# Bessel function equation

I have to prove that equation: $$\tag{1} \operatorname{J}_2(xy)\ \operatorname{Y}_0 (x) - \operatorname{J}_0(x) \operatorname{Y}_2(xy) =0$$ (where $\operatorname{J}_k$ and $\operatorname{Y}_k$ are Bessel functions of first and second kind) has a solution curve lying in the region $\Omega:=]j_{0,1}, \infty[\times]0,1[$, which "starts" from the point $(x_0,y_0)=(j_{0,1},0)$ (where $j_{0,1}$ is the first zero of $\operatorname{J}_0$).

Numerical simulations show there indeed exists a solution curve satisfying the requirement (among many others: in fact, it seems there is a solution curve of (1) "starting" from all the points $(j_{0,n},0)$). You can visualize this running the code:

ImplicitPlot[BesselJ[2, x y] BesselY[0, x] - BesselY[2, x y] BesselJ[0, x] == 0, {x, 2.404825557695773, 31}, {y, 0.01, 1}, AspectRatio -> 1]

on Wolfram's Mathematica.

The problem is that Implicit Function Theorem does not seem to apply, for the "starting point" $(x_0,y_0)$ is not a legit solution of (1); and, if also it were a solution, both partial derivatives of (1)'s lefthand side are singular in that point, cause they involve the singular functions $\operatorname{Y}_k$ for $k=1,2,3$.

Any hints?

Thanks a lot in advance, guys. ;)

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Let $F(x,y) = {{\rm J}_2\left(xy\right)}{{\rm Y}_0\left(x\right)}- {{\rm J}_0\left(x\right)}{{\rm Y}_2\left(xy\right)}$. $F(x_0, t) = J_2(x_0 t) Y_0(t) > 0$ for sufficiently small positive $t$. On the other hand, $F(x_0 + t, t) \sim - \frac{4 J_1(x_0)}{\pi x_0^2 t}$ as $t \to 0+$, and thus is negative for sufficiently small positive $t$. Thus for $t$ in some interval $(0, \epsilon)$ the Intermediate Value Theorem says there is $x(t) \in (x_0,x_0+t)$ with $F(x(t),t) = 0$. In fact it looks to me like as $t \to 0+$, $$x(t) =x_{{0}}+\frac{1}{32}\,{\frac {{x_{{0}}}^{4}Y_{{0}} \left( x_{{0}} \right) \pi }{J_{{1}} \left( x_{{0}} \right) }}{t}^{4}-{\frac {1}{96}}\,{ \frac {Y_{{0}} \left( x_{{0}} \right) {x_{{0}}}^{6}\pi }{J_{{1}} \left( x_{{0}} \right) }}{t}^{6}+O \left( {t}^{8} \ln(t)\right)$$
Great idea! But I have two questions: how could you find the asymptotic expansion of $x(t)$? and, can I use the asymptotic to infer that $x(t)$ is one-to-one? –  Bubbi Feb 11 '13 at 0:06
I got the series from Maple. Presumably $x'(t)$ can be obtained from term-by-term differentiation, which would imply that $x$ is one-to-one on some interval $(0,\epsilon)$. –  Robert Israel Feb 11 '13 at 2:07