# Existence and Attractivity for Fractional Evolution Equations

Authors: Yong Zhou; Jia Wei He; Bashir Ahmad; Ahmed Alsaedi
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

We study the existence and attractivity of solutions for fractional evolution equations with Riemann-Liouville fractional derivative. We establish sufficient conditions for the global attractivity of mild solutions for the Cauchy problems in the case that semigroup is compact.


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## Body

## 1. Introduction

Fractional differential equations have gained considerable importance due to their application in various sciences, such as physics, mechanics, chemistry, and engineering. In recent years, there has been a significant development on ordinary and partial differential equations involving fractional derivatives; see the monographs of Podlubny [1], Kilbas et al. [2], Diethelm [3], Tarasov [4], and Zhou [5, 6] and a series of papers [7–28] and the references cited therein.Recently, Zhou [7], Chen et al. [19], Losada et al. [20], and Banas´ and O’Regan [21] investigated the attractivity of solutions for fractional ordinary differential equations and integral equations. On the other hand, the existence theory of solutions for time fractional evolution equations has been investigated intensively by many authors; for example, see Kim et al. [16], Bazhlekova [22], Wang et al. [23], Zacher [24], and Zhou et al. [25]. However, to the best of our knowledge, there are no results on the attractivity of solutions for fractional evolution equations in the literature.Consider fractional evolution equation with Riemann-Liouville derivative:(1)DL0+αxt=Axt+ft,xt,t∈0,∞,I0+1-αx0=x0,where DL0+α is Riemann-Liouville fractional derivative of order 0<α<1, I0+1-α is Riemann-Liouville fractional integral of order 1-α, A is the infinitesimal generator of a C0-semigroup of bounded linear operators {Q(t)}t≥0 in Banach space X, f:[0,∞)×X→X is a given function satisfying some assumptions, and x0 is an element of the Banach space X.In this paper, we initiate the question of the attractivity of solutions for Cauchy problem (1). We establish sufficient conditions for the global attractivity for mild solutions of (1) in the case that semigroup is compact. These results essentially reveal the characteristics of solutions for fractional evolution equations with Riemann-Liouville derivative. More precisely, integer order evolution equations do not have such attractivity.


## 2. Preliminaries

In this section, we firstly recall some concepts on fractional integrals and derivatives and then give some lemmas which are useful in next sections.Letα∈(0,1) and u∈L1([0,∞),X). The Riemann-Liouville fractional integral is defined by (2)I0+αut=gαt∗ut=∫0tgαt-susds,t>0,where ∗ denotes the convolution, (3)gαt=tα-1Γα,and in case α=0, we set g0(t)=δ(t), the Dirac measure is concentrated at the origin. For u∈C([0,∞),X), the Riemann-Liouville fractional derivative is defined by (4)DL0+αut=ddtg1-αt∗ut.The Wright functionMα(θ) is defined by (5)Mαθ=∑n=1∞-θn-1n-1!Γ1-αn.It is known that Mα(θ) satisfies the following equality: (6)∫0∞θδMαθdθ=Γ1+δΓ1+αδ,forδ≥0.We give the following definition of the mild solution of (1).Definition 1 (see [5]).
By the mild solution of the Cauchy problem (1), we mean that the function x∈C([0,∞),X) satisfies (7)xt=tα-1Pαtx0+∫0tt-sα-1Pαt-sfs,xsds,t>0,where(8)Pαt=∫0∞αθMαθQtαθdθ.Definition 2.
The mild solutionx(t) of the Cauchy problem (1) is attractive if x(t) tends to zero as t→∞.Suppose thatA is the infinitesimal generator of a C0-semigroup {Q(t)}t≥0 of uniformly bounded linear operators on Banach space X. This means that there exists M≥1 such that (9)M=supt∈0,∞⁡QtBX<∞,where B(X) be the space of all bounded linear operators from X to X with the norm QB(X)=sup{Qx:x=1}, where Q∈B(X) and x∈X.Proposition 3 (see [5]).
For any fixedt>0, Pα(t) is linear and bounded operator, that is, for any x∈X, (10)Pαtx≤MΓαx.Proposition 4 (see [5]).
{
P
α
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t
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0 is strongly continuous, which means that, for ∀x∈X and t′′>t′>0, we have (11)Pαt′′x-Pαt′x⟶0ast′′⟶t′.Proposition 5 (see [5]).
Assume that{Q(t)}t>0 is compact operator. Then {Pα(t)}t>0 is also compact operator.LetX be a real Banach space, J=[0,∞): (12)E=y∈CJ,X:limt→∞⁡yt1+t=0,with the norm y=supt∈[0,∞)yt/(1+t). It is easy to see that (E,·) is a Banach space.We need also the following generalization of Ascoli-Arzela theorem, which one can find in [29].Lemma 6.
The setH⊂E is relatively compact if and only if the following conditions hold: (i)
For anyb>0, the function in U={y:y(t)=x(t)/(1+t),x∈H} is equicontinuous on [0,b].(ii)
For anyt∈[0,∞), U(t)=H(t)/(1+t) is relatively compact in X.(iii)
limt→∞xt/(1+t)=0 uniformly for x∈H.Theorem 7 (Schauder fixed point theorem).
LetS be a nonempty, closed, and convex subset of the Banach space X and let F:S→S be completely continuous; then F has a fixed point in S.Lemma 8 (see [26]).
Ifα,μ,ω>0, then (13)∫0tt-sα-1sμ-1e-ωsds≤Ctα-1,t>0,where C=max⁡1,21-αΓ(μ)(1+μ(μ+1)/α)ω-μ>0.


