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Numerical Methods for Differential Equations

Chapter 4: Two-point boundary value problems

Gustaf Soderlind and Carmen Arevalo

Numerical Analysis, Lund University

Textbooks: A First Course in the Numerical Analysis of Differential Equations, by Arieh Iserles

and Introduction to Mathematical Modelling with Differential Equations, by Lennart Edsberg

c© Gustaf Soderlind, Numerical Analysis, Mathematical Sciences, Lund University, 2008-09

Numerical Methods for Differential Equations – p. 1/86

Chapter 4: contents

◮ Finite difference approximation of derivatives

◮ Finite difference methods for the 2p-BVP

◮ Newton’s method

◮ Sturm–Liouville problems

◮ Toeplitz matrices

◮ Convergence: Lax’ equivalence theorem

◮ Differential operators

◮ From finite differences to finite elements


1. Approximation of derivatives ( y′ = dy/dx)

First order approximations

Forward difference

y′(x) =y(x + ∆x) − y(x)

∆x+ O(∆x)

Backward difference

y′(x) =y(x) − y(x − ∆x)

∆x+ O(∆x)


Approximation of derivatives . . .

Second order approximations

Symmetric difference quotients

y′(x) =y(x + ∆x) − y(x − ∆x)

2∆x+ O(∆x2)

y′′(x) =y(x + ∆x) − 2y(x) + y(x − ∆x)

∆x2+ O(∆x2)


Derivatives → finite differences → matrices

Matrix representation of forward difference

y′(x) =y(x + ∆x) − y(x)

∆x+ O(∆x)

Introduce vectors y = {y(xi)} and y′ = {y′(xi)} :

y′0

y′1

...

y′N

≈ 1

∆x

−1 1

−1 1. . . . . .

−1 1

y0

y1

...

yN+1


Derivatives . . . matrices

Note Forward difference ∼ (N + 1) × (N + 2) matrix

y′0

y′1

...

y′N

≈ 1

∆x

−1 1

−1 1. . . . . .

−1 1

y0

y1

...

yN+1

Nullspace spanned by y = (1 1 1 . . . 1)T

Compare nullspace of d/dx : y = 1 ⇒ y′ ≡ 0

Analogous result for backward difference



Central difference

y′(x) ≈ y(x + ∆x) − y(x − ∆x)

2∆x

Matrix representation

y′1

y′2

...

y′N

≈ 1

2∆x

−1 0 1. . . . . . . . .

−1 0 1

y0

y1

...

yN+1



Note N × (N + 2) matrix

y′1

y′2

...

y′N

≈ 1

2∆x

−1 0 1. . . . . . . . .

−1 0 1

y0

y1

...

yN+1

Nullspace is two-dimensional:y = (1 1 1 . . . 1)T and y = (1 −1 1 −1 . . . 1)T



“False” nullspace y = (1 −1 1 −1 . . . 1)T does notconverge to a C1 function!

Compare difference equation yn+1 − yn−1 = 0, withcharacteristic equation

z2 − 1 = 0 ⇒ z = ±1

and solutions yn = 1 and yn = (−1)n


2nd order derivatives → matrices

Central difference

y′′(x) ≈ y(x + ∆x) − 2y(x) + y(x − ∆x)

∆x2

y′′1

y′′2

...

y′′N

≈ 1

∆x2

1 −2 1. . . . . . . . .

1 −2 1

y0

y1

...

yN+1

Note N × (N + 2) matrix with nullspace y = (1 1 . . . 1)T

and y = (0 1 2 3 . . . N + 1)T


2nd order derivatives . . .

Nullspace of d2/dx2 :y = 1 and y = x both have y′′ ≡ 0

Compare difference equation yn+1 − 2yn + yn−1 = 0, withcharacteristic equation

z2 − 2z + 1 = 0 ⇒ z = 1, 1

and solutions yn = 1 and yn = n , respectively


Numerical differentiation

First and second derivatives of y = sin πx

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1Data function sin(pi*x) and sampled points for h=1/6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−4

−2

0

2

4First derivative, fwd diff (+), bwd diff (x), symmetric diff (o)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−10

−5

0Second derivative, symmetric 2nd diff (o)


2. Finite difference methods for 2p-BVP

Consider simplest problem

y′′ = f(x, y)

y(0) = α; y(1) = β

Introduce equidistant grid with ∆x = 1/(N + 1)

