How do I show the Wronskian of $(J_{a}(x),Y_{a}(x)) = \dfrac {2} {\pi x}$

Question

Based of using my undergrad class notes. I know that the wronskian of $(J_{a}(x),Y_{a}(x))$ is $ W(J_{a}(x),Y_{a}(x)) = \left| \begin{matrix} J_{a}(x) & Y_{a}(x) \\ J_{a}'(x) & Y_{a}'(x) \end{matrix} \right| $ = $J_{a}(x) Y_{a}'(x)-J_{a}'(x)Y_{a}(x)$. Where $J_{a}(x)$ is the bessel function of the first kind and is defined as $J_{a}(x) = |\dfrac{x} {2}|^a \sum_{k=0} ^{\infty} \dfrac{(-x^2/4)^k} {k! \Gamma(a+k+1)}$. While $Y_{a}$ is bessels function of the second kind and is defined as $ Y_{a}(x) =\dfrac{J_{a}(x) \cos(\pi a)-J_{-a}(x)} {\sin(\pi a).}$.I am using the some fact from my class that the wronskian satisfies $\dfrac{dW} {dx}=\dfrac{-1} {x} W$ which implies $ W(x) = \dfrac{c} {x}$ for some constant C. (I have no idea where this comes from, I tried to "google" Abel's Identity and found nothing like that. So I am supposed to find $C = lim_{x \rightarrow 0} \; xW(x)$. Then I suppose I divide the limit I get for $c$ to then divide by $x$ to get the wronskian $W(x)$. Am I thinking about this right? I could not figure out the limit of the Wronskian of this series.How should I proceed? I am looking for mathematical manipulations to solve this, no graduate theory will help since I have no background for this. Just a reference sheet of bessel functions and gamma properties.

Answer 1

What you're looking for in the first place is a solution to the equation $$ W'(x) = \frac {-1}x W(x). $$ Note that you can re-write this equation as $$ (xW(x))' = W(x) + xW'(x) = 0, $$ which means $$ xW(x) = C, $$ hence $W(x) = c/x$. At this point all you need is to be able to compute the limit of $xW(x) = xW(J_a(X), Y_a(x))$ at only one value $\alpha$ of $\mathbb R$ and/or $\mathbb C$ depending on where you work. If $\alpha = 0$ works fine for you, then so be it ; but another value would work fine too. After you've computed this limit, you know its value is $C$, so you're done.

Hope that helps,

Answer 2

Take the differential equation that defines them and u get your result from there immediately!