FOR MILD SOLUTIONS OF A SECOND-ORDER DIFFERENTIAL INCLUSION
AURELIAN CERNEA
We consider a Cauchy problem for a semilinear second-order differential inclusion in separable and non separable Banach spaces under Filippov type assumptions and prove the existence of solution.
AMS 2000 Subject Classification: 34A60.
Key words: cosine family of operators, mild solution, Lusin measurable multifunc- tion.
1. INTRODUCTION
In this paper we study second-order differential inclusions of the form (1.1) x00∈Ax+F(t, x), x(0) =x0, x0(0) =y0,
where F : [0, T]×X → P(X) is a set valued map and A is the infinitesimal generator of a strongly continuous cosine family of operators {C(t); t ∈ R}
on a Banach space X.
The existence of mild solutions of problem (1.1) has been proved in [1, 2, 4] via fixed point techniques. The aim of this note is to prove a re- sult analogous to the Filippov theorem concerning the existence of solutions to a Lipschitz differential inclusion (see [7]). More precisely, we show that Filippov’s ideas can be suitably adapted in order to prove the existence of mild solutions of problem (1.1).
The first result is obtained in the case when X is a separable Banach space and its proof follows the same pattern as in [8], where a similar result is obtained for mild solutions of semilinear differential inclusions of the form (1.2) x0 ∈Ax+F(t, x), x(0) =x0,
where A is the infinitesimal generator of a strongly continuous semigroup {T(t); t≥0}on a separable Banach space X.
On the other hand, De Blasi and Pianigiani [5] established the existence of mild solutions for problem (1.2) on an arbitrary, not necessarily separable,
REV. ROUMAINE MATH. PURES APPL.,54(2009),1, 1–11
Banach spaceX. Even if the ideas of Filippov are still present, the approach in [5] has a fundamental difference which consists in the construction of measu- rable selections of the multifunction. This construction does not use classical selection theorems as in Kuratowsky and Ryll-Nardzewski [9] or Bressan and Colombo [3]. Our second result extends the above mentioned result to pro- blem (1.1).
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 results.
2. PRELIMINARIES
Let I denote the interval [0, T], T > 0, and let X be a real Banach space with the norm | · | and the corresponding metric d(·,·). Denote by B the closed unit ball in X and by B(X) the Banach space of bounded linear operators from X intoX.
We recall that a family{C(t); t∈R} of operators inB(X) is said to be a strongly continuous cosine family if
(i)C(0) =I, whereI is the identity operator inX;
(ii)C(t+s) +C(t−s) = 2C(t)C(s) ∀t, s∈R;
(iii) the mapt→C(t)y is strongly continuous ∀y∈X.
The strongly continuous sine family {S(t); t ∈ R} associated with a strongly continuous cosine family {C(t); t∈R} is defined by
S(t)y:=
Z t 0
C(s)yds, y∈X, t∈R.
Let M ≥ 0 be such that |C(t)| ≤ M ∀t ∈ I (e.g., [10]). Note that
|S(t)| ≤M t∀t∈I.
The infinitesimal generatorA:X →X of a cosine family{C(t); t∈R}
is defined by
Ay= d2
dt2
C(t)y|t=0.
Fore more details on strongly continuous cosine and sine family of operators we refer to [6, 10].
In what follows,A is the infinitesimal generator of a cosine family{C(t);
t ∈ R} on a real separable Banach space X and F(·,·) : I ×X → P(X) is a set-valued map with nonempty closed values, which define the Cauchy problem associated to a second-order differential inclusion, namely,
(2.1) x00∈Ax+F(t, x), x(0) =x0, x0(0) =y0.
A continuous mappingx(·)∈C(I, X) is called amild solutionof problem (2.1) if there exists a (Bochner) integrable function f(·)∈L1(I, X) such that (2.2) f(t)∈F(t, x(t)) a.e.(I)
(2.3) x(t) =C(t)x0+S(t)y0+ Z t
0
S(t−u)f(u)du ∀t∈I,
i.e., f(·) is a (Bochner) integrable selection of the set-valued map F(·, x(·)) and x(·) is the mild solution of the Cauchy problem
(2.4) x00=Ax+f(t), x(0) =x0, x0(0) =y0.
