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).
PDEPE: Unable to meet integration tolerances
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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
Akzeptierte Antwort
Grzegorz Knor
am 15 Sep. 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
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu PDE Solvers finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!