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MATLAB PDE BC'S

2 vues (au cours des 30 derniers jours)
Mr.DDWW
Mr.DDWW le 25 Mar 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.

Réponses (2)

Torsten
Torsten le 25 Mar 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 commentaires
Mr.DDWW
Mr.DDWW le 26 Mar 2022
Well, I am supposed to obtain the numerical solution from the image by changing the size of the mesh
Torsten
Torsten le 26 Mar 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.

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Mr.DDWW
Mr.DDWW le 27 Mar 2022
It is a slab. So the symmetry (m) = 1

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