= Solution
Choose spin orientation so that $0<a<M$; the printed $M/a$ threshold assumes nonzero positive $a$. Put $H=2Mr/(r^2+a^2)$ and $F=1-H$. The axial metric has components $(-F,H,1+H)$ and determinant $-1$. Conserved energy and timelike normalization give
$$
\varepsilon=F\dot{\tilde t}-H\dot r,\qquad-F\dot{\tilde t}^{\,2}+2H\dot{\tilde t}\dot r+(1+H)\dot r^2=-1.
$$
Eliminating $\dot{\tilde t}$ gives \b[the <Kerr axial timelike effective potential>]
$$
\boxed{\dot r^2+V_{\rm eff}(r)=\varepsilon^2,\qquad V_{\rm eff}=1-\frac{2Mr}{r^2+a^2}.}
$$
It extends through the zeros of $F$ in the regular original coordinates. Since $V_{\rm eff}(0)=1$, reaching zero requires $\varepsilon^2\geq1$. At equality the infalling orbit reaches zero with zero radial velocity and turns back, because the nearby negative sheet is forbidden. Along the axis zero is regular, unlike the off-axis ring.
For $r<0$, $V'_{\rm eff}=2M(r^2-a^2)/(r^2+a^2)^2$ and the global maximum is $V_{\rm eff}(-a)=1+M/a$. If $\varepsilon^2>1+M/a$, an inward particle has no negative-sheet turning point. \b[It passes through zero and continues toward $\boxed{r\to-\infty}$], another asymptotically flat end, with $\dot r\to-\sqrt{\varepsilon^2-1}$ and infinite <proper time> to infinity. It first crosses the outer and inner horizons. Perturbative Cauchy-horizon instability can invalidate this ideal <Kerr axial analytic extension> physically.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-58-kerr-axis.png]
{title=Conformal blocks of the Kerr axis with an energetic timelike orbit continuing to negative-r infinity}
This is the two-dimensional axis restriction. The dashed $r=0$ curve is regular; the four-dimensional ring lies off-axis and is not an axial singular boundary.
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