Abstract

This paper is concerned with a nonlocal Cauchy problem for fractional integrodifferential equations in a separable Banach space X. We establish an existence theorem for mild solutions to the nonlocal Cauchy problem, by virtue of measure of noncompactness and the fixed point theorem for condensing maps. As an application, the existence of the mild solution to a nonlocal Cauchy problem for a concrete integrodifferential equation is obtained.

1. Introduction

Nonlocal Cauchy problem for equations is an initial problem for the corresponding equations with nonlocal initial data. Such a Cauchy problem has better effects than the normal Cauchy problem with the classical initial data when we deal with many concrete problem coming from engineering and physics (cf., e.g., [110] and references therein). Therefore, the study of this type of Cauchy problem is important and significant. Actually, as we have seen from the just mentioned literature, there have been many significant developments in this field.

On the other hand, fractional differential and integrodifferential equations arise from various real processes and phenomena appeared in physics, chemical technology, materials, earthquake analysis, robots, electric fractal network, statistical mechanics biotechnology, medicine, and economics. They have in recent years been an object of investigations with much increasing interest. For more information on this subject see for instance [9, 1118] and references therein.

Throughout this paper, 𝑋 is a separable Banach space; 𝐿(𝑋) is the Banach space of all linear bounded operators on 𝑋; 𝐴 is the generator of an analytic and uniformly bounded semigroup {𝑇(𝑡)}𝑡0 on 𝑋 with 𝑇(𝑡)𝐿(𝑋)𝑀 for a constant 𝑀>0, and 𝐶([𝑎,𝑏],𝑋) is the space of all 𝑋-valued continuous functions on[𝑎,𝑏] with the supremum norm as follows:𝑥[𝑎,𝑏][][]=max{𝑥(𝑡)𝑡𝑎,𝑏},forany𝑥𝐶(𝑎,𝑏,𝑋).(1.1)

Let 0<𝑞<1, 𝑇>0, Δ={(𝑡,𝑠)[0,𝑇]×[0,𝑇]𝑡𝑠}, 𝑓[0,𝑇]×𝐶([0,𝑇],𝑋)𝑋, and Δ×𝐶([0,𝑇],𝑋)𝑋. The nonlocal Cauchy problem for abstract fractional integrodifferential equations, with which we are concerned, is in the following form:𝑐D𝑞𝑥(𝑡)=𝐴𝑥(𝑡)+𝑓(𝑡,𝑥(𝑡))+𝑡0[],𝑘(𝑡,𝑠)(𝑡,𝑠,𝑥(𝑠))𝑑𝑠,𝑡0,𝑇𝑥(0)=𝑔(𝑥)+𝑥0,(1.2) where 𝑘 and 𝑔 are given functions to be specified later and the fractional derivative is understood in the Caputo sense, this means that, the fractional derivative is understood in the following sense:𝑐𝐷𝑞𝑥(𝑡)=𝐿𝐷𝑞(𝑥(𝑡)𝑥(0)),𝑡>0,0<𝑞<1,(1.3) and where𝐿𝐷𝑞1𝑥(𝑡)=𝑑Γ(1𝑞)𝑑𝑡𝑡0(𝑡𝑠)𝑞𝑥(𝑠)𝑑𝑠,𝑡>0,0<𝑞<1(1.4) is the Riemann-Liouville derivative of order 𝑞 of 𝑥(𝑡), where Γ() is the Gamma function.

Our main purpose is to establish an existence theorem for the mild solutions to the nonlocal Cauchy problem based on a special measure of noncompactness under weak assumptions on the nonlinearity 𝑓 and the semigroup {𝑇(𝑡)}𝑡0 generated by 𝐴.

2. Existence Result and Proof

As usual, we abbreviate 𝑢𝐿𝑝([0,𝑇],𝐑+) with 𝑢𝐿𝑝, for any 𝑢𝐿𝑝([0,𝑇],𝐑+).

As in [16, 17], we define the fractional integral of order 𝑞 with the lower limit zero for a function 𝑓𝐴𝐶[0,) as𝐼𝑞1𝑓(𝑡)=Γ(𝑞)𝑡0(𝑡𝑠)𝑞1𝑓(𝑠)𝑑𝑠,𝑡>0,0<𝑞<1,(2.1) provided the right side is point-wise defined on [0,).

Now we recall some very basic concepts in the theory of measures of noncompactness and condensing maps (see, e.g., [19, 20]).

Definition 2.1. Let 𝐸 be a Banach space, 2𝐸 the family of all nonempty subsets of 𝐸, (𝒜,) a partially ordered set, and 𝛼2𝐸𝒜. If for every Ω2𝐸: 𝛼co(Ω)=𝛼(Ω),(2.2) then we say that 𝛼 is a measure of noncompactness in 𝐸.

Definition 2.2. Let 𝐸 be a Banach space, and 𝔉𝑌𝐸𝐸 is continuous. Let 𝛼 be a measure of noncompactness in 𝐸 such that(i)for any Ω0,Ω12𝐸 with Ω0Ω1, 𝛼Ω0Ω𝛼1;(2.3)(ii) for every 𝑎0𝐸, Ω2𝐸, 𝛼𝑎0Ω=𝛼(Ω).(2.4) If for every bounded set Ω𝑌 which is not relatively compact, 𝛼(𝔉(Ω))<𝛼(Ω),(2.5) then we say that 𝔉 is condensing with respect to the measure of noncompactness 𝛼 (or 𝛼-condensing).

