A general Lipschitz uniqueness criterion for scalar ordinary differential equations

The classical Lipschitz-type criteria guarantee unique solvability of the scalar initial value problem ẋ = f (t, x), x(t0) = x0, by putting restrictions on | f (t, x)− f (t, y)| in dependence of |x− y|. Geometrically it means that the field differences are estimated in the direction of the x-axis. In 1989, Stettner and the second author could establish a generalized Lipschitz condition in both arguments by showing that the field differences can be measured in a suitably chosen direction v = (dt, dx), provided that it does not coincide with the directional vector (1, f (t0, x0)). Considering the vector v depending on t, a new general uniqueness result is derived and a short proof based on the implicit function theorem is developed. The advantage of the new criterion is shown by an example. A comparison with known results is given as well.


Introduction
We consider the scalar initial value problem and assume throughout the paper that f : D → R is a continuous function on an open neighborhood D of the point (t 0 , x 0 ) ∈ R 2 .Problem (1.1) is called locally uniquely solvable if there exists an open interval I containing t 0 such that (1.1) has exactly one solution on I.

A general Lipschitz uniqueness criterion
Theorem 2.1.Let v(t) = (ϕ(t), ψ(t)) be a continuously differentiable vector on an open neighborhood of t 0 with real entries ϕ and ψ such that whenever the arguments of f are well-defined and belong to D.
Proof.Peano's theorem guarantees that (1.1) has at least one solution x : [t 0 − α 0 , t 0 + α 0 ] → R for some α 0 > 0. By assumption (i) there exists α ∈ (0, α 0 ] with ψ(t) = f (t, x(t))ϕ(t) for all t ∈ (t 0 − α, t 0 + α).To prove that (1.1) is locally uniquely solvable with solution x on I := (t 0 − α, t 0 + α) assume to the contrary that there exists a solution y : I → R of (1.1) and x ≡ y on [t 0 , t 0 + α) (the case x ≡ y on (t 0 − α, t 0 ] is treated similarly).For x 1 by continuity and also We show that the equation is uniquely solvable with respect to k = k(t) on a subinterval of I.The problem suggests to apply the implicit function theorem.Let This function is defined in an open set containing (t 1 , 0) with the property we get with assumption (2.2) The implicit function theorem (cf., e.g., [8,Theorem 9.28]) now yields that there exists a unique continuously differentiable function k = k(t) on an open interval I 1 ⊂ I containing t 1 such that k(t 1 ) = 0 and F(t, k(t)) = 0 for all t ∈ I 1 .
We show that k(t) ≡ 0 on a subinterval of I 1 with t 1 ∈ I 1 .Due to (2.2), there exist a constant η > 0 and an open interval Moreover, there exists a constant M such that which is equivalent to Since u(t 1 ) = k 2 (t 1 ) = 0, we get u(t) = k 2 (t) ≡ 0 and hence from (2.3), x(t) ≡ y(t) on I 2 , which contradicts the definition of t 1 .

Concluding remarks and comparison with known results
The function k(t) in the proof of Theorem 2.1 measures in the case when v(t) is a unit vector the distance between the points (t, x(t)) and (t + k(t)ϕ(t), y(t + k(t)ϕ(t))) on the graphs of the solutions x and y because dist (t, x(t)), (t + k(t)ϕ(t), y(t By the specification v(t) = (ϕ(t), ψ(t)) = (0, 1) we get the well-known Lipschitz condition.The specification v(t) = (ϕ(t), ψ(t)) = (1, 0) yields the result by Mortici cited above.The latter case contains the following special uniqueness criterion which is given in [7].It was already known by Peano.Finally, the choice v(t) = (ϕ(t), ψ(t)) = (d t , d x ) turns our result into the following criterion published in German by Stettner and Nowak [9].Theorem 3.2.Let D be an open neighborhood of the point (t 0 , x 0 ) and f : D → R be continuous on D. Let d t , d x be real numbers such that i ) iii ) for a constant L ≥ 0 and every k ∈ R the inequality is satisfied whenever the arguments of f are in D.
Then (1.1) has at most one solution.
Now we illustrate the advantage of Theorem 2.1.

Example 1. Consider the initial value problem
where It is easy to check that f is not Lipschitz continuous with respect to x in any neighborhood of (0, 0), and the problem cannot be treated by Theorem 3.2 using a constant vector v = (d t , d x ).
Consider the theoretically possible cases and note that β) is impossible.Then condition (2) of the form where without loss of generality we can assume k = 0.

Corollary 3 . 1 .
If f : R → R + is continuous and positive then the equation ẋ = f (x) has uniqueness, i.e. exactly one solution passes through every point of R 2 .