Stationary sheets and charge conservation

By [A1], $\mathrm{curl}H=J_v+K{\delta }_S+{\mathrm{\partial }}_{t}D$ and $\mathrm{div}D={\rho }_v+\sigma {\delta }_S$ hold in distributions. For any vector distribution $H$, the coordinate expression for $\mathrm{div}\mathrm{curl}H$ is ${\mathrm{\partial }}_1{\mathrm{\partial }}_2{H}_3-{\mathrm{\partial }}_1{\mathrm{\partial }}_3{H}_2+{\mathrm{\partial }}_2{\mathrm{\partial }}_3{H}_1-{\mathrm{\partial }}_2{\mathrm{\partial }}_1{H}_3+{\mathrm{\partial }}_3{\mathrm{\partial }}_1{H}_2-{\mathrm{\partial }}_3{\mathrm{\partial }}_2{H}_1=0,$ because distributional partial derivatives commute by [F6], so the terms cancel in pairs. Taking the divergence of the first equation and substituting the second gives the distributional source-compatibility identity ${\mathrm{\partial }}_t{\rho }_v+{\mathrm{\partial }}_t(\sigma {\delta }_S)+\mathrm{div}{J}_v+\mathrm{div}(K{\delta }_S)=0.$

Let $\psi \in C_c^{\mathrm{\infty }}(U)$ and choose a compact interval $J\subset I$ containing its time support in its interior. The function $\mathrm{\Phi }(x,t):=\sigma (x,t)\psi (x,t)$ is $C^1$ on the compact set $S\times J$. Its time difference quotients converge uniformly in $x$ on compact subintervals to ${\mathrm{\partial }}_t\mathrm{\Phi }$, by uniform continuity of ${\mathrm{\partial }}_t\mathrm{\Phi }$; since $S$ has finite surface measure, $G(t):={\int }_S\mathrm{\Phi }(x,t)dS(x)$ is $C^1$ and $G^{\mathrm{\prime }}(t)={\int }_S(({\mathrm{\partial }}_t\sigma )\psi +\sigma {\mathrm{\partial }}_t\psi )dS$. Also $G$ vanishes near the endpoints of $J$. Using the surface-sheet pairing and one-dimensional integration by parts [F9], $⟨{\mathrm{\partial }}_t(\sigma {\delta }_S),\psi ⟩=-{\int }_J{\int }_S\sigma {\mathrm{\partial }}_t\psi dSdt={\int }_J{\int }_S({\mathrm{\partial }}_t\sigma )\psi dSdt\mathrm{.}$ Hence ${\mathrm{\partial }}_t(\sigma {\delta }_S)=({\mathrm{\partial }}_t\sigma ){\delta }_S$, with no transport term because $S$ is fixed.

For any locally integrable scalar or vector field $f$ that is $C^1$ on the two fixed sides of $S$ and has the stated joint one-sided $C^1$ extensions, its distributional time derivative is the regular distribution of its sidewise classical derivative. Indeed, on a compact spacetime cylinder, $f={\mathbf{1}}_{{\mathrm{\Omega }}_{-}}{f}_{-}+{\mathbf{1}}_{{\mathrm{\Omega }}_{+}}{f}_{+}$ almost everywhere because $S$ is three-dimensionally null by [F8]. Pairing with a compactly supported test $\psi$, Fubini gives $-{\sum }_{\pm }{\int }_{{\mathrm{\Omega }}_{\pm }}{\int }_J{f}_{\pm }(x,t){\mathrm{\partial }}_t\psi (x,t)dtd{\lambda }_3(x)$. For almost every fixed $x\in {\mathrm{\Omega }}_{\pm }$, one-dimensional integration by parts in $t$ has no endpoint term because $\psi$ is compactly supported in $I$; the joint one-sided extensions make the sidewise derivatives locally bounded near $S$, and ordinary $C^1$ regularity gives local integrability away from $S$. Fubini therefore gives ${\int }_U\{{\mathrm{\partial }}_{t}f\}\psi d{\lambda }_{3}dt$. For complex-valued tests the same calculation is applied to real and imaginary parts. The indicators of the two sides are independent of time, so no spatial-interface delta occurs. This applies componentwise to $D,B$, and ${\rho }_v$.

The four spatial jumps follow from the proof argument of [A1], not by identifying its narrower published Statement with the present hypotheses. That supplier's proof applies the spatial divergence/curl jump formulas at each time, compares the volume and surface coefficients, and then localizes the continuous surface coefficients. These steps use that $S$ is fixed, that $D,B,E,H$ have the stated joint one-sided $C^1$ extensions, and that the sheet densities are continuous; they do not differentiate $\sigma$ or $K$ in time. Here $\sigma ,K$ are jointly $C^1$, hence continuous and locally integrable on $S\times I$; step 1.3 supplies the same regular time-derivative premise for $D,B$. Fubini on compactly supported spacetime tests is justified by [F12]. Thus the proof-level jump argument applies to this time-dependent data, giving $n\cdot [D]=\sigma$, $n\cdot [B]=0$, $n\times [E]=0$, and $n\times [H]=K$ as surface-distribution identities. Their coefficients are continuous on $S\times I$, so the supplier's coefficient localization gives the stated pointwise identities. This is the broader consequence of its proof, not its published Statement.

