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 D!u ?

|!| = 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

D!u ="3u

"x31

+"3u

"x32

+"3u

"x33

+"3u

"x21"x2

+"3u

"x21"x3

+"3u

"x22"x3

+"3u

"x1"x22

+"3u

"x1"x23

+"3u

"x2"x23

+"3u

"x1"x2"x3.

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 | D!u is continuous in ! for all |!| $ k},Ck(!) := {u : ! " R | D!u is continuous in ! for all |!| $ k},

C!(!) := {u : ! " R | D!u 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 )u)Ck(!) is defined by

)u)Ck(!) :=!

|!|"k

supx#!

|D!u(x)|.

Example 4.1.5. Let I = (0, 1), u(x) := x, u ! C(I). Then, supx#I |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]. Define#j(x) : [0, 1] " R by

#j(x) :=

#$$%

$$&

x+ xj$1

hx ! (xj$1, xj),

xj+1 + x

hx ! (xj , xj+1),

0 elsewhere.

Then we see #j ! C(I) and support#j = [xj$1, 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

)v)Lp(!) =*(

!|v(x)|pdx

+1/p

.

Especially the space L2(!) is a Hilbert space with inner product -·, ·.L2(!) definedby

-u, v.L2(!) :=(

!u(x)v(x)dx

and norm ) ·) L2(!) defined by )u)L2(!) :=,-u, v.L2(!).

31

We have Minkowski’s inequality as follows. For u, v ! Lp(!), 1 $ p < ,

)u + v)Lp(!) $ )u)Lp(!) + )v)Lp(!).

We also have Holder’s inequality. For u ! Lp(!) and v ! Lq(!), 1 $ p, q < , with1/p + 1/q = 1

--(

!u(x)v(x)dx

-- $ )u)Lp(!))v)Lq(!).

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 ! C!0 (!). Then we

see by integration by parts(

!D!u(x)v(x)dx = (+1)|!|

(

!u(x)D!v(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 # ! C!0 (!).

Then the weak derivative of u of order ! denoted by D!u is defined by D!u = 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 # ! C!0 (!), 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 # ! C!

0 (!) 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(!) | D!u ! Lp(!) &|!| $ k}

34

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

)u)W k,p(!) :=

.

/!

|!|"k

)D!u)pLp(!)

0

11/p

.

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

-u, v.Hk(!) :=!

|!|"k

-D!u, D!v.L2(!).

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

-u, v.H1(!) = -u, v.L2(!) + -Du,Dv.L2(!)

=( b

au(x)v(x)dx +

( b

aDu(x)Dv(x)dx.

-u, v.H2(!) = -u, v.L2(!) + -Du,Dv.L2(!) + -D2u, D2v.L2(!)

=( b

au(x)v(x)dx +

( b

aDu(x)Dv(x)dx +

( b

aD2u(x)D2v(x)dx.

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

|-u, v.Hk(!)| $!

|!|"k

|-D!u, D!v.L2(!)|

$!

|!|"k

)D!u)L2(!))D!v)L2(!)

$2 !

|!|"k

)D!u)2L2(!)

2 !

|!|"k

)D!v)2L2(!)

= )u)Hk(!))v)Hk(!).

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 that3+ 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 su"ciently 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

=4+p(x)

du(x)dx

#(x)5x=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 the

test 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 term6! u#dx to make sense, we need u, # ! L2(!). For the derivatives du/dx, d#/dx

to make sense, we take this further, so u, # ! H1(!).

2) Let us chooseV :=

7# ! H1(!) | v(0) = v(1) = 0

8,

36

whereH1(!) =

7# ! L2(!) | D# ! L2(!)

8.

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-Milgram’s theorem.

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

|l(#)| = |(

!f#dx|

$ )f)L2(!))#)L2(!)

$ )f)L2(!) ()#)2L2(!) + )D#)2L2(!))

= cl)#)V ,

where we have set cl := )f)L2(!). 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(#, %)| $ |(

!pD#D%dx| + |

(

!q#%dx|

$ maxx#!

|p(x)|(

!|D#D%|dx + max

x#!|q(x)|

(

!|#%|dx

$ maxx#!

|p(x)|)D#)L2(!))D%)L2(!) + maxx#!

|q(x)|)#)L2(!))%)L2(!)

$ C()D#)L2(!))D%)L2(!) + )#)L2(!))%)L2(!))

$ C,)D#)2L2(!) + )#)2L2(!)