We shall call (x(·), f(·)) a trajectory-selection pairof (2.1) iff(·) verifies (2.2) and x(·) is a mild solution of (2.4).
For the solution sets of (2.1) we shall use the notation (2.5)
S(x0, y0) ={(x(·), f(·)); (x(·), f(·)) is a trajectory-selection pair of (2.1)}.
In what follows, u0, v0 ∈ X, g(·) ∈ L1(I, X) and y(·) ∈ C(I, X) is a mild solution of the Cauchy problem
(2.6) y00 =Ay+g(t), y(0) =u0, y0(0) =v0.
Hypothesis 2.1. i) A is an infinitesimal generator of a given strongly continuous bounded cosine family {C(t); t ∈ R} on the separable Banach space X.
ii) F(·,·) : I ×X → P(X) has nonempty closed values and F(·, x) is measurable for every x∈X.
iii) There exist β >0 and L(·) ∈L1(I,(0,∞)) such that F(t,·) is L(t)- Lipschitz on y(t) +βB for almost allt∈I, to mean that
dH(F(t, x1), F(t, x2))≤L(t)|x1−x2| ∀x1, x2∈y(t) +βB, where dH(A, C) is the Hausdorff distance
dH(A, C) = max{d∗(A, C), d∗(C, A)}, d∗(A, C) = sup{d(a, C);a∈A}.
iv)The function t→γ(t) :=d(g(t), F(t, y(t))is integrable on I.
Setm(t) = eM TR0tL(u)du,t∈I.
OnC(I, X)×L1(I, X) we consider the norm
|(x, f)|C×L=|x|C+|f|1 ∀(x, f)∈C(I, X)×L1(I, X), where, as usual, |x|C = supt∈I|x(t)|, x ∈ C(I, X) and |f|1 = RT
0 |f(t)|dt, f ∈L1(I, X).
The technical results summarized in the next lemma are well known in the theory of set-valued maps. We refer, for example, to [8].
Lemma 2.2([8]). Let X be a separable Banach space, letH :I → P(X) be a measurable set-valued map with nonempty closed values and g, h:I →X, L:I →(0,∞) measurable functions. Then
i)the function t→d(h(t), H(t) is measurable;
ii) if H(t)∩(g(t) +L(t)B) 6= ∅ a.e. (I) then the set-valued map t → H(t)∩(g(t) +L(t)B) has a measurable selection.
Moreover, if Hypothesis 2.1 is satisfied and x(·) ∈ C(I, X) with ||x− y||C ≤β then the set-valued mapt→F(t, x(t))is measurable.
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 inY with centeryand radiusrwill be denoted byBY(y, r) (resp.BY(y, r).
A multifunctionF :Y → P(X) with closed bounded nonempty values is said to be dH-continuous at y0 ∈Y if for every ε > 0 there existsδ >0 such thatdH(F(y), F(y0))≤εfor anyy ∈BY(y0, r)F is calleddH-continuous if it is so at each point y0∈Y.
LetA∈ L withµ(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 thatF restricted toKε is dH-continuous.
It is clear that ifF, G:A→ P(X) andf :A→X are Lusin measurable, then so areF restricted toB(measurableB⊂A),F+Gandt→d(f(t), F(t)).
Moreover, the uniform limit of a sequence of Lusin measurable multifunctions also is Lusin measurable.
If X is not separable, a continuous mapping x(·) ∈ C(I, X) is called a mild solution of problem (2.1) if there exists a Lusin measurable function f(·)∈L1(I, X), (Bochner) integrable such that (2.2) and (2.3) are satisfied.
In what follows, X is a real Banach space and we assume the following hypotheses.
Hypothesis 2.3. i)A is the infinitesimal generator of a given strongly continuous bounded cosine family {C(t); t∈R}.
ii) F(·,·) : I ×X → P(X) has nonempty closed bounded values and F(·, x) is Lusin measurable on I for any x∈X.
iii)There exists l(·)∈L1(I,(0,∞)) such that∀t∈I, dH(F(t, x1), F(t, x2))≤l(t)|x1−x2|, ∀x1, x2∈X.
iv)There exists q(·) ∈L1(I,(0,∞)) such that ∀t∈I we have F(t,0)⊂ q(t)B.
