# Green's function for the Laplace equation: why it is not zero?

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 reason is that the fundamental solution does not solve the homogeneous Laplace equation. Commented Jul 22, 2017 at 10:16
• Why? I mean, $\Delta(G)=\Delta(E) + \Delta(g)$, both $E$ and $g$ are harmonic in both the variables, and by the argument above, i can fix one of them at a time to show that $G$ has to be $0$ everywhere. Then I have that $\Delta(G)=0$. (I know that my argument is wrong, but I don't know why) Commented Jul 22, 2017 at 10:27
• $E$ is not a harmonic function (otherwise it would not be a fundamental solution). Commented Jul 22, 2017 at 13:14
• The definition of harmonic function i'm using is to be a function with continuous partial derivatives and satisfying Laplace equation. Now $E$ is an harmonic function in both of the variables. I'm editing my question with an explicit definition of what is the fundamental solution Commented Jul 22, 2017 at 19:44

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.
• And, more pointedly, in all these examples $\Delta_x E(x,y)=\delta(x-y)$, with Dirac's $\delta$, where the derivative is in an entirely legitimate distributional sense. Yes, it is pointwise $0$ off a set of measure $0$, but measure theory is not to-the-point here. Commented Jul 22, 2017 at 20:50