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Contents

1 Classification of PDE Models 4

1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Elliptic PDEs 16

2.1 Maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.2 Finite Difference Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3 Sobolev Spaces and Variational Formulation . . . . . . . . . . . . . . . . . 36

2.4 Finite Element Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3 Parabolic PDEs 66

3.1 Initial-boundary value problems . . . . . . . . . . . . . . . . . . . . . . . . 66

3.2 Finite difference methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

3.3 Stability analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

3.4 Semidiscretisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4 Hyperbolic PDEs of Second Order 91

4.1 Wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

4.2 Finite difference methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

4.3 Methods of Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

5 Hyperbolic Systems of First Order 114

5.1 Systems of two equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

5.2 Conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

5.3 Numerical methods for linear systems . . . . . . . . . . . . . . . . . . . . . 131

5.4 Conservative methods for nonlinear systems . . . . . . . . . . . . . . . . . 148

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Chapter 1

Examples and Classification

1

This lecture deals with the numerical solution of partial differential equa-tions(PDEs). The exact solution depends on several independent variables,which are often the time and the space coordinates. Different types of PDEmodels exist even in the linear case. Each class exhibits certain propertiesand thus requires corresponding numerical methods. Initial and/or bound-ary conditions appear.

In contrast, systems ofordinary differential equations(ODEs) can be writ-ten in the general form

y(x) =f(x, y(x)) (y: n, f : n n).The independent variable x often represents the time. Thus initial valueproblems y(x0) = y0 are the most important task. An analytical solutionis not feasible in general. Hence we need numerical methods to achieve an

approximate solution. Nevertheless, a convergent numerical method canresolve an arbitrary system of ODEs.

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1.1 Examples

We present three important examples, which illustrate the three classes ofPDE models.

Poisson equation

We consider an open bounded domain 2. For example, we can choose = (0, 1) (0, 1). Foru C2(), the Laplace operator is defined via

u:=2u

x2 +2u

y2 . (1.1)

The Poisson equation (in two space dimensions x, y) reads

u= fwith a predetermined function f :

. The special case f 0 re-

produces the Laplace equation. Hence the solution of the Poisson equation(1.1) is stationary, i.e., it does not change in time.

Now we specify boundary value problems. Let be the boundary of .Boundary conditions of Dirichlet type read

u(x, y) =g(x, y) for (x, y) with a given function g :

. Boundary conditions of Neumann type

specify the derivative of the solution perpendicular to the boundary, i.e.,

u

(x, y) := (x, y), u(x, y) =h(x, y) for (x, y) with the normal vector (2 = 1) and a given function h : .Often mixed boundary conditions

u(x, y) =g(x, y) for (x, y) D, u

(x, y) =h(x, y) for (x, y) N

appear with D N= and D N= .

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We derive the Poisson equation for electric fields in case of three space dimensions (x =(x1, x2, x3)). Let E :

3 3 be the electric field and : 3 the correspondingpotential. It holds

E(x) =(x)with the gradient = (

x1, x2

, x3

). It follows

divE(x) = (x).

Let : 3 describe the charge distribution and >0 be the permittivity. The firstMaxwells equation (Gauss law) reads

divE(x) =(x)

.

Comparing the two relations, we obtain the Poisson equation

(x) = (x)

,

whereis given and is unknown.

A connection to complex analysis is given for holomorphic functions. Letg : beholomorphic and be bounded, connected and open. On the one hand, Cauchysintegral formula yields

g(z) = 1

2i

g()

z d for z .

It follows that g is already determined uniquely inside by its values on the boundary .On the other hand, the formulas of complex differentiation (Cauchy-Riemann-PDEs) imply

(Reg) = 0 and (Img) = 0,

i.e., real and imaginary part are solutions of the Laplace equation in two dimensions. The

values ofg on specify Dirichlet boundary conditions. It follows a unique solution for the

real and the imaginary part in , respectively. Thus the two theoretical concepts agree.

Wave equation

In a single space dimension, the wave equation reads

2u

t2 =c2

2u

x2, (1.2)

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where the real constant c > 0 is the wave speed. The solutionu dependson space as well as time. We solve the wave equation using dAlemberts

method. New variables are introduced via

=x ct, =x + ct.It follows

x =

x+

x =

+

,

2

x

2 =

2

2+ 2

2

+ 2

2,

t =

t+

t = c

+

,

2

t2 = c2

2

2 2

2

+

2

2

.

We obtain the transformed PDE

c22

2 2

2

+

2

2u(, ) =c2

2

2+ 2

2

+

2

2u(, )and thus

2u

= 0.

It is straightforward to verify that the general solution is given by

u(, ) = () + ()

with arbitrary functions , C2( ). A special case is (only for

u

t (x, 0) 0). It followsu(x, t) = (x ct) + (x+ ct).

The functions , follow from initial conditions. As an interpretation, werewrite

(x ct) = (x+ ct c(t + t)) = (x ct)with x := x+ ct and t := t+ t. In the period t, the informationtravels from the point x to the point x with x = ct. Hence the term

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(xct) represents a wave moving at speedc. Likewise, the term (x + ct)yields a wave moving at speed

c. The solution u is the superposition of

two waves travelling at opposite speeds.

For initial values u(x, 0) = u0(x), ut (x, 0) = cu1(x) at t= 0, a formula for

the exact solution is available, i.e.,

u(x, t) =1

2

u0(x + ct) + u0(x ct) +

x+ctxct

u1(s) ds

. (1.3)

It follows that the solutionu at a point (x, t) fort>0 depends on initial

values at t= 0 in the interval x[x ct, x+ct] only. Hence the waveequation includes a transport of information at a finite speed.The linear PDE (1.2) of second order can be transformed into a corre-sponding system of PDEs of first order. We definev1 :=

ut and v2 :=

ux .

Assumingu C2, the theorem of Schwarz yieldsv1x

= 2u

xt=

v2t

.

The PDE (1.2) implies v1t

=c2v2x

.

It follows the system

t

v1v2

+

0 c21 0

x

v1v2

=

00

. (1.4)

The resulting matrix exhibits the eigenvalues +c andc, i.e., the wavespeeds. To obtain the solution u of (1.2), an integration using v1, v2 stillhas to be done.

Heat equation

In a single space dimension, the heat equation reads

u

t =

2u

x2 (1.5)

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with a constant > 0. To demonstrate some solutions of this PDE, wechoose = 1 and consider the finite domain x

[0, ] without loss of

generality. We arrange homogeneous boundary conditions

u(0, t) = 0, u(, t) = 0 for all t 0. (1.6)The functions

vk(x, t) := ek2t sin(kx) for k (1.7)

satisfy the heat equation (1.5) and the boundary conditions (1.6). Giveninitial values u(x, 0) =u0(x) for x[0, ] with u0(0) = u0() = 0, we can

apply a Fourier expansion

u0(x) =k=1

aksin(kx)

with coefficients ak . Since the heat equation (1.5) is linear, we obtainthe solution as a superposition of the functions (1.7)

u(x, t) =

k=1 akek2t sin(kx)

fort 0.Alternatively, boundary conditions of Neumann type can be specified. Ho-mogeneous conditions

u

x(0, t) = 0,

u

x(, t) = 0 for all t 0

imply that there is no heat flux through the boundaries.

Given initial conditionsu(x, 0) =u0(x) in the whole space domain, it followsa formula for the exact solution of the heat equation (= 1)

u(x, t) = 1

2

t

+

e2/4tu0(x ) d (1.8)

provided that the integral exists. We recognise that the solution in somepoint (x, t) depends on the initial values u0() for all . Hence the

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transport of information proceeds at infinite speed. However, the magnitudeof the information is damped exponentially for increasing distances.

We derive the heat equation in three space dimensions. LetT : 3 be the temperature, F : 3 3 the heat flux and > 0 the diffusionconstant. It follows

F = T.For the energy E:

3 , we obtain

E

t = div F =divT =T.

Let := ET be a constant material parameter. It holds Et =

ET

Tt . Con-

sequently, the heat equation

T

t =

T

is achieved with= .

Black-Scholes equation

The above examples are motivated by physics and technical applications.We discuss shortly an example from financial mathematics. LetSbe theprice of a stock and V be the fair price of a European call option basedon this stock. It follows a PDE with solution V, where the independentvariables are the timet 0 and the valueS 0. The famous Black-Scholesequation reads

V

t + 12

2S22V

S2 + rS

V

S rV = 0, (1.9)

with constants r, > 0. Although the Black-Scholes equation (1.9) lookscomplicated, it can be transformed into the heat equation (1.5) by trans-formations in the domain of dependence. Thus the properties of the Black-Scholes equation are the same as for the heat equation.

Remark: The wave equation (1.2), the heat equation (1.5) and the Black-Scholes equation (1.9) are relatively simple such that formulas for corre-sponding solutions exist, see (1.3) and (1.8). Hence numerical methods are

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not required necessarily. However, an analytical solution is often not feasibleany more if a source term appears, i.e.,

2u

t2 =c2

2u

x2+ f(x,t,u) or

u

t =

2u

x2+ f(x,t,u).

Now we need numerical schemes to obtain an approximative solution. Nev-ertheless, the fundamental properties of the PDEs do not change by addinga source term.

1.2 Classification

We consider a linear PDE of second ordern

i,j=1

aij2u

xixj=f

x1, . . . , xn, u,

u

x1, . . . ,

u

xn

(1.10)

with n 2 independent variables. Let the solution u : satisfyu C2() for some open domain

n. We apply the abbreviationsx= (x1, . . . , xn) and

u= ( ux

1

, . . . , uxn

). The types of PDEs with respect

to the degree of linearity read:

linear PDE: The coefficientsaij are constants or depend on x only andthe right-hand side is linear (f=b(x)+c(x)u+d1(x)

ux1

+ +dn(x) uxn). semi-linear PDE: The coefficientsaij are constants or depend on xonly

and the right-hand side f is nonlinear.

quasi-linear PDE: The coefficients aij depend on u and/or

u. (The

right-hand sidefcan be linear or nonlinear.)

The definition of well-posed problems is as follows.

