# On the Dimension of the Solution Set for Semilinear Fractional Differential Inclusions

Authors: Ravi P. Agarwal; Bashir Ahmad; Ahmed Alsaedi; Naseer Shahzad
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate the existence and dimension of the solution set for a nonlocal problem of semilinear fractional differential inclusions. The main tools of our study include some well-known results on multivalued maps.


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## Body

## 1. Introduction

The subject of fractional calculus has recently emerged as an important and popular field of research due to its numerous applications in science and engineering. Examples can be found in various disciplines such as mechanics, electricity, signal and image processing, thermodynamics, biophysics, blood flow phenomena, aerodynamics, economics, and fitting of experimental data [1–4] whereas the theoretical development of the subject can be found in [5, 6]. For some recent results on fractional differential equations and inclusions, see [7–16].In this paper, we study the following problem for semilinear fractional differential inclusion with nonlocal condition:(1.1)cDqx(t)∈A(t)x(t)+F(t,x(t)),t∈[0,T](T>0),x(0)+g(x)=x0,x0∈Rn,
where cDq denote the Caputo fractional derivative of order q∈(0,1] [5], A(t) is a bounded linear operator on [0,T] (the function t→A(t) is continuous in the uniform operator topology), g:ℭ([0,T],ℝn)→ℝn, and F:[0,T]×ℝn→P(ℝn), where P(ℝn) is the family of all nonempty subsets of ℝn.


## 2. Terminology and Preliminary Results

In this section, we discuss some basic concepts of multivalued analysis and recall some results involving multivalued maps.Letℭ([0,T],ℝn) denote the Banach space of continuous functions from [0,T] into ℝn with the norm ∥x∥∞=supt∈[0,T]∥x(t)∥. Let L1([0,T],ℝn) be the Banach space of measureable functions x:[0,T]→ℝn that are Lebesgue integrable and normed by ∥x∥L1=∫0T∥x(t)∥dt.For a nonempty subsetC of a complete metric space X, let P(C)={Y⊆C:Y≠∅}, Pcl(C)={Y∈P(C):Yisclosed}, Pb(C) = {Y∈P(C):Yisbounded}, Pb,cl(C) = {Y∈P(C):Yisboundedandclosed}, and Pcp(C) = {Y∈P(C):Yiscompact}. If C is a nonempty subset of a Banach space X, then we set Pc,cl(C) = {Y∈P(C):Yisclosedandconvex}, and Pc,cp(C) = {Y∈P(C):Yiscompactandconvex}.A multivalued mapℱ:C→P(X) is closed (resp., compact) valued if ℱ(x) is closed (resp., compact) for all x∈C. The map ℱ is bounded on bounded sets if ℱ(𝔹)=∪x∈𝔹ℱ(x) is bounded in X for all 𝔹∈Pb(C) (i.e., supx∈𝔹{sup{∥y∥:y∈ℱ(x)}}<∞). The map ℱ is called upper semicontinuous (u.s.c.) if {x∈C:ℱ(x)⊂V} is open in C whenever V⊂X is open. ℱ is called lower semi-continuous (l.s.c.) if the set {y∈C:ℱ(y)∩V≠∅} is open for any open set V⊂X. ℱ is called continuous if it is both l.s.c. and u.s.c. ℱ is said to be completely continuous if ℱ(𝔹) is relatively compact for every 𝔹∈Pb(C). A mapping f:C→X is called a selection of ℱ:C→P(X) if f(x)∈ℱ(x) for every x∈C. We say that the mapping ℱ has a fixed point if there is x∈X such that x∈ℱ(x). The fixed points set of the multivalued operator ℱ will be denoted by Fix(ℱ). A multivalued map ℱ:[0,T]→Pcl(ℝn) is said to be measurable if, for every y1∈ℝn, the function(2.1)t⟼d(y1,F(t))=inf⁡{‖y1-y2‖:y2∈F(t)}
is measurable.Definition 2.1.
Let(X,d) be a metric space. Consider H:P(X)×P(X)→ℝ∪{∞} given by
(2.2)H(A,B)=max⁡{sup⁡a∈Ad(a,B),sup⁡b∈Bd(b,A)},
where d(a,B)=infb∈Bd(a,b). H is the (generalized) Pompeiu-Hausdorff functional. It is known that (Pb,cl(X),H) is a metric space and (Pcl(X),H) is a generalized metric space (see [17]).Definition 2.2.
A multivalued operatorℱ:X→Pcl(X) is called a k-contraction if there exists 0<k<1 such that
(2.3)H(F(x),F(y))≤kd(x,y),foreachx,y∈X.
It is known that ℱ:X→Pcp(X) is continuous on X if and only if ℱ is continuous on X with respect to the Hausdorff metric. Also, if ℱ:X→Pcp(X) is a k-contraction, then ℱ is continuous with respect to Hausdorff metric.Further details of multivalued maps can be found in ([18, 19]).For the forthcoming analysis, we need the following results on multivalued maps.Lemma 2.3 (Covitz and Nadler [20]).
Let(X,d) be a complete metric space. If Φ:X→Pcl(X) is a k-contraction, then, Fix(Φ)≠∅.Lemma 2.4 (Dzedzej and Gelman [21]).
LetF:[0,α]→Pc,cp(ℝn) be a measurable map such that the Lebesgue measure μ of the set {t:dimF(t)<1} is zero. Then there are arbitrarily many linearly independent measurable selections x1(·),x2(·),…,xm(·) of F.Lemma 2.5 (Saint-Raymond [22]).
LetK be a compact metric space with dimK<n, X a Banach space, and Ω:K→Pc,cp(X) a lower semicontinuous map such that 0∈Ω(x) and dimΩ(x)≥n for every x∈K. Then, there exists a continuous selection f of Ω such that f(x)≠0 for each x∈K.Lemma 2.6 (Michael’s selection theorem [23]).
LetC be a metric space, X a Banach space and Ω:C→Pc,cl(C) a lower semicontinuous map. Then, there exists a continuous selection f:C→X of Ω.Lemma 2.7 (see Dzedzej and Gelman [21] and Petrusel [24]).
LetC be a nonempty closed convex subset of a Banach space X. Suppose that Ω:C→Pc,cp(C) is a k-contraction. If f:C→C is a continuous selection of Ω, then Fix(f) is nonempty.


