[SciPy-User] integrate.ode sets t0 values outside of my data range

[Warren Weckesser]

On Wed, Jul 30, 2014 at 4:05 AM, camille chambon <camillechambon at yahoo.fr>
wrote:

>
> camille chambon <camillechambon <at> yahoo.fr> writes:
>
> >
> >
> > Hello,
> >
> > I would like to solve the ODE dy/dt = -2y + data(t), between t=0..3, for
> y(t=0)=1.
> >
> > I
> >  wrote the following code:
> >
> > import numpy as npfrom scipy.integrate import odeintfrom
> scipy.interpolate
> import interp1dt = np.linspace(0, 3, 4)data = [1, 2, 3,
> 4]linear_interpolation = interp1d(t, data)def func(y, t0):
> >     print 't0', t0    return -2*y + linear_interpolation(t0)soln =
> odeint(func, 1, t)
> >
> > When I run this code, I get several errors:
> >
> >
> > ValueError: A value in x_new is above the interpolation
> range.odepack.error: Error occurred while calling the Python function
> named func
> >
> > My interpolation range is between 0.0 and 3.0.
> >
> > Printing the value of t0 in func, I realized that t0 is actually
> sometimes
> above my interpolation range: 3.07634612585, 3.0203768998, 3.00638459329,
> ...
> >
> > I have a few  questions:
> >
> > - how does integrate.ode makes t0 vary? Why does it make t0 exceed the
> infimum (3.0) of my interpolation range?
> >
> > - in spite of these errors, integrate.ode returns an array which seems to
> contain correct value. So, should I just catch and ignore these
> >  errors?
> >
> >
> > - if I shouldn't ignore these errors, what is the best way to avoid them?
> >
> >
> > 2 suggestions for the last question:
> >
> > - in interp1d, I could set bounds_error=False and fill_value=data[-1]
> since the t0 outside of my interpolation range seem to be closed to t[-1]:
> > linear_interpolation = interp1d(t, data, bounds_error=False,
> fill_value=data[-1])
> > But first I would like to be sure that with any other func and any other
> data the t0 will always remain closed to t[-1]. For example, if
> integrate.ode chooses a t0 below my interpolation range, the fill_value
> would be still data[-1], which would not be correct. Maybe to know how
> integrate.ode makes t0 vary would help me to be sure of that (see my first
> question).
> >
> > - in func, I could enclose the linear_interpolation call in a try/except
> block, and, when I catch a ValueError, I recall  linear_interpolation but
> with t0 truncated:
> >
> > def func(y, t0):
> >     try:
> >  interpolated_value = linear_interpolation(t0)
> >     except ValueError:
> >
> >         interpolated_value = linear_interpolation(int(t0)) # truncate t0
> >
> >
> >     return -2*y + interpolated_value
> >
> >
> >
> >
> >
> > At least this solution permits linear_interpolation to still raise an
> exception if integrate.ode makes a t0 above 4.0 or below -1.0. I can then
> be
> alerted of incoherent behavior. But it is not really readable and the
> truncation seems to me a little arbitrary by now.
> >
> >
> > Maybe I'm just overthinking about these errors. Please let me know.
> >
> > Thanks in advance.
> >
> > Cheers,
> >
> > Camille
> >
> >
> >
> > _______________________________________________
> > SciPy-User mailing list
> > SciPy-User <at> scipy.org
> > http://mail.scipy.org/mailman/listinfo/scipy-user
> >
>
> Sorry, I made a typo in the range of t: I want to solve the ODE dy/dt = -2y
> + data(t) between t=0..3, and not between t=0..4. I corrected my original
> message.
> Cheers,
> Camille
>


Camille,

I added an answer to your question on StackOverflow:
http://stackoverflow.com/questions/25031966/integrate-ode-sets-t0-values-outside-of-my-data-range/

Summary:
* It is normal for the ODE solver to evaluate your function at points
beyond the last requested time value.
* One way to avoid the interpolation problem is to extend the interpolation
data linearly, using the last two data point.


Warren



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