Setn(t) =Rt
0l(u)du,t∈I. The technical results summarized in the next lemma are essential in the proof of our result. For the in proofs we refer to [5].
Lemma 2.4([5]). 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 2.3 is 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. THE MAIN RESULTS We are ready now to prove our main results.
Theorem 3.1. Consider δ ≥0, assume that Hypothesis2.1 is satisfied, and set, η(t) =m(t)(δ+M TRt
0γ(s)ds).
If η(T)≤β, then for any x0, y0 ∈X with M(|x0−u0|+T|y0−v0|)≤δ and any ε >0 there exists (x(·), f(·))∈ S(x0, y0) such that
|x(t)−y(t)| ≤η(t) +εM T tm(t), ∀t∈I,
|f(t)−g(t)| ≤L(t)(η(t) +εM T tm(t)) +γ(t) +ε a.e.(I).
Proof. Let ε > 0 such that η(T) +εM T2m(T) < β and set χ(t) = δ+M T Rt
0 γ(s)ds+εM T t,x0(t)≡y(t),f0(t)≡g(t),t∈I.
We claim that is enough to construct sequencesxn(·)∈C(I, X),fn(·)∈ L1(I, X),n≥1, with the properties below:
(3.1) xn(t) =C(t)x0+S(t)y0+ Z t
0
S(t−s)fn(s)ds, ∀t∈I, (3.2) |x1(t)−x0(t)| ≤χ(t), ∀t∈I,
(3.3) |f1(t)−f0(t)| ≤γ(t) +ε a.e. (I), (3.4) fn(t)∈F(t, xn−1(t)) a.e. (I), n≥1,
(3.5) |fn+1(t)−fn(t)| ≤L(t)|xn(t)−xn−1(t)| a.e. (I), n≥1.
Indeed, from (3.1), (3.2) and (3.5), for almost allt∈I we have
|xn+1(t)−xn(t)| ≤ Z t
0
|S(t−t1)| · |fn+1(t1)−fn(t1)|dt1
≤M T Z t
0
L(t1)|xn(t1)−xn−1(t1)|dt1
≤M T Z t
0
L(t1) Z t1
0
|S(t1−t2)| · |fn(t2)−fn−1(t2)|dt2
≤(M T)2 Z t
0
L(t1) Z t1
0
L(t2)|xn−1(t2)−xn−2(t2)|dt2dt1
≤(M T)n Z t
0
L(t1) Z t1
0
L(t2)· · · Z tn−1
0
L(tn)|x1(tn)−y(tn)|dtn· · ·dt1
≤χ(t)(M T)n Z t
0
L(t1) Z t1
0
L(t2)· · · Z tn−1
0
L(tn)dtn· · ·dt1
=χ(t)(M TRt
0L(s)ds)n
n! .
Therefore {xn(·)} is a Cauchy sequence in the Banach space C(I, X).
Thus, from (3.5) for almost all t ∈ I, the sequence {fn(t)} is Cauchy in X.
Moreover, from (3.2) and the last inequality we have (3.6) |xn(t)−y(t)| ≤ |x1(t)−y(t)|+
n−1
X
i=2
|xi+1(t)−xi(t)|
≤χ(t)
1 +M T Z t
0
L(s)ds+ (M TRt
0L(s)ds)2 2! · · ·
≤χ(t)eM TR0tL(s)ds=η(t) +εM T tm(t), and taking into account the choice of εwe get
(3.7) |xn(·)−y(·)|C ≤β, ∀n≥0.
On the other hand, from (3.3), (3.5) and (3.6) for almost allt∈Iwe have (3.8) |fn(t)−g(t)| ≤
n−1
X
i=1
|fi+1(t)−fi(t)|+|f1(t)−g(t)|
≤L(t)
n−2
X
i=1
|xi(t)−xi−1(t)|+γ(t) +ε≤L(t)(η(t) +εtm(t)) +γ(t) +ε.
