Material response and stationary interfaces — Examples

Multiplying the dimension of ${\epsilon }_0$ from [L1] by the dimension $\mathrm{C}{\mathrm{m}}^{-2}$ of $D$ from [L2] gives ${\mathrm{C}}^3{\mathrm{s}}^2{\mathrm{k}\mathrm{g}}^{-1}{\mathrm{m}}^{-5}$, which differs from the dimension $\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-2}{\mathrm{C}}^{-1}$ of $E$ in the exponent of $\mathrm{C}$ and of $\mathrm{m}$; the proposed law $E={\epsilon }_{0}D$ is therefore excluded.

Solving $D={\epsilon }_{0}E+P$ for $E$ gives $E=(D-P)\mathrm{/}{\epsilon }_0$, so in vacuum $E={\epsilon }_0^{-1}D$; a law $E=\eta D$ with constant $\eta >0$ of dimension $\mathrm{V}\mathrm{m}{\mathrm{C}}^{-1}$ has the correct dimension for every such $\eta$, so the unit check alone leaves $\eta$ free and says nothing about locality, linearity or instantaneity.

Give the witness physical units and the required vector type: fix $\tau >0$ in seconds and $E_{\ast }>0$ in $\mathrm{V}{\mathrm{m}}^{-1}$, let $E:\mathbb{R}\to {\mathbb{R}}^3$ be $E(t)=E_{\ast }\min (1,\max (0,-t\mathrm{/}\tau ))e_1$, and set $K(s)={\tau }^{-1}{\mathbf{1}}_{[0,\tau ]}(s)I_3$, where $I_3$ is the identity matrix. This real matrix kernel has units ${\mathrm{s}}^{-1}$ and integrable entries; the history is bounded and continuous, with $E(0)=0$ and $E(-s)=E_{\ast }(s\mathrm{/}\tau )e_1$ for $0\le s\le \tau$. The polynomial $s^2\mathrm{/}(2{\tau }^2)$ has derivative $s\mathrm{/}{\tau }^2$ by expanding its difference quotient, so [L3] evaluates the integral. Therefore $P(0)={\epsilon }_0{E}_{\ast }{e}_1{\int }_0^{\tau }s\mathrm{/}{\tau }^{2}ds={\epsilon }_0{E}_{\ast }{e}_1\mathrm{/}2\mathrm{\ne }0$, while every instantaneous law $P=CE$ with a fixed matrix $C$ gives $P(0)=0$. Thus a memory response need not be an instantaneous coefficient law; replacing its harmonic symbol by a time-domain multiplier changes this declared model.

Steps 1.1-1.3 verify (a)-(c): the identification $E={\epsilon }_{0}D$ fails dimensionally, dimensionally admissible laws do not determine $\eta$ or the response type, and the frequency-domain symbol is defined by its action on harmonic histories rather than as an instantaneous coefficient, as the explicit history shows. The scope (d) holds: the example uses only the framework's unit and definition conventions, asserts no measurement, no material property and no empirical claim, and the material constant $\epsilon$ or susceptibility ${\chi }_e$ remains fixed by the declared model.

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.

On $r\mathrm{\ne }0$ the functions $z$ and $z{r}^{-3}$ are smooth and $r^{-3}$ is smooth away from the origin; the sum and scalar rules of [L1] give $\mathrm{\Delta }z=0$ and $\mathrm{\Delta }(z{r}^{-3})=z\mathrm{\Delta }(r^{-3})+2\mathrm{\nabla }z\cdot \mathrm{\nabla }(r^{-3})$, with ${\mathrm{\partial }}_i{r}^{-3}=-3x_i{r}^{-5}$ and $\mathrm{\Delta }(r^{-3})={\sum }_i{\mathrm{\partial }}_i(-3x_i{r}^{-5})=-9r^{-5}+15r^{-5}=6r^{-5}$; hence $\mathrm{\Delta }(z{r}^{-3})=6z{r}^{-5}+2\cdot (-3z{r}^{-5})=0$, so $\mathrm{\Delta }{\phi }_{\mathrm{i}\mathrm{n}}=0$ on $r<a$ and $\mathrm{\Delta }{\phi }_{\mathrm{o}\mathrm{u}\mathrm{t}}=0$ on $r>a$; the interior potential is linear on the entire closed ball, including the origin, while the exterior expression and its gradient are smooth up to $S$ because $r\ge a>0$ on that closed exterior side.