Discretization

yi+1 − 2yi + yi−1

∆x2= f(xi, yi)

y0 = α; yN+1 = β


Discrete 2pBVP

yi+1 − 2yi + yi−1

∆x2= f(xi, yi) i = 1 : N

y0 = α; yN+1 = β

This is a (nonlinear) system of equations F (y) = 0 for theN unknowns y1, y2, . . . , yN

Solve F (y) = 0 using Newton’s method


Equation system F (y) = 0

F1(y) =α − 2y1 + y2

∆x2− f(x1, y1)

Fi(y) =yi−1 − 2yi + yi+1

∆x2− f(xi, yi)

FN (y) =yN−1 − 2yN + β

∆x2− f(xN , yN)

Note how boundary values enter


Jacobian matrix

F ′(y) = tridiag (1/∆x2 , −2/∆x2 +∂f

∂yi

, 1/∆x2)

is tridiagonal , and

◮ Super- and subdiagonal elements 1/∆x2

◮ Diagonal elements −2/∆x2 − ∂f/∂yi

◮ Sparse LU decomposition runs in O(N) time

◮ Solution effort moderate even when N is large


3. Newton’s method (recap)

Let y(k) be an approximation to the root y

Taylor series expansion

0 = F (y) ≈ F (y(k)) + F ′(y(k)) · (y − y(k))

Define y(k+1) by

0 =: F (y(k)) + F ′(y(k)) · (y(k+1) − y(k))


Newton’s method for F (y) = 0

Newton iteration

1. Compute Jacobian F ′(y(k)) = {∂Fi/∂yj}2. Factorize Jacobian matrix F ′(y(k)) → LU

3. Solve linear system LUδy(k) = −F (y(k))

4. Update y(k+1) := y(k) + δy(k)

Newton’s method is quadratically convergent


Quadratic convergence

Newton’s method converges if

1. ‖F ′(y(k))−1‖ ≤ C ′

2. ‖F ′′(y(k))‖ ≤ C ′′

3. ‖y(0) − y‖ < ε (close enough starting value)

Then convergence is quadratic

‖y(k+1) − y‖ ≤ C · ‖y(k) − y‖2


Boundary Conditions come in many shapes

◮ “Dirichlet” boundary conditionsy(0) = α straightforward to implement

◮ “Neumann” boundary conditionsy′(0) = γ requires special attention

◮ “Robin” conditionsy(0) + c · y′(0) = κ requires same attention

for the method’s convergence order to be preserved


Neumann problem

Exampley′′ = f(x, y)

y(0) = α; y′(1) = β

Equidistant grid, with x = 1 between grid points!

xN + ∆x/2 = 1 = xN+1 − ∆x/2

y′(1) = β → yN+1 − yN

∆x= β

⇒ yN+1 := β ∆x + yN is of second order at x = 1


Robin problem

Exampley′′ = f(x, y)

y(0) = α; y(1) + c · y′(1) = κ

Equidistant grid, with x = 1 between grid points!

y(1) + cy′(1) = κ → yN+1 + yN

2+ c

yN+1 − yN

∆x= κ

⇒ yN+1 :=(2c − ∆x)yN + 2κ∆x

2c + ∆x


4. Sturm–Liouville eigenvalue problems

Diffusion problem

∂u

∂t=

∂

∂x

(

p(x)∂u

∂x

)

= 0 ; y(a) = 0 , y(b) = 0

Separation of variables (one space dimension)

u(t, x) := y(x) · v(t) ⇒ v

v=

(p(x) y′(x))′

y=: λ

Sturm–Liouville eigenvalue problem

d

dx

(

p(x)dy

dx

)

− λy = 0 ; y(a) = 0 , y(b) = 0


Sturm–Liouville eigenvalue problem. . .

Wave equation

∂2u

∂t2= c2 ∂2u

∂x2; y(a) = . . . , y(b) = . . .

Express solution as u(t, x) = y(x) eiωt ⇒

−ω2y = c2y′′

Sturm–Liouville eigenvalue problem

y′′ = λy with λ = −ω2/c2


Sturm–Liouville eigenvalue problem. . .

Find eigenvalues λ and eigenfunctions y(x) with

d

dx

(

p(x)dy

dx

)

= λy ; y(a) = 0 , y(b) = 0

Discretization Matrix eigenvalue problem

T∆x y = λ∆x y

Note Analytic eigenvalue problem converts to algebraic!


