Solving second order PDE

Hi, I am trying to solve the following pde with initial condition CA(0,r)=0 and boundary conditions CA(t,0)=F(t) and CA(t,5)=0.

, where D_A and gamma_A are known constants.

I tried using pdepe but was told that left boundary condition would be ignored when m=1 (cylindrical symmetry).

Then I tried discretizing space variable r before using ode15s, but was confused about how to construct the equation exactly.

Can anybody help?

[darova]

Can you please exaplain more about your equation?

Is it

or

[Lotuny Lee]

@darova Hi, it is a reaction-diffusion equation. And I believe the middle part should be .

Answer 1

[Bill Greene]

The reason that pdepe imposes a boundary condition of the flux equal zero at the

center is that this is required for the problem to be mathematically well-posed.

Imposing a prescribed temperature at the center would require that the flux go to

infinity there.

An easy way to understand this is to solve the problem with the left end a small distance

from the center and with a fine mesh. I have attached a short script below that shows this.

function matlabAnswers_6_27_2020
r0=1e-6;
x = linspace(r0,1,1000);
tf=1;
t = linspace(0,tf,40);

pdeFunc = @(x,t,u,DuDx) heatpde(x,t,u,DuDx);
icFunc = @(x) heatic(x);
bcFunc = @(xl,ul,xr,ur,t) heatbcDirichlet(xl,ul,xr,ur,t);

m=1;
sol = pdepe(m, pdeFunc,icFunc,bcFunc,x,t);

figure; plot(t, sol(:,end)); grid on; title 'Temperature at outer surface'
figure; plot(t, sol(:,1)); grid on; title 'Temperature at center'
figure; plot(x, sol(end,:)); grid; title 'Temperature at final time'

end

function [c,f,s] = heatpde(x,t,u,DuDx)
c = 1;
f = DuDx;
s = 0;
end

function u0 = heatic(x)
u0 = 0;
end

function [pl,ql,pr,qr] = heatbcDirichlet(xl,ul,xr,ur,t)
pl = ul-1;
ql = 0;
pr = 0;
qr = 1;
end

[Lotuny Lee]

Thanks a lot! This looks like a feasible alternative.

Answer 2

[J. Alex Lee]

Does this help: https://www.mathworks.com/help/matlab/ref/pdepe.html

I believe that pdepe is available with base matlab.

It appeas to be able to do the space discretization automatically for you if you

[Lotuny Lee]

Thanks for answering, but my issue with pdepe is that my boundary condition would be ignored.