A short introduction to Sobolev-spaces andapplications for engineering students

Dag Lukkassen

November, 2004

Contents

1 Preliminary denitions and results . . . . . . . . . . . . . . . . . . . . . 31.1 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 The precise denition of completeness (optional for the stu-dents) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Sobolev- and Lebesgue-spaces . . . . . . . . . . . . . . . . . . . . 41.3 An example of a Sobolev-function . . . . . . . . . . . . . . . . . . 61.4 Example of a function which is not a Sobolev-function . . . . . . 61.5 Schwarz inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 81.6 Poincars and Friedrichs inequalities . . . . . . . . . . . . . . . . 91.7 Other useful inequalities . . . . . . . . . . . . . . . . . . . . . . . 11

2 The heat conduction problem . . . . . . . . . . . . . . . . . . . . . . . 122.1 Dirichlet problem (u = 0 on the boundary) . . . . . . . . . . . . . 142.2 Existence and uniqueness of the Dirichlet problem . . . . . . . . . 162.3 Equivalent minimum energy formulation . . . . . . . . . . . . . . 172.4 The Neumann problem (@u=@n = 0 on the boundary) . . . . . . . 20

3 Abstract formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.1 Comparison with minimum problems on the real line . . . . . . . 243.2 The abstract theory for the Dirichlet problem . . . . . . . . . . . 26

4 Eective properties of periodic structures . . . . . . . . . . . . . . . . . 305 Finite Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.1 Example 1 (dimension 1) . . . . . . . . . . . . . . . . . . . . . . . 405.2 Example 2 (dimension 2) . . . . . . . . . . . . . . . . . . . . . . . 475.3 Example 3 (Eective conductivity of periodic structures) . . . . . 53

6 Further exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567 Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607.1 Solution to further exercises . . . . . . . . . . . . . . . . . . . . . 80

2

1. Preliminary denitions and results

1.1. Hilbert spaces

Let V be a vector-space (i.e. x; y 2 V ) x + y 2 V , and kx 2 V for all k 2 R)with a scalar product denoted (also called inner product). This scalar productinduces a norm kxkV = (x x)

12 . If V is complete with respect to this norm

then V is called a Hilbert-space. Roughly speaking, we say that a space V iscomplete if for any sequence fxhg of elements in V converging to an element x;i.e. kxh xkV ! 0, we have that x also belongs to V .

Example 1.1. Let

x =

26664x1x2...xn

37775and

y =

26664y1y2...yn

37775 :The scalar product () between vectors of this type is usually dened by

x y = x1y1 + x2y2 + :::+ xnynand the norm of x is given by

kxk =qx21 + x

22 + :::x

2n ):

It is easy to check that the set of all such vectors forms a vector space. This spaceis usually denoted Rn: It is possible to show that this space is complete withrespect to the given scalar product. Hence Rn is a Hilbert space. In particular,R = R1 is a Hilbert space.

Example 1.2. An example of a set which is not complete is the set Q of rationalnumbers (numbers which can be written on the form a=b where a and b areintegers). This set is not complete with respect to the norm kxk = px2. For

3

example we can nd elements in Q which converge to the numberp2 = 1: 4142::,

(take e.g. the sequence 1; 1:4; 1:41; 1:414; :::) but this number is not a member ofQ:

From now on we will reserve the symbol () for the usual scalar product

x y = x1y1 + x2y2 + :::+ xnyn ; x 2 Rn and y 2 Rn:

Remark 1. Let V be a m-dimensional vector space (with a scalar product ),i.e. there exist m vectors v1; :::vm (so called basis vectors) such that each vectorw 2 V can be written (uniquely) as:

w = a1v1 + :::+ amvm

for some real constants a1; :::am: Then it is possible to prove that V is a Hilbertspace. Hence vector spaces which are not Hilbert spaces must be innitely dimen-sional.

Exercise 1.1. The set Q is not complete and is therefore not a Hilbert space.There is also another reason why Q is not a Hilbert space, what is that?

1.1.1. The precise denition of completeness (optional for the students)

Let V be a vector-space with a scalar product : The corresponding norm is thendened by kxkV = (xx)

12 :We say that fxhg ; h = 1; 2; 3::: , is a Cauchy sequence if

the dierence kxk xmkV tends to 0 as k andm goes to1 (or even more precisely:for any number " > 0 there exists an integer N such that kxk xmkV < " for allk > N and m > N). The space V is said to be complete (with respect to thegiven scalar product) if every Cauchy sequence converges to an element x 2 V; i.e.kxh xkV ! 0 as h!1:

1.2. Sobolev- and Lebesgue-spaces

Sobolev- and Lebesgue-spaces are important examples of Hilbert-spaces. Let

be a bounded domain in Rn:L2() = all functions v that has the property so thatZ

jvj2 dx

W 1;2() = all functions u where u 2 L2() and @u=@xi 2 L2(): (The number1 is used to indicate that the partial derivatives of order 1 should belong to L2().The number 2 refers to power of the integrand in (1.1)). Functions belonging toW 1;2() do not have to be dierentiable at every point; e.g. it is enough if theyare continuous with piecewise continuous partial derivatives in the domain ofdenition and satisfy the above conditions.Let Y be a cell in Rn (a rectangle if n = 2; a box if n = 3): W 1;2per(Y ) =

fu 2 W 1;2(Y ) : u is Y periodicg : (We read it like this: W - periodic, 1,2, of Yis the set of all functions u which is an element of the space W ,1,2 of Y where uis Yperiodic.) Recall that u is Yperiodic means that u have the same valueson opposite faces of the cell YW 1;20 () = fu 2 W 1;2() : u is 0 on the boundary @g : (We read it like this:

W ,0,1,2 of Y ) is the set of all functions u which is an element of the space W ,1,2of Y where u is zero on the boundary @.In detail we see that:

W 1;20 (Y ) W 1;2per(Y ) W 1;2(Y ) L2(Y )and

W 1;20 () W 1;2() L2():The W - spaces are often called Sobolev-spaces. The space L2() is called aLebesgue-space.The scalar product () of W 1;2() is dened by

u v =Z

uv dx+

Z

gradu grad v dx

and hence the norm is dened as

kukW 1;2() =sZ

u2 dx+

Z

jgraduj2 dx:

We use the same norm for the other Sobolev-spaces. Moreover the scalar product() on L2() is dened by

u v =Z

uv dx:

The corresponding norm is then given by

kukL2() =sZ

u2 dx:

5

1.3. An example of a Sobolev-function

If Rn; we will always let jj denote the value

jj =Z

1 dx

(which equals the length, area or volume of if n = 1; 2 or 3; respectively).Let be the square Y = [0; 2]2 and let

u(x) = u(x1; x2) = sinx1 + cosx2:

We see that @u=@x1 = cos x1 and @u=@x2 = sin x2:Moreover, we observe thatZY

juj2 dx =ZY

(sinx1 + cosx2)2 dx

ZY

j1 + 1j2 dx =ZY

4 dx = 4 jY j

=1

4

Z 10

(ln 1 ln 0)dx2 =1

This shows that u =2 W 1;2(Y ); i.e. u is not a Sobolev-function. Remark that forthis example both

RY

@u@x2 2 dx

Figure 1.1: in Exercise 1.7

Exercise 1.7. Let R2 be the set which consists of all points above the linex2 = 1 and below the curve x2 =

p4 x21 (see Figure 1.1). Determine whether

the functions v and w; dened by

v(x1; x2) = (x21 + x

22 4) (2x2 2) ;

w(x1; x2) = x132

are in some of the Lebesgue and Sobolev spaces L2();W 1;2();W 1;20 ():

1.5. Schwarz inequality

If V is a vector space with a scalar product () then the Schwarz inequality takesthe following form

jx yj kxkV kykV = (x x)12 (y y) 12 :

Let V be the space of vector valued functions with n components with scalarproduct dened by

u v =Z

u v dx; (1.2)

where is a positive bounded function, such that

0 < (x) +

where and + are constants. In this case the Schwarz inequality takes the formZ

u v dx Z

juj2 dx 1

2Z

jvj2 dx 1

2

: (1.3)

In particular we obtain thatZ

uv dx

Z

juj2 dx 1

2Z

jvj2 dx 1

2

; (1.4)

where u and v are real functions.

Exercise 1.8. Recall the denition of a scalar product (inner product). Let V bethe space of continuous real valued functions dened on the interval = [1; 1],let be a function such that 1 (x) 2 and let be dened by

u v =Z 11(x)u(x)v(x)dx:

1. Show that V is a vector space and show that denes a scalar product onV .

2. Let kukV be the norm kukV = (u u)12 : Find the number kukV in the case

when u(x) = x and = 1:

3. Let uh be dened by

uh(x) =

8

Figure 1.2: 3 examples of sets with Lipschitz continuous boundaries. Note thatthe smallest angle on the boundary is larger than 0.

Figure 1.3: 3 examples of sets which boundaries are not Lipschitz continuous.

10

has this property and also boundaries of convex domains. See also Figure 1.2 andFigure 1.3.Let be a domain whose boundary is Lipschitz continuous. Then the Friedrichs

inequality: Z

v2 dx C0Z

jgrad vj2 dx; 8v 2 W 1;20 ()is valid: If, in addition, is connected then the Poincars inequality is also valid:Z

u2 dx C0(Z

u dx

2+

Z

jgraduj2 dx)

8u 2 W 1;2():

(the symbol 8 means: for all). Note that C0 > 0 in both cases, and C0 is aconstant which only dependent of (not the functions u or v):

Exercise 1.9. Use the inequality (1.4) and show thatZ

u dx

2 K

Z

juj2 dx

(1.5)

for some positive constant K which is independent of u. Prove Friedrichs inequal-ity in the case = [0; 1] (Hint: use that

jv(t) v(0)j =Z t0

v0(x) dx

Z t0

jv0(x) j dx

Z 10

jv0(x) j dx =Z

jgrad vj dx

and then use (1.5))

1.7. Other useful inequalities

We also have the following inequalityZ

f(x) dx

Z

jf(x) j dx;

which is valid for any real valued or vector valued function f (which is possible tointegrate). In addition we sometimes use the triangle inequality

ja+ bj jaj+ jbj ;

11

which is valid for any real number a and b: More generally, for elements a; b ininner product spaces with norm k()k ; the triangle inequality takes the form

ka+ bk kak+ kbk :

2. The heat conduction problem

We start with the strong formulation of the stationary heat conduction problem:

div ((x) gradu(x)) = f(x); x 2 ; (2.1)where we have certain information of u on the boundary @ of the connectedbody , see Figure 2.1 (f(x) is the heat source in the system and (x) is theconductivity). We will assume that f 2 L2() and that

0 < (x) +

Figure 2.1: Body

gradu =

@u

@x1; ; @u

@xn

;

we obtain that

div ( gradu) =@

@x1@u

@x1+ + @

@xn@u

@xn:

In the two-dimensional case (n = 2) this means that 2.1 can be written as:

@

@x1@u

@x1+

@

@x2@u

@x2= f

We multiply with a test function on both sides of (2.1), and then integrate.This leads to the weak formulation of (2.1):Z

div ( gradu) dx = Z

f dx: (2.2)

From the following version of Greens formula (see Exercise 2.1):Z

divwdx = Z

grad w dx+Z@

w n ds (2.3)

we obtain (putting w = gradu = [@u=@x1; ::; @u=@xn]) that

13

Z

div ( gradu) dx =

= Z

grad ( gradu) dx+Z@

( gradu) n ds:

The equation (2.2) can therefore be written as

Z

(grad) gradu dx+Z@

( gradu) n ds = Z

f dx: (2.4)

We remark that the strong formulation implies the weak formulation.

