TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedaudited 2026-09-06

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Local fully nonlinear Charpit graph construction

Statement

Let $O \subseteq R^{n} \times R \times R^{n}$ be open and $F \in C^{2} (O)$ . Let $V \subseteq R^{n - 1}$ be open, let $y_{0} \in V$ , and let $γ : V \to R^{n}$ , $ϕ : V \to R$ , and $p_{0} : V \to R^{n}$ be $C^{1}$ maps such that $(γ (y), ϕ (y), p_{0} (y)) \in O$ , $F (γ (y), ϕ (y), p_{0} (y)) = 0$ , and $D ϕ (y) = p_{0} (y) D γ (y)$ for every $y \in V$ . Suppose also that $rank [F_{p} (γ (y_{0}), ϕ (y_{0}), p_{0} (y_{0})), D γ (y_{0})] = n .$ Then the Charpit strip through $(γ, ϕ, p_{0})$ projects locally to a classical graph $u$ satisfying $F (x, u, D u) = 0$ . It is unique while that projection is locally invertible among graphs obtained by inverse-projecting this fixed Charpit strip.

Facts & Assumptions

Given: The stated $C^{2}$ equation, compatible $C^{1}$ strip data, and full-rank condition at $y_{0}$ .

Proof

technique · direct

1.1

The Charpit vector field is $C^{1}$ because $F \in C^{2}$ , hence locally Lipschitz. Continuous dependence gives a unique common local strip through the $C^{1}$ initial data $(γ, ϕ, p_{0})$ .

givenconstruct

2.1

The $C^{1}$ -dependence theorem makes this strip $C^{1}$ in $(s, y)$ . Its projected derivative at $s = 0$ has columns $[F_{p}, D γ]$ , hence is invertible by the rank hypothesis.

step 1.1givenalgebra

3.1

The inverse function theorem supplies a local inverse of $(s, y) \mapsto X (s, y)$ ; define $u (X (s, y)) : = Z (s, y)$ .

step 2.1construct

4.1

Constraint preservation gives $F (X, Z, P) = 0$ , while contact preservation gives $D_{y} Z = P \cdot D_{y} X$ ; the $s$ identity is $Z_{s} = P \cdot X_{s}$ . Since $D_{(s, y)} X$ is invertible, these identities imply $D u (X) = P$ .

step 3.1givenalgebra

5.1

Substitution in the preserved constraint gives $F (x, u (x), D u (x)) = 0$ . The fixed strip and its local inverse determine this inverse-projected graph uniquely while the projection remains locally invertible.

step 4.1given∎

Depends on

Used by

Characteristic initial data need not determine a fully nonlinear solution Counterexample

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Part I: Explicit methods — Lecture notes for MA342H (standard reference, not scraped)