Lesson 14 Ultraspherical spectral methods · 2 • We ﬁnished last lecture constructing a value...

Lesson 14Ultraspherical spectral methods

1

2

• We finished last lecture constructing a value space Chebyshev spectral method

– This is the method used by Chebfun

• Today, we are going to construct a coefficient space Chebyshev spectral method

• Like Taylor/Fourier coefficient space methods, his method achieves:

– High accuracy

– Efficient solver

• This approach is a very recent development by SO and Alex Townsend (Oxford)

Value space Chebyshev spectral method

3

50 100 150 200

10-1210-1010-810-610-40.01

Error

u

00(x) + exu(x) = 0, u(�1) = 0, u(1) = 1

Matrix sparsity

4

�

��

�2�

0

2

�

��

L[a] =

�

��

a0 a�1 a�2 a�3

a1 a0 a�1 a�2

a2 a1 a0 a�1

a3 a2 a1 a0

�

��

Fourier Taylor Chebyshev

Differentiation

Multiplication

5

�

��

�2�

0

2

�

��

�

��

0 12

34

5

�

��

L[a] =

�

��

a0 a�1 a�2 a�3

a1 a0 a�1 a�2

a2 a1 a0 a�1

a3 a2 a1 a0

�

��

T [a] =

�

��

a0

a1 a0

a2 a1 a0

a3 a2 a1 a0

�

��


Differentiation

Multiplication

6

�

��

�2�

0

2

�

��

�

��

0 12

34

5

�

��

�

��

0 1 3 5 · · ·4 8 12

6 108 12

1012

�

��

L[a] =

�

��

a0 a�1 a�2 a�3

a1 a0 a�1 a�2

a2 a1 a0 a�1

a3 a2 a1 a0

�

��

T [a] =

�

��

a0

a1 a0

a2 a1 a0

a3 a2 a1 a0

�

��


Differentiation

Multiplication ?

7


Differentiation

Multiplication

Why is bandedness good?

• An m-banded matrix is one with m nonzero sub/super diagonals, i.e., the (i,j)th entry satisfies

Ai,j = 0 JSV i � j > m

• Banded matrices have two good properties:• Low memory cost• Fast solvers for A�1b

First order differentiation

9

10

• We want to modify our method for Chebyshev polynomials so that differentiation is banded

• The idea: represent the derivative in a different basis

• We use the Chebyshev U polynomials

– We define these by differentiating the Chebyshev T polynomials:

Uk(x) =T �

k+1(x)

k + 1, JSV k = 0, 1, 2, . . .

11

� 8LYW�MJ

a(x) =��

k=0

akTk(x)

[I�VITVIWIRX�XLI�HIVMZEXMZI�EW

a�(x) =��

k=0

ak+1T�k+1(x) =

��

k=0

ak+1(k + 1)Uk(x)

� -R�QEXVM\�JSVQ�[I�KIX�XLI�HMJJIVIRXMEXMSR�STIVEXSV�JVSQ�8 XS�9

D =

�

��

0 12

34

� � �

�

��

• We need to convert Chebyshev T to U

• Recall

12

�Tk(x) x =

Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C


• Recall

13

�Tk(x) x =

Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

• Thus we have

T0(x) = U0(x), T1(x) =U1(x)

2,

Tk(x) =x

�Tk(x) x

=x

�Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

�=

T �k+1(x)

2(k + 1)�

T �k�1(x)

2(k � 1)

=Uk(x)

2� Uk�2(x)

2


• Recall

14

�Tk(x) x =

Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

• Thus we have

T0(x) = U0(x), T1(x) =U1(x)

2,

Tk(x) =x

�Tk(x) x

=x

�Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

�=

T �k+1(x)

2(k + 1)�

T �k�1(x)

2(k � 1)

=Uk(x)

2� Uk�2(x)

2


• Recall

15

�Tk(x) x =

Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

• Thus we have

T0(x) = U0(x), T1(x) =U1(x)

2,

Tk(x) =x

�Tk(x) x

=x

�Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

�=

T �k+1(x)

2(k + 1)�

T �k�1(x)

2(k � 1)

