PDE Laplace equation. Integral representation form and Green function 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
$$ u(y) = - \int_{\Omega} G \Delta u dx - \int_{\partial \Omega} \partial_v G(x,y) u(x) d\sigma(x) $$
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:
$$
u(y) = - \int_{\Omega} E \Delta u dx + \int_{\partial \Omega} E(x,y) \partial_vu(x) - \partial_vE(x,y) u(x) d\sigma(x)
$$

It follows from the divergence theorem applied to the vector field $h_y\nabla u - \nabla h_y u$. 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 $u$. 
