# Existence and Attractivity for Fractional Evolution Equations.

1. IntroductionFractional 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:

[mathematical expression not reproducible], (1)

where [sup.L][D.sup.[alpha].sub.0+] is Riemann-Liouville fractional derivative of order 0 < [alpha] < 1, [I.sup.1-[alpha].sub.0+] is Riemann-Liouville fractional integral of order 1 - [alpha], A is the infinitesimal generator of a [C.sub.0]-semigroup of bounded linear operators [{Q(t)}.sub.t[greater than or equal to]0] in Banach space X, f : [0, [infinity]) x X [right arrow] X is a given function satisfying some assumptions, and [x.sub.0] 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.

Let [alpha] [member of] (0,1) and u [member of] [L.sup.1]([0, [infinity]),X). The Riemann-Liouville fractional integral is defined by

[I.sp.[alpha].sub.0+] u (t) = [g.sub.[alpha]] (t) * u(t) = [[integral].sup.t.sub.0] [g.sub.[alpha]] (t-s) u (s) ds, t > 0, (2)

where * denotes the convolution,

[g.sub.[alpha]] (t) = [t.sup.[alpha]-1]/[GAMMA]([alpha]), (3)

and in case [alpha] = 0, we set [g.sub.0](t) = [delta](t), the Dirac measure is concentrated at the origin. For u [member of] C([0, [infinity]), X), the Riemann-Liouville fractional derivative is defined by

[sup.L][D.sup.[alpha].sub.0+] u (t) = d/dt ([g.sub.1-[alph,a]] (t) * u(t)). (4)

The Wright function [M.sub.[alpha]]([theta]) is defined by

[M.sub.[alpha]]([theta]) = [[infinity].summation over (n=1)] [(-[theta]).sup.n-1]/(n - 1)![GAMMA] (1 - [alpha]n). (5)

It is known that [M.sub.[alpha]]([theta]) satisfies the following equality:

[[integral].sup.[infinity].sub.0] [[theta].sup.[delta]] [M.sub.[alpha]] ([theta]) d[theta] = [GAMMA](1 + [delta])/ [GAMMA](1 + [alpha][delta]), for [delta] [greater than or equal to] 0. (6)

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 [member of] C([0, [infinity]),X) satisfies

[mathematical expression not reproducible], (7)

where

[P.sub.[alpha]] (t) = [[integral].sup.[infinity].sub.0] [alpha] [theta] [M.sub.[alpha]] ([theta]) Q ([t.sup.[alpha][theta])/d[theta]. (8)

Definition 2. The mild solution x(t) of the Cauchy problem (1) is attractive if x(t) tends to zero as t [right arrow] [infinity].

Suppose that A is the infinitesimal generator of a [C.sub.0]-semigroup [{Q(t)}.sub.t[greater than or equal to]0] of uniformly bounded linear operators on Banach space X. This means that there exists M [greater than or equal to] 1 such that

[mathematical expression not reproducible], (9)

where B(X) be the space of all bounded linear operators from X to X with the norm [[parallel]Q[parallel].sub.B(X)] = sup{[absolute value of Q(x)]: [absolute value of x] = 1}, where Q [member of] B(X) and x [member of] X.

Proposition 3 (see [5]). For any fixed t > 0, [P.sub.[alpha]](t) is linear and bounded operator, that is, for any x [member of] X,

[absolute value of [P.sub.[alpha]] (t) x] [less than or equal to] M/[GAMMA]([alpha]) [absolute value of x]. (10)

Proposition 4 (see [5]). [{[P.sub.[alpha]](t)}.sub.t>0] is strongly continuous, which means that, for [for all]x [member of] X and t" > t' >0, we have

[absolute value of [P.sub.[alpha]] (t")x--[P.sub.[alpha]] (t')x] [right arrow] 0 as t" [right arrow] t'. (11)

Proposition 5 (see [5]). Assume that [{Q(t)}.sub.t>0] is compact operator. Then [{[P.sub.[alpha]](t)}.sub.t>0] is also compact operator.