Exercise 2.1. Prove (2.3) (Hint: Use the product rule

@ (wi)

@xi=@

@xiwi +

@wi@xi

and the Divergence theorem of Gauss:Z

div v dx =

Z@

v n ds)

2.1. Dirichlet problem (u = 0 on the boundary)

In the case of the Dirichlet problem we let = 0 on the boundary. HenceZ@

( gradu) n ds = 0;

and we end up with the weak formulationZ

gradu grad dx =Z

f dx:

For the case u = 0 on the boundary @, the precise formulation of the weakproblem takes the following form:Find u 2 W 1;20 () such thatZ

gradu grad dx =Z

f dx

14

for all 2 W 1;20 (): In other words

a(u; ) = L () ; 8 2 W 1;20 (); (2.5)

where

a(u; ) =

Z

gradu grad dx

and

L () =

Z

f dx:

It it possible to prove that a(:; :) is a bilinear form onW 1;20 (); i.e. that for allfunctions u; v; w in W 1;20 () and k 2 R; it holds that a(u; v) is a real number and

a(u+ v; w) = a(u;w) + a(v; w);

a(ku; w) = ka(u;w);

a(w; u+ v) = a(w; u) + a(w; v);

a(u; kw) = ka(u;w):

Exercise 2.2. Show that

a(u; v) =

Z

gradu grad v dx

denes a bilinear form on W 1;20 ().

Exercise 2.3. Determine whether a(:; :) is a bilinear form on V :

1. a(u; v) = u+ v; V = R

2. a(u; v) = uv; V = R

3. a(u; v) = u(0)v(0); V = C([0; 1]); i.e. the space of continuous functions onthe interval [0; 1].

4. a(u; v) =R 10u(x)v(x)dx; V = C([0; 1]);

5. a(u; v) = (u(0)v(0))2 ; V = C([0; 1]);

15

2.2. Existence and uniqueness of the Dirichlet problem

As we know a given problem may not have a solution (e.g. the problem: Findx 2 R such that x2 + 1 = 0). However the existence of a solution of the weakformulation follows by a well-known result in mathematics called the Lax-MilgramLemma (see Theorem 3.1 and the discussion in the same section). Moreover thesolution of the weak formulation is unique. This fact is proved as follows: Considertwo solutions u1 and u2 of (2.5)Z

gradu grad dx =Z

f dx; 8 2 W 1;20 ():

Then, Z

gradu1 grad dx =Z

f dx; 8 2 W 1;20 ()

and Z

gradu2 grad dx =Z

f dx; 8 2 W 1;20 ():

Thus, we get Z

grad(u1 u2) grad dx = 0; 8 2 W 1;20 ():

Since this holds for 8 2 W 1;20 ; it particularly holds for = u1 u2, which givesus that Z

jgrad(u1 u2)j2 dx = 0: (2.6)

Since (x) > 0 for all x 2 and

jgrad(u1 u2)j2 0; 8x 2 ;

(2.6) gives thatgrad(u1 u2) = 0; x 2 ;

and hence(u1 u2) = constant in :

Because u1 = u2 = 0 on the boundary @, the constant is 0, i.e. u1 = u2 for allx 2 . This shows that the solution u in (2.5) is unique.

16

2.3. Equivalent minimum energy formulation

A function F dened on the set K has a minimum at x0 with the minimum valueF (x0) if

F (x0) = minx2K

F (x):

This is the same as the statement:

F (x0) F (x) 8x 2 K:

[For a simple analogy let us consider the function

F (x) = x2:

( F (x) is shown in the curve below). We easily see that

F (0) = minx2R

F (x) = 0

mF (0) F (x) for 8x 2 R .]

We will now show that (2.5) is equivalent with the following minimum. energyformulation: Find u 2 W 1;20 () such that

F (u) = min2W 1;20 ()

F ():

which is the same as nding u 2 W 1;20 () such that

F (u) F (); 8 2 W 1;20 (); (2.7)

whereF () =

1

2a(; ) L(): (2.8)

17

52.50-2.5-5

25

20

15

10

5

0

x

y

x

y

Curve for x2

To prove that (2.5) () (2.7), we must check if the implication is true in theboth ways. First we prove the implication (2.5) ) (2.7):We note that

a(w;w) =

Z

gradw gradw dx =

=

Z

jgradwj2 dx 0:Letting w = u ; we get that

a(u ; u ) 0:Since a is bilinear and a(u; ) = a(; u) we obtain that

a(u ; u ) = a(u; u ) a(; u ) == a(u; u) a(u; ) [a(; u) a(; )] == a(u; u) a(u; ) a(; u) + a(; ) =

= a(u; u) 2a(; u) + a(; ):Thus,

a(u; u) 2a(; u) + a(; ) = a(u ; u ) 0: (2.9)i.e. (the tricky part)

1

2a(; ) a(; u) 1

2a(u; u) a(u; u):

18

Since u is a solution we have that a(; u) = L(), and a(u; u) = L(u): Hence,

1

2a(; ) L() 1

2a(u; u) L(u);

which is the same asF () F (u):

This completes the proof of (2.5) ) (2.7).Proof of (2.7) ) (2.5): Let 2 W 1;20 () and x 2 R be arbitrary. Then

(u+ x) 2 W 1;20 (); and since u is a minimum point, we obtain that

F (u) F (u+ x) 8x 2 R:Put g(x) = F (u+x); x 2 R. Observe that g(0) = F (u+0) = F (u): Therefore,we have that

g(0) g(x) 8x 2 R;i.e. g has a minimum at x = 0: Hence g0(0) = 0 if the derivative g0(x) exists atx = 0: But

g(x) =1

2a(u+ x; u+ x) L(u+ x) =

=1

2a(u; u+ x) +

1

2a(x; u+ x) L(u) L(x) =

=1

2a(u; u) +

1

2a(u; x) +

1

2a(x; u) +

1

2a(x; x) L(u) L(x) =

=1

2a(u; u) +

x

2a(u; ) +

x

2a(; u) +

x2

2a(; ) L(u) xL() =

=1

2a(u; u) L(u) + xa(u; ) xL() + x

2

2a(; );

where we used the symmetry of a(:; :). Thus

g0(x) = a(u; ) L() + xa(; )and it follows that

0 = g0(0) = a(u; ) L();i.e.

a(u; ) = L()

This proves that (2.7)) (2.5).

19

2.4. The Neumann problem (@u=@n = 0 on the boundary)

In the case of the Neumann problem we put no restriction on the boundary valuesof the test function in (2.4), and instead of using the space W 1;20 () we use thewhole space W 1;2(). Thus using ( gradu) n = @u=@n = 0 in (2.4) we obtainthe following weak formulation of the Neumann problem: Find u 2 W 1;2() suchthat

a(u; ) = L () ; 8 2 W 1;2(); (2.10)where (as before)

a(u; ) =

Z

gradu grad dx

and

L () =

Z

f dx:

If, in addition, f 2 L2(); R

f dx = 0 and the boundary of is Lipschitz contin-

uous, then the solution of (2.10) exists and is unique within an additive constant(see Exercise 3.6).

3. Abstract formulation

There exists many problems in physics and mechanics and in order to avoid thenecessity to give an innitenumber of concrete examples we will give an abstract(i.e. more general) formulation in the case of so-called elliptic problems. Ellipticequations model for example stationary heat conduction and static problems inelasticity. A formulation in the abstract way will also make it easier to understandthe basic structure of the nite element method.Let V be a Hilbert space with scalar product and a corresponding norm kkV

(dened as usual as kvkV =pv v). Assume that a(:; :) is a bilinear form on V;

i.e. for all u; v; w in V and k 2 R; it holds that

a(u+ v; w) = a(u;w) + a(v; w);

a(ku; w) = ka(u;w);

a(w; u+ v) = a(w; u) + a(w; v);

a(u; kw) = ka(u;w):

20

Moreover, assume that L a linear form on V: This means that

L(u+ v) = L(u) + L(v);

L(ku) = kL(u);

In addition assume that

1. a(:; :) is symmetric, i.e. a(; v) = a(v; ); 8; v 2 V2. a(:; :) is continuous, i.e. there is a constant > 0 such that ja(; v)j

kkV kvkV 8; v 2 V;

3. a(:; :) is V -elliptic, i.e. there is a constant > 0 such that a(; ) kk2V 8 2 V:

4. L is continuous, i.e. there is a constant > 0 such that jL()j kkV8 2 V:

We will now consider the following abstract minimization problem: Find u 2 Vsuch that

F (u) = min2V

F () (3.1)

(i.e. F (u) F (), 8 2 V ) where

F () =1

2a(; ) L();

and we will also consider the following abstract weak formulation problem: Findu 2 V such that

a(u; ) = L(), 8 2 V: (3.2)

Theorem 3.1. [Lax-Milgram Lemma] If the conditions 2, 3 and 4 are satisedthen there exists a unique solution u 2 V of the problem (3.2).

Theorem 3.2. If the conditions 1, 2, 3 and 4 are satised then there exists aunique solution u 2 V of the problem (3.1). In addition, the problems in (3.1)and (3.2) are equivalent, i.e. u 2 V satises (3.1) if and only if u satises (3.2).

21

Proposition 3.3. If the conditions 2 and 4 are satised then the function F ()is continuous, i.e.

