Differential Inclusion Approach for Mixed Constrained Problems Revisited

Differential Inclusion Approach for Mixed Constrained Problems Revisited Report FEUP/DEEC/MRP1/2013 Maria do Rosário de Pinho, Igor Kornienko 1 2013 1The support of the European Union Seventh Framework Programme [FP7-PEOPLE-2010-ITN] under grant agreement n. 64735-SADCO is greatly acknowledged. This work was also supported by FEDER funds through COMPETE and by Portuguese funds through the Portuguese Foundation for Science and Technology, within projects PTDC/EEA-CRO/116014/2009 and PTDC/EEI-AUT/1450/ 2012. Abstract Properties of control systems described by differential inclusions are well established in the literature. Of special relevance to optimal control problems are properties concerning measurability, convexity, compactness of trajectories and Lipschitz continuity of the multifunctions mapping defining the differential inclusion of interest. In this work we concentrate on dynamic control systems coupled with mixed state-control constraints. We characterize a class of such systems that can be described by an appropriate differential inclusion so as exhibit “good” properties of the multifunction. We also illustrate the importance of our findings by treating some applications scenarios. 1 Introduction Control systems described in terms by differential inclusions have been extensively studied in the literature (see, e.g., [1, 2, 5, 7, 12, 13, 16, 17] to name but a few). Differential inclusions appear in control theory when dynamical systems are expressed as ˙x(t)∈F(t, x(t)),(1) where t∈I⊂R,x∈Rnand Fis a multifunction (or set-valued mapping) with closed values in Rn. Such systems make it possible to study in a uniform way a large number of control problems (in this respect see for example [5]). Indeed, differential inclusions have proved to be a useful framework for optimal control problems. They are convenient to state conditions under which existence of solution is ensured and an useful tool to derive optimality conditions. It is commonly accepted that differential inclusions are a natural framework” to study dynamical systems with mixed statecontrol constraints (see [17], pp. 38). Such approach has been used for example in [7], [10] and, recently, in [8] and [9]. The usefulness of differential inclusions for optimal control problems requires that differential inclusion have certain properties. In particular, it is essential to establish under which conditions the trajectories of control systems described in terms of ordinary differential equations are the functions satisfying (1). In this respect many questions arise as those on the measurability of the multifunction defining the differential inclusion (so existence of measurable selections is guaranteed), compactness of trajectories, convexity properties (two subjects relevant for the existence of solution to optimal control problems), etc. Although such aspects are clearly and concisely treated in the literature for control systems of the form ˙x(t) = f(t, x(t), u(t)) a.e. t∈[a, b], u(t)∈U(t)a.e. t∈[a, b], (see for example Chapter 2 in [17]), the same cannot be said when control systems are coupled with mixed