# Existence Results for Fractional Semilinear Integrodifferential Equations of Mixed Type with Delay

Authors: Xue Wang; Bo Zhu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we discuss a class of fractional semilinear integrodifferential equations of mixed type with delay. Based on the theories of resolvent operators, the measure of noncompactness, and the fixed point theorems, we establish the existence and uniqueness of global mild solutions for the equations. An example is provided to illustrate the application of our main results.


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## Body

## 1. Introduction

Fractional calculus can be used to describe some nonclassical phenomena in natural science and engineering applications. Fractional differential equations have been applied in different fields ranging from engineering, finance, and physics in the past few decades. Researchers have conducted extensive explorations on this subject and have achieved fruitful results for the fractional differential equations [1–13]. Zhu and Han [10] and Chadha and Pandey [11] studied the fractional integrodifferential equations and discussed the existence of mild solutions. Based on the theory of the resolvent family and fixed point theorems, Chen et al. [14–17] analyzed nonautonomous evolution equations in a Banach space. Moreover, some researchers considered sufficient conditions on the existence of mild solutions for fractional differential equations by the measure of noncompactness [4, 18, 19]. The initial boundary value problem for the fractional integrodifferential equations with delay has been investigated by using fixed point theorems [4, 5, 18, 20]. In [3, 21–24], differential equations of mixed type have been studied and some results have been concluded.Chen [22] studied the fractional nonautonomous evolution equations of mixed type:
(1)cDtβut+Atut=ft,ut,Tut,Sut,t∈0,a,u0=A−10u0,where
(2)Tut=∫0tKt,susds,Sut=∫0aHt,susds,where the kernels K and H are linear functions. The operator T is an integral with a variable upper limit, and the operator S is an ordinary definite integral; accordingly, problem (1) is called fractional semilinear integrodifferential equations of mixed type.Li and Jia [25] investigated the existence of mild solutions for abstract delay fractional differential equations:
(3)cDtβut=Aut+Jt1−βft,ut,t∈0,T,ut=φt,t∈−r,0,where β∈0,1, Jt1−β is the Riemann-Liouville fractional integral, the linear operator A is independent on t, and the Lipschitz coefficient of f is constant.To the best of our knowledge, there are no results on the fractional integrodifferential equations of mixed type with delay. Motivated by this idea, we consider the following problem:(4)cDtβxt=Atxt+Jt1−βft,xt,Kxt,Hxt,t∈0,T0,xt=ϕt,t∈−r,0,where β∈0,1, cDtβ is the Caputo fractional derivative of order β, At is a closed and linear operator with domain DA defined on a Banach space E, Jt1−β is the Riemann-Liouville fractional integral of order 1−β, K and H are defined by
(5)Kxt=∫0tKt,s,xsds,Hxt=∫0T0Ht,s,xsds,where K:D×C−r,0;E⟶E and H:D0×C−r,0;E⟶E are continuous and nonlinear functions, D=t,s∈R2:0≤s≤t≤T0, J=0,T0, D0=t,s∈R2:0≤t,s≤T0, ϕ∈C−r,0, f is to be specified later, and xt means the element of C−r,0;E defined by xtθ=xt+θ,−r≤θ≤0, for x∈C−r,T0;E,t∈J.We demonstrate the existence and uniqueness of global mild solutions for problem (4) under the conditions of the compact resolvent operator and noncompact resolvent operator, respectively. The kernels K and H of the operators K and H are nonlinear functions. In addition, the operator At is dependent on t. The rest of this paper is organized as follows. Basic definitions and auxiliary results are presented in Section 2. In Section 3, we prove the existence and uniqueness of mild solutions via various fixed point theorems, the measure of noncompactness, and the Banach contraction mapping principle. An example is provided to illustrate the main theorems in Section 4. Finally, Section 5 is the summary of our results.