F (h) F ()! 0; whenever kh kV ! 0:

Proposition 3.4. If the conditions 3 and 4 are satised then the function F ()has the property:

F ()!1 if kkV !1:

We will not prove Theorem 3.1. The equivalence in Theorem 3.2 has beenproved for the Dirichlet problem earlier. The proof in the general case is thesame. The fact that the problem (3.1) has a unique solution then follows byTheorem 3.1.Let us prove Proposition 3.3: We have that

F (h) F () = (1

2a(h; h) L(h)) (

1

2a(; ) L())

The trick is to add the 0-term

12a(; h) +

1

2a(; h) = 0;

which give us

F (h) F () =1

2(a(h; h) a(; h))+

+1

2(a(; h) a(; )) + (L() L(h)):

Next we use that a is bilinear and that L is linear:

F (h) F () =1

2(a(h ; h) +

1

2(a(; h )) + L( h):

Thus,

jF (h) F ()j 1

2ja(h ; h)j+

1

2ja(; h )j+ jL( h)j :

Condition 2) gives the inequalities

1

2ja(h ; h)j

1

2

kh kV khkV

22

12ja(; h )j

1

2

kkV kh kV

and condition 4) gives the inequality

jL( h)j k hkV :

Thus

jF (h) F ()j 1

2ja(h ; h)j+

1

2ja(; h )j+ jL( h)j

12

kh kV khkV +

1

2

kkV kh kV + k hkV =

1

2

khkV +

1

2

kkV +

kh kV ! 0

as kh kV ! 0; i.e.jF (h) F ()j ! 0

as kh kV ! 0:We turn to the proof of Proposition 3.4: Condition 3 gives that

a(; ) kk2Vand condition 4 gives that

jL()j kkSince

F () =1

2a(; ) L() 1

2a(; ) jL()j ;

we obtain that

F () 12a(; ) jL()j 1

2 kk2V kkV :

Thus,F ()

kkV 2kkV !1

as kkV !1: Thus F ()!1 as kkV !1:

23

Figure 3.1: The functions f1 and f2:

3.1. Comparison with minimum problems on the real line

In Figure 3.1 we see the graphs of the function

f1(x) =

2x if x > 0

2x+ 1 if x 0

(to the left) and the functionf2(x) = e

x2 :

Note that non of these two functions has any minimum points. The rst functionf1 has the badproperty that is not continuous, but has the goodpropertythat f1(x) grows as jxj grows. The second function f2 has the opposite properties:the goodproperty is that it is continuous at every point, but the badpropertyis that f2(x) does not grow as jxj grows.The function

f3(x) =x

p22

(see Figure 3.2) possesses both these goodproperties and has a minimum pointon the real line.We see that the function F (u) has similar goodproperties when the condi-

tions 2) 3) and 4) are satised. The continuity property follows by Proposition3.3 and the growth property follows by Proposition 3.4Note the minimum point for the function f3 (which equals

p2) is not a rational

number. If we had minimized over the set Q = fall rational numbersg instead ofR; then the function f3 would not have had any minima. This motivates why

24

Figure 3.2: The function f3

the space V; on which we are seeking the minimizer (the solution), has to be acomplete space.

Exercise 3.1. 1. Show that all vectors u and v in R2

u =

u1u2

; v =

v1v2

satisfy the inequality

ju1v1 + 2u2v2j u21 + 2u

22

12v21 + 2v

22

12 (3.3)

2. Let V = R2 with the usual scalar product. Consider the problem: Findu 2 V such that

a(u; v) = L(v) for all v 2 V;where a(u; v) = u1v1+2u2v2 and L(v) = 2v1+4v2: Use Lax Milgrams Lemmaand (3.3) to verify that this problem has a unique solution.

3. The purpose of this exercise is to give an example of a problem where thestudents easily may nd the solution themselves (in contrast to many of theproblems involving partial dierential equations). Calculate explicitly thesolution of the above problem.

4. Consider the minimum problem: Find u 2 V such thatF (u) F (v) for all v 2 V;

where F (v) = 12a(v; v)L(v): Calculate explicitly the solution of this prob-

lem. Compare with the solution you found above. Any comments?

25

Exercise 3.2. The purpose of this exercise is to show what might happen if notall the conditions 2-4 are satised. As in Exercise 3.1, let V = R2: Moreover leta(u; v) = u1v1 + u1v2 + u2v1 + u2v2:

1. Let L(v) = 0 for all v 2 V: Show that both vectors

u =

00

and u =

11

are solutions to the problem:

a(u; v) = L(v) for all v 2 V:

2. Let L(v) = v1 for all v 2 V: Show that there are no solutions u to theproblem:

a(u; v) = L(v) for all v 2 V:3. Apparently, the above two problems did not satisfy the conclusion of Lax-Milgrams Lemma. Show that a(u; v) is not V -elliptic, i.e. that condition 3is not satised.

3.2. The abstract theory for the Dirichlet problem

Let us now show that the bilinear form a (:; :) and L dened for the Dirichletproblem in (2.5) satises the conditions 1), 2), 3) and 4) Proof: Condition 1) isobvious because

a(u; ) =

Z

gradu grad dx =Z

grad gradu dx = a(; u):

Proof: Condition 2): By Schwarz inequality (1.3)

ja(; v)j =Z

(grad) grad v dx

Z

jgradj2 dx 1

2Z

jgrad vj2 dx 1

2

Z

+ jgradj2 dx 1

2Z

+ jgrad vj2 dx 1

2

=

26

=+ 12

Z

jgradj2 dx 1

2 + 12

Z

jgrad vj2 dx 1

2

=

= +Z

jgradj2 dx 1

2Z

jgrad vj2 dx 1

2

+Z

jgradj2 + jj2 dx 12 Z

jgrad vj2 + jvj2 dx 12 == + kkW 1;20 () kvkW 1;20 () :

By putting = + and V = W 1;20 () we get that

ja(; v)j kkV kvkV 8; v 2 V;

This proves 2).Proof: Condition 3):

a(; ) =

Z

jgradj2 dx Z

jgradj2 dx = Z

jgradj2 dx:

By Friedrichs inequality we have thatZ

2 dx C0Z

jgradj2 dx:

Thus Z

2 dx+

Z

jgradj2 dx C0Z

jgradj2 dx+Z

jgradj2 dx =

= (C0 + 1)

Z

jgradj2 dx;

which gives that

1

C0 + 1

Z

2 dx+

Z

jgradj2 dxZ

jgradj2 dx;

i.e.

1

C0 + 1

Z

2 dx+

Z

jgradj2 dx

Z

jgradj2 dx:

27

Hence we obtain that

a(; ) Z

jgradj2 dx 1C0 + 1

Z

2 dx+

Z

jgradj2 dx=

= 1

C0 + 1kk2W 1;20 () :

By putting

= 1

C0 + 1

andV = W 1;20 ()

we obtain thata(; ) kk2V ;

which proves 3).Proof of Condition 4):

L() =

Z

f dx

jL()j =Z

f dx

Z

j f j dx Z

jj2 dx 1

2Z

jf j2 dx 1

2

Z

jj2 + jgradj2 dx 1

2Z

jf j2 dx 1

2

= kkW 1;20 ()

where =R

jf j2 dx 12 (In the second inequality from the left we have used

Schwarz inequality (1.4)). Let W 1;20 () = V , then jL()j kkV , which proves4).

Exercise 3.3. In this task we consider the following partial dierential equation(strong formulation):8>>>:

@2u(x)

@x21+@2u(x)

@x22 @u(x)

@x1 @u(x)

@x2= sin x1; x = (x1; x2) 2 ;

u = 0 on @;

28

where is the open square = h0; 1i2. Derive (this means: verify) the followingweak formulation of the above problem: Find u 2 W 1;20 () such that

a(u; v) = L(v) for all v 2 W 1;20 (); (3.4)

where

a(u; v) =

Z

gradu(x) grad v(x) dx +Z

(@u(x)

@x1+@u(x)

@x2)v(x) dx;

L(v) =

Z

v(x) sinx1dx:

Next, prove that L(v) is continuous.

Exercise 3.4. In this task we consider the following Dirichlet problem (weakformulation): Find u 2 W 1;20 () such that

a(u; v) = L(v) for all v 2 W 1;20 (); (3.5)

where

a(u; v) =

Z

4@u

@x1; 2@u

@x2

grad v dx (=

Z

4@u

@x1

@v

@x1+ 2

@u

@x2

@v

@x2dx);

L(v) =

Z

vdx

and = h0; 1i2 :Show that there exists a unique solution of (3.5).

Exercise 3.5. Given the fact that W 1;2() is a Hilbert space (this fact will notbe proven in this report). Let V be the space:

V =

u 2 W 1;2() :

Z

u dx = 0

(with the same norm as that for W 1;2()):

1. Show that V is a vector space.

2. Show that V is complete and conclude that V is a Hilbert space (this taskis more di cult, hint: use the inequality (1.5)).

29

3. Assume f 2 L2() and that the boundary of is Lipschitz continuous.Consider the modiedNeumann problem: Find u 2 V such that

a(u; ) = L () ; 8 2 V; (3.6)where (as before)

a(u; ) =

Z

gradu grad dx

and

L () =

Z

f dx:

Show that (3.6) has a unique solution.

Exercise 3.6. Prove that ifR

f dx = 0 and the boundary of is Lipschitz

continuous, then the solution of the Neumann problem (2.10) exists and is uniquewithin an additive constant (Hint: Observe rst that if 2 W 1;2() then thefunction v = ( 1jj

R

dx) belongs to the space V dened in the previous

exercise, where jj is the areaof , i.e. jj = R

dx).

Exercise 3.7. It turns out that the conditionR

f dx = 0 in the previous exercise

is a necessary condition to obtain a solution.

1. show this fact theoretically

2. give a physical explanation of this fact

4. Eective properties of periodic structures

Consider the stationary heat-conduction equation for a periodic structure of theform

div gradu = f; x 2 R2;where (x) is periodic relative to a small cell Y and bounded between the positiveconstants and + (see Figure 4.1).When the structure is symmetric it is possibleto use so-called G-convergence results to show that the eective behavior of theabove partial dierential equation is close to the homogeneous equation

div

1;e 00 2;e

gradu = f;

30

Figure 4.1: Periodic structure

where

i;e =1

jY jZY

(x)

@uper@xi

+ 1

dx; (4.1)

where uper is the solution of the following periodic problem, called the cell-problem:Find uper 2 W 1;2per(Y ) such thatZ

Y

graduper grad dx = ZY

ei grad dx; 8 2 W 1;2per(Y );

where ei is the usual basis-vector in the i-th direction. The parameter i;e iscalled the eective conductivity in the i-th direction.

Theorem 4.1. The cell-problem has a solution uper 2 W 1;2per(Y ): This solution isunique within an arbitrary constant, i.e. if uper;1 and uper;2 are solutions thenuper;1 uper;2 = constant:Proof: We put

a(uper; ) =

ZY

graduper graddx (4.2)

and

L() = ZY

ei grad dx: (4.3)

31

Thus the cell problem can be written as: Find uper such that

a(uper; ) = L(); 8 2 W 1;2per(Y ):We rst let V be the space

V =

2 W 1;2per(Y ) :

ZY

dx = 0

with the usual norm in W 1;2(Y ) and show that the problem

a(uper; ) = L(); 8 2 V (4.4)has a unique solution. This is done by rst proving that a(uper; ) and L() havethe properties 2), 3) and 4). Since V is a Hilbert space (see Exercise 3.5), theexistence and uniqueness will then follow by Theorem 3.1.Proof of condition 2): By Schwarz inequality (1.3)

ja(; v)j =Z

(grad) grad v dx

Z

jgradj2 dx 1

2Z

jgrad vj2 dx 1

2

Z

+ jgradj2 dx 1

2Z

+ jgrad vj2 dx 1

2

=

=+ 12

Z

jgradj2 dx 1

2 + 12

Z

jgrad vj2 dx 1

2

=

= +Z

jgradj2 dx 1

2Z

jgrad vj2 dx 1

2

+Z

jgradj2 + jj2 dx 12 Z

jgrad vj2 + jvj2 dx 12 == + kkV kvkV :

Thus we get thatja(; v)j kkV kvkV 8; v 2 V;

This proves 2).