Definition 1 A PDE or system of PDEs with corresponding initial and/orboundary conditions is well-posed if and only if a unique solution exists andthe solution depends continuously on the input data. Otherwise, the problemis called ill-posed.

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PDEs ofsecond order

of first order

elliptic PDE parabolic PDE

hyperbolic PDE

hyperbolic systems

Figure 1: Classification of PDEs.

The coefficientsaij form a matrixA = (aij) nn. Without loss of gener-ality, we assume that the matrix A is symmetric due to uC2. Thus A isdiagonalisable and all eigenvalues (EVs) 1, . . . , n are real. The classifica-tion of PDEs (1.10) is based on the definiteness ofA. (The classification isindependent of the right-hand side f.) In case of two dimensions (n = 2),it holds det(A) = 12, i.e., the definiteness follows from the sign of the

determinant. The quadratic form

q(z) :=zAz (A

22, z

2)

can be investigated for a geometrical interpretation of the definiteness. Thisyields the nomenclature of the types of PDEs. Fig. 1 illustrates the corre-sponding classes.

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Case 1: A (pos. or neg.) definiteelliptic PDE(all EVs ofAare positive or all EVs ofA are negative)

Elliptic PDEs describe stationary solutions. Boundary conditions yield well-posed problems.

Forn= 2, the set{z 2 :zAz= 1} represents an ellipse.Example: Poisson equationInn dimensions, the Poisson equation reads

u= 2u

x12

2u

xn2 =f(x1, . . . , xn). (1.11)

It follows that the matrix A

nn is diagonal an all diagonal elements(eigenvalues) are equal to1. Consequently, the PDE (1.11) is elliptic.

Case 2: A indefinite, det A= 0hyperbolic PDE(at least two EVs have opposite sign, all EVs are non-zero)Hyperbolic PDEs model transport processes. Initial conditions (possiblyadditional boundary conditions) result in well-posed problems.

Forn= 2, the set{z 2 :zAz= 1} represents a hyperbola.Example: Wave equationInn space dimensions, the wave equation is given by

2u

t2 c2u=

2u

t2 c2

2u

x12+ +

2u

xn2

= 0 (1.12)

with wave speed c > 0. Again the coefficient matrix A

(n+1)(n+1) isdiagonal. It follows a simple EV +1 and a multiple EV

c2. Hence the

wave equation (1.12) is a hyperbolic PDE.

Often one EV exhibits a different sign than all other EVs (the time differsqualitatively from the space coordinates). If two pairs of eigenvalues havean opposite sign, then the PDE is called ultrahyperbolic(n4 necessary).However, we will not consider ultrahyperbolic PDEs in this lecture.

In the casen= 2, a hyperbolic PDE of second order can be transformed intoan equivalent hyperbolic PDE of first order. An example has been given

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in (1.4). Vice versa, not each hyperbolic PDE of first order is equivalent toa PDE of second order.

Case 3: det A= 0parabolic PDE(at least one EV is equal to zero)Parabolic PDEs describe diffusion, for example. Initial conditions in addi-tion to boundary conditions yield well-posed problems.

For n = 2, the set{z

2 : zAz = 1} corresponds to straight lines. Iflinear terms are added to the quadratic form (q(x) :=zAz+ bz), then a

parabola can appear.Example: Heat equationInn space dimensions, the heat equation reads

u

t u= u

t

2u

x12+ +

2u

xn2

= 0 (1.13)

including a constant > 0. The coefficient matrix A (n+1)(n+1) isdiagonal. A simple EV zero and a multiple eigenvalue

appears. It

follows that the PDE (1.13) is parabolic.

The classification is unique in case of constant coefficients aij. Foraij(x),the same PDE (1.10) may exhibit different types in different domains .For aij(u), the type of the PDE may even depend on the correspondingsolution u. However, this happens rather seldom in practice.

Scaling

Multiplying the PDE (1.10) by a coefficient = 0 changes the matrix Ainto A. The differences in the signs of the eigenvalues remain the same.Thus the type of the PDE is invariant with respect to this scaling.

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Basis transformations

Now we investigate the invariance of the type of a PDE (1.10) with respectto basis transformations in the domain of dependence . We consider con-stant coefficients aij, since a generalisation to non-constant coefficients isstraightforward. Let y = Bx using a regular matrix B = (bij) nn. Inthe new coordinatesy , the solution is u(y) = u(Bx) =u(x). The chain ruleof multivariate differentiation implies

u(Bx)

xi=

n

k=1u(Bx)

ykbki,

2u(Bx)

xjxi=

nk=1

nl=1

2u(Bx)

ylykbkiblj.

It followsn

i,j=1

aij2u(x)

xjxi=

ni,j=1

aij

nk,l=1

2u(Bx)

ylykbkiblj =

nk,l=1

ni,j=1

aijbkiblj

2u(y)

ylyk.

Let A= (akl) be the coefficients in the new basis. It holds

A= BAB. (1.14)

The matrix A is always symmetric. However, the eigenvalues ofA are in-variant just for orthogonal matrices B, i.e., B1 = B. Hence orthogonaltransformations do not change the type of the PDE. Each symmetric ma-trix A can be diagonalised via D = SAS using an orthogonal matrix S.Thus each PDE (1.10) can be transformed into an equivalent equation withdiagonal coefficient matrix of the same type.

Non-orthogonal basis transformations may change the type of the PDE incase of n 3. Nevertheless, the type is invariant for an arbitrary basistransformation in case of n = 2. Thereby, the type depends just on thesign of det(A), i.e., elliptic for det(A) > 0, hyperbolic for det(A) < 0 andparabolic for det(A) = 0. The transformation (1.14) yields

det(A) = det(B) det(A) det(B) = (det(B))2 det(A).Hence the sign of det(A) is identical to the sign of det(A).

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Chapter 2

Elliptic PDEs

2

In this chapter, we discuss the numerical solution of boundary value prob-lems of second-order elliptic PDEs. Thereby, the Poisson equation repre-sents a benchmark problem. Two classes of numerical methods exist: finitedifference methods and finite element methods.

2.1 Maximum principle

We consider the Poisson equation

u(x) :=ni=1

2u(x)

x2i=f(x1, . . . , xn) (2.1)

in n2 space dimensions. Let n be an open and bounded domain.Boundary conditions of Dirichlet type read

u(x) =g(x) for x (2.2)with a predetermined function g : . We assume the existenceof a solution u C2()C0(). (Corresponding theorems on existencerequire some assumptions and are hard to prove.) The uniqueness as wellas the continuous dependence on the input data follows from the maximumprinciple.

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Theorem 1 (maximum principle) Letu C2() C0(). It holds:

(i) maximum principle: Ifu = f 0 in, then u exhibits its maxi-mum on the boundary.

(ii) minimum principle: Ifu= f 0in, thenuexhibits its minimumon the boundary.

(iii) comparison: Ifv C2() C0() andu v in anduvon, then it followsu v in.

Proof:

We show the property (i) first. We assume f sup

xu(x),

thenis also a local maximum. It followsu() = 0 and the Hesse matrix2u() = (uxi,xj()) is negative semi-definite. In particular, the entries onthe diagonal are not positive. Thus it holds

(ux1,x1() + + uxn,xn()) 0.This is a contradiction tou= f supx

u(x).

We defineh(x) := (1 x1)2 + + (n xn)2 andw(x) :=u(x) + h(x)using a real number >0. Since hC2() C0() holds, the function wexhibits its maximum inside for sufficiently small . It follows

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The property (ii) follows from (i) applying the maximum principle for thefunction v:=

u.

To verify the property (iii), we define w:= v u. It followsw= v+ u 0

due to the assumptions. It holds w 0 on the boundary. The minimumprinciple implies w(x) 0 for all x . Using this maximum principle, we achieve the following estimate.

Theorem 2 Foru C2() C0(), it holds|u(x)| sup

z|u(z)| + c sup

z|u(z)|. (2.3)

for eachx with some constantc 0.

Proof:

The bounded domain is situated inside a circle of radiusR with its center

atx= 0. We define

w(x) :=R2 ni=1

x2i .

It follows wxi,xj =2ij. It holdsw = 2n and 0w R2 in . Nowwe arrange

v(x) := supz

|u(z)| + w(x) 12nsupz

|u(z)| 0.

Due to this construction, we havev |u| in and v |u| on .Theorem 1 (iii) impliesv(x)u(x)+v(x) in . Since w R2 holds,it follows (2.3) withc:= R

2

2n .

Let u1 and u2 be two solutions of the boundary value problem (2.1),(2.2),i.e., it holdsu1 = f1,u2 = f2 in and u1 = g1, u2 = g2 on .Inserting the differenceu1 u2 into (2.3) yields

|u1(x) u2(x)| supz

|g1(z) g2(z)| + c supz

|f1(z) f2(z)| (2.4)

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for allx . Hence the solution is Lipschitz-continuous with respect to theinput data. Moreover, the solution of a boudary value problem of Dirichlet

type (2.1),(2.2) is unique (choosef1 f2,g1 g2).Boundary value problems of Dirichlet type are well-posed, see Definition 1,due to (2.4) (just existence is not shown but assumed). We give an examplethat initial value problems are ill-posed. We consider the Laplace equationu = 0 in the domain ={(x, y) 2 : y 0}. Let initial values bespecified aty= 0

u(x, 0) = 1nsin(nx), uy (x, 0) = 0.

It follows the unique solution

u(x, y) = 1n

cosh(ny)sin(nx),

which grows like eny. It holds|u(x, 0)| 1n , whereas u becomes larger andlarger at y = 1 for n . Considering the limit case u(x, 0) = 0, thesolution does not depend continuously on the initial data.

Now we consider a general differential operator of elliptic type.

Definition 2 The linear differential operatorL: C2() C0()

L:= n

i,j=1

aij(x) 2

xixj(2.5)

is called elliptic, if the matrixA= (aij) is positive definite for eachx,i.e., it holds A(x) > 0 for all

n. The operator (2.5) is called

uniformly elliptic in n, if a constant >0 exists such thatA(x) 22 for all n and all x . (2.6)

The maximum principle given in Theorem 1 also holds with Linstead ofin case of a general elliptic operator (2.5). The estimate from Theorem 2 isvalid withL instead of for uniformly elliptic operators (2.5).

