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Asymptotic expansion method for the two point boundary value problem with small periodic structure
Applied Mathematics and Computation 161 (2005) 667–678 www.elsevier.com/locate/amc
Asymptotic expansion method for the two point boundary value problem with small periodic structure Wen-ming He
a,*
, Jun-zhi Cui
b
a
b
Department of Mathematics, Wenzhou Normal College, Wenzhou, Zhejiang 325027, PR China Academy of Mathematics and System Science, Chinese Academy of Science, Beijing 100080, PR China
Abstract For the two point boundary value problem with small periodic structure, we propose an asymptotic expansion method to get its solution. Ó 2004 Elsevier Inc. All rights reserved.
1. Introduction Consider the following one-dimensional Dirichlet problem with rapidly oscillating coefficient 8 e d ¼ f ðxÞ; a xe du < dx dx e u ðcÞ ¼ u0 ; : e u ðdÞ ¼ u1
x 2 ðc; dÞ;
and
*
Corresponding author. E-mail address: [email protected] (W.-m. He).
0096-3003/$ - see front matter Ó 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.12.069
ð1:1Þ
668
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
8 d x due < dx a e dx ¼ f ðxÞ; ue ðcÞ ¼ u0 ; : d due ðdÞ a e dx ¼ u1 ;
x 2 ðc; dÞ; ð1:2Þ
where aðxeÞ is 1-periodic function. When e is very small and the smoothness of aðxeÞ is very weak, it is very difficult to get numerical solution of Eqs. (1.1) or (1.2) using finite element method directly (when the smoothness of aðxeÞ is arbitrary bad, Babuska find that finite element method can perform arbitrary badly in [2]). In [1], homogenization method is introduced to solve the equation (1.1). The procedures are presented as follows: (1) Let a0 ¼ R 1
1
1 dn 0 aðnÞ
.
(2) Get the solution u of the following equation: 8 d ¼ f ðxÞ; x 2 ðc; dÞ; a0 du < dx dx uðcÞ ¼ u0 ; : uðdÞ ¼ u1 :
ð1:3Þ
Then u is the homogenization solution of ue . The homogenization method to obtain the approximate solution of Eq. (1.2) is similar. For Eq. (1.1), in [1], it is proved that e!0
ue * u
in H 1 ðc; dÞ
but error estimate is not given. In this paper, we propose an asymptotic expansion method to get the solution of Eq. (1.1) on basis of [1]. By virtue of it, we obtain the solution of Eq. (1.2).
2. Asymptotic expansion method for the equation (1.1) Now Let us consider Eq. (1.1), Suppose f 2 C 1 ðc; dÞ. First we consider the simplest case: d ¼ c þ meðm 2 N Þ. We divide ½c; d into ½c; c þ e ; . . . ; ½c þ ie; c þ ði þ 1Þe ; . . . ; ½c þ ðm 1Þe; d . For Eq. (1.1), there exists ^ a1 , ^ a2 , fi , satisfying the following equality ue ðc þ ieÞ ¼ ^ a1 ue ðc þ ði 1ÞeÞ þ ^ a2 ue ðc þ ði þ 1ÞeÞ þ fi ; ^2 have no relation with fi Þ: ð^ a1 and a a2 , there exists the following Lemma 2.1. About ^ a1 and ^
ð2:1Þ
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
669
Lemma 2.1. ^ a1 ¼ ^ a2 ¼ 12. Proof. Let f 0; ue ðcÞ ¼ 0; ue ðdÞ ¼ 1, Then we have Z x 1 1 y dy: ue ðxÞ ¼ R d 1 a dy c e c aðy Þ e
ð2:2Þ
So we have ue ðc þ jeÞ ¼
j m
ðj ¼ 0; . . . ; m 1Þ:
Therefore ue ðc þ ieÞ
ue ½c þ ði 1Þe þ ue ½c þ ði þ 1Þe ¼ 0: 2
Lemma 2.1 is proved. h Now we consider how to obtain the simplest expression of fi . Let E ¼ ½c þ ði 1Þe; c þ ði þ 1Þe ;
f ð0Þ ðxÞ ¼ f ðxÞ:
Using Taylor expansion, we get f ðxÞ ¼ f ðc þ ieÞ þ f ð1Þ ðc þ ieÞðx c ieÞ þ f ðnÞ ðc þ ieÞ n ðx c ieÞ þ n! 1 X f ðnÞ ðc þ ieÞ n ðx c ieÞ ; x 2 E; ¼ n! n¼0 þ
then we obtain e X 1 d x du f ðnÞ ðc þ ieÞ n a ðx c ieÞ ; ¼ dx e dx n! n¼0
ð2:3Þ
x 2 E:
ð2:4Þ
