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For a single positive-density component, staticity requires $\rho +3P=0$ from the Friedmann acceleration equation. Among the listed components, only the cosmic string network has $w=-1/3$. Put $a(t)=a_E>0$ and use the Friedmann equation to obtainAll other components are absent. This is the static closed cosmic-string universe.

There is a normalization condition hidden in this expression. String conservation fixes $C_s={\rho }_s{a}^2$, and a static solution exists only whenOnce this condition is imposed, every positive radius $a_E$ is a static solution, with its density determined as above. One cannot independently prescribe an arbitrary string normalization and then choose a radius to compensate: both the string density term and curvature term scale as $a^{-2}$.

The distinction between the evolution equation and the constraint is essential. For strings alone, the exact equations reduce toLinearizing the acceleration equation gives $\delta a=A+Bt$. At the exactly tuned static $C_s$, however, the full Friedmann equation requires ${\dot{a}}^2=0$, so $B=0$ for an admissible exact perturbation.

A small nonzero velocity $v$ can be admissible if its initial density also changes:Arbitrarily small $v$ and the corresponding second-order density change therefore produce a secular departure. There is no Lyapunov stability against all nearby constrained initial data for the static closed cosmic-string universe. Neutral radius shifts at fixed normalization and drift when the normalization changes are different perturbation classes.