# Orthogonality with Bessel functions of rational order

Given a set of functions $f_{mv}(r,\phi)=J_{v}(k_{mv}r)\cos(v \phi)$ in polar coordinates, where $J_{v}$ denotes Bessel's function of the first kind of $v$th order and $k_{mv}$ denotes the $m$th root of $J'_{v}(k_{mv}a)=0$ at boundary $r=a$, how do we prove that these functions are orthogonal iff $v$ is integer?

[Note that they are tested for orthogonality over $v$ and $m$, as they will be later used to express functions by expansions like $\sum\limits_{v}\sum\limits_{m}c_{mv}f_{mv}$].

And then, for a case where we have rational $v$ (ratio of two integers, $v=M/N$), is there a way to construct a new orthogonal set that is of similar form to the one above (or linearly re-adjust/combine the terms in the original one above to make it orthogonal under such condition)?

• @paul It is intended: if and only if (=iff). Apologies. – user135626 Apr 3 '17 at 11:35
• @paul Question was updated to fix this. – user135626 Apr 3 '17 at 11:41
• @Francois Done. – user135626 Apr 3 '17 at 16:20

Your functions $r\mapsto J_\nu(kr)$ are orthogonal because they are eigenfunctions of a self-adjoint Sturm-Liouville problem. This follows from Lagrange's (a.k.a. Green's) identity, or most explicitly from the relation (e.g. Lebedev (5.14.3, 5.14.9)) $$\int_0^a J_\nu(\alpha r)J_\nu(\beta r)\,r\,dr = \frac{a\beta J_\nu(\alpha a)J_\nu'(\beta a)-a\alpha J_\nu(\beta a)J_\nu'(\alpha a)}{\alpha^2-\beta^2}.$$ For non-integral $\nu$, this still works but the product with $\cos(\nu\phi)$ won't be single-valued in polar coordinates...
• If we could ignore or avoid the angular multi-valuedness for a moment (e.g. by limiting the chosen domains of $\phi$ in practice), would the product itself be orthogonal for different values of $v$ when $v=M/N$? – user135626 Apr 3 '17 at 11:57
• @user135626 Rather the opposite: if $\mu, \nu$ have common denominator $N>1$ (e.g. $\mu=0$, $\smash{\nu=\frac12}$), then $\smash{\int_0^b e^{i(\nu-\mu)\phi}d\phi=0}$ generally requires $b$ to be a multiple of $2N\pi$, not just $2\pi$, i.e. you must expand the domain to an $N$-fold cover of the circle. – Francois Ziegler Apr 3 '17 at 14:35