Context
Bessel functions of the first kind, denoted as $J_\alpha(x)$, are solutions of Bessel's differential equation [1]. For appropriate boundary conditions the solutions satisfy an orthogonality relationship [1]. In particular:
$$
\int_0^1 x J_\alpha\left(x u_{\alpha,m}\right) J_\alpha\left(x u_{\alpha,n}\right) \,dx = \frac{\delta_{m,n}}{2} \left[J_{\alpha+1} \left(u_{\alpha,m}\right)\right]^2 .
$$
The Bessel functions of the second kind, denoted as $Y_\alpha(x)$, are solutions of the Bessel differential equation too.
Questions
-
Is it true that, for appropriate boundary conditions the solutions satisfy an orthogonality relationship:
$$
\int_0^1 x Y_\alpha\left(x u_{\alpha,m}\right) Y_\alpha\left(x u_{\alpha,n}\right) \,dx = \frac{\delta_{m,n}}{2} \left[Y_{\alpha+1} \left(u_{\alpha,m}\right)\right]^2 \,\,?
$$ -
For appropriate boundary conditions, do the solutions satisfy an orthogonality relationship:
$$
\int_0^1 x J_\alpha\left(x u_{\alpha,m}\right) Y_\alpha\left(x u_{\alpha,n}\right) \,dx \,\,?
$$
Bibliography
[1] Wikipedia contributors. "Bessel function." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 16 Mar. 2021. Web. 17 Mar. 2021.
Best Answer
I know that $$\int_0^\infty J_\alpha(z) J_\beta(z) \frac {dz} z= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2}. $$
I also know that for integral $\alpha$ and $\beta$, I can rewrite in terms of a Neuman fucntion as $$\int_0^\infty J_\alpha(z) \left[\frac{\sin{(\pi\,\beta)}\,N_\beta(z) + J_{-\beta}(z) }{\cos{(\pi\beta)}} \right] \frac {dz} z= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2}. $$ With a minor adjustment, $$ \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2} = \int_0^\infty J_\alpha(z) \left[\frac{\sin{(\pi\,\beta)}\,N_\beta(z) + (-1)^\beta\,J_{\beta}(z) }{\cos{(\pi\beta)}} \right] \frac {dz} z. $$ Thus, $$ \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2} = \int_0^\infty J_\alpha(z) \left[\frac{\sin{(\pi\,\beta)}\,N_\beta(z) + }{\cos{(\pi\beta)}} \right] \frac {dz} z + \int_0^\infty J_\alpha(z) \left[\frac{ (-1)^\beta\,J_{\beta}(z) }{\cos{(\pi\beta)}} \right] \frac {dz} z $$ Further, $$ \int_0^\infty J_\alpha(z) \left[\frac{\sin{(\pi\,\beta)}\,N_\beta(z) + }{\cos{(\pi\beta)}} \right] \frac {dz} z + \frac{(-1)^\beta}{\cos{(\pi\beta)}}\int_0^\infty J_\alpha(z) \left[ J_{\beta}(z) \right] \frac {dz} z = \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2} $$ And $$ \int_0^\infty J_\alpha(z) \, N_\beta(z) \frac {dz} z = \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2} \,\left[1 - \frac{(-1)^\beta}{\cos{(\pi\beta)}}\right]\frac{\cos{(\pi\beta)}}{\sin{(\pi\beta)}} $$ Thus, I find a orthogonality condition: $$ \boxed{ \int_0^\infty J_\alpha(z) \, N_\beta(z) \frac {dz} z = \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2} \,\frac{\left[\cos{(\pi\beta)} - (-1)^\beta \right] }{\sin{(\pi\beta)}} .} $$ I state without proof that the integral evaluates to zero for (i) $\alpha \neq \beta$ and (ii) $\alpha = \beta$ and $\alpha \neq 0$. Thus, I conclude that the Bessel of the first and second kind are orthogonal with respect to each other for $\alpha \neq \beta$ and for $\alpha = \beta$ and $\alpha \neq 0$. Using this approach, I cannot conclude whether the Bessel of the first and second kind are orthogonal with respect to each other when $\alpha = \beta =0$.