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Re: Power series solution to differential equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg41571] Re: [mg41533] Power series solution to differential equations
  • From: Selwyn Hollis <selwynh at earthlink.net>
  • Date: Tue, 27 May 2003 01:47:32 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Okay, here's a good way to get Frobenius series solutions.

Let's use a Bessel equation with this operator:

   T[y_] := t^2*D[y,t,t] + t*D[y,t] + (t^2 - 1/4)*y

The plan is to find solutions of the form y[t]:= t^q (1+w[t]).

    n:=7;

    eqns = 
Thread[CoefficientList[Simplify[t^(1-q)*T[t^q*(1+w[t]+O[t]^n)]], t] == 
0]

    t^q (1+w[t]+O[t]^n) /.
     Solve[eqns, Prepend[Table[D[w[t], {t,j}], {j,1,n-1}], q]/. t->0]

In this example, we get a spurious third solution involving w'''[0], 
which can easily be discarded using something like

    Flatten[If[MemberQ[#1, Derivative[_][w][0], Infinity], {}, #1] & /@ 
%]

Looks like it's time for me to simplify the code in my DETools package 
:-)

-----
Selwyn Hollis
http://www.math.armstrong.edu/faculty/hollis




On Monday, May 26, 2003, at 05:46  AM, Dr. Wolfgang Hintze wrote:

> Given a differential equation of the form
>
> diffeq = a[x] u''[x] + b[x] u'[x] + f[x, u[x]] == 0
>
> where ' means d/dx we assume that u[x] has a power series expansion
> about x0 of the form (t = x-x0)
>
> u[t] = Sum[ c[k] t^(k+z) , {k, 0, Infinity }]
>
> We have to determine z and the coefficients c[k].
>
> Question: what is the best way to tackle this problem in Mathematica?
>
> Any hint is greatly appreciated.
>
> Wolfgang
>
>
>


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