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### Transcript of Variational Formulation of Boundary Value Problems · PDF fileChapter 4 Variational...

• Chapter 4

Variational Formulation ofBoundary Value Problems

4.1 Elements of Function Spaces

4.1.1 Space of Continuous Functions

N is a set of non-negative integers.

1) An n-tuple = (1, , n) Nn is called a multi-index.

2) The length of is

|| :=n

j=1

j.

3) 0

= (0, , 0).

Set D := (/x1)1 (/xn)n .

Example 4.1.1. Assume n = 3, = (1, 2, 3) N3, u(x1, x2, x3) : R3 R.

29

What is

||=3 Du ?

|| = 3 =3

j=1

j = 3.

= = (3,0,0), (0,3,0), (0,0,3), (2,1,0), (2,0,1), (0,2,1),(1, 2, 0), (1, 0, 2), (0, 1, 2), (1, 1, 1).

=

||=3

Du =3u

x31+

3u

x32+

3u

x33+

3u

x21x2+

3u

x21x3+

3u

x22x3

+3u

x1x22+

3u

x1x23+

3u

x2x23+

3u

x1x2x3.

This sort of list can get very long. Hence D is useful notation.

Definition 4.1.2. Let be an open set in Rn. Let k N. Define spaces Ck(),Ck() and C() by

Ck() := {u : R | Du is continuous in for all || k},Ck() := {u : R | Du is continuous in for all || k},

C() := {u : R | Du is continuous in for all Nn},

where is the closure of . If is bounded, = , where is the boundaryof . We denote C() = C0() and C() = C0().

Example 4.1.3. Set I := (0, 1) and u(x) := 1/x2 x I. Then clearly for allk 0 u Ck(I). However, in I = [0, 1] u is not continuous at 0. Thus, u / C(I).

Definition 4.1.4. For a bounded open set Rn, k N and u Ck(), thenorm uCk() is defined by

uCk() :=

||k

supx

|Du(x)|.

Example 4.1.5. Let I = (0, 1), u(x) := x, u C(I). Then, supxI |u(x)| = 1.

Definition 4.1.6. For an open set Rn and u C() the support of u denotedby supportu ( Rn) is defined by

supportu := the closure of {x | u(x) *= 0}.

Remark. The support of u is the smallest closed subset of such that u = 0 in\supportu.

30

• Example 4.1.7. 1) Let 0 = x0 < x1 < < xn = 1 be a patition of [0, 1]. Definej(x) : [0, 1] R by

j(x) :=

x xj1h

x (xj1, xj),xj+1 x

hx (xj , xj+1),

0 elsewhere.

Then we see j C(I) and supportj = [xj1, xj+1].

2) Define w(x) : Rn R by

w(x) :=

e

11 |x|2 |x| < 1,

0 otherwise.

Then we see w C(Rn) and supportw = {x Rn | |x| 1}.

Definition 4.1.8. Define Ck0 () ( Ck()) by

Ck0 () := {u Ck() | supportu is a bounded subset of }.

4.1.2 Spaces of Integrable Functions

Definition 4.1.9. Let denote an open subset of Rn and assume 1 p < . Wedefine a space of integrable functions Lp() by

Lp() :={

v : R |

|v(x)|pdx < +

}.

The space Lp() is a Banach space with norm Lp() defined by

vLp() =(

|v(x)|pdx

)1/p.

Especially the space L2() is a Hilbert space with inner product , L2() definedby

u, vL2() :=

u(x)v(x)dx

and norm L2() defined by uL2() :=u, vL2().

31

We have Minkowskis inequality as follows. For u, v Lp(), 1 p <

u + vLp() uLp() + vLp().

We also have Holders inequality. For u Lp() and v Lq(), 1 p, q < with1/p + 1/q = 1

u(x)v(x)dx

uLp()vLq().

Now, any two integrable functions are equal if they are equal almost everywhere,that is, they are equal except on a set of zero measure. Strictly speaking, Lp()consists of equivalent classes of functions.

Example 4.1.10. Let u, v : (1, 1) R be

u(x) ={

1 x (0, 1),0 x (1, 0], v(x) =

{1 x [0, 1),0 x (1, 0),

The functions u and v are equal almost everywhere, since the set {0} where u(0) *=v(0) has zero measure in the interval (1, 1). So u and v are equal as integrablefunctions in (1, 1).

Suppose that u Ck(), where is an open set of Rn. Let v C0 (). Then wesee by integration by parts

Du(x)v(x)dx = (1)||

u(x)Dv(x)dx,

where || k.

Definition 4.1.11. A function u : R is locally integrable if u L1(U) forevery bounded open set U such that U .

Definition 4.1.12. Suppose u : R is locally integrable and there is a locallyintegrable function w : R such that

w(x)(x)dx = (1)||

u(x)D(x)dx

for all C0 ().

Then the weak derivative of u of order denoted by Du is defined by Du = w.

32

• Note that at most only one w satisfies (4.1.12) so the weak derivative of u is well-defined. Indeed, the following DuBois-Raymond lemma shows such w is unique.

Lemma 4.1.13. (DuBois-Raymond) Suppose is an open set in Rn and w : Ris locally integrable. If

w(x)(x)dx = 0

for all C0 (), then w(x) = 0 for a.e x .

We will useD for both classical and weak derivatives.

Example 4.1.14. Let = R. Set u(x) = (1 |x|)+, x , where

(x)+ :={

x x > 0,0 x 0.

