Hamiltonian Mechanics and Completely Integrable Systems — Examples

Here $H_p=p\mathrm{/}m$ and $H_q=0$, so [F1] gives $\dot{q}=p\mathrm{/}m$ and $\dot{p}=0$.

The second equation gives $p(t)=p_0$; substituting into the first and integrating gives $q(t)=q_0+t{p}_0\mathrm{/}m$. These formulas exist for all real time and directly satisfy the initial condition.

By [F1], $\dot{q}=p\mathrm{/}m$ and $\dot{p}=-m{\mathrm{\Omega }}^{2}q$, hence $\ddot{q}+{\mathrm{\Omega }}^{2}q=0$. Thus $q(t)=A\cos (\mathrm{\Omega }t)+B\sin (\mathrm{\Omega }t)$ and $p(t)=m\dot{q}(t)$. Every nonconstant solution has period $2\pi \mathrm{/}\mathrm{\Omega }$.

Conservation follows directly from $\frac{d}{dt}H=(p\mathrm{/}m)\dot{p}+m{\mathrm{\Omega }}^{2}q\dot{q}=0$. Setting the constant value to $E>0$ and dividing its level equation by $E$ gives the displayed ellipse. For $E=0$, positivity forces $(q,p)=(0,0)$, the stationary equilibrium.

The level equation is $p^2=2(E-1+\cos q)$. Its critical points solve $p=0$ and $\sin q=0$: $(0,0)$ is a minimum of energy $0$, while $(\pi ,0)$ is a saddle of energy $2$.

Assume $0<E<2$. The allowed angles form a proper interval about $q=0$; the positive and negative square-root branches meet at two turning points $p=0$, producing a closed oscillatory curve. By [F1], the sign of $p$ is the direction of angular travel and reverses at those endpoints.

Assume $E>2$. The right-hand side is strictly positive for every $q$. The two graphs $p=\pm \sqrt{2(E-1+\cos q)}$ are disjoint circles over $S^1$, and [F1] gives rotation with fixed sign.

Assume $E=2$. The two branches meet at the saddle and form the homoclinic separatrix; the level is singular and is not a regular Liouville torus.

Finally $E=0$ gives only the stable equilibrium, while $E<0$ gives the empty level because $H\ge 0$. These alternatives and steps 2.1--2.3 exhaust all energies.

For $L(q,v)=\frac{1}{2}{g}_q(v,v)$, [F2] gives $\mathbb{F}L=g^{\mathrm{♭}}$ and $H=\frac{1}{2}{g}^{-1}(p,p)$. The Euler–Lagrange equation of this kinetic-energy Lagrangian is the coordinate equation ${\ddot{q}}^k+{\mathrm{\Gamma }}_{ij}^k{\dot{q}}^i{\dot{q}}^j=0$ for the Levi–Civita connection in [F1], hence is ${\mathrm{\nabla }}_{\dot{q}}\dot{q}=0$.

By [F3], $(q,\dot{q})$ solves that geodesic equation exactly when $(q,p)=(q,g^{\mathrm{♭}}\dot{q})$ is a Hamiltonian trajectory of $H$. Thus the two flows correspond wherever their maximal trajectories exist; no completeness of $g$ is asserted.

Insert $Y_a(q)=a\times q$ into [F1]. The scalar triple-product identity gives $H_{Y_a}(q,p)=p\cdot (a\times q)=a\cdot (q\times p)$.

Hence the infinitesimal cotangent-lifted rotation is $X_{H_a}$. Varying $a$ shows that the vector-valued observable is $q\times p$, the usual angular momentum.

The only nonzero derivatives are $\mathrm{\partial }{q}^i\mathrm{/}\mathrm{\partial }{q}^k={\delta }_k^i$ and $\mathrm{\partial }{p}_j\mathrm{/}\mathrm{\partial }{p}_k={\delta }_{jk}$. Substitution in [F1] gives zero for the $q$--$q$ and $p$--$p$ brackets and ${\sum }_k{\delta }_k^i{\delta }_{jk}={\delta }_j^i$ for $\{q^i,p_j\}$.

