solution u

a'

Regularity

220

can be written as part (c) xøL for i 1.

In part (b) of the proof we will treat xou, and in

(b) We set

(i=O,l,...,N)

d1(u,v) :=a(Xgu,v)—a(u,Xsv) and wish to show the estimate (u E

Ido(u, v)l CdIUImIVlm_a for

8

VE

8

J dx.

= XoDatj+ lower denvatives of u, one has

Since

=

—

161

(9.1.24)

= 1. Each suminand of do(u,v) has the form —

with

1)

with

161 2m—1. One integrates

161

= m by parts, and obteins a bound CIUImIVIm_i, whence (24) follows.

a way that the correThe coefficients Denote the resulting bilinear form sponding condition (22) Ia satisfied on v). Since Xo E C°°(.Q) has a support supp(xo) C U°, the extension of by poses no probleres. We may formally Xou through Xou(x) =0 for x since only the restriction of v to U° is of define do(u,v) for v any consequence. By (24) do(u, v) can be written, for a fixed u E in the form

do(u,v) =

With d0

Idols_rn

CdlUlm.

Xou is the weak solution of

v) = a(xou, v) = a(tt, xov) + do(u, v) = (f, xov)o + (d0, v)o for ally Hm(R"). = (thus too

Theorem 3 (resp. 8J proves xou IXoulrn÷.

CEIXOII_m+. +

l4I-m+. + iXOt&Irnl


wheresisstillrestrictedto8

+ lUlmI, (9.1.25a)

1.

(i= 1,...,N)shows

(c) The same reasoning as for

= (xJ+di,v)o By assumption the maps a1

:

for V E

U1 —'

1d11_rn+s S CdIUIm.

and their inverses belong tI(x) for i = a1(x), that is

Put u(i) ü=uo(a')'. Inasimilar way define to

[reap.

and

In

9.1 Solutions of the Boundary Value Problem in H'(I1), 8 > m

=

221

>J

one can replace the derivatives with derivatives with respect to the new coordinates, since t 0, and thus obtain the form

= which again is and whose new coefficients &ap satisfy the conditions corresponding to (22). As in (b), Ôap can be continued to such that the resulting bilinear form is One can apply Theorem 11 to

for all ii E

= Which yields

and

E

Oa[IJI_m+e +

+

Transforming back (ci. Theorems 6.2.17,

C,

+

yields

I-m+. +

C, Ill I—m+. + CdIuIm + CIUIm

and thus 1

i s N,

$

1.

(9.1.25b)

(d) (25a,b) hold for $

1. Since Hm+. = estimate (23) has been proved for a 1. If the conditions of the theorem allow an s E (1,2], one proves (23) as follows. Since U E has been proved already, one can estimate the forms after further integration by parts via fr4(u,v)I CdItiIm+lIVIm...,. Accordingly, = with d1 Hm+'((?) and CdJUIm+l. If one inserts the above estimate (23) for s = 1, one obtains (25a,b) and hence also (23) for 1 <s 2. FUrther induction yields (23) for admissible a E (k, k + 1]. (e) The case of inhomogeneous boundary values is treated as in Theorem 14. U Analogously one may prove

Theorem 9.1.17. (Natural boundary conditions) Let 1?

with t 0.

Exercise 9.1.18. Transfer Corollaries 5,6, and 7 to the situation of Theorems 16 and 17. What conditions on a must be added?

222

9 Regularity

Corollary 9.1.19. Let 12 and the coefficients of a(.,.) satisftj the conditions in Theorem 16, rvsp. 17. An e genft*nction, i.e., a solulsonu E V (V = Hr((i) of Theorem 17) of in the cose of Theorem 16, V = H'(Il) in the

a(u,v)=0 forallvEV,

(9.1.26)

PROOF. Use Theorem 16(17) forf =O(andçp=O). According to Theorem 6.2.30 (Sobolev's lemma) one obtains sufficient conditions via for u to be a classical solution from D The minimal conditions for U E result from a different theoretical approach, which essentially goes back to Schauder. The following theorem, for example, can be found in Miranda [1, It shows the regularity of the operator L in (5.1.1).

Theorem 9.1,20. Let k 2, 0 < A < 1. Let .0 E

be a bounded operutor L = E a be uniformly elliptic in (2 (i.e., (5.1.3a) holds). Let and E Then the boundanj value problem Lu = E f domain. Let the

in (2, u =

f

on F either has a unique (classical) solution u there exists a finite-dimensionaL eigenspace {O} such that for EC all e E the following holds: Le 0 in .0, e = 0 on F. If a 0, the first 'p

alternative always holds.

The condition (2€ in Theorem 16 is stronger than necessary. For the Dirichiet problem the Lipechitz continuity of F is already sufficient to obtain the following result.

Theorem 9.1.21. (Ne&s 111) Let Si bilinear form (5) be

1/2 t > The coefficients

E

weak solution u a(u, v) with I E

C°'1 be a bounded domain. Let the

Let the following hold:

=

a

>

0.

if t$I = m. Then the

must belong to of the problem

/ f(x)v(x) dx

belongs to

for dlv

E

and satisfies the estimate (23).

The condition of can be replaced by that of uniform ellipticity (7.2.3) (cf. Theorems 7.2.11, 7.2.13). The statement of Theorem 21 cannot be extended to 8 1/2 since then u Hr'(Si) would contain another boundary condition.

9.1 Solutions of the Boundary Value Problem in H'($2), 8> m

223

(cf. The proof of Theorem 21 uses an isomorphism R = R related to and also between Hr(fl) and proof of Theorem 8) between such that the form b(u,v) := a(Ru,Rv) is and It is necessary to prove that b(u, v) := a(u, R2v) is also

/

We know f H_rn+8(.11) implies R2f E a(u, v) = v)o is also a solution of a(u, = b(u, so that u E follows.

Eath solution of = (1,

(f,

9.1.5 Regularity for Convex Domains and Domains with Corners A domain .0 is convex if with x', x" E.0, x' + t(x" — x') belongs to I? for all 0 t 1 Convex domains in particular belong to C°", but permit stronger regularity statements than Theorem 21. .

Theorem 9.1.22. (Kadlec [1]) Let I? be bounded and convex. Let the bilinear form (5) be Lipschitz-continuous:

Let the coefficients of the principal part be

for all

=

=

1;

for the remaining ones let the following hol&

for all a,f3,-y withy

E

Then every weak solution u E a(u, v)

of the problem

clx for all v E = j f(x)v(x)

L2(Q) belongs to

with f

IaI+I/31 <1.

fl HJ(Q) and satisfies the estimate

C1[IfIo + luli).

(9.1.27)

The constant C1 depends only on the diameter of.0. In Section 8.4.2 IP-regularity was required. A generalisation of (27) in the form of for the biharinonic differential equation with m 2 is known for convex polygons (cf. Blum—Rannather [1)). For the Poisson equation

the inequality (27) can be reformulated as follows.

Corollary 9.1.23. For the solution of the

Poisson equation —Llu =

L2(Q) in a convex domain.0 with u =0 on F, the following holds:

fE

/ Iflo.

(9.1.28)

tal=2 As

112(Q)

in Lemma 8.4.1 one shows that the left side of (28) is a norm of

fl

which is equivalent to f

12•

Thus, (27) follows. As a model

9 Regularity

224

for the proof of Theorem 22 we carry out the proof for Corollary 23 for the case (2 C PROOF. (a) First let us aesume that the convex domain is smooth: (2E C°°. According to Theorem 16 L2(.11) has a solution u E H3(i?) n = H0'(Q). We want to show

/

:= >

for all u E

(9.1.20)

IaI=2

it suffices to prove (29) that for all u in the dense subset {u E C°°(D): u=Oon F) Integration by parts yields JIz'u12&rdv = =

=

+4k,

where n = (nm, is the normal vector. The tangent direction is given by t = (—n11, n1). Then is the negative tangential derivative Now and can be expreseed in terms of and Since both is and

2j(uxzn*, -

=

=

+

Integration by parts of the second suinmand yields

-

j

-

=

/

-

The bracketed expresauxi in the last display is the curvature in x I', which for a convex domain is always 0. (28) has thus been proved. (b) Every convex domain (2can be approximated monotonically by convex Weinterpret V,, :={uE supp(u) C Th,} as Ritz-Galerkin space c HJ(a). Each U E Cr(.f?) lies in for sufficiently large gi. With C000(Q), is therefore also a dense subset of Fbr every v, the RAtz-Galerkin problem provides the solution of E = 0 on 13OV. In = in may be continued by = 0. Theorem 8.2.2, which is also applicable in the case dim = oo, proves -. 0, where is is the solution of

f

-

9.1 Solutions of the Boundary Value Problem In H(Q), > m

= on

will show that for every f in Q. Theorem 9.1.26 C V,,, v For each yE

to

if

fl5

the restriction of u we have

=

= lim a,,

this one infers El*.2 Ill

il/ iiL2(fl) (ef. (29)), and obtains (28). on asubspace

What role the H2-regulazity plays for the VhC

Exercise 9.1.24. Let the subepace Vh C inf{Iu —

vlj:v

Vh}
for all u

satisfy fl

(cf. (8.4.14)). Let the Poisson problem be 112-regular (aaxrding to Theorem 21 convexity of 11 is sufficient). Let Qv: be the orthogonal —, Vh C Show that there exists a Ci such that projection on Vp, with respect to —

CihluIi

for all u

Hints: (1) With the Poiseon problem the boundary value problem

f

= in Q and u =0 on F are also H2-regular (cf. Remark 2). The corresponding bilinear form a(.,.) is the scalar product in (2) Qv agrees with the Ritz projection Sh for a(.,.). (3) Use Corollary 8.4.12.

In connection with finite elements one often considers polygonal domains a. Since polygons belong to C°", the Dirichiet problem, according to Theorem 21, is with 0 8<1/2. If the polygon is convex (i.e., if the inner angles are then as in Theorem 22 one has H2-regularity (m = 1). One obtains results between H3/2 and H2 if the maximal inner angle of the polygon lies between and 2ir (cf. Schatz-Wahlbin [1)). If a has a reentrant corner the boundary value problem can no longer be H2-regular (cf. Example 2.1.4; with an inner angle a the solution belongs to for 8< 7r/a). Stronger regularity properties may be obtained, however, if special compatibility conditions are satisfied in the corners (cf. Kondrat'ev [1J).

Example 9.1.25. Let u be the solution of the Poiseon equation =f in the rectangle a = (0,1) x (0,1) with u = 0 on F. Only for 8 <3 does

f

H—2(a) lead to u E H'(fl). Under the additional compatibility condition that f vanish at all corners, f(0,0) = 1(0,1) = /(1,0) = 1(1,1) = 0, however, one can also conclude, for I with 3 < 8 <4, that u E H'(a).

226

9 Regulanty

9.1.6 Regularity In the Interior Up to now all regularity statements have referred to the entire domain (1. Instead one can also investigate the regularity of the solution u in ho Cc h The following theorem shows that the answer depends neither on the smoothness of the boundary nor on the type of boundary condition.

Theorem 9.1.26. Let ho CC hi C (1 and 8

0. Let the bilinear form (5) be condition (22) with £7 be replaced

For the coefficients

byhj and withtsEt4 ort>s. LetuEVCHm(Q) beaweaksolution of the problem a(u,v) = for all v E V, where the restriction belongs to Then the restriction of u to belongs to and satisfies


J?i,

IH—"'+(fl1)

+

I fr'I

PROOF. A special covering of .0 is given by U° = U1 = obtain the assertion from Part b of the proof of Theorem 16.

9.2

Thus we U

Regularity Properties of Difference Equations

The convergence estimates for difference equations in Section 4.4 read, for under the condition that u example, Vu — uhlko or u This regularity assumption is frequently not satisfied (cf. Ex-

arnples 2.1.3-4). A comparison with the error estimate lu — S Ch21u12 for the finite-element method suggests that similar estimates also exist for difference solutions. To obtain the latter, one needs to replace the stability estimate C (or C), whith corresponds to L2(fl)), by stronger estimates whith correspond to E L(H1(fl), or

L(L2(h), H2(h)).

9.2.1 Discrete H'-REgularlty For £7 C Ut's the following grids are defined: Qh

:= {x

Z},

v,h,

A grid function vh defined on

oh := i7flQh.

(9.2.1)

is extended toQ,, by vh =0:

Vh(X) = 0

for x

Qh\Qh.

(9.2.2)

The Eudidean norm is now called the Li-norm: r

Ivalo :=

:= [hm

ItJh(X)12J

xEQk

(9.2.3a)

9.2 Regularity Properties of Difference Equations

227

It comes from the scalar product (9.2.3b)

vh(x)wh(x).

XEQh

with the norm

The discrete analogue of H0'(fl) is n

1/2

+

:= IIVhUH1 :=

(vh

is the forward difference in the

where

Iv,d_i :=

= 0 on Qh\a4)

(9.2.3c)

direction. The dual norm reads

:= sup{I(vh,wh)oI/IwhIj:wh = 0 in QhV?h}.

The associated matrix norms

=

(9.2.3d)

=

ILhI1.—_1,

etc., are defined by := sup{ILhvhL/Ivhlj:0

satisfies (2))

for

i,j

{—1,0,

1}.

(9.2.3d')

Exercise 9.2.1. Show that (a) is the spectral norm of Lh (cf. §4.3). (b) The following inverse estimates hold: for

(9.2.3e)

The matrix Lh yields the bilinear form

ah(uh,vh) := is said

to be

(9.2.4a)

if a CE >0 exists such that

uh) CHIUhI?

for all uh and all h >0.

Correspondingly, 04(.,.) Is said to be Hg-coercive if there exist CK E IR with

ah(uh,uh)

for all

—

and all h >0.

(9.2.4b)

>0 and (9.2.4c)

As defined in Sect. 4.5, L,, (reap. ah(•,.)) Is said to be Li-stable if

0. We call Lh

if

C1 Exercise 9.2.2. (b) = (c) If

is

(9.2.4d)

(a)

for all h > 0.

implies Li-stability. for i,j E {—1,0, 1).

then so is

(9.2.4e)

9 Regularity

228

are stable with respect to (4d) follows with Co :=

(d) If Lh and then

C1,

i.e., IlLalLo

implies

(e)

The following statement resembles the alternative in Theorem 6.5.15.

Theorem 9.2.3. If

is

and If

then Lh is

is

also

PROOF. (a) Let

= fh such that ah(uh,uh)

=

(uh,fh)o.

Coercivity provides Eah(uh, uh) +

= I(uh, fh)o + C1uh109/CE

+ CIuhIol(uhIo. On

the basis of

the stability estimate IUhIO From thisone infers C (cf. Exercise 2b).

(b) Now let Lhuh

=

we obtain

C and hence

to Part a, One has CelfhI_1. = I(fh,uh)oI lfhI_lIUhIl through +

fh. Aecording

By estimating

one obtams the coercivity

C1 with

and hence

:=

U

+

Instead of the Li-stability in Theorem 3 one can also aseume the solvability of the continuous problem and a consistency condition (cf. Corollary

11.3.5).

By analogy with Lemma 7.2.12 the following may be proved.

Exercise 9.2.4. If

is and if öLh with IaoI, (cii conat, contains at most first differences, then Lh + öLh is also Aecording to Exercise 4 it is sufficient to investigate the principal part of a difference operator as to its coercivity.

Example 0.2.5. Let i7 C JR2 be bounded. On 0h let Lh be given difference

(Lhuh)(x, y) =

for (x,y)

by the

method (5.1.18) (with a = 0):

ô;afl

+

y) +

y+ (9.2.5)

.17h, where u,4(x,y) = 0 for (x,y)

e >0 in Th. Then

ah(.,.)

is

and

Qh\Qh. Here, let —a11, is

229

9.2 Regularity Properties of Difference Equations

PROOF. For arbitrary vh, VJh defined on Qh the following rules of summation by parts hold; (9.2.6)

=

= Hence it follows that ah(tLh,uh) = (Lhuh,uh)O

=

—

(a11

(.+

•_

+

(an

101.

and in Lemma 6.2.11 one proves that for bounded $2 the norms are equivalent (uniformly with respect to h) + results from Exercise 2e. so that (4b) follows. The As

I

Exercise 9.2.6. Let

be bounded. Let the equation

1? C

+

+

+

=I

Let by the difference stars (5.1.18/19), where Uk = 0 in the differential equation be uniformly elliptic in <0, allan — a12a21 > 0. 1\irther, let C°(TJ) hold. Show that for sufficiently —e(aij + for all 4 > 0, is small h the associated matrix Lh is Hint: For sufficiently small 4 the following holds: be given on

—all

(

li

(

h\2

for all di, d2 E JR.

The difference methods constructed so far remain if differences of lower order are added (ci. Exercise 4) or if the principal term + ... is replaced by + ... with Oh E C1(Th). The above difference methods are described by the same difference operator regardless whether the grid points are close to or far from the boundary. The homogeneous Dirichiet boundary condition is discretised by (2): =0 on If one wants to approximate the boundary condition more aceurately, one needs to select special discretisations in the points near the boundary of Qh (cf. Section 4.8.1, 4.8.2). One thus obtains an irregularity which makes a proof of as in Example 5, difficult. We begin with the one-dimensional case.

Lemma 9.2.7. Let L,, be the matrix of the one-dimensional Shortley- Welter diacretisation.s of —u" = on .0, u =0 on OS?:

f

9 RegularIty

230

2

h2 .(..!flh(X) —

Sr(Sr + 3j)

SiSr

I.

forz

(Ih, where

0<

Uh(X+ 8rh)

2

—

+

)

if

= 1(x) (9.2.7)

811. For

< 1 (of. (4.8.7)), u(e) =

arbitrury 5? C JR there holds Thus for bounded 5?, Lh is

PROOF. (a) First, let

u,,(x — Srh)1

0 on

be aasuined to be connected. Let the grid points of

5?hbexi:=xo+lhE5?forl=O,".,k>O.Lettheboundaryporntsof5?be —

s1,0h

and Zk ÷

with 81,0, 8r,k E (0, 1). The other factors of Equation

= 1. If one takes into account vh (7) in x = x1 are obtains the following identity:

0

on IR\I? one

(vh,Lhvh)O :=

+ h1 {vh(xo) [(! _2) vh(xO) + (i —

=

+ vh(xk) [(i- —2) Vh(xk) +

(i —

i..)

1

2

)

Vh(X1)]

Vh(Xk_1)]}i

and s,. := Since vh(x0) = and vh(z1) = because of vh(x_1) = 0, the first sununand inside + the bracse can be written as where

:=

2) Vh(ZO) +

—

=h

l—8j [8j+2 1+81 1

For a :=

(i

2

I ÷ SI )

Vh(Z1)] (9.2.8b)

8 +Vh(x_1)2—c vh(x_1)D + vh(zO)

— use the inequality —a/3 with = (Si + 2)/SI and note that the function s(1 — .)/L4(1 + 8)(2+ is 1)/2). The second summand inside bounded by 0.018 (maximum at 8= the braces is treated analogously and yleith

(3 :=

(vh, Lhvh)O

+

—

(9.2.8c)

(b) In Part (a) it was aesumed that k > 0. For k =0 one obtains (vh, LhVh)0 =

SjS,.

BISr

that (8c) also holds. (c) Let 5? be arbitrary. Let the components of connection be = (a1, (i Z) with b1 Let be the discretisatiori matrix aseociated with with support in S1h = QhflQcan be written where

so

9.2 RegularIty Properties of Difference Equations

=

0

Let

outside

:= 11h fl

231

be the first and last grid points of

and

The statement follows from :S

=

2(vh, Lhvh)O.

Theorem 9.2.8.