,)D%)2L2(!) + )%)2L2(!)

= C)#)H1(!))%)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

= C)#)2V ,

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

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

(P) Find u ! V such that(

!(pDuD# + qu#)dx =

(

!f#dx,

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 exists C > 0 such that(

!|#|2dx $ C

(

!|D#|2dx for &# ! V.

By using this inequality we can prove the unique existence of the solution solving(P) with q 0 0 in the same way as above.

4.3 One Dimensional Problem: Neumann condition

Let ! = (0, 1), p(x), q(x) ! C(!) and f(x) ! L2(!). We consider the followingproblem.

Find u : ! " R such that#$%

$&

+ d

dx

*pdu

dx

++ qu = f, x ! !,

d

dxu(x) = 0, x ! "!.

(NBVP)

38

Specifying the value of du/dx at boundary points is said to be a Neumann boundarycondition. We assume the same conditions for p, q as before, i.e, p(x) ' p0 > 0,q(x) ' q0 > 0 in !. Let us derive the variational form. Take a su"cient smoothtest function #, multiply (NBVP) by # and integrate.

(

!f(x)#(x)dx =

(

!

*+ d

dx

*p(x)

du(x)dx

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

+dx

=4+p(x)

du(x)dx

#(x)5x=1

x=0

+(

!

*p(x)

du(x)dx

d#(x)dx

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

dx

=(

!

*p(x)

du(x)dx

d#(x)dx

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

dx.

We have eliminated the term [p(x)du(x)/dx#(x)]x=1x=0 by taking into account the Neu-

mann boundary conditions du(x)/dx = 0 for x = 0, 1. Let us choose the functionalspace V := H1(!) in this case and define

a(u, v) :=(

!(pDuDv + quv)dx,

l(v) :=(

!fv dx.

The corresponding variational problem is that:-

(P) Find u ! V such that(

!(pDuD# + qu#)dx =

(

!f#dx,

for any # ! V .

Again the linear form l(·) : V " R and the bilinear form a(·, ·) : V / V " R satisfythe assumptions of Lax-Milgram’s theorem, hence the problem (P) has the uniquesolution.

Remark. Consider (NBVP) with q 0 0. Then we see(

!f(x)dx = +

(

!

d

dx(p(x)

du(x)dx

)dx

=4p(x)

du(x)dx

5x=1

x=0

= 0.

39

Thus, we need to assume6! f(x)dx = 0 as a compatibility condition in this case. In

order to prove the unique existence of the solution, we need to modify the functionalspace. Let us define H1

m(!) by

H1m(!) := {# ! H1(!) |

(

!#dx = 0},

and equip the same inner product as H1(!). Again Poincare’s inequality is availablefor this space H1

m(!), i.e, for all # ! H1m(!)

(

!|D#|2dx $ C

(

!|#|2dx,

where C > 0 is a constant. By using this inequality we can prove the uniqueexistence of the solution solving (P) with q 0 0 and V = H1

m(!) in the same wayas above on the assumption

6! f(x)dx = 0.

4.4 One Dimensional Problem: Robin/Newton Condi-tion

Let ! = (0, 1), p(x), q(x) ! C(!), f(x) ! L2(!), &, g0, g1 ! R be constants. Weconsider the following problem. Find u : ! " R such that

#$$$$$%

$$$$$&

+ d

dx

*pdu

dx

++ qu = f, x ! !,

+pd

dxu(0) + &u(0) = g0,

pd

dxu(1) + &u(1) = g1.

(RNBVP)

Let us derive the variational form. Take su"ciently smooth #. This kind of boundarycondition is said to be a Robin/Newton boundary condition. We assume the sameconditions for p, q as before, i.e, p(x) ' p0 > 0, q(x) ' q0 > 0 in ! and that & ' 0.

(

!f(x)#(x)dx =

(

!

*+ d

dx

*p(x)

du(x)dx

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

+dx

=4+p(x)

du(x)dx

#(x)5x=1

x=0

+(

!

*p(x)

du(x)dx

d#(x)dx

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

dx

= (+g1 + &u(1))#(1)+ (g0 + &u(0))#(0)

+(

!

*p(x)

du(x)dx

d#(x)dx

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

dx,

40

which is equal to(

!

*p(x)

du(x)dx

d#(x)dx

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

dx + &u(1)#(1) + &u(0)#(0)

=(

!f(x)#(x)dx+ g1#(1) + g0#(0).

