[proofplan] We fix $x \in U$ and excise a small ball $B(x, \varepsilon) \subset U$. On the punctured domain $U_\varepsilon := U \setminus \overline{B}(x, \varepsilon)$, the Green's function $G(x, \cdot)$ is smooth, so [Green's second identity](/theorems/27) applies to the pair $(u, G(x, \cdot))$. The PDE $-\Delta u = f$, the boundary condition $G(x, \cdot) = 0$ on $\partial U$, and the asymptotic behaviour of $G$ near its singularity reduce the identity to three contributions: a volume integral of $fG$, a boundary integral of $g \, \partial_\nu G$ on $\partial U$, and surface integrals over $\partial B(x, \varepsilon)$. Passing $\varepsilon \to 0$, the surface integrals converge to $u(x)$ and the volume integral converges to $\int_U fG \, d\mathcal{L}^n$, yielding the representation formula. [/proofplan] [step:Record the distributional identity for $G$ and the PDE for $u$] Fix $x \in U$. By hypothesis (ii), $-\Delta_y G(x, \cdot) = \delta_x$ in $\mathcal{D}'(U)$, meaning that for every $\varphi \in C_c^\infty(U)$, \begin{align*} \varphi(x) = \delta_x(\varphi) = (-\Delta_y G(x, \cdot))(\varphi) = \int_U G(x, y)\, \Delta \varphi(y) \, d\mathcal{L}^n(y). \end{align*} The solution $u \in C^2(U) \cap C(\overline{U})$ satisfies $-\Delta u = f$ in $U$ and $u = g$ on $\partial U$. [/step] [step:Excise the singularity and apply Green's second identity on the punctured domain] Since $G(x, \cdot)$ has a singularity at $y = x$ (its distributional Laplacian is $-\delta_x$, not an $L^1$ function), we cannot apply [Green's second identity](/theorems/27) directly on $U$. Instead, choose $\varepsilon > 0$ small enough that $\overline{B}(x, \varepsilon) \subset U$, and define the punctured domain \begin{align*} U_\varepsilon := U \setminus \overline{B}(x, \varepsilon). \end{align*} On $U_\varepsilon$, the function $G(x, \cdot)$ is $C^2$ (the singularity at $y = x$ has been removed) and satisfies $\Delta_y G(x, y) = 0$ for $y \in U_\varepsilon$, since $-\Delta_y G(x, \cdot) = \delta_x$ is supported at $\{x\}$. Both $u$ and $G(x, \cdot)$ are $C^2$ on $U_\varepsilon$, and $U_\varepsilon$ is a bounded domain with piecewise $C^1$ boundary $\partial U_\varepsilon = \partial U \cup \partial B(x, \varepsilon)$. Therefore [Green's second identity](/theorems/27) applies on $U_\varepsilon$: \begin{align*} \int_{U_\varepsilon} \bigl( u(y)\, \Delta_y G(x, y) - G(x, y)\, \Delta u(y) \bigr) \, d\mathcal{L}^n(y) = \int_{\partial U_\varepsilon} \bigl( u(y)\, \partial_{\nu_\varepsilon} G(x, y) - G(x, y)\, \partial_{\nu_\varepsilon} u(y) \bigr) \, d\mathcal{H}^{n-1}(y), \end{align*} where $\nu_\varepsilon$ denotes the outward unit normal to $U_\varepsilon$. On $\partial U$, the outward normal $\nu_\varepsilon$ coincides with $\nu$. On $\partial B(x, \varepsilon)$, the outward normal to $U_\varepsilon$ points *inward* toward $x$, so $\nu_\varepsilon(y) = -\frac{y - x}{|y - x|} = -\frac{y - x}{\varepsilon}$. [guided] Why can we not apply Green's second identity directly on $U$? The identity requires both functions to be $C^2$ on the domain. The function $u \in C^2(U)$ satisfies this, but $G(x, \cdot)$ does not: its distributional Laplacian $-\Delta_y G(x, \cdot) = \delta_x$ is a Dirac mass at $x$, meaning $G(x, \cdot)$ has a non-removable singularity at $y = x$. (In $\mathbb{R}^n$ with $n \geq 2$, the fundamental solution $\Phi(x - y)$ has a singularity of order $|x - y|^{2-n}$ for $n \geq 3$ or $\log|x - y|$ for $n = 2$, and $G$ inherits this singular behaviour.) The standard technique is **excision**: we remove a small ball $\overline{B}(x, \varepsilon) \subset U$ around the singularity. On the punctured domain $U_\varepsilon := U \setminus \overline{B}(x, \varepsilon)$, the function $G(x, \cdot)$ is $C^2$ because the only singularity has been excised. Moreover, $\Delta_y G(x, y) = 0$ for $y \in U_\varepsilon$, since the distributional equation $-\Delta_y G = \delta_x$ tells us that $G(x, \cdot)$ is harmonic away from $x$. The boundary of $U_\varepsilon$ has two components: the original $\partial U$ and the sphere $\partial B(x, \varepsilon)$. The outward normal to $U_\varepsilon$ on $\partial B(x, \varepsilon)$ points toward $x$ (inward with respect to the ball, outward with respect to the punctured domain), giving $\nu_\varepsilon(y) = -\frac{y - x}{\varepsilon}$. With both functions $C^2$ on $U_\varepsilon$ and $\partial U_\varepsilon$ piecewise $C^1$, [Green's second identity](/theorems/27) yields: \begin{align*} \int_{U_\varepsilon} \bigl( u(y)\, \Delta_y G(x, y) - G(x, y)\, \Delta u(y) \bigr) \, d\mathcal{L}^n(y) = \int_{\partial U_\varepsilon} \bigl( u(y)\, \partial_{\nu_\varepsilon} G(x, y) - G(x, y)\, \partial_{\nu_\varepsilon} u(y) \bigr) \, d\mathcal{H}^{n-1}(y). \end{align*} [/guided] [/step] [step:Evaluate the volume integral on $U_\varepsilon$ using $\Delta_y G = 0$ and $-\Delta u = f$] Since $\Delta_y G(x, y) = 0$ for all $y \in U_\varepsilon$ (the singularity at $x$ lies outside $U_\varepsilon$), the first term in the volume integral vanishes. Using $-\Delta u = f$: \begin{align*} \int_{U_\varepsilon} \bigl( u(y)\, \Delta_y G(x, y) - G(x, y)\, \Delta u(y) \bigr) \, d\mathcal{L}^n(y) = \int_{U_\varepsilon} G(x, y)\, f(y) \, d\mathcal{L}^n(y). \end{align*} [/step] [step:Evaluate the boundary integrals on $\partial U$ using $G(x, \cdot) = 0$ and $u = g$] The boundary $\partial U_\varepsilon = \partial U \cup \partial B(x, \varepsilon)$ splits the right-hand side into integrals over $\partial U$ and $\partial B(x, \varepsilon)$. On $\partial U$, the outward normal $\nu_\varepsilon = \nu$. By hypothesis (i), $G(x, y) = 0$ for $y \in \partial U$, so the $G \, \partial_\nu u$ term vanishes there. Since $u = g$ on $\partial U$, the $\partial U$ contribution is \begin{align*} \int_{\partial U} g(y)\, \partial_\nu G(x, y) \, d\mathcal{H}^{n-1}(y). \end{align*} The boundary integrals involving $\partial_\nu G(x, \cdot)$ on $\partial U$ are finite by hypothesis (iii), and $u, \partial_\nu u$ are [continuous](/page/Continuity) on $\overline{U}$ hence bounded on the compact set $\partial U$. [/step] [step:Evaluate the surface integrals over $\partial B(x, \varepsilon)$ and pass $\varepsilon \to 0$] On $\partial B(x, \varepsilon)$, the outward normal to $U_\varepsilon$ is $\nu_\varepsilon(y) = -\frac{y - x}{\varepsilon}$. The sphere $\partial B(x, \varepsilon)$ has $\mathcal{H}^{n-1}$-measure $n \alpha_n \varepsilon^{n-1}$, where $\alpha_n = \mathcal{L}^n(B(0,1))$ is the volume of the unit ball in $\mathbb{R}^n$. The surface integrals over $\partial B(x, \varepsilon)$ are \begin{align*} I_\varepsilon &:= \int_{\partial B(x, \varepsilon)} u(y)\, \partial_{\nu_\varepsilon} G(x, y) \, d\mathcal{H}^{n-1}(y) - \int_{\partial B(x, \varepsilon)} G(x, y)\, \partial_{\nu_\varepsilon} u(y) \, d\mathcal{H}^{n-1}(y). \end{align*} **Second integral.** Since $u \in C^2(U)$, the function $\partial_{\nu_\varepsilon} u$ is bounded on $\overline{B}(x, \varepsilon)$ by some constant $M > 0$ independent of $\varepsilon$ (for $\varepsilon$ small enough). The Green's function satisfies $G(x, y) = O(|y - x|^{2-n})$ as $y \to x$ for $n \geq 3$ (respectively $G(x, y) = O(\log|y - x|)$ for $n = 2$). On $\partial B(x, \varepsilon)$, $|y - x| = \varepsilon$, so \begin{align*} \left| \int_{\partial B(x, \varepsilon)} G(x, y)\, \partial_{\nu_\varepsilon} u(y) \, d\mathcal{H}^{n-1}(y) \right| &\leq M \cdot \sup_{y \in \partial B(x, \varepsilon)} |G(x, y)| \cdot \mathcal{H}^{n-1}(\partial B(x, \varepsilon)) \\ &= O(\varepsilon^{2-n}) \cdot O(\varepsilon^{n-1}) = O(\varepsilon) \to 0 \quad \text{as } \varepsilon \to 0 \end{align*} for $n \geq 3$. For $n = 2$, the bound becomes $O(|\log \varepsilon|) \cdot O(\varepsilon) \to 0$. **First integral.** Write $G(x, y) = \Phi(x - y) - \phi^x(y)$, where $\Phi$ is the fundamental solution of $-\Delta$ and $\phi^x \in C^2(\overline{U})$ is the corrector (harmonic in $U$ with $\phi^x = \Phi(x - \cdot)$ on $\partial U$). Then $\partial_{\nu_\varepsilon} G = \partial_{\nu_\varepsilon} \Phi(x - \cdot) - \partial_{\nu_\varepsilon} \phi^x$. The corrector term satisfies $|\partial_{\nu_\varepsilon} \phi^x| \leq C$ on $\partial B(x, \varepsilon)$, so its integral is $O(\varepsilon^{n-1}) \to 0$. For the fundamental solution, with $\nu_\varepsilon(y) = -\frac{y-x}{\varepsilon}$: \begin{align*} \partial_{\nu_\varepsilon} \Phi(x - y) = \nabla_y \Phi(x - y) \cdot \nu_\varepsilon(y) = \frac{1}{n \alpha_n \varepsilon^{n-1}} \quad \text{for } y \in \partial B(x, \varepsilon), \end{align*} using the standard computation $\nabla_y \Phi(x - y) = \frac{1}{n \alpha_n} \frac{y - x}{|y - x|^n}$. Therefore \begin{align*} \int_{\partial B(x, \varepsilon)} u(y)\, \partial_{\nu_\varepsilon} \Phi(x - y) \, d\mathcal{H}^{n-1}(y) = \frac{1}{n \alpha_n \varepsilon^{n-1}} \int_{\partial B(x, \varepsilon)} u(y) \, d\mathcal{H}^{n-1}(y) = \frac{1}{\mathcal{H}^{n-1}(\partial B(x, \varepsilon))} \int_{\partial B(x, \varepsilon)} u(y) \, d\mathcal{H}^{n-1}(y). \end{align*} This is the spherical average of $u$ over $\partial B(x, \varepsilon)$. Since $u$ is continuous at $x$, this average converges to $u(x)$ as $\varepsilon \to 0$. Combining, $I_\varepsilon \to u(x)$ as $\varepsilon \to 0$. [guided] This is the heart of the argument. The Green's function $G(x, \cdot)$ is singular at $y = x$, and we must understand precisely how the surface integrals over $\partial B(x, \varepsilon)$ behave as the excised ball shrinks. The key structural fact is that $G(x, y) = \Phi(x - y) - \phi^x(y)$, where $\Phi$ is the fundamental solution of $-\Delta$ and $\phi^x$ is a smooth corrector (harmonic in $U$, chosen so that $G(x, \cdot) = 0$ on $\partial U$). The singular behaviour of $G$ near $x$ comes entirely from $\Phi$, and $\phi^x$ is smooth up to and including $x$. For the second integral (the one involving $G \, \partial_{\nu_\varepsilon} u$): $\partial_{\nu_\varepsilon} u$ is bounded near $x$ since $u \in C^2$, and $G(x, y) = O(|y - x|^{2-n})$ for $n \geq 3$. The surface area of $\partial B(x, \varepsilon)$ is $n \alpha_n \varepsilon^{n-1}$. So the integral is bounded by $C \varepsilon^{2-n} \cdot \varepsilon^{n-1} = C\varepsilon \to 0$. The $n = 2$ case gives $O(|\log \varepsilon| \cdot \varepsilon) \to 0$. For the first integral (the one involving $u \, \partial_{\nu_\varepsilon} G$): the corrector contributes $O(\varepsilon^{n-1}) \to 0$. The fundamental solution contributes a constant flux through $\partial B(x, \varepsilon)$. Explicitly, $\nabla_y \Phi(x - y) = \frac{1}{n \alpha_n} \frac{y - x}{|y - x|^n}$, and dotting with $\nu_\varepsilon = -\frac{y - x}{\varepsilon}$ gives $\partial_{\nu_\varepsilon} \Phi = \frac{1}{n \alpha_n \varepsilon^{n-1}}$. Integrating $u \cdot \frac{1}{n \alpha_n \varepsilon^{n-1}}$ over $\partial B(x, \varepsilon)$ (which has measure $n \alpha_n \varepsilon^{n-1}$) yields exactly the spherical average of $u$. By continuity of $u$ at $x$, this average converges to $u(x)$. Thus $I_\varepsilon \to u(x) - 0 = u(x)$ as $\varepsilon \to 0$. [/guided] [/step] [step:Combine all contributions and pass $\varepsilon \to 0$ to obtain the representation formula] Assembling the identity on $U_\varepsilon$: the volume integral equals $\int_{U_\varepsilon} G(x, y) f(y) \, d\mathcal{L}^n(y)$, the $\partial U$ contribution is $\int_{\partial U} g(y) \, \partial_\nu G(x, y) \, d\mathcal{H}^{n-1}(y)$, and the $\partial B(x, \varepsilon)$ contribution is $I_\varepsilon$. Green's second identity gives \begin{align*} \int_{U_\varepsilon} G(x, y)\, f(y) \, d\mathcal{L}^n(y) = \int_{\partial U} g(y)\, \partial_\nu G(x, y) \, d\mathcal{H}^{n-1}(y) + I_\varepsilon. \end{align*} As $\varepsilon \to 0$: the volume integral $\int_{U_\varepsilon} G(x, y) f(y) \, d\mathcal{L}^n(y) \to \int_U G(x, y) f(y) \, d\mathcal{L}^n(y)$ by the dominated convergence theorem (since $G(x, \cdot) f \in L^1(U)$ by hypothesis (i) and continuity of $f$), and $I_\varepsilon \to u(x)$. The $\partial U$ integral is independent of $\varepsilon$. Therefore \begin{align*} \int_U G(x, y)\, f(y) \, d\mathcal{L}^n(y) = \int_{\partial U} g(y)\, \partial_\nu G(x, y) \, d\mathcal{H}^{n-1}(y) + u(x). \end{align*} Rearranging: \begin{align*} u(x) = -\int_{\partial U} g(y)\, \partial_\nu G(x, y) \, d\mathcal{H}^{n-1}(y) + \int_U f(y)\, G(x, y) \, d\mathcal{L}^n(y), \end{align*} which is the claimed representation formula. [/step]