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

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.

Hadamard instability despite analytic solvability

Statement refuted

For positive integers k the harmonic analytic functions $u_{k} (x, t) = e^{- \sqrt{k}} \sinh (k t) \cos (k x) / k$ have zero value data and normal data $e^{- \sqrt{k}} \cos (k x)$ tending to zero in every $C^{j}$ seminorm on compact x-intervals, but $u_{k} (0, t) \to + \infty$ for every fixed t>0. Thus analytic solvability gives no continuous solution map from that smooth-data topology to pointwise evaluation at any positive time.

Facts & Assumptions

Given: For each positive integer k define $u_{k} (x, t) = e^{- \sqrt{k}} \sinh (k t) \cos (k x) / k$ . Its PDE, data, and limiting behavior are to be verified.

[F1]

Sine and cosine differentiate into one another with the stated signs. (The derivatives of sine and cosine are cosine and minus sine).

[F2]

The exponential equals each of its derivatives. (The exponential function is smooth and $(\exp)^{'} = \exp$ ).

[F3]

Sine and cosine are bounded in modulus by one. (Parity and the Pythagorean identity for sine and cosine).

[F4]

Every fixed polynomial is dominated by a positive exponential. (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ).

[F5]

The exponential power series converges absolutely for every real argument. (The exponential series converges absolutely for every real argument).

[F6]

The sine and cosine defining power series converge absolutely everywhere. (The sine and cosine power series converge absolutely for every real argument).

[F7]

Products and substitutions by zero-constant inner series preserve convergent analyticity on a sufficiently small neighborhood. (Operations preserving coefficient majorisation).

Counterexample

1.1

Use $\sinh s = (e^{s} - e^{- s}) / 2$ . F1 and F2 give $(u_{k})_{t t} = k e^{- \sqrt{k}} \sinh (k t) \cos (k x)$ and $(u_{k})_{x x} = - k e^{- \sqrt{k}} \sinh (k t) \cos (k x)$ , so their sum is zero. At t=0, $\sinh 0 = 0$ and its derivative is one, giving $u_{k} (x, 0) = 0$ and $(u_{k})_{t} (x, 0) = e^{- \sqrt{k}} \cos (k x)$ . The exponential and trigonometric series make each u_k analytic.

givenF1F2algebraF5F6F7

2.1

For every nonnegative integer j, each x-derivative of order j of the normal data has modulus at most $k^{j} e^{- \sqrt{k}}$ by F1 and F3. With $z = \sqrt{k}$ this is $z^{2 j} e^{- z} \to 0$ by F4. The bound is uniform on the entire real line, hence on every compact interval. All value-data derivatives are already zero.

step 1.1F1F3F4

3.1

Fix t>0. Eventually $\sinh (k t) \geq e^{k t} / 4$ and $\sqrt{k} \leq k t / 2$ . Hence $u_{k} (0, t) \geq e^{k t / 2} / (4 k) \to \infty$ by F4. The zero data give the zero solution, whereas these data converge to zero in every displayed seminorm and their solution values diverge. This disproves the asserted continuous dependence even though the Laplace principal coefficient on u_tt is one and each function is analytic.

step 1.1step 2.1F4∎

Source notes

Ageno, §2.4.1 Hadamard example, PDF pp. 28–29. The factor exp(-sqrt(k)) is a local strengthening making all fixed derivative seminorms tend to zero.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

34 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

Ageno, Part III: Analysis of Partial Differential Equations (standard reference, not scraped)