Heat and viscous entropy specialization

Differentiating $g_{ab}{u}^a{u}^b=-1$ covariantly and using [F5] gives $u^d{\mathrm{\nabla }}_c{u}_d=0$ for every $c$. Also $u^a{h}_a{}^c=u^c+(u_a{u}^a)u^c=0$, and similarly $u^b{h}_b{}^d=0$. Therefore both indices of $h_a{}^c{h}_b{}^d{\mathrm{\nabla }}_{(c}{u}_{d)}$ are spatial, and subtracting the spatial tensor $\theta h_{ab}\mathrm{/}3$ preserves spatiality. Since $h_{ab}$ and ${\mathrm{\nabla }}_{(c}{u}_{d)}$ are symmetric, their two-index projection and hence ${\sigma }_{ab}$ are symmetric.

Expanding $h$ and using $g(u,u)=-1$ gives $g^{ab}{h}_a{}^c{h}_b{}^d=g^{cd}+u^c{u}^d$ and $g^{ab}{h}_{ab}=3$. Hence $g^{ab}{\sigma }_{ab}=(g^{cd}+u^c{u}^d){\mathrm{\nabla }}_{(c}{u}_{d)}-\theta =\theta +u^c(u^d{\mathrm{\nabla }}_c{u}_d)-\theta =0$, where the middle term vanishes by step 1.1.

By [F4], each pointwise rest space $u^{\perp }$ is a finite-dimensional positive-definite inner-product space. Choose an algebraic basis there and apply the finite Gram–Schmidt construction; each nonzero residual has positive norm and can be normalized. Since $\sigma$ is spatial and symmetric by step 1.1, in that basis ${\sigma }^{ab}{\sigma }_{ab}={\sum }_{i,j=1}^3{\sigma }_{ij}^2\ge 0$, and the sum is zero exactly when every component, hence $\sigma$, is zero. The same positive-definite metric gives $g(Y,Y)\ge 0$ for every spatial vector, with equality only for $Y=0$.

[F1] fixes the state and $q=cQ$ convention, while [F2] supplies the Gibbs and Euler identities required by [F3]. By [F5], the closure ${\pi }^{ab}=-2\eta c{\sigma }^{ab}$ is symmetric, spatial and trace-free, so it satisfies [F3]'s stress hypotheses; $Q$ is spatial by [F4], and $V=0$. Since $u_b{f}^b=0$, the remaining balance hypotheses of [F3] also hold. Its entropy current reduces to $S^a=s{u}^a+Q^a\mathrm{/}T$. Because $Q$ is spatial, $Q^a({\mathrm{\nabla }}_{a}T+T{a}_a)=Q^a{X}_a$, and therefore ${\mathrm{\nabla }}_a{S}^a=-\frac{\mathrm{\Pi }\theta }{T}-\frac{{\pi }^{ab}{\sigma }_{ab}}{T}-\frac{Q^a{X}_a}{T^2}\mathrm{.}$

From $Q^a=-\kappa X^a\mathrm{/}c$, $\mathrm{\Pi }=-{\zeta }_{\mathrm{b}}c\theta$, and ${\pi }^{ab}=-2\eta c{\sigma }^{ab}$, the three terms in step 1.1 become, respectively, ${\zeta }_{\mathrm{b}}c{\theta }^2\mathrm{/}T$, $2\eta c{\sigma }^{ab}{\sigma }_{ab}\mathrm{/}T$, and $\kappa X^a{X}_a\mathrm{/}(c{T}^2)$. The vector $X$ is spatial by [F4], so [F5]'s positive-definite rest metric gives $X^a{X}_a\ge 0$; the same fact in [F5] gives ${\sigma }^{ab}{\sigma }_{ab}\ge 0$. Since $c,T>0$ and the coefficient functions are nonnegative, their sum is the stated identity and is nonnegative.

Put $\alpha =A_0\mathrm{/}{C}_V>0$. The extension ${\tilde{\vartheta }}_0$ is bounded and uniformly continuous because it is continuous and periodic. Apply [F5] with time variable $\tau =\alpha t_K$ to define $\vartheta (t_K,x)=\int \mathrm{\Gamma }(x-y,\alpha t_K){\tilde{\vartheta }}_0(y)dy$ for $t_K>0$ and $\vartheta (0,x)={\vartheta }_0(x)$. The chain rule gives ${\mathrm{\partial }}_{t_K}\vartheta =\alpha {\mathrm{\Delta }}_x\vartheta$, and local uniform initial recovery on the compact set $\bar{\mathrm{\Omega }}$ gives continuity to the stated initial trace.

