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

= rsin
y

z

= rcostjJ

the Laplace operator is given by 02

2 0 1 [ 1 02 + r2 sin2 tjJ 0
Ll = 8r2 + ; or

0 02 ]

+ cot tjJ otjJ + OtjJ2

.

(2.1.4)

Note. In the n-dimensional case the transformation to polar coordinates leads to 02 n -1 0 1 (2.1.5) L l = 2- + - - - + - B 8r r or r 2 ' where the Beltrami operator B contains only derivatives with respect to the angle variables.

2.2 Singularity Function The singularity function is defined by s(x,y) :=

{

- ';2 log Ix -

yl

Ix _ yl2-n (n - 2)wn

for n

= 2,

for n

> 2.

(2.2.1a)

for X,y E IRn , where Wn

= 2r(1/2t /r(n/2)

,

W2

= 27r,

W3

= 47r,

(2.2.1b)

with r the Gamma function, is the surface of the n-dimensional unit sphere. The Euclidean norm of x in IRn is denoted by ..

Ixl

=

1/2

(LX;) .

(2.2.2)

i=1

Lemma 2.2.1. For fixed y E IRn the potential equation in IRn\{y} is solved by s(x,y).

2.2 Singularity Function

15

The proof can be carried out directly. It is simplest to introduce polar coordinates with Y as origin and to use (1.5), since s(x, y) depends only on r= Ix-yl. For the next theorem we need to introduce the normal derivative 8/ 8n. Let nbe a domain with smooth boundary r. Let n(x) E m.n denote the outer normal direction at x E r, i.e. n is a unit vector perpendicular to the tangential hyperplane at x and points outwards. The normal derivative of u at x E r is defined as

8u(x)/8n = (n(x), Vu(x»,

(2.2.3a)

where (2.2.3c) is the gradient of u and n

(x,y)

= 2:XiYi

(2.2.3c)

i=l

is the scalar product in m.n . In the case of the sphere KR(y) (cf. (7» the normal direction is radial, and Bu/8n becomes 8u/8r with respect to r = Ix - yl, if one uses polar coordinates with the origin at y. It follows from 8s(x, y)/8r = -Ix - yll-n /wn , that

8s(x,y)/8n = _R1-

n

/w n

for x E 8KR (y).

(2.2.4)

The first Green formula reads (cf. Green [1])

r uL1vdx = - lnr (Vu, Vv) + lan r u 88vn dr

ln

(2.2.5a)

and holds for u E Clem, v E C2(Q) if the domain n satisfies suitable conditions. Here Ian' .. dr denotes the surface integral. Domains for which Equation (5a) holds are called normal domains. To see sufficient conditions for this refer to Kellogg [1, Chapter IV) and Hellwig [1, 1-1.2]. Functions u, v E C 2 (Q) in a normal domain n satisfy the second Green formula (2.2.5b)

Theorem 2.2.2. Let n be a normal domain, and let u there. Then u(y)

E

C2(Q) be harmonic

= far [s(X, y) :n u(x) - u(x) 8~ s(x, y)] drx

for all yEn. Here 8/8nx and drx refer to the variable x.

(2.2.6)

16

2 The Potential Equation

PROOF. By

(2.2.7) we denote the ball with centre y and radius r. Since the singularity function s(·,y) is not differentiable on x = y, Green's formula (5b) is not directly applicable. Let n" := n\K,,(y), with € so small that K,,(y) en. Since n" is again a normal domain, it follows from Llu = Lls = 0 (cf. Lemma 1) and (5b) with v = s that

r

Jan"

[u(X)aa s(x,y) - s(X'Y)aa U(X)] drx = nx n

o.

(2.2.8a)

We have an" = anu aKE(y). However, at x E aKE(y), the normal directions of an" and of aK,,(y) differ in their signs. The same holds for the normal derivatives, so that the integral in (8a) can be decomposed into fan • ... = Jan' •. - J8K.(y) . ". The assertion of the theorem would be proved if we could show that J8K.(y)··· - -u(y) for € - O. The normal derivative au/an is bounded on aK,,(y), and J8K.(Y) s(x,y) dr converges like O(€llog€1) or O(€) towards zero, as can be seen from (1) and f8K.(y) dr = €n-1 Wn . Thus, we obtain

r

s(X'Y)aa u(x)drx- O

JOK.(y)

From J8K.(t/) dr

= €n-1 Wn

r

(2.2.8b)

n

and (4) one infers

u(y)aa s(x,y)drx = -u(y).

nx

JOK.(y)

(2.2.8c)

The continuity of u in y yields

r

J8K.(y)

(u(x)-u(y»aa s(x,y)drx :$ max lu(x) - u(y)l- 0 as € - O. xE8K.(y) nx (2.2.8d)

Equations 8b,c,d show that

J8K.(y)

ruins - sinuJdr -

that (8a) proves the theorem.

-u(y) (€ - 0), so •

Any function of the form

1'(x, y) = s(x, y) + ~(x, y)

(2.2.9)

is called a fundamental solution, of the potential equation, in n if for fixed yEn the function ~(.,y) is harmonic in n and belongs to C2(il).

Corollary 2.2.3. Under the conditions of Theorem 2 for every fundamental solution in

n the following holds:

2.3 The Mean Value Property and Maximum Principle

u(y) = Ian ['Y(X, y) :n u(x) - u(x)

8=x 'Y(x, y)] drx.

PROOF. (5b) implies fan[iP8uj8n - u8iPj8n] dr

17

(2.2.10)

= o.

•

For a possible weakening of the condition iP = 'Y - s E C 2 (ll) for iP(·, y) E

Cl(ll) n C 2 (il) refer to Hellwig [1, 1-1.4]. Exercise 2.2.4. Let il = KR(Y)' Define 'Y(x, e)

={

(nJ)wn [IX - el 2- n - (KJPlx - (,I)

-2~

[log Ix - el-Iog

(~.rllx -

2-n]

ell)]

for n > 2 for n

(2.2.lla)

=2

with x, e E il, e' = Y + R 2 1e - yl- 2(e - y) and show: (a) 'Y is a fundamental solution in il, (b) 'Y(x, e) = 'Y(e, x), (c) on the surface r = 8KR (y) the following holds:

~'Y(x,e) = ~'Y(e,x) = 8n{

8n{

__I_(R2 -Ix - yl2)lx _ €I-n

Rwn

(e E r).

(2.2.llb)

2.3 The Mean Value Property and Maximum Principle Definition 2.3.1. A function u has the mean value property in il if u E CO(.a) and if for all x Eiland all R > 0 with KR(X) c il the following equation holds

(2.3.1) Since f8KR(X) dr = wn~-l the right side in (1) is the mean value of u taken over the surface of the sphere. An equivalent characterisation results if one averages over the sphere KR(X).

u E Co(ll) has the second mean value property in il if fKR(X) u(e) d€ for all x E il, R> 0 with KR(X) C il. Show that

Exercise 2.3.2.

u(x)

= R,!lwn

this mean value property is equivalent to the mean value property (1). Hint:

fKR(X) u(e) d€ = fOR faKr(x) u(e)dr{ dr. Functions with the mean value property satisfy a maximum principle, as is known from the function theory for holomorphic functions:

18

2 The Potential Equation

Theorem 2.3.3. (Maximum-minimum principle) Let n be a domain and let U E CO(n) be a nonconstant function which has the mean value property. Then u takes on neither a maximum nor a minimum in n.

PROOF. (a) It suffices to investigate the case of a maximum since a minimum of u is a maximum of -u, and -u also has the mean value property. (b) For an indirect proof we assume that there exists a maximum in yEn: u(y)

=M

~

u(x)

for all x E n.

In (c) we will show u(y') = M for arbitrary y' En, i.e. u == M in contrast to the assumption u ¢ const. (c) Proof of u(y') = M. Let y' E n. Since il is connected, there exists a path connecting y and y' running through n, i.e. there exists a continuous cp: (0, 1]- n with cp(O) = y, cp(l) = y'. We set

I := {s

E

[0, l]:u(cp(t»

=M

for all 0 ~ t ~ s}.

I contains at least 0, and is closed since u and cp are continuous. Thus there exists s* = max{s E I}, and the definition of I shows that I = [0, s*]. In (4) it is proved that s* = 1 so that y' = cp(l) E I and hence u(y') = M follows. (d) Proof of s* = 1. The opposite assumption s* < 1 can be made and shown to be contradicted by proving that u(x) = M in a neighbourhood ofx* := cp(s*). Since x* E il, there exists R > 0 with KR(X*) C n. Evidently, it follows that u = M in KR(X)* if it is shown that u = M on aKr(x)* for all 0 < r ~ R. (e) Proof of u = M on aKr(x*). Equation (1) in x* reads M

= u(x*) =

r

u(~) drd(wnrn-l).

JaKr(x*)

In general we have u(~) ~ M . If one had u(~') < M for~' E aKr(x*) and thus also u < M in a neighbourhood of ~', one would have on the right side a mean value smaller than M. Therefore, u = M on aKr(x*) has been proved. • Simple deductions from Theorem 3 are contained in Corollary 2.3.4. Let n be bounded. (a) A function with the mean value property takes its maximum and its minimum on an. (b) If two functions with the mean value property coincide on the boundary an, they are identical.

n

PROOF. (a) The extrema are assumed on the compact set = n U an. According to Theorem 3, the extremum cannot be in n if u is not constant on a connected component of n. But in this case the assertion is also obvious. (b) If u and v with u = v satisfy the mean value property on an then the latter is also satisfied for w := u - v. Since w = 0 on an, part (a) indicates that maxij w = mini) W = O. Thus u = v in n. •

2.3 The Mean Value Property and Maximum Principle

19

Lemma 2.3.5. Harmonic functions have the mean value property.

PROOF. Let u be hannonic in nand Y E KR(y) c n. We apply the representation (2.6) for n = KR(Y). The value s(x,y) is constant on8KR(Y): let it be denoted by a(R). Because of (2.4), Equation (2.6) becomes

The equation agrees with (1) if the first integral vanishes. The latter follows from Lemma 2.3.6. Let u E 0 2 (0) be harmonic in a normal domain we have 8u -8 dr=o. ail n

i

n.

Then

(2.3.2)

PROOF. In Green's formula (2.5a) substitute 1 and u for u and v respectively.

•

Lemma 5, Theorem 3, and Corollary 4 together imply Theorems 7 and 8:

Theorem 2.3.7. (The maximum-minimum principle for harmonic functions) Let u be harmonic in the domain n and nonconstant. There exists no maximum and no minimum in n. Theorem 2.3.8. (Uniqueness). Let n be bounded. A function harmonic in n assumes its maximum and its minimum on 8n and is uniquely determined by its values on 8n. The representation (1) of u(y) by the values on 8KR(Y) is a special case of the following formula which will be proved at the end of this section and which provides Equation (1) for x = y.

Theorem 2.3.9. (Poisson's integral formula) Assume we have

E

(2.3.3) is given by the function

u(x)

= R2 -Ix RMln

YI2

r laKR(Y)


Ix - ~In

which belongs to OOO(KR(y)) n CO(KR(Y) ).

dre for x E KR(y),

(2.3.4)

20

2 The Potential Equation

The mean value property only assumes 1.£ E COCa), while harmonic functions belong to C2(n)nCO(.a). This makes the following assertion surprising: Theorem 2.3.10. A function is harmonic in n if and only if it has the mean value property there.

PROOF. Because of Lemma 5 it remains to be shown that a function v with the mean value property is harmonic. Let x E KR(X) C n be given arbitrarily. According to Theorem 9 there exists a function 1.£ harmonic in K R(X) with

According to Lemma 5, U has the mean value property, as does v, and Corollary 4b proves that 1.£ = v in KR(X), i.e. v is harmonic in KR(X), Since KR(X) C n is arbitrary, v is harmonic in n. • An important application of Theorem 10 is Theorem 2.3.11. (Harnack) Let 1.£1. 1.£2, ••• be a sequence of functions harmonic in n and converging uniformly in li. Then u = limk-+oo Uk is harmonic inn.

PROOF. The limit function is continuous: 1.£ E CO(li). The limit process, applied to Uk(X) = I8KR(x) Uk (e) drd(wnR!'-l) yields Equation (1) for Uj i.e. 1.£ has the mean value property. According to Theorem 10, 1.£ is also harmonic inn.

•

Theorems 3 and 7 on the maximum-minimum principle pertain to global extrema. The proof of Theorem 3 does not yet exclude local extrema in the interior. It merely shows that 1.£ is then always constant in a circle KR(Y) C n. As is known from function theory one can derive from this u(x) = M in :0, if 1.£ is analytic, i. e., there is a convergent power series expansion in a neighbourhood of any x En. Indeed, the following theorem holds whose proof can be found, for example, in Hellwig [1,III-§1.5]: Theorem 2.3.12. A function harmonic in

n

is analytic there.

The proof of the Poisson formula still needs to be carried out. (a) First we must show that u in (4) is a function harmonic in KR(y), i. e., it satisfies ..11.£ = O. Since the integrand is twice continuously differentiable and the domain of integration r = 8KR(Y) is compact, the Laplace operator commutes with the integral sign:

Llu(x)

= (Rwn)-l

1


el-n] dre

for x E KR(Y)'

(2.3.5)

2.3 The Mean Value Property and Maximwn Principle

21

According to Exercise 2.2.4 there exists a fundamental solution 'Y(x, e) such that

eE r,x E KR(Y)· (2.3.6) From Llx~ = ~Llx'Y(x, e) = 0 and (5), one infers that Llu = O. (b) The expression (4) defines u(x) at first only for x E KR(Y)' It still needs to be shown that u has a continuous extension on KR(Y) = KR(Y) u r (Le., u E CO(KR(Y))) and that the continuously extended values agree with the boundary values cp:

lim

u(x)

x-+z

= cp(z)

for

Z E

r.

(2.3.7)

xEKR(Y)

By Equation (6), putting u R2

Let

Z E

r

-Ix _Y12 f

== 1 in Corollary 2.2.3 gives the identity

Jr Ix -el-ndr~ = 1

Rwn

for all

x E KR(Y)'

(2.3.8)

be arbitrary. Due to Equation (8) one can then write: (2.3.9a)

R

Figure 2.3.1. Construction of ra, n

We define ro = r n Kp(z), n = r\ro (see Figure 1) and split the expression (9a) into u(x) - cp(z) = 10 + 11. where 1. = R2 ,

-Ix - YI2 f cp(e) - cp(z) dJ', Rwn Jr. Ix-eln ~,

i=O,l.

2 The Potential Equation

22

Since

Iirof

dr~1 :S max l
re~ l
dr~

iro Ix - el n


~~In,

it follows from Equation (8) that (2.3.9b) Because of the continuity of

> 0 such that for given 10 > 0 (2.3.9c)

10 :S 10/2. Set C

m~Er

l
1 zi :S 0(10) := 210 (p)n 2 4G.

and Ix - zl :S p/2. The last inequality implies Ix - el ~ ~

for

eE Tt

(see Figure 1).

Together with R2_lx_yI 2 = (R+lx-yl)(R-lx-yl) :S 2R(R-lx-yl) :S 2Rlz - xl :S 2RO(€) one obtains

I

f
= R2 -Ix &In

yI 2

irl

II11:S

10

2·

(2.3.9d)

Thus for every 10 > 0 there exists a 0(10) > 0 such that Ix - zl :s:; oCt:) implies the estimate Iu(x) -
r

2.4 Continuous Dependence on the Boundary Data

23

2.4 Continuous Dependence on the Boundary Data Definition 2.4.1. An abstract prdblem of the form A(x)

= y,

x E X, Y E Y,

is said to be well-posed if for all y E Y it has a unique solution x E X and if the latter depends continuously on y. It is important to recognise whether a mathematical problem is well-posed since otherwise essential difficulties may occur in its numerical solution. In the case of the boundary value problem (1.1a,b) Xc C 2 (n) nCO(n) is the space of functions harmonic in nand Y = CO (r) is the set of continuous boundary data on r = an. The topologies of X and Yare given by the supremum norms:

lIulioo := sup lu(x)1 xED

and

1I
(2.4.1)

xEr

The question of the existence of a solution of (1.1a,b) will have to be postponed (see Section 7). The uniqueness, however, has been confirmed already in Theorem 2.3.8, if n is bounded. That the boundedness of n cannot be dropped without further ado is shown in

Example 2.4.2. The functions U(X1.X2) = Xl in n = (O,oo) x m, u(X1. X2) = log{x~ + x~) in n = m,2\Kl{O) U{X1.X2) = sin{xl)sinh(x2) in n = (O,7r) X (O,oo)

(2.4.2a) (2.4.2b) (2.4.2c)

and also the trivial u = 0, are solutions of the boundary value problem Llu = 0 in n, u = 0 on r = an. For bounded n the harmonic functions {solutions of (1.1a,b)) depend not only continuously but also Lipschitz-continuously on the boundary data:

Theorem 2.4.3. Let

n be bounded. If uI and uH

are solutions of

then

(2.4.3) PROOF. v := u I - u ll is a solution of Llv = 0 in n and v =
24

2 The Potential Equation

-

The definition (1) of 11·1100 implies (3).

If lI
3 shows that the associated solutions satisfy lIu n - ulloo -

Theorem 2.4.4. Let D be bounded. Let
PROOF. Since according to Theorem 3 lIun - umll oo ~ lI
b)

c)

Figure 2.4.1. Approximation of a by a' and

a"

Let iln be a sequence of domains with rn = aDn. We say that rn - r if dist(rn • r) - O. Here, dist(rn , r) := sup{dist(x, r) : x Ern}, dist(x, r) := inf{lx - yl: Y E r}. Further, one must specify when 0, there exist numbers N(E) and O(E) > 0 such that the following implication holds: n ~ N(E),

x E r,

Y Ern,

Ix-YI ~ O(E) => l
f.

The sequence Un E CO(Dn) converges uniformly to u E CO(D) if

(2.4.4a)

2.4 Continuous Dependence on the Boundary Data

lim sup{lun(x) - u(x)l: x E iln n il}

n-+oo

= O.

25

(2.4.4b)

Remark 2.4.5. (a) Let K be a set which is compact (Le. complete and bounded) with r C K, and rn C K for all n. Let 4>n E COCK) converge uniformly on K to 4>. If CPn = 4>n on rn and cP = 4> on r then (4a) is satisfied. (b) Let iln C il hold for all n and let Un be the following (not continuous) continuation of Un onto fi: Un = Un on fin, Un = u on fi\li n . Then (4b) is aquivalent to uniform convergence Un -+ u on fi in the usual sense. Theorem 2.4.6. Let fln C il, with fl bounded, and let rn functions Un, which are harmonic in iln , be solutions of Llun = 0

in iln ,

Un

= CPn

-+

on rn.

r. Let the (2.4.5a)

Let CPn E CO(rn ) converge uniformly in the sense of (4a) to cP E CO(r). Then the following assertions hold: (a) if there exists a solution u E C2(fl) n cO(fi) of Llu

=0

in il,

u

= cP

on r

(2.4.5b)

then Un -+ U holds in the sense of (4b). (b) if conversely Un -+ U E co(n) is satisfied in the sense of (4b), then u is the solution of (5b).

PROOF. (a) Let the continuation Un be defined as in Remark 5b. Since u is uniformly continuous on n, there exists liu (f) > 0 for all f > 0 such that (2.4.6a) Set Ii* (f) := min {liu (~) , Ii (~)} with Ii from (4a). Because rn -+ r there exists Nr(f), so that dist(rn , r) S 8*(f) for n ~ Nr(f). For n ~ N*(f) := max{Nr(f), N(f/2)} (N from (4a» we want to show that Iun(x) -u(x)1 S f, for x E For x E the estimate is trivial because Un(x) = u(x). For all x E fin C il, however, there holds

n.

n\nn

IUn(x) - u(x)1

= lun(x) - u(x)1 S

IIJ,~ IU n - ul

(2.4.6b)

(cf. Theorem 2.3.8), because Un -u is harmonic in iln . It remains to estimate IUn(x) -u(x)1 for x Ern. For x Ern with n ~ N*(f) there exists y E r with

Ix - yl S li(f/2). Thus we obtain lun(x) - u(x)1

= ICPn(x) -

u(x)1 S ICPn(x) - cp(y)1 + Icp(y) - u(x)1 S

S ICPn(x) - cp(y)1 + lu(y) - u(x)1

s 2f + 2f = f

from (4a) and (6a). Since x E rn is arbitrary it follows that IUn -ul Sf on rn, and (6b) proves the uniform convergence Un -+ u on fi. Hence, from Remark 5b it follows that (4b) is satisfied.

26

2 The Potential Equation

(b) Let K C Q be a compact set. Since r .. --+ r there exists a N(K) such that K C Q.. for n ~ N(K). Thus, the sequence {u .. : n ~ N(K)} converges uniformly in the usual sense on K to u so that one can apply Theorem 2.3.11: consequently u is harmonic in K. Since K C Q may be chosen arbitrarily, it follows that u E C2(Q). By assumption, we already have u E c°(.?i). That the boundary value u =

r.. =


=
'Pn

= 1 on 8K1/ .. (0),


=1

satisfy the condition
Remark 2.4.7. In Q = K1(0)\{0} c m.2 the potential equation has no solution u E C 2 (Q) n CO(n) which assumes the boundary values u(x) = 0 on 8K1 (0) and u(x) = 1 in x = O.

3 The Poisson Equation

3.1 Posing the Problem The Poisson equation reads Llu

= f in il

(3.1.la)

with given f E CO(il). In the physical interpretation f is the source term [for example, the charge density in the case of an electrical potential u). To determine the solution uniquely one needs a boundary value specification, for example, the Dirichlet condition u = cp on F.

(3.1.lb)

Definition 3.1.1. The function u is called the classical solution of the boundary value problem (la,b) if u E C2(il) n CO(D) satisfies the equations (la,b) pointwise. Until we introduce weak solutions in Section 7, "solution" will always mean "classical solution" . The solution of the boundary value problem (la,b) will in general no longer satisfy the mean value property and the maximum principle. But these properties still hold for the differences of two solutions Ul and U2 of the Poisson equation, since Ll(Ul -U2) = f - f = o. Thus the uniqueness of the solution of problem (la,b) immediately follows and Theorem 2.4.3 can be brought over: Theorem 3.1.2. Let il be bounded. (a) The solution of (la,b) is uniquely determined. (b) If u I and u II are solutions of the Poisson equation for boundary values cpI and cpII, then we have (3.1.2)

PROOF. (b) The proof of Theorem 2.4.3 can be repeated verbatim here. (a) If u I and u II are two solutions of (la,b) then (2) shows that \luI -uIIlloo ::; IIcp - cplloo = o. Thus u I = u II . • Theorems 2.4.4 and 2.4.6 can be transferred likewise.

W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_3, © Springer-Verlag Berlin Heidelberg 2010

27

28

3 The Poisson Equation

3.2 Representation of the Solution by the Green Function Lemma 3.2.1. Let the solution of (1.1a,b) belong to C2(n), where normal domain. Then u may be represented as

u(x) =

-In

1(e,x)f(e) de +

l

n

is a

[1(e,X) :n u(e) - u(e) a!e 1(e,X)] dre (3.2.1)

for every fundamental solution 1 in (2.2.9). The proof is the same as in Theorem 2.2.2 or in Corollary 2.2.3. The term = n\KE(x) becomes IOe 1f £leo Since the singularity of 1(e, x) is integrable at e = x, IO e 1 f de converges to 10 1/ £le as € -+ o.

IOE1..:1ude with n E

Exercise 3.2.2. (a) Let n c lRn be bounded, Xo E n, f E CO(n\{xo}), and I/(x)l :::; Clx - XoI- s for s < n. Show that 10 /(x) dx exists as an improper integral. (b) Let n c rn.n be bounded and let Xo(e) E n depend continuously on ED, with D compact. Let f(x, e) be continuous in (x, e) E .0 x D with x -=I xo(e) and let 1/(x,e)l:::; Clx-Xo(e)I-S, s < n. Show that F(e):= 10 f(x,e) dx is continuous: FE CO(D).

e

In the boundary integral in (1) one may replace u(e) bY'P(e) (cf. (l.1b». The function auj an on r , however, is unknown and cannot be specified arbitrarily either, since the boundary values (1.1b) already determine the solution uniquely (cf. Theorem 3.1.2). To make Ir18ujande vanish one must select the fundamental solution so that 1(e, x) = 0 for e E r, x E n. Definition 3.2.3. A fundamental solution 9 in (2.2.9) is called a Green function (of the first kind) if g(e,x) = 0 for all E r, x En.

e

The existence of a Green function is closely related to the solvability of the boundary value problem for the potential equation: Remark 3.2.4. The Green function exists if and only if for all x E n the boundary value problem .14> = 0 in nand 4> = -s(·, x) on has a solution 4> E C 2 (n).

r

The above consideration results in Theorem 3.2.5. Let n be a normal domain. Let the boundary value problem (1.1a,b) have a solution u E 02(.0). Assume the existence of a Green function of the first kind. Then one can express u explicitly by

3.2 Representation of the Solution by the Green Function

u(x)

=-

r g(e, x)/(e) d(e) -1rune cp(e) !.l0 g(e, x) dre.

in

29

(3.2.2)

In the following we reverse the implication. Let the existence of the Green function be assumed. Then, does function u defined by Equation (2) represent the classical solution of the boundary value problem (1.1a,b)? Here it must be proved, in particular, that u E C2(il) and Llu = I. Firstly, it is not even clear yet whether the function u(x) defined by Equation (2) depends continuously on x since the definition of a fundamental solution 'Y(e, x) does not require continuity with respect to the second argument x. Despite that, the Green function is g(e, x) with respect to x in C2(il\ {e})' as the following result shows (cf. Leis [1, p. 67)). Exercise 3.2.6. Let il be a normal domain. Let the Green function exist for il, and for fixed y E il let g(.,y) E C2(il\{y}) (weaker conditions are possible I). Now prove that g(x,y)

= g(y,x)

(3.2.3)

for X,y E il.

Hint: Apply the Green formula (2.2.5b) with ilE = il\[KE(x') U KE(X")], x',x" E il, u(x) := g(x,x'), v(x) := g(x, x"), and use (2.2.10). If one tries to reverse the assertion of Theorem 5, one encounters the surprising difficulty of having to set precise conditions on the source term I. The natural requirement 1 E C°(il) is necessary for u E C2(ll), but it is not sufficient, as the following theorem, whose proof will be appended at the end of this section, shows.

Theorem 3.2.7. Even il the boundary r and the boundary values cp are sufficiently smooth and il the Green function exists, there are lunctions 1 E C°(11) to which no solutions u E C2(ll) correspond. Theorem 7 shows that Equation (2) need not represent a classical solution for 1 E CoO?). However, a sufficient condition for 1 to do so is Holder continuity. Definition 3.2.8. 1 E CO(ll) is said to be Holder continuous in II with the exponent>. E (0,1) if there exists a constant C = CU) such that I/(x) - l(y)1 ~ Clx - yl>'

for all x, YEll.

(3.2.4a)

We write 1 E C>'(Q) and define the norm II/lIc>.(.Q) as the smallest constant C which satisfies (4a):

_._

II/lIc>.(o) .- sup

{I/(X) - l(y)l. Ix _ yl>'

. x, Y E il, x

}

:f y .

(3.2.4b)

30

3 The Poisson Equation

The function 1 E Ck(n) is said to be k-fold Holder continuously differentiable in n (with the exponent 'x), if DV 1 E CA(n) for alllvi ~ k. Here v

= (Vb' •• ,vn ) with Vi

E

Z,

Vi

~

0,

Ivi

= Vl + ... + Vn

(3.2.5a)

is a multi-index and (3.2.5b) a lvi-fold partial derivative operator. The k-fold HOlder continuously differentiable functions form the linear space ck+A(n) with the norm (3.2.4c)

= k + ,x

one also writes C 8 (n) for ck+A(n). The k-fold Lipschitz continuously differentiable functions 1 E Ck,l(n) are the result of the choice ,x = 1 in (4a, b). For reasons of completeness let us add that

If s

(3.2.4d) is the norm in Ck(n) for integer k ~ O. Exercise 3.2.9. (a) 1 is said to be locally Holder continuous in il iffor each x E n there exists a neighborhood Ke(x) such that 1 E CA(Ke(x) nil). Prove that if n is compact then 1 E CA(n) follows from the local Holder continuity in n. Formulate and prove corresponding statements for ck+A(n) and Ck,l(n). (b) Let s > O. Show that IxI8 E C 8 (KR (0», if s ¢ IN, otherwise IxI8 E C 8 - l ,l(KR (0». Hint: 1 - t 8 ~ (1 - t)B for 0 ~ t ~ 1, s ~ O. The function u from Equation (2) can be decomposed into Ul +U2 where Ul = - J(J gl cJe and U2 = - Jr <pagl an dr; U is the solution of the boundary value problem (1.1a,b) if we are able to show that Ul and U2 are solutions of .dUl

= 1 in

n,

Ul

= 0 on r,

.du2

= 0 in il,

U2

=

Theorem 3.2.10. If the Green function exists and satisfies suitable conditions then U(x)

= - f
is a classical solution of .du = 0 in il, with U

(3.2.6)

=

The proof goes in principle just as for Theorem 2.3.9 (cf. Leis [1, p.69]).

3.2 Representation of the Solution by the Green Function

31

Theorem 3.2.11. Suppose the Greenfunctiong(·, x) E C2(n\{x}) for x E

n

exists, and let it be f E c>.(n). Then u(x) =

-10 f(t;)g(t;, x) df.

(3.2.7)

is a classical solution of Llu = f in n, with u = 0 on r. PROOF. The boundary condition u(x) = 0 for x E r follows easily from g(x, t;) = 0 and (3). The property u E C 1 (n) and the representation uz.(x) = - fn!(t;)gz.(t;,x)df. result from the

n n:

m.

n be bounded and A := {(t;, x) E Exercise 3.2.12. Let n c x t; =I- x}. For the derivatives of f with respect to x assume D~f E CO(A) and ID~f(t;,x)1 ~ Clx - t;1-s with s < n for any Ivl ~ k. Prove that then F(x) := fn f(t;, x) df. E Ck(!]) and DV F(x) = fn D~f(t;, x) df., Ivl ~ k.

To prove u E C2(n) this step cannot be repeated since gz.Zj (t;, x) = O(It; - xl- n ) has a singularity which is not integrable. We write the derivative in the form

uz.(x) = -

fa

[f(t;) - !(x)]gz.(t;, x) df. - f(x)

fa

gz.(t;, x) df..

(3.2.8a)

Let 8j F(x) be the difference quotient (F(xE ) - F(x»/f., with xj = Xj + f. and x~ = Xi for i =I- j. The product rule 8j (FG)(x) = G(x)8j F(x) + F(xE )8j G(x) applied to Equation (8a) gives

8j u z• (x) = -

kf [f(t;) -

f(x E)]8j gz• (t;, x) df. - f(x E)8j 88

f

~Jn

g(t;, x) df..

Since [f(t;) - f(x)]gz,Zj (t;, x) = O(It; - xl>.-n) is integrable, the limit f. -+ 0 results in the formula

uz.zj(x)=-

kf l!(t;)-f(x)]gz.Zj(t;,x)df.-f(X)8~8;~kf g(t;,x)df.. (3.2.8b)

Equation (8b) implies

Llu =

10 [f(t;) - f(x)]Llg df. - f(x)Ll 10

9 df. = - f(x)Ll

10 g(t;, x) df.,

so that it remains only to show that Llfngdf. = -1. Choose KR(Z) so that x E KR(Z) c n. The Green function has the form (2.2.9): 9 = s+!P. The first two terms in

f g(t;, x) df. = f g(t;,x)df.+ f !p(t;,x)df.+ f s(t;,x)df. Jn Jn\KR{Z) JKR{Z) JKR(Z) are harmonic in KR(Z), so that Ll fKR(Z) s(t;,x) df. = -1 is what has to be proved.

32

3 The Poisson Equation Let a(r) be defined by

s(~,x)

=

a(l~

- xl) (cf. (2.2.1)). For fixed r > 0

set

1n _ 1 f s(~, z) dre. (3.2.8c) wnr 18Kr(z) For all x fj 8KR(Z) (Le., Ix-zl =I r) v is harmonic, since s(~, x) is nonsingular on 8Kr (z) and satisfies .dxs(~,x) = O. Since s(·,x) is harmonic in Kr(z) for r < Iz - xl, the mean-value property (2.3.1) holds, which can now be written vex) :=

vex)

= s(z, x) = a(lz -

xl)

for Iz - xl > r.

(3.2.8d)

Using Exercise 3.2.2b we see that vex) is continuous in IRn , so that we also have (3.2.8e) vex) = a(r) for Iz - xl = r. Thus v is harmonic in Kr(x) with the constant boundary values (8e). The unique solution is therefore

vex)

= a(r)

for Iz - xl ~ r.

(3.2.8f)

The equations (8c,d,f) yield

f

s(~, x) drt; = wnrn-1a(max{r, Iz -

xl})

18Kr(z)

and then, since 0 < Iz - xl < R,

s({,x)~=

f 1 KR(Z)

Wn

l

iz-xi

fR f

10

s({,x) dre dr

18Kr (z)

rn-1a(/z-xl)dr+wn

lR

o

rn-1a(r)dr

Iz-xl

xln

lR

rn IR -Wn rn a(lz-xD+wn-a(r) -a'(r)dr n n Iz-xl Iz-xl n Wn Rna(R) _ Wn fR rn (_ r 1- n ) dr = Wn Rna(R) + fR ::.. dr n liz-xi n Wn n liz-xi n =Wn

Iz

-

= Wn Rna(R) + R2 _ Iz - x1 2 . n 2n 2n From this we see that, independently of R, n, and z, there results

.d

f

s(~,x)~=-l.

lKR(z)

From Theorems 10 and 11 follows

•

Theorem 3.2.13. Under the same assumptions as for Theorems 10 and 11 Equation (2) gives a representation for the classical solution of the boundary value problem (l.la,b).