=Uk(x)

2� Uk�2(x)

2


• Recall

16

�Tk(x) x =

Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

• Thus we have

T0(x) = U0(x), T1(x) =U1(x)

2,

Tk(x) =x

�Tk(x) x

=x

�Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

�=

T �k+1(x)

2(k + 1)�

T �k�1(x)

2(k � 1)

=Uk(x)

2� Uk�2(x)

2


• Recall

17

�Tk(x) x =

Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

• Thus we have

T0(x) = U0(x), T1(x) =U1(x)

2,

Tk(x) =x

�Tk(x) x

=x

�Tk+1(x)

2(k + 1)� Tk�1(x)

2(k � 1)+ C

�=

T �k+1(x)

2(k + 1)�

T �k�1(x)

2(k � 1)

=Uk(x)

2� Uk�2(x)

2

18

� ;I�LEZI�XLI�GSRZIVWMSR�VIPEXMSRWLMTW

T0(x) = U0(x), T1(x) =U1(x)

2, Tk(x) =

Uk(x)

2� Uk�2(x)

2

� ;I�YWI�XLIWI�XS�GSRWXVYGX�XLI�8 XS�9 GSRZIVWMSR�STIVEXSV �

U =

�

��

1 0 � 12

12 0 � 1

212 0 � 1

2� � �

� � �

�

��

� -R�SXLIV�[SVHW� MJ

f(x) =��

k=0

fkTk(x) ERH

�

��u0

u1��

�

�� = U

�

��f0

f1��

�

��

XLIR

f(x) =��

k=0

ukUk(x)

• Example:

19

u� + u = f ERH u(�1) = 1

• Example:

19

u� + u = f ERH u(�1) = 1

• We represent u in Chebyshev T series and f in Chebyshev U series

• Thus the ODE becomes

(D + U) u = U f

• Example:

19

u� + u = f ERH u(�1) = 1



(D + U) u = U f

u(�1) =��

k=0

ukTk(�1) =��

k=0

uk k (�1)

=��

k=0

uk k� =��

k=0

uk(�1)k

• The boundary conditions are

• Example:

19

u� + u = f ERH u(�1) = 1



(D + U) u = U f

u(�1) =��

k=0

ukTk(�1) =��

k=0

uk k (�1)

=��

k=0

uk k� =��

k=0

uk(�1)k

• The boundary conditions are

• Thus we have

Bu = 1 JSV B = (1, �1, 1, �1, . . .)

20

• Example: u� + u = f ERH u(�1) = 1

�B

D + U

�u =

�

��

1 �1 1 �1 1 · · ·1 1 � 1

212 2 � 1

212 3 � 1

2. . .

. . .. . .

�

��u =

�1

U f

�

21

• Example: u� + u = f ERH u(�1) = 1

�Bn

Dn + Un

�un =

�

��

1 �1 1 �1 1 · · · (�1)n�1

1 1 � 12

12 2 � 1

212 3 � 1

2. . .

. . .. . .

12 n � 2 � 1

212 n � 1

�

��

un =

�1

UnCfn

�

Truncation

23

Multiplication of Chebyshev series

24

L[a] =

�

��

a0 a�1 a�2 a�3

a1 a0 a�1 a�2

a2 a1 a0 a�1

a3 a2 a1 a0

�

��

T [a] =

�

��

a0

a1 a0

a2 a1 a0

a3 a2 a1 a0

�

��


Relationship

Operator ?

zjzk = zj+kj� k� = (j+k)� Tj(x)Tk(x) = #

25

� ;I�KS�XS�XLI�GMVGPI�

Tj(J(z))Tk(J(z)) =zj + z�j

2

zk + z�k

2

=zj+k + zj�k + zk�j + z�j�k

4

=zj+k + z�j�k

4+

zj�k + zk�j

4

=1

2

zj+k + z�j�k

2+

1

2

z|j�k| + z�|j�k|

2

=Tj+k(J(z))

2+

T|j�k|(J(z))

2

� -R�SXLIV�[SVHW�

Tj(x)Tk(x) =Tj+k(x)

2+

T|j�k|(x)