Skew-symmetry, also visible by reversing the two terms in the coordinate formula, gives $\{p_j,q^i\}=-{\delta }_j^i$.

Contraction gives ${\iota }_X\omega =dy$, which is closed; hence [F1] shows that $X$ is symplectic.

On the closed loop $\gamma (t)=[(0,t)]$, $0\le t\le 1$, one has ${\int }_{\gamma }dy=1$. Every exact one-form has zero integral around a closed curve by the fundamental theorem of calculus, so $dy$ is not exact. Its class is nonzero, and [F2] proves that $X$ is not Hamiltonian.

Differentiating $L(q,v+sw)$ at $s=0$ gives $g_q(v,w)$, so its fibre derivative is $p=g_q^{\mathrm{♭}}(v)$. Positive definiteness makes $g_q^{\mathrm{♭}}$ invertible, with inverse $g_q^{\mathrm{♯}}$, and [F1] gives hyperregularity.

The energy is $p(v)-L=g(v,v)-\frac{1}{2}g(v,v)+V=\frac{1}{2}g(v,v)+V$. Substituting $v=g^{\mathrm{♯}}p$ gives $H(q,p)=\frac{1}{2}{g}^{-1}(p,p)+V(q)$, as asserted.

Differentiating the displayed substitution gives $dq\wedge dp=d\varphi \wedge dJ$; its Jacobian coefficient is ${\cos }^2\varphi +{\sin }^2\varphi =1$. Also direct substitution gives $H=\mathrm{\Omega }J$.

Since $d\theta \wedge dI=(d\varphi \mathrm{/}(2\pi ))\wedge (2\pi dJ)=d\varphi \wedge dJ$, [F1] applies. Moreover $H=\mathrm{\Omega }I\mathrm{/}(2\pi )$, so the library Hamiltonian convention gives $\dot{\theta }=\mathrm{\Omega }\mathrm{/}(2\pi )$, equivalently $\dot{\varphi }=\mathrm{\Omega }$. Thus the displayed substitution explicitly supplies the action–angle coordinates. If angle has period $2\pi$ instead, the corresponding action is $J=H\mathrm{/}\mathrm{\Omega }$; the factor $2\pi$ is purely normalization.

The Hamiltonian flow of $J$ is the axial circle action, so its orbit cycle $b$ is unchanged by transport around the source's positively oriented loop $\gamma$ enclosing $(1,0)$.

By [F2], the two energy-level pieces used to cross the critical energy have Chern numbers $c_1=c_2+1$. The source's solid-torus gluing sends the basis by $T(c)=\left(\begin{array}{cc}1& c\\ 0& 1\end{array}\right)$; hence its complete calculation gives $M_{\gamma }=T(c_1)T(c_2{)}^{-1}=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ in the basis $(a,b)$. This step invokes, rather than reproves, the source's Takens index theorem and gluing argument.

The matrix is not the identity, so [F1] rules out a single global action–angle chart on the regular torus bundle around the puncture. Reversing the loop or changing the integral basis can invert or conjugate the matrix but cannot make its monodromy trivial.

Here $dH=qdq+pdp$, so it is nonzero on ${\mathbb{R}}^2\setminus \{0\}$, an open dense set. Thus $H$ supplies a completely integrable one-degree-of-freedom system in the sense of [F1].

But $H^{-1}(0)=\{(0,0)\}$ because it is a sum of squares, and $d{H}_{(0,0)}=0$. The fibre is singular and zero-dimensional, hence cannot be diffeomorphic to $S^1$. This shows why the regular-value hypothesis in the torus theorem is essential.

Both functions depend only on $q^1$. Their Hamiltonian fields are scalar multiples of ${\mathrm{\partial }}_{p_1}$, so their symplectic pairing and hence $\{F_1,F_2\}$ vanish.

Nevertheless $d{F}_2=2q^{1}d{F}_1$ at every point, so the integral map $(F_1,F_2)$ has rank at most one instead of two and has no regular locus of the required rank. By [F1], it is not a completely integrable system, and Liouville–Arnold cannot furnish action–angle coordinates for it.