LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Factorisation of the one-dimensional wave operator

Statement

Let $c > 0$ and let $u$ be $C^{2}$ on an open subset of $R^{2}$ (Wave equation, Cauchy data and wave speed). Then $\partial_{t}^{2} u - c^{2} \partial_{x}^{2} u = (\partial_{t} - c \partial_{x}) (\partial_{t} + c \partial_{x}) u = (\partial_{t} + c \partial_{x}) (\partial_{t} - c \partial_{x}) u .$ In the characteristic coordinates $ξ = x - c t$ , $η = x + c t$ one has $\partial_{t}^{2} u - c^{2} \partial_{x}^{2} u = - 4 c^{2} \partial_{ξ} \partial_{η} u,$ so $u$ solves the homogeneous one-dimensional wave equation exactly on the open set where $u_{ξ η} = 0$ .

Facts & Assumptions

Given: a speed $c > 0$ and a $C^{2}$ function $u$ on an open subset of $R^{2}$ , with coordinates $(x, t)$ and characteristic coordinates $ξ = x - c t$ , $η = x + c t$ .

[F1]

If $f$ is $C^{2}$ on an open subset of $R^{m}$ , then $\partial_{i} \partial_{j} f = \partial_{j} \partial_{i} f$ for every pair of coordinate indices (Clairaut--Schwarz theorem for continuous second partial derivatives).

[F2]

If $f : U \to V \subseteq R^{n}$ is totally differentiable at $a$ and $g : V \to R^{p}$ is totally differentiable at $f (a)$ , then $g \circ f$ is totally differentiable at $a$ with $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ). The required total differentiability follows from continuous coordinate partial derivatives (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[F3]

Sums, constant multiples of differentiable functions are differentiable with the usual rules (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 $g (c) \neq 0$ ).

Proof

1.1F1F3algebra

Expanding the two compositions and using [F1] for the mixed terms, $(\partial_{t} - c \partial_{x}) (\partial_{t} + c \partial_{x}) u = \partial_{t}^{2} u + c \partial_{t} \partial_{x} u - c \partial_{x} \partial_{t} u - c^{2} \partial_{x}^{2} u = \partial_{t}^{2} u - c^{2} \partial_{x}^{2} u$ , and $(\partial_{t} + c \partial_{x}) (\partial_{t} - c \partial_{x}) u = \partial_{t}^{2} u - c \partial_{t} \partial_{x} u + c \partial_{x} \partial_{t} u - c^{2} \partial_{x}^{2} u = \partial_{t}^{2} u - c^{2} \partial_{x}^{2} u$ ; hence both factorisations equal the operator applied to $u$ .

1.2F2F3algebra

Write $U (ξ, η) : = u (x, t)$ with $ξ = x - c t$ , $η = x + c t$ . By [F2] the chain rule for the substitution $(x, t) \mapsto (ξ, η) = (x - c t, x + c t)$ gives $\partial_{x} u = U_{ξ} + U_{η}$ and $\partial_{t} u = - c U_{ξ} + c U_{η}$ , with the right-hand sides evaluated at $(x - c t, x + c t)$ , hence $\partial_{t} - c \partial_{x} = - 2 c \partial_{ξ}$ and $\partial_{t} + c \partial_{x} = 2 c \partial_{η}$ as operators on $U$ ; composing, $\partial_{t}^{2} u - c^{2} \partial_{x}^{2} u = (\partial_{t} - c \partial_{x}) (\partial_{t} + c \partial_{x}) u = (- 2 c \partial_{ξ}) (2 c \partial_{η}) U = - 4 c^{2} U_{ξ η}$ .

2.1algebra∎

Since $c > 0$ the factor $- 4 c^{2}$ is nonzero, so $\partial_{t}^{2} u = c^{2} \partial_{x}^{2} u$ at a point if and only if $U_{ξ η} = 0$ there; this proves the claimed equivalence and completes the factorisation identities.

Depends on

Used by

Data on one characteristic line do not determine a one-dimensional wave Counterexample
Right- and left-travelling waves Example
General solution of the one-dimensional wave equation Lemma

Dependency tree · two levels

33 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

Victor Ivrii, Partial Differential Equations (University of Toronto, 2018, CC BY-SA) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (2025 archived author manuscript, AMS Graduate Studies in Mathematics) (standard reference, not scraped)
Per Kristen Jakobsen, An Introduction to Partial Differential Equations (arXiv:1901.03022) (standard reference, not scraped)