Letx(·)∈C(I, X) be the limit of the Cauchy sequence xn(·). By (3.8), the sequence fn(·) is bounded in the mean and we have already proved that the sequence {fn(t)}is Cauchy in X for almost all t∈I. Let f(·)∈L1(I, X) with f(t) = limn→∞fn(t).
By Hypothesis 2.1.2 iii), for almost all t ∈ I the set Q(t) = {(x, v);
v ∈F(t, x), |x−y(t)| ≤ β} is closed. In addition, (3.4) and (3.7) imply that (xn−1(t), fn(t))∈Q(t) forn≥1 andt∈I. So, passing to the limit, we deduce that (2.2) holds for almost all t∈I.
Moreover, passing to the limit in (3.1) and using the Lebesgue dominated convergence theorem, we get (2.3). Finally, passing to the limit in (3.6) and (3.8) we obtain the desired estimations.
It remains to construct the sequences xn(·), fn(·) with the properties (3.1)–(3.5). The construction will be done by induction. First, we apply Lem- ma 2.2 and deduce that the set-valued map t→F(t, y(t)) is measurable with closed values and
F(t, y(t))∩ {g(t) + (γ(t) +ε)B} 6=∅ a.e. (I).
From Lemma 2.2 we find a measurable selection f1(·) of the set-valued map H1(t) :=F(t, y(t))∩{g(t)+(γ(t)+ε)B}. Obviously,f1(·) satisfies (3.3). Define x1(·) as in (3.1) withn= 1. Therefore, we have
|x1(t)−y(t)| ≤ |C(t)(x0−u0)|+|S(t)(y0−v0)|+
Z t 0
S(t−s)(f1(s)−g(s))ds
≤δ+M Z t
0
(γ(s) +ε)ds≤η(t) +M T εt≤β.
Assuming that for some N ≥1 we have already constructed xn(·) ∈C(I, X) and fn(·) ∈ L1(I, X), n = 1,2, . . . , N, satisfying (3.1)–(3.5), we define the set-valued map
HN+1(t) :=F(t, xN(t))∩ {fN(t) +L(t)|xN(t)−xN−1(t)|B}, t∈I.
By Lemma 2.2 the set-valued map t → F(t, xN(t)) is measurable and since F(t,·) is Lipschitz, HN+1(t) 6= ∅ for almost all t ∈ I. We apply Lemma 2.2 and find a measurable selection fN+1(·) ofF(·, xN(·)) such that
|fN+1(t)−fN(t)| ≤L(t)|xN(t)−xN−1(t)| a.e. (I)
We define xN+1(·) as in (3.1) withn=N+ 1. The proof is complete.
The next corollary of Theorem 3.1 shows the Lipschitz dependence of the mild solutions on the initial conditions.
Corollary 3.2. Let (y, g) be a trajectory-selection of(2.1)and assume that Hypothesis 2.1 is satisfied. Then there exists K > 0 such that for any η = (η1, η2) in a neighborhood of (y(0), y0(0)) we have
dC×L((y, g),S(η1, η2))≤K(|η1−y(0)|+|η2−y0(0)|).
Proof. Take 0 < ε < 1. We use Theorem 3.1 and deduce the exis- tence of δ >0 such that for anyη = (η1, η2) ∈B((y(0), y0(0)), δ) there exists a trajectory-selection (xε, fε) of (2.1) with xε(0) = η1 and x0ε(0) = η2 and such that
|xε−y|C ≤m(T)(M|η1−y(0)|+M T|η2−y0(0)|) +εM T2m(T)
and
|fε−g|1≤m(T)(M|η1−y(0)|+M T|η2−y0(0)|) +ε(M T2m(T) + 1).
Since ε >0 is arbitrary, the proof is complete.
We now consider the case whenXis not separable. For simplicity,F(·,·) is assumed to be (globally) Lipschitz instead of locally Lipschitz as in Theo- rem 3.1.
Theorem 3.1. We assume that Hypothesis 2.3 is satisfied. Then for every x0, y0 ∈X the Cauchy problem (1.1)has a solutionx(·) :I →X.