Sturm–Liouville problem. Discretization

pi−1/2yi−1 − (pi−1/2 + pi+1/2)yi + pi+1/2yi+1

∆x2= λ∆xyi

y0 = yN+1 = 0

Symmetric tridiagonal N × N eigenvalue problem

T∆x y = λ∆x y

There are N eigenvalues λ∆x,n = λn + O(∆x2)


Sturm–Liouville problem. Simple example

Exampley′′ = λy

y(0) = y(1) = 0

Analytic solution

y(x) = A sin√−λx + B cos

√−λx

Boundary values ⇒ B = 0 and

A sin√−λ = 0

A 6= 0 ⇒√

−λn = nπ n = 1, 2, . . .


Simple Sturm–Liouville example . . .

y′′ = λy

y(0) = y(1) = 0

Eigenvalues λn = −n2π2 , n = 1, 2, . . .

Eigenfunctions yn(x) = sin nπx


Discrete Sturm–Liouville. Same example

Discretization of y′′ = λy with BVs ⇒

yi−1 − 2yi + yi+1

∆x2= λ∆xyi

y0 = yN+1 = 0 ; ∆x = 1/(N + 1)

Tridiagonal N × N matrix formulation

1

∆x2

−2 1

1 −2 1. . .

1 −2

y1

y2

...

yN

= λ∆x

y1

y2

...

yN


Discrete Sturm–Liouville . . .

Algebraic eigenvalue problem

T∆x y = λ∆x y

Smallest eigenvalue λ∆x = −π2 + O(∆x2)

The first few eigenvalues are well approximated, but theapproximation gradually gets worse


Discrete Sturm–Liouville. ComputationFirst three (N =19) eigenvectors of T∆x

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.4

−0.2

0

0.2

0.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.4

−0.2

0

0.2

0.4


Discrete Sturm–Liouville. Eigenvalues N = 19

Discrete eigenvalues λ∆x −9.8493 −39.1548 −87.1948

Exact eigenvalues λ −9.8696 −39.4784 −88.8264

Relative errors 0.21% 0.82% 1.63%

◮ Lowest eigenvalues are more accurate

◮ Good approximations for√

N first eigenvalues

(Here approximately first 4 – 5 modes)


Discrete Sturm–Liouville. High modesEigenvectors 7, 13, 19 (N =19) of T∆x

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.4

−0.2

0

0.2

0.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.4

−0.2

0

0.2

0.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.4

−0.2

0

0.2

0.4


5. Toeplitz matrices

A Toeplitz matrix is constant along diagonals

Example (symmetric)

T∆x =1

∆x2

−2 1 0 . . .

1 −2 1

1 −2 1. . .

. . . 0 1 −2


Toeplitz matrices . . .

Much is known about Toeplitz matrices

◮ Eigenvalues

◮ Norms

◮ Inverses

◮ etc.

They can be generated in MATLAB using the built-infunction toeplitz


Eigenvalues of Toeplitz matrices

Example Solve the eigenvalue problem Ty = λy for

T =

−2 1 0 . . .

1 −2 1

1 −2 1. . .

. . . 0 1 −2

Note λ[T ] = −2 + λ[S]


Eigenvalues . . .

. . . the problem gets simplified

Sy =

0 1 0 . . .

1 0 1

1 0 1. . . 1

. . . 0 1 0

y1

y2

...

yN

= λy

Find the eigenvalues of S!


Eigenvalues . . . and difference equations!

Consider the nth equation of Sy = λy :

yn+1 + yn−1 = λyn

Linear difference equation with boundary values

y0 = 0; yN+1 = 0

Characteristic equation

z2 − λz + 1 = 0


Eigenvalues . . . characteristic equation

Roots of z2 − λz + 1 = 0 are z and 1/z (product 1)

General solution yn = αzn + βz−n





Boundary condition y0 = 0 = α + β ⇒

Solution yn = α(zn − z−n)







Boundary conditionyN+1 = 0 = α(zN+1 − z−(N+1)) ⇒ z2(N+1) = 1







Boundary conditionyN+1 = 0 = α(zN+1 − z−(N+1)) ⇒ z2(N+1) = 1

Roots zk = exp( kπi

N + 1

)

k = 1 : N


Eigenvalues . . .

