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Regular versus hyperregular: the bounded-gradient Lagrangian $\sqrt{1 + v^{2}}$ and the degenerate zero Lagrangian

Example

Take $n = 1$ , $J = U = R$ , and consider the three fibre-full $C^{2}$ Lagrangians $L_{1} (v) = \sqrt{1 + v^{2}}$ , $L_{2} (v) = 0$ and $L_{3} (v) = \frac{1}{2} v^{2}$ on $J \times U \times R$ (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity). Then:

$L_{1}$ is regular on the chart but not hyperregular: its velocity Hessian is $(1 + v^{2})^{- 3 / 2} > 0$ at every $v$ , while the fibre Legendre map $v \mapsto L_{1}^{'} (v) = v / \sqrt{1 + v^{2}}$ has range $(- 1, 1)$ , so it is not surjective onto $R$ ;
$L_{2}$ is neither regular nor hyperregular: $L_{2}^{'} \equiv 0$ and $L_{2}^{''} \equiv 0$ , so the velocity Hessian is singular everywhere and the fibre map is constant, hence not injective;
$L_{3}$ is both regular and hyperregular: $L_{3}^{''} \equiv 1$ and the fibre map $v \mapsto v$ is a $C^{\infty}$ bijection of $R$ onto itself.

Thus regularity does not imply hyperregularity, and neither condition is automatic.

Facts & Assumptions

Given: The three $C^{2}$ Lagrangians $L_{1} (v) = \sqrt{1 + v^{2}}$ , $L_{2} (v) = 0$ , $L_{3} (v) = \frac{1}{2} v^{2}$ , their fibre maps $v \mapsto L_{j}^{'} (v)$ , and their second derivatives.

[F1]

Power rule and chain rule (Continuity and derivatives of positive-base real powers, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , $C^{k}$ Euclidean maps and diffeomorphisms, Absolute value in an ordered field, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length): $L_{1}^{'} (v) = \frac{1}{2} (1 + v^{2})^{- 1 / 2} \cdot 2 v = v (1 + v^{2})^{- 1 / 2}$ and $L_{1}^{''} (v) = (1 + v^{2})^{- 3 / 2} > 0$ for every real $v$ ; $L_{2}^{'} = L_{2}^{''} = 0$ ; $L_{3}^{'} (v) = v$ and $L_{3}^{''} (v) = 1$ .

[F2]

Monotonicity and injectivity (On an interval $I$ , for $f$ continuous on $I$ and differentiable at every interior point: $f^{'} \geq 0$ throughout gives $f$ nondecreasing, $f^{'} > 0$ gives $f$ increasing, $f^{'} \leq 0$ and $f^{'} < 0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{'} \geq 0$ and a nonincreasing $f$ has $f^{'} \leq 0$ wherever it is differentiable, and no strict converse is claimed, A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant, Invertible Euclidean linear maps, Linear map between vector spaces over the same field): since $L_{1}^{''} > 0$ on $R$ , the function $φ_{1} : = L_{1}^{'}$ is strictly increasing, hence injective; since $L_{2}^{'} \equiv 0$ is constant, it is not injective; since $φ_{3} = i d$ is the identity, it is bijective with inverse itself.

[F3]

Elementary estimates: $0 < \sqrt{1 + v^{2}}$ and $∣ v (1 + v^{2})^{- 1 / 2} ∣ = ∣ v ∣ / \sqrt{1 + v^{2}} < 1$ for every real $v$ , because $∣ v ∣ < \sqrt{1 + v^{2}}$ follows from squaring; in particular $p = 1$ is not in the range of $φ_{1}$ , since $v / \sqrt{1 + v^{2}} = 1$ would give $v \geq 0$ and $v^{2} = 1 + v^{2}$ , i.e. $0 = 1$ .

Verification

technique · direct; compute the derivatives, compare the ranges with $\mathbb R$, and check regularity and hyperregularity against the definition

1.1F1algebra

By [F1] the three Lagrangians are $C^{2}$ , and the velocity Hessians at $v$ are $(1 + v^{2})^{- 3 / 2}$ , $0$ and $1$ respectively. Hence $L_{1}$ and $L_{3}$ are regular on the chart, while $L_{2}$ is regular at no point, because regularity means a nonsingular one-by-one velocity Hessian (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity, Nonsingular velocity Hessian is equivalent to invertibility of the Legendre derivative, hence gives a local C^1 inverse).

2.1F2F3step 1.1

By [F2], $φ_{1}$ is injective and $φ_{3}$ is bijective $R \to R$ , while $φ_{2}$ is constant. Moreover, by [F3] the range of $φ_{1}$ is contained in $(- 1, 1)$ and omits $p = 1$ ; for every $p \in (- 1, 1)$ , setting $v = p / \sqrt{1 - p^{2}}$ gives $1 + v^{2} = 1 / (1 - p^{2})$ and $v / \sqrt{1 + v^{2}} = p$ . Thus the range is exactly $(- 1, 1)$ , and $φ_{1}$ is not surjective onto $R$ . Therefore $L_{1}$ is regular but not hyperregular, $L_{3}$ is hyperregular, and $L_{2}$ fails hyperregularity by non-injectivity (and fails regularity as well).

3.1step 1.1step 2.1∎

The three clauses of the Example are proved: clause 1 by the regularity of $L_{1}$ from step 1.1 and the failure of surjectivity from step 2.1; clause 2 by the vanishing Hessian and constancy from steps 1.1 and 2.1; clause 3 by the invertibility of the identity map and the nonvanishing Hessian.

Remarks

The first example is the standard witness that global hyperregularity is a genuine extra hypothesis: the local inverse branch exists near every velocity by Nonsingular velocity Hessian is equivalent to invertibility of the Legendre derivative, hence gives a local C^1 inverse, but no global momentum variable can be solved over the whole fibre. The third example is the model case supplied by A mechanical chart Lagrangian with positive definite kinetic matrix is hyperregular, with the explicit kinetic-plus-potential Hamiltonian with constant positive definite $G$ , and the second is the degenerate case used in Scope of the Legendre map: regular versus hyperregular, the meaning of the covector datum, and the limits of the branch description.
The choice $L_{1} = \sqrt{1 + v^{2}}$ is a mathematical example, not a claim about any physical Lagrangian; no interpretation of the velocities or of the momentum variable is asserted.

Regular versus hyperregular: the bounded-gradient Lagrangian $\sqrt{1 + v^{2}}$ and the degenerate zero Lagrangian

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

Regular versus hyperregular: the bounded-gradient Lagrangian 1+v21+v2​ and the degenerate zero Lagrangian

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

Regular versus hyperregular: the bounded-gradient Lagrangian $\sqrt{1 + v^{2}}$ and the degenerate zero Lagrangian