[proofplan] Apply Green's second identity on the excised domain $\Omega_\varepsilon := \Omega \setminus \overline{B(x,\varepsilon)}$ with $v(y) := \Phi(x - y)$ and $u$. Since $\Delta_y \Phi(x - y) = 0$ for $y \neq x$, the volume integral reduces to $\int_{\Omega_\varepsilon} \Phi(x-y)\,\Delta u(y) \, d\mathcal{L}^n(y)$. The boundary integrals on $\partial B(x,\varepsilon)$ are evaluated using the same computation as in the [Fundamental Solution theorem](/theorems/566), extracting $u(x)$ in the limit $\varepsilon \to 0$. [/proofplan] [step:Apply Green's second identity on the excised domain $\Omega \setminus B(x,\varepsilon)$] Fix $x \in \Omega$ and let $\varepsilon > 0$ be small enough that $B(x,\varepsilon) \subseteq \Omega$. On $\Omega_\varepsilon := \Omega \setminus \overline{B(x,\varepsilon)}$, both $u$ and $v(y) := \Phi(x - y)$ are $C^2$. Green's second identity gives: \begin{align*} \int_{\Omega_\varepsilon} (v\,\Delta u - u\,\Delta v) \, d\mathcal{L}^n = \int_{\partial\Omega} \left(v\frac{\partial u}{\partial\nu} - u\frac{\partial v}{\partial\nu}\right) d\mathcal{H}^{n-1} - \int_{\partial B(x,\varepsilon)} \left(v\frac{\partial u}{\partial\nu} - u\frac{\partial v}{\partial\nu}\right) d\mathcal{H}^{n-1}, \end{align*} where $\nu$ is the outward normal to $\Omega_\varepsilon$ (outward on $\partial\Omega$, inward on $\partial B(x,\varepsilon)$). Since $\Delta v = \Delta_y \Phi(x - y) = 0$ for $y \neq x$ (by the [Fundamental Solution theorem](/theorems/566)): \begin{align*} \int_{\Omega_\varepsilon} \Phi(x-y)\,\Delta u(y) \, d\mathcal{L}^n(y) = \int_{\partial\Omega} \left(\Phi\frac{\partial u}{\partial\nu} - u\frac{\partial\Phi}{\partial\nu}\right) d\mathcal{H}^{n-1} - \int_{\partial B_\varepsilon} \left(\Phi\frac{\partial u}{\partial\nu} - u\frac{\partial\Phi}{\partial\nu}\right) d\mathcal{H}^{n-1}. \end{align*} [/step] [step:Evaluate the boundary integrals on $\partial B(x,\varepsilon)$ and take $\varepsilon \to 0$] The computations are identical to those in the proof of the [Fundamental Solution theorem](/theorems/566), with $u$ replacing $\varphi$: The term $\int_{\partial B_\varepsilon} \Phi\,\frac{\partial u}{\partial\nu} \, d\mathcal{H}^{n-1}$ is bounded by $C\varepsilon$ (or $C\varepsilon|\log\varepsilon|$ for $n = 2$) and vanishes as $\varepsilon \to 0$. The term $\int_{\partial B_\varepsilon} u\,\frac{\partial\Phi}{\partial\nu} \, d\mathcal{H}^{n-1}$ equals the average of $u$ over $\partial B(x,\varepsilon)$ (since $\frac{\partial\Phi}{\partial\nu}(x - y) = \frac{1}{n\omega_n\varepsilon^{n-1}}$ on $\partial B(x,\varepsilon)$ with the inward-pointing normal), which converges to $u(x)$ by continuity. The volume integral converges by dominated convergence since $\Phi(x - \cdot) \in L^1_{\mathrm{loc}}$ and $\Delta u$ is bounded on compact sets. [/step] [step:Rearrange to obtain the representation formula] Taking $\varepsilon \to 0$: \begin{align*} \int_\Omega \Phi(x-y)\,\Delta u(y) \, d\mathcal{L}^n(y) = \int_{\partial\Omega} \left(\Phi(x-y)\frac{\partial u}{\partial\nu}(y) - u(y)\frac{\partial\Phi}{\partial\nu_y}(x-y)\right) d\mathcal{H}^{n-1}(y) - u(x). \end{align*} Rearranging: \begin{align*} u(x) = -\int_{\partial\Omega} \Phi(x-y)\frac{\partial u}{\partial\nu}(y) \, d\mathcal{H}^{n-1}(y) + \int_{\partial\Omega} u(y)\frac{\partial\Phi}{\partial\nu_y}(x-y) \, d\mathcal{H}^{n-1}(y) - \int_\Omega \Phi(x-y)\,\Delta u(y) \, d\mathcal{L}^n(y). \end{align*} [/step]