The Henon-Heiles potential is

$$ U(x,y ) = \frac{1}{2} (x^2 + y^2 + 2 x^2 y - \frac{2}{3} y^3) .$$

This is a two degree-of-freedom system. The full Hamiltonian is

$$ H = p_x^2 + p_y^2 + U(x,y ) . $$

It is shown by numerics that it is non-integrable. But can one prove it rigorously analytically? The problem boils down to proving the non-existence of a second first-integral/integral-of-motion.

If this problem is too difficult, is there any simpler model whose non-integrability can be proven analytically?


General analytic methods to prove non-integrability are discussed in e.g. this Phys.SE post. In this answer, we will sketch how to apply the following Poincare corollary.

Poincare corollary: If an autonomous Hamiltonian Liouville-integrable system has a periodic solution $z_0(t)=z_0(t+T)$, then the monodromy matrix for the linearized system along $z_0$ can only have 1 as a (generalized) eigenvalue.

A proof of the Poincare corollary is given in my Phys.SE answer here.

$\uparrow$ Fig. 1. Poincare maps in the $(y,\dot{y})$ plane with $(x,\dot{x})=(0,0)$ for various fixed energy-levels $E$ of the Henon-Heiles (HH) system. We consider below the outermost periodic orbit for $E=1/6$ and $E=1/12$.

Sketched analytic proof that the HH system is not integrable: The HH system has Lagrangian$^1$ $$ L ~=~T-V, \qquad T ~=~\frac{1}{2}(\dot{x}^2+\dot{y}^2) ~=~\frac{1}{2}(\dot{r}^2+r^2\dot{\theta}^2), $$ $$V~=~\frac{1}{2}(x^2+y^2)+x^2y-\frac{1}{3}y^3 ~=~\frac{r^2}{2}+\frac{r^3}{3}\sin 3\theta, \qquad x+iy ~\equiv~re^{i\theta}. \tag{1}$$ The EL equations are $$ -\ddot{x}_0~=~x_0 +2x_0y_0, \tag{2}$$ $$ -\ddot{y}_0~=~y_0 +x_0^2-y_0^2.\tag{3} $$ We seek a periodic solution $y_0(t)=y_0(t+T)$ with $x_0(t)\equiv 0$. Then eq. (2) is automatically satisfied. The energy $$ E~=~ \frac{1}{2}\dot{y}_0^2+\frac{1}{2}y_0^2 -\frac{1}{3}y_0^3 \tag{4}$$ is conserved. Consider $(x,y)=(x_0,y_0)+ (\xi,\eta)$, where $(\xi,\eta)$ is an infinitesimal fluctuation. The linearized EL equations $$ \ddot{\xi}~=~-(1+2y_0)\xi ,\tag{5} $$ $$\ddot{\eta}~=~(2y_0-1)\eta,\tag{6}$$ decouple. [The eq. (6) corresponds to dimensional reduction to the 1D subsystem in the $y$-direction alone. This is manifestly integrable, so we already know that the corresponding 2 generalized eigenvalues for the monodromy matrix in the $y$-direction is $1$. Hence eq. (5) will be the actual integrability test.]

First attempt: $E=\frac{1}{6}$. The first integral (4) becomes $$\dot{y}_0~=~\pm (1-y_0)\sqrt{y_0+1/2}, \qquad -\frac{1}{2}~\leq~y_0~\leq~1. \tag{7}$$ It has solution $$ y_0(t)~=~ \frac{3}{2}\tanh u-\frac{1}{2} ~=~1-\frac{3/2}{\cosh^2u},\qquad u~\equiv~\frac{t}{2}.\tag{8} $$ Unfortunately the period is infinite $T=\infty$. For the record, the linearized eqs. (5) & (6) become Legendre's DE $$ \frac{d^2\xi}{du^2}~=~- 12\xi \tanh^2 u , \tag{9}$$ $$ \frac{d^2\eta}{du^2}~=~\left(4 - \frac{12}{\cosh^2 u}\right) \eta. \tag{10}$$

Second attempt: $E=\frac{1}{12}$. The first integral (4) becomes $$\dot{y}_0~=~\pm \sqrt{\frac{2}{3}(y_0-1)(y_0^2-y_0-1/2)}, \qquad \frac{1-\sqrt{3}}{2}~\leq~y_0~\leq~\frac{1}{2}.\tag{11}$$ It has a solution in terms of Jacobi elliptic functions $$ y_0(t)~=~ \sqrt{3}{\rm sn}^2u +\frac{1-\sqrt{3}}{2}, \qquad {\rm sn} u~\equiv~{\rm sn}(u|m\!=\!2), \qquad u~\equiv~\frac{t}{2\sqrt[4]{3}}.\tag{12} $$ The linearized eqs. (5) & (6) become Lamé's DE on Jacobi form$^2$ $$ \frac{d^2\xi}{du^2}~=~\left(12-8\sqrt{3}- 24 {\rm sn}^2u \right) \xi \tag{13},$$ $$ \frac{d^2\eta}{du^2}~=~\left(24 {\rm sn}^2u - 12\right) \eta \tag{14}.$$ Eqs. (13) & (14) may in principle be solved analytically. At this stage we admittedly just plugged eq. (13) into Mathematica, and checked numerically that the corresponding 2 generalized eigenvalues for the monodromy matrix in the $x$-direction are far from 1. Hence the HH system is not integrable, cf. the Poincare corollary. $\Box$


  1. H. Goldstein, Classical Mechanics, 3rd edition, Section 11.6.


$^1$ Although the $4\times 4$ monodromy matrix refers to the 4D phase space, it is convenient to perform most of analysis in the 2D configuration space, and postpone the Legendre transformation.

$^2$ One solution to eq. (14) is $$\eta_1(u)~:=~{\rm sn}u~{\rm cn}u~{\rm dn}u. \tag{15}$$ Another independent solution can in principle be found from the formula $$ \eta_2(u)~:=~\eta_1(u)\int^u\! \frac{du^{\prime}}{\eta_1(u^{\prime})^2}. \tag{16}$$

  • $\begingroup$ Presumably the same strategy would hold to show that the 3-particle Toda system - which can be reduced to 2 positions and 2 momenta and is closely related to HH - is integrable? (Of course we know it is as a second invariant of motion is known). $\endgroup$ – ZeroTheHero Nov 15 '18 at 15:34

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