Constrained smooth vacuum Maxwell initial-value theory, finite-energy conservation and finite propagation

Setting and construction. Put ${\dot{E}}_0=c^2\mathrm{curl}{B}_0$ and ${\dot{B}}_0=-\mathrm{curl}{E}_0$, both $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^3$; for each $i\in \{1,2,3\}$ apply clause (a) of [F1] on the slab $[t_0,T]$ with $h=0$ and scalar data $(E_{0,i},{\dot{E}}_{0,i})$ to obtain a $C^{\mathrm{\infty }}$ function $E_i$ with $W{E}_i=0$ on ${\mathbb{R}}^3\times (t_0,T)$, $E_i(\cdot ,t_0)=E_{0,i}$, ${\mathrm{\partial }}_t{E}_i(\cdot ,t_0)={\dot{E}}_{0,i}$, and likewise $B_i$ with data $(B_{0,i},{\dot{B}}_{0,i})$; by clause (a) each $E_i$ has the spherical-mean representation ${\mathrm{\partial }}_s{H}_a+H_b$ with mean the $M_r$ of [F13], and the construction is legitimate because all six data functions are $C^{\mathrm{\infty }}$ and [F1] carries ${\mathrm{A}\mathrm{C}}_{\omega }$ through [A1].

Defects and their wave equations. Define the jointly $C^{\mathrm{\infty }}$ fields $R:={\mathrm{\partial }}_{t}E-c^2\mathrm{curl}B$ and $Q:={\mathrm{\partial }}_{t}B+\mathrm{curl}E$ on ${\mathbb{R}}^3\times [t_0,T]$; the displayed derivatives exist and are continuous (indeed $C^{\mathrm{\infty }}$) because the components are $C^{\mathrm{\infty }}$, and $W{E}_i=W{B}_i=0$ by step 1.1. Then each component satisfies $W{R}_i=0$ and $W{Q}_i=0$ on ${\mathbb{R}}^3\times (t_0,T)$: for $R_i$, ${\mathrm{\partial }}_t^2{R}_i={\mathrm{\partial }}_t^3{E}_i-c^2(\mathrm{curl}{\mathrm{\partial }}_t^{2}B{)}_i=c^2\mathrm{\Delta }{\mathrm{\partial }}_t{E}_i-c^4(\mathrm{curl}\mathrm{\Delta }B{)}_i=c^2\mathrm{\Delta }({\mathrm{\partial }}_t{E}_i-c^2(\mathrm{curl}B{)}_i)$ using ${\mathrm{\partial }}_t^2{E}_i=c^2\mathrm{\Delta }{E}_i$, ${\mathrm{\partial }}_t^2{B}_i=c^2\mathrm{\Delta }{B}_i$ from step 1.1, commutation of the mixed partial derivatives and of $\mathrm{\Delta }$ with the first derivatives [F12], and the same computation with $E$ and $B$ exchanged gives $Q$; equivalently $R_i,Q_i$ are $C^{\mathrm{\infty }}$ in the extension sense on the closed slab.

Initial values of the defects. At $t_0$, $R_i(\cdot ,t_0)={\mathrm{\partial }}_t{E}_i(\cdot ,t_0)-c^2(\mathrm{curl}B(\cdot ,t_0){)}_i={\dot{E}}_{0,i}-c^2(\mathrm{curl}{B}_0{)}_i=0$ and $Q_i(\cdot ,t_0)={\dot{B}}_{0,i}+(\mathrm{curl}{E}_0{)}_i=0$ by step 1.1 and the definitions of ${\dot{E}}_0,{\dot{B}}_0$. For the time derivatives, the component wave equations of step 1.1 give ${\mathrm{\partial }}_t^2{E}_i(\cdot ,t_0)=c^2\mathrm{\Delta }{E}_{0,i}$ and ${\mathrm{\partial }}_t^2{B}_i(\cdot ,t_0)=c^2\mathrm{\Delta }{B}_{0,i}$, so ${\mathrm{\partial }}_t{R}_i(\cdot ,t_0)=c^2\mathrm{\Delta }{E}_{0,i}-c^2(\mathrm{curl}{\dot{B}}_0{)}_i=c^2\mathrm{\Delta }{E}_{0,i}+c^2(\mathrm{curl}\mathrm{curl}{E}_0{)}_i=c^2(\mathrm{\Delta }{E}_0+\mathrm{curl}\mathrm{curl}{E}_0{)}_i=c^2(\mathrm{\nabla }\mathrm{div}{E}_0{)}_i=0$ by the curl-curl identity [F12] and the constraint $\mathrm{div}{E}_0=0$ of the Statement, and symmetrically ${\mathrm{\partial }}_t{Q}_i(\cdot ,t_0)=c^2\mathrm{\Delta }{B}_{0,i}+c^2(\mathrm{curl}\mathrm{curl}{B}_0{)}_i=c^2(\mathrm{\nabla }\mathrm{div}{B}_0{)}_i=0$.