Sum of the roots of z2 − λz + 1 = 0 are

λk = zk + 1/zk ⇒

λk[S] = exp( kπi

N + 1

)

+ exp(

− kπi

N + 1

)

= 2 coskπ

N + 1


Eigenvalues . . .

Sum of the roots of z2 − λz + 1 = 0 are

λk = zk + 1/zk ⇒

λk[S] = exp( kπi

N + 1

)

+ exp(

− kπi

N + 1

)

= 2 coskπ

N + 1

Hence

λk[T ] = − 2 + 2 coskπ

N + 1= −4 sin2 kπ

2(N + 1)


Eigenvalues of Toeplitz matrices

Theorem The N × N Toeplitz matrix

T =

−2 1 0 . . .

1 −2 1

1 −2 1. . .

. . . 0 1 −2

has N real eigenvalues ( k = 1 : N )

λk[T ] = −4 sin2 kπ

2(N + 1)∈ (−4, 0)


Eigenvalues of Toeplitz matrices . . .

Consider the operator approximation

d2

dx2↔ 1

∆x2T

on x ∈ [0, 1], with ∆x = 1/(N + 1)

Corollary The eigenvalues of T∆x := T/∆x2 are

λk[T∆x] = −4(N + 1)2 sin2 kπ

2(N + 1)≈ −k2π2

for k ≪ N


What are the eigenvalues of d2/dx2 on [0, 1]?

Consider the Sturm–Liouville problem

u′′ = λu ; u(0) = u(1) = 0

Solutions u(x) = A sin√−λ x + B cos

√−λ x

Boundary conditions B = 0 and A sin√−λ = 0


What are the eigenvalues of d2/dx2 on [0, 1]?

Consider the Sturm–Liouville problem

u′′ = λu ; u(0) = u(1) = 0

Solutions u(x) = A sin√−λ x + B cos

√−λ x

Boundary conditions B = 0 and A sin√−λ = 0

Theorem The eigenvalues are λk[d2/dx2] = −k2π2

Note k ∈ Z+


What are the norms of T?

Lemma For a symmetric matrix A, it holds

‖A‖2 = maxk

|λk|


What are the norms of T?


‖A‖2 = maxk

|λk|


µ2[A] = maxk

λk

(Both results actually hold for normal matrices)


Proofs: Norm

Definition

‖A‖22 = max

xTx6=0

xTATAx

xTx

Find stationary points of the Rayleigh quotient of ATA,given by ρ(x) = xTATAx/xTx

gradxρ(x) = (2ATAxxTx − 2xxTATAx)/(xTx)2 := 0

ATAx = ρ(x)x ⇒ A2x = ρ(x)x


Proofs: Norm . . .

So ρ(x) = λ2, where λ is an eigenvalue of A

Therefore ‖A‖22 = max |λ[A]|2 or

‖A‖2 = max |λ[A]|

when A is symmetric


Proofs: Logarithmic norm

Definition

µ2[A] = maxxTx6=0

xTAx

xTx

Find stationary points of the Rayleigh quotient of A, givenby ρ(x) = xTAx/xTx

gradxρ(x) = [(A + AT)xxTx − 2xxTAx]/(xTx)2 := 0

1

2(A + AT)x = ρ(x)x ⇒ Ax = ρ(x)x


Proofs: Logarithmic norm . . .

So ρ(x) = λ, where λ is an eigenvalue of A

Therefore µ2[A] = maxλ[A] when A is symmetric.

For symmetric matrices we have proved

‖A‖2 = maxk

|λk| µ2[A] = maxk

λk


What are the norms of T∆x?

Eigenvalues of T∆x = T/∆x2 are

λk[T∆x] = −4(N + 1)2 sin2 kπ

2(N + 1)

So ‖T∆x‖2 = |λN | and µ2[T∆x] = λ1

Theorem The Euclidean norms of T∆x are

‖T∆x‖2 ≈4

∆x2µ2[T∆x] ≈ −π2


The norm of T−1∆x

Recall that µ[A] < 0 ⇒ ‖A−1‖ ≤ −1/µ[A]

Approximate y′′ = f(x) with y(0) = y(1) = 0 by

T∆xu = q

Note that µ2[T∆x] ≈ −π2 then implies that this problem hasa unique solution, as

‖T−1∆x‖2 /

1

π2


What norms to use. RMS and Euclidean norms

The norm of a function is measured in the L2 norm

‖l‖22 =

∫ 1

0

l(x)2dx

A corresponding discrete function (vector) is thenmeasured in the root mean square (RMS) norm