Let y ¼ x c ie , suppose u0 , a0 and an ðn P 1Þ to be the solution of Eqs. (2.5)– e (2.7), respectively. 8
0 > < dyd aðyÞ du ¼ 0; y 2 ð 1; 1Þ; dy e ð2:5Þ u ð 1Þ ¼ u ðc þ ði 1ÞeÞ; > : 0 u0 ð1Þ ¼ ue ðc þ ði þ 1ÞeÞ; (
da0 d aðyÞ ¼ 1; y 2 ð 1; 1Þ; dy dy ð2:6Þ a0 ð 1Þ ¼ a0 ð1Þ ¼ 0;
670
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
(
n n aðyÞ da ¼ yn! ; y 2 ð 1; 1Þ; dy an ð 1Þ ¼ an ð1Þ ¼ 0:
d dy
ð2:7Þ
By virtue of (2.4)–(2.7), we conclude ue ðxÞ ¼ u0 ðyÞ þ
1 X
enþ2 an ðyÞf ðnÞ ðc þ ieÞ:
ð2:8Þ
n¼0
Therefore ue ðc þ ieÞ ¼
1 ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ X þ enþ2 an ð0Þf ðnÞ ðc þ ieÞ: 2 n¼0
ð2:9Þ Let an ¼ an ð0Þ, we have ue ðc þ ieÞ ¼
1 ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ X þ enþ2 an f ðnÞ ðc þ ieÞ: 2 n¼0
ð2:10Þ Assume that ~v satisfies 8 2 > < ddx~2v ¼ 0; x 2 ðc; dÞ; ~vðcÞ ¼ ue ðcÞ; > : ~vðdÞ ¼ ue ðdÞ:
ð2:11Þ
Then we have ~vðc þ ieÞ ¼
~v½c þ ði 1Þe þ ~v½c þ ði þ 1Þe : 2
Suppose v0 to be the solution of the following equation 2 d v0 ¼ f ðxÞ; x 2 ðc; dÞ; dx2 v0 ðcÞ ¼ v0 ðdÞ ¼ 0:
ð2:12Þ
Using Taylor expansion to right item f ðxÞ of Eq. (2.12), we obtain 1 d2 v0 X f ðkÞ ðc þ ieÞ k ðx c ieÞ ; ¼ k! dx2 k¼0
~0 ðyÞ satisfy Further let a ( 2 d ~ a0 ¼ 1; y 2 ð 1; 1Þ; dy 2 ~ a0 ð1Þ ¼ 0: a0 ð 1Þ ¼ ~ For k P 1, let ~ ak ðyÞ satisfy
x 2 E:
ð2:13Þ
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
(
d2 ~ ak dy 2
671
k
¼ yk! ; y 2 ð 1; 1Þ; ~ ak ð1Þ ¼ 0: ak ð 1Þ ¼ ~
ð2:14Þ
It follows from (2.12)–(2.14) that v0 ðc þ ieÞ ¼
v0 ðc þ ði 1ÞeÞ þ v0 ðc þ ði þ 1ÞeÞ 2 1 X ~ þ ak ð0Þekþ2 f ðkÞ ðc þ ieÞ:
ð2:15Þ
k¼0
Let ~ ak ¼ ~ ak ð0Þ, then we get ~ v0 ðc þ ieÞ ¼
1 v0 ðc þ ði 1ÞeÞ þ v0 ðc þ ði þ 1ÞeÞ X ~ak ekþ2 f ðkÞ ðc þ ieÞ: þ 2 k¼0
Let vj ðj P 1Þ satisfy ( d2 vj ¼ f ðjÞ ðxÞ; dx2 vj ðcÞ ¼ vj ðdÞ ¼ 0:
ð2:16Þ
Using Taylor expansion to right item of Eq. (2.16), we get 1 d2 vj ðxÞ X f ðkþjÞ ðc þ ieÞ k ðx c ieÞ ; ¼ dx2 k! k¼0
x 2 ðc þ ði 1Þe; c þ ði þ 1ÞeÞ:
Using the same way as above, we deduce vj ðc þ ieÞ ¼
1 vj ðc þ ði 1ÞeÞ þ vj ðc þ ði þ 1ÞeÞ X ~ak ekþ2 f ðkþjÞ ðc þ ieÞ: þ 2 k¼0
Let 0
~ a0 B~ a B 1 B~ B a2 B... A¼B B... B B~ B an @... ...
0 ~ a0 ~ a1 ... ... ~ an 1 ... ...
0 0 ~ a0 ... ... ... ... ...
0 0 0 ... ... ~ a2 ... ...
... ... ... ... ... ~ a1 ... ...
0 0 0 ... ... ~ a0 ... ...
1 ... ...C C ...C C ...C C; ...C C ...C C ...A ...
F ¼ ða0 ; a1 ; a2 ; . . . ; an ; . . .Þ: Assume that b ¼ ðb0 ; b1 ; . . . ; bn 1 ; bn ; . . .Þ is the solution of the following linear equation: Ab ¼ F :
ð2:17Þ
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W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
Then we have 1 X 1 ue ðc þ ieÞ ¼ ½ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ þ enþ2 an f ðnÞ ðc þ ieÞ 2 n¼0
1 ¼ ½ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ 2 1 n X X þ enþ2 bj ~ an j f ðnÞ ðc þ ieÞ n¼0
j¼0
1 ¼ ½ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ 2 1 1 X X þ bj ej ekþ2 ~ ak f ðkþjÞ ðc þ ieÞ: j¼0
k¼0
Let uS ¼ ~v þ
1 X
b j e j vj :
ð2:18Þ
j¼0
By virtue of above analysis, we obtain 8 u ðc þ ieÞ ¼ 12 ½u þ ði 1ÞeÞ þ uS ðc þ ði þ 1ÞeÞ > > PS ðc < S P1 1 þ j¼0 bj ej k¼0 ekþ2 ~ ak f ðkþjÞ ðc þ ieÞ; e u ðcÞ ¼ u ¼ u ðcÞ; > S 0 > : uS ðdÞ ¼ u1 ¼ ue ðdÞ:
ð2:19Þ
We obtain from (2.19) that ue ðc þ ieÞ ¼ uS ðc þ ieÞ. Finally, we get the following Lemma 2.2. Lemma 2.2. Suppose that b ¼ ðb0 ; b1 ; b2 ; . . . ; bn ; . . .Þ and uS to be defined as (2.17) and (2.18) respectively, then ue ðc þ ieÞ ¼ uS ðc þ ieÞ. Now let us consider the case that x 2 ðc þ ði 1Þe; c þ ieÞð1 6 i 6 n 1Þ, let y¼
