Research article

Existence of solutions to a class of nonlinear boundary value problems with right and left fractional derivarives

  • Received: 24 December 2019 Accepted: 11 May 2020 Published: 01 June 2020
  • MSC : 34A08, 34B15

  • We discuss the existence of solutions for a boundary value problem involving both left Riemann-Liouville and right Caputo types derivatives. For this, we convert the posed problem to a sum of two integral operators then we apply Krasnoselskii's fixed point theorem to conclude the existence of nontrivial solutions.

    Citation: Hamza Moffek, Assia Guezane-Lakoud. Existence of solutions to a class of nonlinear boundary value problems with right and left fractional derivarives[J]. AIMS Mathematics, 2020, 5(5): 4770-4780. doi: 10.3934/math.2020305

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  • We discuss the existence of solutions for a boundary value problem involving both left Riemann-Liouville and right Caputo types derivatives. For this, we convert the posed problem to a sum of two integral operators then we apply Krasnoselskii's fixed point theorem to conclude the existence of nontrivial solutions.


    Differential equations of noninteger order can represent the dynamics of various memory systems and arise from a variety of applications, including several fields of science and engineering such as geology, physics, optics, chemistry, biology, economics, signal and image processing, ... Although the literature on fractional differential equations is now vast, more studies are needed. Recently, the investigation of the qualitative properties of solutions to fractional initial and boundary value problems has attracted the attention of many authors [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17], and different tools are used in these researches, such as the method of upper and lower solutions, the variational method, the coincidence degree theory, the fixed point theorems ...

    The aim of this work is the study of the existence of solutions, for the following nonlinear boundary value problem (P) involving both the right Caputo and the left Riemann-Liouville fractional derivatives:

    CDα1Dβ0+u(t)+ω2u(t)+f(t,u(t))=0, tJ=[0,1]. (1.1)
    u(0)=0,Dβ0+u(1)=0, (1.2)

    where 0<α,β<1, α+β>1, ωR, CDα1 and Dβ0+ denote respectively the right Caputo derivative and the left Riemann Liouville derivative, u is the unknown function and f:J×RR is a Caratheodory function. Let us mention that if α and β tend to one, then problem (P) is a classical oscillator boundary value problem that is investigated in [2]. Note that problem (P) is studied in [13] by lower and upper solutions method, the authors proved the existence of solution under some specific conditions on the nonlinear term f. In the present study, we prove the existence of solution for problem (P) under Lipschitz type condition on the nonlinear term f and by using Krasnoselskii's fixed point theorem.

    Different methods are used in the study of differential equations involving mixed type fractional derivatives. By the help of operational method and the successive approximations, some linear differential equations containing left and right fractional derivatives that may appear in fractional variational calculus, are studied in [8,9].

    Recently, the method of upper and lower solutions is applied in [13,15,16] to solve nonlinear differential equations containing mixed fractional derivatives.

    In [3], the authors considered a coupled system of nonlinear differential equations involving mixed type fractional derivatives

    CDα1Dβ0+x(t)=f(t,x(t),y(t))=0,CDp1Dq0+y(t)=g(t,x(t),y(t))=0, 0<t<1,

    with nonlocal boundary conditions

    x(0)=x(0)=0,  x(1)=γy(η),0<η<1,y(0)=y(0)=0,  y(1)=δx(θ),0<θ<1,

    here 1<α,p<2, 0<β,p<1, γ,δR. The existence and uniqueness of solution is proved by the help of Leray-Schauder alternative and Banach fixed point theorem.

    By Krasnoselskii's fixed point theorem, the authors in [12,14], investigated some boundary value problems involving mixed type fractional derivatives. In particular in [12], proved, under Lipschitz type condition on the nonlinear term, the existence of solution in a weighted space, for the following boundary value problem

    CDα1Dβ0+u(t)+f(t,u(t))=0, 0<t<1,limt0+t1βu(t)=u(1)=u(η)

    where 0<α,β<1, 1<α+β<2.

    In [14], the authors studied by the help of Krasnoselskii's fixed point theorem and Arzela Ascoli's theorem, the existence of solution for the problem

    CDα1Dβ0+u(t)+f(t,u(t))=0,0<t<1
    u(0)=u(0)=u(1)=0

    where 0<α1, 1<β2, CDα1 denotes right Caputo derivative, Dβ0+ denotes the left Riemann–Liouville and f: [0,1]×RR satisfies Lipschitz type condition.

