A NOTE ON CONSTRAINED SECOND-ORDER DIFFERENTIAL INCLUSIONS
WITHOUT CONVEXITY
AURELIAN CERNEA
We prove the existence of viable solutions to the Cauchy problemx∈F(x, x) + f(t, x, x), x(0) = x0, x(0) = y0,x(t) ∈ K, where K ⊂ Rn is a closed set, F a set-valued map contained in the Clarke subdifferential of a uniformly regular function, andfa Carath´eodory map.
AMS 2000 Subject Classifications: 34A60.
Key words: uniform regular function, Clarke subdifferential, viable solution, dif- ferential inclusion.
1. INTRODUCTION
In this note we consider the second order differential inclusions of the form
(1.1) x∈F(x, x) +f(t, x, x), x(0) =x0, x(0) =y0,
where F(·,·) : D ⊂ Rn × Rn → P(Rn) is a given set-valued map, f(·,·,·) :D1⊂R×Rn×Rn → P(Rn) is a given function and x0, y0 ∈Rn.
Existence of solutions of problem (1.1) that satisfy a constraint of the formx(t)∈K,∀t, well known as viable solutions, has been studied by many authors, mainly in the case when the multifunction is convex valued andf ≡0 ([2], [7], [8], [9] etc.).
Recently [1], the situation where the multifunction is not convex valued has been considered. More exactly, in [1] is proved the existence of viable solutions of problem (1.1) when F(·,·) is an upper semicontinuous, compact valued multifunction contained in the subdifferential of a proper convex func- tion.
The aim of this note is to improve on the result in [1]. Namely, we prove existence of viable solutions of problem (1.1) in the case where the set- valued map F(·,·) is upper semicontinuous compact valued and contained in the Clarke subdifferential of a uniformly regular function. Note that an alternative improvement of the result in [1] is provided in [6]. More precisely,
MATH. REPORTS9(59),2 (2007), 175–181
the convexity of the function that contains the set-valued map in [1] is relaxed in [6] in the sense that this function is assumed to beφ-convex of order two.
The proof of our result follows general ideas in [1] and [3]. We note that in the proof we only pointed out the differences that appear with respect to the other approaches.
The paper is organized as follows. In Section 2 we recall some preliminary facts that we need in the sequel while in Section 3 we prove our main result.
2. PRELIMINARIES
We denote by P(Rn) the set of all subsets of Rn and by R+ the set of all positive real numbers. For > 0 we putB(x, ) = {y ∈Rn;y−x< } andB(x, ) ={y∈Rn;y−x ≤}. By B we denote the unit ball inRn, by cl(A) the closure of the set A⊂Rn, by co(A) the convex hull of A, and put A= sup{a;a∈A}.
Consider a locally Lipschitz continuous functionV :Rn→R. For every direction v ∈ Rn its Clarke directional derivative at x in the direction v is defined by
DCV(x;v) = lim sup
y→x, t→0+
V(y+tv)−V(y)
t .
TheClarke subdifferential (generalized gradient) of V at x is defined by
∂CV(x) ={q∈Rn, q, v ≤DCV(x;v)∀v∈Rn}. We recall that theproximal subdifferential ofV(·) is defined by
∂PV(x) ={q ∈Rn, ∃δ, σ >0 such that V(y)−V(x) +σy−x2≥ q, y−x ∀y∈B(x, δ)}.
Definition 2.1 ([3]). Let V :Rn →R∪ {∞} be a lower semicontinuous function and letO ⊂dom(V) be a nonempty open subset. We say that V is uniformly regularover O if there exists a positive numberβ >0 such that for allx∈O and for all ξ∈∂PV(x) one has
ξ, y−x ≤V(y)−V(x) +βy−x2 ∀y∈O.
We say that V is uniformly regular over the closed set K if there exists an open setO containing K such thatV is uniformly regular over O.
In [3] there are several examples and properties of such maps. According to [3], any lower semicontinuous proper convex functionV is uniformly regular over any nonempty compact subset of its domain with β = 0; any lower-C2 functionV is uniformly regular over any nonempty convex compact subset of its domain.
The next results will be used in the proof of our result.
Lemma2.2 ([3]). Let V :Rn→R be a locally Lipschitz function and let
∅ =K⊂dom(V) a closed set. If V is uniformly regular over K, then i) ∂CV(x) =∂PV(x) ∀x∈K;
ii) if z(·) : [0, τ] → Rn is absolutely continuous, g(·) : [0, τ] → Rn is measurable and g(t)∈∂CV(z(t)) a.e. ([0, τ]), then
(V ◦z)(t) =z(t)2, a.e.([0, τ]).
