本部分运用Nystrom 方法求解Fredholm integral equations of the second kind，对应书上P100-103。
# -*- coding: utf-8 -*-
"""
Created on Wed May 30 21:16:58 2018
@author: shaowu
本代码主要运用Nystrom 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)cos(t)和sqrt(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》P100-103.
"""
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(x):
"""
定义精确解
"""
return scp.exp(x)
def function_k(s,t):
"""
定义核函数
"""
return scp.exp(s*t)
def simpson_solve_A(a,b,lamda,n):
"""
用Simpson求积公式求解（4.1.5）式子中的系数矩阵
"""
h=(b-a)/n
t=[a+i*h for i in range(0,n+1)]
A=[]
for i in t:
A.append([-(b-a)/6*function_k(a,i),-(b-a)/6*4*function_k((a+b)/2,i),\
-(b-a)/6*function_k(b,i)])
A=np.array(A)
for i in range(A.shape[0]):
A[i][i]=lamda+A[i][i]
return A
def simpsom_sovle_y(a,b,lamda,n):
"""
由给定的精确解，计算Simpson点对应的右端项y
"""
h=(b-a)/n
t=[a+i*h for i in range(0,n+1)]
return [lamda*function_x(i)-scp.integrate.quad(lambda s:(function_k(s,i)*function_x(s)),a,b)[0]\
for i in t]
def simpsom_solve_xn(A,y):
"""
求解xn
"""
return np.linalg.solve(A,y)
def simpsom_error(xn,a,b,n):
"""
计算误差
"""
h=(b-a)/n
t=[a+i*h for i in range(0,n+1)]
return function_x(t)-xn
#########################################################################################################
def gauss_xw(n=2):
"""
默认用3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(n+1)
return x,w
def gauss_sovle_y(x,a,b,lamda):
"""
由给定的精确解，计算高斯点对应的右端项y
"""
t=(b-a)/2*x+(a+b)/2 #对区间[a，b]做变化
return [lamda*function_x(i)-scp.integrate.quad(lambda s:(function_k(s,i)*function_x(s)),a,b)[0]\
for i in t]
def gauss_solve_A(x,w,a,b,lamda):
"""
计算系数矩阵
"""
c=(b-a)/2
t=(b-a)/2*x+(a+b)/2 #把区间[a,b]变化到[-1,1]
A=[]
for i in t:
A.append([-c*w[j]*function_k(t[j],i) for j in range(len(t))])
A=np.array(A)
for i in range(A.shape[0]):
A[i][i]=lamda+A[i][i]
return A
def gauss_solve_xn(A,y):
"""
计算xn
"""
return np.linalg.solve(A,y)
def gauss_error(xn,x,a,b):
"""
计算误差
"""
t=(b-a)/2*x+(a+b)/2
return function_x(t)-xn
if __name__ == '__main__':
print('******************************程序入口*******************************')
lamda=int(input('pleas input lambda:'))
n=int(input('please input Simpson node 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...')
#下面是使用3个Simpson节点（输入时n=2的情况）：
y=simpsom_sovle_y(a,b,lamda,n)
A=simpson_solve_A(a,b,lamda,n)
xn=simpsom_solve_xn(A,y)
E1=simpsom_error(xn,a,b,n)
print(E1)
#############################################################################
#下面是使用m个高斯勒让德节点（输入时m相应少1）：
x,w=gauss_xw(m)
y=gauss_sovle_y(x,a,b,lamda)
A=gauss_solve_A(x,w,a,b,lamda)
xn=gauss_solve_xn(A,y)
E2=gauss_error(xn,x,a,b)
print(E2)
print('all_cost_time:',time.time()-start_time)
运行结果：
******************************程序入口*******************************
pleas input lambda:2
please input Simpson node n:2
please input the left value of interval:0
please input the right value of interval:1
please input the node of Gauss_Legendre:5
计算中...
[-0.0047043 -0.00796678 -0.01640422]
[ 2.72448730e-13 2.98427949e-13 3.51274565e-13 4.72955008e-13
8.05577827e-13 1.36868294e-12]
all_cost_time: 35.219531774520874