constraints. The system of interest, herein denoted as (Σ), comprises a differential equation ˙x(t) = f(t, x(t), u(t)) a.e. t∈[a, b], and mixed constraints (x(t), u(t)) ∈S(t)a.e. t∈[a, b],(2) together with boundary conditions (x(a), x(b)) ∈E. (3) The data comprises a fixed interval [a, b], a function f: [a, b]×Rn×Rk→Rn, a multifunction S: [a, b]→Rn×Rk and a set E⊂Rn×Rn. For such system, a pair (x, u)comprising an absolutely continuous function x(the state trajectory) and a measurable function u(the control), will be called throughout an feasible process if it satisfies all the constraints of the above system. Our aim is to gather together properties of multifunctions relevant to optimal control problems involving control systems of the form (Σ). To highlight the required properties, while keeping exposition as simple as possible, we work under somewhat strong assumptions. Conditions under which the state trajectories for (Σ) coincide with the trajectories of a certain differential inclusion ˙x(t)∈Fm(t, x(t)) a.e. t∈[a, b], (where Fmis a multifunction to be defined shortly) satisfying the boundaries constraints (3) will be central in our analysis. We shall pay particular attention to the case where S(t) := {(x, u)∈Rn×U:g(t, x(t), u(t)) ⩽0},(4) where U⊂Rkand g: [a, b]×Rn×Rk→Rm. However, we do not limit our discussion to this case. We emphasize that the contribution of this paper does not reside on the novelties of the results (since most of them appear to be known) but rather on gathering them together. A remarkable fact is that our work highlights the importance of a bounded slope condition imposed on the mixed constraints in the vein [7]. This paper is organized in the following way. In section II we introduce a definition and an important result due to Clarke (in [7]) that will be essential in our setting. Before engaging in our main theme we briefly report 1 on two optimal control problems where inequality mixed constraint appear naturally. Next, in section IV, we present the main assumptions. Our main results appear in Section V. In the last section we present a sketch of proof of our last (and probably the most important) theorem. Notations: If g∈Rm, the inequality g⩽0is interpreted component-wise. Define Rm −={ξ∈Rm:ξ⩽0}and likewise for Rm +. The closed ball centred at xwith radius δis denoted ¯ B(x, δ)and likewise for the open ball regardless of the dimension of the underlying space. On the other hand, ¯ Band Bdenote the closed and open unit ball centred at the origin. Also | · | is the Euclidean norm or the induced matrix norm on Rp×q. If Ω⊂Rpand F: Ω →Rqis a multifunction (or set-valued mapping), then the graph of Fis defined as Gr F:= {(x, y)∈Ω×Rq:y∈F(x)}. We say that a set S⊂R×Rn×Rmis L×B -measurable when we refer to measurability relative to the σ-field generated by the products of Lebesgue measurable subsets in Rand Borel measurable subsets in Rn×Rm. Consider now a function h: [a, b]→Rp. We say that h∈W1,1([a, b]; Rp)if and only if it is absolutely continuous; h∈L1([a, b]; Rp)iff his integrable; and h∈L∞([a, b]; Rp)iff his essentially bounded. The norm of L1([a, b]; Rp)is denoted by k·k1and the norm of L∞([a, b]; Rp)is k·k∞. Take A⊂Rnto