‖l(x)‖2∆x =

N∑

1

l(xi)2∆x =

1

N + 1

N∑

1

l(xi)2 =

1

N + 1‖l(x)‖2

2

Important! Note that ‖T−1∆x‖∆x ≡ ‖T−1

∆x‖2


Solution and error bounds

Approximate y′′ = f(x) with y(0) = y(1) = 0 by

T∆xu = q

Convergence analysis will show

Theorem µ2[T∆x] ≈ −π2 implies a unique solution with

‖u‖∆x /‖q‖∆x

π2‖e(x)‖∆x /

‖l(x)‖∆x

π2

Solution–data Global–local error


6. Convergence of finite difference methods

Basic problem

y′′ = f(x, y)

y(0) = α; y(1) = β

Equidistant discretization

yi+1 − 2yi + yi−1

∆x2= f(xi, yi)

y0 = α; yN+1 = β


Error definitions

Local error

Insert exact solution y(x) into discretization:

y(xi−1) − 2y(xi) + y(xi+1)

∆x2= f(xi, y(xi))−l(xi)

Taylor expansion, using f(xi, y(xi)) = y′′(xi):

−l(xi) = 2

(

∆x2

4!y(4)(xi) +

∆x4

6!y(6)(xi) + . . .

)

Only even powers of ∆x due to symmetry


Global error

Definition The global error is defined e(xi) = yi − y(xi)

Convergence Will show that e(x) → 0 as ∆x → 0, ormore specificallly

e(xi) = c1∆x2 + c2∆x4 + . . .

Again only even powers due to symmetry


Convergence

Assume f is a linear function, ⇒ F is linear ⇒

F (y) = 0 ⇔ T∆xy = q

with T∆x tridiagonal. Then

Numerical solution T∆xy = q

Exact solution T∆xy(x) = q − l(x)


Convergence . . .

Solve T∆xy = q formally to get

Numerically y = T−1∆x · q

Exact y(x) = T−1∆x · (q − l(x))

Global error e(x) = T−1∆x · l(x)

Error bound ‖e(x)‖∆x ≤ ‖T−1∆x‖2 · ‖l(x)‖∆x


Matching norms: RMS and Euclidean norms

Note how root mean square (RMS) norm of vector

‖l(x)‖2∆x =

N∑

1

l(xi)2∆x =

1

N + 1

N∑

1

l(xi)2 =

1

N + 1‖l(x)‖2

2

is matched to the Euclidean matrix norm, by way of

‖T−1∆x‖∆x ≡ ‖T−1

∆x‖2

(Prove this relation!)


Convergence . . .

Recall µ2[T∆x] ≈ −π2 ⇒ ‖T−1∆x‖2 / 1/π2 (∆x→0)

Since ‖e(x)‖∆x ≤ ‖T−1∆x‖2 · ‖l(x)‖∆x and

‖l‖∆x = γ1∆x2 + γ2∆x4 . . . we have

‖e‖∆x ≤ C · ‖l‖∆x = c1∆x2 + c2∆x4 + . . .

. . . and we have convergence as ∆x→0! :)


Convergence: Lax’ principle

Conclusion

Consistency: local error l → 0 as ∆x → 0

Stability: ‖T−1∆x‖2 ≤ C as ∆x → 0

Convergence: global error e → 0 as ∆x → 0

Theorem (Lax’ Principle)

Consistency + Stability ⇒ Convergence

(“Fundamental theorem of numerical analysis”)


7. Differential operators. Integration by parts

Logarithmic norm of matrix:

µ2[A] = maxx6=0

xTAx

xTx⇒ xTAx ≤ µ2[A] · xTx

For d2/dx2 , introduce the inner product

〈u, v〉 =

∫ 1

0

u(x)v(x) dx ⇒ ‖u‖22 = 〈u, u〉


The logarithmic norm of d2/dx2

Can we find a constant µ2[d2/dx2] such that

〈u, u′′〉 ≤ µ2[d2/dx2] · ‖u‖2

2

for all functions u ∈ C20 [0, 1]?

