Distributional divergence and curl across a stationary interface

By [F1] the components of $F$ and of its one-sided first derivatives are locally integrable on $\mathrm{\Omega }$, the piecewise fields $\{{\mathrm{\partial }}_i{F}_j\}$ are locally integrable, and $[F_j]$ is continuous on $S$; hence all the distributions displayed in the Statement are defined, and $n_i[F_j]{\delta }_S$ is the surface distribution of the continuous density $n_i[F_j]$ of [F1].

Reduction to finite chart boxes. At a point of $S$ use the $C^2$ slice diffeomorphism $\varphi$ from [F1], with $\varphi (S)=\{s=0\}$ locally. Choose a closed rectangular product box inside its open image, centred at the image point, and write $B$ for its open inverse image. Shrink the box so its closure is compactly contained in the chart and in $\mathrm{\Omega }$. The two sets $B_{\pm }=B\cap {\mathrm{\Omega }}_{\pm }$ are inverse images of the two open half-boxes: the local side designation is determined by the normal convention and continuity, and the half-boxes are connected. Each has a specified finite piecewise $C^1$ presentation, with the interface rectangle at $s=0$ and the finitely many outer rectangular faces transported by the $C^2$ chart inverse. Faces are regular parametrized compact patches. Their seams are transported line segments and vertices, hence null in ambient volume: the chart inverse has bounded first derivatives on the compact box, so a subdivision of each segment into $O({\delta }^{-1})$ pieces has images covered by balls of radius $C\delta$, whose total three-dimensional content is $O({\delta }^2)\to 0$. For the required surface-null condition, parametrize each face by its two free box coordinates. Its seam preimage is a finite union of line segments, covered by $O({\delta }^{-1})$ parameter squares of area $O({\delta }^2)$; thus it has two-dimensional Lebesgue measure zero. The chart surface density is bounded on the compact face, so its surface integral on those seams is zero. This verifies the finite-face and seam obligations in [F3], without asserting any property of a spherical chart cut. Compactness of the test support and [F4] now give finitely many such $B_1,\dots ,B_N$ covering $\mathrm{supp}\psi \cap S$, with compactly supported smooth cutoffs ${\chi }_k$ inside $B_k$ summing to one near that intersection. Write $\psi ={\sum }_k{\chi }_k\psi +(1-{\sum }_k{\chi }_k)\psi$; the last test vanishes near $S$, where the identity is the ordinary classical integration-by-parts identity. Thus it suffices to prove the displayed component identity for a test supported in one $B_k$.

Local computation: let $\phi \in \mathcal{D}(\mathrm{\Omega })$ have support in a chart box $B=B_k$ as in step 1.2. By [F2], $⟨{\mathrm{\partial }}_i{F}_j,\phi ⟩=-{\int }_{\mathrm{\Omega }}{F}_j{\mathrm{\partial }}_i\phi =-{\int }_{{\mathrm{\Omega }}_{+}}{F}_j{\mathrm{\partial }}_i\phi -{\int }_{{\mathrm{\Omega }}_{-}}{F}_j{\mathrm{\partial }}_i\phi$. On each side apply the divergence theorem [F3] to $G^{\pm }:=F_j\phi e_i$. Its values and first spatial derivatives extend continuously to the closed half-box by the one-sided trace hypothesis and the product rule, so it belongs to $C^1(\overline{B_{\pm }})$ in the stated up-to-boundary convention, which requires no ambient extension. It vanishes near every outer box face because $\mathrm{supp}\phi \subseteq B$. For complex $\phi$, apply the real divergence theorem separately to its real and imaginary parts; the outward normal of ${\mathrm{\Omega }}_{+}$ on the interface face is $-n$ and that of ${\mathrm{\Omega }}_{-}$ is $+n$, so ${\int }_{{\mathrm{\Omega }}_{+}}{\mathrm{\partial }}_i(F_j\phi )=-{\int }_{S\cap B}(F_j{)}_{+}\phi n_{i}dS$ and ${\int }_{{\mathrm{\Omega }}_{-}}{\mathrm{\partial }}_i(F_j\phi )={\int }_{S\cap B}(F_j{)}_{-}\phi n_{i}dS$. Substituting ${\int }_{{\mathrm{\Omega }}_{\pm }}{F}_j{\mathrm{\partial }}_i\phi ={\int }_{{\mathrm{\Omega }}_{\pm }}{\mathrm{\partial }}_i(F_j\phi )-{\int }_{{\mathrm{\Omega }}_{\pm }}({\mathrm{\partial }}_i{F}_j)\phi$ gives $⟨{\mathrm{\partial }}_i{F}_j,\phi ⟩={\int }_{\mathrm{\Omega }}\{{\mathrm{\partial }}_i{F}_j\}\phi +{\int }_S((F_j{)}_{+}-(F_j{)}_{-})\phi n_{i}dS={\int }_{\mathrm{\Omega }}\{{\mathrm{\partial }}_i{F}_j\}\phi +⟨n_i[F_j]{\delta }_S,\phi ⟩$, which is the componentwise identity.

Summing step 2.1 over $i$ (and using the definitions of Divergence and curl of a $C^1$ vector field and of the piecewise fields) gives $⟨\mathrm{div}F,\phi ⟩={\int }_{\mathrm{\Omega }}\{\mathrm{div}F\}\phi +⟨n\cdot [F]{\delta }_S,\phi ⟩$ for every test supported in some $B_k$, hence, by the finite partition of step 1.2, for every $\psi \in \mathcal{D}(\mathrm{\Omega })$; this is the divergence identity on $\mathrm{\Omega }$.

Antisymmetrizing step 2.1 gives the curl identity: by [F5], $(\mathrm{curl}F{)}_k={ϵ}_{kij}{\mathrm{\partial }}_i{F}_j={ϵ}_{kij}\{{\mathrm{\partial }}_i{F}_j\}+{ϵ}_{kij}{n}_i[F_j]{\delta }_S=\{\mathrm{curl}F{\}}_k+(n\times [F]{)}_k{\delta }_S$ for each $k$, since ${ϵ}_{kij}\{{\mathrm{\partial }}_i{F}_j\}$ is the classical $k$-th component of the piecewise curl and ${ϵ}_{kij}{n}_i[F_j]=(n\times [F]{)}_k$ in the stated convention; hence, for every $\psi \in \mathcal{D}(\mathrm{\Omega })$, $\mathrm{curl}F=\{\mathrm{curl}F{\}}_{\mathrm{p}\mathrm{i}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}}+n\times [F]{\delta }_S$.