INTEGRODIFFERENTIAL INCLUSIONS
AURELIAN CERNEA
We prove the Lipschitz dependence on the initial condition of the solution set of innite horizon integrodierential inclusions by using an appropiate norm on the space of solutions and by applying the contraction principle in the space of selections of the multifunction instead of the space of solutions.
AMS 2000 Subject Classication: 34A60.
Key words: integrodierential inclusion, xed point, contractive set-valued map.
1. INTRODUCTION
In this paper we study nonlinear integrodierential inclusions of the form (1.1) x0∈F(t, x, V(x)(t)), x(0) =x0,
where X is a separable Banach space, F : [0,∞)×X ×X → P(X) is a set-valued map and V : C([0,∞), X) → C([0,∞), X) is a nonlinear Volterra integral operator. Qualitative properties and structure of the set of solutions of this problem have been studied by many authors. See, e.g., [2, 4, 5, 6, 7, 8, 9, 12, 14].
Here we study the properties of the map that associates with a given initial condition (0, x0) the set of solutions of (1.1) starting from x0, and our main purpose is to prove that this solution map depends Lipschitz-continuously on the initial condition in the case of innite horizon problems. The proof is based on an idea of Kannai and Tallos [11], who apply the contraction principle in the space of the selections of the multifunction instead of the space of solutions. Our result extends the result obtained in [11] for dierential inclusions (i.e., systems with no nonlinear Volterra integral operator). Since the Volterra integral operator plays a certain role in our problem, the approach to problem (1.1) is not at all straightforward and requires special computations.
The paper is organized as follows. In Section 2 we present the notation, denitions and the preliminary results to be used in the sequel while in Sec- tion 3 we prove the main result.
REV. ROUMAINE MATH. PURES APPL., 53 (2008), 4, 277283
2. PRELIMINARIES
Let denote by I the interval [0,∞) and let X be a real separable Ba- nach space with the norm k · k and the corresponding metric d(·,·). Let V : C(I, X) → C(I, X) be the nonlinear Volterra integral operator dened by V(x)(t) =Rt
0k(t, s, x(s))ds.
In what follows we assume the following hypotheses.
Hypothesis 2.1. i)F(·,·,·) :I×X×X → P(X)has nonempty closed values and F(·, x, y) is measurable for every x, y∈X.
ii) There exists L(·) ∈ L1loc(I,R+) such that L(t) ≥ 1 and F(t,·,·) is L(t)-Lipschitz for almost all t∈I, to mean that
dH(F(t, x1, y1), F(t, x2, y2))≤L(t)(kx1−x2k+ky1−y2k)∀x1, x2, y1, y2∈X, where dH(A, B) is the Hausdor distance
dH(A, B) = max{d∗(A, B),d∗(B, A)}, d∗(A, B) = sup{d(a, B);a∈A}.
iii) d(0, F(t,0, V(0)(t)))≤L(t) a.e. (I).
Hypothesis 2.2. i) k(·,·,·) : I×X×X → X is a function such that (t, s)→k(t, s, x) is measurable ∀x∈X.
ii) kk(t, s, x)−k(t, s, y)k ≤L(t)kx−yk a.e.(t, s)∈I×I,∀x, y∈X. Putm(t) =Rt
0L(u)duand letS(x0) be the solution set of problem (1.1), i.e., the set of all absolutely continuous functions that satisfy (1.1) a.e. on I.
Letα >2 be given and introduce the vector space Lα=
f(·)∈L1loc(I, X);
Z ∞
0
e−αm(t)kf(t)kdt <∞
of locally integrable functions. Lα is a Banach space with the norm kfk1 =
Z ∞
0
e−αm(t)kf(t)kdt, f ∈Lα. Consider also the vector space
ACα={x(·) :I →X; x(·) is absolutely continuous and x0(·)∈Lα} equipped with the norm
kxkα=kx(0)k+ Z ∞
0
e−αm(t)kx0(t)kdt.
As is well known, ACα is a Banach space.
We show rst that all solutions of (1.1) belong toACα.
Proposition 2.3. If α > 2 and Hypotheses 2.1 and 2.2 are satised, then S(x0)⊂ACα for every x0∈X.
Proof. Letx(·) be a solution of (1.1) and consider the map g(t) =
Z t
0
e−m(s)kx0(s)kds, t≥0.