2

26

� ;I�RS[�GSRWXVYGX�XLI�QYPXMTPMGEXMSR�STIVEXSVW�MR GSIJ½GMIRX�WTEGI XIVQ�F]�XIVQ

� 7MRGI T0(x) = 1� XLI�½VWX�GEWI�MW�XVMZMEP�

M[T0] =

�

��1

1� � �

�

��

� 2S[�JSV T1(x) [I�LEZI

M[T1] =1

2

�

��

0 11 0 1

1 0 1� � �

� � �

�

��+

1

2

�

��

0 0 · · ·1 0 · · ·0 0 · · ·��

��

�

��

27

�

M[T2] =1

2

�

��

0 0 10 0 0 11 0 0 0 1

1 0 0 0 1� � �

� � �

�

��+

1

2

�

��

00 11

�

��

�

M[T3] =1

2

�

��

0 0 0 10 0 0 0 10 0 0 0 0 11 0 0 0 0 0 1

1 0 0 0 0 0 1� � �

� � �

�

��

+1

2

�

��

00 0 10 11

�

��

� ;I�RS[�WII�E TEXXIVR

28

M[a] =1

2

2

666666664

0

BBBBBBBB@

2a0 a1 a2 a3 . . .

a1 2a0 a1 a2. . .

a2 a1 2a0 a1. . .

a3 a2 a1 2a0. . .

.... . .

. . .. . .

. . .

1

CCCCCCCCA

+

0

BBBBBBB@

0 0 0 0 . . .a1 a2 a3 a4 . . .

a2 a3 a4 a5 . ..

a3 a4 a5 a6 . ..

... . ..

. ..

. ..

. ..

1

CCCCCCCA

3

777777775

.

Multiplication operator (Toeplitz + Hankel) a(x) =��

k=0

akTk(x)

29

Multiplication operator (Toeplitz + Hankel)

Because approximation of a by its Chebyshev series converges spectrally fast, this matrix is in practice banded

M[a] =1

2

!

"""""""#

$

%%%%%%%&

2a0 a1 a2a1 2a0 a1 a2

a2 a1 2a0 a1. . .

a2 a1 2a0. . .

. . .. . .

. . .

'

((((((()

+

$

%%%%%%&

0 0 0 0 . . .a1 a2a2

'

(((((()

*

+++++++,

.

a(x) =m�

k=0

akTk(x)

30

• Our operator equation is now

�B

D + UM[a]

�u =

�1

U f

�

• We discretize this as�

Bn

Dn + UnMn[a]

�un =

�1

UnCfn

�

• This is banded

31

Computed Solution

u� + x3u = 100 sin(20,000x2), u(�1) = 0

Error at one

5000 10000 15000 20000 25000

10-10

10-8

10-6

10-4

0.01

1

32

u� + sin(2,000x2)u = 0, u(�1) = 1

−1 −0.5 0 0.5 10.96

0.965

0.97

0.975

0.98

0.985

0.99

0.995

1

1.005

1.01

x

u(x)

Solution

0 2000 4000 6000 8000 10000 1200010−30

10−25

10−20

10−15

10−10

10−5

100

||un−u n

−1|| 2

n

Cauchy Error

33

Second order ODEs

34

• We replicate the approach of first order equations:

– We construct a new basis so that second order differentiation is banded

– We find a conversion relationship between the Chebyshev T basis and the new basis

– We use the bandedness of multiplication in coefficient space

35

� 7XIT�� HI½RI�E�RI[�FEWMW� XLI YPXVEWTLIVMGEP�TSP]RSQMEPW

C(2)k (x) =

U �k+1(x)

2

� 8LIR�WIGSRH�SVHIV�HMJJIVIRXMEXMSR�MW

D2 = 2

�

��

0 11

1� � �

�

��D = 2

�

��

0 11

1� � �

�

��

�

��

0 12

34

� � �

�

��

= 2

�

��

0 0 23

4� � �

�

��

36


C(2)k (x) =

U �k+1(x)

