TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Conditional continuous residual-to-error certificate

Statement

Assume Countable Choice. Let $d \in {2, 3}$ , $T > 0$ , and $Q_{d} = R^{d} / (2 π Z)^{d}$ with normalized measure $d μ = (2 π)^{- d} d x$ . Use $∥ w ∥_{2} = (\int_{Q_{d}} ∣ w ∣^{2} d μ)^{1 / 2}$ and let $P$ act on vector Fourier coefficients by $\hat{P w} (0) = \hat{w} (0)$ and $\hat{P w} (k) = (I - k k^{T} / ∣ k ∣^{2}) \hat{w} (k)$ for $k \neq 0$ . For a smooth spatial vector field $w : Q_{d} \to R^{d}$ , define $div w (x) = \sum_{j} \partial_{j} w_{j} (x)$ ; for a smooth time-dependent vector field $w : [0, T] \times Q_{d} \to R^{d}$ , define ${div}_{x} w (t, x) = \sum_{j} \partial_{j} w_{j} (t, x)$ , with derivatives taken only in $x$ . Here divergence-free means $div w = 0$ at every $x$ for a spatial field and ${div}_{x} w = 0$ at every $(t, x)$ for a time-dependent field. For a tensor field $F$ , define $(div F)_{i} = \sum_{j} \partial_{j} F_{i j}$ ; define $(w \otimes z)_{i j} = w_{i} z_{j}$ and $Δ = \sum_{j} \partial_{j}^{2}$ . Let $ν \geq 0$ , let $f \in C^{\infty} ([0, T] \times Q_{d}; R^{d})$ be arbitrary, with no divergence condition imposed on $f$ , let $u_{0} \in C^{\infty} (Q_{d}; R^{d})$ satisfy $div u_{0} = 0$ on $Q_{d}$ , and let $u, v \in C^{\infty} ([0, T] \times Q_{d}; R^{d})$ satisfy ${div}_{x} u = {div}_{x} v = 0$ on $[0, T] \times Q_{d}$ . Thus only $u_{0}, u, v$ are required to be divergence-free. Suppose $u (0) = u_{0}$ and

$\partial_{t} u + P div (u \otimes u) - ν Δ u = P f,$

and define $r_{v} = \partial_{t} v + P div (v \otimes v) - ν Δ v - P f$ . For a smooth time-dependent vector field $b$ , define its spatial Jacobian by $(\nabla_{x} b (t, x))_{i j} = \partial_{j} b_{i} (t, x)$ , and for $a \in R^{d}$ define $(a \cdot \nabla_{x}) b = (\nabla_{x} b) a$ . For a smooth scalar field $ϕ : [0, T] \times Q_{d} \to R$ , let $\nabla_{x} ϕ = (\partial_{j} ϕ)_{j = 1}^{d}$ . For $M \in R^{d \times d}$ , define $∥ M ∥_{o p} = \sup_{z \in R^{d}, ∣ z ∣ = 1} ∣ M z ∣$ , using the Euclidean norm. Set $A (t) = \int_{0}^{t} \sup_{x \in Q_{d}} ∥ \nabla_{x} u (s, x) ∥_{o p} d s$ . Then for every $t \in [0, T]$ ,

$∥ v (t) - u (t) ∥_{2} \leq e^{A (t)} (∥ v (0) - u_{0} ∥_{2} + \int_{0}^{t} e^{- A (s)} ∥ r_{v} (s) ∥_{2} d s) .$

This is a conditional estimate for an already given smooth exact solution; it proves no existence theorem and gives no error guarantee from finitely many collocation residual values alone. All quantities are mathematical, with no physical units assigned to $ν$ .

Statement references

Tanveer, Energy Estimate, Uniqueness and ν dependence of Smooth Solutions, Ohio State lecture notes.

Carlo Morosi and Livio Pizzocchero, On approximate solutions of the incompressible Euler and Navier-Stokes equations.

Facts & Assumptions

Given: The torus, Fourier multiplier, smooth fields, common forcing and viscosity, exact equation, and residual convention stated in the Statement and in the continuous periodic solenoidal surrogate residual Definition.

[A1]

$u, v$ are smooth and divergence-free on the compact periodic domain, $ν \geq 0$ , $r_{v} \in C ([0, T]; L^{2})$ , and $u (0) = u_{0}$ .

[F1]

Continuous periodic solenoidal surrogate residual: $r_{v} = \partial_{t} v + P div (v \otimes v) - ν Δ v - P f$ , with the same normalized $L^{2}$ norm and tensor-divergence convention used here.

[F2]

Periodic Leray and pressure maps: $P$ is self-adjoint as an orthogonal Fourier projection, commutes with derivatives, and fixes divergence-free fields mode by mode.

[F3]

Holder's inequality for integrals, including the endpoint cases: for measurable $a, b \in L^{2} (Q_{d})$ , $\int_{Q_{d}} ∣ a b ∣ d μ \leq ∥ a ∥_{2} ∥ b ∥_{2}$ .

[F4]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every sequence of nonempty sets has a choice function; this is the explicit choice premise carried by the periodic Fourier multiplier framework.

