这是《The Numerical Solution of Integral Equations of the Second Kind》书上的例题代码实现（Galerkin方法），P66-P68,P74-77
# -*- coding: utf-8 -*-
"""
Created on Fri Jun 1 19:45:59 2018
@author: shaowu
本代码主要运用Galerkin method实现Fredholm integral equations of the second kind的求解,方程如下：
lamda*x(t)-integrate(K(t,s)*x(s))ds=y(t),a=0<=t<=b,K(t,s)取exp(s*t)
首先，我们给定两个精确解exp(t)，求出其对应的y(t)，然后再来反解x(t).更详细说明可
参见《The Numerical Solution of Integral Equations of the Second Kind》P66-P68,P74-77.
"""
import sympy as sp
import scipy as scp
import numpy as np
import pandas as pd
import numpy.linalg as LA
import matplotlib.pyplot as plt
from scipy.special.orthogonal import p_roots
import time
start_time=time.time()
def function_x(t):
return scp.exp(t)
def function_k(s,t):
return scp.exp(s*t)
def gauss_xw(m=100):
"""
默认用100个点求高斯——勒让德节点xi和权weight，并返回x和w数组
"""
x,w=p_roots(m+1)
return x,w
def gauss_solve_f(x,w,lamda,a,b,n): #参数n>=1
"""
求解课本3.3.67式中的右端项,记为f
参数：
x,w为高斯点和对应的权,
a,b 对应于区间[a,b],
n 子区间的个数
返回的是一个n+1维的列表
"""
c=(b-a)/2
s=(b-a)/2*x+(a+b)/2 #把区间[a,b]变化到[-1,1]
h=(b-a)/(n)
t=[a+i*h for i in range(0,n+1)] #等距划分a=t0<t1<...<tn=b
return [sum([c*w[k]*lamda*function_x(s[k])*hat_f(i,s[k],a,b)-
c*w[k]*sum([c*w[j]*function_k(s[j],s[k])*function_x(s[j])*hat_f(i,s[k],a,b)\
for j in range(len(s))]) for k in range(len(s))])\
for i in range(len(t))]
def hat_f(j,x,a,b):
"""
定义hat函数
"""
h=(b-a)/(n)
t=[a+i*h for i in range(0,n+1)]
if j==0:
if x>=t[0] and x<=t[1]:
return 1-abs(x-t[0])/h
else:
return 0
if j==n:
if t[n-1]<=x and x<=t[n]:
return 1-abs(x-t[n])/h
else:
return 0
if j>0 and j<n:
if t[j-1]<=x and x<=t[j+1]:
return 1-abs(x-t[j])/h
else:
return 0
def elements_of_matrix(x,w,a,b,lamda,n):
"""
求方程（3.3.67）中的系数矩阵；
返回的是一个数，即对应系数矩阵中的一个元素
"""
c=(b-a)/2
s=(b-a)/2*x+(a+b)/2 #把区间[a,b]变化到[-1,1]
h=(b-a)/(n)
t=[a+i*h for i in range(0,n+1)] #等距划分a=t0<t1<...<tn=b
A=[]
for i in range(len(t)):
A.append([scp.integrate.quad(lambda x: lamda*hat_f(i,x,a,b)*hat_f(j,x,a,b),a,b)[0]-\
sum([c*w[k]*sum([c*w[e]*function_k(s[e],s[k])*hat_f(i,s[e],a,b)*hat_f(j,s[k],a,b)\
for e in range(len(s))]) for k in range(len(s))]) for j in range(len(t))])
return np.array(A)
def solve_c(A,f):
"""
计算c
"""
return np.linalg.solve(A,f)
def solve_xn(c,a,b,n):
"""
计算xn
"""
h=(b-a)/(n)
t=[a+i*h for i in range(0,n+1)] #等距划分a=t0<t1<...<tn=b
xn=[sum([c[j]*hat_f(j,i,a,b) for j in range(len(t))]) for i in t]
error=function_x(t)-xn
#print(LA.norm(error,np.inf))
return xn,LA.norm(error,np.inf)
if __name__ == '__main__':
print('******************************程序入口*******************************')
lamda=int(input('pleas input lambda:'))
n=int(input('please input n:'))
a=int(input('please input the left value of interval:'))
b=int(input('please input the right value of interval:'))
m=int(input('please input the node of Gauss_Legendre:'))
print('计算中...')
x,w=gauss_xw(m)
f=gauss_solve_f(x,w,lamda,a,b,n)
A=elements_of_matrix(x,w,a,b,lamda,n)
c=solve_c(A,f)
xn,error=solve_xn(c,a,b,n)
print('the error is:',error)
print('all_cost_time:',time.time()-start_time)
运行结果：
******************************程序入口*******************************
pleas input lambda:50
please input n:8
please input the left value of interval:0
please input the right value of interval:1
please input the node of Gauss_Legendre:400
计算中...
the error is: 0.00360146251855
all_cost_time: 188.2648663520813