    Motivated by the above papers, we study the existence of solutions for problem (P). For this, we convert the problem (P) into an integral equation which we write as a sum of two integral operators, including a contraction and a completely continuous operator, then we apply Riesz compactness criteria and Krasnoselskii fixed point theorem to prove the existence of solution.

    We give some background on fractional calculus that can be found in [18,19,20]. Let g be a real function defined on [0,1] and α>0. Then the left and right Riemann-Liouville fractional integrals of order α of g are defined respectively by

    Iα0+g(t)=1Γ(α)t0g(s)(ts)1αds,
    Iα1g(t)=1Γ(α)1tg(s)(st)1αds.

    The left Riemann-Liouville fractional derivative of order α>0, of a function g is

    Dα0+g(t)=dndtn(Inα0+g(t))

    where n=[α]+1.

    The left and right Caputo fractional derivatives of order α>0 of a function g are defined respectively as

    CDα0+g(t)=Inα0+(dndtng(t)),
    CDα1g(t)=(1)nInα1(dndtng(t)).

    Proposition 2.1. Let fCn([0,1]). Then

    Iα0+CDα0+f(t)=f(t)n1k=0f(k)(0)k!tk,
    Iα1CDα1f(t)=f(t)n1k=0(1)kf(k)(1)k!(1t)k.

    Theorem 2.1 (Riesz compactness criteria). [11]. Let F be a bounded set in Lp(0,1), 1p<. Assume that

    (i) limh0τhffLp=0 uniformly on F, where τhf(t)=f(t+h).

    (ii)limε011ε|f(t)|pdt=0 uniformly on F. Then F is relatively compact in Lp(0,1).

    Theorem 2.2 (Krasnoselskii fixed point Theorem). [21]. Let Ω be a closed bounded convex nonempty subset of a Banach space E. Suppose that A and B map Ω into

    E such that

    (i) A is completely continuous,

    (ii) B is a contraction mapping,

    (iii) x,yM implies Ax+ByΩ. Then there exists zΩ with z=Az+Bz.

    To study the nonlinear problem (P), we consider first, the associated linear problem

    CDα1Dβ0+u(t)+y(t)=0, 0<t<1, (3.1)
    u(0)=0,Dβ0+u(1)=0.

    Lemma 3.1. Assume that yLp(0,1), p>1, then u is a solution for the linear boundary value problem (3.1)–(1.2) if and only if u satisfies the integral equation

    u(t)=10G(t,τ)y(τ)dτ,

    where

    G(t,τ)=1Γ(α)Γ(β){τ0(ts)β1(τs)α1ds,  0τt1,t0(ts)β1(τs)α1ds,  0tτ1.

    Proof. Applying the right-hand side fractional integral Iα1 to equation (3.1), we get

    Dβ0+u(t)=Iα1y(t)+a,  aR

    The boundary condition Dβ0+u(1)=0, gives a=0, then applying the fractional integral Iβ0+ to the obtained equation, it yields

    u(t)=Iβ0+Iα1y(t)+ctβ1,  cR (3.2)

    Multiplying the equation (3.2) by t1β, then using the condition u(0)=0, we obtain c=0, thus

    u(t)=Iβ0+Iα1y(t)=1Γ(α)Γ(β)t0(ts)β11s(τs)α1y(τ)dτds.

    Finally, by Fubini theorem, we get

    u(t)=1Γ(α)Γ(β)t0τ0(ts)β1(τs)α1y(τ)dsdτ+1Γ(α)Γ(β)1tt0(ts)β1(τs)α1y(τ)dsdτ.

    Lemma 3.2. The function G is continuous, nonnegative and

    G(t,τ)1(α+β1)Γ(α)Γ(β),for all  t,τJ.

    Remark 3.1. Let us mention the case α+β1+. Since α+β>1 and 0<α,β<1, then α>12 or β>12. If α>12, then α+β1+implies (α1 and β0) or (α12+ and β12), then the problem (P) is reduced respectively to

    u(t)+ω2u(t)+f(t,u(t))=0, tJ=[0,1].(P1)u(0)=0

    and

    CD121D120+u(t)+ω2u(t)+f(t,u(t))=0, tJ=[0,1].(P2)u(0)=0,D120+u(1)=0,

    For problem (P2), let us fix α=1/2, then we have,

    G(1,1)=1Γ(α)Γ(β)10(1s)β+α2ds=1Γ(α)Γ(β)(β+α1)=1Γ(1/2)Γ(β)10(1s)β32ds=1Γ(1/2)Γ(β)(β12),as β12

    thus the Green function is not bounded.