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x

y

1

1

h

h

Figure 2: Grid in finite difference method.

2.2 Finite Difference Methods

We introduce the class of finite difference methods for boundary value prob-lems of elliptic PDEs in two space dimensions.

Laplace operator on unit square

As benchmark, we consider a boundary value problem of Dirichlet type onthe unit square :={(x, y) : 0< x, y


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holds foru C4()

u(x + h, y) = u(x, y) + hux(x, y) + h2 12uxx(x, y) + h

3 16uxxx(x, y)

+ h4 124uxxxx(x + 1h, y)

u(x h, y) = u(x, y) hux(x, y) +h2 12uxx(x, y) h3 16uxxx(x, y)+ h4 124uxxxx(x 2h, y)

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Figure 3: Five-point star (left) and nine-point star (right).

We obtain a linear systemAhUh=Fhof dimensionn = M2. The matrixAh

is a band matrix

Ah= 1

h2

C II C . . .

. . . . . . II C

with C=

4 11 . . . . . .

. . . . . . 11 4

. (2.11)

The matrix Ah is sparse, since each row includes at most five non-zeroelements. Obviously, the matrix Ah is symmetric. It can be shown that Ahis always positive definite. Hence the matrix is regular. The corresponding

solution Uh = A1h Fh represents an approximation of the solution u of thePDE in the grid points.

Laplace operator on general domain

Now we consider an arbitrary open and bounded domain

2, see Fig-ure 4. The application of a finite difference method requires the constructionof a grid. We define an auxiliary (infinite) grid

Gh:= {(x, y) = (ih,jh) :i, j }.Now the used (finite) grid reads

h:= Gh .Let h ={z1, . . . , z R} be the grid points. Boundary conditions appear inthe points of

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x

y

h

h

Figure 4: Grid in a general domain .

E

S

W

N

(xh ,y)W

(x,yh )S

(x,y+h )N

(x+h ,y)EZ

Figure 5: Five-point star for variable step sizes.

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To arrange the difference formulas near the boundary, variable step sizeshave to be considered.

We apply a five-point star again, see Figure 5. Taylor expansion yields

2u

x2

Z

= 2

hE(hE+ hW)uE 2

hEhWuZ+

2

hW(hE+ hW)uW+ O(h)

2u

y2

Z

= 2

hN(hN+ hS)uN 2

hNhSuZ+

2

hS(hN+ hS)uS+ O(h)

provided that u

C3(). Hence four (possibly) different step sizes are in-volved. Leth:= max{hE, hW, hN, hS}. This scheme to include the bound-ary data is also called the Shortley-Weller star.

In general, a five-point star and a nine-point star are specified via theircoefficients

N

W Z ES

and

NW N NEW Z E

SW S SE

,

respectively, cf. Figure 3. A discretisation of the Poisson equation (2.7)using arbitrary five point stars reads

l=Z,E,S,W,N

lUl=fZ

for eachZ h. A general difference formula can be written in the formLhU :=l lUl,

where the sum is over all non-zero coefficients l. The discrete operator Lhdepends on the step sizes. We apply the notation

Lhu:=l

lu(Zl),

where a function u : (usually a solution of the PDE problem) is

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We outline the algorithm of the finite difference method for the Dirichletproblem of the Poisson equation on a general domain .

Algorithm: FDM for Dirichlet problem

1. Choose a step size h >0 and determine h as well as h.

2. Choose a numbering of the unknown UZ forZ h.3. Arrange the difference formulas

ZUZ+ EUE+ WUW+ NUN+ SUS=fZ

for eachZ h.4. If boundary values UB with B h appear in the left-hand side of

a difference formula, then replace UB bygB and shift this term to theright-hand side.

5. Arrange and solve the linear system

AhUh=Fh

with the chosen numbering of the unknowns Uh= (UZi) forZi h.

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General differential operator

Difference schemes for mixed derivatives also exist. For example, it holds2u

xy(xi, yj) =

ui+1,j+1 ui1,j+1 ui+1,j1+ ui1,j14h2

+ O(h2). (2.12)

We verify this formula via (multivariate) Taylor expansions of the neigh-bouring points around the central pointu ui,j.

ui+1,j+1 = u+ hux+ huy+ 12h

2(uxx+ 2uxy+ uyy)

+ 1

6

h3(uxxx+ 3uxxy+ 3uxyy + uyyy) +O

(h4)

ui1,j+1 = u hux+ huy+ 12h2(uxx 2uxy+ uyy)+ 1

6h3(uxxx+ 3uxxy 3uxyy + uyyy) + O(h4)

ui+1,j1 = u+ hux huy+ 12h2(uxx 2uxy+ uyy)+ 1

6h3(uxxx 3uxxy+ 3uxyy uyyy) + O(h4)

ui1,j1 = u hux huy+ 12h2(uxx+ 2uxy+ uyy)+ 1

6h3(

uxxx

3uxxy

3uxyy

uyyy) +

O(h4)

ui+1,j+1 ui1,j+1 ui+1,j1+ ui1,j1= 4h2uxy+ O(h4),Now we can discretise an arbitrary differential operator of second order

Lu:= a2u

x2+ 2b

2u

xy+ c

2u

y2 (2.13)

with a,b,c for n = 2. The operator (2.13) is elliptic, if and onlyif ac > b2 holds. The derivatives

2ux2 ,

2uy2 are replaced by the difference

formulas in h and the mixed derivative 2uxy is substituted by (2.13). Itfollows the (discrete) difference operator

Lhu = a

h2 [ui1,j 2ui,j+ ui+1,j]+ b2h2 [ui+1,j+1 ui1,j+1 ui+1,j1+ ui1,j1]+ ch2 [ui,j1 2ui,j+ ui,j+1].

(2.14)

The difference formula represents a nine-point star, see Figure 3.

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Definition 5 (consistency) A difference operatorLh is called consistentwith respect to a differential operatorL: V

W, if the local error satisfies

limh0

(h) = 0 uniformly on h

for all functionsu V. The method is consistent of orderp (at least), if(h) = O(hp) uniformly on h

for all functionsu V.

For example, the difference operator (2.14) is consistent of order p = 2 withrespect to the differential operator (2.13) for V =C4().

We analyse the convergence of the Dirichlet problem of the Poisson equation,i.e., the discretisation h of the Laplace operator . It holds

AhUh=Fh, AhUh=Fh+ Rh,

where Uh = (u(zi)) represents the data of the exact solution in the gridpoints. ThusRh is a vector, which contains the local errors (h). Since the

difference formula is consistent of order p = 2, it holdsRh =O(h2).We apply the maximum norm, since the size of the vectors depends on h.Assuming thatAh is regular for all h >0, we obtain

Uh Uh= A1h Fh A1h (Fh+ Rh) = A1h Rhand thus

Uh Uh

A1h

Rh C

A1h

h2

with some constant C > 0 for sufficiently small h. However, to guarantee

the convergence, we require a condition likeA1h K or A1h Khfor all h < h0 uniformly with some constant K > 0. Such a conditioncorresponds to the stability of the finite difference method.

We obtain a more general stability criterion by the following theoreticalresult. Thereby, we have to assume that the grid h is connected.

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It holds l < 0 and UlUZ 0 for all l in the original sum. Thus allterms of the original sum are non-negative. It follows that each term must

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Theorem 4 We consider an elliptic PDE Lu = f on with Dirichletboundary conditions u = g on . Let AhUh = Fh be the linear system

from a finite difference operator satisfying the assumptions of the discretemaximum principle. It follows that the matrixAh is regular and thus has aunique solution.

Proof:

The homogeneous linear system AhUh = 0 represents the discretisation ofthe PDE forf 0 andg 0. LetUhbe a solution. The discrete maximumprinciple implies UZ 0 in each Z h, whereas the discrete minimumprinciple yieldsUZ 0 in eachZ h. It followsUh= 0. Hence the matrixis not singular.

It remains to show that the finite difference method is convergent. In thefollowing, we restrict to the Laplace operator Lu =u. We apply thefive-point star, which is consistent of order at least one and satisfies thediscrete maximum principle.

Lemma 1 LetuC2() C0() be the solution of the Poisson equationu = fwith Dirichlet boundary conditionsu = g on. LetLh be thedifference operator of the five-point star on a gridh. Then the local errorand global error of the finite difference method fulfill the estimate

maxZh

|u(Z) UZ| KmaxZh

|(Z)| (2.15)

with a constantK >0, which is independent of the step sizeh.

Proof:

We investigate the local and global errors

(Z) = u(Z) Lhu(Z), (Z) =u(Z) UZ for Z h.It follows due to the linearity of the operators

Lh(Z) =Lhu(Z) LhUZ=Lhu(Z) f(Z) =Lhu(Z) + u(Z) = (Z).31


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The global error vanishes at the boundary h, since the solution is equalto the predetermined boundary values. We analyse the problem

Lh= in h, = 0 on h.The scaling

:=

, :=

with := max

Zh|(Z)|

yields the problem

Lh= with 1 (Z) 1 for all Z h.It still holds = 0 on h. Let {(x, y) 2 : x2 +y2 < R2}. Wedefine the auxiliary function

w(x, y) := 14(R2 x2 y2) 0.

Since all third derivatives of w are identical zero, the local errors of thefive-point star vanish. It follows Lhw =w = 1 in h. Since the five-point star satisfies the discrete maximum principle, the discrete comparisonimplies

w 14R2 for all Z h.Usingw instead ofw, the discrete comparison yields

14R

2 w for all Z h.Hence it follows

maxZ

h

|(Z)| 14R2 maxZh

|(Z)|

and we can choose the constant K := R2

4 .

In the proof of Lemma 1, both the consistency and the discrete maximumprinciple corresponding to the five-point star are applied. The discretemaximum principle guarantees the stability of the finite difference method.Now we can conclude the convergence.

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Theorem 5 LetuC3() be the solution satisfying the Poisson equation

u = fwith Dirichlet boundary conditionsu = g on. LetLh be the

difference operator of the five-point star. Then the numerical solution of thefinite difference method converges to the exact solution and it holds

maxZh

|u(Z) UZ| = O(h).