be a closed set with and consider x∗∈A. Also let f:Rk→R∪ {+∞} be a a lower semicontinuous function. With respect to f,x∗∈Rkwill be such that f(x∗)<+∞. Concerning nonsmooth analysis we use the following notation: NP A(x∗)is the proximal normal cone to Aat x∗,NL A(x∗)is the limiting normal cone to Aat x∗,NC A(x∗)is the Clarke normal cone to Aat x∗,∂Lf(x∗)is limiting subdifferential of f at x∗and ∂Cf(x∗)is (Clarke) subdifferential of fat x∗. If fis Lipschitz continuous near x∗, the convex hull of the limiting subdifferential, co ∂Lf(x∗) = ∂Cf(x). 2 Auxiliary Result Before proceeding we state an adaptation of Theorem 3.5.2 in [7] that will be important in the forthcoming analysis. Consider a multifunction Γ:[a, b]×Rn→Rk. For each t∈[a, b], consider the multifunction x→Γ(t, x)and suppose that the graph G(t)of this multifunction is closed. Suppose that u∗∈Γ(t, x∗)and that the following condition holds: [BS’] There exist ε > 0,R > 0and K>0such that, for almost all t, x∈B(x∗, ε), u ∈B(u∗, R),(α, β)∈NP G(t)(x, u) =⇒ |α|⩽K|β|. The following theorem asserts that a multifunction satisfying [BS’] is pseudo-Lipschitz. Theorem 2.1 (adaptation of Theorem 3.5.2 in [7]) Let x→Γ(t, x)satisfy [BS’] near (x∗, u∗)∈G(t). Then for any ξ∈]0,1[ and any x1, x2∈B(x∗,¯ε)the following holds Γ(t, x1)∩¯ B(u∗,(1 −ξ)R)⊂Γ(t, x2) + K|x1−x2|¯ B, where ¯ε= min{ε, ξR/3K}. 3 Main Assumptions Mixed constraints, also known as state dependent control constraints, can be written in the general form (see [8]) as (2). We associate with Sthe multifunction Sm: [0,1] ×Rn→Rkdefined as Sm(t, x) = u∈Rk: (x, u)∈S(t). For each t∈[a, b]the set S(t)is the graph of x→Sm(t, x), that is, (x, u)∈S(t)⇐⇒ u∈Sm(t, x). 2 Let us also introduce two more multifunctions F: [0,1] ×Rn→Rn×R×Rmand Fm: [0,1] ×Rn→Rn×R defined as F(t, x) : = {(f(t, x, u), g(t, x, u)) : u∈U},(5) Fm(t, x) : = {f(t, x, u) : u∈Sm(t, x)}(6) The multifunction (5) will be of importance when we concentrate on Sas in (4). Consider now an absolutely continuous function x∗: [a, b]→Rnsuch that ˙x∗(t)∈Fm(t, x∗(t)) a.e..(7) Take ε > 0and define X(t) := x∗(t) + ε¯ Band S∗ ε(t) := S(t)∩(x∗(t) + εB)×Rk.(8) We now state several assumptions that will be use in the forthcoming analysis. Let φ: [a, b]×Rn×Rk→Rp be a general function (φmay then replaced by for g). (B1) The function t→φ(t, x, u)is L-measurable for all (x, u)∈Rn×Rk. (B2) The multifunction Sis L-measurable and, for each t∈[a, b],S(t)is closed. (B3) The set Eis closed. (B4) For each t∈[a, b]and x∈X(t), there exists u∈Rksuch that (x, u)∈S(t). Furthermore, each t∈[a, b] there exists a constant σsuch that (x, u)∈S(t) =⇒ |u|< σ. (BS) There exists a constant K>0such that, for almost all t∈[a, b]and all (x, u)∈S∗ ε(t), (α, β)∈NP S(t)(x, u) =⇒ |α|⩽K|β|. (CA) For all t∈[a, b]and x∈X(t),Fm(t, x)is convex. (LC) There exist constants kφ xand kφ usuch that, for almost all t∈[a, b]and all (xi, ui)∈S∗ ε(t)(i= 1,2), we have |φ(t, x1, u1)−φ(t, x2, u2)|⩽kφ x|x1−x2|+kφ u|u1−u2|. Some of the above assumptions could (in some situations) be stated in weaker forms. Such strengthening of the assumptions include the second part of (B4) and also the second part of nature of the constant in (BS) and (LC). Indeed, the parameters Kin (BS) and kφ xand kφ uin (B1) are constants while in [8] such parameters are considered to be merely measurable functions. These strengthening of the hypotheses however allow us to avoid some technical details in the forthcoming analysis. 