In what follows we assume
Hypotheses 2.3. i) Ω = K ×O, where K ⊂ Rn is a closed set and O⊂Rn is a nonempty open set.
ii) F(·,·) : Ω → P(Rn) is upper semicontinuous (i.e., ∀z ∈ Ω, ∀ > 0 there exists δ > 0 such that z−z < δ implies F(z) ⊂ F(z) +B) with compact values.
iii)f(·,·,·) :R×Ω→Rn is a Carath`eodory function, i.e., ∀(x, y)∈Ω, t→f(t, x, y) is measurable, for all t∈Rf(t,·) is continuous and there exists m(·)∈L2(R,R+) such thatf(t, x, y) ≤m(t) ∀(t, x, y)∈R×Ω.
iv) For all (t, x, v) ∈R×Ω, there exists w∈F(x, v) such that lim inf
h→0
1 h2d
x+hv+h2 2 w+
t+h
t f(s, x, v)ds, K
= 0.
v) There exists a locally Lipschitz function V : Rn → R∪ {+∞} uni- formly regular overO such that
F(x, y)⊂∂CV(y), ∀(x, y)∈Ω.
3. THE MAIN RESULT
In order to prove our result we need the following lemmas.
Lemma3.1 ([1]). Assume that Hypotheses 2.3 i)–iv) are satisfied. Con- sider (x0, y0)∈Ω, r >0 such that B(x0, r)⊂O, M := sup{F(t, x); (t, x)∈ Ω0 := [K×B(y0, r)]∩B((x0, y0), r)}, T1 >0such thatT1
0 (m(s) +M+ 1)ds <
3r,T2= min{3(Mr+1),3(y2r
0+r)}andT ∈(0,min{T1, T2}). Then for every >0 there exist η ∈ (0, ) and p ≥ 1 such that for all i= 1, . . . , p−1 there exists (hi,(xi, yi), wi)∈[η, ]×Ω0×Rn for which
xi =xi−1+hi−1yi−1+h2i−1
2 wi−1+
hi−2+hi−1
hi−2
f(s, xi−1, yi−1)ds∈K, yi =yi−1+hi−1wi−1, wi ∈F(xi, yi) +
TB,
(xi, yi)∈Ω0, p−1
i=0
hi < T ≤ p
i=0
hi.
Moreover, for >0 small enough we have p−1
i=0 h2i
2 ≤p−1
i=0 hi < T.
Fork≥1 and q= 1, . . . , pdenote byhkq the real number associated with = 1k and (t, x, y) = (hkq−1, xq, yq) given by Lemma 3.1.
Define t0k = 0, tpk =T, tqp = hk0 +· · ·+hkq−1 and consider the sequence xk(·) : [tq−1k , tqk]→Rn,k≥1, defined by
xk(0) =x0, xk(t) =xq−1+(t−tq−1k )yq−1+1
2(t−tq−1k )2wq−1+ t
tq−1k (t−s)f(s, xq−1, yq−1)ds.
Lemma 3.2 ([1]). Assume that Hypotheses 2.3 i)–iv) are satisfied and consider the sequencexk(·) constructed above. Then there exist a subsequence, still denoted byxk(·), and an absolutely continuous function x(·) : [0, T]→Rn such that
i)xk(·) converges uniformly to x(·);
ii) xk(·) converges uniformly to x(·);
iii)xk(·) converges weakly in L2([0, T],Rn) to x(·);
iv)the sequence(p
q=1
tqk
tq−1k < xk(s), f(s, xk(tq−1k ), xk(tq−1k ))>ds)kcon- verges toT
0 < x(s), f(s, x(s), x(s))>ds;
v)for every t∈(0, T) there existsq∈ {1, . . . , p} such that
k→∞lim d((xk(t), xk(t), xk(t)−f(t, xk(tq−1k ), xk(tq−1k ))),graph(F)) = 0;
vi)x(·) is a solution of the convexified problem
x∈coF(x, x) +f(t, x, x), x(0) =x0, x(0) =v0. We are now able to prove our result.
Theorem3.3. Assume that Hypotheses2.3are satisfied. Then for every (x0, y0) ∈ Ω there exist T > 0 and a solution x(·) : [0, T] → Rn of problem (1.1) that satisfiesx(t)∈K ∀t∈[0, T].
Proof. Let (x0, y0)∈Ω and considerr >0,T >0 as in Lemma 3.1 and xk(·) : [0, T] → Rn, x(·) : [0, T] → Rn as in Lemma 3.2. Let β > 0 be the constant appearing in Definition 2.1.
It follows from the statement vi) in Lemma 3.2 and Hypothesis 2.3 v) that
(3.1) x(t)−f(t, x(t), x(t))∈∂CV(x(t))
for almost allt∈[0, T].
