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Research article

New results for nonlinear fractional jerk equations with resonant boundary value conditions

  • Received: 26 May 2020 Accepted: 30 June 2020 Published: 14 July 2020
  • MSC : 26A33, 34B15

  • A novel fractional-order jerk equation with resonant boundary value conditions is proposed. Using coincidence degree theory, we obtain the existence of solutions of nonlinear fractional jerk equation with two-point boundary conditions. This paper enriches some existing literatures. Finally, an example is given to demonstrate the effectiveness of our main result.

    Citation: Lei Hu, Cheng Wang, Shuqin Zhang. New results for nonlinear fractional jerk equations with resonant boundary value conditions[J]. AIMS Mathematics, 2020, 5(6): 5801-5812. doi: 10.3934/math.2020372

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  • A novel fractional-order jerk equation with resonant boundary value conditions is proposed. Using coincidence degree theory, we obtain the existence of solutions of nonlinear fractional jerk equation with two-point boundary conditions. This paper enriches some existing literatures. Finally, an example is given to demonstrate the effectiveness of our main result.


    It is well known that a three-dimensional dynamical system:

    {˙x(t)=y,˙y(t)=z,˙z(t)=f(x,y,z),

    can be transformed into a third-order differential equation of the form:

    x(t)=f(x,˙x,¨x).

    The above equation involves a third derivative of the variable x, which in a mechanical system is a rate of change of acceleration, it is called a jerk equation. Jerk equations are widely used in applied science and engineering, such as it can describe a three-order dynamic vibration model. In recent years, the discussion of jerk equations have attracted the attention of scholars and some results have been obtained. For recent publication on jerk equations, we refer the reader to see [1,2,3,4,5,6,7].

    In 2017, by using classical contraction mapping theorem, Elsonbaty and El-Sayed [1] obtained the existence and uniqueness of the solution of the system:

    {˙x1(t)=x2,˙x2(t)=x3,˙x3(t)=bx1cx2x3x31,

    where the dot denotes differentiation with respect to time, a and c are positive parameters, and cR.

    In [2], a second approximate solution of nonlinear jerk equation:

    {x(t)+f(x,˙x,¨x)=0,x(0)=0,˙x(0)=B,¨x(0)=0,

    was obtained by using modified harmonic balance method.

    Very recently, Liu and Chang [3] developed two iterative algorithms to determine the periodic solutions of the following nonlinear jerk equation:

    {x(t)=J(x,˙x,¨x),x(0)=x0,˙x(0)=˙x0,¨x(0)=¨x0,

    where the initial values x0,˙x0 and ¨x0 for the periodic solution of the above are unknown and the period is denoted by T>0.

    What is interesting to us is the appearance of the article [4]. Prakash and Singh presented a new fractional order jerk equations which do not have equilibrium point:

    {Dα0+x(t)=y(t),Dβ0+y(t)=z(t),Dγ0+z(t)=f(x,y,z),

    where f=y(t)+3y2(t)x2(t)x(t)z(t)+βW(x(t))y(t), α,β,γ are fractional order and 0<α,β,γ1. In addition, the authors successfully designed a fractional-order backstepping controller to stabilise the chaos in the system. This proposed system is novel in the sense that the jerk systems involving fractional order. It enlightens us to think the problem: how to observe the existence result of fractional jerk differential equation.

    Differential equations with fractional order arise from a variety of applications in various fields of science and engineering, such as electromagnetic, mechanics, biology, chemistry etc. (see [8]). Over last two decades, there has been significant progress in the area of fractional differential equations. Though many works have been done dealing with the existence of solutions to fractional differential equation (see [9,10,11,12,13,14,15,16,17,18,19,20,21]), no results on boundary value problem of fractional jerk equation are reported in the literature. To the best of the authors' knowledge, this is the first paper dealing with resonant boundary value problem of fractional jerk equations.

    In this paper, we are interested in the solutions to the following fractional jerk equations:

    {Dα0+u(t)=y(t),Dβ0+y(t)=z(t),Dγ0+z(t)=f(t,u(t),u(t),u(t)), (1.1)

    with boundary value conditions given as

    u(0)=u(ξ),Dα0+u(0)=(Dβ0+(Dα0+u))(0)=0, (1.2)

    where t(0,1), 0<α,β,γ1, 0<ξ1, Dα0+,Dβ0+,Dγ0+ denote the Caputo fractional derivatives and f:[0,1]×R3R is continuous.

