Completeness and periodicity of the negative-energy Kepler orbit, with $T=2\pi \mu ab\mathrm{/}J$ and $T^2=4{\pi }^2\mu a^3\mathrm{/}\kappa$

By [L1] and [L2], for $t$ in the interior of $I$ the function $\phi (t):=\theta (t)$ satisfies $\rho (t)=\varrho (\phi (t))\in [r_{\min },r_{\max }]$ and, by [L4], ${\rho }^{\mathrm{\prime }}(t)={\varrho }^{\mathrm{\prime }}(\phi (t)){\theta }^{\mathrm{\prime }}(t)$ with $\mathrm{\mid }{\theta }^{\mathrm{\prime }}\mathrm{\mid }=J\mathrm{/}(\mu {\rho }^2)\le J\mathrm{/}(\mu r_{\min }^{})$; since ${\varrho }^{\mathrm{\prime }}$ is continuous and $2\pi$-periodic, it is bounded on $\mathbb{R}$ by $M_1:=\max \mathrm{\mid }{\varrho }^{\mathrm{\prime }}\mathrm{\mid }$, so $\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }\le M:=M_{1}J\mathrm{/}(\mu r_{\min }^{})$ on the interior of $I$; by continuity the bounds for $\rho$ and ${\rho }^{\mathrm{\prime }}$ hold on all of $I$.

Consider the state $z:=(\rho ,{\rho }^{\mathrm{\prime }},\theta )$ in the open set $U_0:=(0,\mathrm{\infty })\times \mathbb{R}\times \mathbb{R}$ with right-hand side $F(x):=(x_2,-{\mu }^{-1}(U_J^{\mathrm{e}\mathrm{f}\mathrm{f}}{)}^{\mathrm{\prime }}(x_1),J\mathrm{/}(\mu x_1^2))$; the radial equation of [L1] and ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)$ give $z^{\mathrm{\prime }}=F(z)$. Here $(U_J^{\mathrm{e}\mathrm{f}\mathrm{f}}{)}^{\mathrm{\prime }}(s)=\kappa \mathrm{/}{s}^2-J^2\mathrm{/}(\mu s^3)$, so $F$ is locally Lipschitz: on a state box with $x_1\ge \delta >0$, each $x_1^{-k}$ for $k=2,3$ satisfies $\mathrm{\mid }{x}_1^{-k}-y_1^{-k}\mathrm{\mid }\le k{\delta }^{-k-1}\mathrm{\mid }{x}_1-y_1\mathrm{\mid }$ by factoring the difference of reciprocal powers, while the first component is linear. Along every solution of this system, $H(\rho ,{\rho }^{\mathrm{\prime }}):=\frac{1}{2}\mu {\rho }^{\mathrm{\prime }2}+U_J^{\mathrm{e}\mathrm{f}\mathrm{f}}(\rho )$ is constant, since $H^{\mathrm{\prime }}={\rho }^{\mathrm{\prime }}(\mu {\rho }^{\mathrm{\prime }\mathrm{\prime }}+(U_J^{\mathrm{e}\mathrm{f}\mathrm{f}}{)}^{\mathrm{\prime }}(\rho ))=0$; its value is the given $E<0$. With $p=J^2\mathrm{/}(\mu \kappa )$, direct algebra gives $E-U_J^{\mathrm{e}\mathrm{f}\mathrm{f}}(s)=\frac{\kappa (1-e^2)}{2p{s}^2}(s-\frac{p}{1+e})(\frac{p}{1-e}-s)\mathrm{.}$ Thus every such solution has $r_{\min }=p\mathrm{/}(1+e)\le \rho \le r_{\max }=p\mathrm{/}(1-e)$; its energy identity bounds ${\rho }^{\mathrm{\prime }}$, and ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)$ is bounded as well. On a bounded time interval, $\theta$ is consequently bounded, so near every finite endpoint $z$ lies in a compact subset of $U_0$. Compact continuation [L3] extends it past every finite right endpoint. For a finite left endpoint apply [L3] to $\widehat{z}(s):=z(-s)$, which solves the time-reversed locally Lipschitz system ${\widehat{z}}^{\mathrm{\prime }}=-F(\widehat{z})$. Hence the maximal extension of $z$ is defined on all of $\mathbb{R}$. For $n(\theta ):=\cos \theta e_1+\sin \theta e_2$ and $n_{\perp }(\theta ):=-\sin \theta e_1+\cos \theta e_2$, differentiating $r=\rho n(\theta )$ gives $r^{\mathrm{\prime }\mathrm{\prime }}=({\rho }^{\mathrm{\prime }\mathrm{\prime }}-\rho {\theta }^{\mathrm{\prime }2})n(\theta )+(2{\rho }^{\mathrm{\prime }}{\theta }^{\mathrm{\prime }}+\rho {\theta }^{\mathrm{\prime }\mathrm{\prime }})n_{\perp }(\theta )\mathrm{.}$ Since ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)$, ${\theta }^{\mathrm{\prime }\mathrm{\prime }}=-2{\rho }^{\mathrm{\prime }}{\theta }^{\mathrm{\prime }}\mathrm{/}\rho$ and the second coefficient vanishes; the radial equation then gives $\mu ({\rho }^{\mathrm{\prime }\mathrm{\prime }}-\rho {\theta }^{\mathrm{\prime }2})=-U^{\mathrm{\prime }}(\rho )$. Thus the reconstructed $r$ satisfies the central-force equation. For any other continuation, the product rule and the central-force equation give $j^{\mathrm{\prime }}=\mu (r^{\mathrm{\prime }}\times r^{\mathrm{\prime }}+r\times r^{\mathrm{\prime }\mathrm{\prime }})=0$, so its angular momentum vector agrees with the one on the overlap; cross-product orthogonality places it in the same plane. By [L1] it has polar data there, and shifting its angle lift by an integer multiple of $2\pi$ makes the angle agree at a common time. The resulting state solves the same locally Lipschitz system with the same initial state, so uniqueness [L3] gives the same $\rho ,\theta$ and hence the same trajectory. This is clause 1.