The local flow of $\xi ={\mathrm{\partial }}_{x^0}$ is ${\mathrm{\Phi }}_t(x^0,x^1,x^2,x^3)=(x^0+t,x^1,x^2,x^3)$. Since $g$ is independent of $x^0$, ${\mathrm{\Phi }}_t^{\ast }g=g$, so [F12] gives ${\mathcal{L}}_{\xi }g=0$. The components give $g(\xi ,\xi )=-N^2$, $u^0=N^{-1}$, $u^i=0$, $u_0=-N$, and $u_i=0$; hence $g(u,u)=-1$ and $g(u,\xi )=-N<0$, so $u$ is future directed. The Christoffel formula [F9] gives ${\mathrm{\Gamma }}^0{}_{0i}={\mathrm{\Gamma }}^0{}_{i0}={\mathrm{\partial }}_{i}N\mathrm{/}N$, ${\mathrm{\Gamma }}^i{}_{00}=N{\mathrm{\partial }}_{i}N$, and all other coefficients vanish. In particular ${\mathrm{\Gamma }}^a{}_{a0}=0$ and ${\mathrm{\Gamma }}^a{}_{ai}={\mathrm{\partial }}_i\log N$. Since ${\mathrm{\partial }}_a{u}^a=0$, $u^i=0$, and ${\mathrm{\Gamma }}^a{}_{a0}=0$, the coordinate formula ${\mathrm{\nabla }}_a{u}^a={\mathrm{\partial }}_a{u}^a+{\mathrm{\Gamma }}^a{}_{ac}{u}^c$ gives $\theta =0$. Directly, ${\mathrm{\nabla }}_0{u}_i={\mathrm{\partial }}_{i}N$, ${\mathrm{\nabla }}_i{u}_0={\mathrm{\nabla }}_0{u}_0={\mathrm{\nabla }}_i{u}_j=0$, and ${\mathrm{\Delta }}_0{}^a=0$, ${\mathrm{\Delta }}_i{}^a={\delta }_i{}^a$; the projected spatial symmetric derivative vanishes and ${\sigma }_{ab}=0$. Also $a_i=u^0{\mathrm{\nabla }}_0{u}_i={\mathrm{\partial }}_{i}N\mathrm{/}N={\mathrm{\partial }}_i\log N$ and $a_0=0$. The chosen material state map is smooth because $T=\vartheta \mathrm{/}N>0$ for $t_K>0$; differentiation of $e$ gives $T=e_{{\sigma }_p}=T_{\mathrm{r}\mathrm{e}\mathrm{f}}{e}^{({\sigma }_p-{\sigma }_{\mathrm{r}\mathrm{e}\mathrm{f}})\mathrm{/}{c}_v}$ and $p=-e_v=0$, whence $ϵ=n_{0}e=C_{V}T+n_0{e}_0$ and $s=n_0{\sigma }_p=n_0{\sigma }_{\mathrm{r}\mathrm{e}\mathrm{f}}+C_V\log (T\mathrm{/}{T}_{\mathrm{r}\mathrm{e}\mathrm{f}})$. Finally $J=n_{0}u$ has zero divergence, while $V=\mathrm{\Pi }=\pi =0$ and $Q=q\mathrm{/}c$ give precisely the stress in the Statement and the specialization of [F2].

Positivity and unit mass in [F6] give $m_0\le \vartheta \le M_0$. For a period translation $P_i(x)=x+2L_i{e}_i$, the substitution $y=P_{i}z$ in [F7] has absolute Jacobian $1$; periodicity of the datum and radiality of $\mathrm{\Gamma }$ yield $\vartheta (P_{i}x,t_K)=\vartheta (x,t_K)$. For reflection $R_i(x)=(x_1,\dots ,-x_i,\dots ,x_3)$, the same change-of-variables formula and the evenness of the datum and kernel give $\vartheta (R_{i}x,t_K)=\vartheta (x,t_K)$. Periodicity plus reflection makes $\vartheta$ even about each plane $x_i=L_i$ as well as $x_i=0$. Smoothness for $t_K>0$ then forces the normal derivative to vanish on each face; $N^{2}q=-A_0{\mathrm{\nabla }}_x(NT)=-A_0{\mathrm{\nabla }}_x\vartheta$, so this is exactly the insulated condition.