3.2 Representation of the Solution by the Green Function

33

Finally we put forward two inequalities for the Green function as exercises: Exercise 3.2.14. In il, and ill C il2, respectively, let the Green functions 9, 91 and 92 exist. Show (a) 0:59(X,Y):5s(x,y) forx,yEilCIRn , n~3 (3.2.9) What is the inequality for n=2? (b) 91(X,y):5 92(X,y) for X,y E!h C il2 (3.2.10) Hint: Exercise 2.3.14. Supplement: Proof of Theorem 7. If we make use of a later theorem (Theorem 6.1.13) then Theorem 7 follows from Theorem 3.2.15. The solution u does not depend continuously on I, il the supremum norm (2.4.1) is used as the norm in CO(n), and that from (4d) is used as the norm in C 2 (n). PROOF. Let {} = Kl(O) C IR2 und cp = o. The disk {} is a normal region for which the Green function is known (cf. Theorem 3.3.1). By Theorem 11 there exist solutions un E Q2(n) of Llun = In in il, un = 0 on r for the functions 2 _x21Pn(r), r:= lxi, In(x) = x 2 r2

I rl-

Pn(r):= min ( n· r, log '2

1)

'

which belong to CO(n) and are uniformly bounded: II/nlloo = 1jlog2. By Theorem 11 we have un(x) = 9(~, x)ln(~) de. Since I/n(~)I :5 nl~l, it follows from Exercise 12 that

In

U:1Z1 =

-fa

9z1zJn de =

-fa

ipzlzJn de

-

fa

sZlzJn de at x = 0,

where 9 = ip+s. The first integral is bounded since ip E C 2 (n). The derivative of the singularity function is SZlZl (~, 0) = (~~ - ~i)jl~14. The special choice of In gives for x = 0 U:1Z1 (0) =

-fa

ipzlzl (~, O)/n(~) de + fa[~~

The surface integral K := IaKr(O)[~~

-~~]21~1-6Pn(I~l)de.

- ~iI21~1-5 d~

> 0 does not depend on

I:

r E (0,1], so that the second integral takes on the form In := K r-1pn(r) dr. Since t[rllog(rj2)1)-1 dr diverges as f -+ 0, we deduce In -+ 00 as n -+ 00. Since lIunllc2(.a) ~ IU:1Z1 (0)1, it follows that the map I f-+ U is not bounded, and thus not continuous. •

34

3 The Poisson Equation

3.3 The Green Function for the Ball Theorem 3.3.1. The Green function for the ball KR(y) is given by the function in (2.2.1180). For f E C>'(ll) and cP E C2+>'(F) with 0 < .A < 1, the representation formula (2.2) defines a solution u E C2+>'(ll) of the boundary value problem ~u = f in n, u = cp on F. The proof of the theorem follows from a result of Schauder, which is cited in Theorem 9.1.20. In the case n = 2 the plane IR2 can be identified with GJ by the correspondence (:c, y) +-+ z = :c + iy. The following considerations are based on Exercise 3.3.2. Let the map cp: z =:c + iy En ....... be holomorphic. Show

C= e+ i11 = p(z)

E n'

(3.3.1)

Equation (1) shows, in particular, that a holomorphic transformation of coordinates maps harmonic functions into harmonic functions. An arbitrary simply connected region with at least two boundary points can, by the Riemann mapping theorem, be mapped by a conformal mapping Pzo: zEn ....... P.zo (z) E Kl (0) onto the unit disk such that PZo (zo) = 0 for any given ZO E n. Let g(C,C /) be the Green function for Kl(O). One may check that G(z, zo) := g(Pzo (z), 0) is again a fundamental solution. Now z E an implies pzo(z) E aKl(O), i.e., G(z,zo) = O. Thus G(z,zo) is the Green function in n. This proves Theorem 3.3.3. Let n c IR2 be simply connected with at least two boundary points. Then there exists a Green function of the first kind for n. The explicit forms of various Green functions can be found, for example, in the book by Wloka. [I, Exercises 21.1-21.8]. Of numerical interest might be the fact that with conformal mapping one may remove comers which are disturbing (e.g., reentrant comers) (cf. Gladwell-Wait [I, p.70]). Example 3.3.4. Let n be the L-shaped region in Example 2.1.4. Choose P = z2/3: n --+ n'. Then P is conformal in n. The sides of the angle Fo C an (cf. Fig. 2.1.1) are mapped into a single line segment, so that n' has no more comers sticking in. The Poisson equation ~u = f in n corresponds to the equation ~v(C) = ~IClf(C) in n'.

3.4 The Neumann Boundary Value Problem

35

3.4 The Neumann Boundary Value Problem In (1.1b) and in (2.1.1b) the boundary values u =

o

on u(x) =
onr.

(3.4.1)

In physics this second boundary condition, as it is also called, occurs more frequently than the Dirichlet condition. For example, if u is the velocity potential of a gas, then ~ = 0 means that the gas can only move tangentially at the boundary r. Except in some unusual cases, the boundary value problem Lu = f in nand ou/on =

n

Theorem 3.4.1. Let be a normal region. The Poisson equation Llu with the Neumann boundary condition (1) is only solvable if

£

cp(e) dr~ =

In

=f

(3.4.2)

f(x) d:x..

If a solution u does exist, then u + c, with c any constant, is also a solution.

PROOF. (1) One may repeat the proof of Lemma 2.3.6 for Llu = f. (2) Obviously u + c satisfies the same equation.

•

Later, in Example 7.4.8, we will show that the Neumann boundary value problem for the Poisson equation has a solution if and only if (2) is satisfied, and that two solutions can differ only by a constant. In the representation (2.1) both the values u(e), for E r, and the normal derivative ou/on occur. The Green function of the first kind was chosen in such a way that g(e, x) = 0 for E In the case of the second boundary conditions (1) one makes the assumption that o'Y(e,x)/on~ = c (c: constant), i.e. MJ(e,x)/on~ = c - os(e,x)/on~

e

e r.

for ifJ

= 'Y -

s. The Corollary 2.2.3 with u == 1 and 'Y

= s gives

L:=

£dr.

Since ifJ must be harmonic (Le., f := LlifJ = 0), from Equation (2) we see that cL + 1 = 0 is a necessary condition for the existence of ifJ. Thus the condition on the Green Function of the Second Kind for the potential equation is

o'Y(x,e)/onx = -1/

£dr.

36

3 The Poisson Equation

Thus the term fr u8"{/8ndr in (2.1) becomes const· frudr. Since u is only determined up to a constant (cf. Theorem 1), one can fix this constant with the additional condition fr u dr = o. This gives the following result, if we write 9 for "(:

u(x) =

-1 f(~)g(~,

x) ~ +

1'P(~)g(~,

x) dr~.

The Green function of the second kind for the ball KR(O) C IR3 can be found in Leis [1, p. 79].

3.5 The Integral Equation Method In the representation (2.1) of the Poisson solution the singularity function s can, in particular, be chosen to be "(. If in addition one imposes the given Neumann data (4.1), one obtains

u(x)

=

£

k(x, ~)u(~) dr~ + g(x) for x

with the kernel function k(x,·) :=

-8s(~,x)/n~

E

n,

(3.5.1)

and the functions

g(x) := gl(X) + g2(X), gl(X):= g2(X) :=

£s(~,x)'P(~)dr~

1s(~, x)f(~) dr~.

The right-hand side in Equation (1) with the unknown boundary value r, can be used as an ansatz solution:

u(~),

~ E

4>(x) =

£

k(x, ~)u(~) dr~ + g(x).

(3.5.2)

The first summand on the right of (2) is called the double-layer potential (dipole potential); gl is the single-layer potential, while g2 is a volume potential. For each u E CO(r) the 4> in (2) is a solution of the Poisson equation (1.1a) in n. However, 4> is also defined for an argument x in the exterior domain m.n\n. A closer look at the kernel function k(x,e) shows that it is in fact only weakly singular for the case of smooth boundaries r. Thus 4> is also defined for x E r. The function 4> which is now defined on all IRn is not continuous at points of the boundary r. At Xo E r there exists both an interior limit 4>_(Xo) for x -+ Xo, x E n and an exterior limit 4>+(Xo) for x -+ Xo, x E IRn\n. In addition we have the third function value 4>(Xo) of (2). Their connection is given by the following jump discontinuity relation (cf. Hackbusch [7, Satz 8.2.8]):

3.5 The Integral Equation Method

4'+(Xo) - 4'_(XO) 4'+(Xo) + 4'_(xo)

= 2u(xo), = 24'(xo),

37

(3.5.3a) for Xo E r.

(3.5.3b)

In order that the ansatz (2) does indeed give the solution u in (1), the boundary value 4'_, continued from the interior, must agree with the function u which is put in the integral: 4'_ = u. Now one can solve Equation (3a,b) for !P_: 4'_(Xo) = 4'(Xo) - u(Xo). The equation 4'_ = u thus leads to

u(Xo)

= 44'(Xo) =

l k(xo,e)u(~") dr{ +

g(Xo) for Xo

E

r.

(3.5.4)

Equation (4) is called a Fredholm integral equation of the second kind for the unknown function u E CO(r). The original Neumann boundaryvalue problem (l.la), (4.1) and the integral eqation (4) are equivalent in the following sense: (a) If u is the solution of the Neumann boundary-value problem, then the boundary values, u(e), E r, satisfy the integral equation (4). (b) If u E CO (r) is a solution of the integral equation (4), then the expression (1) gives a solution of (l.la), (4.1) in the entire domain n. The transformation of a boundary-value problem into an integral equation, and the subsequent solution of the integral equation is referred to as the integral equation method. It allows, for example, a new approach to existence statements, in that one shows the solvability of (4). The integral equation (4) can also be attacked numerically. If methods similar to the finiteelement method described in Section 8 are used, then the result is called the boundary-element method (BEM). One can find references to the integral equation method in, e.g., Hackbusch [7, §§7-9] and Kress[l].

e

4 Difference Methods for the Poisson Equation

4.1 Introduction: The One-Dimensional Case Before developing difference methods for the partial differential Poisson equation, let us first recall the discretisation of ordinary differential equations. The equation a(x)u"(x) + b(x)u'(x) + c(x)u(x) = !(x) can be supplemented with initial conditions U(Xl) = Ul. U'(Xl) = u~ or with boundary conditions U(Xl) = Ul, U(X2) = U2. The ordinary initial value problems correspond to the hyperbolic and parabolic initial value problems, while an ordinary boundary value problem may be viewed as an elliptic boundary value problem in one variable. In particular one can view - u"(x)

u(O)

= lex)

= 'Po,

for x E (0,1), u(l) = CPI

(4.1.1a) (4.1.1b)

n

= ! in the domain = (0,1) with Dirichlet conditions on the boundary r = {O,l}. Difference methods are characterised by the fact that derivatives are replaced by difference quotients (divided differences), in the following called, for short, "differences" . The first derivative u'(x) can be approximated by several (so-called "first") differences, for example, by the forward or right difference: as the one-dimensional Poisson equation -Llu

(8+u)(x) := [u(x + h) - u(x)Jlh, the backward or left difference

(4.1.2a)

(8-u)(x) := [u(x) - u(x - h)Jlh, or the symmetric difference

(4.1.2b)

(OOu)(x) := [u(x + h) - u(x - h)l/(2h), (4.1.2c) where h > 0 is called the step size. An obvious second difference for ul/(x)

is (-8-8+u)(x) := [u(x + h) - 2u(x) +u(x - h))/h 2 •

(4.1.3)

ao,

One also calls 8+ , 8-, and 8-8+ index/ D7 / emphdifference operators. The product 8-8+ may be viewed as 8- 08+ or as a+ oa-, Le .. (a+ a- )u( x) = 8+(8-u(x».

W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_4, © Springer-Verlag Berlin Heidelberg 2010

38

4.1 Introduction: The One-Dimensional Case

Lemma 4.1.1. Let[x - h, x

+ h] en.

39

Then

a±u(x) = u'(x) + hR with

IRI ~ 4I1uIlC2(Si)

if 1.1. E C 2 (n)

(4.1.4a)

aou(x) = u'(x) + h 2 R with

IRI ~ ~lIuIlC3(.a)

if 1.1. E C 3(n)

(4.1.4b)

a-a+u(x)

= u"(x) + h 2 R with IRI ~ 112"uIC4(0) ifuEC4 (n).(4.1.4c)

PROOF. We give the proof only for (4c). If one applies Taylor's formula u(x ± h)

= u(x) ± hu'(x) + h 2 u"(x)/2 ± h3u"'(x)/6 + h4R 4,

R4 = h- 4

l

z ±h

z

u""(e)(x ± h - e)3/3!dI;. = u""(x ± ~h)/4!,

with ~ E (0,1), to Equation (3), the result is (4c) because R u""(x - ~2h)]/24.

• • • • • • •

(4.1.5a) (4.1.5b)

= [u""(x+~lh)+

•

Figure 4.1.1

• · • • · · · · · li h We replace il = (0,1) and ilh

Grid for h = 1/8

n = [0,1] by the grids

= {h, 2h,···, (n -l)h = 1- h},

(4.1.6a)

nh = {O, h, 2h, ... , 1 -

(4.1.6b)

h, I}

of step size h = l/n. For x E ilh, a-a+u(x) only contains the values of 1.1. at x, x ± h E h • Under the assumption that the solution 1.1. of Equations (la,b) belongs to C 4 (n), (4c) yields the equations

n

(4.1.7) If one neglects the remainder term O(h2) in Equation (7), one obtains - a-a+Uh(X)

== h-2 [-Uh(X - h) + 2Uh(X) - Uh(X + h)] = f(x) for x E ilh . (4.1.8a)

These are n-1 equations in n+1 unknowns {un(x),x E equations are supplied by boundary conditions (lb):

nh }. The two missing (4.1.8b)

Uh is a grid function defined on Uh

nh. Its restriction to ilh yields the vector

= (uh(h),uh(2h), ... ,Uh(1-h))T.

40

4 Difference Methods for the Poisson Equation

If in (Sa) one eliminates the components Uh(O) and uh(l) with the aid of

Equation (Sb), one gets the system of equations (4.1.9a) with 2 -1

-1

2

-1 2

-1

-1

-1

(4.1.9b)

2 -1

-1 2

qh = (f(h) + h- 2<po, f(2h), f(3h), .. " f(I- 2h), f(I- h) + h- 2
4.2 The Five-Point Formula First we select the unit square

n={(x,y):O<x
nh := {(x,y) E n:x/h, y/h E Z} for step size h

=

(4.2.1a)

~ (n E IN). The discrete boundary points form the set

n:= {(x,y) E r:x/h,y/h E Z}.

(4.2.Ib)

nh := nh U rh = {(x,y) E nh:x/h, y/h E Z}.

(4.2.Ic)

As in (1.Ib) we set

y

Figure 4.2.1 A two-dimensional grid

0

0

0

• • • • • • • • • • • • • • •

0

0

• • • • • • • • • •

0 0 0

.

:

Points of n h

0:

Points of rh

0 0 0

x

4.2 The Five-Point Formula

41

In the Poisson equation -Llu == -U",,,, - ulIlI U

IP on

=

= f in il,

r,

(4.2.2a) (4.2.2b)

the second derivatives u"'''' and UuIl can each be replaced by the respective differences (l.4c) in the x and y directions: (LlhU)(X,y):= (8;8;% +8;8t)u(x,y) = h-2[u(x - h,y) +u(x + h, y) + u(x,y - h) + u(x,y + h) - 4u(x,y)]. (4.2.3) Since on the right side of (3) the function u is evaluated at five points, Llh is also called the Five-Point Formula. The discretisation of the boundary

value problem (2a,b) using Llh leads to the difference equations -LlhUh(X) Uh{X)

= f(x)

= IP{x)

for x E ilh,

(4.2.4a)

for x E n.

(4.2.4b)

Through (4a,b) one obtains one equation per grid point x = (x,y) E il h , and hence one equation per component of the grid function Uh = (Uh(X»xEGh. Except in the one-dimensional case, there exists no natural arrangement of grid points, thus one cannot immediately obtain a matrix representation as in (1.9b). The only natural indexing of Un is that through x E ilh or the pair (i,j) E 71? with x = (x,y) = (ih,jh). Let the matrix elements be given by Lxe:=

{

4h-2 if x = ~ E il h , _h- 2 ifxEilh,~Eilh'x-~=

o

(:h) orx-~= (±t),

otherwise.

(4.2.5) For x = ~, Lxx is a diagonal element; in the second case, Lxe = _h-2, we say that x and ~ are neighbours. If one eliminates the components Uh(X), x En, with the aid of Equation (4b), then Equation (4a) assumes the following form: (4.2.6a) LxeUh(e) = %(x) for x E ilh,

I:

eE!2h

where qh

:= A

+ IPh,

A(x):= f(x),

IPh(X):= -

I: LxeIP(~)·

(4.2.6b)

eErh For the proof split the sum -LlhUh(X)

=

I: LxeUh(~) eEGh

into L:eEnh ... and L:eErh .... The second partial sum is -IPh and is moved to the right side of the equation.

42

4 Difference Methods for the Poisson Equation

Remark 4.2.1. fh is the restriction of f to the grid ilh . For all points far from the boundary we have 'Ph(X) = 0; here x E ilh is said to be far from the boundary if all neighbours x± (0, h), x± (h, 0) belong to ilh. In the case of homogeneous boundary values 'P = 0 we have % = fh. The system of equations (6a) can be expressed in the form (1.9a):

where the matrix Lh = (L~)x,~En" and the grid functions Uh = (Uh(X))xEn" and % = (qh(X))xEO" are described by their components (cf. (5), (6b)). Strictly speaking Lh is not a matrix in the usual sense but a linear mapping since the indices x E ilh are not ordered. Exercise 4.2.2. (a) An N x N matrix P is called a permutation matrix if w := Pv, for all v E IRN, has the coefficients Wi = v1f(i), 1 ~ i ~ N, where 7r is a permutation of indices {I, .. ·, N}. Show that P is unitary, Le., p-l = pT. (b) Let I be an index set with N elements. Let the "matrix" coefficients Aa ,8, 01, f3 E I, be given. To the arrangement OIl. ... ,OIN of indices corresponds the N x N matrix A = (aij )i,j=l, ...• N with Qjj = Aa,a;' Let A be the matrix which belongs to a second arrangement all"', aN of I. Prove that there exists a permutation matrix P such that A = PAPT . (c) Let Aa,8 = A,8a for all 01, f3 E I. For each arrangement of the indices the corresponding matrix is symmetric. A possible linear enumeration of indices x E ilh is lexicographical ordering (h, h), (h,2h),

(2h, h), (2h,2h),

(3h, h), (3h,2h),

(1- h,h), (1 - h,2h),

(h,1 - h),

(2h, 1 - h),

(3h,1 - h),

(1 - h, I - h).

(4.2.7)

Generally, the point x = (Xli"', Xd) precedes the point y = (Yl,"" Yd) in lexicographical order, if for a j E {I, ... , d} the conditions Xi = Yi (i > j) and Xj < Yj hold. Each line in (7) corresponds to a so-called x-row in the grid th. A vector Uh whose (n - 1)2 components are enumerated in the series (7) thus separates into n - I blocks (so-called x-blocks). The block decomposition of the vectors generates a block decomposition of the matrix Lh which is given in (8). Exercise 4.2.3. (a) With the lexicographical numbering of grid points the matrix Lh has the form

IT=[~I ~I I 4.2 The Five-Point Formula

Lh

=h

-2 [

T -1

-1 T

-1 .... -1

..

T -1

i '

.

-I

-1

43

~1 ~1

'

(4.2.8) where T is an (n - 1) x (n - 1) matrix and Lh contains (n - 1)2 blocks. 1 is the (n - 1) x (n - 1) identity matrix. (b) Let il be the rectangle il

= (0, a)

x (0, b)

= {(x,y):O < x < a, 0 < y < b}.

Let the step size h satisfy the conditions a = nh and b = mho Show that the discretisation (4a,b) in the corresponding grid ilh leads to a matrix which also has the form (8). But here Lh contains (m - 1)2 blocks of the size (n - 1) x (n -1). Another frequently used arrangement is the chequer-board ordering (or red-black ordering). To this end one chessboard pattern divides ilh into "red" and "black" fields: ilhr := {(x, y) E ilh : (x ilhb := {(x,y) E ilh: (x

+ y)jh odd}, + y)jh even}.

(4.2.9)

First one numbers the red squares (x,y) E ilh r lexicographically, and then those of ilh b • The partition (9) induces a partition of vectors into 2 blocks and a partition of the matrix Lh into 2 . 2 = 4 blocks.

Exercise 4.2.4. With respect to the chequer-board ordering, the matrix Lh assumes the form 4

A 4

(4.2.10)

4

4

where, in general, A is a rectangular block matrix because for n even, ilh r and ilh b contain a different number of points. The complete (n - 1)2 x (n - 1)2 matrix Lh in (8) or (10) is needed neither for the theoretical investigation of the system of equations Lh Uh = qh nor for its numerical solution. All properties of Lh considered in the following are invariant with respect to re-numbering of the grid points. Even though numerical methods for the solution of LhUh = qh implicitly use an arrangement of grid points (with the exception of special algorithms for parallel computers), they never employ the complete (n - 1)2 x (n - 1)2 matrix L h . Every usable

44

4 Difference Methods for the Poisson Equation

algorithm must take into account that Lh is sparse, Le., it has substantially more zero than nonzero elements. In the following we again return to indexing by x E nh. Nevertheless we will continue to refer to the Lh defined by (5) as a matrix. The difference operator Llh is also described by the star

<1.

~h

-2

[1 ~4 1]

(4.2.11)

The general definition of a difference star (with variable coefficients) reads

C-l,l(X,y) C-l,O(X, y) C-l,-l(X,y)

CO,l(X,y) CO,o(x, y) CO,-l(X,y)

Cl,l(X,y) Cl,O(X, y) Cl,-l(X,y)

00

=

L

t;j(x,y)uh(x+ih,y+jh),

(4.2.12)

',j=-oo

in which the zero coefficients have not been written out. Attention. The star (11) does not represent a submatrix of Lh! The coefficients of the star appear in each row of Lh.

Remark 4.2.5. Note that the difference operator -Llh cannot be equated with the matrix Lh since Llh does not contain information on the type or place of the boundary conditions. We shall say the matrix Lh belongs to a difference star (12) if the system of equations LhUh = Qh results from the difference equations in x E nh after elimination of the Dirichlet boundary values Uh(X) = cp(x), x E n. Even if the vector Uh in LhUh = qh contains only components Uh(X), x E nh , one occasionally equates Uh with the grid function on which assumes the prescribed boundary values (4b) on n.

nh

4.3 M-matrices, Matrix Norms, Positive Definite Matrices The elements of the matrix A are denoted by aa{3, Ot, f3 E I. Here A and the index set I assume the places of Lh and nh . We write

A

~

B if aa{3

and define analogously A byO.

~

B, A

~ ba {3

for allOt, f3 E l,

> B, A <

B. The zero matrix is denoted

4.3 M-matrices, Matrix Norms, Positive Definite Matrices

45

Definition 4.3.1. A is called an M-matrix if aaa

>0

for all a E I,

aa{3::;

0

for all a

i:- {3,

(4.3.1a)

A nonsingular and A -1 ~ O.

(4.3.1b)

The inequalities (la) can immediately be proved for Lk (cf. (2.5)). However we still need criteria and auxiliary results to prove (lb). The index a E I is said to be directly connected with {3 E I if aa{3 i:- O. We say that a E I is connected with {3 E I if there exists a "connection" (chain of direct connections) (4.3.2)

The index set I together with the direct connections form the graph of A (cf. Fig. 1). Frequently A has a symmetrical structure, i.e., aa{3 i:- 0 holds if and only if a{3a i:- O. In this case a is (directly) connected with {3 if and only if {3 is (directly) connected with a. Definition 4.3.2. A matrix A is said to be irreduci ble if every a E I is connected with every {3 E I.

CD, @, CD: Indices _

: Connections

Figure 4.3.1. Graph of the irreducible matrix A

In the case of the matrix A = Lk, two indices x, y E ilk are connected if and only if y = x or if y is a neighbour of x. Arbitrary x, y E ilk can evidently be connected by a chain x = x(O) , x(l), ... ,x(k) = y of neighbouring points. Thus Lk is irreducible. Exercise 4.3.3. Prove that A is irreducible if and only if there is no ordering of the indices such that the resulting matrix has the form A

=

[All

o

Al2] , A22

where All and A22 are respectively square nl x nl and (nl ~ 1, n2 ~ 1), and A12 is an nl x n2 submatrix.

n2

x

n2

matrices

46

4 Difference Methods for the Poisson Equation

The important question as to whether A = Lh is nonsingular can be treated as a special case of the following statement. 4.3.~ershgorin) Let Kr{z) denote the open disk {( E CD: Izand let Kr{z) := {( E CD: Iz - (I ~ r} denote the closed disk. (a) All eigenvalues of A lie in

Criterion

(I < r},

UKrQ(aaa)

with ra =

aEI

L

laa,8l·

~¢Q ~El

(b) If A is irreducible, the eigenvalues even lie in

aEI

aEI

PROOF. (a) Let A be an eigenvalue of A and u a corresponding eigenvector which, without loss of generality, satisfies /lu/l co = 1, where

(4.3.3) is the maximum norm. There exists (at least) one index 'Y E I with lu-yl

= 1.

Assertion 1. lu-y I = 1 implies

From (*) follows A E K r ., (an) and hence the statement. To prove the assertion use the equation from Au = AU associated to the index 7:

Av.-y =

L a-y,8u,8, ~I

that is

(A - an )u-y =

L a-y,8u,8' ~~

From lu-yl = 1 follows IA - a-y-yl = I(A - a-y-y)u-yl ~ II:,8¥-y a-y,8u,8l. By taking the modulus into the sum and by using IU,81 ~ /luI/co = 1, (*) follows. (b) Let A be irreducible and let A be an arbitrary eigenvalue of A with associated eigenvector u which in tum is again normalised by /luI/co = 1. The case A E UaE/ Kr,,(aaa) immediately leads to the statement. Therefore let A ¢ UaEIKrQ(aaa) be assumed.

f:. 0, i.e., 7 is directly connected with {3; then lu-yl implies IU,81 = 1 and IA - a,8,81 = r,8'

Assertion 2. Let aa,8 and IA - a-y-yl

= r-y

=1

Part (a) proves the existence of a 'Y E I with Iv.-yl = 1 and IA - ani ~ r-y. According to the assumption, IA - ani = r-y must hold so that assertion 2 is

4.3 M-matrices, Matrix Norms, Positive Definite Matrices

47

applicable to "I. Since A is irreducible, for an arbitrary {3 E f there exists a connection (2) of "I with {3: ao = "I, a1. ... ,ak = {3, aCto_lao f; O. Assertion 2 shows lua,1 = 1 and l,x - aa,a, I = ra, for all i = 0"" ,k; in particular, ,x E 8Kr~(af3f3) for {3 = ak. Since {3 was chosen arbitrarily, it follows that ,x E naEl 8Kr",(aaa), and the statement is proved. Proof of Assertion 2. Besides the inequality chain (*) there also holds l,x an I = r -y so that all the inequalities in (*) become equations. In particular

L: 1a-yf3ll uf31 = L: 1a-yf31 f3~-y

f3~-y

must hold. Since lu.al ~ lIuli oo = 1, the identity la-y.al lu.al = la-y.al must be satisfied for each summand. Hence a-y.a =I- 0 implies lu.a I = 1. The application of Assertion 1 to {3 yields l,x - a.a.al ~ r.a. The assumption ,x
Definition 4.3.6. (a) A is said to be diagonally dominant if

L: laa.al < laaal

(4.3.4a)

.a~a

for all a E f. (b) A is said to be irreducibly diagonally dominant if A is irreducible, the inequality (4a) holds for at least one index a E f and

L: laa.al ~ laaal

for all a E f.

(4.3.4b)

~¢",

~EI

Note that while an irreducible and diagonally dominant matrix is irreducibly diagonally dominant, the reverse need not hold. The matrix Lh from Section 4.2, while not diagonally dominant, is irreducibly diagonally dominant, for Lh is irreducible and satisfies (4b). At all points near the boundary-Le., those x E ilh, which have a boundary point yEn as a neighbour-however, (4a) holds: L~¢'" laa.al ~ 3h- 2 < 4h- 2 = ~EI aaa· The spectral radius p(A) of a matrix A is given by the eigenvalue that is largest in modulus: p(A) := max{I,xI:,x eigenvalue of A}.

(4.3.5)

48

4 Difference Methods for the Poisson Equation

In the following we split A into

A

=D -

B,

D

= diag

{a",,,,: a E I},

(4.3.6a)

where D is the diagonal part of A: d",,,,

= a",,,,,

d"'f3

=0

for a

f- {3.

(4.3.6b)

B := D - A is the off-diagonal part: (4.3.6c) Criterion 4.3.1. Let (6a-c) hold. Sufficient conditions for

p(D-1B) < 1

(4.3.7)

are the diagonal dominance or the irreducible diagonal dominance of A.

PROOF. (a) The coefficients of C := D-IB read caf3

= -aaf3 / aaa (a f- {3), COla = O.

From the diagonal dominance (4a) follows that ra := Lf3¥'" IC"'f31 < 1 for all a E I. By the Gershgorin Criterion 4a all eigenValues oX of C lie in UaEI KrQ(caa ) = UaEI KrQ(O) so that P,I ::; maxr", < 1 and hence p(C) = p(D-1B) < 1 also follows. (b) If A is irreducibly diagonally dominant then rf3 ::; 1 for all {3 E I and r", < 1 for at least one a. According to Criterion 4b all eigenvalues of C lie in Uf3E I K r ,8 (0) U (n f3E I 8Kr ,8 (0)) . This set lies in K1 (0) if n pEl 8Kr,8 (0) C K1(0). At first let us assume that all rf3 agree: rf3 = r for all {:J. Since ra < 1 for one a E I, it follows that r < 1 and nf3 8Kr,8(0) = 8Kr (0) C Kl(O). But if all rf3 are not equal then nf38Kr,8(0) is empty. Thus in both cases oX E K1(0) holds and (7) is proved. • Exercise 4.3.8. (a) Weaken irreducible diagonal dominance as follows: Let A satisfy the inequalities (4b) and for all {3 E I let the connection (2) exist for an index a E I for which the strict inequality (4a) holds. Prove that even under this assumption p(D-IB) < 1 holds. Hint. Use Exercise 5. (b) Show that the geometric series S = L~o C" converges if and only if p( C) < 1. Then the following holds: S = (I - C) -1. Hint: Represent C in the form QRQ T (Q a unitary, and R an upper triangular matrix) and show IiC"lioo ::; K[p(C))". (c) Let u be a vector. We define lui as the vector (!) with the entries lui a := lu",l. For two vectors one writes v ::; w if Va ::; w'" (a E I). Show that: (1) AB 2: 0, if A 2: 0, B 2: 0; AB > 0, if A > 0, B > 0; (2) AD> 0 if A> 0, and D 2: 0 is a nonsingular diagonal matrix; (3) Av ::; Aw if A 2: 0 and v ::; w; IIvll oo IIwll oo if 0 S v S w;

s

4.3 M-matrices, Matrix Norms, Positive Definite Matrices

49

(4) Au ~ IAul ~ Alul if A ~ O. The importance of inequality (7) results from Lemma 4.3.9. Let A satisfy (la). Let D and B be defined by (6a-c). A is an M-matrix if and only if p(D-1B) < 1. PROOF. (a) Let C := D-1B satisfy p(C) < 1. Then the geometric series S := E:'o Cv converges (cf. Exercise 8b). From D-1 ~ 0 and B ~ 0 one infers C ~ 0, C v ~ 0, and S ~ O. Since I = S(I - C) = SD-1 D - B = (SD)-1 A, A has the inverse A -1 = SD- 1. D-1 ~ 0 and S ~ 0 result in A -1 ~ O. From this (lb) also results, i.e., A is an M-matrix. (b) Let A be an M -matrix. For an eigenvalue>. of D-1 B select an eigenvector u =I- O. According to Exercise 8c we have 1>'1 lui = I>,ul = ID-1Bul ~ D- 1Blul·

Because A -lD ~ 0 (cf. (la,b)) one obtains -A -lDD-1Blul ~ -A -lDI>'llul so that lui = A -l(D - B)lul = A -1D(I - D- 1B)lul ~ A -1Dlul- A -1 D I>'llul = (1 -I>'DA -1Dlul

follows. For 1>'1 ~ 1 we would get the inequality, lui ~ 0, i.e., u = 0, in contradiction to the assumption u =I- O. From this follows 1>'1 < 1 for every • eigenvalue of C = D- 1B, thus p(D-1B) < 1. Criterion 7 and Lemma 9 imply Criterion 4.3.10. If a matrix A with the property (la) is diagonally dominant or irreducibly diagonally dominant, then A is an M -matrix. Theorem 4.3.11. An irreducible M -matrix A has an element-wise positive inverse: A -1 > O. PROOF. Let a,/3 E I be selected arbitrarily. There exists a connection (2): a = ao, at,···, ak = /3. Set C:= D-1B. Since ca'_la, > 0, it follows that

"n,"','Yk-IEI

According to Lemma 9, p(C) < 1 holds, so that S := E:'o CV converges. Since Sa{3 ~ (C k )a{3 > 0 and a,p E I are arbitrary, S > 0 is proved. The assertion results from A -1 = SD -1 > 0 (cf. proof of Lemma 9). • In the following we derive norm estimates for A -1.

50

4 Difference Methods for the Poisson Equation

Definition 4.3.12. Let V be a linear space (vector space) over the field of real numbers (K := JR) or complex numbers (K := GJ). The functional 11·11 is called a norm in V if lIuli = 0 only for u = 0, lIu + vII ~ lIull + IIvll for all u, v E V, IIAull = IAI lIuli for all A E K, u E V.

(4.3.8a) (4.3.8b) (4.3.8c)

Example. Let V = JR#1 where #1:= is the number of elements of the index set I. The maximum norm defined in (3) satisfies the norm axioms (8a-c). If one views the elements u E V as vectors, 11·11 is called a vector norm. But the matrices also form a linear space. In the latter case one calls II . II a matrix norm. A special class of matrix norms is contained in Definition 4.3.13. Let V be the vector space with vector norm 11·11. Then one calls (4.3.9) IIIAIiI := sup{IIAull/liull: 0 ::J u E V} the matrix norm associated with the vector norm

11·11.

Exercise 4.3.14. Let 111·111 be defined by (9). Show that: (a) (b) the following holds:

111·111 is a norm;

IIIABIlI ~ IIIAIIIIIIBIII,

(4.3.10a) (4.3.10b) (4.3.10c) (4.3.lOa)

1111111 = 1 (I: unit matrix),

IIAull ~ IIIAlllliull, IIIAIiI ~ peA).

Example. The matrix norm associated with the maximum norm II ·1100 (cf. (3» is called the row sum norm and is also denoted by II . 1100. It has the explicit representation (4.3.11) Exercise 4.3.15. (a) Prove (11). (b) For matrices 0

~

B

~

C there holds

IIBlloo ~ IIClloo. In the next theorem we denote by 1 the vector having only ones as components: :ao< = 1 for all a E I. For the notation v

~

w see Exercise 8c.

Theorem 4.3.16. Let A be an M -matrix and let a vector w exist with Aw ~ 1. Then IIA-11l 00 ~ IIwll oo .