This suggests that we should define a(·, ·), l(·) and the functional space V as follow-ing.

a(u, v) :=(

!(pDuDv + quv)dx + &u(1)v(1) + &u(0)v(0),

l(v) :=(

!fvdx+ g1v(1) + g0v(0),

V := H1(!).

As usual we need to show that the (bi)linear forms a, l are bounded and a is coer-cive to establish the unique solvability of (RNBV P ). Let us assume the followinginequality holds true for a while.

|#(x)| $ C)#)H1(!) (4.1)

for all # ! H1(!). Then it is easy to see that a(·, ·), l(·) are bounded. Now

a(#, #) ' min(p0, q0))#)2V + &(#(1)2 + #(0)2)

' min(p0, q0))#)2V= !)#)2V ,

where ! = min(p0, q0). Hence the bilinear form a becomes coercive and Lax-Milgram’s theorem assures the unique existence of the solution.

4.5 One Dimensional H1 inequalities

Here we derive some inequalities in the one dimensional case ! = (a, b). We definea function space H1

"0(!) by

H1"0(!) := {# ! H1(!) | #(a) = 0}.

The following inequality is one example of Poincare-Friedrichs inequality.

41

Proposition 4.5.1. For all # ! H1"0(!),

)#)L2(!) $112(b+ a))D#)L2(!).

Proof. We can write that for a $ &x $ b

#(x) =( x

aD#(')d'.

Then we see that

)#)2L2(!) =( b

a#(x)2dx

=( b

a

*( x

aD#(')d'

+2

dx

$( b

a

*( x

a12dx

+ *( x

aD#(')2d'

+dx

=( b

a(x+ a)

( x

aD#(')2d'dx

$( b

a(x+ a)

( b

aD#(')2d'dx

=12(b+ a)2)D#)2L2(a,b).

Example 4.5.2. We can apply this inequality to prove the unique solvability ofthe Dirichlet problem with a functional space V , a bilinear form a(·, ·) and a linearfunctional l(·) defined by

V := {# ! H1(0, 1) | #(0) = #(1) = 0},

a(u, v) :=( 1

0pDuDvdx for &u, v ! V,

l(u) :=( 1

0fudx for &u ! V.

We only check that the bilinear form a is bounded and coercive. We see that

|a(u, v)| $ supx#(0,1)

|p(x)|)Du)L2(!))Dv)L2(!) $ supx#(0,1)

|p(x)|)u)H1(!))v)H1(!)

42

and

a(v, v) ' p0)Dv)2L2(!)

=p0

2)Dv)2L2(!) +

p0

2)Dv)2L2(!)

' p0

2

9)Dv)2L2(!) +

2)v)2L2(!)

(b+ a)2

:

' !)v)2V ,

where ! := p0 min(1, 2/(b+ a)2)/2.

Proposition 4.5.3. The following Agmon’s inequality holds. For all # ! H1"0(!)

maxx#!

|#(x)|2 $ 2)#)L2(!))#)H1(!).

Proof.

#(x)2 =( x

a

d#(')2

d'd'

= 2( x

a#(')D#(')d'

$ 2*( x

a#(')2d'

+1/2 *( x

aD#(')2d'

+1/2

$ 2*( b

a#(')2d'

+1/2 *( b

aD#(')2d'

+1/2

$ 2)#)L2(!))#)H1(!),

which gives the inequality.

Noting that)#)L2(!) $

1b+ a max

x#[a,b]|#(x)|,

this Agmon’s inequality yields

maxx#!

|#(x)|2 $ 21

b+ a maxx#[a,b]

|#(x)|)#)H1(!),

ormaxx#!

|#(x)| $ 21

b+ a)#)H1(!)

for any # ! H1"0(!).

43

4.6 Weak Solutions to Elliptic Problems

The simplest elliptic equation is Laplaces’ equation:

#u = 0, (4.2)

where # :="n

j=1#2

#x2j

is the Laplace operator. A general second order ellipticequation is: given a bounded open set ! ( Rn find u such that:

+n!

i,j=1

"

"xj

*aij(x)

"u

"xi

++

n!

i=1

bi(x)"u

"xi+ c(x)u = f(x) x ! !, (4.3)

where classically aij ! C1(!), i, j = 1, . . . , n; bi ! C(!), i = 1, . . . , n; c ! C(!); f !C(!). For the equation to be elliptic we require

n!

i,j=1

aij(x)(i(j ' Cn!

i=1

(2i & ( = ((1, . . . , (n) ! Rn, (4.4)

where C > 0 is independent of x, (. Condition (4.4) is called uniform ellipticity.