4 On S(t)defined by (4) Let us analyse the case where the set S(t)has a specific structure given by (4). First we introduce additional assumptions on Uand F. (ICU) The set Uis compact and for each x∈Rnthere exists a u∈Usuch that g(t, x, u)⩽0. (IMC) There exists a constant Msuch that, for almost every t, all (x, u)∈S∗ ε(t),η∈NL U(u),γ∈Rm +with hγ, g(t, x, u)i= 0, we have (α, β −η)∈∂L x,uhγ, g(t, x, u)i=⇒ |γ|⩽M|β|. (ICC) For all t∈[a, b]and x∈X(t),F(t, x)is convex. Next we shall relate these assumptions with previous ones imposed on S(t)and Fm. Lemma 4.1 Consider S(t)as defined by (4). Assume that gsatisfies (B1) and (LC) and that (ICU) holds. Then Sis a Lebesgue measurable multifunction and for each t,S(t)is closed. 3 Remark: For Sas defined by (4) Lemma 4.1 states conditions on gand Uimplying that (B2) and (B4) hold. Proof. For each t∈[a, b],S(t)is nonempty by (ICU). By (LC) we know that gis a Carathéodory function. Then Proposition 14.33 in [15] asserts that S(t)is a closed set for each tand t→S(t)is Lebesgue measurable. Now we investigate the relation between (IMC) and (BS). We fist characterize (α, β)∈NL S(t)(x, u). Lemma 4.2 Consider S(t)as defined by (4) and assume that gsatisfies (B1) and (LC) and that (ICU) and (IMC) hold. Then for almost every t∈[a, b], for all (x, u)∈S∗ ε(t)and all (α, β)∈NL S∗ ε(t)(x, u), there exists an γ⩾0with hγ, g(t, x, u)i= 0 such that (α, β)∈∂L (x,u)hγ, g(t, x, u)i+{0} × NL U(u).(9) Proof. (Proof of Lemma 4.2) Let t∈[a, b]be such that (IMC) and (LC) hold. Let ϕ(x, u) = g(t, x, u) and set C1(t) = ϕ−1(Rm −)and C2(t) = X(t)×U. Now take any (x, u)∈S(t), x ∈x∗(t) + εB and (α, β)∈NL S(t)(x, u). Observe that x∈X(t)but it is not on the boundary of X. This will be of importance in what follow since (IMC) will be invoked. Our next task is to characterize NL C1(t)(ϕ(x, u)) in terms of ∂L (x,u)hγ, ϕ(x, u)i. Corollary 10.50 in [15] is essential here. To do so, we first claim that if γ∈NL Rm − (ϕ(x, u)) such that (0,0) ∈∂L (x,u)hγ, ϕ(x, u)i, then γ= 0. Take any such γ. Since γ∈NL Rm − (ϕ(x, u)), we have hγ, ϕ(x, u)i= 0, γ ⩾0. Since (IMC) holds, we deduce from (0,0) ∈∂L (x,u)hγ, ϕ(x, u)iand the fact that 0∈NL U(u), that |γ|⩽0. It follows that γ= 0. Then Corollary 10.50 in [15] holds yielding NL C1(t)(x, u)⊂[n∂L (x,u)hγ, ϕ(x, u)i:γ∈NL D(ϕ(x, u))o.(10) This means that there exists a γ⩾0,hγ, ϕ(x, u)i= 0 such that if (v1, v2)∈NL C1(t)(x, u), then (v1, v2)∈ ∂Lhγ, ϕ(x, u)i. Our next step is to prove that NL C1(t)(x, u)and NL C2(t)(x, u)are transversal in (x, u), i.e., (ξ, ζ)∈ −NL C1(t)(x, u)∩NL C2(t)(x, u) =⇒(ξ, ζ) = (0,0).(11) Since NL C2(t)(x, u) = NL X(t)(x, u)×NL U(x, u),x∈intX(t), we have ζ∈NL U(x, u)and ξ= 0. By (10) we have, for some γwith the required properties, (0,−ζ)∈∂Lhγ, ϕ(x, u)i. Invoking (IMC) with α= 0,β= 0 and η=ζ, we deduce that γ= 0. But then (0, ζ) = (0,0), proving (11). Then we can apply Theorem 6.42 in [15] to conclude that NS∗ ε(t)(x, u)⊂NL C1(t)(x, u) + NL C2(t)(x, u). It follows from the above that (9) holds, proving the Lemma.  