The defects vanish and the evolution equations hold. By step 2.1 and step 3.1 each $R_i$ is $C^2$ in the extension sense on the closed slab with $W{R}_i=0$ on ${\mathbb{R}}^3\times (t_0,T)$ and zero Cauchy data at $t_0$; clause (b) of [F1] applied with $(a,b,h)=(0,0,0)$ therefore gives $R_i=0$ on the whole slab, and the same argument gives $Q_i=0$. Hence ${\mathrm{\partial }}_{t}E=c^2\mathrm{curl}B$ and ${\mathrm{\partial }}_{t}B=-\mathrm{curl}E$ hold pointwise on ${\mathbb{R}}^3\times [t_0,T]$, the two evolution equations of the smooth branch with $J=0$.

The Gauss constraints propagate: all four equations hold. The fields $E,B$ of step 1.1 are jointly $C^2$ and $\rho =0,J=0$ are $C^1$ on the laboratory region, the two evolution equations hold by step 4.1, and the continuity equation ${\mathrm{\partial }}_t\rho +\mathrm{div}J=0$ is trivially satisfied; the initial defects vanish because $\mathrm{div}E(\cdot ,t_0)=\mathrm{div}{E}_0=0$ and $\mathrm{div}B(\cdot ,t_0)=\mathrm{div}{B}_0=0$ by the constraints of the Statement. Apply clause (a) of [F2] on the open interval $(t_0,T)$: each divergence is constant there at every fixed $x$. Continuity up to $t_0$ identifies that constant with its zero initial value, and continuity up to $T$ gives the same value at $T$. Thus $\mathrm{div}E\equiv 0$ and $\mathrm{div}B\equiv 0$ on ${\mathbb{R}}^3\times [t_0,T]$, so with step 4.1 all four equations $\mathrm{div}E=0$, $\mathrm{div}B=0$, $\mathrm{curl}E=-{\mathrm{\partial }}_{t}B$, $\mathrm{curl}B=c^{-2}{\mathrm{\partial }}_{t}E$ of the source-free smooth branch [F11] hold, as does the initial-value clause ${\mathrm{\partial }}_{t}E(\cdot ,t_0)={\dot{E}}_0$, ${\mathrm{\partial }}_{t}B(\cdot ,t_0)={\dot{B}}_0$ from step 1.1; this proves clauses (a) and the equation part of the Statement.

Uniqueness. Let $E^{\mathrm{\prime }},B^{\mathrm{\prime }}$ be jointly $C^2$ on ${\mathbb{R}}^3\times [t_0,T]$ in the extension sense, satisfy the four source-free equations of [F11] and have $E^{\mathrm{\prime }}(\cdot ,t_0)=E_0$, $B^{\mathrm{\prime }}(\cdot ,t_0)=B_0$. Then the Ampère and Faraday equations at $t_0$ give ${\mathrm{\partial }}_t{E}^{\mathrm{\prime }}(\cdot ,t_0)=c^2\mathrm{curl}{B}^{\mathrm{\prime }}(\cdot ,t_0)=c^2\mathrm{curl}{B}_0={\dot{E}}_0$ and ${\mathrm{\partial }}_t{B}^{\mathrm{\prime }}(\cdot ,t_0)=-\mathrm{curl}{E}_0={\dot{B}}_0$; by [F3] each component of $E^{\mathrm{\prime }}$ and of $B^{\mathrm{\prime }}$ satisfies $Wv=0$ on the whole slab, and clause (b) of [F1] identifies each component with the corresponding component constructed in step 1.1 and verified in step 5.1, so $(E^{\mathrm{\prime }},B^{\mathrm{\prime }})=(E,B)$.