Definition 2.3. Let 𝜛𝑞(1𝜎)=𝜋𝑛=1(1)𝑛1𝜎𝑞𝑛1Γ(𝑛𝑞+1)𝑛!sin(𝑛𝜋𝑞),𝜎(0,)(2.6) be a one-sided stable probability density, and 𝜉𝑞1(𝜎)=𝑞𝜎11/𝑞𝜛𝑞𝜎1/𝑞,𝜎(0,).(2.7) For any 𝑧𝑋, we define operators {𝑌(𝑡)}𝑡0 and {𝑍(𝑡)}𝑡0 by 𝑌(𝑡)𝑧=0𝜉𝑞(𝜎)𝑇(𝑡𝑞𝜎)𝑧𝑑𝜎,𝑍(𝑡)𝑧=𝑞0𝜎𝑡𝑞1𝜉𝑞(𝜎)𝑇(𝑡𝑞𝜎)𝑧𝑑𝜎.(2.8) If a continuous function 𝑥[0,𝑇]𝑋 satisfies 𝑥(𝑡)=𝑌(𝑡)𝑔(𝑥)+𝑥0+𝑡0[][],𝑍(𝑡𝑠)𝑓(𝑠,𝑥(𝑠))+𝑎(𝑥)(𝑠)𝑑𝑠,𝑡0,𝑇(2.9) then the function 𝑥 is called a mild solution of (1.2).
Our main result is as follows.

Theorem 2.4. Assume that (1)𝑓(,𝑤) and (,,𝑤) are measurable for each 𝑤𝐶([0,𝑇],𝑋); 𝑘(𝑡,) is measurable for each 𝑡[0,𝑇];(2)𝑓(𝑡,) is continuous for a.e. 𝑡[0,𝑇]; 𝑔 is completely continuous; (𝑡,𝑠,) is continuous for a.e. (𝑡,𝑠)Δ; the map 𝑡𝑘𝑡=𝑘(𝑡,) is continuous from [0,𝑇] to 𝐿([0,𝑇],𝐑);(3)there exist two positive functions 𝜇(),𝜂()𝐿𝑝(0,𝑇,𝐑+)(𝑝>1/𝑞>1) and two positive functions 𝑚(,) and 𝜁(,) on Δ with sup[]𝑡0,𝑇𝑡0𝑚(𝑡,𝑠)𝑑𝑠=𝑚<,sup[]𝑡0,𝑇𝑡0𝜁(𝑡,𝑠)𝑑𝑠=𝜁<,(2.10) such that [](𝑓(𝑡,𝑤)𝜇(𝑡)𝑤(a.e.𝑡0,𝑇),𝑡,𝑠,𝑤)𝑚(𝑡,𝑠)𝑤(a.e.(𝑡,𝑠)Δ),(2.11) for all 𝑤𝐶([0,𝑇],𝑋), and []𝜒𝜒(𝑓(𝑡,𝐷))𝜂(𝑡)𝜒(𝐷),(a.e.𝑡0,T),((𝑡,𝑠,𝐷))𝜁(𝑡,𝑠)𝜒(𝐷),(a.e.(𝑡,𝑠)Δ),(2.12) for any bounded set 𝐷𝐶([0,𝑇],𝑋), where 𝜒 is the Hausdorff measure of noncompactness: 𝜒(Ω)=inf{𝜀>0Ωhasanite𝜀-net}.(2.13)(4)𝑔() satisfies []𝑔(𝑥)𝑏,𝑥𝐶(0,𝑇,𝑋),(2.14) for a positive constant 𝑏, and 𝑘||𝑘||(𝑡)=esssup(𝑡,𝑠),0𝑠𝑡(2.15) is bounded on [0,𝑇].
Then the mild solutions set of problem (1.2) is a nonempty compact subset of the space 𝐶([0,𝑇],𝑋), in the case of 𝑞𝑀Γ(1+𝑞)𝑝1𝑝𝑞1(𝑝1)/𝑝𝑇𝑞1/𝑝𝜇𝐿𝑝<1.(2.16)

