Proving the Divergence Theorem and Green's Identities from Stokes Theorem

Divergence Theorem: Let $n \in \mathbb{N}$ and $\Omega \subset \mathbb{R}^n$ a bounded domain with smooth boundary. Let $N$ be the outwardpointing normal vector field. Let $U$ be an open neighborhood of $\bar \Omega$ and $X:= \sum_{j=1}^n F_j \frac{\partial}{\partial x_j}$ be a $C^1$ vector field on $U$. Then \begin{equation} \int_\Omega \sum_{j=1}^n \partial_j F_j d V = \int_{\partial \Omega} \langle N, X\rangle dS. \end{equation} Proof: Let $§$ denote contraction in the first slot. Let $\omega := X § dV$. Then \begin{equation} \omega = \sum_{j=1}^n (-1)^{j-1} F_j dx_1 \wedge \cdots \wedge \widehat{dx}_j \wedge \cdots \wedge dx_n. \end{equation} From here it follows by applying the definition of $d$ that \begin{equation} d \omega = \sum_{j=1}^n \partial_j F_j d V . \end{equation} Therefore by Stokes theorem \begin{equation} \int_\Omega \sum_{j=1}^n \partial_j F_j d V = \int_\Omega d \omega = \int_{\partial \Omega} \iota^* \omega, \end{equation} where $\iota : \partial \Omega \to \bar \Omega$ is the embedding. By definition $dS = N §\iota^* dV$. Now let \begin{equation} X^\perp := \langle N, X\rangle N \end{equation} and $Y := X - X^\perp$. Then by linearity \begin{equation} \iota^* \omega = \iota^* (X § dV) = \iota^* (X^\perp §d V) + \iota^* (Y § dV) \end{equation} and \begin{equation} \iota^* (X^\perp § dV) = \langle N, X\rangle N § \iota^* dV = \langle N, X \rangle dS. \end{equation} Let $v_1, \dots, v_{n-1}$ be any tangent vectors of $\partial \Omega$. Then \begin{equation} \iota^* (Y § dV) (v_1, \dots, v_{n-1}) = dV ( Y, v_1, \dots, v_{n-1}) =0, \end{equation} because $Y$ is tangent to $\partial \Omega$ (since $\langle Y, N\rangle =0$) and $n$ vectors from a $n-1$ dimensional vector space must be linearly dependent (as well as $dV$ being a top form). In total this shows that \begin{equation} \int_\Omega \sum_{j=1}^n \partial_j F_j d V = \int_\Omega d \omega = \int_{\partial \Omega} \iota^* \omega = \int_{\partial \Omega} \langle N, X \rangle dS, \end{equation} which is the divergence theorem. $\square$

Green’s First Identity: In the situation of the divergence theorem: Let $\varphi, \psi \in C^2(U)$. Then \begin{equation} \int_{\Omega} \varphi \Delta \psi + \sum_{j=1}^n \partial_j \psi \partial_j \varphi dV = \int_{\partial \Omega} \varphi \bigg\langle \sum_{j=1}^n \partial_j \psi \frac{\partial }{\partial x_j}, N \bigg\rangle dS. \end{equation} Proof: Take $X:= \sum_{j=1}^n \varphi \partial_j \psi \frac{\partial }{\partial x_j}$ in the divergence theorem and use the product rule. $\square$

Green’s Second Identity: In the situation of the divergence theorem: Let $\varphi, \psi \in C^2(U)$. Then \begin{equation} \int_{\Omega} \varphi \Delta \psi - \Delta \varphi \psi dV = \int_{\partial \Omega} \bigg\langle \sum_{j=1}^n \varphi \partial_j \psi \frac{\partial }{\partial x_j} - \psi \partial_j \varphi \frac{\partial }{\partial x_j} , N \bigg\rangle dS. \end{equation} Proof: This follows from applying Green’s first identity twice and by noticing that the symmetric terms cancel.$\square$