Put $q:=z\mathrm{/}r$ for the fixed directional factor along a radial line. At $r=a$: ${\phi }_{\mathrm{i}\mathrm{n}}(a)=-C{E}_{0}aq$ and ${\phi }_{\mathrm{o}\mathrm{u}\mathrm{t}}(a)=(-E_{0}a+d\mathrm{/}{a}^2)q$, whose equality for every angle is the scalar equation $C=1-d\mathrm{/}(E_0{a}^3)$; the radial derivatives are ${\mathrm{\partial }}_r{\phi }_{\mathrm{i}\mathrm{n}}=-C{E}_{0}q$ and ${\mathrm{\partial }}_r{\phi }_{\mathrm{o}\mathrm{u}\mathrm{t}}=(-E_0-2d\mathrm{/}{a}^3)q$, whose equality after the weights ${\epsilon }_{\mathrm{i}\mathrm{n}},{\epsilon }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ is ${\epsilon }_{\mathrm{i}\mathrm{n}}C={\epsilon }_{\mathrm{o}\mathrm{u}\mathrm{t}}(1+2d\mathrm{/}(E_0{a}^3))$. Substituting the displayed $C$ and $d$ satisfies both equations (the determinant of the pair is ${\epsilon }_{\mathrm{i}\mathrm{n}}+2{\epsilon }_{\mathrm{o}\mathrm{u}\mathrm{t}}>0$), so the two matching identities hold. By the chain rule along any smooth surface chart, equal traces have equal tangential derivatives and hence equal tangential electric fields; conversely, equality of those derivatives makes the trace difference constant along every piecewise smooth path on the sphere, whose points are joined by great-circle arcs, without fixing its value. The displayed trace computation fixes it to zero. The weighted radial identity is exactly normal-$D$ continuity because $E=-\mathrm{\nabla }\phi$ and $D=\epsilon E$. The asymptotics (c) follow from $\mathrm{\mid }z\mathrm{\mid }\le r$: $\mathrm{\mid }{\phi }_{\mathrm{o}\mathrm{u}\mathrm{t}}+E_{0}z\mathrm{\mid }=\mathrm{\mid }d\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }{r}^{-3}\le \mathrm{\mid }d\mathrm{\mid }{r}^{-2}$ and $\mathrm{\nabla }(dz{r}^{-3})=d(r^{-3}{e}_3-3z{r}^{-5}x)$ has squared norm $d^2{r}^{-6}(1+3q^2)\le 4d^2{r}^{-6}$ because $\mathrm{\mid }q\mathrm{\mid }\le 1$, hence norm at most $2\mathrm{\mid }d\mathrm{\mid }{r}^{-3}$.

For the declared static completion, [L3] gives $\mathrm{div}D=\{\mathrm{div}D\}+n\cdot [D]{\delta }_S$ and $\mathrm{curl}E=\{\mathrm{curl}E\}+n\times [E]{\delta }_S$, with analogous formulas for $B$ and $H$; by step 1.1 the piecewise classical divergences and curls of $D$ and $E$ vanish (each side is a constant multiple of the gradient of a harmonic function, so $\mathrm{div}E=0$ and $\mathrm{curl}E=0$ there, the latter by the mixed-partial commutation of [L1]), and by step 1.2 the normal jump $n\cdot [D]$ and tangential jump $n\times [E]$ vanish; since $B=H=0$ and ${\rho }_f=J_f=0$ with ${\sigma }_f=K_f=0$, all four distributional macroscopic equations hold on the local region and the hypotheses of Stationary macroscopic interface jump conditions for D, B, E and H ($S=\{r=a\}$ a compact embedded $C^2$ hypersurface in $\mathrm{\Omega }$ because ${\bar{B}}_a\subseteq \mathrm{\Omega }$, locally integrable fields, $C^1$ off $S$, joint one-sided $C^1$ extensions, stationary free sheet data) are verified.