Proof. First of all, let us prove our definition of the mild solution to problem (1.2) is well defined and reasonable. Actually, the proof is basic. We present it here for the completeness of the proof as well as the convenience of reading.
Write 𝑎(𝑥)(𝑡)=𝑡0𝑘(𝑡,𝑠)(𝑡,𝑠,𝑥(𝑠))𝑑𝑠,̂𝑥(𝜆)=0𝑒𝜆𝑡𝑓𝑥(𝑡)𝑑𝑡,(𝜆)=0𝑒𝜆𝑡𝑓(𝑡,𝑥(𝑡))𝑑𝑡,̂𝑎(𝜆)=0𝑒𝜆𝑡𝑎(𝑥)(𝑡)𝑑𝑡.(2.17) Clearly, the nonlocal Cauchy problem (1.2) can be written as the following equivalent integral equation: 𝑥(𝑡)=𝑔(𝑥)+𝑥0+1Γ(𝑞)𝑡0(𝑡𝑠)𝑞1[][],𝐴𝑥(𝑠)+𝑓(𝑠,𝑥(𝑠))+𝑎(𝑥)(𝑠)𝑑𝑠,𝑡0,𝑇(2.18) provided that the integral in (2.18) exists. Formally taking the Laplace transform to (2.18), we have 1̂𝑥(𝜆)=𝜆𝑔(𝑥)+𝑥0+1𝜆𝑞1𝐴̂𝑥(𝜆)+𝜆𝑞.𝑓(𝜆)+̂𝑎(𝜆)(2.19) Therefore, if the related integrals exist, then we obtain ̂𝑥(𝜆)=𝜆𝑞1(𝜆𝑞𝐴)1𝑔(𝑥)+𝑥0+(𝜆𝑞𝐴)1𝑓(𝜆)+̂𝑎(𝜆)=𝜆𝑞10𝑒𝜆𝑞𝑠𝑇(𝑠)𝑔(𝑥)+𝑥0𝑑𝑠+0𝑒𝜆𝑞𝑠𝑇(𝑠)𝑓(𝜆)+̂𝑎(𝜆)𝑑𝑠=𝑞0(𝜆𝑡)𝑞1𝑒(𝜆𝑡)𝑞𝑇(𝑡𝑞)𝑔(𝑥)+𝑥0+𝑑𝑡0𝑒(𝜆𝜏)𝑞𝑞𝜏𝑞1𝑇(𝜏𝑞)0𝑒𝜆𝑡1(𝑓(𝑡,𝑥(𝑡))+𝑎(𝑥)(𝑡))𝑑𝑡𝑑𝜏=𝜆0𝑑𝑒𝑑𝑡(𝜆𝑡)𝑞𝑇(𝑡𝑞)𝑔(𝑥)+𝑥0+𝑑𝑡00𝑒𝜆𝜏𝜎𝑞𝜏𝑞1𝜛𝑞(𝜎)𝑇(𝜏𝑞)0𝑒𝜆𝑡=(𝑓(𝑡,𝑥(𝑡))+𝑎(𝑥)(𝑡))𝑑𝑡𝑑𝜎𝑑𝜏00𝑒𝜆𝑡𝜎𝜎𝜛𝑞(𝜎)𝑇(𝑡𝑞)𝑔(𝑥)+𝑥0𝑑𝜎𝑑𝑡+𝑞00𝑒𝜆𝜃𝜃𝑞1𝜎𝑞𝜛𝑞𝜃(𝜎)𝑇𝑞𝜎𝑞0𝑒𝜆𝑡=(𝑓(𝑡,𝑥(𝑡))+𝑎(𝑥)(𝑡))𝑑𝑡𝑑𝜎𝑑𝜃0𝑒𝜆𝑡0𝜛𝑞(𝑡𝜎)𝑇𝑞𝜎𝑞𝑔(𝑥)+𝑥0𝑑𝜎𝑑𝑡+𝑞00𝑡𝑒𝜆𝜏(𝜏𝑡)𝑞1𝜎𝑞𝜛𝑞(𝜎)𝑇(𝜏𝑡)𝑞𝜎𝑞=(𝑓(𝑡,𝑥(𝑡))+𝑎(𝑥)(𝑡))𝑑𝜏𝑑𝑡𝑑𝜎0𝑒𝜆𝑡0𝜛𝑞𝑡(𝜎)𝑇𝑞𝜎𝑞𝑔(𝑥)+𝑥0𝑑𝜎𝑑𝑡+𝑞00𝜏0𝑒𝜆𝜏(𝜏𝑡)𝑞1𝜎𝑞𝜛𝑞(𝜎)𝑇(𝜏𝑡)𝑞𝜎𝑞(=𝑓(𝑡,𝑥(𝑡))+𝑎(𝑥)(𝑡))𝑑𝑡𝑑𝜏𝑑𝜎0𝑒𝜆𝑡0𝜛𝑞𝑡(𝜎)𝑇𝑞𝜎𝑞𝑔(𝑥)+𝑥0+𝑑𝜎𝑑𝑡0𝑒𝜆𝑡𝑞𝑡00(𝑡𝑠)𝑞1𝜎𝑞𝜛𝑞(𝜎)𝑇(𝑡𝑠)𝑞𝜎𝑞(𝑓(𝑠,𝑥(𝑠))+𝑎(𝑥)(𝑠))𝑑𝜎𝑑𝑠𝑑𝑡.(2.20) Now using the uniqueness of the Laplace transform (cf., e.g., [21, Theorem  1.1.6]), we deduce that 𝑥(𝑡)=0𝜛𝑞𝑡(𝜎)𝑇𝑞𝜎𝑞𝑔(𝑥)+𝑥0𝑑𝜎+𝑞𝑡00(𝑡𝑠)𝑞1𝜎𝑞𝜛𝑞(𝜎)𝑇(𝑡𝑠)𝑞𝜎𝑞𝑓(𝑠,𝑥(𝑠))𝑑𝜎𝑑𝑠+𝑞𝑡00(𝑡𝑠)𝑞1𝜎𝑞𝜛𝑞(𝜎)𝑇(𝑡𝑠)𝑞𝜎𝑞=𝑎(𝑥)(𝑠)𝑑𝜎𝑑𝑠0𝜉𝑞(𝜎)𝑇(𝑡𝑞𝜎)𝑔(𝑥)+𝑥0𝑑𝜎+𝑞𝑡00𝜎(𝑡𝑠)𝑞1𝜉𝑞(𝜎)𝑇((𝑡𝑠)𝑞𝜎)𝑓(𝑠,𝑥(𝑠))𝑑𝜎𝑑𝑠+𝑞𝑡00𝜎(𝑡𝑠)𝑞1𝜉𝑞(𝜎)𝑇((𝑡𝑠)𝑞𝜎)𝑎(𝑥)(𝑠)𝑑𝜎𝑑𝑠.(2.21) Consequently, we see that the mild solution to problem (1.2) given by Definition 2.3 is well defined.
