I am using an ode45 with options but my time vector still doesn't have the 1st step as I defined. I should have t=[0 5 15 25 35...] But I have t=[0 10 20 30...]
tspan=[1 10000]
options = odeset('NormControl','on','RelTol',simul.RelTol,'AbsTol',simul.AbsTol,'InitialStep',5,'Refine',1,'stats','off','MaxStep',MaxStp,'OutputFcn',@IntegrationFcn);
[ t , y ] = ode45( @dy , tspan , y_0 , options );

2 commentaires

madhan ravi
madhan ravi le 21 Oct 2018
Modifié(e) : madhan ravi le 21 Oct 2018
Upload your code for dy
Francesco Fortunato
Francesco Fortunato le 21 Oct 2018
Sorry, I am not sure I understand what you mean

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 Réponse acceptée

Walter Roberson
Walter Roberson le 21 Oct 2018

1 vote

When you pass in a tspan that has two elements then ode45 will automatically pick times to evaluate at as needed to maintain accuracy.
You can pass a vector of times instead. ode45 will output for those times. It will continue to evaluate at whichever points it wants to maintain accuracy, but it will calculate for the times given and report only for those.

2 commentaires

Francesco Fortunato
Francesco Fortunato le 21 Oct 2018
Thank you for your reply, Could I try to get this a bit straighter? If my tspan is a vector tspan=[5 10 20 30 …] ODE45 will select the point to maintain the accuracy but will output the solution at the instants in tspan; If my tspan has two elements ODE45 will select the points and output the solution at those points. Is it correct?
Right. But you would want tspan starting from zero in most conditions, as the first element of tspan tells it the time at which the boundary conditions apply. So for example,
tspan = [0 5 10:10:10000];

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