    Lemma 3.3. The function uLp(0,1) is a solution of the integral equation

    u(t)=10G(t,τ)f(τ,u(τ))dτ+ω210G(t,τ)u(τ)dτ.

    if and only if u is a solution of the fractional boundary value problem (P).

    Now we define the operators A and B on Lp(0,1) as

    Au(t)=10G(t,τ)f(τ,u(τ))dτ,Bu(t)=ω210G(t,τ)u(τ)dτ.

    Obviously, the problem (P) has a solution if and only if the operator A+B has a fixed point in Lp(0,1). Before stating and proving the main results, we introduce the following hypotheses.

    (H1) M=sup0t1|f(t,0)|<, and there exists a constant k, 0<k(α+β1)Γ(α)Γ(β)12, such that

    |f(t,u)f(t,v)|k|uv|,0t1,u,vR.

    (H2) ω2(α+β1)Γ(α)Γ(β)<12.

    Theorem 3.1. Assume that (H1)–(H2) hold, then the fractional boundary value problem (P) has a nontrivial solution in Lp(0,1).

    To prove Theorem 3.1, we need the following lemmas.

    Lemma 3.4. Under the hypotheses (H1)–(H2), the operator A is completely continuous on Lp(0,1).

    Proof. Let

    Ω={uLp(0,1),uLpR}

    such that

    RM(α+β1)Γ(α)Γ(β)(k+ω2) (3.3)

    Clearly, Ω is a nonempty, bounded and convex subset of the Banach space Lp(0,1). We should prove that A is continuous and relatively compact on Lp(0,1).

    Claim 1. The mapping A is continuous on Ω. In fact, consider the sequence (un)nΩ, such that unu in Lp(0,1), then from Lemma 3.2, hypothesis (H1) and Hölder inequality, we get

    |Aun(t)Au(t)|10G(t,τ)|f(τ,un(τ))f(τ,u(τ))|dτk(α+β1)Γ(α)Γ(β)10|un(τ)u(τ)|dτk(α+β1)Γ(α)Γ(β)un(.)u(.)Lp(0,1).

    Hence

    AunAuLp(0,1)k(α+β1)Γ(α)Γ(β)un(.)u(.)Lp(0,1)0,as n.

    Claim 2. (Au) is bounded in Lp(0,1). Indeed, let uΩ, then by condition (H1) and Hölder inequality, it yields

    |Au(t)|1(α+β1)Γ(α)Γ(β)10|f(τ,u(τ))|dτ1(α+β1)Γ(α)Γ(β)(k(10|u(τ)|dτ)+10|f(τ,0)|dτ)1(α+β1)Γ(α)Γ(β)(k(10|u(τ)|pdτ)1p+10|f(τ,0)|dτ)kR+M(α+β1)Γ(α)Γ(β),

    thus

    AuLpkR+M(α+β1)Γ(α)Γ(β).

    Claim 3. (Au) is relatively compact. In fact, let uΩ, and p>1, we have

    |Au(t+h)Au(t)|10|G(t+h,τ)G(t,τ)||f(τ,u(τ))|dτ10|G(t+h,τ)G(t,τ)|(k|u(τ)|+|f(τ,0)|)dτ(kR+M)(10|G(t+h,τ)G(t,τ)|pdτ)1p
    (kR+M)(t0|G(t+h,τ)G(t,τ)|pdτ+t+ht|G(t+h,τ)G(t,τ)|pdτ+1t+h|G(t+h,τ)G(t,τ)|pdτ)1pkR+MΓ(α)Γ(β)(t0(τ0((ts)β1(t+hs)β1)(τs)α1ds)pdτ+1t(t0((ts)β1(t+hs)β1)(τs)α1ds)pdτ+t+ht(τt(t+hs)β1ds)pdτ)1p=(kR+M)Γ(α)Γ(β)(I1+I2+I3)1p,

    hence

    |Au(t+h)Au(t)|(kR+M)Γ(α)Γ(β)(I1+I2+I3)1p. (3.4)

    Let us calculate Ii, i=1,2,3.

