Let $\widehat{\cal O}$ be an $n$-dimensional differential operator i.e. for $n=1$ it leads to an ordinary differential equation and for $n>1$ to a partial differential equation. The Greensfunction $G$ obeys $$ \widehat{\cal O} \, G\left(\vec{x},\vec{x}'\right) = \delta\left(\vec{x}-\vec{x}'\right) $$ where the operator acts on $\vec{x}$. I'm wondering if $G$ in general is symmetric with respect to its arguments. If the Operator is translational invariant, then the Greensfunction is of the form $G\left(\vec{x}-\vec{x}'\right)$. If it is not I do not see why it should be symmetric. If $\Psi\left(\vec{x}\right)$ is a solution of the inhomogenous equation $$ \widehat{\cal O} \, \Psi\left(\vec{x}\right) = \omega\left(\vec{x}\right) $$ then it seems to be sufficient, that $G$ solves the first equation only with respect to $\vec{x}$, because $$ \Psi\left(\vec{x}\right) = \int_V G\left(\vec{x},\vec{x}'\right) \omega\left(\vec{x}'\right) \, {\rm d}V' $$ and no solution with respect to $\vec{x}'$ is required. Are there other criteria? For the laplacian the Greensfunction $G=-\frac{1}{4\pi\left|\vec{x}-\vec{x}'\right|}$ indeed seems to be symmetric, but the laplacian in cartesian coordinates is translationally invariant.
Or must this eigendecomposition $$ G\left(\vec{x},\vec{x}'\right) = \sum_n \frac{\psi_n\left(\vec{x}\right)\psi_n\left(\vec{x}'\right)}{\lambda_n} $$ always hold, where $\psi_n$ is normalized and solves $$ \widehat{\cal O} \, \psi_n = \lambda_n \psi_n \, . $$ Are there Greensfunction which are not representable like this? Why should any Greensfunction be precisely of this form?