Let $n \in \mathbb{N}$ with $n\geq 3$. Let $\alpha (n)$ be the volume of the unit ball $B(0,1)$. Then the surface area of $\partial B(0,1)$ is $n \alpha(n)$. Define the fundamental solution $\Phi: \mathbb{R}^n \setminus \{0\}\to \mathbb{R}$ of $-\Delta$ by \begin{equation} \Phi(x):= \frac{1}{(n-2) n\alpha(n)} \frac{1}{\| x \|^{n-2}}. \end{equation} The fundamental solution has the following easily verified (for 5. integrate in a ball centered at $0$ and use spherical coordinates) properties:

$\Phi \in C^\infty (\mathbb{R}^n \setminus \{0\})$,
$\forall j \in \{1,\dots,n \}, x \in \mathbb{R}^n\setminus \{0\}$: \begin{equation} \partial_j \Phi (x) = - \frac{1}{n \alpha (n)} \frac{x_j}{\| x\|^n}, \end{equation}
$\forall j ,k \in \{1,\dots,n \}, x \in \mathbb{R}^n\setminus \{0\}$: \begin{equation} \partial_k \partial_j \Phi (x) = - \frac{1}{n \alpha (n)} \bigg( \frac{\delta_{j,k}}{\| x\|^n} -n \frac{x_k x_j}{ \| x \|^{n+2}} \bigg), \end{equation}
$- \Delta \Phi =0$,
$\Phi \in L^1_{\mathrm{loc}} (\mathbb{R}^n)$.

Proposition 1 (Green’s representation formula): Let $\Omega \subset \mathbb{R}^n$ be a bounded domain with smooth boundary. Let $u \in C^2(\bar \Omega)$. Then for all $x \in \Omega$: \begin{equation} u(x) = -\int_{\Omega} \Delta u \Psi dV + \int_{\partial \Omega} \Psi \langle \nabla u , N \rangle - u \langle \nabla \Psi , N \rangle dS , \end{equation} where $\Psi := \Phi (\bullet-x)$.

Proof: Let $\varepsilon \in (0, \infty)$ so that $\bar B(x,\varepsilon) \subset \Omega$. Let $\Omega_\varepsilon := \Omega \setminus \bar B (x, \varepsilon)$. By the chain rule for all $j \in \{1,\dots, n\}$ and $y \in \mathbb{R}^n \setminus \{ x\}$: $\partial_j \Psi ( y) = \partial_j \Phi (y-x)$ and $\Delta \Psi = \Delta \Phi =0$. By Green’s second identity: \begin{equation} \begin{split} \int_{\Omega_\varepsilon} \Delta u \Psi dV &= \underbrace{\int_{\Omega_\varepsilon} u \Delta \Psi dV}_{=0} + \int_{\partial \Omega_\varepsilon} \Psi \langle \nabla u , N \rangle - u \langle \nabla \Psi , N \rangle dS,\\ &=\int_{\partial \Omega} \Psi \langle \nabla u , N \rangle - u \langle \nabla \Psi , N \rangle dS + \int_{\partial B(x, \varepsilon)} \Psi \langle \nabla u , N \rangle - u \langle \nabla \Psi , N \rangle dS, \end{split} \end{equation} where $N$ is the outer surface normal vectorfield and $dS$ the surface measure of $\partial \Omega_\varepsilon$. Now, using the Cauchy-Schwarz inequality and spherical coordinates: \begin{equation} \bigg|\int_{\partial B(x, \varepsilon)} \Psi \langle \nabla u , N \rangle dS \bigg| \leq \sup_{y \in \Omega} \| \nabla u \| \int_{\partial B(0, \varepsilon)} \Phi dS \end{equation} and \begin{equation} \int_{\partial B(0, \varepsilon)} \Phi dS = \frac{1}{(n-2) n\alpha(n)} \frac{1}{ \varepsilon^{n-2}} n \alpha(n) \varepsilon^{n-1} = \frac{1}{n-2} \varepsilon. \end{equation} By definition of $N$ (pointing inwards!) and the formula for $\nabla \Phi$: \begin{equation} \int_{\partial B(x, \varepsilon)} - u \langle \nabla \Psi , N \rangle dS = - \frac{1}{n \alpha (n) \varepsilon^{n-1}} \int_{\partial B(x, \varepsilon)} u dS \to -u (x) \ (\varepsilon \to 0). \end{equation} The limit follows from the fact that $u$ is continuous and that $n \alpha (n) \varepsilon^{n-1}$ is the surface area of $\partial B(x,\varepsilon)$. Therefore by combining the preceding \begin{equation} \int_{\Omega} \Delta u \Psi dV = \lim_{\varepsilon \to 0} \int_{\Omega_\varepsilon} \Delta u \Psi dV = \int_{\partial \Omega} \Psi \langle \nabla u , N \rangle - u \langle \nabla \Psi , N \rangle dS - u(x). \quad \square \end{equation}

