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

Electronic Journal of Qualitative Theory of Differential Equations 2018, No. 13, 1–14; https://doi.org/10.14232/ejqtde.2018.1.13 www.math.u-szeged.hu/ejqtde/ A Lipschitz condition along a transversal foliation implies local uniqueness for ODEs José Ángel CidB1and F. Adrián F. Tojo2 1Departamento de Matemáticas, Universidade de Vigo, Campus de Ourense, Spain 2Departamento de Análise Matemática, Universidade de Santiago de Compostela, Spain Received 28 December 2017, appeared 16 February 2018 Communicated by Josef Diblík Abstract. We prove the following result: if a continuous vector field Fis Lipschitz when restricted to the hypersurfaces determined by a suitable foliation and a transversal condition is satisfied at the initial condition, then Fdetermines a locally unique integral curve. We also present some illustrative examples and sufficient conditions in order to apply our main result. Keywords: uniqueness, Lipschitz condition, foliation, modulus of continuity, rotation formula. 2010 Mathematics Subject Classification: 34A12. 1 Introduction Uniqueness for ODEs is an important and quite old subject, but still an active field of research [7–9], 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⊂R2be an open neighborhood of (t0,x0)and f:U⊂R2→Rbe continuous and consider the scalar initial value problem x0(t) = f(t,x(t)),x(t0) = x0. (1.1) It was proved, independently by Mortici, [12], and Cid and Pouso [4,5], that local uniqueness holds provided that the following conditions are satisfied: •f(t,x)is Lipschitz with respect to t, •f(t0,x0)6=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⊂R2is an open neighborhood of (t0,x0), f:U⊂R2→Ris continuous and (u1,u2)∈R2such that BCorresponding author. Email: [email protected] 2J. Á. Cid and F. A. F. Tojo •|f(t,x)−f(t+ku1,x+ku2)| ≤ L|k|on D, •u26=f(t0,x0)u1, then the scalar problem (1.1) has a unique local solution. By taking either (u1,u2) = (0,1) or (u1,u2)=(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 (u1,u2)to depend on t. Let us now consider the autonomous initial value problem for a system of differential equations z0(t) = F(z(t)),z(t0) = p0, (1.2) where n∈N,F:U⊂Rn+1→Rn+1and p0∈U. Trough the paper we shall need the following definition: if g:D⊂Rn+1→E, where Eis a normed space, we will say that gis Lipschitz in D when fixing the first variable if there exists L>0 such that for all (s,x1,x2, . . . , xn),(s,y1,y2, . . . , yn)∈Dwe have that kg(s,x1,x2, . . . , xn)−g(s,y1,y2, . . . , yn)kE≤Lk(x1,x2, . . . , xn)−(y1,y2, . . . , yn)k, and where k·k stands for any norm in Rn. Moreover, for any function gwith values in Rn+1 we denote g= (g1,g2, . . . , gn+1). The following alternative version of Lipschitz uniqueness theorem for systems was proved by Cid in [3]. Theorem 1.1. Let U ⊂Rn+1an open neighborhood of p0and F :U→Rn+1continuous. If moreover • F is Lipschitz in U when fixing the first variable, • F1(p0)6=0, then there exists α>0such that problem (1.2)has a unique solution in [t0−α,t0+α]. Remark 1.2. The classical Lipschitz theorem is included in the previous one. In order to see this, let n∈N,U⊂Rn+1be an open set, f:U→Rnand (t0,x0)∈Uand consider the non-autonomous problem x0(t) = f(t,x(t)),x(t0) = x0. (1.3) As it is