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

[Oleksandr Huziy]

Salut Camille:


Just don't ask for the solution in the last point, since in order to
calculate a value of the solution at that point it has to go a bit
beyond (y should be defined from both sides of ti for the derivative
to exist in ti (this is a necessary condition)), but interpolation
does not really define values beyond the right limit.

Here I've modified your example:
http://nbviewer.ipython.org/github/guziy/PyNotebooks/blob/master/ode_demo.ipynb

Cheers



2014-07-30 4:05 GMT-04:00 camille chambon <camillechambon at yahoo.fr>:
>
> 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
>
>
> _______________________________________________
> SciPy-User mailing list
> SciPy-User at scipy.org
> http://mail.scipy.org/mailman/listinfo/scipy-user



-- 
Sasha