Let (1 C 1R2 be boundeL Let Lh be the matrix associated with the Shortley- Weller discretisation of the Poisson equation (cf. Section 4.8.1). Then Lh is

PROOF. Let be the portion of the z differences Ii' differencceJ such that Lh = The restriction of to a "grid row" {(zih,y) i?:v + Z} (y is fixed) corresponds to the matrix Lh in Lemma 7. Thus one obtains (vh, one

With the analogous inequality (vh, obtains 2(vh,LhVh)O

I

I

li——i

If

Theorem 9.2.9. Let fl C JR2 be bounded. Let the Poisson equation be dixcretised with the aid of the five-point formula (4.8.14e) at points far from the boundary and by inteipolation (4.8.16) at points near the boundary. The associated matrix is PROOF. The corresponding, one-dimensional formulae, except for the scaling factor ÷ 2, agree with (7) so that the proof of Theorem 8 can easily be carried over.

Exercise 9.2.10.

Show that for a convex domain 4? C JR2 the matrix Lh of the Shortley-Weller discretisation satisfies the inequality

(vh,Lhvh)o

+

Theorems 8 and 9 can be strengthened in the following way:

Corollary 9.2.11. Let Lh be as in Theorem 8 or9. Let Dh = diag {d(x):x be a diagonal matrix with d(x) :=

1}

where

E

(x E 4?h),

s,., sc,, come from (4.8.7), respectively (4.8.16). For points far from the boundary, x E S?h, evidently d(x) = 1 hoLds. The matrix

LI

.

h

h,

232

Regularity

9

which belongs to the re-sa4led system of equations

=

:= Dhfh, is also

C. PROOF. In the case of the Shortloy-Weller method one needs to combine the correction terms (8b) for the x and y directions. The result is that they remain

sothat

? e

> 0. One obtains an analogous estimate for the difference method in

I

Section 4.8.2.

9.2.2 Consistency In the followrng we carry over the estimate lUll C'hlfk, ChIul2 which holds for finite-element solutions, to difference methods. To this end

one needs to prove the consistency condition := flL11R,1 —

—

CKh

(9.2.9)

in

E2u E

for suitable restrictions L2(Q)

R11: H2(S?) fl

To construct the restrictions we first continue u According to Theorem 6.2.40c one assumes

E2u = for all U with

on (1,

(9.2.lOa)

11E2u11H2(n.2)

An analogous continuation E0L2(a)

E

f := Eof = f

L2(1R2)

I E L2(fl)

on (1,

(9.2.1Db)

7 = 0 on 1R2\Q. Let the averaging operators

is given, for example, C0(Wt

) be defined by

= h1

= h1

nV2 I

J—h/2

The restrictions R,,, :=

(9.2.11)

are chosen as follows:

y) =

i.e.,

h2 I

I

J—h/2 J—h/2

with

=

F,2u

for (x, y) E

U(z +

y+

dir7

(9.2.12a)

i.e., (&11f)(x, y)

h4

/1112 /11/2 /11/2 fh/2 —11/2

with

—h/2

7 = E(Jf

—11/2

J—h/2

for (x,y)

(9.2.12b)

9.2 Regularity Properties of Difference Equations

The characteristic properties of the convolutions

Exercise 9.2.12. Let & It4x + h/2, y) — u(x

be

233

4 are the subject of

the symmetric difference operator (&ti)(z, y)

h/2, y)]/h;, is defined analogously. Show that

—

C, II4IIHk(1t2)+—Hk(I(2) C (in particular for

k=O,±l,2). :=

(c)

< 11v11L2(R2) for all v

(d) For a E C°"(1R2) holds

—

=

Here (e)

L2(1R2).

=

and

for ii,

—

E IN.

juj IHk+1(Ita) for u E

(f) flu — Consider the differential operator

L=

+ 1,3=1

a4(x)O/8x4 + a(x).

(9.2.13a)

4=1

First we asswne that S? =

and discretise L with the regular difference

JR'S

operator

+a(x)

Lh =

(9.2.13b)

with an arbitrary combination of the ±-signs.

Lemma 9.2.13. Let £2 = IR". Let

Let Land Lh be

given by (13a,b). Then the consistency estimate (9) holds.

PROOF. To simplify the notation let us assume that n =2. Let

=

= with responding to shows that

—81

(cf. Exercise 12e,b). Let the term Corin (13b) be, for example, all(x; Exercise 12a

= = with &2(z, y) := ux(X,y)] and 12c,b). Since approximates

=

y) =

h,

—

—

=

y) —

h, y) — u(x,

y)J =

— 4, y) — (cf. Exercise

the error term 52 does not appear if one also by

= +

Finally one obtains

+

— —

uZju

=

—

234

9 Regularity

< — = Exercise 12c,f). Putting this altogether one obtains with

(cf.

= Si +

—

+ 63).

For the first error term the following holds: 1161(1ff_i

For arbitrary Vh (vh,

(9.2.14a)

C'hJuI2.

we have (63

=

+

(9.2. 14b)

[alIvhi, 63+

C°"(Jt2) it follows that 80 that IIOIJCO(R3)(IVPIIIH1

From a

+

+ Ss)II,çi = suni Kvs,

+

VkEHL

Cfl63 +

(9.2.14c)

C'hIuI2.

From (14a.,c) we have

Ch.

—

(9.2.14<1)

Analogously, one shows 110

—

-

For

norms . larger norm

Ch.

14a

(9.2.14e)

similar reasoning result8 in an 0(h)-estimate for the and

Both are upper bounds for the

1IH1,_H2(R2) such that

Ch.

(9.2.14f)

Ch.

(9.2.14g)

Likewise — RhaIIj,_1I..ff2(Ra)

Statement (9) follows from (14e,f,g).

U

When generalising the consistency estimate to more general domains 11 C

the following difficulty arises. The entries of the matrix Lh according to

(4.8.7) or (14.8.16) are not bounded by Ch-2. Rather, at points near the boundary the inverse of the distance to the boundary point enters, and this di8tance may be arbitrarily &nall. One way around this, would be to formulate

the discretisation so that the distances between boundary points and points near near the boundary remain, for example, h/2. A second possibility would be a suitable definition of Ra so that the product appearing in

9.2 Regularity Properties of Difference EquatIons

235

(9) can be estimated (cf. Hackbusch [2)). Here we choose a third option: Lh is = DhLh from Corollary 11. replaced by the re-scaled matrix

Theorem 9.2.14. Let .0 E C2 (or convex) and bOtind&L For the discreti,ation ofLu = forL = —4 on.0 withu = 0 on F use the discretisation Lhuh according to (4.8.7) or (4.8.16). Let = DhLh be defined as in Corollar!J 11.

f

Then the consistency estimate (9.2.15)

ç Ch

—

holds.

Here, the matrix L4 from (4.8.7/16) is only taken as an example. The proof will show that the estimate (9), respectively (15), also holds for other x near the boundary, represents a second difference. First, Lh if two lenunas are needed. be the set of points near the boundary. If vh isa grid function Let 'yh C defined on the function then we denote by (vhf,,.j(x)

vh(x)

for x E l'h,

(vhi.Th)(x) = 0

for XE

Lenuna 9.2.15. Let .0 E Ce" be bound&L Then there exists a C = C(.0) independent of 4, such that IVhlmIO ChIvhII.

PROOF.

(9.2.16)

.0 E C°'1 follows: there exist numbers K E IN and h0 >0 such with h h0, not all grid points {x+ (vh, ash): —K i',

that for all x

K}lieinQ.Foreachx€ .0aselectapair(&'o,p0)EZ2with—K and x + poh) K) are defined by

At this grid point x the functions (—K v, p := vh(x), := 0 for (v,p) VhI.r,, is a decomposition with = Vh(X) or

Therefore

=0. we

Without Ices of generality let us assume that v > 0 and p >0. For x E Qh define the chain X0 = X,

X1

= x +(h,0),

...

= x+ (vh,h),

X

=

,

According to the definition, either

=

0

holds or

.0, i.e.

0 (ci. (2)). In both cases we have <

=4

+

—

+ ... +

+ +

—

+ . .. +

+ ..

9 Regulanty

236

hflvj

and thus

that (16)

= of x 'Yh This holds in particular for convex

already has a direct neighbour in

In most cases x E such

Summation

+ Ii')

yields estunate (16) for

over

with C =

follows

domains.

Lemma 9.2.16. Let

be defined by

(12a). Let E2 satisfy (lOs). Then we

have

f',u

for all

+

:=

where

PROOF. (a) Let Q =

x€ 1R2x E (—k, (—n) x

J v(x)dxI

)v(O) —

H01((I) fl 112(11), (9.2.17)

x 112(Q) one shows

Fbr v

(—i,

+

C i/j(vz2x +

since the left-hand side vanishes for linear functions u(x, y) =

+ fix +

Theproofissiniilartotbeonefor(8.4.2).TransformingfromQtoQh x

gives

Ih2v(O) —

(b) Let

f

v(x)

+

+

dxdy.

(9.2.18)

=0.

U

F. Statement (17) follows from (18) with

v(x) :=

÷ hx),

+ hx) =

since

PROOF. (of Theorem 14) (a) Inequality (15) is proved if j(vh,

for all vh E

Ch

—

and u

with IVhIl =

(9.2.19a) = 1. lb this end

is split into In part (b) we show EL'hRh —

Cih,

(9.2.lOb)

C2h,

(9.2.19c)

The other steps of the proof, (c) and (d), yield —

such that (19a) with C = C1 + C2 follows. (b) Lemma 15 shows Ivklo

= C3h.

9.2 Regularity For

one obtains

of Difference Equations

= 1 + fr4j1. The inverse estimate (3e)

+ Iv;111

yields Ivji Ch11vj0 CC3,

237

thus

C4.

(9.2.19e)

Let Lh be the (regular) difference operator on the infinite grid

=

ph): ii, p

Z}.

Since the support of is Qh\'yh, (vi', L'5Wh)O = Lhwh)o holds for all wh. Furthermore, Dhwh)o = (vi', wh)o. This proves the first equality in

—DhRhLJU)oJ =

—RhLJii)oI C5hC4C6 := C1h,

S

:= E2u is the continuation of u to 1R2. Further inequalities result from Lemma 13, (lOe), and where

fltIjIH2(R3)

C6jJnflJp(fl) = C6

(9.2.19f)

(cf. (lOa)). Theorem 6.2.40c guarantees the existence of an extension U with

(19f) if 11

C2. Another sufficient condition for (19f) is the convexity of (2.

(c) The left side of (19c) splits into (va, and The first part is estimated in part (d). Exercise 12c and (19d) yield the second term CahIu(2 =

CGh.

(9.2.19g)

(d) We set wh :=

Since the support of is contained in 'Yh, we have = (vi, w,j0. contains differences with respect to the x and y directions. Accordingly we write wh = + i4. In the following we lunit ourselves (i) to the Shortley-Weller diacretisation, (ii) to the term and (ii) to the case that Ii,,

(i.e., 8r=

1),

x'

The other cases should be treated analogously. We set

:=

E H2(E2).

The Shortley-Weller difference in the x direction reads

=

a,)

where d,,, is the diagonal element of

variables has

F (for example, the form

Since in the equation L'huh fh the = x'), have already been eliminated,

h2

9 Regularity

238

The factor

by 4h2.

+

due

to the definition of Dh, remams bounded one infers c

From Lemma 16 and Kh/2(x')

(9.2.19h)

—

The second divided difference

in x = (x, y) can be

by

at

LIx) h3r

with

g(t) =

(2(t—h)/(h2+h2sj), —2(sjh + t)/(h2si(1 + sj)),

for —sjh t 0, otherwise.

From this one infers 11/21

I

/

2h'

J

{jh

11/2

/ (9 2

+

(cf. (6.2.5a)). From (19h,i) and the corresponding estimate for

one obtains

such that

lwh(x)I

=

1s2

IWh(X)12 C? xE7h

(3C7C6)2

From this follows C34C8

(19g) and (19j) yield the required inequality (19c).

C9h.

(9.2.19j)

U

Remark 9.2.17'. The proof steps for Theorem 14 can be carried out in the same manner for more general difference equations (for example, with variable coefficients, as in Lemma 13).

9.2.3 Optimal Error Estimates In the following we compare the discrete solution uh = with the striction := Rhti of the exact solution ii = From the representation

= = one

— Rh!)

immediately obtains

(fh —

9220 —

9.2 Regularity Properties of Difference Equations

239

f. Let the = fh be chosen so that

Theorem 9.2.18. Let u E Jf2(Q) hold for the solution of Lu nght-hand side fh of the discrete equation Ifh — Rail_i C,h.

(9.2.21)

and if the consistency condition (9) holds, If, furthermore, Lh is satisfies the ermr estimate then IUh —

ufl1

C1(C1 + CxIuI2)h.

PROOF. l"h

(9.2.22)

U

—Rail_i +ILaRa —

Corollary 9.2.19. (a) !neqsiality (21) is satisfied in particular if one chooses

fa := R,1f. (b) The choice fa(x) := f(x) forx E (ci (4.2.6b)) leads to (21) if f E In these eases the following even holds: or I E —

Rail0

reap.

—

Raflo Ch2Jf

—

(9.2.23)

12.

(c) In Theorem 18 one ean replace the H)-regularity of DhLh (cf Corollary 11), (9) by (15), and (21) by IDa/h — DaR,,f I_i

C1h.

The proof of (23) is based on the inequality (18).

by that of L',, (9.2.21')

I

Error eatimate8 of order 0(h2) can be derived in the same way if one has consistency conditions of second order. These are, for example, (24a) or (24b): ILaRa — lLhRh — R,1L1_i4-s


(9.2.24a)

Ch2.

(9.2.24b)

Remark 9.2.20. If

or 1? = (z',z") x (y',y"), the = Qa (i.e., I? = inequalities (24a,b) can be shown in a way similar to Lemma 13.

Example 9.2.21. The difference method in Example 4.5.8 shows quadratic convergence. This case can be analysed as follows. Using Remark 20 one shows (24a). In the following section we prove the Re-regularity IL;1 124-0 C which for the symmetric matrix under discussion, Lh, is equivalent with IL;'104-_2

C (cf. (6.3.3)). Expression (20) leads to —

The corresponding estimate

uj0

Ch21u12.

(9.2.25a)

9 Regularity

240

(9.2.25b)

—

which is based on (24b), fails

due to the fact that the solution in Example

4.5.8 does not belong to H3(O) (cf. Example 9.1.25). The verification of (24b) m the presence of irregular discretisationa at the boundary becomes more complicated.

9.2.4 Under suitable conditions on (1, Lu = f L2(f1) has a solution u H2(.O) n lo (cf., for example, Theorem 9.1.22). For the discrete soHd(S?): 1u12 lution of Lhuh = fh the corresponding question arises: does CIfhJo of hold? First one has to define the norm or if the boundary V coincides with grid lines, one can define If Q = 1/2

+

+ h2

iuhl2

XEQh

in general, however, one has to use irregular differences at the boundary. At points far from the boundary x (Ih\-yh, let = for example, with a grid point near the boundary x = (x,y) E At F, xr := (x + h,y) E (hI It appears practical to use the x' := (x — violate the inverse divided difference (4.8.5). In general, however, this become arbitrarily small. Therefore may estimate 1uh12 with the aid of the grid points (x — sjh,y), (z + h,y), one defines

(x+2h,y): 2

fUh(X+2h,y)—uh(X+h,V)

(2+ &j)h

h —

uh(x+h,y) —245(X —sih,y)1 s1h

y) =0. where u,(x — If, in addition, F passes between (x + h, y) and (x + 2h, y), one needs

toreplacex+2hbyx+h+srh.If(x+h,y)aieonolongerbelongstoQ,

set := h2ua(x). The second differenos is defined accordingly. The mixed difference in the Interior reads := Near the boundary several choices for Dm1, are possible. The definition of reads (

+

+

+

xEfl,,

ExercIse 9.2.22. Prove the inverse estimate

call Lh He-regular if
We 11sh12

S

E

12

Ii.

C. An equivalent formulation is

9.2 Regularity Properties of Difference Equations

241

and let Lh + 6Lh be + 6Lh is also Hz-regular. Hint: Remark

Exercise 9.2.23. Let Lh be


9.1.2.

one can —in analogy to the proof of CorolFor the proof of lary 9.1.23— partially sum the scalar product (LhUh, Lhtth)o in order to show This technique, however, can ILhuhPo and the equivalence of tfhIg only be applied to a rectangle Si. Here we use a simpler proof which is also applicable to general domains. L2(a) be the following piecewise constant interpolation: Let Ph:

(Phuh)(X) :=

X

PhUh(X)

(9.2.26a)

{

Phuh(x,y) := uh(x',y'), if (x',y') E

(9.2.26b)

and

x'—h/2<xx'+h/2, where uh(x)

0

for x E Qh\Qh (cf. (2)).

Lemma 9.2.24. Let Si E

Then we have

— I1i#.0

Ch,

(9.2.27a)

C, C.

(9.2.27b) (9.2.27c)

PROOF. (a) The estimate (27a) is equivalent to (6.3.3)). Thus we must show IWhIO ChIuhll for wh :=

Ch (cf.

— —

I)uh. From

the following Exercise 25 we obtain the expression tuh(X) =

for wh =

—

—

I)uh,

'C

Exercise 12c shows 111h10 — I)P,,uh. For every 1WIL2(J(2) for w := E (0,1) x (0,1) define the grid function with := w(x for x E Qh. One may check that y) is a weighted sum of first differences uh(x,y) — uh(x',y/) where (z',y') E {(x± h,y), (x,y ± h), (z± h,y ± h)}. Thus we have f for all from which one infers

=

11(2 w(x)12

h2

Jw(x +

= J(0,1)2 thus ItIhIo 11010 h(tsh$1.

=

=

=

9 Regularity

242

= {x E

with

fl V $

Qh:

one infers

Iwh —tiJhIO

(cf. part b of the proof and Lemma 15, in which -Ih niay be replaced by Together with hIuhIl one obtains Iwhlo (C' + l)hIuhIl. (b) (27b) with C = 1 follows from lPhuhlo = have already been estimated in (19i). (191) (c) Second differences of = implies

C1u12, i.e. (27c).

ExercIse 9.2.25. L2(1R2) —i

Show that the adjoint operators for Ph and Rh are and R: L2(1R2) with =

=

for x E Qh,

= uh(x) for all x E Qh.

Furthermore, the following holds: The identity = RhL —' Ph

—

[(LhRh —

RhL)L'Ph + (RhPh

1)1

yields the estimate

+

x

IILhRh — RhLIL4_21L'124-oIPhIo,_o + lRhPh

— uI—14—OL

The inverse estimate from Exercise 22 yields 11124—I Ch1 for the identity I: -s Thgether with the Inequalities (27a—c) follows

Theorem 9.2.26. Let Lh be Hz-regular and satisfy the consistency condition (9). Let Ii be convex or/rem C2 and let L be (i.e., C). Then Lh is also He-regular.

a

would be a sufficient condition upon 1? guarantee (27a-c). On E the other hand, the condition given agrees with the conditions for Theorem 14 and with the 112-regularity of L.

Corollary 9.2.27. In Theorem 26 one con replace (9) and the of L,, by (15) and the of L'h. The restdt is the bothL4

of

PROOF. For the proof of the x and note

I

Exercise 9.2.28. If Si E lit" (n <

of L'h define Phuh(X) := 0 for

3)

is bounded and Lh is

then Lh is stable with respect to the row-sum norm:

C'iUhlO S

C. Hint: Use

9.2 Regularity

of Difference Equations

243

For general Lipechitz domains (1, such as, for example, the L..shaped domain in Example 2.1.4, L Theorem cannot be 112-regular 9.2.21, however, guarantees for s [0, 1/2). For Lh one can define and prove analogously (cf. HaCkbUSCh [lJ).