    It is easy to check that the following fractional jerk equation:

    {(Dγ0+(Dβ0+(Dα0+u)))(t)=f(t,u(t),u(t),u(t)),u(0)=u(ξ),Dα0+u(0)=(Dβ0+(Dα0+u))(0)=0, (1.3)

    is equivalent to (1.1). Due to conditions (1.2), Eq (1.3) happens to be at resonance in the sense that the associated linear homogeneous equation:

    {(Dγ0+(Dβ0+(Dα0+u)))(t)=0,u(0)=u(ξ),Dα0+u(0)=(Dβ0+(Dα0+u))(0)=0,

    has u(t)=c,cR as nontrivial solutions.

    In addition, we can see if α=β=γ=1, the Eq (1.3) can be rewritten as

    {u(t)=f(t,u,˙u,¨u),u(0)=u(ξ),˙u(0)=¨u(0)=0,

    which is a standard nonlinear jerk equation. Furthermore, if ξ1, then the above equation is a nonlocal initial value problems of nonlinear jerk equation; if ξ=1, then it is a nonlinear jerk equations with periodic boundary conditions. If u(t),t[0,1] satisfies u(0)=u(1), then boundary value condition u(0)=u(1) is called periodic boundary conditions. According to different values of ξ,α,β,γ, we will get diversiform types of boundary value conditions. Thus, this paper enriches and extends the existing literatures, such as [4,5,6,7].

    The remaining part of this article is organized as follows. In Section 2, we will state some necessary notations, definitions and lemmas. In Section 3, by applying the coincidence degree theory due to Mawhin [22], we obtain the existence of solutions of (1.3). In the last section, an example is given to illustrate our results.

    We present the necessary definitions and lemmas from fractional calculus theory that will be used to prove our main theorems.

    Definition 2.1 ([8]). The Riemann-Liouville fractional integral of order α>0 of a function f:(0,)R is given by

    Iα0+f(t)=1Γ(α)t0(ts)α1f(s)ds,

    provided that the right-hand side is pointwise defined on (0,).

    Definition 2.2 ([8]). The Caputo fractional derivative of order α>0 of a continuous function f:(0,)R is given by

    Dα0+f(t)=1Γ(nα)t0f(n)(s)(ts)αn+1ds,

    where n1<αn, provided that the right-hand side is pointwise defined on (0,).

    Lemma 2.3 ([8]). Let n1<αn, uC(0,1)L1(0,1), then

    Iα0+Dα0+u(t)=u(t)+c0+c1t++cn1tn1,

    where ciR, i=0,1,,n1.

    Lemma 2.4. If β>0,α+β>0, then the equation

    Iα0+Iβ0+f(x)=Iα+β0+f(x)

    is satisfied for continuous function f.

    Firstly, we briefly recall some definitions on the coincidence degree theory. For more details, see [22].

    Let Y,Z be real Banach spaces, L:domLYZ be a Fredholm map of index zero and P:YY, Q:ZZ be continuous projectors such that

    KerL=ImP,ImL=KerQ,Y=KerLKerP,Z=ImLImQ.

    It follows that

    L|domLKerP:domLKerPImL

    is invertible. We denote the inverse of this map by KP. If Ω is an open bounded subset of Y, the map N will be called L-compact on ¯Ω if QN(¯Ω) is bounded and KP,QN=KP(IQ)N:¯ΩY is compact.

    Theorem 2.5 ([22]). Let L be a Fredholm operator of index zero and N be L-compact on ¯Ω. Suppose that the following conditions are satisfied:

    (1)LxλNx for each (x,λ)[(domLKerL)Ω]×(0,1);

    (2)NxImL for each xKerLΩ;

    (3)deg(JQN|KerL,ΩKerL,0)0, where Q:ZZ is a continuous projection as above with ImL=KerQ and J:ImQKerL is any isomorphism.

    Then the equation Lx=Nx has at least one solution in domL¯Ω.

    In this section, we will discuss the existence of solutions to Eq (1.3).

    To obtain our main results, we impose the following condition:

    (H0)α+β+γ>2.

    We define E=C[0,1] with the norm u=max0t1|u(t)| and X={u(t):u(t),u(t),u(t)E} with the norm uX=u+u+u(t). It is clear that (E,) and (X,X) are Banach spaces.