Next, we define the operator 𝐶([0,𝑇],𝑋)𝐶([0,𝑇],𝑋) as follows: (𝑥)(𝑡)=𝑌(𝑡)𝑔(𝑥)+𝑥0+𝑡0[][].𝑍(𝑡𝑠)𝑓(𝑠,𝑥(𝑠))+𝑎(𝑥)(𝑠)𝑑𝑠,𝑡0,𝑇(2.22) It is clear that the operator is well defined.
The operator can be written in the form =1+2, where the operators 𝑖,𝑖=1,2 are defined as follows: 1𝑥𝑔(𝑡)=𝑌(𝑡)(𝑥)+𝑥0[],,𝑡0,𝑇2𝑥(𝑡)=𝑡0𝑍[𝑓][].(𝑡𝑠)(𝑠,𝑥(𝑠))+𝑎(𝑥)(𝑠)𝑑𝑠,𝑡0,𝑇(2.23) The following facts will be used in the proof. (1)0𝜉𝑞(𝜎)𝑑𝜎=1,(2.24) which implies that 𝑌(𝑡)Const;(2.25)(2)0𝜎𝜈𝜉𝑞(𝜎)𝑑𝜎=01𝜎𝑞𝜈𝜛𝑞(𝜎)𝑑𝜎=Γ(1+𝜈)],Γ(1+𝑞𝜈),𝜈(0,1(2.26) which implies that𝑍(𝑡)𝑞𝑀𝑡Γ(1+𝑞)q1,𝑡>0.(2.27) Let {𝑥𝑛}𝑛𝐍𝐶([0,𝑇],𝑋) such that lim𝑛𝑥𝑛𝑥[0,𝑇]=0,(2.28) for an 𝑥𝐶([0,𝑇],𝑋). Then by the assumptions, we know that for almost every 𝑡[0,𝑇] and (𝑡,𝑠)Δ: lim𝑛𝑓𝑡,𝑥𝑛(𝑡)=𝑓(𝑡,𝑥(𝑡)),lim𝑛𝑡,𝑠,𝑥𝑛(𝑠)=(𝑡,𝑠,𝑥(𝑠)).(2.29) Therefore, for sufficiently large 𝑛, we have 𝑓𝑡,𝑥𝑛(𝑡)𝑓(𝑡,𝑥(𝑡))𝜇(𝑡)1+2𝑥[0,𝑇],𝑡,𝑠,𝑥𝑛(𝑠)(𝑡,𝑠,𝑥(𝑠))𝑚(𝑡,𝑠)1+2𝑥[0,𝑇],𝑡0𝑘(𝑡,𝑠)𝑡,𝑠,𝑥𝑛(𝑠)𝑑𝑠𝑡0𝑘(𝑡,𝑠)(𝑡,𝑠,𝑥(𝑠))𝑑𝑠𝑚𝑘1+2𝑥[0,𝑇],(2.30) where 𝑘=sup[]𝑡0,𝑇𝑘(𝑡).(2.31) Hence, lim𝑛𝑡0𝑘(𝑡,𝑠)𝑡,𝑠,𝑥𝑛(𝑠)𝑑𝑠𝑡0𝑘(𝑡,𝑠)(𝑡,𝑠,𝑥(𝑠))𝑑𝑠=0.(2.32) Thus, 2𝑥𝑛2𝑥[0,𝑇]0,as𝑛,(2.33) since (2.27) implies that 𝑡0𝑓𝑍(𝑡𝑠)𝑠,𝑥𝑛(+𝑠)𝑠0𝑘(𝑠,𝜏)𝑠,𝜏,𝑥𝑛(𝜏)𝑑𝜏𝑓(𝑠,𝑥(𝑠))+𝑠0𝑘(𝑠,𝜏)(𝑠,𝜏,𝑥(𝜏))𝑑𝜏𝑑𝑠𝑞𝑀Γ(1+𝑞)𝑡0(𝑡𝑠)𝑞1𝑓𝑠,𝑥𝑛+(𝑠)𝑓(𝑠,𝑥(𝑠))𝑠0𝑘(𝑠,𝜏)𝑠,𝜏,𝑥𝑛(𝜏)(𝑠,𝜏,𝑥(𝜏))𝑑𝜏𝑑𝑠.(2.34) By (2.33) and our assumptions, we see that is continuous.
Since 𝜒 is the Hausdorff measure of noncompactness in 𝑋, we know that 𝜒 is monotone, nonsingular, invariant with respect to union with compact sets, algebraically semiadditive, and regular. This means that(i)for any Ω0,Ω12𝐸 with Ω0Ω1, 𝜒Ω0Ω𝜒1;(2.35)(ii)for every 𝑎0𝐸, Ω2𝐸, 𝜒𝑎0Ω=𝜒(Ω);(2.36)(iii)for every relatively compact set 𝐷𝐸, Ω2𝐸, 𝜒({𝐷}Ω)=𝜒(Ω);(2.37)(iv)for each Ω0, Ω12𝐸, 𝜒Ω0+Ω1Ω𝜒0Ω+𝜒1;(2.38)(v)𝜒(Ω)=0 is equivalent to the relative compactness of Ω.Noting that for any 𝜓𝐿1([0,𝑇],𝑋), we have lim𝐿+sup[]𝑡0,𝑇𝑡0𝑒𝐿(𝑡𝑠)𝜓(𝑠)𝑑𝑠=0.