Domain of dependence and the cone reading. Let $(E_0^{\mathrm{\prime }},B_0^{\mathrm{\prime }})$ be constrained data agreeing with $(E_0,B_0)$ and with $\mathrm{curl}{E}_0,\mathrm{curl}{B}_0$ on the closed ball ${\bar{B}}_{c(t_{\ast }-t_0)}(x_{\ast })$, and let $(E^{\mathrm{\prime }},B^{\mathrm{\prime }})$ be the corresponding constructed solution of steps 1.1 and 5.1, whose scalar data for the $i$-th component of $E$ are $(E_{0,i},{\dot{E}}_{0,i})$, respectively $(E_{0,i}^{\mathrm{\prime }},{\dot{E}}_{0,i}^{\mathrm{\prime }})$, with ${\dot{E}}_0^{\mathrm{\prime }}=c^2\mathrm{curl}{B}_0^{\mathrm{\prime }}$. On the closed ball the scalar data agree: $E_{0,i}=E_{0,i}^{\mathrm{\prime }}$ and $c^2(\mathrm{curl}{B}_0{)}_i=c^2(\mathrm{curl}{B}_0^{\mathrm{\prime }}{)}_i$; the sources are $h=\tilde{h}=0$ on the closed backward cone $K(x_{\ast },t_{\ast })$, which is exactly the cone object of [F10]. Clause (c) of [F1] applied to each of the six scalar problems therefore gives $E(x_{\ast },t_{\ast })=E^{\mathrm{\prime }}(x_{\ast },t_{\ast })$ and $B(x_{\ast },t_{\ast })=B^{\mathrm{\prime }}(x_{\ast },t_{\ast })$ for every $x_{\ast }\in {\mathbb{R}}^3$ and $t_{\ast }\in [t_0,T]$. For the model reading, put the event $q=(x_{\ast },t_{\ast })$ in the inertial chart $x^0=ct$ of [F10]; for an event $p=(y,v)$ with $t_0\le v\le t_{\ast }$ the displacement $q-p$ has components $(c(t_{\ast }-v),x_{\ast }-y)$, and by the closure description of the future cone in [F10] one has $p\le q$ if and only if $c(t_{\ast }-v)\ge \mathrm{\mid }{x}_{\ast }-y\mathrm{\mid }$, that is, $(y,v)\in K(x_{\ast },t_{\ast })$; hence $K(x_{\ast },t_{\ast })$ is the model causal past $J^{-}(q)$ intersected with the slab, and an event in the slab outside it cannot be joined to $q$ by a piecewise $C^1$ future causal path, since such a path has displacement in the closed future cone by [F10]. This is the modelling statement of clause (c) and carries no empirical claim.

Compact support. Assume $E_0,B_0$ vanish off a compact $K$ and choose $\rho >0$ with $K\subseteq {\bar{B}}_{\rho }(0)$, which exists because compact subsets of ${\mathbb{R}}^3$ are bounded [F9]. For $x\mathrm{\notin }K$ the fields $E_0,B_0$ vanish on a neighbourhood of $x$, so ${\dot{E}}_0=c^2\mathrm{curl}{B}_0$ and ${\dot{B}}_0=-\mathrm{curl}{E}_0$ vanish at $x$ as well: each component of the scalar data of step 1.1 vanishes off $K$, and $h=0$ vanishes off the compact set $\{0\}$. Fix $t\in [t_0,T]$ and $x$ with $\mathrm{\mid }x\mathrm{\mid }>\rho +c(t-t_0)$; then for every $y\in {\bar{B}}_{c(t-t_0)}(x)$ the reverse triangle inequality gives $\mathrm{\mid }y\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-\mathrm{\mid }y-x\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-c(t-t_0)>\rho$, so $y\mathrm{\notin }K$ and all six scalar data vanish at $y$, and $(y,v)\in K(x,t)$ likewise has $v\in [t_0,t]$, $\mathrm{\mid }y-x\mathrm{\mid }\le c(t-v)$ and hence $\mathrm{\mid }y\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-c(t-v)\ge \mathrm{\mid }x\mathrm{\mid }-c(t-t_0)>\rho$, so $h(y,v)=0$; clause (d) of [F1] applied to each component with $\rho$ and the compact $\{0\}$ therefore gives $E(x,t)=B(x,t)=0$, which is clause (d).

