The Matlab event function doesn't stop the integration

Hi,

I have a ode system that I solve with Matlab. I want to find the steady state of the system and to do so, I use the event function as described here .

But some times, the solver doesn't stop even if the criterium is achieved. For example, if you solve the following system with x0 = 10 the solver will stop before 2000 but with x0 = 0.0001 it won't.

The event function (eventfun_t.m)

function [x,isterm,dir] = eventfun_t(t,y)
    dy = test_systeme(t,y);
    x = norm(dy) - 1e-3;

      isterm = 1;
      dir = 0;  %or -1, doesn't matter
  end

The system (test_systeme.m)

function dx = test_systeme(t,x)
    v = x./(x+1);
    dx = -v;
end

Solve the system

x0 = 10;
eventfonction = @(t,y) eventfun_t(t,y);
optionsode=odeset('Events',eventfonction);
[t x]=ode15s(@(t,x) test_systeme(t,x),[0 2000],x0,optionsode);

I suppose it's because with x0 = 0.0001 norm(dy) is already lower than 1e-3 but in that case, how can I stop the solver without checking by myself ?

[Jan]

Did you consider Andrew Newell's comment?

 You'd need to check that the norm is not less than the
 tolerance before the integration starts.

[Cécile Moulin]

Hi, thanks for the answer. I didn't read that comment. Where is it ? But I thought about it. And I wonder if there is a way to do it only with the event function. I even tried this way :

function [x,isterm,dir] = eventfun_t(t,y)
    dy = test_systeme(t,y);
    if t~= 0 && x < 1e-3
       x = 0;
    else
       x = 1;
    end
      isterm = 1;
      dir = 0;  %or -1, doesn't matter
  end

to obtain a fisrt value different from 0, but it doesn't work.

[Jan]

This is the first comment to the answer in the thread you've posted a link to.

    dy = test_systeme(t,y);
    if t~= 0 && x < 1e-3
      x = 0

This cannot work, because x is undefined. Even if you mean dy instead of x, this is not a useful event function. Remember, that the event is triggered, when the 1st output of this functions becomes 0. Because it is very unlikely, that 0 is hit exactly, the integrators detect a change of the sign of this value and adjust the step size, to locate the exact zero crossing. Then "if t~= 0 && x < 1e-3, x = 0;" is not useful, because this is not a zero-crossing at t==0 even for tiny dy.

I recommend to read the instructions again: doc: ode-event-location

I did not test it, but Andrew's comment sounds reasonable: The ODE integrators do not detect an event at the initial step. So run the event function manually before starting the integration. This is cheap and easy, so I do not see a reason to implement any further tricks.

[Cécile Moulin]

Ok thanks for the answer.

Answer 1

[Cécile Moulin]

The anwser is in the comments of the question : there is no trick and I have to check before the integration if the conditions are already achieved.