积分方程数值解

例题：

y t =g t +γ

t −s r

t 0

y s ds

取r =1,γ=1 则原例题可变为：

y t =g t + t −s t 0y s ds 设y t =t 则

g t =t − t −s sds =t −t 3

6

t

y t − t −s t

0y s ds =t −t 3

6

使用方法：配置法

详细算法：

第一步剖分：[0,T] 0=t 0

I n := V ∈L 2 I :V

∈2n P m −1 n

=1,⋯,N

t ni =t n −1+c i h n 0≤c 1

第三步配置方程：U h t ni =f t ni + K t ni ,s t

ni 0u h s ds n =1,⋯,N i =1,⋯,m K t ni ,s t ni 0u h s ds = K t ni ,s t n −10u h s ds + K t ni ,s t

n −1

+c i h n

t n −1

u h s ds

= K t ni ,s t l

t

l −1

u h s ds +n −1l=1 K t ni ,s t

n −1+c i h n

t n −1

u h s ds

= h l K t ni ,t e +vh e u h t l −1+vh l 1

n −1l=1dv +h n K t ni ,t l +vh e u h t l −1+vh e dv c

i 0

U nj L nj t ni =f t ni +F n t ni m j=1+h n K t ni ,t l +vh e u h t l −1+vh e dv c

i 0

=f t ni + h l v lj K t ni ,t l −1+vh l L j v 1

m j=1n −1l=1dv +h n V nj m j=1 K t ni ,t n −1+vh n L j v c

i 0

dv B n

l ≔ K t ni ,t l −1+vh l L j v 1

0dv i,j =1,⋯,m 0≤l

B n ≔ K t ni ,t n +vh n L j v c

i 0dv

i,j =1,⋯,m

I m −h n B n U n =g n +G n n =0,1,⋯,N −1

G n ≔ F n t n,1 ,⋯,F n t n,m T

= h l B n l

n −1

l=0U l

U n,i =f t n,i + K t n,i ,s U n s T

0ds =f t n,i + K t n,i ,s N l=1U n s ds

= f t n,i + U lj K t ni ,s L lj s ds l

l −1

m j=1N l=1 =f t n,i + h l U lj K t ni ,t l −1+vh l L lj v dv 10

m j=1N l=1 U −U n ∞=max 1≤n ≤N U t

=max 1≤n ≤N max t ∈I n U t −U n t =max 1≤n ≤N max t ∈N U t ni −U n t ni

分片一次误差图

分片两次误差图