## 2. Preliminaries

Definition 1 [6, 26].
The Riemann-Liouville fractional integralJtβ and derivative Dtβ of a function f:0,∞⟶R of order β>0 are defined by
(6)Jtβft=1Γβ∫0tt−sβ−1fsds,Dtβft=1Γn−βddtn∫0tt−sn−β−1fsds,n−1<β≤n,where ft∈L10,T0;E, Γ· denotes the gamma function, and n∈N.Remark 2 [25].
Dtβft=DtmJtm−βft, where Dtm=dm/dtm and Jtm−βft∈Wm,10,T0;E.Definition 3 [26, 27].
The Caputo fractional derivative of orderβ>0 of a function f:0,∞⟶R is given by
(7)cDtβft=1Γn−β∫0tt−sn−β−1fnsds,n−1<β<n.Remark 4 [25].
For the Riemann-Liouville fractional integral operator and the Caputo fractional derivative operator, the following conclusions are obtained:(8)cDtβft=Dtβft−∑k=0m−1tkk!fk0,cDtβJtβft=ft,JtβcDtβft=ft−∑k=0m−1tkk!fk0.Definition 5 [28, 29].
LetAt be a closed and linear operator with domain DA defined on a Banach space E and β>0. Let ρAt be the resolvent set of At. At is called the generator of a β-resolvent family if there exist ω≥0 and a strongly continuous function Uβ:R+2⟶BE such that λβ:Reλ>ω⊂ρA and
(9)λβI−As−1x=∫0∞e−λt−sUβt,sxdt,Reλ>ω,x∈E.
In this case,Uβt,s is called the β-resolvent family generated by At.Remark 6 [29, 30].
Uβt,s satisfies the following properties:
(1)
Uβs,s=I and Uβt,s=Uβt,rUβr,s, for 0≤s≤r≤t≤a(2)
t,s⟶Uβt,s is strongly continuous for 0≤s≤t≤a(3)
IfUβt,s is compact for t,s>0, then Uβt,s is continuous in the uniform operator topologyLemma 7 [21].
LetB⊂CJ,E be equicontinuous and bounded; then, Co¯B⊂CJ,E is also equicontinuous and bounded.Lemma 8 [24].
LetB⊂CJ,E be equicontinuous and bounded; then, αBt is continuous on J and
(10)α∫JBsds≤∫JαBsds,αB=maxt∈JαBt,where α denotes the measure of noncompactness.Lemma 9 [21].
LetE be a Banach space and D⊂E be bounded; then, there exists a countable set D0⊂D such that αD≤2αD0.Lemma 10 [31].
LetE be a Banach space and D⊂E be a bounded closed and convex set. Assume that Q:D⟶D is a strict set contraction mapping; then, Q has at least one fixed point in D.Definition 11.
A functionx∈C−r,T0;E is a mild solution of problem (4), if x satisfies the following equations:
(11)xt=Uβt,0ϕ0+∫0tUβt,sfs,xs,Kxs,Hxsds,t∈0,T0,ϕt,t∈−r,0.


