Global smooth periodic linear acoustics

The mass residual at $({\rho }_{\ast }+ϵz,ϵv)$ is $ϵ(z_t+{\rho }_{\ast }\mathrm{div}v)+{ϵ}^2\mathrm{div}(zv)$. The velocity residual is $ϵv_t+{ϵ}^{2}v\cdot \mathrm{\nabla }v+ϵ[P^{\mathrm{\prime }}({\rho }_{\ast }+ϵz)\mathrm{/}({\rho }_{\ast }+ϵz)]\mathrm{\nabla }z$. Differentiating at zero gives exactly the two displayed linear equations because $P^{\mathrm{\prime }}({\rho }_{\ast })=c_{\ast }^2$. Thus the tangent interpretation follows from the defined residual derivative, without assuming that perturbations solve the nonlinear system for nonzero $ϵ$.

Write Fourier coefficients with kernel $e^{-ik\cdot x}$. For each $k\in {\mathbb{Z}}^d$, the PDE is the finite-dimensional ODE $\dot{\widehat{z}}=-i{\rho }_{\ast }k\cdot \widehat{v}$, $\dot{\widehat{v}}=-i(c_{\ast }^2\mathrm{/}{\rho }_{\ast })k\widehat{z}$. For $k\mathrm{\ne }0$, differentiating the first equation and substituting the second gives $\ddot{\widehat{z}}+c_{\ast }^2\mathrm{\mid }k{\mathrm{\mid }}^2\widehat{z}=0$. With ${\omega }_k=c_{\ast }\mathrm{\mid }k\mathrm{\mid }$, its unique solution is $\widehat{z}(t,k)={\widehat{z}}_0(k)\cos ({\omega }_{k}t)-i{\rho }_{\ast }(k\cdot {\widehat{v}}_0(k))\sin ({\omega }_{k}t)\mathrm{/}{\omega }_k$, and integrating the second mode equation gives $\widehat{v}(t,k)={\widehat{v}}_0(k)-i(c_{\ast }^2\mathrm{/}{\rho }_{\ast })k{\int }_0^t\widehat{z}(s,k)ds$. At $k=0$ both coefficients are constant.

Differentiation of these formulas recovers both mode equations and their initial coefficients. For each $k$, differentiating $E_k=\mathrm{\mid }\widehat{v}{\mathrm{\mid }}^2+(c_{\ast }^2\mathrm{/}{\rho }_{\ast }^2)\mathrm{\mid }\widehat{z}{\mathrm{\mid }}^2$ gives zero: the two cross terms are $-i(c_{\ast }^2\mathrm{/}{\rho }_{\ast })[(k\widehat{z})\cdot \bar{\widehat{v}}+(k\cdot \widehat{v})\bar{\widehat{z}}]$, whose bracket is real, so its real part vanishes. Thus $E_k(t)=E_k(0)$ for every mode. The mode formulas also preserve conjugate symmetry because their coefficients at $-k$ are conjugates of those at $k$.

For each scalar component $f$ of the smooth initial data and every integer $N\ge 0$, periodic integration by parts gives $(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^N\widehat{f}(k)=\widehat{(1-\mathrm{\Delta }{)}^{N}f}(k)$, where $\mathrm{\Delta }={\sum }_{j=1}^d{\mathrm{\partial }}_{x_j}^2$; opposite periodic-face terms cancel. Since $(1-\mathrm{\Delta }{)}^{N}f$ is continuous on the compact torus, its $L^1$ norm $C_{f,N}$ is finite, and $\mathrm{\mid }\widehat{f}(k)\mathrm{\mid }\le C_{f,N}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-N}$. For any integer $m\ge 0$, choose $N$ with $4N-2m>d$. The shell $j\le \mathrm{\mid }k\mathrm{\mid }<j+1$ contains at most $C_d(j+1{)}^{d-1}$ lattice points, by placing disjoint unit cubes at its lattice points inside a shell of fixed enlarged width, so ${\sum }_k(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{m-2N}<\mathrm{\infty }$ by comparison with ${\sum }_{j\ge 1}{j}^{d-1+2m-4N}$. Thus every initial component belongs to every $H^m$ under [F1]. Multiplying the mode identity $E_k(t)=E_k(0)$ from step 2.1 by $(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^m$ and summing these nonnegative finite-energy series proves, by [F2], the stated conserved weighted energy. In particular both coefficient families remain in every $H^m$, uniformly for all real time.

Fourier synthesis [F2], under the stated Countable Choice assumption [A1], now defines real fields $z(t),v(t)$. The uniform $H^m$ bounds and embeddings [F3] imply convergence of the Fourier series with every spatial derivative, locally uniformly in time. The mode equations pass through these convergent series and give the PDE. Write $q=(z,v)$ and let $\mathcal{L}$ be the constant-coefficient first-order spatial operator in that system; recursively ${\mathrm{\partial }}_t^{n}q={\mathcal{L}}^{n}q$, which is a finite sum of spatial derivatives of order at most $n$. Thus all mixed derivatives are continuous, so $q\in C^{\mathrm{\infty }}(\mathbb{R}\times {\mathbb{T}}^d)$.

For any two smooth solutions with the same data, each Fourier-mode difference solves the same linear ODE with zero initial value. Uniqueness of the finite-dimensional linear ODE gives zero difference at each $k$, and Fourier synthesis then gives equality of the fields. Together with the energy conservation and smooth synthesis above, this proves the global existence, conservation, and uniqueness assertions.