Finite-range scattering: the simple outer turning point, the two global branches and the free tails

Square the energy identity of [L1]: $\frac{1}{2}\mu ({\rho }^{\mathrm{\prime }}{)}^2=E-V(\rho )$, which is clause 1's first display. At a time with $\rho =r_{\ast }$ the right side is $E-V(r_{\ast })=0$, so ${\rho }^{\mathrm{\prime }}=0$; then $\mu {\rho }^{\mathrm{\prime }\mathrm{\prime }}(t_{\ast })=-V^{\mathrm{\prime }}(r_{\ast })>0$, so $t_{\ast }$ is a strict local minimum of $\rho$, and in particular $\rho (t)>\rho (t_{\ast })$ for $t$ near but different from $t_{\ast }$.

On the open set where $\rho >r_{\ast }$, the right side of the energy identity is strictly positive, so ${\rho }^{\mathrm{\prime }}\mathrm{\ne }0$; hence the sign of ${\rho }^{\mathrm{\prime }}$ is constant on every connected component of $\{t:\rho (t)>r_{\ast }\}$ (a union of open intervals on which the continuous nonvanishing function ${\rho }^{\mathrm{\prime }}$ has a sign).

Global extension: take the state $z:=(\rho ,{\rho }^{\mathrm{\prime }},\theta )$ in $(0,\mathrm{\infty })\times \mathbb{R}\times \mathbb{R}$ with right-hand side $F(x):=(x_2,-{\mu }^{-1}{V}^{\mathrm{\prime }}(x_1),J\mathrm{/}(\mu x_1^2))$. This field is locally Lipschitz: near any $x_1>0$, continuity of $V^{\mathrm{\prime }\mathrm{\prime }}$ bounds it by some $M$ on a small closed interval; the fundamental theorem for continuous $V^{\mathrm{\prime }\mathrm{\prime }}$ gives $V^{\mathrm{\prime }}(a)-V^{\mathrm{\prime }}(b)={\int }_b^a{V}^{\mathrm{\prime }\mathrm{\prime }}$, and the Darboux definition with $\mathrm{\mid }{V}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }\le M$ gives $\mathrm{\mid }{V}^{\mathrm{\prime }}(a)-V^{\mathrm{\prime }}(b)\mathrm{\mid }\le M\mathrm{\mid }a-b\mathrm{\mid }$. Also $x_1^{-2}$ is locally Lipschitz by factoring the difference of reciprocal powers. Along every solution, $H(\rho ,{\rho }^{\mathrm{\prime }}):=\frac{1}{2}\mu ({\rho }^{\mathrm{\prime }}{)}^2+V(\rho )$ is constant by the radial equation. Also $\rho \ge r_{\ast }$ on every continuation: if it entered $\rho <r_{\ast }$, continuity from $t_{\ast }$ would give a boundary time with $\rho =r_{\ast }$ and values below $r_{\ast }$ arbitrarily nearby on one side; there $H=E$ gives ${\rho }^{\mathrm{\prime }}=0$ and the equation gives ${\rho }^{\mathrm{\prime }\mathrm{\prime }}=-V^{\mathrm{\prime }}(r_{\ast })\mathrm{/}\mu >0$, contradicting the strict local-minimum property. Thus $V(\rho )\le E$ and ${\rho }^{\mathrm{\prime }2}=(2\mathrm{/}\mu )(E-V(\rho ))$. The function $V$ is bounded below on $[r_{\ast },\mathrm{\infty })$ because $U$ is continuous and vanishes beyond $R$, so ${\rho }^{\mathrm{\prime }}$ is bounded; also $0<{\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)\le J\mathrm{/}(\mu r_{\ast }^2)$. On every bounded time interval, $\rho$ and $\theta$ are therefore bounded, so near every finite right endpoint $z$ lies in a compact subset of the state domain and [L2] continues it. For a finite left endpoint apply [L2] to $\widehat{z}(s):=z(-s)$, which solves ${\widehat{z}}^{\mathrm{\prime }}=-F(\widehat{z})$. Hence the maximal extension of $z$ is defined on $\mathbb{R}$. Set $n(\theta ):=\cos \theta e_1+\sin \theta e_2$ and $n_{\perp }(\theta ):=-\sin \theta e_1+\cos \theta e_2$. The relation ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)$ gives ${\theta }^{\mathrm{\prime }\mathrm{\prime }}=-2{\rho }^{\mathrm{\prime }}{\theta }^{\mathrm{\prime }}\mathrm{/}\rho$; differentiating gives $r^{\mathrm{\prime }\mathrm{\prime }}=({\rho }^{\mathrm{\prime }\mathrm{\prime }}-\rho {\theta }^{\mathrm{\prime }2})n(\theta )+(2{\rho }^{\mathrm{\prime }}{\theta }^{\mathrm{\prime }}+\rho {\theta }^{\mathrm{\prime }\mathrm{\prime }})n_{\perp }(\theta ),$ whose second coefficient vanishes. Using $\mu {\rho }^{\mathrm{\prime }\mathrm{\prime }}=-V^{\mathrm{\prime }}(\rho )$ then yields $\mu ({\rho }^{\mathrm{\prime }\mathrm{\prime }}-\rho {\theta }^{\mathrm{\prime }2})=-U^{\mathrm{\prime }}(\rho )$, so the reconstructed curve satisfies the central-force equation. Any other continuation has the same angular momentum on the overlap: along it $j^{\mathrm{\prime }}=\mu (r^{\mathrm{\prime }}\times r^{\mathrm{\prime }}+r\times r^{\mathrm{\prime }\mathrm{\prime }})=0$ by the central-force equation, so it remains in the same plane. By [L1] it has polar data there; shift its angle lift by an integer multiple of $2\pi$ to match $\theta$ at a common time. The resulting state solves the same locally Lipschitz system with the same initial state, so uniqueness [L2] gives the same $\rho ,\theta$ and hence the same trajectory.

