Filter löschen
Filter löschen

Unable to meet integration tolerances without reducing the step size problem.

18 Ansichten (letzte 30 Tage)
Hello Everyone.
I am facing '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, error for the following code. How can I fix it.
clc
clear all
A =1.4036e-18;
B = 3.4302e-18;
e=1.60217663e-19;
T_e=3;
f = @(t,x) [(sqrt(A/x(1)))*(-x(4)-(0.424)+(1.6977)*exp(-(e/T_e)*(x(1))));
(sqrt(A/x(1)))*(x(4)-(0.663)+(2.6540)*exp(-(e/T_e)*(x(2))));
-(2.994*10^5)*x(4);
((x(1))-(x(2))+x(3)+60*(cos(t)))-2.55*10^7*x(4)];
[t,xa] = ode15s(f,[0, 80],[0,0,-62,0]);
figure(1)
plot(t,xa(:,1),'Color','blue'),grid on;
title('xa1')
xlabel('t'), ylabel('Qsp (Powered Electrode Sheath Voltage)')

Antworten (1)

Torsten
Torsten am 7 Feb. 2023
You define an initial condition for x(1) as 0, but you divide by x(1) in equations 1 and 2.
This won't work.
  3 Kommentare
Md. Golam Zakaria
Md. Golam Zakaria am 7 Feb. 2023
@Steven Lord, If I chose nonzero initial conditions will it solve my problem. I didnt do it because It takes too much time to solve
Torsten
Torsten am 7 Feb. 2023
If I chose nonzero initial conditions will it solve my problem.
Nobody knows. Just try it.

Melden Sie sich an, um zu kommentieren.

Produkte


Version

R2018b

Community Treasure Hunt

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

Start Hunting!

Translated by