4.3 M -matrices, Matrix Norms, Positive Definite Matrices

51

PROOF. As in the proof of Lemma 9, let lui be the vector with the components luo,!. For each u we have lui ~ lIull oo l ~ IlullooAw. Since A-1 ~ 0, we obtain

lA-lUi ~ A-llul ~ IlullooA -1 Aw = Ilulloow (cf. Exercise 8c) and IIA-lulloo/liulloo ~ Ilwll oo . Definition 13 implies IIA -11100 ~ Ilwlloo.

that

•

How to estimate with the aid of a majorising matrix is shown in Theorem 4.3.17. following holds

Let

A

and

A'

be M-matrices with

A'

~

A.

Then the

(4.3.12)

PROOF. A-1 ~ 0,

A'-l ~ A-1 follows from A-1 - A,-l = A-l(A' - A)A,-l and A' - A ~ 0, A,-l ~ 0. The remainder follows from Exercise 15b.

•

Exercise 4.3.18. Prove (12) under the following weaker assumptions: A is an M-matrix, A' satisfies (la) and A' ~ A. Hint: Repeat the considerations from the first part of the proof of Lemma 9 with the matrices D' and B' associated to A'. Exercise 4.3.19. Let B a principal submatrix of A, that is, there exists a subset I' c I such that B is given by the entries bal' = aal', a, f3 E I'. Prove that if A is an M-matrix, then so is Band 0 ~ (B-1 )al' ~ (A -1 )al' holds for all a, f3 E I'. Hint: Apply Exercise 18 to the following matrix A': a~1' = aal' (a, f3 E I'), a~a = aaa (a E IV'), a~1' = 0 otherwise. Another well-known vector norm is the Euclidean norm

lIull2 := [c L IUa12]l/2

(4.3.13)

aEI

with fixed scaling constant c > 0 (for example, the choice c = h 2 in connection with the grid functions from Section 4.2 results in the fact that c LaEI represents an approximation to the integration J(1). The matrix norm associated to 11·112 is independent of the factor c. It is called the spectral norm and is also denoted by II . 112. The name derives from the following characterisation. Exercise 4.3.20. Prove: (a) for symmetric matrices there holds peA) (cf. (5)). (b) For each real matrix holds:

(c)

IIAI12 = [p(AT AW/ 2= [maximal eigenvalue of AT Aj1/2. For each matrix holds IIAII~ ~ IIAlbIlATI12. Hint: (b) and (lOd).

IIAII2

=

52

4 Difference Methods for the Poisson Equation

For the proof in the exercise use the scalar product (u, v) := C

L u",v",

(4.3.14a)

"'EI

(c as in (13)) and its properties

(u, u)

= lIull~, (Au, v) = (u, AT v), I(u, v)1

~ lIulbllvll2.

(4.3.14b)

Here the case K = lR is always used as basis., i.e., all matrices and vectors are real. Definition 4.3.21. symmetric and

A matrix A is said to be positive definite if it is

(Au, u) > 0 for all u

# o.

(4.3.15)

Exercise 4.3.22. Prove that (a) a symmetric matrix is positive definite if and only if all eigenvalues are positive. (b) All principal submatrices of a positive definite matrix are positive definite (cf. Exercise 19). (c) The diagonal elements a",,,, of a positive definite matrix are positive. (d) A is called positive semi-definite if the inequality (15) holds with "~" instead of ">". A positive semidefinite matrix A has a unique positive semidefinite square root B = A 1/2, which has the property B2 = A. If A is positive definite, then so is A 1/2. A corollary to Exercise 22a is Lemma 4.3.23. A positive definite matrix A is nonsingular and has a positive definite inverse. The property "A -1 is positive definite" is neither necessary nor sufficient to ensure the property "A -1 ~ 0" of an M-matrix. In both cases, however, (irreducible) diagonal dominance is a sufficient criterion (cf. Criterion 10). Criterion 4.3.24. If a symmetric matrix with positive diagonal entries is diagonally dominant or irreducibly diagonally dominant then it is positive definite. PROOF. Since r", < a"'Cl<> resp. r", ~ a",,,,, the Gershgorin circles which occur in Criterion 4 do not intersect the semi-axis ( -00,0], so that all the eigenvalues must be positive. By Exercise 22a then A is positive definite. • Lemma 4.3.25. Let Amin and Amax be respectively the smallest and largest eigenvalues of a positive definite matrix A. Then there holds

IIA -1112 = 1/ Amin.

(4.3.16)

4.4 Properties of the Matrix Lh

53

PROOF. Exercise 20a shows that IIAII2 = p(A) and IIA -1112 = p(A -1). From (5) then result p(A) = Amax and p(A -1) = l/A min, since Amin > O. •

4.4 Properties of the Matrix Lh Theorem 4.4.1. The matrix Lh (five-point formula) defined in (2.5) has the following properties: Lh is an M-matrix, Lh is positive definite,

IILhlloo ~ 8h- 2 ,

II Lhl12

(4.4.1a) (4.4.1b) (4.4.1c) (4.4.1d)

IILh"ll1oo ~ 1/8, < 8h- 2 ,

~ 8h- 2 cos2(7rh/2)

II Lh"1112 ~ ~h2 sin- 2

t 2h )

=

2!2

+ O(h2) ~ 1~'

(4.4.1e)

PROOF. (a) In Section 4.3 we already noticed that Lh is irreducibly diagonally dominant and satisfies the inequality (3.1a). By Criterion 4.3.10 then Lh is an M-matrix. (b) Since Lh is symmetric and irreducibly diagonally dominant, (lb) follows from Criterion 4.3.24. (c) That IILhlloo ~ 8h- 2 can be read from (2.5) and (3.11). To estimate Lh"l one uses Theorem 4.3.16 with w(x, y) = x(l - x)/2. Then we have Lhw ~ 1 (even that (LhW)(X, y) = 1 unless y = h and y = 1 - h) and IIwlloo ~ w(1/2, y) = 1/8. (d) The inequalities (ld,e) result from Lemma 4.3.25 and

Lemma 4.4.2. The (n _1)2 eigenvectors of Lh are u"'" (1 ~ V,j.L ~ n -1):

u""'(X, y) = sin(v7rx) sin (J-L7rY )

(4.4.2a)

The corresponding eigenvalues are l~v,j.L~n-1.

(4.4.2b)

PROOF. Let n~D be the one-dimensional grid (1.6a) and let u"(x) .sin(v7rx). For each x E nkD there holds 8-8+u"(x) = h-2 [sin(v7r(x - h)) + sin(v7r(x + h)) - 2sin(v7rx)] = 2h- 2 sin(v7rx)[cos(v7rh) -1]

since sin(v7r(x ± h)) = sin(v7rx)cos(v7rh) ± cos(v7rx)sin(v7rh). The identity 1 - cos~ = 2sin2(~/2) then implies

54

4 Difference Methods for the Poisson Equation (4.4.2c)

Let L~D be the matrix (1.9b). Note that (a-a+u)(h) also involves the boundary value u(O) which (L~Du)(h) does not; similarly (a-a+u)(1 - h) depends on u(1). However, since u(O) = sin(O) = 0 and u(1) = sin(v7r) = 0 we have L~Duv = -a-a+uv, and (2c) can be brought over:

1$v$n-1.

(4.4.2c')

The two-dimensional grid function uVP- in (2a) can be written as the (tensor) product UV(x)uP-(y). Now we have that (LhUVP-)(X, y) is equal to the sum uP-(y)(L1DuV)(x) +uV(x)(L1DuP-)(y), so that (2b) follows from (2c'). • In the sequel we want to show the analogies between the properties of the Poisson equation (2.2) and the discrete five-point formula (2.4a,b). The analogue of the mean-value property (2.3.1) is the equation 1 Uh(X,y) = '4[u h(x-h,y)+u h(x+h,y)+uh(x,y-h)+uh(x,y+h)]. (4.4.3)

From (2.3) and (2.4a) with

f

= 0 we obtain the

Remark 4.4.3. The solution Uh of the discrete potential equation (2.4a) with f = 0 satisfies Equation (3) at all grid points (x,y) E {lh. As in the continuous case the mean-value property (3) implies the maximum-minimum principle.

Remark 4.4.4. Let Uh be a non-constant solution of the discrete potential equation (2.4a) with f = O. The extrema max{ Uh(X): x E nh} and min{uh(x):x E nh} are assumed not on n h but on n.

PROOF. If Uh were maximal in (x, y) E nh, then because of Equation (3), all neighbouring points (x± h, y) and (x, y± h) would have to carry the same values. Since every pair of points can be linked by a chain of neighbouring • points, it follows that Uh = const, in contradiction to the assumption. The last proof indirectly uses the fact that Lh is irreducible. The irreducibility of Lh corresponds to the assumption in Theorem 2.3.7 that n is a domain, i.e., connected. The result of carrying over Theorems 2.4.3 and 3.1.2 reads as follows:

Theorem 4.4.5. Let ul and u~ be two solutions of (2.4a): -L1hu~ = f for different boundary values u~ = !pi (i = 1,2). Then the following holds: (4.4.4a) (4.4.4b)

4.4 Properties of the Matrix Lh

55

PROOF. Let wh:= u~ -uk. (1) In the case that O. (2) Let M be the right side of (4a). -M ::; Wh ::; M on implies the • inequalities -M ::; Wh ::; M on rh.

n

The discrete analogue of the Green function 9(X,e) is h- 2 L-,;l. Let oe be the scaled unit vector

_{h-

oe(x) -

2

0

if x if x

=e

of e

(4.4.5a)

The column of the matrix h- 2 L-,;l with index

eE ilh is given by (4.4.5b)

For e E ilh fixed, 9h(·,~) is a grid function defined on ilh. The domain of definition is extended to ilh x nh: (4.4.5c) The 9h(X, e) are entries of h- 2 L-,;l: 9h(X,~) Lh implies Remark 4.4.6. Uh(X,e)

= h- 2 (L-,;l)xe. The symmetry of

= Uh(e, x) for all x,e E nh (cf. (3.2.3)).

The representation (3.2.7) is recalled by: Remark 4.4.7. The solution Uh of the system of equations (2.4a) with boundary values

Uh(X) = h2

L

Uh(e,x)i(e),

x E ilh.

(4.4.6)

eEah

Equation (6) is the component-wise representation of the equation Uh = L-,;l ih. The factor h 2 compensates for h- 2 in (5a). It was introduced so that the summation h2 I: in (6) approximates the integral fa. Since in Section 3.2 we consider the Poisson equation L1u = i, but here the equation -L1u = i, the right-hand sides in (3.2.7) and (6) differ in their signs. The discrete Green function is positive also (cf. (3.2.9)): Remark 4.4.8. 0 < 9h(X, e) ::; h- 2/8 for x,e E ilh . PROOF. The upper bound follows from

56

4 Difference Methods for the Poisson Equation

lIo{lIoo = h- 2, and

(lc). gh > 0 can be inferred from L-;;l > 0 (cf. Theorem

4.3.11).

•

The bound gh(X,~) :::; h- 2 /8 is too pessimistic and can be improved considerably.

Remark 4.4.8'. 0< h(X~) < 9

,

! _ 10g(lx -

- 4

~12 + 2h2) < !log hi.

log 16

-

log 4

(4.4.7)

This estimate via 0(1 log hI) reflects the logarithmic singularity of the singularity ftmction s(x,~) = -log(lx - ~12)/41T. Exercise 4.4.9. For the proof of inequality (7) define Sh(X,~) := 1/4 -log(lx - ~12 + 2h2)/ log 16,

'Uh(X):= Sh(X, 0)

and carry out the following steps: (a) Sh(X,~) ~ 0 for all x E h, ~ E n h, (b) -Llh'Uh(O) = h- 2 , (c) -Llh'Uh(X) ~ 0 for all x E }R2 (longer calculation I), (d) -Llh(Sh(-,~) - gh(·,e» ~ 0 for fixed ~ E nh, (e) 9h(X,e):::; Sh(X,e).

n

Let
L

9h(~,X)
X

E n h·

{En"

Since
r h:= {~ E n: ~ is not a corner point

(0,0), (1,0), (0, 1), or (1, I)}

instead of over the points near the boundary, then the definition of
= -h L

8;;gh(~' x)
x E ilh,

(4.4.8)

{Er;.

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

4.4 Properties of the Matrix Lh

57

8;;Uh(', x) = h- 1IUh(', x) - Uh(' - hn, x)] (note that Uh(',X) = 0 for' E r~ points next to the boundary.

c

rh). The variable, - hn ranges over all

Remark 4.4.10. Equation (8) corresponds to the representation in Theorem 3.2.10. The summation h EeEr~ approximates the integral frFinally we want to take a closer look at the estimate for the solution Uh = L-;;l% through lIuhl/oo ::; I/L-;;ll1ool/%IIoo ::; IIqh1l00/8 (cf. (lc)). According to (2.6b), qh = Ih +

58

4 Difference Methods for the Poisson Equation

for x E r{.. Let the mapping <.Ph 1-+ Uh = L-,;lcph be given by the rectangular matrix A: Uh = Acph. According to Equation (8) the entries of A read

= -h8;;Yh(e, x) = 9h(e - hn,x) = 9h(X,e - hn), x E ilh, eE r~. Since A ~ 0 (cf. Remark 8), one obtains the row sum norm IIAlloo := maxx"E~ a~ as IIAcphlloo for the choice of CPh(X) = 1 in all x E r~. The ax~

solution Vh

= ACPh

then reads Vh

= 1 (why?), so it follows that

(2) The column sums of A are s(e) := "ExEah ~ = "ExEah 9h(X,e - hn) for E r~. The grid function Vh := h-2Lh"ln at the points hn near the boundary has the values

e

e-

s(e)

= Vh(e - hn),

eE r~,

n normal direction,

e

as is implied by Remark 7. Let = (6, ~2) E r~ be a point of the left or right = 0 or 1). As mentioned in the proof of (lc), LhWh ~ 1 boundary (i.e., holds for Wh(X,y) := x(l- x)/2. Since L-,;1 ~ 0, then so is Wh 2:: L-,;ln and hence Vh :$ h-2wh. In particular we have the following estimate

el

s(e)

= Vh(e -

hn) :$ h-2wh(e - hn)

= h- 2h(1 -

e-

h)/2 :$ h- 1/2

e

at the point near the boundary hn = (h,6) or (1 - h, e2) . For a point from the upper or from the lower boundary one obtains the same estimate if one uses Wh(X, y) := y(l- y)/2. Since the column sums s(e) are the row sums of AT, we have proved

IIATlloo = max{s(e):e E r~}:$ h- 1/2. (3) We have IIATAII2 = p(ATA) :$ IIATA1100 :$ IIATIloolIAlloo :$ h- 1 /2 (cf. Exercise 4.3.20a, (3.lOd), (3.10a)) so that the solution Uh = Lh"lcph = Aih satisfies the following estimate

= h2

L <.Ph (x)(A A<.Ph)(x) T

xEr~

Since IIcpll2,r~

= l14>h Ib,r~, the assertion follows.

•

4.5 Convergence

59

4.5 Convergence Let Uk be the vector space of the grid functions on n k . The discrete solution e Uk and the continuous solution U e CO(n) cannot be compared directly because of the different domains of definition. In difference methods it is customary to compare both functions on the grid nk. To this end one must map the solution U by means of a "restriction"

Uk

U 1-+ Rku

e Uk

(u: continuous solution)

(4.5.1)

to Uk. In the following we choose Rk as restriction on nk: (4.5.2) The limit h -+ 0 is made precise as follows. Let H C m+ be a subset with accumulation point zero: 0 e H. For example, the step sizes considered so far form the set H = {1/n:n e IN}. For each he H let Uk be equipped with the norm II . Ilk. Definition 4.5.1. (Convergence) The discrete solutions Uk (with respect to the system of norms II· Ilk, hE H) to u if

e Uk converge (4.5.3a)

We have convergence of order k if (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 e H} is said to be stable with respect to 11·1100 if sup

IILhllioo < 00.

(4.5.4)

kEH

For the discretisation defined in Section 4.2, the stability has been proved in (4.1c) with respect to the row sum norm, and in (4.1d) with respect to the spectral norm. The grid function Ih in -L1huh = Ih is the restriction Ih

= Rhi

(4.5.5a)

of I, where in (2.6b) Rh was chosen as the restriction to ilh: (Rhf)(X)

= I(x)

for all x

e nh .

(4.5.5b)

The formulation of consistency becomes simpler if instead of the matrix Lh one considers the difference operator -L1h on which it is based (cf. Remark 4.2.5). Let Dh be a general difference operator, and let -DhUh = Ih discretise -Llu = I. The discretisation described by the triple (Dh, R h, Rh) is said to be

60

4 Difference Methods for the Poisson Equation

consistent of order k with respect to operator), if

11·1100

(consistent with the Laplace (4.5.6)

for all U E Ck+2(O). Here K is independent of hand u. Remark 4.5.2. Let Rh and Rh be given by (2) and (5b). The five-point formula L1h is consistent of order 2: estimate (6) holds with k = 2 and K = 1/6. PROOF. The expansion (l.4c) can be applied in the x and y directions and yields L1hRhU(X, y)

= L1u(x, y) + h2(Rr,; + R,,)

with IR:J;I, IR"I ~ (1/12)lIuII C4(i1)' (4.5.7)

•

Theorem 4.5.3. Let the discretisation (Dh' R h , R h ) be consistent of order k. Let the matrix Lh associated to the difference opemtor Dh (cf. Remark 4.2.5) be stable with respect to II ·1100' Then the method is converyent of order k if U E Ck+2Ci'i).

PROOF. Wh := Uh - RhU satisfies the difference equations -DhWh

=

-DhUh

+ DhRhU = !h + DhRhu

= Rhf + DhRhU = DhRhU -

RhLlu

in flh.

Because of the homogeneous boundary condition Wh = 0 on rh we have that = L,/(DhRhu - RhL1U) (cf. Remark 4.2.1). (4) and (6) imply (3b). •

Wh

If one substitutes the concrete values IILh"llloo ~ 1/8 and K obtains

= 1/6 one

Corollary 4.5.4. Let the continuous solution of the boundary value problem (2.2a,b) belong to C4(ii). Let Uh be the discrete solution defined in (2.4a,b). Then the converyence of Uh to

U

is of second order:

h2 IIUh - RhUlioo ~ 48I1uIIC4(i1).

(4.5.8)

The assumption U E C4(1i) can be weakened to Corollary 4.5.5. Under the condition U E C 3 ,1(1i) we also have IIUh - RhUlioo ~ (h 2/48) Ilull c 3.1 (ii)" PROOF. The remainder term R4 in (1.5b) may also be written as

(4.5.8')

4.5 Convergence

61

The Lipschitz estimate lu"'(e) - u"'(x)l ::; Ie - xlliull c 3.1Cfi) implies R.t ::; Ilullc3.1(?i/4! so that in (7), (6), and (8) the norm Ilullc4(?i) can be replaced • by lIullc3.1Cfi)'

If, however, one further weakens u E e 3 •1 (n) to one obtains a weaker order of convergence. Corollary 4.5.6. Under the condition of order s - 2:

U

U

E

eB(n), 2 < s < 4,

E eB(n), 2 < s ::; 4, Uh converges (4.5.8")

where Ks := 1/[2s(s - 1)] for 2 ::; s ::; 3 and KB := 1/[2s(8 - 1)(8 - 2)] for 3 ::; 8 ::; 4.

The proof results from

Even though the proofs of convergence are simple, the results remain unsatisfactory. AB 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 e 2 (li), although one needs at least U E e s (n) with s > 2 in Corollary 6 for convergence. Stronger results can be obtained by an analysis which will be discussed in Section 9.2. That errors of the order of magnitude of O(h2) occur even under weaker conditions, is shown in Example 4.5.8. If one solves the difference equation (2.4a, b) for -Llu = 1 in [l = (0,1) X (0,1) and u = 0 on r, one obtains at the centre x = y = 1/2 the values Uh(!, i), which are shown in the first column of Table 1. The exact solution u(!,!) = 0.0736713· ., results from a representation that one can find in Example 8.1.11. The quotients €h/€2h of the errors €h = u(!,!) - Uh(!,!) approximate 1/4. This proves Uh(!,!) = u(!, !)+O(h2), although U ¢ e 2(n). At (!, !) Uk has furthermore the asymptotic expansion Uh = u(!, + h2 + O(h4). The error term e(!,!) independent of h, is eliminated by using the Richardson extrapolation

e(1.1)

The extrapolated values are already very accurate for h column of Table 1).

(!, !)

= 1/16

!)

(cf. the last

62

4 Difference Methods for the Poisson Equation

'!able 4.5.1 Difference solutions for Example 8 h

Uh(!,!)

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

0.0727826 0.07344576 0.0736147373 0.0736571855

Uh,2h(!,

!)

0.0736668 0.07367106 0.07367133

Uh(!,!) Quo-Uh(!, tient

fh =

!)

8.89xlO- 4 2.26xlO- 4 5.66xlO- 5 1.41 x 10-5

0.250 0.251 0.249

fh,2h

Quotient

4.5xlO- 6 0.06 2.7xlO-7

4.6 Discretisations of Higher Order The five-point formula (2.11) is of second order. Even if the solution U belongs to C 8 with s > 4, no better bound for ..1hRhu - Rh..1u than O(h2) would result. An obvious method for constructing difference methods of higher order is the following. As an ansatz for the discretisation of the one-dimensional equation U" = f choose

cm

k

(DhUh)(X)

= h- 2

'2: CIIUh(X + vh). II=-k

The Taylor expansion provides ~

(DhRhU)(X)

= '2: a"h,,-2u" (x) +O(h2k-l),

k

a"

=

,,=0

('2: cllv")/j.LL II=-k

The 2k + 1 equations ao = al = a3 = a4 = ... = a2k = 0 and a2 = 1 form a linear system for the 2k + 1 unknown coefficients CII. For k = lone obtains the usual difference formula (l.4c); for k = 2 a difference of fourth order results: 2

-1

4

5

h (DhUh)(X) = 12[uh(x-2h)+uh(x+2h)]+3[uh(x-h)+uh(x+h)!-2uh(x).

If one applies this approximation for U" to the x and y coordinates, one obtains for -..1 the difference star

2 -h12

[

1 -16

-1~

60 -16 1

-16

1

1

(4.6.1)

(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 ilh, for example, one needs the values Uh( -h, h) and uh(h, -h) outside nh (cf. Figure 1). One possibility would be to use scheme (1) only at points 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 this one means difference methods (2.12) which are characterised by "ca {3 # 0 only for -1 S a,/3 S 1" (cf. Fig. 1). An ansatz with the nine free parameters ca {3, -1 S a, /3 S 1, leads, however, to a negative result: there is no compact nine-point formula with (DhU)(X) = -Llu(x) + O(h3 ). In this sense the five-point formula is indeed optimal. Nevertheless, nine-point procedures of fourth order can be obtained if one also selects the right side fh of the system of equations (2.6a,b) in a suitable manner.

..

.

II

•

•

•••

II

•

X

•••

.. : Difference Scheme (1) .: Compact nine-point formula

• Figure 4.6.1. Scheme (1) and the compact nine-point formula If one applies the compact nine-point scheme

Dh := _

h~2 [=! ;~ -1

to

U

-4

=! ]

(4.6.2)

-1

e eS(n), the Taylor expansion results in

[8

8

~

4 4 h2 h4 -Dh'U = -Llu - 12Ll2u - 360 8x4 + 4 8x2 8y2

]

+ 8y4 Llu + O{h6 ). (4.6.3)

Here it is crucial that the error term can be expressed by - Llu and hence by

f·

For the special choice of the restriction Rh via 1/2 4

1/2

]

I,

(4.6.4)

1/2 i.e.,

fh(X,y) =(Rhf)(x,y) 1 :=12 If(x - h,y) + I(x + h,y)+ f(x, y - h)

+ I{x, y + h) + 8/(x, y)], (4.6.4')

64

4 Difference Methods for the Poisson Equation

one obtains the expansion (4.6.5) which, because 1 = -Llu, agrees with (3) up to O(h4). The matrix Lh of the system of equations which results after the elimination of the boundary values Uh(X) = cp(x), x E r h, has the entries L

~

= h- 2 {

e, e e

20/6 if x = -1/6 if x - = (±h, ±h) or (±h, =Fh), -4/6 ifx - = (±h,O) or (O,±h), o otherwise.

The right side of the system of equations LhUh

(4.6.6)

= qh is

qh := Ih+CPh, Ih := ~I according to (4'), CPh(X) := -

L

L~cp(e). (4.6.7)

~Er,.

The discretisation (Dh , Rh, Rh) with Dh from (2), Rh from (5.2), Rh from (4') is called the mehrstellen method (cf. Collatz [1]).

Exercise 4.6.1. Let Dh and Lh be defined respectively by (2) and (6). Prove that (a) Lh is an M-matrix; (b) IILh'llloo ~ 1/8 (stability), IILhiloo :5 Wh- 2 /3; (c) IIDhRhu - RhLlulioo :5 ifCo·llullc&(o), if u E CS(Ji) (consistency).

Theorem 4.6.2. Let ilh be defined as in Section 4.2. Let Uh be the solution provided by the mehrstellen method LhUh = % with Lh from (6), and qh from (7). Let the solution u of the boundary value problem (2.2a,b) belong to C6(lJ) or C5.1(.G). Then the following error estimates hold in the respective cases: llh4 IIUh - RhUlioo :5 14401Iullc5.1(o)· (4.6.8a) In the case 01 the potential equation, i.e. 1 = 0, we even have resp.

IIUh - RhUlioo ~ Kh6I1ullc8(O)'

IIUh - RhUIloo ~ Kh6I1ullc7.1(O)' (4.6.8b)

ilu E CS(n), resp. u E c.,·l(n). PROOF. As in Theorem 4.5.3, (8a) results from Exercise (lb,c). In the case that 1 = 0 the O(h4 )-term also vanishes in (3). •

4.7 The Discretisation of the Neumann Boundary Value Problem

65

4.7 The Discretisation of the Neumann Boundary Value Problem The Dirichlet boundary condition u{x) = !p{x) was used directly in the difference method; a discretisation was not necessary. A different situation arises for the Neumann boundary value problem

-.1u = finn = (O, 1) X

au/an =!p

(0,1),

on

r.

(4.7.1)

The normal derivative, which reads explicitly

au -=-u an y au an = -Ua;

au -=U an a; au -=u an y

for X= (x, 0) E r, for x={O,Y)Er,

for x = (x, 1) E r, for x = (l,y) E

r,

(4.7.2)

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

4.7.1 One-sided Difference for

Bu/on

The Poisson equation leads to the (n _1)2

= (l/h -

1)2 equations

(4.7.3) which require the values of Uh(X) for all x

n~ := nh U r~,

r~:=

E

n~, where

n \ {(O, 0), (0,1), (1,0), (1, I)}.

To obtain a further 4(n - 1) equations for {Uh(X): x E rh}, we replace, at all x E r~, the normal derivative au/an =!p by the backward difference

(4.7.4)

-.- .. .. I

• -

Discretisation of the Neumann condition Discretisation of the Poisson equation

-i-- · 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 Poisson Equation

h- 1 (Uh(X,0) - Uh(X, h») =
} for x

h- 1 [Uh(0, y) - uh(h, y)) =
} for y = h, 2h, ... , 1 - h.

= h, 2h, ... , 1 - h,

(4.7.4') Equations (3) and (4) yield (n + 1)2 - 4 equations for as many unknowns.

Exercise 4.7.1. After a rescaling of equations (4) to h- 18;;Uh(X) E r~, these equations, together with (3), form a system Lhuh that Lh is symmetric and satisfies (3.1a).

x

= h- 1
As in the Dirichlet problem the variables Uh(X), x E r~, can be eliminated with the aid of (4) in (3). At the point (h, y) near the boundary, for example, Equation (3) becomes h-2(3uh(h, y) - uh(h, Y - h) -uh(h, y+ h) -uh(2h, y») The star h- 2

[ -1 -1

4 -1

h- 2

-1] thus becomes

0

-1 3 -1

[-1

0 3 -1

h-2 [

= f(h, y) + h- 1
-1] ,

h-2 [

-1] ,

h-2 [

-1

-1 3 -1

0].

-1

-1 3 0

-1] ,

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

(h, h) E fl. the star reads h-2 [0 -

~

-1

l

Except for the special case

h = 1/2, one obtains for the (n - 1)2 values Uh(X), x E ilh, the system of equations

with

(4.7.5a) if x E ilh is far from the boundary, if x = (h, h), (h, 1 - h), (1 - h, h), (1 - h, 1 - h), otherwise (4.7.5b) if x, E are neighbours, otherwise for x =1= (4.7.5c) fh(x) = (Rhf)(X) := f(x),
e nh

and

e

4.7 The Discretisation of the Newnann Boundary Value Problem

67

Remark 4.7.2. (a) Lh is symmetric and satisfies the sign condition (3.1a). (b) With lexicographical arrangement of the grid points of ilh, Lh has the form T-I -I

Lh =h -2

[

-I

T-I ..

.... -I

The matrix Lh is singular because the system LhUh = %, like the continuous 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 solvable if and only if _h2

L

f(x) = h

L


(4.7.6)

xEr~

xES'h

Any two solutions of (5a) can only differ by a constant: uJ. - u~

= ell, c E IR.

PROOF. Evidently, Lhll = 0 holds, i.e., 11 E kernel(Lh)' Furthermore, Theorem 4 will then imply dim(kernel (Lh)) = 1. This proves

kernel(L h ) = {ell: e E IR}

(4.7.7)

and thus the second part of the assertion. (5a) is solvable if and only if the scalar product (v, qh) vanishes for all v E kernel (LI) = range (Lh).L. Because = Lh and (7) of

LI

(ll,%) = 0

or

L

%(x) = 0

(4.7.6')

XEnh

is sufficient and necessary. According to Definition (5c), (6) and (6') agree. • Let condition (6) be satisfied. System (5a) can be solved as follows. Select an arbitrary Xo E ilh and normalise the solution Uh (determined except for one constant) by

(4.7.8) Let Uh be the vector Uh without the component Uh(Xo). Let Lh be the submatrix of Lh in which the row and column with index Xo have been left out. Let Q.h be constructed likewise. Then

68

4 Difference Methods for the Poisson Equation £hU = qh

(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, £h is a symmetric M -matrix. Under condition (6), Uh = L;:lqh' supplemented by (8), yields the solution Uh of system (5a).

PROOF. (a) As a principal submatrix of Lh, £h is symmetric. In {}h \{Xo} any two grid points can be connected by a chain of neighbouring points so that Lh is irreducible. For all x E ilh \ {Xo} there holds (3Ab); at neighbouring points ofXo we even have (3Aa), so that £h is irreducibly diagonally dominant. According to Criterion 4.3.10, Lh is an M-matrix, thus nonsingular. (b) If Uh is the solution of (5a), one can assume (8) without loss of generality, so that Uh restricted to nh \ {xo} also solves Equation (9) and has to agree • with the unique solution Uh.

As a corollary of Theorem 4 one obtains that rank(Lh) ~ rank(Lh) = (n - 1)2 -1, i.e., dim(kemel (L h = 1 and hence (7) holds. Another possibility for the solution of Equation (5a) is to pass to an extended system of equations

»

(4.7.lOa)

with where

(T

(4.7. lOb)

can be prescribed arbitrarily.

Theorem 4.7.5. Equation (lOa) is always solvable. If for the last component of the solution Uh we have A = 0, then condition (6) is satisfied and Uh represents the solution of system (5a) which is normalised by nTUh = EXE S1h Uh(X) = (T. However, if A f. 0 holds one can interpret Uh as solution of Lhuh = ilh where ilh = qh - An belongs to j(x) := I(x) - A and j and cp satisfy condition (6).

PROO F. 11 is linearly independent of the columns of Lh so that rank IL h , n] = rank (L h ) + 1 = (n _1)2. Likewise, (R T , 0) is linearly independent of the rows of ILh, R] so that rank (L h ) = (n - 1)2 + 1, i.e., Lh is nonsingular. The other statements can be read from (lOb). • Attention. One should either use Equation (lOa,b) or Equation (9), after first replacing % by ilh := % - (IT %/IT n)l. As a justification of this recommendation note that the condition for solvability of the continuous problem is JS1 I dx + Jr cpdT = 0 and that this does not at all imply the discrete solvability condition (6). For smooth functions f and cp Equation (6) can be shown to hold up to a remainder of order O(h). Thus it is generally unavoidable to

4.7 The Discretisation of the Neumann Boundary Value Problem

69

replace ih and qh by ih -All and qh -All (A = llT %/IT 1). In the case of Equation (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 of LhUh = qh with qh(X) = %(x) for x '" Xo and qh(Xo) := - Ex#Xo %(x). That is, here too, an implicit correction of qh is carried out, with the difference that the correction is not distributed over all components as before, but is concentrated on qh(XO). If Equation (6) is satisfied up to order O(h), then qh(XO) and qh(XO) differ by O(h-l). Therefore the solution Uh of Equation (9) contains a singularity at the point Xo.

Theorem 4.7.6. (Convergence) Let u E C3.1(1i) be the solution of the Neumann problem (1). Let Uh

= (u;) be the solution of Equation

(lOa).

Then there exists acE IR and constants C, C' independent of u and h such that IAI ~ C'hllullco.1(i'i)'

IIUh - Rhu - dll oo

~ C[hllulb,l(i'i) + h21Iullc3,1(i'i))' (4.7.11)

PROOF. (a) We have A=

nT qh/nTn = (h2 LJ(x) + h L
Because f fl f dx + f r 'P dr = 0, the first factor consists only of the quadrature error O(hllullco,l(i'i»' (b) According to Theorem 5, Uh is the solution of LhUh = qh := qh - All. This corresponds to the difference equations -LlhUh

= ih :=!h -

Al

in ilh,

a;;Uh

=
on r~.

The difference Wh := Uh - RhU satisfies -L1hWh

= -LlhUh + LlhRhu = LlhRhu + ih - Al = L1hRhu - RhLlu - XI =: Ch in ilh,

au a;;Wh = a;;Uh-a;; RhU =
(4.7.12a)

on r{ (4.7.12b)

The errors of consistency are

IIChiioo ~ ~h2I1ullc3'1(i'i) 1

It/J(x) I ~ '2hllulb,l(O) + IAI

(cf. Remark 4.5.2), (cf. Lemma 4.1.1).

