# Normalization of Bessel functions of the first kind

1. Nov 20, 2013

### ShayanJ

Before stating the main question,which section should the special functions' questions be asked?

Now consider the Bessel differential equation:

$\rho \frac{d^2}{d\rho^2}J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})+\frac{d}{d\rho}J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})+(\frac{\alpha_{\nu m}^2 \rho}{a^2}-\frac{\nu^2}{\rho})J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})=0$

and a similar equation but with $\alpha_{\nu m}$ replaced by $\alpha_{\nu n}$ where $\alpha_{\nu s}$ is the $s$th root of $J_{\nu}(x)$.

Now if one multiplies the first equation by $J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})$ and the second by $J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})$ and then subtracts the second from the first,the following will be found upon integration of the whole equation from 0 to a:

$\int_0^a J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})\frac{d}{d\rho}[\rho\frac{d}{d\rho}J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})]d\rho-\int_0^a J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})\frac{d}{d\rho}[\rho\frac{d}{d\rho}J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})]d\rho=\frac{\alpha_{\nu n}^2-\alpha_{\nu m}^2}{a^2}\int_0^a J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})J_{\nu}(\alpha_{\nu m} \frac{\rho}{a}) \rho d\rho$

Integrating the LHS by part and cancelling gives:

$J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})\rho\frac{d}{d\rho}J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})|_0^a-J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})\rho\frac{d}{d\rho}J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})|_0^a=\frac{\alpha_{\nu n}^2-\alpha_{\nu m}^2}{a^2}\int_0^a J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})J_{\nu}(\alpha_{\nu m} \frac{\rho}{a}) \rho d\rho$

Using $\frac{d}{dx}J_n(x)=\frac{n}{x}J_n(x)-J_{n+1}(x)$:

$J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})\rho[\frac{\nu a}{\alpha_{\nu m}\rho}J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})-J_{\nu+1}(\alpha_{\nu m}\frac{\rho}{a})]|_0^a-J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})\rho[\frac{\nu a}{\alpha_{\nu n}\rho}J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})-J_{\nu+1}(\alpha_{\nu n}\frac{\rho}{a})]|_0^a=\frac{\alpha_{\nu n}^2-\alpha_{\nu m}^2}{a^2}\int_0^a J_{\nu}(\alpha_{\nu n} \frac{\rho}{a})J_{\nu}(\alpha_{\nu m} \frac{\rho}{a}) \rho d\rho$

After placing $\alpha_{\nu n}=\alpha_{\nu m}+\varepsilon$ and taking the limit as $\varepsilon\rightarrow 0$ and using $\frac{d}{dx}J_n(x)=\frac{n}{x}J_n(x)-J_{n+1}(x)$ to replace terms involving Js with $\varepsilon$ in their arguments and calculating the terms in the boundaries:
$-J_{\nu+1}(\alpha_{\nu m})\varepsilon[-aJ_{\nu+1}(\alpha_{\nu m})](\alpha_{\nu m})=\frac{2\alpha_{\nu m} \varepsilon}{a^2}\int_0^a J^2_{\nu}(\alpha_{\nu m} \frac{\rho}{a})\rho d\rho$

Which gives:

$\int_0^a [J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})]^2\rho d\rho=\frac{a^3}{2\alpha_{\nu m}}[J_{\nu+1}(\alpha_{\nu m})]^2$

But the correct equation is:

$\int_0^a [J_{\nu}(\alpha_{\nu m} \frac{\rho}{a})]^2\rho d\rho=\frac{a^2}{2}[J_{\nu+1}(\alpha_{\nu m})]^2$
(This is what you find about normalization of Bessel functions everywhere)
What's wrong in my calculations?
Thanks

2. Apr 2, 2018

### JavierMarmol

Hello,

the mistake is in the relationship where you change Jv for Jv+1. The x derivate changes to (d ro / dx)(d / d ro) and (d ro / dx) = alpha / a

I was stuck ages with this too! Can't believe there wasn't a reply for 5 years!