In a physics problem I need to solve an equation which involves eigenvalues of the differential operator $L$ applied to a function $f:x\mapsto f(x), x\in \mathbb{R}^+$. $L$ is given by \begin{equation} L = x^2 {d\over dx} + x{d\over dx} + x^2. \label{a} \end{equation} You can see that the eigen-problem $L[\tilde{f}_\nu] = \nu^2 \tilde{f}_\nu$ is simply the Bessel equation which has solutions given by Bessel functions of order $\nu$.
I want to decompose the general solution $f$ into the "modes" $f_\nu$ using the scalar product $$ \left<f,g\right> = \int_0^{+\infty} {dx\over x} f(x)\overline{g(x)}$$ This is the same idea as usual Fourier decomposition of the operator ${d^2\over dx^2} = -k^2$. A very helpful aspect of the Fourier basis $e^{ikx}$ is that it is orthogonal as $$ \int_\mathbb{R} dx e^{i(k-q)x} = \delta(k-q).$$
However, the Bessel functions are not strictly orthogonal since it gives the cardinal sine $$\left<J_\alpha,J_\beta\right>= \int_0^{+\infty} {dx\over x}J_\alpha(x)J_\beta(x) = {\sin_c({\pi\over 2 }(\alpha-\beta))\over \alpha+\beta}$$ for $\text{Re}(\alpha+\beta) > 0$. Does anyone know if
- there is a way to obtain an orthogonal basis, for example by changing the definition of the scalar product (by adding boundary terms?)
- a decomposition into a sum of modes is still possible with a non-orthogonal basis?
Thanks a lot.