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Given S-L equation $\dfrac{1}{x}[\dfrac{d}{dx}(xy')+(\dfrac{-m^2}{x})y]=-\lambda y$

Say $\mathcal{L}$ is the Sturm-Liouville operator, $y_k$ is eigenfunction $J_m(j_{mk}x)$ where $J_m$ is Bessel function and $j_{mk}$ is kth zero of $J_m(x)$. Associated eigenvalues $\lambda_k=(j_{mk})^2$. Let $y(x,\lambda)=J_m(\sqrt{\lambda}x)$.

How can I evaluate $(\mathcal{L}y_k)y-y_k(\mathcal{L}y)$ in two different ways, then form a suitable integral to show $\int^1_0 xJ_m(\sqrt{\lambda}x)J_m(\sqrt{\lambda_k}x)dx=\dfrac{\sqrt{\lambda_k}J'_m(\sqrt{\lambda_k})J_m(\sqrt{\lambda})}{\lambda-\lambda_k}$

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  • $\begingroup$ Do you have a question? $\endgroup$ – DisintegratingByParts Feb 25 '15 at 15:58
  • $\begingroup$ I don't see how you can evaluate the expression in two different ways and arrive at that identity @T.A.E. $\endgroup$ – The Problem Feb 25 '15 at 16:00
  • $\begingroup$ Please rephrase your post so that it becomes a question. :) $\endgroup$ – DisintegratingByParts Feb 25 '15 at 16:02
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One way is just a straight computation. i.e $$( \mathcal{L} y_k )y - y_k ( \mathcal{L}(y)) = \frac{y}{x} \left [ (xy_k')' - \frac{m^2 y_k}{x} \right ] -\frac{y_k}{x} \left [ (xy')' - \frac{m^2 y}{x} \right ] =\frac{W_{1/x}[y,y_k]'}{x}$$ where $W_p[y_1,y_2]$ is the $1/p$ weighted Wronskian. The other is to notice that $$ ( \mathcal{L} y_k )y - y_k ( \mathcal{L}(y)) = (\lambda - \lambda_k) y y_{k} $$ Thus, the identities together give us that $$ x y y_k = \frac{ W_{1/x} [y,y_k]' }{\lambda- \lambda_k} $$

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  • $\begingroup$ Could you explain the computation method please, I don't understand what the operator L does/is $\endgroup$ – The Problem Feb 25 '15 at 16:06
  • $\begingroup$ It's hard to explain from your notation, but have a look at the first step $\endgroup$ – Jeb Feb 25 '15 at 16:09
  • $\begingroup$ Ah that clears things up, this might be a silly question but what does $W_{1/x}$ mean? $\endgroup$ – The Problem Feb 25 '15 at 16:17
  • $\begingroup$ Work out the line before it, and you'll see it is the derivative of the weighted Wronskian, i.e. $ W_p[y_1,y_2] = y_1 ( py_2)' - (py_1)' y_2 $ $\endgroup$ – Jeb Feb 25 '15 at 16:19

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