As a consequence of Lemma 4.2 we relate (IMC) and (BS). Corollary 4.3 Under the assumptions of Lemma 4.2, (BS) holds. Proof. Take any (α, β)∈NP S(t)(x, u). Since S∗ ε(t)⊂S(t)we have NP S(t)(x, u)⊂NP S∗ ε(t)(x, u). On the other hand, we also have NP S∗ ε(t)(x, u)⊂NL S∗ ε(t)(x, u). Thus (α, β)∈NL S∗ ε(t)(x, u)and it follows from Lemma 4.2 and (IMC) that for γ⩾0with hγ, g(t, x, u)i= 0, η∈NL U(u),(α, β −η)∈∂L (x,u)hγ, g(t, x, u)i=⇒ |γ|⩽Mkβ|. By (LC) we also conclude that (x, u)→ hγ, g(t, x, u)iis Lipschitz continuous with constant |γ|max{kg x, kg u}. We then conclude that |α|⩽|(α, β −η)|⩽max{kg x, kg u}|γ|⩽max{kg x, kg u}M|β|. But then (BS) holds with K= max{kg x, kg u}M. 4 5 Convexity of Fm(t, x) Let us now concentrate on (CA) and (ICC). Although for a general S(t), (CA) may be difficult to check, the same is not true when S(t)is as defined by (4). In this case, (ICC) is a condition easier to verify and, as we will show soon, implies (CA). However, the opposite implication does not hold in general. Given the structure of S(t)one may be tempted to think that other easier verifiable condition would involve the multifunctions Ff(t, y) = {f(t, y, u) : u∈U}, Gg(t, y) = {g(t, y, u) : u∈U}. Next we investigate the existence of any relations between convexity properties of F,Fm,Ffand Ggwhen S(t) is defined by inequalities mixed constraints (4). Lemma 5.1 Consider any t∈[a, b]and x∈X(t)such that Sm(t, x)6=∅where S(t)is defined as S(t) := {(x, u)∈Rn×U:g(t, x(t), u(t)) ⩽0}. The following relations hold: 1. F(t, x)convex =⇒Fm(t, x)convex, but the opposite implication does not hold. 2. F(t, x)convex =⇒Ff(t, x)and Gg(t, x)are convex, but the opposite implication does not hold. 3. The convexity of Fm(t, x)does not imply the convexity of Ff(t, x)and Gg(t, x)and the opposite implication does not hold. Proof. 1. F(t, x)convex =⇒Fm(t, x)convex. Take any v1, v2∈Fm(t, x). Then there exist u1, u2∈Usuch that v1=f(t, x, u1),v2=f(t, x, u2), g(t, x, u1)⩽0and g(t, x, u2)⩽0. Set zi=g(t, x, ui),i= 1,2. We have (vi, zi)∈F(t, x),i= 1,2. Since F(t, x)is convex, for any β∈[0,1], there exists u∈Usuch that (v, z) = β(v1, z1) + (1 −β)(v2, z2) = (f(t, x, u), g(t, x, u)). But z=βz1+ (1 −β)z2=g(t, x, u)⩽0. Thus v∈Fm(t, x)proving convexity of Fm(t, x). If Fm(t, x)convex 6=⇒F(t, x)convex. Take U= [−1,1],f(t, x, u) = uand g(t, x, u) = −(u+ 1)2. Then for any x, the set Fm(t, x) = [−1,1] is convex. However F(t, x) = (u, −(u+ 1)2) : u∈U is not. 2. F(t, y)convex =⇒Ff(t, y)and Gg(t, y)are convex. Fix yand take any v1, v2∈F(t, x). Then there exist u1, u2∈Usuch that v1=f(t, x, u1)and v2=f(t, x, u2). Set z1=g((t, x, u1)and z2=g(t, x, u2). Then, for any β∈[0,1] (v, z) = β(v1, z1) + (1 − β)(v2, z2)is such that (v, z)∈F(t, x), i.e, there exists u∈Usuch that (v, z) = (f(t, x, u), g(t, x, u)). It follows that v∈Ff(t, x)and z∈Gg(t, x)proving convexity of Ff(t, x)and Gg(t, x). If Ff(t, x)and Gg(t, x)convex 6=⇒F(t, x)convex. To see this it is enough to define U= [−1,1],f(t, x, u) = u2, and g(t, x, u) = u. Then Ff(t, x) = [0,1], Gg(t, x)[−1,1] and F(t, x) = {(u2, u) : u∈[−1,1]}. 