Since the mapping x(·) is absolutely continuous, from (3.1) and Lem- ma 2.2 we obtain
(3.2) (V(x(t))) =x(t), x(t)−f(t, x(t), x(t)) a.e.[0, T].
From (3.2) we have (3.3) V(x(T))−V(y0) =
T
0 x(s)2ds− T
0 x(s), f(s, x(s), x(s) ds.
On the other hand, for q= 1, . . . , pand t∈[tq−1k , tqk),
xk(t)−f(t, xk(tq−1k ), xk(tq−1k ))∈F(xk(tq−1k ), xk(tq−1k )) + 1 kTB and, therefore,
xk(t)−f(t, xk(tq−1k ), xk(tq−1k ))∈∂CV(xk(tq−1k )) + 1 kTB.
We deduce the existence ofbqk∈B such that xk(t)−f(t, xk(tq−1k ), xk(tq−1k ))− bqk
kT ∈∂CV(xk(tq−1k )).
By Definition 2.2 we have V(xk(tqk))−V(xk(tq−1k ))≥
xk(t)−f(t, xk(tq−1k ), xk(tq−1k ))−
−bqk kT,
tq
k
tq−1k xk(s)ds
−βxk(tqk)−xk(tq−1k )2. Using the fact that xk(·) is constant on [tq−1k , tqk], we can write
V(xk(tqk))−V(xk(tq−1k ))≥ tq
k
tq−1k xk(s), xk(s) ds−
− tq
k
tq−1k xk(s), f(s, xk(tq−1k ), xk(tq−1k )) ds− tq
k
tq−1k
xk(s), bqk kT
ds−
−βxk(tqk)−xk(tq−1k )2. By summing over q the last inequalities we get (3.4) V(xk(T))−V(y0)≥
T
0 xk(s)2ds−
− p q=1
tq
k
tq−1k xk(s), f(s, xk(tq−1k ), xk(tq−1k ))ds+a(k) +b(k),
where a(k) =−
p q=1
1 kT
tq
k
tq−1k xk(s), bqk ds, b(k) =−β p q=1
xk(tqk)−xk(tq−1k )2. On the other hand, we have
|a(k)| ≤ 1 kT
p q=1
bqk tq
k
tq−1k xk(s)ds≤
≤ 1 kT
T
0 xk(s)ds≤ 1 kT
T
0 M + 1
T +m(s)
ds and
|b(k)| ≤β p q=1
tq
k
tq−1k xk(s)ds 2 ≤β
p q=1
1 k
tp
k
tp−1k xk(s)2ds≤
≤ β k
T
0 xk(s)2ds≤ β k
T
0 M+ 1
T +m(s) 2
ds.
We conclude that
k→∞lim a(k) = lim
k→∞b(k) = 0.
Hence, using statement iv) in Lemma 3.2 and the continuity of the function V(·) and letting k→ ∞in (3.4) yield
(3.5) V(x(T))−V(y0)≥
≥lim sup
k→∞
T
0 xk(s)2ds− T
0 x(s), f(s, x(s), x(s)) ds.
Using (3.2) we deduce that lim sup
k→∞
T
0 xk(t)2dt≤ T
0 x(t)2dt
and, sincexk(·) converges weakly inL2([0, T],Rn) tox(·), by the lower semi- continuity of the norm in L2([0, T],Rn) (e.g. Proposition III.30 in [4]) we obtain that
k→∞lim T
0 xk(t)2dt= T
0 x(t)2dt,
i.e., xk(·) converges strongly in L2([0, T],Rn). Hence there exists a subse- quence (still denoted)xk(·) that converges pointwise tox(·).
It follows from statement v) in Lemma 3.2 that
d((x(t), x(t), x(t)−f(t, x(t), x(t))),graph(F)) = 0 a.e. ([0, T]) and, since by Hypothesis 2.4 graph(F) is closed, we obtain
x(t)∈F(x(t), x(t)) +f(t, x(t), x(t)) a.e.([0, T]).
In order to prove the viability constraint satisfied byx(·), fixt∈[0, T].
There exists a sequence (tqk)k such that t= limk→∞tqk. But limk→∞x(t)− xk(tqk)= 0 andxk(tqk)∈K. So, the fact thatK is closed givesx(t)∈K and the proof is complete.
Remark3.4. If V(·) :Rn →R is a proper lower semicontinuous convex function, then∂cV(x) =∂V(x), where∂V(·) is the subdifferential in the sense of Convex Analysis ofV(·), and Theorem 3.3 yields Theorem 2.1 in [1].
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Received 27 February 2006 University of Bucharest
Faculty of Mathematics and Computer Science Str. Academiei 14
010014 Bucharest, Romania