Acoustic and inertial finite-mode solutions

Let ${\rho }_{ϵ}={\rho }_0+ϵr$ and $u_{ϵ}=ϵv$ with constant $V$. The mass residual expands as ${\mathrm{\partial }}_t{\rho }_{ϵ}+\mathrm{div}({\rho }_{ϵ}{u}_{ϵ})=ϵ(r_t+{\rho }_0\mathrm{div}v)+{ϵ}^2\mathrm{div}(rv)$. The momentum residual has ${\mathrm{\partial }}_t{u}_{ϵ}+(u_{ϵ}\cdot \mathrm{\nabla })u_{ϵ}=ϵv_t+{ϵ}^2(v\cdot \mathrm{\nabla })v$, while the chain rule gives $\mathrm{\nabla }P({\rho }_{ϵ})=ϵP^{\mathrm{\prime }}({\rho }_0+ϵr)\mathrm{\nabla }r$. At each point ${\rho }_{ϵ}>0$ for sufficiently small $\mathrm{\mid }ϵ\mathrm{\mid }$, and $P^{\mathrm{\prime }}({\rho }_0+ϵr)\mathrm{/}{\rho }_{ϵ}\to \chi \mathrm{/}{\rho }_0$. Thus the coefficients of $ϵ$ in the two residuals are exactly the stated linear acoustic equations; this first-variation calculation supplies no error estimate for nonzero $ϵ$. For regularity, [F1] gives differentiability and hence continuity of $S,C$, and differentiating its identities yields $S^{\mathrm{\prime }\mathrm{\prime }}=C^{\mathrm{\prime }}=-S$ and $C^{\mathrm{\prime }\mathrm{\prime }}=-S^{\mathrm{\prime }}=-C$, both continuous. Each displayed spatial mode is a time-dependent coefficient times $S$ or $C$ composed with an affine function of $x$; twice applying [F3] gives its first and second spatial partial derivatives, which are continuous by these identities. The time-dependent coefficients $e^{\lambda t}$, $z(t)$, and $\beta (t)$, when present, have continuous first derivatives by [F1], [F3], and [F4] (in particular, ${\beta }^{\mathrm{\prime }}(t)=e\cdot (2\mathrm{\Omega }\times z^{\mathrm{\prime }}(t))$). Therefore the displayed first time partials are continuous as well.

Put $\theta =k\cdot x-\omega t$. By [F1] and [F3], $r_t=A\omega S(\theta )$, $\mathrm{\nabla }r=-AkS(\theta )$, and $\mathrm{div}v=-A\omega S(\theta )\mathrm{/}{\rho }_0$, so $r_t+{\rho }_0\mathrm{div}v=0$.

Let $e=k\mathrm{/}\mathrm{\mid }k\mathrm{\mid }$. At least one standard coordinate unit vector $d$ is not parallel to $e$, since the three coordinate vectors are linearly independent by [F8]. Put $y=d-⟨d,e⟩e$; then $y\mathrm{\ne }0$ and $⟨y,e⟩=0$. Set $a=y\mathrm{/}\mathrm{\mid }y\mathrm{\mid }$ and $b=e\times a$. By [F6], $b$ is orthogonal to $e$ and $a$, and $\mathrm{\mid }b{\mathrm{\mid }}^2=\mathrm{\mid }e{\mathrm{\mid }}^2\mathrm{\mid }a{\mathrm{\mid }}^2-⟨e,a{⟩}^2=1$. The three orthonormal vectors are linearly independent; since ${\mathbb{R}}^3$ has dimension three by [F8], they form a basis.

Also $v_t=A{\omega }^{2}kS(\theta )\mathrm{/}({\rho }_0\mathrm{\mid }k{\mathrm{\mid }}^2)$, while $-(\chi \mathrm{/}{\rho }_0)\mathrm{\nabla }r=\chi AkS(\theta )\mathrm{/}{\rho }_0$. Since $k\mathrm{\ne }0$, varying $x$ makes $\theta$ range over $\mathbb{R}$. From the displayed definition of $C$ and $E(0)={\sum }_{n=0}^{\mathrm{\infty }}{0}^n\mathrm{/}n!=1$, we have $C(0)=\mathrm{Re}((1+1)\mathrm{/}2)=1$; [F1] then gives $S^{\mathrm{\prime }}(0)=C(0)=1$, so $S$ is not identically zero. Thus for $A\mathrm{\ne }0$ these expressions agree for all $x,t$ exactly when ${\omega }^2=\chi \mathrm{\mid }k{\mathrm{\mid }}^2$; if $A=0$ both fields vanish. Finally [F2] gives ${\rho }_0+r\ge {\rho }_0-\mathrm{\mid }A\mathrm{\mid }>0$ under the stated amplitude bound.

