0 个投票

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. :)

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Grzegorz Knor
Grzegorz Knor 2011-9-23

0 个投票

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 个评论
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Alex
Alex 2011-9-23
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 2011-9-23
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.
Bjorn Gustavsson
Bjorn Gustavsson 2011-9-23
@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 2011-10-3
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 ?
Mohammed Thaiki
Mohammed Thaiki 2016-12-1
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

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