## 3. Main Results

Let us introduce the operatorΨ:C−r,T0;E⟶C−r,T0;E by
(12)Ψxt=Uβt,0ϕ0+∫0tUβt,sfs,xs,Kxs,Hxsds,t∈0,T0,ϕt,t∈−r,0.Theorem 12.
Assume that the following conditions hold:
(H1). The resolvent operator Uβt,s is compact for all t,s>0,M⋆=maxUβt,s<+∞,0≤s≤t≤T0.
(H2). K:D×C−r,0;E⟶E and H:D0×C−r,0;E⟶E are continuous; there exist nonnegative Lebesgue integrable functions pi∈LJ,R+i=1,2 such that ∥Kt,s,x∥≤p1t∥x∥C−r,0;E and ∥Ht,s,x∥≤p2t∥x∥C−r,0;E, for all t,s∈D,t,s∈D0,x∈C−r,0;E.
(H3). f:J×C−r,0;E×C−r,0;E×C−r,0;E⟶E is continuous; there exist nonnegative Lebesgue integrable functions a,Li∈LJ,R+i=1,2,3 such that ∥ft,x1,x2,x3∥≤at+∑i=13Lit∥xi∥C−r,0;E, for all t∈J,xi∈C−r,0;E.
Then, problem (4) has at least one mild solution x∈C−r,T0;E.Proof.
Let us set the notationR1>0 such that
(13)R1≥M⋆ϕ0+M⋆∫0T0asds1−M⋆∫0T0L1sds+∫0T0L2s∫0T0p1νdνds+∫0T0L3s∫0T0p2νdνds,where ϕ0=ϕ0 and ∫0T0L1sds+∫0T0L2s∫0T0p1νdνds+∫0T0L3s∫0T0p2νdνds−1>M⋆.
First of all, we consider the setBR1=x∈C−r,T0;E:xC−r,T0;E≤R1 and show that ΨBR1⊂BR1. By using conditions (H2) and (H3), for all x∈BR1, we have
(14)Ψxt≤Uβt,0ϕ0+∫0tUβt,sfs,xs,Kxs,Hxsds≤M⋆ϕ0+M⋆∫0tfs,xs,Kxs,Hxsds≤M⋆ϕ0+M⋆∫0tas+L1sxs+L2sKxs+L3sHxsds≤M⋆ϕ0+M⋆∫0tasds+M⋆∫0tL1sds+∫0tL2s∫0sp1νdνds+∫0tL3s∫0T0p2νdνdsxC−r,0;E≤M⋆ϕ0+M⋆∫0T0asds+M⋆∫0T0L1sds+∫0T0L2s∫0T0p1νdνds+∫0T0L3s∫0T0p2νdνdsxC−r,T0;E≤R1.
So, we conclude thatΨ maps BR1 into itself.
Second, we prove thatΨ:BR1⟶BR1 is continuous.
Letxn0∞⊂C−r,T0;E, with xn⟶xn⟶∞,x∈C−r,T0;E. Using the fact that K:D×C−r,0;E⟶E, H:D0×C−r,0;E⟶E, and f:J×C−r,0;E×C−r,0;E×C−r,0;E⟶E are continuous, we obtain
(15)ft,xnt,Kxnt,Hxnt⟶ft,xt,Kxt,Hxtn⟶∞,for any t∈J uniformly. That is, for any ε>0, there exists a natural number N0, for n>N0,t∈J, such that
(16)ft,xnt,Kxnt,Hxnt−ft,xt,Kxt,Hxt≤εM⋆T0,which implies that
(17)Ψxnt−Ψxt=∫0tUβt,sfs,xns,Kxns,Hxnsds−∫0tUβt,sfs,xs,Kxs,Hxsds≤M⋆∫0tfs,xns,Kxns,Hxns−fs,xs,Kxs,Hxsds≤M⋆T0εM⋆T0=ε.
In consequence,Ψ:BR1⟶BR1 is continuous.
Furthermore, we prove thatΨBR1 is equicontinuous.
To do this, letLs=L1s+L2s∫0T0p1νdν+L3s∫0T0p2νdν. Obviously, it is a nonnegative Lebesgue integrable function. For all x∈BR1,t1,t2∈Jt1<t2, we have
(18)Ψxt2−Ψxt1∥≤∥Uβt2,0−Uβt1,0ϕ0+∫t1t2Uβt2,sfs,xs,Kxs,Hxsds+∫0t1Uβt2,s−Uβt1,sfs,xs,Kxs,Hxsds≤ϕ0Uβt2,0−Uβt1,0+M⋆∫t1t2fs,xs,Kxs,Hxsds+sups∈JUβt2,s−Uβt1,s∫0t1fs,xs,Kxs,Hxsds≤ϕ0Uβt2,0−Uβt1,0+M⋆∫t1t2as+LsR1ds+sups∈JUβt2,s−Uβt1,s∫0t1as+LsR1ds≕I1+I2+I3.
In view of conditionH1, compactness of the resolvent operator Uβt,st,s>0 implies the continuity in the uniform operator topology. That is, for any ε>0, there exists δ1>0, for any t2−t1<δ1,t1,t2∈J, such that I3<ε/3. Hence, for the above ε>0, by using properties of Uβt,s and the above inequalities, there exists δ>0 (δ<δ1) such that Ψxt2−Ψxt1<ε, for anyt2−t1<δ,t1,t2∈J. Consequently, ΨBR1 is equicontinuous.
In the end, we prove thatΨBR1 is precompact.
For any fixedtt∈−r,T0 and 0<ε<t, the operator Ψεxt is defined by
(19)Ψεxt=Uβt,0ϕ0+∫0t−εUβt,sfs,xs,Kxs,Hxsds,t∈0,T0,ϕt,t∈−r,0.
SinceUβt,st,s>0 is a compact resolvent operator, then the set Yεt=Ψεxt:x∈BR1 is relatively compact in E for any ε (0<ε<t).
Moreover, for anyx∈BR1, one can find that
(20)Ψxt−Ψεxt=∫t−εtUβt,sfs,xs,Kxs,Hxsds≤M⋆∫t−εtfs,xs,Kxs,Hxsds≤M⋆∫t−εtas+LsR1ds≤M⋆as+LsR1ε.Thus, Yt=Ψxt:x∈BR1 is totally bounded. Hence, Yt is relatively compact in E, and so, based on the Arzelà-Ascoli theorem, Ψ:BR1⟶BR1 is completely continuous. As all the assumptions of the Schauder fixed point theorem are satisfied, the conclusion implies that the operator Ψ has a fixed point x in C−r;T0,E, which is a global mild solution of problem (4). This completes the proof.Next, we develop the existence of global mild solutions for problem (4) via the measure of noncompactness and fixed point theorem. Furthermore, we employ the notations: TR=x∈C−r,T0;E:xC−r,T0;E≤R, k0=supKt,s,xs:t,s,xs∈D×C−r,0;E, h0=supHt,s,xs:t,s,xs∈D0×C−r,0;E, and R≥maxT0k0,T0h0.Theorem 13.
