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Re: Mathematica and Mathieu DEQ Results?

  • To: mathgroup at
  • Subject: [mg91190] Re: Mathematica and Mathieu DEQ Results?
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at>
  • Date: Fri, 8 Aug 2008 07:18:27 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <g7ed1f$2fh$>

amzoti wrote:

> 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]
>  2. However, according to <
> 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)
>  a. I am confused why I am getting these totally different results.
>  b. The paper I am reading < 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?
>  Why can't Mathematica get the same result (or is it that the DEQ does
> not have a unique solution based on some properties of Mathieu
> functions)?
>  Any insights are appreciated!

It seems you have misread or interpreted what the arguments of Mathieu 
differential equation are and how they relate to the general solution.

If the cosine has for argument two times the complex variable z, i.e. 
cos(2z) (Cos[2 z] in Mathematica syntax), then the solution is as follows:

In[1]:= DSolve[y''[z] + (a - 2 q Cos[2 z]) y[z] == 0, y, z]

Out[1]= {{y ->
    Function[{z}, C[1] MathieuC[a, q, z] + C[2] MathieuS[a, q, z]]}}

In your case the cosine does match this pattern, so the expected 
solution cannot be so straightforwardly deduced as above.

However, note that the solution returned by Mathematica is correct.

In[2]:= mathieu = x''[t] +
          (4*Pi^2 - (2/5)*((-Pi^2)*Cos[(2*(Pi*t))/100]))*x[t] == 0;
sol = DSolve[mathieu, x, t]
mathieu /. sol[[1]] // Simplify

Out[3]= {{x ->
    Function[{t}, C[1]*MathieuC[40000, -2000, (Pi*t)/100] +
            C[2]*MathieuS[40000, -2000, (Pi*t)/100]]}}

Out[4]= True

The following tutorial might be of some help:

-- Jean-Marc

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