x c ði 1Þe ; e
x ^ aðyÞ ¼ að Þ; e
uðyÞ ¼ ue ðxÞ:
Then uðyÞ is the solution of the following problem: 8
P ðkÞ 1 > ; y 2 ð0; 1Þ; aðyÞ du ¼ k¼0 ekþ2 y k f ðcþði 1ÞeÞ < dyd ^ dy k! e uð0Þ ¼ u ðc þ ði
1ÞeÞ; > : uð1Þ ¼ ue ðc þ ieÞ: Assume that ~ a0 and ~ a1 satisfy
ð2:20Þ
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
8
a0 > ¼ 0; < dyd aðyÞ d~ dy ~ a ð0Þ ¼ 0; > : 0 ~ a0 ð1Þ ¼ 1;
673
y 2 ð0; 1Þ; ð2:21Þ
and 8
a1 > < dyd aðyÞ d~ ¼ 0; dy ~ ð0Þ ¼ 1; a > : 1 ~ a1 ð1Þ ¼ 0:
y 2 ð0; 1Þ; ð2:22Þ
^k ðyÞðk ¼ 0; . . . ; n; . . .Þ to be the solution of the following problem Suppose a (
k d^ ak ðyÞ d aðyÞ ¼ yk! ; y 2 ð0; 1Þ; dy dy ð2:23Þ ^ ak ð1Þ ¼ 0: ak ð0Þ ¼ ^ From (2.20)–(2.23), we obtain uðyÞ ¼ ~ a0 ðyÞue ðc þ ði 1ÞeÞ þ ~ a1 ðyÞue ðc þ ieÞ 1 X þ ekþ2 ^ ak ðyÞf ðkÞ ðc þ ði 1ÞeÞ k¼0
a1 ðyÞuS ðc þ ieÞ ¼~ a0 ðyÞuS ðc þ ði 1ÞeÞ þ ~ 1 X þ ekþ2 ^ ak ðyÞf ðkÞ ðc þ ði 1ÞeÞ k¼0
a1 ðyÞuS ðc þ ieÞ ¼~ a0 ðyÞuS ðc þ ði 1ÞeÞ þ ~ 1 1 XX þ ekþ2 ^ ak ðyÞð yÞj f ðkþjÞ ðxÞ: k¼0 j¼0
Let ak ¼
k X
^ aj ðyÞð yÞk j :
j¼0
Then we have uðyÞ ¼ ~ a0 ðyÞuS ðc þ ði 1ÞeÞ þ ~ a1 ðyÞuS ðc þ ieÞ þ
1 X ak ðyÞekþ2 f ðkÞ ðxÞ: k¼0
For x 2 ðc þ ði 1Þe; c þ ieÞ, ue can be expressed as ue ðxÞ ¼ ~ a0 ðyÞuS ðc þ ði 1ÞeÞ þ ~ a1 ðyÞuS ðc þ ieÞ þ
1 X k¼0
From
ak ðyÞekþ2 f ðkÞ ðxÞ:
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W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
uS ðc þ ði 1ÞeÞ ¼ uS ðxÞ þ
1 X
dj uS ; dxj
ej ð yÞj
j¼1
uS ðc þ ieÞ ¼ uS ðxÞ þ
1 X
ej ð1 yÞj
j¼1
dj uS ; dxj
we have 1 X
e
u ðxÞ ¼ ~ a0 ðyÞ uS ðxÞ þ
j¼1
dj uS ðxÞ e ð yÞ dxj
1 X
þ~ a1 ðyÞ uS ðxÞ þ
j¼1
dj uS ðxÞ e ð1 yÞ dxj 1 X
d ~vðxÞ þ~ a1 ðyÞð1 yÞ þ dxj þ
bk ek vk ðxÞÞ þ
k¼0 j
j
1 X
!
j
j
¼ ð~ a0 ðyÞ þ ~ a1 ðyÞÞð~vðxÞ þ "
!
j
j
þ
1 X
ak ðyÞekþ2 f ðkÞ ðxÞ
k¼0 1 X
ej ½~a0 ðyÞð yÞ
j
j¼1 1 X k¼1
j
d vk ðxÞ b k ek dxj
#
2
ak ðyÞekþ2
k¼0
d vk ðxÞ : dx2
ð2:24Þ
a1 ðyÞ, ~v, vk ðk P 0Þ to be defined as above. For Lemma 2.3. Suppose y, ~ a0 ðyÞ, ~ x 2 ðc þ ði 1Þe; c þ ieÞ, we have ue ðxÞ ¼ ð~a0 ðyÞ þ ~ a1 ðyÞÞð~vðxÞ þ "
1 X k¼0
j
d ~vðxÞ þ~ a1 ðyÞð1 yÞ þ dxj j
þ
1 X k¼0
bk ek vk ðxÞÞ þ
1 X
ej ½~a0 ðyÞð yÞ
j
j¼1 1 X k¼1
j
d vk ðxÞ bk e dxj
#
k
2
ak ðyÞekþ2
d vk ðxÞ : dx2
ð2:25Þ
Actually, Lemma 2.3 has not obtained complete asymptotic expansion for the two point boundary value problem with small periodic structure. But from Lemmak 2.3, we can obtain that ue can be expressed by y, u0 , u1 , d v0 ðxÞ dk v0 ðcÞ dk v0 ðdÞ v0 ðxÞ, v0 ðcÞ, v0 ðdÞ, dxk , dxk , dxk completely. The result is the following Theorem 2.1:
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
675
Theorem 2.1. Set d 0 v0 ¼ v0 ;
^ a1 ðyÞ ¼ b1 þ ½ ae0 ðyÞð yÞ þ ~a1 ðyÞð1 yÞ b0 :
^ a0 ðyÞ ¼ b0 ;
When k P 2, let ^ ak ðyÞ ¼
k X ½~ a0 ðyÞð yÞj þ ~ a1 ðyÞð1 yÞj bk j þ ak 2 ðyÞ þ bk : j¼1
Then ue can be expressed as follows: k
0 1 k d v0 ðdÞ v0 ðcÞ 1 k
d dx ðx cÞ X k k dx ðu
u Þðx
cÞ d v ðcÞ 1 0 0 A
þ bk ek @ ue ðxÞ ¼ u0 þ k d c dx d c k¼1 k
1 0 d v0 ðdÞ dk v0 ðcÞ 1
X k k dx dx u1 u0 A
a1 ðyÞð1 yÞ @e bk ekþ1 þ ½ ae0 ðyÞð yÞ þ ~ d
c d c k¼1 þ
1 X
^ ak ðyÞ
k¼0
dk v0 ðxÞ : dxk
Proof. By virtue of the definition of ~v, we have ~vðxÞ ¼ u0 þ
ðu1 u0 Þðx cÞ ; d c
d~vðxÞ u1 u0 ¼ ; dx d c
We obtain from the definition of v0 and vk that d2 vk ðxÞ dkþ2 v0 ðxÞ ¼ : dx2 dxkþ2 Suppose that ~vk ðxÞ satisfies 8 2 d ~vk ðxÞ > > < dx2 ¼ 0; x 2 ðc; dÞ; k v0 ðcÞ ~vk ðcÞ ¼ d dx ; k > k > d : ~vk ðdÞ ¼ v0 ðdÞ : dxk
We obtain k
~vk ðxÞ ¼ From
d v0 ðcÞ þ dxk
dk v0 ðdÞ dxk
k v0 ðcÞ
d dx ðx cÞ k d c
