ON THE EXISTENCE OF VIABLE SOLUTIONS FOR A CLASS OF SECOND ORDER
DIFFERENTIAL INCLUSIONS Aurelian Cernea
Faculty of Mathematics, University of Bucharest Academiei 14, 70109 Bucharest, Romania
Abstract
We prove the existence of viable solutions to the Cauchy problem x
00∈ F (x, x
0), x(0) = x
0, x
0(0) = y
0, where F is a set-valued map defined on a locally compact set M ⊂ R
2n, contained in the Fr´echet subdifferential of a φ-convex function of order two.
Keywords: viable solutions, φ-monotone operators, differential inclusions.
2000 Mathematics Subject Classification: 34A60.
1. Introduction
In the present paper, we consider the second order differential inclusion of the form
(1.1) x
00∈ F (x, x
0), x(0) = x
0, x
0(0) = y
0,
where F (., .) : M ⊂ R
n× R
n→ P(R
n) is a given set-valued map and x
0, y
0∈ R
n.
Existence of viable solutions to problem (1.1) has been studied by many authors, mainly in the case when the multifunction is convex valued ([2], [6], [8], [10] etc.).
Recently in [9], the situation when the multifunction is not convex val-
ued is considered. More exactly, in [9] the existence of viable solutions to
the problem (1.1) is proved when F (., .) is an upper semicontinuous, com-
pact valued multifunction contained in the subdifferential of a proper convex
function.
The aim of this paper is to extend the result of [9] to the case when the multi- function F is contained in the Fr´echet subdifferential of a φ-convex function of order two. Since the class of proper convex functions is strictly contained in the class of φ-convex functions of order two, our result generalizes the one in [9].
On the other hand, our result may be considered as an extension of our previous existence result ([5]) obtained for first order differential inclusion of the form
(1.2) x
0∈ F (x), x(0) = x
0to the more general problem (1.1). The proof of our main result follows the general ideas of [5] and [9].
The paper is organized as follows: in Section 2, we recall some prelim- inary facts that we need in the sequel and in Section 3, we prove our main result.
2. Preliminaries
We denote by P(R
n) the set of all subsets of R
nand by R
+the set of all positive real numbers. For ² > 0 we put B(x, ²) = {y ∈ R
n; ky − xk < ²}.
With B we denote the unit ball in R
n. By cl(A) we denote the closure of the set A ⊂ R
n, by co(A) we denote the convex hull of A and we put kAk = sup{kak; a ∈ A}.
Let Ω ⊂ R
nbe an open set and let V : Ω → R ∪ {+∞} be a function with domain D(V ) = {x ∈ R
n; V (x) < +∞}.
Definition 2.1. The multifunction ∂
FV : Ω → P(R
n), defined as:
∂
FV (x) = {α ∈ R
n, lim inf
y→x
V (y) − V (x)− < α, y − x >
ky − xk ≥ 0} if V (x) < +∞
and ∂
FV (x) = ∅ if V (x) = +∞ is called the Fr´echet subdifferential of V . We also put D(∂
FV ) = {x ∈ R
n; ∂
FV (x) 6= ∅}.
According to [4] the values of ∂
Fare closed and convex.
Definition 2.2. Let V : Ω → R∪{+∞} be a lower semicontinuous function.
We say that V is a φ-convex of order 2 if there exists a continuous map
φ
V: (D(V ))
2× R
2→ R
+such that for every x, y ∈ D(∂
FV ) and every α ∈ ∂
FV (x) we have
(2.1) V (y) ≥ V (x)+ < α, x−y > −φ
V(x, y, V (x), V (y))(1+kαk
2)kx−yk
2. In [4] there are several examples and properties of such maps.
For M ⊂ R
nand x ∈ M we recall that the contingent cone to M at x is defined by
K
M(x) = {v ∈ R
n; lim inf
h→0+
1
h d(x + hv, M ) = 0}
and the second-order contingent set to M at (x, y) ∈ M × R
nis defined by:
K
M2(x, y) = {v ∈ R
n; lim inf
h→0+
d(x + hy +
h22, M ) h
2/2 = 0}.
