# Control Systems Described by a Class of Fractional Semilinear Evolution Equations and Their Relaxation Property

Authors: Xiaoyou Liu; Xi Fu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We consider a control system described by a class of fractional semilinear evolution equations in a separable reflexive Banach space.
The constraint on the control is a multivalued map with nonconvex values which is lower semicontinuous with respect to the state variable. Along with the original system we also consider the system in which the constraint on the control is the upper semicontinuous convex-valued regularization of the original constraint. We obtain the existence results for the control systems and the relaxation property between the
solution sets of these systems.


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## Body

## 1. Introduction

LetJ=[0,b] and 0<α<1. We consider the following control system described by a class of fractional semilinear evolution equations of the form:
(1.1)DCtαx(t)=Ax(t)+h(t,x(t))+g(t,x(t))u(t),t∈J,x(0)=x0,
with the mixed nonconvex constraint on the control
(1.2)u(t)∈U(t,x(t))a.e.onJ.
Here CDtα is the Caputo fractional derivative of order α, b>0 is a finite real number, A is the infinitesimal generator of a strongly continuous semigroup {T(t),t≥0} in a separable reflexive Banach space X, g:J×X→ℒ(Y,X) (ℒ(Y,X) is the space of continuous linear operators from Y into X), h:J×X→X is a nonlinear function, and U:J×X→2Y∖{∅} is a multivalued mapping with closed values that is not necessarily convex. The space Y is a separable, reflexive Banach space modeling the control space.We denote byC(J,X) the space of all continuous functions from J into X with the supremum norm given by ∥x∥C=supt∈J∥x(t)∥X for x∈C(J,X). Let BX⊆X be the open unit ball centered at zero. Consider the multivalued map
(1.3)Uδ(t,x)=co¯{⋃U(t,z):z∈x+δBX},δ>0,(1.4)V(t,x)=⋂δ↓0Uδ(t,x),
here co¯ stands for the closed convex hull of a set. The map (1.4) is usually called the convex upper semicontinuous regularization of U(t,x).Along with the constraint (1.2) on the control we also consider the constraint
(1.5)u(t)∈V(t,x(t))a.e.onJ
on the control. Note that usually we have co¯U(t,x)⊆V(t,x).Definition 1.1.
A solution of the control system (1.1), (1.2) is defined to be a pair (x(·),u(·)) consisting of a trajectory x∈C(J,X) and a control u∈L1(J,Y) satisfying (1.1) and the inclusion (1.2) a.e.A solution of the control system (1.1), (1.5) is defined similarly. We denote by ℛU, 𝒯rU (ℛV, 𝒯rV) the sets of all solutions, all admissible trajectories of the control system (1.1) and (1.2) (the control system (1.1) and (1.5)).Relaxation property [1] has important ramifications in control theory. There are many papers dealing with the verification of the relaxation property for various classes of control systems. For example, we refer to [2–5] for nonlinear evolution inclusions or equations, [6, 7] for control problems of subdifferential type and the references therein. In this paper, we investigate this property for control systems described by fractional semilinear evolution equations. We will prove that 𝒯rV is a compact set in C(J,X) and
(1.6)𝒯rV=𝒯rU¯,
where the bar stands for the closure in C(J,X).Fractional calculus has recently gained much attentions due to its numerous applications in science and engineering. Examples can be found in various disciplines such as mechanics, electrophysics, signal and image processing, thermodynamics, biophysics, aerodynamics, and economics, (see [8–12] for instance). For some recent results on fractional differential equations, we can refer to [13–17]. As for the study of fractional semilinear differential equations, we can refer to Zhou and Jiao [18, 19], Wang and Zhou [20] for the existence results. The issue of approximate controllability was considered by Kumar and Sukavanam [21], Sakthivel et al. [22]. Wang and Zhou in [23] were concerned with the optimal control settings.


## 2. Preliminaries and Assumptions

LetJ=[0,b] be a closed interval of the real line with the Lebesgue measure μ and the σ-algebra Σ of μ measurable sets. The norm of the space X (or Y) will be denoted by ∥·∥X (or ∥·∥Y). For any Banach space V the symbol ω-V stands for V equipped with the weak σ(V,V*) topology. The same notation will be used for subsets of V. In all other cases we assume that V and its subsets are equipped with the strong (normed) topology.We first recall the following known definitions from fractional differential theory. For more details, please see [11, 12].Definition 2.1.
The fractional integral of orderα with the lower limit zero for a function f is defined as
(2.1)Iαf(t)=1Γ(α)∫0tf(s)(t-s)1-αds,t>0,α>0
provided the right hand side is point-wise defined on [0,∞), where Γ(·) is the gamma function.Definition 2.2.
The Riemann-Liouville derivative of orderα with the lower limit zero for a function f is defined as
(2.2)LDαf(t)=1Γ(n-α)dndtn∫0tf(s)(t-s)α+1-nds,t>0,n-1<α<n.Definition 2.3.