## 3. Some Lemmas

In this paper, we always suppose that the operatorA generates a compact C0-semigroup {Q(t)}t>0 on X; that is, the operator Q(t) is compact for t>0.Let(14)CαJ,X=x∈C0,∞,X:limt→0+t1-αxtexists  and  is  finite,limt→∞⁡t1-αxt1+t=0,with the norm (15)xα=supt∈0,∞⁡t1-αxt1+t.Then (Cα(J,X),·α) is a Banach space by the similar proof of [27, Lemma 3.2].For anyx∈Cα(J,X), consider the operator F defined by (16)Fxt=F1xt+F2xt,where (17)F1xt=tα-1Pαtx0,fort∈0,∞,F2xt=∫0tt-sα-1Pαt-sfs,xsds,fort∈0,∞.It is easy to see limt→0+t1-α(Fx)(t)=x0/Γ(α). It is clear that x is a mild solution of (1) in Cα(J,X) if and only if there exists a fixed point x∗∈Cα(J,X), such that x∗=Fx∗ holds.For anyy∈E, let (18)xt=tα-1yt,fort∈0,∞.Then, x∈Cα(J,X). Define an operator F as follows: (19)Fyt=F1yt+F2yt,where (20)F1yt=t1-αF1xt,fort∈0,∞,x0Γα,fort=0,F2yt=t1-αF2xt,fort∈0,∞,0,fort=0.In this section, we always suppose that the following condition holds:(H1)
There existK≥0 and β1∈(α,1), such that ft,xt≤Kt-β1, for any x∈C((0,∞),X).Letγ1∈(0,β1). Then there exists T1>0 such that (21)Mx0ΓαT1γ1-1+MKΓ1-β1Γ1+α-β1T1γ1-β1≤1.Let (22)S1=y:y∈E,tγ1yt1+t≤1,fort≥T1.Before obtaining our main results, we firstly prove some lemmas as follows.Lemma 9.
Assume that (H1) holds. Then,{Uy:Uy(t)=(Fy)(t)/(1+t),y∈S1} is equicontinuous and limt→∞⁡Uyt=0 uniformly for y∈S1.Proof.
It is clear thatS1 is a nonempty, closed, and convex subset of E.
Claim I. {Uy:Uy(t)=(F1y)(t)/(1+t),y∈S1} is equicontinuous.
Fort1=0, t2∈(0,∞), we have (23)F1yt21+t2-F1y0≤11+t2Pαt2x0-x0Γα≤11+t2Pαt2-1Γαx0⟶0,ast2⟶0.
For anyt1,t2∈(0,∞) and t1<t2, we have (24)F1yt21+t2-F1yt11+t1≤Pαt2x01+t2-Pαt1x01+t1≤Pαt2x01+t2-Pαt2x01+t1+Pαt2x01+t1-Pαt1x01+t1≤Pαt2x0t2-t11+t21+t1+Pαt2-Pαt1x011+t1⟶0,ast2⟶t1.Hence, {Uy:Uy(t)=(F1y)(t)/(1+t),y∈S1} is equicontinuous.
Claim II. {Uy:Uy(t)=(F2y)(t)/(1+t),y∈S1} is equicontinuous. For any y∈S1, let x(t)=tα-1y(t), t∈(0,∞). Then x∈S~1, where (25)S~1=x∈CαJ,X:tγ1t1-αxt1+t≤1,fort≥T1,which is nonempty, closed, and convex.
Sinceα<β1<1 and ε>0 is given, then there exists T′>0 enough large, such that (26)MKΓ1-β1Γ1+α-β1T′1-β11+T′<ε2.
