Bessel Functions of the First Kind $J_n(x)$ II: Orthogonality

To accommodate boundary conditions for a finite interval $[0,a]$, we need to consider Bessel functions of the form $J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)$. For $x=\frac{\alpha_{\nu m}}{a}\rho$, Bessel’s equation (9) in here can be written as
$$\label{eq:bessel10}\rho^2\frac{d^2}{d\rho^2}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)+\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)+\left(\frac{\alpha_{\nu m}^2\rho}{a^2}-\frac{\nu^2}{\rho}\right)J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)=0.$$ Changing $\alpha_{\nu m}$ to $\alpha_{\nu n}$, $J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)$ satisfies
$$\label{eq:bessel11}\rho^2\frac{d^2}{d\rho^2}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)+\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)+\left(\frac{\alpha_{\nu n}^2\rho}{a^2}-\frac{\nu^2}{\rho}\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)=0.$$
Multiply \eqref{eq:bessel10} by $J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)$ and \eqref{eq:bessel11} by $J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)$ and subtract:
\begin{align*}
J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\frac{d}{d\rho}&\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]-J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\frac{d}{d\rho}\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\right]\\&=\frac{\alpha_{\nu n}^2-\alpha_{\nu m}^2}{a^2}\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right).\end{align*}
Integrate this equation with respect to $\rho$ from $\rho=0$ to $\rho=a$:
\begin{aligned}\int_0^\rho J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\frac{d}{d\rho}&\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]d\rho\\&-\int_0^\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\frac{d}{d\rho}\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\right]d\rho\\&=\frac{\alpha_{\nu n}^2-\alpha_{\nu m}^2}{a^2}\int_0^\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho d\rho.\end{aligned}\label{eq:bessel12}
Using Integration by Parts, we have
\begin{align*}
\int_0^\rho &J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\frac{d}{d\rho}\left[\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]d\rho\\&=\left[\rho J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]_0^a-\int_0^a \rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)dJ_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right).\end{align*}
Thus \eqref{eq:bessel12} can be written as
\begin{aligned}\left[\rho J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]_0^a-\left[\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\rho\frac{d}{d\rho}J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\right]_0^a\\=\frac{\alpha_{\nu n}^2-\alpha_{\nu m}^2}{a^2}\int_0^\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho d\rho.\end{aligned}\label{eq:bessel13}
Clearly the LHS of \eqref{eq:bessel13} vanishes at $\rho=0$. (Here we consider only $\nu=\mbox{integer}$ case.) It also vanishes at $\rho=a$ if we choose $\alpha_{\nu n}$ and $\alpha_{\nu m}$ to be $n$-th and $m$-th zeros of $J_\nu$. Therefore, for $m\ne n$
$$\label{eq:bessel14}\int_0^\rho J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)J_\nu\left(\frac{\alpha_{\nu n}}{a}\rho\right)\rho d\rho=0.$$
\eqref{eq:bessel14} gives us orthogonality over the interval $[0,a]$.

For $m=n$, we have the normalization integral
$$\int_0^a\left[J_\nu\left(\frac{\alpha_{\nu m}}{a}\rho\right)\right]^2\rho d\rho=\frac{a^2}{2}[J_{\nu+1}(\alpha_{\nu m})]^2.$$