Warning message, what does it mean?

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Thanasis Hou
Thanasis Hou 2017-4-12
回答： Somad 2023-10-16
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.
  1 个评论
Bill Greene
Bill Greene 2017-4-13
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.

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回答（1 个）

Somad
Somad 2023-10-16
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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