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For the ODE $$u''(x) + x^{-n}u(x) = 0$$ can I use Frobenius method for $n > 2$? I think not since we need $x^2 x^{-n}$ to be analytic but is there something else I can use to find a solution? Some generalisation?

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2 Answers 2

up vote 2 down vote accepted

Assume you only want to know the cases that $n$ is integer:

Try let $u(x)=\sum\limits_{k=0}^\infty a_kx^{k+r}$ ,

Then $u'(x)=\sum\limits_{k=0}^\infty(k+r)a_kx^{k+r-1}$

$u''(x)=\sum\limits_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-2}$

$\therefore\sum\limits_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-2}+x^{-n}\sum\limits_{k=0}^\infty a_kx^{k+r}=0$

$\sum\limits_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-2}+\sum\limits_{k=0}^\infty a_kx^{k+r-n}=0$

$\sum\limits_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-2}+\sum\limits_{k=2-n}^\infty a_{k+n-2}x^{k+r-2}=0$

$\sum\limits_{k=2-n}^{-1}a_{k+n-2}x^{k+r-2}+\sum\limits_{k=0}^\infty((k+r)(k+r-1)a_k+a_{k+n-2})x^{k+r-2}=0$

which fails to solve by the conventional version of Frobenius method as we can't get any indicial equations.

But when we try let $u(x)=\sum\limits_{k=0}^\infty a_kx^{r-k}$ ,

Then $u'(x)=\sum\limits_{k=0}^\infty(r-k)a_kx^{r-k-1}$

$u''(x)=\sum\limits_{k=0}^\infty(r-k)(r-k-1)a_kx^{r-k-2}$

$\therefore\sum\limits_{k=0}^\infty(r-k)(r-k-1)a_kx^{r-k-2}+x^{-n}\sum\limits_{k=0}^\infty a_kx^{r-k}=0$

$\sum\limits_{k=0}^\infty(r-k)(r-k-1)a_kx^{r-k-2}+\sum\limits_{k=0}^\infty a_kx^{r-k-n}=0$

$\sum\limits_{k=0}^\infty(r-k)(r-k-1)a_kx^{r-k-2}+\sum\limits_{k=n-2}^\infty a_{k-n+2}x^{r-k-2}=0$

$\sum\limits_{k=0}^{n-3}(r-k)(r-k-1)a_kx^{r-k-2}+\sum\limits_{k=n-2}^\infty((r-k)(r-k-1)a_k+a_{k-n+2})x^{r-k-2}=0$

which can solve by this ''modified version'' of ''Frobenius method'' .

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@doraemonpaul By looking at your answers, I think you'd love the \begin{align} ... \end{align} environment :) –  Pragabhava Nov 15 '12 at 3:23
    
Thank you! Appreciate it –  markclops Nov 15 '12 at 11:24
    
@markclops So the indicial equations gives $r= 0 $and $r = 1$. How can that be? –  soup Nov 18 '12 at 21:33
    
Sorry, last comment should have gone to @doraemonpaul –  soup Nov 19 '12 at 9:49

If $n>2$, then $x=0$ is not a regular singular point. However, you can still get a solution in terms of Bessel functions,

$$ y \left( x \right) ={\it c_1}\,\sqrt {x}\, {{\rm J_{ \left( 2-n \right) ^{-1}}}\left(2\,{\frac {{x}^{-\frac{n}{2}+1}}{n-2}}\right)} +{\it c_2}\,\sqrt {x}\, {{\rm Y_{(2-n)^{-1}}}\left(2\,{\frac {{x}^{-\frac{n}{2}+1}}{n-2}}\right)}\,,$$

where $J$ is Bessel function of the first kind and $Y$ is Bessel function of second kind.

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1  
I know that the general solution of this ODE can be found from eqworld.ipmnet.ru/en/solutions/ode/ode0202.pdf, but the OP wants to know whether Frobenius method can find the general solution of this ODE or not, rather than just want to know the general solution of this ODE. Please stop confusing the OP! –  doraemonpaul Nov 15 '12 at 0:03
    
@doraemonpaul: Stop judging people and take care of yourself. I just posted a solution and it is up to him to accept it or not. –  Mhenni Benghorbal Nov 15 '12 at 2:22
    
Thanks, I appreciate your answer. Take it easy everyone! –  markclops Nov 15 '12 at 11:25
    
@markclops: You are welcome. –  Mhenni Benghorbal Nov 15 '12 at 19:14
    
@downvoter: What's the downvote for? –  Mhenni Benghorbal Nov 26 '12 at 8:20

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