INCLUSIONS IN NON SEPARABLE BANACH SPACES
AURELIAN CERNEA
We consider a Cauchy problem for a Sturm-Liouville type differential inclusion in non separable Banach spaces under Filippov type assumptions and prove the existence of solutions.
AMS 2000 Subject Classification: 34A60.
Key words: Lusin measurable multifunction, selection, solution set.
1. INTRODUCTION
In this paper we study second-order differential inclusions of the form (1.1) (p(t)x0(t))0 ∈F(t, x(t)) a.e. ([0, T]), x(0) =x0, x0(0) =x1, where F : [0, T]×X→ P(X) is a set-valued map,X a non separable Banach space, x0, x1∈X and p(·) : [0, T]→(0,∞) is continuous.
In some recent papers ([5, 9]), several existence results for problem (1.1) are obtained using fixed point techniques. In [4] it is shown that Filippov’s ideas ([7]) can be suitably adapted in order to prove the existence of solutions to problem (1.1). All these approaches are valid provided that the Banach space X is separable.
De Blasi and Pianigiani [6] established the existence of mild solutions for semilinear differential inclusions on an arbitrary, not necessarily separable, Banach space X. Even if the ideas of Filippov are still present, the approach in [6] has a fundamental difference which consists in the construction of the measurable selections of the multifunction. This construction does not use classical selection theorems as those of Kuratowsky and Ryll-Nardzewski [8]
or Bressan and Colombo [1].
The aim of this paper is to obtain an existence result for problem (1.1) similar to that in [6]. We will prove the existence of solutions for problem (1.1) in an arbitrary space X under assumptions on F of Filippov type. We note that similar results for other classes of differential inclusions were obtained in [2, 3].
MATH. REPORTS10(60),3 (2008), 205–211
The paper is organized as follows. In Section 2 we present the notation, definitions and preliminary results to be used in the sequel while in Section 3 we prove the main result.
2. PRELIMINARIES
Consider an arbitrary real Banach space X with norm | · | and the cor- responding metric d(·,·). Let P(X) be the space of all bounded nonempty subsets of X endowed with the Hausdorff pseudometric
dH(A, B) = max{d∗(A, B),d∗(B, A)}, d∗(A, B) = sup
a∈A
d(a, B), where d(x, A) = infa∈A|x−a|,A⊂X,x∈X.
LetL be the σ-algebra of the (Lebesgue) measurable subsets of R and, for A∈ L, let µ(A) be the Lebesgue measure of A.
LetX be a Banach space and Y a metric space. An open (resp. closed) ball in Y with center y and radius r is denoted by BY(y, r) (resp. BY(y, r).
In what follows B=BX(0,1).
A multifunction F :Y → P(X) with closed bounded nonempty values is said to be dH-continuous at y0 ∈ Y if for every ε > 0 there exists δ > 0 such that for any y ∈ BY(y0, r) we have dH(F(y), F(y0)) ≤ ε. F is called dH-continuous if it is dH-continuous at each point y0 ∈Y.
LetA∈ L, with µ(A)<∞. A multifunctionF :Y → P(X) with closed bounded nonempty values is said to be Lusin measurable if for every ε > 0 there exists a compact set Kε ⊂A with µ(A\Kε)< ε such that F restricted toKε is dH-continuous.
It is clear that ifF, G:A→ P(X) andf :A→X are Lusin measurable then the same property is enjoyed by F restricted toB (B ⊂A measurable), F +G and t → d(f(t), F(t)). Moreover, the uniform limit of a sequence of Lusin measurable multifunctions also is Lusin measurable.
Let denote byIthe interval [0, T],T >0. Consider a set-valued mapping, F :I×X → P(X) andx0, x1 ∈X.
A continuous mappingx(·)∈C(I, X) is called a solution of problem (1.1) if there exists a Lusin measurable function f(·) ∈ L1(I, X), (Bochner) inte- grable such that
(2.1) f(t)∈F(t, x(t)) a.e.(I),
(2.2) x(t) =x0+p(0)x1 Z t
0
1 p(s)ds+
Z t 0
1 p(s)
Z s 0
f(u)duds ∀t∈I.
Note that if we denoteS(t) :=Rt 0
1
p(s)ds,t∈I, then (2.2) may be rewrit- ten as
(2.3) x(t) =x0+p(0)x1S(t) + Z t
0
S(t−u)f(u)du ∀t∈I,
We shall call (x(·), f(·)) a trajectory-selection pair of (1.1) if (2.1) and (2.2) are satisfied.
