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I have this 1D code in MATLAB I want to use the same code but convert it to 2D there is an error in this code I tired many time to figure it out put I can’t. I replaced the 3 here ( f ＝ 3 * DuDx;) with ( k/(rho*cp) :
% Define parameters and discretization
m ＝ 0; % Coefficient for a 1D simulation
k ＝ 1.4; % Thermal conductivity
rho ＝ 1.225; % Density
cp ＝ 1.005; % Specific heat capacity
x ＝ linspace(0, 10, 50); % Spatial grid
t ＝ linspace(0.1, 10, 50); % Time grid
% Solve the PDE using pdepe
sol ＝ pdepe(m, @pdex4pde, @pdex4ic, @pdex4bc, x, t);
% Extract the solution
u1 ＝ sol(:,:,1);
% Plotting the solution
figure;
surf(x, t, u1);
xlabel(′Space (x)′);
ylabel(′Time (t)′);
zlabel(′Heat Density u(x,t)′);
title(′Heat Equation Solution′);
function [c, f, s] ＝ pdex4pde(x, t, u, DuDx)
% Define the coefficients of the PDE: c * dudt ＝ x * dfdx + s
Z ＝ k / (rho * cp);
c ＝ 1; % Coefficient multiplying the time derivative
f ＝ Z * DuDx; % Coefficient multiplying the spatial derivative
s ＝ 0; % Source term (if any)
end
function u0 ＝ pdex4ic(x)
% Initial condition function: u0(x) at t ＝ 0
u0 ＝ 300; % Initial heat density
end
function [pl, ql, pr, qr] ＝ pdex4bc(xl, ul, xr, ur, t)
% Boundary condition function
% Left boundary (Dirichlet condition): u(xl, t) ＝ ul
pl ＝ ul; % Value of the solution at the left boundary
ql ＝ 0; % Flux condition at the left boundary (unused here)
% Right boundary (Neumann condition): du/dx ＝ ur – 1
pr ＝ ur – 1; % Flux condition at the right boundary
qr ＝ 0; % Value of the solution at the right boundary (unused here)
end
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