The integral form of the vacuum Maxwell equations and its equivalence with the smooth differential form

For the volume identities of (a), fix $t\in I$ and a volume test geometry $(V,{\nu }_V)$; the fixed-time fields $E(\cdot ,t),B(\cdot ,t)$ are $C^1$ on a neighbourhood of $\bar{V}$ by the said regularity, and the divergence theorem [F4] applies to that presentation: ${\oint }_{\mathrm{\partial }V}E(\cdot ,t)\cdot {\nu }_{V}dS={\int }_V\mathrm{div}E(\cdot ,t)d{\lambda }_3$ and likewise for $B$. Substituting the Gauss laws of [F1] gives ${\oint }_{\mathrm{\partial }V}E(\cdot ,t)\cdot {\nu }_{V}dS=(1\mathrm{/}{\epsilon }_0){\int }_V\rho (\cdot ,t)d{\lambda }_3$ and ${\oint }_{\mathrm{\partial }V}B(\cdot ,t)\cdot {\nu }_{V}dS=0$.

For the patch identities of (a), fix a patch test geometry $(D,r)$ and put $S=r[D]$, with parameter derivatives $r_u,r_v$ and oriented area vector $r_u\times r_v$. The flux functions ${\mathrm{\Phi }}_B(t):={\int }_D⟨B(r(u,v),t),r_u(u,v)\times r_v(u,v)⟩dudv$ and ${\mathrm{\Phi }}_E$ are differentiable on $I$ with ${\mathrm{\Phi }}_B^{\mathrm{\prime }}(t)={\int }_S{\mathrm{\partial }}_{t}B(\cdot ,t)\cdot n_{S}dS$ and ${\mathrm{\Phi }}_E^{\mathrm{\prime }}(t)={\int }_S{\mathrm{\partial }}_{t}E(\cdot ,t)\cdot n_{S}dS$: the integrands are the continuous functions $⟨B(r(u,v),t),r_u\times r_v⟩$ with continuous $t$-derivative $⟨{\mathrm{\partial }}_{t}B(r(u,v),t),r_u\times r_v⟩$ on a neighbourhood of the compact Jordan region $D$ times a compact interval around $t$, [F6] identifies the Riemann integral with the Lebesgue integral, and [F7] differentiates under the integral. Stokes' theorem [F5] applied to the $C^1$ fields gives ${\oint }_{\mathrm{\partial }S}E(\cdot ,t)\cdot d\mathbf{r}={\int }_S\mathrm{curl}E(\cdot ,t)\cdot n_{S}dS$ and the corresponding identity for $B$, in the induced orientation of the patch; substituting the Faraday law $\mathrm{curl}E=-{\mathrm{\partial }}_{t}B$ of [F1] gives ${\oint }_{\mathrm{\partial }S}E(\cdot ,t)\cdot d\mathbf{r}=-{\int }_S{\mathrm{\partial }}_{t}B(\cdot ,t)\cdot n_{S}dS=-{\mathrm{\Phi }}_B^{\mathrm{\prime }}(t)$, and substituting the Ampere law gives ${\oint }_{\mathrm{\partial }S}B(\cdot ,t)\cdot d\mathbf{r}={\mu }_0{\int }_{S}J(\cdot ,t)\cdot n_{S}dS+c^{-2}{\mathrm{\Phi }}_E^{\mathrm{\prime }}(t)$.

For clause (b), fix a closed axis-parallel box $K={\prod }_{i<3}[a_i,b_i]\subseteq \mathrm{\Omega }$ with $a_i<b_i$ and let $V$ be its open interior, a volume test geometry with $\bar{V}=K$ by [F2]. The volume identity hypothesis and the divergence theorem give $0={\oint }_{\mathrm{\partial }V}E(\cdot ,t)\cdot {\nu }_{V}dS-(1\mathrm{/}{\epsilon }_0){\int }_V\rho (\cdot ,t)d{\lambda }_3={\int }_{V}hd{\lambda }_3$ for the continuous function $h:=\mathrm{div}E-\rho \mathrm{/}{\epsilon }_0$, where linearity [F9] combines the two finite integrals and [F4] identifies the flux with ${\int }_V\mathrm{div}E$; by [F2] and [F9] this equals ${\int }_{K}hd{\lambda }_3$, because $K\setminus V$ is ${\lambda }_3$-null. Since every closed axis-parallel box $K\subseteq \mathrm{\Omega }$ with positive sides arises this way, [F8] with $n=3$ gives $h=0$ on $\mathrm{\Omega }$, that is $\mathrm{div}E=\rho \mathrm{/}{\epsilon }_0$. The same computation with the second volume identity and ${\int }_V\mathrm{div}B=0$ gives $\mathrm{div}B=0$.

For the curl equations fix a point $p\in \mathrm{\Omega }$ and let $(j,k)$ be the cyclic pair with $e_j\times e_k=e_i$ for a coordinate index $i$. Put $W:=\{(u,v)\in {\mathbb{R}}^2:p+u{e}_j+v{e}_k\in \mathrm{\Omega }\}$, an open set, and ${\phi }_i(u,v):=(\mathrm{curl}E+{\mathrm{\partial }}_{t}B)(p+u{e}_j+v{e}_k)\cdot e_i$ on $W$, a continuous function. For every closed axis-parallel square $Q\subseteq W$ with side $\mathrm{\ell }>0$ the affine patch $r(u,v):=p+u{e}_j+v{e}_k$ over $Q$ is a patch test geometry with $r_u\times r_v=e_i$ and image in $\mathrm{\Omega }$; Stokes [F5], the differentiability of its flux function and the Faraday identity of the hypothesis give ${\int }_Q{\phi }_{i}dudv={\int }_Q⟨(\mathrm{curl}E)\circ r,r_u\times r_v⟩dudv+\frac{d}{dt}{\int }_Q⟨B\circ r,r_u\times r_v⟩dudv={\oint }_{\mathrm{\partial }S}E\cdot d\mathbf{r}+\frac{d}{dt}{\int }_{S}B\cdot n_{S}dS=0$. Hence the continuous function ${\phi }_i$ has vanishing integral over every closed axis-parallel square contained in $W$, so the planar clause of [F8] gives ${\phi }_i=0$ on $W$. The point $p\in \mathrm{\Omega }$ was chosen arbitrarily, and $(0,0)\in W$ because $p\in \mathrm{\Omega }$, so evaluating at $(0,0)$ gives $(\mathrm{curl}E+{\mathrm{\partial }}_{t}B)(p)\cdot e_i=0$ for every $p\in \mathrm{\Omega }$; letting $i$ run over the three coordinate indices, $\mathrm{curl}E=-{\mathrm{\partial }}_{t}B$ on $\mathrm{\Omega }$. The same argument with the Ampere identity in place of Faraday gives $\mathrm{curl}B={\mu }_{0}J+c^{-2}{\mathrm{\partial }}_{t}E$, the flux derivative producing $c^{-2}{\mathrm{\Phi }}_E^{\mathrm{\prime }}$.

Steps 1.1 and 1.2 prove that the smooth-branch equations imply all the integral identities of (a) for every volume and patch test geometry, and steps 1.3 and 1.4 prove the converse for the same exact class; the recovery uses the open interiors of all closed axis-parallel boxes and all affine square patches, so no single selected volume or patch suffices, the marked ${\mathrm{A}\mathrm{C}}_{\omega }$ is exactly the convention of the divergence theorem and of the surface-measure interface used in [F4], [F2] and [F5], and the fields, sources and constants throughout are those of the framework definitions recorded in [F10].