[proofplan] All three identities follow from the [integration-by-parts formula](/theorems/29) applied componentwise. Identity (i) applies integration by parts to each second partial derivative of $u$ with test function $v \equiv 1$, then sums over coordinates to recover the [Laplacian](/page/Laplace%27s%20Equation) and normal derivative. Identity (ii) applies the same formula to each product $u \cdot \partial_{x_i} v$, then sums to obtain the gradient inner product, the Laplacian term, and a boundary integral. Identity (iii) subtracts two instances of (ii) with the roles of $u$ and $v$ interchanged, cancelling the gradient terms and leaving only Laplacian and boundary data. [/proofplan] [step:Apply integration by parts componentwise and sum to obtain the first Green identity] For each $i \in \{1, \dots, n\}$, apply the [integration-by-parts formula](/theorems/29) to the function $\partial_{x_i} u \in C^1(\overline{U})$ with test function $w \equiv 1$. Since $U$ is bounded with $C^1$ boundary and $u \in C^2(\overline{U})$, the hypotheses of the integration-by-parts formula are satisfied. This gives \begin{align*} \int_U \frac{\partial^2 u}{\partial x_i^2}(x) \, d\mathcal{L}^n(x) = \int_{\partial U} \frac{\partial u}{\partial x_i}(x) \, \nu_i(x) \, d\mathcal{H}^{n-1}(x), \end{align*} where $\nu_i(x)$ denotes the $i$-th component of the outward unit normal $\nu(x)$ at $x \in \partial U$. Summing over $i = 1, \dots, n$ and using the definitions $\Delta u = \sum_{i=1}^n \partial_{x_i}^2 u$ and $\frac{\partial u}{\partial \nu} = \nabla u \cdot \nu = \sum_{i=1}^n \partial_{x_i} u \cdot \nu_i$: \begin{align*} \int_U \Delta u(x) \, d\mathcal{L}^n(x) &= \sum_{i=1}^n \int_{\partial U} \frac{\partial u}{\partial x_i}(x) \, \nu_i(x) \, d\mathcal{H}^{n-1}(x) \\ &= \int_{\partial U} \frac{\partial u}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \end{align*} This is identity (i). $\square$ [guided] The first Green identity says that integrating the Laplacian of $u$ over $U$ equals the flux of $\nabla u$ through the boundary. How do we prove this from integration by parts? The Laplacian $\Delta u = \sum_{i=1}^n \partial_{x_i}^2 u$ is a sum of second-order partial derivatives, so the strategy is to handle each coordinate direction separately and then sum. For each $i \in \{1, \dots, n\}$, we apply the [integration-by-parts formula](/theorems/29) to the function $\partial_{x_i} u \in C^1(\overline{U})$ paired with the constant test function $w \equiv 1$. We must verify the hypotheses: the formula requires $U \subset \mathbb{R}^n$ to be a bounded open set with $C^1$ boundary and the functions to lie in $C^1(\overline{U})$. Since $u \in C^2(\overline{U})$, each partial derivative $\partial_{x_i} u$ is in $C^1(\overline{U})$, and $w \equiv 1 \in C^1(\overline{U})$. The formula yields \begin{align*} \int_U \frac{\partial^2 u}{\partial x_i^2}(x) \, d\mathcal{L}^n(x) = \int_{\partial U} \frac{\partial u}{\partial x_i}(x) \, \nu_i(x) \, d\mathcal{H}^{n-1}(x), \end{align*} where $\nu_i(x)$ denotes the $i$-th component of the outward unit normal $\nu(x)$ at $x \in \partial U$. Now sum over $i = 1, \dots, n$. On the left-hand side, the sum of $\partial_{x_i}^2 u$ over all $i$ is exactly $\Delta u$ by definition. On the right-hand side, the sum $\sum_{i=1}^n \partial_{x_i} u(x) \, \nu_i(x) = \nabla u(x) \cdot \nu(x) = \frac{\partial u}{\partial \nu}(x)$ is the outward normal derivative. Therefore \begin{align*} \int_U \Delta u(x) \, d\mathcal{L}^n(x) &= \sum_{i=1}^n \int_{\partial U} \frac{\partial u}{\partial x_i}(x) \, \nu_i(x) \, d\mathcal{H}^{n-1}(x) \\ &= \int_{\partial U} \frac{\partial u}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \end{align*} This completes the proof of identity (i). $\square$ [/guided] [/step] [step:Apply integration by parts to each product $u \cdot \partial_{x_i} v$ and sum to