CounterexampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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CFL endpoint can have linear growth

Physical scope

Pure mathematics; no physical premise.

Statement refuted

For the staggered leapfrog algorithm, replacing the strict condition $h ∥ K ∥_{2} < 2$ by $h ∥ K ∥_{2} \leq 2$ guarantees a constant $C < \infty$ , independent of the finite horizon $N$ and initial data, such that $\max_{0 \leq n \leq N} ∥ (e_{n}, b_{n - 1 / 2}) ∥_{2} \leq C ∥ (e_{0}, b_{- 1 / 2}) ∥_{2}$ , where the pair norm is Euclidean on $R^{n_{e} + n_{b}}$ .

Statement refuted references

CFL energy stability and finite termination

Staggered compatible leapfrog

Facts & Assumptions

Given: $n_{e} = n_{b} = 1$ , $M_{e} = M_{b} = C = [1]$ , $S = 0$ , $D = G = [0]$ , $h = 2$ , zero forcing, and an arbitrary finite horizon $N \geq 0$ .

[F1]

The staggered algorithm updates $b_{n + 1 / 2} = b_{n - 1 / 2} - h C e_{n}$ and $e_{n + 1} = e_{n} + h M_{e}^{- 1} C^{T} M_{b} b_{n + 1 / 2}$ (Staggered compatible leapfrog).

[F2]

The matrix in the strict CFL condition is $K = M_{b}^{1 / 2} C M_{e}^{- 1 / 2}$ and $∥ K ∥_{2}$ is its induced Euclidean operator norm (CFL energy stability and finite termination).

Counterexample

1.1givenF2algebra

These one-dimensional data satisfy the matrix assumptions of the algorithm. Their positive mass roots equal $1$ , so [F2] gives $K = [1]$ and $h ∥ K ∥_{2} = 2$ , exactly the endpoint in the refuted extension.

2.1givenF1step 1.1algebra

Write $z_{n} : = (e_{n}, b_{n - 1 / 2})^{T}$ . By [F1], $z_{n + 1} = A z_{n}$ with $A = (\begin{array}{cc} - 3 & 2 \\ - 2 & 1 \end{array})$ . Set $J : = A + I = (\begin{array}{cc} - 2 & 2 \\ - 2 & 2 \end{array})$ . Direct multiplication gives $J \neq 0$ and $J^{2} = 0$ , so $A = - (I - J)$ and the binomial formula yields $A^{n} = (- 1)^{n} (I - n J)$ for every integer $n \geq 0$ .

3.1givenstep 2.1algebra∎

For the nonzero initial state $z_{0} = (1, 0)^{T}$ , one has $J z_{0} = (- 2, - 2)^{T} \neq 0$ . The reverse triangle inequality therefore gives $∥ z_{n} ∥_{2} = ∥ z_{0} - n J z_{0} ∥_{2} \geq n ∥ J z_{0} ∥_{2} - ∥ z_{0} ∥_{2} = 2 \sqrt{2} n - 1$ . These outputs occur in the algorithm for each finite horizon $N \geq n$ and their norm ratio to $∥ z_{0} ∥_{2}$ is unbounded as $n$ grows. No constant $C$ independent of $N$ can satisfy the refuted estimate.

CFL endpoint can have linear growth

Statement refuted

Statement refuted references

Facts & Assumptions

Counterexample

Depends on

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