Fort1,t2>T′, in virtue of (H1) and (26), we get (27)F2yt21+t2-F2yt11+t1≤t21-α1+t2∫0t2t2-sα-1Pαt2-sfs,xsds+t11-α1+t1∫0t1t1-sα-1Pαt1-sfs,xsds≤MKΓαt21-α1+t2∫0t2t2-sα-1s-β1ds+MKΓαt11-α1+t1∫0t1t1-sα-1s-β1ds≤2MKΓ1-β1Γ1+α-β1T′1-β11+T′<ε,ast2⟶t1.
Fort1=0,0<t2≤T′, we have (28)F2yt21+t2-F2y0=t21-α1+t2∫0t2t2-sα-1Pαt2-sfs,xsds≤MKΓαt21-α1+t2∫0t2t2-sα-1s-β1ds≤MKΓ1-β1Γ1+α-β1t21-β11+t2⟶0ast2⟶0.For 0<t1<t2≤T′, we have (29)F2yt21+t2-F2yt11+t1≤t11-α1+t1∫t1t2t2-sα-1Pαt2-sfs,xsds+t11-α1+t1∫0t1t2-sα-1-t1-sα-1Pαt2-sfs,xsds+t11-α1+t1∫0t1t1-sα-1Pαt2-s-Pαt1-sfs,xsds+t21-α1+t2-t11-α1+t1∫0t2t2-sα-1Pαt2-sfs,xsds≤I1+I2+I3+I4,where (30)I1=MKΓαt11-α1+t1∫0t2t2-sα-1s-β1ds-∫0t1t1-sα-1s-β1ds,I2=2MKΓαt11-α1+t1∫0t1t1-sα-1-t2-sα-1s-β1ds,I3=t11-α1+t1∫0t1t1-sα-1Pαt2-s-Pαt1-sfs,xsds,I4=t21-α1+t2-t11-α1+t1MKΓ1-β1Γ1+α-β1t2α-β1.
It can deduce thatI1→0 as t2→t1 directly. Indeed, (31)I1=MKΓ1-β1Γ1+α-β1t11-αt2α-β1-t1α-β11+t1≤MKΓ1-β1Γ1+α-β1t11-β1t2α-β1t2-t1β1-α1+t1⟶0,ast2⟶t1.
Note that(32)t1-sα-1-t2-sα-1s-β1≤t1-sα-1s-β1,fors∈0,t1,and the map s→(t1-s)α-1s-β1 is integrable for s∈[0,t1] and t1∈[0,T′], then by Lebesgue dominated convergence theorem, we have (33)∫0t1t1-sα-1-t2-sα-1s-β1ds⟶0,ast2⟶t1,which implies that I2→0 as t2→t1.
For givenε>0 small enough, from the condition (H1), we get (34)I3≤t11-α1+t1∫0t1-εt1-sα-1Pαt2-s-Pαt1-sfs,xsds+t11-α1+t1∫t1-εt1t1-sα-1Pαt2-s-Pαt1-sfs,xsds≤t11-α1+t1∫0t1t1-sα-1fs,xsdssups∈0,t1-ε⁡Pαt2-s-Pαt1-s+t11-α1+t12MΓα∫t1-εt1t1-sα-1fs,xsds≤I31+I32+I33,where (35)I31=t11-α1+t1∫0t1t1-sα-1fs,xsdssups∈0,t1-ε⁡Pαt2-s-Pαt1-sBX,I32=t11-α1+t12MKΓα∫0t1-εt1-ε-sα-1-t1-sα-1s-β1ds,I33=t11-α1+t12MKΓα∫0t1t1-sα-1s-β1ds-∫0t1-εt1-ε-sα-1s-β1ds.
By Proposition4, we have that I31→0 as t2→t1. Similar to the proof that I1, I2 tend to zero, we get I32→0 and I33→0 as ε→0. Thus, we get that I3 tend to zero independently of y∈S1 as t2→t1, ε→0.
Similar to (24), it is easy to proof that I4→0 as t2→t1.
For0≤t1<T′<t2, note that if t2→t1, then t2→T′ and t1→T′, from the above argument, we obtain (36)F2yt21+t2-F2yt11+t1≤F2yt21+t2-F2yT′1+T′+F2yT′1+T′-F2yt11+t1⟶0,ast2⟶t1.