Finite-energy setup and the flux bound. Let $u$ and $S$ be the energy density and Poynting vector of [F5] for the fields of step 5.1; $u$ is a nonnegative $C^{\mathrm{\infty }}$ function and $S$ a $C^{\mathrm{\infty }}$ vector field, and by [F4] with $J=0$ the local Poynting identity ${\mathrm{\partial }}_{t}u+\mathrm{div}S=0$ holds on the whole slab. For every $x$ and $v$, the coordinate identity $\mathrm{\mid }E\times B{\mathrm{\mid }}^2=\mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{\mid }B{\mathrm{\mid }}^2-(E\cdot B{)}^2\le \mathrm{\mid }E{\mathrm{\mid }}^2\mathrm{\mid }B{\mathrm{\mid }}^2$ of [F14] gives $\mathrm{\mid }S\mathrm{\mid }={\mu }_0^{-1}\mathrm{\mid }E\times B\mathrm{\mid }\le {\mu }_0^{-1}\mathrm{\mid }E\mathrm{\mid }\mathrm{\mid }B\mathrm{\mid }=cpq$ with $p=\sqrt{{\epsilon }_0}\mathrm{\mid }E\mathrm{\mid }$ and $q=\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\sqrt{{\mu }_0}$, because $c\sqrt{{\epsilon }_0{\mu }_0}=1$ by [F11]; since $2pq\le p^2+q^2$ for real $p,q$, $\mathrm{\mid }S\mathrm{\mid }\le c(p^2+q^2)\mathrm{/}2=cu$.

Moving-ball identity. Fix $x_0\in {\mathbb{R}}^3$, $r_0>0$ and $t_0\le s\le t\le T$, put $r(v):=r_0+c(t-v)$ for $v\in [s,t]$ (so $r\ge r_0>0$ and $r^{\mathrm{\prime }}=-c$), and put $F(v):={\int }_{B_{r(v)}(x_0)}u(x,v)dx$. For each $v$, the map $T_v(y)=x_0+r(v)y$ is a $C^1$ diffeomorphism of ${\mathbb{R}}^3$ with $\det D{T}_v=r(v{)}^3$, and $T_v(B_1(0))=B_{r(v)}(x_0)$; by the change-of-variables lemma [F7] applied to the nonnegative Borel function $u(\cdot ,v)$, $F(v)={\int }_{B_1(0)}f(y,v)dy$ with $f(y,v):=u(x_0+r(v)y,v)r(v{)}^3$, and the boundary sphere $S^2$ is Lebesgue null by [F9], so the integral may equivalently be taken over the compact set ${\bar{B}}_1(0)$. On an open neighbourhood of ${\bar{B}}_1(0)\times [s,t]$ the function $f$ and its partial derivative ${\mathrm{\partial }}_{v}f=\mathrm{\nabla }u(x_0+ry,v)\cdot r^{\mathrm{\prime }}y{r}^3+{\mathrm{\partial }}_{v}u(x_0+ry,v)r^3+3r^2{r}^{\mathrm{\prime }}u(x_0+ry,v)$ are continuous, because $u$ is $C^{\mathrm{\infty }}$ in the extension sense and $r$ is $C^1$; the parameter-integral lemma [F6] with the compact finite-measure set ${\bar{B}}_1(0)$ therefore gives the one-sided derivatives at the endpoints and $F^{\mathrm{\prime }}(v)={\int }_{{\bar{B}}_1(0)}{\mathrm{\partial }}_{v}f(y,v)dy$ for $v\in (s,t)$. For the signed continuous integrands below, substitution follows by applying [F7] to their positive and negative parts separately; each part has a finite integral on the compact ball, so the equalities may be subtracted. Expanding and substituting back, the first and third terms are ${\int }_{{\bar{B}}_1}[r^3{r}^{\mathrm{\prime }}\mathrm{\nabla }u(x_0+ry,v)\cdot y+3r^2{r}^{\mathrm{\prime }}u(x_0+ry,v)]dy=\frac{r^{\mathrm{\prime }}}{r}{\int }_{B_r(x_0)}\mathrm{div}((x-x_0)u(x,v))dx$, and the second term is $r^3{\int }_{{\bar{B}}_1}{\mathrm{\partial }}_{v}u(x_0+ry,v)dy={\int }_{B_r(x_0)}{\mathrm{\partial }}_{v}u(x,v)dx$; the divergence theorem [F8] applies on the ball $B_r(x_0)$, a bounded $C^1$ domain with outward normal $n=(x-x_0)\mathrm{/}r$, giving ${\int }_{B_r}\mathrm{div}((x-x_0)u)=r{\int }_{\mathrm{\partial }{B}_r}udS$ and, with ${\mathrm{\partial }}_{v}u=-\mathrm{div}S$ from the Poynting identity of step 6.4, ${\int }_{B_r}{\mathrm{\partial }}_{v}u=-{\int }_{\mathrm{\partial }{B}_r}S\cdot ndS$. Hence $F^{\mathrm{\prime }}(v)=-{\int }_{\mathrm{\partial }{B}_r(x_0)}(S\cdot n-r^{\mathrm{\prime }}(v)u)dS=-{\int }_{\mathrm{\partial }{B}_r(x_0)}(S\cdot n+cu)dS$ with $r=r(v)$, and the integrand is nonnegative because $S\cdot n+cu\ge -\mathrm{\mid }S\mathrm{\mid }+cu\ge 0$ by step 6.4; the surface integral of a nonnegative continuous function is nonnegative by the definition of the surface integral [F8], so $F^{\mathrm{\prime }}\le 0$ on $(s,t)$.