Remark 2.3 ([2], [6], [10]). If F (., .) is upper semicontinuous, compact convex valued and x(.) : [0, T ] → R
nis a solution to the Cauchy problem (1.1) such that x(t) ∈ M , ∀t ∈ I, then
(x(t), x
0(t)) ∈ graph(K
M(.)), ∀t ∈ [0, T ].
In what follows, for a multifunction F : M ⊂ R
n× R
n→ P(R
n) and for any (x
0, y
0) ∈ M we consider the problem (1.1) under the following assumptions.
Hypothesis 2.4.
(i) M = M
1× M
2⊂ R
n× R
nis a locally compact set such that graph(K
M1(.)) ⊂ M.
(ii) F is upper semicontinuous (i.e., ∀z ∈ M, ∀² > 0 there exists δ > 0 such that kz − z
0k < δ implies F (z
0) ⊂ F (z) + ²B) with compact values and such that
F (x, y) ∩ K
M21(x, y) 6= ∅, ∀(x, y) ∈ M.
(iii) There exists a proper lower semicontinuous φ-convex function of order two V : R
n→ R ∪ {+∞} such that
F (x, y) ⊂ ∂
FV (y), ∀(x, y) ∈ M.
Finally, by a viable solution to problem (1.1) we mean an absolutely con- tinuous function x(.) : [0, T ] → R
nwith an absolutely continuous derivative x
0(.) such that x(0) = x
0, x
0(0) = y
0,
x
00(t) ∈ F (x(t), x
0(t)) a.e. ([0, T ]) and
(x(t), x
0(t)) ∈ M, ∀t ∈ [0, T ].
3. The main result
In order to prove our main result we need the following lemma.
Lemma 3.1 ([9]). Assume that Hypothesis 2.4 is verified and let M
0⊂ M be a compact subset such that (x
0, y
0) ∈ M
0. Then, for every k ∈ N there exist h
0k∈ (h
0(k),
1k] and u
0k∈ R
nsuch that
x
0+ h
0ky
0+ (h
0k)
22 u
0k∈ M
1, (x
0, y
0, u
0k) ∈ graph(F ) + 1
k (B × B × B).
Our main result is the following:
Theorem 3.2. Consider F : M → P(R
n) and V : R
n→ R∪{∞} satisfying Hypothesis 2.4. Then, for every (x
0, y
0) ∈ graph(K
M(.)) there exist T > 0 and x(.) : [0, T ] → R
na viable solution to problem (1.1).
P roof. Let (x
0, y
0) ∈ graph(K
M(.)). Since M is locally compact, there exists r > 0 such that M
0= M ∩ B
r(x
0, y
0) is a compact set. Moreover, by the upper semicontinuity of F and Proposition 1.1.3 in [1], the set
F (B
r(x
0, y
0)) = ∪
(x,y)∈Br(x0,y0)
F (x, y) is compact, hence there exists L > 0 such that
sup{kvk; v ∈ F (x, y); (x, y) ∈ B
r(x
0, y
0)} ≤ L < +∞.
Let φ
Vbe the continuous function appearing in Definition 2.2.
Since V (.) is continuous on D(V ) (e.g. [7]), by possibly decreasing r one can assume that for all y ∈ B
r(y
0) ∩ D(V )
|V (y) − V (y
0)| ≤ 1.
Set
S := sup{φ
v(y
1, y
2, z
1, z
2); y
i∈ B
r(y
0), z
i∈ [V (y
0) − 1, V (y
0) + 1], i = 1, 2}, T = min
½ r
2(L + 1) ,
r r
L + 1 , r 2(ky
0k + 1)
¾ .
By Lemma 3.1, for (x
0, y
0) ∈ M
0there exist h
0k∈ (h
0(k),
k1] and u
0k∈ R
nsuch that
x
0+ h
0ky
0+ (h
0k)
22 u
0k∈ M
1, (x
0, y
0, u
0k) ∈ graph(F ) + 1
k (B × B × B).
We define
x
1k:= x
0+ h
0ky
0+ (h
0k)
22 u
0k, y
k1:= y
0+ h
0ku
k0.
According to the choice of T and if h
0k< T we have kx
1k− x
0k ≤ h
0kky
0k + (h
0k)
22 ku
0kk
≤ h
0kky
0k + (h
0k)
22 (L + 1) < r, ky
k1− y
0k ≤ h
0kku
k0k ≤ h
0k(L + 1) < r.
Hence (x
1k, y
k1) ∈ M
0and applying again Lemma 3.1 we find h
1k∈ (h
0(k),
1k] and u
1k∈ R
nsuch that
x
1k+ h
1ky
1k+ (h
1k)
22 u
1k∈ M
1, (x
1k, y
k1, u
1k) ∈ graph(F ) + 1
k (B × B × B).
By recurrence, for each k ∈ N , there exist m
k∈ N
∗and h
pk, x
pk, y
pk, u
pksuch
that for every p = 2, . . . , m
k− 1 we have:
(a) Pmj=0k−1h
jk≤ T < Pmj=0k h
jk,
h
jk,
(b) x
pk= x
0k+ ( Pp−1i=0h
ik)y
0+
12P
p−1i=0(h
ik)
2u
ik+ Pp−2i=0P
p−1j=i+1h
ikh
jku
ik, (c) y
kp= y
0k+ Pp−1i=0h
iku
ik,
P
p−1j=i+1h
ikh
jku
ik, (c) y
kp= y
0k+ Pp−1i=0h
iku
ik,
(d) (x
pk, y
pk) ∈ M
0,
(e) (x
pk, y
pk, u
pk) ∈ graph(F ) +
1k(B × B × B).
Assume that h
qk, x
qk, y
kq, u
qkhave been constructed for q ≤ p satisfying (a) – (e) and we construct h
p+1k, x
p+1k, y
kp+1, u
p+1k.
By Lemma 3.1, since (x
pk, y
pk) ∈ M
0there exist h
pk∈ (h
0(k),
1k] and u
pk∈ R
nsuch that
x
pk+ h
pky
pk+ (h
pk)
22 u
pk∈ M
1, (x
pk, y
kp, u
pk) ∈ graph(F ) + 1
k (B × B × B).
If h
0k+ h
1k+ . . . + h
pk≥ T , then we set m
k= p. If h
0k+ h
1k+ . . . + h
pk< T we define
x
p+1k:= x
pk+ h
pky
pk+ (h
pk)
22 u
pk, y
p+1k:= y
pk+ h
pku
kp.
From (b) and (c) it follows
x
p+1k= x
pk+ h
pky
kp+ (h
pk)
22 u
pk= x
0k+ ³ Xp i=0
h
ik´ y0+ 1 2
X
p i=0(h
ik)
2u
ik+
p−1
X
i=0
X
p j=i+1h
ikh
jku
ik,
y
p+1k:= y
kp+ h
pku
kp= y
0k+ Xp i=0
h
iku
ik.
We deduce that
kx
p+1k− x
0k ≤ ³ Xp i=0
h
ik´ ky0k + L + 1 2
³ X
pi=0
h
ik´
2< r, ky
p+1k− y
0k ≤
X
p i=0h
ikku
i0k ≤ (L + 1) ³ Xp i=0
h
ik´ < r,
hence (x
p+1k, y
p+1k) ∈ M
0.
Let us note that this iterative process is finite because h
pk∈ (h
0(k),
1k], implies the existence of m
k∈ N such that
h
0k+ h
1k+ . . . + h
mkk−1≤ T < h
0k+ h
1k+ . . . + h
mkk.