Therefore, it is clear that(37)F2yt21+t2-F2yt11+t1⟶0,ast2⟶t1,independently of y∈S1. Thus, {Uy:Uy(t)=(F2y)(t)/(1+t),y∈S1} is equicontinuous..
Claim III. limt→∞Uyt=0 uniformly for y∈S1. For any y∈S1, by (H1) and Proposition 3, we have (38)Fyt≤Pαtx0+t1-α∫0tt-sα-1Pαt-sfs,xsds≤Mx0Γα+MKΓα∫0tt1-αt-sα-1s-β1ds≤Mx0Γα+MKΓ1-β1Γ1+α-β1t1-β1.We multiply the above inequalities at both sides with the factor 1/(1+t), then (39)Fyt1+t≤Mx0Γα11+t+MKΓ1-β1Γ1+α-β1t1-β11+t,and, therefore, (40)Fyt1+t≤Mx0Γα11+t+MKΓ1-β1Γ1+α-β1t1-β11+t⟶0,ast⟶∞,which shows that limt→∞Uyt=0 uniformly for y∈S1. The proof is completed.Lemma 10.
Assume that (H1) holds. ThenF maps S1 into itself and F is continuous.Proof.
Claim I. F maps S1 into S1. For any y∈S1, let x(t)=tα-1y(t), t∈(0,∞). Then x∈S~1.
Fort≥T1, together inequity (21) and 0<γ1<β1 yield that (41)Mx0Γαtγ11+t+MKΓ1-β1Γ1+α-β1t1-β1+γ11+t≤Mx0Γαtγ1-1+MKΓ1-β1Γ1+α-β1tγ1-β1≤Mx0ΓαT1γ1-1+MKΓ1-β1Γ1+α-β1T1γ1-β1≤1.Then, for t≥T1, from (39), we have (42)tγ1Fyt1+t≤Mx0Γαtγ11+t+MKΓ1-β1Γ1+α-β1t1-β1+γ11+t≤1,which deduce that FS1⊂S1 for t≥T1.
Claim II. F is continuous. For any ym,y∈S1,  m=1,2,…,with limm→∞ym=y, let x(t)=tα-1y(t), xm(t)=tα-1ym(t),t∈(0,∞). Then xm,x∈S~1.
In addition, we have(43)limm→∞⁡ymt=yt,limm→∞⁡tα-1ymt=tα-1yt,fort∈0,∞.By (H1) and the continuity of f on any compact subsets uniformly, then we get (44)limm→∞⁡ft,xmt=limm→∞⁡ft,tα-1ymt=ft,tα-1yt=ft,xt.
Therefore, for a.e.t∈[0,T], (t-s)α-1fs,xms-fs,xs≤2L(t-s)α-1s-β1 via (H1). In addition, the function s→2L(t-s)α-1s-β1 is integrable for s∈[0,t] and t∈[0,T]. Hence, by Lebesgue dominated convergence theorem, we obtain (45)∫0tt-sα-1fs,xms-fs,xsds⟶0,asm⟶∞.Then, for t∈[0,T], (46)Fymt1+t-Fyt1+t≤t1-α1+t∫0tt-sα-1Pαt-sfs,xms-fs,xsds≤MΓαt1-α1+t∫0tt-sα-1fs,xms-fs,xsds,which yields that Fym-Fy→0 as m→∞.
On the other hand, letε>0 be given, fixed enough large T>0 with replaced T′ such that (26) holds. Then, for t>T, by virtue of (26) and (27), we have (47)Fymt1+t-Fyt1+t≤MΓαt1-α1+t∫0tt-sα-1fs,xms+fs,xsds≤2MKΓ1-β1Γ1+α-β1T1-β11+T<ε.Therefore, it is obvious that Fym-Fy→0 as m→∞. Combined with the above statement, It can imply that Fym→Fy uniformly on [0,∞) as m→∞; that is, F is continuous.


