= Solution
To distinguish <Eulerian and Lagrangian fluid perturbations>, use $\boldsymbol\xi=(\xi_x,\xi_y,\xi_z)$ for the <fluid displacement>, $\phi=\Phi'$ for the Eulerian potential perturbation, and $\delta\rho$ for the Eulerian <mass density> perturbation. The equilibrium <mass density> $\rho(z)$ is the profile from part (a); subscript $z$ below denotes differentiation. Linearized <mass conservation> and the <isothermal equation of state> give
$$
\delta\rho=-ik\rho\xi_x-(\rho\xi_z)_z,\qquad
\delta p=c_s^2\delta\rho.
$$
The <velocity> perturbation is $-i\omega\boldsymbol\xi$. Linearized momentum initially has the form
$$
-\omega^2\rho\boldsymbol\xi=-\nabla\delta p-\delta\rho\nabla\Phi_0-\rho\nabla\phi.
$$
Since $\nabla\Phi_0=-c_s^2\nabla\ln\rho$, the <pressure> and perturbed <mass density> gravity terms combine into $-\rho\nabla(c_s^2\delta\rho/\rho)$. Define $W=c_s^2\delta\rho/\rho+\phi$. The closed linearized system is
$$
\boxed{\omega^2\xi_x=ikW,\qquad \omega^2\xi_z=W_z,\qquad
\omega^2\xi_y=0,\qquad
\phi_{zz}-k^2\phi=4\pi G\delta\rho,\qquad
\delta\rho=-ik\rho\xi_x-(\rho\xi_z)_z}.
$$
The $y$ component is a neutral transverse sector; it vanishes for nonzero-frequency modes.
To prove reality rather than assume it, multiply momentum by $\rho\boldsymbol\xi^*$ and integrate over $z$. With $I=\int\rho|\boldsymbol\xi|^2dz>0$, <integration by parts> and the conjugate <continuity equation> give
$$
\omega^2I=\left[\rho\xi_z^*W\right]_{-\infty}^{\infty}
+c_s^2\int\frac{|\delta\rho|^2}{\rho}\,dz+\int\delta\rho^*\phi\,dz.
$$
Using the conjugate <Poisson equation for Newtonian gravity> in the last integral gives
$$
\int\delta\rho^*\phi\,dz
=\frac1{4\pi G}\left[\phi\phi_z^*\right]_{-\infty}^{\infty}
-\frac1{4\pi G}\int(|\phi_z|^2+k^2|\phi|^2)\,dz.
$$
Take disturbances with finite $I$, finite <pressure> and gravitational integrals, and vanishing <pressure> work and gravitational boundary terms. For $k\ne0$, require the potential to decay at both infinities and the <fluid displacement> to make $\rho\xi_z^*W\to0$ as well. In a source-free outer region the decaying potential is proportional to $e^{-|k||z|}$. The <isothermal displacement energy of a self-gravitating slab> then gives
$$
\boxed{\omega^2 I=c_s^2\int\frac{|\delta\rho|^2}{\rho}\,dz
-\frac1{4\pi G}\int(|\phi_z|^2+k^2|\phi|^2)\,dz\in\mathbb R}.
$$
Since $I$ is strictly positive for a nonzero displacement, \b[$\omega^2$ is real]. Positive values describe oscillations; negative values give pure growth and decay. The result is the <energy> identity for the pressure-gravity displacement operator, a <self-adjoint operator> in the <mass density>-weighted <inner product>. The identity excludes an oscillatory growing eigenmode under these <boundary conditions>. At $k=0$ the same identity applies when its boundary terms vanish and its integrals remain finite, with the gravitational integral involving only $|\phi_z|^2$.
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