Solving Bessel Equation: Indicial Equation & Frobenius Solution

[Telemachus]

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

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

Well, this is what I did:

At first I normalized the equation:
$$y''+\frac{y'}{x}+y=0$$
Then
$$P(x)=\frac{1}{ x} \rightarrow xP(x)=1$$
$$Q(x)=1 \rightarrow x^2Q(x)=x^2$$
So x=0 is regular singular point.Then the indicial equation is: $$\alpha(\alpha-1)+p_0\alpha+q_0=0 \rightarrow \alpha^2-alpha+\alpha=0 \rightarrow \alpha=0$$

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

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

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

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

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

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

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

But I should get $$\sum \frac{(-1)^n}{ 2^{2n}(n!)^2 }a_0$$ 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!

 

[HallsofIvy]

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

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

Well, this is what I did:

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


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

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

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

You have not changed the lower limits when you changed n in each sum.
In the last sum you have $a_nx^{n+ 2}$ with n going from 0 to infinity. If your new "n" is the old n+ 2, then you have $a_{n- 2}x^n$ with n going from 2 to infinity. Also, the first sum should have $x^{n-2}$ but you have written $x^n$. If you let the new n be the old n-2, then the term in the first sum is $a_{n+2}(n+2)(n+1)x^n$ with the sum going from 0 to infinity. Finally, in the middle sum you have, again, $x^n$ when it should be $x^{n-1}$. If you let the new n be the old n-1 then you should have $a_{n+1}(n+1)x^n$
That is, you should have
$$\sum_{n = 0}^\infty a_{n+2} (n+2)(n+1) x^n+ \sum_{n = 1}^\infty a_{n+1} (n+1) x^n+\sum_{n = 2}^\infty a_{n-2}x^n$$

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

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

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

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

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

But I should get $$\sum \frac{(-1)^n}{ 2^{2n}(n!)^2 }a_0$$ 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!

 

[Telemachus]

Thanks HallsOfIvy.