Fröbenius method

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I'm trying to solve the differential equation:

[tex]x^2y''+2xy'+(x^2-2)y=0[/tex]

using the Fröbenius method.

So I want a solution on the form

[tex]y=\sum_{n=0}^\infty a_{n}x^{n+s}[/tex]

After finding derivatives of y, inserting into my ODE, and after some rearranging:

[tex]\sum_{n=0}^\infty ((n+s)(n+s-1)+2(n+s)-2)a_{n}x^{n+s} + \sum_{n=2}^\infty a_{n-2}x^{n+s}[/tex]

Looking at n=0, assuming [tex]a_{0}[/tex] different from zero, I get two possible values for s:

[tex]s_{1}=1[/tex]
[tex]s_{2}=-2[/tex]

Both giving [tex]a_{1}=0[/tex]

Choosing s=1, I get

[tex]a_{n}=\frac{-a_{n-2}}{(n+s)(n+s-1)+2(n+s)-2}=\frac{-a_{n-2}}{n(n+3)}[/tex]

I know [tex]a_{n}=0[/tex] for odd n, so I'm interrested in finding [tex]a_{n}[/tex] for even n, expressed by [tex]a_{0}[/tex]. Best thing I can come up with is

[tex]a_{n}=\sum_{n=2,4,6,..}^\infty \frac{3(-1)^{n/2}a_{0}}{(n+1)!(n+3)}[/tex]

which seems overly complicated given the simple recursive formula. Any ideas?


Second problem; I want to find another linear independent solution, but not quite sure where to start given my ugly expression for [tex]a_{n}[/tex].
 
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Answers and Replies

  • #2
Your equation is almost certainly some kind of Bessel equation. You should seek solutions in terms of Bessel functions.

Edit:
In fact, it is a Bessel function. Specifically, a spherical Bessel function apparently. There are also general solutions on the page. Remember that in order to obtain the correct expansion, you must use the boundary values of the problem.
 
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  • #3
HallsofIvy
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I'm trying to solve the differential equation:

[tex]x^2y''+2xy'+(x^2-2)y=0[/tex]

using the Fröbenius method.

So I want a solution on the form

[tex]y=\sum_{n=0}^\infty a_{n}x^{n+s}[/tex]

After finding derivatives of y, inserting into my ODE, and after some rearranging:

[tex]\sum_{n=0}^\infty ((n+s)(n+s-1)+2(n+s)-2)a_{n}x^{n+s} + \sum_{n=2}^\infty a_{n-2}x^{n+s}[/tex]

Looking at n=0, assuming [tex]a_{0}[/tex] different from zero, I get two possible values for s:

[tex]s_{1}=1[/tex]
[tex]s_{2}=-2[/tex]

Both giving [tex]a_{1}=0[/tex]

Choosing s=1, I get

[tex]a_{n}=\frac{-a_{n-2}}{(n+s)(n+s-1)+2(n+s)-2}=\frac{-a_{n-2}}{n(n+3)}[/tex]

I know [tex]a_{n}=0[/tex] for odd n, so I'm interrested in finding [tex]a_{n}[/tex] for even n, expressed by [tex]a_{0}[/tex]. Best thing I can come up with is

[tex]a_{n}=\sum_{n=2,4,6,..}^\infty \frac{3(-1)^{n/2}a_{0}}{(n+1)!(n+3)}[/tex]

which seems overly complicated given the simple recursive formula. Any ideas?


Second problem; I want to find another linear independent solution, but not quite sure where to start given my ugly expression for [tex]a_{n}[/tex].
I don't see how you could get a sum like that. Starting with
[tex]a_n= \frac{-a_{n-2}}{n(n+3)}[/tex]
You get, for the first few terms, [itex]a_2= -a_0/(2(5))[/itex], [itex]a_4= a_0/(2(5)(4)(7))[/itex], [itex]a_6= -a_0/(2(4)(5)(7)(6)(9)[/itex]

Okay, it looks like
[tex]a_{2n}= \frac{3(-1)^n a_0}{(2n)!(2(n+1)+ 1)}[/tex]
but there is no sum!
 
  • #4
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I don't see how you could get a sum like that. Starting with
[tex]a_n= \frac{-a_{n-2}}{n(n+3)}[/tex]
You get, for the first few terms, [itex]a_2= -a_0/(2(5))[/itex], [itex]a_4= a_0/(2(5)(4)(7))[/itex], [itex]a_6= -a_0/(2(4)(5)(7)(6)(9)[/itex]

Okay, it looks like
[tex]a_{2n}= \frac{3(-1)^n a_0}{(2n)!(2(n+1)+ 1)}[/tex]
but there is no sum!
You're absolutely right. My mistake.

After reading up on Bessel functions (thanks ObsessiveMathsFreak!!), I've rewritten my recursion formula as

[tex]a_{2n}=\frac{(-1)^{n}a_{0}\Gamma(1+3/2)}{2^{2n}n!\Gamma(n+5/2)}[/tex]

which seems to expand beautifully.
 

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