MATLAB PDE BC'S

조회 수: 1 (최근 30일)
Mr.DDWW
Mr.DDWW 2022년 3월 25일
답변: Mr.DDWW 2022년 3월 27일
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.

답변 (2개)

Torsten
Torsten 2022년 3월 25일
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
  댓글 수: 4
Mr.DDWW
Mr.DDWW 2022년 3월 26일
Well, I am supposed to obtain the numerical solution from the image by changing the size of the mesh
Torsten
Torsten 2022년 3월 26일
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 2022년 3월 27일
It is a slab. So the symmetry (m) = 1

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