obtain the second Green identity] For each $i \in \{1, \dots, n\}$, apply the [integration-by-parts formula](/theorems/29) to the pair $(u, \partial_{x_i} v)$, where $u \in C^2(\overline{U}) \subset C^1(\overline{U})$ and $\partial_{x_i} v \in C^1(\overline{U})$ (since $v \in C^2(\overline{U})$). The formula gives \begin{align*} \int_U \frac{\partial u}{\partial x_i}(x) \, \frac{\partial v}{\partial x_i}(x) \, d\mathcal{L}^n(x) = -\int_U u(x) \, \frac{\partial^2 v}{\partial x_i^2}(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, \frac{\partial v}{\partial x_i}(x) \, \nu_i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Sum over $i = 1, \dots, n$. On the left-hand side, $\sum_{i=1}^n \partial_{x_i} u \cdot \partial_{x_i} v = \nabla u \cdot \nabla v$. On the right-hand side, $\sum_{i=1}^n \partial_{x_i}^2 v = \Delta v$ and $\sum_{i=1}^n \partial_{x_i} v \cdot \nu_i = \frac{\partial v}{\partial \nu}$. This yields \begin{align*} \int_U \nabla u(x) \cdot \nabla v(x) \, d\mathcal{L}^n(x) = -\int_U u(x) \, \Delta v(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, \frac{\partial v}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Since the Euclidean dot product is commutative, $\nabla u \cdot \nabla v = \nabla v \cdot \nabla u$, giving identity (ii). $\square$ [guided] Identity (ii) relates the $L^2$ inner product of the gradients to a bulk Laplacian term and a boundary term. The idea is the same as for (i) --- apply integration by parts componentwise --- but now we pair $u$ with each partial derivative $\partial_{x_i} v$ rather than using a constant test function. For each $i \in \{1, \dots, n\}$, we apply the [integration-by-parts formula](/theorems/29) to $u$ and $\partial_{x_i} v$. We verify the hypotheses: $u \in C^2(\overline{U}) \subset C^1(\overline{U})$ and $\partial_{x_i} v \in C^1(\overline{U})$ because $v \in C^2(\overline{U})$. The domain $U$ is bounded with $C^1$ boundary by hypothesis. Integration by parts gives \begin{align*} \int_U \frac{\partial u}{\partial x_i}(x) \, \frac{\partial v}{\partial x_i}(x) \, d\mathcal{L}^n(x) = -\int_U u(x) \, \frac{\partial^2 v}{\partial x_i^2}(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, \frac{\partial v}{\partial x_i}(x) \, \nu_i(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Why does the negative sign appear on the bulk term? This is the signature of integration by parts: differentiating $u$ in the $x_i$-direction "transfers" a derivative onto $\partial_{x_i} v$, producing a second derivative $\partial_{x_i}^2 v$, at the cost of a sign flip and a boundary correction. Now sum over $i = 1, \dots, n$. The left-hand side becomes \begin{align*} \sum_{i=1}^n \int_U \frac{\partial u}{\partial x_i}(x) \, \frac{\partial v}{\partial x_i}(x) \, d\mathcal{L}^n(x) = \int_U \nabla u(x) \cdot \nabla v(x) \, d\mathcal{L}^n(x), \end{align*} since $\nabla u \cdot \nabla v = \sum_{i=1}^n \partial_{x_i} u \cdot \partial_{x_i} v$ by definition of the Euclidean inner product. On the right-hand side, the bulk term collects the Laplacian $\sum_{i=1}^n \partial_{x_i}^2 v = \Delta v$, and the boundary term collects the normal derivative $\sum_{i=1}^n \partial_{x_i} v \cdot \nu_i = \nabla v \cdot \nu = \frac{\partial v}{\partial \nu}$. We obtain \begin{align*} \int_U \nabla u(x) \cdot \nabla v(x) \, d\mathcal{L}^n(x) = -\int_U u(x) \, \Delta v(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, \frac{\partial v}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Since the Euclidean dot product is commutative, $\nabla u \cdot \nabla v = \nabla v \cdot \nabla u$, and the left-hand side may equivalently be written $\int_U \nabla v \cdot \nabla u \, d\mathcal{L}^n$. This is identity (ii). $\square$ [/guided] [/step] [step:Subtract two instances of identity (ii) with $u$ and $v$ interchanged to obtain the third Green identity] Write identity (ii) as established above: \begin{align*} \int_U \nabla v(x) \cdot \nabla u(x) \, d\mathcal{L}^n(x) = -\int_U u(x) \, \Delta v(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, \frac{\partial v}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \tag{A} \end{align*} Apply identity (ii) again with the roles of $u$ and $v$ interchanged (both $u, v \in C^2(\overline{U})$, so the hypotheses remain satisfied): \begin{align*} \int_U \nabla u(x) \cdot \nabla v(x) \, d\mathcal{L}^n(x) = -\int_U v(x) \, \Delta u(x) \, d\mathcal{L}^n(x) + \int_{\partial U} v(x) \, \frac{\partial u}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \tag{B} \end{align*} Since $\nabla v \cdot \nabla u = \nabla u \cdot \nabla v$, the left-hand sides of (A) and (B) are equal. Subtracting (B) from (A) cancels the gradient terms and yields \begin{align*} 0 &= -\int_U u(x) \, \Delta v(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, \frac{\partial v}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) \\ &\quad + \int_U v(x) \, \Delta u(x) \, d\mathcal{L}^n(x) - \int_{\partial U} v(x) \, \frac{\partial u}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Rearranging: \begin{align*} \int_U \bigl( u(x) \, \Delta v(x) - v(x) \, \Delta u(x) \bigr) \, d\mathcal{L}^n(x) = \int_{\partial U} \left( u(x) \, \frac{\partial v}{\partial \nu}(x) - v(x) \, \frac{\partial u}{\partial \nu}(x) \right) d\mathcal{H}^{n-1}(x). \end{align*} This is identity (iii), completing the proof of all three Green identities. $\blacksquare$ [guided] Identity (iii) is the most powerful of the three: it relates the asymmetry between $u \Delta v$ and $v \Delta u$ inside the domain to boundary data. The proof is a pure algebraic consequence of identity (ii). Write identity (ii) in the form already established: \begin{align*} \int_U \nabla v(x) \cdot \nabla u(x) \, d\mathcal{L}^n(x) = -\int_U u(x) \, \Delta v(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, \frac{\partial v}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \tag{A} \end{align*} Now apply identity (ii) again, but with the roles of $u$ and $v$ swapped. Both functions lie in $C^2(\overline{U})$, so all hypotheses remain valid. This gives \begin{align*} \int_U \nabla u(x) \cdot \nabla v(x) \, d\mathcal{L}^n(x) = -\int_U v(x) \, \Delta u(x) \, d\mathcal{L}^n(x) + \int_{\partial U} v(x) \, \frac{\partial u}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \tag{B} \end{align*} The key observation is that the left-hand sides of (A) and (B) are identical: $\nabla v \cdot \nabla u = \nabla u \cdot \nabla v$ by commutativity of the Euclidean inner product. Therefore the right-hand sides must also be equal. Setting them equal and rearranging: \begin{align*} -\int_U u(x) \, \Delta v(x) \, d\mathcal{L}^n(x) + \int_{\partial U} u(x) \, \frac{\partial v}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) &= -\int_U v(x) \, \Delta u(x) \, d\mathcal{L}^n(x) \\ &\quad + \int_{\partial U} v(x) \, \frac{\partial u}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Moving the Laplacian terms to one side and the boundary terms to the other: \begin{align*} \int_U \bigl( u(x) \, \Delta v(x) - v(x) \, \Delta u(x) \bigr) \, d\mathcal{L}^n(x) = \int_{\partial U} \left( u(x) \, \frac{\partial v}{\partial \nu}(x) - v(x) \, \frac{\partial u}{\partial \nu}(x) \right) d\mathcal{H}^{n-1}(x). \end{align*} This is identity (iii). Note that (iii) encodes a symmetry principle: the Laplacian is formally self-adjoint with respect to the $L^2$ inner product, and (iii) measures the failure of exact self-adjointness by boundary terms. When both $u$ and $v$ vanish on $\partial U$, the right-hand side is zero and we recover $\int_U u \, \Delta v \, d\mathcal{L}^n = \int_U v \, \Delta u \, d\mathcal{L}^n$, the symmetry of the [Dirichlet](/page/Laplace%27s%20Equation) Laplacian. $\blacksquare$ [/guided] [/step]