# Liouville dynamical system

In classical mechanics, a Liouville dynamical system is an exactly solvable dynamical system in which the kinetic energy T and potential energy V can be expressed in terms of the s generalized coordinates q as follows:
    $T={\frac {1}{2}}\left\{u_{1}(q_{1})+u_{2}(q_{2})+\cdots +u_{s}(q_{s})\right\}\left\{v_{1}(q_{1}){\dot {q}}_{1}^{2}+v_{2}(q_{2}){\dot {q}}_{2}^{2}+\cdots +v_{s}(q_{s}){\dot {q}}_{s}^{2}\right\}$
    $V={\frac {w_{1}(q_{1})+w_{2}(q_{2})+\cdots +w_{s}(q_{s})}{u_{1}(q_{1})+u_{2}(q_{2})+\cdots +u_{s}(q_{s})}}$
The solution of this system consists of a set of separably integrable equations
    ${\frac {\sqrt {2}}{Y}}\,dt={\frac {d\varphi _{1}}{\sqrt {E\chi _{1}-\omega _{1}+\gamma _{1}}}}={\frac {d\varphi _{2}}{\sqrt {E\chi _{2}-\omega _{2}+\gamma _{2}}}}=\cdots ={\frac {d\varphi _{s}}{\sqrt {E\chi _{s}-\omega _{s}+\gamma _{s}}}}$
where E = T + V is the conserved energy and the $\gamma _{s}$ are constants. As described below, the variables have been changed from q$_s$ to φ$_s$, and the functions u$_s$ and w$_s$ substituted by their counterparts χ$_s$ and ω$_s$. This solution has numerous applications, such as the orbit of a small planet about two fixed stars under the influence of Newtonian gravity. The Liouville dynamical system is one of several things named after Joseph Liouville, an eminent French mathematician.

## Example of bicentric orbits

In classical mechanics, Euler's three-body problem describes the motion of a particle in a plane under the influence of two fixed centers, each of which attract the particle with an inverse-square force such as Newtonian gravity or Coulomb's law. Examples of the bicenter problem include a planet moving around two slowly moving stars, or an electron moving in the electric field of two positively charged nuclei, such as the first ion of the hydrogen molecule H$_2$, namely the hydrogen molecular ion or H$_2$$^+$. The strength of the two attractions need not be equal; thus, the two stars may have different masses or the nuclei two different charges.

### Solution

Let the fixed centers of attraction be located along the x-axis at ±a. The potential energy of the moving particle is given by
    $V(x,y)={\frac {-\mu _{1}}{\sqrt {\left(x-a\right)^{2}+y^{2}}}}-{\frac {\mu _{2}}{\sqrt {\left(x+a\right)^{2}+y^{2}}}}.$
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,
    $x=a\cosh \xi \cos \eta ,$
    $y=a\sinh \xi \sin \eta ,$
the potential energy can be written as
    $V(\xi ,\eta )={\frac {-\mu _{1}}{a\left(\cosh \xi -\cos \eta \right)}}-{\frac {\mu _{2}}{a\left(\cosh \xi +\cos \eta \right)}}={\frac {-\mu _{1}\left(\cosh \xi +\cos \eta \right)-\mu _{2}\left(\cosh \xi -\cos \eta \right)}{a\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)}},$
and the kinetic energy as
    $T={\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)\left({\dot {\xi }}^{2}+{\dot {\eta }}^{2}\right).$
This is a Liouville dynamical system if ξ and η are taken as φ$_1$ and φ$_2$, respectively; thus, the function Y equals
    $Y=\cosh ^{2}\xi -\cos ^{2}\eta$
and the function W equals
    $W=-\mu _{1}\left(\cosh \xi +\cos \eta \right)-\mu _{2}\left(\cosh \xi -\cos \eta \right)$
Using the general solution for a Liouville dynamical system below, one obtains
    ${\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)^{2}{\dot {\xi }}^{2}=E\cosh ^{2}\xi +\left({\frac {\mu _{1}+\mu _{2}}{a}}\right)\cosh \xi -\gamma$
    ${\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)^{2}{\dot {\eta }}^{2}=-E\cos ^{2}\eta +\left({\frac {\mu _{1}-\mu _{2}}{a}}\right)\cos \eta +\gamma$
Introducing a parameter u by the formula
    $du={\frac {d\xi }{\sqrt {E\cosh ^{2}\xi +\left({\frac {\mu _{1}+\mu _{2}}{a}}\right)\cosh \xi -\gamma }}}={\frac {d\eta }{\sqrt {-E\cos ^{2}\eta +\left({\frac {\mu _{1}-\mu _{2}}{a}}\right)\cos \eta +\gamma }}},$
gives the parametric solution
    $u=\int {\frac {d\xi }{\sqrt {E\cosh ^{2}\xi +\left({\frac {\mu _{1}+\mu _{2}}{a}}\right)\cosh \xi -\gamma }}}=\int {\frac {d\eta }{\sqrt {-E\cos ^{2}\eta +\left({\frac {\mu _{1}-\mu _{2}}{a}}\right)\cos \eta +\gamma }}}.$
Since these are elliptic integrals, the coordinates ξ and η can be expressed as elliptic functions of u.