Assume that (H1) and the following conditions hold:
(H4). The function f:J×TR×TR×TR⟶E is bounded and continuous, which satisfies
(21)limR→∞supMRR<1T0M⋆,where MR=supft,x1,x2,x3:t,x1,x2,x3∈J×TR×TR×TR.
(H5). For any R, there exist nonnegative Lebesgue integrable functions qi∈LJ,R+, i=1,2,3,4,5 such that for any equicontinuous and countable set Di⊂TRi=1,2,3, αft,D1,D2,D3≤∑i=13qitαDi, αKt,s,D2≤q4tαD2, and αHt,s,D3≤q5tαD3.
(H6). 2M⋆∫0T0q1s+q2s∫0T0q4νdν+q3s∫0T0q5νdνds<1.
Then, problem (4) has at least one mild solution.Proof.
By (H4), there exists 0<μ<1/T0M⋆ and R0>0, for any R≥R0, such that
(22)MR<μR.
LetR⋆=maxR0,M⋆ϕ01−M⋆T0μ−1; we first consider the set BR⋆=x∈C−r,T0;E:xC−r,T0;E≤R⋆ and show that ΨBR⋆⊂BR⋆. From the above inequality, for all x∈BR⋆, we have
(23)ΨxC−r,T0;E≤Uβt,0ϕ0+∫0tUβt,sfs,xs,Kxs,Hxsds≤M⋆ϕ0+M⋆T0MR⋆≤M⋆ϕ0+M⋆T0μR⋆≤R⋆.
Meanwhile, applying the arguments employed in the proof of Theorem12, we conclude that Ψ is a continuous and bounded operator on BR⋆.
Then, we prove thatΨBR⋆ is equicontinuous. For any x∈BR⋆,t1,t2∈Jt1<t2, we have
(24)Ψxt2−Ψxt1≤Uβt2,0−Uβt1,0ϕ0+∫t1t2Uβt2,sfs,xs,Kxs,Hxsds+∫0t1Uβt2,s−Uβt1,sfs,xs,Kxs,Hxsds≤ϕ0Uβt2,0−Uβt1,0+M⋆t2−t1MR⋆+sups∈JUβt2,s−Uβt1,sMR⋆t1.
By (H1), the compactness of Uβt,s, for t,s>0, implies the continuity in the uniform operator topology. Namely, for any ε>0, there exists δ1>0, for any ∣t2−t1∣<δ1,t1,t2∈J, such that
(25)sups∈JUβt2,s−Uβt1,sMR⋆t1<ε3.
Therefore, for the aboveε>0, there exists δ>0 (δ<δ1) such that Ψxt2−Ψxt1<ε, for all x∈BR⋆,t2−t1<δ,t1,t2∈J, which shows that ΨBR⋆ is equicontinuous. In view of Lemma 7, Co¯ΨBR⋆⊂BR⋆ is bounded and equicontinuous.
Finally, we prove thatΨ:Co¯ΨBR⋆⟶Co¯ΨBR⋆ is a condensing operator. By Lemma 9, for any D⊂Co¯ΨBR⋆, there exists a countable set D0=xn⊂D such that
(26)αΨD≤2αΨD0.
By using condition (H5) and Lemma 8, we obtain
(27)αΨD0t=α∫0tUβt,sfs,D0s,KD0s,HD0sds≤M⋆∫0tαfs,D0s,KD0s,HD0sds≤M⋆∫0tq1sαD0s+q2sαKD0s+q3sαHD0sds≤M⋆∫0tq1s+q2s∫0sq4νdν+q3s∫0T0q5νdνdsαD.
In addition, using Lemma8, we have
(28)αΨD0=maxt∈JαΨD0t.
Consequently,(29)αΨD≤2M⋆∫0T0q1s+q2s∫0T0q4νdν+q3s∫0T0q5νdνdsαD.
By (H6), we obtain that Ψ is a condensing operator on Co¯ΨBR⋆. By Lemma 10, there exists at least one fixed point x∈Co¯ΨBR⋆⊂C−r,T0;E for Ψ. In conclusion, problem (4) has at least one global mild solution. This completes the proof.Remark 14.
Theorems12 and 13 above are concluded under the conditions that Uβt,s is compact for t,s>0 and the functions f, K, and H satisfy corresponding conditions; in contrast, when the resolvent operator Uβt,s is noncompact, we could obtain Theorem 15 if f, K, and H meet the Lipschitz conditions.Theorem 15.
Assume that the following conditions hold:
(H7). f:J×C−r,0;E×C−r,0;E×C−r,0;E⟶E is continuous; there exist nonnegative Lebesgue integrable functions gi∈LJ,R+i=1,2,3, for all t∈J,ui,vi∈E, such that
(30)ft,u1,u2,u3−ft,v1,v2,v3≤∑i=13gitui−viC−r,0;E.
(H8). K:D×C−r,0;E⟶E and H:D0×C−r,0;E⟶E; there exist nonnegative Lebesgue integrable functions g4,g5∈LJ,R+, for all u,v∈E,t,s∈D,t,s∈D0 such that
(31)Kt,s,u−Kt,s,v≤g4tu−vC−r,0;E,Ht,s,u−Ht,s,v≤g5tu−vC−r,0;E.
(H9). M⋆∫0T0g1s+g2s∫0T0g4νdν+g3s∫0T0g5νdνds<1.
Then, problem (4) has a unique mild solution.Proof.
For anyu,v∈C−r,T0;E,
(32)Ψut−Ψvt≤M⋆∫0tfs,us,Kus,Hus−fs,vs,Kvs,Hvsds≤M⋆∫0tg1sus−vs+g2sKus−Kvs+g3sHus−Hvsds≤M⋆∫0T0g1s+g2s∫0T0g4νdν+g3s∫0T0g5νdνdsu−vC−r,T0;E.
By (H9), we have Ψu−ΨvC−r,T0;E<u−vC−r,T0;E. These arguments enable us to conclude that the operator Ψ is a contraction mapping. Hence, the operator Ψ has a unique fixed point x⋆∈C−r,T0;E, which implies that problem (4) has a unique global mild solution. This completes the proof.Remark 16.
In Theorem15, we develop the uniqueness of the mild solution for problem (4) via the Banach contraction mapping principle. In conditions (H7) and (H8), gi∈LJ,R+i=1,2,3,4,5 turn out to be nonnegative Lebesgue integrable functions instead of constants.