Since the solution Wh exists, condition (6) is satisfied with Ch and t/J:

70

4 Difference Methods for the Poisson Equation

Then Wh := Wh - cl with c = nTWh/nT n is the solution of (12a,b) normalised by nTWh = O. The application of the following Theorem 4.7.7 to Equation (12a,b) provides the inequality (11). • Theorem 4.7.7. (Stability) Let condition (6) be satisfied. Let the solution Uh of (3), (4) be normalised by IT Uh = O. Then there exist constants Ct, C 2 independent of U and h such that (4.7.13)

The proof of this theorem, which corresponds to Theorem 4.4.11, will be supplied in Sect. 4.7.4. 4.7.2 Symmetric Difference for

au/an

AB can be seen from Theorem 6, the one-sided difference a;; causes the error term O(h). It seems obvious to replace a;; by a symmetric difference. To this end the five-point discretisation is set up at all points x E h = [Jh u n (cf. (2.1c)): -LlhUh = fh := Rhf in'lih , (4.7.14a) where Rh is the restriction to 'li h . The symmetric difference &,! is defined by

n

(~Uh)(X) := (2h)-1[Uh(X + hn) - Uh(X - hn)] = !f1(x) in x E

n.

(4.7.14b)

Here, 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 comer x = (0,0) one has, for example, the normals

The corresponding equations (14b) also contain different (!) values IP(O+, 0) = limz -+o!f1(x,O) and !f1(0,0+) = limy-+o!f1(O,y). For x E rh the difference formula (14a) needs the values in points x + hn outside of 'li h • These can be eliminated with the aid of (14b) so that a system of equations Lhuh = qh remains for the (n + 1)2 components Uh(X), x E h .

n

Exercise 4.7.8. (a) If the grid points of'li h are arranged in lexicographical order, Lh has the form

Lh = h- 2

[

T

-21

-/

T

-/ -/

'. I'

T -/ -21 T

T

=

[_i -;

-1..

-1

4

-1

-2

4

I.

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

71

=

The analogue of Theorem 3 reads: Theorem 4.7.9. Equation (14a,b) is solvable if and only if nT Dhqh = 0 for Dh in Exercise 8b. Any two solutions may differ by only a constant. The formulation of nT DMh = 0 with the aid of f and cp reads: _h2

2:

d(x)d(y)f(x, y) = 2h

2:

(4.7.15)

d(x)d(y)cp(x, y),

where the summand for the corner points occurs twice in the second sum, and both the different limits for cp are taken into account.

Remark 4.7.10. The sums in (15) represent summed trapezoidal formulae. fdx+ cpdr = 0 follows Equation (15), except for a remainder Thus, from O( h2l1ullcl.l (0)'

In

Ir

Theorems 4 and 5 can be transferred without difficulty. The convergence theorem, Theorem 6, becomes Theorem 4.7.11. Let u E the solution of LhUh

=

c3 •1 (n)

(D~% )

be a solution of (1). Let Uh

with Lh

= [D~.fh ~].

=

("f: ) be

We have conver-

gence of second order:

The proof is essentially the same as for Theorem 6. An additional technical difficulty is the fact that the consistency error LlhRh u - Rh Llu must also be Instead of treating the determined in x E n although u is only defined in difference equation and the boundary discretisation separately as in (12a,b) one should directly analyse the equations LhUh = % from which the values Uh(X + hn) outside of li have already been eliminated.

n.

4.7.3 Symmetric Difference for

au/an on an Offset

Grid

If we offset the above grid by h/2 in the x and y directions, we obtain the grid

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

flh

are at a distance h/2 from

72

4 Difference Methods for the Poisson Equation

n:= {(X,y) E r: ~ - ~ E Z or

*-~

E

z}

(cf. Figure 2). To each point near the boundary x-hn/2 (x E rh) corresponds an outlying neighbour x + hn/2. o

0

0

0

0,r--"--"'--•••• +0 ot • • • • ,0

0: Points in ilh

ot •••• *0 0, ...••••-to

.. : Points in rh

I

I

I

I

I

I

~

o

-~-.-

0

0

. 0

Figure 4.7.2. Offset grid

The discretisation of the Poisson equation (1) is -LlhUh(X) = f(x)

in x E ilh,

h- 1[Uh(X + hn/2) - Uh(X - hn/2») = cp(x)

in x

(4.7.17a)

En.

(4.7.17b)

The difference (17b) is symmetric with respect to the boundary point x and nevertheless agrees with the backward difference 8;; at the grid point x+hn/2.

Remark 4.1.12. After eliminating the values Uh(X + hn/2), x E n, we obtain a system of equations LhUh = Qh, where Remark 2 is also valid for this matrix L h . In contrast to Section 4.7.1, Lh is of size n 2 x n 2. The Theorems 3, 4, 5, and 7 hold analogously. Theorem 6 holds with the inequality (16) instead of (11).

4.1.4 Proof of the Stability Theorem 1 In the case of Dirichlet boundary values the stability statement of Theorem 4.4.11 follows immediately from the maximum principle and the bound for L-,;1. The corresponding statement of Theorem 7 for Neumann boundary values, 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 additional factor Ilog hi. However, this factor is not to be avoided if one uses Equation (9), LhUh = qh, without condition (6) being satisfied. Let the discrete Green function 9h(X,€), x E ilh Un, € E ilh be defined by -Llh9h(X, €) = 6(x, €) :=

{h- 2 o

e} _{h- /(4 - 4h)

for x = otherwise

1

0

(4.7.18a) } if x is near the boundary if x is far from the boundary ,

4.7 The Discretisation of the Neumann Boundary Value Problem

73

(4.7.18b) where Llh and a;; act on x. 9h exists since LXE!h 8(x,~) solvability condition (6').

Lemma 4.1.13. For arbitrary % with :o.T% Uh(X) := h2

L

= 0 (c/.

=

0 proves the

(6')),

qh(~)9h(X'~)

(4.7.19)

{EO"

represents a solution

0/ LhUh = %.

PROOF. At points far from the boundary X E ilh, (LhUh)(X) = -Llhuh(x) = qh(X), At points near the boundary x E ilh, from a;;Uh = 0 and (6'), one has the identity (LhUh)(X)

= (-Llhuh)(x) = Qh(x) -

h- 1 4 _ 4h

L

Qh(~)

= qh(X),

•

{EO"

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

Theorem 4.1.14. The Green function

9h(X,~)

can be so selected that

(4.7.20) This inequality corresponds to the bound (4.7) in the Dirichlet case. Before Theorem 14 is proved, we want to show that Theorem 7 follows from it. Analogously to Jo[l + jlog Ix-~111 dx ~ Kl and Jr[l + liog Ix-~1I1 drx ~ K2 we obtain h2

L [1 + jlog(lx - ~I + h) 11 ~ K~, h L [1 + Ilog(lx - ~I + h)1l ~ K~. xEr:.

From (19), (20), and qh

nT

= ih +
If Uh := Uh - nTuh/1T 1 is the solution of LhUh = qh normalised by Uh = 0, one can see that

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

•

74

4 Difference Methods for the Poisson Equation

e

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

e

e

e,

(4. 7. 21a) a;;Gh

=0

on r~,

(4. 7. 21 b)

IGh(x)1 ~ hG'/(lx - el + h), where G' does not depend on e and Theorem 7.

(4. 7. 21 c)

e. Then Theorem 14 holds, and thus also

PROOF. For e,e' E nh let 9h(X,e,e') be defined as a solution of a;;gh(·,e,e') =0 on

r~,

-Llh9h(X,e,e')=h- 2 {_\

o

!~~:~,}

in Dh.

otherwise

For e' = e+he, gh(·,e,e') agrees with Gh in Lemma 15. For arbitrary e, e', one finds a connection e = e 1 , ... ,e' = e' with ek+l - e k = ±hek , e k = (1,0) or (0,1). Since 9h(·,e,e') = 'E~=19h(·,ek-l,ek), (21c) implies the estimate

eo,

,

Igh(X,e,e')I ~

hI: G'/Ox -

e k- 11+ h).

k=l If one considers first the case Xl = el = e~, X2 ~ e2 < e2 and one applies 'E~kl 11k ~ const . (1 + log(k2h) -log(klh», one obtains Igh(X, e,e')1 ~ G"[1

+ pog(lx -

el + h)l + pog(lx - e'l + h)ll·

In the general case, this bound also comes out, but one must choose the connection e k such that Ix - ekl ~ min{lx - el, Ix - e'I}. Since we know that h 'Ee/Er~ Ilog(lx - e'l + h)l ~ const for all x E Dh, gh(X,e):= 4.!:4h

I: gh(X,e,e') e/Er~

satisfies the inequality (20). Because 'Er 1 = (4-4h)jh the equations (18a,b) h are also satisfied, i.e., gh(X, e) is the Green function. • We shall construct the function Gh needed in Lemma 15 explicitly. The basic building block is the discrete singularity function Sh(X,e) on the grid in m.2 Qh:= {(x,y) E m.2 :xlh,ylh E Z}. Lemma 4.7.16. and

The singularity function defined by Sh(X, e) := ah(x - e)

4.7 The Discretisation of the Newnann Boundary Value Problem

75

j1f j1f +a:2112)/h - 1 d'T]l d'T/2 2 sin ('T]d2) + sin2('T/2/2) = __1_ j1f j1f sin 2«xl'T]l + x2'T/2)/(2h)) d'T]l d'T/2 2 1

ei(a:1111

O"h(X) := - 1611"2

-1f

-1f

811"2

-1f

-1f

Sin ('T]1/2) + sin2('T/2/2)

for all x, eE Qh has the property -Llhsh(X, e) = h- 2 for x = o otherwise.

PROOF. Let e(x, 'T]) := exp(i(xl'T]l

eand -Llhsh =

+ x2'T/2)/h). Note that

-Llh(e(x, 'T]) - 1) = 4h- 2e(x, 'T])[sin2('T]d2) + sin2('T/2/2)]

1:1:

and

e(x,'T])d'T] =

•

{4~2 ~~~~;~}.

For multi-indices a = (aba2) E 7l.2 with al ~ 0, a2 ~ 0 one defines the partial difference operators of order lal = al + a2 by (4.7.22) Starting with the representation in Lemma 16, Thomee [1] proves

For the construction of the function Gh in Lemma 15 we select, without loss of generality, e = (1,0) and keep E ilh with + he E ilh . The function

e

e

(4.7.23) satisfies (21a) but not the boundary condition (21b). Let the symmetrisation operators Sa: and Sy be defined by 1

1

Sa:Uh(X,y):= '2[Uh(X,y)+Uh(h-x,y)], SlIUh(x,y):= '2[Uh(X,y)+Uh(X,h-y)] for (x,y) E Qh. The function (4.7.24) is symmetric with respect to the axes y = h/2 and x = h/2. Thus we have (21b) on the left and lower boundary: a;G~(O,y) = a;;G~(x,O) = 0

(4.7.25)

Furthermore, G h satisfies condition (21a) just as Gh does. For each f3 E 7l.2 we define the operator P{3 by

76

4 Difference Methods for the Poisson Equation

41 [Uh(X + f31L, y + f32L) + Uh(X +Uh(X + f31L, y - f32L) + Uh(X - f31L, y -

(P,8Uh)(X, y) =

f31L,y + f32L) f32L)],

where L = 2 - 2h, and set G~(x) := (P,8G~)(x) - (P,8G~)(O).

Lemma 4.7.18. For f3 = (f31.f32) with 18QG~(x)1 ~ hK/If31 3 Here, K is independent of the

(4.7.26)

1If3l1oo ~ 2 there holds

allial ~ 2, x E ilh. choice of the ~,~ + he E ilh.

(4.7.27)

for

PROOF. According to the definition we have G~(O,O)

= O.

(4.7.28a)

The operator P,8 preserves the symmetry, i.e., Uh = SzUh implies P,8Uh SzP,8Uh. Thus we have that G~ = SzG~ = SlIG~ and consequently

=

(4.7.28b) G~(x) is the linear combination of h8;ah(X-e) for differentl with Ix-el+h ~ K'If3I, if x E ilh . According to Lemma 17,

(4.7.28c) so that (27) follows for form

lal = 2. Let lal = 1. 8QGh(X) can be written in the

aaG~(O)

I

+ L[8QG~(Xk) -

aaG~(xk-l)]

k=l

with x O = x, x' = x, IIxk - Xk-1ll oo = h. Each summand has the form ±h8'YG~ with hi = 2 so that (28b,c) lead to the estimate (28d) (4.7.28d) Likewise one infers from (28a) and (28d) the inequality (27) for proves Lemma 18. Since

L:,8EZ2\{O}

lal =

0, which •

1f31-3 < 00, Gh(x) :=

L

G~(x)

(4.7.29)

,8EZ2

exists. Since -LlhG~(X) = 0 in ilh for all f3 #- (0,0), Gh also satisfies Equation (21a). As already mentioned in the proof of (28b), Gh = SzG h = SliGh

4.7 The Discretisation of the Newnann Boundary Value Problem

77

holds so that G h also satisfies Equation (25): 8;;Gh = 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: Gh(X, y) = Gh(X + L, y) = Gh(X, y + L).

PROOF. Let G~(x+f31L, y+f32L)-G~(f31L, f32L) be abbreviated to 'Y{3(x, y). Definition (29) says that (4.7.30a) Now'Y{3 can be written as a sum over the differences h8t G'h (X+f31 L, f32L+J-th) from J-th = 0 to p.h = x, and over h8tG'h(f31L + vh,f32L) from vh = 0 to vh = y. For 1f311 ~ 2 the distance between the arguments of ~ and ~ + he is always::; (1f311 - 1)L. Each summand is thus, by Lemma 17, bounded by O(h2j(lf311-1)2) = O(h2jf3f). As a sum of such terms then 'Y{3 can be estimated by (4.7.30b) 1'Y/31 ::; O(hjf3~) for 1f311 ~ 2. We want now to show that the following equation (30c) also holds:

Gh

= k-+oo lim

"L.J "'Y{3 = k-+oo lim "L.J "L.J 'Y{3. L.J 1{31-11::;k 1/3219 1-k::;.Bl 9+k 113219

(4.7.30e)

The sums (30a) and (30c) differ by Dh := L:1{321::;kl'Y(1+k,{32) - 'Y(-k,{32l The values 1 + k and -k that can be given to f31 satisfy 1f311 ::; 2k. The absolute value of the sum Dh is then bounded, from (30b), by L:1{321::;k O(hj k 2 ) = (2k+ 1)O(hjk2) = O(hjk), so that the limits of the double sums in (30a) and (30e) are the same. Changing the variable f31 to f31 -1 then transforms (30e) into

(4.7.3Od) The first part of the identity k

L

[G'h(f31 L, f32L) - G~(f31L + L, f32L)]

/31,f32=-k k

=

L

[G~(-kL,f32L)-G~(kL+L,f32L)]

f32=-k

k

=

L

[G'h(h + kL, f32L) - G'h(kL + L, f32L)]

(4.7.30e)

(32=-k

is elementary; the second results from the symmetry G~ = S",G~. As above, the summands of the last sum can be bounded by O(hjk2) so that (30e) vanishes for k ..... 00. Together with (3Od) one obtains

78

4 Difference Methods for the Poisson Equation

= k->oo lim The proof of Gh{x, y) Since L

=2-

Gh(l, y)

'Yf3{x + L, y)

"L..J

= Gh{X + L, y).

-k~f31.f329

= Gh{x, y + L) is analogous.

2h, the symmetry Gh

= Gh(h -

1, y)

= 811:Gh and the periodicity yield

= Gh(h -

1 + L, y)

= Gh(l -

•

h, y),

i.e., 8;;Gh{1,y) = 0 on the upper boundary (l,y) E rho Likewise we show that 8;;Gh{X, 1) = 0 on the right boundary. Thus (21b) is also proved. It remains to show (21c): IGh{X)1 ~ hC/{lx - ~I + h) in nh . The G~ for 11.81100 ~ 1 are linear combinations of h8; O"h(X for = and other ~ nh, which are generated by 811:, 8 11 , and Pf3. For all there holds Ix - el ~ Ix - el, so that from Lemma 17 follows

-e)e e e

e

On the other hand, Lemma 18 shows that

I

L

G~{x)1 ~ h

1If3l1co~2

L

K/I.813 = hK' ~ hK"/(lx - ~I + h).

1If3l1oo~2

The assumptions of Lemma 15 are hence satisfied: Gh with (21a-c) exists. Thus theorems 14 and 7 are proved. •

4.8 Discretisation in an Arbitrary Domain 4.8.1 Shortley-Weller Approximation Let the boundary value problem

-.L1u =

f in

be given for an arbitrary domain over n one obtains

n,

u = cp on

r

(4.8.1)

n. If one places a square grid of step size h

flh := {(x,y) E fl: x/h E 71.

and y/h E 7l.}

(4.8.2)

as the set of grid points. By contrast, the set r h of boundary points must be defined differently from the case of the square. The left neighbour point of (x, y) E flh reads (x - h, y). If the connecting segment {(x - t9h, y): 19 E {O,l]} does not lie completely in fl, there exists a boundary point (x - sh, y) E

r

with (x, y) E flh' S E (O, 1), (x - t9h, y) En for all 19 E [0, s), (4.8.3a)

4.8 Discretisation 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, lower, and upper neighbour points can be boundary points of the following form: (x + sh, y) E r

with (x, y) E {}h, s E (0,1], (x + 'I9h, y) E {} for all '19 E [0, s), (4.8.3b) (x, y - sh) E r with (x, y) E {}h, S E (0,1], (x, y - 'I9h) E {} for all '19 E [0, s), (4.8.3c) (x, y + sh) E r with (x, y) E {}h, S E (0,1], (x, y + 'I9h) E {} for all '19 E [0, s). (4.8.3d) We set

n

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

(4.8.4)

A grid point (x, y) E {}h possessing a neighbour from n, is said to be near the boundary. All other points of {}h are said to be far from the boundary. As can be seen in Figure 1, a point may be near the boundary although (x ± h, y) and (x, y ± h) belong to {}h (namely if not all connecting segments lie completely in il). • E .fJh near the boundary • E {}h far from boundary DEn on boundary

Figure 4.8.1. [Jh and rh

Exercise 4.8.1. For a convex domain .fJ show that

n

= {(x, y) E

r:

x/h E 71. or y/h E 7l.}.

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: u"(x)

= 2[u(x") -

u(x) _ u(x) - u(xl)]/(x" _ x') x"-x x-x'

+ Rem.

(4.8.5)

Exercise 4.8.2. Show that (a) the Taylor expansion implies

IRem I ~

1 (x" - x)2 + (x - x ' )2 3 x" _ x' lIu llc3([z',zll])

max{x" - x x - x'} ~ 3 ' lI u llcs([z',zll)) for the remainder in Equation (5) if u E C 3 ([x' , x"]).

(4.8.6)

80

4 Difference Methods for the Poisson Equation

(b) In Equation (6) one can replace the norm of C 3 ([x /, x"]) by that of C2,1 ([x', x"]). (c) If x" = x + h and x' = x - h the difference in (5) agrees with the usual second difference a-a+u(x). To set up the difference equation for -Llu = fat (x,y) E ilh we use the four neighbouring points (x - Sth, y), (x + Srh, y), (x, y - suh), (x, y + soh) E n h U rh, 0

< s.

$ 1,

as defined above. The subscripts to the S are suggest respectively left, right, under and over. For points far from the boundary s. = 1 holds; for points near the boundary (x, y) at least one neighbour lies on r h and the corresponding distance s.h may be smaller than h. Equation (5) with x' = x - Sth and x" = x + srh provides an approximation for u..,..,. Analogously one can replace Uyy by a divided difference. One obtains the difference scheme of Shortley and Weller [1]:

-DhU(X,Y):=h-2{(~+_2_)u(x,y)(2 )u(x+srh,y) StSr SoSu Sr Sr + St (

2

St Sr + St

_ Su

)u(x-sth,y)-

(

2

So So + Su

)u(x,y+soh)

(2 ) u(x, y - su h )} . So + Su (4.8.7)

Remark 4.8.3. If St = Sr =

Su

= So = 1, Dh agrees with Ll h .

The discrete boundary value problem assumes the form -DhUh =!h:= Rhf on n h with (Rhf)(X) = f(x) for x E ilh , Uh =!P on rho

(4.8.8a) (4.8.8b)

The five coefficients on the right side of Equation (7) define the matrix elements Lxe for = x and for the four neighbours of x. Otherwise we set Lxe = o. The right side of the system of equations

e

e

(4.8.9a)

in which Uh is interpreted as a grid function on nh (not nh Un) is given by %

= !h + !Ph,

!Ph(X):= -

L

Lx~!p(e)·

~Er,.

Again CPh(X) = 0 holds for points x E ilh far from the boundary.

(4.8.9b)

4.8 Discretisation in an Arbitrary Domain

81

Theorem 4.8.4. Let n be bounded and contained in a strip (xo, xo+d) x IR or IR x (Yo, Yo +d) of width d. For the matrix L,. belonging to the Shortley- Weller discretisation the following holds: (a) L,. is genemlly not symmetric; (b) L,. is M -matrix with (4.8.10) PROOF. (a) Let x = (x,y) E n,. be near the boundary, but its neighbour x, = (x + h, y) E n,. be far from the boundary. Then it holds that LXXi = -2h- 2/( 1 + Bt) f. _h- 2 = LXiX' if Bt < 1. Other than in Exercise 4.7.1, in general no scaling can be found so that D,.Lh (Dh diagonal) becomes symmetric. (b) Lh need not necessarily be irreducible and hence irreducibly diagonally dominant. But the weaker condition in Exercise 4.3.8 is satisfied and proves the M -matrix property. (c) For the proof of (10) we use Theorem 4.3.16. If the domain nlies in the strip (xo, Xo + d) x IR we select Wh(X, y) := Rhw, W := (x - xo)(xo + d - x)/2. The remainder in Equation (5) contains only third derivatives that vanish for w. Thus D,.w,. agrees with Llw = -1:

The corresponding system of equations reads L,.w,. = q,. := J,. +
Exercise 4.8.5. Prove the analogue of the derivative (4.4.9):

With (10) stability has been proved. The order of consistency however is only 1, for at points near the boundary eh := DhR,.u - il,.Llu

is of the order of magnitUde O(hl). Here, Rh is the restriction to n,. u n. il,. has been defined in (Sa). We want to show that nevertheless there is convergence of order 2. The difference w,. := u,. - R,.u between the discrete solution u,. = Lh"1q,. and the solution u E C 3•1(a) of (1) satisfies -D,.w,. = - D,.u,. + D,.RhU = ilhf + DhRhU (4.8.11a) (4.8.11b)

82

4 Difference Methods for the Poisson Equation

so that Wh = L/:1ch follows. Ch can be written as c'K + cK, where c'K (resp. cK) is the error of discretisation of the x difference (resp. y difference). In turn, c'K . spl't' -z = C2:,1 + C2:,2 : IS 1 Into (.;h h h

C2:,2(X,y):= {c'K(x,y) h 0 Analogously one defines

if SI = ~r = otherwIse

1},

C",,1 ' - J: h . - Ch

_

c",,2

h'

cK,l and cK,2 and sets

+ Ch, _11,1

Ch1..= C"',1 h

The errors ~ are described by (5.7):

IIw~lIoo S IIL/:1I1oollc~lIoo, Ilc~lIoo S ~h2I1ullc,s.1(n).

(4.8.12a)

With K := ih3I1ullc2.1(n) define

Uh = Kl in ilh,

Uh

=0

For x E ilh far from the boundary Ch(X) however, we have

Ch(X)

=K

L

L~

~En"

= -K L

on rh,

= 0; for x

L~

Ch:= LhUh. E ilh near the boundary,

(x E ilh near boundary).

~Er"

Consider, for example, the case x = (x,y) E ilh near the boundary with (x - SI h, y) En. According to Exercise (2a, b) the x difference has the error

e=

Ic2:,1{x)1 < h

-

2+ 2 h 2+ 2 Sr St lIull 21 - = h- 2Sr St K < K 3 Sr + St C • (0) 8 r + St -

2h-2 St{Sr

+ St)

= -KL

~

.

The analogous estimate for cK,1{x) gives Ic~(x)l S Ch{X). Because c~(x) = Ch{X) = 0 at the x E ilh far from the boundary one has -Ch S c~ S Ch. Since L/:1 ~ 0 it follows that -Uh S wl S Uh, i.e., (4.8.12b) Equations (12a,b) together with wl

+ w~ = Wh = Uh -

Rhu prove

n

Theorem 4.8.6. Let satisfy the assumption of Theorem 4. The conue1Yence for the Shonley- Weller method is second order: IIUh - RhUlioo S

S if u E C 3,1(i'i).

~h3I1ullc2.1(n) + IIL/:1I1oo~h2I1ullcs.l(n) ( 1 h3

'3

d 2 2) + 48 h lI u ll c3.1(i'i)'

(4.8.13)

4.8 Discretisation in an Arbitrary Domain

83

Exercise 4.8.1. 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 of third order: IIUh - RhUlloo = O(h3 ).

4.8.2 Interpolation at Points near the Boundary

Instead of discretising the Poisson equation at grid points x E ilh near the boundary, one could also try to determine Uh(X) by interpolation from the neighbouring points. If, for example, x = (x,y) E ilh is near the boundary, (x - slh, y) E n and (x + srh, y) E ilh Un, then linear interpolation yields Uh(X, y)

=

[SlUh(X + srh, y)

+ SrUh(X -

slh, y)JI(sr

+ s!).

Thus, at the point x we set up the equation (4.8.14a)

e

= (x - slh, y) should be a boundary point, one can replace Uh(e) by cp(e). If, however, (x, y + soh) or (x, y - suh) is a boundary point, we choose interpolation in the y direction:

Since

(4.8. 14b) At any x E

nh

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

-(..1 hUh)(X) = fh(x) = Rhf(x) = f(x)

(x far from boundary). (4.8.14c)

Theorem 4.8.8. Let il satisfy the assumptions of Theorem 4. Let the discretisation be given by (14a-c) with the choice between (14a) and (14b) being made in such a way that always (at least) one boundary point is used for the interpolation. Let the system of equations resulting after the elimination of the be Lhuh = %. Lh is an (in geneml unboundary values Uh(e) = cp(e), E symmetric) M -matrix which satisfies the estimate (10). The discrete solutions Uh converge with the orner 2, if U E C 3,l(n):

e n

(4.8.15)

PROOF. The M-matrix property and (10) are proved as in Theorem 4.2. Let x E ilh be a point near the boundary at which (14a) is used. The error of interpolation is ck(x,y):= (Sl

+ Sr)RhU(X,y) -

slRhU(X + srh,y) - SrRhU(X - Slh,y),

1 I ~ '2SrSl(Sl 1 ICh(x,y)

+ sr)h2 IIUIICl,l(ii)'

84

4 Difference Methods for the Poisson Equation

With this c~ and K := h2I1ulb.l(ii)' one can essentially just repeat the proof of Theorem 6. • By rescaling and adding the equations (14a,b), one obtains

su)

h-2 {(s/ + Sr + So + Uh(X, y) _ ..!.Uh(X _ s,h, y) S'Sr SOSU s, 1 1 - -Uh(X + srh, y) - -Uh(X, y + soh) Sr So

--.!...Uh(X,y - Suh)} = O.

(4.8.16)

Su

Using this device one can obtain a symmetric matrix Lh, even at arbitrary n. Exercise 4.8.9. Show that the discretisation (14c), (16) leads to a symmetric M-matrix. The estimate of convergence reads

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

U(X, y) = u(x - s,h, y) = rp(x - s,h, y),

if (x, y) E n h, (x - s,h, y) E

n,

is less desirable since it only provides first-order convergence: IIUh -Rhulloo = O(h). By contrast, interpolation of higher order is very applicable indeed (cf. Pereyra-Proskurowski-Widlund [1]). However, it is described by an equation which also contains points at an distance of ~ 2h. Higher boundary approximations are in particular then necessary if one wants to apply extrapolation methods (cf. Marchuk-Shaidurov [1, p. 162 ff.]).

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: (5.1.1a) Lu=f in

n

with (5.1.1b) We mentioned that, without loss of generality, one can assume tlij(X)

= aji(x)

(5.1.1c)

so that the matrix (5.1.1d)

A(x) := (aij(x))i,j=l .... ,n

is symmetric. A differential operator apparently more general than (1 b) is

~[I

a2

i,j=l

'3

L = LJ aij ax.ax'

III]

ana a2 ~[Ia a II] + ax' aij ax' + ax.ax. aij +LJ ai ax' + ax' ai +a. 3

,

i=l

'3

'

•

(5.1.2) But since, for example, (aU u) = aUu"',"'j + (aau/aXj)u"", the operator (2) can be described in the form (lb), provided the coefficients are sufficiently often differentiable. According to Definition 1.2.3, Equation (1) is elliptic in n if all eigenvalues of A(x) have the same sign. One can assume without loss of generality that all eigenvalues are positive so that A(x) is positive definite (cf. Exercise 4.3.22a). Thus, L is elliptic in n if

a!j

a!,

n

L: tlij(X)ei~j > 0

for all x E

n,

0

#- e E m,n.

(5.1.3a)

i,j=l

W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_5, © Springer-Verlag Berlin Heidelberg 2010

85

86

5 General BOWldary Value Problems

For any x E n there exists c(x) := min{aij(x)eiej : lei = I} and it must be positive (c(x) is the smallest eigenvalue of A(x)!). Hence one can also write (3a) in the form (3a'): n

L: aij(X)eiej ~ c(x)leI

c(x) > 0 for all x E

2,

n,

e E JRn .

(5.1.3a')

i,j=l

Definition 5.1.1. The equation (la), or the operator L, is defined to be uniformly elliptic in if

n

inf{c(x):x En} On

r

=

> 0 (c(x) from (3a')).

(5.1.3b)

an we impose the following Dirichlet boundary value condition: 11.

=
on

r.

(5.1.4)

5.1.2 Maximum Principle In general, the maximum principle does not hold for the equation Lu = f, nor is the solution of the boundary value problem (la), (4) uniquely determined. Example 5.1.2. Let n = (0,11") x (0,11"),

L atj(x*)ux•x; (x*) + a(x*)u(x*), i,j

(5.1.5a)

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations

87

and the Hesse matrix B := (u xiXj (X)*))i,j=l, ... ,n must be positive semidefinite (here, a matrix B is said to be positive semidefinite if it is synunetric and satisfies 2:: 0 for all E IRn). From Exercise 4 follows

<e, Be)

e

n

tr(A(x*)B)

=

L aij(x*)UxiXj(x*) 2:: o.

(5.1.5b)

i,j=l

If the minimum u(x*) were negative, one would obtain from (5a,b) the result I(x*) 2:: a(x*)u(x*) > 0 in contradiction to the assumption I::; o. Part (b) is proved analogously. • Exercise 5.1.4. The trace is defined by tr(A) := I:~1 au. Prove that: (a) ~i 2:: 0 and tr(A) 2:: 0 if A is positive semidefinite; (b) tr(AB) = tr(BA) = I:~j=l ~jbji; (c) tr(AB) 2:: 0, if A and B are positive semidefinite. Hint for (c): Bl/2AB1/2 is positive semidefinite; Exercise 4.3.22d . .AI:, in the case of the potential equation, from Theorem 3 follows the

n

Corollary 5.1.5. Let be bounded (not necessarily a domain); furthermore, assume that the conditions of Theorem 3 hold. If I::; 0 in n [I 2:: 0 in n] and il there exists a negative minimum [positive maximum] of u in it must lie on the boundary an.

n

Remark 5.1.6. The continuity of the coefficients ~j,~, and a of L in Theorem 3 and Corollary 5 can be replaced by the assumption a < 0 in n or by the stipulation: In every compact set Ken let ~j, ~ and a be bounded and let L be uniformly elliptic.

5.1.3 Uniqueness of the Solution and Continuous Dependence

Lemma 5.1.7. Let n be bounded, let the coefficients of L be continuous and let (3a) be valid and a ::; 0 in n. Let Ul, U2 E e 2(n) n eO(n) be solutions of the boundary value problems LUi

If It 2::

= Ii

h in nand
in

n,

::; IP2 on

Ui

r,

=
(5.1.6)

then also Ul ::; U2 in D.

n

PROOF. v := U2 - Ul satisfies the equations Lv = h - It ::; 0 in and v =
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 in il and u = cP on r is uniquely determined. PROOF. Let Ut, U2 be two solutions. Lemma 7 with 11 = cP shows that Ul $ U2 as well as U2 $ Ul. Thus, Ul

on

f

= U2.

12 = f,

CPl =

CP2

= •

The next theorem states that the solution depends Lipschitz-continuously and cpo

Theorem 5.1.9. Let L be uniformly elliptic in il. Under the conditions of Lemma 7 the following holds: (5.1.7)

for solutions Ul and U2 of (6). In this inequality the number M depends only on K := sup{lllij(x)l, lai(x)l, la(x)1 :x E il}, on the ellipticity constant defined by m := inf{c(x):x E il} > 0 (cf. (3b)) and on the diameter of the domain il.

PROOF. Let il C KR(Z). For all x E il we have Zl - R $ Xl $ Zl select a

~

+ R.

We

0 so that

ma2

-

K(a + 1) ~ 1

and we define

We compare w with the solution v := Ul-U2 of Lv = 11 on From the choice of KR(Z) we have

r.

w(x) ~ v(x)

for x E

12 in il, v =

CPl-CP2

r.

Furthermore, the selection made of a ensures: (Lw)(x)

=

,a(x)llcpl...-

CP21100, +

$0 {ae2Ra - ~a(Xl~zl+R)lal1(x) a 2+al(x) a+a(x)lHII1 - 121100 '-v-'

'-v-'

~

~

1 ~ m ~ -K ~-K $ K(a + 1)}1I11 - 121100 $ -1111 - 121100 $ l1(x) - h(x) = (Lv)(x). $ 0

~

-{ma2 -

If one uses Lemma 7 with Ul

= v, U2 = w,

the result is v $ w in

n, i.e.,

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations

89

where M := e2Ra • Analogously one proves -w $ v so that (7) follows.

•

Exercise 5.1.10. (a) Let {l be bounded; let the coefficients of L be continuous in {l. Now show that L is uniformly elliptic in (l if and only if L is elliptic inn. (b) Theorem 9 holds for the special case !1 = h without the assumption of uniform ellipticity. (c) Let a strip in rn.n be described by

s = {x E rn.n:O $

("I, x - x*) $ S},

where x* E rn.n is a boundary point of s, "I E rn.n with 1"11 = 1 is a unit vector, and S is the strip width. Show that inequality (7) holds with M := e6a , if n c S and a are as in the proof of Theorem 9. (d) Under the conditions of Theorem 9 show that (5.1.8)

n

Exercise 5.1.11. Let the coordinate transformation P: x E 1--+ ~ E {l' and the inverse p-1: {l' -+ be continuously differentiable. To the operator L (in the x coordinates) let L' correspond in the ~ coordinates. Show that if L satisfies the conditions of Lemma 7, or of Theorem 9, then so does L'.

n

The right side f and the boundary values cp are not the only parameters on which the solution u depends. We next investigate how the solution depends on the coefficients Uij, Ui, and a of the differential operator L. Theorem 5.1.12. Let the coefficients of LI and LII be a!j, aL ai, resp. aU, a~I, all. Let u I and u ll be solutions of

LIu I = LII,uIl = f

in

(l,

u l = u ll = cp on

r.

