Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 20 (Hamiltonian Dynamical Systems)

Use the standard symplectic form ${\omega }_0={\sum }_{j=1}^{n}d{q}_j\wedge d{p}_j$. A canonical transformation is a diffeomorphism $f$ preserving this symplectic form:Thus, writing $f(q,p)=(Q,P)$, the defining identity is ${\sum }_{j}d{Q}_j\wedge d{P}_j={\sum }_{j}d{q}_j\wedge d{p}_j$. For a local canonical transformation, require this on its open domain. In particular preservation of the symplectic form, rather than just of phase-space volume, is the condition in dimensions greater than two.

Throughout these solutions choose the Hamiltonian vector field convention ${\iota }_{X_F}\omega =dF$. It gives $\dot{q}=F_p$, $\dot{p}=-F_q$ and agrees with the positive coordinate Poisson bracket used below. The opposite convention for Hamiltonian vector fields changes signs in the Lie bracket of vector fields identity but none of the commuting or integrability conclusions.

The Poisson bracket in the original coordinates isFor $z=(q,p)$, setThen ${\omega }_0(u,v)=u^{T}Jv$ and $\{F,G\}=\nabla F^{T}J\nabla G$. The pullback definition of a canonical transformation gives the pointwise matrix conditionSince $A$ is invertible, this is equivalent to $AJ{A}^T=J$. Indeed, invert $A^{T}JA=J$ and use $J^{-1}=-J$ to obtain $A^{-1}J{A}^{-T}=J$, then multiply by $A$ and $A^T$. Applying the same calculation to the inverse implication gives equivalence. These are the two equivalent symplectic matrix identities that connect the symplectic form with Poisson brackets.

The chain rule yieldsIf $f$ is a canonical transformation, the symplectic matrix identity $AJ{A}^T=J$ makes this expression $\{F,G{\}}_{Q,P}\circ f$. Thus preservation of the symplectic form implies preservation of every Poisson bracket, for all $C^1$ functions.

Conversely, assume this Poisson bracket identity for every $F,G$. Take $F(Z)=u\cdot Z$ and $G(Z)=v\cdot Z$, for arbitrary constant vectors $u,v\in {\mathbb{R}}^{2n}$. It gives $u^{T}AJ{A}^{T}v=u^{T}Jv$. Since this holds for every $u,v$, $AJ{A}^T=J$, hence $A^{T}JA=J$. Therefore preservation of all Poisson brackets is equivalent to canonicity. Only first derivatives of $f,F,G$ enter this argument.

Apply preservation of Poisson brackets to the coordinate functions. It givesConversely, the entries of $AJ{A}^T$ are exactly the Poisson brackets between all transformed coordinate functions. The displayed relations therefore say $AJ{A}^T=J$, which implies the canonical transformation condition and, by the chain rule, preservation of every Poisson bracket. Thus all three conditions are equivalent:In particular the original coordinate bracket $\{q_i,p_j\}$ is ${\delta }_{ij}$ with the chosen Poisson bracket convention.

Under the given rank hypothesis, the necessary and sufficient local condition isVanishing only at the single point would not suffice: an extension preserves the coordinate Poisson brackets throughout its domain. Necessity follows from the coordinate criterion just proved.

For sufficiency, put $Q_i=a_i$ and $Y_i=X_{Q_i}$. Because the symplectic form is a nondegenerate bilinear form at every point, the independent differentials $d{Q}_i$ give independent fields $Y_1,\dots ,Y_n$. They commute, sinceMoreover $d{Q}_j(Y_i)=\{Q_j,Q_i\}=0$, so they span the tangent spaces of the $n$-dimensional fibres of the submersion $Q$. Choose a local section $s(Q)$ transverse to those fibres. Its joint local flow defines coordinates $(Q,t)$ byThe inverse function theorem makes $\Psi$ a local diffeomorphism, and commuting flows give ${\Psi }_{\ast }{\partial }_{t_i}=Y_i$.

