# Question on deriving Bessel's equation

1. May 17, 2013

### yungman

$$\hbox { Bessel's equation of order p: }\;x^2y''+xy'+(x^2-p^2)y=0\;\hbox { and let}\;y=\sum_0^{\infty}C_nx^{n+r}$$
$$\Rightarrow\; \left[\sum_0^{\infty}C_k(k+r)(k+r-1)x^k+\sum_0^{\infty}C_k(k+r)x^k+\sum_2^{\infty} C_{k-2}x^k-p^2\sum_0^{\infty}C_kx^k\right]x^r\;=\;0$$
$$\Rightarrow\; x^r\left[(r^2-p^2)C_0+[(r+1)^2-p^2]C_1 x+\sum_2^{\infty}\left([(k+r)(k+r)-p^2]C_k+C_{k-2}\right)x^k\right]\;=\;0$$

According to the book, all the terms have to be zero for the equation to be zero.
$$(r^2-p^2)C_0=0\;\Rightarrow r=^+_-p$$
The book also say
$$[(r+1)^2-p^2]C_1=0\;\Rightarrow C_1=0\;\hbox { because from above, }\;r=^+_-p$$

My problem with this assumption $C_1=0$ is I can just as easy say $(r+1)^2-p^2=0$ and claim $C_0=0$!!! This will change the whole equation of the Bessel's equation!!!

Why do I have to follow the book to let $C_1=0$ which result in all $C_{2n+1}=0$? Is it just because it is simpler and more convenient this way?

Also, the book just said let $C_0=\frac {1}{2^p\Gamma(1+p)}$ without explaining why. Why?

I know it is for fitting the formula of:

$$J_p=\sum_0^{\infty} \frac{(-1)^k}{k! \Gamma(k+p+1)}\left(\frac {x}{2}\right)^{2k+p}$$

But can you just let $C_0$ to be anything?

Thanks

Last edited: May 17, 2013
2. May 17, 2013

### fzero

You can make either choice. You will still find that the leading term in $x^{k+r}$ in the solution has degree $p$. Choosing $C_0\neq 0$ makes $r=p$ and the sum over even powers $k$, while choosing $C_1\neq 0$ makes $r=p-1$ and the sum over odd powers. In either case the series starts with $x^p$ and multiplies this by a polynomial in even powers. So the choice corresponds to shifting the definitions of $k$ and $r$ by $\pm 1$ in such a way that they cancel out in the complete solution.

Yes, since the equation is linear, any multiple of a solution is still a solution. So we can choose any convenient normalization that we want when we define the Bessel functions.

3. May 17, 2013