Let X be a real Banach space, J = [0, [infinity]):

[mathematical expression not reproducible], (12)

with the norm [parallel]y[parallel] = [sup.sub.t[member of][0[infinity])] [absolute value of y(t)|/(1 + t)]. It is easy to see that (E, [parallel] x [parallel]) is a Banach space.

We need also the following generalization of Ascoli-Arzela theorem, which one can find in [29].

Lemma 6. The set H [subset] E is relatively compact if and only if the following conditions hold:

(i) For any b > 0, thefunctioninU = {y: y(t) = x(t)/(1 + t), x [member of] H} is equicontinuous on [0, b].

(ii) For any t [member of] [0, [infinity]), U(t) = H(t)/(1 + t) is relatively compact in X.

(iii) [lim.sub.t[right arrow][infinity]] [absolute value of x(t)]/(1 +t) = 0 uniformly for x [member of] H.

Theorem 7 (Schauder fixed point theorem). Let S be a nonempty, closed, and convex subset of the Banach space X and let F : S [right arrow] S be completely continuous; then F has a fixed point in S.

Lemma 8 (see [26]). If [alpha], [omega] > 0, then

[mathematical expression not reproducible], (13)

where C = max{[1,2.sup.1-[alpha]]}[GAMMA]([mu])(1 + [mu]([mu] + 1)/[alpha])[[omega].sup.-[mu]] > 0.

3. Some Lemmas

In this paper, we always suppose that the operator A generates a compact [C.sub.0]-semigroup [{Q(t)}.sub.t>0] on X; that is, the operator Q(t) is compact for t > 0.

Let

[mathematical expression not reproducible], (14)

with the norm

[mathematical expression not reproducible]. (15)

Then ([C.sub.[alpha]](J, X), [[parallel] x [parallel].sub.[alpha]]) is a Banach space by the similar proof of [27, Lemma 3.2].

For any x [member of] [C.sub.[alpha]](J, X), consider the operator F defined by

(Fx) (t) = ([F.sub.1]X) (t) + ([F.sub.2]x) (t), (16)

where

[mathematical expression not reproducible]. (17)

It is easy to see [mathematical expression not reproducible]. It is clear that x is a mild solution of (1) in [C.sub.[alpha]](J, X) if and only if there exists a fixed point [x.sup.*] [member of] [C.sub.[alpha]](J, X), such that [x.sup.*] = F[x.sup.*] holds. For any y [member of] E, let

x (t) = [t.sup.[alpha]-1]y (t), for t [member of] (0,[infinity]). (18)

Then, x [member of] [C.sub.[alpha]](J, X). Define an operator F as follows:

(Fy)(t) = ([F.sub.1]y)(t) + ([F.sub.2]y)(t), (19)

where

[mathematical expression not reproducible]. (20)

In this section, we always suppose that the following condition holds:

(H1) There exist [mathematical expression not reproducible].

Let [[gamma].sub.1] [member of] (0, [[beta].sub.1]). Then there exists [T.sub.1] > 0 such that

[mathematical expression not reproducible]. (21)

Let

[mathematical expression not reproducible]. (22)

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 [member of] [S.sub.1]) is equicontinuous and [lim.sub.t[right arrow][infinity]] [absolute value of Uy(t)] = 0 uniformly for y [member of] [S.sub.1].

Proof. It is clear that [S.sub.1] is a nonempty, closed, and convex subset of E.

Claim I. {Uy: Uy(t) = ([F.sub.1]y)(t)/(1 + t), y [member of] [S.sub.1]} is equicontinuous.

For [t.sub.1] = 0, [t.sub.2] [member of] (0, [infinity]), we have

[mathematical expression not reproducible]. (23)

For any [t.sub.1],[t.sub.2] [member of] (0, [infinity]) and [t.sub.1] < [t.sub.2], we have

[mathematical expression not reproducible]. (24)

Hence, {Uy: Uy(t) = ([F.sub.1]y)(t)/(1 + t), y [member of] [S.sub.1]} is equicontinuous.

