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Re: Power series solution to differential equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg41572] Re: [mg41533] Power series solution to differential equations
  • From: Selwyn Hollis <selwynh at earthlink.net>
  • Date: Tue, 27 May 2003 01:47:33 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Oops, forgot to mention one thing:

    w[0]:=0

--Selwyn

On Monday, May 26, 2003, at 01:34  PM, Selwyn Hollis wrote:

> 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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