# Existence and Uniqueness of Mild Solutions to Impulsive Nonlocal Cauchy Problems

Authors: Mohamed Hannabou; Khalid Hilal; Ahmed Kajouni
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, a class of nonlocal impulsive differential equation with conformable fractional derivative is studied. By utilizing the theory of operators semigroup and fractional derivative, a new concept on a solution for our problem is introduced. We used some fixed point theorems such as Banach contraction mapping principle, Schauder’s fixed point theorem, Schaefer’s fixed point theorem, and Krasnoselskii’s fixed point theorem, and we derive many existence and uniqueness results concerning the solution for impulsive nonlocal Cauchy problems. Some concrete applications to partial differential equations are considered. Some concrete applications to partial differential equations are considered.


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## Body

## 1. Introduction

Fractional differential equations have gained popularity due to their applications in many domains of science and engineering[1–3]. In consequence, many researchers pay attention to form a simple and best definition of fractional derivative. Recently, a new definition of fractional derivative named conformable fractional derivative has been introduced in [4]. This novel fractional derivative is very easy and satisfies all the properties of the standard one. In short time, many studies and discussion related to conformable fractional derivative have appeared in several areas of applications [1–10].Motivated by the abovementioned works, we consider the following impulsive differential equation with conformable fractional derivative:(1)dαxtdtα=Axt+ft,xt,α∈0,1,t∈J=0,b,t≠tk,xtk+=xtk−+yk,k=1,2,…,n,x0=x0∈X,where dα./dtα is the so-called fractional conformable derivative [4]. This novel fractional derivative attracts the attention of many authors in various domains of science [1–10]. A:DA⊆X⟶X is the generator of a C0-semigroup Tt,t≥0 on a Banach space X, f:J×X⟶X is continuous, x0, yk are the element of X, and 0=t0<t1<t2<…<tn<tn+1=b, xtk+=limh⟶0+xtk+h and xtk−=xtk represent respectively the right and left limits of xt at t=tk.One of the main novelties of this paper is the concept on a mild solution for system (1). Then, using some fixed point theorems such as Banach contraction mapping principle and Schauder’s fixed point theorem, we derive many existence and uniqueness results concerning the mild solution for system (1) under the different assumptions on the nonlinear terms.As a second problem, we discuss in Section4, a nonlocal impulsive differential equation with conformable fractional derivative(2)dαxtdtα=Axt+ft,xt,α∈0,1,t∈J=0,b,t≠tk,xtk+=xtk−+yk,k=1,2,…,n,x0=x0+gx,where A, f, yk are defined as above, g is a given function and constitutes a Cauchy problem. The condition x0=x0+gx represents the nonlocal condition [11]. For good effect of this condition, we refer to [12, 13]. We adopt the ideas given in [14–16] and obtained some new existence and uniqueness results for system (2) under the different assumptions on the nonlocal terms.The content of this paper is organized as follows. In Section2, we recall some preliminary facts on the conformable fractional calculus. Sections 3 and 4 are devoted to prove the main result. At last, some interesting examples are presented to illustrate the theory.