Claim II. {Uy: Uy(t) = ([F.sub.2]y)(t)/(1 + t), y [member of] [S.sub.1]} is equicontinuous. For any y [member of] [S.sub.1], let x(t) = [t.sup.[alpha]-1]y(t), t [member of] (0, [infinity]). Then x [member of] [[??].sub.1], where

[mathematical expression not reproducible], (25)

which is nonempty, closed, and convex.

Since [alpha] < [[beta].sub.1] < 1 and [epsilon] > 0 is given, then there exists [T.sub.1] > 0 enough large, such that

[mathematical expression not reproducible]. (26)

For [t.sub.1], [t.sub.2] > [T.sub.1] , in virtue of (H1) and (26), we get

[mathematical expression not reproducible]. (27)

For [t.sub.1] = 0, 0 < [t.sub.2] [less than or equal to] [T.sub.1], we have

[mathematical expression not reproducible]. (28)

For 0 < [t.sub.1] < [t.sub.2] [less than or equal to] T', we have

[mathematical expression not reproducible], (29)

where

[mathematical expression not reproducible]. (30

It can deduce that [I.sub.1] [right arrow] 0 as [t.sub.2] [right arrow] [t.sub.1] directly. Indeed,

[mathematical expression not reproducible]. (31)

Note that

[mathematical expression not reproducible], (32)

and the map [mathematical expression not reproducible] is integrable for s [member of] [0, [t.sub.1]] and [t.sub.1] [member of] [0, T'], then by Lebesgue dominated convergence theorem, we have

[mathematical expression not reproducible], (33)

which implies that [I.sub.2] [right arrow] 0 as [t.sub.2] [right arrow] [t.sub.1].

For given [epsilon] > 0 small enough, from the condition (H1), we get

[mathematical expression not reproducible], (34

where

[mathematical expression not reproducible]. (35)

By Proposition 4, we have that [I.sub.31] [right arrow] 0 as [t.sub.2] [right arrow] [t.sub.1]. Similar to the proof that [I.sub.1], [I.sub.2] tend to zero, we get [I.sub.32] [right arrow] 0 and [I.sub.33] [right arrow] 0 as [epsilon] [right arrow] 0. Thus, we get that [I.sub.3] tend to zero independently of y [member of] [S.sub.1] as [t.sub.2] [right arrow] [t.sub.1], [member of] [right arrow] 0.

Similar to (24), it is easy to proof that [I.sub.4] [right arrow] 0 as [t.sub.2] [right arrow] [t.sub.1]. For 0 [less than or equal to] [t.sub.1] < T' < [t.sub.2], note that if [t.sub.2] [right arrow] [t.sub.1], then [t.sub.2] [right arrow] T' and [t.sub.1] [right arrow] T', from the above argument, we obtain

[mathematical expression not reproducible]. (36)

Therefore, it is clear that

[mathematical expression not reproducible], (37)

independently of y [member of] Sx. Thus, {Uy: Uy(t) = ([F.sub.2]y)(t)/(1 + t), y [member of] [S.sub.1]} is equicontinuous..

Claim III. [lim.sub.t[right arrow][infinity]] [absolute value of Uy(t)] = 0 uniformly for y [member of] [S.sub.1]. For any Y [member of] [S.sub.1], by (H1) and Proposition 3, we have

[mathematical expression not reproducible]. (38)

We multiply the above inequalities at both sides with the factor 1/(1 +t), then

[mathematical expression not reproducible], (39)

and, therefore,

[mathematical expression not reproducible], (40)

which shows that [lim.sub.t[right arrow][infinity]] [absolute value of Uy(t)] = 0 uniformly for y [member of] [S.sub.1]. The proof is completed.

Lemma 10. Assume that (H1) holds. Then F maps [S.sub.1] into itself and F is continuous.

Proof.

Claim I. F maps [S.sub.1] into [S.sub.1].For any y [member of] [S.sub.1], t [member of] (0, [infinity]). Then x [member of] [[??].sub.1].