## 3. Main Results

Definition 3.1.
A functionx∈ℭ([0,T],ℝn) is a solution of the problem (1.1) if there exists a function f∈L1([0,T],ℝn) such that f(t)∈F(t,x(t)) a.e. on [0,T] and
(3.1)x(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)f(s)ds.
LetSx0([0,α]) denote the set of all solutions of (1.1) on the interval [0,α], where 0<α≤T.Lemma 3.2.
Assume that(H1)
F:[0,T]×ℝn→Pcp(ℝn) is such that F(·,x):[0,T]→Pcp(ℝn) is measurable for each x∈ℝn,(H2)
H(F(t,x),F(t,x¯))≤κ1(t)∥x-x¯∥ for almost all t∈[0,T] and x,x¯∈ℝn with κ1∈ℭ([0,T],ℝ+) and ∥F(t,x)∥=sup{∥v∥:v∈F(t,x)}≤κ1(t) for almost all t∈[0,T] and x∈ℝn,(H3)
g:ℭ([0,T],ℝn)→ℝn is continuous and ∥g(x)-g(y)∥≤κ2∥x-y∥∞ for all x,y∈ℭ([0,T],ℝn) and some κ2>0. Then, the Cauchy problem (1.1) has at least one solution on [0,T] if
(3.2)κ2+TqΓ(q+1)(A1+‖κ1‖∞)<1,
where A1=maxt∈[0,T]∥A(t)∥.Proof.
For eachy∈ℭ([0,T],ℝn), define the set of selections of F by
(3.3)SF,y∶={v∈L1([0,T],Rn):v(t)∈F(t,y(t))fora.e.t∈[0,T]}.
Observe that, by assumptions (H1) and (H2), F(·,x(·)) is measurable and has a measureable selection v(·) (see [25, Theorem III.6]). Also κ1∈ℭ([0,T],ℝ+) and
(3.4)‖v(t)‖≤‖F(t,x(t))‖≤κ1(t).
Thus, the set SF,x is nonempty for each x∈ℭ([0,T],ℝn). Now we show that the operator Ω defined by
(3.5)Ω(x)={h∈C([0,T],Rn):h(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)f(s)ds,f∈SF,x}
satisfies the assumptions of Lemma 2.3. To show that Ω(x)∈Pcl(ℭ([0,T],ℝn)) for each x∈ℭ([0,T],ℝn), let {un}n≥0∈Ω(x) be such that un→u(n→∞) in ℭ([0,T],ℝn). Then, u∈ℭ([0,T],ℝn) and there exists vn∈SF,x such that, for each t∈[0,T],
(3.6)un(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)vn(s)ds.
As F has compact values, we pass to a subsequence to obtain that vn converges to v in L1([0,T],ℝn). Thus, v∈SF,x and, for each t∈[0,T],
(3.7)un(t)⟶u(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)v(s)ds.
Hence, u∈Ω(x).
Next we show that there existsk>0 such that
(3.8)H(Ω(x),Ω(x¯))≤k‖x-x¯‖∞foreachx,x¯∈C([0,T],Rn).
Let x,x¯∈ℭ([0,T],ℝn) and h1∈Ω(x). Then, there exists v1(t)∈SF,x such that, for each t∈[0,T],
(3.9)h1(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)v1(s)ds.
By (H2), we have
(3.10)H(F(t,x),F(t,x¯))≤κ1(t)‖x(t)-x¯(t)‖.
So, there exists w∈F(t,x¯(t)) such that
(3.11)‖v1(t)-w‖≤κ1(t)‖x(t)-x¯(t)‖,t∈[0,T].
Define V:[0,T]→P(ℝn) by
(3.12)V(t)={w∈Rn:‖v1(t)-w‖≤κ1(t)‖x(t)-x¯(t)‖}.
Since the nonempty closed valued operator V(t)∩F(t,x¯(t)) is measurable [25, Proposition III.4], there exists a function v2(t) that is a measurable selection for V(t)∩F(t,x¯(t)). So v2(t)∈F(t,x¯(t)) and, for each t∈[0,T], we have ∥v1(t)-v2(t)∥≤κ1(t)∥x(t)-x¯(t)∥.
For eacht∈[0,T], let us define
(3.13)h2(t)=x0-g(x¯)+∫0t(t-s)q-1Γ(q)A(s)x¯(s)ds+∫0t(t-s)q-1Γ(q)v2(s)ds.
Thus,
