Planar dielectric interface with prescribed free surface charge

Localize a test by a smooth partition of unity [L3] to finitely many rectangular boxes compactly contained in $U$; for boxes meeting the plane choose the height interval to straddle $z=0$. It suffices to prove the identity for each localized test and sum. Let $F\in \{E,B,D,H\}$ and let $\psi$ be supported in one such box. For fixed $(x,y,t)$ with $z\mathrm{\ne }0$ the map $z\mapsto F(x,y,z,t)$ is $C^1$ on each part of that height interval and has the stated one-sided $C^1$ extensions at zero; applying [L1] on the two closed height subintervals and summing gives ${\int }_{\mathbb{R}}F{\mathrm{\partial }}_z\psi dz=-[F](x,y,0,t)\psi (x,y,0,t)-{\int }_{\mathbb{R}}\{{\mathrm{\partial }}_{z}F\}\psi dz$, because the outer boundary terms vanish by compact support; integrating over $x,y,t$ by [L2], with the plane $\{z=0\}$ null by [L4] so that the split into the two half-spaces is exact, gives $⟨{\mathrm{\partial }}_{z}F,\psi ⟩={\int }_U\{{\mathrm{\partial }}_{z}F\}\psi +⟨[F]{\delta }_S,\psi ⟩$ componentwise, i.e. $\mathrm{div}F=\{\mathrm{div}F\}+e_z\cdot [F]{\delta }_S$ and $\mathrm{curl}F=\{\mathrm{curl}F\}+e_z\times [F]{\delta }_S$ by [L6], the tangential derivatives contributing no sheet term.

The same per-line integration by parts in $t$, with $\psi$ of compact $t$-support and the plane stationary and null by [L4], gives $⟨{\mathrm{\partial }}_{t}F,\psi ⟩={\int }_U\{{\mathrm{\partial }}_{t}F\}\psi$ for every test $\psi$: time differentiation at the fixed plane contributes no sheet term, and the piecewise derivatives are locally bounded by the joint one-sided $C^1$ hypothesis.

Substituting the spatial identities of step 1.1 and the temporal identity of step 1.2 into the assumed equations [A2] gives four distributional identities of the form $u_g+a{\delta }_S=0$ on the local region, with locally integrable volume parts $g$: $u_{\{\mathrm{div}D\}-{\rho }_v}+(n\cdot [D]-{\sigma }_f){\delta }_S=0$, $u_{\{\mathrm{div}B\}}+n\cdot [B]{\delta }_S=0$, $u_{\{\mathrm{curl}E\}+\{{\mathrm{\partial }}_{t}B\}}+n\times [E]{\delta }_S=0$ and $u_{\{\mathrm{curl}H\}-J_v-\{{\mathrm{\partial }}_{t}D\}}+n\times [H]{\delta }_S=0$, where the two half-space fields are $C^1$ with locally bounded derivatives and ${\rho }_v,J_v$ are locally integrable.

[Coefficient extraction] Let $b\in C_c^{\mathrm{\infty }}((-1,1))$, extended by zero to $\mathbb{R}$, with $b(0)=1$, $0\le b\le 1$ [L3] and put ${\psi }_h:=\psi b(z\mathrm{/}h)$ for a test $\psi$; testing an identity $u_g+a{\delta }_S=0$ of step 2.1 with ${\psi }_h$ gives $0={\int }_{U}g{\psi }_h+{\int }_{\{z=0\}}a{\psi }_h$, whose second term tends to ${\int }_{\{z=0\}}a\psi$ because $b(z\mathrm{/}h)\to b(0)=1$ everywhere on the plane and the integrands are dominated by $\mathrm{\mid }a\psi \mathrm{\mid }$ on the compactly supported part of the plane, while the first term tends to $0$ because $\mathrm{\mid }{\int }_{U}g{\psi }_h\mathrm{\mid }\le {\int }_U\mathrm{\mid }g\psi \mathrm{\mid }{\mathbf{1}}_{\{\mathrm{\mid }z\mathrm{\mid }\le h\}}$ with integrands converging to $0$ almost everywhere, dominated by the integrable $\mathrm{\mid }g\psi \mathrm{\mid }$, the exceptional set $\{z=0\}$ being null [L4]. Hence ${\int }_{\{z=0\}}a\psi =0$ for every test $\psi$; taking $\psi (x,y,z,t)=\eta (x,y,t)\beta (z)$ with $\eta$ supported in the tangential/time part of one of those local boxes and $\beta$ supported in its height interval with $\beta (0)=1$ reduces this to $\int a(x,y,0,t)\eta (x,y,t)dxdydt=0$ for every $\eta$, so the continuous function $(x,y,t)\mapsto a(x,y,0,t)$ vanishes identically by [L5]. All four coefficients of step 2.1 therefore vanish pointwise on the plane.

Applying step 3.1 and the constitutive relations [A1]: (a) $n\times [E]=0$, so the tangential parts agree, $E_{2t}=E_{1t}$; (b) $n\cdot [D]={\sigma }_f$ with $D_j={\epsilon }_j{E}_j$ on the two sides gives ${\epsilon }_2{E}_{2n}-{\epsilon }_1{E}_{1n}={\sigma }_f$ and, since ${\epsilon }_2>0$, $E_{2n}=({\epsilon }_1\mathrm{/}{\epsilon }_2)E_{1n}+{\sigma }_f\mathrm{/}{\epsilon }_2$; (c) if ${\sigma }_f=0$ the previous identity reads ${\epsilon }_2{E}_{2n}={\epsilon }_1{E}_{1n}$, so $E_{2n}=({\epsilon }_1\mathrm{/}{\epsilon }_2)E_{1n}$ has the same sign as $E_{1n}$ and vanishes with it, and $E_{2n}=E_{1n}$ holds if and only if $({\epsilon }_1-{\epsilon }_2)E_{1n}=0$, i.e. when $E_{1n}=0$ or ${\epsilon }_1={\epsilon }_2$; (d) $n\cdot [B]=0$ and $n\times [H]=0$, and in the nonmagnetic case $H={\mu }_0^{-1}B$ on both sides the two identities give $[B_n]=0$ and $[B_t]=0$, hence $B_{2n}=B_{1n}$ and $B_{2t}=B_{1t}$ by the decomposition of [L6]. The scope (e) holds: the computation uses only the assumed local equations and the declared linear branch, asserts no global configuration, no existence, no further constitutive law and no empirical claim.