how to solve PDE with derivative boundary conditions ?

Question

Mohammad Adeeb 2021-4-1

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/789969-how-to-solve-pde-with-derivative-boundary-conditions

评论： darova 2021-4-3

hey all

im trying to solve PDE with derivative boundary condition , so i tend to use the imaginary node method , could i have another way to solve it without any built in function

this is the qustion:

𝜕𝑇𝜕𝑡=𝜕2𝑇𝜕𝑥2+𝑞(𝑥) (1)

With 𝑞(𝑥)=100sin(𝜋𝑥) (2)

1)

𝑇(𝑥,0)=0 (3)

2)

𝜕𝑇𝜕𝑡(0,𝑡)=𝑇(0,𝑡)−10 (4)

3)

𝜕𝑇𝜕𝑡(1,𝑡)=10−𝑇(1,𝑡) (5)

clear all;
close all;
clc;
%% Demo program for parapolic pde
dt = 0.25;
dx = 0.1*dt;
alpha=1;
t = 0:dt:15;
x = 0:dx:4;
q_x=(100*sin(pi*x));
N = length(x)-1;
T=[]; %Dynamic size
T(1,:) = zeros(1,5) ; %Initial condition
for j=1:length(t)-1
    T(1,N-1) = T(j+1,N) + (2*dx*(T(j+1,N+1)-10));
    for i=2:N
        T(j+1,i) = T(j,i)+alpha*(dt/(dx^2))*(T(j,i+1)+ T(j,i-1)-2*T(j,i))+q_x;
    end
     T(2,N+2) =  T(j+1,N) + (2*dx*(10-T(j+1,N+1)));
end
mesh(t,x,T)
colorbar;

the code isn't evaluated , what is the proplem?

3 个评论
显示 1更早的评论隐藏 1更早的评论

Mohammad Adeeb 2021-4-2

...

Mohammad Adeeb 2021-4-2

...

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

darova 2021-4-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/789969-how-to-solve-pde-with-derivative-boundary-conditions#answer_665389

在 MATLAB Online 中打开

Try these corrections

T = zeros(length(t),length(x));
for j=1:length(t)-1
    T(j+1,1) = T(j,1) + dt*(T(j,1)-10);
    T(j+1,N) = T(j,N) + dt*(10-T(j,N));
    for i=2:N-1     % changed
        T(j+1,i) = T(j,i)+alpha*(dt/(dx^2))*(T(j,i+1)+ T(j,i-1)-2*T(j,i)) + q_x(i); % note: q_x(i)
    end
end
mesh(t,x,T)

2 个评论
显示无隐藏无

Mohammad Adeeb 2021-4-2

编辑：Mohammad Adeeb 2021-4-2

it's worked but the mesh result is totally wrong

darova 2021-4-3

在 MATLAB Online 中打开

I made some change sto your code. Some notes:

should be larger than ( should be small )
should be small too
i changed boundary conditions 𝜕𝑇𝜕𝑡(0,𝑡)=𝑇(0,𝑡)−10 and 𝜕𝑇𝜕𝑡(1,𝑡)=10−𝑇(1,𝑡)

clc,clear
%% Demo program for parapolic pde
dt = 0.25;
dx = 5*dt;
alpha=1;
t = 0:dt:5;
x = 0:dx:20;
q_x = sin(pi*x/max(x));
N = length(x);
r = alpha*dt/dx^2;
T = zeros(length(t),length(x));
for j=1:length(t)-1
    T(j+1,1) = T(j,1) + dt*(T(j,1)-1/10);   % changed these
    T(j+1,N) = T(j,N) + dt*(1/10-T(j,N));
    for i=2:N-1     % changed
        T(j+1,i) = T(j,i)+r*diff(T(j,i-1:i+1),2) + q_x(i); % note: q_x(i)
    end
end
surf(x,t,T)