The equation is usually supplemented with boundary conditions - Dirichlet, Neu-mann, Robin, or a mixed Dirichlet/Neumann boundary.

In the case of the homogeneous Dirichlet problem (u = 0 on "!) u is said to bea classical solution provided u ! C2(!) 2 C(!). Elliptic theory tells us that thereexists a unique classical solution provided aij , bi, c, f and "! are su"ciently smooth.However we are only intersted in problems where the data is not smooth, for examplef = sign(1/2 + |x|),! =( +1, 1). This problem can’t have u ! C2(!) because #uhas a jump discontinuity are |x| = 1/2. With the help of functional analysis theexistence/uniqueness theory for ‘weak’, ‘variational’ solutions turn out to be easyand is good for FEM.

4.7 Variational Formulation of Elliptic Equation: Neu-mann Condition

Let ! be a bounded domain in Rn with smooth boundary "!. Let p, q ! C(!) suchthat

p(x) ' p0 > 0, q(x) ' q0 > 0 &x ! !,

44

and f ! L2(!).

Find u : ! " R such that3 +3 · (p3u) + qu = f, x ! !,

"u

"n= 0, x ! "!,

(NBVP)

where n is the unit outward normal to the boundary "!. Note that

3 · (p3u) =n!

i=1

"

"xi

*p

"u

"xi

+

=n!

i=1

*p"2u

"x2i

+"p

"xi

"u

"xi

+

= p#u +3p ·3u,

"u

"n= 3u · n.

So we have a second order PDE. In one dimensional problem, in order to derive thevariational formulation we used integration by parts. Let us revise some formulaerelated to the integration by parts.

Notation:

3v =*

"v

"x1, . . . ,

"v

"x1

+T

3 ·3 v = 32v = #v

÷ ·A =n!

i=1

"Ai

"xi

(D2v)ij ="2v

"xi"xj

Tr(D2v) = #v.

Theorem 4.7.1 (Divergence theorem). Let A : ! " Rn be a C1 vector field. Thefollowing equality holds. (

!÷Adx =

(

#!A · nds.

45

Remark. Suppose A = fei with the coordinate vectorei = (0, · · · , 1, · · · , 0)T , i.e, the jth component is {ei}j = &i,j . Then we see

÷A ="

"xif.

So by the Divergence theorem(

!

"f

"xidx =

(

#!fnids.

In one dimensional case where ! = (a, b), "! = {a, b}, Divergence theorem becomes

( b

a

"f

"xdx = f(b)+ f(a).

Let us derive the integration by parts formula.

Proposition 4.7.1 (Integration by parts).For A ! C1(!; Rn), g ! C1(!),

(

!A ·3gdx =

(

#!gA · nds+

(

!g ÷ Adx.

Proof. By Divergence theorem we see that(

!÷(Ag)dx =

(

#!A · ngds.

Alternatively,÷(Ag) = g ÷ A + A ·3g.

By combining these equality we get the desired formula.

For example, if A = 3u and g = v, we have by integration by parts formula that(

!3u ·3vdx =

(

#!v3u · nds+

(

!v ÷ (3u)dx.

Noting that

3u · n ="u"n

and ÷ (3u) = !u,

46

where # is Laplacian, we obtain(

!v#udx =

(

#!v"u

"nds+

(

!3u ·3vdx.

Looking at our boundary value problem +÷ (p3u) + qu = f , we have that

+(

!÷(p3u)vdx =

(

!p3u ·3vdx+

(

#!p"u

"nvds.

Let v be a su"ciently smooth test function. Multiply (NBVP) by v and integrateusing Divergence theorem.

(

!fvdx =

(

!(+3 · (p3u) + qu)vdx

=(

!p3u3vdx+

(

#!p"u

"nvds +

(

!quvdx.

Since "u/"n = 0 on "!, we do not need to place a restriction on the test functionv. So if u solve (BVP), then

(

!(p3u3v + quv)dx =

(

!fvdx,

for any su"ciently smooth function v.

