A Lipschitz condition along a transversal foliation implies local uniqueness for ODEs

We prove the following result: if a continuous vector field $F$ is Lipschitz when restricted to the hypersurfaces determined by a suitable foliation and a transversal condition is satisfied at the initial condition, then $F$ determines a locally unique integral curve. We also present some illustrative examples and sufficient conditions in order to apply our main result.


Introduction
Uniqueness for ODEs is an important and quite old subject, but still an active field of research [7][8][9], being Lipschitz uniqueness theorem the cornerstone on the topic. Besides the existence of many generalizations of that theorem, see [1,6,10], one recent and fruitful line of research has been the searching for alternative or weaker forms of the Lipschitz condition. For instance, let U ⊂ 2 be an open neighborhood of (t 0 , x 0 ) and f : U ⊂ 2 → be continuous and consider the scalar initial value problem x (t) = f (t, x(t)), x(t 0 ) = x 0 . (1.1) • f (t 0 , x 0 ) = 0.
A more general result had been proved before by Stettner and Nowak, [14], but in a paper restricted to German readers. They proved that if U ⊂ 2 is an open neighborhood of (t 0 , x 0 ), f : U ⊂ 2 → is continuous and (u 1 , u 2 ) ∈ 2 such that • u 2 = f (t 0 , x 0 )u 1 , then the scalar problem (1.1) has a unique local solution. By taking either (u 1 , u 2 ) = (0, 1) or (u 1 , u 2 ) = (1, 0) this result covers both the classical Lipschitz uniqueness theorem and the previous alternative version. Moreover this result has been remarkably generalized in [8] by Diblík, Nowak and Siegmund by allowing the vector (u 1 , u 2 ) to depend on t.
The following alternative version of Lipschiz uniqueness theorem for systems has been proved by Cid in [3].
Recently, Diblík, Nowak and Siegmund have obtained in [13] a generalization of both [3] and [14]. Their result reads as follows: The previous theorem has the following geometric meaning: uniqueness for the autonomous system (1.2) follows provided that the continuous vector field F is Lipschitz when restricted to a family of parallel hyperplanes to that covers U and that the vector field at the initial condition F (p 0 ) is transversal to .
Our main goal in this paper is to extend Theorem 1.3 from the linear foliation generated by the hyperplane to a general n-foliation. The paper is organized as follows: in Section 2 we present our main result which relies on an appropriate change of coordinates and Theorem 1.1. We will show by examples that our result is in fact a meaningful generalization of Theorem 1.3. In Section 3 we present some useful results about Lipschitz functions, including the definition of a modulus of Lipschitz continuity along a hyperplane that will be used in Section 4 for obtaining explicit sufficient conditions on F for the existence of a suitable n-foliation. Another key ingredient for that result shall be a general rotation formula proved too at Section 4.
Through the paper 〈·, ·〉 shall denote the usual scalar product in the Euclidean space. with 0 ∈ J and a family of differentiable functions {g s : V → U} s∈J such that g 0 (0) = p 0 ∈ U and Φ : (s, y) ∈ J × V → g s ( y) ∈ U is a diffeomorphism. Then we say {g s } s∈J is a local n-foliation of U at p 0 . Remark 2.2. An observation regarding notation. If Φ : n+1 → n+1 is a diffeomorphism, we denote by Φ its derivative and by Φ −1 its inverse. Also, we write (Φ −1 ) for the derivative of the inverse. Observe that Φ takes values in n+1 ( ) so, although we cannot consider the functional inverse of Φ , we can consider the inverse matrix, whenever it exists, of every Φ (x) for x ∈ n+1 . We denote this function by (Φ ) −1 . Clearly, the chain rule implies that The following is our main result. • (C1) Transversality condition: Since {g s } s∈J is a foliation, Φ is a diffeomorphism. Then, considering y = (s, y 1 , . . . , y n ), differentiating (2.2) with respect to t and taking into account equation (1.2), By definition of g s , Φ(0) = p 0 , so we can consider the problem Now, by (C2) we have that h is the product of locally Lipschitz functions when fixing the first variable. Furthermore, if e 1 = (1, 0, . . . , 0) ∈ n and taking into account (C1), Hence, we can apply Theorem 1.1 to problem (2.4) and conclude that problem (1.2) has, locally, a unique solution.

