CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Characteristic analytic data may be nonunique or incompatible

Statement refuted

Analytic Cauchy data on a characteristic analytic surface need not determine a unique analytic solution or even admit a solution. For $u_{x} = 0$ in $(x, t)$ on the surface t=0, zero trace data have infinitely many analytic solutions, while trace $u (x, 0) = x$ admits no differentiable solution near zero.

Facts & Assumptions

Given: The equation $u_{x} = 0$ with initial surface $t = 0$ , and the candidate traces and solutions specified in the statement.

[F1]

A conormal is characteristic when the principal symbol vanishes there. (Characteristic covectors, hypersurfaces, and noncharacteristic data).

Counterexample

1.1

The principal symbol is $p (ξ_{x}, ξ_{t}) = ξ_{x}$ , and the nonzero conormal to t=0 is dt=(0,1). Thus p(dt)=0, so the surface is characteristic by F1. For every real c, $u_{c} (x, t) = c t$ is analytic, satisfies $(u_{c})_{x} = 0$ , and has zero initial trace. Distinct c give distinct germs and distinct normal derivatives.

givenF1algebra

2.1

If a differentiable u satisfied $u_{x} = 0$ near zero and $u (x, 0) = x$ , differentiating the trace along x would give $u_{x} (x, 0) = 1$ , contradicting the equation value zero. Therefore these analytic trace data are incompatible.

givenalgebra∎

Source notes

Gantumur, §5 transport discussion after Exercise 24, printed p. 14.

Depends on

Used by

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Sources

Gantumur, Math 580 Lecture Notes 2: The Cauchy-Kovalevskaya Theorem (standard reference, not scraped)