In what follows X is a real Banach space and we assume the following hypotheses.
Hypothesis 2.1. (i) F(·,·) : I ×X → P(X) has nonempty closed bounded values and F(·, x) is Lusin measurable on I for any x∈X.
(ii)There exists l(·)∈L1(I,(0,∞))such that
dH(F(t, x1), F(t, x2))≤l(t)|x1−x2| ∀t∈I,∀x1, x2∈X.
(iii) There existsq(·)∈L1(I,(0,∞)) such that∀t∈I we have F(t,0)⊂q(t)B.
Setm(t) =Rt
0 l(u)du,t∈I, andM := supt∈I p(t)1 . Note that|S(t)| ≤M t
∀t∈I.
The next technical results summarized in Lemma 2.2 are essential in the proof of our result. For the proof we refer to [6].
Lemma2.2 ([6]). (i)LetFi :I → P(X),i= 1,2, be two Lusin measurable multifunctions and let εi>0, i= 1,2 be such that
H(t) := (F1(t) +ε1B)∩(F2(t) +ε2B)6=∅, ∀t∈I.
Then the multifunction H :I → P(X) has a Lusin measurable selection h:I →X.
(ii)Assume that Hypothesis 2.1is satisfied. Then, for any x(·) :I →X continuous, u(·) :I →X measurable and ε >0,
a)the multifunction t→F(t, x(t))is Lusin measurable on I;
b) the multifunctionG:I → P(X) defined by
G(t) := (F(t, x(t)) +εB)∩BX(u(t),d(u(t), F(t, x(t))) +ε) has a Lusin measurable selection g:I →X.
3. MAIN RESULT
We are ready now to prove our main result.
Theorem 3.1. We assume that Hypothesis 2.1 is satisfied. Then for every x0, x1∈X the Cauchy problem(1.1)has a solution x(·) :I →X.
Proof. Let us note first that, ifz(·) : I → X is continuous, then every Lusin measurable selection u:I →X of the multifunctiont→F(t, z(t)) +B is Bochner integrable on I. More exactly, for any t∈I we have
|u(t)| ≤dH(F(t, z(t)) +B,0)≤dH(F(t, z(t)), F(t,0))+
+ dH(F(t,0),0) + 1≤l(t)|z(t)|+q(t) + 1.
Let 0 < ε < 1, εn = 2n+2ε . Consider an arbitrary Lusin measurable, Bochner integrable function f0(·) :I →X and define
x0(t) =x0+p(0)S(t)x1+ Z t
0
S(t−u)f0(u)du, t∈I.
Since x0(·) is continuous, by Lemma 2.2 (ii) there exists a Lusin measurable function f1(·) :I →X satisfying
f1(t)∈(F(t, x0(t)) +ε1B)∩B(f0(t),d(f0(t), F(t, x0(t))) +ε1), t∈I.
Obviously, f1(·) is Bochner integrable onI. Define x1(·) :I →X by x1(t) =x0+p(0)S(t)x1+
Z t 0
S(t−u)f1(u)du, t∈I.
By induction, we construct a sequence xn:I →X, n≥2 given by (3.1) xn(t) =x0+p(0)S(t)x1+
Z t 0
S(t−u)fn(u)du, t∈I, where fn(·) :I →X a Lusin measurable function such that (3.2)
fn(t)∈(F(t, xn−1(t)) +εnB)∩B(fn−1(t),d(fn−1(t), F(t, xn−1(t)))+εn), t∈I.
At the same time, as we saw at the begining of the proof,fn(·) also is Bochner integrable.
From (3.2), forn≥2 and t∈I we obtain
|fn(t)−fn−1(t)| ≤d(fn−1(t), F(t, xn−1(t))) +εn≤
≤d(fn−1(t), F(t, xn−2(t))) + dH(F(t, xn−2(t)), F(t, xn−1(t))) +εn≤
≤εn−1+l(t)|xn−1(t)−xn−2(t)|+εn. Since εn−1+εn< εn−2 we deduce that
(3.3) |fn(t)−fn−1(t)| ≤εn−2+l(t)|xn−1(t)−xn−2(t)|, n≥2.