Ifu C4() and all step sizes are equidistant, then it holdsmaxZh

|u(Z) UZ| = O(h2).

Proof:

The consistency of the five-point star yields (Z) =O(h) for all Z h.The estimate (2.15) from Lemma 1 implies

maxZh

|(Z)| 14R2Ch for all 0< h < h0

with some constantC >0. In case ofu C4() and equidistant step sizes,it follows (Z) = O(h2) and thus convergence of order p= 2.

A further analysis shows that a convergence of order p= 2 is also given forvariable step sizes. Moreover, the convergence can also be shown in case ofu C2().

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Generalisations

We outline the generalisation of the above finite difference method to furtherproblems.

von-Neumann boundary value problem: We consider the Poisson equa-tion u= f in 2 with boundary conditions u =g(x, y) on.For example, let = (0, 1) (0, 1). We apply the grid points (2.8) fori, j = 0, 1, . . . , M , M +1. In comparison to Dirichlet problems, 4M ad-ditional unknowns appear (the four edges of the square are not used).

Hence we arrange 4Mequations using difference formulas to replacethe derivative u

:

g(xi, 0) = uy (xi, 0) .= 1

h[ui,0 ui,1] for i= 1, . . . , M ,

g(xi, 1) = u

y (xi, 1) .= 1h [ui,M+1 ui,M] for i= 1, . . . , M ,

g(0, yj) = ux (0, yj) .= 1h [u0,j u1,j] for j = 1, . . . , M ,

g(1, yj) = u

x (1, yj) .= 1h [uM+1,j uM,j ] for j = 1, . . . , M .

Techniques for Neumann boundary conditions on arbitrary domains

also exist. Remark that a solution of a pure von-Neumann boundaryvalue problem is not unique and thus requires an additional condition.

space-dependent coefficients: Given the elliptic PDE

Lu:= a(x, y)2u

x2+ 2b(x, y)

2u

xy+ c(x, y)

2u

y2 =f(x, y),

the coefficients a,b,c : depend on space. Appropriate finite

difference methods can be constructed in this case. For proving the

convergence, a uniformly elliptic operator, see Definition 2, has to beassumed.

right-hand side includes the solution: We consider the (nonlinear) PDEu = f(x,y,u) with a nonlinear function f. Let be the unitsquare. Homogeneous boundary condition u = 0 on are applied.The five-point star with equidistant step sizes yields the equations

1h2 [4ui,j ui1,j ui+1,j ui,j1 ui,j+1] =f(xi, yj, ui,j)

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for i, j = 1, . . . , M . We obtain a nonlinear system for the unknowns ui,j.Newtons method yields an approximation of the corresponding dis-

crete solution. In the special case f(x,y,u) = b(x, y) +c(x, y)u, itfollows a linear system again.

three-dimensional space: The Laplace operator in three space dimen-sions reads u=uxx+uyy + uzz . We consider an open and boundeddomain

3 for the Poisson equationu= f. The above theoryof finite difference methods can be repeated in this case. The sameresults appear concerning consistency, stability and convergence.

Iterative solution of linear systems

The finite difference methods yield large and sparse linear systems. Gaus-sian elimination becomes expensive, since many fill-ins appear in the fac-torisation. In contrast, iterative methods allow for efficient algorithms.

The types of iterative solvers are:

stationary methods: Jacobi method, Gauss-Seidel method, SOR, etc. instationary methods: conjugate gradient method, GMRES, etc. multigrid methods.

An introduction to iterative methods for solving linear systems can be foundin J. Stoer, R. Bulirsch: Introduction to Numerical Analysis. (2nd ed.)Springer, New York, 1993. (Chapter 8)

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2.3 Sobolev Spaces and Variational Formulation

In this section, we introduce weak solutions of elliptic PDEs. Finite elementmethods represent convergent techniques for weak solutions, whereas finitedifference methods fail.

Classical solutions

Classical solutions of a PDE are sufficiently smooth in some sense.

Definition 7 (classical solution) Let n be open and bounded. Foran elliptic PDELu= f, a functionu: is called a classical solution,if it holds

for Dirichlet problems: u C2() C0(), for von-Neumann problems: u C2() C1().

As an example forn = 2, we consider the Laplace equation on three quarterof the unit disc as domain

:= {(x, y)

2 :x2 + y2 0.

It holds u C2() C0(), i.e., u is a classical solution of the Dirichletproblem. However, due to w(z) = 23z

1/3 and w(z) =29z4/3, both thefirst and the second derivative of u is not bounded in a neighbourhood

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of z = 0. It follows that u / C2(). Hence we cannot guarantee theconvergence of the finite difference method constructed in Sect. 2.2. An

alternative numerical method is required.

Weak Derivatives and Sobolev Spaces

We consider an open domain

n. We define the space of test functions

C0 () :={ C() : supp() , supp() is compact},where supp() :={x :(x) = 0}. If is bounded, it follows that

(x) = 0 for x . Furthermore, we apply the Hilbert space L2

(),which has the inner product

f, gL2 :=

f(x) g(x) dx

for eachf, g L2(). The corresponding norm reads

fL2 =

f(x)2 dx .

The set C0 () L2() is dense. A multi-index is given by

:= (1, . . . , n) n0 , || :=ni=1

i.

For u Ck() with k =||, an elementary differential operator can bedefined via

Du:= ||u

x11

xnn

.

Foru Ck(), Du represents a usual (strong) derivative.

Definition 8 (weak derivative) Given a functionfL2(), a functiong L2() is called the weak derivativeDf off, if it holds

g(x) (x) dx= (1)||

f(x) D(x) dx

for all C0 (). We writeDf=g.37


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The property of this definition can also be written as

g, L2 = (1)|

|f, D

L2 for all C0 ().It can be shown that a weak derivative of a function is unique, since thespaceC0 () is dense in L

2().

Example: In the special case n= 1, a function f C1(a, b) leads to ba

f(x)(x) dx= [f(x)(x)]x=bx=a ba

f(x)(x) dx= ba

f(x)(x) dx

for each

C0 (a, b). Hence f

is a weak derivative off.

In case ofn 2, we apply Greens formula

v wxi

dx=

v w ids

v

xi wdx (2.16)

for i {1, . . . , n} andv, w C1(), where i is the ith component of theouter normal vector on the boundary . The formula (2.16) does not holdfor arbitrary domains . A domain with a (so-called) smooth boundary issufficient for the application of the formula. For simplicity, we restrict to

domains, where the formula (2.16) holds.

The concept of the weak derivative is used to define the Sobolev spaces.

Definition 9 (Sobolev space) Form0, Hm() is the set of all func-tions u L2(), where a weak derivative Du L2() exists for all|| m. The spaceHm() exhibits an inner product

u, v

Hm :=

||mDu, Dv

L2.

The setHm() is called a Sobolev space. Hm is a Sobolev norm.

Hence Hm() can be interpreted as a generalisation of the space Cm(),which is not a Hilbert space. The spaces (Hm(), Hm) are Hilbert spaces,i.e., they are complete. Remark that it holdsHm() L2() for m 1andH0() =L2().

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To investigate PDE problems with homogeneous Dirichlet boundary condi-tions ( is bounded), we define the subsets Hm0 () :

C0 () as the closure

ofC0 () Hm() with respect to the norm Hm. More precisely, itholds

Hm0 () =

u Hm() :(vi)i C0 () with limi

u viHm = 0

.

It follows that the subspace (Hm0 , Hm) is also a Hilbert space.Furthermore, we obtain a semi-norm on Hm() via

|u|Hm := ||=m

Du2L21/2

.

If is inside a cube of edge length s, then it holds the equivalence

|u|Hm uHm (1 + s)m|u|Hm for all u Hm0 (). (2.17)The above construction can also be done using the spaces Lp() instead ofL2() with 1 p . It follows the Sobolev spaces Wmp (), where itholds Wm2 () =H

m().

Symmetric operators and bilinear forms

In the following, we assume homogeneous Dirichlet boundary conditions.Given an elliptic PDELu = fin andu= g on. Letu0be a sufficientlysmooth function with u0 = g on . The function w := u u0 satisfiesLw = f in with f := f Lu0 and w = 0 on . Hence we havetransformed the problem to homogeneous boundary conditions.

We consider a general linear differential operator of the form

Lu:=

ni,j=1

aij2u

xixj

+

nj=1

aju

xj

+ a0u. (2.18)

Thereby, we assumeaij C2(),aj C1(),a0 C0(). The correspond-ing adjoint operator L is defined by the property

Lu,vL2 = u, LvL239


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foru, v C2() withu= 0, v= 0 on . It follows

Lv= ni,j=1

2(aijv)xixj

nj=1

(ajv)xj

+ a0v.

A symmetric (also: self-adjoint) operator satisfies L= L. It can be shownthat a symmetric operator exhibits the form

Lu:=

ni,j=1

xi

aij

u

xj

+ a0u. (2.19)

In particular, each operator (2.18) is self-adjoint in case of constant coef-ficients aij and a1 = = an = 0, for example. Greens formula (2.16)implies

Lu,v =

ni,j=1

aiju

xj

v

xidx

+

a0uvdx. (2.20)

We recognise that the right-hand side is symmetric in u and v. Hence itholdsLu,v = u,Lv. Furthermore, just derivatives of first order appear.

Definition 10 LetH be a Hilbert space. A bilinear forma:H H is symmetric, ifa(u, v) =a(v, u) holds for allu, vH. A bilinear formais continuous , if a constantC >0 exists such that

|a(u, v)| C u v for all u, v H.A symmetric, continuous bilinear forma is calledH-elliptic (also: coercive),if a constant >0 exists with

a(u, u) u2 for all u H.

In particular, each H-elliptic bilinear form is positive, i.e., a(u, u) > 0 foru= 0.Each H-elliptic bilinear form a induces a norm

ua:=a(u, u) for u H,40


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which is called the energy norm. The energy norm is equivalent to the normof the Hilbert space.

The relation (2.20) motivates the definition of a symmetric bilinear formcorresponding to a uniformly elliptic differential operator.