## 2. Preliminary

In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper, and we recall some concepts on conformable fractional calculus.LetLbX be the Banach space of all linear and bounded operators on X. For a C0-semigroup Tt,t≥0 on X, we set M=supt∈JTtα/αLbX. Let CJ,X be the Banach space of all X-valued continuous functions from J=0,b into X, endowed with the norm xC=supt∈Jxt. We also introduce the set of functions(3)PCJ,X=x:J⟶X|xis continuous att∈J\t1,t2,…,tn,xis continuous from left and has right hand limits att∈t1,t2,…,tn,endowed with the norm(4)xPC=maxsupt∈Jxt+0,supt∈Jxt−0,where it is easy to see that PCJ,X,.PC is a Banach space.Theorem 1.
(Krasnoselskii’s fixed point theorem). Assume thatK is a closed bounded convex subset of a Banach space X. Furthermore, assume that Γ1 and Γ2 are mappings from K into X such that(1)
Γ1u+Γ2v, for all u,v∈K(2)
Γ1 is a contraction(3)
Γ2 is continuous and compact
Then,Γ1+Γ2 has a fixed point in K.Definition 1.
(see [4]). Let α∈0,1. The conformable fractional derivative of order α of a function x. for t>0 is defined as(5)dαxtdtα=limε⟶0xt+εt1−α−xtε.
Fort=0, we adapt the following definition:(6)dαx0dtα=limt⟶0+dαxtdtα.
The fractional integralIα. associated with the conformable fractional derivative is defined by(7)Iαxt=∫0tsα−1xsds.Theorem 2.
(see [4]). If x. is a continuous function in the domain of Iα., then we have(8)dαIαxtdtα=xt.Definition 2.
(see [2]). The Laplace transform of a function x. is defined by(9)ℒftλ≔∫0+∞e−λtftdt,λ>0.
It is remarkable that the above transform is not compatible with the conformable fractional derivative. For this, the adapted transform is given by the following definition.Definition 3.
(see [5]). The fractional Laplace transform of order α∈0,1 of a function x. is defined by(10)ℒαxtλ≔∫0+∞tα−1e−λtα/αxtdt,λ>0.
The following proposition gives us the actions of the fractional integral and the fraction Laplace transform on the conformable fractional derivative, respectively.Proposition 1.
(see [5]). If x. is a differentiable function, then we have the following results:(11)Iαdαx.dtαt=xt−x0,ℒαdαxtdtαλ=λℒαxtλ−x0.
According to [6], we have the following remark.Remark 1.
For two functionsx. and y., we have(12)ℒαxtααλ=ℒxtλ,ℒα∫0tsα−1xtα−sααysdsλ=ℒxtλℒαytλ.Lemma 1.
A measurable functionf:J⟶X is Bochner integrable, if f is Lebesuge integrable.Lemma 2.
Forσ∈0,1 and 0<a≤b, we have a−bσ≤b−aσ.


