TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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A regular index-one DAE has a local reduced ODE branch

Statement

Let $d, m \geq 1$ , let $Ω \subseteq R \times R^{d} \times R^{m}$ be open, and let $f : Ω \to R^{d}$ , $g : Ω \to R^{m}$ be $C^{k}$ for $k \geq 1$ . Suppose $g (t_{*}, x_{*}, z_{*}) = 0$ and $g_{z} (t_{*}, x_{*}, z_{*})$ is invertible. Then there are neighborhoods $U$ of $(t_{*}, x_{*})$ and $V$ of $z_{*}$ , and a unique $C^{k}$ map $ψ : U \to V$ , such that, for $(t, x) \in U$ and $z \in V$ ,

g (t, x, z) = 0 ⟺ z = ψ (t, x) .

Its derivatives satisfy

ψ_{t} = - g_{z}^{- 1} g_{t}, ψ_{x} = - g_{z}^{- 1} g_{x},

with the derivatives of $g$ evaluated at $(t, x, ψ (t, x))$ . The reduced vector field $F (t, x) = f (t, x, ψ (t, x))$ is $C^{k}$ on $U$ . Thus solutions of the DAE that stay in this selected chart are exactly solutions of $x^{'} = F (t, x)$ paired with $z = ψ (t, x)$ .

Facts & Assumptions

Given: The dimensions, open domain, $C^{k}$ maps, root, and invertible algebraic Jacobian in the Statement.

[F1]

Semi-explicit index-one DAEs and consistent initial data: a semi-explicit DAE is $x^{'} = f (t, x, z)$ , $g (t, x, z) = 0$ ; a consistent pair satisfies $g = 0$ .

[F2]

The parametrized implicit function theorem with $C^{k}$ regularity: for a $C^{k}$ map whose derivative in the unknown block is invertible at a zero, there are neighborhoods and a unique $C^{k}$ implicit branch, with derivative $- D_{z} g^{- 1} D_{(t, x)} g$ .

[F3]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : the derivative of a composite is the composition of the derivatives.

Proof

1.1givenF2algebra

Treat $(t, x)$ as the parameter block and $z$ as the unknown block in [F2]. Its hypotheses hold because $g$ is $C^{k}$ , $g (t_{*}, x_{*}, z_{*}) = 0$ , and $g_{z} (t_{*}, x_{*}, z_{*})$ is invertible. Therefore [F2] gives neighborhoods $U, V$ and a unique $C^{k}$ function $ψ$ with the stated equivalence and derivative formula. Splitting that combined derivative into its $t$ and $x$ blocks gives the two displayed identities.

1.2F2F3algebra

Differentiating $g (t, x, ψ (t, x)) = 0$ in the $t$ direction using [F3] gives $g_{t} + g_{z} ψ_{t} = 0$ ; multiplying by $g_{z}^{- 1}$ gives $ψ_{t} = - g_{z}^{- 1} g_{t}$ . Differentiating in each $x$ direction gives $g_{x} + g_{z} ψ_{x} = 0$ , hence $ψ_{x} = - g_{z}^{- 1} g_{x}$ .

2.1F1F3step 1.1∎

The map $(t, x) \mapsto (t, x, ψ (t, x))$ is $C^{k}$ , so [F3] applied to its composition with the $C^{k}$ map $f$ shows $F (t, x) = f (t, x, ψ (t, x))$ is $C^{k}$ . On the selected chart the equivalence in step 1.1 forces every DAE solution to have $z (t) = ψ (t, x (t))$ , and substitution gives $x^{'} = F (t, x)$ ; conversely, a reduced solution paired with $z = ψ (t, x)$ satisfies both DAE equations.

Depends on

Used by

Cited to discharge well-definedness by Semi-explicit index-one DAEs and consistent initial data.

Dependency tree · two levels

14 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

Ernst Hairer and Gerhard Wanner, Solving Ordinary Differential Equations II, Chapter VI §VI.1, Solving Index 1 Problems, pp. 372–381 (standard reference, not scraped)
Kendall Atkinson, Weimin Han, and David Stewart, Numerical Solution of Ordinary Differential Equations, Chapter 10 §10.3 (standard reference, not scraped)