Remark 2.4. 1) Condition (2.1) can be easily interpreted geometrically: the vector
is normal to the hypersurface given by g 0 (V ) at p 0 . So, condition (2.1) means that the vector F (p 0 ) is not tangent to that hypersurface, and therefore it is called the 'transversality condition'.
2) Notice that from [3, Example 3.1] we know that if the transversality condition (2.1) does not hold then the Lipschitz condition along the foliation, that is (C2), is not enough to ensure uniqueness. On the other hand, by [3, Example 3.4] we also know that (C1) and a Lipschitz condition along a local (n-1)-foliation do not imply uniqueness. So, in some sense, conditions (C1) and (C2) are sharp. [13], where only foliations consisting of hyperplanes are considered. In the next example we show the limitations of linear (or affine) coordinate changes which are used in [13].

Theorem 2.3 generalizes the main result in
Is there a linear change of coordinates Φ such that F • Φ is Lipschitz in a neighborhood of zero when fixing the first variable? The answer is no. Any linear change of variables Φ will be given by two linearly independent vectors v, w ∈ 2 as Φ(z, t) = zw + t v. If F • Φ is Lipschitz in a neighborhood of zero when fixing the first variable, that is, z, that implies that the directional derivative of F at any point of the neighborhood in the direction of v, whenever it exists, is a lower bound for any Lipschitz constant. To see that this cannot happen, take S = {(x, y) ∈ 2 : y = x 2 } and realize that F is differentiable in 2 Now consider a neighborhood N of 0. In particular, we can consider the points of the form This quantity is unbounded in N \S unless the numerator is 0 for every λ ∈ (−ε, ε), but that means that v = 0, so v and w cannot be linearly independent. Hence, no linear change of coordinates Φ makes F • Φ Lipschitz in a neighborhood of zero when fixing the first variable.
, which is clearly Lipschitz when fixing the first variable.
In the figure you can see the parabolas g z (t) foliating the plane, where g 0 (t) is the thicker one. EXAMPLE 2.6. With what we learned from Example 2.5, it is easy to see that uniqueness for the scalar initial value problem

Some results about Lipschitz functions
We will now establish some properties of Lipschitz functions that will be useful for checking condition (C2) in Theorem 2.3. Before that, consider the following Lemma.

Lemma 3.1. Let A, B, C ∈ n ( ), A and C invertible. Then
where · is the usual matrix norm.
Proof. It is enough to observe that

If g is locally Lipschitz and g −1 (the inverse matrix function) is locally bounded, then g −1 is
locally Lipschitz.
2. If g is locally Lipschitz when fixing the first variable and g −1 is locally bounded, then g −1 is locally Lipschitz when fixing the first variable.
Proof. 1. Let K be a compact subset of U, k 1 be a Lipschitz constant for g in K and k 2 a bound for g −1 in K. Then, for x, y ∈ K, using Lemma 3.1, Hence, g(x) −1 − g( y) −1 ≤ k 1 k 2 2 x − y in K and g −1 is locally Lipschitz. 2. We proceed as in 2. Let K be a compact subset of U, (t, x), (t, y) ∈ K, k 1 be a Lipschitz constant for g in K when fixing t and k 2 a bound for g −1 in K. Then, Hence, g(t, x) −1 − g(t, y) −1 ≤ k 1 k 2 2 x − y and g −1 is locally Lipschitz when fixing the first variable. Proof. 1. Just apply Lemma 3.2.1 to g = f .

Notice that
and that ( f ) −1 is locally Lipschitz by the previous claim. On the other hand, since f is locally continuous we have that f is locally a 1 -diffeomorphism, and thus f −1 is locally Lipschitz. Therefore ( f −1 ) is locally Lipschitz since it is the composition of two locally Lipschitz functions.
3. Just apply Lemma 3.2.2 to g = f .