## 3. Main Results

Now, we give the main contribution results.To obtain the uniqueness of mild solution, we will need the following assumption.Case 1.
We suppose thatf is Lipschitz.
Let us list the following hypotheses:
HA: A is the infinitesimal generator of a compact semigroup Tt,t≥0 in X.
HF1: f:J×X⟶X is continuous and there exists a constant q1∈0,α and a real-valued function Lft∈L1/q1J,R+ such that(13)ft,x−ft,y≤Lftx−y,t∈J,x,y∈X.
For brevity, let us take(14)T∗=α−11−q1tk+1α−1/1−q11−q1LfL1/q1tk,tk+1,R+.
By using the following Duhamel formula (see [7]), we can introduce the following definition of the mild solution for system (1).Definition 4.
We say that a functionx∈PC0,b,X is called a mild solution of Cauchy problem,(15)dαxtdtα=Axt+ft,xt,α∈0,1,t∈J=0,b,t≠tk,xtk+=xtk−+yk,k=1,2,…,n,x0=x0,if x satisfies(16)xt=Ttααx0+∫0tsα−1Ttα−sααfs,xsds,t∈0,t1,Ttααx0+Ttα−t1ααy1+∫0tsα−1Ttα−sααfs,xsds,t∈t1,t2,⋮Ttααx0+∑i=1nTtα−t1ααyi+∫0tsα−1Ttα−sααfs,xsds,t∈tn,b.Theorem 3.
IfTtt>0 is compact and HA−HF1 are satisfied, then Cauchy problem (1) has a unique mild solution on J, provided that(17)0<MT∗<1.Proof.
Letx0∈X be fixed. Define an operator Γ:PCJ,X⟶PCJ,X by(18)Γxt=Ttααx0+∫0tsα−1Ttα−sααfs,xsds,t∈0,t1,Ttααx0+Ttα−t1ααy1+∫0tsα−1Ttα−sααfs,xsds,t∈t1,t2,⋮Ttααx0+∑i=1nTtα−t1ααyi+∫0tsα−1Ttα−sααfs,xsds,t∈tn,b.
By our assumptions and Lemma1, Γ is well defined on PCJ,X.Step 1.
We prove thatΓx∈PCJ,X for x∈PCJ,X.Claim 1.
For0≤τ<t≤t1, taking into account the imposed assumptions and applying Lemma 2, we obtain(19)Γxt−Γxτ≤Ttααx0−Tτααx0+∫τtsα−1Ttα−sααfs,xsds+∫0τsα−1Ttα−sαα−Tτα−sααfs,xsds≤Ttααx0−Tτααx0+MfPCαtα−τα+MfPCαt1α≤Ttααx0−Tτααx0+MfPCαtα−τα+MfPCαt1α,where we use the inequality tα−τα≤t−τα. The first and second terms tend to zero as t⟶τ. Moreover, it is obvious that the last terms tend to zero too as t⟶τ. Thus, we can deduce that Γx∈PC0,t1,X.Claim 2.
Fort1≤τ<t<t2, keeping in mind our assumptions and applying Lemma 2 again, we have(20)Γxt−Γxτ≤Tτααx0−Tτααx0+Ttα−t1αα−Tτα−t1ααy1+∫τtsα−1Ttα−sααfs,xsds+∫0τsα−1Ttα−sαα−Tτα−sααfs,xsds≤Tτααx0−Tτααx0+Ttα−t1αα−Tτα−t1ααy1+MfPCαtα−τα+MfPCαt2α.
Ast⟶τ, the right-hand side of the above inequality tends to zero. Thus, we can deduce that Γx∈PCt1,t2,X.
Similarly, we can also obtain thatΓx∈PCt1,t2,X,…,Γx∈PCtn,b,X. That is, Γx∈PCJ,X.Step 2.
We show thatΓ is the contraction on PCJ,X.Claim 1.
For eacht∈0,t1, it comes from our assumptions that(21)Γxt−Γyt≤∫0tsα−1Ttα−sααfs,xs−fs,ysds≤supt∈0,τTtαα∫0tsα−1Lfsx−yds≤∫0tsα−1/1−q11−q1dsLfL1/q10,t1,R+Mx−yPC≤α−11−q1t1α−1/1−q11−q1LfL1/q10,t1,R+Mx−yPC.
In general, for eacht∈tk,tk+1, using our assumptions again,(22)Γxt−Γyt≤Mα−11−q1tk+1α−1/1−q11−q1LfL1/q1tk,tk+1,R+x−y,thus(23)Γx−ΓyPC≤T∗Mx−yPC.
Hence, condition (17) allows us to conclude in view of the Banach contraction mapping principle that Γ has a unique fixed point x∈PCJ,X which is just the unique mild solution of system (1).Case 2.