(3.14)‖h1(t)-h2(t)‖≤‖g(x)-g(x¯)‖+∫0t|t-s|q-1Γ(q)‖A(s)(x-x¯(s))‖ds+∫0t|t-s|q-1Γ(q)‖v1(s)-v2(s)‖ds.
Hence,
(3.15)‖h1-h2‖∞≤κ2‖x-x¯‖∞+TqΓ(q+1)(A1+‖κ1‖∞)‖x-x¯‖∞=(κ2+TqΓ(q+1)(A1+‖κ1‖∞))‖x-x¯‖∞.
Analogously, interchanging the roles of x and x¯, we obtain
(3.16)H(Ω(x),Ω(x¯))≤k‖x-x¯‖∞,foreachx,x¯∈C([0,T],Rn),
where k=(κ2+(Tq/Γ(q+1))(A1+∥κ1∥∞))<1. Since Ω is a contraction, it follows by Lemma 2.3 that Ω has a fixed point x that is a solution of (1.1). This completes the proof.Lemma 3.3.
LetF:[0,T]×ℝn→Pc,cp(ℝn) satisfy (H1), (H2), and (H3) and suppose that Ω:ℭ([0,T],ℝn)→P(ℭ([0,T],ℝn)) is defined by
(3.17)Ω(x)={h∈C([0,T],Rn):h(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)f(s)ds,f∈SF,x}.
Then, Ω(x)∈Pc,cp(ℭ([0,T],ℝn)) for each x∈ℭ([0,T],ℝn).Proof.
First we show thatΩ(x) is convex for each x∈ℭ([0,T],ℝn). For that, let h1,h2∈Ω(x). Then. there exist f1,f2∈SF,x such that, for each t∈[0,T], we have
(3.18)hi(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)fi(s)ds,i=1,2.
Let 0≤λ≤1. Then, for each t∈[0,T], we have
(3.19)[λh1+(1-λ)h2](t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)[λf1(s)+(1-λ)f2(s)]ds.
Since SF,x is convex (F has convex values), it follows that λh1+(1-λ)h2∈Ω(x).
Next, we show thatΩ maps bounded sets into bounded sets in ℭ([0,T],ℝn). For a positive number r, let Br={x∈ℭ([0,T],ℝn):∥x∥∞≤r} be a bounded set in ℭ([0,T],ℝn). Then, for each h∈Ω(x),x∈Br, there exists f∈SF,x such that
(3.20)h(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)f(s)ds,
and, in view of (H1), we have
(3.21)‖h(t)‖≤‖x0‖+sup⁡x∈Br‖g(x)‖+∫0t|t-s|q-1Γ(q)‖A(s)x(s)‖ds+∫0t|t-s|q-1Γ(q)‖f(s)‖ds≤‖x0‖+sup⁡x∈Br‖g(x)‖+TqΓ(q+1)(A1r+‖κ1‖∞).
Thus,
(3.22)‖h‖∞≤‖x0‖+sup⁡x∈Br‖g(x)‖+TqΓ(q+1)(A1r+‖κ1‖∞).
Now we show that Ω maps bounded sets into equicontinuous sets in ℭ([0,T],ℝn). Let t′,t′′∈[0,T] with t′<t′′ and x∈Br, where Br is a bounded set in ℭ([0,T],ℝn). For each h∈Ω(x), we obtain
(3.23)‖h(t′′)-h(t′)‖=‖∫0t′′(t′′-s)q-1Γ(q)(A(s)x(s)+f(s))ds-∫0t′(t′-s)q-1Γ(q)(A(s)x(s)+f(s))ds‖≤‖∫0t′[(t′′-s)q-1-(t′-s)q-1]Γ(q)(A(s)x(s)+f(s))ds‖+‖∫t′t′′(t′′-s)q-1Γ(q)(A(s)x(s)+f(s))ds‖.
Obviously the right-hand side of the above inequality tends to zero independently of x∈Br′ as t′′-t′→0. By the Arzela-Ascoli theorem, Ω:ℭ([0,T],ℝn)→P(ℭ([0,T],ℝn)) is completely continuous. As in Lemma 3.2, Ω is closed valued. Consequently, Ω(x)∈Pc,cp(ℭ([0,T],ℝn)) for each x∈ℭ([0,T],ℝn).
For0<α≤T, let us consider the operator
(3.24)Ω(x)={h∈C([0,α],Rn):h(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)f(s)ds,f∈SF,x}.
It is well known that Fix(Ω)=Sx0([0,α]) and, in view of Lemma 3.2, it is nonempty for each 0<α≤T.Theorem 3.4.
Suppose thatF:[0,α]×ℝn→Pc,cp(ℝn) satisfies (H1), (H2), and (H3) and that the Lebesgue measure μ of the set {t:dimF(t,x)<1forsomex∈ℝn} is zero. Then, for each α, 0<α<min{((1-κ2)Γ(q+1)/(A1+∥κ1∥∞))1/q,T}, the set Sx0([0,α]) of solutions of (1.1) has an infinite dimension for any x0.Proof.
