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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2021; 61(1): 139-153

Published online March 31, 2021

### Existence of Positive Solutions for a Class of Conformable Fractional Differential Equations with Parameterized Integral Boundary Conditions

Department of Physics, University of Sciences and Technology of Oran-MB, El Mnaouar, BP 1505, 31000 Oran, Algeria Laboratory of Fundamental and Applied Mathematics of Oran (LMFAO), University of Oran 1, 31000 Oran, Algeria

Received: June 10, 2019; Revised: June 2, 2020; Accepted: June 2, 2020

### Abstract

In this paper, we study the existence of positive solutions for a class of conformable fractional differential equations with integral boundary conditions. By using the properties of Green's function with the fixed point theorem in a cone, we prove the existence of a positive solution. We also provide some examples to illustrate our results.

Keywords: conformable fractional derivatives, integral boundary value problems, positive solutions, fixed point theorems, cone.

### 1. Introduction

Fractional calculus and fractional differential equations are recently experiencing rapid development. There are several notions of fractional derivatives, some classical, such as the Riemann-Liouville or Caputo definitions, and some novel, such as conformable fractional derivatives [18], β-derivatives [9], or others [12, 20]. Recently, the new definition of a conformable fractional derivative, given by [1, 2, 18], has drawn much interest from many researchers [6, 7, 17, 22, 23, 24, 26]. Recent results on conformable fractional differential equations can also be found in [3, 8, 11].

In 2017, X. Dong et al.[15] studied the existence and multiplicity of positive solutions for the following conformable fractional differential equation with p-Laplacian operator

Dα(ϕp(Dαu(t)))=f(t,u(t)),0<t<1, u(0)=u(1)=Dαu(0)=Dαu(1)=0.

Here, 1<α2 is a real number, Dα is the conformable fractional derivative, ϕp(s)=|s|p2s, p>1, ϕp1=ϕq, 1/p+1/q=1, and f:[0,1]×[0,+)[0,+) is continuous. Using an approximation method and fixed point theorems on the cone, some existence results are established.

In [10], the authors considered the following three-point boundary value problem for a conformable fractional differential equation

Dα(D+λ)x(t)=f(t,x(t)),t[0,1], x(0)=0,x(0)=0,x(1)=βx(η).

Here Dα is the conformable fractional derivative of order α(1,2], D is the ordinary derivative, f:[0,1]× is a known continuous function, λ and β are real numbers, λ>0, and η(0,1). The authors proved their results using the classical Banach fixed point theorem and Krasnoselskii's fixed point theorem.

In [5], D. R. Anderson et al., considered the following conformable fractional-order boundary value problem with Sturm-Liouville boundary conditions

DβDαx(t)=f(t,x(t)),0t1, γx(0)δDαx(0)=0=ηx(1)+ζDαx(1).

Here α,β(0,1] and the derivatives are conformable fractional derivatives, with γ,δ,η,ζ0, and d=ηδ+γζ+γη/α>0. Employing a functional compression expansion fixed point theorem due to Avery, Henderson, and O'Regan, they proved the existence of a positive solution.

In a recent paper [16], using the well-known topological transversality theorem, L. He et al., obtained the existence of solutions for the fractional differential equation

Dαx(t)=f(t,x(t),Dα1x(t)),t[0,1],

with one of the following boundary value conditions

x(0)=A,Dα1x(1)=B;orDα1x(0)=A,x(1)=B,

where α(1,2] is a real number, Dαx(t) is the conformable fractional order derivative of a function x(t), and f:[0,1]×2 is a continuous function. The existence results of solutions to the problem are obtained under f satisfying some sign conditions.

In the same year, Q. Song et al. [21] investigated the following fractional Dirichlet boundary value problem

Dαx(t)=f(t,x(t),Dα1x(t)),t[0,1], x(0)=A, x(1)=B,

where 1<α2, Dαx(t) is the conformable fractional derivative, and f:[0,1]×2 is a continuous function. The existence results of solutions to the problem are obtained under certain sign conditions on the nonlinearity f.