3. If Fm(t, x)convex 6=⇒Ff(t, x)and Gg(t, x)convex. Take U= [−1,1],f(t, x, u) = uand g(t, x, u) = (−u, u3−u). Then Sm(t, x) = [0,1] and consequently Fm(t, x) = [0,1] is convex. On the other hand, although Ff(t, x) = [0,1] is convex, we do not have convexity of Gg(t, x) = {(−u, u3−u) : u∈[−1,1]}. If Ff(t, x)and Gg(t, x)are convex 6=⇒Fm(t, x)convex. Take U= [−1,1],f(t, x, u) = uand g(t, x, u) = −u2+ 1/4. Then Ff(t, x)=[−1,1] and Gg(t, x) = [−3/4,1/4] are both convex. However, Fm(t, x) = [−1,−1/2] ∪[1/2,1] is not convex. 5 We summarize our findings: Ff,g(t, y)convex =⇒ 6⇐=F(t, y), G(t, y)convex w ~ w −~ w −w  − Fm(t, y)convex ⇐⇒ Fm(t, y)convex 6 Properties of the Multifunctions Here we establish important properties of the multifunctions Sand Fm. We We first focus on a general Sand later on we turn to (4). Lemma 6.1 Assume that (B2) and (B4) hold and that fsatisfies (B1) and (LC). Then 1. For each t∈[a, b]and each x∈X(t),Sm(t, x)and Fm(t, x)are nonempty and compact. 2. The multifunction Fmis L×B-measurable. 3. The graph of (t, x)→Sm(t, x)is a L×B-measurable set. 4. For almost all t∈[a, b]and all x(t)∈X(t)there exists an integrable function csuch that for all γ(t)∈ Fm(t, x(t)) we have |γ(t)|⩽c(t). Proof. The first part of the Lemma follows from (B4) and (LC). For each t∈[a, b]and each x∈X(t), (B4) guarantees that the sets Sm(t, x)and Fm(t, x)are nonempty. The set Sm(t, x)is compact since it is closed by (B2) and bounded by (B4). Taking into account that u→f(t, x, u)is continuous by (LC), we get the compactness of the set Fm(t, x). We now turn to 2. of the Lemma. Take any open set A⊂Rn. We want to prove that {(t, x)∈[a, b]×Rn:Fm(t, x)∩A6=∅} is L×B-measurable. Since, by (B1) and (LC), t→f(t, x, u)is measurable for each (x, u)and (x, u)→f(t, x, u) is continuous for each t, Proposition 2.3.6 in [17] asserts that fis an L×B-measurable function. It follows that the set f−1(A) = (t, x, u)∈[a, b]×Rn×Rk:f(t, x, u)∈A is L×B measurable. On the other hand, the multifunction t→S(t)is L-measurable and closed valued by (B4). It follows from, for example, Theorem 2.3.7 in [17], that its graph Υ := {(t, x, u)∈[a, b]×Rn×Rk: (x, u)∈S(t)},(12) is a L×B measurable set. Consequently, f−1(A)∩Υis a L×B-measurable set. Taking into account that (B4) guarantees that for any (t, x)∈ {(t, x)∈[a, b]×Rn:Fm(t, x)∩A6=∅} there exists a u∈Rksuch that (t, x, u)∈f−1(A)∩Υ we conclude the L×Bmeasurability of Fm. Statement 3. of the Lemma follows from the L×B measurability of the set (12) and the fact that (x, u)∈S(t) is equivalent to u∈Sm(t, x). It remains to prove 4. Take t∈[a, b]such that ˙x∗(t)∈Fm(t, x∗(t)) (see (7)). Let u∗be such that u∗∈Sm(t, x∗(t)) and ˙x∗(t) = f(t, x∗(t), u∗(t)). Take xsuch that x∈X(t). Since by (B4) we have Fm(t, x)6=∅, take any γ∈Fm(t, x). By definition of Fmthere there exists a u∈Sm(t, x)such that γ=f(t, x, u). Appealing to (LC) we now have |γ|⩽|f(t, x∗(t), u∗(t))|+ 2kf xε+ 2kf uσ=|˙x∗(t)|2kf xε+ 2kf uσ. Set c(t) = |˙x∗(t)|+ 2kf xε+ 2kf uσ. Observe that upper bound does not depend on the choice of xor uand it holds for almost every t. Since ˙x∗is an integrable function we conclude that c∈L1proving our claim.  