(2.39) So, there exists a positive constant 𝐿 such that 𝑞𝑀Γ(1+𝑞)sup[]𝑡0,𝑇𝑡0(𝑡𝑠)𝑞1𝜂(𝑠)𝑒𝐿(𝑡𝑠)𝑑𝑠=𝐿1<13,𝑞𝑀𝑘𝜁Γ(1+𝑞)sup[]𝑡0,𝑇𝑡0(𝑡𝑠)𝑞1𝑒𝐿(𝑡𝑠)𝑑𝑠=𝐿2<13,𝑞𝑀Γ(1+𝑞)sup[]𝑡0,𝑇𝑡0(𝑡𝑠)𝑞1𝜇(𝑠)+𝑚𝑘𝑒𝐿(𝑡𝑠)𝑑𝑠=𝐿3<13.(2.40) For every bounded subset Ω𝐶([0,𝑇],𝑋), we define mod𝑐(Ω)=lim𝛿0sup𝑣Ωmax|𝑡1𝑡2|𝛿𝑣𝑡1𝑡𝑣2,Ψ(Ω)=sup[]𝑡0,𝑇𝑒𝐿𝑡,𝜒(Ω(𝑡))𝛼(Ω)=Ψ(Ω),mod𝑐.(Ω)(2.41) Then mod𝑐(Ω) is the module of equicontinuity of Ω, and 𝛼 is a measure of noncompactness in the space 𝐶([0,𝑇],𝑋) with values in the cone 𝐑2+.
Let Ω𝐶([0,𝑇],𝑋) be a nonempty, bounded set such that 𝛼((Ω))𝛼(Ω).(2.42) By the assumptions and the continuity of 𝑇(𝑡) in the uniform operator topology for 𝑡>0, we get mod𝑐1Ω=0.(2.43) Clearly, 𝜇𝑓(𝑠,𝑥(𝑠))+𝑎(𝑥)(𝑠)(𝑠)+𝑚𝑘𝑥[0,𝑇].(2.44) Let 𝛿>0, 𝑡1,𝑡2(0,𝑇] such that 0<𝑡1𝑡2𝛿 and 𝑥Ω. Then 𝑡10𝑍𝑡1[]𝑠𝑓(𝑠,𝑥(𝑠))+𝑎(𝑥)(𝑠)𝑑𝑠𝑡20𝑍𝑡2[]𝑠𝑓(𝑠,𝑥(𝑠))+𝑎(𝑥)(𝑠)𝑑𝑠𝑥[0,𝑇]𝑡20𝑍𝑡1𝑡𝑠𝑍2𝑠𝜇(𝑠)+𝑚𝑘𝑑𝑠+𝑡1𝑡2𝑍𝑡1𝑠𝜇(𝑠)+𝑚𝑘𝑑𝑠𝑞𝑥[0,𝑇]𝑡200|||𝑡1𝑠𝑞1𝑡2𝑠𝑞1|||𝜎𝜉𝑞𝑇𝑡(𝜎)1𝑠𝑞𝜎𝜇(𝑠)+𝑚𝑘+𝑑𝜎𝑑𝑠𝑡200𝑡2𝑠𝑞1𝜎𝜉𝑞𝑇𝑡(𝜎)1𝑠𝑞𝜎𝑡𝑇2𝑠𝑞𝜎×𝜇(𝑠)+𝑚𝑘+𝑑𝜎𝑑𝑠𝑞𝑀𝑥[0,𝑇]Γ(1+𝑞)𝑡1𝑡2𝑡1𝑠𝑞1𝜇(𝑠)+𝑚𝑘𝑑𝑠𝑞𝑀𝑥[0,𝑇]Γ(1+𝑞)𝑡20|||𝑡1𝑠𝑞1𝑡2𝑠𝑞1|||𝜇(𝑠)+𝑚𝑘+𝑑𝜎𝑑𝑠𝑡1𝑡2𝑡1𝑠𝑞1𝜇(𝑠)+𝑚𝑘+𝑑𝑠𝑡200𝑡2𝑠𝑞1𝜎𝜉𝑞(𝑇𝑡𝜎)1𝑠𝑞𝜎𝑡𝑇2𝑠𝑞𝜎𝜇(𝑠)+𝑚𝑘𝑑𝜎𝑑𝑠.(2.45) It is not hard to see that the right-hand side of (2.45) tend to 0 as 𝑡2𝑡1. Thus, the set {(2𝑥)()𝑥Ω} is equicontinuous, then mod𝑐(2Ω)=0. Combining with (2.43), we have mod𝑐(Ω)=0, which implies mod𝑐(Ω)=0 from (2.42). Next, we show that Ψ(Ω)=0.
It is easy to see that Ψ1Ω=0.(2.46) For any 𝑡[0,𝑇], we define 2(Ω)(𝑡)=𝑡0.𝑍(𝑡𝑠)𝑓(𝑠,𝑥(𝑠))𝑑𝑠𝑥Ω(2.47) We consider the multifunction 𝑠[0,𝑡]𝐺(𝑠): 𝐺(𝑠)={𝑍(𝑡𝑠)𝑓(𝑠,𝑥(𝑠))𝑥Ω}.(2.48) Obviously, 𝐺 is integrable, that is, 𝐺 admits a Bochner integrable selection 𝔤[0,]𝐸, and []𝔤(𝑡)𝐺(𝑡),fora.e.𝑡0,.(2.49) From (2.27) and our assumptions, it follows that 𝐺 is integrably bounded, that is, there exists a function 𝜚𝐿1([0,],𝐸) such that [].𝐺(𝑡)=sup{𝔤𝔤𝐺(𝑡)}𝜚(𝑡),a.e.𝑡0,(2.50) Moreover, we have the following estimate for a.e. 