Let LI satisfy the conditions of Theorem 9 and let u II belong to C 2 (n). With M from (7) we then have

lIu I - uIIlloo $

(.t lIa!j - aU 1100) lIuIl II + (~lIa~ - a: Ill oo) lI lb(i)

M{

C 2(ii)

',3=1

uIl

+ lIaI -

all 1100

II ItJ a~. '3 = aH '3' the condition u

then just

u ll E

(5.1.9)

lIuII 1100 }.

E C1(n) CO(n) n C2({l) will do.

n C2({l)

is sulhcient· =. a~I : U ' , zf also a~,

90

5 General Boundary Value Problems

PROOF. Set f' := L1u II . Then 11/'-11100 = II(LI-LII)uIIlloo can be bounded by the right side of (9) without the factor M. Theorem 9 applied to L1u II = I' and LIul = I implies lIul - ulIlioo =::; Mill' - 11100' •

5.1.4 Difference Methods for the General Differential Equation of Second Order For notational reasons we limit ourselves to the two-dimensional case n = 2. E m.2 require special discretisations at the boundary as General domains explained in Section 4.8. Here we only want to discuss the difference formulae at points far from the boundary. Therefore it will suffice to base our comments on the unit square n= (0,1) x (0,1).

n

Let L be given by (lb). If one wishes to obtain a difference method of consistency order 2, the following choice suggests itself, which will later be modified for reasons of stability.

au (x, y )8: 8; + 2a12(X, y)~~ + a22(x, y)8t 8;+ al(x,y)~ + ~(x,y)8~ + a(x,y) =h- 2 [

+

-a12 (X,y)/2 all(x,y) a12(x,y)/2

(5.1.10)

a22(x,y) -2[au(x,y)+a22(x,y)] a22(x,y)

a12 (X,y)/2] au(x,y) -a12(x,y)/2

~h-l [-al~X'Y) ~(~,y) al(~'Y)] + [~ o

-a2(x,y)

0

0

a(:y) 0

~].

0

Remark 5.1.13. The difference method (10) is a nin&-point scheme of consistencyorder 2. Let Lh be the matrix of the system of difference equations in ated to (10) (cf. Remark 4.2.5):

nh

associ(5.1.11)

The solvability of Equation (11), for arbitrary qh, is equivalent to uniqueness. In the continuous case, uniqueness was essentially given by the condition a=::;O and the ellipticity condition alla22 > a~2 (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 01 L let (3a,b) hold, and assume that (5.1.12)

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations

91

and a(x, y) ~ 0 in n. Then there exists for L a seven-point difference method of consistency order 1 such that the associated matrix Lh is an M -matrix except for the sign. In particular, the resulting Equation (11) is solvable.

= a22 > o.

Note that Condition (12) follows from (3a) only if all

PROOF. At the grid point (x, y) E nh (cf. (4.2.1a)) we abbreviate the coefficients ilij(X,y), ili(X,y), and a(x,y) to O-ij, ili, and a. The principal part ,,£aijE)2/8xi8Xj (Xl = X,X2 = y) is discretised by the following difference stars:

~ h- 2 8xr 82

8Xl8x2

[

0 1 -2 0

!h-

2

~h-2

1] ,

[ -:

[-1 ~

-1 2 -1 1 -2 1

~

[

-1].

if al2 ;::: 0,

8xfh-2

q,

0

1 -2 1

0]. (5.1.13)

if al2 < o.

-1

For al2 ;::: 0, we thus obtain

a22 - a12 2(al2 - all - a22) a22 - a12 If one introduces at2 := max{al2'0} and a12 := min{al2'0} then both for al2 ;::: 0 and for a12 < 0 we get the seven-point star

(5.1.13')

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 ili8/8x. are replaced by the forward and backward differences, respectively, ~8;' if ai ~ 0:

tili .=1

8~i + a: h- l [-a1 -,ali~ la21 -a 2

at'

at]

+ [0

~0 a

0],

(5.1.14)

is defined in the same way as ar2. Because ~ 0, the diagonal where element is negative, the others nonnegative. If one adds these terms to (13'),

92

5 General Boundary Value Problems

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 on&-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, X2 ± p.h) for fixed V,p. E 71.. (but in general Ivl,Ip.1 > 1) (cf. Brambl&-Hubbard [1]). Layton-Morley [1) point 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 CLi{}/ {}x;. as in (10). The following corollary shows that -Lh is also an M-matrix when hCLi is sufficiently small. Corollary 5.1.16.

In addition to the assumptions of Theorem 14 let the

following hold:

(5.1.15) Then the discretisation of'ECLi{}/{}x, from (10) together with (13') leads to a seven-point difference method of second order of consistency Buch that -Lh is an M -matrix.

PROOF. -Lh satisfies (4.3.1a) and is irreducibly diagonally-dominant.

•

Exercise 5.1.17. The condition CLi;. ~ la121 + hiCLil/2 instead of (15) is not sufficient. Construct a counterexample with au = ~2 = 1, a12 = 0, h = 1/3, a;, (x, y) variable, and lasl = 6 so that Lh is singular. Considerably weaker conditions for the nonsingularity of Lh than in Theorem 14 and Corollary 16 are needed in Section 9.2 (cf. Exercise 9.2.6, Corollary 11.3.5). In general, Lh is not a symmetric matrix. Symmetry of Lh is to be expected only if L is also symmetric: L = L'. Here the formally adjoint differential operator L' which is associated to L in (2), is defined by L

,=

~

[III {}2

L....t a;'j {} .{} . ',j x, x J

{} II {}

{}2 I] - L....t ~ [ II {} {} I] a, ~ + ~ai +a.

+ ~a,ja-: + {}x,.{}X J.aij x, xJ

,=1

X.

x,

(5.1.16) It is easy to see that a symmetric operator can always be written in the form n

L=

{}

{}

L ~a;'j{x)a-:x + a(x), i,j=l x,

asj =

aji·

(5.1.17)

J

A difference method for this is given for the case fiv&-point star

n

= 2 and a12 = 0 by the

5.1 Dirichlet Boundary Value Problems for Linear Differential Equations

+ [ 00

0 0] a(x,y) 0 . 000

93

(5.1.18)

Theorem 5.1.1B. Let a1l, a22 E 0 2.1(0). The difference method (18) is consistent of order 2. The associated matrix L", is symmetric. If aii > 0 (ellipticity) and a:::; 0, then -L", is a positive definite M-matrix. PROOF. (a) For the proof of consistency, expand

v(x + h/2) := a1l(x + h/2,y)[u(x + h,y) - u(x,y)]

(y fixed)

around x + h/2, and then expand v(x + h/2) - v(x - h/2) around x. (b) The symmetry results from L:c~ = L~:c, for E il", (cf. Exercise 4.2.2c). (c) L", is irreducibly diagonally dominant so that Criteria 4.3.10 and 4.3.24 are applicable. •

x,e

The mixed term ~a12(x'Y):1I 1 [-a12 (XA) - a12(xB) 2h-2 2a12(XA)

o

+ /ya12(x,y)/x

a12(xc) + a12(xB) -2[a12(xA) + a12(xc)] adxA) + a12(xD)

(x - ~,y), XB = (x - ~,y (x+ ~,y - h). Another way of writing (19) is:

with

XA

=

+ h),

Xc

=

can be discretised by 0 ] 2a12(XC) -a12(xc) - a12(xD) (5.1.19) (x + ~,y), and XD =

4[a:b(. -~, .)a; + atb(. -~, .)a;; + a;;b(. +~, .)at + a;b(. +~, .)a:] , (5.1.19') where b := a12. Exercise 5.1.19. (a) For the coefficients of Lin (17) let the following hold: au > 0, a22 > 0, a12 = a2l :::; 0, au + a12 > 0, and also 2a22 (x, y + h/2) + a12(x - h/2, y + h) + a12(x + h/2, y) > 0, and a :::; O. Show that the difference scheme which is described by (18) and (19) has consistency order 2 and that the associated matrix - L", is a symmetric, irreducibly diagonally-dominant and positive definite M-matrix. (b) What is the suitable discretisation for the case a12 ~ 0 ? Exercise 5.1.20. The difference formula from Theorem 14 for the operator L reads au at a; + a12 (at a: + a; a;;) + a22a: a;; + a1 at + a2a: + a when

94

5 General Boundary Value Problems

a12 ~ 0, a1 ~ 0, a2 ~ o. Let the associated matrix be L h . Then prove that the transposed matrix LX describes a difference method for the adjoint operator L' and also possesses consistency order 1.

In general it is possible to show for regular difference methods that L~ is a discretisation of L'. The role of regularity is demonstrated in Example 5.1.21. Let Lu := u" + au' in n = (-1,1) with a(x) ~ 0 for x ~ 0 and a(x) ~ 0 for x > o. According to (14) au' is discretised for x ~ 0 by a(x)a+u(x) and for x> 0 by a(x)a-u(x). Let the associated matrix be Lh = L h,2 + L h,1t where L h,2 and L h,1 correspond to the terms u" and au' respectively. According to the above, 1Vh should be a discretisation of -(av)'. But the differences Lr,1Vh at x = 0 ~nd x = h are

Lr

h-1[a( -h)Vh(-h) - a(O)vh(O) - a(h)vh(h)] = -(av)' - h- 1a(0)vh(0) + O(h), h-l[a(O)vh(O) + a(h)vh(h) - a(2h)vh(2h)]

= -(av)' + h-1a(0)vh(0) + O(h);

LI

thus they are not consistent. Nevertheless, is a possible discretisation of L'v = v" - (av)', for it can be shown that the error Vh - RhV has order of magnitude O(h). To prove stability one has to show IIL;;11100 ~ const. Obviously it is sufficient to prove this inequality for sufficiently small h. In the proof of Theorem 9 we used the fact that -Lw

~

1 in

n,

W ~

0 on

r

for w(x) := exp(2Ra) - exp(a(xl - %1 + R)). Let DhUh(X) be the difference equations from which LhUh results after elimination of the boundary values. We set Wh := 2RhW; i. e., Wh(X) = 2w(x) for x E li h . The following holds: -DhWh

= -2(DhRh -

RhL)w - 2RhLw ~ 2 - 2(DhRh - RhL)W.

Each consistent difference method satisfies II (DhRh -RhL)wlloo ~ O. For sufficiently small ho we thus have

This inequality agrees for x E nh far from the boundary with -LhWh(X) ?: 1. For x E ilh close to the boundary, -DhWh(X) also contains the sum - E~Erh Lx~Wh(e), which is not contained in -LhWh(X). For the discretisations from Theorem 14, Corollary 16, Theorem 18, and Exercise 19, however, L~ ?: 0 (x E ilh, E n) holds, so that because Wh ?: 0 on n,

e

-LhWh ?: -DhWh ?:

n

(h ~

holds for all grid points. Theorem 4.3.16 yields

ho)

5.2 General Boundary Conditions

95

Theorem 5.1.22. The discretisations from Theorem 14, Corollary 16, Theorem 18, and Exercise 19 are stable under the conditions posed there, i.e., IILh"1I1oo :$ const for all h E H = {l/n: n E IN}. According to Theorem 4.5.3 the methods converge. The order of convergence agrees with the corresponding order of consistency.

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

e

Lxg(€,x)

Leg(e,x)

= 0,

= 0

for x,€ E il, x =I-

ej

g(€,x)=O forxEror€Er. Here, L' is the adjoint differential operator (16). If L =I- L', then 9 is no longer symmetric: g(x,€) =I- g(€, x). Under suitable conditions the solution of (la-c), (4) can be represented as

u(x) =

-In

g(e, x)f(€) de

-£
x) dr~

at

where B = B~ = 2:~j=1 nj Uij is a boundary differential operator (nj are the components of the normal vector n = n(e), € E r). Only when the principal part of L agrees with .d is B the normal derivative. In the discrete case the inverse Lh"1 again corresponds to the Green function g(., .).

5.2 General Boundary Conditions 5.2.1 Formulating the Boundary Value Problem Let the differential equation be given by (l.la,b). The Dirichlet boundary condition (1.4) can be written in the form

Bu

=
on

r

(5.2.1a)

where B is the identity (to be precise: the trace on r). In more general settings B can be an operator - a so-called boundary differential operator of order 1: n B =

L b (x)8/8x + bo(x), i

i

x

E

r.

(5.2.1b)

i=1

If one introduces the vector b(x) the form

= (b 1 (x),··· ,bn(x»T,

Bu can be written in

96

5 General Boundary Value Problems

Bu = (b(x), Vu(x)} + bo(x)u(x) or B = bTV + boo

(5.2.1b')

Example 5.2.1. (a) From b = 0, bo(x) =I- 0 there results what is known as the Dirichlet condition u =

')'3:u

=

';(1- x)/2

il:Llu =0 (a)

(b)

Figure 15.2.1. (a) Boundary value problem with changing boundary condition type (b) Dirichlet problem in a disk with a cut

The normal derivative B = a/an (Le. b = n) is important in connection with L = -Ll. For the general operator L in (1.1b) the so-called conormal derivative B with

b

= An (A = (a.jkj=l, ... ,n, a.j as in (Ub»

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 = -Ll, B = I (Dirichlet condition) uniqueness is guaranteed (cf. Theorem 3.1.2), while the problem associated to L = -Ll, B = nTV = a/an 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, Le., b(x) = must not occur for any x E r. But it is possible that =1= 0) in 'Y C r and b(x) =1= 0 in r\'Y. Then there is a Dirichlet condition u =

o and bo(x) = 0, b(x) = 0 (and bo

Example 5.2.3. Let n be the upper semicircle around x = y = 0 with radius 1. Let the differential equation and boundary conditions be given as in Figure la. The boundary condition changes its order at x = y = O. The solution in polar coordinates reads: u = r l / 2 sin(

The same singularity as in Example 3 occurs in the problem described in Figure lb; the solution is also r1/2 sin(

Figure 5.2.2. A domain symmetric with respect to 'Y

Example 5.2.4. Let n = nl U !"h U 'Y be as in Figure 2: let the reflection of n1 in 'Y result in n2. If one seeks a solution of Lu = f in n, Bu =

While the boundary condition Bu =

98

5 General Boundary Value Problems

Then, in addition to BtL boundary condition

=

r\bl u 'Y2), we can require the periodic

(5.2.2) on 'Yl and 'Y2. The solution is periodically continuable in the x-direction (with period X2 - xt}. The origin of periodic boundary conditions is discussed in Example 5. Example 5.2.5. (a) Let il' be an annulus which is described by the polar coordinates r E (rb r2),

5.2.2 Difference Methods for General Boundary Conditions The boundary conditions posing the least difficulties are the periodic boundary conditions. Define ilh as the set of grid points in il and on 'Yl (but not on 'Y2; cf. Figure 3). 'Yl

•

o

Figure 5.2.3. Discretisation for periodic boundary conditions

5.2 General Boundary Conditions

99

The difference equation at a grid point (Xl> y) E 1'1 has a left neighbour h,y) outside Replace Uh(X1 - h,y) in the difference equation by Uh(X2 - h, y). By doing so we have transferred the periodicity onto the difference solution without explicitly discretising Equation (2). Of course, the step size must be chosen so that X2 - Xl is a multiple of h. (Xl -

n.

Exercise 5.2.6. For periodic boundary conditions the structure of the matrix Lh changes. Let L = -.1 in the square n = (0,1) x (0,1). Assume Dirichlet 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 Lh for a lexicographical arrangement of the grid points. In Section 4.7 we described the discretisation of the boundary condition B = n TV' = a/an for the case that n is a rectangle and r coincides with the grid. In the same situation one can discretise the general boundary condition (1a,b) as follows. outer

r

... ......

inner ---, (x+h,y+h) L·lne ... ~.I< ......h ...-) b2 (x - x) - b1 (y - y) ... ,. \x+ ,y • (x + h, y) (.::) X,y I

=

°

---. I I

Figure 5.2.4. Discretisation of Bu =
Let (x, y) E n = nh n r be a grid point on r. The point of intersection (x + h, y) drawn in Figure 4 is given by y = y + h~(x, y)/bl (x, Y). Note that b1 1= 0, since b TV is not to be a tangential derivative. For u E Q2(D) the directional derivative bTV'u = (b, V'u) in (x,y) is approximated by

[u(x,y) -u(x+ h,Y)]/Vh2 + (y - ij)2 = (bTVu)(x,y) + O(h).

(5.2.3a)

On the other hand, one obtains u(x + h, y) under the same conditions by interpolation up to order O(h2 ):

100

5 General Boundary Value Problems

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

(BhUh)(X, y) :

{Uh(X, y) - y+~-tiuh(X + h, y) - ?Uh(X + h, y + h) }

.jh2 + (ii _ y)2

+ bo(x, Y)Uh(X, y)

(5.2.3c)

=
Lemma 5.2.7. Let Rh be the restriction to the grid points U E C 2(n) one has

nh . For a/unction

Bh(Uh - RhU)(X, y) =
0/ the boundary condition has consistency order 1.

Equation (3c) has the form

(BhUh) (x, y) = COUh(X, y) - CIUh(X + h, y) - C2Uh(X + h, y

+ h).

(5.2.3c')

We assume that the difference equations L(En" Lx(uh(~) = fh(x) (before the elimination of the values Uh(~), ~ E n) satisfy the inequalities Lx( $ 0

(x 1: ~),

Lxx

=-

LLx(

(x E nh,~ E

nh ).

(5.2.4a)

(;o!x

Here Lxx > 0 and Lx( $ 0 correspond to the sign condition (4.3.1a). The second inequality in (4a) agrees with (4.3.4b). The corresponding inequalities for (3c') read: CI ~ 0, C2 ~ 0, CO ~ CI + C2. (5.2.4b) Let the difference equations in x E ilh and the boundary equations (3c') given for (x, y) E rh be combined into AhUh = gh. (4a,b) implies ~ $ 0, axx ~ - L(;o!x~(X,~ E nh). Therefore, Ah is an M-matrix if Ah is irreducible and (4.3.4a) holds for at least one x E nh . The above considerations explain why the boundary discretisation should satisfy the conditions (4b). (4b) holds for the case bo = 0 if and only if b2 /bl E [0,1). If ~/bl E [-1,0], (4b) can be satisfied if one interpplates between y and y - h (instead of y and y + h). If, however, I~I > Ibll, i.e., if the tangential component is greater, one could interpolate between (x + h, y) and (x + h, y ± kh), where Ib 2/b l l $ k. For the case b2 /b 1 ~ 1, however, it is more practical to interpolate at the point (x, y + h) between (x, y + h) and (x + h, Y + 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 il = (0,1) x (0,1). If n is a general region, two discretisation techniques offer themselves (cr. Sections 4.8.1 and 4.8.2).

5.2 General Boundary Conditions

101

(x - 81h,y) = (x, 17). ...

• : Points in f1h .: Points in

n

Figure 5.2.5. First boundary discretisation

First option for discretisation: Let f1h and n be chosen as in Section 4.8.1. At near-boundary points the differential equation is approximated by a difference scheme which in the case of L = -..d corresponds to the Shortley-Weller method. For this one needs the values of Uh at the boundary points E As in Figure 5 let (x, y) = (x - 8th, y) E be a boundary point. Again we can set up an equation analogous to (3a-c). In Figure 5 (x, y) has the same position as (x + h, y) in Figure 4. In general one has to use the point of intersection of the line which is also presented by (x, y) +tb (t E IR), and the dotted straight line in Figure

e n.

n

5. Second option for discretisation: Let ilh be the grid f1h used above. Now let ilh consist of all points of ilh that are far from the boundary. For all (x, Y) E ilh difference equations (with equidistant step size) are declared which involve {Uh(e):e E ilh}. For each point (x, y) E n := ilh\ilh a boundary discretisation must be found (cf. Figure 6). Equation (3a) can be set up with (x, y) and (x + h, y) instead of (x, y) and (x + h, y). The arguments of the coefficients b1 and b2 are (x, y). This point results implicitly from ~(x, y)(x

- x)

= b1(x, y)(y - y),

(x, y) E r.

(5.2.5)

Remark 5.2.8. (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

r .. - - -. " ,

(X,Y~

,,

,

Line ' (Ax,yA) +tb

,,*,I (x + h, ii)

,.(X,y).I (x+h,y)

.--- .. I I

• Points in

nh

• Points in n

Figure 5.2.6. Second boundary discretisation

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 possible to simplify the boundary conditions by using coordinate transformations. Let Bu = cp be prescribed on 1 cr. If one finds a transformation (x, y) +-+ (e,1]) such that the equations = 0 and b2X( = blY( are satisfied on 1, one obtains for the transformed problem Bu == u8u/8n + 1101.1. = cp on a vertical boundary piece (8/8n = -alae). The discretisation may be carried out as described in Section 4.7. The same reasoning as in the preceding sections proves s tab iii t y and convergence:

e

Remark 5.2.9. Let the difference method DhUh = fh satisfy the conditions of Theorem 5.1.22 so that -DhWh ;::: 1 holds for Wh := K1 + Wh (K > o is constant; Wh is as in the proof of Theorem 5.1.22). Let the boundary discretisation (3c') satisfy CO - Cl - C2

= bo(x) ;::: € > 0 for all x E rho

For sufficiently large K, one then obtains (Bhwh)(x) ;::: 1 for x En. Hence it follows that AhWh ;::: 1, where Ah is the M-matrix defined immediately after Equation (4b). Since IIAh"ll1oo ~ const (cf. Theorem 4.3.16), stability has been proved. The order of convergence is the minimum of 1 (order of consistency of Bh ; cf. Lemma 7) and the order of consistency of DhUh = fh. In general it is not possible to approximate b TV by symmetric differences. To construct a discretisation of Bu = cp with consistency order 2 despite this fact, set: 3

Bhuh(x, Y) := COUh(X, y) -

L CiUh(Xi, yd = g(x, y), i=l

(x, y) E n,

5.3 Bowtdary Problems of Higher Order

103

where (Xi, y,) are three points in nh. If, in order to remove the first two tenns in the Taylor expansion, one uses the differential equation Lu = f in (x,17) and the tangential derivative of Bu =
5.3 Boundary Problems of Higher Order 5.3.1 The Biharmonic 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) L12u

= f in n

(5.3.1)

where Ll2 = a 4/ ax4 + 2a 4/ ax2ay2 + a 4/ ay4. 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

=
au/an

=
(5.3.2)

with
=
and

Llu

= I{J2 on r

(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 L1v = f in n, r and, subsequently, L1u = v in nand u =
v =
Remark 5.3.2. The solutions of L1 2 u = 0 do not satisfy a maximumminimum principle (counterexample: u = x2 + y2 in n = KR(O».

104

5 General Boundary Value Problems

5.3.2 General Linear Differential Equations of Order 2m The partial derivative Dcz (a E zn : multi-index) of order lal = al + ... + an is defined in (3.2.5). A differential operator of order 2m has the form

L

=

L

acz(x)DCZ

(x E

.a)

(5.3.4a)

Iczl$2m and defines the differential equation of order 2m:

Lv. = f

(5.3.4b)

in.o.

Ellipticity has been explained thus far only for equations of second order (cf. Definition 1.2.3). Definition 5.3.3. The differential operator L (with real-valued coefficients acz) is said to be elliptic (of order 2m) at x E .a if

L

(5.3.5a)

acz(x)eCZ '" 0 for all 0 '" e E IRn.

Iczl=2m

..

Here, ecz is an abbreviation for the polynomial ef 1 e~2 ··e~n of degree lal. We set P(x, e) := E 1czl=2m acz(x)ecz. Evidently (5a) is equivalent to P(x, e) '" o for all e E IRn with lei = 1. For reasons of continuity either P(x, e) > 0 or P(x, e) < 0 must hold. Without loss of generality, we may assume that p(x,e) > 0; otherwise we scale with the factor -1 (changing from Lv.(x) = f(x) to -Lv.(x) = - f(x)). Since the set{e E IRn : lei = I} is compact it follows that c(x) := min{P(x,e): lei = I} > 0 and this justifies the formulation of (5a) as

L

acz(x)eCZ ~ c(x)leI 2m

for all e E IRn with c(x)

> O.

(5.3.5b)

Iczl=2m Definition 5.3.4. The differential operator L is said to be uniformly elliptic in .a if inf{c(x):x E .a} > 0 for c(x) from (5b). Exercise 5.3.5. (a) Translate L from (1.1b) into the notation of (4a). What are the coefficients acz for L = Ll2? (b) Prove that the biharmonic operator Ll2 is uniformly elliptic. (c) Let acz be real-valued. Why are there no elliptic operators L of odd order? (d) If the coefficients acz are sufficiently smooth one can write L from (4a) in the form L = E1czlSm EI.8ISm( -1)lfJIDfJaczfJ(x)DCZ (cf. (1.1b) and (1.2)). For m = 1 (equation of order 2) we have used one boundary condition; for the biharmonic equation (m = 2) two boundary conditions occur. In general one needs m boundary conditions

5.3 Boundary Problems of Higher Order Bju:=

I:

bjaDau =
(j

= 1,2,· .. ,m) on r

105 (5.3.6)

lal:=;m;;

with boundary differential operators B j of order 0 ::; mj < 2m. Remark 5.3.6. The boundary operators B j cannot be chosen arbitrarily, but must be independent of each other (they must form a so-called "normal system"; cf. Lions-Magenes (1, p. 113] and Wloka (1, Definition 14.1]). In particular the orders mj must differ pairwise. We have a Dirichlet boundary condi tion if B j = 8 j - 1/8n j - 1 = = 1,···, m (cf. (2)). 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 = ..12 in rn?, for example, the singularity function is 8~ Ix - Yl2log Ix - yl (cf. Wloka (1, Exercise 21.9]). ({)/8n)j-1 for j

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

Dh = ..1~ =

= ..12 is the 13-point difference

2 [ h- 4 1 -8 2

. 1 -8 20 -8 1

2 -8 2

(5.3.7)

which can be represented as the square of the five-point star ..1h from (4.2.3). Let the grid ilh C il and the boundaries rh C r, "th C rn?\!1 be defined as in Figure 1. The difference equation (5.3.8a)

n

requires the values of Uh in h := ilh U rh U"th. These are determined by the boundary conditions (cf. (2)) for x E r, 8~Uh(X) := (Uh(X + hn) - Uh(X - hn)I/(2h) =
Uh(X) =
(5.3.8b) (5.3.8c)

106

5 General Boundary Value Problems 0

0

0

0

0

0

0

0

IJ

IJ

C

C

IJ

C

C

0

0

• •

0

IJ

0

c

c c c c

0

IJ

• • • • • • • • • • • • • • • • • • • •

0

0

c c

•

IJ

0

0

C

IJ

C

C

IJ

C

C

0

0

0

0

0

0

0

0

• •

0

0 0

Figure 5.S.1. Grid for the biharmonic equation in

a=

(0,1) x (0,1)

As in Equation (4.7.14b) we assign to the comer points two nonnal directions each. Correspondingly, to each comer point belong two equations (Be). Note that two (m = 2) boundary layers occur, r h and "rh, and that two boundary conditions are given.

Remark 5.3.7. If with the aid of (8b,c) one eliminates in Equation (8a) the U"rh}, one obtains a system of equations LhUh = % unknowns {Uh(e):e for {Uh(e): E nh}. Lh is not an M-matrix since the sign condition (4.3.1a) is violated. But L,/ also is in general not ~ 0 (cf. (4.3.1b)).

e

En

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.B. Let Uh be the discrete solution of the biharmonic equation in the square n = (0, 1) x (0, 1) defined by (8a-c). The matrix Lh described in Remark 7 is symmetric and positive definite. It satisfies

(5.3.9a) If IP1

=0

on n (cf. (8b)), then Uh can be estimated by

lI u hlln"

1

:5 25611fhlln"

h- 1/ 2

+ 16.j2IIP2Ir",

(5.3.9b)

where the norms are defined as follows:

Let the sum EXEr" in Il{J2lr" contain the corner points twice, where, for example, the value IP2(X) at x = (0,0) should be interpreted once as IP2(0, 0+), and once aSI{J2(O,O+) (cf. comment on (Be)).

5.3 BOlmdary Problems of Higher Order

107

PROOF. We limit ourselves to the most important part, the inequality (9b). In the infinite grid Qh:= {(x,y) E JR2:x/h,y/h E Z} we set Uh(X) := Uh(X)

for x E ilh,

Uh(X):= 0

for x E Qh \ilh.

After partial summation one obtains

L

uh(x)lh(x)

=

XEah

L

uh(x)Ll~Uh(X) =

XEQh

L

(LlhUh)(X)(LlhUh)(X)

XEQh

(First consider the identities Ev(x)a+a-u(x) = - E(a-v)(x)(a-u)(x) = E(a+a-v)(x)u(x) in the one-dimensional case.) From the definition of Uh it follows that LlUh = 0 on Qh \(ilh un). On one has Uh = Uh = 0, and

n

Uh(X - hn) = Uh(X - hn), Uh(X + hn) = 0, Uh(X + hn) = Uh(X - h(n) + 2h
for x E n. If one sets a';Uh(X) := [Uh(X) - Uh(X - hn)]/h = -Uh(X - hn)/h at x E rh , then

for x E

rh, implies that

(uh,LhUh)ah

= (uh,lh)ah = IILlhuhll~h +2h- 1 Ia;Uhl}h -2h- 1(a;Uh,
where h ·)ah and (., ·)rh are the scalar products belonging to the norms (9c). According to (5.3.10) one estimates (a;Uh,
+ 1
Thus one obtains

IILlhuhll~h

+ 2h- 1 Ia;Uhl}h

::; lIuhllahllfhllah

::; (Uh' fh)rh

+ 2h- 1(a;Uh, CP2)rh

(5.3.11)

+ 2h- 1 Ia;Uhl}h + h- 11
Theorem 4.4.12 shows that Iluhllah ::; IILlhuhllnh/16. Thus for
=

0

(5.3.12a)

= 0, (11) implies the inequality IILlhuhll~h ::; (2h)-1Icp21}h' therefore

lI uhllah

1 1 ::; 1611Llhuhllah ::; 16(2h)-1/21
(5.3.12b)

In the general case we write Uh as uk +ukI , where
108

5 General Boundary Value Problems

Theorem 5.3.9. (Convergence) Let the solution of the biharmonic boundary value problem (1), (2) in il = (0,1) x (0,1) satisfy u E c5,1(a). Then the discrete solution of equations (8a-c) is converyent (at least) of order

i:

Iluh - RhUlioh ::; Clh2I1ullc5.1(ii)

+ C2h3/2I1ullc2.1(ii)'

(5.3.13)

where Rh is the restriction to ilh U rho PROOF. The Taylor expansion shows that

h2 LlUhRhU = Llu + 41 (uz,n:z

1 IRI::; 360 Ilull c 5.1 (ii)in ilh. (5.3.14) Outside ilh one can specify RhU arbitrarily since these values do not appear in (13). In 'Yh we set

+ UgSlSlSl) + h4 R,

RhU(X + hn) := u(x - hn)

h3

+ 2h~2(X) + "'3Unnn -

h4 41 Unnnn

2h5 8

+ T!(8n)

5

u(x),

x En·

The choice is made so that (14) also holds for x E rho Applying Llh to LlhRhU yields 2 2 2 . LlhRhU = DhRhU = RhLl U + O(h lIull c 5.1(ii) m ilh

(Rh is the restriction to ilh). The difference Wh := Uh - RhU (defined in satisfies

nh)

DhWh = DhUh - DhRhU

= Rh(f Wh(X) ~Wh(X)

.&u) + O( h2 I1 u ll c 5.1(ii» =: 9h,

(5.3.15a)

= Uh(X) -

(RhU)(X) = ~l(X) - u(x) = 0 for x E rh,

(5.3.15b)

= 8~Uh(X) - 8~RhU(X) = ~2(X) - [~2(X) + O(h21Iullc2.1(ii»] =: "ph in x En.

(5.3.15c)

If one sets 9h = O(h2I1ullc5.1(ii» and"ph = O(h2I1ullc2.1(ii» instead of fh and ~2 in Theorem 8 (note that ~1 = Or) then (9b) implies assertion (13). • By using more complicated methods one can replace the factor h- 1/ 2 in (13) by C(f)h- E (f > 0 arbitrary) so that the order of convergence is O(h2-E), i.e., almost 2.

Remark 5.3.10. Inequality (9a) shows cond2(Lh) := IILh1l211Lh"1112 = O(h-4). In general a difference method for a differential equation of order 2m leads to the condition (5.3.16)

5.3 Boundary Problems of Higher Order

109

This indicates greater sensitivity to roundoff errors at higher orders 2m (cf. §4.4 in Stoer[l] and Bulirsch-Stoer[l]). Difference methods for boundary value problems of fourth order with variable coefficients and general domains n are discussed in an paper by Zlamal [1]. There too convergence of order O(h3/2) is shown. By contrast, O(h2)convergence was proved by Bramble [1] for the 13-point star (7) in a general domain n 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 Hilbert Spaces 6.1.1 Normed Spaces Let X be a linear space (alternative term: vector space) over K where K = R or K = C. In the following the normal case K = R is always intended. K = C occurs only in connection with Fourier transforms. The notion of a norm II·II:X -+ [0,00) is explained in Definition 4.3.12. The linear space X equipped with a norm is called a normed space and is denoted by the pair (X, 11·11). Whenever it is clear which norm belongs to X, this norm is called 1I·lIx and one writes X instead of (X, 1I·lIx). Example 6.1.1. (a) The Euclidean norm 1·1 from (4.3.13) and the maximum norm (4.3.3) are norms on Rn. (b) The continuous functions form the (infinite-dimensional) space CO(n). If n is bounded, all u e CO(n) are bounded so that the supremum norm (2.4.1) is defined and satisfies the norm axioms. If is unbounded, the bounded, continuous functions form a proper subset (CO(n), 1I'lI c o(il» of CO(n). Instead of II 'lIco(o)' we use the traditional notation 11·1100' (c) The HOlder-continuous functions introduced in Definition 3.2.8 form the normed space (CB(n), 1I·lIco co».

n

The norm defines a topology on X: A C X is open if for all x e A there exists an € > 0 so that the "ball" Ke(x) = {y e X: IIx - yll < €} is contained in A. We write Xn -+

x

or x

= n-+oo lim X n ,

if

IIxn - xII -+ O.

