Finite-slab energy control and subsidiary propagation for the linearized electrovacuum system

The metric principal block in [F1] is symmetric with positive time matrix on each spacelike slice. The Maxwell block in [F2] is symmetric with positive time matrix; conversion from the geometric field to the state variables is multiplication by a fixed nonzero scalar and leaves those matrices unchanged. The Maxwell stress is algebraic in $g^{-1}$ and $\mathcal{F}$, so its variation contributes only a zeroth-order term. The field-strength reduction [F2] puts derivatives of the metric and frame into lower-order terms; after linearization, derivatives of $h$ are represented by the state components $p,q$. Thus the full principal matrices are direct sums, the time matrix is positive, and the spatial matrices are symmetric. Smoothness on the compact slab gives uniform $a,b$ bounds and bounded coefficient derivatives.

Put $U:=df$ and $V:=d\delta (\star \mathcal{F})$. Since the background Maxwell residuals vanish, variation of the normal-slot equations gives ${\iota }_{n}U={\iota }_{n}V=0$; in the chosen chart $n={\mathrm{\partial }}_t$. Also $dU=d^{2}f=0$ and $dV=d^2\delta (\star \mathcal{F})=0$. Hence each is spatial and has the form $U={\rho }_U(t,\theta )d{\theta }^1\wedge d{\theta }^2\wedge d{\theta }^3,V={\rho }_V(t,\theta )d{\theta }^1\wedge d{\theta }^2\wedge d{\theta }^3\mathrm{.}$ Their exterior derivatives are ${\mathrm{\partial }}_t{\rho }_{U}dt\wedge d{\theta }^1\wedge d{\theta }^2\wedge d{\theta }^3$ and ${\mathrm{\partial }}_t{\rho }_{V}dt\wedge d{\theta }^1\wedge d{\theta }^2\wedge d{\theta }^3$. Closedness makes both coefficients constant in $t$; zero initial restrictions therefore give $U=V=0$ throughout, which propagates both linearized Gauss constraints.

Put ${\mathcal{R}}_{ab}:={\mathrm{Ric}}_{ab}-{\mathrm{\nabla }}_{(a}{H}_{b)}-\mathrm{\Lambda }{g}_{ab}-2Q_{ab}$ and ${\mathcal{E}}_{ab}:=G_{ab}+\mathrm{\Lambda }{g}_{ab}-2Q_{ab}$. Direct contraction of $Q_{ab}$ gives $g^{ab}{Q}_{ab}=0$, so $g^{ab}{\mathcal{R}}_{ab}=S-{\mathrm{\nabla }}_c{H}^c-4\mathrm{\Lambda }$. Substituting $\mathrm{Ric}=\mathcal{R}+{\mathrm{\nabla }}_{(a}{H}_{b)}+\mathrm{\Lambda }g+2Q$ into $\mathcal{E}$ yields ${\mathcal{E}}_{ab}={\mathcal{R}}_{ab}-\frac{1}{2}{g}_{ab}{g}^{cd}{\mathcal{R}}_{cd}+\frac{1}{2}({\mathrm{\nabla }}_a{H}_b+{\mathrm{\nabla }}_b{H}_a-g_{ab}{\mathrm{\nabla }}_c{H}^c)\mathrm{.}$ At the background $\mathcal{R}=0$ and $H=0$, so the traced identity gives $S=4\mathrm{\Lambda }$ and hence $\mathcal{E}=0$. The linearized reduced Einstein equation gives $\delta \mathcal{R}=0$. Differentiating this identity then gives $\delta {\mathcal{E}}_{ab}=\frac{1}{2}({\mathrm{\nabla }}_a{\dot{H}}_b+{\mathrm{\nabla }}_b{\dot{H}}_a-g_{ab}{\mathrm{\nabla }}_c{\dot{H}}^c),$ since every variation multiplying a zero background residual vanishes.

