= Solution
Consider the nontrivial gravitational mode with $k\ne0$. At $\omega^2=0$, horizontal momentum requires $ikW=0$, hence $W=0$ and
$$
\delta\rho=-\frac{\rho}{c_s^2}\phi.
$$
Substituting into the <Poisson equation for Newtonian gravity> and using $4\pi G\rho_0/c_s^2=2/H^2$ gives
$$
\phi_{zz}+\left[\frac2{H^2}\operatorname{sech}^2(z/H)-k^2\right]\phi=0.
$$
Set $\tau=\tanh(z/H)$ and $\nu=kH$. Since $\partial_z=(1-\tau^2)\partial_\tau/H$, this becomes the <associated Legendre equation> of degree one and order $\nu$:
$$
\boxed{\frac d{d\tau}\left[(1-\tau^2)\frac{d\phi}{d\tau}\right]
+\left[2-\frac{\nu^2}{1-\tau^2}\right]\phi=0}.
$$
To verify the two explicit solutions, put $\eta=z/H$ and $T=\tanh\eta$. They are
$$
F_+=e^{\nu\eta}(\nu-T),\qquad F_-=e^{-\nu\eta}(\nu+T),
$$
because $e^{2\eta}=(1+\tau)/(1-\tau)$. Differentiating $F_+$ twice gives
$$
(F_+)_{\eta\eta}=e^{\nu\eta}\left[\nu^2(\nu-T)-2\nu(1-T^2)+2T(1-T^2)\right],
$$
which cancels identically against $[2(1-T^2)-\nu^2]F_+$. Reflection $\eta\mapsto-\eta$ proves the same for $F_-$. Their <Wronskian> is $2\nu(1-\nu^2)$, evaluated at $\eta=0$; thus they are generally independent but coincide or become dependent at the exceptional orders $0,\pm1$.
Use $n=|k|H>0$ to analyze decay, which is unaffected by the sign of $k$. The solution decaying at $+\infty$ is proportional to $e^{-n\eta}(n+\tanh\eta)$. At $-\infty$ its leading term is $(n-1)e^{n|\eta|}$, so it also decays there only if $n=1$. At this value
$$
e^{-\eta}(1+\tanh\eta)=e^{\eta}(1-\tanh\eta)=\operatorname{sech}\eta.
$$
Therefore the <marginal fragmentation mode of an isothermal slab> has
$$
\boxed{|k|=\frac1H,\qquad \Phi'=\phi\propto\operatorname{sech}(z/H),\qquad
\delta\rho\propto\operatorname{sech}^3(z/H)}.
$$
At $n=1$ the second independent solution obtained by <reduction of order> grows at infinity, so the coincidence of the two explicit formulas does not supply an extra admissible mode. At $k=0$ a uniform vertical translation gives another neutral disturbance, and mass-preserving relabeling displacements can have $\delta\rho=\phi=0$. Neither is the nonzero-<wavenumber> fragmentation threshold just calculated.
\b[The unstable fragmentation modes have wavelengths greater than $2\pi H$.] Gas-<pressure> support becomes less effective relative to <self-gravity> at long wavelength. The side of the threshold can also be checked from the <energy> identity: at $|k|=1/H$ the marginal <mass density> disturbance has zero quadratic <energy>. Keep that <mass density> disturbance and decrease $|k|$; the <pressure> integral stays fixed, while the gravitational term becomes more negative because the positive inverse operator $(-\partial_z^2+k^2)^{-1}$ increases as $k^2$ decreases. Thus the quadratic <energy> becomes negative, giving an unstable mode on the longer-wavelength side. This argument concerns the gravitational sector and its stated decay conditions.
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