Example 6.1.2. Let In. I e CO(n). The limit process In -+ I (with respect to 1I·1l00) denotes the uniform convergence known from analysis.

W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_6, © Springer-Verlag Berlin Heidelberg 2010

110

6.1 Banach Spaces and Hilbert Spaces

Exercise 6.1.3. The nonn 11·11: X reversed triangle inequality holds

-+

111

[0,00) is continuous; in particular the

IlIxll -IIYIII :s; IIx - yll

(6.1.1)

for x, y E X.

As Example 1a shows, several norms can be defined on X. Two norms are said to be equivalent if there exists a number

II ·11 and III . III on X o < C < 00 such that 1

C

IIxll :s; IIlxlll :s; Cllxll

(6.1.2)

for all x E X.

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

6.1.2 Operators Let X and Y be normed spaces with the norms II . IIx resp. II . lIy. A linear mapping T: X -+ Y is called an operator. If the operator norm IITlly<-x := sup{IITxlly/llxllx: O f x E X}

(6.1.3)

is finite, T is said to be bounded (cf. (4.3.9)). Exercise 6.1.5. (a) T is bounded if and only if it is continuous. (b) 11·lly.....x is a nonn. With the addition (Tl + T2)X = T1X + T2X and the multiplication by scalars, the bounded operators fonn a linear space which is denoted by L(X, Y). (L(X, Y), II . lIy<-x) is a nonned space. With the additional multiplication (T1T2)x = T1(T2X), L(X,X) even forms an algebra with unit I, since IIIlIx .....x = 1. I always denotes the identity: Ix = x. Exercise 6.1.6. Prove that IITxlly

:s; IITlly<-xllxllx for all x

E X, T E L(X, Y).

(6.1.4a)

If Tl E L(Y, Z) and T2 E L(X, Y), then T1T2 E L(X, Z) and (6.1.4b)

We write Tn -+ T for T, Tn E L(X, Y) if liT - Tnlly .....x the operator nonn).

-+

0 (convergence in

112

6 Tools from Functional Analysis

6.1.3 Banach Spaces A sequence {xn E X: n ~ I} is said to be Cauchy convergent, or is called a Cauchy sequence, if sup{llxn - Xm IIx: n, m ~ k} -+ 0 for k -+ 00. A space X is said to be complete if any Cauchy-convergent sequence converges to an x EX. A Banach space is a complete normed space.

Example 6.1.7. (a) m.n is complete, but 0 as well as (Ck.l(D), 1I·lb.,lCD» are complete, thus Banach spaces. PROOF of (b). If {Un} is Cauchy convergent then there exists a limit u* (x) := limn--+ooUn(x) for all xED. Since Un converges uniformly to u·, u· must be continuous, i.e., u* E CO(D). •

Denote by Loo(D) the set of functions that are bounded and locally integrable on D. Here we do not distinguish between functions which agree almost everywhere. In this case the supremum norm is defined by IIuliLOOCD) := inf{sup{lu(X) I: X E D\A}:A set of measure zero}}

(Loo(D), II 'IILooCD» 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 subspace. The quotient space XI Z has as elements the classes x = {x + z: z E Z}. Show that X/Z with the norm IIxil = inf{IIx + zllx: z E Z} is a Banach space.

A set A is said to be dense in (X, II· IIx) if A c X and A = X, i.e., every x E X is the limit of a sequence au E A. If (X, II· IIx) is a normed, but not complete, space, one calls (X, 1I·lIx) the completion of X, if X is dense in X and IIxllx = IIxlix for all x EX. The completion is uniquely determined up to isomorphism, and can be constructed in the same way as one constructs m. as the completion of
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 opemtor To E L(Xo, Y) defined on the subspace Xo with IITolly<-xo = sup{IIToxllylllxllx:O =1= x E Xo} has a unique continuation T E L(X, Y), i.e., Tx = Tox for all x E Xo. (b) For Xn -+ x (xn E X o, x E X) holds Tx = lim,.-+oo Toxn. (c) IITlly+-x

=

IIToIIT+-xo·

PROOF. For x E Xo, T is defined by Tx = Tox, while for x E X\Xo there exists a sequence Xn -+ x, Xn E Xo and Tx = limn -+oo TXn = lim,.-+oo Toxn. It remains to show that lim Toxn exists and is independent of the choice of the sequence Xn E Xo. Since IIToxn - TOxmlly ~ IITolly+-xllxn - xmllx, the sequence Toxn is Cauchy convergent. Because of the completeness of Y there exists y E Y with Toxn -+ y. Similarly, for a second sequence x~ E Xo with x~ -+ x, there exists for the same reason a y' E Y such that Tox~ -+ y'. Since y' - y = lim(Tox~ - Toxn) and IITox~ - TOxnlly ~ IITo(x~ - xn)lIy ~ IITolly<-xollx~ - xnllx -+ 0, Tx is well defined by (b). Part (c) follows from IITxliy IlIxllx = lim IIToxnl/y\lIxnllx for Xn -+ x, Xn E Xo. •

Exercise 6.1.12. Let Xo be dense in (X, 11·11). Let 111·111 be a second norm on Xo, that is equivalent with 11·11 on Xo. Show that the completion of Xo with respect to 111·111 results in (X, III· liD and 11·11 and 111·111 are also equivalent onX. A corollary of the "open mapping theorem" (cf. Yosida [1, §II.5]) is the following, important result:

Theorem 6.1.13. Let X, Y be Banach spaces. Let T E L(X, Y) be injective and surjective. Then T-l 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) Yo := range(T) := {Tx: x E X} with 1I·lly as norm is a Banach subspace ofY. (b) T- 1 exists on Yo and is bounded: T- 1 E L(Yo, X). Let X and Y be Banach spaces with X C Y. Obviously the inclusion denoted by I : x E X 1-+ X E Y is a linear mapping. If it is bounded, i.e.,

IE L(X, Y)

that is, IIxlly ~ Cllxllx

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, II . lIy), then X is said to be densely and con tin uously embedded in Y.

114

6 Tools from Functional Analysis

Exercise 6.1.15. Let X eYe Z be Banach spaces. Show that if X is [densely and] continuously embedded in Y and Y is [densely and] continuously embedded in Z, then X is [densely and] continuously embedded in Z.

6.1.4 Hilbert Spaces

A mapping (., .): X x X -+ K (K = IR or K = CIJ) is called a scalar product on X if (x, x)

> 0 for all 0 ::J. x

E

X,

(6.1.6a)

(Ax +y,z) = A(X,Z) + (y,z)

(6.1.6b)

(x,y) = (y,x)

(6.1.6c)

for all A E K,x,y,z E X, for all x,y E X,

where ~ denotes the complex conjugate of A. Exercise 6.1.16. Show that (a) proof use (b) the Schwarz inequality:

Ilxll

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

l(x,y)1 ~ IIxlillyll for all x,y E X. Hint: consider IIx - Ayll ~ 0 for A= IIxll 2 /(x, y), if(x, y) ::J. o.

(6.1.7)

(c) (-, .): X x X -+ K is continuous.

A Banach space X is called a Hilbert space if there exists a scalar product (., ·)x on X such that IIxlix = (x,x)lt 2 for all x EX. x E X and y E X are said to be orthogonal (x .1 y) if (x, y)x = o. If A c X is a subset of the Hilbert space X then the orthogonal space A1. := {x EX: (x, a)x = 0 for all a E A} defines a closed subspace of X.

Lemma 6.1.17. Let U be a closed subspace of a Hilbert space X. Then X can be decomposed into the direct sum X = U $ U 1., i. e., any x E X has a unique decomposition x = u + v with u E U, v E U 1.. Purthermore, IIxll~ = lIull~ + Ilvll~· Exercise 6.1.18. Let A be a subset of the Hilbert space X. Prove the equivalence of the following statements: (a) A is dense in X; (b) A1. = {a}; (c) x E X and (a,x)x = 0 for all a E A imply x = 0; (d) for every O::J. x E X there exists an a E A with (a,x)x ::J. O.

6.2 Sobolev Spaces

115

6.2 Sobolev Spaces In the following il is always an open subset of JRn .

L2(il) consists of all Lebesgue-measurable functions whose squares on il are Lebesgue-integrable. Two functions u, v E L2(il) are considered to be equal (u = v) if u(x) = v(x) for almost all x E il, i.e., for all x E n\A, where the exceptional set A has Lebesgue measure J.t(A) = o. Theorem 6.2.1. L2(n) forms a Hilbert space with the scalar product (u,v)o:= (U,V)L2(S1):=

L

u(x)v(x)dx

(6.2.1)

and the norm

(6.2.2)

Lemma 6.2.2. The spaces COO(il)nL2(il) and C8"(il) are dense in L2(il). Here C[f(il) := {u E COO(il): supp(u) compact, supp(u) cc n}.

(6.2.3)

supp(u) := {x E il:u(x) # O} denotes the support ofu. The double inclusion il' cc il indicates that il' lies in the interior of il (i.e., il' C il and TI' n

an = 0).

Let Da be the partial derivative operator (3.2.5b). In the following we need so-called weak derivatives Dau which are defined in a nonclassical way. Definition 6.2.3. u E L2(il) has a (weak) derivative v := Dau E L2(il) if for the latter v E L2(il) holds: (w,v)o

= (-l)la l(D aw,u)o for all w E C8"(il).

(6.2.4)

Exercise 6.2.4. Show the following: (a) Let u have a weak derivative Dau E L2(il). If the classical derivative nau exists in il' C il, it coincides there (almost everywhere) with the weak one. (b) If u has the weak derivative Va = Dau E L2(n), and if Va has the weak derivative v a+J3 = DJ3va E L2(il) then v a+J3 is also the weak Da+J3-derivative of u; i.e., Da+J3u = DJ3(Dau ). (c) Let il C JRn be bounded and 0 E il. u(x) := Ixl 17 (0" E JR) has weak first derivatives in L2(il) if 0" = 0 or 20" + n > 2.

116

6 Tools from Functional Analysis

(d) For u" E COO(n) let u" -+ L2(n) nonn. Then v = nau.

U

E L2(n) and Da u" -+ v E L2(n) in the

For further applications we consider -

N-

Example 6.2.5. Let n = Ui=l ni, where the bounded subdomains n, are disjoint and have piecewise smooth boundaries. Let k E IN. A function u E Ck-l(a) with restrictions u In.E ck(lii) (1 ~ i ~ N) has a (weak) kth derivative Va = nau E L2(n), lal ~ k, which coincides with the classical one in i.

un

PROOF. The (k - 1)st derivatives exist as classical derivatives so that the assertion need only be shown for k = 1. Let Da = a/ax;, W E C{f(n). The assertion is obtained via integration by parts:

-(Daw,u)o

= -(wz.pu)o = -

In wziudx

I:ln, wzjudx = I: [rIn, wuzi dx - lan. r wun~i) dr] = I:l dx = I: r wvadx = r wvadx = (w,va)o, n, In. ln

=-

i

i

WU Zi

i

i

since every x E ani \ an also belongs to ank \ an for some k =f. i with opposite • nonnal direction n(k) = _n(i). The inequality I(u,v)ol ~ lulolvlo (cf. (1.7)) reads explicitly I

For v I

In

u(x)v(x) dxl

~

[In

lu(x)12 dx

In

Iv(x)12 dx]1/2.

(6.2.5a)

= 1 the result is

In

u(x) dxl

~ J1.(n)1/2[1n lu(x)l2 dx]1/2

(J1.(n): measure of n).

(6.2.5b)

For a E LOO(n),u,v E L2(n) one now has (6.2.5c)

6.2.2 Hk(n) and HG(n) Let k E IN U {o}. Let Hk(n) c L2(n) be the set of all functions having weak derivatives Da u E L2(n) for lal ~ k:

6.2 Sobolev Spaces

117

The Sobolev space denoted here by Hk(il) is denoted by W;(il) in some other places. Usually, there Hk(il) is a space which only coincides for sufficiently smoothly bounded il with W;(il). Theorem 6.2.6. Hk(il) forms a Hilbert space with the scalar product (U,V)k:= (U,V)Hk(n) :=

L

(Dau, D a V)£2(n)

(6.2.6)

lal9

and the (Sobolev) norm lulk := lIuIlHIo(n) :=

L

II Da u IlL2(n)"

(6.2.7)

lal~k

Lemma 6.2.7. O=(il)

n Hk(il)

lies densely in Hk(il).

Lemma 7, whose proof can be found, for example, in Wloka [1, Theorem 3.5] permits a second definition of the Sobolev space Hk(il): Remark 6.2.8. Let Xo := {u E 0= (il): lulle < co}. The completion of Xo in £2(il) with respect to norm (7) results in Hk(il). Definition 6.2.9. The completion of 08"(il) in £2(il) with respect to the norm (7) is denoted by HG(il). Theorem 6.2.10. The Hilbert space HG(il) is a subspace of Hk(il) with the same scalar product (6) and same norm (7). 08"(il) is dense in HG(il). For k = 0 the1l~ holds (6.2.8)

PROOF. (a) 08"(il) C Xo (cf. Remark 8) implies HG(il) C Hk(il) C £2(il). (b) According to the definition 08"(il) is dense in HG(il). (c) For k = 0 the norms (2) and (7) coincide. Lemma 2 proves Hg(il) = £2(il). Because of HG(il) C Hk(il) C £2(il) (k arbitrary) (8) follows. • Lemma 6.2.11. For il bounded,

II· IIH"(n)

and (6.2.9)

are equivalent norms in HC(il).

PROOF. (a) Evidently,

1·lk,o ~ 1I·IIHk(n).

118

6 Tools from Functional Analysis

(b) Let u E Coca). For an a with lal = k -1 set v := DOtu E Coca). There exists an R with a c KR(O). For each x E a, Xl E [-R, Rj; thus

(cf. (5b». Integration over x E a yields Ivl~ ~ 4R2Iv:C11~ ~ 4R21ul~ 0' since V:Cl is k-fold derivative. Summation over all a with lal = k-l yields luiLl,O ~ Ck-llul~,o. Now luIJ-l,O ~ Cj-liullo follows likewise for alII ~ j ~ k, and

thus lullo ~ CjCHl· .... Ck-dul~,o· Since lul~ = 2:~=0 lullo ~ Clul~,o' for all u E Coca), the statement follows from Exercise 6.1.12. •

Exercise 6.2.12. Show that (a) for bounded a and k 2:: 1, Hk(a) and H3(n) are different. Hint: Consider the constant function u(x) = 1 and use Lemma 11. (b) Lemma 11 holds even if n is bounded in one direction, i.e., if II lies on a strip {x E lRn : IXkl < R} (k E {I,···, n}). Theorem 6.2.13. Let m 2:: 1. There exist constants C 7](10, m) such that

= C(m)

lulk ~ Clul~mlul~m-k)/m for all 0 ~ k ~ m, u E Ho(ll),

lulk

~ flulm + 7](E)lulo for all E > 0, 0 ~ k

PROOF. (a) Partial integration for

< m, u E HO(ll).

and 7](10)

=

(6.2.lOa) (6.2. lOb)

lal = 1 shows that

IDOtul~ = (DOtu, DOtu)o = _(D 2Ot u, u)o ~ ID 20t ui0lui0 ~ lul2lulo.

Since also lul~ ~ lul2lulo, it follows that lul~ ~ (n + 1)luI2Iulo. If one replaces u by DP u with 1.81 = I, one obtains in the same way lul,2 ~ Clull+llub-l

(1 ~ l

< m; u

E

Ho(ll».

(6.2.lOc)

(b) Let u =I 0 be fixed. Set VI := log lub, WI := [lv m + {m -1)vo1/m and ZI := V, - W,. (lOc) leads to 2z1 - ZI-l - ZI+l ~ e:= loge, where Zo = Zm = O. For z = {Zl.··., zm_d T one obtains Az ~ en. Here A is the M-matrix (4.1.9b) for h = 1. A -1 2:: 0 proves that z ~ c:= eA- 1 n. VI = ZI+WI ~ CI+WI (I ~ 1 < m) implies lull = exp(vl) ~ exp(ci + WI), i.e., (lOa) with C:= exp(ellA -111100)' (c) Elementary calculation shows that for each E > 0, 0 ~ ~ eo < 1 there exists an 7]{ 10, 8 0 ) such that

e

aeb l - e ~ Ea+7](E)b for all a,b 2:: O. Formula (lOb) is a corollary of (lOa,d).

(6.2.1Od)

•

6.2 Sobolev Spaces

119

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

{DOtu,DPu)o :;; €Iul~ + {4€)-1Iul~ (€ l> 0, lal :;; m, I\PII:;; k:;; m, '11. E Hm{n)), {DOtu,DPu)o :;; €Iul~ +ll{€)lul~ (€ > 0, lal:;; m, IPI < m, '11. E H[{'(n)).

Remark 6.2.14. The set {u E COO{n):supp{u) compact, 1'11.1", in H"'{il). .

(6.2.10e) (6.2.lOf)

< co} is dense

PROOF. Let '11. E H"'(n), € > O. According to Lemma 7 there exists a function COO{n) with 1'11. - uEI", :;; €/2. There exists a E coo{m,n) with a{x) = 1 for Ixl :;; 1, a{x) = 0 for Ixl ~ 2. For sufficiently large R, one also has luE{x)-a{x/ R)uE{x)l", :;; €/2. Thus there exists v{x) = a{x/R)uE{x) E c8"{n)

UEE

with 1'11. -

vi", :;; €.

•

Since "supp(u) compact" already implies "supp(u)

cc m,n"

we obtain

The Leibniz rule for derivatives of products proves

Theorem 6.2.16. E H"'(n).

l\auI\Hk(SJ):;; C",l\al\ck(.i'i)lIuIlHk(SJ) for all a

E

C"'(Q),

'11.

Theorem 16 together with the substitution rule for volume integrals shows

Theorem 6.2.17. (Transformation theorem). LetT:n -+ n ' be a oneto-one mapping onto n ' with T E cmax(""l)(n) and IdetdT/dxl ~ 0 > 0 in n. We write v = '11. 0 T for vex) = u(T(x)). Then '11. E H"'(n' ) ['11. E H~(n')] also implies '11. 0 T E H"'{n) [E H~(n)l and (6.2.11)

6.2.3 Fourier 'Iransformation and H"'{m,n) For

'11.

E C8"(rn.n ) one defines the Fourier-transformed function

u by (6.2.12)

Note that bounded.

u is

described by a proper integral since the support of

'11.

is

120

6 Tools from Functional Analysis

Lemma 6.2.18. Let

U

E Co:'(ffin). For R

~ 00

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

PROOF. It suffices to discuss the case n tion with respect to { results in

IR(u; y)

= (I/,K)

= 1 (by Fubini's theorem). Integra-

L

(x - y)-l sin(R(x - y»u(x) dx.

Then IR(I;y) = (1/'n") IRt-1 sin(t) dt = 1 for all R > O. Since U E coo(ffin), then also w(x, y) := [u(x)-u(y»)/(x-y) E coo(ffi2n). The estimates w(x, y) = O(I/lxl) and wa:(x, y) = 0(I/x2 ) hold uniformly for y E supp(u). Partial integration yields

IR(u(,) - u(y);y) = IR«' - y)w(·, y); y) = (1/7r) fIR sin(R(x - y»w(x,y) dx

=

-(I/7r)

L

cos(R(x -y»Wa:(x,y)dx/R = O(I/R).

The statement follows from

IR(u; y)

= U(y)IR(I;y) + IR(u(,) -

u(y);y)

= u(y) + O(I/R).

•

PROOF. Lemma 18 shows

Lemma 6.2.20. The inverse Fourier transformation 9="- l u = u is defined for u E Co:'(ffin) by (13):

(9="-lu)(x) := (27r)-n/2 {

1IR"

ei({,x)u({)d{

:= lim (27r)-n/2 [ R-+oo

11{loo~R

ei({,x)u({)d{.

(6.2.13)

6.2 Sobolev Spaces

121

PROOF. We have from Lemma 18

f

(21T)-n/2

f

ei({,x)u(e)£le = (21T)-n

J 1{loo5,.R

f

ei({.x-y)u(y) dy£le

J 1{loo5,.R JJRR

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

•

Theorem 6.2.21. !1',!1'-1 E L(L2(IRn), L2(IRn» with 1I!1'1I£2(JRR)t-L2(JRR) = 1I!1'-1IlL2(IRR)t-L2(RR) = 1, i.e., !1' is an isometric mapping of L2(IRn) onto itself The scalar product satisfies (u,v)o = (u,v)o for all u,v E L2(IRn). PROOF. Since Cg:'(IRn) is dense in L2(IRn) (cf. Lemma 2), !1' can be continued to!1': L2(IRn) ..... L2(IRn) (cf. Theorem 6.1.11). The norm estimate follows from Lemma 19. The roles of !1' and :r- 1 are interchangeable (cf. (12) and (13»; thus !1'-1 E L(L2(IRn), L2(IRn» also holds. The second statement results from

(u, v)o =

~(Iu + vl5 -lul5 -lvI5) = ~(Iu + vl5 -lul5 -lvI5) = (u, v)o . •

Exercise 6.2.22. Prove that: (a) With e a

= ell .. ·e~R, there holds (6.2.14)

(b) There exists C for all e E IRn.

= C(k) such that iJ(1+leI 2)k ~ Llal91eal2

~ C(1+leI 2)k

Lemma 6.2.23. (a) lulk = h/LaI9IeaI2u(e)lo holds for all u E Hk(IRn). (b) A norm on Hk(IRn) equivalent to 1·lk is (6.2.15)

PROOF. (a) It suffices to show the statement for u E Cg:'(IRn) (cf. Corollary 15, Theorem 6.1.11): 2

lul~ =

E lal9

IDau l5 =

E lal9

I!1'D au15 =

E

leau (e)15

lal9

(b) The statement follows from Exercise 22b.

=

E lal9

lea I2u(e)

.

o

•

Lemma 6.2.24. Let 8h.j be the difference operator8h•j u(x) := [u(x+~ej)­ u(x- ~ej)}fh, where ej is jth unit vector. Ifu E Hk(IRn) and 18h,julk ~ C for all h > 0, 1 ~ j ~ n, then u E Hk+1(IRn) holds. Conversely, 18h•j ulk ~ lulk+l holds for all u E Hk+1(IRn), h > o.

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6 Tools from Functional Analysis

PROOF. (a) From !f(u(· + 8ej))(e) = e-i6~iu(e) follows !f(8h ,ju)(e) = sin(ejhI2)u(e). Hence, since 4h- 2sin2(ejhI2) ~ ej for h ~ 1/1el, follows 1!f(8h ,ju)(e)1 2 = ejlu(e)l2 for lei ~ 1/h. Summation over j = 1, ... , n and integration over then gives

¥

e

r .. (1 + leI 2)k+1Iu(e)1 2de 1m.

(lul~+d2 =

~ ~

r ... + (lul~)2 + 11~19/h { (1 + lel 2)klel 2lul 2de

11~1?1/h

1

1~I?l/h

~ (

11~1?1/h

The inteliral over

A2 A2 ... + (Iulk) +~ .LJ(18h ,j u lk) j=l

'" + Cklul~ + nC2 •

lei ~ 1/h vanishes for

(b) For the converse use (8h ,ju)(x)

=

h -+ 0 so that the statement follows. f~~~2u:l)i(x+thej)dt. •

n

Let 8 ~ O. For = m.n one can define the following scalar product (16a) and the Sobolev norm (16b) for all u e C8"(m.n): (6.2.16a) (6.2.16b) The completion in L2(m.n) also defines the Sobolev space HB(m.n) for noninteger order 8. On the basis of Lemma 23 and Exercise 6.1.12 the newly defined HB(m.n ) for 8 e IN U {O} agree with the Sobolev spaces used until this point. Let c m.n . The number 8 ~ 0 can be decomposed as s = k + A with k e INu {O} and 0 < A < 1. We define

n

(u,v)s:=

~ lal::>k

+

lul s

Jlxn

[1

Dau(x)Dav(x)dx

n

[Dau(x) -

D~:(~)~[I~:~(X) -

:= IIUIiHaCS1) := J(u,u)s

(8 = k +A, 0

Dav(y)) dxdY] , (6.2.17a)

< 8 < 1).

(6.2.17b)

The norm I·IB is called the Sobolev-Slobodeckir norm. One can define the Hilbert spaces H8(il) and H3(n) in the same way as in the case 8 = k E IN. The properties of these spaces are summarised in the following theorem (cf. Adams [1), Wloka [1]):

6.2 Sobolev Spaces

123

Theorem 6.2.25. Let s ~ O. (a) For n = IRn the norms (I6b) and (I7b) are equivalent, i.e., both norms define the same space HB(IRn) = H3(IRn). (b) {u E COO(n): supp(u) compact, lulB < oo} is dense in HB(n). (c) CO"(n) is dense in H3(n). (d) aDa(bu) E Hs-Ial(n), if lal ~ s, u E HB(n), a E ct-ial(n), bE Ct(n), where t = s E lN U {O} or t > s. (e) HB(n) c Ht(n), H&(n) c H&(n) for s ~ t. (f) In (IOa,b) k and m can be real. (g) In (11) replace IITllc"Co) by IITllc"co)' with t > k if k rt IN. Exercise 6.2.26. Check, using the norm (I6b), as well as (I7b), that the characteristic function u(x) = 1 in [-1,1], u(x) = 0 for Ixl > 1 belongs to HB(IR) if and only if 0 ~ s < 1/2.

6.2.5 Trace and Extension Theorems

The nature of boundary value problems requires that one can form boundary values ul r (restriction of u to r = an, or trace of u on r = an) in a meaningful way. As can be seen easily, a Holder-continuous function u E CS (n) has a restriction ul r E CB(r) if only r is sufficiently smooth. But, from u E HS(r) does not necessarily follow ul r E HS(r). Since the equality u = v on HS (.0) only means that u(x) = vex) almost everywhere in .0, and r is a set of measure zero, u(x) =f vex) may hold everywhere on r. Also, the boundary value u(x) (x E r) cannot be defined by a continuous extension either since, for example, u E Hl(n) need not be continuous (cf. Exercise 4c). The inverse problem for the definition of ul r is extension: does there exist, for a given boundary value r.p on r, a function u E HS(n) such that r.p = ul r? If the answer is negative there exists no solution u E HB (.0) to the Dirichlet boundary value problem. First we study these problems on the half-space

n=

IR~ := {(Xl.··· ,xn ) E IRn:x n

Functions u E H8(IR~) restricted to r.

r = an =

IRn- 1 x {O}. (6.2.18) will first be continued to it E HB(IRn) , and then it

> O} with

Theorem 6.2.27. Let s ~ O. There exists an extension operator ifJB E L(HS(IR~), HS(IRn» such that for all u E HS(IR~) the extension it = ifJsu and u coincide on IR~. PROOF. For s = 0, set it = u on IR~ and it = 0 otherwise. Since lIuIlL2(Rn) = lIuIlL2(R+), the mapping defined by ifJou := it is bounded: lIifJolIL2(Rn)<-L2(R+) = 1. For s ~ 1 define ifJsu = it by

124

6 Tools from Functional Analysis

(Le., by reflection on r). For s = lone obtains, for example, luh = J2lu/t, i.e., ~s E L(H8(m.~), HB(m.n)). For larger s one uses higher interpolation formulae for u( ... , -xn ) (cf. Exercise 9.1.13, cf. Wloka [1, p. 101)). • In the following the restriction ul r is written in the form 'YU. At first, 'Y is defined only on CO'(m.n ): 'Y: CO'(m.n ) -

cO'(r) c L 2 (m.n -

1 ),

hu)(x):= u(x)

We write x = (x', xn) with x' = (Xl."·, xn-t) E m.n - 1 : HB(r) = HB(m.n- 1).

m.n- 1 . r

for all (x) E r. (6.2.19) is identified with

Theorem 6.2.28. Let s > 1/2. Then '"Y from (19) can be continued to 'Y E L(HB(m.n), Hs-l/2(m.n- 1 )). Thus we have in particular !'YUl s -l/2 S cslul s for u E H8(m.n ). In the case n = 1, i.e., 'Yu = u(O), we have hul S Cslul B.

PROOF. It suffices to show l'Y'ILI:-l/2 s C~I'UI: for u E CO'(m.n ) (cf. Theorem 6.1.11 and Theorem 25a). Let the Fourier transforms of 1£ E CO'(m.n ) and W := 'YU E CO(m.n - 1) be u = !fnu and w = !fn-lw (!flc:k-dimensional Fourier transform). !fn can be written as product !flo !fn-1 where !fn-1 acts on x' E m.n - 1 and !f1 on Xn . Therefore W("x n) := !fn-1U("Xn) has the properties u(e',,) = !flW(e',·) and w = W(-,O). According to Lemma 20, w(e') = W(e', 0) = [!f11u(e', ')I\~n=O has the representation

w(e')

= (271')-1/2

L

u(e',en)den

for all e' E

m.n- 1.

(6.2.20)

Since u E HS(m.n ), O(e', en) := (1 + 1e'12 + e~)B/2u(e', en) lies in L2(m.n) (cf. Lemma 23b). Inequality (5a) yields

The first factor may be computed as K B (l + le'1 2)1/2-B with KB = fIR(! + x2)-s dx < 00, since 8 > 1/2. The second is U(e') := 1I0(e', ')IIL2(IR) E L2(Rn-l) because IIUIIL2(Rn-1) = liOIIL2(Rn) (Fubini's theorem). Together we have:

Integration over e' E

m.n- 1 results in

6.2 Sobolev Spaces

125

Thus Iwl:-l/2 ::; C~lul: is proved with C~ = (Ks/(27r))1/2 (cf. (16b)). In the case n = 1, w(e') already represents 'Yu = u(O), and the integration over e' is not required. • Theorem 28 describes the restriction u(·, 0) = 'YU to Xn = o. Evidently, similarly we have lu(.,xn )ls-l/2 ::; Cslul s for any other Xn e m., with the same constant Cs. The mapping Xn 1-+ u(·,x n ) is continuous [resp. Holdercontinuous] in the following sense.

Theorem 6.2.29. For s

> 1/2 the following

statements hold:

lim lIu(·,x n ) - u(·,Yn)IIH.-l/2(lRn-1) = 0 for all Xn

'IIn--t:r:n

e IR,u e HS(IRn), (6.2.21a) (6.2.21b)

IIU(·,X n ) -U(-,Yn)llH8-1/2->'(lRn-1)::; Ks.~lxn -Ynl~IIUIlH8(lRn) for all u e HS(m.n ), 0::; >. < 1, >.::; s -1/2 (and for>. = 1 if s > 3/2).

PROOF. (a) Let Uv e C~(IRn) be a sequence with Uv -+ U e HS(IRn) and set IPv(x):= lIuv(-,x)IIHo-l/2(lRn-1). The function IPv is continuous in m. and converges uniformly to lIu(.,x)IIH.-l/2(lRn-1) since luv(-,x) - u(.,X)ls-l/2 ::; Cslu v - uls for all x e IlL Thus (21a) follows. (b) ue(·,xn ) := u(·,xn + €) - u(·,xn ) has the Fourier transform ue(e) = ue(e',en) = [exp(ien€) - l]u(e) so that lue (e)1 2 = 4sin2(en€/2)lu(e)1 2 . As in the proof for Theorem 28 set W(.,x n ) := :7n-1Ue(·,X n ), W = W(.,O). The first integral in the estimate of 27rlw(OI2 now reads

1m.(1 + 1e'12 + e;)-S sin2(en€/2) den = (1 + le'1 2) 1/2-s fIR (1 + e)-S sin2(1]t) dt with 1] = ~(1 + le'1 2)1/2. The decomposition of the last integral into subintegrals over It I ::; 1/1] and It I ~ 1/1] shows flR(1 + t 2)-S sin2 (1]t) dt ::; Cs.~1]2~. The remainder of the argument follows the same lines as in the proof of The• orem 28. Up to this point we have obtained HS(IRn) by completion of CQ'(IRn) in L2(m.n). The next theorem shows that for sufficiently large s one can also complete in CO(IRn) n L2(IRn) so that HS(IRn) contains only classical functions (Le. continuous, HOlder-continuous, [HOlder-] continuously differentiable functions).

Theorem 6.2.30. (Sobolev's lemma) Hs(m.n) c Ck(IRn) holds for k IN U {O}, s > k + n/2 and HB(IRn) c ct(m.n) for 0 < t ¢. IN, s ~ t + n/2.

e

PROOF. (a) Let s ~ t + n/2, 0 < t < 1, u e HS(IRn). For given x, y e IRn we want to show lu(x) - u(y)1 ::; Clx - ylt with C independent of x,y. The coordinates of the m.n can be rotated so that x = (Xl! 0, .. ·,0), y = (Yl. 0, ... ,0). An (n-l)-fold application of Theorem 28 to u( .), u(-, 0), u(·, 0,0)

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6 Tools from Functional Analysis

, etc., results in 1£(·,0,0, .. ·,0) E HB-(n-l)/2(1R). Theorem 29 provides the desired estimate with C = KB-(n-l)/2,tlU(" 0,···, 0)IB-(n-l)/2 :5 C'luls ; thus 1£ E ct(lRn) and II 'IICt(Rn) :5 CII ·IIHo(Rn). Then 1£ E CO(lRn) follows from Ct(lRn) c CO(lRn). (b) Let s ~ t + n/2, 1 < t < 2, 1£ E HB(lRn). Part (a) is applicable to Datu E Hs-l(lRn) (cf. Theorem 25d,e) with lal :5 1: Datu E Ct-l(lRn ) for lal :5 1. Thus 1£ E Ct(lRn), etc. • We return to the statements of Theorems 27 and 28. For all 1£ E C8"(IR+') the restriction 'Yu = 1£(·,0) agrees with 'Y~BU. Completion in HS(IR+') yields

This proves the following corollary: Corollary 6.2.31. Let s > 1/2. For the restriction 'Yu := 1£(.,0) we have 'Y E L(H"(IR+.), H,,-1/2(lRn- 1

».

°

With the restriction to Xn = one evidently loses half an order of differentiability. Conversely one gains half an order if one continues w E Hs- 1/ 2 (lRn - 1 ) suitably in IRn. Theorem 6.2.32. Let s > 1/2, wE Hs-l/ 2(lRn - l ). There exists a function 1£ E H"(lRn ) [1£ E H"(IR+.») such that luis :5 C"lwl,,-1/2 and 'YU = w, i.e., W = 1£(.,0).

