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Computational Differential Equations

http://www.math.unibas.ch/ ∼beilina

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Course requirements

1. Lectures

2. Assignments (homework)

3. Project

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Lectures

Mathematical Institute, grossen Horsaal

1. Wednesday, 12.00-14.00

2. Friday, 14.00-16.00

Project and Assignments:

Thursday, 14.00-16.00 Mathematical Institute, Seminarraum/Matlabraum

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Literature

1. S.C.Brenner, L.R.Scott. The Mathematical theory of finite element

methods, Springer-Verlag, 1994.

2. K.Eriksson, D.Estep and C.Johnson, Computational Differential

Equations, Studentlitteratur, Lund, 1996.

3. Romanov V.G. and Kabanikhin S.I. Inverse Problems for Maxwell’s

Equations. VSP, Utrecht, The Netherlands, 1994.

4. Kabanikhin S.I. and Lorenzi A. Identification Problems of Wave

Phenomena (Theory and Numerics). VSP, Utrecht, The Netherlands,

1999.

5. L.Beilina, Adaptive Finite Element/Difference Methods for

Time-Dependent Inverse Scattering Problems.

www.phi.chalmers.se/preprints/ 

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Course plan

1. Basic Concepts :

(a) Weak formulation of Boundary value Problem;

(b) The Finite Element Method (FEM);

(c) The Finite Difference Method (FDM);

(d) Computer implementation of FEM and FDM.

2. Sobolev Spaces

(a) Review of Lebesgue Integration Theory;

(b) Weak Derivatives;

(c) Sobolev norms and associated spaces;

(d) Trace Theorems.

3. Variational formulation of boundary value problems

(a) Hilbert Spaces ;

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(b) Riesz Representation Theorem;

(c) The Lax-Milgram Theorem;

4. The construction of a Finite Element Spaces

(a) Definition of a Finite Element. Triangular Finite Elements.

Rectangular Finite Elements;

(b) The Interpolant.

5. n-dimensional variational problems and adaptive error control

(a) Variational Formulation of Poisson’s Equation. Treatment of 

different boundary conditions;

(b) Adaptive error control;

(c) The Heat Equation. A FEM for the Heat Equation. Error

estimates and adaptive error control;

(d) The Wave Equation. A FEM for the Wave Equation. Error

estimates and adaptive error control.

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6. Application of FEM to inverse scattering problems

(a) The FEM for Inverse Acoustic Scattering;

(b) The FEM for Inverse Elastic Scattering;

(c) The FEM for Inverse Electromagnetic scattering.

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Weak formulation of Boundary Value Problem

Let us consider two-point boundary-value problem

−

∂ 2u

∂x2 = f, in (0, 1),u(0) = 0,

u(1) = 0.

(1)

if u is the solution and v is any regular function such that v(0) = 0, then

we can integrate by parts to get

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(f, v) := 10

f (x)v(x)dx

=

 1

0

−u(x)v(x)dx

=

 1

0

u(x)v(x),

a(u, v) :=  1

0

u(x)v(x)

(2)

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Let us define also

V  := {v ∈ L2(0, 1) : a(v, v) < ∞, v(0) = 0}. (3)

Then the variational formulation of (1) is: find u ∈ V  such that

a(u, v) = (f, v) ∀v ∈ V. (4)

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Theorem. Suppose f  ∈ C 0([0, 1]) and u ∈ C 2([0, 1]) satisfy (4). Then u

solves (1).

Proof Let v ∈ V  ∩ C 1

([0, 1]), then integration by parts yields

(f, v) = a(u, v) =

 1

0

−u(x)v(x)dx + u(1)v(1) (5)

Thus, (f  − (−u(x)), v) = 0 ∀v ∈ V  ∪ C 1([0, 1]) : v(1) = 0. We can

choose w = f  − (−u(x)) ∈ C 0([0, 1]). If w = 0, then w is of one sign

on [x0, x1] ⊂ [0, 1] with x0 < x1 (continuity). Choose

v(x) = (x − x0)2

(x − x1)2

in [x0, x1], and v = 0 else. But in this case(w, v) = 0 - contradiction. Thus −u = f . Now we can apply (6) with

v(x) = x to find u(1) = 0. u ∈ V  implies u(0) = 0 and u solves (1). 2

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Conventions for boundary conditions

Boundary

ConditionVariational Name Proper Name

u(x) = 0 essential Dirichlet

u

(x) = 0 natural Neumannu(x) + γu

(x) = 0 combination of both Robin

Dirichlet b.c. appears in the variational formulation explicitly and it is

called essential. Neumann b.c. we call “natural” because it is

incorporated implicitly.

