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