By (e), for every k ∈ N and any p ∈ {0, 1, . . . , m
k} there exists (a
pk, b
pk, v
kp) ∈ graph(F ) such that
kx
pk− a
pkk < 1
k , ky
pk− b
pkk < 1
k , ku
pk− v
kpk < 1 k . It follows that
kx
pkk ≤ kx
pk− x
0k + kx
0k ≤ 1 + kx
0k, ky
pkk ≤ ky
pk− x
0k + ky
0k ≤ 1 + ky
0k, ku
pkk ≤ ku
pk− v
pkk + kv
pkk ≤ 1 + L.
For every k ∈ N
∗, by the upper semicontinuity of F (.) in (x
p−1k, y
kp−1) ∈ M
0, there exists δ = δ(k) such that kx − x
p−1kk < δ, ky − y
kp−1k < δ implies F (x, y) ⊂ F (x
p−1k, y
p−1k) +
1kB. One may take δ ≥
1k, hence ka
p−1k− x
p−1kk <
δ, kb
p−1k− y
p−1kk < δ and then
(3.1)
F (a
p−1k, b
p−1k) ⊂ F (x
p−1k, y
kp−1) + 1 k B F (a
p−1k, b
p−1k) + 1
k B ⊂ F (x
p−1k, y
kp−1) + 2 k B u
p−1k∈ F (x
p−1k, y
p−1k) + 2
k B
We define t
pk= h
0k+ h
1k+ . . . + h
p−1k, t
0k= 0. Obviously, ∀ k ∈ N , p ∈ {1, . . . , m
k}
t
pk− t
p−1k< 1
k and t
mkk≤ T < t
mkk+1. For k ≥ 1, p ∈ {1, . . . , m
k} we define I
kp:= [t
p−1k, t
pk] and
x
k(t) = x
p−1k+ (t − t
p−1k)y
p−1k+ 1
2 (t − t
p−1k)
2u
p−1k, t ∈ I
kp.
Hence
x
0k(t) = y
p−1k+ (t − t
p−1k)u
p−1k, t ∈ I
kp. x
00k(t) = u
p−1k, t ∈ I
kp,
thus, for all t ∈ [0, T ]
(3.2)
kx
00k(t)k ≤ ku
p−1kk < L + 1,
kx
0k(t)k ≤ ky
p−1kk + (t − t
pk)ku
pkk < ky
0k + L + 2, kx
k(t)k ≤ kx
0k + ky
0k + L + 3.
On the other hand,
kx
k(t) − x
pkk ≤ 1
k (ky
0k + L + 2), ∀t ∈ [0, T ], kx
0k(t) − y
pkk ≤ 1
k (L + 1), ∀t ∈ [0, T ], hence from (e) we find that
(x
k(t), x
0k(t), x
00k(t)) ∈ graphF + ²(k)(B × B × B), where ²(k) → 0 as k → ∞.
Then by (3.2) we find that x
00k(.) is bounded in L
2([0, T ], R
n), x
k(.), x
0k(.) are bounded in C([0, T ], R
n), and equi-Lipschitzean; therefore, apply- ing Theorem 0.3.4 in [1] we obtain the existence of a subsequence (again denoted by x
k(.)) and an absolutely continuous function x(.) : [0, T ] → R
nsuch that
x
k(.) converges uniformly to x(.), x
0k(.) converges uniformly to x
0(.),
x
00k(.) converges weakly in L
2([0, T ], R
n) to x
00(.).
Taking into account Hypothesis 2.4 and Theorem 1.4.1 in [1] we obtain that x
00(t) ∈ coF (x(t), x
0(t)) ⊂ ∂
FV (x
0(t)) a.e. ([0, T ]).
Since the mapping x(.) is absolutely continuous and x
00(t) ∈ ∂
FV (x
0(t))
almost everywhere on [0, T ], we apply Theorem 2.2 in [4] and we deduce
that there exists T
1> 0 such that the mapping t → V (x
0(t)) is absolutely continuous on [0, min{T, T
1}] and
(3.3) (V (x
0(t)))
0=< x
00(t), x
00(t) > a.e. [0, min{T, T
1}].
