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
�
������
Fourier Taylor Chebyshev
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
�
������
Fourier Taylor Chebyshev
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
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
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
• We need to convert Chebyshev T to U
• 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
• We need to convert Chebyshev T to U
• 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
• We need to convert Chebyshev T to U
• 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
• We need to convert Chebyshev T to U
• 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)
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)
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)
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
• 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
• We represent u in Chebyshev T series and f in Chebyshev U series
• Thus the ODE becomes
(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
• We represent u in Chebyshev T series and f in Chebyshev U series
• Thus the ODE becomes
(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
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
�
������
Fourier Taylor Chebyshev
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
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
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
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
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
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 · · ·���
���� � �
�
����
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
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
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
� 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� � �
�
����
� 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� � �
�
����
37
� 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� � �
�
����
38
� 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� � �
�
����
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
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
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
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� � �
� � �
�
����
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
�
�
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
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)
���
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