:
d2~vðxÞ ¼ 0: dx2
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W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
d2 ðvk ðxÞ þ ~vk ðxÞÞ dkþ2 v0 ðxÞ ¼ ; dx2 dxkþ2 dk v0 ðcÞ vk ðcÞ þ ~vk ðcÞ ¼ ; dxk k d v0 ðdÞ ; vk ðdÞ þ ~vk ðdÞ ¼ dxk we have dk v0 ðxÞ dk v0 ðxÞ dk v0 ðcÞ k ~
v ðxÞ ¼
vk ðxÞ ¼ dxk dxk dxk
dk v0 ðdÞ dxk
k v0 ðcÞ
d dx ðx cÞ k d c
:
From Lemma 2.3, we conclude k
3 2 d v0 ðdÞ dk v0 ðcÞ 1 k k ðx cÞ X k dxk dx ðu
u Þðx cÞ d v ðxÞ d v ðcÞ 1 0 0 0 5 þ ue ðxÞ ¼ u0 þ bk ek 4
d c d c dxk dxk k¼1 ! dk v0 ðdÞ dk v0 ðcÞ 1
dxk u1 u0 X kþ1 dxk
þe½~ a0 ðyÞð yÞþ~ a1 ðyÞð1 yÞ be d c k¼1 k d c þ þ
1 1 X X dkþj v0 ðxÞ ½~ a0 ðyÞð yÞj þ~ a1 ðyÞð1 yÞj bk ekþj dxkþj j¼1 k¼0 1 X
ak ðyÞekþ2
k¼0
dkþ2 v0 ðxÞ : dxkþ2
From the definition of ^ ak , finally we have k
1 0 d v0 ðdÞ dk v0 ðcÞ 1 k
ðx
cÞ X k k dx dx ðu1 u0 Þðx cÞ d v0 ðcÞ A þ þ bk ek @ ue ðxÞ ¼ u0 þ k d c dx d
c k¼1 k
1 0 d v0 ðdÞ dk v0 ðcÞ 1
X k k dx dx u1 u0 A þ ½~ a0 ðyÞð yÞ þ ~
a1 ðyÞð1 yÞ @e bk ekþ1 d
c d c k¼1 þ
1 X k¼0
^ ak ðyÞ
dk v0 ðxÞ : dxk
Theorem 2.1 is proved.
ð2:26Þ
h
Theorem 2.1 obtained a complete asymptotic expansion for the two point boundary value problem with small periodic structure. Above, we consider the special case: d ¼ c þ me. when d ¼ c þ me þ deð0 < d < 1Þ, we can take the following steps to get the solution of Eq. (1.1).
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
(1) Solve the following equation: 8 2 P j ðjÞ < ddxw2 ¼ 1 j¼0 bj e f ðxÞ; wðcÞ ¼ 0; : wðc þ meÞ ¼ 0:
677
ð2:27Þ
(2) From (2.27), we obtain ue ðc þ ði 1ÞeÞ ¼ wðc þ ðm 1ÞeÞ þ
1 m u0 þ ue ðc þ meÞ: m mþ1
For x 2 ðc þ ðm 1Þe; dÞ, let y¼
x c ðm 1Þe ; e
^ aðyÞ ¼ a
x
e
;
F ðyÞ ¼ e2 f ðxÞ;
uðyÞ ¼ ue ðxÞ:
(3) Obtain the solution of the following problem: 8
> aðyÞ du ¼ F ðyÞ; < dyd ^ dy
y 2 ð0; 1 þ dÞ;
1 m u0 þ mþ1 u1 ; uð0Þ ¼ wðc þ ðm 1ÞeÞ þ mþ1 > : uð1 þ dÞ ¼ u1 :
ð2:28Þ
Let ~ u1 ¼ uð0Þ: (4) By virtue of the same way as above, solve the following problem: 8 e d ¼ f ðxÞ; x 2 ðc; c þ ðm 1ÞeÞ; a xe du < dx dx e ð2:29Þ u ðcÞ ¼ u0 ; : e u ðc þ ðm 1ÞeÞ ¼ ~ u1 : For Eq. (1.2), suppose d ¼ c þ me þ deð0 < d < 1Þ, we can take following steps to obtain its solution: (1) Solve the following equation 8 1 P > d2 w j ðjÞ > < dx2 ¼ bj e f ðxÞ; j¼0
ð2:30Þ
> wðcÞ ¼ 0; > : wðc þ meÞ ¼ 0: (2) From (2.30), we obtain ue ðc þ ðm 1ÞeÞ ¼ wðc þ ðm 1ÞeÞ þ
1 m u0 þ ue ðc þ meÞ: mþ1 mþ1
678
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
(3) For x 2 ðc þ ðm 1Þe; dÞ, let y¼
x c ðm 1Þe ; e
^ aðyÞ ¼ a
x
e
;
F ðyÞ ¼ e2 f ðxÞ;
uðyÞ ¼ ue ðxÞ:
Obtain the solution of the following problem: 8
aðyÞ du ^ d > > ¼ F ðyÞ; y 2 ð0; 1 þ dÞ; < dy dy 1 m u0 þ mþ1 u1 ; uð0Þ ¼ wðc þ ðm 1ÞeÞ þ mþ1 > > duð1þdÞ :^ að1 þ dÞ dy ¼ u1 : Let ~ u1 ¼ uð0Þ: (4) By virtue of the same way as above, solve the following problem: 8 e d a xe du ¼ f ðxÞ; x 2 ðc; c þ ðm 1Þe; < dx dx e u ðcÞ ¼ u0 ; : e u ðc þ ðm 1ÞeÞ ¼ ~ u1 : By virtue of Theorem 2.1, we maybe can study the following problem 8 e d þ ke ue ¼ 0; x 2 ðc; dÞ; a xe du < dx dx e ð2:31Þ u ðcÞ ¼ 0; : e u ðdÞ ¼ 0:
Acknowledgements This work is supported by National Natural Science Foundation of China. Subsidized by Special Funds for Major State Basic Research Projects.
References [1] O.A. Oleinik, A.S. Shamaev, G.A. Yosifian, Mathematical Problems in Elasticity and Homogenization, North-Holland, Amsterdam, 1992. [2] Ivo Babuska, J.E. Osborn, Can a finite element method perform arbitrarily badly? Mathematics of Computation 69 (2000) 443–462.