f is not Lipschitz.
We make the following assumptions:
HF2: f:J×X⟶X is continuous and maps a bounded set into a bounded set.
HF3: the function ft,.:X⟶X is continuous, and for all r>0, there exists a function μr∈L∞0,b,ℝ+ such that supx≤rft,x≤μrt, for all t∈0,b.
C1: for each x0∈X, there exists a constant r>0 such that(24)Mx0+∑i=1nyk+bααsups∈J,ϕ∈YΓfs,ϕs≤r,where(25)YΓ=ϕ∈PCJ,Xϕ≤r,t∈J.Theorem 4.
Suppose thatHA, HF2,HF3, and C1 are satisfied. Then, for every x0∈X, system (1) has at least a mild solution on J.Proof.
Letx0∈X be fixed. We introduce that map Γ:PCJ,X⟶PCJ,X by(26)Γxt=Γ1xt+Γ2xt,where(27)Γ1xt=Ttααx0+∫0tsα−1Ttα−sααfs,xsds,t∈J\t1,…,tnΓ2xt=0,t∈0,t1∑i=1kTt−tiαyi,t∈tk,tk+1,k=1,2,…,n.
For eacht∈0,t1, x∈ϒΓ,(28)Γxt≤Γ1xt+Γ2xt≤Mx0|+bααsups∈J,ϕ∈YΓfs,ϕs.
For eacht∈tk,tk+1, x∈YΓ,(29)Γxt≤Γ1xt+Γ2xt≤Mx0+∑i=1nyk+bααsups∈J,ϕ∈YΓfs,ϕs.
Noting the conditionC1, we see that Γ:ϒΓ⟶ϒΓ.Step 1.
We prove thatΓ is a continuous mapping from ϒΓ to ϒΓ.
In order to derive the continuity ofΓ, we only check that Γ1 and Γ2 are all continuous.
For this purpose, we assume thatxn⟶x in ϒΓ; it comes from the continuity of f that sα−1fs,xns⟶sα−1fs,xs, as n⟶∞.
Noting that(30)sα−1fs,xns−fs,xs≤sα−1sups∈J,ϕ∈YΓfs,ϕs,fors∈0,t,t∈J,by means of Lebesgue dominated convergence theorem, we obtain that ∫0tsα−1fs,xns−fs,xsds⟶0, as n⟶∞. It is easy to see that, for each t∈J,(31)Γ1xnt−Γ1xt≤M∫0tsα−1fs,xns−fs,xsds⟶0,n⟶∞.
Thus,Γ1 is continuous. On the contrary, it is obvious that Γ2 is continuous. Since Γ1 and Γ2 are continuous, Γ is also continuous.Step 2.
We show thatΓ is a compact operator, or Γ1 and Γ2 are compact operators.
The compactness ofΓ2 is clear since it is a constant map.
Compactness ofΓ1:Claim 1.
We prove thatΓ1xt|x∈Br is relatively compact in X.
For some fixedt∈0,b, let ε∈0,t, x∈Br, and define the operator Γ1ε by(32)Γ1εxt=∫0tα−εα1/αsα−1Ttα−sααfs,xsds.
We can writeΓ1ε as follows:(33)Γ1εxt=Tεαα∫0tα−εα1/αsα−1Ttα−sα−εααfs,xsds.
According to the compactness ofTtt>0, the set Γ1εxt|x∈Br is relatively compact in X. Using HF3, we have(34)Γ1εxt−Γ1xt≤μrL∞0,b,ℝ+Mεαα.
Therefore,Γ1xt|x∈Br is relatively compact in X. It is clear that Γ1x0|x∈Br is compact. Finally, Γ1xt|x∈Br is relatively compact in X, for all t∈0,b.Claim 2.
We prove thatΓ1Br is equicontinuous.
Lett1,t2∈0,b such that t1<t2. We have(35)Γ1xt2−Γ1xt1=∫0t1sα−1Tt2α−sαα−Tt1α−sααfs,xsds+∫t1t2sα−1Tt2α−sααfs,xsds=Tt2α−sαα−I∫0t1sα−1Tt1α−sααfs,xsds+∫t1t2sα−1Tt2α−sααfs,xsds.
Hence,(36)Γ1xt2−Γ1xt1≤MμrL∞0,b,ℝ+αt2α−t1α+ταTt2α−t1αα−I.
We conclude thatΓ1x, x∈Br are equicontinuous at t∈0,b. By using the Arzela–Ascoli theorem, we obtain that Γ2 is compact. Now, Schauder’s fixed point theorem implies that Γ has a fixed point, which gives rise to a mild solution.


