Response derivatives and linear acoustic sign

Let $U\subset (0,\mathrm{\infty })\times \mathbb{R}$ be open and let $e:U\to \mathbb{R}$ have continuous total derivatives through order two. Fix $(v_b,s_b)\in U$ with $T_b=e_s(v_b,s_b)>0$ and $e_{ss}(v_b,s_b)>0$, set ${\rho }_b=1\mathrm{/}{v}_b$ and $p=-e_v$, and use the local chart $s=s(v,T)$ from the unique branch satisfying $e_s(v,s(v,T))=T$. Define $c_v:={\frac{\mathrm{\partial }}{\mathrm{\partial }T}e(v,s(v,T))\mid }_{v=v_b,T=T_b},c_s^2:=v_b^2{e}_{vv}(v_b,s_b)\mathrm{.}$ Because $U$ is open, choose an open interval $J_{\rho }\subset (0,\mathrm{\infty })$ containing ${\rho }_b$ such that $(1\mathrm{/}\rho ,s_b)\in U$ for every $\rho \in J_{\rho }$, and define $P:J_{\rho }\to \mathbb{R}$ by $P(\rho ):=p(1\mathrm{/}\rho ,s_b)=-e_v(1\mathrm{/}\rho ,s_b)$. Then $P\in C^1(J_{\rho })$ and the chain rule gives $P^{\mathrm{\prime }}({\rho }_b)=v_b^2{e}_{vv}(v_b,s_b)=c_s^2$. This is the local isentropic pressure-response coefficient; its sign is not assumed.

Let $I\subset \mathbb{R}$ be a nonempty open time interval. On $I\times {\mathbb{R}}^3$, let $\rho :I\times {\mathbb{R}}^3\to J_{\rho }$ and $u:I\times {\mathbb{R}}^3\to {\mathbb{R}}^3$ have continuous partial derivatives through order two. Use the Euclidean dot product $x\cdot y:={\sum }_{i=1}^3{x}_i{y}_i$ and norm $\mathrm{\mid }x\mathrm{\mid }:=\sqrt{x\cdot x}$ on ${\mathbb{R}}^3$. Use $\mathrm{\nabla }h=({\mathrm{\partial }}_{x_i}h{)}_{i=1}^3$, $\mathrm{div}u={\sum }_{i=1}^3{\mathrm{\partial }}_{x_i}{u}_i$, and $(u\cdot \mathrm{\nabla })u_i={\sum }_{j=1}^3{u}_j{\mathrm{\partial }}_{x_j}{u}_i$. Consider the mathematical barotropic system ${\mathrm{\partial }}_t\rho +\mathrm{div}(\rho u)=0,\rho ({\mathrm{\partial }}_{t}u+(u\cdot \mathrm{\nabla })u)+\mathrm{\nabla }P(\rho )=0.$ Here $\mathrm{\Delta }r:={\sum }_{i=1}^3{\mathrm{\partial }}_{x_i{x}_i}r$. For $r:I\times {\mathbb{R}}^3\to \mathbb{R}$ and $w:I\times {\mathbb{R}}^3\to {\mathbb{R}}^3$ with continuous partial derivatives through order two, the pointwise first variation about $({\rho }_b,0)$ is defined by differentiating at $\epsilon =0$ after substituting $\rho ={\rho }_b+\epsilon r$ and $u=\epsilon w$, with $\epsilon$ sufficiently close to zero locally at each point so the perturbed density lies in $J_{\rho }$. It is $r_t+{\rho }_b\mathrm{div}w=0,{\rho }_b{w}_t+c_s^2\mathrm{\nabla }r=0,r_{tt}=c_s^2\mathrm{\Delta }r\mathrm{.}$ If $c_s^2>0$, then with $c_s=\sqrt{c_s^2}$, for every nonzero $k\in {\mathbb{R}}^3$ and every $A\in \mathbb{R}$ the following are plane-wave solutions: $r=A\cos (k\cdot x-\omega t),w=\frac{\omega A}{{\rho }_b\mathrm{\mid }k{\mathrm{\mid }}^2}k\cos (k\cdot x-\omega t),\omega =c_s\mathrm{\mid }k\mathrm{\mid }\mathrm{.}$ If $c_s^2<0$, then for every nonzero $k\in {\mathbb{R}}^3$ and every $A\in \mathbb{R}$, set $a=\sqrt{-c_s^2}\mathrm{\mid }k\mathrm{\mid }>0$. The fields $r=A{e}^{at}\cos (k\cdot x),w=-\frac{aA}{{\rho }_b\mathrm{\mid }k{\mathrm{\mid }}^2}k{e}^{at}\sin (k\cdot x)$ solve the same linearized system and have exponential time factor $e^{at}$; for $A\mathrm{\ne }0$, their amplitudes grow exponentially. These are exact statements about the linearized mathematical system; they provide no finite-amplitude error or lifespan estimate and make no empirical claim about a material.

Rohan Abeyaratne, Continuum Mechanics, Volume II, Chapter 12 §12.1, Problem 12.4, printed pp. 348–350