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Continuity of the force map alone does not give unique motions: two distinct Newtonian motions through one datum for $F = λ \sqrt{∣ x_{1} ∣ / ℓ_{*}} e_{1}$

Physical scope

A refutation, inside the framework's equation of motion, of the claim that a merely continuous force map suffices for uniqueness through an initial datum: the unit-mass model in three dimensions with force $F (x) = λ \sqrt{∣ x_{1} ∣ / ℓ_{*}} e_{1}$ , where $[λ] = μ ℓ τ^{- 2}$ and $[ℓ_{*}] = ℓ$ , is continuous but not state-locally-Lipschitz at the origin and exhibits two distinct $C^{2}$ motions through the rest datum. The item refutes only the weakened variant of the force postulate in which continuity replaces the local-Lipschitz clause; it asserts no motion of any physical body, no measurement and no empirical claim.

Statement refuted

Consider the framework's equation of motion with one label of unit mass ( $m = 1 k g$ , Particle labels, worldlines, the frame-time parameter condition and fixed inertial masses clause (v)), time window $I_{0} = R$ , configuration domain $Q = R^{3}$ , and a force map $F : R \times R^{3} \times R^{3} \to R^{3}$ with force components of dimension $μ ℓ τ^{- 2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework) that is merely continuous — that is, the standing hypotheses of clause (iii) of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with the state-local-Lipschitz requirement dropped. Then for every initial datum $(t_{0}, x_{0}, v_{0})$ there is at most one motion of the model through that datum, on every interval containing $t_{0}$ .

Facts & Assumptions

Given: The unit-mass model above and the datum $(t_{0}, x_{0}, v_{0}) = (0 s, 0, 0)$ .

[L1]

A motion of the model on an interval $J$ with nonempty interior is a $C^{2}$ curve $x : J \to R^{3}$ with $\ddot{x} (t) = F (t, x (t), \dot{x} (t))$ for all $t \in J$ (The associated first-order system, motions, maximal motions and the Newtonian flow of a force model clause (c) with $m = 1$ and $N = 1$ ); differentiability and second differentiability are those of 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, and an interval is that of Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length.

[L2]

The map $u \mapsto ∣ u ∣$ is continuous on $R$ (Absolute value in an ordered field) and the map $s \mapsto \sqrt{s}$ is continuous on $[0, \infty)$ as the continuous inverse of squaring on that set (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 $f$ ); compositions, sums, scalar multiples and coordinate projections are continuous (A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs, 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, 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).

[L3]

Derivative rules: $(t^{n})^{'} = n t^{n - 1}$ on $R$ for $n \geq 1$ (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); sums and scalar multiples are differentiated termwise (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 $g (c) \neq 0$ ); a difference quotient is read through the definition of the derivative (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).

[L4]

Units (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework): position has dimension $ℓ$ , time has dimension $τ$ , mass has dimension $μ$ , and force has dimension $μ ℓ τ^{- 2}$ ; thus $λ \sqrt{∣ x_{1} ∣ / ℓ_{*}}$ has force dimension when $[λ] = μ ℓ τ^{- 2}$ and $[ℓ_{*}] = ℓ$ .

Counterexample

Proof technique: direct; exhibit two distinct $C^{2}$ motions through the rest datum for the continuous force map $F (t, x, v) = (λ \sqrt{∣ x_{1} ∣ / ℓ_{*}}, 0, 0),$ where $λ > 0$ is a declared force scale and $ℓ_{*} > 0$ is a declared length (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework).

1.1givenL2L4

The map $F (t, x, v) = (λ \sqrt{∣ x_{1} ∣ / ℓ_{*}}, 0, 0)$ is continuous on $R \times R^{3} \times R^{3}$ : its first component is the composition of the continuous coordinate map $x \mapsto x_{1}$ , absolute value, division by the positive constant $ℓ_{*}$ , and the square root on $[0, \infty)$ ([L2]); its other components are constant. By [L4] its force components have dimension $μ ℓ τ^{- 2}$ , so it is a dimensionally consistent continuous force map for the weakened model.

