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For $c(y)=1-y^2$ in the strip $|y|<1$, the travel-time coordinate is $s(y)=\mathrm{artanh}y$. The characteristic curves areequivalently $y=\tanh (C\mp x)$. Outside the strip the same separated ordinary differential equation gives hyperbolic cotangent branches, and the equilibrium curves $y=\pm 1$ are also characteristic.

Put $u=\tanh (\lambda t)=e^{\varphi /{\varphi }_0}$. The hyperbolic cotangent double-angle identity givessoAlsoSolving the Friedmann equation for the scalar potential,therefore gives