10 Special Differential Equations

If the boundary value problems have special properties, one often uses special discretisations for them. We give two examples of this in Sections 10.1 and 10.2. For the first example the variational formulation is of decisive irnportance.

10.1 Differential Equations with Discontinuous Coefficients 10.1.1 Formulation The seif-adjoint differential equation

+ a(x)n(x) = 1(x) in Si

(10.1.la)

(cf. (5.1.17)) occurs often in physics. It can also be written as

with A(x)

so

that for a = 0 and f =

conservation law In the applications In physics the coeffidents

are,

0 we have the

in general, constants

can be varying, if the constitution of the of the material. The functions material depends on poertion. As soon as we have several materials in contact may be discontinuous on the boundary with oath other, the coefficients of contact (see Figure 1). The equation (Ia) can be understood in a classical sense only if E C1(ii). For discontinuous therefore one uses the variational formulation. If one supplements Equation (is) with the Dirichiet condition

u=0 onl'

(1O.l.lb)

then the weak formulation is written u

with a(u,v) =

vE

(1O.l.2a)

10.1 Differential Equations with Discontinuous Coefficlent8

245

where

a(u,v)

(1O.1.2b)

—

Note that Equations (2a,b) are defined for arbitrary

FIgure 10.1.1. An interior boundary

E

=

fl

In the sequel we shall assume the situation to be as in Figure 1: i? is by the b mdar into two subregions and Let the coefficients

be piecewise smooth: for k = 1,2. Along 'y the E coefficients may be discontinuous so that the one-sided boundary values lim

:= lim

(x E i')

may be different. In addition let us assume that the solution u is continuous, u E C°(Th), but only that it Is piecewise smooth, u E C1(.01) and u

C'($?2). The one-sided boundary values of the derivatives are denoted and (x E With these assumptions iraegration by parts of dx gives the result dx — f1, dl'. Since — ii = (ni, n2,...) in Figure 1 is the outgoing normal to but the ingoing normal with respect to in £12 there results dx = dx + J dl'. Putting this together one has f0

-

a(ts, v)

=

j

{au

v

+

/

dl'.

Thus from the variational equation (2a) there follows in addition to the

differential equation (la) also the transition equation

iJ=1

=

Thus we have shown the following lemma.

on

(10.1.3)

246

10 Special

Differential Equations

C'((4), k = 1,2. If the weak solution u of Lemma 10.1.1. Assume and a2, then it is (2a) in a is continuous and piecewise dsfferentiable in In U (12 = the dassicol solution of the differential equation (la) in addition to fulfilling the boundary condition (ib) the solution also satisfies the transition condition (3) on

Corollary 10.1.2. If the coefficients are not continuous then the solution of the equation (2a) does not, in general, belong to C2(fl), but will have disconHowever, the tangential derivative along 'y may tinuous derivatives along be continuous.

Example 10.1.3. Suppose the coefficient of a(u, v) := f a(x)u'v1 dx are given by a(x) = 1 on the interval and a(x) = 2 on this one1). dimensional example the point

the equation (2a) with /(x) =

u(x) =

1

.

on —0) = 2 .

plays the role of the curve y. The solution of 1

Is —

x2]

(CE)

1

+ 0).

As mentioned at the beginning, the differential equation can be written

inthefonndiv4,=fon(11US?2, whereq4:=A(x)graduon (1andu=Oon F. The transition condItion (3) means that (4,, a) is continuous on Since (4,, t) is also continuous for any tangential direction t, 4, is continuous on 'y and thus throughout 0. The regularity proofs of Section 9.1 can be carried over to the present situation. The smoothnees upon the coefficients of a(.,.) are in each case to be required piecewise on i and in addition the dividing line (or hypersurface) must be sufficiently smooth. Then there results the piecewise regularityu and u E instead of u Bringing in the transition conditIon (3) one obtaIns 4' = A(x)gradu on the whole region.

10.1.2 Dlscretisatlon When one discretises using finite elements there are the following difficulties:

Remark 10.1.4. Linear or bilinear elements give a finite-element solution with the error estimate Iu — = 0(h1/2). The L2(i?) error bound is in general no better than (ti — u"(O = 0(h).

PROOF. (1,:={tEr: tfl'y}consistainallflniteelementsthat&ejn contact with 'y. Since u has discontinuous derivatives on i', one can have no

10.2 A Singular Perturbation Problem

247

better estimate on fl.1 than Vu — Vv = 0(1) (v E Vh). The 8Urface area of £2, is of the order 0(h), so that there results lu — 0(h'/2). lbr the bound on lu ii'(o

Example 10.1.5. In Example 3 choose e (1 + h)/2 and discretise using piecewise linear elements of length h. Then there results an error at x = 1/2 of u(x) —1z"(z) = ah+0(h2) with a 0.00463. Since u—ui' behaves linearly in (0,1/2), one obtains the error 0(h) not only for the V norm — u1'jo but also for the maximum norm IlL — and for the L1 norm Ilu — ti'IItl(ol). The usual bounds on the errors fu — uhIi = 0(h) and fu

—

= 0(h2)

however be attained. To do this one must adjust the geometry of the triangulation to the curve If the dividing line -y is piecewise linear, one must choose the triangulation such that -y concides with sides of (interior) triangles. If 'y is curvilinear then it may be approximated by isoparametric can

elements (cf. §8.6). Similar aseertions hold for the difference scheme (5.1.18).

Example 10.1.6. Let the discontinuity position e in Example 3 be a grid point, i.e., form

E

N Then the difference scheme (5.1.18), which here takes the

h2

—us)

=1

for 1 ii

h-1 —1, is suited to the equation from Example 3. In general the error is of the order 0(h2). Since the solution of the differential equation is piecewise quadratic here, it is in fact exactly given by the difference solution. On the other hand, is not a grid point then the error is of the order 0(h). In the two-dimensional case one obtains 0(h2) difference solutions if 'y coincides with lines of the grid. Otherwise the error worsens to 0(h). Another poesible form of difference approximation consists in approximating the differential equations separately in the regions and .172, and then, to handle the unknown values on to discretise the transition condition (3).

10.2 A Singular Perturbation Problem 10.2.1 The Convection-Diffusion Equation In the following we shall consider the boundary value problem

-e4u +

=

f

in .17,

u = 0 on 1

(10.2.1)

fore >0. One calls the diffusion term and = (c,Vu) the convection term. For small e the convection term dominates.

248

Differential Equations

10

Exercise 10.2.1. Let the coefficients transformed by v(x) elliptic equation

=I

+

Equation (1) is elliptic for all

be

constants. Equation (1) can be into the symmetric Hd(fl)-

in a, v = 0 on 1.

(10.2.1')

> 0. That there is a unique solution

follows from Exercise 1 (cf. also Theorem 5.1.8). Denote the solution by

Remark 10.2.2. ug(x) and

cannot converge uniformly on Th for

—40.

PROOF. If

and 8u,/8x4 were continuous in one would be able to take the limit —+0 in Equation (1) and one would obtain the first-order differential equation

intl

(10.2.2)

:= This equation is of hyperbolic type, but not compatible with the boundary condition u =0 on I' (ci. Section 1.4), i.e., Equation (2) has, in general, no solution with =0 on 1'. U for

Equation (2) is called the "reduced equation". Equations (1) and (2) differ in the "perturbation term" -€ilu. Since the equations, (1) and (2), are of different types one speaks of a "singular" perturbation. The following example will show that *10 = limc..oue exists and satisfies Equation (2), but that the boundary condition mio =0 is only satisfied on a part Vr, of the boundary r.

Example 10.2.3.

(a) The solution of the ordinary boundary value problem

—at" + u'

=

1

in (0,1),

ti(0) = u(1) = 0

(10.2.3a)

= x_(ea/( 1). On [0,1), converges to *10(z) =x. This function satisfies the reduced equation (2), u' = 1, and the left boundary condition =0, but not uo(1) 0. (b) The solution of —at" — u'

= 0 in (0, 1),

u(0) = 0, *41) =

1

(l0.2.3b)

is Ue(X) = (1 — e_x/e)/(1 — In this case, u0(x) := = 1 satisfies Equation (2), —u' = 0, and uo(1) = 1, but not the boundary condition

at z =

0.

Which boundary condition is fulfilled depends for Equations (3a,b) on the sign of the convection term ±tt'. In the many-dimensional case the decisive factor is the direction of the vector c = (Cl,...

249

10.2 A Singular Perturbation Problem

tAO

1-.

1

1x 0-

0

I-

0

1

FIgure 10.2.1. (a) Solution of Example 3a; (b) Solution of Example 3b

In Figure la,b the solutions of Example 3 are sketched. In the interior Ue only in the neighbourhood of x = 1 (Figure la) [resp. x = 0 close to in order to satisfy the second boundary differ from (Figure lb)i does attain the condition. These neighbourhoods, in which the derivatives of is

order O(1/€), are called boundary layers. For Example 3 the thidmees of the boundary layer is of order Exercise 10.2.4. The interval [1 — ij, in which the function —1) exceeds the value <1, has the thi&ness =

An analysis of smgularly perturbed problems is to be found, e.g., in Kevorkian-Cole [1J and Coering et a!. [1J.

10.2.2 Stable Difference Schemes The special case of Equation (1) for c = (1,0) is

—E41L+u=f ma,

(10.2.4)

The symmetric difference formula (5.1.10) for Equation (4) is —€

= h2 —€ — h/2 For a fixed follows

+ h/2

(10.2,5)

L, is consistent of order 2. From Corollary 5.1.16 there

Remark 10.2.5. As soon as h

the difference scheme (5) leads to an

M-matrix. The property of being an M-matrix was used in Section 5.1.4 to demonstrate the solvability of the discrete equation. It turns out that the difference

10 Special Differential Equations

250

However, with increasing h/€ there develops an instability which is made clear in Table 1.

equations are also solvable for h

Thble 10.2.1

2/32 3/32

4/32

...

30/32

31/32

0.93 0.97 0.99 0.99

0.87 0.95 0.97 0.99

0.75 0.89 0.94 0.98

0.04 0.22 0.47 0.82

0.02 0.13

1.00 1.03

0.99 0.99 0.95 0.94 0.94

... ... ... ... ... ... ... ... ...

0.87 0.23 0.08 0.06 0.06

1.11

z 0 1/32 0.5 0.1

1

0.05 0.02

1

1

1 1 1

i04

1 1

1

4.95 48.6 4859.5

0.81

0.92 0.96 0.98 0.99 1.04 5.00 48.7 4859.6

0.99 0.99 0.90 0.88 0.88

1

0

0 0

0.31 0.73

0

1.89 5.88 49.6 4860.4

0 0 0 0

0

of Table 1 shows the values Ujt,(x, 1/2) of the difference solution Equation (4) in the unit square ii = (0,1) x (0,1) with f = 0 and w(x,y) h/2 = 1/64 = 0.015625, Lh (1 —x)sin(iry). The grid step is h = 1/32. Fbr

is an (irreducibly diagonally dominant) M-matrix. Because of the maximum prinaple (cf. Remark 4.4.4) the values lie between =0 and maxw =1. The solution of the reduced equation (2) is y) = sin v) = Accordingly the values 1/2) from the first part of Table 1 approach the limiting value sin ir/2 =1.

In the second part of the table we have h/2. Thus Lh is not an M-matrix. Notice that 1/2) > 1, i.e., even the maximum principle does not hold. In addition one can see that Ue,h(X, 1/2) converges to 1 — x for the even multiples x = iih, and thus not to uo(x,y) = sin icy. For the intervening odd multiples x = vh an oscillation of amplitude O(h2/2€) develops.

The resulting difficulty is similar to that for initial value problems for stiff (ordinary) differential equations: If one keeps constant and leta it go to zero, the convergence assertions of Section 5.1.4 hold. However, if is very small, the condition it < 2€, without which one does not obtain a reasonable solution, cannot in practice be satisfied. One way out of this difficulty has already been described in Section 5.1.4. One must apprsdmate the convection term by suitable one-sided differences (5.1.14)

-t +

= h2 —t— h4 4€ + hIcjI + hIc2I

-€-h4

wherect :=max{0,4}, c

—

+ hcj

(10.2.6)

10.2 A Singular Perturbation Problem

251

Remark 10.2.6. If one discretises Equation (1) using (6) then one obtains, for all e > 0 and h > 0, an M-matnx Lh. For fixed e, the scheme has consistency order 1.

Exercise 10.2.7. The discretisation of Equation (3b) corresponding to (6) is 2e +h — hi and this gives the discrete solution uh(x) = = — (1 + — (1 + If one applies the difference operator (6) to a smooth function one obtains the Taylor series

= Lu — { The

0(h) term hIcjIu

+ Ic2luyp}h + 0(h2).

(10.2.7)

+ hIc2Iu..fl, is called the numerical viscosity (or

the numerical ellipticity) since it amplilles the principal part. A second remedy consists in replacing the parameter e by a discretisation If one chooses using Ch :=

hIcjI/2, hlc2I/2} or Ch

e

+

(10.2.8)

max

then the symmetric difference method

-1 =

—1

—1 +

4

1

02

0

cj

(10.2.9)

C2

leads

to an M-matrix. It is true that the convection term has been discretised

to a second order of consistency, but the error of the diffusion term is O(eh —c),

which is relevant in practice, amounts to

which, for the case h > 0(h). Instead of (7) one has Lu

The difference

—

—

—

c)zlu + 0(h2).

(10.2. 10)

is called the artificial viscosity.

In the one.dimensional case the methods using numerical and artificial viscosities do not differ:

Remark 10.2.8. If, in the one-dimensional case, on chooses

according to the second alternative in (8), then the difference formulae (6) and (9) coincide.

10.2.3 Finite Elements The difficulties described in the previous section are not restricted to difference methods.

Exercise 10.2.9. Show that

10 Special Differential Equations

252

(a) Linear finite elements on a square grid triangulation applied in the case of Equation (4) give a discretisation that is identical with the difference method —1

L,, =

4

—1

+h

—1

—1

0 —1/3 —1/6

—1/6 0

1/6 1/3

1/6

0

(1O.2.lla)

(cf. Exercise 8.3.13). (b) For bilinear elements (cf. Exercise 8.3.17) one obtains

=

1

—1

—1

—1

8

—1

—1

—1

—1

h

+— 12

0 0

+1

—4 —1

a

+1

1

+4

(c) One-dimensional linear elements for —eu" + u' = difference formula

=

th' [—1

2

—11+1—1

0

.

(10.2.llb)

f lead to the central (10.2.llc)

1).

The Exercises 9b,c show that finite-element methods correspond to central difference formulas, and thus can equally well lead to instability. The method of artificial viscosity corresponds to the finite-element solution of the equation (c, grad *4 = for appropriate As in Exercise 9b one may show

f

Remark 10.2.10. If one sets

:= IciIh, Ic2Ih} and uses bilinear elements then the discretisation of Equation (1) leads to an M-matrix.

On the other hand the matrix (ha) has different signs in the sub-diagonal and in the super-diagonal, so that it is not possible to have an M-matrix for any value of €h. The analogues of one-sided differences are more difficult to construct. One

approach is to combine a finite-element method for the diffusion term with a (one-sided) difference method for the convection term (cf. Thomasset [1, §2.4]).

A second possibility is the generalisation of the Galerkin method to the

Petrov-Galerkin method, in which the discrete solution of the general equation (8.1.1) is defined by problem (12): Find u E V,1, so that a(u,v) = 1(v) for ally e W,,

(cf. Fletcher [1, §7.2), Thomassetll, §2.2]). Here we have

dimVh = dimWh (but in general Vh

Wh).

(10.2.12)

11 Eigenvalue Problems

11.1 Formulation of Elgenvalue Problems The classical formula*;ion of an eigenva)ue problem reads Le

Xe

in .17,

B,e = 0 on F (j

1,•• ,m).

(11.1.1)

Here L is an elliptic differential operator of order 2rn, and B, are boundary operators. A solution e of (1) is called an eigenfunction if e 0. In this case, A is the elgenvalue associated with e. As in Section 7, one can replace the classical representation (1) by a variational formulation, with a suitable bilinear form a(., .): V xV —+ IR taking the place of {L,B,}: Find e

V with a(e,v) = A(e, v)0

for all v

V.

(11.1.2a)

dx is the L2($?)-scalar product. Strictly speaking one ought = where U is the Hubert space of the Gelfand triple to replace (., .)o by (., V c U c V1 (cf. Section 6.3.3). But here we limit ourselves to the standard case U = L2(fl). The adjoint eigenvalue problem is formulated as (ti,

Find e E V with a(v, e) =

for all V E V.

(11.1.2b)

Definition 11.1.1. Let A E e

C. By E(A) one denotes the subepace of all E(A) is called the eigenspace V which Satisfy Equation (1)[ resp.

for A. With E(A) one denotes the corresponding eigenspace of Equation (2b). A is called an eigenvalue if dim E(A) 1.

Theorems 6.5.15 and 7.2.14 already contain the following statements:

Theorem 11.1.2. Let V C L2(.17) be continuously, densely and compactly embedded (for example, let V = with bounded.1?). Let a(.,.) be Vcoercive. Then the problems (2a,b) have countably many eigenvalues A E C which may only have an accumulation point at oo. For all A C we have dimE(A) = dimE(A)
(11.1.3)

254

11

Elgenvalue Problems

Exercise 11.1.3. In addition to the assumptions of Theorem 2 let a(.,.) be symmetric. Show that all eigenvalues are real and problems (2a) and (2b) are identical so that E(A) = 1) For ordinary differential equations of second order (i.e., (1 C 1R1, m it is known that all eigenva.lues are simple: dim E(A) = 1. This statement is incorrect for partial differential equations as the following example shows.

Example 11.1.4. The Poisson equation (0, b), with Dirichiet boundary values e =

0

= Ac in the rectangle (0, a) x on F, has the eigenvalues A =

(vir/a)2 + (or/b)2 with 1.', E I'J. The associated eigenfunction reads e(x, y) a = bone obtains

e"4(x, y) :=

eigenvalues A = A11"4 = A"1', which have multiplicity at least 2 since e1"4 and e14'1' are linearly independent eigenfunctions for the same elgenvalue. A triple eigenvalue, for example, exists for a = b, A = 50ir2/a2: for &'

E(A) = span{e1'7,e7'1,e55}.

The eigenhunctions e E E(A), by definition, belong to V. The regularity investigations of Section 9.1 immediately result in a stronger regularity.

Theorem 11.1.5. Let V = Hr(Q) with m 1, or V =

m=

1.

with

Under the aastsmptions of Theorems 9.1.16 reap. 9.1.17] we have

E(A) c Hm+s(j7).

PROOF. Along with a(.,.), asswnptions. Since aA(e, v) = from Corollary 9.1.19.

0

for e

:= a(u,v) E(A), V

—

A(u,v)o also satisfies the V, the statement follows

I

Besides the standard form (2a) there are generalised elgenvalue problems. An example is the Steklov problem

—& =0

in .1?,

tIe/an =

Ac

on 1',

whose variational formulation reads e E H'(Il), = (v H'(Q)). One can show that all eigenvalues are real and that the statements of Theorem 2 hold.

11.2 Finite Element Discretisation 11.2.1 Discretisation Let Vh C V be a (finite-element) subapace. The Ritz-Galerkin (resp. finiteelement) discretisations of the eigenvalue problems (1 .2a,b) read Vh,

= Ah(e",v)O for all v E

Vh,

(11.2.la)

11.2 Finite Element Discretisation v

Vh.