In these coordinates ${\iota }_{{\partial }_{t_i}}{\Psi }^{\ast }{\omega }_0=d{Q}_i$. Each flow preserves the symplectic form, by Cartan's magic formula. Thus its coefficients are independent of $t$, andwhere $\beta =s^{\ast }{\omega }_0$ is a closed differential two-form on the base. On a small ball, the Poincaré lemma gives $\beta =d\alpha$, with $\alpha ={\sum }_i{\alpha }_i(Q)d{Q}_i$. DefineThenConsequently the original coordinates are extended to the required canonical transformation $(Q,P)$, with the prescribed first half $Q=a$. Its Jacobian matrix is invertible by construction. This is the canonical completion of independent commuting functions. For the printed $C^2$ data, the flows and resulting canonical transformation can be taken $C^1$, which is sufficient to preserve the symplectic form; smooth data give smooth coordinates.

A function $F$ is an integral of motion of $H$ when it is constant along every trajectory of the Hamiltonian flow. Infinitesimally,The antisymmetry of a Lie bracket of the Poisson bracket gives $X_{F}H=\{H,F\}=0$. Hence $H$ is also an integral of motion of $F$.

For the commuting assertion, ${\mathcal{L}}_{X_F}\omega =d({\iota }_{X_F}\omega )=d^{2}F=0$, because $\omega$ is closed. The contraction identity for a Lie bracket of vector fields then givesSince the symplectic form is a nondegenerate bilinear form at every point, the Hamiltonian vector fields commute:With the chosen convention the general identity is $[X_F,X_H]=X_{\{H,F\}}$. Equivalently, it follows by applying the Jacobi identity for the Poisson bracket to an arbitrary test function. Local flows commute whenever both compositions are defined; no global completeness assumption is needed.

Take the separate oscillator energiesThey depend on disjoint canonical coordinate pairs, soThus they are first integrals in involution. Their differentialsare nonzero precisely when $(x_i,y_i)\ne (0,0)$ and have disjoint supports. Hence they are functionally independent on the specified open set.

To understand all further global integrals of motion, use the actionsDirect differentiation gives $d{x}_i\wedge d{y}_i=d{\theta }_i\wedge d{R}_i$, so these are action-angle variables withWhen ${\alpha }_2/{\alpha }_1$ is irrational, every orbit is dense in its fixed-action two-dimensional flat torus. One elementary proof samples the orbit at times $2\pi j/{\alpha }_1$: the first angle returns, and the second undergoes an irrational rotation. These samples are dense in the second circle. Allowing a fixed additional time supplies any desired first angle, proving density in the whole flat torus.

A continuous, globally defined integral of motion must have the same value on an orbit and its closure. Therefore it is constant on every fixed-action flat torus. A $C^1$ such integral has the form $F=\varphi (R_1,R_2)$, with $\varphi$ $C^1$, by evaluating $F$ at a fixed choice of angles. HenceIt cannot be independent of $E_1,E_2$. This proves there are at most two globally independent differentiable integrals. If a third global integral were independent at a point on an excluded coordinate plane, independence would persist in a neighbourhood and hence at nearby points of the regular open set, again a contradiction.

The word global matters: on an angular chart, ${\alpha }_2{\theta }_1-{\alpha }_1{\theta }_2$ is a local integral of motion independent of the actions. It is not single-valued on the flat torus when the frequency ratio is irrational. Thus the assertion would be false for arbitrary local, rather than global, integrals.

The Hamiltonian perturbation preserving one regular level is explicit:Linearity of the Hamiltonian vector field in the differential of a smooth map givesIt vanishes on ${\Sigma }_0$. At a point of ${\Sigma }_c$, its contraction with the symplectic form isSince the differentials are independent there, this vanishes exactly when $c=0$. Thus on every nonempty nonzero regular fibre the two Hamiltonian vector fields differ at every point, not merely as functions somewhere:Mutual Poisson commutation of the $G_j$ is unnecessary; the given first-integral property does not by itself make the perturbed flow preserve every common level, nor is that requested.

There is a genuine missing nonemptiness hypothesis in the literal statement. Independent differentials on an empty set is a vacuous condition, and any two vector-field restrictions to an empty set are equal. For example, on ${\mathbb{R}}^2$ let $r=1$, $G_1=H=e^q$, and $\rho =1$. Every nonempty ${\Sigma }_c$, $0<c<1$, is regular, while ${\Sigma }_c=\varnothing$ for $-1<c\le 0$. In particular, the restrictions at $c=-1/2$ can never be unequal for any $K$. Therefore the printed universal inequality requires either all the compared fibres to be nonempty, or the pointwise formulation displayed above. The explicit $K$ fully proves either intended qualification.

On a $2n$-dimensional symplectic manifold, a completely integrable Hamiltonian system has $n$ first integrals in involution $F_1=H,F_2,\dots ,F_n$ whose differentials are independent almost everywhere, hence on a dense open regular set:An equivalent formulation allows $H$ to be a function of the $n$ commuting independent integrals. The independence requirement is on the regular set, permitting exceptional critical points and singular fibres.

For example, on ${\mathbb{R}}^{2n}$ take $H={\sum }_{j=1}^n{E}_j$ with $E_j=(p_j^2+{\alpha }_j^2{q}_j^2)/2$ and ${\alpha }_j>0$. The disjoint coordinate pairs make all Poisson brackets $\{E_i,E_j\}$ zero, and the differentials are independent whenever no pair $(q_j,p_j)$ vanishes. Replacing $E_1$ by $H$ is an invertible linear map of this list. Thus $F_1=H,F_j=E_j$ for $j\ge 2$ give the required example. Irrational frequency ratios affect orbit density, but not complete integrability.

The Arnold-Liouville theorem concerns a connected compact component $L$ of a regular common level of $n$ commuting independent smooth functions $F=(F_1,\dots ,F_n)$ on a $2n$-dimensional symplectic manifold. It states that $L$ is a Lagrangian torus, and that a neighbourhood of $L$ has action-angle variablesin which every $F_i$ depends only on $I$. If $H=F_1$ or, more generally, $H=h(F)$, its Hamiltonian flow satisfiesAngles have period $2\pi$ here. The theorem is local near this regular compact component; it does not assert globally defined action-angle variables across singular levels or over an entire base with monodromy.

First, $L$ has dimension of a manifold $n$ by the submersion theorem. The Hamiltonian vector fields $X_{F_i}$ are independent, tangent to $L$, and commute. They span $TL$. Moreoverso $L$ is a Lagrangian submanifold. Compactness makes these restricted vector fields complete. Their joint flow is an ${\mathbb{R}}^n$-action on $L$. Its orbits are open because the fields span $TL$, so connectedness gives one orbit. The stabilizer $\Lambda$ of a point is discrete by the inverse function theorem, andCompactness forces $\Lambda$ to be a full-rank Euclidean lattice: otherwise an unbounded linear coordinate transverse to its span would descend to the quotient. Thus $L$ is an $n$-torus.

Next choose a small ball of regular values near $F(L)$ and a neighbourhood $N$ of this component which is a product family of compact tori $L_c$. This local trivialization follows directly by choosing transverse vector fields $Z_j$ with $d{F}_i(Z_j)={\delta }_{ij}$ and lifting short radial paths in the base; compactness of $L$ gives a uniform neighbourhood where their flows exist. Hence $N$ deformation retracts onto $L$. The commuting joint flows on nearby fibres have smoothly varying full-rank Hamiltonian period lattices. A basis $t^{(j)}(c)\in {\mathbb{R}}^n$ of their periods can be chosen smoothly on this small ball: continue the return equations from a basis at the central fibre, using their nonsingular vertical flow derivatives and the implicit function theorem. No global choice of lattice basis is needed.

