CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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Backward heat amplifies small high-frequency errors

Statement refuted

Assume Countable Choice. The claim refuted is that the backward heat problem on $(0, π)$ depends continuously on its terminal data in the $L^{2} (0, π)$ norm: arbitrarily small terminal perturbations do not, in general, produce small perturbations of the initial state.

Facts & Assumptions

Given: Countable Choice and $T > 0$ and, for every integer $k \geq 1$ , the $k$ -th decaying sine mode ${\tilde{u}}_{k} (x, t) = e^{- k^{2} t} \sin (k x)$ on $(0, π)$ .

[F1]

Each ${\tilde{u}}_{k}$ is a classical solution of $u_{t} = u_{x x}$ on $(0, π) \times (0, T]$ with zero Dirichlet data, smooth up to $t = 0$ , with amplification factor $e^{k^{2} T}$ between terminal and initial data (Sine modes decay under Dirichlet heat flow, The backward heat solution map is unbounded).

[F2]

The $L^{2} (0, π)$ inner product is $⟨ f, g ⟩ = \int_{0}^{π} f g$ on the quotient space of The space $L^{p} (μ)$ as the quotient by null functions ( $L^{2}$ with the integral pairing is a Hilbert space), and $∥ \sin (k \cdot) ∥_{2} = \sqrt{π / 2}$ (L2 normalisation of the sine modes on an interval); the derivative identities for sine are those of The derivatives of sine and cosine are cosine and minus sine.

[F3]

$e^{- k^{2} T} \to 0$ faster than every polynomial as $k \to \infty$ (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ).

Counterexample

Given: Countable Choice and $T > 0$ and the modes ${\tilde{u}}_{k} (x, t) = e^{- k^{2} t} \sin (k x)$ .

1.1F1F2given

For every $k$ the function ${\tilde{u}}_{k}$ is a classical solution of $u_{t} = u_{x x}$ on $(0, π) \times (0, T]$ with ${\tilde{u}}_{k} (0, t) = {\tilde{u}}_{k} (π, t) = 0$ , terminal data $g_{k} (x) : = e^{- k^{2} T} \sin (k x)$ and initial state ${\tilde{u}}_{k} (x, 0) = \sin (k x)$ .

2.1step 1.1F1F2given

By [F2] the terminal data have norm $∥ g_{k} ∥_{2} = e^{- k^{2} T} ∥ \sin (k \cdot) ∥_{2} = \sqrt{π / 2} e^{- k^{2} T}$ , while the initial states satisfy $∥ {\tilde{u}}_{k} (\cdot, 0) ∥_{2} = ∥ \sin (k \cdot) ∥_{2} = \sqrt{π / 2}$ for every $k$ ; moreover ${\tilde{u}}_{k} (\cdot, 0) = e^{k^{2} T} g_{k}$ , so the terminal-to-initial amplification factor of the $k$ -th mode is exactly $e^{k^{2} T}$ .

3.1step 2.1F3given∎

By [F3] the terminal norms $∥ g_{k} ∥_{2} = \sqrt{π / 2} e^{- k^{2} T}$ tend to $0$ while the initial norms remain the fixed constant $\sqrt{π / 2}$ , so terminal perturbations of arbitrarily small $L^{2}$ norm come from solutions whose initial states have order-one norm; this refutes continuous dependence on the terminal data, and the amplification is a failure of stability rather than of existence (the solution pair is explicit for every $k$ ).

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Sources

John K. Hunter, Notes on Partial Differential Equations (revised 18 June 2014, UC Davis) (standard reference, not scraped)
Victor Ivrii, Partial Differential Equations (University of Toronto, 2018, CC BY-SA) (standard reference, not scraped)