well known, problem (1.3) is equivalent to the autonomous one (1.2), where F(z1,z2, . . . , zn+1):= (1, f(z1,z2, . . . , zn+1)), and p0:= (t0,x0). Now, if f(t,x)is Lipschitz with respect to xthen F(z1,z2, . . . , zn+1)is Lipschitz when fixing the first variable and moreover F1(p0) = 16=0, so Theorem 1.1 applies. Recently, Diblík, Nowak and Siegmund obtained in [13] a generalization of both [3] and [14]. Their result reads as follows. Theorem 1.3. Let U ⊂Rn+1be an open neighborhood of p0, F :U→Rn+1be continuous and Va linear hyperplane in Rn+1such that • F is Lipschitz continuous along V, that is, there exists L >0such that if x,y∈U and x −y∈ V, then kF(x)−F(y)k ≤ Lkx−yk, Lipschitz condition, foliations and uniqueness for ODEs 3 and the transversality condition • F(p0)6∈ V holds. Then there exists α>0such that problem (1.2)has a unique solution in [t0−α,t0+α]. The previous theorem has the following geometric meaning: uniqueness for the autonomous system (1.2) follows provided that the continuous vector field Fis Lipschitz when restricted to a family of parallel hyperplanes to Vthat covers Uand that the vector field at the initial condition F(p0)is transversal to V. Our main goal in this paper is to extend Theorem 1.3 from the linear foliation generated by the hyperplane Vto 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 Ffor 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 h·,·ishall denote the usual scalar product in the Euclidean space. 2 The main result: a general uniqueness theorem Definition 2.1. Let p0∈Rn+1. Assume there exist open subsets V⊂Rn,U⊂Rn+1, an open interval J⊂Rwith 0 ∈Jand a family of differentiable functions {gs:V→U}s∈Jsuch that g0(0) = p0∈Uand Φ:(s,y)∈J×V→gs(y)∈Uis a diffeomorphism. Then we say {gs}s∈J is a local n-foliation of U at p0. Remark 2.2. An observation regarding notation. If Φ:Rn+1→Rn+1is a diffeomorphism, we denote by Φ0its derivative and by Φ−1its inverse. Also, we write (Φ−1)0for the derivative of the inverse. Observe that Φ0takes values in Mn+1(R)so, although we cannot consider the functional inverse of Φ0, we can consider the inverse matrix, whenever it exists, of every Φ0(x) for x∈Rn+1. We denote this function by (Φ0)−1. Clearly, the chain rule implies that (Φ0)−1(x) = (Φ−1)0(Φ(x)). The following is our main result. Theorem 2.3. Let U ⊂Rn+1, V ⊂Rnbe open sets, p0∈U, F :U⊂Rn+1→Rn+1a continuous function and {gs:V→U}s∈Ja local n-foliation of U at p0which defines the diffeomorphism Φ: J×V→U. If the following assumptions hold, (C1) Transversality condition: * ∂Φ−1 1 ∂z1(p0), . . . , ∂Φ−1 1 ∂zn+1(p0)!,F(p0)+6=0, (2.1) (C2) Lipschitz condition along the foliation: F ◦Φand (Φ0)−1are Lipschitz in a neighborhood of zero when fixing the first variable, then there exists α>0such that problem (1.2)has a unique solution in [t0−α,t0+α]. 4J. Á. Cid and F. A. F. Tojo Proof. Consider the change of coordinates z= (z1, . . . , zn+1) = Φ(s,y1, . . . , yn):=gs(y1, . . . , yn). (2.2) Since {gs}s∈Jis a foliation, Φis a diffeomorphism. Then, considering y= (s,y1, . . . , yn), differentiating (2.2) with respect to tand taking into account equation (1.2), dz dt=Φ0(y)dy dt=F(z) = (F◦Φ)(y). (2.3) Since Φis a diffeomorphism, Φ0(y)is an invertible matrix for every y, so dy dt=Φ0(y)−1(F◦Φ)(y). By definition of gs,Φ(0) = p0, so we can consider the problem dy dt(t) = h(y),y(t0) = 0, (2.4) where h(y) = Φ0(y)−1F(Φ(y)). Now, by (C2) we have that his the product of locally Lipschitz functions when fixing the first variable. Furthermore, if e1= (1, 0, . . . , 0)∈Rnand taking into account (C1), h1(0) = eT 1Φ0(0)−1F(p0) = eT 1(Φ−1)0(p0)F(p0) = * ∂Φ−1 1 ∂z1(p0), . . . , ∂Φ−1 1 ∂zn+1(p0)!,F(p0)+6=0. 