Orthogonality of Bessel Functions
Theorem
Let’s assume that the roots of the first kind Bessel function $\alpha, \beta$ are $J_{\nu}(x)$. Then, in the interval $[0,1]$, $\sqrt{x}J_{\nu}(x)$ forms an orthogonal set.
$$ \int_{0}^{1} x J_{\nu}(\alpha x) J_{\nu}(\beta x)dx = \begin{cases} 0 &\alpha\ne \beta \\ \frac{1}{2}J^{2}_{\nu+1}(\alpha)=\frac{1}{2}J_{\nu-1}^{2}(\alpha)=\frac{1}{2}J_{\nu^{\prime}}^{2}(\alpha) &\alpha=\beta \end{cases} $$
Description
The above content can also be expressed as ‘Bessel function $J_{\nu}(x)$ is orthogonal in the interval $[0,1]$ with respect to the weight function $x$’.
Proof
$\alpha \ne \beta$
The differential equation that satisfies $J_{\nu}(\alpha x)$, $J_{\nu}(\beta x)$ is as follows.
$$ \begin{align*} x(xy^{\prime})^{\prime}+(\alpha^{2}x^{2}-\nu^{2})y &= 0 \\ x(xy^{\prime})^{\prime}+(\beta ^{2}x^{2}-\nu^{2})y &= 0 \end{align*} $$
Here, let’s substitute $J_{\nu}(\alpha x)=u$ for $J_{\nu}(\beta x)=v$
$$ \begin{align} x(xu^{\prime})^{\prime}+(\alpha^{2}x^{2}-\nu^{2})u &= 0 \\ x(xv^{\prime})^{\prime}+(\beta ^{2}x^{2}-\nu^{2})v &= 0 \end{align} $$
Now, calculating $v \cdot (1)-u \cdot (2)$ gives us
$$ \begin{align*} && vx(xu^{\prime})^{\prime}-ux(xv^{\prime})^{\prime}+(\alpha^{2}-\beta^{2})x^{2}uv = 0 \\ \implies && v(xu^{\prime})^{\prime}-u(xv^{\prime})^{\prime}+(\alpha^{2}-\beta^{2})xuv = 0 \\ \implies && (vxu^{\prime}-uxv^{\prime})^{\prime}+(\alpha^{2}-\beta^{2})xuv = 0 \end{align*} $$
Integrating both sides over the interval $[0,1]$ gives
$$ [vxu^{\prime}-uxv^{\prime}]_{0}^{1}+ (\alpha^{2} -\beta^{2})\int_{0}^{1}xuvdx = 0 \tag{3} $$
Since $u(1)=J_{\nu}(\alpha)=0=J_{\nu}(\beta)=v(1)$, the first term is $0$. Therefore,
$$ (\alpha^{2} -\beta^{2})\int_{0}^{1}xuvdx = 0 $$
But since $(\alpha^{2}-\beta ^{2}) \ne 0$,
$$ \int_{0}^{1}xuvdx=\int_{0}^{1}xJ_{\nu}(\alpha x)J_{\nu}(\beta x)=0 $$
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$\alpha = \beta$
Now, let’s assume that $\alpha$ is the root of $J_{\nu}(x)$ but $\beta$ is not. From the proof above, $(3)$ can be derived regardless of whether $\alpha$, $\beta$ are roots of $J_{\nu}(x)$, so let’s start with $(3)$. To summarize,
$$ \begin{align*} && [v(1)u^{\prime}(1)-u(1)v^{\prime}(1)]+(\alpha^{2} -\beta^{2})\int_{0}^{1}xuv dx &=0 \\ \implies && [J_{\nu}(\beta) \alpha J_{\nu}^{\prime}(\alpha)-J_{\nu}(\alpha)\beta J_{\nu}^{\prime}(\beta)]+(\alpha^{2} -\beta^{2})\int_{0}^{1}xuv dx &= 0 \\ \implies && J_{\nu}(\beta) \alpha J_{\nu}^{\prime}(\alpha)+(\alpha^{2} -\beta^{2})\int_{0}^{1}xuv dx &=0 \\ \implies && \int_{0}^{1}xJ_{\nu}(\alpha x)J_{\nu}(\beta x) dx &= \frac{J_{\nu}(\beta)\alpha J_{\nu}^{\prime}(\alpha )}{\beta^{2}- \alpha^{2}} \end{align*} $$
Taking the limit of $\beta \rightarrow \alpha$ on both sides gives
$$ \lim \limits_{\beta \rightarrow \alpha} \int_{0}^{1}xJ_{\nu}(\alpha x)J_{\nu}(\beta x) dx =\lim \limits_{\beta \rightarrow \alpha} \frac{J_{\nu}(\beta)\alpha J_{\nu}^{\prime}(\alpha )}{\beta^{2}- \alpha^{2}}=\frac{ 0 }{ 0 } $$
Therefore, let’s calculate using L’Hospital’s Rule. Differentiating the right side with $\beta$ gives
$$ \begin{align*} \lim \limits_{\beta \rightarrow \alpha} \int_{0}^{1}xJ_{\nu}(\alpha x)J_{\nu}(\beta x) dx &= \lim \limits_{\beta \rightarrow \alpha} \frac{J_{\nu}(\beta)\alpha J_{\nu}^{\prime}(\alpha )}{\beta^{2}- \alpha^{2}} \\ &= \lim \limits_{\beta \rightarrow \alpha} \frac{J_{\nu}^{\prime}(\beta)\alpha J_{\nu}^{\prime}(\alpha )}{2\beta} \\ &= \frac{J_{\nu}^{\prime}(\alpha)\alpha J_{\nu}^{\prime}(\alpha )}{2\alpha} \\ &= \frac{1}{2}J_{\nu}^{\prime}(\alpha) \end{align*} $$
And by the recursive relationship of Bessel’s function $(e)$,
$$ \frac{1}{2}J_{\nu}^{\prime}(\alpha) =\frac{1}{2}J_{\nu-1}(\alpha) =\frac{1}{2}J_{\nu+1}(\alpha) $$
Therefore,
$$ \int_{0}^{1}xJ_{\nu}^{2}(\alpha x)dx=\frac{1}{2}J^{2}_{\nu+1}(\alpha)=\frac{1}{2}J_{\nu-1}^{2}(\alpha)=\frac{1}{2}J_{\nu^{\prime}}^{2}(\alpha) $$
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