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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?

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