For t [greater than or equal to] [T.sub.1], together inequity (21) and 0 < [[gamma].sub.1] < [[beta].sub.1] yield that

[mathematical expression not reproducible]. (41)

Then, for t [greater than or equal to] [T.sub.1], from (39), we have

[mathematical expression not reproducible], (42)

which deduce that c [S.sub.1] [subset] [S.sub.1] for t [greater than or equal to] [T.sub.1].

Claim II. F is continuous. For any [mathematical expression not reproducible].

In addition, we have

[mathematical expression not reproducible]. (43)

By (H1) and the continuity of f on any compact subsets uniformly, then we get

[mathematical expression not reproducible]. (44)

Therefore, for a.e. [mathematical expression not reproducible]. In addition, the function [mathematical expression not reproducible] is integrable for s [member of] [0, t] and t [member of] [0, T]. Hence, by Lebesgue dominated convergence theorem, we obtain

[mathematical expression not reproducible]. (45)

Then, for t [member of] [0,T],

[mathematical expression not reproducible], (46)

which yields that [parallel]F[y.sub.m] - Fy[parallel] [right arrow] 0 as m [right arrow] [infinity].

On the other hand, let [epsilon] > 0 be given, fixed enough large T > 0 with replaced [T.sub.1] such that (26) holds. Then, for t > T, by virtue of (26) and (27), we have

[mathematical expression not reproducible]. (47)

Therefore, it is obvious that [parallel]F[y.sub.m] - Fy[parallel] [right arrow] 0 as m [right arrow] [infinity]. Combined with the above statement, It can imply that F[y.sub.m] [right arrow] Fy uniformly on [0, [infinity]) as m [right arrow] [infinity]; that is, F is continuous.

4. Main Results

Theorem 11. Assume that Q(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 [[??].sub.1] if and only if y is a fixed point of y = F y in [s.sub.1], where x(t) = [t.sup.[alpha]-1] y(t). Thus, it is sufficient to show that y = F y has a fixed point in [S.sub.1]. By Lemma 9, we know that {Uy: Uy(t) = (Fy)(t)/(1 + t), y [member of] [S.sub.1]} is equicontinuous and [lim.sub.t[right arrow][infinity]]Uy(t) = 0 uniformly for y [member of] [S.sub.1]. It remains to verify that, for any t [member of] [0, [infinity]), Uy(t) is relatively compact in X according to Lemma 6. Obviously, Uy(0) is relatively compact in X. Let t [member of] (0, [infinity]) be fixed. For every [epsilon] [member of] (0, t) and [delta] > 0, define an operator [F.sub.[epsilon],[delta]] on [S.sub.1] as follows:

[mathematical expression not reproducible]. (48)

Since Q(t) is compact for t > 0, by Proposition 5, we know that [P.sub.[alpha]](t) is compact. In addition, from the compactness of Q([[epsilon].sup.[alpha]][delta]), we obtain that the set {[U.sub.[epsilon],[delta]] y, y [member of] [S.sub.1]} is relatively compact in X for any [epsilon] [member of] (0, t) and for any [delta] > 0. For every y [member of] [S.sub.1] ,we have

[mathematical expression not reproducible]. (49)

Therefore, the set Uy(t) is closed to an arbitrary compact set. As a result, the set Uy(t) is also relatively compact set in X for t [member of] [0, [infinity]). By Lemma 6, we know that F[S.sub.1] is a relatively compact set. On the other hand, by Lemma 10, we know that F maps [S.sub.1] 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.sup.*] [member of] [S.sub.1] such that [y.sup.*] = F[y.sup.*] holds. Let [x.sup.*](t) = [t.sup.[alpha]-1][y.sup.*](t); then

[mathematical expression not reproducible], (50)

and, therefore, [x.sup.*] is a mild solution of (1).

Noting that, for any x [member of] [[??].sub.1], using the condition (H1), we have

[mathematical expression not reproducible], (51)

which yields that [x.sup.*](t) [right arrow] 0 as t [right arrow] [infinity]. Thus, the solution [x.sup.*] (t) is attractive.

