Krdytkyu

Let $Omega$ be a domain in $mathbb{R}^{d}$ and assume that for any $y in Omega$ there is a function $h_{y} in C^{2}(overline{Omega})$ such that
begin{equation}
label{eq8.1}
begin{cases}
Delta h_{y}(x) = 0 text{ in } Omega \
h_{y}(x) = E(x,y) text{ on } partial Omega
end{cases}
end{equation}

Where $E$ denotes the fundamental solution to $Delta$ in dimension $d$. Under those assumptions we define the Green function $G(x,y) = E(x,y) - h_{y}(x)$

Let $Omega$ be a bounded domain such that one can define the Green function $G$. Then for any $u in C^{2}(overline{Omega})$ and any $y in Omega$, we have
$$ u(y) = - int_{Omega} G Delta u dx - int_{partial Omega} partial_v G(x,y) u(x) dsigma(x) $$

Where $v$ is the outer normal of $partial Omega$. I don't see how the above integral representation is supposed to directly follow from the definition of the Green function and the fact that if $E$ is a fundamental solution of $Delta$ we have:

$$
u(y) = - int_{Omega} E Delta u dx + int_{partial Omega} E(x,y) partial_vu(x) - partial_vE(x,y) u(x) dsigma(x)
$$

It follows from the divergence theorem applied to the vector field $h_ynabla u - nabla h_y u$. Then, the we get the terms term $partial_v u h_y = partial_vE$ and $partial_v h_y u$ which cancels the term from $partial_v G u$, so we recover the original integral representation for $u$.

First of all, remember what $E$ is. In $3$ or more dimensions, it is

$$E(x,y) = frac{1}{|x-y|^{n-2}}$$

If you take the second derivative and sum you obtain $$Delta E(x,y) = sum_{i=1}^{n}frac{partial^2 E(x,y)}{partial x_i^2} = 0 mbox{ for $xneq y$}$$

this is the same as saying $Delta E(x,y) = delta_x(y)$, where $delta_x$ function is the function such that $$delta_x = delta(y-x)$$ Where $delta$ is the Dirac Delta Function:

$$delta(x) = begin{cases}
+infty & x= 0 \
0 & xneq 0 \
end{cases}$$

This isn't a rigorous function definition, because $infty$ in just one point is meaningless. However it can be defined rigorously.

Well, the delta funciton has the nice property that

$$int_{-infty}^{infty} f(x)delta(x-a) dx = f(a)$$

The intuition is that if you have such function $E$ with the property that its laplacian is a delta function, then

$$Delta_x E(x,y) = delta_x(y-x) implies int_{-infty}^{infty}f(x)Delta E(x,y) = int_{infty}^{infty} f(x)delta_x(y-x) = f(x)$$

See that you have $f(x)$ on one side and our magical function $E$ on the other. We're gonna do that for more dimensions, but rigorously.

Consider the second Green identity:

$$int_{Omega} uDelta w - wDelta u dy = int_{partial Omega}left(ufrac{partial w}{partial n}-wfrac{partial u}{partial n}right)$$

So by doing $u=$ and $w=E$ we have:

$$int_{Omega} uDelta E - EDelta u dy = int_{partial Omega}left(ufrac{partial E}{partial n}-Efrac{partial u}{partial n}right)$$

we cannot integrate $E$ on the entire $Omega$ because it is not defined when $x=y$, so we need to separate this integral in the region $Omega_epsilon$, which is a region formed by $Omega$ minus a ball around the point of indefinition like this:

$$Omega_epsilon = Omega-B_x(epsilon)$$

So if we separate $Omega = Omega_epsilon cup B_x(epsilon)$, we know that the integral on $Omega_epsilon$ is $0$ because there's no indefinition there, and we hav to deal with the integral on $Omega_epsilon$. Note that $partialOmega_epsilon = partial Omega cup partial B_x(epsilon)$, so we end up with:

$$int_{Omega_epsilon} uDelta E - EDelta u dy =int_{Omega_epsilon}-EDelta u dy implies \ int_{partial Omega_epsilon}left(ufrac{partial E}{partial n}-Efrac{partial u}{partial n}right) = int_{partial Omega}left(ufrac{partial E}{partial n}-Efrac{partial u}{partial n}right)+ int_{partial B_x(epsilon)}left(ufrac{partial E}{partial n}-Efrac{partial u}{partial n}right)$$

First note that

$$lim_{epsilonto 0}int_{Omega_epsilon}-EDelta u dy = int_{Omega}-EDelta u dy$$

since it does not depend on $epsilon$ and by an additional unexplained argument here, $E$ is integrable on $x=y$, so we can write that integral because it is finite.

Now by another argument I'll not make here,

$$lim_{epsilonto 0} left(int_{partial Omega}left(ufrac{partial E}{partial n}-Efrac{partial u}{partial n}right)+ int_{partial B_x(epsilon)}left(ufrac{partial E}{partial n}-Efrac{partial u}{partial n}right)right) = \ int_{partial Omega} ufrac{partial E}{partial n}-Efrac{partial u}{partial n} dy + u(x)$$

so we end up with

$$int_{partial Omega} ufrac{partial E}{partial n}-Efrac{partial u}{partial n} dy + u(x) = int_{Omega}-EDelta u dy tag{1}$$

Note that it only solves half of the problem. We don't know for example how does $frac{partial u}{partial n}$ looks like on $partial Omega$. This is mitigated by introducing the corrector function $h_x(y)$ that you talked about.

The unexplained parts and the corrector function details can be found here