    Define

    L:domLE,u(Dγ0+(Dβ0+(Dα0+u)))(t), (3.1)
    N:XE,uf(t,u(t),u(t),u(t)), (3.2)

    where

    domL={uX:u(0)=u(ξ),Dα0+u(0)=(Dβ0+(Dα0+u))(0)=0}.

    Then the three-point boundary value problem (1.3) can be written as

    Lu=Nu.

    Lemma 3.1. L is defined as (3.1), then

    KerL={uX:u=c,cR}, (3.3)
    ImL={xE:Iα+β+γ0+x(ξ)=0}. (3.4)

    Proof. It is obvious that the operator L is linear. By Lu=0 and Lemmas 2.3, we have (Dβ0+(Dα0+u))(t)=c0,c0R. In view of (Dβ0+(Dα0+u))(0)=0, we get c0=0. Then, (Dβ0+(Dα0+u))(t)=0. Thus, by Lemmas 2.3 again, we obtain Dα0+u(t)=c1,c1R. In view of Dα0+u(0)=0, we get c1=0. So, Dα0+u(t)=0. Then, according to u(0)=u(ξ), we have u(t)=c,cR. Hence, we obtain (3.3).

    Next, we prove (3.4) hold. Let xImL, so there exists udomL such that x(t)=(Dγ0+(Dβ0+(Dα0+u)))(t). By Lemma 2.3 and the definition of domL, we have

    u(t)=Iα+β+γ0+x(t)+Iβ+α0+c0+Iα0+c1+c2.

    In view of Dβ0+u(0)=(Dβ0+(Dα0+u))(0)=0, we get c0=c1=0. Hence,

    u(t)=Iα+β+γ0+x(t)+c2.

    According to u(0)=u(ξ), we have Iα+β+γ0+x(ξ)=0. On the other hand, suppose x satisfies the above equations. Let u(t)=Iα+β+γ0+x(t), we can prove u(t)domL and Lu(t)=x. Then, (3.4) holds.

    For simplicity of notation, we write p=1+(α+β+γ)2Γ(1+α+β+γ).

    Lemma 3.2. The mapping L:domLYZ is a Fredholm operator of index zero.

    Proof. The linear continuous projector operator P can be defined as

    Pu=u(0).

    Obviously, P2=P. It is clear that

    KerP={u:u(0)=0}.

    It follows from u=uPu+Pu that Y=KerP+KerL. For uKerLKerP, then u=c,cR. Furthermore, by the definition of KerP, we have c=0. Thus,

    Y=KerLKerP.

    The linear operator Q can be defined as

    Qx(t)=Γ(1+α+β+γ)ξα+β+γIα+β+γ0+x(ξ).

    Obviously, QxR. For x(t)E, we have

    Q(Qx(t))=Qx(t)Γ(1+α+β+γ)ξα+β+γ(Iα+β+γ0+1)t=ξ=Qx(t).

    So, the operator Q is idempotent. It follows from x=xQx+Qx that E=ImL+ImQ. Moreover, by KerQ=ImL and Q2=Q, we get ImLImQ={0}. Hence,

    E=ImLImQ.

    Now, IndL=dimKerLcodimImL=0, and so L is a Fredholm mapping of index zero.

    For every uX, we have

    PuX=|u(0)|. (3.5)

    Furthermore, the operator KP:ImLdomLKerP can be defined by

    KPx(t)=Iα+β+γ0+x(t).

    For x(t)ImL, we have

    LKPx(t)=LIα+β+γ0+x(t)=(Dγ0+(Dβ0+(Dα0+Iα+β+γ0+x)))(t)=x(t). (3.6)

    On the other hand, for udomLKerP, according to Lemma 2.3 and the definitions of domL and KerP, we have

    Iα+β+γ0+Lu(t)=Iα+β+γ0+(Dγ0+(Dβ0+(Dα0+u)))(t)=u(t). (3.7)

    Combining (3.6) and (3.7), we have KP=(LdomLKerP)1.

    For xImL and α+β+γ>2, we have

    KPxX=Iα+β+γ0+xX=Iα+β+γ0+x+(Iα+β+γ0+x)+(Iα+β+γ0+x)=[1Γ(1+α+β+γ)+1Γ(α+β+γ)+1Γ(α+β+γ1)]x=[1+(α+β+γ)2Γ(1+α+β+γ)]x=px. (3.8)

    Again for uΩ1, udom(L)Ker(L), then (IP)udomLKerP and LPu=0, thus from (3.8), we have

    (IP)uX=KPL(IP)uX=KPLuX=pNu.