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 ∂Φ−1 1 ∂z1(p0), . . . , ∂Φ−1 1 ∂zn+1(p0)!, is normal to the hypersurface given by g0(V)at p0. So, condition (2.1) means that the vector F(p0)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. Theorem 2.3 generalizes the main result in [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]. Lipschitz condition, foliations and uniqueness for ODEs 5 Example 2.5. Let F(x,y):=1+ (y−x2)2 3. Is there a linear change of coordinates Φsuch that F◦Φis Lipschitz in a neighborhood of zero when fixing the first variable? The answer is no. Any linear change of variables Φwill be given by two linearly independent vectors v,w∈R2as Φ(z,t) = zw +tv. If F◦Φis Lipschitz in a neighborhood of zero when fixing the first variable, that is, z, that implies that the directional derivative of Fat 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)∈R2:y=x2}and realize that Fis differentiable in R2\S, with ∇F(x,y) = 2 3(y−x2)−1 3(−2x,1), for (x,y)∈R2\S. Let v= (v1,v2)∈R2. The directional derivative of Fat (x,y)in the direction of vis DvF(x,y) = h∇F(x,y),vi=2 3(y−x2)−1 3(v2−2v1x), for (x,y)∈R2\S. Now consider a neighborhood Nof 0. In particular, we can consider the points of the form (x,y) = (λ,λ2+µ)∈N\Sfor µ6=0 and λ∈(−e,e), so DvF(x,y) = 2 3 v2−2λv1 µ1/3 . This quantity is unbounded in N\Sunless the numerator is 0 for every λ∈(−e,e), but that means that v=0, so vand wcannot be linearly independent. Hence, no linear change of coordinates Φmakes F◦ΦLipschitz in a neighborhood of zero when fixing the first variable. -2 -1 0 1 2 -2 -1 0 1 2 Figure 2.1: The parabolas gz(t)foliating the plane, where g0(t)is the thicker one. Nevertheless, take (x,y) = Φ(z,t) = gz(t)=(t,z+t2). We have Φ−1(x,y)=(y−x2,x) and both are differentiable, so Φis a diffeomorphism. Now, (F◦Φ)(z,t) = 1+z2 3, which is clearly Lipschitz when fixing the first variable. Example 2.6. With what we learned from Example 2.5, it is easy to see that uniqueness for the scalar initial value problem x0(t) = 1+ (x(t)−t2)2 3,x(0) = 0, (2.5) can not be dealt with [13, Theorem 2] neither with [8, Theorem 1]. However, by using the local 1-foliation associated to diffeomorphism Φgiven in Example 2.5, it is easy to show that conditions (C1) and (C2) of Theorem 2.3 are satisfied. Therefore, we have the local uniqueness of solution. 6J. Á. Cid and F. A. F. Tojo 3 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∈ Mn(R), A and C invertible. Then kABCk ≥ kBk kA−1kkC−1k, where k·k is the usual matrix norm. Proof. It is enough to observe that kBk=kA−1ABCC−1k ≤ kA−1kkABCkkC−1k. Lemma 3.2. Let U be an open subset of Rnand g :U→GLn(R). 1. If g is locally Lipschitz and g−1(the inverse matrix function) is locally bounded, then g−1is locally Lipschitz. 2. If g is locally Lipschitz when fixing the first variable and g−1is locally bounded, then g−1is locally Lipschitz when fixing the first variable. Proof. 