Stationarity of the phase-space action is equivalent to Hamilton's equations, and hyperregular elimination recovers the Lagrangian action

Let $n\ge 1$, let $O_H\subseteq \mathbb{R}\times {\mathbb{R}}^n\times {\mathbb{R}}^n$ be open and nonempty, let $H:O_H\to \mathbb{R}$ be of class $C^2$, let $a<b$, and suppose that for a $C^1$ path $\gamma (t)=(q(t),p(t))$ ($C^k$ Euclidean maps and diffeomorphisms, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral) with $(t,q(t),p(t))\in O_H$ for all $t\in [a,b]$ the phase action

$A(\gamma ):={\int }_a^b[p(t)\cdot q^{\mathrm{\prime }}(t)-H(t,q(t),p(t))]dt$

is defined, the integrand being continuous (The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as ${\sup }_{P}L(f,P)$ and ${\inf }_{P}U(f,P)$, Darboux integrability as their equality, and the notation ${\int }_a^{b}f$, A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$). Then:

1. for every $C^1$ variation pair $(\delta q,\delta p)$ with $\delta q(a)=\delta q(b)=0$ and with $\gamma +s(\delta q,\delta p)$ in the domain of $A$ for small $\mathrm{\mid }s\mathrm{\mid }$,

$\frac{d}{ds}{\mid }_{s=0}A(\gamma +s(\delta q,\delta p))={\int }_a^b\left[\right(q^{\mathrm{\prime }}-H_p)\cdot \delta p+(-p^{\mathrm{\prime }}-H_q)\cdot \delta q]dt;$

2. $\gamma$ is stationary (the displayed derivative vanishes for every such variation) if and only if Hamilton's equations hold on $[a,b]$:

$q^{\mathrm{\prime }}(t)=H_p(t,q(t),p(t)),p^{\mathrm{\prime }}(t)=-H_q(t,q(t),p(t))(t\in [a,b]);$

3. (elimination) let in addition $\mathcal{L}:J\times U\times {\mathbb{R}}^n\to \mathbb{R}$ be a hyperregular $C^2$ chart with inverse branch $v(t,q,p)$ and Hamiltonian $H(t,q,p)=p\cdot v(t,q,p)-\mathcal{L}(t,q,v(t,q,p))$ (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity), such that the phase domain of $H$ contains the points below. For every $C^2$ curve $q:[a,b]\to U$ with $(t,q(t),q^{\mathrm{\prime }}(t))$ in the domain of $\mathcal{L}$, put $p(t):={\mathcal{L}}_v(t,q(t),q^{\mathrm{\prime }}(t))$. Then $A(q,p)=S(q)$, where $S$ is the action of $\mathcal{L}$; consequently $q$ is stationary for $S$ if and only if $(q,p)$ is stationary for the phase action.

• The identification in clause 3 is made on the actual inverse branch: the path in phase space is the Legendre image of the Lagrangian path, so no dynamical assumption such as $p=q^{\mathrm{\prime }}$ is built into the statement; the elimination uses the inverse relation of the hyperregular chart, which is exactly the hypothesis under which the branch exists globally.

• The phase action has free $p$ endpoints and fixed $q$ endpoints. This is why the $p$-variations require no boundary condition and the $q$-variations do: the boundary term produced by integrating $\int p\cdot \delta q^{\mathrm{\prime }}$ vanishes precisely because $\delta q$ vanishes at the endpoints, while no corresponding term arises from $\delta p$.