Positive first-order residual and actual instability/tail counterexamples

Substitute the first-order laws into [F1] under the no-work assumption $u_b{f}^b=0$ of [A3]. They give $-\mathrm{\Pi }\theta \mathrm{/}T=\zeta c{\theta }^2\mathrm{/}T$, $-{\pi }^{ab}{\sigma }_{ab}\mathrm{/}T=2\eta c{\sigma }^{ab}{\sigma }_{ab}\mathrm{/}T$, $-Q^a({\mathrm{\nabla }}_{a}T+T{a}_a)\mathrm{/}{T}^2=\kappa {\mathrm{\Delta }}^{ab}({\mathrm{\nabla }}_{a}T+T{a}_a)({\mathrm{\nabla }}_{b}T+T{a}_b)\mathrm{/}(c{T}^2)$, and $-V^a{\mathrm{\nabla }}_a(\mu \mathrm{/}T)=L{\mathrm{\Delta }}^{ab}{\mathrm{\nabla }}_a(\mu \mathrm{/}T){\mathrm{\nabla }}_b(\mu \mathrm{/}T)$. The rest-space metric induced by $\mathrm{\Delta }$ is positive definite by [F5]; with $T,c>0$ and nonnegative coefficients every summand is nonnegative, proving the stated residual inequality.

In Minkowski coordinates, $T^{00}=ϵ$ and $T^{i0}=Q^i$ for the fixed-rest stress of [A2]. The temporal balance ${\mathrm{\nabla }}_a{T}^{a0}=f^0=0$ is $c^{-1}{\mathrm{\partial }}_tϵ+{\mathrm{\partial }}_i{Q}^i=0$. The spatial components are ${\mathrm{\nabla }}_a{T}^{ai}=c^{-1}{\mathrm{\partial }}_t{Q}^i+{\mathrm{\partial }}_{i}p=f^i$, as required by [A2]. Using $ϵ={ϵ}_0+C\vartheta$ and $q=cQ=-\kappa \mathrm{\nabla }\vartheta$, the temporal balance becomes $C{\mathrm{\partial }}_t\vartheta +\mathrm{\nabla }\cdot q=0$, or $C{\mathrm{\partial }}_t\vartheta -\kappa \mathrm{\Delta }\vartheta =0$. Put $a_h=\kappa \mathrm{/}C>0$ and $U(\mathbf{x},\tau )={\int }_{{\mathbb{R}}^3}\mathrm{\Gamma }(\mathbf{x}-\mathbf{y},\tau ){\vartheta }_0(\mathbf{y})d\mathbf{y}$. The smooth compactly supported datum in [A2] is bounded and uniformly continuous, so [F4] gives a smooth solution of $U_{\tau }=\mathrm{\Delta }U$ for $\tau >0$ with initial trace ${\vartheta }_0$ locally uniformly. Then $\vartheta (\mathbf{x},t)=U(\mathbf{x},a_{h}t)$ solves the fixed-rest equation and has the stated boundary trace. Kernel positivity [F2] and ${\vartheta }_0\ge 0$ give $\vartheta \ge 0$. The unit integral from [F3] and ${\vartheta }_0\le \delta$ give $\vartheta (\mathbf{x},t)\le \delta {\int }_{{\mathbb{R}}^3}\mathrm{\Gamma }(\mathbf{x}-\mathbf{y},a_{h}t)d\mathbf{y}=\delta ,$ so the temperature remains in the specified state interval.

Write $u^0=1+O(\mathrm{\mid }\mathbf{w}{\mathrm{\mid }}^2)$ and $u^i=w^i+O(\mathrm{\mid }\mathbf{w}{\mathrm{\mid }}^2)$ in the homogeneous rest perturbation of [A4]. With $x^0=ct$, $a^i=u^a{\mathrm{\nabla }}_a{u}^i=c^{-1}{\dot{w}}^i$ to first order, so the first-order acceleration heat law gives $Q^i=-\kappa T_0{\dot{w}}^i\mathrm{/}{c}^2$. In ${\mathrm{\nabla }}_a{T}^{ai}=0$, the linearized mixed component is $T^{0i}=h_0{w}^i+Q^i$: the perfect part contributes $h_0{w}^i$, the term $u^0{Q}^i$ contributes $Q^i$, and $Q^0{u}^i$ is second order. All spatial derivatives vanish, so ${\mathrm{\partial }}_0{T}^{0i}=0$ gives $h_0{\dot{w}}^i\mathrm{/}c+{\dot{Q}}^i\mathrm{/}c=0$. Multiplication by $c$ gives $h_0{\dot{w}}^i+{\dot{Q}}^i=0$, the stated homogeneous linearized momentum equation.

Since ${\vartheta }_0$ is continuous, nonnegative, and nonzero, it is bounded below by a positive constant on some closed ball of positive radius. For every fixed $\mathbf{x}$ and $t>0$, the continuous kernel $\mathrm{\Gamma }(\mathbf{x}-\mathbf{y},a_{h}t)$ has a positive minimum on that ball by [F2], so the convolution from step 1.2 is strictly positive. Thus $\mathrm{supp}\vartheta (\cdot ,t)={\mathbb{R}}^3$. For any finite $v\ge 0$, $\mathrm{supp}{\vartheta }_0+{\bar{B}}_{vt}(0)$ is bounded at each fixed $t>0$ because $\mathrm{supp}{\vartheta }_0$ is compact; it cannot contain $\mathrm{supp}\vartheta (\cdot ,t)$. This proves the claimed failure of finite propagation.

Let $\alpha =\kappa T_0\mathrm{/}{c}^2>0$. Substitution of $Q^i=-\alpha {\dot{w}}^i$ into the equation of step 1.3 gives ${\ddot{w}}^i=(h_0\mathrm{/}\alpha ){\dot{w}}^i=\lambda {\dot{w}}^i$. For any ${\mathbf{w}}_{\ast }\mathrm{\ne }0$, $w^i(t)=w_{\ast }^i{e}^{\lambda t}$ satisfies this equation because ${\ddot{w}}^i={\lambda }^2{w}^i=\lambda {\dot{w}}^i$. Since $\lambda >0$, its norm grows exponentially. This proves a linearly growing mode without asserting that the linearized solution remains within the regime of a nonlinear material model.