𝑠[0,𝑡]: 𝜒(𝐺(𝑠))𝑞𝑀Γ(1+𝑞)(𝑡𝑠)𝑞1𝜒(𝑓(𝑠,Ω(𝑠)))𝑞𝑀Γ(1+𝑞)(𝑡𝑠)𝑞1=𝜂(𝑠)𝜒(Ω(𝑠))𝑞𝑀Γ(1+𝑞)(𝑡𝑠)𝑞1𝜂(𝑠)𝑒𝐿𝑠𝑒𝐿𝑠𝜒(Ω(𝑠))𝑞𝑀Γ(1+𝑞)(𝑡𝑠)𝑞1𝜂(𝑠)𝑒𝐿𝑠Ψ(Ω).(2.51) Therefore, since 𝑋 is a separable Banach space, we know by [20, Theorem  4.2.3] that 𝜒2(Ω)(𝑡)=𝜒𝑡0𝐺(𝑠)𝑑𝑠𝑞𝑀Γ(1+𝑞)𝑡0(𝑡𝑠)𝑞1𝜂(𝑠)𝑒𝐿𝑠𝑑𝑠Ψ(Ω).(2.52) So sup[]𝑡0,𝑇𝑒𝐿𝑡𝜒2(Ω)(𝑡)𝑞𝑀Γ(1+𝑞)sup[]𝑡0,𝑇𝑡0(t𝑠)𝑞1𝜂(𝑠)𝑒𝐿(𝑡𝑠)𝑑𝑠Ψ(Ω)=𝐿1Ψ(Ω).(2.53) Similarly, if we set 2(Ω)(𝑡)=𝑡0,𝑍(𝑡𝑠)𝑎(𝑥)(𝑠)𝑑𝑠𝑥Ω(2.54) then we see that the multifunction 𝑠[0,𝑡]𝐺(𝑠), 𝐺(𝑠)={𝑍(𝑡𝑠)𝑎(𝑥)(𝑠)𝑥Ω}(2.55) is integrable and integrably bounded. Thus, we obtain the following estimate for a.e. 𝑠[0,𝑡]: 𝜒𝐺(𝑠)𝑞𝑀𝑘𝜁Γ(1+𝑞)(𝑡𝑠)𝑞1𝑒𝐿𝑠Ψ(Ω),sup[]𝑡0,𝑇𝑒𝐿𝑡𝜒2(Ω)(𝑡)𝑞𝑀𝑘𝜁Γ(1+𝑞)sup[]𝑡0,𝑇𝑡0(𝑡𝑠)𝑞1𝑒𝐿(𝑡𝑠)𝑑𝑠Ψ(Ω)=𝐿2Ψ(Ω).(2.56) Now, from (2.53) and (2.56), it follows that Ψ((Ω))Ψ1(Ω)+Ψ2𝐿(Ω)1+𝐿2Ψ(Ω)=𝐿Ψ(Ω),(2.57) where 0<𝐿<1. Then by (2.42), we get Ψ(Ω)=0. Hence 𝛼(Ω)=(0,0). Thus, Ω is relatively compact due to the regularity property of 𝛼. This means that is 𝛼-condensing.
Let us introduce in the space 𝐶([0,𝑇],𝑋) the equivalent norm defined as 𝑥=sup[]𝑡0,𝑇𝑒𝐿𝑡.𝑥(𝑡)(2.58) Consider the set 𝐵𝑟=([]𝑥𝐶0,𝑇,𝑋)𝑥𝑟.(2.59) Next, we show that there exists some 𝑟>0 such that 𝐵𝑟𝐵𝑟. Suppose on the contrary that for each 𝑟>0 there exist 𝑥𝑟()𝐵𝑟, and some 𝑡[0,𝑇] such that (𝑥𝑟)(𝑡)>𝑟.
From the assumptions, we have 1𝑥𝑟(𝑡)𝑥𝑀𝑏+0.(2.60) Moreover, 2𝑥𝑟(𝑡)𝑡0𝑓𝑍(t𝑠)𝑠,𝑥𝑟(𝑥𝑠)+𝑎𝑟(𝑠)𝑑𝑠𝑞𝑀Γ(1+𝑞)𝑡0(𝑡𝑠)𝑞1𝑥𝜇(𝑠)𝑟(𝑠)+𝑚𝑘𝑒𝐿𝑠𝑥𝑟=𝑑𝑠𝑞𝑀Γ(1+𝑞)𝑡0(𝑡𝑠)𝑞1𝜇(𝑠)𝑒𝐿𝑠𝑒𝐿𝑠𝑥𝑟(𝑠)𝑑𝑠+𝑚𝑘𝑡0(𝑡𝑠)𝑞1𝑒𝐿𝑠𝑥𝑑𝑠𝑟𝑞𝑀𝑟Γ(1+𝑞)𝑡0(𝑡𝑠)𝑞1𝜇(𝑠)+𝑚𝑘𝑒𝐿𝑠𝑑𝑠.(2.61) Therefore, 𝑟<sup𝑡[0,𝑇]𝑒𝐿𝑡𝑥𝑟𝑥(𝑡)𝑀𝑏+0+𝑞𝑀𝑟Γ(1+𝑞)sup[]𝑡0,𝑇𝑡0(𝑡𝑠)𝑞1𝜇(𝑠)+𝑚𝑘𝑒𝐿(𝑡𝑠)𝑑𝑠.(2.62) Dividing both sides of (2.62) by 𝑟, and taking 𝑟, we have 𝑞𝑀Γ(1+𝑞)sup[]𝑡0,𝑇𝑡0(𝑡𝑠)𝑞1𝜇(𝑠)+𝑚𝑘𝑒𝐿(𝑡𝑠)𝑑𝑠1.(2.63) This is a contradiction. Hence for some positive number 𝑟, 𝐵𝑟𝐵𝑟. According to the following known fact.