32

Proof of condition 3):

a(; ) =

ZY

jgradj2 dx ZY

jgradj2 dx = ZY

jgradj2 dx:

By Poincars inequality we have thatZY

2 dx C0(ZY

dx

2+

ZY

jgradj2 dx):

Since V = 2 W 1;2per(Y ) :

RY dx = 0

; we haveZ

Y

jj2 dx C00 +

ZY

jgradj2 dx= C0

ZY

jgradj2 dx;

and soZY

jj2 dx+ZY

jgradj2 dx C0Z

Y

jgradj2 dx+

ZY

jgradj2 dx;

i.e. ZY

jj2 dx+ZY

jgradj2 dx (C0 + 1)Z

Y

jgradj2 dx;

which gives that

1

C0 + 1

ZY

jj2 dx+ZY

jgradj2 dxZY

jgradj2 dx:

Thus,

1

C0 + 1

ZY

2 dx+

ZY

jgradj2 dx

ZY

jgradj2 dx:

Hence we obtain that

a(; ) ZY

jgradj2 dx 1C0 + 1

ZY

jj2 dx+ZY

jgradj2 dx=

= 1

C0 + 1kk2V :

33

By putting = 1C0+1

we obtain that

a(; ) kk2Vwhich proves 3).Proof: Condition 4):

L() = ZY

ei grad dx

jL()j = Z

Y

ei grad dx = Z

Y

ei grad dx

Z

Y

jeij2 dx 1

2Z

Y

jgradj2 dx 1

2

Z

Y

+ jeij2 dx 1

2Z

Y

+ jgradj2 dx 1

2

=

=+ 12+ 12

ZY

jeij2 dx 1

2Z

Y

jgradj2 dx 1

2

=

=+Z

Y

jeij2 dx 1

2Z

Y

jgradj2 dx 1

2

+ZY

jeij2 dx 1

2Z

Y

jgradj2 dx+ZY

jj2 dx 1

2

=+Z

Y

jeij2 dx 1

2

kk2V(in the rst line we have used Schwarz inequality (1.4). By putting

=+Z

Y

jeij2 dx 1

2

;

we obtainjL()j kkV ;

which proves 4). Thus we have proved that the problem

34

a(uper; ) = L(); 8 2 Vhas a unique solution. But note that the solution uper must also be a solution tothe cell problem, because if 2 W 1;2per(Y ) then the function 0; dened by

0 = K;where

K =1

jY jZY

dx;

must belong to V (sinceZY

0dx =

ZY

dxZY

Kdx =

ZY

dxK jY j =ZY

dxZY

dx = 0):

Hence,a(uper; 0) = L(0):

Noting that

a(uper; ) = a(uper; 0 +K) =

Z

(graduper) grad(0 +K) dx

=

Z

(graduper) grad0 dx = a(uper; 0);and similarly that

L() = L(0);

we nd thata(uper; ) = L()

Next, let uper;1 and uper;2 be solutions to (4.4) then

a(uper;1; ) = L();

a(uper;2; ) = L();

soa(uper;1; ) a(uper;2; ) = 0; 8 2 W 1;2per(Y ):

Hence, using the linearity of a(:; :) we obtain that

a ((uper;1 uper;2); ) = 0; 8 2 W 1;2per(Y ):

35

Particularly this holds for = uper;1 uper;2;

i.e.a ((uper;1 uper;2); (uper;1 uper;2)) = 0;

and according to (4.2) we have that

a ((uper;1 uper;2); (uper;1 uper;2)) =ZY

jgrad(uper;1 uper;2)j2 = 0:

Thus ZY

jgrad(uper;1 uper;2)j2 dx = 0;

which gives us that

grad (uper;1 uper;2) = 0; 8 x 2 Y:

Thus uper;1 uper;2 = constant. This completes the proof.We also state the following theorem (without proof):

Theorem 4.2. It holds that

i;e =1

jY j min2W 1;2per (Y )

ZY

jgrad+ eij2 dx:

Exercise 4.1. Show thatZY

jgrad+ eij2 dx = 2F () +ZY

dx

whereF () =

1

2a(; ) L()

a(; ) and L() being given by (4.2) and (4.3). Next, use this, Theorem 3.2 andthe denition of i;e (4.1) to prove Theorem 4.2.

36

Figure 5.1: Finite triangulation of the body :

5. Finite Element Method

In order to formulate the nite element method starting from the weak formulationwe have to dene/construct a nite-dimensional subspace Vh of V . For simplicityand the sake of illustration we consider the Dirichlet problem (where V = W 1;20 ())for the conductivity problem (2.1) and assume that the body is divided intotriangles, where @ is a polygonal curve, see Figure 5.1. We must also introducea mesh-parameter

h = maxK2Th

diam (K); diam (K) = diameter of K = longest side of K;

where Th is the set Th = fK1; :::; Kmg of non-overlapping triangles Ki such that

=

SK2Th

K = K1 [K2::: [Km:

We dene Vh as

Vh = fw j w = 0 on @, continuous on ; wjK is a rst order polynomial for all K 2 ThgNote that wjK denotes the restriction of w to K (i.e. the function which is denedonly on K and which is equal to w on K), i.e. Vh consists of all continuousfunctions which is a rst order polynomial on each triangle K and vanishes on@. Observe that Vh V = W 1;20 ()

37

Figure 5.2: Basis function vj.

Remark 2. More generally V is a function space (Hilbert space) which takesinto account the boundary conditions (in the case of elasticity problems thesefunctions are vector valued), the trianglesin Th may be polygons and wjK maybe a (possibly vector-valued) polynomial..

To describe a function w 2 Vh we will use the values w(Ni) of w as parameters,where Ni; i = 1; :::;M are nodes (see Figure 5.1) of triangles Th. Since w = 0 on@ (i.e. known values) we may exclude the nodes on the boundary @. Considerthe basis function vj 2 Vh illustrated in Figure 5.2, where j = 1; :::;M , dened as

vj(Ni) = ij 1 if i = j0 if i 6= j

i; j = 1; :::;M:

The set of points x for which vj(x) 6= 0 is called the support of vj: For the functionvj this set consists of the triangles with common node Nj (see the shaded area inFigure 5.2). A function w 2 Vh has the representation

w(x) =

MXj=1

jvj(x); j = w(Nj); for x 2 [ @:

We are now in the position to formulate the following nite element method forthe variational formulation of the Dirichlet problem (based on the variational

38

formulation 2.5): Find uh 2 Vh such thata(uh; v) = L(v) 8v 2 Vh; (5.1)

Since uh = 1v1 + ::: + mvm; (5.1) holds and a(:; :) is a bilinear form we obtainthat

L(v1) = a(uh; v1) = a(1v1 + :::+ mvm; v1) = 1a(v1; v1) + :::+ ma(vm; v1)

L(v2) = a(uh; v2) = a(1v1 + :::+ mvm; v2) = 1a(v1; v2) + :::+ ma(vm; v2)...

L(vm) = a(uh; vm) = a(1v1 + :::+ mvm; vm) = 1a(v1; vm) + :::+ ma(vm; vm)

Thus 26664L(v1)L(v2)...L(vm)

37775 =26664a(v1; v1) a(v2; v1) a(vm; v1)a(v1; v2) a(v2; v2)...

. . .a(v1; vm) a(vm; vm)

377752666412...m

37775 ;i.e.

A = b; (5.2)

where A = (aij) is the m m stiness matrixwith elements (aij) = a(vi; vj)and = (i) and b = (bi) are m-vectors with elements i = uh(Ni); bi = L(vi):Thus solving this system with respect to directly gives us the solution

uh = 1v1 + :::+ mvm: (5.3)

Observe that problem (5.1) (which is equivalent with (5.2) has a unique solu-tion. This follows by letting Vh = V in Theorem 3.1. It is important to note thatuh is the best approximation in Vh of the solution u with respect to the energynorm. This means that

ku uhka ku vka 8v 2 Vh (5.4)where

kvka =pa (v; v):

Exercise 5.1. 1. Prove that a (u uh; uh v) = 0 for all v 2 Vh:2. Use that u v = (u uh) + (uh v) to show that

ku vk2a = ku uhk2a + 2a (u uh; uh v) + kuh vk2a3. Use the above exercises to show the inequality (5.4).

39

Figure 5.3: The functions and v:

Figure 5.4: Basis function v1:

5.1. Example 1 (dimension 1)

We will now show how to nd the FEM-solution in the one-dimensional heatconduction problem with Dirichlet boundary condition (V = W 1;20 ()). We put

= [0; 3] and let Th = f[0; 1] ; [1; 2] ; [2; 3]g : In Figure 5.3 we have illustrated ageneral function 2 V and a function v 2 Vh: The (two) basis functions for Vhare illustrated in Figure 5.4 and 5.5.We recall that

a(u; v) =

Z

gradu grad v dx

and

L(v) =

Z

fv dx

where f is an internal heating source which in our case has the form illustratedin Figure 5.6. We put = 1 for simplicity.Since this example is one-dimensional

40

Figure 5.5: Basis function v2:

Figure 5.6: The internal source f:

a(u; v) =

Z

gradu grad v dx =Z

du

dx dvdxdx:

In this case the linear systemA = b

takes the form: a(v1; v1) a(v1; v2)a(v2; v1) a(v2; v2)

12

=

L(v1)L(v2)

We need the expressions for a(v1; v1), a(v2; v1), a(v1; v2), a(v2; v2); L(v1),and L(v2):By considering Figure 5.4 we see that

a(v1; v1) =

Z 30

dv1dx

dv1dxdx =

=

Z 10

dv1dx

dv1dxdx+

Z 21

dv1dx

dv1dxdx+

Z 32

dv1dx

dv1dxdx =

41

=Z 10

1 1 1 dx+Z 21

1 (1) (1) dx+Z 32

1 0 0 dx =

=

Z 10

1 dx+

Z 21

1 dx+

Z 32

0 dx = [x]10 + [x]21 + 0 = 1 + 1 + 0 = 2:

Using both Figure 5.5 and 5.4 we obtain that

a(v2; v1) =

Z 30

dv2dx

dv1dxdx =

=

Z 10

dv2dx

dv1dxdx+

Z 21

dv2dx

dv1dxdx+

Z 32

dv2dx

dv1dxdx =

=

Z 10

1 0 1 dx+Z 21

1 1 (1) dx+Z 32

1 0 (1) dx =

=

Z 10

0 dx+

Z 21

1 dx+Z 32

0 dx = 0 + [x]21 + 0 = 0 1 + 0 = 1:

Using the symmetry fact of a(v1; v2) (or by considering Figure 5.5 and 5.4) wend that

a(v2; v1) = a(v1; v2) =

Z 30

dv2dx

dv1dxdx = 1:

By considering Figure 5.5 we see that

a(v2; v2) =

Z 30

dv2dx

dv2dxdx =

=

Z 10

dv2dx

dv2dxdx+

Z 21

dv2dx

dv2dxdx+

Z 32

dv2dx

dv2dxdx =

=

Z 10

1 0 0 dx+Z 21

1 1 1 dx+Z 32

1 (1) (1) dx =

=

Z 10

0 dx+

Z 21

1 dx+

Z 32

1 dx = 0 + [x]21 + [x]32 = 0 + 1 + 1 = 2:

To nd L(v1) we split the integral in 3 parts:

L (v1) =

Z 30

f v1 dx =Z 10

f v1 dx+Z 21

f v1 dx+Z 32

f v1 dx:

42

We need to know the expression for v1 is in each interval, which is found as follows:From Figure 5.4 we see that

v1 = x; for 0 x 1(equal to the linear line with the slope = 1). Since the slope is -1 in the nextinterval, we use the formula

y y1x x1 = 1

and the fact that the point (x1; y1) = (1; 1) lies on the graph of v1 to obtain

y 1x 1 = 1 () y = x+ 2:

This shows thatv1 = x+ 2; for 1 x 2:

Moreover,v1 = 0; for 2 x 3:

By Figure 5.6 we see that

f = 1; for 0 x 1;f = 1; for 1 x 2

andf = 0; for 2 x 3:

Thus,

L (v1) =

Z 30

f v1 dx =Z 10

1 x dx+Z 21

1 (x+ 2) dx+Z 32

0 0 dx =

=

Z 10

x dx+

Z 21

(x+ 2) dx+Z 32

0 dx =

1

2x210

+

12x2 + 2x

21

+ 0 =

=1

2+1

2+ 0 = 1:

Similarly, we have that

L (v2) =

Z 30

f v2 dx =Z 10

f v2 dx+Z 21

f v2 dx+Z 32

f v2 dx:

43

We must nd v2 in each interval, and this done similarly as above for v1: Summingup we obtain that

v2 = 0; for 0 x 1;v2 = x 1; for 1 x 2;v2 = x+ 3; for 2 x 3:

Thus

L (v2) =

Z 30

f v2 dx =Z 10

1 0 dx+Z 21

1 (x 1) dx+Z 32

0 (x+ 3) dx =

=

Z 10

0 dx+

Z 21

(x 1) dx+Z 32

0 dx = 0 +

1

2x2 x

21

+ 0 = 0 +1

2+ 0 =

1

2:

Now we have got all the parameters we need for solving the system

A = b;

which takes the form 2 11 2

12

=

112

;

The solution is 12

=

2 11 2

1 112

=

5646

:

(

2 11 2

1=1

3

2 11 2

)

(Recall the formula a dc b

1=

1

ab dcb dc a

)

Thus, we have found the approximative solution

uh = 1v1 + 2v2 =5

6v1 +

4

6v2

of the problem, see Figure 5.7. The solution uh is only an approximation of theexact solution u and can be improved by dividing into more intervals. This

44

Figure 5.7: The solution of the problem.

means that we will have to deal with not only two basis functions but severalmore which gives us a more complicated system. We get m equations to solve.The larger m the more accurate the solution will be. In two or three dimensionalproblems the complexity becomes even greater. That is why we are very gratefulfor the powerful computers that can solve these kind of problems numerically veryfast.

Exercise 5.2. Put = 1; = [0; 6] and let Th = f[0; 2] ; [2; 4] ; [4; 6]g : Moreover,let Vh be the two dimensional function space with basis functions v1 and v2 asillustrated in Figure 5.8. Find the best approximation uh in Vh of the exactsolution u for the Dirichlet problem when f is given as in Figure 5.8:

Exercise 5.3. Let ; and f be the same as in the previous exercise, but nowlet

Th = f[0; 1] ; [1; 2] ; [2; 3] ; [3; 4] ; [4; 5] ; [5; 6]gand Vh be the 5-dimensional function space with basis functions fvig as illustratedin Figure 5.9. Formulate the corresponding linear system:

A = b:

45

Figure 5.8: Basis functions and source.

Figure 5.9: Basis functions and source.

46

Exercise 5.4. It is important to remark that for the one dimensional case oneusually can nd the exact solution of the conductivity problem without anyFEM-calculation (the reason why we still do this is usually only for illustrationpurposes). Find the exact solution u for the Dirichlet problem when f = 1;

= [0; 1] and = 1 by solving the strong formulation of the problem directly.

Exercise 5.5. Try to do the same as in Exercise 5.4 but for the Neumann prob-lem. What happens? What if f = sin x; = [0; 2]: Discuss your observationsin view of Exercise 3.7).

5.2. Example 2 (dimension 2)

Now let be the square [0; 1]2 and = 1: Again we consider the Dirichlet problemand let Th be the triangulation illustrated in Figure 5.10 with N N (interior)nodes. Vh consist of all functions which are 0 on the boundary, continuous and arst order polynomial on each K 2 Th. The basis function vi has support in theshaded area in Figure 5.10 and vi (Ni) = 1: We have that

a (vi; vj) =

Z

grad vi grad vj dx =

=

Z

@vi@x1

;@vi@x2

@vj@x1

;@vj@x2

dx =

=

Z

@vi@x1

@vj@x1

+@vi@x2

@vj@x2

dx

Observe that a (vi; vj) = 0 unless when Ni and Nj are nodes of the same triangle.As a concrete example we will use with triangulation and 3 3 = 9 nodes

illustrated in Figure 5.11.To nd the matrix A in 5.2 where the components areAij = a (vi; vj)we need to know the gradients for each triangle in the supportedarea around each node in . All other gradients outside the shaded area are 0.We see from Figure 5.12 that each supported area for one node, which consists ofsix triangles (see Figure 5.12), looks the same for all nodes in . We will nownd the gradients on each triangle in the supported area.Triangle I:

@v1@x1

=114

= 4

@v1@x2

= 0

47

Figure 5.10: Two-dimensional :

Figure 5.11: with nine nodes.

48

Figure 5.12: Supported area for a node.

For the other triangles we argue similarly and obtain the following table:

Triangle @v1@x1

@v1@x2

I 4 0II 0 4III 4 4IV 4 0V 0 4VI 4 4

Now, we can nd each component of A from

Aij = a(vi; vj) =

Z

@vi@x1

@vj@x1

+@vi@x2

@vj@x2

dx:

The area of each triangle is

1

2(N + 1)2=

1

2(3 + 1)2=1

32

Therefore,

A11 = a(v1; v1) =V IXB=I

@v1@x1

B

@v1@x1

B

+

@v1@x2

B

@v1@x2

B

1

32=

=1

32[(4 4 + 0 0) + (0 0 + 4 4) + ((4) (4) + 4 4)+

49

Figure 5.13: The two rst gures shows the support of the basis function corre-sponding to node 1 and node 2 respectively. The last gure shows the intersectionbetween these supports.

+((4) (4) + 0 0) + (0 0 + (4) (4)) + (4 4 + (4) (4))] =

=1

32[16 + 16 + 32 + 16 + 16 + 32] = 4:

To nd A12 we must consider the support of the basis function of the two nodes 1and 2 and multiply the gradients in the intersection of the two shaded areas (seeFigure 5.13) with each other. The intersection ((III1 \ I2) [ (IV1 \ V I2)) is theonly contributing triangles (v1; v2 6= 0) in the element A12:

A12 = a(v1; v2) =

"@v1@x1

III1

@v2@x1

I2

+

@v1@x2

III1

@v2@x2

I2

#1

32+

+

"@v1@x1

IV1

@v2@x1

V I2

+

@v1@x2

IV1

@v2@x2

V I2

#1

32=

=1

32[((4) 4 + 4 0)] + 1

32[(4) 4 + 0 (4)] =

=1

32[16 + 0] + 1

32[16 + 0] =

12

+

12

= 1:

Element A13 = 0 since none of the triangles in node 1 and 3 are in contact witheach other. Element A14 must be checked because the triangles V1 and III4; and

50

the triangles V I1 and II4 is connected to each other. Therefore

A14 = a(v1; v4) =

"@v1@x1

V1

@v4@x1

III4

+

@v1@x2

V1

@v4@x2

III4

#1

32+

+

"@v1@x1

V I1

@v4@x1

II4

+

@v1@x2

V I1

@v4@x2

II4

#1

32=

=1

32[(0 (4) + (4) 4)] + 1

32[4 0 + (4) 4] =

12

+

12

= 1

Element A15 must also be checked because the triangles IV1 and II5; and thetriangles V1 and I5 is connected to each other. We obtain that

A15 = a(v1; v5) =

"@v1@x1

IV1

@v5@x1

II5

+

@v1@x2

IV1

@v5@x2

II5

#1

32+

+

"@v1@x1

V1

@v4@x1

I5

+

@v1@x2

V1

@v4@x2

I5

#1

32=

=1

32[((4) 0 + 0 4)] + 1

32[0 4 + (4) 0] = 0 + 0 = 0:

Element A15 turns out to be 0 even if it these triangles are connected. Note thatwhenever triangles connect we have to examine the element further, because it hasa potential to be dierent from 0. Node 1 has no further triangles from other nodesthat connect. This means that element A13, A16, A17, A18 and A19 are all 0. In thesame way we nd all other elements by examine each supported area at each nodewhere the triangles connect. All elements at the diagonal have the same coe cient,this means A11 = A22 = A33 = A44 = A55 = A66 = A77 = A88 = A99 = 4: We alsosee that A12 = A21 = A14 = A41 = 1: So far we have found the following part ofthe matrix: 26666666666664

4 -1 0 -1 0 0 0 0 0-1 40 4-1 40 40 40 40 40 4

37777777777775:

51

Figure 5.14: Coe sients.

If we now consider node 2 we see that it has triangles that connect with node1,2,3,5 and 6. The elements A12; A21 and A22 are already found, and we see thatA26 = A62 have the same relation to each other with respect to the triangles aselement A15 = 0. We also see that A25 = A52 and A23 = A32 have the samerelation to each other with respect to the triangles as element A14 = 1 andA12 = 1, respectively: If we continue with these kind of comparisons we ndthat the 9x9 matrix will look like this:26666666666664

4 -1 0 -1 0 0 0 0 0-1 4 -1 0 -1 0 0 0 00 -1 4 0 0 -1 0 0 0-1 0 0 4 -1 0 -1 0 00 -1 0 -1 4 -1 0 -1 00 0 -1 0 -1 4 0 0 -10 0 0 -1 0 0 4 -1 00 0 0 0 -1 0 -1 4 -10 0 0 0 0 -1 0 -1 4

37777777777775:

Each node has a connection to four other nodes where the coe cient turns outto be dierent from zero in addition to their own pairingin the system like theone in Figure 5.14.Something dierent happens for the elements that is aectedby the nodes on r(N) and r(N +1), where r describes the rows in ; see Fig.5.10.In our case N = 3 this happens for relations between node 3 and 4, and 6 and 7.This will aect the elements A34 = A43 and A67 = A76; by turning them into 0instead of -1 what we were likely to believe from the Figure 5.14: This happensbecause the shaded areas of the pair of nodes 3 and 4, and 6 and 7 do not connectto each other. If N in Figure 5.10 had been 10, we would have that the elementsA11; A12 6= 0 and the elements A13; :::; A1 10 = 0 while suddenly A1 11 6= 0, and allother elements would be 0.

52

Exercise 5.6. Find the stiness-matrix A when = 1; = [0; 1]2 and N = 2(i.e. 2 2 interior nodes and h = 1=3)

Exercise 5.7. Find the stiness-matrix A when = 1; =0; 5

4

with 4 4

interior nodes (i.e. the shortest side length of each triangle h = 1=4).

5.3. Example 3 (Eective conductivity of periodic structures)

We have in example 1 and 2 considered the Dirichlet problem. Let us now considerthe periodic problem given in 3.4. For simplicity we only consider the two-phaselaminate case, i.e. the case when Y can be described according to Figure 5.15. Let

Figure 5.15: The Y-cell.

us rst nd 2;e. We use the subspace Vh of V = W 1;2per(Y ) consisting of periodicroof-functions (see Figure 5.16) which are 0 at two of the boundaries.According

Figure 5.16: The roof-function.

