I am trying to solve the following equation for days (not a mathematician):
$\sin \phi \;J_n (a \phi) = e^{in\pi} \;\sin \phi' \;J_n (a \phi')$
I need to find such $\phi'$ that the equality is satisfied. For those curious, it comes out of certain boundary conditions in my problem.
Please, at least let me know if there is hope, and if there is, how to approach that.
I have tried expanding in series and equating therm by term; I have looked at analytical continuations (maybe not smart enough to work that out).
If this is not solvable, any analysis, comments are welcome.
Additional information: one may treat $a$ as a small parameter if this is helpful.
If you are working with just integral bessel functions (where $n$ is an integer) you can solve this problem by rewriting it. Let $F_{n}(u) = sin(u) J_n(a u)$ and $e^{i n \pi} = (-1)^n$. Thus your equation becomes:
\begin{equation} F_n(\phi) = (-1)^n F_n(\phi') \end{equation}
Now we look at the even/oddness of $F_n$. If $n$ is even then $F_n$ is odd (sin is odd, bessel is even), if $n$ is odd then $F_n$ is even. Thus some of the solutions are:
\begin{eqnarray} \phi' = \phi \,\, \text{if n is even} \newline \text{a solution can't be determined this way if n is odd} \end{eqnarray}
Of course this is only part of the issue. Since $F_n(u)$ is an oscillatory function there may be multiple values of $\phi'$ which the value $F_n(\phi')$ evaluates to the value of $F_n(\phi)$.
In general the only way to get these values is by using numerical methods. You can use root finding algorithms but things can get tricky (missing roots etc.). I recommend using a numerical package such as Mathmatica (see below) if you can get it.
https://mathematica.stackexchange.com/questions/91784/how-to-find-numerically-all-roots-of-a-function-in-a-given-range