    With the similar proof showed in [13], we have the following lemma.

    Lemma 3.3. KP(IQ)N:YY is completely continuous.

    Theorem 3.4. Assume the following conditions hold:

    (H1) There exist nonnegative functions ψ(t),φ0(t),φ1(t),φ2(t)E, such that for all t[0,1], (u1,u2,u3)R3, one has

    |f(t,u1,u2,u3)|ψ(t)+φ0(t)|u1|+φ1(t)|u2|+φ2(t)|u3|.

    (H2) There exists A>0 such that for (t,u,u,u), if |u|>A for all t[0,1], one has

    QN(u)0.

    (H3) There exists B>0 such that if |c|>B,cR, one has either

    cQN(c)>0

    or

    cQN(c)<0.

    Then, BVP (1.3) has at least a solution in X provided that

    φ<Γ(α+β+γ+1)2+(α+β+γ+1)2 (3.9)

    where φ=max{φ0(t),φ1(t),φ2(t)}.

    Proof. Let

    Ω1={udomLKerL:Lu=λNu,λ(0,1)}.

    For Lu=λNuImL=KerQ, by (3.2) and the definition of KerQ, we have

    Iα+β+γ0+f(t,u,u,u)(ξ)=0.

    According to (H2), there exists t0(0,1) such that |u(t0)|A. By Lu=λNu,udomLKerL, that is (Dγ0+(Dβ0+(Dα0+u)))(t)=λNu, we have

    u(t)=λΓ(α+β+γ)t0(ts)α+β+γ1f(s,u,u,u)ds+c0.

    Substituting t=t0 into the above equation, we get

    u(t0)=λΓ(α+β+γ)t00(t0s)α+β+γ1f(s,u,u,u)ds+c0.

    So, we have

    u(t)u(t0)=λΓ(α+β+γ)t0(ts)α+β+γ1f(s,u,u,u)dsλΓ(α+β+γ)t00(t0s)α+β+γ1f(s,u,u,u)ds.

    Substituting t=0 into the above equation, together with |u(t0)|A and (H1), we have

    u(0)=u(t0)λΓ(α)t00(t0s)α+β+γ1f(s,u,u,u)ds.

    So

    |u(0)||u(t0)|+λΓ(α+β+γ)t00(t0s)α+β+γ1|f(s,u,u,u)|dsA+1Γ(α+β+γ)t00(t0s)α+β+γ1|f(s,u,u,u)|dsA+1Γ(α+β+γ)t00(t0s)α+β+γ1(|ψ|+|φ0||u|+|φ1||u|+|φ2||u|)dsA+1Γ(α+β+γ)(ψ+uXφ)10(1s)α+β+γ1ds=A+1Γ(α+β+γ+1)(ψ+uXφ). (3.10)

    By (3.10) and (H1), we have

    uX=Pu+(IP)uXPuY+(IP)uX|u(0)|+pNuA+1Γ(α+β+γ+1)(ψ+uXφ)+p|f(s,u,u,u)|A+1Γ(α+β+γ+1)(ψ+uXφ)+p(ψ+uXφ)A+1+pΓ(α+β+γ+1)Γ(α+β+γ+1)(ψ+uXφ)=A+1+pΓ(α+β+γ+1)Γ(α+β+γ+1)ψ+uX1+pΓ(α+β+γ+1)Γ(α+β+γ+1)φ=A+2+(α+β+γ)2Γ(α+β+γ+1)ψ+uX2+(α+β+γ)2Γ(α+β+γ+1)φ

    According to (3.9), we can derive

    uXA+2+(α+β+γ)2Γ(α+β+γ+1)ψ12+(α+β+γ)2Γ(α+β+γ+1)φ:=M (3.11)

    Thus, we have Ω1 is bounded.

    Let

    Ω2={uKerL:NuImL}.

    For uKerL, then u(t)=c,cR. In view of NuImL=KerQ, we have Q(Nu)=0, that is

    Iα+β+γ0+f(t,u,u,u)(ξ)=0.

    By (H2), there exist constants t0[0,1] such that

    |u(t0)|=|c|A. (3.12)

    Therefore, Ω2 is bounded.

