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Write $\Lambda =-|\Lambda |$. For $k=0$,The potential is strictly increasing from $-\infty$ to $+\infty$ and crosses the energy line atThe expanding universe reaches this turning point, reverses, and ends in a Big Crunch. Thus a negative cosmological constant forces recollapse in this spatially flat pressureless model.

Let a slowly varying wave have phase $S(\mathbf{x},t)$, wavevector $\mathbf{k}=\nabla S$, and angular frequency $\omega =-{\partial }_{t}S$. The local dispersion relation is the Hamilton-Jacobi equation $\omega =\Omega (\mathbf{k};\mathbf{x},t)$. Equality of mixed derivatives, followed along a curve with velocity $\dot{\mathbf{x}}={\nabla }_{\mathbf{k}}\Omega$, gives the Hamiltonian ray-tracing equationsHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ is the total derivative evaluated along a ray, and $\dot{\mathbf{x}}$ is the group velocity.

For $\Omega =kc(z)$, $\omega$, $k_x$, and $k_y$ are conserved. If $\psi$ is the angle between the ray and the $z$-axis, then $k\sin \psi =(k_x^2+k_y^2{)}^{1/2}$ and $k=\omega /c$. Hence Snell's law is

For the ray launched with $\mathbf{k}=k(\cos \varphi ,0,\sin \varphi )$ at $z=0$, this constant is $|\cos \varphi |/c_0$. Thereforewhich proves

For $0<\varphi <\pi /2$, write $a=\cos \varphi$. Since the ray and wavevector are parallel,Near the upper turning point, put $z=z_m-\zeta$. Expansion givesIf $x=x_m$ at the turn, integration yields the local ray pathThe ray is tangent to the line $z=z_m$, turns smoothly, crosses $z=0$, and repeats symmetrically between $-z_m$ and $z_m$. Increasing $\varphi$ decreases $z_m$, while $\varphi \downarrow 0$ sends the turning level outward.

The work done along the trajectory isBy Newton's second law, $F=m\ddot{x}$, soThis is the work-energy theorem: the work equals the change in kinetic energy.

A force is conservative when its work between two points is independent of the path, equivalently when it is the negative gradient of a potential energy. Here a potential with $V(0)=0$ isConservation of energy for $x>0$ givesThus

If $v_0>v_e$, the particle escapes with asymptotic speed $\sqrt{v_0^2-v_e^2}$; the graph decreases from $v_0$ to this horizontal asymptote. At the escape velocity $v_0=v_e$, it decreases as $v_e{e}^{-x/(2\lambda )}$ and approaches zero only at infinity. If $v_0<v_e$, it reaches the turning pointwhere $v=0$, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.

For oscillation put $w=1-v_0^2/v_e^2$. One quarter of the period iswhere $u=e^{-x/\lambda }$. Therefore

For $\theta >1$, the stationary-point equation would requireso the two saddles leave the real axis. The contour can be deformed through the appropriate complex saddle or saddles, and their contributions become exponentially small or large according to the sign of the real part of $i\Phi$. The real-axis oscillatory saddle contributions present for $0<\theta <1$ therefore disappear across the turning point $\theta =1$. The admissible deformation and the pole's residue must still respect the prescribed branch cuts and the instruction that the contour pass above $z_0$.

For a bound orbit let $E<0$ be the specific energy and $L$ the specific angular momentum. Introduce the dimensionless radius suggested in the question,The radial energy equation becomeswhere $x_{-}$ and $x_{+}$ are the two turning points. Comparing coefficients givesThe time from periapsis to apoapsis isThus the radial orbital period isIt depends on $E$ but not on $L$, which is the defining isochrone property.