Green's Function Divergence

Given a domain $\Omega \in \mathbb{R}^2$, and a PDE of the form $L = a(x) \partial_x^2 + b(x) \partial_y ^2$ for $x \in \Omega$ , the green's function $G(x,y) : \Omega \times \Omega \rightarrow R$ satisfies $L G(p,q) = \delta(q-p)$ and $G(p,q) = 0$ for $q \in \partial \Omega$.

I would like to understand the divergences of the green's function near the diagonal. In particular, what sort of divergent terms will appear? My feeling is that there is only a logarithmic divergences, and a dipole-like term.

Reasoning is that roughly speaking, I should be able to take care of the $\delta$ with a logarithm, and then correct with bounded functions. Precisely for $x_0 \in \Omega$, split off the zeroth order term of the PDE. $$L = L_0 + L_r$$ where $$L_0 = a(x_0) \partial_x^2 + b(x_0) \partial_y ^2 + L_r \\ L_r = ( da \cdot (x-x_0)\partial_x^2 + db \cdot (x-x_0)\partial_x^2) + O((x-x_0)^2)$$ Now we know the Green's function for $L_0$, $G_0(x,y)$, which will be a logarithm. Then one can correct for $L_r$ by adding the term $$L^{-1} \left( L_r G_0(x,y) \right)$$ My intuition is that the linear part of $L_r$ will produce a dipole term: $L_r G_0(x,y)$ itself diverges, but in a small neighborhood $B_\epsilon$, the "total charge" goes to zero. Apart from this, there will be no divergences.

Is this true, and is there some source where this is studied?