MATLAB PDE BC'S

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Mr.DDWW
Mr.DDWW am 25 Mär. 2022
Beantwortet: Mr.DDWW am 27 Mär. 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.

Antworten (2)

Torsten
Torsten am 25 Mär. 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 Kommentare
Mr.DDWW
Mr.DDWW am 26 Mär. 2022
Well, I am supposed to obtain the numerical solution from the image by changing the size of the mesh
Torsten
Torsten am 26 Mär. 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 am 27 Mär. 2022
It is a slab. So the symmetry (m) = 1

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