Thanks for the clarifications,
On 04/13/2015 06:28 PM, Aaron Meurer wrote:
> On Fri, Apr 10, 2015 at 3:54 AM, Carsten Knoll <[email protected]> wrote:
>>
>> 1. What does an object like
>>
>> Integral(F(t), (t, 0))
>>
>> mean?
>
> It means the integral of F(t), evaluated at t = 0. It's the same as
> Integral(F(t), (t, a, 0)) except we don't want to create a new
> arbitrary symbol a.
>
> Normally this is not a well-defined concept, but for solutions from
> dsolve, it is, because there is always a + C arbitrary constant factor
> in the solution that absorbs the term with a.
OK. But anyway: wouldn't it be simpler for dsolve (from user point of
view) to return a Integral object with two boundary values the lower
being zero and the upper being t. If I see it right, this formulation of
the solution has the same level of generality but the advantage that it
is possible to perform useful calculation with the result. E.g. I want
to determine C1 in terms of the intial value x(0).
>
>>
>> 2. Is the behavior described below intended or is it kind of a bug?
>
> If you mean Integral(F(t), t).subs(t, 0) returning Integral(F(t), (t,
> 0)) instead of vanishing, that is intended. SymPy can't know that
> Integral(F(t), (t, 0)) is supposed to be 0. The only way to get
> constants to absorb into C1 is to use constantsimp.
Here I do not understand how I could "use" constantship? If evaluate
Integral(F(t), (t, 0)) (i.e. I choose t=0) than the integral should be
constant.
In Fact I get
In [22]: sp.Integral(F(t), (t, 0)).is_constant()
Out[22]: True
But this Term is not absorbed into C1. (Should it be?)
>
>>
>> 3. What would be a (generic) workaround? (I was thinking in replacing
>> all Integrals occurring in the rhs-result of dsolve and replacing them
>> by new definite integrals like Integral(F(tau), (tau, 0, t)). But maybe
>> this should happen in dsolve itself.)
>
> If you want to do this, it shouldn't be hard to use replace() to do it.
Thanks for the tip. I was thinking in using subs, but replace seems much
more elegant.
I interpret your suggestion to use replace in my private code (not
altering the behavior of sympy.dsolve), right?
Thanks,
Carsten.
>
> Aaron Meurer
>
>>
>> Tanks and best regards,
>> Carsten.
>>
>>
>> On 04/08/2015 06:18 PM, Carsten Knoll wrote:
>>> I have the following problem
>>>
>>>
>>> In [1]: from sympy import *
>>>
>>> In [3]: x, F, t = symbols('x, F, t')
>>>
>>> In [4]: ode = x(t).diff(t) - F(t)
>>>
>>> In [5]: sol = dsolve(ode).rhs
>>>
>>> In [7]: sol
>>> Out[7]: C1 + Integral(F(t), t)
>>>
>>> In [8]: sol.subs(t, 0)
>>> Out[8]: C1 + Integral(F(t), (t, 0))
>>>
>>>
>>> I want the last expression to evaluate to C1, i.e.
>>>
>>> Integral(F(t), t).subs(t, 0)
>>>
>>> should vanish instead of returning a "strange" object with one bound
>>> specified.
>>>
>>> The application of .doit() does not help either.
>>>
>>>
>>> BTW: IMHO dsolve should return something like
>>>
>>> C1 + Integral(F(tau), (tau, 0, t))
>>>
>>> where tau is a dummy variable
>>>
>>>
>>> Thanks in advance,
>>>
>>> Carsten
>>>
>>
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