To derive the divergence identity, in a positively oriented chart put ${\epsilon }_{abcd}:=({\mathrm{v}\mathrm{o}\mathrm{l}}_g{)}_{abcd}$. The Christoffel formula ${\mathrm{\Gamma }}^a{}_{bc}=\frac{1}{2}{g}^{ad}({\mathrm{\partial }}_b{g}_{dc}+{\mathrm{\partial }}_c{g}_{bd}-{\mathrm{\partial }}_d{g}_{bc})$ gives ${\mathrm{\Gamma }}^a{}_{ae}=\frac{1}{2}{g}^{ad}{\mathrm{\partial }}_e{g}_{ad}$ after the first and third terms cancel; determinant differentiation gives ${\mathrm{\Gamma }}^a{}_{ae}={\mathrm{\partial }}_e\log \sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$. Torsion-freeness gives ${\mathrm{\Gamma }}^p{}_{ep}={\mathrm{\Gamma }}^p{}_{pe}$, so writing ${\epsilon }_{abcd}=\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}{\tilde{\epsilon }}_{abcd}$ with constant alternating symbol ${\tilde{\epsilon }}_{0123}=1$ gives ${\mathrm{\nabla }}_e{\epsilon }_{abcd}=({\mathrm{\partial }}_e\log \sqrt{\mathrm{\mid }\det g\mathrm{\mid }}-{\mathrm{\Gamma }}^p{}_{ep}){\epsilon }_{abcd}=0$. Torsion-freeness also gives $(d{\star }_g\mathcal{F}{)}_{acd}=3{\mathrm{\nabla }}_{[a}({\star }_g\mathcal{F}{)}_{cd]}$. Using [F5] and ${\epsilon }^{bacd}{\epsilon }_{cdef}=-2({\delta }_e^b{\delta }_f^a-{\delta }_f^b{\delta }_e^a)$ yields $\frac{1}{3!}{\epsilon }^{bacd}(d{\star }_g\mathcal{F}{)}_{acd}=\frac{1}{4}{\epsilon }^{bacd}{\epsilon }_{cdef}{\mathrm{\nabla }}_a{\mathcal{F}}^{ef}={\mathrm{\nabla }}_a{\mathcal{F}}^{ab}$. The background equation $d{\star }_g\mathcal{F}=0$ thus gives ${\mathrm{\nabla }}_a{\mathcal{F}}^{ab}=0$. Differentiating this identity at the background, the variation of $\epsilon$ multiplies the zero background $d{\star }_g\mathcal{F}$, so $d\delta (\star \mathcal{F})=0$ gives $\delta ({\mathrm{\nabla }}_a{\mathcal{F}}^{ab})=0$. Metric compatibility and the product rule give ${\mathrm{\nabla }}_a{Q}^{ab}=({\mathrm{\nabla }}_a{\mathcal{F}}^{ac}){\mathcal{F}}^b{}_c+{\mathcal{F}}^{ac}{\mathrm{\nabla }}_a{\mathcal{F}}^b{}_c-\frac{1}{4}{\mathrm{\nabla }}^b({\mathcal{F}}_{cd}{\mathcal{F}}^{cd})\mathrm{.}$ Contracting $d{\mathcal{F}}_{adc}={\mathrm{\nabla }}_a{\mathcal{F}}_{dc}+{\mathrm{\nabla }}_d{\mathcal{F}}_{ca}+{\mathrm{\nabla }}_c{\mathcal{F}}_{ad}$ with ${\mathcal{F}}^{ac}$ gives $2{\mathcal{F}}^{ac}{\mathrm{\nabla }}_a{\mathcal{F}}_{dc}-\frac{1}{2}{\mathrm{\nabla }}_d({\mathcal{F}}_{ac}{\mathcal{F}}^{ac})={\mathcal{F}}^{ac}(d\mathcal{F}{)}_{adc}\mathrm{.}$ Thus ${\mathcal{F}}^{ac}{\mathrm{\nabla }}_a{\mathcal{F}}_{dc}=\frac{1}{2}{\mathcal{F}}^{ac}(d\mathcal{F}{)}_{adc}+\frac{1}{4}{\mathrm{\nabla }}_d({\mathcal{F}}_{ac}{\mathcal{F}}^{ac})$; raising $d$ cancels the final scalar-gradient term in the first display, leaving ${\mathrm{\nabla }}_a{Q}^{ab}=({\mathrm{\nabla }}_a{\mathcal{F}}^{ac}){\mathcal{F}}^b{}_c+\frac{1}{2}{g}^{bd}{\mathcal{F}}^{ac}(d\mathcal{F}{)}_{adc}\mathrm{.}$ Since the background also has $d\mathcal{F}=0$, variation of this last display gives $\delta ({\mathrm{\nabla }}_a{Q}^{ab})=0$. The background $\mathcal{E}$ vanishes by step 1.3, so [F4] and ${\mathrm{\nabla }}^a{\mathcal{E}}_{ab}=-2{\mathrm{\nabla }}^a{Q}_{ab}$ imply ${\mathrm{\nabla }}^a\delta {\mathcal{E}}_{ab}=0$.