Yes and µ2[d2/dx2] = −π2


Integration by parts

∫ 1

0

uv′ dx = [uv]10 −∫ 1

0

u′v dx

Because u(0) = u(1) = 0, this can be written

〈u, v′〉 = −〈u′, v〉

Apply to d2/dx2 ⇒

〈u, u′′〉 = −〈u′, u′〉 = −‖u′‖22


Sobolev’s lemma

Lemma For all functions u with u(0) = u(1) = 0 it holdsthat

‖u′‖2 ≥ π‖u‖2

Proof Fourier analysis (Parseval’s theorem)

u =√

2∞

∑

k=1

ck sin kπx ⇒ u′ = π√

2∞

∑

k=1

kck cos kπx

implies ‖u′‖2 ≥ π‖u‖2


Logarithmic norm of d2/dx2 on [0, 1]

〈u, u′′〉 = −〈u′, u′〉 = −‖u′‖22 ≤ −π2‖u‖2

2

Theorem The logarithmic norm of d2/dx2 on C20 [0, 1] is

µ2[d2/dx2] = −π2

Corollary The 2pBVP u′′ = f(x); u(0) = u(1) = 0; has aunique solution with ‖u‖2 ≤ ‖f‖2/π

2


Self-adjoint operators

Definition

〈v,Au〉 = 〈A∗v, u〉

defines the adjoint operator A∗

Example For vectors and matrices

〈v,Au〉 = vTAu = (ATv)Tu = 〈A∗v, u〉

AT is the adjoint of A

A matrix is self-adjoint (“symmetric”) if A = AT


Self-adjoint differential operators . . .

Example d2/dx2 on C20 [0, 1]

Integrate by parts

〈v, u′′〉 = −〈v′, u′〉 = 〈v′′, u〉

So d2/dx2 is a self-adjoint operator

More generally,

d

dx

(

p(x)d

dx

)

+ q(x)

is self-adjoint on C20 [0, 1]


Eigenvalues of self-adjoint operators . . .

. . . are real: let Au = λu

λ‖u‖22 = 〈u, λu〉 = 〈u,Au〉 = 〈A∗u, u〉

= 〈Au, u〉 = 〈λu, u〉 = λ‖u‖22

So λ = λ implies real eigenvalues


Eigenvectors of self-adjoint operators . . .

. . . are orthogonal; let Au = λu and Av = µv

λ〈v, u〉 = 〈v,Au〉 = 〈A∗v, u〉= 〈Av, u〉 = µ〈v, u〉

So (λ − µ)〈v, u〉 = 0 ⇒ 〈v, u〉 = 0 implying that

eigenvectors are orthogonal


Elliptic operators

Definition An operator is elliptic if for all u 6= 0

〈u,Au〉 > 0

Example −d2/dx2 on C20 [0, 1]

Integrate by parts

−〈u, u′′〉 = 〈u′, u′〉 ≥ π2〈u, u〉

by Sobolev’s lemma


Elliptic operators . . .

More generally,

− d

dx

(

p(x)d

dx

)

+ q(x)

is elliptic if p(x) > 0 and q(x) ≥ 0

Example Laplace’s equation

∆u =∂2u

∂x2+

∂2u

∂y2= 0

−∆ is an elliptic operator


Positive definite operators

Definition An operator is positive definite if it isself-adjoint and elliptic

µ2[−A] < 0

Example −d2/dx2 as µ2[d2/dx2] = −π2 on C2

0 [0, 1]

Negative Laplacian −∆ (leads to FEM theory)


8. From Finite Difference to Finite Element

Start with linear differential equation

Au = f + boundary conditions

Finite Difference Method (FDM). The main idea

Replace functions u and f by vectors and differentialoperator A by matrix to get a linear system of equations

Exampled2

dx2u = f(x) ⇒ T∆xu = f


Finite Elements and the Galerkin method

Galerkin Method. The main idea

Approximate function u by polynomial and keep differentialoperator A as is!

Take some polynomial v satisfying boundary conditions.Insert into original equation to get Av ≈ f

Question Which polynomial gives the best approximation?

Answer Choose v to minimize the residual ‖Av − f‖2


Best approximation = least squares

Let {ϕj} be a polynomial basis, and make the ansatz

v(x) =

N∑

1

cjϕj(x)

Minimizing ‖Av − f‖2 is equivalent to requiring thatthe residual is orthogonal to each and every ϕi

〈ϕi,Av − f〉 = 0 ∀i

This is the least-squares approximation!


