1D Heat equation using an implicit method

5 vues (au cours des 30 derniers jours)
krayem youssef
krayem youssef le 14 Avr 2019
hi guys,
so i made this program to solve the 1D heat equation with an implicit method.
i have a bar of length l=1
the boundaries conditions are T(0)=0 and T(l)=0
and the initial conditions are 1 if l/4<x<3*l/4 and 0 else.
i plot my solution but the the limits on the graph bother me because with an explicit method i have a better shape for the same problem.
if someone can help me solve the problem
clear, clc, close all
%% parametres
Nx = 40;
Nt = 1000;
dx = 1/(Nx-1);
c = 5;
l=1;
cfl = (2*c)/(2*c+1);
dt = cfl*(dx^2)/(2*c);
alpha = c*dt/dx^2;
beta = (1+2*alpha);
x = 0:dx:l;
%% construction de la matrice A
A = full(spdiags(ones(Nx,1)*[-alpha,beta,-alpha],[-1,0,1],Nx,Nx));
%% construction de la matrice b
b = ones(Nx,1);
b(1:Nx/4,1)=0; % initial conditions
b(3*Nx/4:Nx,1)=0; % initial conditions
%% %% %% %%t
time = 0;
for t = 0:dt:Nt
T = linsolve(A,b);
T(1) = 0; % boundary conditions
T(Nx) = 1; % boundary conditions
b = T;
time = time+dt;
figure(1), clf % Solution plot
plot(x,T);
xlabel('x [m]')
ylabel('Temperature [^oC]')
ylim([0 3])
drawnow
end
  1 commentaire
Youssef Benmoussa
Youssef Benmoussa le 12 Avr 2020
bonjour est ce que tu peut m'envoyer le code pour l'implicite ?

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