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Suppose I have a linear operator $$ \frac{\mathrm{d}^2}{\mathrm{d}r^2}+\frac{1}{r}\frac{\mathrm{d}}{\mathrm{d}r} $$ and I want to find its eigenfunctions, that is, to solve the ODE $$ \frac{\mathrm{d}^2R}{\mathrm{d}r^2}+\frac{1}{r}\frac{\mathrm{d}R}{\mathrm{d}r}=\lambda R. $$ Suppose, further, that I have boundary conditions $R(0)\neq\pm\infty$ and $R(a)=0$, then the solutions are Bessel's functions of the first kind $R=J_0\left(\frac{j_n}{a}r\right)$, where $j_i$'s are the roots of $J_0$. I want to show directly that these functions are orthogonal with respect to a weight function. Notice that $$ rR''+R'=\lambda rR =\frac{\mathrm{d}}{\mathrm{d}r}(rR')$$ and suppose that $R_m$ and $R_n$ are eigenfunctions with distinct corresponding eigenvalues $\lambda_m,\lambda_n$, so $$ (rR'_m)'=\lambda_m rR_m \\ (rR'_n)'=\lambda_n rR_n $$ Hence, multiplying by $R_n$ and $R_m$ and subtracting, one gets $$ (\lambda_m-\lambda_n)rR_mR_n=((rR'_m)'R_n-(rR'_n)'R_m) $$ At this stage, it seems that $r$ is the weight function and I need to integrate w.r.t. $r$ from $0$ to $a$ $$ (\lambda_m-\lambda_n)\int^a_0 rR_mR_n\:\mathrm{d}r=\int^a_0 ((rR'_m)'R_n-(rR'_n)'R_m)\:\mathrm{d}r $$ It seems that the integral on the RHS should be equal to zero, but I do not see how.

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    $\begingroup$ There seems to be some derivatives missing in your second-to-last equation. $\endgroup$ – Erick Wong Apr 15 '13 at 13:05
  • $\begingroup$ yes, I've corrected it now, thanks! $\endgroup$ – Jimmy R Apr 15 '13 at 13:08
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I think you made a mistake in the next-to-last step. You would be integrating

$$\begin{align}\int_0^a dr \, [R_n (r R_m')' - R_m (r R_n')'] &= [r R_m' R_n - r R_n' R_m]_0^a - \int_0^a dr \, r (R_m' R_n' - R_n' R_m') = 0- 0\end{align}$$

The second step is a result of applying the boundary conditions.

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  • $\begingroup$ thanks! too many derivatives, too easy to leave out some $\endgroup$ – Jimmy R Apr 15 '13 at 13:14

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