Asymptotic expansion method for the two point boundary value problem with small periodic structure Wen-ming He
a,*
, Jun-zhi Cui
b
a
b
Department of Mathematics, Wenzhou Normal College, Wenzhou, Zhejiang 325027, PR China Academy of Mathematics and System Science, Chinese Academy of Science, Beijing 100080, PR China
Abstract For the two point boundary value problem with small periodic structure, we propose an asymptotic expansion method to get its solution. Ó 2004 Elsevier Inc. All rights reserved.
1. Introduction Consider the following one-dimensional Dirichlet problem with rapidly oscillating coefficient 8 e d ¼ f ðxÞ; a xe du < dx dx e u ðcÞ ¼ u0 ; : e u ðdÞ ¼ u1
x 2 ðc; dÞ;
and
*
Corresponding author. E-mail address: [email protected] (W.-m. He).
0096-3003/$ - see front matter Ó 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.12.069
ð1:1Þ
668
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
8 d x due < dx a e dx ¼ f ðxÞ; ue ðcÞ ¼ u0 ; : d due ðdÞ a e dx ¼ u1 ;
x 2 ðc; dÞ; ð1:2Þ
where aðxeÞ is 1-periodic function. When e is very small and the smoothness of aðxeÞ is very weak, it is very difficult to get numerical solution of Eqs. (1.1) or (1.2) using finite element method directly (when the smoothness of aðxeÞ is arbitrary bad, Babuska find that finite element method can perform arbitrary badly in [2]). In [1], homogenization method is introduced to solve the equation (1.1). The procedures are presented as follows: (1) Let a0 ¼ R 1
1
1 dn 0 aðnÞ
.
(2) Get the solution u of the following equation: 8 d ¼ f ðxÞ; x 2 ðc; dÞ; a0 du < dx dx uðcÞ ¼ u0 ; : uðdÞ ¼ u1 :
ð1:3Þ
Then u is the homogenization solution of ue . The homogenization method to obtain the approximate solution of Eq. (1.2) is similar. For Eq. (1.1), in [1], it is proved that e!0
ue * u
in H 1 ðc; dÞ
but error estimate is not given. In this paper, we propose an asymptotic expansion method to get the solution of Eq. (1.1) on basis of [1]. By virtue of it, we obtain the solution of Eq. (1.2).
2. Asymptotic expansion method for the equation (1.1) Now Let us consider Eq. (1.1), Suppose f 2 C 1 ðc; dÞ. First we consider the simplest case: d ¼ c þ meðm 2 N Þ. We divide ½c; d into ½c; c þ e ; . . . ; ½c þ ie; c þ ði þ 1Þe ; . . . ; ½c þ ðm 1Þe; d . For Eq. (1.1), there exists ^ a1 , ^ a2 , fi , satisfying the following equality ue ðc þ ieÞ ¼ ^ a1 ue ðc þ ði 1ÞeÞ þ ^ a2 ue ðc þ ði þ 1ÞeÞ þ fi ; ^2 have no relation with fi Þ: ð^ a1 and a a2 , there exists the following Lemma 2.1. About ^ a1 and ^
ð2:1Þ
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
669
Lemma 2.1. ^ a1 ¼ ^ a2 ¼ 12. Proof. Let f 0; ue ðcÞ ¼ 0; ue ðdÞ ¼ 1, Then we have Z x 1 1 y dy: ue ðxÞ ¼ R d 1 a dy c e c aðy Þ e
ð2:2Þ
So we have ue ðc þ jeÞ ¼
j m
ðj ¼ 0; . . . ; m 1Þ:
Therefore ue ðc þ ieÞ
ue ½c þ ði 1Þe þ ue ½c þ ði þ 1Þe ¼ 0: 2
Lemma 2.1 is proved. h Now we consider how to obtain the simplest expression of fi . Let E ¼ ½c þ ði 1Þe; c þ ði þ 1Þe ;
f ð0Þ ðxÞ ¼ f ðxÞ:
Using Taylor expansion, we get f ðxÞ ¼ f ðc þ ieÞ þ f ð1Þ ðc þ ieÞðx c ieÞ þ f ðnÞ ðc þ ieÞ n ðx c ieÞ þ n! 1 X f ðnÞ ðc þ ieÞ n ðx c ieÞ ; x 2 E; ¼ n! n¼0 þ
then we obtain e X 1 d x du f ðnÞ ðc þ ieÞ n a ðx c ieÞ ; ¼ dx e dx n! n¼0
ð2:3Þ
x 2 E:
ð2:4Þ
Let y ¼ x c ie , suppose u0 , a0 and an ðn P 1Þ to be the solution of Eqs. (2.5)– e (2.7), respectively. 8
0 > < dyd aðyÞ du ¼ 0; y 2 ð 1; 1Þ; dy e ð2:5Þ u ð 1Þ ¼ u ðc þ ði 1ÞeÞ; > : 0 u0 ð1Þ ¼ ue ðc þ ði þ 1ÞeÞ; (