## 4. Existence Results for Impulsive Nonlocal Cauchy Problems

In this section, we extend the results obtained in Section3 to nonlocal problems for impulsive conformable fractional evolution equations. More precisely, we will prove the existence and uniqueness of the mild solutions for system (2). As we all known, the nonlocal conditions have a better effect on the solution and are more precise for physical measurements than the classical initial condition alone.Definition 5.
By a mild solution of system (2), we mean that a function x∈PCJ,X, which satisfies the following integral equation:(37)xt=Ttααx0+gx+∫0tsα−1Ttα−sααfs,xsds,t∈0,t1,Ttααx0+gx+Ttα−t1ααy1+∫0tsα−1Ttα−sααfs,xsds,t∈t1,t2,⋮Ttααx0+gx+∑i=1nTtα−tiααyi+∫0tsα−1Ttα−sααfs,xsds,t∈tn,b.Case 1.
g is Lipschitz.
To end this section, we make the following assumption:
Hg1: g:PCJ,X⟶X and there exists a constant Lg>0 such that
gx−gy≤Lgx−yPC,x,y∈PCJ,X.Theorem 5.
LetHA, HF1, and Hg1 be satisfied. Then, for every x0∈X, system (2) has a unique mild solution on J, provided that(38)0<μ′≔MLg+T∗<1.Proof.
Define the operatorℱ:PCJ,X⟶PCJ,X by(39)ℱxt=Ttααx0+gx+∫0tsα−1Ttα−sααfs,xsds,t∈0,t1,Ttααx0+gx+Ttα−t1ααy1+∫0tsα−1Ttα−sααfs,xsds,t∈t1,t2,⋮Ttααx0+gx+∑i=1nTtα−tiααyi+∫0tsα−1Ttα−sααfs,xsds,t∈tn,b.
It is obvious thatℱ is well defined on PCJ,X.Step 1.
We prove thatℱx∈PCJ,X, for x∈PCJ,X.
For0≤τ<t≤t1, by our assumptions,(40)Ttααgx−Tτααgx≤Ttαα−TτααLgxPC+g0.
Ast⟶τ, the right-hand side of the above inequality tends to zero. Recalling Step 1 in Theorem 3, we know that ℱx∈PCJ,X.Step 2.
ℱ is the contraction.
We only taket∈tk,tk+1, then we have(41)ℱxt−ℱyt≤MLg+T∗x−yPC,so we get(42)ℱx−ℱyPC≤μ′x−yPC,where(43)μ′≔MLg+T∗.
Hence, condition (38) allows us to conclude, in view of the Banach contraction mapping principle again, that ℱ has a unique fixed point x∈PCJ,X which is the mild solution of system (2).Theorem 6.
Suppose thatHA, HF3, and Hg1 are satisfied. If MLg<1/2, then system (2) has at least a mild solution on J.Proof.
Choose(44)σ≥2Mx0+g0+∑ni=1yi+bααμrL∞J,R+.
ConsiderBσ=x∈PCJ,X|xPC≤σ. Define the operators N on Bσ by(45)Nxt=N1xt+N2xt+N3xt,where(46)N1xt=Ttααx0+gx,t∈J,N2xt=∫0tsα−1Ttα−sααfs,xsds,t∈J,and N3 is the same as the operator Γ2, defined in Theorem 4.
It suffices to proceed exactly the steps of the proof in Theorem4, while replacing Br by Bσ to obtain that N2+N3 are continuous and compact. We want to use Krasnoselkii’s fixed point theorem. Thus, to complete the rest proof of this theorem, it suffices to show that N1 is a contraction mapping and that if x,y∈Bσ, then N1x+N2+N3y∈Bσ. Indeed, for any x∈Bσ, we have(47)N1xPC+N2yPC+N3yPC≤Mx0+g0+Lgσ+M∑ni=1yi+MbααμrL∞J,R+.
SinceMLg<1/2, we can deduce that(48)N1x+N2+N3yPC≤σ.
Next, for anyt∈tk,tk+1,x,y∈Ctk,tk+1,X,(49)N1x−N1yCtk,tk+1,X≤MLgx−yCtk,tk+1,X.
Therefore, we can deduce thatN1 is the contraction from MLg<1. Moreover, N2+N3 is compact and continuous. Hence, by the well-known Krasnoselskii’s fixed point theorem, we can conclude that system (2) has at least one mild solution on J.
Case 2:g is not Lipschitz.
Hg2: g:PCJ,X⟶X and maps bounded sets into bounded sets.
C2: for each x0∈X, there exists a constant r′>0 such that(50)Mx0+supϕ∈ϒΓ′ϕt+M∑ni=1yi+Mbααsups∈J,ϕ∈ϒΓ′fs,ϕs≤r′,where(51)ϒΓ′=ϕ∈PCJ,X|ϕ≤r′,fort∈J.Theorem 7.
Suppose thatHA, HF2, C2, and Hg2 are satisfied. Then, for every x0∈X, system (2) has at least a PC-mild solution on J.Proof.
Define an operatorℱ on PCJ,X by(52)ℱxt=ℱ1xt+ℱ2xt,where(53)ℱ1xt=Ttααx0+gx+∫0tsα−1Ttα−sααfs,xsds,t∈J.and ℱ2 is the same as Γ2 defined in Theorem 4. Thus, we need to check that ℱ1 is compact. Observing the expression of the ℱ1, we only check that, for each t∈J, the set Ttx0+gx|x∈ϒΓ′ is precompact in X since Tt,t>0 is compact and Hg2. On the other hand, the equicontinuity of Ttx0+gx|t∈J,x∈ϒΓ′ can be shown using the same idea.
Therefore,ℱ is also a compact operator. By Schauder’s fixed point theorem again, ℱ has a fixed point, which gives rise to a mild solution.