PROOF. Let u = :1'nu and w= :1'n-lw be the Fourier transforms. 'YU = w is equivalent to (20). For

u(e) = u(e', en) := w(e')(1 + le'1 2),,-1/2(1 + le'I 2 + e;)-" / K", K,,:=

L+ (1

t 2 )-"dt,

one checks that (20) and lui: = K;l/2Iwl:_l/2 hold. Restriction of u E HB(lRn) to IR+. proves the parenthetical addition. • If one replaces IR+. with a general domain n c IRn , then IRn- 1 ~ IRn- 1 x {O} = aIR+. becomes r = an, and the necessity arises of defining the Sobolev space H" (r). We begin with the

°

Definition 6.2.33. Let < t E IRU{ oo} [resp. k E 1NU{O}]. We write n E C t [n E Ck,l], if for every x E r := an there exists a neighbourhood U C IRn such that there exists a bijective mapping
6.2 Sobolev Spaces cp-l E Ct(Kl(O» [cp E Ck.l(U), cp-l E Ck.l(Kl(O))),

127

cP E Ct(U),

cp(U n r) = {e E Kl(O):en = O}, cp(U nil) = {e E Kl (0): en > O}, cp(U n (llln\il» = {e E Kl(O): en

(6.2.22a) (6.2.22b) (6.2.22c) (6.2.22d)

< O}.

Here Kl(O) is a circle (ball) if 1·1 is the Euclidean norm. For the maximum norm I . 100, Kl(O) is a square (cube). Likewise, Kl(O) can be replaced by another circle KR(z) or any rectangle (xl,x1) x··· x (x~,x~).

Figure 6.2.1. Covering neighbourhoods of rand

[J

Example 6.2.34. Let il be the circle Kl(O) C lll2. A neighbourhood ofx* = (1,0) is Ul from Fig. 1. The mapping x E Ul 1-+ cp(x) := e E (-1,1) x (-1,1) with Xl = (1 - e2/2)cos(7rel/2), X2 = (1 - e2/2)sin(7ret/2) is bijective and satisfies (22a-d) with t = 00. The same holds for any x E r. Thus il E Coo. Exercise 6.2.35. (a) The rectangle il = (xl, x1) x (x2, x~) and the L-domain from Example 2.1.4 are domains il E CO· l . (b) The cut circle in Figure 5.2.1b does not belong to CO· l . Lemma 6.2.36. Let il E C t [il E C k •l ] be a bounded domain. Then there exists N E lN, Ut (O:5 i :5 N), Ut , ~ (1 :5 i :5 N) with N

U i open, bounded (0 :5 i :5 N),

U Ui :::> 'ii,

UO CC il,

(6.2.23a)

i=O

N

Ui ;=

ui n r

(1:5 i :5 N),

UU = r i

(6.2.23b)

= 1, ... , N,

(6.2.23c)

i=l

at: Ui ~

0

-4

ai(Ui ) C llln-l bijective for i

ajl E Ct(aj(Ui n Uj

»[resp. at

0

ajl E Ck.l(aj(Ui n Uj))]. (6.2.23d)

On Ui (1 :5 i :5 N) are defined mappings CPi with the properties (22a-d).

128

6 Tools from Functional Analysis

PROOF. For every x E r there exist U = U(x) and

l/i} CC n. Let UxErU(x)UU V' be an open covering of the compact set Therefore there exists a finite covering through U' := U(x.) (1 ~ i ~ N) and at most one Vj which is denoted by UO. If one sets Ui := U' n r, 0:; = ax.,
n.

A set of pairs {(U., 0:;): 1 ~ i ~ N} fulfilling the conditions (23b-d) is called a C t _ [resp. C k,l]_ coordinate system for r. In Example 34 one has N = 4. The map all is given by al 1 (€1) = (cos(1ret/2), sin(1r6/2» E U1j and likewise a,41(€1) = 008«3 + 6)1r/2), sin«3 + e1)1r/2) E U\ where in each case -1 < ei < 1. On a1(Ul n U4) one obtains a4(al 1(€1» = €1 + 1. Lemma 6.2.37. (Partition of unity) Let {U':O ~ i ~ N} satisfy (23a). There exist functions O'i E Og>(m.n ), 0 ~ i ~ N, with N

supp(O") CUi,

LO'l(x) = 1

for all x E

n.

(6.2.24)

.=0 The general construction of the O'i can be found, for example, in Wloka [I, §1.2]. In the special case of Figure lone may proceed as follows. Let O'(t) := 0 for It I ~ 1 and O'(t) := exp(1/(t2 - 1» for t E (-1,1). Then 0' E C.r(m.) and supp(O') = [-1,1]. In U' from Figure lone defines, for example, tPo(x) := 0'(9IxI2/4),

tP1{X):= 0'(2r - 2)0'(2

vI:

for x = r

(c~
etc.

These O'i{X) := tPi{X)/ tPl(x) satisfy (24). A function u on r can be written in the form ~ O'lu. Each summand O'lu is parametrisable over o:;(Ui ) C m.n - 1: (O'lu) 0 a; : a.(U.) cc m.n - 1 -+ IR.. This in turn makes possible

Definition 6.2.38. Let n E C t [E Ck,1]. Assume (U., a.) and 0'. satisfy (23b-d) and (24). Let s ~ t E IN [s ~ k+ 1] or s < t ¢ IN, t > 1. The Sobolev space HB(r) is defined as the set of all functions u: r -+ m. such that (O'iU) 0 a;l E H&(m.n - 1) (1 ~ i ~ N). Theorem 6.2.39. (a) HB(r) is a Hilbert space with the scalar product N

(U,V)B:= (U,V)HO(r):= L«UO'i) o a; 1, (vO'.)oa;1)HO(lRn -

1) •

• =1

(b) If {(Ui , a.): 1 ~ i ~ N} is another C t _ [Ok,1:] coordinate system of rand {Ui} another partition of unity, then the space HB(r) defined by this is equal to HB{r) as a set. The n017nS of H8(r) and fI8(r) are equivalent.

6.2 Sobolev Spaces

129

PROOF. For (b): Transformation Theorem 17 [resp. 25g] ; for DE CO,l, 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. "f now denotes the restriction to r = an: "fU = u Ir. Theorem 6.2.40. Let n E C t with 1/2 < s :::; t E IN or 1/2 < s < t [resp. DE e k ,l, 1/2 < s = k + 1 E IN]. (a) The restriction "fU of u E HII(n) belongs to Hs- 1/ 2(r): "f E L(HS(D), HS-l/2(r)). (b) For each W E Hs-l/2(r) there exists an extension u E HS(D) with W = "fU, luis:::; e s lwls-l/2. (c) For each W E HS(n) there exists a continuation Ew E H8(ntn) with E E L(H8 (n), Hs(ntn)). PROOF. The proofs follow the same pattern. Let U i , Ut and
:::; Cslul s .

•

Remark 6.2.41. (a) Under the conditions of Theorem 40 and the additional condition s > lal + 1/2 there exists a restriction "fDau E Hs-la l-1/2(r) of the derivative of u E HS(D). (b) For each u E Hg(n) with s < 1+ 1/2 one has "fDau = 0 if lal :::; l. Theorem 6.2.42. For n E CO,l holds HJ(D)

= {u E Hl(D): ulr = O}.

PROOF. (a) u E HJ(D) is the limit of UII E CO>(D), all of which satisfy ulIJr = O. Since ulli r -+ ul r with respect to Hl/2(r), it follows that u E H (n) and ul r = o.

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6 Tools from Functional Analysis

(b) Conversely, let u E H1(n) and ull' = O. The proof that u E HJ(n) can be divided up as follows: (ba) By using the partition of unity over {Ui: 0:::; i :::; N} (cf. Lenuna 37), the statement reduces to the case n = ffi.+. (bb) Without loss of generality one can assume n = 1: n = H4. (be) According to Remark 14, for each E > 0 there exists a iiE E COO(ffi.+) with finite support such that lu - iieh :::; rJ/2. For each rJ > 0 there exists CPfj E COO(ffi.+) with CPfj(x)

=0

for x:::; rJ/2,

Icp~(x)1 :::; 3/rJ,

cPfj(x)

=1

for x 2:: rJ,

0:::; cPfj(x) :::; 1.

The function U E := iie - (1 - cpt}iiE(O) satisfies U E E COO(H4), uE(O) = 0, supp(uE) finite, and lu - uE/t :::; CEo Thus X := {v E COO(ffi.+):v(O) = 0, supp(v) finite} is dense in {u E Hl(ffi.+):u(O) = a}. (bd) The statement would be proved if Cgo(ffi.+) were also dense in X with respect to I· /t. Let vEX. Evidently, vfj := CPfjv E Cgo(ffi.+) holds for all rJ > O. Set 8(rJ) := IIv'IIL3(o.'1) and note that 8(rJ) -+ 0 for rJ -+ O. Since vex) = v'l(x) for x 2:: rJ, it remains to estimate the variables IIv - v'lIlL3(o.'1) and IIv' - v~IIL3(o.fj)' Because of v~ - v' = cP~v + (cp'l - l)v', one obtains IIv~ - v'IIL3(o.'1) :::; IIcp~IILoo(o.'1)lIvIlL3(o.'1)

:::; (3/rJ)lI v IlL3(o.'1)

+ IIcp'l -

lIlLOO(o.'1)lI v 'IIL2(o.'1)

+ 8(rJ)·

J;

Since vex) = v'(e) cIe, (5b) implies the estimate Iv(x)1 :::; .jij8(rJ) for all o :::; x :::; rJ, and hence IIvIlL3(o.'1) :::; rJ8(rJ). Then the statement follows from Iv - v'll~ :::; IIv - V'llli3(o.'1) + IIv' - v~lIi3(o.'1) :5 C8 2 (rJ) -+ O. • When n E C1, the normal direction n exists at all boundary points. Analogously to Theorem 42 one proves Corollary 6.2.43. For n E C1 and k E IN holds H~(n)

= {u E Hk(n):o'u/on'lr = 0 forO:::; 1:5 k -I} = {u E Hk(Il):DQul r = 0 forO:::; lal:5 k -I}.

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

6.3 Dual Spaces

131

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

Ilx'lIxl ;= IIx'IIR<-X = sup{lx'(x)lIlIxllx:o =I x EX}.

(6.3.1)

The elements x' E X' are called linear functionals on X. Instead of x'(x) (application of x' to x) one also writes (x,X')XXXI, or (x',x)x1xx, and calls (-, ')XXXI the dual form on X x X':

(x,X')XXXI = (x',x)x1xx = x'(x).

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

PROOF. (a) Let y' E Y'. Because Xc Y, y' is defined on X. (b) Since X is a dense subspace ofY, according to Theorem 6.1.11 and (1.5), for each y' E Y' holds:

lIy'IIYI = sup ly'(x)lfllxlly xEX

~ c1 suply'(x)lfllxllx = C1l1y'lIxl, x

•

Le., 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. y' E Y'

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

(Tx,y')yXY ' = (x,X')XXXI

for all x E X

(6.3.2)

defines a unique x' E X'. The linear mapping y' 1-+ x' defines the dual operator -+ X' with T' y' = x'. The following holds: T' E L(Y', X') and

T': Y'

(6.3.3)

PROOF. If one writes (2) in the form y'(Tx) = x'(x), one can see that x' = y' 0 T. (3) follows from the definitions of the norms;

IIT'lIxl<-YI = sup IIT'y'IIx1flly'IIYI = y'#O

sup

x#O,y'#O

l(x,T'y')lxxxlf[llxllxlly'IIYI]

= sup I(Tx, y')YxYllf[llxllxlly'lIy] = sup IITxllY IlIxlix = IITlIy<-x. x#O,y'#O

x#O

• Example 6.3.3. Let n = (0,1), X = (COCi?), II . 1100), and x E n. The mapping cx: u E c°(li) 1-+ u(x) E IR is a functional: Cx E CO(D)' (the socalled delta function). The Laplace operator .d belongs to L(C 2 (D), CO(D».

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6 Tools from Functional Analysis

The dual mapping Ll' E L(oo(O)', Q2(ll)') is applicable to 6x : Ll'6x is characterised by (Ll'6x )u = Llu(x) for all u E C 2 (ll). Exercise 6.3.4. Let S E L(X, Y), T E L(Y, Z). Show that (TS)'

= S'T'.

Exercise 6.3.5. Show that if T E L(X, Y) is surjective, then T' is injective.

6.3.2 Adjoint Operators

Let X be a Hilbert space (over 1R). Every y E X defined by III(x) := (x, y)x is a linear functional III E X, with 1I/lIlixi (cr. Yosida [1, §III.6]).

= lIyllx. The converse also holds

Theorem 6.3.6. (Riesz representation theorem) Let X be a Hilbert space and I E X'. There exists a unique Y/ E X such that I(x)

= (x,y/)x

for all x E X

and IIflixl

= lIy/lix.

Corollary 6.3.1. Let X be a Hilbert space. (a) There exists a one-to-one correspondence (the Riesz isomorphism) J x E L(X,X') with JXY = Ill' J;l I = y/ that presenJes the norm, i.e., IIJxllxl+-x = IIJ;lllx+-xl = 1. (b) X' is a Hilbert space with the scalar product (x', y')xl = (J;lx',J;ly')x. The dual norm IIx'lIxl from (1) agrees with the norm induced by (x',x,)~2. (c) One always identifies X with X" because x(x') := x'(x). Prom this follows JX' = J Xl , Jx = J Til = T lor T E L(X, Y) if Y = Y" is also a Hilbert space. (d) One can identify X and X': X = X', Jx = I.

x,

Let X, Y be Hilbert spaces and T E L(X, Y). The mapping defined by T* := J;lT' Jy E L(Y, X) is called the operator adjoint to T and satisfies (Tx, y)1I = (x, T*y)x for all x E X, Y E Y;

IITlly+-x = IIT*lIx+-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. TE L(X,X) is said to be 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.B. Let Xo be a closed subspace of the Hilbert space X. An orthogonal projection is given by Tx := y E Xo with

6.3 Dual Spaces

IIx - Yllx

= inf{lIx - Ellx:E E X o}.

133

(6.3.5).

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

PROOF. (a) x can be decomposed uniquely into x = y+z (y E Xo, z E Xl) (cf. Lemma 6.1.17). y is the unique solution of (5). x E Xo implies y = x, thus T2 = T. The analogous decomposition x' = y' + z' shows (x, T*x') = (Tx, x') = (y, x') = (y, y' +z') = (y, V') = (y+z, V') = (x, Tx'), hence T = 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' E Xo holds (Tz, V') = (z, T*y') = (z, Ty') = (z, V') = 0, thus Tz E xl. Together with Tz E Xo follows Tz = 0 so that Tx = Ty + Tz = y. (c) Tx = y and Ilxll~ = lIylli + IIzll~ ~ lIyll~ prove IITllx+-x :s; 1. •

6.S.S Scales of Hilbert Spaces We assume:

vcU

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

Lemma 6.S.9. densely in V'.

Under assumption (6) U' is embedded continuously and

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 =I- v' E V' be arbitrary and '1.1. := J y 1v' EVe U. According to the definition, '1.1.' := Juu E U' C V' is characterised by u'(x) = (x, u)u for all x E U. For x:= '1.1. = Jy1v' E V follows

(v', u')v1

= (Jy 1V', Jy1u')v = ('1.1., Jy1u')v = '1.1.'('1.1.) = ('1.1., u)u > O.

•

According to Corollary 7d , U and U' can be identified. By this one obtains the Gelfand triple V cUe V'

(V

c

U continuously and densely embedded).

(6.3.7)

Corollary 6.S.10. In a Gelfand triple (7) V and U are also continuously and 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 = V CU. But it is not possible to identify U with U' and V with V' simultaneously. In the first case one interprets x(y) = (y,x}uxul for X,y E U as (y,x)u (in particular for x, y EVe U), in the second case as (y,x)v. U ' C V'

Exercise 6.3.11.

Let (7) be true. Set W := {Jv1u:u E U} and define (x, y)w := (Jvx, JVY)u as the scalar product on W. Show that (a) (b) (c) (d)

W is a Hilbert space; W C V is a continuous and dense embedding; (v, w)v = (v, Jvw)u for all v E V, wE W; l(x,Y)vl :5l1xllullYllw for all X,y E W.

Because U = U' the scalar product (x, y)u can also be written in the form y(x) = (x,y}uxu1. If x E V, then y(x) = (x,y}vxvi also holds. That means that (x,y)u = (x,y}vxv1 for all x E V,y E U C V'. Likewise one obtains (x,y)u = (x,y}v'xv for all x E U and y E V. The dense and continuous embedding U C V' proves

Remark 6.3.12. Let V cUe V' be a Gelfand triple. The continuous extension of the scalar product (-, ·)u to V x V' [V' x V) results in the dual form (', '}VXVI [(" '}v1xv), Therefore the following notation is practical: (x, y}vxvi = (x, y)u (x,y}v1xv = (x,y)u

for x E V, Y E V', for x E V',y E V.

In connection with Sobolev spaces one always chooses U := L2(O) so that the embeddings read as follows:

c L2(f1) c (Hg(o»' HII(ll) c L2(ll) c (HB(ll»'

Hg(O)

(8 ~ 0),

(6.3.8a)

(8 ~ 0).

(6.3.8b)

Exercise 6.3.13. Show that (8a) and (8b) are Gelfand triples. The dual space of H8(ll) is also denoted by H-S(ll) or HOB(ll): HolI(ll) := H- 8 (ll) := (Hg(ll»,

(8

~

0).

The norm of H-S(ll) according to (1) reads: lul-s := sup{l(u,v)L3(.a)l/lv ls :O

-I v E Ho(ll)},

for

8

~0

where (U,V)L3(.a) is the dual form on Hg(ll) x H-II(O) (cf. Remark 12).

Remark 6.3.14. (a) Let II = m,n. The norm dual to

1'1:,

lul~8 := sup{l(u, v)ol/lvl:: 0 -I v E Hs(m,n)},

6.4 Compact Operators

135

is equivalent to I·I-B and has the representation (2.16b) with -s instead of s. (b) The Fourier transform shows Da E L(HB(m.n), Hs-1al(m.n )) for all s Em.. (c) au E HB(n), if U E HB(J1), a E oteri), where t = lsi E IN U {O} or t > lsi.

6.4 Compact Operators Definition 6.4.1. A subset K of a Banach space is said to be precompact [compact] if each sequence Xi E K (i E IN) contains a convergent subsequence X'1e [and limk-HX) X'1e E K]. 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. Dieudonne [1, (3.16.1)]). For the terms "relatively compact" and "precompact" see also Dieudonne [1, (3.17.5)J.

Remark 6.4.2. (a) K c m.n is precompact [compact] if and only if K is bounded [and complete]. (b) Let X be a Banach space. The unit sphere {x EX: IIxll ::; I} is compact if and only if dim(X) < 00. Definition 6.4.3. Let X and Y be Banach spaces. The mapping T E L(X, Y) is said to be compact if {Tx: x EX, Ilxllx ::; I}, the image of the unit sphere in X, is precompact in Y. Exercise 6.4.4. When is the identity I E L(X, X) compact? Lemma 6.4.5. Let X, Y, and Z be Banach spaces. (a) Let one of the mappings Ti E L(X, Y), T2 E L(Y, Z) be compact. Then T2Ti E L(X, Z) is also compact. (b) T E L(X, Y) is compact if and only ifT' E L(Y', X') is compact. PROOF. (a) Let Ki := {x E X: IIxllx ::; I}. If Ti is compact, i.e., Ti(Kt} is precompact, then T2(Ti(Kt}) is also precompact and thus T2Tl is compact. If, however, T2 is compact, one proves the assertion as follows. Since scaling does not change compactness, IITl lly+-x ::; 1 can be assumed without loss of generality. Hence T i (Kl) is a subset of the unit sphere in Y and therefore T2(Tl(Kt}) is precompact. (b) cf. Yosida [1, §X]. • 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 E L(X, Y), Ix = x is compact.

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6 Tools from Functional Analysis

Together with Definitions 1 and 3 one obtains: X C Y is compactly embedded if every sequence Xl E X with IIXilix :::;; 1 contains a subsequence convergent in Y.

Example 6.4.7. Let nbe bounded. OB(n) C CO(n) is a compact embedding for s > O. PROOF. Functions Ui E oB(Ii) with IIUillc'(Q) :::;; 1 are equi-continuous and uniformly bounded. The assertion is based on the theorem of Arzela-Ascoli (cf. Yosida [1, §III.3]). •

Analogous results can be obtained for Sobolev spaces (cf. Adams [1, p. 1441, Wloka [1, §7J):

Theorem 6.4.8. Let n c Rfl be open and bounded. (a) The embeddings Hg(n) c H&(n) (s, t E R, s > t) are compact. (b) FUrther, let n E CO,l. The embeddings Hk(n) c H'(n) (k, lEN U {O}, k > 1) are compact. (c) Let 0 :::;; t < sand n E or (r > t, r > 1) or n E Ok,l (k + 1 > t). Then the embedding H8(n) c Ht(n) is compact. Remark 6.4.9. In Theorem 8b one can replace n E 0°,1 by the "uniform cone property" (cf. Wloka [1, §2.1]). To ensure n E 0°,1 it is sufficient that the boundary an is piecewise smooth and the inside angles of possible comers are smaller than 211'. Indented comers (cf. Figure 2.1.1) are thus permitted while a cut domain (cf. Figure 5.2.1b) is excluded. In Section 6.S the following situation will arise:

v cUe 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 (1) hold. Let V CUbe a compact embedding. Then T E L(V', V'), T E L(U, U), T E L(V, V), T E L(V', U), and T E L(U, V) are compact. PROOF. As an example, let us do T E L(U, V). Since the inclusion I E L(V, U) is compact we see I E L(U, V') is also compact (cf. Lemma Sb). T E L(U, V), as the product of the compact mapping I E L(U, V') with • T E L(V', V), is compact (cf. Lemma 5a).

Exercise 6.4.11. Let dim X compact.

< 00 or dim Y < 00. Show that T

E L(X, Y) is

6.5 Bilinear Forms

137

The significance of compact operators T E L(X, X) lies in the fact that the equation Tx - AX = y (with X, y EX, Y given, X sought) has properties analogous to the finite-dimensional case. Theorem 6.4.12. (Riesz-Schauder theory) Let X be a Banach spacej let T E L(X, X) be compact. (a) For each A E ClJ\{O} one of the following alternatives holds:

(i) (T - AI)-l E L(X, X)

or

(ii) A is an eigenvalue.

In case (i) the equation Tx - AX = Y has a unique solution X E X for all Y E X. In case (ii) there exists a finite-dimensional eigenspace E(A, T) := kernel (T - AI) # {O}. All X E E(A,T)\{O} solve the eigenvalue problem Tx=Ax, x#O. (b) The spectrum q(T) ofT consists by definition of all eigenvalues and, if not T-l E L(X, X), A = O. There exist at most countably many eigenvalues which can only accumulate at zero. A E a(T) if and only if'). E a(T'). Furthermore,

dim(E(A, T» = dim(E(X, T'» < 00. (c) For A E q(T)\{O}, Tx - Ax = y has at least one solution x only if (y, X'}XXXI = 0 for all x' E E(5., T').

E

X if and

In Lemma 6.5.18 we need Lemma 6.4.13. Let Xc Y c Z be continuously embedded Banach spaces and let X C Y be compactly embedded. Then for every f > 0 there exists a GE such that (6.4.2) IIxlly ~ fllxllx + GEllxllz for all x E X. PROOF. Let f > 0 be fixed. The negation of (2) reads: There exists Xi E X with (lixilly - flixillx)/lIxiliz -+ 00. For Yi := (fllxillx)-lxi E X we thus have (IlYilly -1)/IlYillz -+ 00. From this one infers IIYillz -+ 0 and IIYilly > 1 for sufficiently large i. Since IIYilix ~ 1/f. and Xc Y is a compact embedding, a subsequence Yik converges to y* E Y. Now, IIYilly > 1 implies IIY*lIy ~ 1, i.e., y* # O. On the other hand, Yik also converges in Z to y* since Y c Z is continuously embedded. IIYiliz -+ 0 gives the contradiction sought: y* = o.

•

6.5 Bilinear Forms In the following let us assume that V is a Hilbert space. The mapping a(., .): V x V -+ IR 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)

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6 Tools from Functional Analysis

for (x, y, Z E V, A E IR) (in the complex case one speaks of sesquilinear forms: a(x,Ay) = ~a(x,y)). a(·,·) is said to be continuous (or bounded) ifthere exists a C s such that la(x, y)l ~ Csllxllvllyllv

for all x, y E V.

(6.5.1)

Lemma 6.5.1. (a) To a continuous bilinear form one can assign a unique operator A E L(V, V') such that a(x, y)

= (Ax, y)v'xv

for all x, y E V,

IIAllvl<-v ~ cs.

(6.5.2)

(b) Let VI and V2 be dense in V. Let a(·,·) be defined on VI x V2 and satisfy (1) with '~E Vl,y E V2" instead of '~,y E V". Then a(·,·) can be extended uniquely to V x V so that (1) holds with the same Cs for all X,y E V.

PROOF. (a) Let x E V be fixed.
= sup{la(x, y)l: x, Y E V, IIxliv = lIyllv = 1}.

(6.5.3)

A is called the operator that is associated to a(·, .). The bilinear form a*(-,·) adjoint to a(·,·) is given by a*(x,y):= a(y,x) (x,y E V). The bilinear form is said to be symmetric if a(·,.) = a*c-, .). 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 6.5.8. Let A E L(V, V') be the operator associated to a continuous bilinear form a(·,·). Then the following statements (i), (ii), (iii) are equivalent: (i) A-I E L(V', V) exists; (ii) 10,10' > 0 exist such that inf{sup{la(x, y)l: Y E V, IIYllv

= 1}: x

= 1} = 10 > 0,

(6.5.4a)

inf{sup{la(x, y)l:x E V,lIxliv

= 1}:y E V,lIyllv = 1} =10' >0;

(6.5.4b)

E

V, IIxliv

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

sup{la(x,y)l:x E V, IIxliv If one of the statements (i)-(iii) holds, then

= 1} > O.

(6.5.4c)

6.5 Bilinear Forms

f

= f' = 1/IIA- 1 I1v+-v'

(f,f' from (4a,b».

139

(6.5.4d)

From (4a) follows inf{sup{la(x, v)l: V E V,lIvllv = 1}:x E

v, IIxliv = 1} ~ f > O.

(6.5.4e)

Conversely, (4a) follows from (4e) with a possibly laryer f > O. (4e) and (4c) are also called the Babus ka conditions. (4e) is equivalent to sup{la(x, y)l: y E V, IIYllv = 1} ~ fllxllv

for all x E V,

(6.5.4e')

because (4e) is equal to (4e') for all x E V, IIxliv = 1. The scaling condition IIxliv = 1 can evidently be dropped. The left-hand side in (4e') agrees with the definition of the dual norm of Ax so that (4e) and (4e') are also equivalent to (4e"): (6.5.4e") IIAxll v ' ~ fllxllv for all x E V. PROOF. (a) "(i) ::::} (ii)": Let A-l E L(V', V) exist. Then (4a) follows from . f{ III

} ...

. f =

la(x, y)1 . f I(Ax, y)1 sup II x II V II y II V = ..III eY sup lIeY II X II V II Y II V

..III eY lIeY

.. #0 11#0

=

0:#0 11#0

inf sup I(AA- 1 x', y)1 IIA-1x'lIvllyliv

.. 'eY' lIeY

.. '#0

=

11#0

inf IIA-1x'lIv1llx'lIv'

aI/ev' .. '#0

=

1/IIA- 1 I1v+-v'

=

inf IIA- 1x'II- 1 sup I(x', y)1 ..'eY' V lIeY IIYllv .. '#0

= 1/[

11#0

sup IIA-1x'llv/llx'IIV'] z/EV'

.. '#0

=:

f.

In the same way one shows (4b) with f' = 1/IIA'-1Ilv+-v'. Because A'-l = (A-l)" (3.3), and V" = V, it follows that f = f'. (b) "(ii) ::::} (iii)": (4c) is a weakening of (4b). (c) "(iii) ::::} (i)": f > 0 in (4a) proves that A is injective. We wish to show that the image W := {Ax: x E V} c V' is closed. For a sequence {w v } with IIw· - wvllv' -4 0 we must therefore show that w· E W. According to the definition of W there exists Xv E V with Axv = wv . From (4a) one infers via (4e) and (4e") (with x := Xv - xI') that IIxv - xJ'llv :::; IIwv - wJ'lIv' If. Since {wv } is Cauchy convergent, this property carries over to {xv}. There exists an x* E V with Xv --+ x* in V. The continuity of A E L(V, V') proves Wv = Axv --+ Ax* so that w* = Ax* E W. According to Lemma 6.1.17 one can decompose V' into WEBW.L.1f A were not surjective (thus W i V'), there would exist awE W.L with w i O. Then y:= Jv'w = JV1 E V would satisfy y i 0 (cf. Theorem 6.3.6, Corollary 6.3.7). Since a(x,y) = (Ax,Y)v'xv = (Ax,w)v' = 0 for all x E V, a contradiction to (4c) would result. Therefore, A is also surjective, and Theorem 6.1.13 proves A-l E L(V', V). (d) Statement (4d) has already resulted from Part a of the proof. •

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6 Tools from Functional Analysis

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 (4b) with e' = e and conversely.

< 00, then (4a) implies the statement

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) 2: CEllxll~

for all x

E

(6.5.5)

V.

Exercise 6.5.6. Show that: (a) If W C V is a Hilbert subspace with the norm equal (or equivalent) to V, then a V-elliptic bilinear form is also W-elliptic. (b) Let a(·, .): V x V - m. be continuous. If a(x, x) 2: CEllxll~ holds for all x E Vo where Vo is dense in V, then (5) follows with the same CEo (c) Let a(., .): V x V - m. be continuous, symmetric, nonnegative (a( v, v) 2: 0) and satisfy (4a,c). Then a(.,.) is V-elliptic with CE ~ e2/Cs (e from (4a,b), Cs from (1». Hint: First prove that la(u,v)1 S [a(u,u)a(v,v)J1/2 (cf. Exercise 6.1.16).

Lemma 6.5.7. V-ellipticity (5) implies (1) and (4a,b) with e = e' PROOF. Let x E V,lIxliv = 1. sup{la(x,y)l:y E V,lIyllv CE proves (4a) with e ~ CEo (4b) follows analogously.

=

~

CEo

I} ~ la(x,x)l2:

•

The combination of Lemmata 1, 3, 7 together with IIA'-lllv+--vl IIA-lllv+--vl (cf. Lemma 6.3.2) proves

=

Theorem 6.5.B. Let the bilinear form be V-elliptic [or satisfy (1), (4a,c)]. Then the corresponding operator A satisfies the conditions

= IIAllvl+--v S CS,

A E L(V, V'),

IIA'lIvl+--v

A- l

IIA'-lllv+--vl

E L(V', V),

with Cs from (1) and C'

= I/CE

= IIA-lllv+--vl S c' [resp. C'

(6.5.6)

= l/e = l/e'].

With the help of the bilinear form a(·,·) and a functional f E V' one can formulate the following problem: Find x E V

with

a(x,y)

= fey) for all y E V.

(6.5.7)

According to Lemma lone can write (7) in the form (Ax- f, y}vlxv = 0 for all Y E V, i.e., Ax = f in V'. The equation Ax = f is solvable for f E V' if and only if A-I 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 -elliptic]. Then Problem

6.5 Bilinear Forms

(7) has exactly one solution x := A-l f. This satisfies C = lie = lie' [resp. C = liCE].

IIxliv

141

~ Cllfllv' with

Corollary 6.5.10. Under the assumptions of Theorem 9 the analogous statement holds with the same estimate for the adjoint problem Find x* E V

PROOF. IIA-lllv+-v'

with a*(x*, y)

= f(y)

= IIA,-lllv+-vl

(cf. Exercise 2a, Theorem 8).

(6.5.8)

for all y E V.

•

Exercise 6.5.11. Let a(·, .): Vx V - m. be continuous. Let Vo be dense in V. Show that the solution x E V of problem (7) is already uniquely determined by "a(x, y) = f(y) for all y E Vo". The same holds for (8). Problem (7) may be equivalent to a variational problem:

Theorem 6.5.12. f E V'. Then

Let a(·,·) be V-elliptic and symmetric; furthermore, let J(x) := a(x,x) - 2f(x)

(6.5.9)

(x E 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(x + y) = a(x + y,x + y) - 2f(x + y) = a(x, x) + a(x, y) + a(y, x) + a(y, y) - 2f(x) = J(x) + a(y, y) + 2[a(x, y) - f(y)] = J(x)

+ a(y, y) ~ J(x) + CEllyll} =

one can read J(z) > J(x) for all z

J(x)

2f(y)

+ CEliz - xII},

•

=f 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 cUe V'

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

Definition 6.5.13. Let V cUe V' be a Gelfand triple. A bilinear form a(·, .) is said to be V-coerci ve if it is continuous and if there exists CK E m. and CE > 0 such that

(6.5.10)

142

6 Tools from Functional Analysis

Exercise 6.5.14. Set a(x,y):= a(x,y) +CK(x,Y)u with C K from (10). Let I: V --+ V' be the inclusion. Show that (a) the coercivity condition (10) is equivalent to the V -ellipticity of a. (b) If A E L(V, V') is associated to a(·, .), then so is A := A + C,.I to a{·, .). Why does A E L(V, V') hold?

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 cUe V' be a Gelfand triple with compact embedding V cU. Let the bilinear form a{., .) be V -coercive with corresponding operator A. Let I: V --+ V' be the inclusion. (a) For each AE CU one of the following alternatives holds:

(A - AI)-l E L{V' , V) (ii) A is an eigenvalue. (i)

and (A' - Xl)-l E L{V', V),

In case (i) Ax - Ax = f and A'x· - 'Xx. = f are uniquely solvable for all f E V' (i.e., a(x, y) - A(X, y)u = fey) and a·(x·,y) -X(x., y)u = fey) for all y E V). In case (li) there exist finite-dimensional eigenspaces {O} =f E{A) := kernel (A - AI) and {O} =f E'(A) := kernel (A' - XI) such that Ax = Ax A'x·

= X*

= A{X, y)u

for x E E{A),

i.e.,

a{x, y)

for x* E E'(X),

i.e.,

a·(x*,y)

= X(x*,y)u

for all y E V, (6.5.11a) for all y E V. (6.5.11b)

(b) The spectrum u(A) of A consists of at most countably many eigenvalues which cannot accumulate in CU. A E u(A) if and only if X E U(A'). Pu.rthermore dimE(A) = dimE'(X) < 00. (c) For A E u{A), Ax - Ax = f E V' has at least one solution x E V if and only f..L E'{A), i.e., (f,x*}v'xv = (f,x*)u = 0 for 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 CK from (10) satisfies A + CKI E L(V, V'), (A + CKI)-l E L(V', V) (cf. Exercise 14 a). Lemma 6.5.4 shows that K := (A + CK 1)-1 I: V --+ V is compact. Hence the RieszSchauder theory is applicable to K - 1'1. Since K -1'1= -1'(1 -1'-lK)

= -1'(A+CKI)-l{A+CKI -1'-lI} = -I'(A + CKI)-l(A - AI)

with A = 1/1' - CK, the statements of Theorem 6.4.12 transfer via K - 1'1 to A - AI = -1'-1 (A + CKI) (K - 1'1). •

6.5 Bilinear Forms

143

Remark 6.5.16. The spectrum u(A) has measure zero so that under the conditions of Theorem 15 the solvability of Ax - AX = f is guaranteed for almost all A. Problem (7) is solvable if not "accidentally" 0 E u(A). Lemma 6.5.17. Under the conditions of Theorem 15 the inequalities (4a,b) are equivalent.