Denoteq0(t) := d(f0(t), F(t, x0(t))),t∈I. We next prove by recurrence that for n≥2 andt∈I we have
(3.4) |xn(t)−xn−1(t)| ≤
n−2
X
k=0
Z t 0
εn−2−k
(M T)k+1(m(t)−m(u))k
k! du+
+ε0
Z t 0
(M T)n(m(t)−m(u))n−1
(n−1)! du+
Z t 0
(M T)n(m(t)−m(u))n−1
(n−1)! q0(u)du.
We start with n= 2. By (3.1), (3.2) and (3.3), fort∈I one has
|x2(t)−x1(t)| ≤ Z t
0
|S(t−s)| · |f2(s)−f1(s)|ds≤
≤ Z t
0
M T[ε0+l(s)|x1(s)−x0(s)|]ds≤
≤ε0M T t+ Z t
0
M T l(s) Z s
0
|S(s−u)| · |f1(u)−f0(u)|du
ds≤
≤ε0M T t+ Z t
0
(M T)2l(s) Z s
0
(q0(u) +ε1)du
ds≤
≤ε0M T t+ Z t
0
(M T)2(q0(u) +ε1) Z t
u
l(s)ds
du=
=ε0M T t+ Z t
0
(M T)2(m(t)−m(s))[q0(s) +ε0]ds, i.e, (3.4) is verified for n= 2. Using again (3.2) and (3.3) we have
|xn+1(t)−xn(t)| ≤ Z t
0
|S(t−u)| · |fn+1(u)−fn(u)|du≤
≤ Z t
0
M T[εn−1+l(s)|xn(s)−xn−1(s)|]ds≤
≤εn−1M T t+ Z t
0
l(s) n−2
X
k=0
Z s 0
εn−2−k
(M T)k+2(m(s)−m(u))k
k! du+
+ Z s
0
(M T)n+1(m(s)−m(u))n−1
(n−1)! (q0(u) +ε0)du
ds=
=εn−1M T t+
n−2
X
k=0
εn−2−k
Z t 0
Z s 0
(M T)k+2(m(s)−m(u))k
k! l(s)du
ds+
+ Z t
0
l(s) Z s
0
(M T)n+1(m(s)−m(u))n−1
(n−1)! l(s)[q0(u) +ε0]du
ds=
=εn−1M T t+
n−2
X
k=0
εn−2−k
Z t 0
Z t u
(M T)k+2(m(s)−m(u))k
k! l(s)ds
du+
+ Z t
0
Z t u
(M T)n+1(m(s)−m(u))n−1 (n−1)! l(s)ds
[q0(u) +ε0]du=
=εn−1M T t+
n−2
X
k=0
εn−2−k
Z t 0
(M T)k+2(m(s)−m(u))k+1
(k+ 1)! du+
+ Z t
0
(M T)n+1(m(s)−m(u))n
n! [q0(u) +ε0]du=
=
n−1
X
k=0
εn−1−k
Z t 0
(M T)k+1(m(s)−m(u))k
k! du+
+ Z t
0
(M T)n+1(m(s)−m(u))n
n! [q0(u) +ε0]du, and statement (3.4) holds for n+ 1.
It follows from (3.4) that forn≥2 and t∈I one has (3.5) |xn(t)−xn−1(t)| ≤an,
where an=
n−2
X
k=0
εn−2−k
(M T)k+1m(T)k
k! +(M T)nm(T)n−1 (n−1)!
Z 1 0
q0(u)du+ε0
. Obviously, the series whose nth term is an is convergent. So, from (3.5) we have thatxn(·) converges uniformly onIto a continuous functionx(·) :I →X.
On the other hand, by (3.5) we have
|fn(t)−fn−1(t)| ≤εn−2+l(t)an−1, t∈I, n≥3,
which implies that the sequencefn(·) converges to a Lusin measurable function f(·) :I →X. Sincexn(·) is bounded and
|fn(t)| ≤l(t)|xn−1(t)|+q(t) + 1
we deduce that f(·) also is Bochner integrable. Letting n→ ∞ in (3.1) and using Lebesgue dominated convergence theorem, yield
x(t) =x0+p(0)S(t)x1+ Z t
0
S(t−u)f(u)du, t∈I.
On the other hand, from (3.2) we get
fn(t)∈F(t, xn(t)) +εnB, t∈I, n≥1.
Letting n → ∞ we obtain f(t) ∈ F(t, x(t)), t ∈ I, and the proof is com- plete.
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Received 28 January 2008 University of Bucharest
Faculty of Mathematics and Computer Science Str. Academiei 14
010014 Bucharest, Romania