    Let

    Ω3={uKerL:λu+(1λ)QNu=0,λ[0,1]}.

    For uKerL, then u(t)=c0. By the definition of the set Ω3, we have

    λc0+(1λ)QN(c0)=0. (3.13)

    Next, the proof is divided into three cases.

    Case 1. If λ=1, then c0=0.

    Case 2. If λ=0, similar to the proof of the boundness of Ω2, we have |c0|A.

    Case 3. If λ(0,1), we also have |c0|B. Otherwise, if |c0|>B, in view of the first part of (H3), we obtain

    λc20+(1λ)c0QN(c0)>0,

    which contradicts (3.13). Thus, Ω3 is bounded.

    If the second part of (H3) holds, we can prove the set

    Ω3={uKerL:λu+(1λ)QNu=0,λ[0,1]}

    is bounded.

    Let Ω={xX:xX<max{M,A,B}+1}. It follows from Lemma 3.2 and Lemma 3.3 that L is a Fredholm operator of index zero and N is Lcompact on ˉΩ. By (3.11) and (3.12), we get that the following two conditions of Theorem 2.5 are satisfied

    (1). LuλNu, for every u[(domLKerL)Ω]×(0,1);

    (2). NuImL for every uKerLΩ.

    Let H(u,λ)=±λIu+(1λ)JQNu, where I is the identical operator. By the proof of the boundness of Ω3, we have that H(u,λ)0 for uKerLΩ. Therefore, via the homotopy property of degree, we obtain

    deg(JQN|KerL,ΩKerL,0)=deg(H(,0),ΩKerL,0)=deg(H(,1),ΩKerL,0)=deg(I,ΩKerL,0)=10.

    So, the condition (3) of Theorem 2.5 is satisfied. Applying Theorem 2.5, we conclude that Lu=Nu has at least one solution in domL¯Ω. Then fractional jerk equation (1.1) has at least one solution and the proof is complete.

    Example 4.1. Let us consider the following fractional jerk equation:

    {(D0.60+(D0.70+(D0.80+u)))(t)=f(t,u(t),u(t),u(t)),u(0)=u(13),D0.80+u(0)=(D0.70+(D0.80+u))(0)=0, (4.1)

    where

    f(t,u(t),u(t),u(t))=t20+u(t)10+cosu(t)20+arctan2u(t)5π2.

    Corresponding to BVP (1.3), we have that α=0.8, β=0.7, γ=0.6, α+β+γ=2.1>2. By calculation, we have

    Γ(α+β+γ+1)2+(α+β+γ+1)2=Γ(3.1)11.610.19.

    By a simple proof, we have

    |f(t,u(t),u(t),u(t))|=|t20+u(t)10+cosu(t)20+arctan2u(t)5π2|320+|u(t)|10.

    Choose φ0(t)=320, φ1(t)=110, φ2=φ3=0, then (H1) is satisfied. Furthermore,

    φ=max{φ0(t),φ1(t),φ2(t)}=0.15<0.19.

    Hence, (3.9) holds. In addition, by a simple calculation, by choosing A=B=22, then condition (H2) and (H3) are satisfied. Consequently, By Theorem 3.4, BVP (4.1) has at least one solution.

    In this article, we considered a class of nonlinear fractional jerk equations with boundary value conditions at resonance. By using the theory of Mawhin's continuation theorem, we obtained the existence of solutions of boundary value problems (1.1). An example is presented to illustrate the main results. To the authors' knowledge, the existence of solutions for fractional jerk BVPs at resonance has not been reported. So, it is novel in term of fractional jerk equations and our result enriches and extends some related results in the literature. There is still more work to be done in the future on this interesting problem. For example, seeking the existence of positive solutions for fractional jerk equations with resonant boundary conditions.

    This research was supported by the Research Fund for the Doctoral Program of Shandong Jiaotong University, Shandong Jiaotong University “Climbing” Research Innovation Team Program (No.SDJTUC1803), a project of Shandong Province Higher Educational Science and Technology Program (No.J17KA073, J18KA330), the key research and development project of Shandong Province (No.2018GGX105011, No.2019GGX101055) and University outstanding youth innovation team development plan of Shandong Province (No.2019KJN023). This research was also funded by the National Natural Science Foundation of China (11671181, 61803230) and the Fundamental Research Funds for the Central Universities (2020YJSLX01).

    The authors declare no conflicts of interest in this paper.



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