### Constant of motion

The bicentric problem has a constant of motion, namely,
    $r_{1}^{2}\,r_{2}^{2}{\frac {d\theta _{1}}{dt}}{\frac {d\theta _{2}}{dt}}+2\,c\left(\mu _{1}\cos \theta _{1}-\mu _{2}\cos \theta _{2}\right),$
from which the problem can be solved using the method of the last multiplier.

## Derivation
### New variables

To eliminate the v functions, the variables are changed to an equivalent set
    $\varphi _{r}=\int dq_{r}{\sqrt {v_{r}(q_{r})}},$
giving the relation
    $v_{1}(q_{1}){\dot {q}}_{1}^{2}+v_{2}(q_{2}){\dot {q}}_{2}^{2}+\cdots +v_{s}(q_{s}){\dot {q}}_{s}^{2}={\dot {\varphi }}_{1}^{2}+{\dot {\varphi }}_{2}^{2}+\cdots +{\dot {\varphi }}_{s}^{2}=F,$
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,
    $Y=\chi _{1}(\varphi _{1})+\chi _{2}(\varphi _{2})+\cdots +\chi _{s}(\varphi _{s}),$
the kinetic energy can be written as
    $T={\frac {1}{2}}YF.$
Similarly, denoting the sum of the ω functions by W
    $W=\omega _{1}(\varphi _{1})+\omega _{2}(\varphi _{2})+\cdots +\omega _{s}(\varphi _{s}),$
the potential energy V can be written as
    $V={\frac {W}{Y}}.$

### Lagrange equation

The Lagrange equation for the rth variable $\varphi _{r}$ is
    ${\frac {d}{dt}}\left({\frac {\partial T}{\partial {\dot {\varphi }}_{r}}}\right)={\frac {d}{dt}}\left(Y{\dot {\varphi }}_{r}\right)={\frac {1}{2}}F{\frac {\partial Y}{\partial \varphi _{r}}}-{\frac {\partial V}{\partial \varphi _{r}}}.$
Multiplying both sides by $2Y{\dot {\varphi }}_{r}$, re-arranging, and exploiting the relation 2T = YF yields the equation
    $2Y{\dot {\varphi }}_{r}{\frac {d}{dt}}\left(Y{\dot {\varphi }}_{r}\right)=2T{\dot {\varphi }}_{r}{\frac {\partial Y}{\partial \varphi _{r}}}-2Y{\dot {\varphi }}_{r}{\frac {\partial V}{\partial \varphi _{r}}}=2{\dot {\varphi }}_{r}{\frac {\partial }{\partial \varphi _{r}}}\left[(E-V)Y\right],$
which may be written as
    ${\frac {d}{dt}}\left(Y^{2}{\dot {\varphi }}_{r}^{2}\right)=2E{\dot {\varphi }}_{r}{\frac {\partial Y}{\partial \varphi _{r}}}-2{\dot {\varphi }}_{r}{\frac {\partial W}{\partial \varphi _{r}}}=2E{\dot {\varphi }}_{r}{\frac {d\chi _{r}}{d\varphi _{r}}}-2{\dot {\varphi }}_{r}{\frac {d\omega _{r}}{d\varphi _{r}}},$
where E = T + V is the (conserved) total energy. It follows that
    ${\frac {d}{dt}}\left(Y^{2}{\dot {\varphi }}_{r}^{2}\right)=2{\frac {d}{dt}}\left(E\chi _{r}-\omega _{r}\right),$
which may be integrated once to yield
    ${\frac {1}{2}}Y^{2}{\dot {\varphi }}_{r}^{2}=E\chi _{r}-\omega _{r}+\gamma _{r},$
where the $\gamma _{r}$ are constants of integration subject to the energy conservation
    $\sum _{r=1}^{s}\gamma _{r}=0.$
Inverting, taking the square root and separating the variables yields a set of separably integrable equations:
    ${\frac {\sqrt {2}}{Y}}dt={\frac {d\varphi _{1}}{\sqrt {E\chi _{1}-\omega _{1}+\gamma _{1}}}}={\frac {d\varphi _{2}}{\sqrt {E\chi _{2}-\omega _{2}+\gamma _{2}}}}=\cdots ={\frac {d\varphi _{s}}{\sqrt {E\chi _{s}-\omega _{s}+\gamma _{s}}}}.$