Unable to meet integration tolerances

13 views (last 30 days)
Nishant Gupta
Nishant Gupta on 12 Jul 2020
Commented: Bill Greene on 12 Jul 2020
I am using the 'pdepe' solver in Matlab but I get this error-
Warning: Failure at t=1.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t.
> In ode15s (line 653)
In pdepe (line 289)
In differential (line 5)
Warning: Time integration has failed. Solution is available at requested time points up to t=1.000000e+00.
> In pdepe (line 303)
In differential (line 5)
Here's the code I have-
x = linspace(0,10,20);
t = linspace(1, 1000, 100);
m = 0;
sol = pdepe(m, @pde1pde, @pde1ic, @pde1bc, x, t)
function [c,f,s] = pde1pde(x, t, u, dudx)
rho = 906;
Cp = 0.4365;
MW = 104.15;
dH = -17800;
k = (1.22-0.002*u)/36;
xp = 1-x;
A0 = 1.964*(10^5)*exp(-10040./u);
A1 = 2.57-5.05*u*(10^(-3));
A2 = 9.56-1.76*u*(10^(-2));
A3 = -3.03+7.85*u*(10^(-3));
A = A0*exp(A1*(xp) + A2*(xp^2) + A3*(xp^3));
c = (rho*Cp)/k;
f = dudx;
s = (((rho/MW)*x)^(2.5))*A*(-dH)/k;
function u0 = pde1ic(x)
T0 = 20 + 273.15;
u0 = T0;
function [pl, ql, pr, qr] = pde1bc(xl, ul, xr, ur, t)
pl = 0;
ql = 1;
pr = 0;
qr = 1;
I have tried changing the arguements of t = linspace(1,1000, 100) but I get always get the same error. The solution is provided only for the first value of the vector t and it is equal to T0 irrespective of the value of t. Any hellp is appreciated

Answers (1)

Bill Greene
Bill Greene on 12 Jul 2020
Your equations have a fundamental error. Just calculate the "s" term along the length at the initial temperature and the problem will be obvious.
Bill Greene
Bill Greene on 12 Jul 2020
I suggest you edit your question to describe your equations in mathematical form, defining all symbols, and including boundary and initial conditions.

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!