function $$\alpha(\theta, \phi)=-{1\over \sin\phi \sin\theta}+{1\over \kappa} \cot\phi
\eqn\hamone$$ with symplectic form $$\Omega=\pm  {1\over (m \,\kappa )\, \sin^2\phi \sin\theta}\, 
d\theta\wedge d\phi \ .
\eqn\hamtwo$$ Since $\alpha$ is preserved by the flow, we can rewrite  as $$\sin\theta={\kappa\over \cos\phi-\alpha \kappa \sin\phi}\ ,
\eqn\moned$$ where $\alpha$ is a real constant. The solutions of  in the $x$ parameterisation are $$\eqalign{
\cot\phi\equiv z&={\kappa\over 2 \sqrt{1-\kappa^2}}\exp{\big[\pm m 
\sqrt{1-\kappa^2} (x-x_0)\big]}
\cr & \qquad\qquad
\bigg([\exp{\big[\mp m \sqrt{1-\kappa^2} (x-x_0)\big]}+{\alpha\over 
\sqrt{1-\kappa^2}}]^2+1-\alpha^2\bigg)
\cr
\cos\theta&=\cases{{\sqrt{(z-\alpha\kappa)^2-\kappa^2(1+z^2)}\over 
z-\alpha\kappa}\ , \qquad -\infty< x
<x_{min}
\cr 
-{\sqrt{(z-\alpha\kappa)^2-\kappa^2(1+z^2)}\over z-\alpha\kappa}\ , 
\qquad x_{min}< x <+\infty\ ,}}
\eqn\mtwod$$ where at $x=x_{min}$, $z(x)$ takes its absolute minimum value. To verify that  solves the equations , we differentiate $z$ with respect to $x$ and use the equation for $\phi$ to eliminate the derivative of $\phi$ from the expression. The $x$ dependence can also be eliminated by inverting $z$ to express the exponential of $x$ in terms of $z$. However this equation is quadratic and there are two possible solutions distinguished by a sign. To satisfy the equations , one has to choose for $-\infty< x <x_{min}$ the solution with the plus sign and for $x_{min}< x <+\infty$ the solution with the minus sign together with the corresponding expression for $\cos\theta$ in .