On $\mathbb{R}$, $\rho \in [r_{\min },r_{\max }]$ and ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)\in [J(1-e{)}^2\mathrm{/}(\mu p^2),J(1+e{)}^2\mathrm{/}(\mu p^2)]$, since $r_{\min }=p\mathrm{/}(1+e)$ and $r_{\max }=p\mathrm{/}(1-e)$. Applying the conic supplier [L2] to the global extension gives the same profile $\varrho$: its sine and cosine coefficients agree with the original ones on the nonempty open angle interval $\theta (I^{\circ })$, so their differences vanish identically (evaluate the difference and its derivative at one point). Thus $\rho =\varrho \circ \theta$ on $\mathbb{R}$. Hence $\theta$ is strictly increasing with ${\theta }^{\mathrm{\prime }}(t)\ge c_1>0$, so $\theta (t)\to \pm \mathrm{\infty }$ as $t\to \pm \mathrm{\infty }$ and $\theta (\mathbb{R})=\mathbb{R}$.

Let $t_0\in \mathbb{R}$ and let $T>0$ satisfy $\phi (t_0+T)=\phi (t_0)+2\pi$, i.e. $\theta (t_0+T)=\theta (t_0)+2\pi$ (existence and uniqueness of $T$ come from step 2.1). Then $\rho (t_0+T)=\varrho (\phi (t_0)+2\pi )=\varrho (\phi (t_0))=\rho (t_0)$ by the $2\pi$-periodicity of $\varrho$, and ${\rho }^{\mathrm{\prime }}(t_0+T)={\varrho }^{\mathrm{\prime }}(\phi (t_0)+2\pi ){\theta }^{\mathrm{\prime }}(t_0+T)={\varrho }^{\mathrm{\prime }}(\phi (t_0))\frac{J}{\mu \rho (t_0+T{)}^2}={\varrho }^{\mathrm{\prime }}(\phi (t_0))\frac{J}{\mu \rho (t_0{)}^2}={\rho }^{\mathrm{\prime }}(t_0),$ using [L4], the $2\pi$-periodicity of ${\varrho }^{\mathrm{\prime }}$, and step 2.1. Also $n(\theta (t_0+T))=n(\theta (t_0))$ by the periodicity of sine and cosine [L6], so $r(t_0+T)=r(t_0)$ and $r^{\mathrm{\prime }}(t_0+T)=r^{\mathrm{\prime }}(t_0)$ (differentiate $r=\rho n(\theta )$ and use the equalities just proved).