Theorem 12. Assume that Q(t) (t > 0) is compact, and function f [member of] [C.sub.[alpha]](J, X) satisfies the following condition:

(H2) There exist [mathematical expression not reproducible].

Then the Cauchy problem (1) admits at least one attractive solution.

Proof. Let [[gamma].sub.2] [member of] (0,1). Then there exist constants C = C([alpha], [[beta].sub.2], [mu]) and [T.sub.2] > 0 such that

[mathematical expression not reproducible]. (52)

Let

[mathematical expression not reproducible]. (53)

For any y [member of] [S.sub.2], let x(t) = [t.sup.[alpha]-1] y(t), t [member of] (0,[infinity]). Then x [member of] [[??].sub.2], where

[mathematical expression not reproducible]. (54)

By (H2) and Lemma 8, similar to (39), we get

[mathematical expression not reproducible], (55)

for t [greater than or equal to] [T.sub.2], together inequity (52) and 0 < [[gamma].sub.2] < 1 yield that

[mathematical expression not reproducible]. (56)

Then, for t [greater than or equal to] [T.sub.2], from (55), we have

[mathematical expression not reproducible], (57)

which deduce that F[S.sub.2] [subset] [S.sub.2] for t [greater than or equal to] [T.sub.2].

The remaining part of the proof is similar to that of Theorem 11, and we omit it.

5. Example

Example 1. Let X = [L.sup.2]([0, [pi]], R). Consider the following Cauchy problem for fractional partial differential equations:

[mathematical expression not reproducible], (58)

where 0 < [alpha] < 1, G is a given function, [u.sub.0](z) [member of] X.

We define an operator A by Av = v" with the domain

D (A) = {v [member of] X: v, v absoluty continuous, v" [member of] X, v (0) = v ([pi]) = 0}. (59)

Then A generates a compact, analytic, self-adjoint [C.sub.0]-semigroup [{Q(t)}.sub.t > 0].

Let x(t) = u(t, x); that is, x(t)(z) = u(t,z), t [member of] (0,[infinity]), z [member of] [0, [pi]]. And the function f : (0, [infinity]) x X [right arrow] X is given by

f(t,x(t))(z) = [[partial derivative].sub.z]G(t,u(t,z)). (60)

Then the Cauchy problem (58) can be rewritten as the following format in X:

[sup.L] [D.sup.[alpha].sub.0+] x (t) = Ax (t) + f(t,x(t)), t [member of] (0, [infinity]), [I.sup.1-[alpha].sub.0+] x (0) = [x.sub.0], (61)

We can take [alpha] = 1/4 and f(t, x(t)) = [t.sup.-1/3] sin 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,

[mathematical expression not reproducible], (62)

As the argument described above, (62) can be rewritten as

x' (t) = Ax (t) + [t.sup.-1/3], t [member of] (0, [infinity]), x(0) = 0, (63)

Then (63) has an equivalent representation:

x(t) = [3/2] [t.sup.2/3] + [3/2] [[integral].sup.t.sub.0] A[e.sup.A(t-s)][s.sup.2/3]ds, (64)

with x(t) [right arrow] x as t [right arrow] [infinity]. 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.

https://doi.org/10.1155/2018/1070713

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The work was supported by the National Natural Science Foundation of China (no. 11671339).

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Yong Zhou (iD), (1,2) Jia Wei He, (1) Bashir Ahmad, (2) and Ahmed Alsaedi (2)

(1) Faculty of Mathematics and Computational Science, Xiangtan University, Hunan 411105, China

(2) Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Faculty of Science, King Abdulaziz University, Jeddah 21589, Saudi Arabia

Correspondence should be addressed to Yong Zhou; yzhou@xtu.edu.cn

Received 31 August 2017; Accepted 6 December 2017; Published 10 January 2018

Academic Editor: Francisco R. Villatoro

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Title Annotation: | Research Article |
---|---|

Author: | Zhou, Yong; He, Jia Wei; Ahmad, Bashir; Alsaedi, Ahmed |

Publication: | Discrete Dynamics in Nature and Society |

Date: | Jan 1, 2018 |

Words: | 4273 |

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