The Method of the Green's Function for the Laplacian

Let $n \in \mathbb{N}$ and Let $\Omega \subset \mathbb{R}^n$ a bounded region with smooth boundary. Let $\Phi$ be the fundamental solution of $-\Delta$ (see previous post on the topic). Suppose that for all $x \in \Omega$ there exists (a necessarily unique by a corollary to the maximum principle) $\varphi \in C^2 (\bar \Omega)$ with \begin{equation} \begin{cases} \Delta \varphi = 0 , \ \text{on} \ \Omega, \\ \varphi = \Phi (\bullet-x) , \ \text{on} \ \partial \Omega. \end{cases} \end{equation}

Definition: The Green’s function $G: \Omega \times \bar \Omega \setminus \{ (x,y) \in \Omega \times \bar \Omega : x =y \} \to \mathbb{R}$ of $\Omega$ is defined by \begin{equation} G(x,y) := \Phi (x-y) - \varphi(y) , \end{equation} where $\varphi$ is the unique function $\varphi \in C^2 (\bar \Omega)$ with \begin{equation} \begin{cases} \Delta \varphi = 0 , \ \text{on} \ \Omega, \\ \varphi = \Phi (\bullet-x) , \ \text{on} \ \partial \Omega. \end{cases} \end{equation}

Proposition 1 (representation formula with the Green’s function): Let $u \in C^2(\bar \Omega)$. Then for all $ x\in \Omega$ \begin{equation} u(x) = \int_\Omega -\Delta u G(x,\bullet) dV - \int_{\partial \Omega} u \langle \nabla G(x,\bullet ) , N \rangle dS. \end{equation}

Proof: Let $x \in \Omega$. Let $\varphi$ be the unique function $\varphi \in C^2 (\bar \Omega)$ with \begin{equation} \begin{cases} \Delta \varphi = 0 , \ \text{on} \ \Omega, \\ \varphi = \Phi (\bullet-x) , \ \text{on} \ \partial \Omega. \end{cases} \end{equation} By Green’s representation formula (see my post on the fundamental solution) with $N$ the outer surface normal vector: \begin{equation} u(x) = -\int_{\Omega} \Delta u \Phi (\bullet-x) dV + \int_{\partial \Omega} \Phi(\bullet -x) \langle \nabla u , N \rangle - u \langle \nabla \Phi(\bullet -x) , N \rangle dS . \end{equation} Applying Green’s second formula: \begin{equation} \int_\Omega \phi \Delta u dV = \underbrace{\int_\Omega \Delta \phi u dV}_{=0} + \int_{\partial \Omega} \phi \langle \nabla u, N \rangle - u \langle \nabla \phi , N \rangle dS \end{equation} Since $\phi = \Phi (\bullet-x)$ on $\partial \Omega$: \begin{equation} u(x) = -\int_{\Omega} \Delta u (\Phi (\bullet-x)- \phi) dV - \int_{\partial \Omega} u \langle \nabla (\Phi(\bullet -x)- \phi) , N \rangle dS . \end{equation} Which is the desired conclusion. $\square$

Remark: The representation formula shows if $f \in C(\Omega)$, $g \in C(\partial \Omega)$ and $u \in C^2(\bar \Omega)$ with $-\Delta u = f$ on $\Omega$ and $u = g$ on $ \partial \Omega$. Then $u$ is completly determined by $f$ and $g$. It also gives the perfect candidate for a solution.

Proposition 2: Assume $n>2$. Then $\varphi : \bar B(0,1) \to \mathbb{R}$ \begin{equation} \varphi (y) := \frac{1}{n (n-2) \alpha(n)}, \end{equation} where $\alpha(n)$ is the volume of $B(0,1)$, satisfies $\varphi \in C^2 (\bar \Omega)$ and \begin{equation} \begin{cases} \Delta \varphi = 0 , \ \text{on} \ B(0,1), \\ \varphi = \Phi , \ \text{on} \ \partial B(0,1). \end{cases} \end{equation} Let $x \in B(0,1) \setminus \{0\}$. Let $\tilde{x} := \frac{x}{\|x\|}$. Then $\varphi : \bar B(0,1) \to \mathbb{R}$ \begin{equation} \varphi (y):= \Phi ( \|x\| (y-\tilde{x})) \end{equation} satisfies $\varphi \in C^2 (\bar B(0,1))$ and \begin{equation} \begin{cases} \Delta \varphi = 0 , \ \text{on} \ B(0,1), \\ \varphi = \Phi(\bullet -x) , \ \text{on} \ \partial B(0,1). \end{cases} \end{equation}