Thus,

33

u(x) :=

0 x 1,1 + x 1 x 0,1 x 0 x 1,0 1 x.

!

"y

x##

###

\$\$

\$\$\$

-1 10

1

Clearly we see that u is locally integrable, u C() and u / C1(). However, itmay have a weak derivative. Take any C0 () and = 1. Then,

(1)||

u(x)D(x)dx =

u(x)(x)dx

= 1

1(1 |x|)(x)dx

= 0

1(1 + x)(x)dx

1

0(1 x)(x)dx

= 0

11 (x)dx +

1

0(1)(x)dx

=

w(x)(x)dx,

where

w(x) :=

0 x < 1,1 1 < x < 0,1 0 < x < 1,0 1 < x.

Here we do not worry about the points x = 1, 0, 1, since they have zero measure.Thus, u has its weak derivative Du = w.

Definition 4.1.15. Let k be a non-negative integer and p [0,). The spaceW k,p() defined by

W k,p() := {u Lp() | Du Lp() || k}

34

• is called a Sobolev space. It is a Banach space with the norm

uW k,p() :=

||k

DupLp()

1/p

.

Especially, when p = 2, we denote Hk() as W k,2(). It is a Hilbert space with theinner product

u, vHk() :=

||k

Du, DvL2().

Of special interest are H1() and H2(). If = (a, b), R, we see that

u, vH1() = u, vL2() + Du,DvL2()

= b

au(x)v(x)dx +

b

u, vH2() = u, vL2() + Du,DvL2() + D2u, D2vL2()

= b

au(x)v(x)dx +

b

b

Remark. 1) By using Holders inequality, we can prove Cauchy-Schwarz inequalityfor the inner product of Hk() as follows.

|u, vHk()|

||k

|Du, DvL2()|

||k

DuL2()DvL2()

||k

Du2L2()

||k

Dv2L2()

= uHk()vHk().

2) Let = (a, b) R and u H1(). Then u C(). In higher space dimensionsthis statement is no longer true.

4.2 One Dimensional Problem: Dirichlet condition

Let = (0, 1), p(), q() C() and f() L2(). Note that = {x = 0} { x =1}. We consider the following problem.

35

Find u : R such that{ d

dx(p

du

dx) + qu = f, x ,

u(x) = 0, x .(BVP)

Specifying the value of u at boundary points is said to be a Dirichlet boundarycondition. Now the methodology is

1) multiply the equation by a test function, integrate by parts and use boundaryconditions appropriately,

2) identify V , a(, ) and l(),

3) verify, if possible, the assumptions of Lax-Milgram.

= Unique existence to the variational formulation of the BVP.

Let : R be sufficiently smooth. We will call our test function. Let us followthe methodology.

1)

f(x)(x)dx =

( d

dx(p(x)

du(x)dx

)(x) + q(x)u(x)(x))

dx

=[p(x)du(x)

dx(x)

]x=1

x=0

+

(p(x)

du(x)dx

d(x)dx

+ q(x)u(x)(x))

dx.

We want to eliminate the term [p(x)du(x)/dx(x)]x=1x=0, so we suppose that thetest function satisfies the same Dirichlet conditions as u, i.e, (0) = (1) = 0.Then we have that

(p(x)

du(x)dx

d(x)dx

+ q(x)u(x)(x))

dx =

f(x)(x)dx

for any test function . We want u, v to be from the same space. For the term udx to make sense, we need u, L

2(). For the derivatives du/dx, d/dxto make sense, we take this further, so u, H1().

2) Let us chooseV :=

{ H1() | v(0) = v(1) = 0

},

36

• whereH1() =

{ L2() | D L2()

}.

We equip the inner product , V := , H1(). Let us define

a(u, v) :=

(pDuDv + quv)dx,

l(v) :=

fv dx.

Moreover, assume that p(x) p0 > 0, q(x) q0 > 0 for all x .

3) We will verify the assumptions of Lax-Milgrams theorem.

i) For V and f L2(), we see by Cauchy-Schwarz inequality that

|l()| = |

fdx|

fL2()L2() fL2() (2L2() + D

2L2())

= clV ,

where we have set cl := fL2(). Thus, l : V R is bounded. Clearly l islinear, i.e, l( + ) = l() + l() for any , V and , R.

ii) Obviously a(, ) : V V R is bilinear. Moreover a(, ) is bounded. Indeed,

|a(, )| |

pDDdx| + |

qdx|

maxx

|p(x)|

|DD|dx + max

x|q(x)|

||dx

maxx

|p(x)|DL2()DL2() + maxx

|q(x)|L2()L2()

C(DL2()DL2() + L2()L2())

CD2L2() +

2L2()

D2L2() +

2L2()

= CH1()H1(),

where we have set

C := max{maxx

|p(x)|,maxx

|q(x)|}.

37

The bilinear form a(, ) is coercive, since for all V

a(, ) =

p|D|2dx +

q||2dx

p0

|D|2dx + q0

||2dx

= C2V ,

where we have set C := min{p0, q0}.

We can now apply Lax-Milgrams theorem to see that there uniquely exists asolution to the following problem

(P) Find u V such that

(pDuD + qu)dx =

fdx,

for any V .

Remark. For V = H10 () we can use the norm V defined by

2V =

|D|2dx,

since the following Poincare inequality holds:- there exist