Without loss of generality we may assume that T = min{T, T
1}.
Using the properties of the mapping V (.), the definition of S and (3.3) we find b
p−1k∈ B such that
u
p−1k− 2
k b
p−1k∈ F (x
k(t), x
0k(t)) ⊂ ∂
FV (x
0k(t)) and such that for every t ∈ I
kp−1we have
V (x
0k(t
pk)) − V (x
0k(t
p−1k)) ≥ hu
p−1k− 2 k b
p−1k,
Z
tpk
tp−1k
x
00k(t)dti
− φ
V(x
0k(t
pk), x
0k(t
p−1k), V (x
0k(t
pk)), V (x
0k(t
p−1k)) . ³ 1 + ku
p−1k− 2
k b
p−1kk
2´ kx0k(t
pk) − x
0k(t
p−1k )k
2. If T ∈ (t
mkk−1, t
mkk], we obtain
V (x
0k(T )) − V (x
0k(t
mkk−1)) ≥ hu
mkk−1− 2
k b
mkk−1, Z T
tmk−1k
x
00k(t)dti
− φ
V(x
0k(T ), x
0k(t
mkk−1), V (x
0k(T )), V (x
0k(t
mkk−1)) . ³ 1 + ku
mkk−1− 2
k b
mkk−1k
2´ kx0k(T ) − x
0k(t
mkk−1)k
2. By adding on p the last inequalities we get
V (x
0k(T )) − V (y
0) ≥ Z T
0
kx
00k(t)k
2dt + a(k) + b(k), where
a(k) = −
mk
X
p=1
2
k hb
p−1k, (t
pk− t
p−1k)u
p−1ki b(k) = −
mk
X
p=1
φ
V(x
0k(t
pk), x
0k(t
p−1k)
k, V (x
0k(t
pk)), V (x
0k(t
p−1k)) . ³ 1 + ku
p−1k− 2
k b
p−1kk
2´ kx0k(t
pk) − x
0k(t
p−1k )k
2.
On the other hand, one has
|a(k)| ≤ 2 k
mk
X
p=1
kb
p−1kk.k Z tp
k
tp−1k
x
00k(t)dtk
≤ 2 k
Z
T0
kx
00k(s)kds ≤ 1
k 2T (L + 1) and
|b(k)| ≤
mk
X
p=1
S(1 + L
2)
mk
X
p=1
k Z tp
k
tp−1k
x
00k(t)dtk
2≤ S(1 + L
2)
mk
X
p=1
1 k
Z
tpk
tp−1k
kx
00k(t)k
2dt ≤ S(1 + L
2)
mk
X
p=1
1 k
Z
T0
kx
00k(t)k
2dt
≤ 1
k S(1 + L
2)T (L + 1)
2and thus lim
k→∞a(k) = lim
k→∞b(k) = 0. We deduce that (3.4) V (x
0k(T )) − V (y
0) ≥ lim sup
k→∞
Z
T0
kx
00k(t)k
2dt . Using (3.3) we infer that
lim sup
k→∞
Z
T0
kx
00k(t)k
2dt ≤ Z T
0
kx
00(t)k
2dt
and, since {x
00k(.)}
kconverges weakly in L
2([0, T ], R
n) to x
00(.), by the lower semicontinuity of the norm in L
2([0, T ], R
n) (e.g. Proposition III.30 in [3]) we obtain that
k→∞
lim Z T
0
kx
00k(t)k
2dt = Z T
0
kx
00(t)k
2dt
i.e., x
00k(.) converges strongly in L
2([0, T ], R
n). Hence, there exists a subse- quence (still denoted) x
00k(.) that converges pointwise to x
00(.).
On the other hand, from Hypothesis 2.4 graph(F ) is closed and it follows that
k→∞