Then for almost all twe have
g0(t) = e−m(t)kx0(t)k ≤e−m(t)[d(0, F(t,0, V(0)(t))) +dH(F(t,0, V(0)(t)), F(t, x(t), V(x)(t))]≤e−m(t)L(t)[1 +kx(t)k+kV(x)(t)−V(0)(t)k]≤
≤e−m(t)L(t)
1 +kx(t)k+ Z t
0
L(s)kx(s)kds
≤e−m(t)L(t)
1+
+kx0k+ Z t
0
kx0(s)kds+ Z t
0
L(s)kx0kds+ Z t
0
L(s) Z s
0
kx0(u)kdu
ds
=
= e−m(t)L(t)
1 + (1 +m(t))kx0k+ Z t
0
(1 +m(t)−m(s))kx0(s)kds
≤
≤L(t)
1 +kx0k+ Z t
0
e−m(s)kx0(s)kds
=L(t)[1 +kx0k+g(t)].
According to Bellman-Gronwall's lemma, we have g(t)≤(1 +kx0k)em(t)−(1 +kx0k).
Since L(t)≥1,∀t∈I, we get
kx0(t)k= em(t)g(t)≤(1 +kx0k)L(t)e2m(t), thus
Z ∞
0
e−αm(t)kx0(t)kdt≤(1 +kx0k) Z ∞
0
L(t)e(2−α)m(t)dt≤ 1 +kx0k α−2 ≤ ∞, i.e., x(·)∈ACα.
Finally, we recall some basic results concerning set valued contractions that we shall use in the sequel.
Let(Z, d) be a metric space and consider a set valued map T on Z with nonempty closed values in Z. T is said to be a λ-contraction if there exists 0< λ <1such that dH(T(x), T(y))≤λd(x, y)∀x, y∈Z.
IfZ is complete, then every set valued contraction has a xed point, i.e.
a point z∈Z such thatz∈T(z) ([10]). Denote by Fix(T)the set of all xed point of the multifunction T. Obviously, Fix(T) is closed.
Proposition 2.4 ([13]). Let Z be a complete metric space and suppose that T1, T2 are λ-contractions with closed values in Z. Then
dH(Fix(T1),Fix(T2))≤ 1 1−λ sup
z∈Z dH(T1(z), T2(z)).
3. THE MAIN RESULT
We are now ready to show that the set of all solutions of (1.1) depends Lipschitz-continuously on the initial condition.
Theorem 3.1. Let Hypotheses 2.1 and 2.2 be satised and let α > 2. Then the map x0→ S(x0) is Lipschitz-continuous on X with nonempty closed values in ACα.
Proof. Let us considerx0∈X,f(·)∈Lα and dene the set valued maps (3.1) Mx0,f(t) =F
t, x0+
Z t
0
f(u)du, Z t
0
k
t, s, x0+ Z s
0
f(u)du
ds
, t≥0, (3.2) Tx0(f) ={φ(·)∈Lα; φ(t)∈Mx0,f(t) a.e.(I)}.
We shall rst prove thatTx0(f)is nonempty and closed for everyf ∈Lα. The fact that that the set valued mapMx0,f(·)is measurable is well known. For example, the mapst→x0+Rt
0 f(u)du,t→Rt
0 k(t, s, x0+Rs
0 f(u)du)dscan be approximated by step functions and we can apply Theorem III. 40 in [3]. Since the values ofFare closed andXis separable, the measurable selection theorem (Theorem III.6 in [3]) shows that Mx0,f(·)admits a measurable selection φ(·). The same computations as in Proposition 2.3 yield
kφ(t)k ≤L(t)
1 + (1 +m(t))kx0k+ Z t
0
(1 +m(t)−m(s))kf(s)kds
. Integrating by parts and using the fact thatex≥x+ 1∀x≥0, we have
Z ∞
0
e−αm(t)kφ(t)kdt≤ Z ∞
0
e−αm(t)L(t)
1 + (1 +m(t))kx0k+
+ Z t
0
(1+m(t)−m(s))kf(s)kds
dt≤ Z ∞
0
e−αm(t)L(t)dt+kx0k Z ∞
0
e(1−α)m(t)L(t)dt +
Z ∞
0
Z ∞
s
e−αm(t)L(t)(1 +m(t)−m(s))dt
kf(s)kds≤ 1
α + kx0k α−1+ +
Z ∞
0
Z ∞
s
e(1−αm(t))L(t)dt
e−m(s)kf(s)kds≤ 1
α + kx0k α−1+ 1
α−1kfk1. Hence, ifφ(·) is a measurable selection of Mx0,f(·), thenφ(·)∈Lα, thus Tx0(f)6=∅.
The setTx0(f) is closed. Indeed, ifφn∈Tx0(f)andkφn−φk1 →0, then we can consider a subsequence φnk such that φnk(t)→φ(t) for a.e. t∈I, and we nd that φ∈Tx0(f).
