Regular but not hyperregular: the Lagrangian $e^v$, its branch $H(p)=p\ln p-p$ on $p>0$ and the absent inverse at nonpositive momenta

Take $n=1$, $J=U=\mathbb{R}$, and consider the fibre-full $C^2$ chart Lagrangian

$\mathcal{L}:\mathbb{R}\times \mathbb{R}\times \mathbb{R}⟶\mathbb{R},\mathcal{L}(t,q,v):=e^v$

(The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity, A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations). Then:

1. $\mathcal{L}$ is regular on the chart: ${\mathcal{L}}_v(t,q,v)=e^v$ and ${\mathcal{L}}_{vv}(t,q,v)=e^v>0$ for every $(t,q,v)$, so the velocity Hessian is nonsingular at every state and the Legendre map is a local $C^1$ diffeomorphism at every state (Nonsingular velocity Hessian is equivalent to invertibility of the Legendre derivative, hence gives a local C^1 inverse).
2. $\mathcal{L}$ is not hyperregular on the chart: the fibre velocity map $v\mapsto e^v$ has image $(0,\mathrm{\infty })$ and is not a bijection of $\mathbb{R}$ onto $\mathbb{R}$, so there is no inverse branch defined on the whole momentum line; a momentum $p\le 0$ is not the fibre derivative of any velocity, and no Hamiltonian of a branch is defined there (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity).
3. on the momentum half-line $p>0$ the inverse branch is $v(p)=\ln p$, the unique real number with $e^{v(p)}=p$ (The exponential is a continuous bijection from $\mathbb{R}$ onto $(0,\mathrm{\infty })$), and the Hamiltonian of that branch is

$H:(0,\mathrm{\infty })⟶\mathbb{R},H(p)=pv(p)-e^{v(p)}=p\ln p-p,$

with $H^{\mathrm{\prime }}(p)=\ln p=v(p)$ for every $p>0$; in particular the differential identity $H_p=v$ of The Hamiltonian on an inverse Legendre branch: differential, equivalence with the Euler-Lagrange equation, and the energy identity clause 1 holds on this branch even though the chart is not hyperregular. 4. for every nonempty open interval $I$ and every $C^2$ curve $q:I\to \mathbb{R}$ with $p(t):=e^{q^{\mathrm{\prime }}(t)}$ one has: $\dot{q}=H^{\mathrm{\prime }}(p)$ and $\dot{p}=0$ hold for all $t\in I$ if and only if the Euler-Lagrange expression of $\mathcal{L}$ vanishes, $E_{\mathcal{L}}(q)(t)=0$, for all $t\in I$.

The example is a pure chart computation: $\mathcal{L}=e^v$ is not of the natural mechanical form $\frac{1}{2}{v}^{\mathsf{T}}Gv+A\cdot v-V$, and no physical interpretation of it is claimed.

• Where the global structure stops. The failure is exactly surjectivity of the fibre map: injectivity of $v\mapsto e^v$ holds and gives the unique branch $v(p)=\ln p$ over the image, but momentum zero and negative momenta are not fibre derivatives of any velocity of the chart. The example thus separates the two hypotheses of The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity — nonsingular velocity Hessian and bijectivity onto the momentum fibre — and shows that a local branch at every state need not assemble into a global one.

• Not a natural mechanical Lagrangian. The chart Lagrangian $e^v$ is not of the form $\frac{1}{2}{v}^{\mathsf{T}}Gv+A\cdot v-V$ with positive definite $G$; the regular non-hyperregular separation exhibited here therefore does not contradict the hyperregularity of natural mechanical charts, which rests on the positivity of the kinetic matrix.