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Osgood's criterion gives uniqueness without a Lipschitz bound

Statement

Suppose a continuous vector field has a state modulus $ρ$ satisfying the Osgood divergence condition on a neighborhood of two solution graphs. Then the Osgood divergence condition gives uniqueness of solutions through the same initial value, on both sides of the initial time.

The Osgood divergence condition gives uniqueness of solutions through the same initial value.

Facts & Assumptions

Given: Two solutions $x, y$ through the same initial value and an Osgood state modulus $ρ$ .

[L1]

The Euclidean inner product satisfies $⟨ z, z ⟩ \geq 0$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L2]

Cauchy-Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

[L3]

The chain rule differentiates a differentiable composite without dividing by the inner increment (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L4]

The Osgood condition is the divergence of $\int_{0^{+}} d r / ρ (r)$ for a positive modulus away from zero (Moduli of continuity and the Osgood divergence condition).

[L5]

An integrable derivative satisfies the endpoint-increment formula (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[L6]

The integral function of a continuous function on a nondegenerate interval is a primitive of that function (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

Proof

technique · contradiction

1.1

Put $R (t) = ∥ x (t) - y (t) ∥_{2}^{2}$ ; differentiation by [L3], [L1], [L2], and the modulus estimate give $∣ R^{'} (t) ∣ \leq 2 \sqrt{R (t)} ρ (\sqrt{R (t)})$ .

givenL1L2L3

2.1

Suppose, for contradiction, that $R$ becomes positive after the initial time, and choose a time $t$ before $\sqrt{R}$ can leave the modulus neighborhood. For $ε > 0$ , put $q_{ε} = \sqrt{R + ε}$ . Then step 1.1 and monotonicity of $ρ$ give $q_{ε}^{'} \leq ρ (q_{ε})$ .

step 1.1L3assume-contraalgebra

3.1

By [L6], $H_{ε} (q) = \int_{\sqrt{ε}}^{q} d r / ρ (r)$ has derivative $1 / ρ (q)$ on the positive interval in use. Apply [L3] and [L5]: since $q_{ε} (t_{0}) = \sqrt{ε}$ , step 2.1 gives $\int_{\sqrt{ε}}^{\sqrt{R (t) + ε}} d r / ρ (r) \leq t - t_{0}$ . As $ε ↓ 0$ , the left side diverges by [L4] because $R (t) > 0$ , a contradiction. Reflection proves the backward direction, and repeating the local argument at every agreement time gives uniqueness on the whole common interval.

step 2.1L3L4L5L6discharge-contradiction∎

Depends on

Used by

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

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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)