$J_n(x)=0$ has no repeated root except zero. How to prove it?

Question

Suppose $J_n(x)=0$ has repeated roots. Then it must have at least two equal roots, say $x=x_0(\neq 0)$ i.e., $x_0$ is a double root of $J_n(x)=0$. Therefore $J_n(x_0)=0$ & $J_n’(x_0)=0$

Then what should I do?

Answer 1

This is probably overkill:

$J_n$ satisfies two recurrence relations:

$$\frac{2n}{x}J_n(x)=J_{n-1}(x)+J_{n+1}(x)$$

and

$$2\frac{dJ_n(x)}{dx}=J_{n-1}(x)-J_{n+1}(x).$$

If $x_0\neq 0$ is a repeated root, your reasoning implies $J_{n-1}(x_0)=J_{n+1}(x)=0$

Bourget's hypothesis states that $J_n(x),J_{n+m}(x)$ cannot share zeros other than $x=0$. This was proven by Siegel, and you can find a proof here (Theorem 3.1). The case of $m=1$ should be a bit easier.

Answer 2

$J_n$ solves the Sturm–Liouville equation $$ -(xy')' +\frac{n^2}{x} y = x^2y, $$ or more commonly $$ x^2 y'' + xy' +(x^2-n^2)y = 0. $$ If $J_n(a)$ has a multiple zero at $a \neq 0$, $J_n'(a) = 0$. But then the differential equation implies that $J_n''(a)=0$, and repeated differentiation and substitution implies that $J_n^{(k)}(a)$ also equals zero. Since $J_n$ is analytic away from $0$, it follows that $J_n$ vanishes identically on an open set containing $a$, which is a contradiction.

See here for some other properties of the zeros of $J_n$ and $J'_n$ (most of which apply to any solution of Bessel's equation where it is analytic).