53

Figure 5.17: Basisfunction with height 1.

to the description in 3.4 we have to nd the FEM-solution of the following problem:Find uh 2 Vh such thatZ

Y

graduh grad dx = ZY

e2 grad dx 8 2 Vh:

We continue as in example 1 and 2. Put

a (uh; ) =

ZY

graduh grad dx

and

L () = ZY

e2 grad dx:

As before we have to ndaij = a (vi; vj)

andbi = L (vi)

where vi are basis functions.. In this case we only have one basis function, vi:This function is 1 on the top of the roof, see also Fig5.17. We have to nda11 = a (v1; v1) and b1 = L (v1) : (Thats it.) We put

a (uh; v1) =

ZY

graduh grad v1 dx;

54

and then we obtain that

a (v1; v1) =

ZY

grad v1 grad v1dx =ZY

0;@v1@x2

0;@v1@x2

dx =

=

ZY

@v1@x2

@v1@x2

dx =

ZW

W@v1@x2

@v1@x2

dx+

ZB

B@v1@x2

@v1@x2

dx =

=

ZW

W (2 2) dx+ZB

B ((2) (2)) dx = W 4 12+B 4 1

2= W 2+B 2:

We also have that

L (v1) = ZY

e2 grad v1 dx = ZY

[0; 1]

0;@v1@x2

dx =

ZY

@v1@x2

dx =

= ZW

W@v1@x2

dxZB

B@v1@x2

dx = W 2 12 B (2) 1

2= W B:

In this case where A = [a11] ; = 1 and b = b1 we observe that

A = B =) a111 = b1which is the equation we have to solve. Then

(W 2 + B 2) 1 = W Band

1 =W B

(W 2 + B 2) :

Becauseuh = 1 v1

thus@uh@x2

= 1 @v1@x2

:

Finally from Section 4 we have that

2;e =1

jY jZY

(x)

@uh@x2

+ 1

dx =

=1

jY jZ

W

W (1 2 + 1) dx+ZB

B (1 (2) + 1) dx=

55

= 1

1

2W (1 2 + 1) +

1

2B (1 (2) + 1) dx

=

=1

2W

W + BW + B

+W + BW + B

+1

2B

W BW + B

+W + BW + B

=

=W BW + B

+W BW + B

= 2W BW + B

=1

121W+ 1

21B

;

which is the harmonic mean. Similarly we can derive that

1;e =1

2W +

1

2B;

which is the arithmetic mean.

Remark 3. In this example it turns out that the FEM-solution is the same asthe exact solution.

6. Further exercises

Exercise 6.1. What is a Hilbert space?

Exercise 6.2. Let be the open square Y = h0; 2i2 in R2: Compute theL2(Y ) norm and the W 1;2(Y ) norm of the function u(x1; x2) = cos x1 (recallthat

R 20sin2 t dt =

R 20cos2 t dt = )

Exercise 6.3. Let be the open square Y = h0; 2i2. Determine whether the fol-lowing functions are in some of the Lebesgue and Sobolev spaces L2(Y );W 1;2(Y );W 1;2per(Y );W

1;20 (Y ):

1. u(x1; x2) = x21 + x22

2. u(x1; x2) = x321

3. u(x1; x2) = x121 + x

122

4. u(x1; x2) = x 12

1

5. u(x1; x2) = cosx1 + sinx2

56

6. u(x1; x2) = cos x14

7. u(x1; x2) = sinx1 sin x2

Exercise 6.4. State (i.e. formulate) the Poincars inequality and the Friedrichsinequality

In the exercises below we consider the strong formulation of the followingDirichlet problem:

@2u

@x21+ k

@2u

@x22= f(x); x 2 (6.1)

u = 0 on @

where = h0; 1i2, k is a constant k 1 and f 2 L2():

Exercise 6.5. Show that the strong formulation (6.1) can be written on the form

div

@u(x)

@x1; k@u(x)

@x2

= f(x); x 2 ;

Exercise 6.6. Use this and the Greens formulaZ

divwdx = Z

grad w dx+Z@

w n ds

to verify the following weak formulation of the above problem: Find u 2 W 1;20 ()such that

a(u; v) = L(v) for all v 2 W 1;20 ()where

a(u; v) =

Z

@u(x)

@x1; k@u(x)

@x2

grad v dx =

=

Z

@u

@x1

@v

@x1+ k

@u

@x2

@v

@x2dx

and

L(v) =

Z

fvdx

57

Exercise 6.7. Show that the above weak formulation has a unique solution byusing the Lax-Milgram lemma.

Exercise 6.8. Show that the above weak formulation is equivalent with the fol-lowing minimum energy principle: Find u 2 W 1;20 () such that

F (u) = minu2W 1;20 ()

F (v);

where

F (v) =

Z

1

2

@v

@x1

2+k

2

@v

@x2

2 fv dx:

Exercise 6.9. Let g 2 W 1;2() and consider the following problem: Find w 2W 1;20 () such that

a(w; v) = G(v) for all v 2 W 1;20 ()where

G(v) = L(v) a(g; v)(L(v) and a(u; v) is dened above). Prove that this problem has a unique solution.

Exercise 6.10. Let g 2 W 1;2() and consider the following Dirichlet problem(corresponding to the boundary condition u = g on @): Find u 2 W 1;2(); suchthat u g 2 W 1;20 () and such that

a(u; v) = L(v) for all v 2 W 1;20 (): (6.2)Show that this problem has an unique solution. (Hint: Verify that u = w+ g is asolution, where w is given in Exercise 6.9).

Exercise 6.11. Do the same as in Exercise 5.3, with

Th = f[0; 1] ; [1; 2] ; [2; 3] ; [3; 4] ; [4; 6]gand where Vh is the corresponding 4-dimensional space.

Exercise 6.12. Let = 1=2; = [0; 1] and f = 1. Moreover, let

Th = f[0; h] ; [h; 2h] ; :::; [1 h; 1]gwith the corresponding FEM-space Vh V = W 1;20 (): The FEM-solution uhconverges to a function u as h! 0: Find the function u (without nding uh).

58

References

[1] Johnson, C. (1987) Numerical solution of partial dierential equations by the niteelement method, Studentlitteratur, Lund.

[2] Jikov, V.V., Kozlov, S.M. and Oleinik, O.A. (1994) Homogenization of dierentialoperators and integral functionals, Springer-Verlag, Berlin.

[3] Marti, J.T. (1986) Introduction to Sobolev Spaces and Finite Element Solution ofElliptic Boundary Value Problems. Academic Press, London.

[4] Meidell, A. (1998) Anvendt Homogeniseringsteori, ISBN-82-7823-037-4,HiN-rapport(51s.), Hgskolen i Narvik, Narvik.

[5] Persson, L-E., Persson, L., Svanstedt, N., Wyller, J. (1993) The HomogenizationMethod, Studentlitteratur, Lund.

59

7. Solutions to exercises

Exercise 1.1 Q is not a vector space since e.g. 1 2 Q; but 1 k =2 Q for any scalark which is irrational.

Exercise 1.2

1. ZY

ju(x)j2 dx =Z 10

(x)4 dx Z 10

(1)4 dx = 1

5. Similarly we prove that u 2 L2(Y ) and for u 2 W 1;2(Y ) for u(x) = cos (2x) ;but here we have that u(0) = u(1) = 1;so u 2 W 1;2per(Y ) but u =2 W 1;20 (Y )).

Exercise 1.3:

kukL2(Y ) =Z

Y

u2dx

12

=

Z 10

Z 10

x21 + x

22

2dx1dx2

12

=

r28

45

kukW 1;2(Y ) =Z

Y

u2 + jgraduj2 dx 1

2

=

Z 10

Z 10

x21 + x

22

2+ (2x1)

2 + (2x2)2dx1dx2

12

=

r148

45

Exercise 1.4: For u(x) = u(x1; x2) = x1=21 we have thatZ 1

0

j@u=@x2j2 dx = 0

and similarly as above we obtain that @u=@x2 2 L2(Y ) for all > 12 . If = 0 ,then @u=@x1 = 0 and @u=@x2 2 L2(Y ). Summing up this means that u 2 W 1;2(Y )when > 1

2or = 0:

Exercise 1.6: v 2? :The three functions v,

@v=@x1 =ex1 sin (x1) + ex1 cos (x1) sin (x2) ;

@v=@x2 = ex1 sin (x1) cos (x2)

are all bounded functions (cannot grow innitely large in any point in ). Forexample,

j@v=@x1j e01 + e01 1 1 = 1 + ;

which gives thatZ

@v

@x1

2dx

Z

(1 + )2 dx (1 + )2

are all bounded functions (cannot grow innitely large in any point in ). Forexample,

j@v=@x1j (2 2 (2 2 + 2)) = 24;which gives that Z

@v

@x1

2dx

Z

(24)2 dx (24)2 jj

u v =Z 11(x)u(x)v(x)dx =

Z 11(x)v(x)u(x)dx = v u

u u =Z 11(x) ju(x)j2 dx

Z 111 ju(x)j2 dx 0

and, thus, if u u = 0 then R 11 ju(x)j2 dx = 0 which implies that u = 0.2. kukV =

qR 11 1x

2dx =q

23

3. uh 2 V and uh ! u as h!1, where u is given by

u(x) =

0 if 1 x 01 if 0 < x 1 :

In order to see this we observe that

kuh ukV =sZ 1

1 juh uj2 dx =

=

vuuutZ 01 juh uj2 dx| {z }

0

+

Z 1=h0

juh uj2 dx+Z 11=h

juh uj2 dx| {z }0

=

=

sZ 1=h0

juh uj2 dx sZ 1=h

0

2 j1j2 dx =r1

h2! 0ash!1:

We observe that u =2 V (not continuous at x = 0): Thus V is not completeand V is therefore not a Hilbert space.

Exercise 1.9: We put v = 1 in (1.4) and obtainZ

u1 dx

Z

juj2 dx 1

2Z

1dx

12

:

By taking the square of each side we obtainZ

udx

2 K

Z

juj2 dx

(7.1)

64

where K =R

1dx. By the hint and the fact that v(0) = 0;we have that

jv(t)j Z 10

jgrad vj dx:

Hence, by (7.1) (with u = jgrad vj) we obtain thatZ 10

jv(t)j2 dt Z 10

Z

jgrad vj dx2dt =

Z

jgrad vj dx2 K

Z

jgrad vj2 dx;

i.e. Z

v2dx KZ

jgrad vj2 dx:

Exercise 2.1: We add the following n identities:

@ (w1)

@x1=

@

@x1w1 +

@w1@x1

...@ (wn)

@xn=

@

@xnwn +

@wn@xn

and obtaindiv (w)= grad w+ divw

the Divergence theorem of Gauss (with v =w)Z

div v dx =

Z@

v n ds

then yields that Z

grad wdx+Z

divwdx =

Z@

w n ds;

i.e. that Z

divwdx = Z

grad wdx+Z@

w n ds;

which is (2.3).

65

Exercise 2.2:

a(u+ v; w) =

Z

grad (u+ v) gradw dx =

=

Z

gradu gradwdx+Z

grad v gradwdx= a(u;w) + a(v; w)

Similarly we prove thata(ku; w) = ka(u;w);

a(w; u+ v) = a(w; u) + a(w; v);

a(u; kw) = ka(u;w);

Thus a is a bilinear form on W 1;20 ().