Theorem 6 The bilinear forma: H10() H10() given by

a(u, v) :=

ni,j=1

aiju

xj

v

xidx

+

a0uvdx (2.21)

with aij, a0 C0(), A = (aij) symmetric and positive definite, a0 0is continuous and H10()-elliptic provided that the underlying differentialoperator is uniformly elliptic.

Proof:

We define c := sup{|aij(x)| : x , 1 i, j n}. It follows using theCauchy-Schwarz inequality n

i,j=1

aijuxivxj dx c n

i,j=1

|uxivxj | dx

cn

i,j=1

uxiL2 vxjL2

cn

i,j=1

|u|H1 |v|H1

= cn2 |u|H1 |v|H1.We arrangeb:= sup{|a0(x)| :x }. It follows

a0uvdx

b

|uv| dx b uL2 vL2.

We obtain applyinguL2 uH1 and|u|H1 uH1 withC :=b+ cn2

|a(u, v)| C uH1 vH1.41


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Since the differential operator is uniformly elliptic, see (2.6), it holds usingthe monotonicity of the integral

ni,j=1

aijvxivxj dx

ni=1

(vxi)2 dx

forv H10(). It follows due to a0 0

a(v, v) ni=1

(vxi)2 dx= |v|2H1

for each v

H10(). The equivalence (2.17) implies a(v, v)

K

v

2H1

for v H10() with some constant K > 0 depending just on . Thus thebilinear forma isH10()-elliptic with:=K.

Variational formulation

Now we consider the PDE Lu = fwith a uniformly elliptic operatorL. Letu be a classical solution and Lu, f L2(). It follows

Lu

f = 0

(Lu f)v = 0Lu f, vL2 = 0

Lu,vL2 f, vL2 = 0for eachv L2(). We define the linear mapping

(v) := f, vL2 (2.22)forv

L2() or a corresponding subspace. The Cauchy-Schwarz inequality

yields|(v)| fL2vL2. Hence is bounded on L2() with fL2.It follows thatis also bounded onH1().

Let V, i.e., :V be an arbitrary linear mapping. We apply the

notation, v :=(v), which refers to the bilinear form, :V V .

The right-hand side f yields the linear mapping (2.22), whereas the left-hand sideLu corresponds to the bilinear form (2.21). It follows the conceptof a weak solution.

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Definition 11 (weak solution) A function u H10() is called a weaksolution of the elliptic PDE problem

Lu = f in

u = 0 on ,

if the corresponding bilinear form (2.21) and linear mapping (2.22) satisfy

a(u, v) = , v for all v H10(). (2.23)

Now we show that a classical solution represents also a weak solution of

the problem. For simplicity, we demand the property u C2

() for theclassical solution.

Theorem 7 Letu be a classical solution of

n

i,j=1

xi

aij

u

xj

+ a0u = f in

u = 0 on

withu C2()andaij C1(), a0, f C0() L2(). Thenu representsalso a weak solution of the problem.

Proof:

We apply Greens formula

v wxi

dx=

v w ids

v

xi wdx,

where we choose w:=aijuxj C1() andv C0 (). It follows

v

xi

aij

u

xj

dx=

aijv

xi

u

xjdx.

We apply the bilinear and linear form, respectively,

a(u, v) :=

ni,j=1

aiju

xi

v

xj+ a0uvdx, , v :=

f vdx.

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It follows

a(u, v) , v =

v ni,j=1

xiaij uxj+ a0u f dx

=

v[Lu f] dx = 0

for all vC0 () due toLu= f. The bilinear form a as well as the linearform (v) =f, vL2 are continuous on H10(). Since C0 () H10() isdense, it followsa(u, v), v = 0 for allv H10(). Furthermore, it holdsu

H10() due tou

C0()

H1() andu= 0 on.

Solutions satifying the assumptions of Theorem 7 can be computed by finitedifference methods due to u C2().Now we show an important equivalence of our problem.

Theorem 8 Let V be a linear space and a : VV a symmetric,positive bilinear form and: V

a linear mapping. The function

J(v) := 12

a(v, v) , v (2.24)exhibits a minimum inV atu if and only if

a(u, v) = , v for all v V. (2.25)There exists at most one minimum.

Proof:

A positive bilinear form fulfills a(u, u)>0 for all u= 0. Foru, vV andt , we calculate

J(u+ tv) = 12

a(u+ tv,u+ tv) , u+ tv= J(u) + t [a(u, v) , v] + 1

2t2a(v, v).

Ifu V satisfies the condition (2.25), then it follows using t= 1J(u+ v) =J(u) + 12a(v, v)> J(u)

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forv V withv= 0. Hence u is the unique minimum.

Vice versa, letu Vbe a minimum of the function (2.24). For eachv V,it holds

d

dtJ(u+ tv)

t=0

= 0.

Due tod

dtJ(u+ tv) =a(u, v) , v + ta(v, v),

it follows the condition (2.25).

Theorem 8 yields the uniqueness of a weak solution. If a classical solutionexists satisfying additional properties (like u C2()), then this functionalso represents the unique weak solution.

We obtain an additional characterisation of the weak solution by Theorem 8:The weak solution of the PDE also represents a solution of a minimisationproblem

J(v) := 12a(v, v) , v min. (2.26)and vice versa. The task (2.26) is called a variational formulation (of theproblem) or a variational problem.

Theorem 9 (Lax-Milgram) Let H be a Hilbert space and V H be aclosed convex set. For an H-elliptic bilinear form a : HH and H, the variational problem (2.26) exhibits a unique solution inV.

Proof:

The mappingJis bounded from below, sice it holds

J(v) 12

v2 v = 12(v )2 122 122.

We definec := inf{J(v) :v V}. Let (vn)n Vbe a sequence satisfyinglimn

J(vn) =c.

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It follows

vn vm2

a(vn vm, vn vm)= 2a(vn, vn) + 2a(vm, vm) a(vn+ vm, vn+ vm)= 4J(vn) + 4J(vm) 8J(12(vn+ vm)) 4J(vn) + 4J(vm) 8c,

since 12(vn+vm)V holds due to V convex. The upper bound convergesto zero for n, m . Thusvn vm 0 for n, m , i.e., (vn)n isa Cauchy sequence. SinceV is a closed set, a limit u

Vexists. It follows

J(u) =J

limn

vn

= limn

J(vn) = inf{J(v) :v V}

due to the continuity ofJ. Henceu represents a minimum.

Concerning the uniqueness, letu1, u2 V be two solutions of the variationalproblem (2.26). Then it holds J = c for each component of the sequence(u1, u2, u1, u2, . . .). Due to the above calculations, a Cauchy sequence isgiven. It follows

u1

u2

< for each >0. Hence it holds

u1

u2

= 0

andu1=u2.

We apply Theorem 9 in the special case V = H = H10(), since H10()

is a Hilbert space. It follows that the variational problem (2.26) has aunique solution u H10(). Due to Theorem 8, the solution u of thevariational problem is also a weak solution of the PDE problem accordingto Definition 11. We obtain directly the following result.

Theorem 10 LetL be a uniformly elliptic, symmetric differential operator.Then the homogeneous Dirichlet boundary value problem forLu = fexhibitsa unique weak solution inH10().

Remark: Not all open and bounded domains n are feasible, sinceGreens formula has to be applicable. Nevertheless, Greens formula is validfor nearly all domains in practice.

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In the one-dimensional case (n = 1), we obtain a homogeneous boundaryvalue problem of a second-order ordinary differential equation

(p(x)u(x))+ q(x)u(x) =f(x) forx (a, b)withu(a) =u(b) = 0. Assuming p(x)p0 >0 and q(x)0, the involveddifferential operator is uniformly elliptic. A corresponding variational for-mulation can be constructed as above. The detailed derivation for thisspecial case can be found in, for example, Stoer/Bulirsch: Introduction toNumerical Analysis, Springer (Section 7.5).

Von-Neumann boundary conditionsFor the Poisson equationu = f, we demand u = g on in caseof von-Neumann boundary conditions. Weak solutions are considered inthe space H1() now. For a broad class of domains , a bounded linearmapping

:H1() L2(), (v)L2() CvH1() (2.27)exists satisfying (v) = v| for all v C0() H1(). The linear map-ping is called the trace operator.

The trace operator allows for an alternative characterisation of the Hilbert space H10 ().We defined H10 () : C0 (), i.e., the closure of the test functions with respect to theSobolev norm H1. It can be shown that it holds

H10 () ={u H1() :(u) = 0}.This property has already been used in the proof of Theorem 7, since u C0() H1()and u= 0 on implies u H10 () now.

A variational formulation can be derived also in the case of von-Neumann

boundary conditions. For simplicity, let u C2

(), v C1

(). Greensformula yields foru, vL2

vni=1

2u

x2idx=

ni=1

v2u

x2idx=

ni=1

vu

xii ds

v

xi

u

xidx.

The second term implies the definition of the bilinear form

a(u, v) =

n

i=1u

xi

v

xidx (2.28)

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2.4 Finite Element Methods

Now we apply the theory of the previous section to construct numericalmethods for the determination of weak solutions.

Ritz-Galerkin approach

We consider a homogeneous Dirichlet boundary value problem includinga uniformly elliptic, symmetric differential operator (2.19). It follows theexistence of a unique weak solution. Theorem 8 shows two properties, which

characterise the weak solution. Numerical methods can be based on eachof these properties. Typically, finite-dimensional subspaces Sh H10() arechosen, whereh >0 represents a discretisation step size to be defined later.Typically, it holds dim(Sh) forh 0.We obtain three classes of numerical techniques:

Galerkin method: The definition (2.23) is used. The approximation ofthe weak solution is determined in a finite dimensional subspace Sh.

The condition (2.23) shall be satisfied for all v Sh. Petrov-Galerkin method (or: method of weighted residuals): The defi-

nition (2.23) is applied again. The approximation is situated in somespace Sh. The condition (2.23) shall be satisfied for all v Th withanother subspaceTh of the same dimension. The special case Sh=Thyields the Galerkin method.