Shape of the set $\{\rho >r_{\ast }\}$. This set is open in $\mathbb{R}$, and on each of its connected components ${\rho }^{\mathrm{\prime }}$ has a constant sign (step 2.1). Every bounded component $(t_0,t_1)$ satisfies $\rho (t_0)=\rho (t_1)=r_{\ast }$ (endpoints of a component of the open set $\{\rho >r_{\ast }\}$ lie in its complement, where $\rho \le r_{\ast }$, i.e. $\rho =r_{\ast }$); at both endpoints ${\rho }^{\mathrm{\prime }\mathrm{\prime }}=-V^{\mathrm{\prime }}(r_{\ast })\mathrm{/}\mu >0$ by step 1.1, so ${\rho }^{\mathrm{\prime }}$ changes sign from negative to positive at each of them, whence just inside the component the sign is positive near $t_0$ and negative near $t_1$, contradicting constancy of the sign. Hence $\{\rho >r_{\ast }\}$ has no bounded component. Since $t_{\ast }$ is an isolated zero of $\rho -r_{\ast }$ (a strict local minimum by step 1.1) and points just left and just right of $t_{\ast }$ belong to $\{\rho >r_{\ast }\}$, the two components adjacent to $t_{\ast }$ are $(-\mathrm{\infty },t_{\ast })$ and $(t_{\ast },+\mathrm{\infty })$: an open subset of $\mathbb{R}$ has at most two unbounded components, and each of these two is unbounded on its outer side (a bounded component is impossible, and it cannot end on the inner side at a point other than $t_{\ast }$, since its inner endpoint would be an endpoint of a component at which $r_{\ast }$ is attained, producing a second strict minimum and the same sign contradiction as above). Therefore $\rho >r_{\ast }$ exactly for $t\mathrm{\ne }{t}_{\ast }$, ${\rho }^{\mathrm{\prime }}<0$ on $(-\mathrm{\infty },t_{\ast })$ and ${\rho }^{\mathrm{\prime }}>0$ on $(t_{\ast },\mathrm{\infty })$.

The quadrature: set $q(s):=\sqrt{(2\mathrm{/}\mu )(E-V(s))}$ for $s>r_{\ast }$. The integral defining $\tau$ is finite and continuous at $r_{\ast }$ by [L4], and ${\tau }^{\mathrm{\prime }}(s)=1\mathrm{/}q(s)$ for $s>r_{\ast }$. On the outgoing branch, the chain rule and ${\rho }^{\mathrm{\prime }}=q(\rho )$ give $\frac{d}{dt}\tau (\rho (t))=1$ for $t>t_{\ast }$; taking the limit as $t\downarrow t_{\ast }$ yields $\tau (\rho (t))=t-t_{\ast }$. On the incoming branch, ${\rho }^{\mathrm{\prime }}=-q(\rho )$ gives $\frac{d}{dt}\tau (\rho (t))=-1$, hence $\tau (\rho (t))=t_{\ast }-t$. The function $\tau$ is strictly increasing on $[r_{\ast },\mathrm{\infty })$ and tends to $+\mathrm{\infty }$ by [L4], so it maps this interval onto $[0,\mathrm{\infty })$. Inverting the outgoing identity gives $\rho (t_{\ast }+\tau (r))=r$ for every $r\ge r_{\ast }$; the incoming identity gives the corresponding incoming branch. In particular the unique time with $\rho =r_{\ast }$ is $t_{\ast }$, and $\rho (t)>r_{\ast }$ for $t\mathrm{\ne }{t}_{\ast }$. These are clauses 2 and 3.

Free tails: suppose $t_1\ge t_{\ast }$ with $\rho (t_1)>R$. By step 3.1, $\rho$ is strictly increasing on $[t_{\ast },\mathrm{\infty })$, so $\rho (t)\ge \rho (t_1)>R$ for all $t\ge t_1$. On that regime the potential is identically zero and the Cartesian central-force equation reduces to $\mu r^{\mathrm{\prime }\mathrm{\prime }}=-(U^{\mathrm{\prime }}(\rho )\mathrm{/}\rho )r=0$, since $U^{\mathrm{\prime }}(\rho )=0$ for $\rho >R$; hence $r$ is affine on $[t_1,\mathrm{\infty })$ and $r(t)=r(t_1)+(t-t_1)r^{\mathrm{\prime }}(t_1)$ there. The incoming statement is the same argument on $(-\mathrm{\infty },t_2]$. This is clause 4.