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Please help me prove the following identity
$$ a(J_1(a)Y_0(a)-J_0(a)Y_1(a))=\frac{2}{\pi} $$ for any $a$.
$J$ and $Y$ are bessel functions of the first and second kind respectively.
Thank you.

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    $\begingroup$ what is "*" in your notation? Too much MATLAB recently ? :-) $\endgroup$ Mar 12, 2013 at 1:32

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This is equivalent to computing the Wronskian of $J_0$ and $Y_0$, since $J'_0 = -J_1$ and $Y'_0 = -Y_1$. $$ W(x) = \begin{vmatrix} J_0 & Y_0 \\\ J'_0 & Y'_0 \end{vmatrix} = \begin{vmatrix} J_0 & Y_0 \\\ -J_1 & -Y_1 \end{vmatrix} = J_1 Y_0 - J_0 Y_1. $$ But the zeroth-order Bessel equation is: $$ \frac{d^2y}{dx^2} + \frac{1}{x}\frac{dy}{dx} + y = 0. $$ By Abel's formula, $W(a) = \frac{W(1)}{a}$. You now have to compute any value of $W$.

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  • $\begingroup$ $$W(a)=W(1)e^{-\int_1^a\frac{1}{x}dx}=\frac{W(1)}{a}$$ $\endgroup$ Mar 12, 2013 at 9:09
  • $\begingroup$ Thank you.<br> The proof is almost complete now it is required to prove that: $$ (J_1(1)Y_0(1)-J_0(1)Y_1(1))=\frac{2}{\pi} $$ $\endgroup$ Mar 12, 2013 at 11:18
  • $\begingroup$ You must have finished this by now I am sure, but just in case, to conclude, once you know it is constant, you just have to compute the limit as a->0, very easy with the explicit formulae. $\endgroup$
    – username
    Sep 19, 2013 at 8:33

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