I'm looking at the following S-L ODE:
$\frac{d}{dx} \left[e^{ix} f'(x)\right] + e^{ix} \lambda f(x) = 0$,
where $i$ is the complex number, and $\lambda$ denotes the eigenvalues. However, this can only have real eigenvalues if the problem is regular, i.e., that somehow, one has to show that $e^{ix} > 0$ over some interval. As this is a complex function, I'm having difficulty showing this.
Any ideas?
If homogenous Dirichlet boundary conditions are available, then multiplying by the complex conjugate of $f$ and integrating by parts once yields
$$\lambda = \frac{\displaystyle\int^{b}_{a} \mathrm{e}^{ix} |f'|^2 \, \mathrm{d}x}{\displaystyle\int^{b}_{a} \mathrm{e}^{ix} |f|^2 \, \mathrm{d}x}$$
For $f= \mathrm{e}^{i sx}$, which is the solution to any linear operator, the fraction above turns out to be real. Moreover, $\lambda = |s|^2$.