Hello everyone, I would like to use pdepe for solving a heat equation 1D space, so it looks good. But I don't really understand where the diffusion coefficient is ?
I would like to solve : (1)--> du/dt = d/dx(D(u) du/dx) with D(u) the non linear diffusion coefficient function of u. Is it possible to put D(u) at this location in the equation (1)?
Thanks to the community. :)

 Accepted Answer

Grzegorz Knor
Grzegorz Knor on 23 Sep 2011
Yes it is possible in pdefun :)
Look at example:
function diffusion
m = 0;
x = linspace(0,1,20);
t = linspace(0,2,5);
sol = pdepe(m,@eqtn,@ic,@bc,x,t);
u = sol(:,:,1);
figure;
surf(x,t,u);
xlabel('Distance x');
ylabel('Time t');
% --------------------------------------------------------------------------
function [c,f,s] = eqtn(x,t,u,DuDx)
c = 1;
f = Dfsn(u)*DuDx;
s = 0;
% --------------------------------------------------------------------------
function u0 = ic(x)
u0 = 1:length(x);
% --------------------------------------------------------------------------
function [pl,ql,pr,qr] = bc(xl,ul,xr,ur,t)
pl = ul;
ql = 2;
pr = ur;
qr = 2;
% --------------------------------------------------------------------------
function d = Dfsn(u)
d = sqrt(u+1);
Where Dfsn is your non linear diffusion coefficient function.

5 Comments

Alex
Alex on 23 Sep 2011
Thanks Grzegorz, but could you tell me more the equation you are solving in your example. Because it seems weird to have an increase of u, a diffusion equation must diffuse not increase.
Grzegorz Knor
Grzegorz Knor on 23 Sep 2011
This equation has no physical meaning. I wrote a simple example that shows just how to use any function D in the parabolic partial differential equation.
@Alex: Depending on your heating and cooling rates of course you can get increasing temperatures, and a temperature varying heat conductivity is not unheard of - that should just reflect a variation of how efficiently heat is conducted at different temperatures. In the physical systems I work with D is roughly proportional to T.^(5/2).
Alex
Alex on 3 Oct 2011
Ok, thanks for the answer. I had work on it, but I remain stuck on pl, ql, pr and qr. In a basic heat equation you have to give an initial condition and boundary condition. What is in that case the boundary condition ?
For example, initial condition is u(t=0,x)=sin(Pix) u(t,x=0)=u(t,x=1)=0 for [0,1] in space.
So what are the link with pl,pr,ql,qr and boundary condition ?
Hello ; I have a problem with a heat transfer script, below the script:
clear all; x=linspace(0,.01,50);%We use 50 values from 0 to 0.01 t=linspace(0,1,60);%We used 60 points from 0 to 1 m=0; sol=pdepe(m,@ecuation,@initialcond,@boundary,x,t) u=sol(:,:,1); % Surface plot command and data surf(x,t,u) colormap([gray]) xlabel('\delta (m)') ylabel('L (m)') zlabel('C_a (M)') shading interp figure for j=1:length(t) plot(x,u(j,:),'k') xlabel('\delta (m)') ylabel('C_a (M)') hold on end
and the three files :
1. @ecuation function [c,f,s]=ecuation(x,t,u,DuDx) c=2*0.5*((x/0.01)-0.5*(x/0.01)^2)/2.1e-5;%term (C) f=DuDx;%Flow term (F) s=0;%source term (S)
2. @initialcond %Initial conditions. function u0=initialcond(x) u0=0;
3. @boundary %Boundary conditions. function [pl,ql,pd,qd]=boundary(xl,ul,xd,ud,t) %for y = 0 pl=0; ql=1; %%for y = δ pd=ur-.1; qd=0;
MATLAB gives error in line :
sol=pdepe(m,@ecuation,@initialcond,@boundary,x,t)
I am waiting for your ideas to solve this problem thank you

Sign in to comment.

More Answers (0)

Categories

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!