### 4.1. Example

In this section, some interesting examples are presented to illustrate the theory. Consider the following impulsive fractional differential equations with nonlocal conditions:(54)dαdtαxt,y=∂2∂y2xt,y+ft,xt,y,α∈0,1,y∈0,π,t∈0,t1∪t1,1,xt,0=xt,π=0,xt1+=xt1−+z,t1=12,t∈0,π,x0,y=x0y+gxt,y,t∈0,1,y∈0,π.LetX=L20,π. Define E1Ax=−∂2/∂y2x, for x∈DA, where DA=x∈X|∂x/∂y,∂2x/∂y2∈Xandx0=xπ=0. Then, A is the infinitesimal generator of a strongly continuous semigroup Tt,t≥0 in L20,π. Moreover, T. is also compact and Tt≤e−t≤1=M1, t≥0.Case 1.
Definef:0,1×X⟶XE2ft,xty=e−txt,y/ρ+et1+xt,y,t∈0,t1∪t1,1,ρ>,1,x∈X,y∈0,π,
E3gxty=∑j=12λjxsj,y,0<λ1,λ2,0<s1<s2<1,s1,s2≠t1,x∈PC0,1,X,y∈0,π.
Clearly,f: [0, 1]  × X ⟶ X are continuous functions,
ft,x−ft,y≤Lfx−y, with Lf=1/ρ+1∈L1/q10,1,R+,q1∈0,α.
It is obvious thatg:PC0,1,X⟶X satisfies gx−gy≤Lgx−yPC with Lg=∑j=12λj.
E1+E2+E3 makes the assumptions in Theorem 5 satisfied. Therefore, equation (54) has a unique mild solution on 0,1, provided that(55)∑2j=1λj+1ρ+11−q1α−q11−q1<1.
Case 2. DefineE4ft,xty=e−tsinxt,y/1+tet+e−t+e−t,t∈0,t1∪t1,1,x∈X,y∈0,π.
Clearly,ft,xty≤e−t/e−t+et+e−t=mt, with mt∈L∞0,1,R+.
E1+E3+E4 makes the assumptions in Theorem 6 satisfied. Therefore, equations (54) has at least one mild solution on 0,1, provided that ∑2j=1λj<1/2.
Case 3. DefineE5ft,xty=c1sinxt,y,c1>0,t∈0,t1∪t1,1,x∈X,y∈0,π,.
E6gxty=∫01lsln1+xs,y1/2ds,l∈L10,1,R,x∈PC0,1,X,y∈0,π.
Clearly,f and g are continuous and map a bounded set into a bounded set.
E1+E5+E6 makes the assumptions in Theorem 7 satisfied, for large r′>0. Therefore, equation (54) has at least one mild solution.


## 4.1. Example

In this section, some interesting examples are presented to illustrate the theory. Consider the following impulsive fractional differential equations with nonlocal conditions:(54)dαdtαxt,y=∂2∂y2xt,y+ft,xt,y,α∈0,1,y∈0,π,t∈0,t1∪t1,1,xt,0=xt,π=0,xt1+=xt1−+z,t1=12,t∈0,π,x0,y=x0y+gxt,y,t∈0,1,y∈0,π.LetX=L20,π. Define E1Ax=−∂2/∂y2x, for x∈DA, where DA=x∈X|∂x/∂y,∂2x/∂y2∈Xandx0=xπ=0. Then, A is the infinitesimal generator of a strongly continuous semigroup Tt,t≥0 in L20,π. Moreover, T. is also compact and Tt≤e−t≤1=M1, t≥0.Case 1.
Definef:0,1×X⟶XE2ft,xty=e−txt,y/ρ+et1+xt,y,t∈0,t1∪t1,1,ρ>,1,x∈X,y∈0,π,
E3gxty=∑j=12λjxsj,y,0<λ1,λ2,0<s1<s2<1,s1,s2≠t1,x∈PC0,1,X,y∈0,π.
Clearly,f: [0, 1]  × X ⟶ X are continuous functions,
ft,x−ft,y≤Lfx−y, with Lf=1/ρ+1∈L1/q10,1,R+,q1∈0,α.
It is obvious thatg:PC0,1,X⟶X satisfies gx−gy≤Lgx−yPC with Lg=∑j=12λj.
E1+E2+E3 makes the assumptions in Theorem 5 satisfied. Therefore, equation (54) has a unique mild solution on 0,1, provided that(55)∑2j=1λj+1ρ+11−q1α−q11−q1<1.
Case 2. DefineE4ft,xty=e−tsinxt,y/1+tet+e−t+e−t,t∈0,t1∪t1,1,x∈X,y∈0,π.
Clearly,ft,xty≤e−t/e−t+et+e−t=mt, with mt∈L∞0,1,R+.
E1+E3+E4 makes the assumptions in Theorem 6 satisfied. Therefore, equations (54) has at least one mild solution on 0,1, provided that ∑2j=1λj<1/2.
Case 3. DefineE5ft,xty=c1sinxt,y,c1>0,t∈0,t1∪t1,1,x∈X,y∈0,π,.
E6gxty=∫01lsln1+xs,y1/2ds,l∈L10,1,R,x∈PC0,1,X,y∈0,π.
Clearly,f and g are continuous and map a bounded set into a bounded set.
E1+E5+E6 makes the assumptions in Theorem 7 satisfied, for large r′>0. Therefore, equation (54) has at least one mild solution.



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*Source: 5729128-2020-11-12.xml*