1. Let Kbe a compact subset of U,k1be a Lipschitz constant for gin Kand k2a bound for g−1in K. Then, for x,y∈K, using Lemma 3.1, k1kx−yk ≥ kg(x)−g(y)k=kg(x)(g(y)−1−g(x)−1)g(y)k ≥ kg(y)−1−g(x)−1k k2 2 . Hence, kg(x)−1−g(y)−1k ≤ k1k2 2kx−ykin Kand g−1is locally Lipschitz. 2. We proceed as in 2. Let Kbe a compact subset of U,(t,x),(t,y)∈K,k1be a Lipschitz constant for gin Kwhen fixing tand k2a bound for g−1in K. Then, k1kx−yk ≥kg(t,x)−g(t,y)k=kg(t,x)(g(t,y)−1−g(t,x)−1)g(t,y)k ≥kg(t,y)−1−g(t,x)−1k k2 2 . Hence, kg(t,x)−1−g(t,y)−1k ≤ k1k2 2kx−ykand g−1is locally Lipschitz when fixing the first variable. Corollary 3.3. Let U be an open subset of Rn, f :U→f(U)⊂Rnbe a diffeomorphism (notice that, in that case, f0:U→GLn(R)). 1. If f0is locally Lipschitz and (f0)−1is locally bounded, then (f0)−1is locally Lipschitz. 2. If f0is locally Lipschitz and (f0)−1is locally bounded, then (f−1)0is locally Lipschitz. 3. If f0is locally Lipschitz when fixing the first variable and (f0)−1is locally bounded, then (f0)−1 is locally Lipschitz when fixing the first variable. Lipschitz condition, foliations and uniqueness for ODEs 7 Proof. 1. Just apply Lemma 3.2.1 to g=f0. 2. Notice that (f−1)0(x) = (f0)−1(f−1(x)), and that (f0)−1is locally Lipschitz by the previous claim. On the other hand, since f0is locally continuous we have that fis locally a C1-diffeomorphism, and thus f−1is locally Lipschitz. Therefore (f−1)0is locally Lipschitz since it is the composition of two locally Lipschitz functions. 3. Just apply Lemma 3.2.2 to g=f0. 3.1 A modulus of continuity for Lipschitz functions along a hyperplane Let Ube an open subset of Rn+1,p0∈Uand consider the tangent space of Uat p, which can be identified with Rn+1. Consider now the real Grassmannian Gr(n,n+1), that is, the manifold of hyperplanes of Rn+1. We know that Gr(n,n+1)∼ =Gr(1, n+1) = Pn, that is, we can identify unequivocally each hyperplane with their perpendicular lines, which are elements of the projective space Pn. Definition 3.4. Consider Bn+1(p,δ)⊂Rn+1to be the open ball of center pand radius δ. Then, for a function F:U→Rn+1and every p∈U,v∈Pnand δ∈R+we define the modulus of continuity ωF(p,v,δ):=sup x,y∈Bn+1(p,δ) x−p,y−p⊥v x6=y kF(x)−F(y)k kx−yk∈[0, +∞]. We also define ωF(p,v):=lim δ→0ωF(p,v,δ) = lim δ→0sup x,y∈Bn+1(p,δ) x−p,y−p⊥v x6=y kF(x)−F(y)k kx−yk =lim (x,y)→(p,p) x−p,y−p⊥v x6=y kF(x)−F(y)k kx−yk∈[0, +∞]. Remark 3.5. If ωF(p,v)<+∞, then there exist δ,e∈R+such that kF(x)−F(y)k ≤ (ωF(p,v) + e)kx−yk,x,y∈Bn+1(p,δ),x−p,y−p⊥v. Equivalently, kF(x+p)−F(y+p)k ≤ (ωF(p,v) + e)kx−yk,x,y∈Bn+1(0, δ),x,y⊥v. Let Abe a orthonormal matrix such that its first column is parallel to v. In that case, since A is orthogonal, x⊥e1implies that Ax ⊥v. Then, kF(Ax +p)−F(Ay +p)k ≤ (ωF(p,v) + e)kA(x−y)k,x,y∈Bn+1(0, δ),x,y⊥e1. That is, taking into account that kAk=1, kF(A(0, x) + p)−F(A(0, y) + p)k ≤ (ωF(p,v) + e)kx−yk,x,y∈Bn(0, δ). Hence, if ϕ(x) = Ax +pthen F◦ϕis locally Lipschitz in an neighborhood of the origin when the first variable is equal to zero. 8J. Á. Cid and F. A. F. Tojo The following lemma illustrates the relation between the modulus of continuity ωFand the partial derivatives of F. Lemma 3.6. Assume F is continuously differentiable in a neighborhood N of p. Then ωF(p,v) = sup