Fix a spatial multi-index $\alpha$ with $\mathrm{\mid }\alpha \mathrm{\mid }\le k$ and put $w_{\alpha }={\mathrm{\partial }}^{\alpha }u$. Differentiating the system gives $A^0{\mathrm{\partial }}_t{w}_{\alpha }+{\sum }_i{A}^i{\mathrm{\partial }}_i{w}_{\alpha }+C{w}_{\alpha }={\mathrm{\partial }}^{\alpha }s-R_{\alpha },$ where $R_{\alpha }=[{\mathrm{\partial }}^{\alpha },A^0]{\mathrm{\partial }}_{t}u+{\sum }_i[{\mathrm{\partial }}^{\alpha },A^i]{\mathrm{\partial }}_{i}u+[{\mathrm{\partial }}^{\alpha },C]u$ and $[P,A]v:=P(Av)-A(Pv)$. The equation and $\mathrm{\parallel }(A^0{)}^{-1}\mathrm{\parallel }\le a^{-1}$ give $\mathrm{\parallel }{\mathrm{\partial }}_{t}u{\mathrm{\parallel }}_{H^{k-1}}\le C(\mathrm{\parallel }u{\mathrm{\parallel }}_{H^k}+\mathrm{\parallel }s{\mathrm{\parallel }}_{H^{k-1}})$ for $k\ge 1$. Expanding the commutators by the finite product rule and using bounded coefficient derivatives yields ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le k}\mathrm{\parallel }{R}_{\alpha }{\mathrm{\parallel }}_{L^2}\le C(\mathrm{\parallel }u{\mathrm{\parallel }}_{H^k}+\mathrm{\parallel }s{\mathrm{\parallel }}_{H^k})$; when $k=0$, every commutator is zero.