PROOF. (4a) proves that A is injective, Le., 0 f/. u(A). Theorem 15a shows A-l E L(V', V) so that (4b) follows from Lemma 3. • Evidently a(·, .) remains V-coercive if one adds a multiple of (., ·)u. Generally, there holds

Lemma 6.5.18. Let a(·,·) be V-coercive where V cUe V'. Then a(·,·) + b(., .) is also V -coercive if the bilinear form b(., .) satisfies one of the following conditions: (a) for every € > 0 exists Ce such that (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: Ib(x,x)1 ~ CBllxllxllxlly

for all X E V.

(6.5.12b)

(c) Let the embeddings V eX, V c Y be continuous. Let (12b) hold. For II . IIx or II . lIy assume that for every f. > 0 there exists a C~ such that

IIxlix ~ €lIxliv + C~llxliu

or

IIxlly ~

€lIxllv + C~lIxllu

for x E V. (6.5.12c)

= CE/2 with CE from (10). Then a(·,·) +b(.,.) satisfies the V-coercivity condition with C E /2 > 0 and C K + Ce instead of C E 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 IIxllY ~ Cyllxllv. Choose €' = €/(2CB Cy) in (12c): PROOF. (a) Select €

Ib(x,x)1

~ CB(€'lIxliv +C~,"xllu)CYllxllv ~ ~lIxll~ +Kllxllvllxllu

with K = CBCyC~,. Since Kllxllvllxllu ~ ~"xll~ (5.3.10)), (12a) follows.

+ !K2€-1I1xll& (cf. •

7 Variational Formulation

7.1 Historical Remarks In the preceding chapters it was not possible to establish even for the Dirichlet problem of the potential equation (2.1.1a,b) whether, or under what conditions, a classical solution 1.£ E C 2 (fl) n CO(fl) exists. Green [I] 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 Dirichlet offered a different line of reasoning. The Dirichlet integral 1(1.£):=

f

Ja

IVu(x)1 2 dx =

f tu~.(x)dx

Ja i=1

(7.1.1)

describes the energy in physics. With boundary values 1.£ = II' on r given, one seeks to minimise 1(1.£). This variational problem is equivalent to

1(1.£, v):=

fa

(Vu(x), Vv(x»dx = 0 for all v with v

= 0 on r.

(7.1.2)

The proof of the equivalence results from 1(1.£+ v) = 1(u)+21(u, v)+l(v) and l(v) ~ 0 for all v (cf. Theorem 6.5.12). Green's formula (2.2.5a) provides 1(1.£, v) = fa v.::1udx = 0 for all v with v = 0 on r such that .::1u = 0 follows. Thus, like (2), the variational problem 1(1.£) = min is equivalent to the Dirichlet problem .::1u = 0 in n, 1.£ = II' on r. The so-called Dirichlet principle states that l(u), since it is bounded from below by l(u) 2 0, must take a minimum for some u. According to the above considerations this would ensure the existence of a solution of the Dirichlet problem. In 1870, WeierstraB argued against this line of reasoning, stating that while there may exist an infimum of l(u) over {u E C2(fl) n CO(fl):u = II' on r} it need not necessarily be in this set. For example, the integral J(u):= fo1u 2(x)dx in {u E CO ([0, 1]):1.£(0) = 0,1.£(1) = I} never takes the value inf J(u) = O. Further, the following example due to Hadamard shows that no finite infimum of the Dirichlet integral need exist. Let r and II' be the polar coordinates in the circle fl = Kl(O). The function u(r, 11') = :E~=1 r n1 n- 2sin(n!
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_7, © Springer-Verlag Berlin Heidelberg 2010

144

7.2 Equations with Homogeneous Dirichlet Boundary Conditions

145

The above difficulties disappear if one seeks the solutions in the more suitable Sobolev spaces instead of in C 2 (il) n c<'(il).

7.2 Equations with Homogeneous Dirichlet Boundary Conditions In the following we investigate the elliptic equation (7.2.1a)

Lu=g in il,

L

:L :L (-1)I.8ID.8 aa.8(x)Da

=

(7.2.1b)

lal:=;m 1.8I:=;m

of order 2m (cf. Section 5.3; Exercise 5.3.5d). The principal part of Lis

Lo = (_1)m

:L

D.8aa,8{x)Da .

(7.2.2)

lal=I.8I=m

According to Definition 5.3.4, L is uniformly elliptic in il if there exists such that

:L

aa.8(x)ea+.8 ~

flel 2m

for all x E il, e E

f

>0

(7.2.3)

ffin.

lal=I.8I=m

Attention. In the case that only aa,8 E LOO{il) is assumed, one needs to replace "for all x E il" by "for almost· all x E il" . We assume the homogeneous Dirichlet boundary conditions ou= 0 ( - 0 U=o -

'on

'on

)2

U=O ...

( 0 )m-l -

"on

U=O

onT

'

(7.2.4)

which are only meaningful if r = oil is sufficiently smooth. Note that in the standard case m = 1 (an equation of second order) condition (4) becomes u=o.

Since with u = 0 on r 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:

Dau{x)

=0

in x E

r

for

lal ~ 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 formulated as uE HQ{il).

(7.2.4")

Let u E C2m{il)nH(j{il) be a classical solution of (1a) and (4). To derive the variational formulation we take an arbitrary v E Cgo(il) and consider

146

7 Variational Formulation

Since v E C{f(.fl), the integrand vanishes in the proximity of r so that one can integrate by parts:

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

L

1

aa.B(x)(Dau(x))(D.Bv(x)) dx =

lal.I.BI:S;m [J

1

g(x)v(x) dx

(7.2.5)

[J

for all v E

co(n),

since Lu = g. If conversely u E c2m{n) with boundary conditions (4) satisfies condition (5), then the partial integration can be reversed and f[J{g - Lu)vdx = 0 for all v E C{f{n) 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 equivalent with respect to classical solutions. We introduce the bilinear form a(u, v):=

L

1

aa.B(x){Dau)(D.Bv)dx

(7.2.6)

fa

g(x)v(x)dx.

(7.2.7)

lal.I.BI:S;m [J

and the functional f(v) :=

AB remarked above, the boundary condition (4) for classical solutions u E c 2m(n)nHm(n) means that u E H{{'(.fl). Thus the "variational formulation" or "weak formulation" of the boundary value problem (1), (4) reads as follows: find

u E Hrr(.fl)

with a(u, v)

= f(v) for all v

E Coo{n).

(7.2.8)

A solution of problem (8) which, according to the definition, lies in H{{'(n) but not necessarily in c2m(n), is called a weak solution.

Exercise 7.2.1. (a) Let n be bounded. Show that any classical solution u E c2m(n) n cm(.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 aa.B E LOO{n). The bilinear form defined by (6) is bounded on H{{'(n) x H{{'(n).

7.2 Equations with Homogeneous Dirichlet Boundary Conditions

147

PROOF. Let u,v E Goo(n). The inequality (6.2.5c) yields la(u, v)1 ::;

L Ilaa~IILOO(a)IDauloID~vlo ::; const lulmlvlm. a,~

Since Goo(n) is dense in H{f(n) (cf. Theorem 6.2.10), a{-,.) has an extension to H{f(n) xH{f(n» and is bounded by the same constant (cf. Lemma 6.5.1b) .

•

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

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

for all v E HO'(n).

(7.2.9)

One can regain the form Lu = f by applying Lemma 6.5.1. Let L E L(H{f(n), H-m(n» and f E H-m(n) = (H{f(n»' be defined by a(u,v) = (Lu, V)H-m(n)XH(J'(a) and f(v) = (f, V)H-m(a)XH(J'(a) for all v E HO'(n). Equation (9) states that Lu = f. (7.2.9') While (la) represents an equation Lu = 9 in CO(n) (Le., for a classical solution), (9') is an equation in H-m(n). Theorem 6.5.9 guarantees unique solvability of Equation (9) if a(·,·) is H{f(n)-elliptic. We first investigate the standard case m = 1 (equations of order 2m = 2). Theorem 7.2.3. Let n be bounded, m = 1, aa~ E LOO(n). Let L satisfy (3) (uniform ellipticity) and be equal to the principal part Lo, i.e., aa~ = 0 for lal + 1.81::; 1. Then the form a(·,·) is HJ(n)-elliptic: a(u,u);::: t'lul~,

f'

> O.

(7.2.10)

PROOF. Since lal = 1.81 = lone can identify a and.8 according to Da = 8/8xi, D~ = 8/8xj with indices i,j E {I,,,,, n}. For fixed x E n use (3) with ~ = Vu(x):

L lal,I~I=l

n

aa~(x)~a+~ =

L

aij(x)~i~j ;::: fl~12 = fIVu(x)12.

i,j=l

Integration over n yields a(u, u) ;::: fIn IVul 2 dx. Since In IVul 2 dx ;::: Gnluli (cf. Lemma 6.2.11), (10) follows with f' = fGn. • Corollary 7.2.4. The condition "n bounded" may be dropped if for a .8 = 0 one assumes aoo(x) ;::: 1J > 0 (instead of aoo = 0). Example 7.2.5. The Helmholtz equation -Llu the bilinear form

+u

=

=

f in n leads to

148

7 Variational Formulation

a(u, v):=

1rf: a

uz.(x)vz.(x) + u(x)v(x)] dx =

i=1

1 a

[(Vu, Vv)

+ uv] dx.

a(u,v) is the scalar product in HJ(D) (and H1(D)). The fact that a(u,u) = lul~ proves the HJ(n)-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 0.0{3 (Ial = IPI = 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 oj the principal part be constants: aa{3 = const Jor lal = IPI = m. FUrthermore, assume that aa{3 = 0 Jor 0 < lal + IPI :$ 2m - 1, aoo ~ 0 Jor a = P = O. Let L be uniJormly elliptic (cJ. (3)). F'u.rther let either n be bounded or aoo ~

'1 > O.

Then a(·,·) is Hlf(D)-elliptic.

PROOF. We continue u E Hlf(n) through '1.£ = 0 onto JRn • Theorem 6.2.21, Exercise 6.2.22, and inequality (3) show that a(u, '1.£)

-1

aoou2 dx =

a

L

1

L

aa{3

aaf3DauIJf3udx

lal,I{3I=m a lal,I{3I=m =

L

J.

D a uD{3udx

R"

aaf3

L)(ie )au (e)][(i e ){3u(e)]d{

lal,If3I=m

=

J. [ L

aa{3ea+{3l1u(e) 12 d{

R" lal,If3I=m

~ E r leI 2m lu(e)1 2 d{. lIt" Let 0.00 ~ '1 > O. There exists an E' > 0, so that Elel 2m ~ E' E~al::;m lea l2 - '1 for all e E JRn . From this follows EJ leI 2m lu(e)1 2d{ ~ E'lul m - '11'1.£16 and a(u,u) ~ E'lul~ (cf. Lemma 6.2.23). If n is bounded, use Lemma 6.2.11. • Having shown the Hlf(n)-ellipticity of the form a(·,·), we are now able to apply the general proofs after Theorem 6.5.9.

Theorem 7.2.8.

(Existence and uniqueness of weak solutions)!J a(·,·) is H{f(n)-elliptic then there exists a solution u E Hlf(D) oj Problem (9) which satisfies 1

lul m:$ CE IJI-m

(CE from (6.5.5)).

(7.2.11)

7.2 Equations with Homogeneous Dirichlet Boundary Conditions

Since (11) holds for all f E H-m(a) and u is equivalent to

= L-1 f

149

(cf. (9')), inequality (11)

(7.2.11')

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

Theorem 7.2.9. Let a(·,·) be an H{f(a)-elliptic and symmetric bilinear form. Then (9) is equivalent to the variational problem find where

u E HQ'(a),

such that

J(u)::; J(v)

for all v E HQ'(a), (7.2.12a)

1 J(v) := 2a(v, v) - f(v).

(7.2.12b)

Attention. If a(·,·) is either not Ho(a)-elliptic or not symmetric, Problem (9) remains meaningful although the solution does not minimize the functional J(u). Example 7.2.10. The Poisson equation -Llu = f in a, u = 0 on leads to a(u,v) = In{Vu(x), Vv (x» dx. For a bounded domain a, a(.,.) is HJ(n)-elliptic (cf. Theorem 3), so that for any f E H-1(a) there exists exactly one (weak) solution u E HJ(a) of the Poisson equation. This is also the solution of the variational problem! In IVul 2 dx - feu) = min.

r,

A weaker condition than Ho(a)-ellipticity is the Ho(a)-coercivity: a(u,u) ~ €Iul~ - Glula.

= 1, and let the coefficients a"'{3 E LOO(a) satisfy condition (3) of uniform ellipticity. Then a(·,·) is HJ(a)-coercive.

Theorem 7.2.11. Let m

PROOF. We write L as L = LI + L Il , with LI satisfying the conditions of Theorem 3, resp. Corollary 4 if a is not bounded, and LII containing only • derivatives of order ::; 1. Then we can apply the following lemma.

Lemma 7.2.12. Let a(·, .) = a' (-, .) + a" (-, .) be decomposed such that a' (', .) is HQ(ll)-elliptic, or perhaps only H{f(ll)-coercive, while a"(u,u)

=

L lal.I.BI$m lal+I.BI<2m

1

a",{3D"'uD{3udx

n

with a",{3 E LOO(.o) contains only derivatives of order::; 2m - 1. Then a(·,·) is also H{f(.o)-coercive.

150

7 Variational Formulation

PROOF. (6.5.12c) follows from (6.2.lOb) with X = Hbal(n), Y = H)fl(n), V = H{f(n), and U = L2(n), so that Lemma 6.5.18c proves the assertion. • The generalisation of Theorem 11 to arbitrary m conditions on the coefficients of the principal part.

~

1 requires stronger

Theorem 1.2.13. (Garding's theorem) Let L be uniformly elliptic (cf. (3)) and assume aa{3 E LOO(n). Purthermore, let the coefficients aa{3 with lal = 1,81 = m be uniformly continuous in n. Then a(·,·) is H[{'(n)-coercive. If conversely aa{3 E C(n) holds for lal = 1,81 = m and aa{3 E LOO(n) otherwise, then from the H{j(n)-coercivity follows uniform ellipticity (3). Details of the proof can be found in Wloka [1, Theorem 19.2]. The proof given there also holds for unbounded n, 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 1.2.14. Let n be bounded and a(·,·) be H{j(n)-coercive. Then one of the following alternatives holds: (i) Problem (9) has exactly one (weak) solution 11. E H{j(n). (ii) The kernels E = kernel(L) and E* = kernel(L') are k-dimensional for a k E IN, i.e., a(e,v) = 0,

a(v,e*) = 0

for all e E E,e* E E*,v E H{j(n).

Purther, the eigenvalue problem a(e, v) = .>.(e, V)L~(a)

for all v E HO'(n)

(7.2.13)

has countably many eigenvalues which do not accumulate in CIJ.

PROOF. Since for bounded n the embedding V:= H{j(n) c U := L2(n) is compact (cf. Theorem 604.8b), Theorem 6.5.15 is applicable. •

7.3 Inhomogeneous Dirichlet Boundary Conditions Next, we consider the boundary value problem Lu = g in

n,

11.

=
on

r,

(7.3.1)

where L is a differential operator of second order (Le., m = 1). The variational formulation of the boundary value problem reads: find

11.

E

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

11.

=
on r

for all v

E

such that HJ(n).

(7.3.2a) (7.3.2b)

7.3 Inhomogeneous Dirichlet Boundary Conditions

151

But according to Section 6.2.5 the restriction ulr of u E Hl(ll) on r is well defined as a function in Hl/2(r). Thus "u = cp on r" must be understood as the equality ulr = cp in Hl/2(r). In contrast to the preceding section one uses a(·,·) in (2b) as a bilinear form on Hl(ll) x HJ(ll). 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) it is necessary that: there exists a UO E Hl(ll)

with uolr

= cpo

(7.3.3)

If a function Uo with Property (3) is known, a second characterisation of the

weak solution results: Let Uo

satisfy (3);

find

W

a(w, v) = f'(v) := f(v) - a(uo, v)

E

HJ(ll), such that

(7.3.4a)

for all v E HJ(ll).

(7.3.4b)

Remark 7.3.2. The variational problems (2a,b) and (4a,b) are equivalent. If and ware the solutions of (4a,b), then u = Uo +w is a solution of Problem (2a,b).1f u is a solution of (2a,b), then, for example, UO = u and w = 0 satisfy Problem (4a, b).

Uo

Exercise 7.3.3. Show that

f'

E H-1(ll) for

f' from (4b) and

If'1-1 ~ Ifl-1 + Gsluoh

(7.3.5)

with Gs from la(u,v)\ ~ Gsluhlv\1 (cf. (6.5.1)). Remark 7.3.4. The problem (1) and the variational formulation (2a,b) have the same classical solutions if such exist. PROOF. It suffices to assume v E C~(ll) in (2b). Integration by parts can • be carried out as in Section 7.2 and proves the assertion.

Theorem 7.3.5. (Existence and uniqueness). Let problem (2.9) (with homogeneous boundary values) be uniquely solvable for all f E H-1(ll). Then Condition (3) is sufficient, and necessary, for the unique solvability of Problem (2a,b).

PROOF. If there exists a solution u E H1(ll) 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). • Remark 7.3.6. Under the condition

n E CO,l, (3) is equivalent to

cp E Hl/2(r).

(7.3.6)

152

7 Variational Formulation

PROOF. If Uo satisfies Condition (3) then Theorem 6.2.40 shows that cp E H1/2(r). If conversely cp E H1/2(r), then the same theorem guarantees an extension UO E Hl(n) to fl with Uolr = cp and (7.3.7)

•

Let inequality (2.11') hold in the case of homogeneous boundary values. Equation (5) shows lult 5 luoll + Iwlt 5 luoh + (1/1-1 + C'luoh)/f for the solution of Problem (4a,b). The estimate (7) proves Theorem 7.3.7. Let n E CO· 1. Let the bilinear lorm be restricted to Hl(n) x HJ(fl) and let it satisfy (2.11'). Then lor every IE H-1(n) and cp E H 1/2(r) there exists exactly one solution u E Hl(fl) 01 Problem (2a,b) with (7.3.8)

Exercise 7.3.8. Let a(·,·) be symmetric and HJ(n)-elliptic. Show that Problem (2a,b) with I = 0 is equivalent to the variational problem: Find u E H1(fl) with ulr = cp such that a(u, 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(fl) x Hm(fl). In analogy to Theorem 7.2.2 there holds Theorem 7.4.1. Assume aal3 E LOO(n). The bilinear lorm defined by (2.6) is bounded on Hm(n) X Hm(.fl): la(u, v)1 5 Ea.l3l1aaI3I1Loo(o)IUlmIVlm lor all u,v E Hm(fl). Now let I be a functional from (Hm(fl))'. Equation (2.7) with U E L2(n), 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 U E L2(n), cp E L2(r) hold. Show that f(v):= iau(x)v(x)dx+

i

cp(x)v(x) dr,

(7.4.1a)

defines a functional in (Hl(fl»' with IIfll(Hl(O»1 5 C(lIullp(S1) + IIcpIlL2(r». This implies 1 E (Hm(fl»' for all m > 1. More precisely, the following also holds:

7.4 Natural Boundary Conditions

IIfll(Hl(O»' :::; C(lIglI(Hl(O»'

+ IIcpIlH-l/2(r»·

153

(7.4.1b)

Frequently, variational problems with a physical background have the form: find u E Hm(n),

a(u,v) = f(v)

such that

for all v E Hm(n).

(7.4.2a) (7.4.2b)

In contrast to the condition u E Hlf(n) from Section 7.2, u E Hm(n) contains no boundary condition. Nevertheless, Problem (2a,b) has a unique solution if a(·,·) is Hm(n)-elliptic. This condition is easy to satisfy. Theorem 7.4.3. Under the conditions of Theorem 7.2.3 or Corollary 7.2.4 a(·,·) is Hl(Q)-elliptic: a(u,u) 2: flul~ for all u E Hl(n). Problem (2a,b) (with m = 1) has exactly one solution which satisfies the estimate (3): (7.4.3)

PROOF. The same as for Theorem 7.2.3 or Corollary 7.2.4, and Theorem 7.2.8. • Corollary 7.4.4. (a) A unique solution which satisfies the estimate (3), also exists if instead of Hl(n)-ellipticity one assumes: a(·,·) is H1(n)-coercive, n E CO,1 is bounded, A = 0 is not an eigenvalue (i.e., a(u, v) = 0 for all v E H1(n) implies u = 0). (b) The combination of inequalities (3) and (1b) results in (7.4.3')

for the solution of (2a,b), if f is defined by (1a) with g E (H1(n))', cp E H-1/2(r). PROOF. (a). According to Theorem 6.4.8b, H1(n) is compactly embedded in £2(Q) so that the statement of Theorem 7.2.14 can be transferred. If A = 0 is not an eigenvalue then £-1 E £«Hl(n»)', H1(n» holds (cf. Theorem 6.4.12). • To find out which classical boundary value problem corresponds to the variational formulation (2a,b), we assume that (2a,b) has a classical solution U E Hm(n) n c2m(Q). Further, v E COO(n) can be assumed also (cf. Lemma 6.5.1b). For reasons of simplicity we limit ourselves to the case m = 1. Under the assumption aa{3 E C1(n) and under suitable conditions on n the following general Green formula is applicable:

154

7 Variational Formulation

a( u, v)

= =

r [t G.tjUill, Villi + t i,j=l

Jn

In [+

aoiUVill,

i=1

+

t

G.toUa;, V + aoouv] dx

,=1

,t1(G.tjUill .)illi - t(lZoiU)ill'

+ t£lioUilli +aooU]Vdx

l [~njG.tjUill' + ~~tzoiu]vdr. s"

(17.4.4)

s

Here, the ~ are the components of the normal direction n define the boundary differential operator

= n(x), x e r. We

nan njG.tj- + L~lZoi, , '-1 ax, s-'-1 ',3-

B:= L

(7.4.5)

when L is described by (2.1b). Equation (4) becomes a(u,v) = InvLudx+ IrvBudr. By the formulation of the problem, a(u, v) agrees with f(v) from (Ia). If we first choose V e HJ(.!J) c Hl(.!J) the boundary integrals drop out and we obtain Lu = 9 as in Section 7.2. By this the identity a(u,v) = f(v) reduces to IrvBudr = Ir«pVdr for all v e H1(.!J). According to Theorem 6.2.40b, vir runs over the set H1/2(r) if v runs over Hl(.!J), so that we have Ir 1/J(Bu - 'P) dr = 0 for all1/J e H1/2(r); thus Bu = 'P. This proves

Theorem 7.4.5. Let r be sufficiently smooth. A classical solution of the problem (2a,b) with f from (1a) is also the classical solution of the boundary value problem (7.4.6) Lu = 9 in n, Bu = 'P on r, and conversely.

The condition Bu = 'P 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 -Llu = 9 has as the natural boundary condition the Neumann condition -au/On = o. Theorem 7.4.7. Let n e CO,l be bounded and a(·,·) be W"'(.!J)-coercive. Then the statements of Theorem 7.2.14 holds with Hm(.!J) instead of H{?(.!J). Example 7.4.8. Let n e CO,l be a bounded domain. The bilinear form a(u, v) = In( (,Vu, Vv) + cuv)dx, associated to the Helmholtz equation -Llu + cu = f

in

n

(c

> 0), au/an = 0 on r,

is H1(.!J)-elliptic since a(u, u) ~ min(1, c)lul~. For c = 0, however, a(·,·) is only H1(.!J)-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 eigenspace E = kernel(L). u E E satisfies a(u, u) = 0, thus Vu = O. Since fl is connected, it follows that u(x) = const 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) = f(v) (v E Hl(fl)) is solvable if and only if f .L E, i.e., f(l) = O. If f(v) := Ing(x)v(x)dx, f(l) = 0 reads as In gdx = O. If, however, f is given by (la), the integrability condition reads f(l) = In 9 dx + Ir cpdr = 0 (this is Equation (3.4.2), in which f should be replaced by -g).

Remark 7.4.9. While the classical formulation of a boundary condition such as au/an = 0 requires conditions on the boundary r, the problem (2a,b) can ba formulated for arbitrary measurable fl. In the following, we proceed in the opposite direction: does there exist, for a classically formulated boundary value problem Lu = g in fl, Bu = cp on r, with given Land 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

t; bj(x) a + bo(x) n

B

=

aXi

(x E r).

(7.4.7)

With b T = (bl, .. ·, bn) one can also write B = bTV +bo (cf. (5.2.1b,b')). Here bTV is not allowed to be a tangential derivative (cf. Remark 5.2.2):

(b(x), n(x)}

=f 0

for all xE

r.

(7.4.8)

Remark 7.4.10. Let m = 1. Let (8) hold. Let A(x) be the matrix A = (aij) (cf. (5.1.1d)). By passing from Bu = cp to the equivalent scaled equation u Bu = ucp with u(x) = (n(x), A(x)n(x)} / (n(x), b(x)}, one can ensure that (n,ub) = (n, An). Thus in the following it is always assumed that b already satisfies (n, b) = (n, An). PROOF. Because of (8) u is well defined. For Bu = cp and uBu = ucp to be equivalent, u =f 0 is required. This is guaranteed by the uniform ellipticity: (n, An) ~ elnl 2 = e.

Theorem 7.4.11.

(Construction of the bilinear form). Let m = 1.

Let Land B be given by (2.1b) and (7), with b satisfying condition (8). Then there exists a bilinear form a(.,.) on Hl(fl) x Hl(fl) such that to the variational problem (2a,b) co'TTesponds the classical formulation Lu = g in fl, Bu = cp on r.

156

7 Variational Formulation

PROOF. The bilinear form we seek is not uniquely detennined . 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 P(x) := (P1(X),· . " Pn{x)) E C1(0) be arbitrary. The differential operator

maps every u e C1(n) into zero: L11.1. = O. 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 + L 1 . Equation (5) shows that the boundary operator associated with a(·,·) has the absolute term (7.4.9) i=1

If bo = 0, the choice of Pi. = -aoi is successful. Otherwise, two other options are available. (aa) Select Pi such that on r the following holds: (3;(x) = bo(x)n,(x) -aoi(X). Since Inl = 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 n of the boundary values Pi(X), x e r . (ab) Set Pi = -aoi and add a suitable boundary integral:

a(u, v) :=

In {~atju:r:.v:r:j+ ~(ato '", +

i

- ao,)u:r:. v +

bo{x)uv dr.

[aoo - ~(ao,):r:, ]uv} dx ,

(7.4.1O)

The integration by parts described above shows that the boundary operator associated to (10) reads (7.4.11) (b) The operator (11) can be written in the form iJ = j)TV +bo with j) = An. Since (n, b) = (n, An) = (n, b) has been assumed already, d := b - b is orthogonal to n. To change j) to b there are again two options. (ba) Define the n x n-matrix A 8 on r by A8 = dnT - ndT , Le., a:j = dtnj - nidj. This A8(x) is skew-symmetric: A8T = _A8. Continue AS(x), which is at first only defined on r, to a skew-symmetric matrix AS E C1(n). [Here we have the same practical difficulty as in step (aa).] The entries of AS define

7.4 Natural Boundary Conditions

Again, L 2 u

157

= 0 holds for all u E G 2 {il), since

Thus L can be replaced by L + L2 without changing the boundary value problem. The coefficients belonging to L + L2 result in a boundary operator B whose derivative terms read Li,j nj{a;,j + aij)a/axi = [(A + A 8 )n]TV. By the construction of AS we have ASn = (dnT - ndT)n = d = b - b , since (n, n) = 1 and (d, n) = O. Since we also have An = b, the derivative term in B gives bTV as desired. The transition L ~ L + L2 does not change the absolute term in 8, so that from (11) follows B = bTV + boo (bb) Let iJ = bTV + bo 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 ;::: 3]. If r is sufficiently smooth then the restriction vir of v E H1(il) is an element of H1/2(r). By Remark 6.3.14a, one can show that T E L(H1/2(r), H-1/2(r)). Since T(ulr) E H-1/2(r), fr .,pT(ulr) dr) is well-defined for.,p E Hl/2(r), in particular, for .,p = vir with v E H1(il). Thus b(u,v):= £(v1r)T(u1r)dr is a bilinear form bounded on H1(il) x Hl(il). We add b(.,·) to a(·,·) in (10). Integration by parts yields the boundary operator iJ + T = bTV + bo+ (b - b)TV = bTV + bo = B. • Remark 7.4.12. Let the bilinear form (2.6) be Hl(il)-coercive (cf. Theorem 3). Let its coefficients, as well as the boundary r, be sufficiently smooth (E 0 1 ). Then the constructions of the preceding proof again result in an B7jH1 (il)-coercive 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) := fa Li)aijuz.vzj + (atj)zjvuz.ldx. Here Lemma 7.2.12 is also applicable, since the skew symmetry of AS results in b(u, u) = fa L(atj)zjuuz,dx. (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 r be described by {(X1(S),X2(S)): 0 :5 s :5 I}. If X1,X2 E G1([O,1]) and x~(O) = x~(1), and if we have d E Gl(r) for the d in T = dTV, then b(.,.) has the representation b(u,v) = f01v(s)r{s)u'(s)ds, where u(s) = U(X1(S),X2(S)), v(s) = V(X1(S),X2(S)), r E G1{[0,1]). Thanks to periodicity, integration by parts yields in b{u, v) = - f~ (rv)'u ds without boundary terms, so that b( u, u) =

! [f01uru' ds - f~ (ru)'u dS] = -! f01 r'u2 ds. This implies

158

7 Variational Formulation Ib(u, u)1 !S

~lIrIlOl([0.1)) lIulrlll~(r) !S 01Irllol([0.1]) lIullt-'(n)

for all 8 E (1/2,1) (cf. Theorem 6.2.4080). Since lul~ !S apply Lemma 6.5.18c.

flul1- Clul~, one can •

The case m ~ 2 has been excluded in this section (except for Theorem 1). Boundary value problems of order 2m require m boundary conditions Bju = CPj on r (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 19.3], Lions-Magenes [1, p. 210]). The resulting complications can be seen with the aid of the biharmonic equation. Example 7.4.13. (a) To the variational problem: find u E H2(fl) with

a(u, v) := L L1uL1vdx = f(v) := LgvdX+

£

(7.4.1280)

(CP1:: -CP2 v )dr

for all v E JI2(fl), corresponds the classical formulation

.12 u

= 9 in fl,

L1u = CP1

and

a L11£ = CP2 an

(7.4. 12b)

on r.

But the bilinear form a(·,·) is not H 2 (fl)-coercive. (b) To the variational problem: find 1£ E H2(n) n HJ(fl) with

a(1£, v) = f(v):= L gvdx+

£

CP:: dr for all v E H2(fl)

n HJ(fl) (7.4.12c)

(a( " .) as in (1280» corresponds the classical formulation Ll 2u

=9

in fl,

1£ = 0 and .11£ = cP on

r.

(7.4.12d)

The bilinear form is H2(fl) n HJ(fl)-coercive. (c) The boundary conditions in

L1 21£

=9

in fl,

a an

1£ = 0 and -L11£ = cP on

r

(7.4.12e)

are admissible. Nevertheless this boundary value problem cannot be written in the present form as a variational problem. A variational formulation for (12e) reads: Find u E H2(fl) n HJ(fl) such that

a(1£,v)=f(v):= LgvdX- £cpvdr

forallvEH 2(fl) with

::lr=O (7.4.12f)

7.4 Natural Boundary Conditions

159

(a(.,.) as in (12a». But this does not agree with the present concept since and v belong to different spaces.

U

PROOF. The equivalence of the variational and the classical formulation can be shown via integration by parts:

a(u, v)

=

In

vLl 2udx +

1

[LlU:: - 8!u v] dr.

The noncoercivity in part (a) results as follows. Let il c IRn be bounded. For all a E IR, Ua(Xl,··· ,xn ) = sin(axl)exp(ax2) lies in H2(il) and satisfies Llua = 0, hence also a(u a , u a ) = o. If a(·,·) were coercive, there would exist a C with 0 = a(ua,ua ) ~ Elua l2 - Clualo for all a, i.e., lua l2 ::; (C/E)lualo. The contradiction results from lua l2 ~ 182ua/ax~lo = a 21ua lo for sufficiently • large a. Natural and Dirichlet conditions can occur together. In Example 13b, u

=

o is a Dirichlet condition and (aLlu/8n) = cp a natural boundary condition. Even in the case m = 1 both sorts of boundary conditions can occur.

Example 7.4.14. Let "( be a nonempty, proper subset of r. The boundary value problem -Llu = 9

in il,

u = 0 on ,,(,

au/8n

= cp

on r\,,(

(7.4.13)

in the variational formulation reads as follows: find u E H~(il) such that

a(u, v):=

r('\Iu, '\Iv) dx

Jn

= f(v):=

rgvdx + Jr

Jn

cpvdr

r\"r

for all v E H~(il), where H~(il) := {u E Hl(il): u = 0 on "(}. Equation (13) is occasionally termed a Robin problem.

Exercise 7.4.15. Let a(u,v) := In [(V'u, V'v} +cuv] dx with c > 0 on V x V with V:= {u E Hl(il):u constant on r} be defined. Show that (a) a(·, .) is V-elliptic. (b) The weak formulation: u E V, a( u, v) = In gv dx + Ir cpv dr for all v E V corresponds to the problem -Llu+ cu

=9

in .0,

u constant on

r,

l ~: = l dr

cpdr.

(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 9 and cp of the differential equation and the boundary condition are combined in the functional f. In the variational formulation the components 9 and cp are

160

7 Variational Formulation

indistinguishable! u E HI(n) has first derivatives in L2(a), whose restrictions to r do not have to make sense. That is why Bu cannot be defined in general; Bu =