# Bessel function p=0

• May 7th 2011, 01:50 PM
Ulysses
Bessel function p=0
Hi there. Well, I'm stuck with this problem, which says:
When p=0 the Bessel equation is: $\displaystyle x^2y''+xy'+x^2y=0$

Show that its indicial equation only has one root and find the Frobenius solution correspondingly. (Answer: $\displaystyle y=\sum \frac{(-1)^n}{ 2^{2n}(n!)^2 }x^{2n}$

Well, this is what I did:

At first I normalized the equation:
$\displaystyle y''+\frac{y'}{x}+y=0$
Then
$\displaystyle P(x)=\frac{1}{ x} \rightarrow xP(x)=1$
$\displaystyle Q(x)=1 \rightarrow x^2Q(x)=x^2$
So x=0 is regular singular point.

Then the indicial equation is: $\displaystyle \alpha(\alpha-1)+p_0\alpha+q_0=0 \rightarrow \alpha^2-alpha+\alpha=0 \rightarrow \alpha=0$

$\displaystyle y=\sum_{n = 0}^\infty a_n x^n \rightarrow y'=\sum_{n = 1}^\infty a_n n x^{n-1} \rightarrow y''=\sum_{n = 2}^\infty a_n n(n-1) x^{n-2}$

Then $\displaystyle x^2y''+xy'+x^2y=\sum_{n = 2}^\infty a_n n(n-1) x^n+ \sum_{n = 1}^\infty a_n n x^n+\sum_{n = 0}^\infty a_n x^{n+2} =$
$\displaystyle =\sum_{n = 2}^\infty a_n n(n-1) x^n+ \sum_{n = 1}^\infty a_n n x^n+\sum_{n = 0}^\infty a_{n-2}x^n$

So from here I took
$\displaystyle a_n n(n-1)+a_{n-2}=0$

$\displaystyle a_n=\frac{-a_{n-2}}{n(n-1)},n=2k \rightarrow a_{2k}=\frac{-a_{2k-2}}{2k(2k-1)}$

Then I've made some iterations, but I can't find the form that the problem gives as the answer, some of the iterations:

$\displaystyle a_2=\frac{-a_0}{2 },a_4=\frac{a_0}{4.3.2 },a_6=\frac{-a_0}{6.5.4.3.2 }$

So the answer I seem to get is $\displaystyle a_{2k}=\frac{(-1)^k}{(2k)!}a_0$

But I should get $\displaystyle \sum \frac{(-1)^n}{ 2^{2n}(n!)^2 }a_0$ or something like that, which is how the answer the problem gives looks like.

I'm probably doing something wrong, but I couldn't figure it out what it is.

Bye there, thanks for helping!
• May 7th 2011, 02:48 PM
topsquark
Quote:

Originally Posted by Ulysses
Hi there. Well, I'm stuck with this problem, which says:
When p=0 the Bessel equation is: $\displaystyle x^2y''+xy'+x^2y=0$

Show that its indicial equation only has one root and find the Frobenius solution correspondingly. (Answer: $\displaystyle y=\sum \frac{(-1)^n}{ 2^{2n}(n!)^2 }x^{2n}$

Well, this is what I did:

At first I normalized the equation:
$\displaystyle y''+\frac{y'}{x}+y=0$
Then
$\displaystyle P(x)=\frac{1}{ x} \rightarrow xP(x)=1$
$\displaystyle Q(x)=1 \rightarrow x^2Q(x)=x^2$
So x=0 is regular singular point.

Then the indicial equation is: $\displaystyle \alpha(\alpha-1)+p_0\alpha+q_0=0 \rightarrow \alpha^2-alpha+\alpha=0 \rightarrow \alpha=0$

$\displaystyle y=\sum_{n = 0}^\infty a_n x^n \rightarrow y'=\sum_{n = 1}^\infty a_n n x^{n-1} \rightarrow y''=\sum_{n = 2}^\infty a_n n(n-1) x^{n-2}$

Then $\displaystyle x^2y''+xy'+x^2y=\sum_{n = 2}^\infty a_n n(n-1) x^n+ \sum_{n = 1}^\infty a_n n x^n+\sum_{n = 0}^\infty a_n x^{n+2} =$
$\displaystyle =\sum_{n = 2}^\infty a_n n(n-1) x^n+ \sum_{n = 1}^\infty a_n n x^n+\sum_{n = 0}^\infty a_{n-2}x^n$

So from here I took
$\displaystyle a_n n(n-1)+a_{n-2}=0$

You missed the term from the middle summation.
$\displaystyle a_n n( n - 1) + a_n n + a_{n - 2} = 0$

-Dan
• May 7th 2011, 03:28 PM
Ulysses
Thanks! its done now.