da0 d aðyÞ ¼ 1; y 2 ð 1; 1Þ; dy dy ð2:6Þ a0 ð 1Þ ¼ a0 ð1Þ ¼ 0;
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W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
(
n n aðyÞ da ¼ yn! ; y 2 ð 1; 1Þ; dy an ð 1Þ ¼ an ð1Þ ¼ 0:
d dy
ð2:7Þ
By virtue of (2.4)–(2.7), we conclude ue ðxÞ ¼ u0 ðyÞ þ
1 X
enþ2 an ðyÞf ðnÞ ðc þ ieÞ:
ð2:8Þ
n¼0
Therefore ue ðc þ ieÞ ¼
1 ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ X þ enþ2 an ð0Þf ðnÞ ðc þ ieÞ: 2 n¼0
ð2:9Þ Let an ¼ an ð0Þ, we have ue ðc þ ieÞ ¼
1 ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ X þ enþ2 an f ðnÞ ðc þ ieÞ: 2 n¼0
ð2:10Þ Assume that ~v satisfies 8 2 > < ddx~2v ¼ 0; x 2 ðc; dÞ; ~vðcÞ ¼ ue ðcÞ; > : ~vðdÞ ¼ ue ðdÞ:
ð2:11Þ
Then we have ~vðc þ ieÞ ¼
~v½c þ ði 1Þe þ ~v½c þ ði þ 1Þe : 2
Suppose v0 to be the solution of the following equation 2 d v0 ¼ f ðxÞ; x 2 ðc; dÞ; dx2 v0 ðcÞ ¼ v0 ðdÞ ¼ 0:
ð2:12Þ
Using Taylor expansion to right item f ðxÞ of Eq. (2.12), we obtain 1 d2 v0 X f ðkÞ ðc þ ieÞ k ðx c ieÞ ; ¼ k! dx2 k¼0
~0 ðyÞ satisfy Further let a ( 2 d ~ a0 ¼ 1; y 2 ð 1; 1Þ; dy 2 ~ a0 ð1Þ ¼ 0: a0 ð 1Þ ¼ ~ For k P 1, let ~ ak ðyÞ satisfy
x 2 E:
ð2:13Þ
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
(
d2 ~ ak dy 2
671
k
¼ yk! ; y 2 ð 1; 1Þ; ~ ak ð1Þ ¼ 0: ak ð 1Þ ¼ ~
ð2:14Þ
It follows from (2.12)–(2.14) that v0 ðc þ ieÞ ¼
v0 ðc þ ði 1ÞeÞ þ v0 ðc þ ði þ 1ÞeÞ 2 1 X ~ þ ak ð0Þekþ2 f ðkÞ ðc þ ieÞ:
ð2:15Þ
k¼0
Let ~ ak ¼ ~ ak ð0Þ, then we get ~ v0 ðc þ ieÞ ¼
1 v0 ðc þ ði 1ÞeÞ þ v0 ðc þ ði þ 1ÞeÞ X ~ak ekþ2 f ðkÞ ðc þ ieÞ: þ 2 k¼0
Let vj ðj P 1Þ satisfy ( d2 vj ¼ f ðjÞ ðxÞ; dx2 vj ðcÞ ¼ vj ðdÞ ¼ 0:
ð2:16Þ
Using Taylor expansion to right item of Eq. (2.16), we get 1 d2 vj ðxÞ X f ðkþjÞ ðc þ ieÞ k ðx c ieÞ ; ¼ dx2 k! k¼0
x 2 ðc þ ði 1Þe; c þ ði þ 1ÞeÞ:
Using the same way as above, we deduce vj ðc þ ieÞ ¼
1 vj ðc þ ði 1ÞeÞ þ vj ðc þ ði þ 1ÞeÞ X ~ak ekþ2 f ðkþjÞ ðc þ ieÞ: þ 2 k¼0
Let 0
~ a0 B~ a B 1 B~ B a2 B... A¼B B... B B~ B an @... ...
0 ~ a0 ~ a1 ... ... ~ an 1 ... ...
0 0 ~ a0 ... ... ... ... ...
0 0 0 ... ... ~ a2 ... ...
... ... ... ... ... ~ a1 ... ...
0 0 0 ... ... ~ a0 ... ...
1 ... ...C C ...C C ...C C; ...C C ...C C ...A ...
F ¼ ða0 ; a1 ; a2 ; . . . ; an ; . . .Þ: Assume that b ¼ ðb0 ; b1 ; . . . ; bn 1 ; bn ; . . .Þ is the solution of the following linear equation: Ab ¼ F :
ð2:17Þ
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W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
Then we have 1 X 1 ue ðc þ ieÞ ¼ ½ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ þ enþ2 an f ðnÞ ðc þ ieÞ 2 n¼0
1 ¼ ½ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ 2 1 n X X þ enþ2 bj ~ an j f ðnÞ ðc þ ieÞ n¼0
j¼0
1 ¼ ½ue ðc þ ði 1ÞeÞ þ ue ðc þ ði þ 1ÞeÞ 2 1 1 X X þ bj ej ekþ2 ~ ak f ðkþjÞ ðc þ ieÞ: j¼0
k¼0
Let uS ¼ ~v þ
1 X
b j e j vj :
ð2:18Þ
j¼0
By virtue of above analysis, we obtain 8 u ðc þ ieÞ ¼ 12 ½u þ ði 1ÞeÞ þ uS ðc þ ði þ 1ÞeÞ > > PS ðc < S P1 1 þ j¼0 bj ej k¼0 ekþ2 ~ ak f ðkþjÞ ðc þ ieÞ; e u ðcÞ ¼ u ¼ u ðcÞ; > S 0 > : uS ðdÞ ¼ u1 ¼ ue ðdÞ:
ð2:19Þ
We obtain from (2.19) that ue ðc þ ieÞ ¼ uS ðc þ ieÞ. Finally, we get the following Lemma 2.2. Lemma 2.2. Suppose that b ¼ ðb0 ; b1 ; b2 ; . . . ; bn ; . . .Þ and uS to be defined as (2.17) and (2.18) respectively, then ue ðc þ ieÞ ¼ uS ðc þ ieÞ. Now let us consider the case that x 2 ðc þ ði 1Þe; c þ ieÞð1 6 i 6 n 1Þ, let y¼
x c ði 1Þe ; e
x ^ aðyÞ ¼ að Þ; e
uðyÞ ¼ ue ðxÞ:
Then uðyÞ is the solution of the following problem: 8
P ðkÞ 1 > ; y 2 ð0; 1Þ; aðyÞ du ¼ k¼0 ekþ2 y k f ðcþði 1ÞeÞ < dyd ^ dy k! e uð0Þ ¼ u ðc þ ði
1ÞeÞ; > : uð1Þ ¼ ue ðc þ ieÞ: Assume that ~ a0 and ~ a1 satisfy
ð2:20Þ
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
8
a0 > ¼ 0; < dyd aðyÞ d~ dy ~ a ð0Þ ¼ 0; > : 0 ~ a0 ð1Þ ¼ 1;
673
y 2 ð0; 1Þ; ð2:21Þ
and 8
a1 > < dyd aðyÞ d~ ¼ 0; dy ~ ð0Þ ¼ 1; a > : 1 ~ a1 ð1Þ ¼ 0:
y 2 ð0; 1Þ; ð2:22Þ
^k ðyÞðk ¼ 0; . . . ; n; . . .Þ to be the solution of the following problem Suppose a (
k d^ ak ðyÞ d aðyÞ ¼ yk! ; y 2 ð0; 1Þ; dy dy ð2:23Þ ^ ak ð1Þ ¼ 0: ak ð0Þ ¼ ^ From (2.20)–(2.23), we obtain uðyÞ ¼ ~ a0 ðyÞue ðc þ ði 1ÞeÞ þ ~ a1 ðyÞue ðc þ ieÞ 1 X þ ekþ2 ^ ak ðyÞf ðkÞ ðc þ ði 1ÞeÞ k¼0
a1 ðyÞuS ðc þ ieÞ ¼~ a0 ðyÞuS ðc þ ði 1ÞeÞ þ ~ 1 X þ ekþ2 ^ ak ðyÞf ðkÞ ðc þ ði 1ÞeÞ k¼0
a1 ðyÞuS ðc þ ieÞ ¼~ a0 ðyÞuS ðc þ ði 1ÞeÞ þ ~ 1 1 XX þ ekþ2 ^ ak ðyÞð yÞj f ðkþjÞ ðxÞ: k¼0 j¼0
Let ak ¼
k X
^ aj ðyÞð yÞk j :
j¼0
Then we have uðyÞ ¼ ~ a0 ðyÞuS ðc þ ði 1ÞeÞ þ ~ a1 ðyÞuS ðc þ ieÞ þ
1 X ak ðyÞekþ2 f ðkÞ ðxÞ: k¼0
For x 2 ðc þ ði 1Þe; c þ ieÞ, ue can be expressed as ue ðxÞ ¼ ~ a0 ðyÞuS ðc þ ði 1ÞeÞ þ ~ a1 ðyÞuS ðc þ ieÞ þ
1 X k¼0
From
ak ðyÞekþ2 f ðkÞ ðxÞ:
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W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
uS ðc þ ði 1ÞeÞ ¼ uS ðxÞ þ
1 X
dj uS ; dxj
ej ð yÞj
j¼1
uS ðc þ ieÞ ¼ uS ðxÞ þ
1 X
ej ð1 yÞj
j¼1
dj uS ; dxj
we have 1 X
e
u ðxÞ ¼ ~ a0 ðyÞ uS ðxÞ þ
j¼1
dj uS ðxÞ e ð yÞ dxj
1 X
þ~ a1 ðyÞ uS ðxÞ þ
j¼1
dj uS ðxÞ e ð1 yÞ dxj 1 X
d ~vðxÞ þ~ a1 ðyÞð1 yÞ þ dxj þ
bk ek vk ðxÞÞ þ
k¼0 j
j
1 X
!
j
j
¼ ð~ a0 ðyÞ þ ~ a1 ðyÞÞð~vðxÞ þ "
!
j
j
þ
1 X
ak ðyÞekþ2 f ðkÞ ðxÞ
k¼0 1 X
ej ½~a0 ðyÞð yÞ
j
j¼1 1 X k¼1
j
d vk ðxÞ b k ek dxj
#
2
ak ðyÞekþ2
k¼0
d vk ðxÞ : dx2
ð2:24Þ
a1 ðyÞ, ~v, vk ðk P 0Þ to be defined as above. For Lemma 2.3. Suppose y, ~ a0 ðyÞ, ~ x 2 ðc þ ði 1Þe; c þ ieÞ, we have ue ðxÞ ¼ ð~a0 ðyÞ þ ~ a1 ðyÞÞð~vðxÞ þ "
1 X k¼0
j
d ~vðxÞ þ~ a1 ðyÞð1 yÞ þ dxj j
þ
1 X k¼0
bk ek vk ðxÞÞ þ
1 X
ej ½~a0 ðyÞð yÞ
j
j¼1 1 X k¼1
j
d vk ðxÞ bk e dxj
#
k
2
ak ðyÞekþ2
d vk ðxÞ : dx2
ð2:25Þ
Actually, Lemma 2.3 has not obtained complete asymptotic expansion for the two point boundary value problem with small periodic structure. But from Lemmak 2.3, we can obtain that ue can be expressed by y, u0 , u1 , d v0 ðxÞ dk v0 ðcÞ dk v0 ðdÞ v0 ðxÞ, v0 ðcÞ, v0 ðdÞ, dxk , dxk , dxk completely. The result is the following Theorem 2.1:
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
675
Theorem 2.1. Set d 0 v0 ¼ v0 ;
^ a1 ðyÞ ¼ b1 þ ½ ae0 ðyÞð yÞ þ ~a1 ðyÞð1 yÞ b0 :
^ a0 ðyÞ ¼ b0 ;
When k P 2, let ^ ak ðyÞ ¼
k X ½~ a0 ðyÞð yÞj þ ~ a1 ðyÞð1 yÞj bk j þ ak 2 ðyÞ þ bk : j¼1
Then ue can be expressed as follows: k
0 1 k d v0 ðdÞ v0 ðcÞ 1 k
d dx ðx cÞ X k k dx ðu
u Þðx
cÞ d v ðcÞ 1 0 0 A
þ bk ek @ ue ðxÞ ¼ u0 þ k d c dx d c k¼1 k
1 0 d v0 ðdÞ dk v0 ðcÞ 1
X k k dx dx u1 u0 A
a1 ðyÞð1 yÞ @e bk ekþ1 þ ½ ae0 ðyÞð yÞ þ ~ d
c d c k¼1 þ
1 X
^ ak ðyÞ
k¼0
dk v0 ðxÞ : dxk
Proof. By virtue of the definition of ~v, we have ~vðxÞ ¼ u0 þ
ðu1 u0 Þðx cÞ ; d c
d~vðxÞ u1 u0 ¼ ; dx d c
We obtain from the definition of v0 and vk that d2 vk ðxÞ dkþ2 v0 ðxÞ ¼ : dx2 dxkþ2 Suppose that ~vk ðxÞ satisfies 8 2 d ~vk ðxÞ > > < dx2 ¼ 0; x 2 ðc; dÞ; k v0 ðcÞ ~vk ðcÞ ¼ d dx ; k > k > d : ~vk ðdÞ ¼ v0 ðdÞ : dxk
We obtain k
~vk ðxÞ ¼ From
d v0 ðcÞ þ dxk
dk v0 ðdÞ dxk
k v0 ðcÞ
d dx ðx cÞ k d c
:
d2~vðxÞ ¼ 0: dx2