Very recently, in 2018, W. Zhong and L. Wang [25] discussed the existence of positive solutions of the conformable fractional differential equation

Dαx(t)+f(t,x(t))=0,t[0,1],

subject to the boundary conditions

x(0)=0,x(1)=λ01x(t)dt,

where the order α belongs to (1,2], Dαx(t) denotes the conformable fractional derivative of a function x(t) of order α, and f:[0,1]×[0,)[0,) is a continuous function. Employing a fixed point theorem in a cone, they established some criteria for the existence of at least one positive solution.

Inspired and motivated by the above recent works, we intend in the present paper to study the existence of positive solutions for the boundary value problem of conformable fractional differential equation

Dαx(t)+f(t,x(t))=0,t[0,1], x(0)=0,x(1)=λ0ηx(t)dt,

where Dαx(t) denotes the conformable fractional derivative of x at t of order α, α(1,2], η(0,1], fC([0,1]×[0,),[0,)), and the parameter λ is a positive constant.

For the case of η=1, the problem (1.1) and (1.2) reduces to the problem studied by Zhong and Wang in [25]. Our approach is similar to that used in [25], i.e., fixed point theorem in a cone, lower and upper bounds for Green's function are employed as the main tool of analysis. It should be noticed that our results seem more natural than those in [25], and in this case, the results in [25] are special cases of those in this paper. Our work extends and complements the results in [25]. It is worth pointing out that the obtained Green's function in this work is singular at s=0.

In Section 2, we present the necessary definitions and we give some lemmas in order to prove our main results. In particular, we state some properties of Green's function associated with BVP (1.1) and (1.2). In Section 3, some sufficient conditions are established for the existence of positive solution to our BVP when f is superlinear or sublinear. Finally, two examples are also included to illustrate the main results.

### 2. Preliminaries

In this section, we give some definitions and results concerning conformable fractional derivative which can be found in recent literature, see [1, 15, 18].

### Definition 2.1.

([1, 18]) Let α(n,n+1] and f be n-differentiable function at t>0. Then the fractional conformable derivative of order α at t>0 is given by

Dαf(t)=limϵ0f(n)(t+ϵtn+1α)f(n)(t)ϵ,

provided the limits of the right side exists.

If f is α-order differentiable on (0, a), a>0, and limt0+Dαf(t) exists, then define

Dαf(0)=limt0+Dαf(t).

### Definition 2.2.

([1, 18]) Let α(n,n+1] and set β=αn. Then the fractional derivative of a function f:[0,) of order α, where f(n)(t) exists, is defined by

Dαf(t)=Dβf(n)(t).

### Lemma 2.1.

([15, 18]) Let α(n,n+1] and t>0. The function f(t) is (n+1)-differentiable if and only if f is α-differentiable, moreover, Dαf(t)=tn+1αf(n+1)(t).

### Definition 2.3.

([1]) Let α be in (n,n+1]. The fractional integral of a function f:[0,) of order α is defined by

Iαf(t)=1n!0t (ts)nsαn1f(s)ds.

### Lemma 2.2.

([1, 15, 18]) Let α be in (n,n+1]. If f is a continuous function on [0,), then, for all t>0, DαIαf(t)=f(t).

### Lemma 2.3.

([15]) Let α(n,n+1], f be a α-differentiable function at t>0, then Dαf(t)=0 for t(0,) if and only if f(t)=a0+a1t+...+an1tn1+antn, where ak, for k =0,1,...,n.}

### Lemma 2.4.

([1, 15]) Let α be in (n,n+1]. If Dαf(t) is continuous on [0,), then IαDαf(t)=f(t)+c0+c1t+...+cntn for some real numbers ck , k = 0,1,...,n.

In order to study the boundary value problem (1.1-1.2), we consider first the linear equation

Dαx(t)+h(t)=0,t[0,1],

where α(1,2] and hC([0,1]).

### Lemma 2.5.

If λη22, then the unique solution of (2.1) subject to the boundary conditions (1.2) is given by

x(t)=01K(t,s)h(s)ds,

where

K(t,s)=G(t,s)+λt2λη2H(η,s), G(t,s)=(1t)sα1,0st1;t(1s)sα2,0ts1,

and

H(t,s)=(2tt2s)sα1,0st1;t2(1s)sα2,0ts1.

Proof. From Lemma 2.4, we may reduce (2.1) to an equivalent integral equation,

x(t)=Iαh(t)+c0+c1t  =0 t (ts)sα2h(s)ds+c0+c1t,

for some c0,c1. By 1.2, we get c0=0 and c1=Iαh(1)+x(1). Hence

x(t)=Iαh(t)+tIαh(1)+tx(1)=0t (ts)sα2h(s)ds+t01 (1s)sα2h(s)ds+tx(1)=0t (ts)sα2h(s)ds+t0t (1s)sα2h(s)ds+tt1 (1s)sα2h(s)ds+tx(1)=0t (1t)sα1h(s)ds+t1t(1s)sα2h(s)ds+tx(1).

So

x(t)=01G(t,s)h(s)ds+tx(1).

Moreover, in checking the second boundary condition, we get

x(1)=λ0ηx(t)dt=λ0η[Iαh(t)+tIαh(1)+tx(1)]dt=λ0η(0t (ts)sα2h(s)ds)dt+λη2 2Iαh(1)+λη2 2x(1)=λ20η (ηs) 2 sα2h(s)ds+λη2 2Iαh(1)+λη2 2x(1),

which implies

x(1)=λ2λη20η (ηs) 2 sα2h(s)ds+λη22λη2Iαh(1).

Substituting the value of x(1) in (2.5), we get

x(t)=01G(t,s)h(s)dsλt2λη20η(ηs)2sα2h(s)ds+λη2t2λη201(1s)sα2h(s)ds=01G(t,s)h(s)dsλt2λη20η(ηs)2sα2h(s)ds+λη2t2λη20η(1s)sα2h(s)ds+λη2t2λη2η1(1s)sα2h(s)ds=01G(t,s)h(s)ds+λt2λη20ηsα2[η2(1s)(ηs)2]h(s)ds+λt2λη2η1η2(1s)sα2h(s)ds=01G(t,s)h(s)ds+λt2λη20ηsα1(2ηη2s)h(s)ds+λt2λη2η1η2(1s)sα2h(s)ds=01G(t,s)h(s)ds+λt2λη201H(η,s)h(s)ds.

The proof is therefore complete.

We point out here that (2.3)-(2.4) become the usual Green's function when α=2.

### Lemma 2.6.

Let θ(0,12) be fixed. For G(t,s) and H(t,s) given in (2.3-2.4,) we have the following bounds.

• (i) θ2G(s,s)G(t,s)G(s,s), for all (t,s)[θ,1θ]×(0,1];

• (ii) ρ(t)G(s,s)H(t,s)G(s,s), for all (t,s)(0,1]×(0,1], where G(s,s)=(1s)sα1, and ρ(t)=min{t2,t(1t)}=t2,t12;t(1t),t12.

• (iii) θ2G(s,s)H(t,s)G(s,s), for all (t,s)[θ,1θ]×(0,1].

Proof. (i) From Lemma 2.5 in [25], we have

t(1t)G(s,s)G(t,s)G(s,s),(t,s)(0,1]×(0,1].

Therefore, if θ(0,12), then G(t,s) satisfies

θ2G(s,s)G(t,s)G(s,s),(t,s)[θ,1θ]×(0,1].

(ii) If st, then from (2.4) we have

H(t,s)=(2tt2s)sα1=[(t22t)s]sα1  =([(t1)21]s)sα1=[(1s)(1t)2]sα1  (1s)sα1.

On the other hand, we have

H(t,s)=(2tt2s)sα1  =[t(1t)+(ts)]sα1t(1t)sα1(1s)sα1t(1t).

If ts, from (2.4), we have

H(t,s)=t2(1s)sα2t(1s)sα2=ts(1s)sα1(1s)sα1,

and

H(t,s)=t2(1s)sα2t2(1s)sα2s=t2(1s)sα1.

From (2.6), (2.7), (2.8) and (2.9), we have

ρ(t)(1s)sα1H(t,s)(1s)sα1,forall(t,s)(0,1]×(0,1].

(iii) It follows immediately from (ii).

### Lemma 2.7.

Let θ(0,12) be fixed and 0λ<2/η2. If h(t)C([0,1],[0,)), then the unique solution of (2.1) subject to the boundary conditions (1.2) is nonnegative and satisfies

mint[θ,1θ]x(t)θ2x.

Proof. From Lemma 2.5 and Lemma 2.6, x(t) is nonnegative for t[0,1], and we get

x(t)=01K(t,s)h(s)ds  =01G(t,s)h(s)ds+λt2λη2 01H(η,s)h(s)ds  01G(s,s)h(s)ds+λ2λη2 01H(η,s)h(s)ds.

Then

x01G(s,s)h(s)ds+λ2λη2 01H(η,s)h(s)ds.

On the other hand, from (2.10) and Lemma 2.6 for any t[θ,1θ], we have

x(t)=01G(t,s)h(s)ds+λt2λη2 01H(η,s)h(s)ds  θ201G(s,s)h(s)ds+λt2 2λη2 01H(η,s)h(s)ds  θ201G(s,s)h(s)ds+λθ2 2λη2 01H(η,s)h(s)ds  =θ201G(s,s)h(s)ds+λ2λη2 01H(η,s)h(s)ds  θ2x.

From (2.11), we obtain

mint[θ,1θ]x(t)θ2x.

In order to prove our main results, the following well known fixed point theorem is needed in the forthcoming analysis [4, 13, 19].

### Lemma 2.8.

Let B be a Banach space, and let PB, be a cone, and Ω1, Ω2 two bounded open balls of B centered at the origin with Ω¯1Ω2. Assume that A:P(Ω¯2\Ω1)P is a completely continuous operator such that

• (C1) Axx, xPΩ1

• (C2) There exists φP\{0} such that xAx+λφ for xPΩ2 and λ>0.

Then A has a fixed point in P(Ω¯2\Ω1). The same conclusion remains valid if (C1) holds on PΩ2 and (C2) holds on PΩ1.

### 3. Existence Results

Throughout this section, we assume that

• (H) fC([0,1]×[0,),[0,)), and the parameter λ[0,2η2).

Let E=C([0,1],) be the Banach space endowed with the sup norm

x=supt[0,1]|x(t)|.

Let θ(0,12), define the cone P in E by

P=xE:x(t)0,t[0,1],mint[θ,1θ]x(t)θ2x.

Given a positive number r, define the subset Ωr of E by

Ωr={xE:x<r},

and also, define the operator A:EE by

(Ax)(t)=01K(t,s)f(s,x(s))ds.

### Lemma 3.1.

If the hypothesis (H) holds, then A(P)P.

Proof. By (3.1) and Lemma 2.7, we have A(P)P.

In order to discuss the complete continuity of the operator A, denote the operator A by

A=A1+A2,

where the operators A1 and A2 are defined, respectively by

(A1x)(t)=01G(t,s)f(s,x(s))ds,

and

(A2x)(t)=λt2λη201H(η,s)f(s,x(s))ds.

By Lemma 2.7, it follows that A1(P)P, and the complete continuity of the operator A1 was verified in [14, 15]. Also, due to Lemma 2.7, we have the invariance property A2(P)P. Furthermore, the kernel λt2λη2H(η,s) of A2 is continuous on [0,1]×[0,1], and using a standard argument, we can easily check that the operator A2 is also completely continuous. Thus, we get the following lemma:

### Lemma 3.2.

If the hypothesis (H) holds, then the operator A:PP is completely continuous.

The following lemma transforms the boundary value problem (1.1) and (1.2) into an equivalent fixed point problem.

### Lemma 3.3.

If the hypothesis (H) holds, then the problem of nonnegative solutions of (1.1) and (1.2) is equivalent to the fixed point problem x=Ax, xP.

Proof. It follows easily by using the same argument as for the proof of [25, Lemma 3.3].

For convenience, we introduce the following notations

f0=limx0+mint[0,1]f(t,x)x,f=limx+maxt[0,1]f(t,x)x,f0=limx0+maxt[0,1]f(t,x)x,f=limx+mint[0,1]f(t,x)x,Λ1=(θ4θ1θ(G(s,s)+λ2λη2H(η,s))ds)1,Λ2=((1+λ2λη2)01G(s,s)ds)1.

Now, we will state and prove our main results.

### Theorem 3.1.

Assume that the hypothesis (H) holds. If f0>Λ1 and f<Λ22, then the problem (1.1) and (1.2) has at least one positive solution.

Proof. By Lemma 3.2, we get that the operator A:PP is completely continuous.

Since f0>Λ1, there exists ρ1>0 such that f(t,x)Λ1x, for 0<xρ1 and t[0,1]. Thus

f(t,x(t))Λ1x(t)fort[0,1]andxPΩρ1.

By choosing φ1, it is obvious that φP\{0}. Now, we show that for the specified φ, the condition (C2) in Lemma 2.8 is verified. Assume that there exist a function x0PΩρ1 and a positive number λ0 such that

x0=Ax0+λ0φ.

Then, by Lemma 2.6 and Lemma 2.7, for each t[θ,1θ], we have

x0(t)=01K(t,s)f(s,x0(s))ds+λ0  =01G(t,s)f(s,x0(s))ds+λt2λη2 01H(η,s)f(s,x0(s))ds+λ0  θ2θ1θG(s,s)Λ1x0(s)ds+λθ2 2λη2 θ1θH(η,s)Λ1x0(s)ds+λ0  θ2θ1θG(s,s)Λ1θ2x0ds+λθ2 2λη2 θ1θH(η,s)Λ1θ2x0ds+λ0  =x0Λ1(θ4θ1θ(G(s,s)+λ2λη2 H(η,s))ds)+λ0  =x0+λ0.

Thus, x0x0+λ0. This is a contradiction. Hence, the operator A satisfies the condition (C2) in Lemma 2.8.

We next show that the operator A satisfies the condition (C1) in Lemma 2.8. The fact that f<Λ22 says us that there exists a constant γ1>0 such that

f(t,x)Λ22xfort[0,1]andxγ1.

Define now

γ2=max{f(t,x):0t1,0xγ1}.

So, by virtue of (3.4), we get

f(t,x)Λ22x+γ2fort[0,1]andx0.

Set ρ2=max{2ρ1,2γ2Λ21} and xPΩρ2. Then, by Lemma 2.6 and (3.5), we obtain

Ax=maxt[0,1]01K(t,s)f(s,x(s))ds=maxt[0,1]{01G(t,s)f(s,x(s))ds+λt2λη2 01H(η,s)f(s,x(s))ds}01G(s,s)(Λ2 2x(s)+γ2)ds+λ2λη2 01G(s,s)(Λ2 2x(s)+γ2)ds(Λ2 2x+γ2)(1+λ2λη2 )01G(s,s)ds=x2+γ2Λ21x2+x2=x.

Hence, the condition (C1) in Lemma 2.8 is satisfied. By Lemma 2.8 and Lemma 3.3, the operator A has at least one fixed point xP(Ω¯ρ2\Ωρ1), which is a positive solution of the boundary value problem (1.1) and (1.2). The proof is complete.

### Theorem 3.2.

Assume that the hypothesis (H) holds. If f0<Λ2 and f>Λ1, then the problem (1.1) and (1.2) has at least one positive solution.

Proof. We first note that, in virtue of Lemma 3.2, the operator A is completely continuous.

Since f0<Λ2 and f>Λ1, there exist two positive numbers ρ1>0 and γ1>0 such that

f(t,x)Λ2x,fort[0,1]and0<xρ1, f(t,x)Λ1x,fort[0,1]andxγ1.

By (3.6) and Lemma 2.6, for xPΩρ1, we get

Ax=maxt[0,1]01K(t,s)f(s,x(s))dsΛ2(1+λ2λη2 )01G(s,s)x(s)dsΛ2x(1+λ2λη2 )01G(s,s)dsx.

Thus the operator A satisfies the condition (C1) in Lemma 2.8.

Now, we show that the operator A also satisfies the condition (C2) in Lemma 2.8.

Let ρ2=max{2ρ1,γ1θ2}, then by Lemma 2.7, for xPΩρ2, we have

x(t)θ2ρ2γ1,fort[θ,1θ].

Hence, by (3.7), we have

f(t,x(t))Λ1x(t),fort[θ,1θ]andxPΩρ2.

We now choose the function φ1, and clearly, φP\{0}. We then show that

xAx+λφ,forxPΩρ2andλ>0.

If the above fact is not true, then there exist a function x0PΩρ2 and a positive number λ0 such that

x0=Ax0+λ0φ.

Then, by Lemma 2.6 and Lemma 2.7, for each t[θ,1θ], we have

x0(t)=01K(t,s)f(s,x0(s))ds+λ0  θ2θ1θG(s,s)Λ1x0(s)ds+λθ2 2λη2 θ1θH(η,s)Λ1x0(s)ds+λ0  x0Λ1(θ4θ1θ(G(s,s)+λ2λη2 H(η,s))ds)+λ0  =x0+λ0.

Thus, x0x0+λ0. This is a contradiction. Hence, the operator A satisfies the condition (C2) in Lemma 2.8.

By Lemma 2.8 and Lemma 3.3, the operator A has at least one fixed point xP(Ω¯ρ2Ωρ1), which is a positive solution of the boundary value problem (1.1) and (1.2). The proof is complete.

From Theorem 3.1 and Theorem 3.2, we can obtain the following corollary.

### Corollary 3.1.

Suppose that the hypothesis (H) holds. If f0= and f=0 or if f0=0 and f=, then the boundary value problem (1.1) and (1.2) has at least one positive solution.

### Example 4.1.

Consider the following boundary value problem

Dαx(t)+t+ex=0,t[0,1], x(0)=0,x(1)=2013x(t)dt,

where α(1,2], λ=2, η=13, and f(t,x)=t+exC([0,),[0,)), so λη2=29<2. We have

f0=limx0+exx=,f=limx1+exx=0.

Thus, by Corollary 3.1, the fractional boundary value problem (4.1)-(4.2) has at least one positive solution.

### Example 4.2.

As a second example, we consider the fractional boundary value problem

D32x(t)+t+4xe2x/5e2x+ex999500=0,t[0,1], x(0)=0,x(1)=85012x(t)dt,

where α=32, λ=85, η=12, and f(t,x)=t+4xe2x/5e2x+ex999500C([0,),[0,)), so λη2=25<2. We have

f0=limx0+mint[0,1]f(t,x)x=limx0+4e2x/5e2x+ex999500=400,f=limx+maxt[0,1]f(t,x)x=limx+(1x+4e2x/5e2x+ex999500)=45.

By simple calculations, we find that

Λ21=(1+λ2λη2)01G(s,s)ds=2α(α+1)=815.

Hence, we get

Λ2=158>2f=85.

Λ11=θ4θ1θ(G(s,s)+H(12,s))ds  =θ4(θ1θ(1s)s12ds+θ12H(12,s)ds+121θH(12,s)ds)  =θ4(45θ2θ76θθ+12(1θ)1θ25(1θ)21θ  +121θ4152)  =θ430((24θ35)θθ+3(6+3θ4θ2)1θ42).

Using Mathematica software, we easily check that Λ1<400=f0, for all θ[1950,2150]. Therefore, all conditions of Theorem 3.1 are fulfilled. Hence, the problem (4.3)-(4.4) has at least one positive solution.

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