Exercise 2.3:

1. If w 6= 0; then a(u+v; w) = u+v+w 6= (u+ w)+(v + w) = a(u;w)+a(v; w)i.e. a(u+ v; w) 6= a(u;w) + a(v; w): Thus a is not a bilinear form

2. Is a bilinear form

3. Is a bilinear form

4. Is a bilinear form

5. Is not a bilinear form (if u(0)v(0) 6= 0 and k > 1 then a(ku; v) = ((ku(0)) v(0))2 =k2 (u(0)v(0))2 6= k (u(0)v(0))2 = ka(u; v); i.e. a(ku; v) 6= ka(u; v))

Exercise 3.1

1. Using the scalar product u v = u1v1 + 2u2v2 we obtain from the Schwarzinequality

ju vj (u u) 12 (v v) 12that

ju1v1 + 2u2v2j u21 + 2u

22

12v21 + 2v

22

12

66

2. We observe rst that a(u; v) is a bilinear form and want to check if theconditions 2-4 are satised: Condition 2: We must prove that ja(u; v)j

kuk kvk ; for some constant > 0; which in this case means that

ja(u; v)j u21 + u22 12 v21 + v22 12By (3.3) we have that

ja(u; v)j = ju1v1 + 2u2v2j u21 + 2u

22

12v21 + 2v

22

12

2u21 + 2u22 12 2v21 + 2v22 12 = (2) 12 u21 + u22 12 (2) 12 v21 + v22 12= 2

u21 + u

22

12v21 + v

22

12 = 2 kuk kvk :

Thusja(u; v)j 2 kuk kvk ;

so the condition 2 is satised for = 2: We turn to condition 3:

ja(v; v)j = v1v1 + 2v2v2 = v21 + 2v22 v21 + v22 = kvk2 :

Thusja(v; v)j kvk2 ;

so the condition 3 is satised for = 1:Finally we prove condition 4:

jL(v)j = j2v1 + 4v2j = j[2; 4] [v1; v2]j

j[2; 4]j j[v1; v2]j =22 + 42

12 kvk = 2

p5 kvk ;

where the last inequality follows by Schwarz inequality. Thus, jL(v)j 2p5 kvk ; so the condition 4 is satised for = 2p5: Hence, conditions 2-4

are satised, and the problem has a unique solution by Lax Milgram lemma.

3. We have thata(u; v) = L(v) for all v 2 V;

where a(u; v) = u1v1 + 2u2v2 and L(v) = 2v1 + 4v2: Putting

v =

v1v2

=

10

and v =

v1v2

=

01

;

67

we obtain the two equations

u1 1 + 2u2 0 = 2 1 + 4 0;u1 0 + 2u2 1 = 2 0 + 4 1;

which gives the solution

u =

u1u2

=

22

:

4. We have that

F (v) =1

2a(v; v) L(v) = 1

2(v1v1 + 2v2v2) (2v1 + 4v2) =

=

1

2v21 2v1

+v22 4v2

:

In this simple case we can easily nd the minimum point of F (v) by ndingthe minimum points of each of the two terms 1

2v21 2v1 and v22 4v2; sepa-

rately. This is done by dierentiating with respect to v1 and v2 in the rstand second term, respectively:

d12v21 2v1

dv1

= v1 2 = 0) v1 = 2;d (v22 4v2)

dv2= 2v2 4 = 0) v2 = 2:

Thus, the solution (which we as usual call) u is

u =

u1u2

=

22

:

Comments: This is the same solution as we found above, which agreeswith the minimum theorem (Theorem 3.2) since a(u; v) and L(v) satisfyall conditions 1-4. Indeed, above we have veried that the conditions 2,3,4are satised, but we see directly that even condition 1 is satised, sincea(u; v) = u1v1 + 2u2v2 = v1u1 + 2v2u2 = a(v; u); i.e. a(u; v) = a(v; u).

Exercise 3.2

68

1. For u = 0 we obtain that

a(u; v) = 0 v1 + 0 v2 + 0 v1 + 0 v2 = 0for all v 2 V: Since L(v) = 0 for all v 2 V; we see that a(u; v) = L(v) for allv 2 V; so u = 0 is therefore a solution: For the other vector

u =

11

we obtain that

a(u; v) = u1v1 + u1v2 + u2v1 + u2v2

= 1 v1 + 1 v2 + (1) v1 + (1) v2 = 0;so that a(u; v) = L(v) for all v 2 V; which means that u is a solution.

2. Let L(v) = v1 and assume that the problem a(u; v) = L(v) has a solution u(as we soon will see, this will lead to a contradiction): Then, a(u; v) = L(v);particularly for

v =

10

and v =

01

;

which gives the two equations

u1 1 + u1 0 + u2 1 + u2 0 = 1;u1 0 + u1 1 + u2 0 + u2 1 = 0:

In other words,

u1 + u2 = 1;

u1 + u2 = 0;

which is impossible. Thus there are no solutions u to the above problem.

3. Suppose that condition 3 is satised (as we soon will see, this will lead to acontradiction) , i.e. that there is a constant > 0 such that

a(v; v) kvk2V 8v 2 V:In our case this means that

v1v1 + v1v2 + v2v1 + v2v2 v21 + v

22

;

69

Since this inequality holds for all v 2 V; it also holds for example whenv1 = 1 and v2 = 1: But this implies that

1 1 + 1 (1) + (1) 1 + (1) (1) 12 + (1)2 ;i.e. that

0;which contradicts the assumption that > 0: Thus we can conclude thatcondition 3 is not satised.

Exercise 3.3:Solution: We replace

@2u

@x21+@2u

@x22

with div gradu, multiply with v 2 W 1;20 () on both sides of the strong formulationand integrate:Z

v div gradudxZ

(@u(x)

@x1+@u(x)

@x2)v(x) dx =

Z

v(x) sinx1dx: (7.2)

Greens formula with w =gradu givesZ

v div gradudx = Z

grad v gradudx+Z@

v gradu n ds:

Inserting this into (7.2) we obtain

Z

grad vgradudx+Z@

v gradun dsZ

(@u(x)

@x1+@u(x)

@x2)v(x) dx =

Z

v(x) sinx1dx

The second term on the left side vanishes because v = 0 on @; and we obtainZ

grad v gradudx+Z

(@u(x)

@x1+@u(x)

@x2)v(x) dx =

Z

v(x) sinx1dx;

which is the same as (3.4).Next, we prove that L(v) is continuous:

jL(v)j =Z

v(x) sinx1dx

|{z}Schwartz

sZ

(v(x))2 dx

sZ

(sinx1)2 dx

70

sZ

(v(x))2 dx

sZ

(1)2 dx = 1 sZ

(v(x))2 dx

1 sZ

(v(x))2 + jgrad vj2 dx = 1 kvk :

Hence, jL(v)j 1 kvk for all v 2 W 1;20 (), which means that L is continuous.

Exercise 3.4:We observe rst that a(u; v) is a bilinear form and want to check ifthe conditions 2-4 are satised: Condition 2:

ja(u; v)j =Z

4@u(x)

@x1; 2@u(x)

@x2

grad v dx

Scwarz ineq.

sZ

4@u(x)@x1 ; 2@u(x)@x22 dx

sZ

jgrad vj2 dx =

=

sZ

42@u(x)

@x1

2+ 22

@u(x)

@x2

2dx

sZ

jgrad vj2 dx

since 2242

sZ

42@u(x)

@x1

2+ 42

@u(x)

@x2

2dx

sZ

jgrad vj2 dx =

= 4

sZ

@u(x)

@x1

2+

@u(x)

@x2

2dx

sZ

jgrad vj2 dx =

= 4

sZ

jgraduj2 dxsZ

jgrad vj2 dx

4sZ

jgraduj2 + juj2 dxsZ

jgrad vj2 + jvj2 dx = 4 kuk kvk

i.e.ja(u; v)j 4 kuk kvk

Condition 3:

a(v; v) =

Z

4@v(x)

@x1; 2@v(x)

@x2

grad v dx

=71

Z

4

@v(x)

@x1

2+ 2

@v(x)

@x2

2dx

since 42Z

2

@v(x)

@x1

2+ 2

@v(x)

@x2

2dx

= 2Z

@v(x)

@x1

2+

@v(x)

@x2

2dx

= 2

Z

jgrad v j2 dx 2 1C0 + 1

kvk2

[The last inequality follows by Friedrichs inequality:Z

v2 dx C0Z

jgrad vj2 dx:

which givesZ

v2 dx+

Z

jgrad vj2 dx C0Z

jgrad vj2 dx+Z

jgrad vj2 dx =

= (C0 + 1)

Z

jgrad vj2 dx:

Thus1

C0 + 1

Z

v2 dx+

Z

jgrad vj2 dxZ

jgrad vj2 dx

i.e.1

C0 + 1

kvk2 Z

jgrad vj2 dx:]

Thus ja(v; v)j kvk2 where = 2 1C0+1

:Condition 4:

L(v) =

Z

v dx

jL(v)j =Z

v 1 dx Z

jv 1j dx |{z}Schwarz

Z

jvj2 dx 1

2Z

j1j2 dx 1

2

Z

jvj2 + jgrad vj2 dx 1

2Z

j1j2 dx 1

2

= kvkW 1;20 ()

where =R

j1j2 dx 12 = 1. Thus jL(v)j kvkV which proves condition 4.

72

All conditions are satised, and hence, existence and uniqueness of the solu-tion follow by Lax-Milgrams lemma

Exercise 3.5:

1. V is a vector space: Indeed if u 2 V and v 2 V and k 2 R we havethat u + v 2 W 1;2() and ku 2 W 1;2() (simply because W 1;2() is avector space). Moreover,

R

u + v dx =

R

udx +

R

v dx = 0 + 0 = 0 andR

ku dx = k

R

u dx = k0 = 0; i.e. u+ v 2 V and ku 2 V:

2. Next we show that V is complete: Let fuhg be a sequence in V convergingto a function u: Then u 2 W 1;2() since W 1;2() itself is complete. By theinequality (1.5) we obtain thatZ

(u uh) dx2 K

Z

ju uhj2 dx

KZ

ju uhj2 dx+Z

jgrad (u uh)j2 dx= K ku uhk2W 1;2() ;

i.e. 0BB@Z

udxZ

uhdx| {z }=0

1CCA2

K ku uhk2W 1;2() ! 0;

ThusR

udx = 0 and we obtain that u 2 V: This shows that V is complete.

3. We prove that a(:; :) and L satises the conditions 2, 3 and 4. This is doneexactly as for the Dirichlet problem, except that we must use the Poincarsinequality instead of Friedrichs inequality. Thus existence of a solutionfollows by Lax-Milgram lemma (Theorem 3.1).