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W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
d2 ðvk ðxÞ þ ~vk ðxÞÞ dkþ2 v0 ðxÞ ¼ ; dx2 dxkþ2 dk v0 ðcÞ vk ðcÞ þ ~vk ðcÞ ¼ ; dxk k d v0 ðdÞ ; vk ðdÞ þ ~vk ðdÞ ¼ dxk we have dk v0 ðxÞ dk v0 ðxÞ dk v0 ðcÞ k ~
v ðxÞ ¼
vk ðxÞ ¼ dxk dxk dxk
dk v0 ðdÞ dxk
k v0 ðcÞ
d dx ðx cÞ k d c
:
From Lemma 2.3, we conclude k
3 2 d v0 ðdÞ dk v0 ðcÞ 1 k k ðx cÞ X k dxk dx ðu
u Þðx cÞ d v ðxÞ d v ðcÞ 1 0 0 0 5 þ ue ðxÞ ¼ u0 þ bk ek 4
d c d c dxk dxk k¼1 ! dk v0 ðdÞ dk v0 ðcÞ 1
dxk u1 u0 X kþ1 dxk
þe½~ a0 ðyÞð yÞþ~ a1 ðyÞð1 yÞ be d c k¼1 k d c þ þ
1 1 X X dkþj v0 ðxÞ ½~ a0 ðyÞð yÞj þ~ a1 ðyÞð1 yÞj bk ekþj dxkþj j¼1 k¼0 1 X
ak ðyÞekþ2
k¼0
dkþ2 v0 ðxÞ : dxkþ2
From the definition of ^ ak , finally we have k
1 0 d v0 ðdÞ dk v0 ðcÞ 1 k
ðx
cÞ X k k dx dx ðu1 u0 Þðx cÞ d v0 ðcÞ A þ þ bk ek @ ue ðxÞ ¼ u0 þ k d c dx d
c k¼1 k
1 0 d v0 ðdÞ dk v0 ðcÞ 1
X k k dx dx u1 u0 A þ ½~ a0 ðyÞð yÞ þ ~
a1 ðyÞð1 yÞ @e bk ekþ1 d
c d c k¼1 þ
1 X k¼0
^ ak ðyÞ
dk v0 ðxÞ : dxk
Theorem 2.1 is proved.
ð2:26Þ
h
Theorem 2.1 obtained a complete asymptotic expansion for the two point boundary value problem with small periodic structure. Above, we consider the special case: d ¼ c þ me. when d ¼ c þ me þ deð0 < d < 1Þ, we can take the following steps to get the solution of Eq. (1.1).
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
(1) Solve the following equation: 8 2 P j ðjÞ < ddxw2 ¼ 1 j¼0 bj e f ðxÞ; wðcÞ ¼ 0; : wðc þ meÞ ¼ 0:
677
ð2:27Þ
(2) From (2.27), we obtain ue ðc þ ði 1ÞeÞ ¼ wðc þ ðm 1ÞeÞ þ
1 m u0 þ ue ðc þ meÞ: m mþ1
For x 2 ðc þ ðm 1Þe; dÞ, let y¼
x c ðm 1Þe ; e
^ aðyÞ ¼ a
x
e
;
F ðyÞ ¼ e2 f ðxÞ;
uðyÞ ¼ ue ðxÞ:
(3) Obtain the solution of the following problem: 8
> aðyÞ du ¼ F ðyÞ; < dyd ^ dy
y 2 ð0; 1 þ dÞ;
1 m u0 þ mþ1 u1 ; uð0Þ ¼ wðc þ ðm 1ÞeÞ þ mþ1 > : uð1 þ dÞ ¼ u1 :
ð2:28Þ
Let ~ u1 ¼ uð0Þ: (4) By virtue of the same way as above, solve the following problem: 8 e d ¼ f ðxÞ; x 2 ðc; c þ ðm 1ÞeÞ; a xe du < dx dx e ð2:29Þ u ðcÞ ¼ u0 ; : e u ðc þ ðm 1ÞeÞ ¼ ~ u1 : For Eq. (1.2), suppose d ¼ c þ me þ deð0 < d < 1Þ, we can take following steps to obtain its solution: (1) Solve the following equation 8 1 P > d2 w j ðjÞ > < dx2 ¼ bj e f ðxÞ; j¼0
ð2:30Þ
> wðcÞ ¼ 0; > : wðc þ meÞ ¼ 0: (2) From (2.30), we obtain ue ðc þ ðm 1ÞeÞ ¼ wðc þ ðm 1ÞeÞ þ
1 m u0 þ ue ðc þ meÞ: mþ1 mþ1
678
W.-m He, J.-z. Cui / Appl. Math. Comput. 161 (2005) 667–678
(3) For x 2 ðc þ ðm 1Þe; dÞ, let y¼
x c ðm 1Þe ; e
^ aðyÞ ¼ a
x
e
;
F ðyÞ ¼ e2 f ðxÞ;
uðyÞ ¼ ue ðxÞ:
Obtain the solution of the following problem: 8
aðyÞ du ^ d > > ¼ F ðyÞ; y 2 ð0; 1 þ dÞ; < dy dy 1 m u0 þ mþ1 u1 ; uð0Þ ¼ wðc þ ðm 1ÞeÞ þ mþ1 > > duð1þdÞ :^ að1 þ dÞ dy ¼ u1 : Let ~ u1 ¼ uð0Þ: (4) By virtue of the same way as above, solve the following problem: 8 e d a xe du ¼ f ðxÞ; x 2 ðc; c þ ðm 1Þe; < dx dx e u ðcÞ ¼ u0 ; : e u ðc þ ðm 1ÞeÞ ¼ ~ u1 : By virtue of Theorem 2.1, we maybe can study the following problem 8 e d þ ke ue ¼ 0; x 2 ðc; dÞ; a xe du < dx dx e ð2:31Þ u ðcÞ ¼ 0; : e u ðdÞ ¼ 0:
Acknowledgements This work is supported by National Natural Science Foundation of China. Subsidized by Special Funds for Major State Basic Research Projects.
References [1] O.A. Oleinik, A.S. Shamaev, G.A. Yosifian, Mathematical Problems in Elasticity and Homogenization, North-Holland, Amsterdam, 1992. [2] Ivo Babuska, J.E. Osborn, Can a finite element method perform arbitrarily badly? Mathematics of Computation 69 (2000) 443–462.