255

(1l.2.lb)

spanned by the solutions of the problems (in) Iresp. (ib)]. As in Theorem 11.1.2, dirnEh(Ah) = dim holds. If a(.,.) is symmetric, then Eh(Ah) = As in Section 8.1, the formulation (la,b) can be transcribed into matrix notation. are

The discrete elgenspaces

= Pe and be the coefficient vectors for Pe (cf. (8.1.6)). The elgenvalue problems (la,b) are equivalent to

Remark 11.2.1. Let e and Le =

LTe*

= 3Me,

(11.2.1')

where the stiffness matrix L is defined as in Theorem 8.1.3 and M by (8.8.7). Since in general M I, (1') involves generalised eigenvalue problems.

Exercise 11.2.2. Show that (a) M is positive definite and possesses a decomposition M = ATA (for example, A M"2 or the Cholesky factor). (b) The first problem in (1') is equivalent to the ordinary eigenvalue problem = AhÔ withL := (AT)1LA', ë = Ae. The second problem in (1') with ë = Ae'. corresponds to LTÔ* =

When investigating convergence, one must watch out for the following difficulties:

(i) A uniform approximation of all the eigenva.lues and eigenfunctions by discrete eigenvalues and eigenfunctions is impossible since the infinitely many eigenvalues of (1 .2a) are set against the only finitely many of (Ia). It is only possible to characterise a fixed elgenvalue A of (1.2a) as an accumulation point of discrete eigenvalues {Ah: h 1, it may well be that the multiple eigenvalue A is approximated by several discrete eigenvalues I = 1,. , k: dimE(A) = dim The error estimates of jA — are then generally worse than for simple aigenvalues. Only for the mean value A := Ar/k does one obtain the usual estimates (ef. .

p. 3381).

Exercise

11.2.3. Let the eigenvalue problem be as in Example 11.1.4 with = b and A = Let V,, consist of linear elements over a square grid triangulation. Show that to the given triple eigenvalue A corresponds a double eigenvalue and a single eigenvalue distinct from it, with = A (i = 1,2). Hint: the nodal values of the discrete eigenfunctions !lmn_.o agree with the continuous eigenfunctions e1'7, and

a

256

11 Elgenvalue Problems

11.2.2 Qualitative Convergence Results e for A and eh The rate of convergence will be discussed in Section 11.2.3. The basic 0. assumptions are the following:

This section concerns the question as to whether h

Let a(.,.): V x V

lit be V-coercive,

(l1.2.2a)

Let V C L2(fl) be continuously, densely, and compactly embedded, (11.2.2b)

so that the

theory is applicable (Theorem 11.1.2). Further-, 0) be given which increasingly more, let a sequence of subapaces approximate V (cf. (8.2.4a)): u E V.

lirninf{IIu —

(11.2.2c)

We define

aA(u,v) :=a(u,v)—

aA(.,.):V xV —' w(A) :=

mi

sup

uEV

vEV

laA(ti,v)I,

(11.2.3a) (11.2.3b)

flvllv=1 ]IvIIvl wh(A) :=

inf

sup

(11.2.3c)

vEVg,

ExercIse 11.2.4. Let L and Lh be the operators associated with a(., .): V x V JR and a(.,.):Va x Vh —, lit (cf. (8.1.lOa)). Show that (1) If A is not an eigenvalue we have

= lI(Lh —At)'1111,

=

(11.2.4)

(cf. Lemma 6.5.3 and ExercIse 8.1.16).

are continuous in A E (ii) w(A) and (iii) If in (3b,c) one replaces OA(u, v) by aA(v, u) one obtains the variables which correspond to the a4joint problem. The following holds: w(A) and

= w(A) and

= wh(A). Hint: Use Lemma 6.5.17.

With the aid of (4) and Theorem 6.5.15 one proves the following connection between wa(A), and the eigenvalue problems.

Remark 11.2.5. Let (2a,b) hold. A

is an elgenvalue of (1.2a) if and only if 0, and Ah is elgenvalue of(la) if and only if wh(Ah)= 0.

Lemm* 11.2.6. Let a(.,.) be V-coercive (ef. (2a)). Then there exists a such that is V-elliptic. In addition, then 1/CR and 1/CE hold with CE >0 from Definition 6.5.13.

it

257

11.2 Finite Element Discretisation

be osmpact. Let (2a-c) hokL Then there exut =0 numbers C > 0 and ,7(h) > 0, independent of A e A with such that

Lemma 11.2.7. Let A C

w(A)

wh(A) Cw(A) —i1(h),

for aU A E A. (11.2.5)

Cc,Jh(A)—rI(h)

PROOF. Let the operators Z = Z(A): V

—*

V

and 4 = 4(A): V —* Vh

defined as follows:

for all v e

z := Z(A)u is the solution of

V,

(11.2.6a)

where

for all v Vh, (1L2.6b)

is the solution of

Zh(A)u

is chosen acooi-ding to Lemma 6. It shows for all A

From aA(u,v) = w(A)IlujIv

S

A.

(11.2.6c)

z,v) and the definition of w(A) one infers

—

sup

IaA(u,v)l =

vEV

8UP

vEV IIvIIv=1

jvIIv=1

— z,v)l CsIIu — zIIv (11.2.6d)

with Cs :

For an arbitrary u E — zh, v), Lemma 6, and (6d) that

sup

laA(u,v)i

= sup

VEVi.

vEVk

IIvIIv=1

IIvIIv=1

— zIlv

liz

—

infers from

—z"llv

—

—

—

HZ

—

Uultv,

thus Wh (A) (CECS)'w(A) — flZ — ZhIIVIV/CE. From this follows the first part of (5) with C = (CECS)' >0 and ,i(h) = HZ — if

urn sup fl Z(A) h—so

—

= 0.

(11.2.6e)

)iEA

The proof of (6e) is earned out indirectly. Its negation reads: there exist e >0, —*0 with H E A, >0. Then there exist E V —

with = 1, Due to (2b) and the Uj V with lixnAj = A5,

—

e/2 > 0.

(1L2.6f)

of A there exists a subsequence A, e A, u' E V' in V'. (2c) and Theorem 8.2.2 show H(Z(A) — Z,,(A)u1$v —50. Together with (6c) obtain

258

11 Elgenvalue Problems

— Zh1(A,)ju,IIv

II[Z(A,) — Z(A*)!uJIIv + II[Zhj(Aj) — Zh,(A))uj$Iv — u*jflv + fl[Z(A) — + —uJHv + IIZh — <2C1A3 1- 2Czflu, — uiIv' + lI[Z(A) — Zh,(A')Ju'jIv

0

in contradiction to (6f).

For the proof of the second part of (5) one estimate by

(6d) and the following

forall U"EVh,

sup yE V,,

SUp

sup

vEV

vEV

IIvIIv=t

11V11v1

— IIZ(A) — Zh(A)IIv,_v111u40v

Let u E V

with

IIuflv

= I be selected such that

sup

inf

sup

oEV IvHv=1

tLEV

vEV

Since sup IaA(u — u's,

all nh e Vh.

for

IaA(u,v)l=w(A).

IIuIIv=1 flvIIv=1 CsIIu

—

it

follows for arbitrary

E Vh

that

— IIZ(A) — Zh(A)IJV,...V — CgJJu —

From (2c) and (6e) follows the second part of (5).

A corollary to Lemma 7 is Theorem 8.2.8.11 Problem (8.1.1) for all! V' is solvable, then A =0 cannot be an eigenvalue, i.e., w(0) > 0. Thus it follows

that CN, WN,(O) c >0 for sufficiently large i. A second corollary concerns the convergence of the eigen values.

Theorem 11.2.8. Let (2a-c) hold. If eigenvdues of(la) with A,11

Ao then ),o

—s 0) are discrete

(i —* 00,

is an

of(1.2a).

PROOF. If A0 were not an eigenvalue then w(A) c-neighbourhood of

since w(A)

is

> 0 would be in the continuous (cf. Exercise 4(u)). There

would exist h0 >0 such that ,1(h) for all 4 h0 (C and from (5)). For all E K1Po) with h0 the contradiction follows from (5):

0=

—

—

=

>0.

Lemma 11.2.9. (Minimum Principle) Let hold. The functions w(A) and w,1(A) in the interior of Ac C have no pmper positive minimum.

11.2 Finite Element Dlscretisation

259

with w(A)> with o(., .). Let PROOF. Let L be the operator A. For sufficiently small e > 0 we 0, be an arbitrary point in the interior of is defined in Al)' Thus (L — > 0 in C A and have and holomorphic. Caucby'a integral formula says (L —

)J)' = 2ws

—

A)'(L — CI)'dC

for all A E Ke(A).

From this one infers —

ES

C)IIvv

A')

(ci. Exercise 4(i)). Thus, w(A) cannot

i.e., w(A) min{w(():C E

For wh(A) the conclusion is the same.

aseume a proper minimum in

The converse of Theorem 8 is contained in

Theorem 11.2.10. Let (2a-c) hold. Let A0 be the eigenvalue of(1.2a). Then Ah = A0. of(la) (for all h) stich that there exist discrete eigenvalnes

PROOF. Let e > 0 be arbitrary. Aceording to Theorem 11.1.2, Ao is an isolated eigenvalue: >0 for 0 < IA — Aol e (e sufficiently small). is compact, we have that is continuous and Since positive. Because of (5) and w(Ao) = 0 one obtains is IA — Aol = for sufficiently small h Wh(A)

Cw(A)

—

Cp, —

> v1(h)/C

must have a proper minimum iii Ke(A0). Thus By Lemma 9 the minimal value is zero. Thus there exists a Ah Ke(A0) which is a discrete eigenvalue, Wh(Ah) 0. for all A E

I

The convergence of the eigenfunctions is obtained from Theorem 11.2.11. Let (2a—c) hold. Let e1'

be discrete eigenfunc= 1 and A0. Then there exists a subsequence e1" which converges in V to an eigenfunction eE(A0): Lions with

e

E(Ao),

— dlv

0

(i

oo),

lIellv

1.

PROOF. The functions

are uniformly bounded in V. Since V C L2(fl) is compactly embedded (ci. (2b)), there exists a subeequence which in L2(Q) to an e L2(Q): lie —

—. 0

(i

oo).

(11.2.7a)

11 Elgenvalue Problems

260

We define z = Z(Ao)e, = Theorem 8.2.2 there exists an h1(e) > 0

<e/2

liz —

for


A combination of

(i.e., (6b) for A = Ao) and (7c) yields

:=

(A0 —

—

v)o — (A,4 — Ao)(e",

because Ao and (7*), there exists an e/(2CEJ (CE from Lemma 6) and

V,4. Since

v

(l1.2.7c)

for all v E

(A,4

=

et", v) =

—

(11.2.Th)

is a solution of

function

=

according to (6a,b). According to

suth that

—' 0

>0 such that

<e/2

—

(11.2.7e)

h2(e). (Th) and (7e) show that liz — min{h1(e), e for thus = z in V. Therefore lime" = z in L2(.Q) C V also holds. (7*) proves z = e E V such that e = z = becomes a(e, v) = Ao(e,v)o. Therefore e = lime"' Is an elgenfunction of (1.2*). In particular, Idly = limlle"Ilv = 1. for

ExercIse 11.2.12. Let (2*-c) hold. Let

and e be as in Theorem

11. Show that (a) If dim E(Ao) =1 then also dim (b) Let dim E(Ao) = 1. Then we have

I(e",e)vi 1/2 and

11.2.3

:=

=1. e

in V for

:=

if

otherwise.

QuantitatIve Convergence Results

The geometric and algebraic multiplicities of an eigenvalue Ao of (1.2*) agree

if

dim Ker(L — Aol)

dim Ker(L — Aol)2.

(11.2.8)

11.2.13. Let (2a1b) and dim E(Ao) = I hoüL Then (8) is equivalent

to (e,e')o

0 forO

e E E(Ao), 0

e'

E(Ao).

PROOF. We have Ker(L — Aol) = E(Ao) = span {e}. dim Ker(L — Aol)2 > 1 holds if and only if there exists a solution v V for (L—AoI)v = e. According to Theorem 6.5.15c this equation has a solution If and only if (e,e)o 0.

•

Let E(Ao) = span {e}, E'(Ao) = span {e'}. Under the aesumptlon (8) e and e can be normalised so that

(e,e')o = 1.

(11.2.8')

261

11.2 FnIte Element Dscretiaation

= {v' e V':(v',e')o We define V := {v E V:(v,e')o = 0}, be the dual norm for = 11 liv. Fbr Problem (9):

For f

V', find u

(f,v)o for all V

aA(u,v)

O}. Let

(11.2.9)

E

one defines the variable corresponding to (3b)

ml

:=

sup

(11.2.10)

laA(u,v)i.

uEc'

Lt"IIv=l 11v11v1

Lemma 11.2.14. Let (2a,b), (8), and dim E(Ao) = 1 hold. Then there exists e. Problem (9) has exactly C > 0 for all IA — an e > 0 such that one solution u E V with ilulIv Ill Ilv'/2z(A), if

> 0.

C. x PROOF. Let L: 1' be the operator aesociated with a(., .): e (e sufficiently small) (L — AI)u = f has a unique solution For 0 < — Aol u E V. From f V' follows

0=

= ([L — A.1]u,e')o = (u, [L — M)'e')o =

—

X)(u, c')0,

— AI)': —+ Thus there exists as the restriction of For A = Ac Problem (9) has a unique solution cording to Theorem 6.5.15c. Then there exists (L — Al)-1 for all A

i.e., u E (L —

)J)-' to V' C V'.

According to Remark 5 (with V instead of V) , c2i(A) must be positive in Kd(Ao). The continuity of proves C > 0. In analogy to (4) one results from The bound by has llullv =

I

Exercise 11.2.15. Show that

Lemrnk 11.2.16. Let (2a-c), dim E(Ao) = 1, and condition (8) hold. Let A,, be discrete eigenvalues with urn,,....0 A,, = According to Exercise 12b there exists an e1' E,,(A,,) and e'1' with e" —' e E E'(7.o), (e,e')o = 1. This enables one to construct the space := {v" E V,,: (v", = 0} and the variable

inf

sup IaA(u,v)l.

UuIIv=1

there exists a C > 0 independent of h and A E ( such that Wh(A) Cw,,(A). For sufficiently smallc >0 and h, Wh(A) irj> Ofor c. Then

PROOF. (a) First statement: there exists h0 >0 and C such that

min{iiv+ae"llv:aEC}IIvIlv/C

262

11 Elgenvalue Problems

The proof is carried out indirectly. The negation reads: there exists a se0. = 1 and with C, hj —4 0, + quence E —* ? in L2(a). Evi—. a', Thus there exist subsequences with must have the limit w' = liznwj = v' + a'e' dently, := v, + = 0, it follows that w = 0, in L2(a). Since limIIw4IL2(n) lim(vj,e*h)o and thus v' = —a'e'. From 0 = = (v',e')o = one infers a' = 0. Thus the contradiction follows from 1 = tim 11v41v = + IIwiIIv = (b) Second statement: L21,,(A) Ccüh(A) with C from (a). Because aj(u,v)

foruEVh we have

=

inf

sup IaA(u,v)I/(IIuIIvIIvIIv)

C inf

sup max(aA(u,v+ae")I/[(ItLIIvIjv+ae"flv] aEC

=C inf

sup

IaA(u,w)I/[IIuhIvIIwUv]

O#ioEV,.

C

in!

sup

(c) Let e > 0 be chosen such that Ao is the only elgenvalue in For sufficiently small h, Ah is the only discrete eigenvalue in Ke(A0). In the proof of

From Part b follows that > 0 for A E According i := to Exercise 12a, aA(u,v) = (f,v)o (v E is solvable for each 1€ and all A E Ke(A0) such that = 0 is excluded. Lemma 9 shows that

I

Exercise 11.2.17. Let (2a,b) hold. The functions u, v V (u, v)o 0. Let d(eA, Vh) be defined as in (8.2.2). Show that if d(u, Vh) is sufficiently small then there exists a u1'

Vh with

IIu_uhllv 2d(u,Vh), Lemmk 11.2.18.

Let (2a—c) hOZCL Let Ao be an eigenvalue with (8) and dimE(Ao) =1. For sufficientty h there exists Eh(Ah) with —

C[IAo — AhI + d(e, Vh)].

PROOF. Let zh := Zh(AO)e be the solution of (6b). Since e = Z(Ao)e, one has n11e—zh

For all v

Vh we have

IIvCId(e,Vh).

(1l.2.lIa)

11.2 Finite Element Dlscretisation

v) =

= so

(.\o

—

(Ao

.&)(e, v)0

—

— (Ah

263

v)0

—

—

)th)(e,v)o + (Ah —

ehv)0 + (Ah —

—

that

=

+(Ah —.p)(e —z',v)o for ally

—

Vh.

(11.2.llb)

According to Lemma 16 we have f(e, — eh can be scaled so that = 0. (lib) corresponds to Problem (9). Lemma —

+ lie

Wh(Ah) 'Cu)10 —

Together with (1 la)

one

Theorem 11.2.19. =

1.

Let

14 proves

z"Iivi C'[

+ d(e, Vh)J.

—

obtains the statement.

Let

(2a—c) hold. Let

e

lAo —

U

Ao be

an esgenvalue with

E'(10), itchy = 1, (e,e)o =

E(A0), e there exist discrete eigenvalues Xh

dimE(Ao)

—

> 0 for sufficiently small h.

(8) and 1.

Then

with

)'hl Cd(e, Vh)d(e', Vh).

(11.2.12)

PROOF. Choose according to &ercise 17 such that 1Ie — uhhlv 2d(e,Vh), (e — u",e)o = 0. Discrete elgenvalues exist by Theo—i

rem 10. From

0=

a,10(eh,es)

=

= aAh(e",e) —(Ao _Ah)(e!l,e*)o —u") —()1.o —

_tiI)_(Ao_Ah)(eh,e*)o = follows

—

exists an

— e,

e — u")

—

— eilv[11e

Ahi

(Ao

e,

—

+

—

—

AhiJ.

e)o]

By Lemma 18 there

Eh(Ah) such that

P10 —

C'C"[(Ao

From this one obtains (12)

—

Ahl + d(e, Vh)J[I)1o — Ahi

with

C = 3C'C"

for

+ 2d(e*, Vh)J.

sufficiently small h, since

U

d(e,Vh)—40.

Theorem 11.2.20.

asswnptions of Theorem 19 there exiet for

e

elgenfunctions

E(Ao), e' E

Under the E()10) discrete

E Ea(Ah),

with lie

—

Cd(e, Vh),

lie*

— e"lIv Cd(e, Vh).

(11.2.13)

PROOF. Insert (12) with d(e, Vh) const into Lemma 18. The second estimate in (13) follows analogously. U

11 Eigenvalue Problems

264

In the following, let V C H'(fl). Theorem 11.1.5 proves C H14'(37).

C

(11.2.14a)

Also, let (14b) hold (cf. (8.4.10)): for all u

d(u, Vh) Ch'IIUIIHl+.(n)

E

U

(l1.2.14b)

be the cigenvalue Corollary 11.2.21. Let (2a), (2c), (14a,b) hold. Let with(8) and dimE(Ao) = 1. Then there exists Ah,e" Eh(Ah), esh E

such

that

IAo

— AhJ Ch2',

Ch',

lie —

lie5 — e*hIlv CM.

(11.2.15)

Occasionally eigenfunctions may have better regularity than is proven for ordinary boundary value problems. For example, let —& = Ac be in the rectangle (1 = (0,1) x (0, 1) with e= 0 on V. First, Theorem 11.1.5 implies thus e C°(17) (cf. Theorem 6.2.30). Thus e = 0 holds e E H2(.Q) ii in the corners of .0. AccordIng to Example 9.1.25 it follows that e (j7)

for s <4. As in Section 8.4.2 one obtains better error estimates for e —

in the

L2-norm. The proof is postponed until after Corollary 29.