1.2givenL1L3

The zero curve $x_{0} (t) : = (0, 0, 0)$ is a motion through the datum on $R$ : it is $C^{2}$ with ${\ddot{x}}_{0} = 0$ and $F (t, x_{0} (t), {\dot{x}}_{0} (t)) = (0, 0, 0)$ .

1.3givenL1L3L4algebra

Let $t_{1} : = 1 s$ and $A : = (λ / (12 m \sqrt{ℓ_{*}}))^{2} > 0$ . Define $x_{1} : R \to R^{3}$ by $x_{1} (t) : = 0$ for $t \leq t_{1}$ and $x_{1} (t) : = (A (t - t_{1})^{4}, 0, 0)$ for $t \geq t_{1}$ . The coefficient $A$ has dimension $ℓ τ^{- 4}$ , so $A (t - t_{1})^{4}$ has dimension $ℓ$ . For $t > t_{1}$ the first component has derivatives $4 A (t - t_{1})^{3}$ and $12 A (t - t_{1})^{2}$ by [L3]. At $t = t_{1}$ its first difference quotient is $A h^{3} \to 0$ from the right and $0$ from the left; the difference quotient of its first derivative is $4 A h^{2} \to 0$ from the right and $0$ from the left. Thus $x_{1}$ is $C^{2}$ on $R$ , with $x_{1} (0 s) = 0$ and ${\dot{x}}_{1} (0 s) = 0$ .

2.1step 1.3L1algebra

$x_{1}$ satisfies the equation: for $t < t_{1}$ both sides vanish; for $t > t_{1}$ its first acceleration component is $12 A (t - t_{1})^{2}$ , while the first component of $F (t, x_{1} (t), {\dot{x}}_{1} (t))$ is $λ \sqrt{A / ℓ_{*}} (t - t_{1})^{2}$ . The definition of $A$ gives $12 m A = λ \sqrt{A / ℓ_{*}}$ , so $m {\ddot{x}}_{1} = F$ ; at $t = t_{1}$ both sides vanish. Hence $x_{1}$ is a motion on $R$ through the datum by [L1].

3.1step 1.2step 2.1algebra∎

The two motions differ at $t_{2} : = 2 s$ : since $t_{2} - t_{1} = 1 s$ , $x_{1} (t_{2}) = A (1 s)^{4} e_{1} \neq 0 = x_{0} (t_{2})$ . Thus the initial value problem for the continuous force map $F$ has two distinct solutions on every interval containing $[0 s, 2 s]$ , and the refuted statement is false.

Remarks

Where the Lipschitz clause fails. For $u \neq 0$ the quotient of the force difference by the state difference at the origin is $\frac{∥ F (t, u e_{1}, 0) - F (t, 0, 0) ∥_{2}}{∥ (u e_{1}, 0) - (0, 0) ∥_{2}} = \frac{λ \sqrt{∣ u ∣ / ℓ_{*}}}{∣ u ∣} = \frac{λ}{\sqrt{ℓ_{*}} \sqrt{∣ u ∣}},$ which is unbounded as $u \to 0$ ; so $F$ admits no finite state-Lipschitz constant on any cylinder containing the rest datum (Local Lipschitz continuity in the state variable, locally uniform in time and parameters). This is exactly the clause that Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance imposes and that the counterexample drops.
The framework theorem is untouched. Since $F$ is not a force map of the adopted branch, Local existence and uniqueness for the Newtonian initial-value problem, with the quantitative Picard-Lindelöf cylinder does not apply to it; the counterexample refutes only the claim that continuity alone suffices for uniqueness, and thereby shows that the local-Lipschitz clause of the force postulate is not redundant.
No physical claim. The counterexample is an abstract computation inside the model's equation of motion; the force map $F$ is not asserted to model any physical interaction.

Continuity of the force map alone does not give unique motions: two distinct Newtonian motions through one datum for $F = λ \sqrt{∣ x_{1} ∣ / ℓ_{*}} e_{1}$

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

Continuity of the force map alone does not give unique motions: two distinct Newtonian motions through one datum for F=λ∣x1∣/ℓ∗ e1F=λ∣x1​∣/ℓ∗​​e1​

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

Continuity of the force map alone does not give unique motions: two distinct Newtonian motions through one datum for $F = λ \sqrt{∣ x_{1} ∣ / ℓ_{*}} e_{1}$