Let ${\vartheta }_1=N{T}_1$ and ${\vartheta }_2=N{T}_2$ be the redshifted temperatures of two solutions in the stated class with the same initial data, and put $w={\vartheta }_1-{\vartheta }_2$. The transformed equation and face condition give $C_V{\mathrm{\partial }}_{t_K}w=A_0{\mathrm{\Delta }}_{x}w$ and ${\mathrm{\partial }}_{\nu }w=0$. For $0<s<t_K$, differentiating the squared integral and integrating by parts on each coordinate interval gives $\frac{d}{d{t}_K}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2=2{\int }_{\mathrm{\Omega }}w{\mathrm{\partial }}_{t_K}wdx=-2(A_0\mathrm{/}{C}_V)\mathrm{\parallel }{\mathrm{\nabla }}_{x}w{\mathrm{\parallel }}_2^2\le 0$, since every boundary term contains ${\mathrm{\partial }}_{\nu }w=0$. Thus $\mathrm{\parallel }w(t_K){\mathrm{\parallel }}_2^2\le \mathrm{\parallel }w(s){\mathrm{\parallel }}_2^2$; continuity on the compact initial cylinder and equal initial data make the latter tend to zero as $s\downarrow 0$, proving uniqueness.

The nonzero stress components are ${\mathsf{T}}^{00}=ϵ\mathrm{/}{N}^2$, ${\mathsf{T}}^{0i}={\mathsf{T}}^{i0}=q^i\mathrm{/}(cN)$, and ${\mathsf{T}}^{ij}=0$. The determinant is $\det g=-N^2$, and [F9] with step 1.2 gives ${\mathrm{\nabla }}_a{\mathsf{T}}^{a0}=N^{-1}{\mathrm{\partial }}_{x^a}(N{\mathsf{T}}^{a0})+{\mathrm{\Gamma }}^0{}_{ac}{\mathsf{T}}^{ac}$: the trace term uses ${\mathrm{\Gamma }}^a{}_{ac}={\mathrm{\partial }}_c\log N$, and the only nonzero term in the last contraction is $2{\mathrm{\Gamma }}^0{}_{0i}{\mathsf{T}}^{0i}$. Since $u_0=-N$, $u_i=0$, and ${\mathrm{\partial }}_{x^0}=c^{-1}{\mathrm{\partial }}_{t_K}$, $u_b{f}^b=-N{\mathrm{\nabla }}_a{\mathsf{T}}^{a0}=-\frac{1}{cN}({\mathrm{\partial }}_{t_K}ϵ+N{\mathrm{div}}_{x}q+2q\cdot {\mathrm{\nabla }}_{x}N)\mathrm{.}$ Here $ϵ=C_{V}T+n_0{e}_0$, so this expression vanishes exactly when $N{C}_V{\mathrm{\partial }}_{t_K}T+{\mathrm{div}}_x(N^{2}q)=0$. The constructed $\vartheta$ satisfies this balance because $N^{2}q=-A_0{\mathrm{\nabla }}_x\vartheta$ and $C_V{\mathrm{\partial }}_{t_K}\vartheta =A_0{\mathrm{\Delta }}_x\vartheta$. By the definition $f^b={\mathrm{\nabla }}_a{\mathsf{T}}^{ab}$ the stress balance holds, and the calculation proves $u_b{f}^b=0$.

For every compact time interval in $t_K>0$, the integrands and their time derivatives are continuous on its product with $\bar{\mathrm{\Omega }}$; the fundamental theorem of calculus in time and [F11] therefore justify differentiating the finite-volume integral. The heat balance and boundary condition give $\frac{d}{d{t}_K}{\int }_{\mathrm{\Omega }}Nϵdx={\int }_{\mathrm{\Omega }}N{C}_V{\mathrm{\partial }}_{t_K}Tdx=-{\int }_{\mathrm{\Omega }}{\mathrm{div}}_x(N^{2}q)dx=0.$ The last equality follows by applying the one-dimensional fundamental theorem of calculus to each coordinate derivative and summing the six face fluxes, each zero by the insulated condition. Continuity at $t_K=0$ extends this constant value to the initial time, proving conservation of ${\mathcal{E}}_K$.

In [F2] set $\kappa ={\kappa }_{\mathrm{t}\mathrm{h}}=A_0\mathrm{/}N$, $\eta ={\zeta }_{\mathrm{b}}=0$, $V=0$, and use $u_b{f}^b=0$ from step 2.3. Since $X_i={\mathrm{\partial }}_{i}T+T{\mathrm{\partial }}_i\log N=N^{-1}{\mathrm{\partial }}_i\vartheta$, the spatial metric is Euclidean and $T=\vartheta \mathrm{/}N$, the identity in [F2] gives ${\mathrm{\nabla }}_a{S}^a={\kappa }_{\mathrm{t}\mathrm{h}}{X}^a{X}_a\mathrm{/}(c{T}^2)=A_0\mathrm{\mid }{\mathrm{\nabla }}_x\vartheta {\mathrm{\mid }}^2\mathrm{/}(cN{\vartheta }^2)$. The coordinate divergence formula from [F9] is ${\mathrm{\nabla }}_a{S}^a=N^{-1}{\mathrm{\partial }}_{x^a}(N{S}^a)$; with $N{S}^0=s$ and $N{S}^i=N{q}^i\mathrm{/}(cT)$ it yields ${\mathrm{\partial }}_{t_K}s+{\mathrm{div}}_x(Nq\mathrm{/}T)=cN{\mathrm{\nabla }}_a{S}^a\mathrm{.}$ Differentiation under the finite-volume integral is justified as in step 3.1. Integrating, the boundary term vanishes because $Nq\cdot \nu \mathrm{/}T=0$ follows from $N^{2}q\cdot \nu =0$ and $N,T>0$. Thus $\frac{d\mathcal{S}}{d{t}_K}=A_0{\int }_{\mathrm{\Omega }}\mathrm{\mid }{\mathrm{\nabla }}_x\vartheta {\mathrm{\mid }}^2\mathrm{/}{\vartheta }^{2}dx\ge 0$.

