# Orthogonality of first and second kinds of solutions of Bessel's equation.

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

1. Is it true that, for appropriate boundary conditions the solutions satisfy an orthogonality relationship: $$\int_0^1 x Y_\alpha\left(x u_{\alpha,m}\right) Y_\alpha\left(x u_{\alpha,n}\right) \,dx = \frac{\delta_{m,n}}{2} \left[Y_{\alpha+1} \left(u_{\alpha,m}\right)\right]^2 \,\,?$$

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.

• I do not think it is true (for both relations). The orthogonality is related to the fact that $J$'s are eigenfunctions of a self-adjoint operator. But $Y$'s have bad behavior at zero which makes it unlikely (to me) that one can choose the boundary conditions as to make the relevant differential operator self-adjoint. A signature of this is that for $|\alpha|\ge 1$ your 1st integral simply diverges on the lower bound $0$. Commented Mar 18, 2021 at 20:15

For integral $$\alpha$$ and $$\beta$$, an orthogonality relation appears to be $$\int_0^\infty J_\alpha(z) \, N_\beta(z) \frac {dz} z = \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2} \,\frac{\left[\cos{(\pi\beta)} - (-1)^\beta \right] }{\sin{(\pi\beta)}}$$
I 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 $$\alpha = \beta =0$$.