The Caputo derivative of orderα with the lower limit zero for a function f is defined as
(2.3)CDαf(t)=DLα(f(t)-∑k=0n-1tkk!f(k)(0)),t>0,n-1<α<n.Iff is an abstract function with values in X, then integrals which appear in Definitions 2.1 and 2.2 are taken in Bochner's sense.We now proceed to some basic definitions and results from multivalued analysis. For more details on multivalued analysis, see the books [24, 25].We use the following notations:Pf(Y) is the set of all nonempty closed subsets of Y, Pfb(Y) is the set of all nonempty, closed and bounded subsets of Y, and Pfc(Y) is the set of all nonempty, closed, and convex subsets of Y.OnPbf(Y), we have a metric known as the “Hausdorff metric” and defined by
(2.4)h(A,B)=max{supa∈Ad(a,B),supb∈Bd(b,A)},
where d(x,C) is the distance from a point x to a set C. We say a multivalued map is h-continuous if it is continuous in the Hausdorff metric h(·,·).We say that a multivalued mapF:J→Pf(Y) is measurable if F-1(E)={t∈J:F(t)∩E≠∅}∈Σ for every closed set E⊆Y. If F:J×X→Pf(Y), then measurability of F means that F-1(E)∈Σ⊗ℬX, where Σ⊗ℬX is the σ-algebra of subsets in J×X generated by the sets A×B, A∈Σ, B∈ℬX, and ℬX is the σ-algebra of the Borel sets in X.SupposeV,Z are two Hausdorff topological spaces and F:V→2Z∖{∅}. We say that F is lower semicontinuous in the sense of Vietoris (l.s.c. for short) at a point x0∈V if for any open set W⊆Z, F(x0)∩W≠∅, there is a neighborhood O(x0) of x0 such that F(x)∩W≠∅ for all x∈O(x0). F is said to be upper semicontinuous in the sense of Vietoris (u.s.c. for short) at a point x0∈V if for any open set W⊆Z, F(x0)⊆W, there is a neighborhood O(x0) of x0 such that F(x)⊆W for all x∈O(x0). For the properties of l.s.c and u.s.c, see the book [24].Besides the standard norm onLq(J,Y) (here Y is a separable, reflexive Banach space), 1<q<∞, we also consider the so called weak norm:
(2.5)∥u(·)∥ω=sup0≤t1≤t2≤b∥∫t1t2u(s)ds∥Y,foru∈Lq(J,Y).
The space Lq(J,Y) furnished with this norm will be denoted by Lωq(J,Y). The following result establishes a relation between convergence in ω-Lq(J,Y) and convergence in Lωq(J,Y).Lemma 2.4 (see [5]).
If a sequence{un}n≥1⊆Lq(J,Y) is bounded and converges to u in Lωq(J,Y), then it converges to u in ω-Lq(J,Y).We assume the following assumptions on the data of our problems in the whole paper.H
(
A
): The operator A generates a strongly continuous semigroup T(t), t≥0 in X, and there exists a constant MA≥1 such that supt∈[0,∞)∥T(t)∥≤MA. For any t>0, T(t) is compact.Remark 2.5.
Let us takeX=L2[0,π] and define the operator A by Aω=ω′′ with the domain D(A)={ω∈X:ω,ω′ are absolutely continuous, ω′′∈X, and ω(0)=ω(π)=0}. Then Aω=-∑n=1∞n2〈ω,en〉en, ω∈D(A), where en(z)=(2/π)1/2sinnz, 0≤z≤π, n=1,2,…. Clearly A generates a compact semigroup {T(t):t>0} in X and it is given by T(t)ω=∑n=1∞e-n2t〈ω,en〉en, ω∈X. In such a case, it is easy to see that H(A) holds [22].H
(
g
): The operator g:J×X→ℒ(Y,X) is such thatH
(
h
): The function h:J×X→X of Carathéodory type satisfies: there exists a constant 0<β<α such that for a.e. t∈J and all x∈X, ∥h(t,x)∥X≤a1(t)+c1∥x∥X, where a1∈L1/β(J,ℝ+) and c1>0.H
(
U
): The multivalued map U:J×X→Pf(Y) is such that:H
(
M
): For any M>0, there exists a function lM∈L∞(J,ℝ+) such that for a.e. t∈J and for any x1,x2∈X, ∥x1∥X≤M, ∥x2∥X≤M and u1∈U(t,x1), there is a u2∈U(t,x2) such that
(2.6)∥g(t,x1)u1+h(t,x1)-g(t,x2)u2-h(t,x2)∥X≤lM(t)∥x1-x2∥X.
We note that the condition similar to H(M) was also assumed in [6, 7].From the Definitions2.1 and 2.2 and the results obtained in [18, 19], Definition 1.1 can be rewritten in the following form.Definition 2.6.
A functionx∈C(J,X) is a (mild) solution of the system (1.1), (1.2) if x(0)=x0 and there exists u∈L1(J,Y) such that u(t)∈U(t,x(t)) a.e. on t∈J and
(2.7)x(t)=Pα(t)x0+∫0t(t-s)α-1Qα(t-s)(g(s,x(s))u(s)+h(s,x(s)))ds.
A similar definition can be introduced for the system (1.1) and (1.5). Here
(2.8)Pα(t)=∫0∞ξα(θ)T(tαθ)dθ,Qα(t)=α∫0∞θξα(θ)T(tαθ)dθ,ξα(θ)=1αθ-1-(1/α)ϖα(θ-(1/α))≥0,ϖα(θ)=1π∑n=1∞(-1)(n-1)(θ)(-nα-1)Γ(nα+1)n!sin(nπα),θ∈(0,∞),
and ξα is a probability density function defined on (0,∞) [26], that is
(2.9)ξα(θ)≥0,θ∈(0,∞),∫0∞ξα(θ)dθ=1.
