CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-21

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An almost-Lipschitz vector field has a unique solution through zero but is not locally Lipschitz there

Statement refuted

Local Lipschitz continuity is necessary for uniqueness through an initial point. Define $f (0) = 0$ and, for $0 < ∣ x ∣ < e^{- 1}$ ,

$f (x) = x (1 + ∣ \log ∣ x ∣ ∣),$

with any continuous extension outside that interval. An almost-Lipschitz vector field has a unique solution through zero but is not locally Lipschitz there.

Facts & Assumptions

Given: The displayed field and the zero IVP $y^{'} = f (y)$ , $y (0) = 0$ .

[L1]

The Osgood divergence condition gives uniqueness of solutions through the same initial value (Osgood's criterion gives uniqueness without a Lipschitz bound).

[L2]

For $x > 0$ , $\log^{'} (x) = 1 / x$ and $\log x = \int_{1}^{x} d t / t$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

[L3]

An Osgood modulus is positive away from zero, nondecreasing, and has a divergent reciprocal integral at zero (Moduli of continuity and the Osgood divergence condition).

Counterexample

technique · direct

1.1givenL2algebra

The quotient $∣ f (x) - f (0) ∣ / ∣ x ∣ = 1 + ∣ \log ∣ x ∣ ∣$ is unbounded as $x \to 0$ by [L2], so $f$ is not locally Lipschitz at zero.

2.1step 1.1L1L2L3algebra∎

Define $ϕ (r) = r (1 - \log r)$ for $0 < r \leq e^{- 1}$ with $ϕ (0) = 0$ , and put $ρ = 2 ϕ$ on $[0, e^{- 1}]$ . Differentiation using [L2] shows that $ϕ$ is increasing and concave on this interval. On one side of zero, concavity with $ϕ (0) = 0$ gives $∣ ϕ (a) - ϕ (b) ∣ \leq ϕ (∣ a - b ∣)$ ; on opposite sides, $∣ a - b ∣ = ∣ a ∣ + ∣ b ∣$ and monotonicity gives $ϕ (∣ a ∣) + ϕ (∣ b ∣) \leq 2 ϕ (∣ a - b ∣)$ . Since $f$ is the odd extension of $ϕ$ near zero, $ρ$ is therefore a state modulus there. It has the properties in [L3], and the substitution $u = 1 - \log r$ gives its reciprocal divergence, so [L1] makes the zero solution unique through the origin.

Depends on

Used by

False: local Lipschitz continuity is necessary for uniqueness of an ODE solution False statement

Dependency tree · two levels

22 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

Gerald Teschl, Ordinary Differential Equations and Dynamical Systems, Ch. 2 (standard reference, not scraped)
Jiri Lebl, Basic Analysis I, Section 6.3 (standard reference, not scraped)