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.