Let the operatorΩ be defined by
(3.25)Ω(x)={h∈C([0,α],Rn):h(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)x(s)ds+∫0t(t-s)q-1Γ(q)f(s)ds,f∈SF,x}.
Lemma 3.3 guarantees that Ω(x)∈Pc,cp(ℭ([0,α],ℝn)) for each x∈ℭ([0,α],ℝn) and as in the proof of Lemma 3.2, it is a contraction if κ2+(αq/Γ(q+1))(A1+∥κ1∥∞)<1 or α<{((1-κ2)Γ(q+1)/(A1+∥κ1∥∞))1/q,T}. We shall show that dimΩ(x)≥m for any x∈ℭ([0,α],ℝn) and arbitrary m∈ℕ. Consider G(t)=F(t,x(t)). By Lemma 2.4, there exist linearly independent measurable selections x1(·),x2(·),…,xm(·) of G. Set
(3.26)yi(t)=x0-g(x)+∫0t(t-s)q-1Γ(q)A(s)xi(s)ds+∫0t(t-s)q-1Γ(q)xi(s)ds∈Ω(x).
Assume that ∑i=1maiyi(t)=0 a.e. in [0,α]. Taking the Caputo derivatives a.e. in [0,α], we have ∑i=1maixi(t)=0 a.e. in [0,α] and hence ai=0 for all i. As a result, yi(·) are linearly independent. Thus, Ω(x) contains an m-dimensional simplex. So, dimΩ(x)≥m. As in Lemma 3.2, Fix(Ω) is nonempty. It is known that every multivalued k-contraction having compact values is condensing with respect to the Hausdorff measure of noncompactness χ [26]. Since Fix(Ω)⊂Ω(Fix(Ω)), we have
(3.27)χ(Fix⁡(Ω))≤χ(Ω(Fix⁡(Ω))).
Since Ω is χ-condensing, Fix(Ω) is compact. Consider a map I-Ω:Fix(Ω)→Pc,cp(ℝn), where I is the identity operator. Assume that dimFix(Ω)<n. Then, Lemma 2.5 guarantees that there is a continuous selection g of I-Ω such that g(x)≠0 for each x∈Fix(Ω). This implies that there exists a continuous selection h of F:Fix(F)→Pc,cp(ℝn) without fixed points. Define Λ:ℝn→Pc,cp(ℝn) by
(3.28)Λ(x)={Ω(x),x∈Rn∖Fix⁡(Ω),h(x),x∈Fix⁡(Ω).
Since Λ is lower semicontinuous, in view of Michael’s selection result (Lemma 2.6), Λ admits a continuous selection f:ℝn→ℝn. Thus f:ℝn→ℝn is a continuous selection of Ω with no fixed points and f=h on Fix(Ω) contradicting Lemma 2.7. As a result, Fix(Ω)=Sx0([0,α]) is infinite dimensional.Definition 3.5.
A metric spaceX is said to be an AR-space if, whenever it is nonempty closed subset of another metric space Y, there exists a continuous retraction r:Y→X, r(x)=x for x∈X. In particular, it is contractible (and hence connected).Theorem 3.6 (see [27]).
LetC be a nonempty closed convex subset of a Banach space X and F:C→Pc,cp(C) a contraction. Then Fix(F) is a nonempty AR-space.The following result is a consequence of Theorems3.4 and 3.6.Corollary 3.7.
Suppose thatF:[0,α]×ℝn→Pc,cp(ℝn) satisfies (H1), (H2), and (H3) and that the Lebesgue measure μ of the set{t:dimF(t,x)<1forsomex∈ℝn} is zero. Then, for each α, 0<α<min{((1-κ2)Γ(q+1)/(A1+∥κ1∥∞))1/q,T}, the set Sx0([0,α]) of solutions of (1.1) is a compact and infinite dimensional AR-space.



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