The last entry in tspan must be different from the first entry. ODE45

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Anil Rajajapantulu
Anil Rajajapantulu 2019-6-11
评论： Anil Rajajapantulu 2019-6-12
Hello,
I am using ODE45 to solve a ODE, where the time steps are very small. while sloving it shows error the last entry in tspan must be different from the first entry. I went back a chek, but my intial and final times are not same. Can someone provide me an alternative way ?
tInitial =0.48965737046814611799910865741481
tFinal =0.48965737046814611799928732454269
tspan = [tInitial tFinal]
[t,N] = ode45(@(t,N) thisObj.diffusionalGrowth(), tspan, oldDia);

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Bjorn Gustavsson
Bjorn Gustavsson 2019-6-11
tInitial and tFinal differ in the ~22nd decimal. That is not much for a double-precision representation. Why not run the simulation from 0 to 1.78e-22? And are that the entire time-scale you want to integrate the DE? If that time is in units of seconds it is a very short time, if it is in hours it is still a very short time, even if it is in units of years it is still about a 5 femtoseconds...
HTH
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Bjorn Gustavsson
Bjorn Gustavsson 2019-6-11
That your system is dynamic is not making it impossible to integrate the system for longer periods of time. You can do things like this:
function dN = your_ode(t,N,additional,arguments)
Dynamic_variations = other_function(t,y,arguments);
dN_additional = additional(1)*cos(additional(2)*t);
dN = N + Dynamic_variations + dN_additional;
end
Then integrate from your start-time for the entire duration:
additional = [2,4*pi];
arguments = {123,'s'};
t_span = [0 exp(pi)];
N0 = 1;
[t,N] = ode45(@(t,N) your_ode(t,N,additional,arguments), t_span, N0);
If you have discrete events it might be possible to use the events-handling capacity - see ballode.m for an example of how to handle this.
HTH

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