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Matlab: How to write the nonlinear term using pdenonlin solver

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Fehaid Alshammari
Fehaid Alshammari le 13 Fév 2015
Clôturé : MATLAB Answer Bot le 20 Août 2021
I'm solving a nonlinear diffusion-advection equation in 2D spatial domain using pdenonlin . It seems straightforward to write the nonlinear term if it's of the form u^2, u^3, .. . The nonlinear term in my equation is of the form 2u/(4+u) . When I use char to pass this term to the f cofficient, the solutions don't seem right! I think I'm doing something wrong here. The equation I'm solving is u_xx+u_yy-2u/(4+u)=0 . What's wrong in my code? Thanks
c = 1;
a = 0;
f = char('-2*u./(4+u)')
d = 1; xmin=0;xmax=0.575;ymin=0;ymax=0.05315;ymax2=0.066;
gdm = [3;4;xmin;xmax;xmax;xmin;ymax;ymax2;ymin;ymin];
g = decsg(gdm, 'S1', ('S1')');
hmax = .1; % element size
[p, e, t] = initmesh(g, 'Hmax', hmax);
[p,e,t] = refinemesh(g,p,e,t);
[p,e,t] = refinemesh(g,p,e,t);
[p,e,t] = refinemesh(g,p,e,t);
[p,e,t] = refinemesh(g,p,e,t);
[p,e,t] = refinemesh(g,p,e,t);
numberOfPDE = 1;
pb = pde(numberOfPDE);
% Create a geometry entity
pg = pdeGeometryFromEdges(g);
bc1 = pdeBoundaryConditions(pg.Edges(1),'u',100);
bc2 = pdeBoundaryConditions(pg.Edges(2),'u',66);
bc3 = pdeBoundaryConditions(pg.Edges(3),'u',11);
bc4 = pdeBoundaryConditions(pg.Edges(4),'g',0);
pb.BoundaryConditions = [bc1,bc2,bc3,bc4];
u = pdenonlin(pb,p,e,t,c,a,f, 'jacobian', 'lumped');
figure;
hold on
pdeplot(p, e, t, 'xydata', u, 'contour', 'off', 'colormap', 'jet(99)');
title 'chemical Diffusion, Steady State Solution'
xlabel 'X-coordinate, cm'
ylabel 'Y-coordinate, cm'
  2 commentaires
Richard Garner
Richard Garner le 14 Fév 2015
I think you want to treat your nonlinear term as an "acoeff*u" term. That is, you use a_coeff=1/(4+u). fcoeff should be zero. Now, that having been said, I don't know what happens if 4+u happens to be zero during the course of the calculation. But that would be an algorithm issue. Or perhaps an inherent mathematical, or even numerical, issue.
Fehaid Alshammari
Fehaid Alshammari le 16 Fév 2015
@Richard Thanks .. My 'u' always positive so that shouldn't be a problem

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