Applied Mathematics E-Notes, 3(2003), 115-123 c
Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
Existence Of Solutions For Nonconvex Second Order
Differential Inclusions ∗
Vasile Lupulescu†
Received 20 April 2002
Abstract
In this paper we prove an existence result for a second order differential inclusion
¡
¢
¡
¢
x ∈ F x, x + f t, x, x , x (0) = x0 , x (0) = y0 ,
where F is an upper semicontinuous, compact valued multifunction, such that
F (x, y) ⊂ ∂V (y), for some convex proper lower semicontinuous function V , and
f is a Carathéodory function.
1
Introduction
For the Cauchy problem
x ∈ F (x) , x (0) = ξ,
where F is an upper semicontinuous, cyclically monotone, compact values multifunction, the existence of local solutions was obtained by Bressan, et al. [4]. For some
extensions of this results we refer to [1], [7], [12] and [13]. On the other hand, for
second order differential inclusions
x ∈ F (x, x ) , x (0) = x0 , x (0) = y0 ,
existence results were obtained by many authors (we refer to [3], [8], [9], [11], [13]).
The case when F is an upper semicontinuous, compact valued multifunction, such
that F (x, y) ⊂ ∂V (y) , for some convex proper lower semicontinuous function V, was
considered in [10].
In this paper we prove an existence result for a second order differential inclusion
x ∈ F (x, x ) + f (t, x, x ) , x (0) = x0 , x (0) = y0 ,
where F is an upper semicontinuous, compact valued multifunction, such that F (x, y) ⊂
∂V (y), for some convex proper lower semicontinuous function V , and f is a Carathéodory
function.
∗ Mathematics
† Universitatea
Subject Classifications: 34G20, 47H04.
“Constantin Brâncuşi”, Bulevardul Republicii Nr. 1, 1400 Târgu-Jiu, Romania
115
116
2
Differential Inclusions
Statement of Result
Let Rm be the m-dimensional Euclidean space with scalar product ., . and norm . .
For x ∈ Rm and ε > 0 let
Bε (x) = {y ∈ Rm : x − y < ε}
be the open ball centered at x with radius ε, and let B ε (x) be its closure. For x ∈ Rm
and for a closed subsets A ⊂ Rm we denote by d(x, A) the distance from x to A given
by
d (x, A) = inf { x − y : y ∈ A} .
Let V : Rm → R be a proper lower semicontinuous convex function. The multim
function ∂V : Rm → 2R defined by
∂V (x) = {ξ ∈ Rm : V (y) − V (x) N ξ, y − x , ∀y ∈ Rm }
is called the subdifferential (in the sense of convex analysis) of the function V.
m
We say that a multifunction F : Rm → 2R is upper semicontinuous if for every
m
x ∈ R and every ε > 0 there exists δ > 0 such that
F (y) ⊂ F (x) + Bε (0) , ∀y ∈ Bδ (x) .
m
For a multifunction F : Ω ⊂ R2m → 2R
Cauchy problem
and for any (x0 , y0 ) ∈ Ω we consider the
x ∈ F (x, x ) + f (t, x, x ) , x (0) = x0 , x (0) = y0 ,
(1)
under the following assumptions:
m
(H1 ) Ω ⊂ R2m is an open set and F : Ω → 2R
valued multifunction.
is an upper semicontinuous compact
(H2 ) There exists a proper convex and lower semicontinuous function V : Rm → R
such that
F (x, y) ⊂ ∂V (y) , ∀ (x, y) ∈ Ω.
(2)
(H3 ) f : R × Rm ×Rm → Rm is a Carathéodory function, i.e. for every x, y ∈ Rm ,
t → f (t, x, y) is measurable, for t ∈ R, (x, y) → f (t, x, y) is continuous and there
exists m (.) ∈ L2 R∗+ such that:
f (t, x, y) ≤ m (t) , (∀) (x, y) ∈ Rm ×Rm , a.e. t ∈ R.
(3)
By a solution of the problem (1) we mean any absolutely continuous function x :
[0, T ] → Rm with absolutely continuous derivative x such that x(0) = x0 , x(0) = y0 ,
and
x (t) ∈ F (x(t), x (t)) + f (t, x (t) , x (t)) , a.e. on [0, T ].
Our main result is the following:
m
THEOREM 1. If F : Ω ⊂ R2m → 2R , f : R × Rm ×Rm → Rm and V : Rm → R
satisfy assumptions (H1 ) , (H2 ) and (H3 ) then for every (x0 , y0 ) ∈ Ω there exist T > 0
and a solution x : [0, T ] → Rm of the problem (1).
V. Lupulescu
3
117
Proof of Our Result
Let (x0 , y0 ) ∈ Ω. Since Ω is open, there exists r > 0 such that the compact set K :=
B r (x0 , y0 ) is contained in Ω. Moreover, by the upper semicontinuity of F in (H1 ) and
by Proposition 1.1.3 in [2], the set
F (x, y)
F (K) :=
(x,y)∈K
is compact, hence there exists M > 0 such that
sup { v : v ∈ F (x, y) , (x, y) ∈ K} ≤ M.
Set
T := min
r
,
M
r
r
,
M 2 ( y0 + 1)
.
By (H3 ) there exists T > 0 such that
T 33
(m (t) + M ) dt < r.
0
We shall prove the existence of a solution of the problem (1) defined on the interval
[0, T ], where 0 < T ≤ min{T , T }.
j
j−1 j
For each integer n ≥ 1 and for 1 ≤ j ≤ n we set tjn := jT
n , In = [tn , tn ] and for
j
t ∈ In we define
1
xn (t) = xjn + (t − tjn )ynj + (t − tjn )2 vnj +
2
t
jT
n
(s − t) f (s, xjn , ynj )ds,
where x0n = x0 , yn0 = y0 , and, for 0 ≤ j ≤ n − 1, vnj ∈ F (xjn , ynj ),