2


D2 = 2

�

��

0 11

1� � �

�

��D = 2

�

��

0 11

1� � �

�

��

�

��

0 12

34

� � �

�

��

= 2

�

��

0 0 23

4� � �

�

��


C(2)k (x) =

U �k+1(x)

2


D2 = 2

�

��

0 11

1� � �

�

��D = 2

�

��

0 11

1� � �

�

��

�

��

0 12

34

� � �

�

��

= 2

�

��

0 0 23

4� � �

�

��

37


C(2)k (x) =

U �k+1(x)

2


D2 = 2

�

��

0 11

1� � �

�

��D = 2

�

��

0 11

1� � �

�

��

�

��

0 12

34

� � �

�

��

= 2

�

��

0 0 23

4� � �

�

��

38


C(2)k (x) =

U �k+1(x)

2


D2 = 2

�

��

0 11

1� � �

�

��D = 2

�

��

0 11

1� � �

�

��

�

��

0 12

34

� � �

�

��

= 2

�

��

0 0 23

4� � �

�

��

39

� 7XIT�� ;I�RIIH�XS GSRZIVX 'LIF]WLIZ�9 TSP]RSQMEPW�XS�YPXVEWTLIVMGEP�' TSP]�RSQMEPW

� 8LI�½VWX�X[S�GER�FI�JSYRH�HMVIGXP]�

U0(x) = C(2)0 (x), U1(x) =

C(2)1 (x)

2

� ;I�RS[�YWI�XLI�WEQI�XVMGO�SJ MRXIKVEXMRK�XLIR�HMJJIVIRXMEXMRK�

Uk(x) =x

�Uk(x) x =

x

Tk+1(x)

k + 1

=1

k + 1 x

�Uk+1(x)

2� Uk�1(x)

2

�

=C(2)

k (x) � C(2)k�2(x)

k + 1

40

� ;I�LEZI�XLI�GSRZIVWMSR�VIPEXMSRWLMTW�

U0(x) = C(2)0 (x), U1(x) =

C(2)1 (x)

2,

Uk(x) =C(2)

k (x) � C(2)k�2(x)

k + 1

� 8LYW�[I�GER�GSRWXVYGX�XLI FERHIH STIVEXSV�XLEX�GSRZIVXW�'LIF]WLIZ�9 XS�9PXVE�WTLIVMGEP�'�

U1 =

�

��

1 0 � 13

12 0 � 1

413 0 � 1

5� � �

� � �

�

��

41

� ;I�[ERX�XS�WSPZI�XLI�3()

u�� + a(x)u = f(x), u(�1) = c�1 ERH u(1) = c1

� ;I�EKEMR�VITVIWIRX�XLI�YRORS[R u MR�'LIF]WLIZ�8 WIVMIW�FYX�RS[�XLI�VMKLX�LERHWMHI�MR�YPXVEWTLIVMGEP�' WIVMIW

� *SV�XLI�QYPXMTPMGEXMSR�STIVEXSV� [I�QYPXMTP]� MR�8 WIVMIW� XLIR�GSRZIVX�XS�9 WIVMIW�XLIR�XS�' WIVMIW

� 8LYW�[I�KIX�

BD2 + U1UM[a]

�u =

�

�c�1

c1

U1U f

�

�

42

u�� = xu,

u(�1) = Ai (�1),

u(1) = Ai (1)

Example:

42

0 0 4 0 0 0 0 0 0 00 0 0 6 0 0 0 0 0 00 0 0 0 8 0 0 0 0 00 0 0 0 0 10 0 0 0 00 0 0 0 0 0 12 0 0 00 0 0 0 0 0 0 14 0 00 0 0 0 0 0 0 0 16 00 0 0 0 0 0 0 0 0 180 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0

D2

u�� = xu,

u(�1) = Ai (�1),

u(1) = Ai (1)

Example:

42

0 0 4 0 0 0 0 0 0 00 0 0 6 0 0 0 0 0 00 0 0 0 8 0 0 0 0 00 0 0 0 0 10 0 0 0 00 0 0 0 0 0 12 0 0 00 0 0 0 0 0 0 14 0 00 0 0 0 0 0 0 0 16 00 0 0 0 0 0 0 0 0 180 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0

