radial diffusion pde boundary conditions

I am writing a spherical diffusion pde and looking at examples for help. I am wondering what some terms mean.

For example, i think icfun is supposed to identitify initial conditions (concentrations at time t=0)? is this correct?

and bcfun i think are the spatial boundary conditions. but pdepe solves in 1-d so what are the four terms pl, ql, pr, and qr?? do these terms change for spherical problems, when m=2 for pdepe?

what is pdefun on the pdepe input values?

any help is appreciated. thank you

Answer 1

[Torsten]

Here is an example for the 1d spherical diffusion equation

dT/dt = 1/r^2 * d/dr (r^2*D*dT/dr)

with boundary conditions

dT/dr = 0 at r = 0

T = 275 at r = 1

and initial condition

T = 0 for 0 <= r < 1

T = 275 for r = 1.

integrated for 1000 time units.

r = linspace(0,1,25);
t = linspace(0,1000,25);
m = 2;
sol = pdepe(m,@heatsphere,@heatic,@heatbc,r,t);
u = sol(:,:,1);
surf(r,t,u)
xlabel('r')
ylabel('t')
zlabel('u(r,t)')
view([150 25])

plot(t,sol(:,1))
xlabel('Time')
ylabel('Temperature u(0,t)')
title('Temperature change at center of sphere')

function [c,f,s] = heatsphere(r,t,u,dudr)
D = 1e-4;
c = 1;
f = D*dudr;
s = 0;
end
%----------------------------------------------
function u0 = heatic(r)
u0 = 0;
if r==1 
  u0 = 275;
end
end
%----------------------------------------------
function [pl,ql,pr,qr] = heatbc(rl,ul,rr,ur,t)
pl = 0.0;
ql = 1.0;
pr = ur - 275;
qr = 0;
end
%----------------------------------------------

[Juliana]

thank you for the example!! this is definitely helpful

however it is still unclear to me what pl, ql, pr, and qr signify, so that i can translate my system into the pdesolver.

thanks again

[Torsten]

"l" refers to the left boundary point, "r" refers to the right boundary point.

ul is T at r=0, ur is T at r=1.

The general boundary condition is

p + q*f = 0.

In the above case, f = D*dT/dr, thus

p + q*D*dT/dr = 0.

Left boundary point (r=0) with boundary condition dT/dr = 0 gives

pl = 0, ql = 1, e. g., because this leads to 0 + 1*D*dT/dr = 0 or dT/dr = 0.

Right boundary point (r=1) with boundary condition T = 275 gives

pr = ur - 275, qr = 0 because this leads to (T - 275) + 0*D*dT/dr = 0 or T = 275.

[Juliana]

wow okay that makes sense. i cannot thank you enough!!