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 $\int^1_0 xJ_m(\sqrt{\lambda}x)J_m(\sqrt{\lambda_k}x)dx=\dfrac{\sqrt{\lambda_k}J'_m(\sqrt{\lambda_k})J_m(\sqrt{\lambda})}{\lambda-\lambda_k}$

  • $\begingroup$ 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

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 $$ x y y_k = \frac{ W_{1/x} [y,y_k]' }{\lambda- \lambda_k} $$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.