Exercise 3.6:If 2 W 1;2(); then the function v = ( 1jj

R

dx) belongs to V (since

1jjR

vdx = 0 ). If u is the solution of (3.6) then

a(u; v) = L (v) : (7.3)

Sincegrad = grad v;

73

we have that

a(u; ) =

Z

( gradu grad)dx =Z

( gradu grad v)dx = a(u; v) (7.4)

Moreover,

L (v) =

Z

fvdx =

Z

f

( 1jj

Z

dx)

dx

=

Z

fdx ( 1jjZ

dx)

Z

fdx| {z }=0

(7.5)

=

Z

fdx = L ()

Thus, (7.3), (7.4) and (7.5) gives that

a(u; ) = L () ;

and this holds for all 2 W 1;2(). This shows that u is a solution of the Neumannproblem (2.10).Next, let u2 be an other solution to (2.10). Then

a(u; ) = L()

a(u2; ) = L()

thena(u; ) a(u2; ) = 0; 8 2 W 1;2(Y ):

Hence, using the linearity of a(:; :) we obtain that

a ((u u2); ) = 0; 8 2 W 1;2(Y ):Particularly this holds for

= u u2i.e.

a ((u u2); (u u2)) = 0;Thus Z

Y

jgrad(u u2)j2 dx = 0;

74

which gives us thatgrad (u u2) = 0; 8 x 2 :

Thus u u2 = constant.

Exercise 3.7:

1. By inserting = 1 in a(u; ) = L() we obtain thatZ

gradugrad| {z }0

dx =

Z

|{z}1

fdx;

i.e. that Z

fdx = 0

2. The Neumann condition @u=@n = 0 on the boundary implies no heat transferthrough the boundary, i.e. the body is insulated. If the total heat deliveredto the body Z

fdx

is larger than 0 then temperature will go to 1; if it is less than 0 thetemperature will never stop to fall. Both these extreme cases of solutionsdo not belong to W 1;2():

Exercise 4.1:ZY

jgrad+ eij2 dx =ZY

(grad+ ei) (grad+ ei) dx =

=

ZY

jgradj2 dx+2ZY

ei grad dx+ZY

jeij2 dx = a(; )2L()+ZY

dx

2F () +

ZY

dx

Thus, we obtain

min2W 1;2p er (Y )

1

jY jZY

jgrad+ eij2 dx = 2jY j min2W 1;2p er (Y )F () +

1

jY jZY

dx

= (according to Theorem 3.2) =

75

=2

jY jF (uper) +1

jY jZY

dx =2

jY j1

2a(uper; uper) L(uper)

+

1

jY jZY

dx

=2

jY j1

2L(uper) L(uper)

+

1

jY jZY

dx =1

jY jL(uper) +

ZY

dx

=

1

jY jZ

Y

ei graduper dx+ZY

dx

=

1

jY jZ

Y

(@uper@xi

+ 1)dx

= i;e:

Exercise 5.1: Since = v uh 2 Vh V , we get thata (u uh; ) = a (u; ) a (uh; ) = L() L() = 0:

Moreover, since a is bilinear and symmetric, we derive that

kx+ yk2a = a (x+ y; x+ y) = a (x; x) + 2a (x; y) + a (y; y) == kxk2a + 2a (x; y) + kyk2a

Thus

ku vk2a = k(u uh) + (uh v)k = ku uhk2a + 2a (u uh; uh v) + kuh vk2a :Since a (u uh; uh v) = 0; we obtain that

ku vk2a = ku uhk2a + kuh vk2aThus,

ku uhk2a = ku vk2a kuh vk2a ku vk2a

Exercise 5.2: 1 1

2121

12

=

21

;

The solution is 12

=

1 1

2121

1 21

=

10383

:

Exercise 5.3:

76

x 6543210

3.5

3

2.5

2

1.5

1

0.5

0

Figure 7.1: The plot of uh in the case of two and ve basis-functions

A =

2666642 1 0 0 01 2 1 0 00 1 2 1 00 0 1 2 10 0 0 1 2

37777526666412345

377775 =26666411112

0

377775The solution is (not a part of the exercise, see also Figure 7.1):266664

12345

377775 =266664

136103728343

377775

Exercise 5.4: Strong formulation:

div 1 gradu = div gradu =d

dx

du

dx

= 1

Integration gives:du

dx= x+ c1

u(x) =1

2x2 + c1x+ c2

77

Since u(0) = u(1) = 0; c2 = 0 and 0 = 1212 + c11; i.e. c1 = 12 : Thus

u(x) =1

2x2 1

2x:

Exercise 5.5: From above we have that

du

dx= x+ c1

Since we are dealing with the Neumann problem we have that

du

dx(0) = 0 + c1 = 0;

(i.e. c1 = 0) anddu

dx(1) = 1 + c1 = 0;

(i.e. c1 = 1) which is a contradiction, so this is impossible, there exists nosolution in this case. On the other hand if f = sin x; = [0; 2]; then

d

dx

du

dx

= sinx;

which gives thatdu

dx= cosx+ c1:

The Neumann condition gives that

du

dx(0) = cos 0 + c1 = 0

( i.e. c1 = 1) anddu

dx(2) = cos 2 + c1 = 0

( i.e. c1 = 1): Thus there are no contradictions as above where f = 1: Integrating cosx+ 1 we obtain that

u = sin x+ x+ c2;

78

where c2 is an arbitrary constant. We note thatZ

fdx =

Z 20

sin xdx = 0:

According to Exercise 3.7) this is in general a necessary condition for obtaining asolution for the Neumann problem. In the previous exerciseZ

fdx =

Z 10

1dx = 1 6= 0:

Thus, we conclude that our observations agrees with the general theory stated inExercise 3.7.

Exercise 5.6:

A =

26644 1 1 01 4 0 11 0 4 10 1 1 4

3775

Exercise 5.7:A =2666666666666666666666666664

4 1 0 0 1 0 0 0 0 0 0 0 0 0 0 01 4 1 0 0 1 0 0 0 0 0 0 0 0 0 00 1 4 1 0 0 1 0 0 0 0 0 0 0 0 00 0 1 4 0 0 0 1 0 0 0 0 0 0 0 01 0 0 0 4 1 0 0 1 0 0 0 0 0 0 00 1 0 0 1 4 1 0 0 1 0 0 0 0 0 00 0 1 0 0 1 4 1 0 0 1 0 0 0 0 00 0 0 1 0 0 1 4 0 0 0 1 0 0 0 00 0 0 0 1 0 0 0 4 1 0 0 1 0 0 00 0 0 0 0 1 0 0 1 4 1 0 0 1 0 00 0 0 0 0 0 1 0 0 1 4 1 0 0 1 00 0 0 0 0 0 0 1 0 0 1 4 0 0 0 10 0 0 0 0 0 0 0 1 0 0 0 4 1 0 00 0 0 0 0 0 0 0 0 1 0 0 1 4 1 00 0 0 0 0 0 0 0 0 0 1 0 0 1 4 10 0 0 0 0 0 0 0 0 0 0 1 0 0 1 4

377777777777777777777777777579

7.1. Solution to further exercises

Exercise 6.2:kukL2 =

qR 20

R 20(cosx1)

2 dx1dx2 =p2,

kukW 1;2 =qR 2

0

R 20

(cosx1)

2 + ( sin x1)2dx1dx2 = 2

Exercise 6.3: The smallest space u belongs to is: 1)W 1;2(Y ) 2)W 1;2(Y ) 3)L2(Y )4)non of these spaces 5)W 1;2per(Y ) 6)W

1;2(Y ) 7)W 1;20 (Y )

Exercise 6.6: We multiply a test function 2 W 1;20 () on both sides, andintegrate Z

div

@u(x)

@x1; k@u(x)

@x2

dx =

Z

f dx:

We do as before except that we use

w =

@u(x)

@x1; k@u(x)

@x2

and obtain the weak formulation.

Exercise 6.7 and 6.8: We observe rst that a(u; v) is a bilinear form and wantto check if the conditions 1-4 are satised: condition 1)

a(u; v) =

Z

@u

@x1

@v

@x1+ k

@u

@x2

@v

@x2dx =

Z

@v

@x1

@u

@x1+ k

@v

@x2

@u

@x2dx = a(v; u):

Condition 2:

ja(u; v)j =Z

@u(x)

@x1; k@u(x)

@x2

grad v dx

Scwarz ineq.

sZ

@u(x)@x1 ; k@u(x)@x22 dx

sZ

jgrad vj2 dx =

=

sZ

@u(x)

@x1

2+ k2

@u(x)

@x2

2dx

sZ

jgrad vj2 dx

80

since 1k

sZ

k2@u(x)

@x1

2+ k2

@u(x)

@x2

2dx

sZ

jgrad vj2 dx =

= k

sZ

@u(x)

@x1

2+

@u(x)

@x2

2dx

sZ

jgrad vj2 dx =

= k

sZ

jgraduj2 dxsZ

jgrad vj2 dx

ksZ

jgraduj2 + juj2 dxsZ

jgrad vj2 + jvj2 dx = k kuk kvk

i.e.ja(u; v)j k kuk kvk

Condition 3:

a(v; v) =

Z

@v(x)

@x1; k@v(x)

@x2

grad v dx

=Z

@v(x)

@x1

2+ k

@v(x)

@x2

2dx

since k1Z

@v(x)

@x1

2+

@v(x)

@x2

2dx

=Z

jgrad v j2 dx 1C0 + 1

kvk2

The last inequality follows by Friedrichs inequality [This is shown earlier in thetext for the Dirichlet problem]. Thus ja(v; v)j kvk2 for some positive constant:Condition 4: The proof is the same as for the Dirichlet problem (see the text).Since conditions 2-4 are satised there exists a unique solution by Lax-Milgram

lemma (Theorem 3.1), moreover since conditions 1-4 are satised the equivalenceof the two formulations follows by Theorem 3.2.

Exercise 6.9: Again we must show that conditions 2-4 are satised. We havealready proved condition 2) and 3) for a(w; v) above.Condition 4: We have that:

jG(v)j = jL(v) a(g; v)j jL(v)j+ ja(g; v)j (7.6)

81

Since L(v) satises condition 4 (jL(v)j c kvk for some positive constant c) anda(g; v) satises condition 2 (ja(g; v)j kgk kvk for some positive constant ) weobtain from (7.6) that

jG(v)j c kvk+ kgk kvk = (c+ kgk) kvk

i.e.jG(v)j kvk ; 8v

for some positive constant (= c+ kgk). Thus the existence and uniqueness ofthe solution follows by Lax-Milgram lemma.

Exercise 6.10: u g = (w + g) g = w 2 W 1;20 (): Moreover,

a(u; v) = a(w + g; v) =bilinear

a(w; v) + a(g; v) = G(v) + a(g; v)

= (L(v) a(g; v)) + a(g; v) = L(v):

Summing up: u g 2 W 1;20 () and a(u; v) = L(v) for all v 2 W 1;2(): Hence u isa solution of the problem. The uniqueness of the solution is proved as follows: Letu1 and u2 be solutions w1 = u1 g and w2 = u2 g are solutions of the problemgiven in Exercise 6.9 (since a(w1; v) = a(u1; v) a(g; v) = L(v) a(g; v) = G(v)and similarly that a(w2; v) = G(v)). But since these are unique, we have thatw1 = w2; i.e. u1 = u2)

Exercise 6.11: uh converges to the exact solution u; which can be found bysolving the problem (as before)

div1

2gradu =

1

2div gradu =

1

2

d

dx

du

dx

= 1

u(x) = x2 x:

82