The volume-current jump formula [F3] gives $\mathrm{div}{J}_v=\{\mathrm{div}{J}_v\}+n\cdot [J_v]{\delta }_S$. For the sheet current, pair its distributional divergence with a test $\psi$: by the distributional derivative definition and $K\cdot n=0$, $⟨\mathrm{div}(K{\delta }_S),\psi ⟩=-{\int }_I{\int }_{S}K\cdot \mathrm{\nabla }\psi dSdt={\int }_I⟨{\mathrm{Div}}_{S}K,\psi {\mathrm{\mid }}_S⟩dt\mathrm{.}$ The surface divergence here is the functional of [F4]; the identity uses only that $K$ is tangential and the interface has no edge. Substitution into step 1.1 yields $\{{\mathrm{\partial }}_t{\rho }_v+\mathrm{div}{J}_v\}+({\mathrm{\partial }}_t\sigma +n\cdot [J_v]){\delta }_S+{\mathrm{Div}}_{S}K=0.$

By step 1.3, ${\mathrm{\partial }}_t{\rho }_v$ is the regular distribution of a locally integrable sidewise derivative and has no ${\delta }_S$ term; the other volume coefficient $\{\mathrm{div}{J}_v\}$ is locally integrable by the one-sided $C^1$ hypotheses. In every open subset disjoint from $S$, the last two terms of step 2.2 vanish, so ${\mathrm{\partial }}_t{\rho }_v+\{\mathrm{div}{J}_v\}$ pairs to zero against every compactly supported test there. By [F8] it vanishes almost everywhere on $U\setminus S$; because [F8] also gives that $S$ is three-dimensionally null, the coefficient vanishes almost everywhere on all of $U$. Removing its regular distribution from step 2.2 leaves $({\mathrm{\partial }}_t\sigma +n\cdot [J_v]){\delta }_S+{\mathrm{Div}}_{S}K=0$ as a surface distribution.

For each $t\in I$, the field $K(\cdot ,t)$ is a $C^1$ tangential field on the compact $C^2$ surface, so [F11] supplies its unique continuous weak-divergence density. In each chart this density is $J_X^{-1}{\mathrm{\partial }}_a(J_X{K}^a)$; since $K$ is jointly $C^1$ and $J_X$ is positive and $C^1$, the chart formula is jointly continuous in $(x,t)$. Uniqueness makes these local formulas agree on overlaps. Write this joint continuous function as $g(x,t)={\mathrm{div}}_{S}K(x,t)$. The definition [F4], integrated against temporal tests, then identifies the surface functional in step 3.1 with the spacetime distribution $g{\delta }_S$.

Put $q={\mathrm{\partial }}_t\sigma +n\cdot [J_v]+g$, which is continuous on $S\times I$ by the stated joint $C^1$ regularity and step 4.1. The surface-distribution identity in step 3.1 says that for every real ambient spatial test $\eta \in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{R})$ and real temporal test $\theta \in C_c^{\mathrm{\infty }}(I;\mathbb{R})$, ${\int }_I\theta (t)({\int }_{S}q(x,t)\eta (x)dS(x))dt=0.$ For fixed $\eta$, the real inner pairing $p(t):={\int }_{S}q(x,t)\eta (x)dS(x)$ is continuous in $t$ by [F10]. If $p(t_{\ast })\mathrm{\ne }0$, continuity gives $r>0$ with $[t_{\ast }-r,t_{\ast }+r]\subset I$ on which $p$ has the same strict sign as $p(t_{\ast })$. Define $\theta (t)=\exp \text{\hspace{0.17em}⁣}(-1\mathrm{/}(1-((t-t_{\ast })\mathrm{/}r{)}^2))$ for $\mathrm{\mid }t-t_{\ast }\mathrm{\mid }<r$ and $\theta (t)=0$ otherwise. This is a nonnegative smooth compactly supported temporal test, positive on that interval, so ${\int }_I\theta (t)p(t)dt$ has the same strict sign as $p(t_{\ast })$, contradicting the displayed identity. Thus $p(t)=0$ for every $t$. Fixing any $t$ and applying [F5] to the continuous density $q(\cdot ,t)$ shows $q(x,t)=0$ at every $x\in S$.

For the vacuum specialization, put $D={\epsilon }_{0}E$ and $H=B\mathrm{/}{\mu }_0$. By [A2], $\mathrm{div}D={\epsilon }_0\mathrm{div}E=\rho$, $\mathrm{div}B=0$, and $\mathrm{curl}E=-{\mathrm{\partial }}_{t}B$; also $\mathrm{curl}H={\mu }_0^{-1}\mathrm{curl}B=J+({\mu }_0{c}^2{)}^{-1}{\mathrm{\partial }}_{t}E=J+{\mathrm{\partial }}_{t}D$ in distributions by [F7]. Thus the adopted vacuum stationary-sheet model supplies the thin-interface solution hypothesis [A1] with ${\rho }_f=\rho$ and $J_f=J$ in the interface neighbourhood, so the preceding steps apply. This is a model specialization, not a derivation of the vacuum postulate.