## 4. Main Results

Theorem 11.
Assume thatQ(t)(t>0) is compact, and the condition (H1) holds. Then the Cauchy problem (1) admits at least one attractive solution.Proof.
Obviously,x is a mild solution of (1) in S~1 if and only if y is a fixed point of y=Fy in S1, where x(t)=tα-1y(t). Thus, it is sufficient to show that y=Fy has a fixed point in S1. By Lemma 9, we know that {Uy:Uy(t)=(Fy)(t)/(1+t),y∈S1} is equicontinuous and limt→∞Uyt=0 uniformly for y∈S1. It remains to verify that, for any t∈[0,∞), Uy(t) is relatively compact in X according to Lemma 6. Obviously, Uy(0) is relatively compact in X. Let t∈(0,∞) be fixed. For every ε∈(0,t) and δ>0, define an operator Fε,δ on S1 as follows: (48)Uε,δyt=Fε,δyt1+t=Pαtx01+t+∫0t-ε∫δ∞αθt1-αt-sα-11+tMαθQt-sαθfs,xsdθds=Pαtx01+t+Qεαδ∫0t-ε∫δ∞αθt1-αt-sα-11+tMαθQt-sαθ-εαδfs,xsdθds.
SinceQ(t) is compact for t>0, by Proposition 5, we know that Pα(t) is compact. In addition, from the compactness of Q(εαδ), we obtain that the set {Uε,δy,y∈S1} is relatively compact in X for any ε∈(0,t) and for any δ>0. For every y∈S1, we have (49)Uyt-Uε,δyt≤t1-α1+t∫0t-ε∫0δαθt-sα-1MαθQt-sαθfs,xsdθds+t1-α1+t∫t-εt∫0∞αθt-sα-1MαθQt-sαθfs,xsdθds≤αMt1-α1+t∫0t-εt-sα-1fs,xsds∫0δθMαθdθ+αMt1-α1+t∫t-εtt-sα-1fs,xsds∫0∞θMαθdθ≤αMKt1-β11+tΓαΓ1-β1Γ1+α-β1∫0δθMαθdθ+αMKt1-α1+t∫1-ε/t11-sα-1s-β1ds∫0∞θMαθdθ⟶0,asε⟶0,δ⟶0.
Therefore, the setUy(t) is closed to an arbitrary compact set. As a result, the set Uy(t) is also relatively compact set in X for t∈[0,∞). By Lemma 6, we know that FS1 is a relatively compact set. On the other hand, by Lemma 10, we know that F maps S1 into itself and F is continuous. Hence, F is a completely continuous operator. Therefore, according to Schauder fixed point theorem, there exists at least one fixed point y∗∈S1 such that y∗=Fy∗ holds. Let x∗(t)=tα-1y∗(t); then (50)x∗t=tα-1Pαtx0+∫0tt-sα-1Pαt-sfs,x∗sds,t>0,and, therefore, x∗ is a mild solution of (1).
Noting that, for anyx∈S~1, using the condition (H1), we have (51)limt→∞⁡xt≤limt→∞⁡Mx0Γαtα-1+limt→∞⁡MKΓ1-β1Γ1+α-β1tα-β1=0,which yields that x∗(t)→0 as t→∞. Thus, the solution x∗(t) is attractive.Theorem 12.
Assume thatQ(t)(t>0) is compact, and function f∈Cα(J,X) satisfies the following condition: (H2)
There existL≥0, μ>0 and β2>0, such that ft,xt≤Ltβ2-1e-μt, for any x∈C((0,∞),X).
Then the Cauchy problem (1) admits at least one attractive solution.Proof.
Letγ2∈(0,1). Then there exist constants C=C(α,β2,μ) and T2>0 such that (52)Mx0ΓαT2γ2-1+MLCΓαT2γ2-1≤1.Let (53)S2=y:y∈E,tγ2yt1+t≤1,fort≥T2.
For anyy∈S2, let x(t)=tα-1y(t), t∈(0,∞). Then x∈S~2, where (54)S~2=x∈CαJ,X:tγ2t1-αxt1+t≤1,fort≥T2.
By (H2) and Lemma8, similar to (39), we get (55)tγ2Fyt1+t≤tγ2Pαtx01+t+∫0tt1+γ2-αt-sα-11+tPαt-sfs,xsds≤Mx0Γαtγ21+t+MLΓαt1+γ2-α1+t∫0tt-sα-1sβ2-1e-μsds≤Mx0Γαtγ2-1+MLCΓαtγ2-1,for t≥T2, together inequity (52) and 0<γ2<1 yield that (56)Mx0Γαtγ2-1+MLCΓαtγ2-1≤Mx0ΓαT2γ2-1+MLCΓαT2γ2-1≤1.Then, for t≥T2, from (55), we have (57)tγ2Fyt1+t≤Mx0Γαtγ2-1+MLCΓαtγ2-1≤1,which deduce that FS2⊂S2 for t≥T2.
The remaining part of the proof is similar to that of Theorem11, and we omit it.