A modulus of continuity for Lipschitz functions along an hyperplane
Let U be an open subset of n+1 , p 0 ∈ U and consider the tangent space of U at p, which can be identified with n+1 . Consider now the real Grassmannian Gr(n, n + 1), that is, the manifold of hyperplanes of n+1 . We know that Gr(n, n + 1) ∼ = Gr(1, n + 1) = n , that is, we can identify unequivocally each hyperplane with their perpendicular lines, which are elements of the projective space n .

Definition 3.4.
Consider B n+1 (p, δ) ⊂ n+1 to be the open ball of center p and radius δ. Then, for a function F : U → n+1 and every p ∈ U, v ∈ n and δ ∈ + we define the modulus of continuity We also define

Remark 3.5.
If ω F (p, v) < +∞, then there exist δ, ε ∈ + such that Equivalently, Let A be a orthonormal matrix such that its first column is parallel to v. In that case, since A is orthogonal, x ⊥ e 1 implies that Ax ⊥ v.Then, That is, taking into account that A = 1, Hence, if ϕ(x) = Ax + p then F • ϕ is locally Lipschitz in an neighborhood of the origin when the first variable is equal to zero.
The following lemma illustrates the relation between the modulus of continuity ω F and the partial derivatives of F . Lemma 3.6. Assume F is continuously differentiable in a neighborhood N of p. Then Proof. Since F (z) is continuous at p, for {ε n } → 0 there exists {δ n } → 0 such that if z ∈ B n+1 (p, δ n ) and w = 1 then F (z)(w) ≤ F (p)(w) + ε n . Hence, using the Mean Value Theorem, Then, taking the limit when n → ∞, we obtain On the other hand, assume w ∈ n and w ⊥ v. Then F (p + t w) = F (p) + t(D w F (p) + g(t)) where g is continuous and lim t→0 g(t) = 0. Therefore, Taking the limit when t tends to zero, D w F (p) ≤ ω F (p, v), which ends the proof. Let (X , d X ) and (Y, d Y ) be two metric spaces and consider F : X → Y and p ∈ X . We say F is strongly absolutely differentiable at p if and only if the following limit exists: However, notice that there some important differences between ω F (·, ·) and F | | when X = n and Y = m . First, since ω(·, ·) is defined with a supremmum, ω(·, ·) is well defined in more cases than F | | . Also, in the definition of ω F (·, v), we are avoiding the direction of a certain vector v. This means that, while strong absolute differentiation implies continuity at the point (see [2, Theorem 3.1]), ω(·, ·) does not.

Sufficient conditions ensuring a Lipschitz condition along a foliation
The next Lemma is a key ingredient in the main result of this section. It gives an alternative expression to the rotation matrix provided by the Rodrigues' Rotation Formula and generalizes it for n-dimensional vector spaces.

Lemma 4.1 (Codesido's Rotation Formula)
. Let x, y ∈ n+1 and define K y x ∈ n+1 ( ) as K y x := y x T − x y T . Now, let u, v ∈ n , v = −u, and define R v u ∈ n+1 ( ) as where Id is the identity matrix of order n + 1.
Then, R v u ∈ SO(n + 1) and R v u u = v, that is, R v u is a rotation in n+1 that sends the unitary vector u to v. Furthermore, the function R : {(u, v) ∈ n × n : u = −v} → SO(n + 1), defined by R(u, v) := R v u , is analytic. Therefore, Clearly, R v u is analytic on S = {(u, v) ∈ n × n : u = −v} and so is the determinant function. Now, we are going to prove that S is a connected set: firstly, define the linear subspaces  Problem 5], and since the projection π : X → S defined as π(z) = (z 1 , z 2 , . . . , z n+1 ) (z 1 z 2 , . . . , z n+1 ) , (z n+2 , z n+3 , . . . , z 2n+2 ) (z n+2 , z n+3 , . . . , z 2n+2 ) , is continuous and onto, we have that S is connected too. Therefore, |R v u | is continuous on the connected set S and takes values in