Does Green's function only work with hermitian operators?

Question

If I was to use the Green's function method to solve the ordinary differential equation $$Ly=f(x)$$ Would $L$ have to be Hermitian, either way please can you explain why?

Answer 1

No, $L$ does not have to be Hermitian (however, being Hermitian would make Green's function symmetric in its two arguments).

For example, $Ly=y''-y'$ is not a Hermitian operator. Its Green's function on the interval $[-1,1]$ with zero boundary values can be found by piecing together two solutions (of the form $A+Be^x$) so that at $x=a$ the derivative jumps up by $1$, and the boundary conditions hold. In this way I got $$ G(x,a) = \frac{e^{1-a}-1}{e^2-1}(e^{x+1}-1) + (e^{x-a}-1)H(x-a) $$
where $H$ is the Heaviside function.

Generally, every linear differential operator $L$ admits Green's function, defined as the solution of $Ly=\delta_{a}$ where $\delta_a$ is the Dirac delta at $a$.