 Solution

For the Einstein static universe with $P=0$ and ${\rho }_0>0$, the acceleration equation gives $\Lambda =4\pi {\rho }_0$. The first Friedmann equation then gives $k/a_0^2=4\pi {\rho }_0$, soKeep $\Lambda ,k$ fixed and use the specified perturbation closure $\delta P=\delta \rho /3$. Linearization of the constraint gives $\delta \rho =-3{\rho }_0\delta a/a_0$. In the acceleration equation the background terms cancel, leavingThusThe growing mode establishes linear instability of the static universe; a specially chosen decaying perturbation does not remove instability. This uses the pressure-density relation explicitly supplied for the perturbation. If zero density is allowed, there is also the degenerate vacuum static case $k=0$, $\Lambda =0$, which is not the positive-density static universe considered here.