Unable to meet integration tolerances
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
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-
clc
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;
end
function u0 = pde1ic(x)
T0 = 20 + 273.15;
u0 = T0;
end
function [pl, ql, pr, qr] = pde1bc(xl, ul, xr, ur, t)
pl = 0;
ql = 1;
pr = 0;
qr = 1;
end
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
0 Kommentare
Antworten (1)
Bill Greene
am 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.
2 Kommentare
Bill Greene
am 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.
Siehe auch
Kategorien
Mehr zu Eigenvalue Problems 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!