w⊥v kwk=1kDwF(p)k. Proof. Since F0(z)is continuous at p, for {en} → 0 there exists {δn} → 0 such that if z∈ Bn+1(p,δn)and kwk=1 then kF0(z)(w)k≤kF0(p)(w)k+en. Hence, using the mean value theorem, sup x,y∈Bn+1(p,δn) x−p,y−p⊥v x6=y kF(x)−F(y)k kx−yk≤sup x,y,z∈Bn+1(p,δn) x−p,y−p⊥v x6=y kF0(z)(x−y)k kx−yk≤sup z∈Bn+1(p,δn) u∈Bn+1(0,2δn) u⊥v u6=0 kF0(z)(u)k kuk =sup z∈Bn+1(p,δn) d∈(0,2δn) w⊥v kwk=1 kF0(z)(dw)k kdwk=sup z∈Bn+1(p,δn) w⊥v kwk=1 kF0(z)(w)k ≤sup w⊥v kwk=1kF0(p)(w)k+en=sup w⊥v kwk=1kDwF(p)k+en. Then, taking the limit when n→∞, we obtain ωF(p,v)≤sup w⊥v kwk=1kDwF(p)k. On the other hand, assume w∈Snand w⊥v. Then F(p+tw) = F(p) + t(DwF(p) + g(t)) where gis continuous and limt→0g(t) = 0. Therefore, kDwF(p)k=    F(p+tw)−F(p) t−g(t)   ≤sup x,y∈Bn+1(p,t) x−p,y−p⊥v x6=y kF(x)−F(y)k kx−yk+|g(t)|. Taking the limit when ttends to zero, kDwF(p)k ≤ ωF(p,v), which ends the proof. Remark 3.7. This definition of the modulus of continuity ωF(·,·)is somewhat similar to the definition of strong absolute differentiation which appears in [2, expression (1)]: Let (X,dX)and (Y,dY)be two metric spaces and consider F:X→Yand p∈X. We say Fis strongly absolutely differentiable at p if and only if the following limit exists: F|0|(p):=lim (x,y)→(p,p) x6=y dY(F(x),F(y)) dX(x,y). However, notice that there some important differences between ωF(·,·)and F|0| when X= Rnand Y=Rm. First, since ω(·,·)is defined with a supremum, ω(·,·)is well defined in more cases than F|0|. Also, in the definition of ωF(·,v), we are avoiding the direction of a certain Lipschitz condition, foliations and uniqueness for ODEs 9 vector v. This means that, while strong absolute differentiation implies continuity at the point (see [2, Theorem 3.1]), ω(·,·)does not. Regarding the similarities, when the partial derivatives of Fexist, F|0| =k∑n k=1∂F ∂xkk(see [2, Theorem 3.6]). Example 3.8. Consider again F(x,y):=1+ (y−x2)2 3and S={(x,y)∈R2:y=x2}. As was stated in Example 2.5, we have that F|R2\S∈ C∞(R2\S)and ∇F(x,y) = 2 3(y−x2)−1 3(−2x,1), for (x,y)∈R2\S. Therefore, ω(p,v)<+∞for every (p,v)∈(R2\S)×P1. On the other hand, for p= (x0,x2 0)∈Sand v= (v1:v2)∈P1, if x= (x1,y1)−p⊥vthen x=λ(−v2,v1) + pfor some λ∈R. Analogously, we take y=µ(−v2,v1) + pfor some µ∈R. Hence, ωF(p,v) = lim (x,y)→(p,p) x−p,y−p⊥v x6=y kF(x)−F(y)k kx−yk=lim (λ,µ)→(0,0) λ6=µ |F(λ(−v2,v1) + p)−F(µ(−v2,v1) + p)| k(λ−µ)(−v2,v1)k =lim (λ,µ)→(0,0) λ6=µ |[λ(2x0v2+v1)−λ2v2 2]2 3−[µ(2x0v2+v1)−µ2v2 2]2 3| |λ−µ|. We now can consider two cases: (v1:v2)=(−2x0: 1)and (v1:v2)6= (−2x0: 1). In the first case, taking into account that z2+z+1≥3/4 for every z∈R, ωF(p,v) = lim (λ,µ)→(0,0) λ6=µ |(−λ2v2 2)2 3−(−µ2v2 2)2 3| |λ−µ|=lim (λ,µ)→(0,0) λ6=µ |µ4 3−λ4 3||v2|2 3 |λ−µ| =|v2|2 3lim (λ,µ)→(0,0) λ6=µ µ1 3+λ µ2 3+µ1 3λ1 3+λ2 3 =|v2|2 3lim (λ,µ)→(0,0) λ6=µ  µ1 3+λ1 31 µ λ2 3+µ λ1 3+1 ≤ |v2|2 3lim (λ,µ)→(0,0) λ6=µ µ1 3+4 3λ1 3 =0. Observe that in this deduction we have assumed λ6=0. It is clear that, when λ=0, the limit is zero as well. In the case (v1:v2)6= (−2x0: 1)the quotient inside the limit is not bounded and ωF(p,v) = +∞. Therefore, ω−1 F([0, +∞)) = (R2\S)×P1∪{((x,x2),(−2x: 1)) ∈R2×P1:x∈R}. 4 How to get 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 Rodrigues’ rotation formula and generalizes it for n-dimensional vector spaces.