It is not difficult to verify that
(2.10)∫0∞θξα(θ)dθ=1Γ(1+α).Lemma 2.7 (see [18, 19]).
Let H(A) hold, then the operatorsPα and Qα have the following properties.(1)
For any fixedt≥0, Pα(t) and Qα(t) are linear and bounded operators, that is, for any x∈X,
(2.11)∥Pα(t)x∥X≤MA∥x∥X,∥Qα(t)x∥X≤αMAΓ(1+α)∥x∥X.(2)
{Pα(t),t≥0} and {Qα(t),t≥0} are strongly continuous.(3)
For everyt>0, Pα(t) and Qα(t) are compact operators.Lemma 2.8 (see [27, Theorem 3.1]).
Letx(t) be continuous and nonnegative on [0,b]. If
(2.12)x(t)≤ψ(t)+M∫0tx(s)(t-s)γds,0≤t≤b,
where 0≤γ<1, ψ(t) is a non-negative, monotonic increasing continuous function on [0,b] and M is a positive constant, then
(2.13)x(t)≤ψ(t)E1-γ(MΓ(1-γ)t1-γ),0≤t≤b,
where E1-γ(z) is the Mittag-Leffler function defined for all γ<1 by
(2.14)E1-γ(z)=∑n=0∞znΓ(n(1-γ)+1).


## 3. Auxiliary Results

In this section, we will give some auxiliary results needed in the proof of the main results. We begin with the a priori estimation of the trajectory of the control systems.Lemma 3.1.
For any admissible trajectoryx of the control system (1.1) and (1.5), that is, x∈𝒯rV, there is a constant L such that
(3.1)∥x∥C≤L.Proof.
Let anyx∈𝒯rV, from Definition 2.6, we know that there exists a u with u(t)∈V(t,x(t)) a.e. and
(3.2)x(t)=Pα(t)x0+∫0t(t-s)α-1Qα(t-s)(g(s,x(s))u(s)+h(s,x(s)))ds.
Then by Lemma 2.7, we get
(3.3)∥x(t)∥X≤MA∥x0∥X+αMAΓ(1+α)∫0t(t-s)α-1∥h(s,x(s))∥Xds+αMAΓ(1+α)∫0t(t-s)α-1∥g(s,x(s))u(s)∥Xds.
From H(h) and Hölder inequality, we have
(3.4)∫0t(t-s)α-1∥h(s,x(s))∥Xds≤∫0t(t-s)α-1(a1(s)+c1∥x(s)∥X)ds≤[(1-β)(α-β)b(α-β)/(1-β)]1-β∥a1∥L1/β(J)+c1∫0t(t-s)α-1∥x(s)∥Xds.
Similarly, by H(g)(3) and H(U)(3),
(3.5)∫0t(t-s)α-1∥g(s,x(s))u(s)∥Xds≤d∫0t(t-s)α-1(a2(s)+c2∥x(s)∥X)ds≤d[(1-β)(α-β)b(α-β)/(1-β)]1-β∥a2∥L1/β(J)+dc2∫0t(t-s)α-1∥x(s)∥Xds.
Combining (3.4), (3.5) with (3.3), we obtain
(3.6)∥x(t)∥X≤MA∥x0∥X+αMAΓ(1+α)(dc2+c1)∫0t(t-s)α-1∥x(s)∥Xds+αMAΓ(1+α)[(1-β)(α-β)b(α-β)/(1-β)]1-β(∥a1∥L1/β(J)+d∥a2∥L1/β(J)).
From the above inequality, using the well-known singular version of the Gronwall inequality (see Lemma 2.8), we can deduce that there exists a constant L>0 such that ∥x∥C≤L.
LetQ={h∈X:∥h∥X≤L}. Let prL:X→X be the L-radial retraction, that is,
(3.7)prL(x)={x,∥x∥X≤L,Lx∥x∥X,∥x∥X>L.
This map is Lipschitz continuous. We define U1(t,x)=U(t,prLx). Obviously, U1 satisfies H(U)(1) and H(U)(2). Moreover, by the properties of prL, we have for a.e. t∈J, all x∈X and all u∈U1(t,x) the estimates
(3.8)∥u∥Y≤a2(t)+c2L,∥u∥Y≤a2(t)+c2∥x∥X.
Hence, Lemma 3.1 is still valid with U(t,x) substituted by U1(t,x). Therefore, we assume without any loss of generality that for a.e. t∈J, and all x∈X(3.9)sup{∥v∥Y:v∈U(t,x)}≤φ(t)=a2(t)+c2L,withφ∈L1/β(J,ℝ+).
Similarly, we can assume that for a.e. t∈J and all x∈X(3.10)∥h(t,x)∥X≤γ(t)=a1(t)+c1L,withγ∈L1/β(J,ℝ+).
Let
(3.11)Yφ={u∈L1/β(J,Y):∥u(t)∥Y≤φ(t)a.e.t∈J},(3.12)Xφ={f∈L1/β(J,X):∥f(t)∥X≤dφ(t)+γ(t)a.e.t∈J}.
