CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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$y^{'} = 2 \sqrt{∣ y ∣}$ has a continuum of delayed-start solutions through the origin

Statement refuted

Continuity of the right-hand side of a first-order IVP is enough for uniqueness. The continuous field $f (y) = 2 \sqrt{∣ y ∣}$ refutes this: $y^{'} = 2 \sqrt{∣ y ∣}$ has distinct delayed-start solutions through the origin.

Facts & Assumptions

Given: For each $c \geq 0$ , define $y_{c} (t) = 0$ for every $t \leq c$ and $y_{c} (t) = (t - c)^{2}$ for $t \geq c$ .

[L1]

For rational $r > 0$ , $0^{r} = 0$ (Rational powers $a^{r}$ of a positive base).

[L2]

Local state-Lipschitz continuity requires one finite constant bounding the state difference quotient near the point (Local Lipschitz continuity in the state variable, locally uniform in time and parameters).

Counterexample

technique · direct

1.1

On the first piece $y_{c}^{'} = 0 = 2 \sqrt{∣ y_{c} ∣}$ by [L1], on the second $y_{c}^{'} = 2 (t - c) = 2 \sqrt{(t - c)^{2}}$ , and at $t = c$ both one-sided derivatives are $0$ , so every $y_{c}$ is a solution through $(0, 0)$ .

givenL1algebra

2.1

Distinct delays give distinct solutions, while $∣ f (y) - f (0) ∣ / ∣ y ∣ = 2 / \sqrt{y}$ for $y > 0$ , so [L2] rules out every finite local Lipschitz constant at zero.

step 1.1L2algebra∎

Depends on

Used by

False: continuity of the right-hand side guarantees unique ODE solutions False statement

Dependency tree · two levels

24 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)

y′=2∣y∣y′=2∣y∣​ has a continuum of delayed-start solutions through the origin