Because $\omega {|}_L=0$ and $N$ retracts onto $L$, its closed symplectic form is exact on $N$. Choose a one-form $\lambda$ with $d\lambda =-\omega$. Let ${\gamma }_j(c)$ be the smoothly continued cycles represented by the period basis and define the action integralsFor a transverse variation $v$ of the fibre, differentiating a cycle integral and using Cartan's magic formula eliminates the integral of the exact term. With the cycle parametrized by the joint-flow time $t^{(j)}$, this givesThusThe period matrix is nonsingular, so $I$ gives coordinates on the base. The corresponding Hamiltonian vector fields satisfyTheir time-$2\pi$ flows return along the basis cycles. They commute because each $I_j$ is a function of the commuting $F_i$. Consequently they give a free torus action on $N$.

Choose a section over the action-coordinate ball and use this torus action to define angles $\theta$. Since $X_{I_j}={\partial }_{{\theta }_j}$, the symplectic form takes the formThe closed base two-form $\beta$ is exact on the ball: write $\beta =d\alpha$, $\alpha ={\sum }_j{a}_j(I)d{I}_j$, by the Poincaré lemma. Changing the angular origins by ${\theta }_j^{\prime }={\theta }_j+a_j(I)$ removes this term, since ${\sum }_{j}d{\theta }_j^{\prime }\wedge d{I}_j=\omega$. This completes the construction of action-angle variables. The functions $F_i$ are constant on the fibres, so are functions of $I$ alone, and Hamilton's equations give the claimed straight-line motion. This proves both the topological and symplectic parts of the Arnold-Liouville theorem.

Take $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and $|k{|}_1={\sum }_j|k_j|$. A Diophantine frequency vector satisfiesChanging the norm only changes the admissible $\gamma$. The bound excludes resonances and quantifies the small divisors in the inverse of ${\mathcal{D}}_{\omega }$.

A periodic solution necessarily has zero-mean right-hand side, because the integral of each angular derivative vanishes. Conversely, if $⟨g⟩=0$, expand its Fourier series:The Diophantine condition makes every denominator nonzero. These are all differentiable periodic solutions: any difference satisfies ${\mathcal{D}}_{\omega }h=0$ and has no nonzero Fourier coefficient. Thus a solution exists exactly for zero-mean $g$, and is unique up to a constant; setting $⟨f⟩=0$ fixes it. The analytic conclusion is on every strictly smaller complex strip, not necessarily the original boundary strip.

Shifting each contour in the Fourier coefficient integral toward the appropriate edge of the complex strip givesHence for $0<\delta <\sigma$,This converges normally, including differentiated series on any still smaller strip, and proves both holomorphic extension and ${\mathcal{D}}_{\omega }f=g$. Real-valuedness follows from $g_{-k}=\overline{g_k}$ and the corresponding identity for $f_k$.

An explicit bound is obtained by counting lattice shells. The number of $k$ with $|k{|}_1=m$ is at mostPut $s=n+\tau$ and $p=s-1\ge 0$. For $p>0$, $m^p{e}^{-\delta m/2}\le (2p/(e\delta ){)}^p$; for $p=0$ use the bound $1$. Since ${\sum }_{m\ge 1}{e}^{-\delta m/2}=(e^{\delta /2}-1{)}^{-1}\le 2/\delta$, this givesUse $0^0=1$ in this constant when $n=1,\tau =0$. The estimate is valid for every positive $\delta$; optimality of the exponent is not needed. For a unit-period flat torus, insert the factors $2\pi$ in both the Fourier exponent and denominator and adjust the constant.

Fix a smaller angular strip and a momentum ball $B^{\prime }⋐B$; the canonical transformation will be defined there and map into the original domain for small $ϵ$. This domain restriction is natural for a near-identity change of coordinates with a nonzero momentum shift.

First solve the scalar cohomological equation on a Diophantine torusThe preceding result gives a real-analytic periodic $u$ on every smaller strip. We need a near-identity angular diffeomorphism $\chi$ and a constant vector $a$ such thatHere is a complete analytic construction of this translated conjugacy of a Diophantine vector field.

