= Solution
On a length scale $\ell$, <pressure> communicates support over a sound-crossing <time> $\ell/v_s$, whereas self-gravity acts over a <time> of order $(G\rho_0)^{-1/2}$. When the former is longer, gravity can amplify an overdensity before <pressure> smooths it. Thus <Jeans instability> occurs at scales of order $\ell\gtrsim v_s/\sqrt{G\rho_0}$; the precise numerical factor depends on whether scale means <wavelength> or inverse <wavenumber>.
Take a uniform static background, small perturbations, an inviscid <barotropic fluid>, and $\delta p=v_s^2\delta\rho$ with $v_s^2=(dp/d\rho)_0>0$. An infinite Newtonian homogeneous density cannot itself have zero background gravitational <acceleration> while satisfying the unmodified <Poisson equation for Newtonian gravity>. Specify the <Jeans swindle>: subtract the homogeneous field, or regard an external prescription as balancing it, and apply Poisson's equation only to perturbations. The linearized <continuity equation> and <Euler equations for an inviscid fluid> are
$$
\partial_t\delta\rho+\rho_0\nabla\cdot\delta\mathbf v=0,\qquad\partial_t\delta\mathbf v=-\frac{v_s^2}{\rho_0}\nabla\delta\rho-\nabla\delta\phi,\qquad\nabla^2\delta\phi=4\pi G\delta\rho.
$$
Taking a <time> <derivative> of the first equation and substituting the second yields $\partial_t^2\delta\rho=v_s^2\nabla^2\delta\rho+4\pi G\rho_0\delta\rho$. A <plane wave> $e^{i(\mathbf k\cdot\mathbf x-\omega t)}$ therefore gives
$$
\boxed{\omega^2=v_s^2k^2-4\pi G\rho_0,\qquad k_J=\frac{\sqrt{4\pi G\rho_0}}{v_s}.}
$$
For $0<k<k_J$ there is exponential growth at rate $\sqrt{4\pi G\rho_0-v_s^2k^2}$. At $k_J$ the density mode is neutral to this linear order, and above it there are acoustic oscillations. The critical <wavelength> is $2\pi/k_J=\sqrt\pi\,v_s/\sqrt{G\rho_0}$. Transverse <velocity> perturbations decouple as zero-frequency vortical modes in this idealization.
For the attractive <Yukawa potential>, assume a real screening parameter $\alpha\geq0$. Its Green function obeys $(\nabla^2-\alpha^2)(-Gm e^{-\alpha r}/r)=4\pi Gm\delta^{(3)}(\mathbf r)$, including the delta source. Thus $\delta\phi_{\mathbf k}=-4\pi G\delta\rho_{\mathbf k}/(k^2+\alpha^2)$. The same fluid elimination gives <Jeans instability with Yukawa gravity>:
$$
\boxed{\omega^2=k^2\left[v_s^2-\frac{4\pi G\rho_0}{k^2+\alpha^2}\right].}
$$
There exists a growing nonzero-wavenumber mode precisely when
$$
\boxed{4\pi G\rho_0>v_s^2\alpha^2,\qquad0<k<\sqrt{k_J^2-\alpha^2}.}
$$
Equality gives no exponentially growing nonzero mode. If screening exceeds this threshold, the finite-range gravitational response is too weak to overcome <pressure> at any <wavelength>. Otherwise all sufficiently long but nonconstant waves remain unstable, with a longer minimum unstable <wavelength> than in Newtonian gravity. For $\alpha>0$ the long-wave gravitational contribution also scales as $k^2$, so the growth rate tends to zero as $k\to0$; screening does not produce a second positive lower cutoff in <wavenumber>. A homogeneous screened background can have the constant <gravitational potential> $-4\pi G\rho_0/\alpha^2$, with zero <force>, so no Newtonian background subtraction is needed for nonzero screening. The limits $k\to0$ and $\alpha\to0$ are nonuniform. An exact $k=0$ density change is not one of the growing finite-wavelength modes. In the pressureless edge case $v_s=0$, every finite $k>0$ is unstable for positive density and finite screening.
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