problem of pdepe!

function pdex11

m = 0;
x = linspace(0,1,20);
t = linspace(0,2,5);

sol = pdepe(m,@pdex1pde,@pdex1ic,@pdex1bc,x,t);
% Extract the first solution component as u.
u = sol(:,:,1);

% A surface plot is often a good way to study a solution.
surf(x,t,u) 
title('Numerical solution computed with 20 mesh points.')
xlabel('Distance x')
ylabel('Time t')

% A solution profile can also be illuminating.
figure
plot(x,u(end,:))
title('Solution at t = 2')
xlabel('Distance x')
ylabel('u(x,2)')
% --------------------------------------------------------------
function [c,f,s] = pdex1pde(x,t,u,DuDx)
c = pi^2;
f = DuDx;
s = 0;
% --------------------------------------------------------------
function u0 = pdex1ic(x)
u0 = sin(pi*x);
% --------------------------------------------------------------
function [pl,ql,pr,qr] = pdex1bc(xl,ul,xr,ur,t)
pl = ul;
ql = 0;
pr = pi * exp(-t);
qr = 1;

And when I change the context of function into (c=1,f=DuDx/pi^2,s=0), the result is totally different.

Why?

Answer 1

[David Goodmanson]

Hi Junyi,

When you made the change, the flux went down by a factor of 1/pi^2. That's fine, but one of your boundary conditions depends on the flux and you need to change that one to compensate. If, in pdex1bc you change pr to

pr = (pi*exp(-t))/pi^2

then the value of u over the x,t plane (I changed t to have 10 points instead of 5) agrees with the original result within 3e-7.