Now to use Lax-Milgram, we have to set up V , a(·, ·) and l(·). In order for the twoinner products on the left hand side to make sense, we take

V = H1(!),

a(u, v) =(

!(p3u3v + quv)dx,

l(v) =(

!fvdx,

for all u, v ! V . Note that V is a real Hilbert space with the norm

)v)V = )v)H1(!) =*(

!|3v|2dx +

(

!v2dx

+1/2

and obviously a(·, ·) : V / V " R is bilinear and l(·) : V " R is linear. Moreover

47

we observe

a(v, v) =(

!(p|3v|2 + qv2)dx

' p0

(

!|3v|2dx + q0

(

!v2dx

' min{p0, q0})v)2H1(!)

= !)v)2H1(!),

where we have put ! := min{p0, q0}. Thus a(·, ·) is coercive.

|a(v, w)| =--(

!(p3v ·3w + qvw) dx

--

$(

!(|p3u ·3w| + |qvw|)dx

$ C

(

!(|3v|2 + v2)1/2(|3w|2 + w2)1/2dx

$ C

*(

!(|3v|2 + v2)dx

+1/2 *(

!(|3w|2 + w2)dx

+1/2

= C)v)H1(!))w)H1(!).

Therefore, a(·, ·) is bounded. Finally let us check the boundedness of l(·).

|l(v)| =--(

!fvdx

--

$(

!|f ||v|dx

$*(

!f2dx

+1/2 *(

!v2dx

+1/2

= )f)L2(!))v)L2(!)

$ )f)H1(!))v)H1(!)

= Cl)v)H1(!).

Thus, l(·) is bounded. Now we can apply Lax-Milgram to prove that there exists aunique solution u ! V to the following problem (P).

(P) Find u ! V such thata(u, v) = l(v)

for all v ! V .

48

4.8 Variational Formulation of Elliptic Equation: Dirich-let Problem

On the same assumptions on !, p, q, f , we consider the following problem.

Find u : ! " R such that'+3 · (p3u) + qu = f, x ! !,u = 0, x ! "!,

(DBVP)

Let us derive the variational form of (DBVP) as in the section 2.5. Multiply (DBVP)by a su"ciently smooth test function v and integrate. Then we see

(

!fvdx =

(

!(p3u3v + quv)dx+

(

#!p"u

"nvdx.

Since we have u 0 0 on "!, we have to force our test function v to satisfy the samecondition; v 0 0 on "!. Then we obtain

(

!(p3u3v + quv)dx =

(

!fvdx

for any su"cient smooth function v with v 0 0 on "!. Set

V := {v ! H1(!) | v = 0 on "!}= H1

0 (!).

Note that V is a real Hilbert space with the inner product

-v, w.V := -3v,3w.L2(!) + -v, w.L2(!)

and )v)V = )v)H1(!). As before we define

a(v, w) :=(

!(p3v3w + qvw) dx,

l(v) =(

!fv dx.

The same argument as the previous section shows that a(·, ·) is a coersive andbounded bi-linear form and l(·) is a bounded linear functional on V . Therefore Lax-Milgram’s theorem tells us that there uniquely exists a solution to the variationalproblem of (DBVP).

49

4.9 Inhomogeneous Boundary Condition

Let V be a Hilbert space and a(·, ·) be a bilinear coercive form on V /V , let l(·) belinear, let V0 be a closed subspace of V and g ! V . Set Vg = {v ! V : v = v0+g, v0 !V0} and consider the problem: find u ! Vg such that a(u, v) = l(v) & v ! V0. Wecan show that there exists a unique solution.

Let u0 = u+ g. Then the problem becomes: find u0 ! V0 such that

a(u0) = l(v)+ a(g, v) & v ! V0. (4.5)

For the finite element method it becomes: find uh such that:

uh = g0#0 +M$1!

j=1

!j#j + g1#M . (4.6)

this means that A is the same as the homogeneous case but b now has contributionsfrom g0, g1.

4.10 Second Order Elliptic Problems

Consider the problem

+n!

i,j=1

"

"xj

*aij(x)

"u

"xi

++

n!

i=1

bi(x)"u

"xi+ c(x)u = f(x) &x ! ! (4.7)

with u = 0 on "!. Multiply by a test funciton and integrate by parts in the secondorder term using the divergence theorem. The result is the weak (variational) formof the BVP: find u ! V such that a(u, v) = l(v) & v ! V where V = H1

0 (!) and

a(w, v) :=n!

i,j=1

(

!aij(x)

"w

"xi

"v

"xj+

n!

i=1

(

!bi(x)

"w

"xiv(x) +

(

!c(x)w(x),

l(v) :=(

!f(x)v(x) = (f, v).

