CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Negative acoustic response yields exponential mode growth

Physical scope

Exact counterexample for the linearized equations of an explicitly adopted smooth barotropic state relation with negative fixed-entropy pressure response. It is not a finite-amplitude solution or an empirical claim about a realized material.

Statement refuted

Let $U \subset (0, \infty) \times R$ be open and let $e \in C^{2} (U)$ , with $T = e_{s}$ and $p = - e_{v}$ (each partial derivative holds the other state coordinate fixed). Fix $(v_{b}, s_{b}) \in U$ with $T_{b} = e_{s} (v_{b}, s_{b}) > 0$ and $e_{s s} (v_{b}, s_{b}) > 0$ , and put $ρ_{b} = 1 / v_{b}$ . Let $P \in C^{1} ((0, \infty))$ agree with $- e_{v} (1 / ρ, s_{b})$ on an open interval $J_{ρ} \subset (0, \infty)$ containing $ρ_{b}$ , with $(1 / ρ, s_{b}) \in U$ for every $ρ \in J_{ρ}$ , and suppose $P^{'} (ρ_{b}) < 0$ . For any nonempty open time interval $I \subseteq R$ , let $δ ρ \in C^{2} (I \times R^{3}, R)$ and $w \in C^{2} (I \times R^{3}, R^{3})$ solve the first-variation equations about the uniform rest state $(ρ, u) = (ρ_{b}, 0)$ , $\partial_{t} δ ρ + ρ_{b} div w = 0, ρ_{b} \partial_{t} w + P^{'} (ρ_{b}) \nabla δ ρ = 0,$ where $div w = \sum_{i = 1}^{3} \partial_{x_{i}} w_{i}$ and $\nabla δ ρ = (\partial_{x_{i}} δ ρ)_{i = 1}^{3}$ . The false assertion is that for every such state and solution, at each fixed $x \in R^{3}$ , $\sup_{t \in I} (∣ δ ρ (t, x) ∣ + \sqrt{\sum_{i = 1}^{3} w_{i} (t, x)^{2}}) < \infty .$ In SI units, $t$ is in seconds, $x$ in metres, $v_{b}$ in $m^{3} k g^{- 1}$ , $s_{b}$ in $J k g^{- 1} K^{- 1}$ , $e$ in $J k g^{- 1}$ , $T_{b}$ in kelvins, $p$ and $P$ in pascals, $ρ_{b}$ and $δ ρ$ in $k g m^{- 3}$ , and $w$ in $m s^{- 1}$ .

Statement references

David Tong, Fluid Mechanics, Part III lecture notes, §4.4.4, printed pp. 150–151

An Introduction to Continuum Mechanics, Second Edition, Chapter 6, §6.5.2, printed pp. 243–244

Facts & Assumptions

Given: A smooth local state relation with $T_{b} > 0$ , $e_{s s} (v_{b}, s_{b}) > 0$ , a positive reference density, and a first-variation system about rest.

[L1]

Positive-response linear sound in adopted barotropic fluid: for the stated fixed-entropy barotropic restriction of a $C^{2}$ local state relation, the first variation is $\partial_{t} δ ρ + ρ_{b} div w = 0$ , $ρ_{b} \partial_{t} w + c_{s}^{2} \nabla δ ρ = 0$ , with $c_{s}^{2} = P^{'} (ρ_{b}) = v_{b}^{2} e_{v v} (v_{b}, s_{b})$ . The current theorem carrier and complete conditional proof were inspected, and its current Step 1 readiness record reports ready; this records authoring readiness, not independent proof approval.

[A1]

The local state relation and the negative-response coefficient below are adopted mathematical model data. No assertion is made that a measured material has this response.

Counterexample

Given: Choose $v_{b} > 0$ , $T_{b} > 0$ , $p_{b} > 0$ , $s_{b} \in R$ , $e_{b} \in R$ , and constants $C > 0$ , $A > 0$ with units chosen so every term below has units $J k g^{- 1}$ . On a sufficiently small open rectangle about $(v_{b}, s_{b})$ with $v > 0$ , define $e (v, s) = e_{b} + T_{b} (s - s_{b}) + \frac{C}{2} (s - s_{b})^{2} - p_{b} (v - v_{b}) - \frac{A}{2} (v - v_{b})^{2} .$ The rectangle can be chosen so that $e_{s} = T_{b} + C (s - s_{b}) > 0$ . At the reference state, $e_{s s} = C > 0$ , $e_{v v} = - A < 0$ , $ρ_{b} = 1 / v_{b} > 0$ , and $c_{s}^{2} = v_{b}^{2} e_{v v} = - A v_{b}^{2} < 0$ . Define the global barotropic pressure law $P : (0, \infty) \to R$ by $P (ρ) = p_{b} + A (1 / ρ - v_{b})$ . It is $C^{\infty}$ on $(0, \infty)$ and agrees with $p (1 / ρ, s_{b}) = - e_{v} (1 / ρ, s_{b})$ whenever $(1 / ρ, s_{b})$ lies in the local rectangle. Thus it supplies the global $C^{1}$ pressure law required by the adopted model while retaining the chosen local state relation; no global sign condition on $P$ is imposed. Let $I = R$ , choose $k \in R^{3} ∖ {0}$ and amplitude $a_{0} \neq 0$ , set $α = \sqrt{- c_{s}^{2}} ∣ k ∣ > 0$ , and define $δ ρ (t, x) = a_{0} e^{α t} \cos (k \cdot x), w (t, x) = - \frac{α a_{0}}{ρ_{b} ∣ k ∣^{2}} k e^{α t} \sin (k \cdot x) .$

1.1A1L1givenalgebra

Differentiation of the chosen state relation gives $T_{b} > 0$ , $e_{s s} = C > 0$ , and $e_{v v} = - A < 0$ at $(v_{b}, s_{b})$ . The global law has $P \in C^{1} ((0, \infty))$ and $P^{'} (ρ) = - A / ρ^{2}$ ; on the local density interval its equality with $p (1 / ρ, s_{b})$ gives $P^{'} (ρ_{b}) = v_{b}^{2} e_{v v} = - A v_{b}^{2} = c_{s}^{2} < 0$ .

2.1L1givenstep 1.1algebra

The derivatives of the stated mode are $\partial_{t} δ ρ = α δ ρ$ , $div w = - α δ ρ / ρ_{b}$ , $\partial_{t} w = - α^{2} a_{0} k e^{α t} \sin (k \cdot x) / (ρ_{b} ∣ k ∣^{2})$ , and $\nabla δ ρ = - a_{0} k e^{α t} \sin (k \cdot x)$ . Therefore the continuity equation holds, and the momentum equation holds because $- α^{2} / ∣ k ∣^{2} - c_{s}^{2} = 0$ by the definition of $α$ .

3.1A1L1step 2.1algebra∎

Since $α > 0$ and $k \cdot x$ cannot have both sine and cosine zero, at every fixed $x$ either $δ ρ (t, x)$ or $w (t, x)$ has a nonzero coefficient multiplying $e^{α t}$ . Thus the pair is unbounded in time on $I = R$ , refuting the stated assertion. This smooth first-order solution concerns only the linearized mathematical system; it supplies no finite-amplitude solution or empirical material claim.

Depends on

Positive-response linear sound in adopted barotropic fluid