2
 xj+1
= xjn + Tn ynj + 12 Tn vnj
n

ynj+1 = ynj +
(4)
(5)
T j
n vn .
j
j
j
Set, for t ∈ (tj−1
n , tn ), j ∈ {1, 2, ..., n}, fn (t) := f (s, xn , yn ).
j
j
We claim that (xn , yn ) ∈ K for each j ∈ {1, 2, ..., n} . By the choice of T one has
x1n − x0 ≤
T
1 T
1
y0 + ( )2 v0 < T y0 + M T 2 < r
n
2 n
2
and
yn1 − y0 ≤ T v0 < r,
hence the claim is true for j = 1.
We claim that for each j > 1 one has

2
 xjn = x0n + j Tn yn0 + 12 Tn [(2j − 1) vn0 + (2j − 3) vn1 + ... + vnj−1 ]

ynj = yn0 +
T 0
n [vn
+ vn1 + ... + vnj−1 ].
(6)
118
Differential Inclusions
The statement holds true for j = 0. Assume it holds for j, with 1 ≤ j < n. Then by
(5) one obtains that
xj+1
n
T j
1 T
yn + ( )2 vnj
n
2 n
jT 0 1 T 2
0
= xn +
y + ( ) [(2j − 1) vn0 + (2j − 1) vn1 + ... + vnj−1 ] +
n n 2 n
T
T
1 T
+ yn0 + ( )2 [vn0 + vn1 + ... + vnj−1 ]vnj + ( )2 vnj
n
n
2 n
T 0 1 T 2
0
0
= xn + (j + 1) yn + ( ) [(2j + 1) vn + (2j − 1) vn1 + ... + vnj ],
n
2 n
= xjn +
and
T j
T
v = yn0 + [vn0 + vn1 + ... + vnj ].
n n
n
Therefore the relations in (6) are satisfied for each j, with 1 ≤ j ≤ n and our claim
was proved.
Now, by (6) it follows easily that
ynj+1 = ynj +
xjn − x0
≤
=
jT
1 T
y0 + ( )2 [(2j − 1) + (2j − 3) + ... + 3 + 1] M
n
2 n
jT
1
1
jT
y0 + M ( )2 < T y0 + M T 2 < r.
n
2
n
2
and
jT
M < r,
n
proving that (xjn , ynj ) ∈ K := Br (x0 , y0 ) , for each j, with 1 ≤ j ≤ n.
By (4) we have that
ynj − y0 ≤
t
xn (t) = ynj + (t − tjn )vnj +
jT
n
fn (s) ds,
xn (t) = vnj + fn (t) , ∀t ∈ Inj ,
hence