Change of variables in the defining relation of $T$: $\theta :[t_0,t_0+T]\to [\theta (t_0),\theta (t_0)+2\pi ]$ is a continuous bijection with ${\theta }^{\mathrm{\prime }}>0$, so its inverse ${\theta }^{-1}$ is differentiable with $({\theta }^{-1}{)}^{\mathrm{\prime }}=1\mathrm{/}{\theta }^{\mathrm{\prime }}$ composed with ${\theta }^{-1}$ [L5]. Hence, using ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)$ and $\rho =\varrho (\theta )$ on the interval, $T={\int }_{t_0}^{t_0+T}\mathrm{d}t={\int }_{\theta (t_0)}^{\theta (t_0)+2\pi }\frac{1}{{\theta }^{\mathrm{\prime }}}\mathrm{d}\theta =\frac{\mu }{J}{\int }_{\theta (t_0)}^{\theta (t_0)+2\pi }\varrho (\theta {)}^2\mathrm{d}\theta =\frac{\mu }{J}{\int }_0^{2\pi }\varrho (\psi {)}^2\mathrm{d}\psi ,$ the last equality by the $2\pi$-periodicity of $\varrho$.

Define $z_T(t):=(\rho (t+T),{\rho }^{\mathrm{\prime }}(t+T),\theta (t+T)-2\pi )$. Since the state field in step 1.2 is independent of $\theta$, $z_T$ solves the same equation as $z(t):=(\rho (t),{\rho }^{\mathrm{\prime }}(t),\theta (t))$. By step 3.1 and the definition of $T$, $z_T(t_0)=z(t_0)$; uniqueness [L3] gives $z_T=z$ on $\mathbb{R}$. Periodicity of sine and cosine then yields $r(t+T)=r(t)$, and differentiation yields $r^{\mathrm{\prime }}(t+T)=r^{\mathrm{\prime }}(t)$. This is clause 2.

By clause 3 of The subcritical Kepler conic is an ellipse with semiaxes $a=p\mathrm{/}(1-e^2)$, $b=a\sqrt{1-e^2}$, focus at the origin and area $\pi ab$; its area is also $\frac{1}{2}{\int }_0^{2\pi }{\varrho }^2$, applied to the unshifted profile $\tilde{\varrho }(\psi ):=p\mathrm{/}(1+e\cos \psi )$, with a translation of the integration variable by ${\theta }_0$ and invariance of a full-period integral under translation, $\frac{1}{2}{\int }_0^{2\pi }\varrho (\psi {)}^2\mathrm{d}\psi ={\lambda }_2(E_{\mathrm{e}\mathrm{l}\mathrm{l}})=\pi ab$, so step 4.1 gives $T=\frac{2\pi \mu ab}{J}$.

Algebra: $p=J^2\mathrm{/}(\mu \kappa )$ and $p=a(1-e^2)$ by the definitions, so $J^2=\mu \kappa a(1-e^2)$ and $J=\sqrt{\mu \kappa a(1-e^2)}$; with $b=a\sqrt{1-e^2}$, $\frac{2\pi \mu ab}{J}=\frac{2\pi \mu a^2\sqrt{1-e^2}}{\sqrt{\mu \kappa a(1-e^2)}}=2\pi \sqrt{\frac{\mu }{\kappa }}{a}^{3\mathrm{/}2},$ and squaring gives $T^2=4{\pi }^2\mu a^3\mathrm{/}\kappa$. This is clause 3.