Green's function not self adjoint for self adjoint ODE

[member 428835]

Find Green's function of $$K(f(x)) = (1-x^2)f''(x)-2xf'(x)+\left(2-\frac{1}{1-x^2}\right)f(x):x\in[cos(\alpha),1]$$
subject to boundary conditions: $$f|_{x=1} < \infty\\
f|_{x=\cos(\alpha)} = 0.$$

Two fundamental solutions are associated Legendre polynomials (after all, this is Legendre's associated ODE). These are notated $P_1^1(x)$ and $Q_1^1(x)$.

To construct a Green's function for $K$, we use variation of parameters. I omit details, but the Green's function I find is

$$ G_L(x,y) = \frac{P_1^1(x)\left( \frac{\tau_2}{\tau_1} P_1^1(y)- Q_1^1(y)\right)}{W(y)}:\cos\alpha<x<y<1\\
G_R(x,y) = \frac{P_1^1(y)\left( \frac{\tau_2}{\tau_1} P_1^1(x)- Q_1^1(x)\right)}{W(y)}:\cos\alpha<y<x<1$$
where $W$ is the Wronskian of $P_1^1,Q_1^1$ and $\tau_1 = P_1^1(\cos\alpha)$ and $\tau_2 = Q_1^1(\cos\alpha)$.

My question is, why isn't this Green's function self adjoint despite the ODE being self-adjoint?

 

[jvicens]

I would like to point out that the Green's function found in the forum post is indeed self-adjoint. This can be seen by taking the inner product of the Green's function with itself, which should result in a real number. Additionally, the boundary conditions are symmetric, which is a necessary condition for the Green's function to be self-adjoint.

The reason for the confusion may be due to the fact that the Green's function is not symmetric, i.e. G(x,y) != G(y,x). However, this does not affect its self-adjointness. The Green's function is self-adjoint as long as it satisfies the following condition:

∫G(x,y)f(y)dy = ∫f(x)G(x,y)dy

This condition is satisfied by the Green's function found in the forum post. Therefore, it is a valid Green's function for the given ODE and boundary conditions.