Lecture 8

Chebyshev collocation method for differential equations

Katarina Gustavsson

MA5251 Spectral Methods and Applications, 2011

Two point boundary value problem Time independent – linear/non-linear

i Time independent boundary value problem in a general form !uxx (x)+ p(x)ux (x)+ q(x)ux (x) = f (x,u), -1" x "1 ! >0 is a fixed parameter and p(x), q(x) and f (x) are given functionsi Boundary conditons on general (mixed) form: #$u($1)+ %$ux ($1) = g$ #+u(1)+ %+ux (1) = g+i Dirichlet boundary conditions: u($1) = g$ u(1) = g+i Neumann boundary conditions: ux ($1) = g$ ux (1) = g+ The coefficients # ± %± g± are known

Time dependent problems - linear/non-linear

i Heat equation ut ! "uxx = 0 with boundary conditions given on u(-1,t) = g! and u(1,t) = g+ and initial condition u(x,0) = f (x)i Linear wave equation ut + ux= 0 boundary conditions given on u(-1,t) = g! and ux (1,t) = g+ and initial condition u(x,0) = f (x)i Burgers equation ut + uux ! "uxx = 0 with boundary conditions given on u(-1,t) = g! and u(1,t) = g+ and initial condition u(x,0) = f (x)

Review on the Chebyshev transform

i A function u(x) can be expanded in Chebyshev series

u(x) = ukTk (x) k=0

!

" uk =2

#cku(x)Tk (x)$ (x)dx

%1

1

& $ (x) = (1% x2 )%1/2 and ck = 2 for k = 0 and ck = 1 for k '1i The Chebyshev polynomials are given by Tk (x) = cos(k cos

%1(x))i In the discrete Chebyshev-Gauss-Lobatto case:

x j = cos# jN, j = 0,1,2…,N and $ j =

#2 !ck

where

!ck = 2 for j = 0,N and !ck = 1 for j = 1,2,…,N %1i The discrete Chebyshev coefficients are given by

!uk =2!ckN

1!cju(x j )

j=0

N

" cos # kjN

()*

+,- =

2!ckN

1!cju(x j )

j=0

N

" . ei kj#N

()*

+,-, k = 0,1,…N

Review on the Chebyshev transform, cont.

i A function u(x) can be expanded in discrete Chebyshev series

u(x j ) = !ukTk (x j ), j = 0,1,…,N k=0

N

!

i Chebyshev interpolant INu = !ukTk (x)k=0

N

! (1)

i Or in terms of Chebyshev lagrangian polynomials INu = u(x j )" jj=0

N

! (x) (2)

" j (x) =(#1) j+1(1# x2 ) $TN (x)!cjN

2 (x # x j )i Note that INu(x j ) = u(x j ) and that (1)% (2)

Review on the Chebyshev transform - derivatives

i Derivative in transform (polynomial) space (discrete)

INu(x j )( )! := DNu( ) j = !uk(1)Tk

k=0

N

" (x j )

INu(x j )( ) !!!!! := D2Nu( ) j = !uk

(2)Tkk=0

N

" (x j )

i The coefficients can be found by a recursive relation ck !uk

(1) = !uk+2(1) + 2(k +1) !uk+1, k = N #1,N # 2,…0

ck !uk(2) = !uk+2

(2) + 2(k +1) !u(1)k+1 k = N #1,N # 2,…0 !uN+1

(1) = !uN(1) = 0 and !uN+1

(2) = !uN(2) = 0

i Cost of O(N logN ) if FFT is used

Typo!!