## 5. Example

Example 1.
LetX=L2([0,π],R). Consider the following Cauchy problem for fractional partial differential equations: (58)DL0+αut,z=∂z2ut,z+∂zGt,ut,z,z∈0,π,t∈0,∞,ut,0=ut,π=0,t∈0,∞,I0+1-αu0,z=u0z,z∈0,π,where 0<α<1, G is a given function, u0(z)∈X.We define an operatorA by Av=v′′ with the domain (59)DA=v∈X:v,v′absoluty  continuous,v′′∈X,v0=vπ=0.Then A generates a compact, analytic, self-adjoint C0-semigroup {Q(t)}t>0.Letx(t)=u(t,·); that is, x(t)(z)=u(t,z), t∈(0,∞), z∈[0,π]. And the function f:(0,∞)×X→X is given by (60)ft,xtz=∂zGt,ut,z.Then the Cauchy problem (58) can be rewritten as the following format in X: (61)DL0+αxt=Axt+ft,xt,t∈0,∞,I0+1-αx0=x0,We can takeα=1/4 and f(t,x(t))=t-1/3sin⁡x(t). Then, (H1) is satisfied. According to Theorem 11, problem (58) has at least one attractive solution.However, for the following integer order differential evolution equations,(62)∂∂tut,z=∂2∂z2ut,z+t-1/3,z∈0,π,t∈0,∞,ut,0=ut,π=0,t∈0,∞,u0,z=0,z∈0,π,As the argument described above, (62) can be rewritten as (63)x′t=Axt+t-1/3,t∈0,∞,x0=0,Then (63) has an equivalent representation:(64)xt=32t2/3+32∫0tAeAt-ss2/3ds,with x(t)→∞ as t→∞. This result shows that fractional evolution equations with Riemann-Liouville derivative have the global attractivity, whereas the integer order evolution equations do not have such attractivity.



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