Hi Leonid,
Thank you for spending time for my problem.
Your answer is very clear.
Regards,
Dabitto
Le jeudi 28 décembre 2017 19:48:05 UTC+1, Leonid Kovalev a écrit :
>
> The problem is that the solver needs to know the sign of a certain
> expression in terms of the coefficients (the discriminant of a polynomial),
> and it cannot determine the sign based on the information given, as the
> discriminant ends up being `D = (1/(c2*r2) - 1/(c1*r1))**2`. Symbols are
> not automatically assumed real; so the square of some expression like that
> could even be negative. It's a good idea to declare symbols as real, by
>
> sym.symbols('i, r1, c1, r2, c2, t', real=True)
>
> but this doesn't solve the problem because it might happen that c1*r1 ==
> c2*r2, and then D is 0.
>
> I just edited the source to tell it which branch to choose, and got the
> result:
>
> [Eq(x1(t), C1*exp(t*(-(c1*r1 - c2*r2)/(2*c1*c2*r1*r2) - (c1*r1 + c2*r2)/(2
> *c1*c2*r1*r2)))/(c1*r1) + C2*exp(t*((c1*r1 - c2*r2)/(2*c1*c2*r1*r2) - (c1*r1
> + c2*r2)/(2*c1*c2*r1*r2)))/(c1*r1) + i*(r1 + r2)), Eq(x2(t), C1*(1/(c1*r1)
> - (c1*r1 - c2*r2)/(2*c1*c2*r1*r2) - (c1*r1 + c2*r2)/(2*c1*c2*r1*r2))*exp(t
> *(-(c1*r1 - c2*r2)/(2*c1*c2*r1*r2) - (c1*r1 + c2*r2)/(2*c1*c2*r1*r2))) +
> C2*(1/(c1*r1) + (c1*r1 - c2*r2)/(2*c1*c2*r1*r2) - (c1*r1 + c2*r2)/(2*c1*c2
> *r1*r2))*exp(t*((c1*r1 - c2*r2)/(2*c1*c2*r1*r2) - (c1*r1 + c2*r2)/(2*c1*c2
> *r1*r2))) + i*r2)]
>
> So at least you have your solution... Unfortunately I cannot think of any
> workaround at user level. I think the handling of D here
> <https://github.com/sympy/sympy/blob/master/sympy/solvers/ode.py#L6640>
> should be changed in two ways:
>
> a) test it with D.is_positive etc instead of D>0 which throws the error
> you got
> b) if the sign can't be determined, return a Piecewise object listing the
> options with appropriate cases:
>
> Piecewise((sol1, D>0), (sol2, D<0), (sol3, Eq(D, 0)))
>
>
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