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

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)')

Answers (1)

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 Comments

Yes. Evaluating the ODE function at the initial conditions:
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)];
init = f(0, [0,0,-62,0])
init = 4×1
Inf Inf 0 -2
A function with an infinite rate of change at the initial point seems like a Bad Thing.
@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
If I chose nonzero initial conditions will it solve my problem.
Nobody knows. Just try it.

Sign in to comment.

Categories

Find more on Numerical Integration and Differential Equations in Help Center and File Exchange

Products

Release

R2018b

Asked:

on 7 Feb 2023

Commented:

on 7 Feb 2023

Community Treasure Hunt

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

Start Hunting!