Theorem 11.2.22. Let (2a-c), (8), dim E(Ao) = 1, and (14a,b) with $ = 1 hold. Let a(.,.) and a(.,.) be H2 -regular, i.e., fan E L2(Q) , a,h(u,v) = (f,v)o and = (v,f)o (v E Vh, 4 from Lemma 6) have solution8 u,u5 E H2(i7). Let e E E(Ao) and e5 Eh(Ah), and e*h E

E(Ao). Then there exist Ah,e" E

with

lie — e"11L2(a)

C'h2,

lies

—

e"liV(D) C'h2.

If (14a,b) also hold with some s> 1, one must replace C'h2 by

11.2.4

Complementary Problems

In Problem (9) we have already encountered a singular equation which neverthelees was solvable. In the following let A0 be the only eigenvalue in the disc Kr(A0). The equation

aA(u,v)=(f,v)o forallv€V

(11.2.16a)

8 singular for A = Ao. For A A0 Equation (16a) is ill-conditioned. In the following we are going to show that Equation (16a) is well-defined and well-

conditioned if the right-hand side f lies in the orthogonal complement of E(Ao):

11.2 Finite Element Discretisation

f

,i.e.,

(f,e')a =0

for all e' E

265

(11.2.16b)

In the case of A = A0

, with it, it + e (e E E(Ao)) is also the solution. The uniqueness of the solution is obtained under the conditions (8) and (16c):

E*(A0)

(11.2.16c)

Remark 11.2.23. Let (2a,b) and (8) hold. Let A0 be the only eigenvalue in Kr(A0). Then (16a,b) has exactly one solution it for all IA — A0I r which satisfies (16c). There exists a C independent of f and A such that IIuIIv CII! liv' PROOF. This follows from Lemma 14 in which the assumption E(Ao) = 1 is not necessary. U The finite element discretisation of Equation (16a) reads:

Find U"

V,,

= (f,v)o for all v E Vh.

with

(11.2.17)

In general, Equation (17) need not be well-defined, even assuming (16b). For the sake of simplicity we limit ourselves in the following to simple eigenvalues: d.imE(A0) = 1. Equation (17) is replaced by (18a):

Find

E

with aA(u",v) =

for all v

V,,

with f(s) .1.

(11.2.18a)

(11.2.18b) (11.2.18c)

Exercise 11.2.24. Show that is equivalent to: Find it" v) = (f(h), v)o for all v V,, with fl

SA(U1',

=

with as in Lemma 16.

Lemma 16 proves the

Remark 11.2.25. Let (2a-c), (8), dim E(A0) = 1 hold. Let A0 be the only eigenvalue in Kr(Ao). Then there exists an h0 >0 auth that) for all h ho and all A Kr(A0), the Problem (18a,b) has a unique solution it" u"(A) which satisfies the additional conditions (18c). Further there exists a C independent of h, A, and 1(h) such that CII! If E'(A0), f from (16b) need not satisfy condition (18b). If Eh(Ah) and E with (e",e"')o = 1 are known, one can define

f satisfies (18b) since Qh represents the projection on

(11.2.19)

(As)'.

11 Ezgenvalue Problems

266

=

=

Exercise 11.2.26. Letu 1.

1,

= 1,

= 1. Show that

d(u,

E

v" J..

+ llullvinf{11e8 —

= i hold. Let dim E(Ao) = dim Let h be sufficiently small such that Ao be the only eigenvatue in (following Remav* 25) the Problem (18a-c) is solvable. For the solutions u and of (i6a-c) and (iSa—c) the ermr estimate

Theorem 11.2.27. Let (2a-c),

llu—u'iiv

Cfd(u, Vh)+llfllvI inf(lle*

Iv': e E

—1 lIv'i

(il.2.20)

holds, with C independent off, f(h), and h. PROOF. Repeat the proof of Theorem 8.2.1 for aA(-,-) instead of a(-, .). Here one must choose w E Vh with w 1. Furthermore, (8.2.3) becomes

u,v) =

—

— f,v)o

for all v E Vh.

CN agrees with

> 0 (A E (cf. Lemma 16). flu — Why is estimated with the aid of Exercise 26, with llullv il/ liv' being added.

I

Corollary 11.2.28. If 1(h) is defined by (19), then inequality (20) becomes flu —

PROOF.

C'[d(u, Vh) + hilly' inf{hle* — e*hIlv: e' E E'(Ao)}J. (i2.2.2ia)

—lily' Cj(f,e")ol = Cl(f,e Cflf liv' hie* —

Corollary 11.2.29. If additionally the assumptions u E (14a), arid then (21a) yields the estimate d(u,Vh) S Chitihi÷1 hold forts E flu

—

Ch'JIulIffl+.(a).

(11.2.21b)

It remains to add the

PROOF. (Theorem 22) For e E(.Ao) there exists e1' Eh(Ah) with f := e — 1. E(Ao) and Ifhi = Il/ liv Ch = Ch. According to Remark 25 the problem OA0(v,w) = (v,f)o has a solution w I E(A0) for all v E V. The assumption of regularity yields w E H2(I?), 1w12 Cl/k such that E Vh exists with I. 1w — Chlwi2 C'hIflo. The value

11.3 Discretisation by Difference Methods

267

=0—aA0(e5,w")

=

(Ao

—

Aa)(e",w")o

—

=

(cf. (15)). From

be bounded by

can

= (f, f)o

—

a.x0(f,w) = a.x0(f,w — w") + (Ao

—


Ifti Ch one infers tf to C'h2. The same method is then applied to

—

eshI

ExercIse 11.2.30. Formulate the conditions for and the proof of the error from Corollary 28). Hint: Decompose and

estimate

in fi +12 with ft with

11.3 Discretisation by Difference Methods In the following we Limit ourselves to the case of a difference operator of the order 2m = 2. The differential equation Lu = with homogeneous Dirichtet boundary condition is replaced, as in Sections 4 arid 5, by the difference equation L5u5 = fh. The eigenvalue equations Le = Ac, Le = Ae are discretised by

f

Lheh=Ahes,

(11.3.1)

is the transposed and complex-conjugate matrix to L5. The general asswnptions of the following analysis are:

V=

C°" bounded,

I?

a(u,v) = (Lu,v)o

(11.3.2b)

is

Consistency condition ILaRa

(11.3.2a)

Ch.

—

(11.3.2c)

Condition (2c) has been discussed in Section 9.2.2. Assume furthermore that is For suitable A E IR :r=

(I : identity matrix)

L5 —

is thus (LU,hvh,vh)o

(11.3.2d)

Furthermore, let L,.1 := L

i.e.,

—

p1

H2(Q)-regular,

(11.3.2e)


C,

1L51_14_i

C.

(11.3.2f)

11 Elgenvalue Problems

268

In (9.2.26a,b) we have introduced prolongations Ph: —+

L2(S?). Now we need a

mapping

L2(1R2)

and

-.

I Phtlh(x) if Kh12(x) C .0, otherwise,

I. 0

(x E.0),

Phtzh(x) :=

where


(9.2.26b) and

C.

(11.3.2g)

Let Rh and R,, be defIned as in (9.2.12a,b). They Satisfy

IRaio._o C,

(11.3.2h)

C.

IR,111._1 C,

ExercIse 11.3.1. Show that:

IRh—Rhio.--1 Ch,

(11.3.2i)

Ch,

—

11— RhPhl 0._i Ch,

Ch.

if —

(11 .3.2j)

The first inequality in (2j) is equivalent to

(11.3.2j)

l(Phuh,Phvh)o — (uh,vh)oJ ChjuhIolvhIi. Hint: Exercise 9.2.12 and Lemma 9.2.15.

LemmA 11.3.2. Let (2a,c,g,h,i)

hold.

(a) it is true that foraliueH01(f1),

(11.3.2k) !im

— Rh)t40 = 0 for all u —

(b)

If u

RaLjui_i = 0

and lilnh,oIRauk

(11.3.21)

L2(.Q),

for

allu E

0 then u =

E C.

(11.3.2m)

0.

PROOF. (a) The proof of (2k,l,m) follows the same pattern, which will be demonstrated for the case of(21). Fbr >0 one has to show I(Rh — Rh)uI for h Since is dense in there exists a ü with By (2i) there such that i(Rh—Rh)(u—ii)k lu—ui0 follows (Rh— e/2 for h h(e) := Altogether this gives l(Rh

—

(b) From (2k) one infers 0 = hmh_.o(Rhu,P,u)o = hmh,o(PhRhu,u)o = (u,u)O thus u = 0.

I

Exercise 11.3.3. LetAh

Showthat

269

11.3 Discretlsation by Difference Methods

= (Ahuh,uh)o, = (Au,u)o, = (v, LAU)O for v = A1LAU E = (vh,LA,huh)o for Vh =

I

hm I(AhRh — k,,A)uI_i = 0 for all u E

The following analysis is tailored to the properties of the difference meth-

ods under discussion. We have tried to avoid a more abstract theory that is applicable to finite elements as well as difference methods. This kind of approach can be found in Stummel [1) and Chatelin [1). is now defined by The variable :=

inf

inf

sup I(LA,huh,vh)oI =

IlahIl=l

IuhIl=l Ivhh=L

(11.3.3)

ILA,huhI_i.

As in Exercise 11.2.4 we have

ifAisaneigenvalueof Lh;

wh(A)=0, =

(11.3.3')

otherwise.

The analogue of Lemma 11.2.7 reads

Lemma 11.3.4. Let A C C be compact. Let (2a-m) hold. Then there exist independent ofAEA, such that variables C >0 wh(A) Cw(A)

w(A) Cwh(A) —

—

for all A

A,

h > 0. (11.3.4)

PROOF. (a) Since A is compact, w(A) is continuous, and wh(A) is equicontinuous in A, it is sufficient to show that Illflh_.owh(A) Cw(A), and w(A)

Climh.o,Jh(A)forallAEAwithC>0. (b) Define for A E A and uh with IuhIl =

u:=Phuh,

1

and ILA,huhj_1 = wh((A)

z,,

with p from (2e). Without loss of generality it can be assumed that p We have —

zli =

— Zh) + P,,zh — zIl

IPhzh — zli = IPhLZh — Rp,zJ —

—

A.

— ZhIl + IPhzh — zIi,

— PhRJ&]zI1 —

+

—

(ci. (2g,j,m)) so that

luh—zhIlCxIu—zIl—o(1),

C1 >0.

(11.3.5a)

11 Eigenvalue Problems

270

As in (2.6d) one obtains lu —

(11.3.5b)

C2 > 0.

zli C24,(A)Iuli,

From

=

ILA,htLhl_1

+

=

+ (p —

uh)o

—

—

uh)o

+ C5

=

+

+

Al

with CA := max{lp — Al: A E A) > 0 follows —wh(A)j.

from which the statement results directly, or wh(A)

Either CE/2 WhiCh yields

with Co :=

luhlo Co = ColuhIl We

(11.3.5c)

want to show that there exists ho > 0 and C,, = C,(Co) such that (11.3.5d)

for all uh with IuhIo Coluhil and h h0.

luali
C0 and 0 and 0 (h —* 0). From IR,M*ualo lPhuhtl 1 —RhPhIo.lIuhll Ch —*0 follows lUhlO —*0 in contrast IUh — RhP,uhlo C0. Thus we have (5(1) tO lUhlO Together with L,,,,1(uh — zh) = LAhuh, (2d) and (5a,b,d) yield the first of the inequalities (4): The negation of (5d) reads: there exists

=

=

with lUhll = 1, lualo

— Zh)I_1

—

zIj — o(l) C5C1C2w(A)IuIj — o(1) Cw(A) — o(1) with C := C5C1C2/C,, >0. —

(c) Let > 0 be arbitrary; u w(A) ILAuI_I — Set uh := (v, LAU)O

RhU.

with luli = 1 can be selected such that According to Exercise 3 it holds that

for v = A'LAU E

A=

I —4

for vh :=

= (Vh,

From Rhv — Vh —

=

—

L,1,hUh

= [ALA —

(v, LAu)o — (Rhv, RhLAU)O = (LI —

for h

w(A)

0

(ci. (2m), (2k)), one infers ILAUI_1

—

+

—

—10

—, 0,

LAU)O —' 0, —'

ILAUI_i and

o(1) Uh(A) — — o(1)

(h —*0)

11.3 Discretisation by Difference Methods

I

has been proved.

for eache>0.

271

Corollary 11.3.5. Under the assumptions of Lemma 4 the following holds: for all A A then there exists an ho > (1 such that LA,h If there exists for CI1AE A andhho is

PROOF.Wehaveamedw(A)>OinAsomax{w(A):AEA}z:v,>O. Choose h0 according to Lemma 4 such that ws(A) Cw(A) — h < ho. Then 2/ (Ct7) for all A E A,h ho.

for

I

The proof of Theorem 11.2.8 can be earned over without change and results in

Theorem 11.3.6. Assume (2a—m). If of ProbLem (1) with

—. 0) are discrete eigenvaiues then A0 is an eigenvaiue of(1.2a).

—.

Lemma 11.2.9 and Theorem 11.2.10 can also be carried over without changes.

Theorem 11.3.7. Let (2a—m) hold. Let A0 be an cigenvalue of(l.2a). Then there exist discrete eigenvalues Ah of(l) [toraH hJ such that = A0.

Theorem 11.3.8. Let (2a—m) hold. Let eh be discrete eigenJunctions with 1 for Ah, where Ao (h —. 0). Then there exists a subsequence such that converges in to an eigenft*nction 0 0 e E E(Ao).

-

we have

—.0.

PROOF. Because

C (cf. (2g) the functions

:= Pheh are uni-

formly bounded. is compactly embedded in L2(fl) (cf. (2a) and Theorem 6.4.8a) such that a subsequence e5' converges in L2(a) to an e E L2(J2): — do — 0. We have in particular

Estimate (2c) ytelds fR,1z

—

for z = (A0 From —

—

—.

0

=

(Ao

IRaz — zsIi ChIeIo — 1s)L'e, := (A0 — —

such that 1R5(e

—

—

0

eh) + (A5 — A0)ea —. 0 in follows 0 (h —. 0) results. Lemma

2b shows that e = z E HJ(a), and 0 = = IlimeaIi = 1.

e

E E(A0) is excluded because of

1

Theorem 11.3.9. Let (2a—m) hold. Let be the solution of the discrete cigenvatuc problem = with = I, A5 = Ao. Then there

11 Elgenvalue Pmbleins

272

such that

exists a subsequence

in

e' E E*(Ao). Furthermore,

tion 0

—

convei9es to on eigenfizncR,4e'Ii —'0.

Proof Sketch. The proof is not analogous to that of Theorem 8 since the consistency condition (2a,m) does not necessarily imply the corresponding statements for the adjoint operators. One has to carry out the following steps: (a)

converges in L2(f1)_for a subsequence h = —*0. the following holds: and Zh := (Ao —

:= Phek_—,

(b) For z =

(A0

—

=

z—

—

+

—

—

For each v E L2(fl) we obtain

=

(v,z —

(Ao

+ ([P, —

—

—*0

—

for h =

(cf. (2i), (2c)). —+0 may be inferred from Lh(zh — —.0. In = (Ao — particular, (v, — —' 0 for each v E L2(S?). — —'0 for each V E L2(Q) (cf. (21)). (d) — = —+

0

(c) Izh —

(e) follows (v,z — e')o = 0, thus z = e'

E'(A0), where

0.

ExercIse 11.3.10. Carry over Exercise 11.2.12 to the present situation. In Lemma 11.2.16 we defined

Now set

:= {uh: (uh,eja = 0} = = := inf sup I(LA,auh,vh)oI/(Iu,,I1 Ivali).

—-.

(11.3.6)

O$UkEch

A basic condition for the following is dim Eh(Ah) =

E

1,

Eh(Ah) = span

= )huh},

Exercise 10 shows that (7) is valid for h h0 if dim E(Ao)

Lemma 11.3.11. Let (2a-m), dimE(Ao) =

=uh}.

E

1,

1.

and (2.8) hol4L Then there

exi8t ho>OandaC>Othat is independent ofh0 andh, >0 for all

PROOF. (a) There exists a C >0 such that

11.3 Discretisation by Difference Methods

273

forallv,1E1?'h,aEC,h>O. Indirect Proof: Assume that there exists a sequence vh with 0. For a subsequence of := vp4 + E €,

—+0,

= 1,

mL2((I), := v +

P,4wh, —+

=

0

converge. From 0 = = (v,JI — + = 0, & = (w*,e*)o/(e*,e*)o = 0, (v,e)o one infers contradiction results from 1 = limIvh4Il = limlwh,I1 = 0. (b) The rest of the proof runs as in Lemma 11.2.16.

Lemma 11.3.12. choose0 eE

—+

v = 0. The

Let (2a—m), dim E(Ao) = 1, and (2.8) hold. One may

and eh

Eh(.Xh)

so that IR,,e—esli

PROOF. We have that e =

Assume also that The inequality (9.2.22) implies jRhe — zhJl ChIeI2. For sufficiently small h we have (e,,, 0 so that one can scale such that (eh — zh,ejo = 0. From ()io —

—

= (Ah one

zh)

=

(Ah

+

—

+ (Ah —

—

—

—

—

Rhe)

+ (R,1 —

infers that Ieh — ZhIl C(IAh — + h) (cf. Lemma 11) since I(Rh — = 0(h). The statement follows from this.

11.3.13. Under the a8sumptsons of Lemma

12 we have

the inequality

PROOF. Choose such that (R5e — eh, — ej0 = 0. For = (eh, the Rayleigh quotient Ah := Rhe*)o/(Rhe, R,1e)o we then have —

IAh

—

eh(h + I.Xo — AhjJ

with Ch := CIRhe — From (2c, j) one infers that

(LhRp,e,R5e')o — (Le,e*)o

=

— RhLJe,

+ (Le,

—

= 0(h)

= 0(h) such that Rh — For sufficiently and P'h — Ch + ehlAh — small h we have Ch (cf. Theorem 9, Exercise 10) thus Rh —Aol <2Ch. U Leminata 12 and 13 give

274

11 Elgenvalue Problems

Theorem 11.3.14. exists eh

= span {e}, and (2.8) hohL There

Let (2a—m),

Eh(Ah) with

—

Ch.

to 0(h), = o(1) or even IRhe — Since IR,1e' — Theorem 11.2.19 one expects that Pio — AhI = o(h) [resp. 0(h2)J. In general, this estimate is false as the following counterexample shows.

Example 11.3.15. The eigenvalue problem —*/' + u' =

Au in (0, 1) with has the solution u(x) = exp(Ax)sin(Rx) with A = 1/2. The eigcnvalue is ).o = + 1/4. One calculates that the discretisation + = Au has the eigenvalue

u(0) = u(1) =

0

= h212 — + h_11c08(,rh)eA'hI — 1 + +

—

—

with A':=log(1—h)/(2h) such that IA0 — AhI turns out no better than 0(h).

12 Stokes Equations

12.1 Systems of Elliptic Differential Equations In Example 1.1 11 we have already stated the Stokes equations for 47 C 1R2:

— —

LIU2

+ Op/Oxi

fi,

+

12,

(12.1.lai) (12.1.182)

= 0.

8u1/Oxt —

(12.Llb)

another equation ÷ Op/0x3 = f3 needs to be In the case of 42 C added, and the left-hand side of (Ib) must be supplemented by —ôusJäz3. A representation independent of the dimension can be obtained if one takes together (u1, Ireep. (Ui, U2, u3)J as a vector u satisfying, in 42, the equations —

4u+ Vp =

—divu=0. Here,

I

(12.1.2a) (12.l.2b)

dlv is the divergence operator divu =

and n is both the dimension of 47 C JR" and the number of components of Only n 3 is of physical interest here. In fluid mechanics the u(x) Stokes equation describes the flow of an incompreesible medium (neglecting is the the inertial terms) and describes the velocity field. With x velocity of the medium in the direction; the function p denotes the Up to now we have not formulated any boundary conditions. In the fol. lowing we limit ourselves to Dirichlet boundary values:

u=0 onE'.