Proof. Let us first note 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 precisely, 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 functionf0(·) : I → X, that is Bochner integrable and define
x0(t) =C(t)x0+S(t)y0+ Z t
0
S(t−u)f0(u)du, t∈I.
Since x0(·) is continuous, by Lemma 2.4 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) =C(t)x0+S(t)y0+
Z t 0
S(t−u)f1(u)du, t∈I.
We construct by induction a sequencexn:I →X, n≥2, given by (3.9) xn(t) =C(t)x0+S(t)y0+
Z t
0
S(t−u)fn(u)du, t∈I, where fn(·) :I →X a Lusin measurable function satisfying (3.10)
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(·) is also Bochner integrable.
From (3.10), for n≥2, andt∈I we have
|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
(3.11) |fn(t)−fn−1(t)| ≤εn−2+l(t)|xn−1(t)−xn−2(t)|, n≥2.
Denote p0(t) :=d(f0(t), F(t, x0(t))),t∈I. Next, we prove by recurrence that for n≥2 and t∈I we have
(3.12) |xn(t)−xn−1(t)| ≤
n−2
X
k=0
Z t 0
εn−2−k
(M T)k+1(n(t)−n(u))k
k! du
+ε0 Z t
0
(M T)n(n(t)−n(u))n−1
(n−1)! du+
Z t 0
(M T)n(n(t)−n(u))n−1
(n−1)! p0(u)du.
We start with n= 2. By (3.9), (3.10) and (3.11), fort∈I we have
|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
(p0(u) +ε1)du
ds
≤ε0M T t+ Z t
0
(M T)2(p0(u) +ε1) Z t
u
l(s)ds
du
=ε0M T t+ Z t
0
(M T)2(n(t)−n(s))[p0(s) +ε0]ds, i.e., (3.12) is verified forn= 2. Using again (3.11) and (3.12) 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(n(s)−n(u))k
k! du
+ Z s
0
(M T)n+1(n(s)−n(u))n−1
(n−1)! (p0(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(n(s)−n(u))k
k! l(s)du
ds +
Z t 0
l(s) Z s
0
(M T)n+1(n(s)−n(u))n−1
(n−1)! l(s)[p0(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(n(s)−n(u))k
k! l(s)ds
du +
Z t 0
Z t u
(M T)n+1(n(s)−n(u))n−1 (n−1)! l(s)ds
[p0(u) +ε0]du
=εn−1M T t+
n−2
X
k=0
εn−2−k
Z t 0
(M T)k+2(n(s)−n(u))k+1
(k+ 1)! du
+ Z t
0
(M T)n+1(n(s)−n(u))n
n! [p0(u) +ε0]du
=
n−1
X
k=0
εn−1−k
Z t 0
(M T)k+1(n(s)−n(u))k
k! du
+ Z t
0
(M T)n+1(n(s)−n(u))n
n! [p0(u) +ε0]du, so that (3.12) holds for n+ 1.
It follows from (3.12) that for n≥2 and t∈I we have (3.13) |xn(t)−xn−1(t)| ≤an,
where an=
n−2
X
k=0
εn−2−k
(M T)k+1n(T)k
k! +(M T)nn(T)n−1 (n−1)!
Z 1 0
p0(u)du+ε0
. Obviously, the series with nth termanis convergent. So, from (3.13) we have that xn(·) converges uniformly on I to a continuous function x(·) : I → X.
On the other hand, by (3.11) 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.
Lettingn→ ∞in (3.9) and using the Lebesgue dominated convergence theorem we obtain
x(t) =C(t)x0+S(t)y0+ Z t
0
S(t−u)f(u)du, t∈I.
On the other hand, from (3.10) we get
fn(t)∈F(t, xn(t)) +εnB, t∈I, n≥1, and letting n→ ∞ we have
f(t)∈F(t, x(t)), t∈I.
The proof is complete.
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Received 29 May 2007 University of Bucharest
Faculty of Mathematics and Computer Science Str. Academiei 14
010014 Bucharest, Romania