    I1=t0(τ0((ts)β1(t+hs)β1)(τs)α1ds)pdτ(h(1β))pt0(τ0(τs)α1ds)pdτ(h(1β)α(α+1))p.
    I2=1t(t0((ts)β1(t+hs)β1)(τs)α1ds)pdτ(h(1β))pt0((1s)α(ts)α)pds(h(1β))pαp+1.
    I3=t+ht(τt(t+hs)β1ds)pdτ1βpt+ht(hβ(t+hτ)β)pdτhβp+1βp.

    Finally, we get

    Au(.+h)Au(.)Lp(kR+M)Γ(α)Γ(β)((h(1β)α(α+1))p+(h(1β))pαp+1+hβp+1βp)1p (3.5)

    By taking the limit in (3.5) as h0, we obtain that Au(.+h)Au(.)Lp0 for any uΩ. On the other hand we have by the help of claim 2

    11ε|Au(t)|pdtε(kR+M(α+β1)Γ(α)Γ(β))p0 as ε0.

    By Theorem 2.1, we conclude that A is relatively compact on Ω. From the above discussion we conclude that A completely continuous on Lp(0,1).

    Lemma 3.5. Under the hypothesis (H2), the mapping B is a contraction on Ω.

    Proof. Let uΩ and tJ, we have

    |Bu(t)|ω210G(t,τ)|u(τ)|dτω2R(α+β1)Γ(α)Γ(β)<R2.

    thus BuLp(0,1)<R2, and consequently B(Ω)Ω. Now for u,vΩ and tJ, we have

    |Bu(t)Bv(t)|ω210G(t,τ)|u(τ)v(τ)|dτω2(α+β1)Γ(α)Γ(β)uvLp,

    hence

    BuBvLpω2(α+β1)Γ(α)Γ(β)uvLp,

    by hypothesis (H2), we conclude that B is a contraction.

    Lemma 3.6. Assume that hypotheses (H1) and (H2) hold, then Au+BvΩ for all u,vΩ.

    Proof. Let u,vΩ, then taking (3.3) into account, it yields

    Au+BvLpAuLp+BvLpR(ω2+k)+M(α+β1)Γ(α)Γ(β)R,

    hence Au+BvΩ.

    Proof of Theorem 3.1. By Lemmas 3.4, 3.5 and 3.6, we conclude respectively that the mapping A is completely continuous, the mapping B is a contraction and Au+BvΩ for all u,vΩ, then all hypotheses of Theorem 2.2 are satisfied. Hence, there exists a nontrivial solution uΩ for problem (P) such that u=Au+Bu. The proof is complete.

    Now, we give an example to illustrate the usefulness of the obtained results.

    Example 1.Consider the problem (P) with

    f(t,x)=etx9+et(1+x2)+et,(t,x)J×R,ω=0.5, α=0.5, β=0.8,M=sup0t1|f(t,0)|=e=2.7183.

    Let us check hypotheses (H1)-(H2). We have for all (t,x)J×R

    |f(t,x)f(t,y)|et9+et|xy|110|xy|,

    then k=110, 0<k=0.112(α+β1)Γ(α)Γ(β)=0.30953. By Theorem 3.1, we conclude that the problem (P) has a nontrivial solution uLp(0,1), such that uLpR, where R10.103 and u=Au+Bu.

    Example 2.Consider the problem (P) with

    f(t,x)=t13sinx+t315,(t,x)J×R,ω=110, α=13, β=34,M=sup0t1|f(t,0)|=115

    We have for all (t,x)J×R

    |f(t,x)f(t,y)|t1315|sin(x)sin(y)|115|xy|,

    and k=115, k(α+β1)Γ(α)Γ(β)=0.2436912, ω2(α+β1)Γ(α)Γ(β)=3.6554×102<12. Thus hypotheses (H1) and (H2) are satisfied.

    By Theorem 3.1, we conclude that the problem (P) has a nontrivial solution uLp(0,1), such that uLpR, where R=1M(α+β1)Γ(α)Γ(β)(k+ω2)=0.33858 and u=Au+Bu.

    In this article, we have proven the existence of non trivial solutions for a boundary value problem containing different type of fractional derivatives. The existence results are obtained via Krasnoselskii's fixed point theorem. For further investigations we propose to study similar problems with different types of fractional derivatives of higher order, by means of some fixed point theorems.

    The authors sincerely thank the reviewers for their valuable suggestions and useful comments that improved this paper.

    All authors declare no conflicts of interest.



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