We seek to apply the Lax-Milgram theorem. Recall (v, w)H10 (!) =

6! vw +3v3w =

(v, w) + (3v,3w). We have three conditions to check to satisfy the theorem.

50

(1) Is l(·) a bounded linear functional? Clearly

l(!v + $w) = (f, !v + $w) = !(f, v) + $(f, w) = !l(v) + $l(w)

so l(·) is a linear functional on V and

|l(v)| =----(

!f(x)v(x) dx

---- $ )f)L2(!))v)L2(!) $ )f)L2(!))v)H10 (!)

where we have used the Cauchy-Schwartz inequality and thus l(·) is bounded.

(2) Is a(·, ·) bounded? Assume that )aij)L!(!), )bi)L!(!), )c)L!(!) are all boundedfor all i, j and that f ! L2(!). Then

|a(w, v)| $

------

n!

i,j=1

(

!aijwxivxj dx

------+

-----

n!

i=1

(

!biwxiv dx

----- +----(

!cwv dx

----

$n!

i,j=1

maxx#!

|aij(x)|(

!|wxi ||vxj | dx +

n!

i=1

maxx#!

|bi(x)|(

!|wxi ||v| dx + max

x#!|c(x)|

(

!|w||v| dx

$ c

.

/n!

i,j=1

(

!|wxi ||vxj | dx +

n!

i=1

(

!|wxi ||v| dx +

(

!|w||v| dx

0

1

$ c

.

/n!

i,j=1

)wxi))vxj)+n!

i=1

)wxi))v)+ )w))v)

0

1

$ c

.

/n!

i,j=1

)w)V )v)V +n!

i=1

)w)V )v)V + )w)V )v)V

0

1

= c1)w)V )v)V

where c = max{maxij max! |aij(x)|,maxi max! |bi(x)|,max! |c(x)|} and c1 = c(n2+n + 1).

(3) Is a(·, ·) coercive? The crucial assumption is that the aij satisfies the ellipticityassumption

n!

i,j=1

aij(x)(i(j ' cn!

i=1

(2i & ((1, . . . , (n) ! Rn, &x ! !, (4.8)

i.e. for all x ! ! we must have

(T A(x)( ' c)()2 = (T (. (4.9)

51

We also assume that

c(x)+ 12

n!

i=1

"bi(x)"xi

' 0&x ! !. (4.10)

Then

a(v, v) =n!

i,j=1

(

!aij(x)vxivxj +

n!

i=1

(

!bi(x)vxiv +

(

!c(x)v(x)2

' c

(

!

n!

i=1

v2xi

+n!

i=1

(

!bi(x)

"v2/2"xi

+(

cv2.

The middle integral here is 12

6! b · 3(v2), which after integration by parts equals

+12

6! v23 · b so that

a(v, v) ' cn!

i=1

(

!v2xi

+(

!v2(c(x)+ 1

23 · b(x))

' cn!

i=1

(

!v2xi

= c)3v)2

Note that we need 3 · b ! L!(!) for this to work. We wish to show that

a(v, v) ' c0)v)2V = c0()v)+ )3v)). (4.11)

Recall the Poincare-Friedrichs inequalities

)v)2 $ c&)3v)2 & v ! H10 (!).

Hence

a(v, v) ' c)3v)2 ' c

c&)v)

12a(v, v) +

12a(v, v) ' c

2)3v)2 +

c

2c&)v)

' c0;)3v)2 + )v)2

<

4.10.1 Remarks on the Lax-Milgram Result

1. Uniqueness: by our usual methods this follows from the linearity of l(·), thebilinearity of a(·, ·) and the coercivity of a(·, ·).

52

2. Stability estimate: we know that

c0)u)2V $ a(u, u) = l(u) $ c2)u)V

so we can deduce that the solution to our BVP satisfies

)u)H1(!) $1c0)f)L2(!). (4.12)

3. Continuity with repsect to l(·). Consider the two problems

u1 ! V s.t. a(u1, v) = l1(v) & v ! V

u2 ! V s.t. a(u2, v) = l2(v) & v ! V.

