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
>
>
>

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>

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

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