Differentiate ${\mathcal{E}}_k$ and substitute the differentiated system from step 2.2. Symmetry and periodic integration by parts give $-2{\int }_{{\mathbb{T}}^3}{w}_{\alpha }^T{A}^i{\mathrm{\partial }}_i{w}_{\alpha }dx={\int }_{{\mathbb{T}}^3}{w}_{\alpha }^T({\mathrm{\partial }}_i{A}^i)w_{\alpha }dx,$ with no boundary term. All coefficient, lower-order, commutator and source terms are bounded by step 2.2 and Cauchy–Schwarz. For smooth $u$, classical derivatives are its weak derivatives in [F6]. Parseval with the normalized measure $d\mu =(2\pi {)}^{-3}dx$ gives ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le k}\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }u{\mathrm{\parallel }}_{L^2(dx)}^2{\asymp }_k\mathrm{\parallel }u{\mathrm{\parallel }}_{H^k}^2$: at frequency $\mathrm{\ell }$, the polynomial ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le k}\mathrm{\mid }{\mathrm{\ell }}^{\alpha }{\mathrm{\mid }}^2$ is bounded above by a constant times $(1+\mathrm{\mid }\mathrm{\ell }{\mathrm{\mid }}^2{)}^k$; for $\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }\le 1$ its $\alpha =0$ term gives a lower bound, and for $\mathrm{\mid }\mathrm{\ell }\mathrm{\mid }>1$ choose $j$ with $\mathrm{\mid }{\mathrm{\ell }}_j\mathrm{\mid }\ge \mathrm{\mid }\mathrm{\ell }\mathrm{\mid }\mathrm{/}\sqrt{3}$ and use the ${\alpha }_j=k$ term. Thus this proof establishes equivalence of the item's Fourier norm and the supplier's weak-derivative norm; the bounds $aI\le A^0\le bI$ therefore make ${\mathcal{E}}_k$ equivalent to the Fourier norm used in the Statement. The differentiated system now gives $\frac{d}{dt}{\mathcal{E}}_k(t)\le C{\mathcal{E}}_k(t)+C{\mathcal{E}}_k(t{)}^{1\mathrm{/}2}\mathrm{\parallel }s(t){\mathrm{\parallel }}_{H^k}\mathrm{.}$ Here $C$ depends on $k$ and the stated coefficient bounds.

Apply step 3.1 to $Y_{\epsilon }(t)=({\mathcal{E}}_k(t)+\epsilon {)}^{1\mathrm{/}2}$. Division by $2Y_{\epsilon }$ gives $Y_{\epsilon }^{\mathrm{\prime }}\le C{Y}_{\epsilon }+C\mathrm{\parallel }s(t){\mathrm{\parallel }}_{H^k}$. Integrating between $0$ and $t$ and letting $\epsilon \downarrow 0$ yields ${\mathcal{E}}_k(t{)}^{1\mathrm{/}2}\le e^{C\mathrm{\mid }t\mathrm{\mid }}({\mathcal{E}}_k(0{)}^{1\mathrm{/}2}+C{\int }_{\min (0,t)}^{\max (0,t)}\mathrm{\parallel }s(\tau ){\mathrm{\parallel }}_{H^k}d\tau )$. For $t<0$, reverse time and multiply the system by $-1$; symmetry and positivity of the time matrix are preserved. Norm equivalence proves the stated estimate after increasing $C_k$.

Taking the divergence of step 1.3, using step 2.1 and the covector commutator identity derived in [F4], gives ${\mathrm{\square }}_g{\dot{H}}_b+{\mathrm{Ric}}_{ba}{\dot{H}}^a=0$, where ${\mathrm{\square }}_g{Z}_b:=g^{cd}{\mathrm{\nabla }}_c{\mathrm{\nabla }}_d{Z}_b$ is the rough wave operator on covector fields. Each of the four coordinate components has principal part $g^{cd}{\mathrm{\partial }}_c{\mathrm{\partial }}_d{\dot{H}}_b$; connection and curvature terms have at most first order and smooth coefficients. Introduce $v_b={\mathrm{\partial }}_t{\dot{H}}_b$ and $w_{bi}={\mathrm{\partial }}_i{\dot{H}}_b$, giving a 20-component first-order state. The wave reduction of [F1], repeated for these four scalar principal wave equations, has time matrix $\mathrm{diag}(I,I,{\gamma }^{ij}I)$ and symmetric spatial matrices; its curvature term is lower order. The source is zero. Initially ${\dot{H}}_b=0$, $v_b=0$, and $w_{bi}=0$ because $\dot{H}$ vanishes as a spatial function on the initial slice. The integration-by-parts estimate from steps 2.2–4.1 applies to this 20-component system by the same calculation, which uses only matrix symmetry and the positive time-matrix bounds; it forces this state to stay zero, so $\dot{H}=0$ throughout the slab.