Liouville dynamical system

Let the fixed centers of attraction be located along the x-axis at ±a. The potential energy of the moving particle is given by

${\displaystyle V(x,y)={\frac {-\mu _{1}}{\sqrt {\left(x-a\right)^{2}+y^{2}}}}-{\frac {\mu _{2}}{\sqrt {\left(x+a\right)^{2}+y^{2}}}}.}$

The two centers of attraction can be considered as the foci of a set of ellipses. If either center were absent, the particle would move on one of these ellipses, as a solution of the Kepler problem. Therefore, according to Bonnet's theorem, the same ellipses are the solutions for the bicenter problem.

Introducing elliptic coordinates,

${\displaystyle x=a\cosh \xi \cos \eta ,}$

${\displaystyle y=a\sinh \xi \sin \eta ,}$

the potential energy can be written as

${\displaystyle V(\xi ,\eta )={\frac {-\mu _{1}}{a\left(\cosh \xi -\cos \eta \right)}}-{\frac {\mu _{2}}{a\left(\cosh \xi +\cos \eta \right)}}={\frac {-\mu _{1}\left(\cosh \xi +\cos \eta \right)-\mu _{2}\left(\cosh \xi -\cos \eta \right)}{a\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)}},}$

and the kinetic energy as

${\displaystyle T={\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)\left({\dot {\xi }}^{2}+{\dot {\eta }}^{2}\right).}$

This is a Liouville dynamical system if ξ and η are taken as φ₁ and φ₂, respectively; thus, the function Y equals

${\displaystyle Y=\cosh ^{2}\xi -\cos ^{2}\eta }$

and the function W equals

${\displaystyle W=-\mu _{1}\left(\cosh \xi +\cos \eta \right)-\mu _{2}\left(\cosh \xi -\cos \eta \right)}$

Using the general solution for a Liouville dynamical system below, one obtains

${\displaystyle {\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)^{2}{\dot {\xi }}^{2}=E\cosh ^{2}\xi +\left({\frac {\mu _{1}+\mu _{2}}{a}}\right)\cosh \xi -\gamma }$

${\displaystyle {\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)^{2}{\dot {\eta }}^{2}=-E\cos ^{2}\eta +\left({\frac {\mu _{1}-\mu _{2}}{a}}\right)\cos \eta +\gamma }$

Introducing a parameter u by the formula

${\displaystyle du={\frac {d\xi }{\sqrt {E\cosh ^{2}\xi +\left({\frac {\mu _{1}+\mu _{2}}{a}}\right)\cosh \xi -\gamma }}}={\frac {d\eta }{\sqrt {-E\cos ^{2}\eta +\left({\frac {\mu _{1}-\mu _{2}}{a}}\right)\cos \eta +\gamma }}},}$

gives the parametric solution

${\displaystyle u=\int {\frac {d\xi }{\sqrt {E\cosh ^{2}\xi +\left({\frac {\mu _{1}+\mu _{2}}{a}}\right)\cosh \xi -\gamma }}}=\int {\frac {d\eta }{\sqrt {-E\cos ^{2}\eta +\left({\frac {\mu _{1}-\mu _{2}}{a}}\right)\cos \eta +\gamma }}}.}$

Since these are elliptic integrals, the coordinates ξ and η can be expressed as elliptic functions of u.

The bicentric problem has a constant of motion, namely,

${\displaystyle r_{1}^{2}r_{2}^{2}\left({\frac {d\theta _{1}}{dt}}\right)\left({\frac {d\theta _{2}}{dt}}\right)-2c\left[\mu _{1}\cos \theta _{1}+\mu _{2}\cos \theta _{2}\right],}$

from which the problem can be solved using the method of the last multiplier.

