# Multiplicity Results to a Conformable Fractional Differential Equations Involving Integral Boundary Condition

Authors: Shuman Meng; Yujun Cui
Journal: Complexity
 (2019)

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## Abstract

In this article, by using topological degree theory couple with the method of lower and upper solutions, we study the existence of at least three solutions to Riemann-Stieltjes integral initial value problem of the typeDαx(t)=f(t,x),t∈[0,1], x(0)=∫01x(t)dA(t), where Dαx(t) is the standard conformable fractional derivative of order α, 0<α≤1, and f∈C([0,1]×R,R). Simultaneously, the fixed point theorem for set-valued increasing operator is applied when considering the given problem.


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## Body

## 1. Introduction

In recent years, fractional differential equations have exerted tremendous influence on some mathematical models of research processes and phenomena in many fields such as electrochemistry, heat conduction, underground water flow, and porous media. A growing number of papers deal with the existence or multiplicity of solutions of initial value problem and boundary value problem for fractional differential equations [1–11]. Recently, the authors [12] give an interesting fractional derivative called the “conformal fractional derivative”, which depends on the limit definition of the function derivative. Moreover, readers can find in [13] the properties of conformable fractional derivatives that are similar to ordinary differential ones. Other related work on conformable fractional differential equation can be found in [12–17] and the references therein.Whether it is ordinary differential equations or fractional differential equations, the existence of solutions to boundary value problems with integral boundary conditions has been studied in the applied sciences and physics. For more important content of the typical theory of differential equations and integral equations with boundary value problems are obtained [18, 19]. Lots of results have been established for differential equations and differential systems with integral boundary conditions by using upper and lower solution, fixed point theory and fixed point theorem; see [20–28].Now there are more and more articles to prove that there are multiple solutions for integral boundary [29–35]. For example, [31] introduced the system of fractional differential equations (1)D0+αut+ft,vt=0,t∈0,1,n-1<α≤n,D0+βvt+ft,ut=0,t∈0,1,m-1<β≤m,where n,m∈N, n,m≥2, D0+α, D0+β are the standard the Riemann-Liouville derivatives of orders α, β. By means of the Guo-Krasnoselskii fixed point theorem and the property of degree to obtain the existence and multiplicity of positive solutions to integral boundary value conditions (2)u0=u′0=⋯=un-20=0,u1=∫01usdHs.and (3)v0=v′0=⋯=vn-20=0,v1=∫01vsdKs.where H,K:[0,1]→R are nondecreasing functions.Inspired by the above work, we consider the existence and multiple solutions of the following fractional differential equation involving integral boundary condition:(4)Dαxt=ft,xt,t∈0,1,x0=∫01xtdAt,where f∈C([0,1]×R,R), ∫01x(t)dA(t) denotes the Riemann-Stieltjes integral with positive Stieltjes measure. Dαf(t) is the standard conformable fractional derivative of order 0<α≤1 of f at t>0, defined by (5)Dαft=limρ→0ft+ρt1-α-ftρ,t∈0,1.If Dαf(t) exists on (0,1), then Dαf(0)=limt→0Dαf(t). Obviously, Dαf(t)=t1-αf′(t) if f is differentiable.In order to study the existence and multiplicity results of the problem (4), this paper is arranged as follows: after this introduction, in Section 2, we briefly show some necessary definitions and lemmas that are used to prove our main results. In Section 3, we shall define a modified bounded function to discuss the existence of solutions for a conformable fractional differential equation with initial value condition. Moreover, in this part we employ two fixed point theorems, which are Schauder’s fixed point theorem and the fixed point theorem for set-valued increasing operator, respectively. In addition, using two pairs of upper and lower solutions and the property of degree theory, we deduce that problem (4) has at least three solutions.


