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Suppose we have a second order elliptic differential operator $$ L(v) = -\text{div}(A(x) \nabla v) $$ $A(x)$ is a bounded and strictly positive definite matrix with Hölder continuous entries. And suppose $\Omega$ is a $C^1$ domain (can be considered more regular if required).

From [2, Theorem 1], we know that there exists a Poisson Kernel $K(x,y)$ for the above operator and domain $\Omega$. That is the solution $u$ such that $Lu=0$ in $\Omega$ can be written as $$ u(x) = \int_{\partial \Omega}u(y) K(x,y) d\mathcal H^{N-1}(y) $$

On the other hand, we have from [1] we have the existence of Green's function $G$ for the operator $L$ in the domain $\Omega$. Natural question to ask is just like we have a integral representation for harmonic functions, can we say that the Poisson Kernel $K(x,y)$ and the co-normal derivative of Green's function on the boundary i.e. $\left ( A(y)\nabla_y G(x,y)\right )\cdot \nu $ are equal on $\partial \Omega$?

In particular, since the Green's function $G$ of $L$ exists for the domain $\Omega$, then for $u$ such that $Lu=0$ is the following expression true? $$ u(x) = \int_{\partial \Omega}u(y) \left ( A(y)\nabla_y G(x,y)\right )\cdot \nu_y\,d\mathcal H^{N-1}(y). $$

More precisely, is the following true?

$$ K(x,y) = \left ( A(y)\nabla_y G(x,y)\right )\cdot \nu_y\; \forall y \in \partial \Omega,\; x\in \Omega $$

The above claim holds true when $A(x)= Id$, i.e. when the operator $L$ is simple Laplacian. Under what condition/s we can say the same for a general second order elliptic operator in divergence form?

[1] Grueter, Michael; Widman, Kjell-Ove, The Green function for uniformly elliptic equations, Manuscr. Math. 37, 303-342 (1982). ZBL0485.35031.

[2] Fabes, Eugene B.; Jerison, David S.; Kenig, Carlos E., Necessary and sufficient conditions for absolute continuity of elliptic- harmonic measure, Ann. Math. (2) 119, 121-141 (1984). ZBL0551.35024.

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  • $\begingroup$ Can you detail how you came to that formula? You integrated by parts against something, then took a limit, probably? $\endgroup$
    – username
    Commented May 26, 2021 at 20:18
  • $\begingroup$ If $Lu=0$, write $u(x)=u(x)+\int_{\Omega} G(x,y) Lu(y)dy$ and integrate by parts twice. Formally you get your formula since $G$ is symmetric and $L_y G(x,y)= \delta_x$. It should be possible to justify it by using the heat kernel, which is more regular for $t>0$ but I am not sure about the regularity on $A$ needed for this. $\endgroup$ Commented May 26, 2021 at 21:16

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