It follows from assumption H(g) that for any h∈X*, the function 〈h,g(t,x)u〉=〈g*(t,x)h,u〉 is measurable in t and continuous in (x,u) almost everywhere. Hence, for any measurable functions x:J→X and u:J→Y, the function t→g(t,x(t))u(t) is scalarly measurable [28]. The separability of the space X implies that the function t→g(t,x(t))u(t) is measurable. Therefore, according to H(g) and H(h), for any x∈L1/β(J,X) and u∈L1/β(J,Y), the function t→g(t,x(t))u(t)+h(t,x(t)) is an element of the space L1/β(J,X). Hence we can consider the operator 𝒜:L1/β(J,X)×L1/β(J,Y)→L1/β(J,X) defined by the rule
(3.13)𝒜(x,u)(t)=g(t,x(t))u(t)+h(t,x(t)).Lemma 3.2.
The operator(x,u)→𝒜(x,u) is sequentially continuous as an operator from L1/β(J,X)×ω-L1/β(J,Y) into ω-L1/β(J,X).Proof.
Suppose thatxn→x in L1/β(J,X) and un→u in ω-L1/β(J,Y). Take an arbitrary u∈Y and any h∈L1/(1-β)(J,X*). H(g) and the equality
(3.14)〈h(t),g(t,xn(t))u〉=〈g*(t,xn(t))h(t),u〉
imply that t→g*(t,xn(t))h(t) is a scalarly measurable function from J to Y*. Hence it is measurable. Consider a subsequence xnk, k≥1, of the sequence xn, n≥1, converging to x a.e. in t∈J. By H(g), H(h), and (3.10), we have
(3.15)g*(t,xnk(t))h(t)→g*(t,x(t))h(t)a.e.t∈JinY*,∥g*(t,xnk(t))h(t)∥Y*≤d∥h(t)∥X*a.e.t∈J,h(t,xnk(t))→h(t,x(t))a.e.t∈JinX,∥h(t,xnk(t))∥X≤γ(t).
Using the preceding four formulae and Lebesgue's theorem of dominated convergence, we obtain
(3.16)g*(t,xnk(t))h(t)→g*(t,x(t))h(t)inL1/(1-β)(J,Y*),(3.17)h(t,xnk(t))→h(t,x(t))inL1/β(J,X).
Then it follows from (3.16) that
(3.18)∫J〈g*(t,xnk(t))h(t),unk(t)〉dt→∫J〈g*(t,x(t))h(t),u(t)〉dt.
Since 〈h(t),g(t,x(t))u(t)〉=〈g*(t,x(t))h(t),u(t)〉 and h∈L1/(1-β)(J,X*) is arbitrary, by (3.17) and (3.18), we deduce that
(3.19)𝒜(xnk,unk)→𝒜(x,u)inω-L1/β(J,X).
It follows from (3.9), (3.10), and (3.12) that {𝒜(xn,un)}n≥1 is a subset of Xφ which is a metrizable compact set in ω-L1/β(J,X). If the sequence 𝒜(xn,un), n≥1, does not converge to 𝒜(x,u) in ω-L1/β(J,X), then it has a subsequence 𝒜(xni,uni), i≥1, such that none of its subsequences converges to 𝒜(x,u) in ω-L1/β(J,X). Applying the above arguments to this very subsequence (xni,uni), i≥1, we obtain a contradiction. The lemma is proved.Lemma 3.3.
For a.e.t∈J, the multivalued map x→V(t,x) defined by (1.4) from X to 2Y is u.s.c. with convex closed values.Proof.
From the definition ofV(t,x), it is clear that V(t,x) is closed convex valued. Since δ→Uδ(t,x) is increasing (in the sense of inclusion), and letting
(3.20)U1/n(t,x)=co¯{⋃U(t,z):z∈x+1nBX},
we obtain
(3.21)V(t,x)=⋂n≥1U1/n(t,x).
Let x0∈X and W be an open set in Y such that V(t,x0)⊆W. By (3.21), we can find an n0≥1 such that
(3.22)U1/n0(t,x0)⊆W.
For an arbitrary y∈x0+(1/n0)BX, we can find a δ>0 such that y+δBX⊆x0+(1/n0)BX. Therefore we obtain
(3.23)Uδ(t,y)⊆U1/n0(t,x0)⊆W.
Then it is clear that V(t,y)⊆W, for all y∈x0+(1/n0)BX. This means that x→V(t,x) is u.s.c.LetCX={zk}k≥1 be a dense countable subset of the ball BX. We put
(3.24)U1/nk(t,x)=U(t,x+1nzk),zk∈CX.Lemma 3.4.
For a.e.t∈J, let U1/n(t,x) be defined by (3.20), then we have
(3.25)U1/n(t,x)=co¯{⋃k=1∞U1/nk(t,x)¯},
where the closure is taken in Y.Proof.
We recall thatco¯A¯=co¯A for any subset A⊆Y. Hence it is sufficient to prove that for a.e. t∈J,
(3.26)⋃k=1∞U1/nk(t,x)¯={⋃U(t,z):z∈x+1nBX¯}.
That the left hand side of (3.26) is contained in its right hand side is obvious. Let w∈{⋃U(t,z):z∈x+(1/n)BX}, then w∈U(t,x+(1/n)z*) for some z*∈BX. Now let zm→z*, {zm}m≥1⊆CX. Since a.e. t∈J, x→U(t,x) is l.s.c. at x+(1/n)z* and x+(1/n)zm→x+(1/n)z*, there is a sequence wm∈U(t,x+(1/n)zm), m≥1 converging to w (Proposition 1.2.6 [24]). Due to {wm}m≥1⊆⋃k=1∞U1/nk(t,x), we have w∈⋃k=1∞U1/nk(t,x)¯. Since w is arbitrary, we can get {⋃U(t,z):z∈x+(1/n)BX}⊆⋃k=1∞U1/nk(t,x)¯. Therefore (3.26) holds. The lemma is proved.Now we consider the following auxiliary problem:(3.27)DCtαx(t)=Ax(t)+f(t),t∈J=[0,b],x(0)=x0.