xn (t) ≤ M + m (t) , ∀t ∈ [0, T ] ,





xn (t) ≤ y0 + 2r, ∀t ∈ [0, T ]





xn (t) ≤ x0 + 2r (T + 1) , ∀t ∈ [0, T ]
(7)
Moreover, for all t ∈ [0, T ] we have
d ((xn (t), xn (t), xn (t) − fn (t) , graph(F )) ≤
2r(T + 1)
.
n
Then, by (7), we have
T
xn (t)
0
2
T
dt ≤
2
(M + m (t)) dt
0
(8)
V. Lupulescu
119
and therefore the sequence (xn )n is bounded in L2 ([0, T ] , Rm ) .
For all τ, t ∈ [0, T ], we have that
t
x (t) − xn (τ ) ≤
t
xn (s) ds ≤
τ
2
(M + m (s)) ds
τ
so that the sequence (xn )n is equiuniformly continuous. Moreover, by (7) we see that
(xn )n is equi-Lipschitzian, hence equiuniformly continuous.
Therefore, (xn )n is bounded in L2 ([0, T ] , Rm ) , (xn )n and (xn )n are bounded in
C ([0, T ] , Rm ) and equiuniformly continuous, hence, by Theorem 0.3.4 in [2] there
exist a subsequence, still denoted by (xn )n , and an absolutely continuous function
x : [0, T ] → Rm such that
(i) (xn )n converges uniformly to x;
(ii) (xn )n converges uniformly to x ;
(iii) (xn )n converges weakly in L2 ([0, T ] , Rm ) to x .
Since (fn (.))n converges to f (., (.)) in L2 ([0, T ] , Rm ), then, by (H2 ), (8) and Theorem 1.4.1 in [2] we obtain
x (t) − f (t, x (t) , x (t)) ∈ coF (x (t) , x (t)) ⊂ ∂V (x (t)) , a.e., t ∈ [0, T ] ,
(9)
where co stands for the closed convex hull.
By (9) and Lemma 3.3 in [5] we obtain that
d
V (x (t)) = x (t) , x (t) − f (t, x (t) , x (t)) , a.e., t ∈ [0, T ] ,
dt
hence,
T
V (x (T )) − V (x (0)) =
x (t)
0
2
T
dt −
x (t) , f (t, x (t) , x (t)) dt.
(10)
0
On the other hand, since
3
xn (t) − fn (t) = vnj ∈ F (xjn , ynj ) ⊂ ∂V (xn (tjn )), ∀t ∈ Inj ,
and so from the properties of the subdifferential of a convex function, it follows that
j
V (xn (tj+1
n )) − V (xn (tn )) ≥
j
xn (t) − fn (t) , xn (tj+1
n ) − xn (tn )
tj+1
n
=
xn (t) − fn (t) ,
tj+1
n
x (t)
=
tjn
2
tjn
33
xn (s) ds
tj+1
n
dt −
tjn
fn (t) , xn (t) dt.
By adding the n inequalities from above, we get
T
V (xm (T )) − V (y0 ) ≥
xn (t)
0
2
T
dt −
fn (t) , xn (t) dt.
0
(11)
120
Differential Inclusions
The convergence of (fn (.))n in L2 -norm and of (xn (.))n in the weak topology of L2
implies that
T
T
lim
n→∞
fn (t) , xn (t) dt =
0
f (t, x(t), x (t)) , x (t) dt.
0
By passing to the limit as n → ∞ in (11) and using the continuity of V we see that
T
V (x (T )) − V (y0 ) ≥ lim sup
n→∞
xn (t)
0
2
T
dt −
f (t, x(t), x (t)) , x (t) dt, (12)
0
hence, by (10) and (12), we obtain
x (t)
2
L2
≥ lim sup xn (t)
n→∞
2
L2
.
Since, by the weak lower semicontinuity of the norm,
x (t)
2
2
L2
≤ lim inf xn (t)
n→∞
2
L2
,
2
we have that x (t) L2 = limn→∞ xn (t) L2 i.e. (xn )n converge to x strongly in
L2 ([0, T ] , Rm ) (Proposition III.30 in [6]). Hence a subsequence again denoted by (xn )n
converge pointwise to x .
Since by (H1 ) the graph of F is closed and, by (8) ,
lim d ((xn (t) , xn (t) , xn (t) − fn (t)) , graph (F )) = 0,
n→∞
we obtain that
x (t) ∈ F (x (t) , x (t)) + f (t, x(t), x (t)) , a.e., t ∈ [0, T ] .
Since x obviously satisfies the initial conditions, it is a solution of the problem (1).
4
An Example
For D ⊂ Rn and x ∈ D, denote by TD (x) the Bouligand’s contingent cone of D at x,
defined by
d(x + hv, D)
TD (x) = v ∈ Rm ; lim inf
=0 .
+
h
h→0
Also, ND (x) is the normal cone of D at x, defined by
ND (x) = {v ∈ Rm ; y, v ≤ 0, (∀) v ∈ TD (x)}.
In what follows we consider D as a closed subset such that θ ∈ D and θ ∈
/ Int(D),
where θ is the zero element of Rm .
We set K = TD (θ), Q = Int (ND (θ)), Ω = B1 (θ) × Q and denote by πK (y) the
projection
πK (y) = {u ∈ K : d(y, u) = d(y, K)}.
V. Lupulescu
121
THEOREM 2. Suppose Int (ND (x)) = ∅ and f : R × Rm × Rm → Rm satisfies
the assumption (H3 ) . Then there exist T > 0 and a solution x : [0, T ] → Rm for the
following Cauchy problem
x ∈ (1 − x ) πK (x ) + f (t, x, x ), (x(0), x (0)) = (x0 , y0 ) .
PROOF. By Proposition 2 in [4] there exists a convex function V such that
πK (y) ⊂ ∂V (y), (∀) y ∈ Q.
We recall (see [4]) that the function V is defined by
V (y) = sup{ϕu (y); u ∈ K},
where
1
u 2 , y ∈ Q.
2
Also, we observe that the following assertions are equivalent:

 (i) u ∈ πK (y) ;
(ii) y − u ≤ y − v , (∀) v ∈ K;

(iii) ϕu (y) ≥ ϕv (y), (∀) v ∈ K.
ϕu (y) = u, y −
(13)
Let (x, y) ∈ Ω be and let z ∈ F (x, y). Then there exists u ∈ πK (y) such that z =
(1 − x ) u. We have that
ϕ(1−
x )u (y)
1
2
(1 − x ) u 2
2
1
≥ (1 − x ) u, y − (1 − x ) u 2
2
1
1
2
= u, y − u − x ( u, y − u 2 )
2
2
= (1 − x ) ϕu (y),
=
(1 − x ) u, y −
hence
ϕ(1−
x )u (y)
≥ (1 − x ) ϕu (y).
(14)
Since u ∈ πK (y) , then ϕu (y) ≥ ϕv (y), for every v ∈ K, and by (14) it follows that
ϕ(1−
x )u (y)
− ϕv (y) ≥ (1 − x ) ϕu (y) − ϕv (y)
≥ (1 − x ) ϕv (y) − ϕv (y) = − x ϕv (y),
hence
ϕ(1−
x )u (y)
− ϕv (y) ≥ − x ϕv (y)
for every v ∈ K.
Since y ∈ Q = Int (ND (θ)) we have that
y, v ≤ 0, for every v ∈ K = TD (θ) ,
(15)
122
Differential Inclusions
hence
ϕv (y) = y, v −
1
v
2
2
≤ 0, for every v ∈ K.
(16)
From (15) and (16), it follows that
ϕ(1−
x )u (y)
≥ ϕv (y), v ∈ K.
(17)
Then (17) and the equivalent assertions in (13) imply that
z = (1 − x ) u ∈ πK (y) ⊂ ∂V (y).
m
If we define the multifunction F : Ω → 2R
by
F (x, y) = (1 − x ) πK (y) ,
then F is with compact valued and upper semicontinuous and there exists a convex
function V : Rm → R such that
F (x, y) ⊂ ∂V (y), (∀) (x, y) ∈ Ω.
Therefore, F and f satisfies assumptions (H1 ) , (H2 ) (H3 ) and our proof is complete.
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123
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