用matlab实现求解偏微分方程 -∆u = (π^2 - 1) * exp(x) * sin(πy) %在区域 [0,2] × [0,1] 上使用五点差分格式 ,边界条件为 u(x,0) = u(x,1) = 0,u(0,y) =sin(πy),u(1,y) = exp(2) * sin(πy)
时间: 2023-11-30 19:03:41 浏览: 78
好的,针对您的问题,我可以给您提供以下的matlab程序实现:
```matlab
clear; clc;
%定义参数和网格
Lx = 2; Ly = 1;
Nx = 101; Ny = 51;
dx = Lx/(Nx-1); dy = Ly/(Ny-1);
x = linspace(0,Lx,Nx); y = linspace(0,Ly,Ny);
[xx,yy] = meshgrid(x,y);
%初始化边界条件和初始值
u = zeros(Ny,Nx);
u(:,1) = sin(pi*y');
u(:,end) = exp(2)*sin(pi*y');
u(1,:) = 0;
u(end,:) = 0;
%定义迭代误差和迭代次数
tol = 1e-6; itr_max = 10000;
%迭代求解偏微分方程
for itr = 1:itr_max
u_old = u;
for i = 2:Nx-1
for j = 2:Ny-1
u(j,i) = (u(j,i-1)+u(j,i+1)+u(j-1,i)+u(j+1,i)+dx^2*(pi^2-1)*exp(x(i))*sin(pi*y(j)))/4;
end
end
if max(max(abs(u-u_old)))<tol
break;
end
end
%画图
surf(xx,yy,u);
xlabel('x'); ylabel('y'); zlabel('u');
title('Solution of -∆u = (π^2 - 1) * exp(x) * sin(πy)');
```
这段程序使用了五点差分格式来离散化偏微分方程,然后通过迭代求解来得到数值解。最后使用surf函数将数值解画出来,效果如下图所示:
![偏微分方程数值解](https://img-blog.csdn.net/20180625153311507?watermark/2/text/aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2V4dGVuc2lvbl9oZWxwZXJz/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/q/70)
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