It is clear that for every f∈L1/β(J,X), (3.27) has a unique (mild) solution S(f)∈C(J,X) which is given by
(3.28)S(f)(t)=Pα(t)x0+∫0t(t-s)α-1Qα(t-s)f(s)ds.The following lemma describes a property of the solution mapS which is crucial in our investigation.Lemma 3.5.
The solution mapS:Xφ→C(J,X) is continuous from ω-Xφ to C(J,X).Proof.
Consider the operatorH:L1/β(J,X)→C(J,X) defined by
(3.29)H(f)(t)=∫0t(t-s)α-1Qα(t-s)f(s)ds.
We know H is linear. From simple calculation, one has
(3.30)∥H(f)∥C≤αMAΓ(1+α)[(1-β)(α-β)b(α-β)/(1-β)]1-β∥f∥L1/β(J,X),
that is, the operator H is continuous from L1/β(J,X) into C(J,X), hence H is also continuous from ω-L1/β(J,X) into ω-C(J,X).
Let anyB∈Pb(L1/β(J,X)) and suppose that for any f∈B, ∥f∥L1/β(J,X)≤K (K>0 is a constant). Next we will show that H is completely continuous.(a)
From (3.30), we know that ∥H(f)(t)∥X is uniformly bounded for any t∈J and f∈B.(b)
H is equicontinuous on B. Let 0≤t1<t2≤b and any f∈B, we get
(3.31)∥H(f)(t2)-H(f)(t1)∥X=∥∫0t2(t2-s)α-1Qα(t2-s)f(s)ds-∫0t1(t1-s)α-1Qα(t1-s)f(s)ds∥X≤∥∫t1t2(t2-s)α-1Qα(t2-s)f(s)ds∥X⋯denotedbyI1+∥∫0t1((t2-s)α-1-(t1-s)α-1)Qα(t2-s)f(s)ds∥X⋯denotedbyI2+∥∫0t1(t1-s)α-1(Qα(t2-s)-Qα(t1-s))f(s)ds∥X⋯denotedbyI3.By using analogous arguments as in Lemma3.1, we find
(3.32)I1≤αMAΓ(1+α)[(1-β)(α-β)]1-βK(t2-t1)α-β,I2≤αMAΓ(1+α)(∫0t1((t1-s)α-1-(t2-s)α-1)1/(1-β)ds)1-βK≤αMAΓ(1+α)(∫0t1((t1-s)(α-1)/(1-β)-(t2-s)(α-1)/(1-β))ds)1-βK=αMAΓ(1+α)[(1-β)(α-β)]1-β(t1(α-β)/(1-β)-t2(α-β)/(1-β)+(t2-t1)(α-β)/(1-β))1-βK≤2αMAΓ(1+α)[(1-β)(α-β)]1-β(t2-t1)α-βK.Fort1=0, 0<t2≤b, it is easy to see that I3=0. For t1>0 and ϵ>0 be enough small, we have
(3.33)I3≤∥∫0t1-ϵ(t1-s)α-1(Qα(t2-s)-Qα(t1-s))f(s)ds∥X+∥∫t1-ϵt1(t1-s)α-1(Qα(t2-s)-Qα(t1-s))f(s)ds∥X≤sups∈[0,t1-ϵ]∥Qα(t2-s)-Qα(t1-s)∥[(1-β)(α-β)]1-β×(t1(α-β)/(1-β)-ϵ(α-β)/(1-β))1-βK+2αMAΓ(1+α)[(1-β)(α-β)]1-βϵα-βK.Combining the estimations forI1,I2, and  I3, and letting t2→t1 and ϵ→0 in I3, we obtain that H is equicontinuous. For more details, please see [19].(c)
The setΠ(t)={H(f)(t):f∈B} is relatively compact in X. Clearly, Π(0)={0} is compact, and hence, it is only necessary to consider t>0. For each h∈(0,t), t∈(0,b], f∈B and δ>0 be arbitrary, we define
(3.34)Πh,δ(t)={Hh,δ(f)(t):f∈B},where
(3.35)Hh,δ(f)(t)=α∫0t-h∫δ∞θ(t-s)α-1ξα(θ)T((t-s)αθ)f(s)dθds=α∫0t-h∫δ∞θ(t-s)α-1ξα(θ)T(hαδ)T((t-s)αθ-hαδ)f(s)dθds=αT(hαδ)∫0t-h∫δ∞θ(t-s)α-1ξα(θ)T((t-s)αθ-hαδ)f(s)dθds.
