Warning message, what does it mean?
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Hello, I am trying to solve a coupled system of PDEs with pdepe solver. However, I do not get the expected results and it pops up a warning.
Warning: Failure at t=0.000000e+00. Unable to meet integration tolerances without reducing the step size below the
smallest value allowed (7.905050e-323) at time t.
> In ode15s (line 668)
In pdepe (line 289)
In BarrierModula (line 10)
Warning: Time integration has failed. Solution is available at requested time points up to t=0.000000e+00.
> In pdepe (line 303)
In BarrierModula (line 10)
Where it is referred to? There is problem with the time interval of simulation, or with the meshing of distance.
Thank you.
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Bill Greene
2017년 4월 13일
That means the ODE solver (ode15s) was unable to converge even the time step in solving the system of ODEs resulting from spatial discretization of the PDE. Usually this indicates a fundamental error in your definition of the PDE, boundary conditions, or initial conditions. Impossible to tell more from just this error message.
답변 (1개)
Somad
2023년 10월 16일
Hi, i got this problem, could anyone help me please?
> In ode15s (line 662)
In pdepe (line 291)
In pdepeads (line 14)
Warning: Time integration has failed. Solution is available at requested time points up to t=0.000000e+00.
> In pdepe (line 305)
In pdepeads (line 14)
this is my code:
clc; clear; close all;
H = 2.5; % length of column [m]
time = 100; % time [s]
x = linspace(0,H,20);
t = linspace(0,time,20);
m = 0;
sol = pdepe(m,@pdefun,@pdeic,@pdebc,x,t);
function [c,f,s] = pdefun(x, t, u, dudx)
epsG = 1;
epsL = 1;
uG = 3.262116752245617;
uL = 1;
aw = 2.858892361301752e+03;
NCO2 = 1.285712213436322e-04;
NH2O = 1.437958044551129e-06;
CCO2g = 3.846789286919412;
CCO2l = 2;
h = 1;
cpCO2 = 38;
Habs = 90e3;
HH20 = 40e3;
c = ones(4,1);
f = [-uG/epsG * dudx(1);
-uL/epsL * dudx(2);
-uG/epsG * dudx(3);
-uL/epsL * dudx(4)];
s = [-aw*NCO2/epsG; %-CCO2g/epsG * dUgdz
aw*NCO2/epsL;
aw/(CCO2g*cpCO2) * h/epsG *(u(4)-u(3));
aw/(CCO2l*cpCO2) * (h*(u(4)-u(3)) - Habs*NCO2 - HH20*NH2O)/epsL];
end
function u0 = pdeic(x)
u0 = zeros(4,1);
end
function [pl,ql,pr,qr] = pdebc(xl,ul,xr,ur,t)
CAin = 10; D = 0.01; U = 0.01;
pl = [ul(1)-CAin; ul(2); ul(3); ul(4)];
ql = [-D/U; 0; 0; 0];
pr = zeros(4,1);
qr = ones(4,1);
end
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