 Solution

Curvature contractions. Use the index contraction stipulated in the question. In three dimensions,The same contraction in four dimensions givesThen the Ricci scalar is $R=4\Lambda$ and the Einstein tensor is $G_{\mu \nu }=-\Lambda g_{\mu \nu }$. Thus $R_{\mu \nu }=\Lambda g_{\mu \nu }$ is precisely the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$.

The static solution. For $A=1+{\lambda }^2{r}^2$, $A^{\prime }=2{\lambda }^{2}r$ and $A^{\prime \prime }=2{\lambda }^2$. The given Ricci tensor components becomeEach equals $-3{\lambda }^2$ times the corresponding metric tensor component. ThereforeThis is Anti-de Sitter spacetime in static coordinates on anti-de Sitter spacetime, with curvature radius $1/\lambda$. The overall metric signature is opposite to the convention used in some descriptions of those coordinates; the geodesic constants below use the question's $(+,-,-,-)$ convention.

Conserved quantities and radial motion. In the equatorial plane, the geodesic Lagrangian is $L=A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2$. Its cyclic coordinates giveFor rest mass $m$, $mk=p_t$ is the conserved Killing energy and $mh=-p_{\varphi }$ is the axial angular momentum. Timelike normalization $L=1$ givesFor purely radial motion $h=0$, differentiating away from a turning point gives $\ddot{r}=-{\lambda }^{2}r$, and continuity extends that equation through the turning point. An outward launch from the origin at $\tau =0$ hasThis is the radial timelike geodesic in anti-de Sitter spacetime. Its first return to the origin isindependent of the outward launch speed for every nontrivial launch $k>1$. The coordinate $r$ is nonnegative, so this sine expression describes the excursion up to that return; subsequent passage through the origin requires changing the radial direction or using a signed diameter coordinate. The speed changes the amplitude, not the harmonic frequency.

Circular periods. The effective radial potential is $V_{\mathrm{eff}}=A(1+h^2/r^2)$. For a circle at $r_0$, $V_{\mathrm{eff}}^{\prime }=0$ gives $h^2={\lambda }^2{r}_0^4$, and normalization gives $k=A(r_0)$. Thus $\dot{t}=1$ and $\dot{\varphi }=\pm \lambda$. A stable circular timelike geodesic in anti-de Sitter spacetime therefore hasindependently of radius. This is the same isochronous behavior as the central harmonic radial equation. A static clock placed at the orbit's radius instead measures $\sqrt{1+{\lambda }^2{r}_0^2}\Delta t$, which does depend on radius; it is a different clock from the orbiting particle's proper time.