Let 𝔐 be a bounded convex closed subset of 𝐸 and 𝔉𝔐𝔐 a 𝛼-condensing map. Then Fix𝔉={𝑥𝑥=𝔉(𝑥)} is nonempty. we see that problem (1.2) has at least one mild solution.
Next, for 𝑐(0,1], we consider the following one-parameter family of maps: [][][]0,1×𝐶(0,𝑇,𝑋)𝐶(0,𝑇,𝑋)(𝑐,𝑥)(𝑐,𝑥)=𝑐(𝑥).(2.64) We will demonstrate that the fixed point set of the family , ]Fix={𝑥(𝑐,𝑥)forsome𝑐(0,1}(2.65) is a priori bounded. Indeed, let 𝑥Fix, for 𝑡[0,𝑇], we have (𝑥𝑡)𝑀𝑔(𝑥)+𝑥0+𝑡0(𝑥𝑍𝑡𝑠)𝑓(𝑠,𝑥(𝑠))+𝑎(𝑥)(𝑠)𝑑𝑠𝑀𝑏+0+𝑞𝑀Γ(1+𝑞)𝑡0(𝑡𝑠)𝑞1𝜇(𝑠)𝑥(𝑠)𝑑𝑠+𝑚𝑘𝑡0(𝑡𝑠)𝑞1sup[]𝜏0,𝑠.𝑥(𝜏)𝑑𝑠(2.66)
Noting that the Hölder inequality, we have 𝑡0(𝑡𝑠)𝑞1𝜇(𝑠)𝑑𝑠𝑝1𝑝𝑞1(𝑝1)/𝑝𝑡(𝑝𝑞1)/𝑝𝜇𝐿𝑝𝑝1𝑝𝑞1(𝑝1)/𝑝𝑇𝑞1/𝑝𝜇𝐿𝑝.(2.67) Therefore, from (2.66), we obtain 𝑥𝑥(𝑡)𝑀𝑏+0+𝑞𝑀Γ(1+𝑞)𝑝1𝑝𝑞1(𝑝1)/𝑝𝑇𝑞1/𝑝𝜇𝐿𝑝sup[]𝑠0,𝑡+𝑥(𝑠)𝑞𝑀𝑚𝑘Γ(1+𝑞)𝑡0(𝑡𝑠)𝑞1sup[]𝜏0,𝑠𝑥(𝜏)𝑑𝑠.(2.68) We denote 𝑦(𝑡)=sup𝑠[0,𝑡]𝑥(𝑠).(2.69) Let ̃𝑡[0,𝑡] such that ̃𝑦(𝑡)=𝑥(𝑡). Then, by (2.68), we can see 𝑥𝑦(𝑡)𝑀𝑏+0+𝑞𝑀Γ(1+𝑞)𝑝1𝑝𝑞1(𝑝1)/𝑝𝑇𝑞1/𝑝𝜇𝐿𝑝+𝑦(𝑡)𝑞𝑀𝑚𝑘Γ(1+𝑞)𝑡0(𝑡𝑠)𝑞1𝑦(𝑠)𝑑𝑠.(2.70) By a generalization of Gronwall’s lemma for singular kernels ([22, Lemma  7.1.1]), we deduce that there exists a constant 𝜅=𝜅(𝑞) such that 𝑀𝑥𝑦(𝑡)𝑏+01(𝑞𝑀/Γ(1+𝑞))((𝑝1)/(𝑝𝑞1))(𝑝1)/𝑝𝑇𝑞1/𝑝𝜇𝐿𝑝+𝑥𝜅𝑀𝑏+0𝑞𝑀𝑚𝑘/Γ(1+𝑞)1(𝑞𝑀/Γ(1+𝑞))((𝑝1)/(𝑝𝑞1))(𝑝1)/𝑝𝑇𝑞1/𝑝𝜇𝐿𝑝2𝑡0(𝑡𝑠)𝑞1𝑀𝑥𝑑𝑠𝑏+01(𝑞𝑀/Γ(1+𝑞))((𝑝1)/(𝑝𝑞1))(𝑝1)/𝑝𝑇𝑞1/𝑝𝜇𝐿𝑝+𝑥𝜅𝑀𝑏+0𝑞𝑀𝑚𝑘𝑇/Γ(1+𝑞)𝑞𝑞1(𝑞𝑀/Γ(1+𝑞))((𝑝1)/(𝑝𝑞1))(𝑝1)/𝑝𝑇𝑞1/𝑝𝜇𝐿𝑝2=𝑤.(2.71) Hence, sup𝑡[0,𝑇]𝑥(𝑡)𝑤.
Now we consider a closed ball: 𝐵𝑅=([]𝑥𝐶0,𝑇,𝑋)𝑥[0,𝑇]([]𝑅𝐶0,𝑇,𝑋).(2.72) We take the radius 𝑅>0 large enough to contain the set Fix inside itself. Moreover, from the proof above, 𝐵𝑅𝐶([0,𝑇],𝑋) is 𝛼-condensing. Consequently, the following known fact implies our conclusion: Let 𝑉𝐸 be a bounded open neighborhood of zero and 𝔉𝑉𝐸 a 𝛼-condensing map satisfying the boundary condition: 𝑥𝜆𝔉(𝑥),(2.73)for all 𝑥𝜕𝑉 and 0<𝜆1. Then, Fix𝔉is nonempty compact.