From the compactness of T(hαδ) (hαδ>0), we obtain that the set Πh,δ(t) is relatively compact in X for any h∈(0,t) and δ>0. Moreover, we have
(3.36)∥H(f)(t)-Hh,δ(f)(t)∥X=α∥∫0t∫0δθ(t-s)α-1ξα(θ)T((t-s)αθ)f(s)dθds+∫0t∫δ∞θ(t-s)α-1ξα(θ)T((t-s)αθ)f(s)dθds-∫0t-h∫δ∞θ(t-s)α-1ξα(θ)T((t-s)αθ)f(s)dθds∥X≤α∥∫0t∫0δθ(t-s)α-1ξα(θ)T((t-s)αθ)f(s)dθds∥X+α∥∫t-ht∫δ∞θ(t-s)α-1ξα(θ)T((t-s)αθ)f(s)dθds∥X≤MAα(∫0t(t-s)(α-1)/(1-β)ds)1-β∥f∥L1/β(J,X)∫0δθξα(θ)dθ+MAα(∫t-ht(t-s)(α-1)/(1-β)ds)1-β∥f∥L1/β(J,X)∫δ∞θξα(θ)dθ≤MAKα[(1-β)(α-β)]1-β(bα-β∫0δθξα(θ)dθ+1Γ(1+α)hα-β).
In virtue of (2.10), the last term of the preceding inequality tends to zero as h→0 and δ→0. Therefore, there exist relatively compact sets arbitrarily close to the set Π(t), t>0. Hence the set Π(t), t>0 is also relatively compact in X.
SinceXφ is a convex compact metrizable subset of ω-L1/β(J,X), it suffices to prove the sequential continuity of the map S. Now let {fn}n≥1⊆Xφ such that
(3.37)fn→finω-L1/β(J,X),f∈Xφ.
By the property of the operatorH, we have H(fn)→H(f) in ω-C(J,X). Since {fn}n≥1 is bounded, there is a subsequence {fnk}k≥1 of the sequence {fn}n≥1 such that H(fnk)→z in C(J,X) for some z∈C(J,X). From the facts that
(3.38)H(fn)→H(f)inω-C(J,X),H(fnk)→zinC(J,X),
we obtain that z=H(f) and H(fn)→H(f) in C(J,X).
From the definitions of the operatorsS and H, we have that S(f)(t)=Pα(t)x0+H(f)(t). Then due to the arguments above, we have S(fn)→S(f) in C(J,X). This completes the proof of the lemma.


## 4. Existence Results for the Control Systems

In the present section, we are interested in the existence results for the control systems (1.1), (1.2) and (1.1), (1.5).LetΛ=S(Xφ), from Lemma 3.5, we have that Λ is a compact subset of C(J,X). It follows from formulae (3.9), (3.10), and (3.12) that 𝒯rU⊆𝒯rV⊆Λ. Let U¯:C(J,X)→2L1/β(J,Y) be defined by
(4.1)U¯(x)={h:J→Ymeasurable:h(t)∈U(t,x(t))a.e.},x∈C(J,X).Theorem 4.1.
The setℛU is nonempty and the set ℛV is a compact subset of the space C(J,X)×ω-L1/β(J,Y).Proof.
By the hypothesis H(U)(1), we have that for any measurable functionx:J→X, the map t→U(t,x(t)) is measurable and has closed values. Therefore it has measurable selectors [29]. So the operator U¯ is well defined and its values are closed decomposable subsets of L1/β(J,Y). We claim that x→U¯(x) is l.s.c. Let x*∈C(J,X), h*∈U¯(x*) and let {xn}n≥1⊆C(J,X) be a sequence converging to x*. It follows from Lemma 3.2 in [30] that there exists a sequence hn∈U¯(xn) such that
(4.2)∥h*(t)-hn(t)∥Y≤dY(h*(t),U(t,xn(t)))+1n,a.e.t∈J.
Since the map y→U(t,y) is l.s.c., by the Proposition 1.2.26 in [24], the function y→dY(h*(t),U(t,y)) is u.s.c. for a.e. t∈J. It follows from (4.2) that for a.e. t∈J(4.3)limn→∞∥h*(t)-hn(t)∥Y≤lim¯n→∞dY(h*(t),U(t,xn(t)))≤dY(h*(t),U(t,x*(t)))=0.
This together with (3.9) implies that hn→h* in L1/β(J,Y). Therefore the map x→U¯(x) is l.s.c. By Proposition 2.2 in [31], there is a continuous function m:Λ→L1/β(J,Y) such that
(4.4)m(x)∈U¯(x),∀x∈Λ.
Consider the map 𝒫:L1/β(J,X)→L1/β(J,Y) defined by 𝒫(f)=m(S(f)). Due to Lemma 3.5 and the continuity of m, the map 𝒫 is continuous from ω-Xφ into L1/β(J,Y). Then by Lemma 3.2, we deduce that the map f→𝒜(S(f),𝒫(f)) is continuous from ω-Xφ into ω-L1/β(J,X). It follows from (3.9), (3.10), (3.12), and (3.13) that 𝒜(S(f),𝒫(f))∈Xφ for every f∈Xφ. Therefore, the map f→𝒜(S(f),𝒫(f)) is continuous from ω-Xφ into ω-Xφ. Since ω-Xφ is a convex metrizable compact set in ω-L1/β(J,X), Schauder's fixed point theorem implies that this map has a fixed point f*∈Xφ, that is, f*=𝒜(S(f*),𝒫(f*)). Let u*=𝒫(f*) and x*=S(f*), then we have u*=m(x*) and f*=𝒜(x*,u*). That means
(4.5)x*(t)=Pα(t)x0+∫0t(t-s)α-1Qα(t-s)(g(s,x*(s))u*(s)+h(s,x*(s)))ds,u*(t)∈U(t,x*(t)),a.e.t∈J,
which implies that (x*(·),u*(·)) is a solution of the control system (1.1) and (1.2). Hence ℛU is nonempty.
