Re: Mathematica and Mathieu DEQ Results?
- To: mathgroup at smc.vnet.net
- Subject: [mg91173] Re: Mathematica and Mathieu DEQ Results?
- From: Albert Retey <awnl at gmx-topmail.de>
- Date: Fri, 8 Aug 2008 07:15:16 -0400 (EDT)
- References: <g7ed1f$2fh$1@smc.vnet.net>
Hi,
> 1. I am using Mathematica, to solve the following DEQ:
> DSolve[x''[t] + (4[Pi]^2 - 2((-Pi^2/5) Cos[2 (Pi t)/100])))x[t] == 0,
> x[t], t] (1)
>
> This results in a solution of even and odd Mathieu functions as
> follows:
>
> x[t] = C[1] MathieuC[40000, -2000, Pi*t/100] + C[2] MathieuS[40000,
> -2000, Pi*t/100]
Looks correct (or at least consistent within Mathematica :-) to me
(crosschecking symbolic results is always a good idea, whatever system
you are using...):
Simplify[ReleaseHold[
Hold[D[x[t], t,
t] + (4 \[Pi]^2 - 2 (-\[Pi]^2/5) Cos[2 \[Pi] t/100]) x[t] ==
0] /. x[t] ->
C[1] MathieuC[40000, -2000, (\[Pi] t)/100] +
C[2] MathieuS[40000, -2000, (\[Pi] t)/100]
]]
> 2. However, according to <http://mathworld.wolfram.com/
> MathieuDifferentialEquation.html,
> the result is:
>
> x[t] = C[1] MathieuC[4 Pi ^2, -Pi^2/5, Pi*t/100] + C[2] MathieuS[4 Pi
> ^2, -Pi^2/5, Pi*t/100] (2)
Can't find that result there, and
FullSimplify[ReleaseHold[
Hold[D[x[t],t,t]+(4\[Pi]^2-2(-\[Pi]^2/5) Cos[2 \[Pi]
t/100])x[t]==0]/.x[t]->C[1] MathieuC[4 Pi^2,-Pi^2/5,Pi*t/100]+C[2]
MathieuS[4 Pi^2,-Pi^2/5,Pi*t/100]
]]
indicates that this is no solution. (Note that in general
differentiating symbolically is much easier than solving differential
equations symbolically, so these checks have proven to be very reliable
for me in the past).
I think you need to be more careful when applying the coordinate
transformation from t -> t Pi/100. It will give a factor (100/Pi)^2 for
the derivative, which can be rearranged to disappear by modifying a and
q. So I think the result from Mathematica is perfectly correct, maybe it
is easier to see what happens with this:
DSolve[{x''[t] + (a - 2 q Cos[2 b t]) x[t] == 0}, x[t], t]
> a. I am confused why I am getting these totally different results.
Hm, I don't think Mathematica has anything to do with this :-)
> b. The paper I am reading <epsppd.epfl.ch/Roma/pdf/P2_091.pdf shows
> three different plots that match with (2) - where I am able to
> duplicate the results/plots of the paper exactly using this method.
>
> Can someone out there shed light on why this is the case?
No, but if they really get the above result I would wonder what was
going wrong in the reviewing process :-)
> Why can't Mathematica get the same result
Hm, you could teach it to get wrong results, but it tries to be correct
by default.
> (or is it that the DEQ does
> not have a unique solution based on some properties of Mathieu
> functions)?
I'm not an expert in Mathieu functions, so I can't tell, but the
following indicates that there are no such properties (this of course
also relies on a correct implementation of them in Mathematica):
Solve[
Flatten@Table[{
C[1] MathieuC[40000, -2000, (\[Pi] t)/100] +
C[2] MathieuS[40000, -2000, (\[Pi] t)/100] ==
C[1] MathieuC[4 Pi^2, -Pi^2/5, Pi*t/100] +
C[2] MathieuS[4 Pi^2, -Pi^2/5, Pi*t/100]
},
{t, 1, 2}
],
{C[1], C[2]}
]
{{C[1] -> 0, C[2] -> 0}}
hth,
albert