3. Example

In this section, let 𝑋=𝐿2([0,𝜋]), we consider the following nonlocal Cauchy problem for an integrodifferential problem:𝜕𝑞𝑡𝜕𝑢(𝑡,𝜉)=2𝜕𝜉21𝑢(𝑡,𝜉)+𝑘𝑘𝑡𝑢(𝑡,𝜉)+1+𝑢(𝑡,𝜉)𝑡0(𝑡𝑠)sin𝑠𝑢(𝑠,𝜉)𝑡]𝑢[]𝑑𝑠,𝑡(0,1(𝑡,0)=𝑢(𝑡,𝜋)=0,𝑡0,1𝑢(0,𝜉)=𝑗𝑖=0𝜋0𝑐𝑖(𝑢𝑡𝜉,𝑦)𝑖,𝑦𝑡1+𝑢𝑖,𝑦𝑑𝑦+𝑢0(𝜉),(3.1) where 𝜕𝑞𝑡 is the Caputo fractional partial derivative of order 0<𝑞<1; 𝜉[0,𝜋]; 𝑘>0 is a constant to be specified later; 𝑢0(𝜉)𝑋;𝑗𝐍+;0<𝑡0<𝑡1<<𝑡𝑗<1;(3.2)𝑐𝑖(,)(𝑖=0,1,,𝑗) are continuous functions and there exists a positive constant 𝑏 such that𝑗𝑖=0𝜋0𝑐𝑖(𝜉,𝑦)𝑑𝑦𝑏.(3.3) For 𝑡(0,1], 𝜉[0,𝜋], we set𝑥(𝑡)(𝜉)=𝑢(𝑡,𝜉),𝑔(𝑥)(𝜉)=𝑗𝑖=0𝜋0𝑐𝑖𝑥𝑡(𝜉,𝑦)𝑖(𝑦)𝑡1+𝑥𝑖𝑘(𝑦)𝑑𝑦,(𝑡,𝑠)=𝑡𝑠,(𝑡,𝑠,𝑥(𝑠))(𝜉)=sin𝑠𝑥(𝑠)(𝜉)𝑡,1𝑓(𝑡,𝑥(𝑡))(𝜉)=𝑘𝑘𝑡𝑥(𝑡)(𝜉).1+𝑥(𝑡)(𝜉)(3.4) On the other hand, it is known that the operator 𝐴 (𝐴𝑢=𝑢 with 𝐷(𝐴)=𝐻2([0,𝜋])𝐻10([0,𝜋])) generates an analytic semigroup and uniformly bounded semigroup {𝑇(𝑡)}𝑡0 on 𝑋 with 𝑇(𝑡)𝐿(𝑋)1. Therefore, (3.1) is a special case of (1.2).

Moreover, we have(1)for all 𝑡(0,1], 1𝑓(𝑡,𝑥)𝑘𝑘𝑡𝑥=𝜇(𝑡)𝑥;(3.5)(2)for any 𝑤,𝑤𝑋, 𝑤1𝑓(𝑡,𝑤)𝑓𝑡,𝑘𝑘𝑡𝑤𝑤,(3.6) that is, for any bounded set 𝐷𝑋, 1𝜒(𝑓(𝑡,𝐷))𝑘𝑘𝑡𝜒(𝐷),(3.7) for a.e. 𝑡[0,1];(3)for almost all (𝑡,𝑠)Δ, (𝑡,𝑠,𝑥)=sin𝑠𝑥(𝑠)(𝜉)𝑡𝑚(𝑡,𝑠)𝑥,(3.8) where 𝑚(𝑡,𝑠)=𝑠/𝑡, and 𝑚=sup[]𝑡0,1𝑡0𝑚(𝑡,𝑠)𝑑𝑠=sup[]𝑡0,1𝑡0𝑠𝑡2𝑑𝑠=3;(3.9)(4)𝑤(𝑡,𝑠,𝑤)𝑡,𝑠,𝑠𝑡𝑤,𝑤(3.10) that is, for any bounded set 𝐷𝑋, 𝜒((𝑡,𝑠,𝐷))𝜁(𝑡,𝑠)𝜒(𝐷),(3.11) where 𝜁(𝑡,𝑠)=𝑠/𝑡, and sup[]𝑡0,1𝑡02𝜁(𝑡,𝑠)𝑑𝑠=3.(3.12) Therefore, Theorem 2.4 implies that the problem (3.1) has at least a mild solution when𝑞((𝑝1)/(𝑝𝑞1))(𝑝1)/𝑝Γ(1+𝑞)𝜇𝐿𝑝<1.(3.13)

Acknowledgment

The work was supported partly by the Chinese Academy of Sciences, the NSF of Yunnan Province (2009ZC054M) and the NSF of China (11171210).