Write $w=ϵ\nabla u$ and, for a current approximation, putAssume $A$ is close to the identity. For a correction $\Delta \chi =Av$, the linearized defect isThe identity follows by differentiating the definition of $e$, so the term $(De)v$ already contains the current error. ChooseThe matrix average is invertible near the identity, and the right-hand side has zero mean. The preceding torus small-divisor estimate therefore solves this vector equation componentwise. Update ${\chi }_{+}=\chi +Av$, $a_{+}=a+\Delta a$. The exact new error isIt is quadratic in the current defect.

For convergence, reserve a fixed outer strip on which $w$ is analytic and work on shrinking inner strips whose total loss is less than half the starting width. Let $s=n+\tau$ and choose loss parameters ${\delta }_j=d{2}^{-j}$ with $0<d<1$ small enough that the fixed multiples of these losses used at each stage fit in the reserved total width. The torus small-divisor estimate and Cauchy estimates give, with constants uniform while $A$ stays close to the identity and $\chi$ remains in the reserved strip,andIndeed, the first remainder is bounded using $\Vert D{e}_j\Vert \le C{\delta }_j^{-1}\Vert e_j\Vert$, and the second using the uniformly bounded second derivatives of $w$ on the reserved strip. Taking a slightly larger exponent $\mu$ covers all these losses. Start with ${\chi }_0=\mathrm{id}$ and $a_0=0$, so $e_0=-w=O(ϵ)$. If a small constant $K$ satisfies $C{d}^{-\mu }K{2}^{2\mu }\le 1$ and $\Vert e_0\Vert \le K$, induction gives $\Vert e_j\Vert \le K{2}^{-2\mu j}$. The estimates on $\Delta {\chi }_j$ and $\Delta a_j$ are summable. Choosing $ϵ$ still smaller keeps the derivatives close to the identity and the images inside the reserved strip, closing the induction. The real periodic analytic limits $\chi ,a$ solve the conjugacy equation. Since $⟨w⟩=0$, the very first constant correction is zero; the first new defect is $O({ϵ}^2)$, and the remaining summed constant corrections are $O({ϵ}^2)$. Also the summed change of $\chi$ is $O(ϵ)$. This proves the claimed estimates as well as existence, without leaving an unsolved linear remainder.

Now set $\eta (x)=ϵ\nabla u(x)+a$ and define the symplectic cotangent lift with a closed momentum shiftThe cotangent lift of a diffeomorphism is symplectic. Translation by $\eta$ is also symplectic because the one-form $\eta \cdot dx=ϵdu+a\cdot dx$ is closed. The constant term need not be exact on the flat torus; closedness is sufficient. Thus their composition is the required near-identity canonical transformation.

Expanding the quadratic kinetic term gives a transformed linear coefficient $D{\chi }^{-1}(\omega +\eta \circ \chi )$, which is exactly $\omega$ by the conjugacy equation. The constant-in-$y^{\prime }$ part issince ${\mathcal{D}}_{\omega }u=⟨V⟩-V$. Consequently defineand, for $ϵ\ne 0$,In the exact identities, squared Euclidean norms mean the real polynomial ${\sum }_j{v}_j^2$, continued bilinearly on complex strips, without complex conjugation. The latter is uniformly bounded and analytic on the retained strip because $a=O({ϵ}^2)$. At $ϵ=0$ use the identity map, $E_0=0$, and $Q_0=|y^{\prime }{|}^2$. We have the exact requested normal formHere $E_{ϵ}=O(ϵ)$, $Q_{ϵ}$ is genuinely homogeneous quadratic in $y^{\prime }$, and its coefficient matrix differs from the identity by $O(ϵ)$. Thus $|Q_{ϵ}-|y^{\prime }{|}^2|\le C|ϵ||y^{\prime }{|}^2$. The remainder depends only on the angular coordinate. If the printed final $\tilde{V}(x)$ uses the old angle $x$, the same term is $\frac{1}{2}|\nabla u(x)+a/ϵ{|}^2$ evaluated at $x=\chi (x^{\prime })$; this is just the corresponding coordinate convention.