Review on the Chebyshev transform - derivatives

i Derivative in "physical" space DNu( ) j = u(l=0

N

! xl ) "# l (x j ) = Djlu(l=0

N

! xl )

i Djl are the entries in the Chebyshev derivative matrix, D

i The matrix entries for the second order derivative matrix, D2, are given by (DD) jl where DD is a matrix-multiplicationi The Chebyshev derivative matrix at quadrature points is given by

Djl =

!cj ($1)j+l

!cl (x j $ xl ) l % j

$ xl2(1$ xl

2 ) 1& l = j & N $1

'

())

*))

Djl =

2N 2 +16

l = j = 0

$ 2N2 +16

l = j = N

'

())

*))

i The matrix approach costs O(N 2 )

Example of a Dirichlet problem

i We wish to solve, by a Chebyshev collocation method,

uxx + xux ! u = (24 + 5x2 )e5x + (2 + 2x2 )cos(x2 )! (4x2 +1)sin(x2 )

f (x )! "######### $######### ,

on !1" x "1 (1)i With boundary conditions

u(!1) = e!5 + sin(1) = g! and u(1) = e5 + sin(1) = g+

i Let x = (x0, x1,…, xN )T where x j = cos

j#N

$%&

'() , j = 0,1,2,…N

are the Chebyshev-Gauss-Lobatto points

i Let f = ( f (x0 ), f (x1),…, f (xN ))T

i Let u = (g+ ,u(x1),u(x2 ),…,u(xN!1),g! )T and

uM = (u(x1),u(x2 ),…,u(xN!1))T be the vector of unknows to be determined

i We will work with the Chebyshev derivative matrices D and D2

i The entries in D are given by Dij , 0 " i, j " N and D2 = DD

Example of a Dirichlet problem, cont.

i The approximation to (1) is given by

DM2 uM ! xM "DMuM ! uM = F

i xM = x(1:N !1), DM = D(1:N !1,1:N !1), DM2 = D2 (1:N !1,1:N !1)

F = f(1:N !1)! D2 (1:N !1,0)+ xM "D(1:N !1,0)#$ %&g+ !

D2 (1:N !1,N )+ xM "D(1:N !1,N )#$ %&g!i The approximation can also be written on the form AuM = F where A is a matrix of size (N !1)' (N !1)i The matrix A is given by A = D2 + (D! I where ( is a diagonal matrix with the values of xM on the diagonal and I is the identity matrixi The approximate solution is given as the solution to the

linear system of equations uM = A!1F

Solution

1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 120

0

20

40

60

80

100

120

140

160

x

u(x)

Exact solutionN=5N=10N=20

0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.85

10

15

20

25

30

35

40

45

50

x

u(x)

Exact solutionN=5N=10N=20

100 101 10210 12

10 10

10 8

10 6

10 4

10 2

100

102

N

err

i The exact solution is u(x) = e5x + sin(x2 )

i The error is defined by err = max1! j!N"1! uj " u(x j )

Convergence

i The error obtained by a second order finite difference approximation with N = 512 is approximately the same as with N = 10 in the spectral method

Code example - MATLAB N=10![D, x] = cheb(N); Chebyshev derivation matrix, D!D2=D*D; Chebyshev derivation matrix, D2!!DM=D(2:N,2:N); For the inner points only!D2M=D2(2:N,2:N); Note that the numbering of the !L=diag(x(2:N),0); elements in a vector/matrix is !I=diag(ones(N-1,1),0); from 1 to N+1 (not 0 to N)! !A=D2M+L*DM-I; A as in Au=F!!f=(24+5*x).*exp(5*x)+(2+2*x.^2).*cos(x.^2);!f=f-(4*x.^2+1).*sin(x.^2);!gminus=exp(-5)+sin(1);!gplus=exp(5)+sin(1);! !F=f(2:N)-(D2(2:N,1)+x(2:N).*D(2:N,1))*gplus!F=F-(D2(2:N,end)+x(2:N).*D(2:N,end))*gminus;! !sol=A\F u=A-1F! !u=[gplus;sol;gminus]; Add the known boundary conditions!!

Remarks •  The collocation method will lead to a full and ill-conditioned

linear system •  Gaussian elimination to solve Au=F is only feasible for

problems with a small number of unknowns (one-dimensional problems)

•  For multi-dimensional problems an iterative method together with an appropriate pre-conditioner should be used

•  We will get back to this when we speak about multi-dimensional problems

100 101 102 103100

102

104

106

108

1010

N

max(||)

Eigenvalues of A are growing as N4