Proof: The first part is obvious. For the second part: Since $\|\tilde{x}\| >1$ it follows that $\varphi$ is well defined and $\varphi \in C^2(\bar B(0,1))$. Let $y \in \bar B(0,1)$. Then \begin{equation} \|x\|^2 \| ( y-\tilde{x})\|^2 = \|x\|^2 \|y\|^2 - 2 \langle x, y\rangle +1. \end{equation} and therefore if $y \in \partial B(0,1)$ then \begin{equation} \|x\|^2 \| ( y-\tilde{x})\|^2 = \| x- y\|^2. \end{equation} This shows $\phi = \Phi (\bullet -x)$ on $\partial B(0,1)$. A simple (but tedious) calculation shows that $\Delta \varphi =0$ on $B(0,1)$. $\square$

Proposition 3 (Green’s function of $B(0,1)$): Assume $n>2$. Let \begin{equation} \begin{split} &G: B(0,1) \times \bar B(0,1) \setminus \{ (x,y) \in B(0,1) \times \bar B(0,1) : x =y \} \to \mathbb{R} \\ &G(x,y):= \Phi (x-y) - \frac{1}{n (n-2)\alpha(n)} \big( \|x\|^2 \|y\|^2 - 2 \langle x,y \rangle +1 \big)^{\frac{2-n}{2}}. \end{split} \end{equation} Then $G$ is the Green’s function of $B(0,1)$ and for all $x\in B(0,1), y \in \partial B(0,1)$: \begin{equation} \langle \nabla G(x, \bullet) (y), N (y) \rangle = -\frac{1- \|x\|^2}{n \alpha (n)} \frac{1}{\| x-y \|^n}, \end{equation} where $N$ is the outer surface normal vector of $B(0,1)$.

Proof: Follows from proposition 2 and an elementary calculation. $\square$

Armed with the Green’s function for $B(0,1)$ and the representation formula it is natural to conjecture:

Proposition 4 (Poisson formula for $B(0,1)$): Let $g\in C(\partial B(0,1))$ and define $u : \bar B(0,1) \to \mathbb{R}$ by \begin{equation} u(x):= \begin{cases} g (x) , \quad \ x \in \partial B(0,1), \\ \frac{1- \|x\|^2}{n \alpha (n)}\int_{\partial B(0,1)}\frac{g(y)}{\| x-y \|^n} dS(y), \quad x \in B(0,1). \end{cases} \end{equation} Then $u \in C(\bar \Omega) \cap C^2 (\Omega)$, $\Delta u =0$ on $\Omega$ and $u=g$ on $\partial B(0,1)$.

Proof: From the representation formula (and equation 16) with the constant $1$ function we get $\forall x \in B(0,1)$: \begin{equation} 1 = \frac{1- \|x\|^2}{n \alpha (n)}\int_{\partial B(0,1)}\frac{1}{\| x-y \|^n} dS(y). \end{equation} This allows to differentiate under the integral sign, showing that $u \in C^2(\Omega)$ and a tedious calculation shows that $\Delta u =0$. It is left to show the continuity of $u$: Let $x_0 \in \partial B(0,1)$ and $\varepsilon \in (0, \infty)$. Let $\delta \in (0, \infty)$ such that for all $y \in \partial B(0,1)$: \begin{equation} \|y-x_0\| <\delta \Rightarrow |g(y)-g(x_0)| < \varepsilon. \end{equation} Let $\tilde{\delta} \in (0, \delta /2)$. Let $x \in B(0,1)$ with $\|x-x_0\|< \tilde{\delta}$. Then \begin{equation} \begin{split} | u(x) - u(x_0) | &\leq \int_{\partial B(0,1)} |g(y)- g(x_0)| \frac{1- \|x\|^2}{n \alpha (n) \|x-y\|^n}dS(y) \\ & \leq \int_{\partial B(0,1) \cap B(x, \delta/2)} |g(y)- g(x_0)| \frac{1- \|x\|^2}{n \alpha (n) \|x-y\|^n}dS(y) \\ &+ \int_{\partial B(0,1) \setminus \bar B(x, \delta/2)} |g(y)- g(x_0)| \frac{1- \|x\|^2}{n \alpha (n) \|x-y\|^n}dS(y) \\ &\leq \varepsilon + (1- \|x\|^2) \sup_{ y \in \partial B(0,1)} |g (y)| (\delta/2)^{-n}. \end{split} \end{equation} From which the proposition follows by choosing $\tilde{\delta}$ appropriatly small. $\square$