Let the Laplace equation in a domain $D$ be:
$\Delta u=0 $
Then, let $E(x,y)$ be the fundamental solution*, and $g(x,y)$ a harmonic function (both in $x$ and $y$).
Then the function $G(x,y)=E(x,y) + g(x,y)$ is called Green's function of the Dirichlet problem for the Laplace equation in $D$ if $G(x,y)=0$ when $x$ or $y$ is in the boundary $S$ of the domain $D$.
My question is Let $y$ be fixed, then we have that $G_y(x)=E(x,y)+g(x,y)=0$, $\forall x \in S$ (also in $x=y$ because $lim_{x \rightarrow y}G(x,y)=0$ if $y \in S$. The function $G$ is harmonic in both $x$ and $y$. Then why we cannot conclude that, by the extremum principle, it will be $0$ everywhere in $D$?
- $E(x,y):= \frac{1}{n-2} |y-x|^{2-n}$ if $n>2$
- $E(x,y):= -ln|y-x| $ if $n=2$
Here $n$ is the dimension of the euclidean space in which we are working
The concept of Green's function involves a domain $\Omega$, a source point $y\in \Omega$, and a function $G(x,y)$ which satisfies a differential equation with respect to $x$ in the domain $\Omega\setminus \{y\}$.
The functions $$E(x,y):= \frac{1}{n-2} |y-x|^{2-n}, \quad n > 2$$ $$E(x,y):= -\ln|y-x|$$ are not harmonic with respect to $x$ in $\Omega$ because they are not even continuous at $y$, let alone differentiable there.