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Ritz-Galerkin Approximation

Let S  ⊂ V  be any finite dimensional subspace. Let us consider

variational formulation (4) with V  replaced by S :

uS  ∈ S  : a(uS , v) = (f, v) ∀v ∈ S. (6)

Theorem. Given f  ∈ L2(0, 1), (6) has a unique solution.

Proof We will write (6) in terms of basis {ϕi : 1 ≤ i ≤ n} of S . Let

uS  =n

j=1 U jϕj and let K ij = a(ϕj , ϕi), F i = (f, ϕi) for

i, j = 1,...,n. Set U  = (U j), K  = (K ij), F  = (F i). Then (6) is

equivalent to solving the square matrix equation

KU  = F. (7)

For square matrix eq. we know that uniqueness is equivalent to existence

- this is a finite dimensional system. Nonuniqueness would imply that ∃V 

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nonzero such that KV  = 0. Write v =

V jϕj and note that equivalence

of (4) and (6) implies that a(v, ϕj) = 0 ∀ j. We mult. this by V j , take

sum over j and get: 0 = a(v, v) = 1

0(v(x))2dx ⇒ we conclude that

v

= 0 ⇒ vis constant. Since

v ∈ S  ⊂ V  ⇒ v(0) = 0and we must have

v = 0. Since {ϕi : 1 ≤ i ≤ n} is a basis of S , this means that V  = 0.

Thus, the solution to (7) must be unique (and hence must exist).

Therefore, the solution uS  to (6) must also exist and be unique. 2

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Error Estimate in energy norm

Orthogonality relation between u and uS : subtructing a(uS , v) froma(u, v) implies:

a(u − uS , w) = 0 ∀w ∈ S. (8)

We define energy norm as

||v||E  = 

a(v, v) ∀v ∈ V. (9)

Relationship between energy norm and inner-product is Schwarz’

inequality:

|a(v, w)| ≤ ||v||E ||w||E  ∀v, w ∈ V. (10)

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Theorem. ||u − uS ||E  = min ||u − v||E  : v ∈ S .

Proof ∀v ∈ S  from Schwarz’ inequality we have:

||u − uS ||2

E  = a(u − uS , u − uS )

= a(u − uS , u − v) + a(u − uS , v − uS )

= a(u − uS , u − v) ≤ ||u − uS ||E ||u − v||E  ∀w ∈ S.

(11)

If ||u − uS ||E  = 0, we can devide by it to obtain

||u − uS ||E  ≤ ||u − v||E  ∀v ∈ S . If ||u − uS ||E  = 0, we take infimum

over v ∈ S  : ||u − uS ||E  ≤ inf {||u − v||E  : v ∈ S }. Since uS  ∈ S  we

have inf {||u − v||E  : v ∈ S } ≤ ||u − uS ||E , therefore

||u − uS ||E  = inf {||u − v||E  : v ∈ S }. 2

This is the basic error estimate for the Ritz-Galerkin method.

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Error estimate in L2-norm

Let us define L2(0, 1)-norm as

||v|| = (v, v)1/2 = (

 1

0

v(x)2dx)1/2 (12)

We want to estimate u − uS  in this norm.

Theorem Assumption inf v∈S  ||w − v||E  ≤ ||w|| implies that

||u − uS || ≤ ||u − uS ||E  ≤ 2||u|| = 2f.

Proof For the proof we use “duality argument”: let w be the solution of 

−w = u − uS  on [0, 1] with b.c. w(0) = w(1) = 0. Integrating by

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parts, we get

||u − uS ||

2

= (u − uS , u − uS )= (u − uS , −w) = a(u − uS , w) ((u − uS )(0) = w(1) = 0)

= a(u − uS , w − v) ∀v ∈ S  (from orthogonality).

(13)

Thus, Schwarz ineq. implies that

||u − uS || ≤ ||u − uS ||E ||w − v||E /||u − uS ||

= ||u − uS ||E ||w − v||E /||w||.(14)

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We take now inf over v ∈ S  to get

||u − uS || ≤ ||u − uS ||E  inf v∈S  ||w − v||E /||w||.

2

We see that L2-norm can be much smaller than the energy norm provided

that w can be approximated well by some function in S .

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