how to solve PDE with derivative boundary conditions ?

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Mohammad Adeeb
Mohammad Adeeb 2021-4-1
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?

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采纳的回答

darova
darova 2021-4-2
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
darova
darova 2021-4-3
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)

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