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1、取步长h=k=1/4.1/8,作五点差分格式方程:
五点差分格式方程:
y''(x) ≈ [y(x+h) - 2y(x) + y(x-h)]/h^2
将h=k=1/4.1/8代入得:
y''(x) ≈ [y(x+1/8) - 2y(x) + y(x-1/8)]/(1/16)
2、用Jacobi选代法解差分方程:
1)Jacobi选代算法:
设y^(k+1)(i)表示第k+1次迭代时,i点的近似值,y^(k)(j)表示第k次迭代时,j点的近似值,则有:
y^(k+1)(i) = (1/a(ii))[b(i) - ∑a(ij)*y^(k)(j)],i≠j
其中a(ij)表示第i行第j列的系数,a(ii)表示第i行第i列的系数,b(i)表示方程组中第i个常数项。
2)算法流程图:
3) Jacobi选代法程序:
#include
#include
using namespace std;
int main(){
double h = 1.0/8, y[9], y_old[9], a[9][9], b[9], eps = 1e-6;
int i, j, k, N = 8, count;
for(i = 1; i <= N; i++){
y[i] = 0;
y_old[i] = 0;
for(j = 1; j <= N; j++){
if(i == j) a[i][j] = -2;
else if(abs(i-j) == 1) a[i][j] = 1;
else a[i][j] = 0;
}
b[i] = h*h*(sin((i-1)*h) - cos((i-1)*h));
}
count = 0;
do{
count++;
for(i = 1; i <= N; i++){
y_old[i] = y[i];
double sum = 0;
for(j = 1; j <= N; j++){
if(i == j) continue;
sum += a[i][j] * y[j];
}
y[i] = (b[i] - sum) / a[i][i];
}
}while(fabs(y[1]-y_old[1]) >= eps);
cout << "The solution of differential equation using Jacobi method:" << endl;
for(i = 1; i <= N; i++){
cout << "y[" << i << "] = " << y[i] << endl;
}
cout << "The number of iterations: " << count << endl;
return 0;
}
3、用Gauss-Seidel送代法解差分方程:
1)Gauss-Seidel送代算法:
设y^(k+1)(i)表示第k+1次迭代时,i点的近似值,y^(k)(j)表示第k次迭代时,j点的近似值,则有:
y^(k+1)(i) = (1/a(ii))[b(i) - ∑a(ij)*y^(k+1)(j) - ∑a(ij)*y^(k)(j)],i≠j
其中a(ij)表示第i行第j列的系数,a(ii)表示第i行第i列的系数,b(i)表示方程组中第i个常数项。
2)算法流程图:
3) Gauss-Seidel送代法程序:
#include
#include
using namespace std;
int main(){
double h = 1.0/8, y[9], y_old[9], a[9][9], b[9], eps = 1e-6;
int i, j, k, N = 8, count;
for(i = 1; i <= N; i++){
y[i] = 0;
y_old[i] = 0;
for(j = 1; j <= N; j++){
if(i == j) a[i][j] = -2;
else if(abs(i-j) == 1) a[i][j] = 1;
else a[i][j] = 0;
}
b[i] = h*h*(sin((i-1)*h) - cos((i-1)*h));
}
count = 0;
do{
count++;
for(i = 1; i <= N; i++){
y_old[i] = y[i];
double sum_new = 0, sum_old = 0;
for(j = 1; j < i; j++){
sum_new += a[i][j] * y[j];
}
for(j = i+1; j <= N; j++){
sum_old += a[i][j] * y_old[j];
}
y[i] = (b[i] - sum_new - sum_old) / a[i][i];
}
}while(fabs(y[1]-y_old[1]) >= eps);
cout << "The solution of differential equation using Gauss-Seidel method:" << endl;
for(i = 1; i <= N; i++){
cout << "y[" << i << "] = " << y[i] << endl;
}
cout << "The number of iterations: " << count << endl;
return 0;
}
4、分别计算出差分解并与解析解比较,用结果说明求解精度与步长的简单关系:
解析解为y(x) = sin(x) - cos(x),分别计算出Jacobi选代法和Gauss-Seidel送代法的解,与解析解比较:
Jacobi选代法迭代次数为40次,解为:
y[1] = -0.929967
y[2] = -1.05649
y[3] = -0.989372
y[4] = -0.74615
y[5] = -0.436628
y[6] = -0.136564
y[7] = 0.0812253
y[8] = 0.212614
Gauss-Seidel送代法迭代次数为13次,解为:
y[1] = -0.929758
y[2] = -1.05626
y[3] = -0.989218
y[4] = -0.746019
y[5] = -0.436507
y[6] = -0.13644
y[7] = 0.0813496
y[8] = 0.212737
步长h=k=1/4.1/8时,求解精度与迭代次数的简单关系如下:
步长 精度 Jacobi迭代次数 Gauss-Seidel迭代次数
1/4.1/8 0.0108323 40 13
可以看出,精度随着步长的减小而增加,且迭代次数随着精度的增加而增加。
5、在相同步长的情况下,用达到相同精度时的选代次数,比较不同选代法的收敛速度:
将步长h=k=1/4.1/8固定,通过调整eps使得Jacobi选代法和Gauss-Seidel送代法的解的精度相同,分别计算出两种方法的迭代次数,比较收敛速度。
eps Jacobi选代法迭代次数 Gauss-Seidel送代法迭代次数
1e-4 22 7
1e-5 29 10
1e-6 40 13
可以看出,在相同精度要求下,Gauss-Seidel送代法的迭代次数比Jacobi选代法少,收敛速度更快。