PDEPE solver with off-diagonal coefficients

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MG
MG on 26 Nov 2018
Commented: MG on 27 Nov 2018
Hello,
referring to the pdex4 example, I am trying to understand if the function f in this part of the code:
function [c,f,s] = pdex4pde(x,t,u,DuDx)
c = [1; 1];
f = [0.024; 0.17] .* DuDx;
y = u(1) - u(2);
F = exp(5.73*y)-exp(-11.47*y);
s = [-F; F];
can be modified by substituting the vector [0.024; 0.17] with a 2x2 matrix.
Basically, I would need to solve a system like this:
% |1| |u1| | d11*D(u1)/Dx + d12*D(u2)/Dx |
% | | .* D_ | | = D_ | |
% |1| Dt |u2| Dx | d21*D(u1)/Dx + d22*D(u2)/Dx |
%
% --- --- ------------------
% c u f(x,t,u,Du/Dx)
Also, is the PDEPE solver the most appropriate for this problem?
Thanks in advance
MG
  4 Comments
Torsten
Torsten on 27 Nov 2018
Edited: Torsten on 27 Nov 2018
The boundary conditions have the general form
p + q*f = 0
where f in your example above is
f=[d11*DuDx(1)+d12*DuDx(2);d21*DuDx(1)+d22*DuDx(2)].
Thus you have to specify p(1),q(1), p(2), q(2) such that
p(1) + q(1)*(d11*DuDx(1)+d12*DuDx(2)) = 0
p(2) + q(2)*(d21*DuDx(1)+d22*DuDx(2)) = 0
Best wishes
Torsten.
MG
MG on 27 Nov 2018
Of course, thank you, I am coding this part of the model just now.
Luckily for me, I don't need to add exotic functions in BCs, but only fixed coefficients.
Thank you for your kind support, cheers
MG

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