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Consider the nontrivial gravitational mode with $k\ne 0$. At ${\omega }^2=0$, horizontal momentum requires $ikW=0$, hence $W=0$ andSubstituting into the Poisson equation for Newtonian gravity and using $4\pi G{\rho }_0/c_s^2=2/H^2$ givesSet $\tau =\tanh (z/H)$ and $\nu =kH$. Since ${\partial }_z=(1-{\tau }^2){\partial }_{\tau }/H$, this becomes the associated Legendre equation of degree one and order $\nu$:

To verify the two explicit solutions, put $\eta =z/H$ and $T=\tanh \eta$. They arebecause $e^{2\eta }=(1+\tau )/(1-\tau )$. Differentiating $F_{+}$ twice giveswhich 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 $0,\pm 1$.

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 valueTherefore the marginal fragmentation mode of an isothermal slab hasAt $n=1$ 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 $\delta \rho =\varphi =0$. Neither is the nonzero-wavenumber fragmentation threshold just calculated.

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.