To eliminate the v functions, the variables are changed to an equivalent set

${\displaystyle \varphi _{r}=\int dq_{r}{\sqrt {v_{r}(q_{r})}},}$

giving the relation

${\displaystyle v_{1}(q_{1}){\dot {q}}_{1}^{2}+v_{2}(q_{2}){\dot {q}}_{2}^{2}+\cdots +v_{s}(q_{s}){\dot {q}}_{s}^{2}={\dot {\varphi }}_{1}^{2}+{\dot {\varphi }}_{2}^{2}+\cdots +{\dot {\varphi }}_{s}^{2}=F,}$

which defines a new variable F. Using the new variables, the u and w functions can be expressed by equivalent functions χ and ω. Denoting the sum of the χ functions by Y,

${\displaystyle Y=\chi _{1}(\varphi _{1})+\chi _{2}(\varphi _{2})+\cdots +\chi _{s}(\varphi _{s}),}$

the kinetic energy can be written as

${\displaystyle T={\frac {1}{2}}YF.}$

Similarly, denoting the sum of the ω functions by W

${\displaystyle W=\omega _{1}(\varphi _{1})+\omega _{2}(\varphi _{2})+\cdots +\omega _{s}(\varphi _{s}),}$

the potential energy V can be written as

${\displaystyle V={\frac {W}{Y}}.}$

The Lagrange equation for the r^th variable ${\displaystyle \varphi _{r}}$ is

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial T}{\partial {\dot {\varphi }}_{r}}}\right)={\frac {d}{dt}}\left(Y{\dot {\varphi }}_{r}\right)={\frac {1}{2}}F{\frac {\partial Y}{\partial \varphi _{r}}}-{\frac {\partial V}{\partial \varphi _{r}}}.}$

Multiplying both sides by ${\displaystyle 2Y{\dot {\varphi }}_{r}}$, re-arranging, and exploiting the relation 2T = YF yields the equation

${\displaystyle 2Y{\dot {\varphi }}_{r}{\frac {d}{dt}}\left(Y{\dot {\varphi }}_{r}\right)=2T{\dot {\varphi }}_{r}{\frac {\partial Y}{\partial \varphi _{r}}}-2Y{\dot {\varphi }}_{r}{\frac {\partial V}{\partial \varphi _{r}}}=2{\dot {\varphi }}_{r}{\frac {\partial }{\partial \varphi _{r}}}\left[(E-V)Y\right],}$

which may be written as

${\displaystyle {\frac {d}{dt}}\left(Y^{2}{\dot {\varphi }}_{r}^{2}\right)=2E{\dot {\varphi }}_{r}{\frac {\partial Y}{\partial \varphi _{r}}}-2{\dot {\varphi }}_{r}{\frac {\partial W}{\partial \varphi _{r}}}=2E{\dot {\varphi }}_{r}{\frac {d\chi _{r}}{d\varphi _{r}}}-2{\dot {\varphi }}_{r}{\frac {d\omega _{r}}{d\varphi _{r}}},}$

where E = T + V is the (conserved) total energy. It follows that

${\displaystyle {\frac {d}{dt}}\left(Y^{2}{\dot {\varphi }}_{r}^{2}\right)=2{\frac {d}{dt}}\left(E\chi _{r}-\omega _{r}\right),}$

which may be integrated once to yield

${\displaystyle {\frac {1}{2}}Y^{2}{\dot {\varphi }}_{r}^{2}=E\chi _{r}-\omega _{r}+\gamma _{r},}$

where the ${\displaystyle \gamma _{r}}$ are constants of integration subject to the energy conservation

${\displaystyle \sum _{r=1}^{s}\gamma _{r}=0.}$

Inverting, taking the square root and separating the variables yields a set of separably integrable equations:

${\displaystyle {\frac {\sqrt {2}}{Y}}dt={\frac {d\varphi _{1}}{\sqrt {E\chi _{1}-\omega _{1}+\gamma _{1}}}}={\frac {d\varphi _{2}}{\sqrt {E\chi _{2}-\omega _{2}+\gamma _{2}}}}=\cdots ={\frac {d\varphi _{s}}{\sqrt {E\chi _{s}-\omega _{s}+\gamma _{s}}}}.}$