The next step of the proof will show thatTx0(·) is a contraction on Lα. Lety, z ∈Lα be given,φ∈Tx0(y)and let ε >0. Consider the set valued map G(t)=Mx0,z(t)∩
x∈X;kφ(t)−xk ≤L(t) Z t
0
(1+m(t)−m(s))ky(s)−z(s)kds+ε
. Since
d(φ(t), Mx0,z(t))≤dH
F(t, x0+ Z t
0
y(u)du, Z t
0
k(t, s, x0+ Z s
0
y(u)du)ds
,
F
t, x0+ Z t
0
z(u)du,Z t 0
k
t, s, x0+ Z s
0
z(u)du ds
≤L(t) Z t
0
ky(u)−z(u)kdu+
+L(t) Z t
0
L(s) Z s
0
ky(u)−z(u)kduds=L(t) Z t
0
(1+m(t)−m(s))ky(s)−z(s)kds, we deduce that G(·) has nonempty closed values. Moreover, according to Proposition III.4 in [3], G(·)is measurable. Letψ(·)be a measurable selection of G(·). It follows thatψ∈Tx0(z) and
kφ−ψk1 = Z ∞
0
e−αm(t)kφ(t)−ψ(t)kdt
≤ Z ∞
0
e−αm(t)L(t)Z t 0
(1 +m(t)−m(s))ky(s)−z(s)kds dt+
Z ∞
0
εe−αm(t)dt
≤ Z ∞
0
hZ ∞
s
e−αm(t)L(t)(1 +m(t)−m(s))dti
ky(s)−z(s)kds+ε Z ∞
0
e−αm(t)dt
≤ Z ∞
0
hZ ∞
s
e(1−αm(t))L(t))dt i
e−m(t)ky(s)−z(s)kds+ε Z ∞
0
e−αm(t)dt
≤ 1
α−1ky−zk1+ε Z ∞
0
e−αm(t)dt.
Since εwas arbitrary, we deduce that d(φ, Tx0(z))≤ 1
α−1ky−zk1. Replacing y(·) by z(·), we obtain
d(Tx0(y), Tx0(z))≤ 1
α−1ky−zk1,
henceTx0(·) is a contraction onLα. Consequently,Tx0(·)admits a xed point f(·)∈Lα. Dene x(t) =x0+Rt
0f(u)du. Since Fix(Tx0) is closed we also get that S(x0)is a closed subset of ACα.
We next prove the inequality (3.3) dH(Tx1(y), Tx2(y))≤ 1
α−1kx1−x2k ∀y∈Lα, x1, x2∈X.
Consider the set valued map
G1(t) =Mx2,y(t)∩ {z∈X; kφ(t)−zk ≤L(t)(1 +m(t))kx1−x2k+ε}, t∈I, whereφ(·) is a measurable selection of Mx1,y(·) andε >0.
With the arguments used in the case of the set valued map G(·), we deduce that G1(·) is measurable with nonempty closed values. Let ψ(·) be a measurable selection of G1(·). It follows thatψ(·)∈Tx2(y)and
kφ−ψk1 = Z ∞
0
e−αm(t)kφ(t)−ψ(t)kdt≤ kx1−x2k Z ∞
0
L(t)e−αm(t)(1+m(t))dt +ε
Z ∞
0
e−αm(t)dt≤ kx1−x2k Z ∞
0
L(t)e(1−αm(t))dt+ε Z ∞
0
e−αm(t)dt
≤ 1
α−1kx1−x2k+ε Z ∞
0
e−αm(t)dt.
Since ε >0 was arbitrary, we deduce that d(φ, Tx2(y))≤ 1
α−1kx1−x2k.
Replacing x2 byx1, we obtain (3.3).
From (3.3) and Proposition 2.4 we have dH(Fix(Tx1),Fix(Tx2))≤ 1
α−2kx1−x2k.
Let x1, x2∈X and x(·)∈ S(x1). In particular, x0(·)∈Fix(Tx1). So, for every ε >0 there existsg(·)∈Fix(Tx2) such that
(3.4) kx0−gk1≤ 1
α−2kx1−x2k+ε.
Put y(t) =S(t)x2+Rt
0g(u)du. One has
kx−ykα =kx1−x2k+kx0−gk1 ≤ kx1−x2k+ 1
α−2kx1−x2k+ε
= α−1
α−2kx1−x2k+ε.
From the last inequality and (3.4) we nd rst that d(x,S(x2))≤ α−1
α−2kx1−x2k and by interchanging x1 andx2 we obtain
d(S(x1),S(x2))≤ α−1
α−2kx1−x2k.
The proof is complete.
Remark 3.2. If the set valued mapF does not depend on the last variable then (1.1) reduces to
x0 ∈F(t, x), x(0) =x0,
and Theorem 3.1 yields known results, namely, Theorem 1 in [11].
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Received 28 January 2008 University of Bucharest
Faculty of Mathematics and Computer Science Str. Academiei 14
010014 Bucharest, Romania
