2
$\begingroup$

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?

2 Answers 2

2

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'' .

  • 0
    @doraemonpaul By looking at your answers, I think you'd love the `\begin{align} ... \end{align}` environment :)2012-11-15
  • 0
    @markclops So the indicial equations gives $r= 0 $and $r = 1$. How can that be?2012-11-18
  • 0
    Sorry, last comment should have gone to @doraemonpaul2012-11-19
0

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.

  • 1
    I know that the general solution of this ODE can be found from http://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!2012-11-15
  • 0
    @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.2012-11-15
  • 0
    Thanks, I appreciate your answer. Take it easy everyone!2012-11-15
  • 0
    @markclops: You are welcome.2012-11-15
  • 0
    @downvoter: What's the downvote for?2012-11-26
  • 0
    @markclops How is this answer helping you to understand the problem and the approach(es) available to solve it? I am curious.2012-12-07