Thena(u1 + u2) = l1(v)+ l2(v) = l(v). (4.13)

Choosing v = u1 + u2:

c0)u1 + u2)2V = l(u1 + u2) $ )l1 + l2)V ")u1 + u2)

# )u1 + u2)V $)l1 + l2)V "

c0

In terms of our original elliptic bvp’s we have that

)u1 + u2)H1(!) $1c0)f1 + f2)L2(!). (4.14)

4. If l is the zero element of V & (i.e. l(v) = 0 & v ! V ) then 0 = a(u, u) #)u)V = 0 by coercivity and u = 0.

4.10.2 Inhomogeneous Boundary Conditions

Consider the elliptic problem

+3 · (p3u) + qu = f x ! ! (4.15)u = g x ! "! (4.16)

where ! is a bounded open subset of R2. We assume that the data p, q, f, g aresu"ciently smooth and that

pM ' p(x) ' p0 > 0 &x ! !qM ' q(x) ' q0 > 0 &x ! !

53

Let v be a test function. Multiply by v and integrate:

0 = +(

!v3 · (p3u) +

(

!quv +

(

!fv

= I1 + I2 + I3

Nov choosing ) = v, f = p3u in:

3 · ()f) = )3 · f +3) · f3 · (vp3u) = v3 · (p3u) +3v · p3u

I1 = +(

!3 · (vp3u)+3v · p3u

=(

!p3v ·3u+

(

#!vp3u · *

Choosing v = 0 on "! we have I1 =6! p3v ·3u. Thus

0 =(

!p3v ·3u + quv + fv & v ! H1

0 (!). (4.17)

Set V0 = H10 (!), a(u, v) =

6! p3v · 3u + quv, l(v) =

6! fv. Note that u *! V0.

However g ! H1(!) so u+ g ! H1(!) and u+ g ! H10 (!) = V0, i.e u ! Vg := {w !

V = H1(!) : w = g + v, v ! V0}.

Thus our variational problem (P ) is to find u ! Vg such that a(u, v) = l(v) & v ! V0.Observe that V0 is a linear space but Vg = g + V0 is an a"ne space. We can’t applyLax-Milgram directly. Consider u& = u+ g ! V0:

a(u& + g, v) = a(u, v) = l(v) & v ! V0

so u& ! V0 solves

a(u&, v) = l(v)+ a(g, v) =: l&(v) & v ! V0

Now we just need to check Lax-Milgram for this problem. Clearly a(·, ·) is bilinear(and symmetric). Coercivity:

a(v, v) =(

!p|3v|2 + qv2 ' p0

(

!|3v|2 + q0

(

!v2 ' min(p0, q0)

(

!|3v|2 + v2 ' c0)v)H1(!).

Boundedness: using the Cauchy-Schwartz inequality we have

|a(w, v)| =----(

!p3w3v + qwv

---- $ pM

(

!|3w||3v| + qM

(

!|w||v|

$ max(pM , qM )()3w))3v)+ )w))v))$ c)w)H1(!))v)H1(!).

54

Clearly l& is linear and

|l&(v)| = |l(v)+ a(g, v)| $| l(v)| + |a(g, v)| $ )f))v)+ c)g)H1(!))v)H1(!) $ L&)v)H1(!).

Thus there exists a unique u& and we conclude therefore that there exists a uniqueu = u& + g.

The bilinear form is symmetric so there is an energy and associated minimisationproblem:

J(v) =12a(v, v)+ l(v)

Find u ! Vg s.t. J(u) $ J(v) & v ! Vg.

Exercise: Prove that these two problems are equivalent.

4.11 Finite Element Method in 2D

Take ! to be a polygon. Let Th be a triangulation of !, Th = {+} and set h$ =diam+ (the length of the longest side), h = maxdiam +. We assume that |Th| < ,.Any triangles in Th must intersect along a complete edge, at a vertex of not at all.

Note that any linear function on R2 is of the form v(x, y) = a+bx+cy and is definedby three parameters. Thus any function

vh ! Vh := {' ! C(!) : vh|$ is linear} (4.18)

is uniquely determined by it values at the vertices of the triangulation:

#i(xj) = &ij i, j = 1, . . . , Nh, xj is a triangle vertex. (4.19)

The support of the basis functions is local, so A will again be sparse.

Example 4.11.1. Find u ! H1(! such that(

!p3u3v + quv dx =

(

!fv dx & v ! H1(!). (4.20)

A finite element method applied to this yields the problem: find uh ! H1(!) suchthat (

!p3uh3vh + quhvh dx =

(

!fvh dx & vh ! Vh. (4.21)

55