PDEPE: Unable to meet integration tolerances

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Abed Alnaif
Abed Alnaif el 14 de Sept. de 2011
Hi,
I am trying to solve a system of 2 PDEs using Matlab's built-in PDE solver, pdepe. I am being returned an "Unable to meet integration tolerances" warning during the ODE15s routine -- see the bottom of this message to see the specific warning. I read that this warning my occur if the equation has singularities, but I didn't notice any such singularities in my equations.
Below is my code, in a self-contained file. I thought that the problem would be either with the p.h parameter being to large, or that it may be due to the presence of the M^2 term. However, I still get the same warning message even after setting p.h = 1, as well as after removing the M^2 term.
Needless to say, any help is greatly appreciated!
Matlab code:
function unable_to_meet_tolerance()
clear all;
% -------- Model Parameters ----------------------
p.x_m = 100;
p.D_M = 10;
* p.D_E = 1;
p.a_M_1 = 1E-1;
p.a_M_2 = 1;
p.a_E = 1E-5;
p.b_E = 1E-2;
p.T_p2 = 1E-3;
p.h = 2;
p.j = 10; % flux
% -------- Model ---------------------------------
% PDE function
function [c,f,s] = pde(~,~,u,dudx)
M = u(1);
E = u(2);
% differential equations
% see Matlab PDE solver help file for definitions of c, f, and s
c = [1; 1];
f = [p.D_M; p.D_E] .* dudx;
s = [ - ( (1+E) * p.a_M_1 * M ) - ( (1+E) * p.a_M_2 * M^2 )
- ( p.a_E * E ) + ( p.b_E/(1+(M/p.T_p2)) )];
end
% boundary conditions
function [pl,ql,pr,qr] = pde_bc(~,~,~,~,~)
pl = [-p.j; 0];
ql = [1; 1/p.D_E];
pr = [0; 0];
qr = [1/p.D_M; 1/p.D_E];
end
% initial conditions
function u0 = pde_ic(~)
u0 = [0; 0]; % zero initial conditions
end
% -------- Simulation ----------------------------
nsteps_x = 1000;
nsteps_t = 5;
t_fin = 1;
x_mesh = linspace(0,p.x_m,nsteps_x+1);
t_mesh = linspace(0,t_fin,nsteps_t+1);
sol = pdepe(0,@pde,@pde_ic,@pde_bc,x_mesh,t_mesh);
end
This is the warning message:
Warning: Failure at t=4.060671e-03. Unable to meet integration tolerances without
reducing the step size below the smallest value allowed (1.387779e-17) at time t.
> In ode15s at 819
In pdepe at 320
In unable_to_meet_tolerance at 53
Warning: Time integration has failed. Solution is available at requested time points
up to t=0.000000e+00.
> In pdepe at 326
In unable_to_meet_tolerance at 53
  1 comentario
Abed Alnaif
Abed Alnaif el 14 de Sept. de 2011
I forgot to mention, as it may or may not be clear from the code above, I am going for zero initial conditions and no-flux boundary conditions (with the exception of M, which has a flux of p.j at x=0).

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Grzegorz Knor
Grzegorz Knor el 15 de Sept. de 2011
Are you sure that boundary conditions are correct?
Maybe one minus is unnecessary?
% boundary conditions
function [pl,ql,pr,qr] = pde_bc(~,~,~,~,~)
pl = [p.j; 0];
ql = [1; 1/p.D_E];
pr = [0; 0];
qr = [1/p.D_M; 1/p.D_E];
end
  1 comentario
Abed Alnaif
Abed Alnaif el 15 de Sept. de 2011
Grzegorz, you are correct (an embarrassing mistake on my part!). Thanks for your help; it works now after I made the change that you suggested!

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