For uniqueness, let $w_{\mathrm{i}\mathrm{n}}={\tilde{\phi }}_{\mathrm{i}\mathrm{n}}-{\phi }_{\mathrm{i}\mathrm{n}}$ on $\overline{B_a}$ and $w_{\mathrm{o}\mathrm{u}\mathrm{t}}={\tilde{\phi }}_{\mathrm{o}\mathrm{u}\mathrm{t}}-{\phi }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ on $\{r\ge a\}$; both are harmonic with $C^2$ regularity on the closed regions, their traces and weighted radial derivatives agree on $S$ by the shared interface conditions, and $\mathrm{\mid }{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}\mathrm{\mid }\le (M+\mathrm{\mid }d\mathrm{\mid })r^{-1}$ for all sufficiently large $r\ge 1$. Apply the first Green identity of [L2] with $u=v=w_{\mathrm{i}\mathrm{n}}$ on the ball $B_a$ and with $u=v=w_{\mathrm{o}\mathrm{u}\mathrm{t}}$ on the annulus $\{a<r<R\}$: since the Laplacians vanish, ${\int }_{B_a}\mathrm{\mid }\mathrm{\nabla }{w}_{\mathrm{i}\mathrm{n}}{\mathrm{\mid }}^2={\oint }_S{w}_{\mathrm{i}\mathrm{n}}{\mathrm{\partial }}_r{w}_{\mathrm{i}\mathrm{n}}$ and ${\int }_{a<r<R}\mathrm{\mid }\mathrm{\nabla }{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}{\mathrm{\mid }}^2={\oint }_{\{r=R\}}{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}{\mathrm{\partial }}_r{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}-{\oint }_S{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}{\mathrm{\partial }}_r{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}$; multiplying by ${\epsilon }_{\mathrm{i}\mathrm{n}}$ and ${\epsilon }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ and adding, the two surface terms on $S$ cancel by $w_{\mathrm{i}\mathrm{n}}=w_{\mathrm{o}\mathrm{u}\mathrm{t}}$ and ${\epsilon }_{\mathrm{i}\mathrm{n}}{\mathrm{\partial }}_r{w}_{\mathrm{i}\mathrm{n}}={\epsilon }_{\mathrm{o}\mathrm{u}\mathrm{t}}{\mathrm{\partial }}_r{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}$, while the term on $\{r=R\}$ is bounded by $(4\pi R^2)\cdot (M+\mathrm{\mid }d\mathrm{\mid })R^{-1}\cdot C{R}^{-2}$ using [L2]'s Cauchy estimate on balls of radius $R\mathrm{/}4$ centred at $\mathrm{\mid }x\mathrm{\mid }=R$, which lie in the exterior for large $R$ and carry the bound $\mathrm{\mid }{w}_{\mathrm{o}\mathrm{u}\mathrm{t}}\mathrm{\mid }\le C\mathrm{/}R$, hence tends to $0$. For each fixed $L>a$, the nonnegative weighted energy on the ball and $a<r<L$ is at most that on the ball and $a<r<R$ for $R>L$, and the latter is bounded by the absolute outer boundary term, which tends to zero. Thus every fixed bounded-region energy is zero; continuity and the positive weights imply that both gradients vanish, so $w_{\mathrm{i}\mathrm{n}}$ is constant on the ball and $w_{\mathrm{o}\mathrm{u}\mathrm{t}}$ is constant on $r>a$; the decay forces that constant to be $0$, hence the common trace is $0$, and then $w_{\mathrm{i}\mathrm{n}}=0$ by the Dirichlet uniqueness of [L2] applied on the ball to $w_{\mathrm{i}\mathrm{n}}$ and the zero function.

The fields follow by differentiation: $E_{\mathrm{i}\mathrm{n}}=-\mathrm{\nabla }{\phi }_{\mathrm{i}\mathrm{n}}=C{E}_0{e}_3$ with $D_{\mathrm{i}\mathrm{n}}={\epsilon }_{\mathrm{i}\mathrm{n}}C{E}_0{e}_3$ on $r<a$; outside, ${\phi }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ is the sum of the applied uniform potential and the potential $dz{r}^{-3}$ of a dipole at the origin with moment $p=4\pi {\epsilon }_{\mathrm{o}\mathrm{u}\mathrm{t}}d{e}_3$, and direct differentiation of the defined dipole potential gives $E_p(x)=[3(p\cdot n)n-p]\mathrm{/}(4\pi {\epsilon }_{\mathrm{o}\mathrm{u}\mathrm{t}}{r}^3)$, so the exterior field has precisely that form; the interior polarisation is $P_{\mathrm{i}\mathrm{n}}=D_{\mathrm{i}\mathrm{n}}-{\epsilon }_0{E}_{\mathrm{i}\mathrm{n}}=({\epsilon }_{\mathrm{i}\mathrm{n}}-{\epsilon }_0)C{E}_0{e}_3$ and the exterior polarisation is $P_{\mathrm{o}\mathrm{u}\mathrm{t}}=({\epsilon }_{\mathrm{o}\mathrm{u}\mathrm{t}}-{\epsilon }_0)E_{\mathrm{o}\mathrm{u}\mathrm{t}}$, so the bound surface charge density is the singular coefficient of ${\rho }_b=-\mathrm{div}P$ at $S$, namely ${\sigma }_b=n\cdot (P_{\mathrm{i}\mathrm{n}}-P_{\mathrm{o}\mathrm{u}\mathrm{t}})$ in $\mathrm{C}{\mathrm{m}}^{-2}$, by applying [L3] to the piecewise smooth $P$ with jump $P_{\mathrm{o}\mathrm{u}\mathrm{t}}-P_{\mathrm{i}\mathrm{n}}$.

Steps 1.1-1.2 verify (a)-(c), step 2.1 verifies (b) and the distributional thin-interface reading, step 2.2 gives the uniqueness statement (d) with existence exhibited by the displayed pair, and step 2.3 gives the fields, polarisation, exterior dipole moment and bound surface charge of (e). The scope (f) holds: the computation is quasistatic, of the declared linear branch at ${\sigma }_f=0$, conditional on the stated constants and regularity; the identification of $p$ with a dynamic wave polarisability requires a separate quasistatic/dipole approximation, and no material data, empirical claim or general existence theorem is supplied beyond this explicit transmission solution.