Proof

Given: The hypotheses in [A1], the common forcing and viscosity, and the supplier conventions in [F1]–[F4].

1.1givenA1F1F4algebra

Set $e = v - u$ . Under the explicit Countable Choice convention [F4], subtracting the exact equation from the residual equation [F1] gives $\partial_{t} e + P div (v \otimes v - u \otimes u) - ν Δ e = r_{v}$ . Since ${div}_{x} u = {div}_{x} v = 0$ , the tensor convention gives $div (v \otimes v) = (v \cdot \nabla_{x}) v$ and $div (u \otimes u) = (u \cdot \nabla_{x}) u$ ; the difference is $(v \cdot \nabla_{x}) e + (e \cdot \nabla_{x}) u$ .

2.1givenA1F2F3step 1.1algebra

Define the real $L^{2}$ pairing by $⟨ a, b ⟩_{2} = \int_{Q_{d}} a (x) \cdot b (x) d μ (x)$ for $a, b \in L^{2} (Q_{d}; R^{d})$ . Pair the resulting equation with $e$ in this pairing. By [F2], $P e = e$ and $P$ is self-adjoint, so each projected transport term paired with $e$ equals its unprojected pairing. Periodicity gives $\int_{Q_{d}} {div}_{x} (v ∣ e ∣^{2}) d μ = 0$ ; since ${div}_{x} v = 0$ , this yields $⟨ (v \cdot \nabla_{x}) e, e ⟩_{2} = \frac{1}{2} \int v \cdot \nabla_{x} ∣ e ∣^{2} d μ = 0$ . The Jacobian convention in the Statement gives $(e \cdot \nabla_{x}) u = (\nabla_{x} u) e$ , so $∣ (e \cdot \nabla_{x}) u ∣ \leq \sup_{x} ∥ \nabla_{x} u ∥_{o p} ∣ e ∣$ pointwise and $∣ ⟨ (e \cdot \nabla_{x}) u, e ⟩_{2} ∣ \leq \sup_{x} ∥ \nabla_{x} u ∥_{o p} ∥ e ∥_{2}^{2}$ . Periodic integration by parts gives $⟨ - ν Δ e, e ⟩_{2} = ν \sum_{i, j} \int_{Q_{d}} ∣ \partial_{j} e_{i} ∣^{2} d μ \geq 0$ . Pointwise Euclidean Cauchy–Schwarz gives $∣ r_{v} \cdot e ∣ \leq ∣ r_{v} ∣ ∣ e ∣$ ; the real scalar functions $∣ r_{v} ∣$ and $∣ e ∣$ belong to $L^{2} (Q_{d})$ , so [F3] applied to them gives $∣ ⟨ r_{v}, e ⟩_{2} ∣ \leq \int_{Q_{d}} ∣ r_{v} ∣ ∣ e ∣ d μ \leq ∥ r_{v} ∥_{2} ∥ e ∥_{2}$ . Using $⟨ \partial_{t} e, e ⟩_{2} = \frac{1}{2} \frac{d}{d t} ∥ e ∥_{2}^{2}$ now gives $\frac{1}{2} \frac{d}{d t} ∥ e ∥_{2}^{2} \leq \sup_{x} ∥ \nabla_{x} u ∥_{o p} ∥ e ∥_{2}^{2} + ∥ r_{v} ∥_{2} ∥ e ∥_{2}$ .

3.1A1step 2.1algebra

For $δ > 0$ put $y_{δ} (t) = (∥ e (t) ∥_{2}^{2} + δ^{2})^{1 / 2}$ and $a (t) = \sup_{x} ∥ \nabla_{x} u (t, x) ∥_{o p}$ . Continuity of $\nabla_{x} u$ on compact $[0, T] \times Q_{d}$ makes $a$ continuous, so the definition of $A$ gives $A^{'} (t) = a (t)$ . The differential inequality in step 2.1 and the bounds $∥ e ∥_{2}^{2} / y_{δ} \leq y_{δ}$ and $∥ e ∥_{2} / y_{δ} \leq 1$ imply $y_{δ}^{'} (t) \leq A^{'} (t) y_{δ} (t) + ∥ r_{v} (t) ∥_{2}$ .

4.1step 3.1algebra∎

The product rule and $A^{'} (t) = \sup_{x} ∥ \nabla_{x} u (t, x) ∥_{o p}$ show that $\frac{d}{d t} (e^{- A (t)} y_{δ} (t)) \leq e^{- A (t)} ∥ r_{v} (t) ∥_{2}$ . Integrating from $0$ to $t$ and multiplying by $e^{A (t)}$ gives $y_{δ} (t) \leq e^{A (t)} y_{δ} (0) + \int_{0}^{t} e^{A (t) - A (s)} ∥ r_{v} (s) ∥_{2} d s$ . As $δ ↓ 0$ , $y_{δ} (t) \to ∥ e (t) ∥_{2}$ and $y_{δ} (0) \to ∥ v (0) - u_{0} ∥_{2}$ , which proves the stated estimate.

Conditional continuous residual-to-error certificate

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources