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MATLAB PDE BC'S

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Mr.DDWW
Mr.DDWW el 25 de Mzo. de 2022
Respondida: Mr.DDWW el 27 de Mzo. de 2022
clc;clear all;close all;
L = 1;
x = linspace(0,L,75);
t = linspace(0,1,75);
m = 1;
sol = pdepe(m,@heatpde,@heatic,@heatbc,x,t);
sol1=1-sol;
figure(1)
surf(x,t,sol1);
xlabel('y/b');
zlabel('(T-T_0)/(T_1-T_0)');
title('Fig 12.1-1'); grid on;
function [c,f,s] = heatpde(x,t,u,dudx)
c = 1;
f = dudx;
s = 0;
end
function u0 = heatic(x)
u0 = 1;
end
function [pl,ql,pr,qr] = heatbc(xl,ul,xr,ur,t)
% left Bc = ul
pl = ul;
ql = 0;
% right BC= ur
pr = ur;
qr = 0;
end
I am a pde code. I am having a problem using the BC's in image. can you please help me.

Respuestas (2)

Torsten
Torsten el 25 de Mzo. de 2022
function [pl,ql,pr,qr] = heatbc(xl,ul,xr,ur,t)
Nu = 1.0;
% left Bc = ul
pl = 0;
ql = 1;
% right BC= ur
pr = Nu*ur;
qr = 1;
end
  4 comentarios
Mr.DDWW
Mr.DDWW el 26 de Mzo. de 2022
Well, I am supposed to obtain the numerical solution from the image by changing the size of the mesh
Torsten
Torsten el 26 de Mzo. de 2022
If it's the equation from the image you are trying to solve, you'll have to set m=0 instead of m=1 in your code.
If you want to solve the problem for different spatial meshes, change the "75" in
x = linspace(0,L,75);
to a different number.

Iniciar sesión para comentar.


Mr.DDWW
Mr.DDWW el 27 de Mzo. de 2022
It is a slab. So the symmetry (m) = 1

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