Rayleigh-Ritz method(or: Ritz method): The solution of the variationalproblem (2.26) is computed approximately. Thereby, a minimum ofJis determined in a finite-dimensional subspace Sh.

We discuss the Galerkin method first. For some subspaceSh, we choose abasis{1, . . . , N}. The approximation is inside this space, i.e.,

uh(x) =N

j=1

jj(x) (2.30)

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with unknown coefficients 1, . . . , N . Replacing the exact solutionuby the approximationuh, the condition (2.23) demands that

a(uh, v) = , v (2.31)for allv H10(). Sinceuh=uin general, this condition cannot be satisfiedfor allv H10(). Alternatively, we demand that the property (2.23) holdsfor allv Sh. Using the basis functions, this condition is equivalent to

a(uh, i) = , i for i= 1, . . . , N .Inserting (2.30), it follows a linear system

Nj=1

ja(j, i) = , i for i= 1, . . . , N

with the unknown coefficients 1, . . . , N. The matrix of this linear systemreads A := (a(j, i)) NN, which is obviously symmetric. Since thebilinear form is positive, it holds for = (1, . . . , N)

= 0

A =N

i,j=1 a(j, i)ji=a N

j=1 jj,N

i=1 ii> 0.Hence the matrix A is positive definite. It follows that a unique solutionexists, which yields the approximation (2.30).

In the Petrov-Galerkin method, we demand that the condition (2.31) issatisfied for all v Th for some other subspace Th H10() satisfyingdim(Sh) = dim(Th). The elements of Th are often called test functions.(However, they do not belong to the set C0 in general.) We select a basis

{1, . . . , N}ofTh. Now the condition (2.31) for all v Th is equivalent toa

Nj=1

jj, i

= , i for i= 1, . . . , N .

It follows the linear system

Nj=1

ja(j, i) = , i for i= 1, . . . , N

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with the matrix A:= (a(j, i)). This matrix is not symmetric in general.It depends on the choice of the subspaces and the bases if the matrix is

regular. In the special caseSh=Th and using the same basis, the approachcoincides with the Galerkin method due to i=i for all i.

For the Rayleigh-Ritz method, we insert the approximation (2.30) into thefunction Jfrom (2.26). It follows (= (1, . . . , N)

)

J

Nj=1

jj

= 1

2

Ni,j=1

jia(j, i) N

j=1

j, j = 12A b

with the same matrix A and right-hand side b as in the Galerkin method.The minimisation in Sh only demands

J

k= 0 for k= 1, . . . , N .

The gradient ofJis J=Ab. It follows the linear systemA= b. Thusthe technique coincides with the Galerkin method in this case. Differentapproaches may appear if the underlying bilinear form a is not symmetricor not positive. For problems, where the Rayleigh-Ritz method and the

Galerkin method are the same, the technique is called the Ritz-Galerkinmethod. The involved matrixAh is also calles stiffness matrix.

The method of weighted residuals can be motivated also in case of smooth solutions. Letuh Sh C2()C0() be an approximation of a classical solution. For a finite-dimensional spaceSh, we choose a basis 1, . . . , N (of ansatz functions) and consider anapproximation (2.30). It follows the residual :

:= Luh f=

N

i=1iLi

f.

We want to determine the coefficients 1, . . . , N such that the residual becomessmall in some sense. In the method of weighted residuals, a space Thof test functions withdimension Nis chosen. We demand that the residual is orthogonal to the space Th withrespect to the inner product ofL2, i.e.,

Luh f, vL2 = 0 for all v Th.Selecting a basis{1, . . . , N} ofTh, this property can be written as

j(x) (x) dx= 0 or , jL2 = 0 for j = 1, . . . , N .

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This expression can be seen as weighted integrals of the residual , where the functions jrepresent the weights. It follows the linear system

Ni=1

iLi, jL2 = f, jL2 for j = 1, . . . , N

for the unknown coefficients. In the special case Sh = Th and choosing the same basis in

each space, it follows the Galerkin method.

Remark: A significant advantage of the Galerkin method is that the matrixof the linear system is symmetric and positive definite for an arbitrarydomain . Hence iterative solvers can be applied efficiently. In the finitedifference method, see Sect. 2.2, the matrix of the linear system is symmetricand positive definite in case of the unit square ( = (0, 1)2). The matrixbecomes unsymmetric for other domains like the unit disc, for example.

Concerning the stability of the Ritz-Galerkin method, we obtain the follow-ing result.

Theorem 11 (stability) Let a : Hm

0 () Hm

0 ()

be a symmetric,continuous and Hm0 ()-elliptic bilinear form. Let : Hm0 () be a

linear, continuous mapping. Then the solution of the Ritz-Galerkin methodsatisfies

uhHm 1. (2.32)independent of the choiceSh Hm0 ().

Proof:

Since is continuous, it holds|(v)| vHm. The Hm0 ()-ellipticityyields

0 uh2Hm a(uh, uh) = , uh uhHm.Foruh= 0, the inequality (2.32) is trivial. For uh= 0, we divide by uhHmand obtain (2.32).

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The stability implies the Lipschitz-continuous dependence of the approximation on theinput data. For example, we consider a perturbation in the right-hand side f. It holds

fL2. Consider two right-hand sidesf1, f2with corresponding weak solutionsu1, u2.The differenceu1u2is a weak solution for the right-hand sidef1f2. The approximationsfollowing from the Galerkin method satisfy according to (2.32)

u1h u2hHm 1

f1 f2L2

due to the linearity. This estimate is independent of the choice ofSh, i.e., it is uniform

in h >0 for a discretisation step size to be defined later.

Concerning the quality of the approximation resulting from the Ritz-Galer-

kin method, the following important theorem holds.

Theorem 12 (Lemma of Cea) LetHbe a Hilbert space, l : H bea linear continuous form and a : HH be a bilinear form, whichis symmetric, continuous and H-elliptic. Then the function u defined bya(u, v) = l, v for allv Hand the approximationuh of the correspondingRitz-Galerkin method using someSh Hsatisfy the estimate

u uh C

infvhSh u vh. (2.33)

Proof:

It holds

a(u, v) = , v for v H, a(uh, v) = , v for v Sh H.By subtraction, we obtain

a(u uh, v) = 0 for all v Sh.For an arbitraryvh Sh, we conclude

u uh2 a(u uh, u uh)= a(u uh, u vh) + a(u uh, vh uh)= a(u uh, u vh) C u uh u vh

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due tovh uh Sh. Dividing this inequality byu uh = 0 yields

u uh C

u vh.

Since vh Sh is arbitrary, it follows the relation (2.33).

Theorem 12 implies already the convergence of the Galerkin method pro-vided that

limh0

infvhSh

u vh = 0.

Hence the subspaces have to be chosen such that the distance to the ex-act solution decreases. However, this is more a question in approximationtheory in our case H=H10(). We apply the Ritz-Galerkin method in thefollowing. It remains to choose the spaces Sh appropriately.

In a finite difference method, the matrix of a linear system is typically sparseor even a band matrix. Thus the computational effort is significantly lowerthan in case of a dense matrix with the same size. We want to achieve alsoa sparse matrix or a band matrix in the Ritz-Galerkin method. We apply

spaces Sh consisting of piecewise polynomial functions. However, it turnsout that the matrix will be sparse just for specific choices of basis functions.

Let supp() := {x :(x) = 0}. The bilinear form (2.21) satifiesa(, ) = 0 if (supp() supp()) = 0

with the Lebesgue measure, since the bilinear form represents an integralin . Hence we will construct a basis such that the supports of the basis

functions overlap only rarely. Of course, it should still holdN

j=1

supp(j) =

for a basis{1, . . . , N}. The domain will be decomposed into smallersubdomains for the construction of the spaceSh as well as the choice of thebasis functions.

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Triangulations

We consider the two-dimensional case (n = 2). Let 2

be an openpolygonal domain. Hence we can divide the domain into triangles.

Definition 12 (triangulation) Let

2 be a domain with a polygonalboundary. A setT ={T1, . . . , T Q}, where the Tj are non-empty closedtriangles, is an admissible triangulation if it holds

(i) =

Q

j=1 Tj,(ii) int(Ti) int(Tj) = fori =j (int: interior),

(iii) Ti Tj fori =j is either an empty set or a corner of both triangles ora complete edge of both triangles.

For eachT T, we define

hT := 1

2diam(T) = 1

2max{x y2:x, y T}(diam: diameter). For a triangulationT, the (global) step size reads

h:= max{hT :T Th}.Each triangleTcontains a (maximal) circle of radiusT. Given a familyThof triangulations for 0 < h < h0, we assume max{hT : T Th} h. AfamilyThof triangulations is called uniform, if a constant >0 exists suchthat T

h

for all T. The familyT

h is called quasi-uniform, ifT

hT

for eachT. Remark that it always holds T hT h. Both properties ex-clude that the angles of the triangles become arbitrarily small. For uniformtriangulations, the size of the triangles is similar for fixed h.

Given an arbitray open and bounded domain 2, the boundary isapproximated by a polygon first. Then the triangulation is applied to thepolygonal domain. For

2, also quadrangles can be used to decomposethe domain. However, triangulations allow for more flexibility.

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Basis functions

We consider an admissible triangulationTh of an open and polygonal do-main 2. We define finite-dimensional function spaces Shconsisting ofall functionsv: satisfying the properties

(i) v Ck() for somek 0,(ii) v|= 0,

(iii) v|Tis a polynomial of degree (at most) l 1 for eachT Th.

Thereby, the choice of the integers k, l is independent ofh >0.

Hence piecewise polynomial functions appear. We apply the case k = 0(globally continuous functions) and l = 1 (piecewise linear functions). Itholds

v|T =T+ Tx+ Ty for each T Thwith coefficients T, T, T .

LetR={(xi, yi) :i = 1, . . . , N } be the set of inner nodes, i.e., the cornersof the triangles inside . Let R= {(xi, yi) :i = N+ 1, . . . , N + K}be theset of boundary nodes, i.e., the corners of the triangles on . We definepiecewise linear basis function i via

i(xj, yj) =

1 if i= j0 if i =j (2.34)

fori= 1, . . . , N andj = 1, . . . , N + K. It holds dim(Sh) =N.