The same compact-time differentiation argument as in step 3.1 and the heat equation give $\frac{d}{d{t}_K}{\int }_{\mathrm{\Omega }}\vartheta dx=(A_0\mathrm{/}{C}_V){\int }_{\mathrm{\Omega }}\mathrm{\Delta }\vartheta dx=0$. The last integral vanishes by the coordinate fundamental theorem of calculus on each pair of opposite faces, whose normal derivatives are zero. Continuity to the initial trace then shows that the mean of $\vartheta (t_K,\cdot )$ is the positive constant $\bar{\vartheta }$.

Put $w=\vartheta -\bar{\vartheta }$, so ${\int }_{\mathrm{\Omega }}w=0$. Expanding the square and using [F11] to integrate the continuous product over $\mathrm{\Omega }\times \mathrm{\Omega }$ gives ${\int }_{\mathrm{\Omega }}{\int }_{\mathrm{\Omega }}(w(x)-w(y){)}^{2}dxdy=2\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }{\int }_{\mathrm{\Omega }}{w}^{2}dx-2{\left({\int }_{\mathrm{\Omega }}wdx\right)}^2=2\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }\mathrm{\parallel }w{\mathrm{\parallel }}_2^2\mathrm{.}$ Define $b_1=w(x_1,x_2,x_3)-w(y_1,x_2,x_3)$, $b_2=w(y_1,x_2,x_3)-w(y_1,y_2,x_3)$, and $b_3=w(y_1,y_2,x_3)-w(y_1,y_2,y_3)$, so $w(x)-w(y)=b_1+b_2+b_3$. For each $i$, the fundamental theorem of calculus expresses $b_i$ as the integral of ${\mathrm{\partial }}_{i}w$ over an interval of length at most $L_i$; [F11] therefore gives $\mathrm{\mid }{b}_i{\mathrm{\mid }}^2\le L_i{\int }_0^{L_i}\mathrm{\mid }{\mathrm{\partial }}_{i}w{\mathrm{\mid }}^{2}d{x}_i$. Integrating this bound over $x,y$ gives ${\int }_{\mathrm{\Omega }\times \mathrm{\Omega }}\mathrm{\mid }{b}_i{\mathrm{\mid }}^{2}dxdy\le \mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }{L}_i^2{\int }_{\mathrm{\Omega }}\mathrm{\mid }{\mathrm{\partial }}_{i}w{\mathrm{\mid }}^{2}dx,$ because the two unused coordinates in direction $i$ contribute $L_i^2$, one unused coordinate in each other direction contributes ${\prod }_{j\mathrm{\ne }i}{L}_j$, and the Cauchy–Schwarz factor contributes the remaining $L_i$. Since $(b_1+b_2+b_3{)}^2\le 3{\sum }_i{b}_i^2$, division by $2\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$ yields $\mathrm{\parallel }w{\mathrm{\parallel }}_2^2\le \frac{3}{2}{\sum }_i{L}_i^2\mathrm{\parallel }{\mathrm{\partial }}_{i}w{\mathrm{\parallel }}_2^2\le C_{\mathrm{\Omega }}\mathrm{\parallel }{\mathrm{\nabla }}_{x}w{\mathrm{\parallel }}_2^2,$ where $C_{\mathrm{\Omega }}=\frac{3}{2}({\max }_i{L}_i{)}^2$. For $t_K>0$, integration by parts and ${\mathrm{\partial }}_{\nu }w=0$ give $\frac{1}{2}\frac{d}{d{t}_K}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2=-\frac{A_0}{C_V}\mathrm{\parallel }{\mathrm{\nabla }}_{x}w{\mathrm{\parallel }}_2^2\le -\frac{A_0}{C_V{C}_{\mathrm{\Omega }}}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2\mathrm{.}$ Multiplying by $\exp (2A_0{t}_K\mathrm{/}(C_V{C}_{\mathrm{\Omega }}))$ and differentiating proves the exponential estimate after taking square roots; continuity at $t_K=0$ supplies the initial norm.