(12.1.3)

This implies that the flow vanishes at the boundary. No boundary condition is given for p. Since both the pairs (u,p) and (U,p + const) satisfy the Stokes equations (2a,b), (3), one obtains: Remark 12.1.1. p is determined only up to a constant by the Stokes equation (2a1b) and the boundary conditions (3).

276

12 Stokes Equations

The Stokes equations have been chosen as an example of a system of differential equations. It remains to investigate whether Equation (2a,b) is elliptic in a sense yet to be defined. Even though the functions U4, for given — in equations p, are solutions of the elliptic = determining p, (2a,b) do not provide an elliptic equation, in any current sense of elliptic. A general system of q differential equations for q functions Ui, ...

can

be written in the form (12.1.4a)

with the differential operators

=

(1

i,j

q).

(12.1.4b)

f where L is the matrix (Li,) of differential operators and u = (Ui,.. •,uq)7, I The order of the operator is at most ks,. Let the numbers ml,••-,mq,m'i,• .•,m'q be chosen so that The equations (4a) are summarised as Lu

(1
k11mj+m As the principal part of

(12.1.5)

one defines

=0.The with

characteristic polynomial

reads:

:=

E

and forms the matrix function

Definition 12.1.2.

(Agmon-Douglis-Nirenberg (1J) Let (5) hold for is said to be elliptic in x E a if

The differential operator L = 0

for all 0

E

(12.1.6a)

£ is said to be uniformly elliptic in a if there exists an E >0 such that for all XE

with 2m := indices 714,

(12.1.6b)

To be more precise one should call £ elliptic with since the definition does depend on In connection with

12.1 Systems of Elliptic Thfferential Equations

277

this problem, as well as for an additional condition for x E F, q = 2, see the original paper of Agmon—Douglas—Nirenberg [lJ. Further information on this subject can be found in Cosner Lii. are not unique. If mj Exercise 12.1.3. (a) Show that the numbers m,, m +k. The definition satisfy the inequality (5), then so do and is independent of k. of (b) Show that for q = 1, i.e., for the case of a single equation one recovers from (6a) the Definitions 1.2.3a Iresp. (5.1.3a)j; (fib) corresponds to (5.i.3a'). (c) For a first-order system (i.e., = 1, mj = 1, m = 0), (6a) coincides with Definition 1.3.2.

In order to describe the Stokes equations in the form (4a,b) we set

q:=n+i, for

The orders are

=

=

2,

= 1 (i n), and

=

0

otherwise. The

numbers

mq=m'q=O

for

satisfy (5).

coincides

=

with L,, and is independent of x:

=

=

i

for

= 0 otherwise.

n,

From this we see 0

IdetL"(e)I =

0

=

with 2m = 2n,

—iei2

... so that (6b) is satisfied, with 1.=

°

—en

Exercise 12.1.4. In elasticity theory the so-called displacement function .0 C JR3

JR3

is described by the Lamé system:

in!?,

on!'

A > 0). Show that this system of three equations is uniformly elliptic: IdetL"(e)I = (&,

For the treatment of Stokes equations we will use a variational formulation

in the next section. For reasons of completeneas we point out the following transformation.

Remark 12.1.5. Let n = 2 and thus u = exists a so-called stream function

(01,02). Because

with uj =

divu = 0 there

02 =

12 Stokes Equations

278

= Insertion in Equation (lat,2) results in the biharmonic equation = 0 on F. This 8f2/Ox1 — Ofj/0x2. The boundary condition (3) means =0 and 845/& =0 on F where ô/8t is the tangential is equivalent to = 0 implies = const on P. Since the constant may be derivative. =0 on P. chosen arbitrarily, one sets =

12.2 Variational Formulation 12.2.1 Weak Formulation of the Stokes Equations Since u = (u1,. •. , u,1) is a vector-valued function, we introduce

Ho'(a) x

x

(n-fold product).

x

A corresponding definition holds for H'(S?), H2(Q), etc. The norm associin the following. will again be denoted by ated with According to Remark 12.1.1 the pressure component p of the Stokes problem is not uniquely determined. In order to determine uniquely the constant in p = fi+const, we standardise p by the requirement p dx =0. That is the reason why in the following p will always belong to the subspace C

:= {p E L2(.O): J p(x) dx = O}. To derive the weak formulation we proceed as in Section 7.1 and assume that u and p are classical solutions of the Stokes problem (1.2a,b). Multiplication of the ith equation in (1.2a) with Vj E Cr(s?) and subsequent integration implies that

j =

f

d(x) =

+

{(Vui(x), Vvj(x)) —

dx

dx for

E

(12.2.la)

1
Summation over i now gives

J{(Vu(x), Vv(x))

—

p(x) div, v(x)} dx

=

L (1(x), v(x)) dx,

(12.2. la')

where the abbreviation

=

(Vu,Vv) i=1

*3=1

used. Equation (1 .2b) is then multiplied with some q grated, giving is

and inte-

12.2 Va atlonal Formulation

_fq(x)divti(x)dx = 0

279

(12.2.lb)

for all q E

With the bilinear forms

J(Vu(x)7Vv(x))dx for u,v E

a(u,v) b(p, v) —

J p(x)div v(x) dx

(12.2.2a) (12.2.2b)

vE

for p

we obtain the lVweak formulation of the Stokes problem as (3a-c):

Find u

such that

and p E

a(u,v) + b(p,v) b(q,u)

1(v) :=

J(f,v)dx

(12.2.3a)

for all v E

(12.2.3b)

for all q

0

In (3b) we first replace "v E

(12.2.3c)

Since both sides of and since Cr(I1) is dense in RW?),

by "v

(3b) depend continuously on v E (3b) follows for all v E

Remark 12.2.1. A classical solution u E

pE

of the Stokes problem (1.2a,b), (1.3) is also a weak solution, i.e., a solution of (3a-c). If conversely (3a-c) has a solution with u C2(Th, p C1 (1?), then it is also the classical solution of the boundary value problem (i.2a,b), (1.3).

PROOF. (a) The above considerations prove the first part. (b) Equation (3c) implies divu = 0. Let i {1,.. . ,n}. In Equation (3b) one can choose v with vj =0 for j i. Integration by parts recovers (la) and hence the ith equation in (1.2a).

I

12.2.2 Saddlepoint Problems The situation in (3a—'c) is a special case of the following problem. We replace the spaces and in (3a—'c) by two Hilbert spaces V and W. Let o(., .): V x V b(., .): W x

—' IR

V

JR

on V x

V,

(12.2.4a)

a continuous bilinear form on W x

V,

(12.2.4b)

a continuous bilinear form

fi

E

V',

12

E W'.

In generalisation of (6.5.1) we call b(., .): W xV

if there exists a Cb

JR

(12.2.4e) JR

continuous (or bounded),

such that CbIjwIIwIIvIIv

for all w E W,v E

The objective of this chapter is to solve the problem (5a-c):

V.

12 Stokes Equations

280

Find v V and w W with a(v,x) + b(w,x) = fi(x) for all x E V, b(y,v) = f2(Y) for all y W.

(12.2.5a)

(12.2.5b) (12.2.5c)

Formally, (5a—c) can be transformed to the form

Find u E X with c(u,z) = f(z) for all z E X

(12.2.6a)

if one sets:

X =V x W,

c(u,z) :=a(v,x)+b(to,x)+b(y,v) and

(x).

1(z) =fl(x)+f2(y) for u=

(12.2.6b)

Exercise 12.2.2. Show that (a) c(., .):X x X —+ IRis a continuous bilinear form.

(b) Problems (5a—c) and (6a,b) are equivalent.

That the variational problems (5) and (6) must be handled differently than in Section 7 is made clear by the following remark.

Remark 12.2.3. The bilinear form c(•,.) in (6b) cannot be X-elliptic.

(0).

PROOF. We have c(u,u)=0 for all tL= In analogy to (6.5.9) we set

J(v,w) := a(v,v) + 2b(w,v) and

—

2fi(v) — 212(w),

therefore J(v, w) = c(u, u) — 21(u) for u =

(t').i'or v E V, w E

W

we know J(v, w) is neither bounded below nor above. Therefore the solution v,w' of (5a—c) does not give a minimum of J; however, under suitable con-

ditions, v,w' may be a saddle point. Theorem 12.2.4. Let (4a-c) hold. Let a(.,.) be symmetrsc and V-elliptic. The pair

E V,

J(v,w)

E W is a solution of the problem (5a-c) if and only if

J(v,w') J(v,w)

for all v E V,w

W.

(12.2.7)

Another equivalent characterisation is

J(v*,w*) = mini(v,w) = max mmJ(v,w). VEV

WEWvEV

PROOF. (aa) Let v',w solve (5). The expreselon in brackets in

(12.2.8)

12.2 VariatIonal Formulation

w) = a(v

J(v, w) —

v,

—

—

281

v)

_v)+b(w*,v* —v) —fi(v —v)J vanishes because of (5b). Since a(v' — v,v — v) >0 for all v v E V, the second inequality in (7) follows. One also proves the converse as for Theorem

6.5.12: If J(v,w) is minimal for v = (ab) If v is a solution of (Sc), then

J(v',w) — J(v', to) =

then (5b) holds.

21b(w*

—

to,?) — f2(W' — to))

vanishes for all to, which proves the first part of (7) in the stronger form J(v,w) = J(v,w). If, however, v is not a solution of (Sc), there exists a WE W such that 6 := — J(v,w) 0. Since J(v',w') — J(v',ii,) = —5 for tis := 2w — to, the first inequality cannot be valid. For the converse = define to. The first part of (7) implies

J(ve,w) _J(v*,w±) =

0

for both signs, and so b(w,v') (Sc).

f2(W)J

= 12(w). Since to W is arbitrary, one obtains

(ba) We set j(w) := mInVEv J(v,w). According to Theorem 6.5.12, j(w) = to), where E V is the solution of (5b). If and are the solutions for to and to', it follows that

—v,,x) = F(s) := b(w—w',x) for all x Since IIFIlv'

— w'Ijy and IIvw

IIvw — vwsflv 5

(bb) By using the definition of

S C'IIFIIv' one obtains

—

— w'IIv

V.

for all w,w" E W.

(12.2.9a)

in (5b) we can write:

J(v*,w*)_J(vv,w)__a(v*,v*)+2b(w*,v*)_211(v)_212(w*) + 2b(w,

—

=

—

+

v*)

—

b(w,

—

—

—

v')

—

2f2(w)J

v'))

to,?) — f2(w — to)] _w,v*)_12(w* w)j.

.—?) + 2jb(w

—

—

(12.2.9b)

(be) Let v', wa be a solution of (5a-c). Because of (Sc) the expression in brackets in (9b) vanishes and we have

J(v*,w*) = (5b) gives

=

+

—

—?)

= j(w).

and so

J(v,w) =j(w') = maxj(w); wEW

(12.2.9c)

282

12 Stokss Equations

i.e., (8) holds. (bd) Now let v', w be a solution of (8). If in (9b) one sets w = w',

=

oneobtainsfromJ(v',w)=j(w')thatv =vw..Henoea(vw_v*,vw_vs)= depends quadratically on — — WIIV (ef. (9a)). The — variation over w := w' — Ày (A IR, y W arbitrary) givee

0=

= 21b(y,v)

—

—

1(y)],

and so (Sc) is proved. (Sb) has already been established with v =

v,L,..

12.2.3 Existence and Uniqueness of the Solution of a Saddlepoint

Problem To make the saddlepoint problem (5a-c) somewhat more transparent we introduce the operators associated with the bilinear forms: A

L(V,V') with a(v,x) =

for v,x

B*EL(V,Wl) with b(w,x) = C E L(X, X') with c(u, z) = (Cu,

V, (12.2.IOa)

BEL(W,V'),

= (w,Bz)wxw', (12.2.lOb) for u, z

Thus problem (6a,b) now has the form Cu =

X.

(12.2.1(k)

while (5a-c) can be written

as

Ày + Bw = B4v = 12.

If one assumes the existence of A1 v= and

fi

(12.2.lla) (12.2.llb)

L(V', V)1 one can solve (ha) for v:

A'(f1

—

Bw)

(12.2.12a)

substitute in (lib):

=

BA'f1

—

12.

(12.2.12b)

The invortibility of A is in no way necessary for the solvability of the saddlepoint problem. However, it does simplify the analysis, and does hold true in the case of the Stokes problem.

Remark 12.2.5. (a) Under the assumptions E L(V',V),

E L(W',W)

(12.2.13)

the saddlepoint problem (5a—c) jresp. Equations (1 la,b)] are uniquely solvable.

(b) A necessary condition for the existence of (BA'B)' is BE

V') is injective.

(12.2.14)

12.2 Variational

Formulation

283

PROOF. (a) Under the assumption of (BA'B)' E L(W',W), (12b) is uniquely solvable for w, and then (12a) yields v. (b) The mjectivity of BA 1B implies (14).

I

Attention. In general, B: W —+ V' is not bijective so that the representation is not possible. The example of the 3x 3 matrix of(BA'B)' as A

=

and B =

=

shows

that a system

of the form (1 la,b) can be solvable even with a singular matrix A. Therefore A closer look reveals the the assumption A—1 E L(V', V) is not subspace V0 := ker

B = {v

V: B'v = O} = {v

V:b(y,v) =

0

for all yE W} C V, (12.2.15)

as we noted before, in general is not trivial. The kernel of a continuous mapping is closed so that V, according to Lemma 6.1.17, can be represented as the sum of orthogonal spaces: which,

V=

V0

:= (V0)1.

with

V1

(12.2.16a)

ExercIse 12.2.6. Let (16a) hold. Show that (a) V' can be represented as where

:= {t/

(12.2.16b)

V':v'(v) = 0 for all V E Vj.}, (12.2.16c)

:= {v' E V':v'(v) = 0 for all v E Vo}.

AsanormonV01andVf oneusesll.IIv'. (b)The Riesz isomorphismJv:V—' V' maps V0onto V0' and V1 ontoVf (d) The following holds:

=

E

E Vf

(12.2.16d)

The decompositions (16a,b) of V and V' define a block decomposition of the operator A:

Aoj1 I

A

L.tilo A00

L(Vo,V01),

A01

L(Vj,Vo'),

A

I)

It11J

A10

L(Vo,VI),

A11

L(V1,Vfl.

Here, for example, A00 is defined as follows:

forvoEVo The corresponding decomposition of B into (Ba, B1) is written as (0, B), since =0 to the definition of V0. Conversely, we have range (B) C so that B = System (lla,b) thus becomes

284

12

Stokes Equations Aoovo

(12.2.17a)

+ A01v1 = ho

A10v0 + Ajjvj + Bw = B*VJ. = /2 where v =

+vJ.1

E V0, vj E Vj,

= ho +

(12.2.1Th) (12.2.17c)

fu, hoE

E

Theorem 12.2.7. Let (4a—c) hold. Let V0 be defined by (15). A necessary and sufficient condition for the unique solvability of the saddkpoint problem (5a-c) for all Ii V' is the aristence of the inverses E

E

L(Vf,W).

(12.2.18)

PROOF. (a) (17a—c) represents a staggered system of equations. (18) implies B*_l = (B—') L(W',V1) so that one can solve (17c) for vj = B''f2. v0 E V0 is obtained from (17a): vo = — Ao.jvj). Finally, w results from (17b). (b) In order to show that (18) is neceesary, we take ho V01 arbitrary, /1.1. = 0 and 12 = 0. By hypothesis here we have a solution (vo, vl, w) K Vo x V1 x W. = 0 because V0 n V1 = {0}. Thus = 0 implies Vj K Vo, so that Aoovo = ho has a unique solution vo E Vo for each 110 V01. Since A00: Vo is bijective and bounded, Theorem 6.1.13 shows that A&J' L(VO', Vo). If arbitrary and ito = 0, 12 = 0, one infers vj = 0 and one takes hi E = 0, so that Bw = has a unique solution w K W. As we did for A00,

Iii

one also infers that B1 E L(Vf, W). The formulation of conditions (18) in terms of the bilinear forms results in the conditions: inf{sup{Ia(vo,xo)I:zo E Vo, IIxoIIv = 1}:vo

Vo,flvoIIv = 1)

? a >0, (12.2.19a)

sup{Ia(xo,vo)I:xo K

Vo, IIxoIIv

= 1} >0 for all 0

V,HxIIv = 1}:w€ W,flwIIw

vO E V0,

(12.2.lOb)

1}fl>0. (12.2.19c)

ExercIse 12.2.8. Show that (19a) [resp. (19c)J are equivalent to (19a') [reap. (19c')j:

sup{ia(vo,zo)I:xo K Vo,IIxoIIv = 1} ajIvoIIv

for all v0

V0,

(12.2. 19a')

sup{Ib(w,x)I:z E V, Dxliv

LemmR 12.2.9.

1} I3Dwllw

for all w€ W. (12.2.19c')

Let (4a,b) hold. Let V0 be defined by (15). Then conditions (18) and (19a—c) ore equivalent. Here we have 1/a, 1/13.

12.2 Variational Formulation

= 0 for z E Vo one can write (19c) in

PROOF. Because of (16d) and the form inf{sup{Ib(w, x)j: x E V1. HxItv = For 0

285

l}:w

W, llwIIw = 1}

> 0. (12.2. 19d)

Vo, therefore according to (15):

x E V1 one has that x

= 1) >0 for all 0

E

xE V1.

(12.2.19e)

As in the proof of Lemma 6.5.3, we obtain the equivalence of (19a,b) with U and of (19d,e) with B' e E

Corollary 12.2.10.

(a) Condition (19b) becomes superfluous if a(.,.) is

symmetric on Vo x Vo or tf Lemma 6.5.17 applies. (b) Each of the following conditions is sufficient for (19a,b) and hence aLso

for

E L(V01, V0):

a(., .): Vo x Vo

JR

is V0-eliiptic:

for all vO E Vo,

a(vo,vo) a(., .): V x V

JR

(12.2.20a) (12.2.20b)

is V-elliptic.

PROOF. (a) As in Lemma 6.5.17. (b) (20b) implies (20a); (20a) yields U

(19a,b).

(4a—c), (18) is also

ExercIse 12.2.11. Show that under the equivalent to the existence of C1

L(X',X) (cf. (lOc)). Find a bound for

flC'flx._x' in terms of flAoo'flv0*_v0', IIAUv'4_v', and

12.2.4 Solvability and Regularity of the Stokes Problem Conditions (19a,b) (i.e., problem:

E L(V0", Vo)) are easy to 8atisfy for the Stokes

Lemma 12.2.12. Let.0 be bounded. Then the forms (2a,b) describing the Stokes problem

the conditions (4a,b) and (19a,b).

PROOF. (4a,b) is self-evident. According to Example 7.2.10, (.0)-elliptic. From this follows the H,?j(Q)-ellipticity of is

lobproves(19a,b).

Vv) dx Corollary U

It remains to prove condition (19c), which for the Stokes problem assumes the form

sup{I f w(x)divu(x) dxl: n E

luli =

l}

I3lwk

for all w E

(12.2.21)

286

12 Stokes Equations

or

(12.2.21')

for all w E

IIVWIIH-1(n)

Lemma 12.2.13. Sufficient and necessary for (21) is that for eachw E such that there exists u E w

PROOF. (a) For w The

=divu,

luli

fl'Iwk.

