Suppose we are given the following elliptic operator:
$$P(u) = -(a^{ij} (x) u(x)_j)_i $$
where $a^{ij}$ is positive, symmetric and bounded (uniformly elliptic) over a smooth bounded domain $\Omega \subset \mathbb{R}^N$. Here, the usual summation convention is used. I'm interested in constructing the parametrix for this operator. Let $p \in \Omega$ be a point. In this article (equation 4.2), the authors give explicit forms for the parametrix as
$$ \Gamma_p(x)= \frac{1}{4 \pi}|a\left(x_p\right)|^{1 / 2} \log (\{ a_{ij} (x_p) (x^i - x_p^i)(x^j-x_p^j) \}) $$ for $N=2$ and
$$ \Gamma_p (x) = \frac{1}{(N-2) \omega_N} |a(x_p)|^{1/2} (a_{ij} (x_p) (x^i - x^i_p)(x^j-x^j_p) ^{- \frac{N-2}{2} } $$
for $N \geq 3$. Here, $a$ is the determinant of $a_{ij}$ and $\omega_N$ is the surface of the $N$-dimensional unit sphere. They later give additional forms (equations 4.25 and 4.26) with further nice properties for the derivative of this function. I've verified that the above coincides with the Green's function for the Laplacian, in which case $a_{ij}$ is just the identity matrix, but I'm struggling to derive these expressions from scratch. These forms do make sense, since this Green's function is known to be proportional to the Green's function of the Laplacian as per this paper, specifically Theorem 7.1.
The usual derivation for the Green's function of the Laplacian with in $2$D that involves a transformation to polar coordinates and solving an ODE in terms of $r = |\boldsymbol{x}|$ doesn't seem to work out. Specifically, we wish to solve $$ -(a^{ij} (x) \Gamma_p (x)_j)_i = \delta (x - x_p) $$ but expanding the derivatives does not give an expression that would be amenable to a polar coordinate transformation reduction to an ODE, as would happen for the Laplacian.
My question is twofold: how do the authors obtain these expressions? and secondly, how is the derivation different for obtaining a parametrix as compared to a Green's function in this scenario?
I would greatly appreciate any commentary or references! Thank you!
I am no expert in PDEs, but I can do the linear transformation part. This could be very standard, but let me write it out in detail.
What I want to show is that if $a^{ij}(x)=a^{ij}$ are constants, then the Green function $\Gamma_p(x)$ satisfying $$ -(a^{ij}(x)\Gamma_p(x)_j)_i = -a^{ij}\Gamma_p(x)_{ji}= \delta(x-x_p) $$ has the desired form (let me just do $N\geq 3$ case) $$ \Gamma_p(x)=\frac{1}{(N-2)\omega_N}|a|^{1/2}(a_{ij}(x^i-x^i_p)(x^j-x_p^j))^{-\frac{N-2}{2}}. $$ Let me also take $x_p=0$ for simplicity.
Since $A=(a^{ij})$ is symmetric and positive-definite, there exists a matrix $B=(b^{ij})$ such that $A=BB^T$, that is, $$ a^{ij}=b^{ik}b^{jk}. $$
Consider the change of variables $$ x^i = b^{ik}y^k, $$ then by the chain rule, \begin{align} \frac{\partial}{\partial y^k}&=\frac{\partial x^i}{\partial y^k}\frac{\partial}{\partial x^i}=b^{ik}\frac{\partial}{\partial x^i},\\ \Delta_y&=\frac{\partial}{\partial y^k}\frac{\partial}{\partial y^k}=b^{ik}\frac{\partial}{\partial x^i}b^{jk}\frac{\partial}{\partial x^j}=a^{ij}\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j}. \end{align}
Let $\tilde \Gamma(y)=\Gamma(x(y))$. Then $\Gamma(x)$ satisfying $$ -a^{ij}\Gamma(x)_{ij}= \delta(x) $$ is the same as $$ -\Delta_y \tilde \Gamma(y) = \frac{1}{|\det B|}\delta(y), $$ where the coefficient comes from the change of volume form $dx=|\det B|\,dy, $ which implies that $\delta(x)=\frac{1}{|\det B|}\delta(y)$.
The standard formula for the fundamental solution of $\Delta_y$ gives
$$ \tilde \Gamma(y)=\frac{1}{(N-2)\omega_N}\frac{1}{\det B}|y|^{-{(N-2)}}=\frac{1}{(N-2)\omega_N}\frac{1}{|\det B|}(y^ky^k)^{-\frac{N-2}2}. $$
Let $A^{-1}=(a_{ij})$, $B^{-1}=(b_{ij})$, so $A^{-1}=(B^{-1})^TB^{-1}$, and $a_{ij}=b_{ki}b_{kj}$. Then from $y^k = b_{ki}x^i$, we see that $$ \Gamma(x)=\frac{1}{(N-2)\omega_N}\frac{1}{|\det B|}(b_{ki}b_{kj}x^ix^j)^{-\frac{N-2}2}=\frac{1}{(N-2)\omega_N}|a|^{1/2}(a_{ij}x^ix^j)^{-\frac{N-2}2}, $$ where $a=\det (a_{ij})=\det A^{-1} = (\det B)^{-2}$ so $|a|^{1/2}=1/|\det B|$.
I believe this is the essence of the calculation. With more understanding of parametrix, you can feel comfortable with their assertions for its formula.