## 4. An Application

In order to show the application of the main results, we consider the following problem:(33)cDtβxz,t=t2∂2∂z2xz,t+Jt1−βt1+t2xz,t+θ+11+t2∫0tasxz,s+θds+11+et∫01bsxz,s+θds,0<t≤1,z∈Ω,θ∈−r,0,xz,θ=φz,θ,z∈Ω,θ∈−r,0,where 0<β<1, cDtβ is the Caputo fractional derivative of order β, Jt1−β is the Riemann-Liouville fractional integral of order 1−β, Ω⊂ℝn is a bounded domain with regular boundary ∂Ω, and φ∈C−r,0;E,E=CΩ¯;ℝ,Ω¯=Ω⋃∂Ω.By settingxt=x·,t, problem (33) can be rewritten as the following abstract form:
(34)cDtβxt=Atxt+Jt1−βft,xt,Kxt,Hxt,t∈0,1,xt=φt,t∈−r,0,where xt=xt+θ, ft,xt,Kxt,Hxt=t/1+t2xt+1/1+t2∫0tasxsds+1/1+et∫01bsxsds, and
(35)DA=x∈CΩ¯,R:x′′∈CΩ¯,R,Atx=x′′,t∈−r,0.It is well known that the operatorAt generates a β-resolvent family Uβt,s [23, 25]. Let equation (34) satisfy the conditions of Theorems 12–15; then, problem (34) has a global mild solution, which means that problem (33) has a mild solution.


## 5. Conclusion

In this paper, we study the existence and uniqueness of the global mild solutions for the fractional integrodifferential equations of mixed type with delay. Under the condition of the compact resolvent operator, we obtain Theorems12 and 13, respectively, via various fixed point theorems and the measure of noncompactness. Theorem 15 is established by using the Banach contraction mapping principle under the condition of the noncompact resolvent operator. Furthermore, an example is provided to illustrate the main theorems. The kernels K and H of the operators K and H are nonlinear functions; meanwhile, the operator At is dependent on t. As a consequence, our main theorems improve and generalize many corresponding results by using different methods.



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