Binet's equation for the attractive inverse-square field: ${\mathrm{d}}^{2}u\mathrm{/}\mathrm{d}{\phi }^2+u=\mu \kappa \mathrm{/}{J}^2$ for $u=1\mathrm{/}\rho$

By [L1], ${\rho }^{\mathrm{\prime }\mathrm{\prime }}=-(U_J^{\mathrm{e}\mathrm{f}\mathrm{f}}{)}^{\mathrm{\prime }}(\rho )\mathrm{/}\mu$ on $I^{\circ }$. Since $U_J^{\mathrm{e}\mathrm{f}\mathrm{f}}$ is $C^2$ and $\rho$ is continuous, the right-hand side is continuous, so $\rho$ is $C^2$ there. Thus ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)$ is $C^1$ and positive. By [L2], $\theta :I^{\circ }\to W$ has a $C^1$ inverse $\sigma$ with ${\sigma }^{\mathrm{\prime }}=1\mathrm{/}({\theta }^{\mathrm{\prime }}\circ \sigma )$; the right-hand side is $C^1$ by [L3], so $\sigma$ is $C^2$. Therefore $u:=1\mathrm{/}(\rho \circ \sigma )$ is a quotient of $C^2$ functions with nonvanishing denominator and is $C^2$ on $W$.

Chain rule along the trajectory: for $t\in I^{\circ }$, $u(\theta (t))=1\mathrm{/}\rho (t)$, so [L3] gives $u^{\mathrm{\prime }}(\theta (t)){\theta }^{\mathrm{\prime }}(t)=-{\rho }^{\mathrm{\prime }}(t)\mathrm{/}\rho (t{)}^2$; multiplying by ${\rho }^2$ and using ${\rho }^2{\theta }^{\mathrm{\prime }}=J\mathrm{/}\mu$ from [L1], ${\rho }^{\mathrm{\prime }}(t)=-\rho (t{)}^2{\theta }^{\mathrm{\prime }}(t)u^{\mathrm{\prime }}(\theta (t))=-\frac{J}{\mu }{u}^{\mathrm{\prime }}(\theta (t)),$ which is clause 3.

Differentiating clause 3 with [L3] and using ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)$ again gives ${\rho }^{\mathrm{\prime }\mathrm{\prime }}(t)=-\frac{J}{\mu }{u}^{\mathrm{\prime }\mathrm{\prime }}(\theta (t)){\theta }^{\mathrm{\prime }}(t)=-\frac{J^2}{{\mu }^2\rho (t{)}^2}{u}^{\mathrm{\prime }\mathrm{\prime }}(\theta (t))(t\in I^{\circ })\mathrm{.}$ Since $W=\theta (I^{\circ })$, every $\phi \in W$ equals $\theta (t)$ for some $t\in I^{\circ }$, so this identity supplies the corresponding value of $u^{\mathrm{\prime }\mathrm{\prime }}$ at every point of $W$.

The reduction equation of [L1] for $U(\rho )=-\kappa \mathrm{/}\rho$ reads $\mu {\rho }^{\mathrm{\prime }\mathrm{\prime }}=-\kappa \mathrm{/}{\rho }^2+J^2\mathrm{/}(\mu {\rho }^3)$. Substituting steps 2.1 and 3.1, with $\rho =1\mathrm{/}u$, gives $-\frac{J^2}{\mu {\rho }^2}{u}^{\mathrm{\prime }\mathrm{\prime }}=-\frac{\kappa }{{\rho }^2}+\frac{J^2}{\mu {\rho }^3}⟹u^{\mathrm{\prime }\mathrm{\prime }}=\frac{\mu \kappa }{J^2}-u,$ that is, $u^{\mathrm{\prime }\mathrm{\prime }}+u=\mu \kappa \mathrm{/}{J}^2$ on $W$, which is clause 2.

Clause 4: the energy identity of [L1] gives $E=\frac{1}{2}\mu {\rho }^{\mathrm{\prime }2}+U_J^{\mathrm{e}\mathrm{f}\mathrm{f}}(\rho )=\frac{1}{2}\mu {\rho }^{\mathrm{\prime }2}-\frac{\kappa }{\rho }+\frac{J^2}{2\mu {\rho }^2}$. Substituting $\rho =1\mathrm{/}u$ and clause 3, $\frac{1}{2}\mu {\rho }^{\mathrm{\prime }2}=\frac{J^2}{2\mu }(u^{\mathrm{\prime }}\circ \theta {)}^2$ and $\frac{J^2}{2\mu {\rho }^2}=\frac{J^2}{2\mu }{u}^2$; hence $E=\frac{J^2}{2\mu }((u^{\mathrm{\prime }}\circ \theta {)}^2+u(\theta {)}^2)-\kappa u(\theta )$ along the trajectory. Since $\theta$ ranges over $W$, the displayed identity between the constant $E$ and the continuous function $\frac{J^2}{2\mu }((u^{\mathrm{\prime }}{)}^2+u^2)-\kappa u$ on $W$ holds on $W$.