## 2. Preliminaries

In the following, letE=C[0,1], then E is a Banach space with the norm x=maxt∈[0,1]|x(t)|.Definition 1 (see [12]).
Letα∈(0,1]; the conformable fractional integral starting from a point 0 of a function f:[0,+∞)→R of order α is defined as (6)Iαft=∫0tsα-1fsds.Lemma 2 (see [14]).
Letf:(0,+∞)→R be differentiable and 0<α≤1. Then, for all t>0 we have (7)IαDαft=ft-f0.Lemma 3 (see [15]).
Letα∈(0,1], k1,k2,c,k∈R, and the function f, g be α-differentiable on [0,+∞); then(i)
Dαc=0 for all constant function f(t)=c;(ii)
Dαtk=ktk-α;(iii)
Dα(k1f(t)+k2g(t))=k1Dαf(t)+k2Dαg(t);(iv)
Dα(fg)=f(t)Dαg(t)+g(t)Dαf(t);(v)
Dα(f/g)=g(t)Dαf(t)-f(t)Dαg(t)/g2 for all function h(t)≠0.Lemma 4 ([15] (mean value theorem)).
Let an interval[a,b]⊂[0,+∞), and the function f:[0,+∞)→R, if it satisfies(1)
f is continuous on [a,b];(2)
f is α-differentiable for some α∈(0,1] on [a,b].
Then there exists a constantξ∈(a,b), such that Dαf(ξ)=f(b)-f(a)/1/αbα-1/αaα.Definition 5.
A functionv∈C([0,1],R) is called a lower solution of problem (4), if it satisfies (8)Dαvt≤ft,vt,t∈0,1,(9)v0≤∫01vtdAt.If inequalities (8), (9) are reversed, then v is an upper solution of problem (4).Definition 6.
A functionv∈C([0,1],R) is said to be a strict lower solution of problem (4), if the inequality (8), (9) is strict for t∈[0,1].Lemma 7 (see [13]).
f : [ a , b ] → R is a given function that satisfies(i)
f is continuous on [a,b];(ii)
f is α-differentiable for some α∈(0,1).
Then we have the following:(1)
ifDαf(x)≥0 for all x∈(a,b), then f is increasing on [a,b];(2)
ifDαf(x)≤0 for all x∈(a,b), then f is decreasing on [a,b].The following lemma is a direct consequence of the application of the definition of conformable fractional derivative and Lemma7.Lemma 8.
Letα∈(0,1]; the function f(t) is continuous and α-differentiable on [a,b].(1)
Iff has an extreme value at t0∈(a,b), then Dαf(t0)=0;(2)
iff has a maximum (minimum) value at t0=a, then (10)Dαft0≤0Dαft0≥0;(3)
iff has a maximum (minimum) value at t0=b, then (11)Dαft0≥0Dαft0≤0.Theorem 9 ([36, 37] (fixed point theorem for set-valued increasing operator)).
LetX be a partially ordered set, M be a nonempty closed set of X, and L:M→2M be a set-valued increasing operator. Assume that(i)
any totally ordered subset ofM is a relatively compact,(ii)
for allx∈M, Lx is a compact set in X,(iii)
there existx0∈M and u∈Lx0 such that x0≤u.
ThenL has a fixed point in M; that is, there exists x∗∈M such that x∗∈Lx∗.Throughout the paper, we list some hypotheses.(H1) f:[0,1]×R→R is continuous.(H2) ∫01dA(t)>0, A(t) is continuous at t=0 and is nondecreasing at [0,1].(H3) Assume that v,w∈C[0,1] is lower and upper solution of problem (4) with v(t)≤w(t).


## 3. Main Results

Based on the above preparations and the assumptions mentioned, we first consider the following initial value problem (IVP)(12)Dαxt=ft,xt,t∈0,1,x0=a,where a∈R. We suppose that (H1) holds. Applying Lemma 2, it is easy to prove that (12) is equivalent to the following integral equation (13)xt=∫0tsα-1fs,xsds+a.Define integral operatorT:E→E by (14)Txt=∫0tsα-1fs,xsds+a.Then, x is a solution of (12) if and only if x∈E is a solution of the operator equation (I-T)x=0, that is, a fixed point of operator T.Theorem 10.
Assume that(H1)-(H3) hold. Then for a fixed number a∈[v(0),w(0)], there exists a solution x(t) of problem (12) such that v(t)≤x(t)≤w(t), t∈[0,1].Proof.
Firstly we define the following modified function:(15)Fvwt,x=ft,wt+x-wt1+x-wt,x>wt,ft,xt,vt≤xt≤wt,ft,vt+vt-x1+x-vt,x<vt.Obviously Fvw is continuous and bounded on [0,1]×R, so there exists M such that max⁡a+1,Fvwt,x≤M.
Secondly, we consider the modified initial value problem(16)IVPvwaDαxt=Fvwt,xt,t∈0,1,x0=a.The above initial value problem (16) is equivalent to the following integral equation: (17)xt=∫0tsα-1Fvws,xsds+a.For any x∈E, we have (18)∫0tsα-1Fvws,xsds+a≤M∫0tsα-1ds+a<M1α+M=M1+αα.Choose M1>M(1+α)/α, and let (19)Ω=x∈E:x<M1.Obviously, Ω is a bounded convex subset in E. From the above argument, the operator T~:Ω→Ω defined as follows is well defined, (20)T~xt=∫0tsα-1Fvws,xsds+a.It is easy to verify that T~:Ω→Ω is compact. By Schauder’s fixed point theorem, T~ has a fixed point x in Ω. Subsequently, x is the solution of problem (16). Also, by the definition of Fvw and Definition 5, we have (21)Fvwt,vt=ft,vt≥Dαvt,t∈0,1,v0≥aand (22)Fvwt,wt=ft,wt≤Dαwt,t∈0,1,w0≤a.Hence v and w are lower and upper solutions of (16).
Our final job is to apply the method in [38] to illustrate that if (16) has upper and lower solutions, hence any solution x of differential equation (16) must satisfy v(t)≤x(t)≤w(t), t∈[0,1] and then is a solution of (12).
We now prove that any solutionx of (16) does satisfy (23)vt≤xt≤wt,t∈0,1.We need to show that v(t)≤x(t), t∈[0,1]. Analogously we can also show that x(t)≤w(t). If v(t)≰x(t) on [0,1], let h(t)=v(t)-x(t), then h(t) has a positive maximum at some t0∈[0,1]. By the initial boundary conditions, we derive h(0)=v(0)-x(0)≤0, so t0≠0. Then there will be two cases:
(i)t0∈(0,1), then by Lemma 8, we have h(t0)>0, that is v(t0)>x(t0) and (24)Dαht0=0;
(ii)t0=1, then by Lemma 8, we have h(1)>0 and Dαh(1)≥0.
For case (i), by the definition ofFvw and Definition 5, we can get (25)Dαht0=Dαvt0-Dαxt0≤ft0,vt0-Fvwt0,xt0=ft0,vt0-ft0,vt0+vt0-xt01+xt0-vt0<0,which contradicts with (24). This implies that h(t) has no positive local maximum on [0,1].
For case (ii), ift0=1, then h(1)>0, h(0)≤0. Notice that h(t) has no positive local maximum on [0,1], and there exists t1∈[0,1) such that (26)ht1=0,ht≤0,t∈0,t1,and (27)Dαht≥0,ht≥0,t∈t1,1.But for every t∈(t1,1), we have (28)Dαht=Dαvt-Dαxt≤ft,vt-Fvwt,xt=ft,vt-ft,vt+vt-xt1+xt-vt<0,again a contradiction. We complete the proof.Theorem 11.
Assume that(H1)-(H3) hold, then problem (4) has a solution x(t) with v(t)≤x(t)≤w(t), ∀t∈[0,1].Proof.
It follows from Theorem10 that for any fixed a∈[v(0),w(0)], IVP (12) have a solution x(t); we can also mark it as xa(t) satisfies v(t)≤xa(t)≤w(t), t∈[0,1]. The following two cases are considered:
(i)v(0)=w(0)=a. It is easy to obtain that ∫01v(t)dA(t)≤∫01xa(t)dA(t)≤∫01w(t)dA(t). Using Definition 5, we can conclude that (29)∫01xatdAt≤∫01wtdAt≤w0=xa0=v0≤∫01vtdAt≤∫01xatdAt,i.e., xa(0)=∫01xa(t)dA(t), then xa(t) is the solution of problem (4).
(ii)v(0)<w(0). Define the set-valued operator on [v(0),w(0)] as follows: (30)La=∫01xatdAt:xatis  the  solution  of  IVP12.It can be seen from Theorem 10 that for any fixed a∈[v(0),w(0)], L(a)⊂[∫01v(t)dA(t),∫01w(t)dA(t)]⊂[v(0),w(0)], this shows that L:[v(0),w(0)]→2[v(0),w(0)]. Moreover, from the definition of L, we can obtain that (4) has a solution x(t) with v(t)≤x(t)≤w(t), ∀t∈[0,1] if L has a fixed point. In order to apply the fixed point theorem for set-valued increasing operator, we shall prove the following:
Step 1. The sequence {zn} of L(a) must have subsequence {znk} converging to z∗∈L(a) for ∀a∈[v(0),w(0)].
Assume thatzn converges to z∗. By using Theorem 10, we know that the solution xn(t) of (12) related to zn satisfies v(t)≤xn(t)≤w(t), that is to say {xn(t)} is uniformly bounded. Moreover, by the continuity of f, we conclude that |f(t,xn(t))| is bounded on [0,1] for all n. Let t1,t2∈[0,1] with t1≤t2. Using Lemma 4, we can conclude that (31)xnt1-xnt2=1αDαxnξt1α-t2α≤1αfξ,xnξt1α-t2α,ξ∈t1,t2.Therefore, {xn(t)} is equicontinuous. It means that {xn(t)} is completely continuous sequences; by Arzela-Ascoli theorems, we get {xn(t)} has subsequences and {xnk}→x∗(t)∈E when k→∞. Note that xnk satisfy (32)xnkt=∫0tsα-1fs,xnksds+a,t∈0,1,znk=∫01xnktdAt,Letting k→∞, we obtain that (33)x∗t=∫0tsα-1fs,x∗sds+a,t∈0,1,z∗=∫01x∗tdAt,This shows that x∗(t) is a solution of problem (4); hence, z∗∈L(a).
Step 2. L is set-valued increasing operator, that is, ∀a,b∈[v(0),w(0)], a≤b and ∀c∈L(a), ∃d∈L(b) such that c≤d.
Without loss of generality, we assume thata<b, for ∀c∈L(a) such that c=∫01xa(t)dA(t) and v(t)≤xa(t)≤w(t), where xa(t) is the solution of (16) related to a. Therefore, we give the modified function Fxaw for xa(t) and w(t) and consider the following modified initial value problem: (34)Dαxt=Fxawt,xt,t∈0,1,x0=b.It is easy to obtain that xa, w are the lower and upper solutions of initial value problem (34). By the help of Theorem 10, we deduce that problem (34) has existence solution xb(t) with xa(t)≤xb(t)≤w(t). Subsequently, ∫01xa(t)dA(t)≤∫01xb(t)dA(t). Let d=∫01xb(t)dA(t), then c≤d. So the operator L is set-valued increasing operator. The theorem is now proved.Above we have studied the solution of (4), and then we present the main results of this paper.Theorem 12.
Suppose that(H1), (H2) and the following conditions hold.
( H 4 )   v 1 , v 2 ∈ E are two lower solutions and w1,w2∈E are two upper solutions of (4) such that v1≤v2≤w2, v1≤w1≤w2, and v2≰w1.
( H 5 )   v 2, w1 are strict lower and upper solutions of problem (4).
( H 6 )   k 1 ≠ 0, where k1=1-∫01dA(t).
Then problem (4) has at least three solutions xi (i=1,2,3), such that v1≤x1≤w1, v2≤x2≤w2, x3≰w1, and x3≱w2 on [0,1].Proof.
Similar to Theorem10, we define the following modified function Fv1w2: (35)Fv1w2t,x=ft,w2t+x-w2t1+x-w2t,x>w2t,ft,xt,v1t≤xt≤w2t,ft,v1t+v1t-x1+x-v1t,x<v1t.Obviously Fv1w2 is continuous and bounded on [0,1]×R, so there exists M>0 such that |Fv1w2(t,x)|≤M.
LetM2>M/α1+k2/k1, where k2=∫01tαdA(t). We consider the modified boundary value problem (36)Dαxt=Fv1w2t,x,t∈0,1,x0=∫01xtdAt.It is clear that the solution x(t) of the above problem satisfies v1(t)≤x(t)≤w2(t) and then is a solution of problem (4).
By Lemma2, it is easy to know that (36) is equivalent to the following integral equation: (37)xt=∫0tsα-1Fv1w2s,xsds+∫01xtdAt.Integrating (37) with respect to A(t) form 0 to 1, (38)∫01xtdAt=∫01∫0tsα-1Fv1w2s,xsdsdAt+1-k1∫01xtdAt.Combining with (37) and (38), we reduce (36) to the following equivalent integral equation: (39)xt=∫0tsα-1Fv1w2s,xsds+k1-1∫01∫0tsα-1Fv1w2s,xsdsdAt.
Now, we prove the existence of at least three solutions of the problem (36). Note that Ω1=x∈C0,1:x<M2 is a bounded convex subset in C[0,1]. Define the following operator T1:E→E(40)T1xt=∫0tsα-1Fv1w2s,xsds+k1-1∫01∫0tsα-1Fv1w2s,xsdsdAt.Then, x is a solution of (36) if and only if x∈E is a solution of the equation (I-T1)x=0, that is, a fixed point of T1. For x∈Ω¯1, we get (41)T1xt≤∫0tsα-1Fv1w2s,xsds+k1-1∫01∫0tsα-1Fv1w2s,xsdsdAt≤Mα1+k2k1<M2.So T1(Ω¯1)⊂Ω1 and T1:Ω1→Ω1 is compact. By the topological degree theory, we have (42)dI-T1,Ω1,θ=1Let (43)Ωv2=x∈Ω1:x>v2and (44)Ωw1=x∈Ω1:x<w1.It follows from the fact v2≰w1 that (45)Ωv2≠∅≠Ωw1,Ω¯v2∩Ω¯w1=∅,then Ω1∖(Ω¯v2∪Ω¯w1)≠∅.
Next, by(H5), we conclude that (36) has no solution on ∂Ωv2∪∂Ωw1. Without loss of generality, we assume that (36) has a solution x^(t) on ∂Ωv2. Thus there exists t0∈[0,1] such that x^(t0)=v2(t0) and x^(t)≥v2(t) for t∈[0,1]. It follows from Lemma 8 that Dαx^(t0)-Dαv2(t0)≤0 when t0∈(0,1]. But we noticed that v2 is a strict lower solution of (36), thus (46)Dαx^t0-Dαv2t0>Fv1w2t0,x^t0-Fv1w2t0,v2t0=Fv1w2t0,x^t0-Fv1w2t0,x^t0=0.This is a contradiction. If t0=1 and x^(t)>v2(t) for t∈(0,1], from the boundary condition and (H2), we have (47)0=x^0-v20≥∫01x^t-v2tdAt>0.This is a contradiction again. So (36) has no solution on ∂Ωv2∪∂Ωw1.
The additivity of degree implies that(48)dI-T1,Ω1,θ=dI-T1,Ωv2,θ+dI-T1,Ωw1,θ+dI-T1,Ω∖Ω¯v2∪Ω¯w1,θ=1.If we prove that (49)dI-T1,Ωv2,θ=dI-T1,Ωw1,θ=1,we can get (50)dI-T1,Ω∖Ω¯v2∪Ω¯w1,θ=-1and there are solutions in Ωv2, Ωw1, Ω∖(Ω¯v2∪Ω¯w1), respectively.
Now we show thatd(I-T1,Ωv2,θ)=d(I-T1,Ωw1,θ)=1. Firstly, we show that d(I-T1,Ωv2,θ)=1.
Define functionsFv2w2 in a similar way, and we consider the modified boundary value problem (51)Dαxt=Fv2w2t,x,t∈0,1,x0=∫01xtdAt.Equation (51) is equivalent to the operator equation (I-T2)x=0, where (52)T2xt=∫0tsα-1Fv2w2s,xsds+k1-1∫01∫0tsα-1Fv2w2s,xsdsdAt.Similar to the above argument, we can conclude that any solution x(t) of (51) satisfies x(t)≥v2(t), which in view of (H5) leads to x(t)≥v2(t) on (0,1); therefore, (53)x∈Ωv2,dI-T2,Ω1∖Ω¯v2,θ=0.Moreover, since T2(Ω1¯)⊆Ω1, we have (54)dI-T2,Ω1,θ=1.It follows from (53) and (54) that (55)1=dI-T2,Ω1,θ=dI-T2,Ωv2,θ+dI-T2,Ω1∖Ω¯v2,θ=dI-T2,Ωv2,θ.Taking into account of Fv1w2=Fv2w2 on Ωv2 we deduce that (56)dI-T1,Ωv2,θ=dI-T2,Ωv2,θ=1Similarly, we can show that d(I-T1,Ωw1,θ)=1. Therefore, there exist three solutions x1∈Ωv2, x2∈Ωw1, x3∈Ω∖(Ω¯v2∪Ω¯w1) of problem (4). The proof is finished.



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