D2

+U1 U M[�x]

u�� = xu,

u(�1) = Ai (�1),

u(1) = Ai (1)

Example:

42

1 0 - 13

0 0 0 0 0 0 0

0 1

20 - 1

40 0 0 0 0 0

0 0 1

30 - 1

50 0 0 0 0

0 0 0 1

40 - 1

60 0 0 0

0 0 0 0 1

50 - 1

70 0 0

0 0 0 0 0 1

60 - 1

80 0

0 0 0 0 0 0 1

70 - 1

90

0 0 0 0 0 0 0 1

80 - 1

10

0 0 0 0 0 0 0 0 1

90

0 0 0 0 0 0 0 0 0 1

10

1 0 - 12

0 0 0 0 0 0 0

0 1

20 - 1

20 0 0 0 0 0

0 0 1

20 - 1

20 0 0 0 0

0 0 0 1

20 - 1

20 0 0 0

0 0 0 0 1

20 - 1

20 0 0

0 0 0 0 0 1

20 - 1

20 0

0 0 0 0 0 0 1

20 - 1

20

0 0 0 0 0 0 0 1

20 - 1

2

0 0 0 0 0 0 0 0 1

20

0 0 0 0 0 0 0 0 0 1

2

0 - 12

0 0 0 0 0 0 0 0

-1 0 - 12

0 0 0 0 0 0 0

0 - 12

0 - 12

0 0 0 0 0 0

0 0 - 12

0 - 12

0 0 0 0 0

0 0 0 - 12

0 - 12

0 0 0 0

0 0 0 0 - 12

0 - 12

0 0 0

0 0 0 0 0 - 12

0 - 12

0 0

0 0 0 0 0 0 - 12

0 - 12

0

0 0 0 0 0 0 0 - 12

0 - 12

0 0 0 0 0 0 0 0 - 12

0

0 0 4 0 0 0 0 0 0 00 0 0 6 0 0 0 0 0 00 0 0 0 8 0 0 0 0 00 0 0 0 0 10 0 0 0 00 0 0 0 0 0 12 0 0 00 0 0 0 0 0 0 14 0 00 0 0 0 0 0 0 0 16 00 0 0 0 0 0 0 0 0 180 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0

D2

+U1 U M[�x]

u�� = xu,

u(�1) = Ai (�1),

u(1) = Ai (1)

Example:

43

1 -1 1 -1 1 -1 1 -1 1 -11 1 1 1 1 1 1 1 1 1

0 - 16

4 1

40 - 1

120 0 0 0

- 14

0 1

166 1

80 - 1

160 0 0

0 - 112

0 1

208 1

120 - 1

200 0

0 0 - 116

0 1

2410 1

160 - 1

240

0 0 0 - 120

0 1

2812 1

200 - 1

28

0 0 0 0 - 124

0 1

3214 1

240

0 0 0 0 0 - 128

0 1

3616 4

63

0 0 0 0 0 0 - 132

0 1

4018

�B

D2 + U1UM[�x]

�=

44

20 40 60 80 100

10-112

10-89

10-66

10-43

10-20u�� = xu,

u(�1) = Ai (�1),

u(1) = Ai (1)

'SQTYXIH u100

Computed coefficients

45

Convergence of derivatives

20 40 60 80 100 120 140

10-13

10-10

10-7

10-4

0.1u�� = xu,

u(�1) = Ai (�1),

u(1) = Ai (1)

n

d = 0, 9, 20, 99

��u(d)n (1/2) � Ai(d)(1/2)

��

46

-1.0 -0.5 0.5 1.0

-8

-6

-4

-2

2

4

u�� + 5002�ex3

� 1�

u = sin(500x2), u(�1) = 1, u(1) = �1

Solution Cauchy error

100 200 300 400 500 600 700

10-11

10-8

10-5

0.01

10

Lesson 14 Ultraspherical spectral methods · 2 • We ﬁnished last lecture constructing a value...

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Transcript of Lesson 14 Ultraspherical spectral methods · 2 • We ﬁnished last lecture constructing a value...