6 Remark: It is important to emphasize that to prove 4. of Lemma 6.1 the fact that our choice of x∗satisfying (7) is essential to assert the integrability of c. A choice of any absolutely continuous function as x∗, not necessarily satisfying (7), would not have been enough to guarantee 4. However, if had we assumed the function fto be uniformly bounded, then we would get 4. with ca constant.  We now investigate Lipschitz properties of x→Sm(t, x)and x→Fm(t, x)for each t. In this respect, (BS) is essential as we shall see. Indeed, conditions (B1), (B2), (B4) and (LC) by themselves, are not enough to guarantee lower semi-continuity of x→Sm(t, x)or x→Fm(t, x), let alone Lipschitz continuity, as the following example shows. Example 6.2 Let us fix t∈[a, b](the interval [a, b]here has no relevance) and set S(t) = {(x, u)∈Rn×R:u∈[−1,1], u|x|⩽0}. Since, for each t, Sm(t, x) = ([−1,1] if x= 0, [−1,0] if x6= 0, we have Fm(t, x) = {x+u:u∈Sm(t, x)}. It is a simple matter to see that (B1), (B2), (B4) hold and that f(x, u) = x+usatisfies (LC). However, both Fmand Smfail to be lower semi-continuous. To see that consider any sequence {xi}such that xi6= 0 and xi→0. Then 1/2∈Sm(t, 0) and 1/2∈Fm(t, 0). But there is no convergent sequence {ui}with limit equal to 1/2, since ui⩽0. Consequently, there is no sequence γi∈Fm(t, xi) converging 1/2. Assumption (BS) excludes this example from our context. Indeed, for any t, we have (1,0) ∈NP S(t)(0,1/2) and for any constant Kwe have 1>0· K. So (BS) is not satisfied. . Remark: It is worth mentioning, for future reference, that the set S(t)in the above example can be defined as in (4) where U= [−1,1] and g(x, u) = u|x|. Under our conditions an appeal to Theorem 2.1 guarantees that x→Sm(t, x)is not merely pseudo-Lipschitz, it is in fact Lipschitz continuous as we show next. Lemma 6.3 Assume that (B2), (B4) and (BS) hold. Then there exist constants kSand εsuch that, for almost every t, x, x0∈x∗(t) + εB =⇒Sm(t, x)⊂Sm(t, x0) + kS|x−x0|¯ B. Proof. Recall that S(t)is the graph of x→Sm(t, x)and, by (B2), it is a closed set. Now take tsuch that (BS) holds. Consider any measurable function u∗such that u∗(t)∈Sm(t, x∗(t)) for almost every t. By (B3) we know that for any u∈Sm(t, x)we have |u|⩽σ. So, for almost every t, any xand usuch that x∈B(x∗(t), ε) (the same εdefining the closed set X(t)) and (x, u)∈S(t)we have u∈B(u∗(t), R)with R= 2σ. Taking into account (BS) we deduce that (BS’) holds. We can then apply Theorem 2.1, where Γ(t, x) = Sm(t, x)and G(t) = S(t), with ξ= 1/2. Observing that, by (B4), for ε= min{ε, σ 3K}and any x∈B(x∗(t), ε)we have Sm(t, x)∩¯ B(u∗(t),(1 −ξ)R) = Sm(t, x)∩¯ B(u∗(t), σ) = Sm(t, x) we get our result.  As an immediate conclusion from the above Lemma we get the following Corollary. Corollary 6.4 Assume that (B2), (B4) and (BS) hold and that fsatisfies (B1) and (LC). Then there exist constant kFmsuch that, for almost every t, x, x0∈x∗(t) + B(0, ε) =⇒Fm(t, x)⊂Fm(t, x0) + kFm|x−x0|¯ B. Proof. Take any x, x0∈x∗(t) + B(0, ε). Take any γ∈Fm(t, x)and γ0∈Fm(t, x0). Let uand u0be such that (x, u)∈Sm(t, x),(x0, u0)∈Sm(t, x0),γ=f(t, x, u)and γ0=f(t, x0, u0). By (LC) and Lemma 6.3 we have |f(t, x, u)−f(t, x0, u0)|⩽kf x|x−x0|+kf u|u−u0| ⩽kf x|x−x0|+kf uK|x−x0| = (kf x+kf uK)|x−x0| and our result follows with kFm=kf x+kf uK. 7