(12.2.21")

select u such that (21") holds and set ii

E

left-hand side in (21) is faw(x)divü(x)dx =

(b) If (21) holds one infers as in Section 12.2.3 the bijectivity of B: Vj —, W B'1w satisfies condition (21"). 1/ft. Therefore u with

Necäs [2j proves

Theorem 12.2.14. Condition (21) is satisfied If 11 E C°" is bounded. Under this assumption, the Stokes problem —4u + Vp = f,

—

div u = g in $1,

has a unique solution (u,p) luli + Info

u =0 on F (12.2.22) forf E H'(Q) andg E (12.2.23)

I/I—i + Igk].

Remark 12.2.15. Under the conditions that n = 2 and that 1? E C2 is a bounded domain1 the existence proof can be carried out as follows. PROOF. We need to prove (21"). Fbr w solve = win .0, go =0 on F. Theorem 9.1.16 shows that go E Since Vgo H1(f1) and n(x) C2(F) it follows that g := ôu/8n H'/2(f') (cf. Theorem 6.2.40a). From (3.4.2) one infers that = fawdx = 0 since w Integration of g over F yields C with OG/Ot = g, where is the tangential derivative. There exists a function %b E with = C and = 0 on F and CIGI3,2 C"Iwk. We set

Clearly, ul,u2

E

H'(.0). Let the nonnal direction at x

(nI(x),n2(x))T. The tangent direction is thus t(x) u = (u1,u2)T one obtains

(u, n) =

=

—cP5fli

(u,t) =

—goxn2

—

+ g011n1 —

F be n(x) =

(n9(x), _ni(x))T. For

= —g-i-ÔC/8t =0,

= —&p/8t

—

= 0,

since go = 0 on F implies Ogo/àt = 0. (u, a) = (u, t) = 0 yields u = 0 on 1' such that u = (ui, u2) E is proved. One verifies that

12.2 VariationaL Formulation

divlh = with

+

=

—

-

=W

+

.

k012 +

IUIi

287

The above proof uses the H2-regularity of the Poisson problem and requires corresponding assumptions on (1. Theorem 14 also assumes Q C°'1 Since the Poisson equation = I is solvable for any domwn .0 which is

contained in the disk KR(0), or at least in a strip {x E < R}, reone might conjecture that a similar sulting in the inequality CRlf result holds for the Stokes problem.

E (0, 1) let = {(x,y): —1 0 such that (21") holds for all 0, w e

Counterexample 12.2.16. For

0
-1

0

1

FIgure 12.2.1. JZ

PROOF. We select w E such that w(s, y) 1 for x > 0, w(x, y) = —1 for x 0. Let the inequality (21") hold for with > 0. Let u with be chosen according to Lemma 13. We continue ti by u = 0 onto Ut . For the restriction on z = 0 we have according to Theorem 6.2.28

CJtzj)1

)IILZ(R)

Let u1(O,

= 1 for 0 < y < and x(v) and Ixto = we have

= Let

.f?(1

0 otherwise. Since ui(0,y) =

lui(O,.)IoIxIo

= {(x,y) E > 0} {(x,y):x = 0,0< y < Because w =1 in and because divtz w we have

=

288

Equations

12

Iw(x)12 dx

1/2

vi div u dx

=J =

=

J

dlv u dx

n) dE' = j (u,

_juidf= _Jui(O,v)dv.

The last two inequalities result in 1/2

from which we infer that

>0 cannot be independent of e.

ExercIse 12.2.17. Construct a domain 1? located on the strip JR. x (0, 1) in which the Stokes equations are not solvable. Hint: Join the domains

(v€N) from Figure 1. As for the case of scalar differential equations one obtains stronger regularity of the Stokes solution if one more than I E H-'(Q).

Theorem 12.2.18. Let £2 be bounded and sufficiently smooth. Let u and p be the (weak) solution of the Stokes problem (22) with I E Hk(fl), 9 E Hk+l(a) fl for k E t4 U {O}. Then we have u E Hk÷2(.Q) fl and there exists a C depending only on £2 such that

fl

pE

PROOF. Cf.

(12.2.23b)

Ia + IgIa+i].

+ LPIk+1

El, Chap.llI, §5]

U

In analogy to Theorem 9.1.22 it is sufficient to £2 in order to obtain u E H2(a) and p H'(s?) from f

the convexity of 112(fl).

Theorem 12.2.19. (Cf. Kellogg-Oaborn [1]) Let £2 C JR2 be bounded and convex. If 1€ L2(S?), then the Stokes equation (1.2a,b) has a unique solution H2(a) n p E H'(a) fl which satisfy the estimate

CIfIo.

1u12 +

For the more general problem (22) with g the solution satisfies 1u12

1ff

+ lpli

L2(a) and g E

(12.2.23c)

0 in a convex polygonal domain

+

If

(12.2.23d)

IIJ(a). Here, HJ(fl) is the sub8pace of H'(a)

with the following (stronger) norm:

:=

+

with

IaI=1

6(x):=min{Ix—eI:e€F corners of thepolygona}.

12.2 Varia*onal Formulation

289

12.2.5 A V0-elliptic Variational Formulation of the Stokes Problem 0). Vo is a closed As kernel for the mapping B = In the i.e., again a Hubert space for the same norm subepace of following we investigate the problem

has been defined in (15) by V0

V0 c

{u

—div

.

Find u

V0

with a(tz,v) = f(v)

(12.2.24)

for all v E V0,

f0(Vu(x), Vv(x)) dx. Problem (24) has the same form as where a(u, v) . , n), only equations the weak formulation of the = (i = has been replaced by Vo. here

Lemma 12.2.20. Let

be a bounded domain (or bounded in one direction;

cf Exercise 6.2.12b). The form a(.,.) is Vo-elliptic. The constant C2 > 0 in depends only on the diameter of 4'?. In particular, problem a(u, u) for 1€ V01. (24) has a unique solution u E V0 with

PROOF. The

of a(.,.) (ci. Lemma 12) carries over to Vo C (cf. Exercise 6.5.6a). This implies the other statements (cf. Theorem

I

6.5.9)

Theorem 12.2.21. LetS? E C°" be a bounded domain. Assume f E H'(Q). Then the solution u Vo of problem (24) coincides with the solution component u of the mixed formulation (3a--c).

PROOF. According to Exercise 6 one can split / E

in such a way that

f—fo+fi, fo€V0', LEV', fo(v)=Oforv€Vj, fj(v)=Ofor vEV0. In (24) one can replace f(v) by fo(v). u V0 c that belongs to Vu is H1(.O). The part of

gj(vj)

results in E VI with

a(u,vj) for all vj. E V1 =

E

(V0)1.

Theorem 14 proves the condition (2!'), which results in the bijectivity of B V (cf. Lemma 13).p B'(fj. —gj), by definition, satisfies

= f1(v1)

b(p,vj) =

=

fj(vi) —a(u,v1)

for all v1 E V1. Furthermore, since div Vi,) = 0 for v0 E V0, it follows that b(p, vij)

For arbitrary v E V1 E V1,

=0

for all V0 E V0.

to be split into v =

+ V1 with

E V0 and

one obtains

b(p, v) = b(p, vo) + b(p, vI) =

fj(vi)

—

a(u,vj.)

f(vj.) — a(u,v) + a(u, v0)

12 Stokes Equations

290

= f(vj)—a(u, v)+f(vo) .= f(v)—a(u,v), because u E 0 for all to E satisfy the variational formulation (3a—c) of the Stokes problem.

by (24). Since also b(w,u) =

V0,

u and p

Note that Problem (24) is solvable for all bounded domains although Problem (3a—c) depends more sensitively on Al (cf. Counterexample 16). Theorem 21 shows that only the component p has a domain-dependent bound ClfoH Gill_i holds for all .0 C KR(0). Ii'io
Strictly speaking, the variational problem (24) is not equivalent to the

Stokes problem since, for example, for A? from Exercise 17 the Stokes equations are not solvable, whereas Problem (24) definitely has a solution. The original formulation (3a—c) may be interpreted as Equation (24), into

which one has introduced the "side condition" div u = function p (cf. Section 8.3.6).

12.3

0

using the Lagrange

Mixed Finite-Element Method for the Stokes Problem

12.3.1 FInite-Element Discretisation of a Saddlepoint Problem One would have an ordinary Ritz-Oalerkin discretisation if in the variational formulation (2.24) one were to replace the space V0 by a finite-dimensional subapace Vp, c V0. But this is not as easy as it sounds.

ExercIse 12.3.1. Let the square (1 = (0,1) x (0,1) be triangulated regularly as in Figure 8.3.2, 5a or decomposed into grid squares. Let c Hd(i7) be the space of the finite triangular elements (cf. (8.3.8)) [resp. of bilinear elements (ef. (8.3.12b))J. Define the corresponding subepace for the Stokes and divu =0} C Vo. Show that problem as Vp, := {u = (ul,u2):uj,u2 E Vp, contains only the null function.

The remaining procedure is thus oriented toward the weak formulation Vh x Wh. The discrete

(2.3a—c). The space X = V x W is replaced with X,, = problem

Find ii'

V,,

+

and

E Wp, with

= fj(z) =f2(y)

for all xE Vp,,

forallyEWp,,

(12.3.la) (12.3.lb) (12.3.lc)

is called a mixed Ritz-Galerkin problem [reap. a mixed finite-element problem if Vp, and Wp, are formed from finite elementsj. To the formulation (2.6a,b) corresponds the equivalent way of writing (la—c)

Find xh

Xh with c(x",z)= f(z) for all z E

Xp,.

(12.3.1')

12.3 Mixed Finite-Element Method for the Stokes Problem

291

In the case of the Stokes equations, (2.3a—c) provides the desired solution which satisfies the side condition div u = 0. However, the finite-element solution of (la-c) generally does not satisfy the condition dlv =0. One can view from (la—c) as the solution of a nonconformal finite-element discretisation of (2.24), as is shown in the following exercise.

Exercise 12.3.2. Let /2

0 in (ic); set := {z E Vh: b(y,z) = Show that each solution v1' in (la-c) is also a solution of

for all y E

Find

= fi(x) for all z

with

Since in general VO,h (2.24).

0

(12.3.2)

V0 (cf. (2.15)), (2) is a nonoonformal discretisation of

Let where

and Nv,h

OfVh and Wh,

dim

(12.3k)

:= dim Wh.

Let the coefficients of v E Vh and w E W,, be v and w: NV.h

v

NW.h

= Pvv :=

w = Pww ;=

(12.3.3b)

As in Section 8.1, we prove

Theorem 12.3.3. The varw.tionoj problem (la—c) is equivalent to the system of equations

fAa BhJ IV] In] 0 j {wJ =

(12.3.4a)

where the matrices and vectors are given by

(1 i,j := f1(bfl,

f2,h :=

1 k Nw,h)

I Nv,h; 1 k

(1

Nw,h).

The connection between (la—c) and (4a-c) is given by v" = Pvv, wh one obtains the system of equations (i), f :=

For u

•. '-)hUL, a-'

-

a

La

a

I'

(12.3.4b) (12.3.4c)

Pww.

(123.4a)

J

which corresponds to the fonnutation (1').

12.3.2 Stability Conditions When selecting the subepaces and one has to be careful, for even seemingly reasonable spaces lead to singular or unstable systems of equations (la-c). The former occurs in the following example.

292

12 Stokes Equations

Example 12.3.4. Let the Stokes problem be posed for the L—shaped domain in Figure 8.3.2. For all three components uI,u2, and p we use piecewise linear triangular elements over the triangulation r which is given by the second or third picture of Figure 8.3.2. Here, let uj = u2 =0 on F and let p Wh satisfy i.e., f0pdx = 0. Then the discrete variational the side condition p problem (la—c) is not solvable.

PROOF. The second ithird) triangulation in Figure 8.3.2 has 5 jlO] inner Thus dim Vp, = nodes x', with each of them carrying values ui(x') and p dic — 0 the 20]. Because of the side condition 2 5 = 10 frlim Vp, dimension of Wp, is smaller by one than the number of inner and boundary nodes: dim Wp, = 21 -1 = 20 fdimWh = 21]. In both cases the statement -

results from

Lemma 12.3.5. It is necessorij for the solvability of (la—c) that NV,,, NW,,,, i.e., dixnVh

dlmWh.

PROOF. According to Theorem 3 the solvability of (la-c) is equivalent to the nonsingularity of the matrix C,, in (4a'). Elementary considerations show that rank (C,,) Nv,,, + rank (B,,). Since B,, is an Nv,,, x Nw,,,-matrix it follows from Nv,,,
I

singular.

Thus an increase in the dimension of W,, does not always lead to an better approximation of w W. The choices of Vp, and W,, must be mutually adjusted. The inequality dim Vp, dimW,, corresponds to the requirement that B in (2.lOb) needs to be injective, but not necessarily to be surjective. In order to formulate the necessary stability conditions, we define V0,,, := {v

Vp,: b(y, v) = 0 for all y E W,,}

(12.3.5)

(cf. Exercise 2). VOJ, is the discrete analogue of the space Vo in (2.15). The read: conditions, which can be traced back to Brezzi inf{sup{ja(vo, xo)I: x0 E

llxollv

>0,

1}: vo E Vo,h, Iivellv = 1}

(12.3.6a)

inf{sup{Ib(w,x)i:x

= 1}:w

Vp,,

W,,,

= 1}

>0. (12.3.6b)

Theorem 12.3.6. Let (2.4a-'c) hold and dim Vh
+

]112

Chill lix' := ChI IjfilIv' +

1112112W,

(12.3.7)

where Ch depends on ap,, Ph and on the bounds Ca and Gb in la(v, x)l CaiIvtivilxllv and Ib(w,x)I Cbi(wi(w1ixiiv. If

12.3 Mixed Finite-Element Method for the Stokes Problem

a >01

293

(12.3.6e)

øh f3> 0

for all parameters h of a sequence of discretisations, then the di8cretisation is said to be stable and C,1 remains botrnded: C,1 C for all h. holds

V,1, Wh PROOF. Theorem 12.2.7 and Lemma 12.2.9 are applicable with (2.19b) follows instead of V0, V, W. (2.19a,c) then are the same as (6a,b), while
I

Condition (6a) is trivial for the Stokes problem: Exercise 12.3.7. Show that Condition (6a) is always satisfied with a constant JR is V-elliptic. czh independent of h, if a(., .): V x V

12.3.3 Stable Finite-Element Spaces for the Stokes Problem 12.3.3.1 Stability Criterion must hold. In a C For the Stokes problem C so that (6a) is satisfied with bounded domain a(•,•) is is somewhat more difficult to prove the conditions (6b,c), which CE > 0 . It the form for the Stokes problem sup{Jb(p,u)I:uE

= 1}J3IpIo

forailpE W,1

(12.3.6d)

wherein $> 0 must be independent of p and h. it is simpler to prove the modified condition sup{Ib(p,

u E V,1, Iuk = 1} /3IpIi

for slip E W,1

(12.3.8)

occurs, the latter > 0 independent oip and h). Since in (8) the norm requires Wh C H1(i7). This excludes, for example, piecewise constant finite elements. Condition (8) is sufficient for staTogether with the usual bility.

Theorem 12.3.8. Let 11 E C°'1 be

Let the Poisson problem be

regular (it is sufficient that 1? be convex; ef. Theorem 9.1.22). Let V,1 satisfy

the approxsmation property inf{ju —

u E 112(11) fl

(12.3.9a)

and the inverse estimate

for all

E V,1.

Then Condition (8) is sufficient for the Brezzi condition (6b,c).

(12.3.9b)

294

12 Stokes Equations

= I and with b(p,u) I31PIO (cf. Theorem 12.2.14 and (2.21)). According to Exercise 9.1.24, on Vh satisfies the conditions the orthogonal there exists u

PROOF. (a) For given p

IuIi=l, From b(p,uh) Chtpli and

b(p,u) — b(p,e) 13(plo — b(p,e)

—

Iplilelo $Ii'lo

—

1 one infers (12.3.10)

E Vh, Iv"li = 1} f3lpJo — ChlpIi.

(b) Because of (8) and (9b) there exists

1 and

E V,, with

=

Ib(p,ti')I From this follows

= 1}

Vh,

with

:=

and (11) with

If one multiplies (10) with

:= //(C +

= 1)

E

(12.3.11)

the sum reads: (12.3.12)

(6b,c)j has been proved.

Since 13 is independent of p and h, (6d)

12.3.3.2 Finite-Element Diacretisatione with the Bubble Function In the following let fl be a polygonal domain and an admissible triangulation. Example 4 shows that linear triangular elements make no sense for and p. We increase the dimension of Vh by adding to it the so-called "bubble functions" or "bulb functions". < 1} the bubble On the reference triangle T = >

function is defined by :=

—

—

i)

in T,

u

0 otherwise.

(12.3.13a)

The name derives from the fact that u is positive only in T and vanishes on (cf. Exercise 8.3.14) OT and outside. The map -. T to a general T results in the expression ü.1(x,y) :=

(12.3.13b)

for the bubble function on

Exercise 12.3.10. Let

be a quasiuniform triangulation. Show that there dzdy. exists a C >0 independent of h such that dx dy

295

12.3 Mixed Finite-Element Method for the Stokes Problem

We set

linear combinations of the linear elements E H0t (fl)

and the bubble functions for f'

Wa: linear elements E

Vh :=

For the side condition have

(12.3.14)

E

see Section 8.3.6. Since ü..p e H0t(fl), we

C

C

be the quasiunifonn tnangulatson on a bounded and Wh be given by (14). Then the stability conpolygonal domain £7. Let and that the dition (8) is satisfieaL Under the further conditions that Si E Pois8on problem be H2(Si)-regt4ar the Brezzi condition (6b,c) also holds.

Theorem 12.3.11. Let rh

PROOF. (a)

On every T E Ta, Vp is constant:

an arbitrary p We set

Vp = v := >

E

so that

ii := v/lyle

(ü.jr. bubble function (13b) on

= 1. Exercise 10 yields

b(p, v)

J (Vp, v) dx = TEm

+

= 0

f

T

Since lVplo and lpI' are equivalent norris on the subapace H'(Q) fl

it follows that lb(p,v)l

and

C'lplilVpk/jvk. In a

similar way one shows that lvlo

C"(VpIo and obtains 43 := C'/C" independent of h. The left-hand side in (8) is

with so

that

(8) follows.

(b) (9a) is satisfied for Vh (cf. Theorem 8.4.5). The same holds for the inverse estimate (9b) (cf. Theorem 8.8.5). Therefore (6b,c) follows from Theorem 8. U

A general result on the stabilisation by bubble functions can be found in Brezzi-Pitkäranta [11.

12 Stokes Equations

296

12.3.3.3 Stable Discretisations with Linear Elements in Vh If one wants to avoid bubble functions, one must increase the dimension of V,, in some other way. In this section we shall consider for Vh and Wh two different triangulations and Th. By decomposing each T rh as in Figure 1, through halving the sides, into four similar triangles, one obtains Th12. We define:

: linear elements for triangulation Th/2,

V,, c

Wh c

H'(fl) fl

(12 3 15)

: linear elements for triangulation

or

V,, c

:

quadratic elements for triangulation Th,

W,, C H'(Q) fl

linear elements for triangulation r11.

FIgure 12.3.1. Theorem 12.3.12.

Th

'12

16

and Th/2

Theorem 11 holds analogowsig for V,, x Wh in (15) or

(16).

PROOF. (a) Let Vh x Wh be given by (15). Fbr each inner triangular side of the triangulation m there exist two triangles T17,T21, E with = (cf. Figure 2). Let 8/Ot be the derivative in the direction of 'y, let 8/On be the directional derivative perpendicular to it. There exists and with

+

=

1,

8/Ox = a78/On + b78/Ot,

O/Oy = b.1O/On

—

a70/Ot.