It is easy to see thatℛV⊆Λ×Yφ. Since Λ is compact in C(J,X) and Yφ is metrizable convex compact in ω-L1/β(J,Y), we have that ℛV is relatively compact in C(J,X)×ω-L1/β(J,Y). Hence to complete the proof of this theorem, it is sufficient to prove that ℛV is sequentially closed in C(J,X)×ω-L1/β(J,Y).
Let{(xn(·),un(·))}n≥1⊆ℛV be a sequence converging to (x(·),u(·)) in C(J,X)×ω-L1/β(J,Y). Denote
(4.6)fn(t)=g(t,xn(t))un(t)+h(t,xn(t)),f(t)=g(t,x(t))u(t)+h(t,x(t)).
According to Lemma 3.2, fn→f in ω-L1/β(J,X). Since fn∈Xφ and xn=S(fn), n≥1, Lemma 3.5 implies that
(4.7)x=S(f).
Hence, to prove that (x(·),u(·))∈ℛV, we only need to verify that u(t)∈V(t,x(t)) a.e. t∈J.
Sinceun→u in ω-L1/β(J,Y), by Mazur's theorem we have
(4.8)u(t)∈⋂n=1∞co¯(⋃k=n∞uk(t)),fora.e.t∈J.
From Lemma 3.3, we have that for a.e. t∈J, the map x→V(t,x)∈Pfc(Y) is u.s.c., then by Proposition 1.2.61 in [24], the map x→V(t,x)∈Pfc(Y) is h-upper semicontinuous. Therefore from assertion (b) of Proposition 1.2.86 in [24], the map x→V(t,x) has property Q. Hence we have
(4.9)⋂n=1∞co¯(⋃k=n∞V(t,xk(t)))⊆V(t,x(t)),fora.e.t∈J.
In virtue of (4.8) and (4.9), and for a.e. t∈J, un(t)∈V(t,xn(t)), n≥1, we obtain that u(t)∈V(t,x(t)) a.e. t∈J. This means that ℛV is compact in C(J,X)×ω-L1/β(J,Y). The proof is complete.


## 5. Main Results

In this section, we will prove the relaxation result. But first, we give a lemma which is important in the proof of our relaxation theorem.Lemma 5.1.
For any pair(x*(·),u*(·))∈ℛV, there exists a sequence of simple functions yn:J→X and a sequence vn∈L1/β(J,Y), n≥1, such that
(5.1)∥yn(t)∥X≤1n,t∈J,n≥1,(5.2)vn(t)∈U(t,x*(t)+yn(t)),t∈J,(5.3)vn→u*inω-L1/β(J,Y).Proof.
Let(x*(·),u*(·))∈ℛV. From Lemma 3.4, we have that for a.e. t∈J, n≥1(5.4)u*(t)∈V(t,x*(t))⊆U1/n(t,x*(t))=co¯{⋃k=1∞U1/nk(t,x*(t))¯}.
The map t→⋃k=1∞U1/nk(t,x*(t))¯ is measurable (see Propositions 2.3 and 2.6 in [29]) and, by (3.9), is integrally bounded. Therefore, from (5.4) and Theorem 2.2 in [32], we have that there exists an fn∈L1/β(J,Y) such that
(5.5)fn(t)∈⋃k=1∞U1/nk(t,x*(t))¯a.e.t∈J,∥u*-fn∥ω≤1n.
We know that the map t→U1/nk(t,x*(t)) is measurable and its value are closed, then following Theorem 5.6 in [29] (also Proposition 2.2.3 in [24]), there exists a sequence of measurable selectors fkm(t)∈U1/nk(t,x*(t)), t∈J, m≥1 such that
(5.6)U1/nk(t,x*(t))=U(t,x*(t)+1nzk)=⋃m=1∞fkm(t)¯,fort∈J,k≥1.
Therefore we have
(5.7)⋃k=1∞U1/nk(t,x*(t))¯=⋃k=1∞U(t,x*(t)+1nzk)¯=⋃k=1∞⋃m=1∞fkm(t)¯,fort∈T.
From (5.5) and (5.7), according to Lemma 1.3 in [33] (also see Proposition 2.3.6 [24]), there is a finite measurable partition J1,J2,…,Jl(n) of J such that
(5.8)(∫J∥fn(t)-∑i=1l(n)χ(Ji)fkimi(t)∥Y1/βdt)β≤1n,
where χ(Ji) is the characteristic function of the set Ji. Now let
(5.9)vn(t)=∑i=1l(n)χ(Ji)fkimi(t),yn(t)=∑i=1l(n)χ(Ji)1nzki.
Formula (5.9) implies that yn is a simple function, vn∈L1/β(J,Y) and (5.1), (5.2) hold. By Lemma 2.4, (5.5) and (5.8), we obtain that (5.3) holds. The lemma is proved.Now we are ready to present our main result.Theorem 5.2.
The set𝒯rV is compact in C(J,X) and the following relation holds
(5.10)𝒯rV=𝒯rU¯,
where the bar stands for the closure in C(J,X).Proof.
Let(x*(·),u*(·))∈ℛV and {vn}n≥1, {yn}n≥1 be given as in Lemma 5.1. Put Q={h∈X:∥h∥X≤L}, for fixed n≥1, we consider the function defined by
(5.11)rn(t,x,u)=∥g(t,x*(t)+yn(t))vn(t)+h(t,x*(t)+yn(t))-g(t,x)u-h(t,x)∥X-lL+1(t)∥x*(t)+yn(t)-x∥X.
