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Local Hamilton-Jacobi solution by invertible characteristics

Statement

Assume ${A C}_{ω}$ . Let $H \in C^{\infty} (U, R)$ on an open set $U \subset R \times T^{*} R^{n}$ , with coordinates $(t, q, p)$ , and let $S_{0} \in C^{\infty} (Q_{0}, R)$ on an open $Q_{0} \subset R^{n}$ . The Hamilton characteristic ODE is ${\dot{q}}^{i} = \partial H / \partial p_{i}$ and ${\dot{p}}_{i} = - \partial H / \partial q^{i}$ . For a covector $p$ and vector $v$ in the canonical coordinates, write $p \cdot v : = \sum_{i} p_{i} v^{i}$ . For each $a_{0} \in Q_{0}$ such that $(0, a_{0}, d S_{0} (a_{0})) \in U$ , there are neighbourhoods of $(0, a_{0})$ and a time interval on which this ODE with $q (0, a) = a$ , $p (0, a) = d S_{0} (a)$ has a smooth solution and $(t, a) \mapsto (t, q (t, a))$ is a diffeomorphism onto its image after restriction. On that image the function

$S (t, q (t, a)) = S_{0} (a) + \int_{0}^{t} [p (u, a) \cdot \dot{q} (u, a) - H (u, q (u, a), p (u, a))] d u$

is smooth, has initial value $S_{0}$ , and satisfies $S_{t} + H (t, q, S_{q}) = 0$ . The claim is local to the invertible projection domain; it asserts no continuation beyond a point where the characteristic projection loses invertibility. For the SI interpretation, assign $S_{0}$ action units $J s$ as well as assigning $t, q, p, H$ SI time, length, momentum, and energy units. Then both $S_{0}$ and the integral have action units, so $[S] = J s$ , $[S_{q}] = [p]$ , and $[S_{t}] = J$ .

Facts & Assumptions

Given: The smooth Hamiltonian and initial exact graph in the statement, with the initial point $(0, a_{0}, d S_{0} (a_{0}))$ in $U$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): assume countable choice ${A C}_{ω}$ only for the cotangent-bundle suppliers; no full axiom of choice is used.

[L1]

Canonical Poisson brackets and local Hamiltonian flow: the current H0 draft states $ι_{X_{H}} ω = d H$ , ${F, G} = F_{q} G_{p} - F_{p} G_{q}$ , local unique Hamilton solutions with smooth dependence on their actual domain, and no completeness claim. Its H25 Step 1 record is ready and hash-current, but that is not independent proof approval; the carrier remains draft/unbound with ai-altered proof provenance. Its sign and local-flow interface match the proof.

[F1]

Smooth dependence of ODE solutions on parameters: a smooth finite-dimensional ODE vector field has jointly smooth local solutions in time and parameters near each initial datum, on a common sufficiently short interval after restricting the parameter neighbourhood.

[F2]

Choice-free smooth inverse function theorem in Euclidean space: a smooth map with invertible derivative at a point is a smooth diffeomorphism between suitable neighbourhoods of that point and its image.

Proof

technique · direct characteristics and endpoint action

Given: The hypotheses of the statement and the exact conditional Hamiltonian-flow interface [L1].

1.1L1F1A1givenalgebra

The map $a \mapsto (0, a, d S_{0} (a))$ is continuous. Since its value at $a_{0}$ lies in the open set $U$ , it stays in $U$ for all $a$ in a sufficiently small neighborhood of $a_{0}$ . The characteristic vector field is $(H_{p}, - H_{q})$ by [L1] and is smooth on $U$ ; hence the ODE existence and parameter theorem [F1] supplies smooth $q (t, a), p (t, a)$ near $(0, a_{0})$ for these initial graph data. At $t = 0$ , $D_{a} q = I$ .

1.2L1A1algebra

Define $\tilde{S} (t, a)$ by the integral in the statement. Differentiating with respect to $a$ and integrating the $p \cdot \partial_{a} \dot{q}$ term by parts yields $\partial_{a} \tilde{S} = d S_{0} (a) + p (t, a) D_{a} q (t, a) - p (0, a) D_{a} q (0, a) + \int_{0}^{t} [- \dot{p} - H_{q}] D_{a} q + [\dot{q} - H_{p}] D_{a} p d u .$ The initial terms cancel because $p (0, a) = d S_{0} (a)$ and $D_{a} q (0, a) = I$ ; both integrands vanish by Hamilton's equations [L1]. Thus $\partial_{a} \tilde{S} = p (t, a) D_{a} q (t, a)$ .

2.1step 1.1F2algebra

The derivative of $(t, a) \mapsto (t, q (t, a))$ at $(0, a_{0})$ is block triangular with diagonal blocks $1$ and $I$ , so is invertible. [F2] gives a smooth local inverse $a = a (t, q)$ ; restrict to its image, which is the actual projection domain.

3.1step 1.2algebragiven∎

Differentiating the endpoint integral in time gives $\partial_{t} \tilde{S} = p \cdot \dot{q} - H$ , by the fundamental theorem of calculus. Put $S (t, q) = \tilde{S} (t, a (t, q))$ . The chain rule and step 1.2 give $S_{q} = p$ ; at fixed $q$ , the inverse relation $q (t, a (t, q)) = q$ gives $D_{a} q \partial_{t} a = - \dot{q}$ . Thus $S_{t} = (p \cdot \dot{q} - H) + p D_{a} q \partial_{t} a = (p \cdot \dot{q} - H) - p \cdot \dot{q} = - H$ . Therefore $S_{t} + H (t, q, S_{q}) = 0$ . At $t = 0$ , the integral is zero and $q = a$ , so $S (0, q) = S_{0} (q)$ .

Local Hamilton-Jacobi solution by invertible characteristics

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources