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

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Acoustic and inertial finite modes — worked interpretation

Example

For the acoustic mode, take $ρ_{0} = 1 k g m^{- 3}$ , $P (ρ) = (4 m^{2} s^{- 2}) ρ$ , $k = (1, 2, 2) m^{- 1}$ , $A = 1 / 4 k g m^{- 3}$ , and $ω = 6 s^{- 1}$ . Then $χ = P^{'} (ρ_{0}) = 4 m^{2} s^{- 2}$ , $∣ k ∣ = 3 m^{- 1}$ , and the fields are

$r (t, x) = \frac{1}{4} C (θ) k g m^{- 3}, v (t, x) = (\frac{1}{6}, \frac{1}{3}, \frac{1}{3}) C (θ) m s^{- 1}, θ = k \cdot x - ω t .$

The rotating example takes $Ω = (0, 0, 3) s^{- 1}$ , $k = (1, 0, 1) m^{- 1}$ , $a = (0, 1, 0)$ , and $W = 2 m s^{- 1}$ . With $e = (1, 0, 1) / \sqrt{2}$ , $b = (- 1, 0, 1) / \sqrt{2}$ , and $α = 3 \sqrt{2} s^{- 1}$ , its theorem-defined vector $z$ and scalar $β$ are

$z (t) = (\frac{S (α t)}{\sqrt{2}}, C (α t), - \frac{S (α t)}{\sqrt{2}}), β (t) = - 3 \sqrt{2} C (α t) s^{- 1} .$

Consequently the rotating fields are

$w (t, x) = 2 z (t) C (k \cdot x) m s^{- 1}, q (t, x) = 6 C (α t) S (k \cdot x) m^{2} s^{- 2} .$

These are exact substitutions into the specified linear systems, not observations or nonlinear fluid solutions.

Example references

Acoustic and rotating finite-mode theorem: Acoustic and inertial finite-mode solutions.

Facts & Assumptions

Given: The formal parameter values and fields in the Example.

[F1]

For $χ > 0$ , the displayed acoustic fields solve the linear acoustic system when $A \neq 0$ and $ω^{2} = χ ∣ k ∣^{2}$ ; the theorem also gives the zero-amplitude case (Acoustic and inertial finite-mode solutions).

[F2]

For constant $Ω$ and nonzero $k$ , the theorem-defined $e, a, b, α, z, β, w, q$ satisfy the rotating linear system when $∣ a ∣ = ∣ e ∣ = 1$ and $a ⊥ e$ (Acoustic and inertial finite-mode solutions).

Verification

1.1givenF1algebra

The acoustic data give $∣ k ∣^{2} = 1 + 4 + 4 = 9 m^{- 2}$ and $χ ∣ k ∣^{2} = 4 \cdot 9 = 36 s^{- 2} = ω^{2}$ . Also $∣ A ∣ = 1 / 4 < ρ_{0} = 1$ . Substitution in [F1] yields exactly the displayed $r$ and $v$ , and the density lower bound from the theorem is $ρ_{0} - ∣ A ∣ = 3 / 4 k g m^{- 3}$ .

2.1givenF2algebra∎

For the rotating data, $∣ k ∣ = \sqrt{2} m^{- 1}$ , $e = k / ∣ k ∣ = (1, 0, 1) / \sqrt{2}$ , and $b = e \times a = (- 1, 0, 1) / \sqrt{2}$ . Thus $∣ a ∣ = ∣ e ∣ = 1$ and $a ⊥ e$ . The theorem gives $α = 2 Ω \cdot e = 3 \sqrt{2} s^{- 1}$ and $z = a C (α t) - b S (α t)$ , which is the displayed vector. Directly, $2 Ω \times z = (- 6 C (α t), 3 \sqrt{2} S (α t), 0) s^{- 1}$ , so $β = e \cdot (2 Ω \times z) = - 3 \sqrt{2} C (α t) s^{- 1}$ . Therefore $- W β / ∣ k ∣ = 6 m^{2} s^{- 2} C (α t)$ and the pressure and velocity fields are exactly those in the Example. The hypotheses of [F2] hold, so its conclusion verifies both rotating equations and zero divergence.

Depends on

Acoustic and inertial finite-mode solutions