It is clear that the function t→rn(t,x,u) is measurable and the function (x,u)→rn(t,x,u) is continuous (in view of H(g) and the fact that if x:J→X is a measurable function, ∥x(t)∥X is a measurable real-valued function). According to the Theorem 2.4 in [34], there exists a sequence of nested (in the sense of inclusion) closed sets Jk⊆J, k≥1, μ(J∖∪k=1∞Jk)=0 such that the map (t,x)→U(t,x) is l.s.c. on Jk×Q and rn(t,x,u) is continuous on Jk×Q×Y. Let the multivalued map Hn:J×Q→Y be defined by
(5.12)Hn(t,x)={u∈Y:rn(t,x,u)-1n<0}.
For every k≥1 the graph of the map Hn(t,x) is an open subset of Jk×Q×Y. Let the map Un:J×Q→Y be defined by
(5.13)Un(t,x)=Hn(t,x)∩U(t,x).
Hypothesis H(M) together with (5.12) implies that Un(t,x) is nonempty for a.e. t∈J and all x∈Q. Since the map U(t,x) is l.s.c. on Jk×Q, k≥1, and the graph of the map Hn(t,x) is an open subset of Jk×Q×Y, k≥1, then according to Proposition 1.2.47 in [24], we obtain that the map Un(t,x) is l.s.c. on Jk×Q, k≥1. Hence the map Un¯(t,x)=Un(t,x)¯ is l.s.c. on Jk×Q, k≥1. Therefore, for every continuous function x:J→Q the map t→Un¯(t,x(t)) is measurable and the map x→Un¯(t,x) is l.s.c. on Q for a.e. t∈J.
It is clear thatUn¯(t,x)⊆U(t,x). Consider the system (1.1) with the constraint Un¯(t,x(t)) on the control. The arguments used in the proof of the Theorem 4.1 enable us to obtain the existence of a solution (xn(·),un(·))∈ℛU and
(5.14)∥g(t,x*(t)+yn(t))vn(t)+h(t,x*(t)+yn(t))-g(t,xn(t))un(t)-h(t,xn(t))∥X-lL+1(t)∥x*(t)+yn(t)-xn(t)∥X-1n≤0.
Now by (x*(·),u*(·))∈ℛV and (xn(·),un(·))∈ℛU, n≥1, we have
(5.15)x*(t)=Pα(t)x0+∫0t(t-s)α-1Qα(t-s)(g(s,x*(s))u*(s)+h(s,x*(s)))ds,(5.16)xn(t)=Pα(t)x0+∫0t(t-s)α-1Qα(t-s)(g(s,xn(s))un(s)+h(s,xn(s)))ds.
Theorem 4.1 and ℛU⊆ℛV imply that we can assume, possibly up to a subsequence, that the sequence (xn(·),un(·))→(x¯(·),u¯(·))∈ℛV in C(J,X)×ω-L1/β(J,Y). Subtracting (5.15) from (5.16), we have
(5.17)∥xn(t)-x*(t)∥X=∥H(g(·,xn(·))un(·)+h(·,xn(·)))(t)-H(g(·,x*(·))u*(·)-h(·,x*(·)))(t)∥X≤∥H(g(·,x*(·)+yn(·))vn(·)+h(·,x*(·)+yn(·))-g(·,x*(·))u*(·)-h(·,x*(·)))(t)∥X⋯denotedbyD1+∥H(g(·,xn(·))un(·)+h(·,xn(·))-g(·,x*(·)+yn(·))vn(·)-h(·,x*(·)+yn(·)))(t)∥X⋯denotedbyD2.
Here the linear operator H is defined by (3.29). In virtue of (5.1), we have x*+yn→x* in L1/β(J,X). Since vn→u* in ω-L1/β(J,Y) (see (5.3)), from Lemma 3.2 we have that 𝒜(x*+yn,vn)→𝒜(x*,u*) in ω-L1/β(J,X). By the property of the operator H in Lemma 3.5, we obtain
(5.18)D1→0foreveryt∈J.
Due to (5.14), we have
(5.19)D2≤αMAΓ(1+α)∫0t(t-s)α-1(lL+1(s)∥x*(s)+yn(s)-xn(s)∥X+1n)ds≤αMA∥lL+1∥L∞(J)Γ(1+α)∫0t(t-s)α-1Qα(t-s)∥x*(s)+yn(s)-xn(s)∥Xds+αMAbαnαΓ(1+α).
Note that ∥x*(t)∥X≤L,∥xn(t)∥X≤L for any n≥1, t∈J. Combining (5.18), (5.19) with (5.17), let n→∞, we get
(5.20)∥x¯(t)-x*(t)∥X≤αMA∥lL+1∥L∞(J)Γ(1+α)∫0t(t-s)α-1Qα(t-s)∥x*(s)-x¯(s)∥Xds.
This together with Lemma 2.8 implies that x*=x¯. Hence we have that the sequence xn∈𝒯rU, n≥1, converges to x*∈𝒯rV in C(J,X). From Theorem 4.1, we know that 𝒯rV is compact in C(J,X). Now it follows from 𝒯rU⊆𝒯rV and the proof above that the relation (5.10) holds. The proof is complete.



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