We have to evaluate the bilinear form (2.21), which can be decomposed into

a(i, j) =

2k,l=1

aklixk

jxl

+ a0ij dx

=TTh

T

2k,l=1

aklixk

jxl

+ a0ij dx

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for i, j = 1, . . . , N . Likewise, the information of the right-hand side isevaluated via

, i =

f(x)i(x) dx=TTh

T

f(x)i(x) dx

fori= 1, . . . , N .

In the case

n, the general definition of finite elements following Ciarletis given now.

Definition 13 (finite elements)A finite element is a triple(T, , ) with the properties:

(i) T n is a polyhedron (it follows thatT is bounded),(ii) C0(T) is a linear space of finite dimensions,

(iii) is a set ofs linear independent mappings : , which defineeach uniquely (generalised interpolation).

Sometimes, just the subdomains T are called the finite elements. Incase of

2, a triangulation implies a corresponding set of finite ele-ments, whereT is a triangle.

Benchmark problem

Given a uniform grid in the square = (0, 1) (0, 1), cf. Figure 2, it isstraightforward to generate a triangulation, see Figure 6. The defined stepsize h is not half of the diameter in this case. We consider the Poissonequationu=f with homogeneous Dirichlet boundary conditions. Thecorresponding bilinear form is given in (2.28).

We apply the piecewise linear basis functions (2.34). For i, letZ= (xi, yi)be the central node. The neighbouring nodes are labelled as shown inFigure 7 (left). We calculate the stiffness matrix Ah = (a(i, j)) in the

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x

y

1

1

h

h

Figure 6: Uniform grid with corresponding triangulation.

Ritz-Galerkin method. Considering (2.28), it follows

a(Z, Z) = (Z)2 dxdy = 2 I,III,IVZx 2

+Zy 2

dxdy

= 2

I,III

Zx

2dxdy+ 2

I,IV

Zy

2dxdy

= 2

h2

I,III

dxdy+ 2

h2

I,IV

dxdy = 2

h2 4 h

2

2 = 4

due to the values of the first derivative, see Figure 7 (right). Furthoremore,we obtain

a(Z, N) =

(Z) (N) dxdy

=

I,IV

Zx

Nx

+Z

y

Ny

dxdy

=

I,IV

1

h

1

hdxdy = 1

h2

I,IV

dxdy

= 1h2

2 h2

2 = 1

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W E

NE

SESW

NW N

S

Z

V

III

IV

I

II

VIII

VII

VI

I II III IV V VI VII VIIIZx

h1 0 h1 0 0 h1 0 h1Zy

h1 0 0 h1 0 h1 h1 0

Figure 7: Basic cell in benchmark problem.

and due to a symmetry alsoa(Z, S) =a(Z, W) =a(Z, E) = 1. It isstraightforward to verify

a(Z, NW) =a(Z, NE) =a(Z, SW) =a(Z, SE) = 0

by observing the supports of the basis functions.

For the right-hand side, we apply an approximation

, i =

f(x, y)i(x, y) dxdy .=h2f(xi, yi),

since it holds f(xj, yj)i(xj, yj) =f(xi, yi)ij for allj and

i(x, y) dxdy=h2.

It follows just the five-point star from the finite difference method, cf. (2.9).Each finite difference method corresponds to some finite element method.

However, not each finite element method is equivalent to a finite differencemethod. Hence finite element techniques allow for more flexibility.

Computation of stiffness matrix

We outline the efficient computation of the stiffness matrix in the Ritz-Galerkin method, where a general admissible triangulation is considered,see Def. 12. The structure of Ah = (a(i, j)) NN suggests to use

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a loop over the inner nodes i = 1, . . . , N to evaluate the bilinear form(node-oriented form). However, it can be shown that this procedure is

inefficient. Alternatively, the loop is arranged over the triangles (element-oriented form).

We consider a polygonal domain 2 with an arbitrary admissibletriangulationTh ={T1, . . . , T Q}. The Poisson equationu = f withhomogeneous Dirichlet boundary conditions is used as benchmark again.The corresponding bilinear form, cf. (2.28),

a

:= a(

,

) = x

x +

y

y dxdy

=

Qq=1

Tq

x

x

+

y

y

dxdy

has to be evaluated for , = 1, . . . , N . We define

aq :=

Tq

x

x

+

y

y

dxdy. (2.35)

It follows forAh NN.

a=

Qq=1

aq and Ah=

Qq=1

Aqh with Aqh:= (a

q).

Leti, j, kbe the index of the corners of the triangle Tq. Hence justi, j, kare non-zero inTqand give a contribution to the integral overTq. We obtainthe structure

Aqh=

... ... ... aqii aqij aqik

... ...

... aqji aqjj aqjk

... ...

... aqki aqkj aqkk

... ...

...

MM,

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x

y

1

1

k

i j

00

Figure 8: Transformation to reference triangle.

where (at most) nine entries are non-zero. The matrix can be written incondensed form

Aqh=

aqii aqij aqikaqji aqjj aqjk

aqki aqkj a

qkk

33. (2.36)

To compute (2.35), we transform each triangle Tq to a reference triangleT = {(, ) 2 : 0 ,, + 1}, see Figure 8. It follows the formula

Aqh= 1

4|Tq|EqEq with Eq:= yj

yk xk

xj

yk yi xi xkyi yj xj xi

,where|Tq| represents the area of the triangle. Recall that the indices i, j, kdepend onq. Thus the entries ofAh follow directly from the coordinates ofthe corners of the triangles.

If one corner ofTqdoes not belong to the inner nodes but to the boundary,say indexi, then a corresponding basis functioniis not defined. It follows

that the first row as well as the first column in (2.36) are omitted. Accord-ingly, two rows and two columns are deleted if two corners are situated onthe boundary. This strategy is in agreement to the homogeneous boundaryconditions.

The matrixAh NN includes N2 entries. SinceAh is the sum ofAqh forq= 1, . . . , Q, we obtain a rough estimate of the non-zero entries in Ah: atmost 9Q entries are non-zero.

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Approximations of higher order

In the previous subsections, we applied piecewise linear polynomials corre-sponding to a triangulationTh ={T1, . . . , T Q} of a polygonal domain .We are able to construct piecewise polynomials of higher degrees. LetPlbe the set of all polynomials up to degree l, i.e.,

Pl :=

p(x, y) =

i,j0,i+jl

cijxiyj

.

It holds dim(Pl) = (l+1)(l+2)

2 , which is also the number of coefficientscij.On each triangle T Th, we choose (l+1)(l+2)2 points zs = (xs, ys) for aninterpolation. Figure 9 illustrates the construction of the points within thereference triangle T. It follows a unique interpolation operator

IT :C0(T) Pl, (ITu)(zs) =u(zs) for s= 1, . . . , (l+1)(l+2)2 .

We obtain a global interpolation operator

Ih:C0

() C0

(), Ih|T =IT.HenceIhuis a piecewise polynomial of degree up to l foru C0(). More-over, Ihu is a globally continuous function. The restiction of the polyno-mial ITu to the edge of the triangle T represents a univariate polynomialof (at most) degree l. Since each edge includes l+ 1 nodes, the univariatepolynomials on the boundary of two neighbouring triangles coincide.

We want to apply the Sobolev spaces Hm(). The theorem of Sobolev

implies H

m

() C0

() for m 2, i.e., each u Hm

() exhibits a con-tinuous representative. It follows that the interpolation operator can beextended to an operator Ih : H

m0 () C0() provided that m 2. We

demand (Ihu)(zs) = 0 for a nodezs due to the homogeneous boundaryconditions.

If the degree l of the piecewise polynomial functions is sufficiently large,then also global interpolants IhuCk() for k 1 can be defined. How-ever, the construction becomes much more complicated. The choice of the

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Figure 9: Nodes for linear (left), quadratic (center) and cubic (right) polynomial interpo-lation in the reference triangle.

interpolation is related to the selection of the finite-dimensional spaces Sh

in the Ritz-Galerkin method.Remark: On a triangulation, we already obtain functions globally Ck()for arbitrary k 0 provided that the degree l of the local polynomials issufficiently large. Thus we do not require more complicated subdomainsof to achieve an approximation of higher order.

Convergence of finite element method

We consider the finite element method for the general problem Lu = fwith a uniformly elliptic differential operator and homogeneous Dirichletboundary conditions in a polygonal domain 2. Let an admissibletriangulationTh ={T1, . . . , T Q} be given. The convergence of the methodfollows from Theorem 12, where we have to discuss approximations resultingfrom interpolation schemes.

A finite element method based on the triangulationTh applies a space

Sh:= v C0( ) : v|= 0, v|T Pl for each T Th . (2.37)We apply the global interpolation operator

Ih:Hm0 () C0(), Ihu|T Pl for each T Th

assumingm2, which has been introduced in the previous subsection. Itholds Ihu Sh foru Hm0 ().

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We define the norm

vm,h:=

TTh v2

Hm

(T)

for functions v : satisfying v|T Hm(T) for each T Th. The

functionsvh Shfrom (2.37) exhibit this property. Remark that v Ck()implies justv Hk+1() even for piecewise polynomial functionsv. It holdsvm,h =vHm for v Hm(). However, the subspace (2.37) fulfills justSh H10().The following theorem holds for general functions, i.e., they are not neces-

sarily the solution of some PDE.

Theorem 13 Lett 2andTh be a quasi-uniform triangulation of. Thecorresponding interpolation by piecewise polynomials of degreet1satisfies

u Ihum,h c htm |u|Ht for u Ht()and0 m t. The constant c 0 depends on , the constant of thequasi-uniform triangulation

Th and the integert.

For the proof, see D. Braess: Finite Elements.

Since the weak solution of our elliptic PDE is defined in H10(), we applythe casem = 1 only. Thereby,Ihu H10() is guaranteed. We assume thatthe unique weak solution satisfies u Ht0() H10() for somet 2. Nowwe achieve the convergence by means of Theorem 12. Due to IhuSh, itholds

infvhSh u vhH1 u IhuH1 c ht1

|u|Ht.We conclude the convergence of order p 1 for the approximation uh Shresulting from the Ritz-Galerkin approach in the norm ofH1() due to

u uhH1 K hp |u|Hp+1 for u Hp+1() (2.38)with K := Cc depending on p. For t = 2, piecewise linear polynomialsare applied. For t >2, we can achieve higher orders of convergence just by

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choosing polynomials of higher degrees in each triangle. The approximationuh is still just continuous globally.