Local energy estimate and the equality. By step 7.1, $F$ is continuous on $[s,t]$ and differentiable on $(s,t)$ with $F^{\mathrm{\prime }}\le 0$, so the mean value theorem [F9] gives $F(t)\le F(s)$, that is, for every $x_0\in {\mathbb{R}}^3$, $r_0>0$ and $t_0\le s\le t\le T$, ${\int }_{B_{r_0}(x_0)}u(x,t)dx\le {\int }_{B_{r_0+c(t-s)}(x_0)}u(x,s)dx$; this is the local inequality of clause (e) for the constructed pair, and the computation of step 7.1 used only the nonnegativity of $u$, the Poynting identity, the bound $\mathrm{\mid }S\mathrm{\mid }\le cu$ and the smoothness of the pair, so the same inequality holds for every smooth source-free Maxwell solution on a slab with its own energy density and flux. Taking $s=t_0$ and the supremum over $r_0>0$ with $x_0=0$ gives $\mathcal{E}(t)\le \mathcal{E}(t_0)$ for every $t\in [t_0,T]$. For the reverse inequality, apply the same computation to the pair $\tilde{E}(x,v):=E(x,t-v)$, $\tilde{B}(x,v):=-B(x,t-v)$ on ${\mathbb{R}}^3\times [0,t-t_0]$: by the chain rule and the four equations of step 5.1, $\mathrm{div}\tilde{E}=\mathrm{div}\tilde{B}=0$, $\mathrm{curl}\tilde{E}=-{\mathrm{\partial }}_v\tilde{B}$ and $\mathrm{curl}\tilde{B}=c^{-2}{\mathrm{\partial }}_v\tilde{E}$, so the pair is a smooth source-free solution on that slab, with energy density $u(x,t-v)$ and Poynting vector $-S(x,t-v)$; applying the local inequality with $s^{\mathrm{\prime }}=0$ and $t^{\mathrm{\prime }}=t-t_0$ gives ${\int }_{B_{r_0}(x_0)}u(x,t_0)dx\le {\int }_{B_{r_0+c(t-t_0)}(x_0)}u(x,t)dx\le \mathcal{E}(t)$ for every $r_0$, so $\mathcal{E}(t_0)\le \mathcal{E}(t)$. Hence if $\mathcal{E}(t_0)<\mathrm{\infty }$ then $\mathcal{E}(t)=\mathcal{E}(t_0)$ for every $t\in [t_0,T]$.