Suppose $\chi <0$, let $\lambda =\mathrm{\mid }k\mathrm{\mid }\sqrt{-\chi }$, and use the stated growing fields. With $\vartheta =k\cdot x$, [F1], [F3], and [F4] give $r_t=\lambda A{e}^{\lambda t}C(\vartheta )$ and $\mathrm{div}v=-\lambda A{e}^{\lambda t}C(\vartheta )\mathrm{/}{\rho }_0$, so the mass equation holds. They also give $v_t=-{\lambda }^{2}A{e}^{\lambda t}kS(\vartheta )\mathrm{/}({\rho }_0\mathrm{\mid }k{\mathrm{\mid }}^2)$ and $-(\chi \mathrm{/}{\rho }_0)\mathrm{\nabla }r=\chi A{e}^{\lambda t}kS(\vartheta )\mathrm{/}{\rho }_0$; these agree because ${\lambda }^2=-\chi \mathrm{\mid }k{\mathrm{\mid }}^2$. Thus the Fourier mode grows as $e^{\lambda t}$ in the linear system.

By [F9], $e\times b=e\times (e\times a)=-a$ and $a\times b=a\times (e\times a)=e$. Write ${\mathrm{\Omega }}_{\perp }=\mathrm{\Omega }-⟨\mathrm{\Omega },e⟩e={\mathrm{\Omega }}_{a}a+{\mathrm{\Omega }}_{b}b$ and $z=z_{a}a+z_{b}b$. Bilinearity and alternation from [F6] give ${\mathrm{\Omega }}_{\perp }\times z=({\mathrm{\Omega }}_a{z}_b-{\mathrm{\Omega }}_b{z}_a)e$. Since $\mathrm{\Omega }=⟨\mathrm{\Omega },e⟩e+{\mathrm{\Omega }}_{\perp }$, it follows that $\mathrm{\Omega }\times z=⟨\mathrm{\Omega },e⟩(e\times z)+({\mathrm{\Omega }}_a{z}_b-{\mathrm{\Omega }}_b{z}_a)e$. The map $\mathrm{\Pi }y=y-⟨y,e⟩e$ fixes $e\times z$ and annihilates the last term, so $\mathrm{\Pi }(\mathrm{\Omega }\times z)=⟨\mathrm{\Omega },e⟩e\times z$.

Differentiating $z(t)=aC(\alpha t)-bS(\alpha t)$ with [F1] gives $z^{\mathrm{\prime }}=-\alpha aS(\alpha t)-\alpha bC(\alpha t)$. The identities $e\times a=b$ and $e\times b=-a$ from steps 1.3 and 2.3 give $e\times z=bC(\alpha t)+aS(\alpha t)$. Hence $z^{\mathrm{\prime }}=-\alpha e\times z=-2\mathrm{\Pi }(\mathrm{\Omega }\times z)$.

By the definition of $\beta$ and step 2.3, $2\mathrm{\Omega }\times z=\alpha e\times z+\beta e$, so step 3.1 gives $z^{\mathrm{\prime }}+2\mathrm{\Omega }\times z=\beta e$. Consequently ${\mathrm{\partial }}_{t}w+2\mathrm{\Omega }\times w=W\beta eC(k\cdot x)$; also $\mathrm{div}w=W{C}^{\mathrm{\prime }}(k\cdot x)z\cdot k=0$, and $-\mathrm{\nabla }q=W\beta eC(k\cdot x)$ by [F1]. Thus the rotating equations hold. Finally $\mathrm{\mid }\alpha \mathrm{\mid }=2\mathrm{\mid }⟨\mathrm{\Omega },e⟩\mathrm{\mid }\le 2\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$ by [F7] and $\mathrm{\mid }e\mathrm{\mid }=1$; when $\alpha =0$ no division by $\alpha$ occurs.