In contrast to Op/On, Op/Otis constant on T17 UT27 U We denote its value by pu,. The mid-pomt of is a node of rh/2. We define the piecewise linear function over by its values at the nodes u.1(x7) =

w,(x') = 0 at the remaining nodes

(12.3.17a)

and set v

:= E (

E Vh,

v/lv$o.

The sum E7 extends over all interior sides of Tj,. In

(12.3 ITh)

U T2-, we have

12.3 Mixed Finite—Element Method for the Stokes Problem

297

(Vp,

so that

J (Vp, (

I

dx =

T17UT2.1

Ch2>1pt1.y12. are the sides of T

Th (r5 is quasiunifonn!), then

From this one infers that b(p, v) 11, and finishes the proof analogously.

dx

i IVpIo, as in the proof of Theorem

(b) In the case of quadratic elements given by (16) one has the same nodes as in (a) (cf. Figure 8.3.8a and Figure 1). Use (17a,b) to define the quadratic function V E Vh and carry out the proof as in (a). U

FIgure 12.3.2.

Ti.,, Ta.-,

12.3.3.4 Error Estimates In the following, the condition (8.1.12) should be replaced by the stability condition (6a.-c). In the place of the approximation property (8.4.6') we now have the inequalities inf{Iv —

inf{Ip —

Vh}

E W,j

Chfi42 for all V E V =

for all p

W=

(12.3.18a)

(12.3.18b)

Condition (18a) is equivalent to condition (9a) in Theorem 8. The following theorem applies to general saddlepoint problems, and can be reduced to Theorem 8.2.1.

Theorem 12.3.13. Let u =

X = V x W be the solution of (2.5a—c) [resp. (6a)1. Let the discrete problem (la-c) with Xh = x C X satiify

12 Stokes Equations

298

the Brezzi condition (6a.-c) and have the solution

= (v's, w"). Then there

exssts a varsable C independent of h such that —

uhjlx Cinf{IIu —

(12.3.19)

Xh}.

PROOF. The Brezzi condition (cf. Theorem 6) yields S lix' for all right1-hand sides f X' in (2.5a--c), in particular for all f E Xh = The above inequality means t' for the operator Lh:Xh —, which belongs to c(., .): Xh x Xh -. IR. According to Exercise 8.1.16 is equivalent to Condition (8.1.12) for c(., .): X x X IR [instead of a(., .): V x V IRj with 1/n. Theorem 8.2.1 yields the statement (19).

For the Stokes problem with (),

instead of u =

inequality (19) is now rewritten as follows: uhi2 + [p

—

C2inf{lu

+

—

_phlo2:uh

(),

=

Vh,p" E Wh}

or lu

+

—

—

uhfi + [p

Wh}.

(12.3.20)

Theorem 12.3.14.

Let the Stokes equation (1.2a,b) have a solution u E (ci. Theorem 12.2.19). For the suhspaces

p

and Wh C l€t the Brezzi condition (6a—c) and the approximation conditions (18a,b) be Then the discrete solution satisfies the estimate C

lu —

+

—

P"lo

C'hflul9 + Ipli].

PROOF. Combine inequalities (20) and (18a,b).

(12.3.21)

N

Using the same reasoning as in the second proof for Theorem 8.4.11 with

c(..) instead of a(.,.) one proves

Theorem 12.3.15. For each f

L2(a), g E

fl H'(fl) let the Stokes

problem ÷ Vp f, —divu = g have a solution u E p H1(Q) fl with 1u12 + C[1f10 + 1gb]. Under the asstImptlon8 of Theorem 14 we then have the estimates —

+ [p — p'i—i

C'h[Ju(i + Iplol,

(12.3.22a)

—

÷ [p

C"h2[Iu12 + [p1']

(12.3.22b)

—

p41.i

for the finite element solutions. Here lpf—t is the dual norm

Corollary 12.3.16. Combining (22b) and (21) one obtains

12.3 MIxed Finite-Element Method for the Stokes Problem —

+

—p"Io

+

299

(12.3.22c)

A new exposition of the mixed finite-element discretisatiori can be found in Brezzi-Fbrtin [1).

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Index

Note: The page numbers given in bold-face refer to occurrences of the term

printed emphatically on the page indicated. Adjoint problem, 141, 190, 196 Adler problem, 159, 183 Agmon, condition of, 158 AnalytIc, 20 condItion, 139 BabuMca-Brezzi condition, 284 Banach space, 110, 111, 112, 113, 131, 135, 137, 161 Basis, 161, 165, 166, 171, 182, 183, 204, 291 hIerarchical, 202 Basis condition, 171, 172, 173, 175, 176, 177, 180, 181, 183, 200

Beltrami operator, 14 Biharmonic equation. Ses Differential equation Bilinear ftrin, 137, 154, 155, 227, 279

adjoint, 138, 190, 191 continuous or bounded, 138, 146, 157, 161, 185, 279, 280 corresponding to a pde of order 2m, 146, 210 corresponding to the Helmholtz equation, 147, 154, 208 corresponding to the Poisson equation, 149, 154, 165, 176, 180, 183

extension of a, 138 operator associated to a. See Operator, 138, 140, 142, 163, 209

symmetrIc, 138, 141, 149, 152, 164, 170, 200, 204, 208, 280, 285

V-elliptic (H"-elliptic, etc.), 140, 141, 147, 148, 149, 152, 153, 154, 159, 164, 170, 183, 200, 204, 212, 228, 280 V-coercive etc.), 141, 142, 143, 149, 150, 153, 154, 157, 158, 166, 171, 185, 210, 211, 212, 213, 218, 219, 222, 223, 226, 227, 228, 229, 256, 267

Block, block matrix, block partitionlog, 283 Boundary condition, 158, 226, 275 Dirichlet, 27, 35, 38, 65, 85, 86, 96, 145, 150, 159, 164, 174, 178, 215, 222, 229 discretisation of the, 100

first, 35,96 mixed, 96 natural, 152, 154, 155, 159, 168, 178, 218, 221 Neumann, 36, 65, 96, 154, 174, 183

periodic, 98 second, 96 third, 96 Boundary element method (BEM),

37 Boundary layer, 249

Index

308

Boundary value, 7, 9 10, 12, 26 Boundary value problem, 13, 19,

27,

28,85

Brezzi condition, 292, 293, 295, 298 Bubble function, 294, 295, 296 Cauchy-Riemann equations, 4, 7 Chequer-board. See Ordering Coefficient vector, 161, 255 Compact (cf. embedding, nine-point formula, operator), 135, 201, 257, 269 Complementary problem, 264 Complete, 24, 112 Completion, 112, 117, 122, 125 CondItion, 108, 204 Cone condition, 136 Conservation form, 244 Convex. See Domain Consistent, consistency, 59, 64, 69, 93, 94, 200, 228, 232, 233, 235, 239, 242, 267 Convection-diffusion equation, 247 Convergence, 168 of the difference method, 59, 60, 61,

69, 83, 84, 95, 102, 108,

239

of the eigenfunction, 255, 259, 260, 271, 272 of the 255, 258, 260 super, 202, 203 Coordinate transformation. See Domain Corner, re-entrant (cf. domaIn), 34, 223, 225 Covering, 127

10, 112, 113, 114, 115, 117, 119, 123, 141, 169, 170, 224

Dense,

Derivative, 104 conormal, 96, 201 normal, 15, 65, 95, tangential,

96,

96, 145

99, 103, 145,

224,

246, 278 weak,

115,

116, 172

Diagonalisable, real, 6, 7 Diagonally dominant, 47, 48, 49, 52 Irreducibly, 47, 48, 49, 52, 53, 92, 93, 250

Difference (divided), 38, 49, 62, 79, 240

backward or left-sided, 38,

269

108, 109 of higher order, 62 Difference operator, 10, 44, 121, 210, 13-point,

213,

228, 233

Difference

star (stencil), 44, 62, 91

Differential equation(s) biharmonic, 74, 95, 103, 104, 105, 106, 108, 158, 194, 218, 278

223,

order, 1, 3, 4, 6, 277 of order 2m, 104, 108, 145, 158 of second order, 2, 5, 8, 12, 85, of

first

90,150

ordinary, 1, 2, 3, 38, 248, 254 partial, 1, 2, 7 system of, 3, 4, 6, 7, 105, 275, 276

Differential operator, 5, 59, 60, 75, 85, 89, 104, 253, 276 formally ad.joint, 92, 95 boundary, 95, 154, 155, 156 Dirichiet integral, 144 Dirichiet principle, 144 Divergence (operator), 275 12 convex, 79, 197, 223, 224, 225, 231, 235, 236, 242, 288, 293 exterior, 36 general, 78, 100, 109, 196, 234, 241, 243 L-shaped, 13, 34, 127, 201, 243 normal, 15, 16, 19, 28, 33 transformation of the, 6, 89 unbounded, 23, 146, 147 Double-layer potential. See Potential Dual form, 131, 134 Dual space, 130 Eigenfunctlon, 222, 253, 254, 259, Domain,

263, 271 5, 6, 7, 10, 47, 51, 52, 53, 137, 142, 150, 258, 254,

Elgenvalue,

255, 256, 258, 259, 260, 261, 262,

264, 265, 271,

Eigenvalue problem, 7, 56, 65,

72, 91, 250

forward or right-sIded, 38, 227, 250 second, 38, 80, 238, 240

symmetric, 38, 70, 71, 102, 233, 249, 251, 252 Difference method, 38, 59, 65, 90, 91, 93, 94, 102, 108, 173, 177, 178, 202, 204, 247, 249, 267,

91,

274 137, 150,

11,

253

adjoint, 253, 256 elliptic, 253, 254, 255, 274 Ethptic,

4,

104,

5, 6, 7, 9, 10, 11, 12, 85,

276

Index

uniformly, 86, 87, 88, 89, 104, 145, 147, 148, 149, 150, 155, 222, 229, 276, 277 V-. See Bilinear form, 248, 256, 280, 285, 289, 293 Embedding compact, 185, 136, 142, 143, 150, 153, 171, 186, 253, 256 continuous, 118, 131, 133, 134, 136, 137, 141, 143, 171, 185, 253, 256 dense, 131, 133, 134, 137, 141, 171, 253, 256

Error estimates for difference methods, 190, 238, 239

for eigenvalue problems, 260, 263, 264

for finite element methods, 185, 190, 193, 196, 246, 266, 297, 298

for Ritz-Galerkln methods, 167 Estimate. See Error estimate, inverse estimate Existence. See Solution Extension. See Bilinear form, operator of a function, 123, 129, 215, 237 Extrapolation (method), 61, 84 Finite elements, 171, 172, 174, 251, 254

constant (bi)cubic, 182, 194, 195, 196, 296 (bi)Ilnear, 172, 174, 175, 178, 179, 182, 183, 185, 190, 194, 199, 204, 246, 247, 252, 295, 296 isoparainetrlc, 198,247 mixed, 290, 299 nonconforjnal, 199, 200, 291 of the serendipity class, 181, 182, 193

quadratIc, 181, 182, 193 Finite element method, 24 Five-point formula, 40, 41, 53, 60, 62, 92, 176, 190, 231 Form. See Bilinear form, dual form, 8esquilinear form Fourier expansion, 9 Fourier transform, 110, 119, 120, 124, 125, 126, 135, 213 FunctIonal, 132, 140, 146, 152 Fundamental solution, 16, 17, 28 Galerkin method. See RJtz-Galerkln, Petrov-Galerkln G6rding, Theorem of, 150

309

Gei'fand triple, 41, 133, 134, 136, 141, 163, 207, 253 (Ierschgorin, Theorem of, 46, 48 Gradient, 15 Green's formula, 15, 29, 144, 153 Green's function of the first kind, 28, 29, 30, 31, 33, 34, 95, 105, 144

discrete, 55, 72, 95 Green's function of the second kind, 35 Grid, 39, 40, 226 Grid function, 39, 44 Grid points close to the boundary, 63, 79, 83 far from the boundary, 42,63, 79 80 neighbouring, 41, 42, Grid size (width, step sIze), 39 Harmonic, 13, 15, 16, 19, 20, 22, 23, 24, 25, 144 Harnack, Theorem of, 20 Heat equation, 3, 4, 5, 7, 8, 10 Hehnholtz equation, 147, 154, 208 hubert space, 110, 114, 115, 117, 122, 132, 134, 279, 289 Hökler continuous, 29, 30, 110, 123, 125

Hyperbolic, 4, 5, 6, 7, 8, 9, 10, 11, 248

Improperly posed. See Well-posed Inclusion, 135, 136, 142 Initial-boundary value problem, 8, 9, 207 Initial value problem, 2, 3, 7, 8, 9, 38

Instationaiy problem, 10, 11 Integral equation method, 36, 37 Interpolation, 84, 124 Hermite, 194, 196, 207 spline, 194, 196, 207 Inverse estimate, 206, 221, 240, 293 Irreducible, 45, 54 Irreducibly diagonally dominant. See Diagonally dominant Lagranglan (6.ctor), 184, 290 Lamé differential equations, 277 Laplace equation (of potential equations), 2, 12 Laplace operator, 12, 14, 131, 176 Lexicographical ordering. See Ordering

Lipschitz continuity, 23, 30 Map, mapping. See Operator Mass matrix, 205

Index

310

Maximum (-minimum) prmciple, 17, 18, 19, 20, 27, 86, 103, 250, 258

Mean-value property, 17, 18, 19, 20,

27,54 Mehrstallen method, 64, 83 M-matnx, 45, 49, 50, 51, 52, 53, 64, 68, 81, 83, 84, 91, 92, 93, 100, 106, 249, 250, 251, 252 Multi-index, 30, 104 Neighbour. See Grid points Neumann condition. See Boundary condition Nine-point formula, 90, 180 compact, 63, 63 Nodal points, 172, 175, 178, 181, 203

Nodal values, 172, 195 Norm, 29, 30, 50 dual, 131, 132, 134, 163, 194, 215, 227, 298 Euclldean, 14, 51, 57, 110, 163, 204, 226 equIvalent, 111, 117, 123, 204, 295

matrix, 50 assocIated, 50, 204, 227 maximum, 46, 110 operator, 111 row sum, 50, 59, 170, 242 Sobolev, 117, 122 Sobolev-Slobodeckil, 122, 247

spectral, 51, 59, 204 supremum, 23, 110, 112 vector, 50 Normal (dIrection), 15, 116, 145, 224

Normal system, 105 Normed space, 110

Operator, ill adjoint, 132, 163, 213, 216, 242 associated. See Bilinear form, 209, 256, 282 compact, 135, 136, 137 difference. See Difference operator differential. See Differential operator

dual, 131 selfadjoint, 132 Ordering (of the grid poInts), 42, 43, 45 lexicographical, 42, 67, 70 chequer-board, 43 Orthogonal (ef. projection), 114 Orthogonal space, 114, 163, 283

Parabolic, 4,5, 5,7,8, 9, 10, 11 Partition of unity, 128, 130, 150, 209, 219

Petrov-Galerkin method, 252 Plate equation (cf. blharmonic), 85, 103 Poisson equation, 27, 34, 35, 38, 41, 57, 103, 149, 155, 177, 201, 225, 231, 254, 276, 287, 289, 293, 295 Poisson's integral formula, 19, 20, 22

Polar coordinates, 13, 14, 15, 97, 98, 144

Positive definite, 5, 52, 53, 93, 106, 164, 204, 255

Positive semidefinite, 52, 87 Potential, double-layer (dipole), 36 sIngle-layer, 36 volume, 36 Potential equation (of. Laplace equatIon), 2, 4, 5, 7, 9, 12, 13, 14, 16, 28, 35 Precompact, 135 Principal part, 6, 12, 95, 145, 147, 148, 150, 223, 228, 251, 276 Projection, 132, 170 orthogonal, 132, 133, 163, 170, 225, 294

Ritz, 170, 189, 191, 192, 193, 225 Rayleigh-Ritz. See Ritz Reduced equation, 248, 250 Reference triangle (reference element), 10, 177, 179, 185, 198, 204, 294

Regularity, 190, 192, 196, 208, 209, 215, 219, 222, 223, 225, 232, 246, 254, 264, 267, 285, 287, 288, 293, 295 discrete, 227, 228, 229, 230, 231, 239, 240, 241, 242, 243, 271 Riesz reprssent.atlon theorem, 132 Riesz isomorphism, 132, 283 Rlesz-Schauder theory, 1ST, 142 Ritz-Galerkln method (cf. solution), 161, 171, 290 Robin problem, 159 Saddle-point (problem), 279, 280, 282, 284, 290, 297 Scaler product, 52, 114, 115, 117, 121, 132, 148, 162, 170, 227 Separation (of variables), 3, 11 Serendipity class. See Finite elements Sesquilinear form, 138

Index

Seven-point formula, 91, 92, 176 Shortley-Weiler discretisation, 78, 80, 81, 83, 229, 231, 237 Side condition, 182, 183, 290, 291 Single-layer potential. See Potential Singular perturbation, 247, 248 Singularity function, 14, 16, 56 discrete Sobolev, lemma of, 125, 212, 222 Sobolev space (cf. norm), 114, 117, 122, 128, 133, 134, 136, 145 Solution classical, 27, 30, 32, 145, 146, 147, 154, 212, 222, 246, 279

existence of a, 9,13,23,29, 32, 35, 37, 67, 137, 142, 144, 151, 155, 161, 286, 292 existence and uniqueness of a, 137, 141, 142, 147, 148, 149, 150, 151, 152, 153, 162, 164, 166, 167, 168, 171, 183, 188, 222, 248, 282, 284, 285, 286, 288, 289, 292 Ritz-Galerkin, 164, 165, 166, 168, 170

uniqueness of the, 18, 19, 26, 27, 35, 86, 87, 88, 148, 149 weak, 27, 146, 150, 159, 190, 208, 209, 210, 212, 213, 218, 219. 222, 223, 226, 246, 279 Sparse matrices, 44, 171, 173 Spectral radIus, 47 Spectrum, 137, 142 Stable, stabilIty, 59, 64, 70, 90, 94, 95, 102, 103, 227, 228, 242, 249, 252, 291, 293 Star (stencil), 178, 180 Stationary problem, 10 Steklov problem, 254 Step size. See Grid size Stiffness matrIx, 102, 163, 165, 171, 177, 191, 200, 202, 203, 255 Stokes equations, 4, 103, 275, 278, 279, 282, 285, 286, 288, 290, 291, 292, 293, 298 Support, 115, 171, 173, 176, 182, 183, 194, 195, 202 Time, 8, 9, 10, 11 Trace (of a function), 3, 123, 129 Trace (of a matrix), 8? Transformation. See Domain, Fourier Transformation theorem, 119, 129 Transition condition, 245, 246, 247 Trefftz'method, 200

311

Triangulation, 174 adaptive, 201 admissible, 174, 175, 179, 185, 190

quasi-unifurm, 188, 190, 204, 208, 294, 295

regular, 188 uniform, 188

Type (of a pde), 1,4,5,6,7,10 Type invariance, 6 Uniqueness. See Solution Variational formulation (weak formulation), 144, 146, 149, 150, 158, 161, 165, 166, 173, 183, 194, 208, 244, 253, 254, 278, 288

Variational problem, 141, 152, 159, 184, 171 dual/complementary, 200 Viscosity, artificial, 251, 252 numerIcal, 251 Wave equation, 2, 3, 5, 8, 10 Weak formulation. See Variational formulation Well-posed (problems), 23 Wilson's rectangle, 199

This book offers a simultaneous treatment of the theory and the numerical treatment of elliptic problems. The subjects dealt with include the classical theory (Green' s function, maximum principle, etc.) as well as the variational formulation. The author describes and analyses finite difference and finite element methods. Specific chapters

are devoted to the eigenvalue problem and the Stokes problem.

ISSN 0179-3632 ISBN 3-540-54822-X

111111

9 783540 548225 http:llwww.springer.de