Remark: In the situation of the preceding: If $\Delta u=0$ then Green’s representation formula shows that the values of $u$ in $\Omega$ are determined only by the values of $u$ and $\nabla u$ on $\partial \Omega$. This is similar to Cauchys integral formula in complex analysis.

Proposition 1 ($\Phi$ is the fundamental solution of $-\Delta$): Let $f \in C_c^2(\mathbb{R}^n)$ and $x \in \mathbb{R}^n$. Then \begin{equation} \int_{\mathbb{R}^n} -\Delta f(y) \Phi(y-x) dV(y) = f(x). \end{equation} Proof: Let $R \in (0, \infty)$ such that the support of $f$ is contained in $B(0,R)$. Then according to Green’s representation formula \begin{equation} \int_{\mathbb{R}^n} -\Delta f(y) \Phi(y-x) dV(y) = \int_{B(0,R)} -\Delta f(y) \Phi(y-x) dV(y) =f (x). \ \square \end{equation}

Proposition 2 (solving Poisson’s equation in $\mathbb{R}^n$): Let $f \in C^2_c (\mathbb{R}^n)$. Let $u : \mathbb{R}^n \to \mathbb{R}$, \begin{equation} u(x):= \int_{\mathbb{R}^n} \Phi(y) f(x-y)d V(y). \end{equation} Then $u \in C^2(\mathbb{R}^n)$, $- \Delta u =f$ and $u \in C_0(\mathbb{R}^n)$.

Proof: Since $\Phi$ is locally integrable $u$ is well defined. By the theorem of differentiation under the integral sign $u\in C^2(\mathbb{R}^n)$ and for all $x \in \mathbb{R}^n$: \begin{equation} - \Delta u (x) = \int_{\mathbb{R}^n} -\Delta f(x-y) \Phi (y) d V(y) = \int_{\mathbb{R}^n} -\Delta f(y) \underbrace{\Phi (x-y)}_{= \Phi (y-x)} d V(y) = f(x), \end{equation} by the preceding proposition. Let $R\in (0, \infty)$ such that the support of $f$ is contained in $B(0,R)$. Let $R^\prime \in (0,\infty)$ with $R^\prime > R$. Let $x \in \mathbb{R}^n$ with $\|x\| \geq R^\prime$. Then \begin{equation} u(x) = \int_{B(0,R )} \Phi(x-y) f(y) dV(y). \end{equation} Inserting the definition of $\Phi$: \begin{equation} |u(x)| \leq \int_{B(0,R )} |f(y)| \frac{1}{n(n-2) \alpha(n)}\frac{1}{\|x-y\|^{n-2}} dV(y) \end{equation} and so \begin{equation} |u(x)|\leq \frac{1}{n(n-2) \alpha(n)}\frac{1}{(R^\prime -R)^{n-2}} \int_{B(0,R )} |f| dV, \end{equation} since by the reverse triangle inequality for all $y \in B(0,R)$: $\|x-y\| \geq |\|x\|- \|y\|| \geq R^\prime - R$. Choosing $R^\prime$ large enough shows that $u \in C^0 (\mathbb{R}^n)$. $\square$