Due to (2.38), we require at least u H2() to obtain convergence of orderp 1. It can be shown that the weak solution satisfies u H2() providedthat f L2() holds and the domain satisfies some basic assumptions.Theorem 13 also yields an estimate corresponding to the norm of L2(),i.e., in the case of m = 0. We expect a convergence of order t in thenorm ofL2(). Unfortunately, Theorem 12 cannot be applied in this case,since the underlying bilinear form is not continuous with respect to the

norm ofL2(). Nevertheless, the strategy of Aubin and Nitsche yields theestimates

u uhL2 K hp+1 |u|Hp+1 for u Hp+1()with constants K >0 depending on p.

For some problems, a uniform convergence can be shown like

supx |

u(x)

uh(x)|

c

h

fL2

with a constant c >0. Such estimates correspond to the spaceL().

We have shown the convergence in a Sobolev norm or the L2-norm, respec-tively. Further estimates can be construced in the energy norm a.

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Chapter 3

Parabolic PDEs

3

Now we consider parabolic PDEs, which are time-dependent problems. Theheat equation represents the benchmark for this class of PDEs. Numericalmethods for initial-boundary value problems of parabolic PDEs will be de-rived and analysed.

3.1 Initial-boundary value problems

Time-dependent parabolic PDEs often exhibit the form

u

t + Lu= f(x1, . . . , xn)

with solution u : D [t0, tend] using some domainD n in space.The linear differential operatorLincludes second-order derivatives ofuwithrespect to space (no derivatives in time) and is often of elliptic type. We

restrict to one space dimension (n= 1) in this chapter.The heat equation reads

v

t =(x)

2v

x2 (3.1)

with a coefficient function : D (D

) and (x) > 0 for each x.

Without loss of generality, we choose(x) 1, i.e.,u

t =

2u

x2. (3.2)

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ba x

0

t

BC BC

IC

Figure 10: Initial-boundary value problem.

Given a solutionuof (3.2), we obtain a solution of (3.1) for constant viathe transformationv(x, t) =u(x,t).

We choose a finite interval [a, b] in space (a < b). Boundary conditions

(BCs) will be specified at x= a andx= b. Initial conditions (ICs) will begiven in the formu(x, t0) =u0(x) for x [a, b] (3.3)

with a predetermined functionu0: [a, b] . Without loss of generality, wedefinet0:= 0. The initial-boundary value problem is sketched in Figure 10.

We distinguish three different types of boundary value problems:

(i) Boundary conditions ofDirichlet typeread

u(a, t) =(t), u(b, t) =(t) for all t 0 (3.4)with predetermined functions , : [0, ) .

(ii) Boundary conditions ofvon-Neumann typedemand

u

x

x=a

=(t), u

x

x=b

=(t) for all t 0 (3.5)

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(iii) Boundary problems ofRobin type, i.e., a mixed problem of the types (i)and (ii), namely

a(t)u(a, t) + a(t)u

x

x=a

=(t), b(t)u(b, t) +b(t)u

x

x=b

=(t)

for allt 0 with predetermined functions , , a, b, a, b.

The initial values (3.3) have to be compatible with the boundary conditions.For example, u0(a) =(0) andu0(b) =(0) is required in case of Dirichletboundary conditions.

Let u be a solution of (3.2) according to homogeneous Dirichlet boundaryconditions (, 0). The function

u(x) := bxba+ xaba

satisfies the inhomogeneous boundary conditions (3.4) for constant values, = 0. It follows that v := u+ u is a solution of (3.2), which fulfillsthe inhomogeneous Dirichlet problem. Hence we consider homogeneous

Dirichlet conditions without loss of generality.

We solve the heat equation (3.2) with homogeneous Dirichlet boundaryconditions analytically fora= 0, b= 1. We assume a separation

u(x, t) =(t)(x).

Inserting this relation into (3.2) gives

(t)(x) =(t)(x)

(t)

(t) =

(x)

(x) =:.

Thereby, represents the separation constant. Solving the two ordi-nary differential equations

(t) =(t), (x) =(x)

yields(t) =Cet, (x) =Ae

x + Be

x.

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We obtain the general solution

u(x, t) = et Aex + Bexwith arbitrary constants A, B . The homogeneous boundary conditionsare satisfied if and only if

= k22 for k= 1, 2, 3, . . . .It follows the family of solutions

vk(x, t) = Ake

k22t

sin(kx)

fork with new coefficients Ak . We use these solutions to construct

a single solution satisfying the predetermined initial conditions (3.3). Itholds u0(0) =u0(1) = 0 due to the homogeneous boundary conditions. Wecan extend the function u0 to an odd function u : [1, 1] by thedefinition u(x) =u0(x) forx 0 and u(x) = u0(x) forx


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Hence we obtain

|u(x, t) u(x, t)| k=1

|ak ak| ek2

2

t.

The Cauchy-Schwarz inequality in2 and Parsevals theorem yield

k=1

|ak ak| ek22t

k=1

|ak ak|2

k=1

|ek22t|2

=

u0

u0L2([0,1])

k=1

e2k22t.

Thereby, we employ that the extensions u,u ofu0,u0 exhibit the period 2in Parsevals theorem and thatu u2L2([1,1]) = 2u0 u02L2([0,1]) due tothe symmetry.

We apply the formula of the limit of a geometric series

k=1 e2k22t =

k=1 e22t

k2

< 1

1 e22t

1 =

e22t

1 e22t

.

It follows

|u(x, t) u(x, t)| u0 u0L2([0,1]) e2t

1 e22t

for allx [0, 1] andt >0. Hence differences in the initial values are dampedout exponentially in time. The condition of this initial-boundary value prob-lem is excellent. Vice versa, an initial-boundary value problem backwards

in time (from t0 = 0 to some tend < 0) is drastically ill-conditioned, sincesmall differences are amplified exponentially.

For the heat equation (3.1), the condition of initial-boundary value problemsdepends on the constant

\{0} as follows:

forward in time backward in time

>0: well-conditioned ill-conditioned


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Initial value problems backwards in time are also called final value problems(the values at the earlier time tend < 0 are unknown, whereas the state at

the final time t0= 0 is given).

We achieve a solution of an initial value problem in the casex (, +),where no boundary appears. It follows

u(x, t) = 1

2

t

+

e(xy)2

4t u0(y) dy. (3.7)

The integrals exist for a bounded measurable function u0 or in the case

of u0 L2

(

), for example. Otherwise, integrability conditions have tobe imposed. The formula (3.7) cannot be evaluated at t = 0. The initialconditions are satisfied in the sense

limt0+

u(x, t) =u0(x) for each x .

Moreover, this convergence is uniform in compact domains D .

Let u0 be continuous, u0 0 and u0 0. Even ifu0 exhibits a compactsupport, it follows u(x, t) > 0 for all x

and each t > 0. Hence the

transport of information proceeds with infinite speed. This also holds incase of initial-boundary value problems within a finite domain x [a, b].

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a x

0

t

BC BC

bIC

Figure 11: Grid in finite difference method.

3.2 Finite difference methods

We want to apply a finite difference method to solve the initial-boundaryvalue problem of the heat equation (3.2) introduced in Sect. 3.1. A grid isconstructed in the (x, t)-domain for x[a, b] and t[0, T], see Figure 11.Without loss of generality, we assumex [0, 1]. The grid points are definedby

xj :=jh for j = 0, 1, . . . , M 1, M , h:= 1M,tn:= nk for n= 0, 1, . . . , N 1, N, k:= TN.

The corresponding step sizes are h = x and k = t in space and time,respectively. Letunj :=u(xj, tn) be the values of the exact solution and U

nj

the corresponding approximations in the grid points. We replace the partialderivatives in the heat equation (3.2) by difference formulas now.

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Classical explicit method

We substitute the time derivative by the common difference formula of firstorder and the space derivative by the symmetric difference formula of secondorder, i.e.,

ut(xj, tn) = 1k(u(xj, tn+1) u(xj, tn)) + k2utt(xj, tn+ k)

uxx(xj, tn) = 1h2(u(xj1, tn) 2u(xj, tn) +u(xj+1, tn))+ h

2

12uxxxx(xj+ h,tn)

with intermediate values (0, 1), (1, 1). Thereby, we assume thatu is sufficiently smooth. The heat equation yields ut(xj, tn) = uxx(xj, tn).It follows

1k(u

n+1j unj ) + O(k) = 1h2(unj1 2unj + unj+1) + O(h2).

Thus the finite difference method reads1k(U

nj+1 Unj) = 1h2(Unj1 2Unj + Unj+1),

Un+1j = Unj +

kh2(U

nj

1

2Unj + U

nj+1).

We define the ratio r:= kh2 . The formula of the technique becomes

Un+1j =rUnj1+ (1 2r)Unj + rUnj+1 (3.8)

for j = 1, . . . , M 1. The scheme is an explicit (one-stage) method.The time layers can be calculated subsequently. The initial values followfrom (3.3), i.e.,

U0j =u0(xj) for j = 0, 1, . . . , M .

In the subsequent layers, the boundary conditions have to be included.Dirichlet boundary conditions yield

Un0 =(tn), UnM=(tn) for each n.

Von-Neumann boundary conditions will be discussed later.

The local discretisation error reads

(k, h) := k2utt(xj, tn+ k) h2

12uxxxx(xj+ h,tn).

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We assume thatutt anduxxxx exist and are continuous on [0, 1] [0, T]. Let

C1:= maxx[0,1],t[0,T] |utt(x, t)| , C2:= maxx[0,1],t[0,T] |uxxxx(x, t)| .It follows

|(k, h)| (k+ h2)(max{12C1, 112C2}) =:C(k+ h2) (3.9)uniformly for all grid points (xj, tn) in [0, 1] [0, T]. Hence the finite dif-ference meth

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