# Orthogonality of eigenfunctions of a linear operator.

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.

• There seems to be some derivatives missing in your second-to-last equation. – Erick Wong Apr 15 '13 at 13:05
• yes, I've corrected it now, thanks! – Jimmy R Apr 15 '13 at 13:08

\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}