Best approximation. . .

As A is linear,

Av = AN

∑

j=1

cjϕj =N

∑

j=1

cjAϕj

Then

〈ϕi,Av − f〉 = 0 ⇔N

∑

j=1

〈ϕi,Aϕj〉cj = 〈ϕi, f〉

This is a linear system of equations∑

aijcj = bi or Ac = b,with matrix and vector elements

aij = 〈ϕi,Aϕj〉 bi = 〈ϕi, f〉


Weak formulation. The problem −u′′ = f

If −u′′ = f then for all v

〈v,−u′′〉 = 〈v, f〉

Integrate by parts and use Dirichlet boundary data to get

Weak formulation 〈v′, u′〉 = 〈v, f〉 ∀v

The weak formulation requires u to be once continuouslydifferentiable, not twice


Weak formulation and energy norm

Definition The energy norm is defined by

a(v, u) = 〈v′, u′〉

and the weak formulation can be written: Find a function u

such that for all test functions v it holds

a(v, u) = 〈v, f〉

What functions? Choose a polynomial space V withbasis {ϕj}, satisfying the boundary conditions, and requirev ∈ V and u ∈ V , all defined on suitable grid {xi}


Example. The Finite Element Method (FEM)

Given grid {xi}, choose piecewise linear basis polynomials

ϕj(xi) = 1 if i = j, otherwise 0

0 0.5 1 1.5 2 2.5 3 3.5 4−0.2

0

0.2

0.4

0.6

0.8

1

1.2

Piecewise linear interpolant v ≈ u can be written

v(x) =N

∑

j=1

cjϕj(x) Note that v(xi) = ci ≈ u(xi)


Galerkin Finite Element Method for −u′′ = f

Best approximation a(v, u) = 〈v, f〉, with u, v ∈ V , leads to

a(ϕi,∑

j=1

cjϕj) = 〈ϕi, f〉

which is equivalent to the finite element equation Kc = b

N∑

j=1

〈ϕ′i, ϕ

′j〉cj = 〈ϕi, f〉 ∀ϕi ∈ V

The stiffness matrix K with elements {〈ϕ′i, ϕ

′j〉}N

i,j=1 can becomputed as soon as the basis {ϕj} has been constructed

The right-hand side vector b depends on the data f


Equation system

Assume an equidistant grid with spacing ∆x. Note that

ϕ′i(x) = 1/∆x on [xi−1, xi]

ϕ′i(x) = −1/∆x on [xi, xi+1]

ϕ′i(x) = 0 elsewhere

Then

〈ϕ′i, ϕ

′i〉 =

∫ xi+1

xi−1

1

∆x2dx =

2

∆x

〈ϕ′i, ϕ

′i+1〉 =

∫ xi+1

xi

−1

∆x2dx =

−1

∆x


Stiffness matrix

For equidistant grid with spacing ∆x the stiffness matrix is

K∆x =1

∆xtridiag(−1 2 − 1)

Note

1) The stiffness matrix is K∆x = −∆x · T∆x

2) It is positive definite, therefore nonsingular

3) Smallest eigenvalue λ1[K∆x] ≈ π2∆x


Mass matrix

Compute RHS integrals using numerical integration

〈ϕi, f〉 ≈ 〈ϕi,N

∑

0

fjϕj〉 =1

∑

k=−1

fi+k〈ϕi, ϕi+k〉

Need to compute 〈ϕi, ϕi+k〉 =∫ xi+1

xi−1ϕi(x)ϕi+k(x) dx

The integrals are 〈ϕi, ϕi〉 = 2∆x/3 and 〈ϕi, ϕi+1〉 = ∆x/6

For equidistant grid with spacing ∆x the mass matrix is

B∆x =∆x

6tridiag(1 4 1)


Assembling the system of equations

Final finite element equations is a tridiagonal linear system

K∆x c = B∆xf

with stiffness matrix

K∆x =1

∆xtridiag(1 − 2 1)

and mass matrix

B∆x =∆x

6tridiag(1 4 1)


Advantages of the Finite Element Method

◮ Produces “continuous solution” not only on grid points

◮ Can also use basis of higher degree splines

◮ Can easily use nonuniform grids

◮ Boundary conditions built into test functions

◮ In PDEs, easy to work with complex geometries

◮ Rich theoretical foundation


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