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A longitudinal acoustic mode with constant transverse flow conserves weighted energy

Example

Assume Countable Choice, $ρ_{*} > 0$ , and $c_{*} > 0$ . On the torus $Q_{2} = R^{2} / (2 π Z)^{2}$ with normalized measure $d μ = (2 π)^{- 2} d x$ , let $e_{1}, e_{2}$ be the standard basis of $R^{2}$ and choose $A, B, W \in R$ . Put $Z (t) = A \cos (c_{*} t) - \frac{ρ_{*} B}{c_{*}} \sin (c_{*} t), V (t) = B \cos (c_{*} t) + \frac{c_{*} A}{ρ_{*}} \sin (c_{*} t),$ $z (t, x) = Z (t) \cos x_{1}, v (t, x) = V (t) \sin x_{1} e_{1} + W e_{2} .$ Then $(z, v)$ solves the linear acoustic tangent system $z_{t} + ρ_{*} div v = 0$ , $v_{t} + (c_{*}^{2} / ρ_{*}) \nabla z = 0$ for all $t \in R$ , with initial data $z (0, x) = A \cos x_{1}$ and $v (0, x) = B \sin x_{1} e_{1} + W e_{2}$ . Its weighted energy is constant and equals $∥ v (t) ∥_{L^{2} (Q_{2})}^{2} + \frac{c_{*}^{2}}{ρ_{*}^{2}} ∥ z (t) ∥_{L^{2} (Q_{2})}^{2} = W^{2} + \frac{1}{2} (B^{2} + \frac{c_{*}^{2}}{ρ_{*}^{2}} A^{2}) .$

Example references

Global smooth periodic linear acoustics

Tong, Fluid Mechanics, Chapter 3, §4.4.4

Facts & Assumptions

Given: $ρ_{*} > 0$ , $c_{*} > 0$ , real amplitudes $A, B, W$ , and the normalized torus conventions in the Example.

[F1]

Global smooth periodic linear acoustics: on the normalized periodic torus the system $z_{t} + ρ_{*} div v = 0$ , $v_{t} + (c_{*}^{2} / ρ_{*}) \nabla z = 0$ has a unique global smooth solution, and $∥ v ∥_{2}^{2} + (c_{*}^{2} / ρ_{*}^{2}) ∥ z ∥_{2}^{2}$ is conserved.

[A1]

Countable Choice: every countable family of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

1.1givenalgebra

Differentiate the amplitudes: $Z^{'} = - ρ_{*} V$ and $V^{'} = (c_{*}^{2} / ρ_{*}) Z$ . Thus $z_{t} = - ρ_{*} V \cos x_{1}$ and $div v = V \cos x_{1}$ , while $v_{t} = V^{'} \sin x_{1} e_{1}$ and $\nabla z = - Z \sin x_{1} e_{1}$ . Substitution gives both equations of the stated linear system. At $t = 0$ , $Z (0) = A$ and $V (0) = B$ , so the initial fields are exactly those in the Example.

2.1givenA1F1step 1.1

By [F1], the fields from step 1.1 are the unique global smooth solution with these data, and their weighted energy is the stated norm sum.

3.1givenstep 2.1algebra

Under the stated normalized measure, $\int_{Q_{2}} \sin^{2} x_{1} d μ = \int_{Q_{2}} \cos^{2} x_{1} d μ = 1 / 2$ and $\int_{Q_{2}} 1 d μ = 1$ . Since $e_{1} ⊥ e_{2}$ , the energy in step 2.1 is $W^{2} + \frac{1}{2} (V^{2} + (c_{*}^{2} / ρ_{*}^{2}) Z^{2})$ . Differentiating the bracket gives $2 V V^{'} + 2 (c_{*}^{2} / ρ_{*}^{2}) Z Z^{'} = 2 (c_{*}^{2} / ρ_{*}) V Z - 2 (c_{*}^{2} / ρ_{*}) Z V = 0$ . Its value at $t = 0$ is $B^{2} + (c_{*}^{2} / ρ_{*}^{2}) A^{2}$ , proving the asserted constant.

4.1givenstep 1.1step 3.1∎

The explicit fields in step 1.1 have only smooth sine, cosine, and constant factors, so they are smooth for every real time; the formula and energy calculation therefore verify the complete Example.

Depends on

Dependency tree · 2 levels

Sources

David Tong, Fluid Mechanics, Chapter 3, §4.4.4