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Continuity alone does not give uniqueness: $x^{'} = \sqrt{∣ x ∣}$ through the origin

Statement refuted

Let $F : R \to R$ be continuous. Then for every $x_{0} \in R$ the initial value problem $x^{'} = F (x)$ , $x (0) = x_{0}$ has at most one solution on every interval containing $0$ .

Facts & Assumptions

Given: The field $F (x) = \sqrt{∣ x ∣}$ and the initial value problem $x^{'} = F (x)$ , $x (0) = 0$ .

[L1]

A solution on an interval $I$ is a differentiable $x : I \to R$ with $x^{'} (t) = F (x (t))$ for every $t \in I$ and $x (0) = 0$ (First-order systems, initial value problems, and solutions on intervals, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set). [given]

[L2]

$u \mapsto ∣ u ∣$ and $u \mapsto \sqrt{u}$ (on $[0, \infty)$ ) are continuous, and sums, products and compositions of continuous functions are continuous; squaring is continuous by the polynomial clause of Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, injective on $[0, \infty)$ by Squaring is monotone on the nonnegatives, and onto $[0, \infty)$ by Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , so $\sqrt{\cdot}$ is its continuous inverse on $[0, \infty)$ (Absolute value in an ordered field, Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ , Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point). [given]

[L3]

$(u^{2})^{'} = 2 u$ , and the chain rule and algebra of derivatives differentiate compositions, products and sums (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ). [given]

Counterexample

technique · direct; exhibit two distinct solutions through the same initial datum

1.1givenL2

The map $F (x) = \sqrt{∣ x ∣}$ is continuous on $R$ as the composition of $u \mapsto ∣ u ∣$ with the square root, by [L2], so the hypothesis of the refuted statement holds.

1.2givenL1

The zero function $x_{0} \equiv 0$ is a solution: it is differentiable with derivative $0 = F (0)$ everywhere.

1.3givenL3algebra

Define $x_{1} : R \to R$ by $x_{1} (t) = 0$ for $t \leq 1$ and $x_{1} (t) = (t - 1)^{2} / 4$ for $t \geq 1$ . Then $x_{1}$ is continuous, and it is differentiable at $t = 1$ with derivative $0$ from both sides: for $h > 0$ , $(h^{2} / 4) / h = h / 4 \to 0$ , while the left difference quotient is $0$ . For $t \neq 1$ the derivative is $0$ (for $t < 1$ ) or $(t - 1) / 2$ (for $t > 1$ ) by [L3]. Hence $x_{1}$ is differentiable on $R$ with $x_{1} (0) = 0$ .

2.1step 1.2step 1.3L1L2algebra∎

For $t > 1$ , $x_{1} (t) = (t - 1)^{2} / 4 \geq 0$ and $(t - 1) / 2 > 0$ , so uniqueness of the nonnegative square root in [L2] gives $\sqrt{∣ x_{1} (t) ∣} = (t - 1) / 2 = x_{1}^{'} (t)$ ; for $t \leq 1$ , $x_{1} (t) = 0$ and $x_{1}^{'} (t) = 0 = \sqrt{∣ 0 ∣}$ . Hence $x_{1}$ is a solution of $x^{'} = F (x)$ through $(0, 0)$ by [L1], and it differs from the zero solution on $(1, \infty)$ : the initial value problem has two distinct solutions on every interval containing $[0, 2]$ , refuting the statement.

Remarks

Why the Lipschitz hypothesis matters. The local uniqueness lemma Continuous and Lipschitz dependence on the initial condition on a Picard-Lindelof cylinder assumes a state-Lipschitz constant; here $F$ fails that hypothesis at the equilibrium: $∣ F (x) - F (0) ∣ / ∣ x ∣ = 1 / \sqrt{∣ x ∣} \to \infty$ as $x \to 0$ . The counterexample shows that continuity alone cannot replace it, so the Lipschitz hypothesis in the oscillator existence and uniqueness results is not redundant.
No physical claim. This is a statement about an abstract scalar ODE; it makes no assertion about any mechanical model.

Continuity alone does not give uniqueness: $x^{'} = \sqrt{∣ x ∣}$ through the origin

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

Continuity alone does not give uniqueness: x′=∣x∣x′=∣x∣​ through the origin

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

Continuity alone does not give uniqueness: $x^{'} = \sqrt{∣ x ∣}$ through the origin