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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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)=bx1−cx2−x3−x31, |
where the dot denotes differentiation with respect to time, a and c are positive parameters, and c∈R.
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)+β−W(x(t))y(t), α,β,γ are fractional order and 0<α,β,γ≤1. 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,γ≤1, 0<ξ≤1, Dα0+,Dβ0+,Dγ0+ denote the Caputo fractional derivatives and f:[0,1]×R3→R 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,c∈R 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 ξ≠1, 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(t−s)α−1f(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(n−α)∫t0f(n)(s)(t−s)α−n+1ds, |
where n−1<α≤n, provided that the right-hand side is pointwise defined on (0,∞).
Lemma 2.3 ([8]). Let n−1<α≤n, u∈C(0,1)∩L1(0,1), then
Iα0+Dα0+u(t)=u(t)+c0+c1t+⋯+cn−1tn−1, |
where ci∈R, i=0,1,⋯,n−1.
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:domL⊂Y→Z be a Fredholm map of index zero and P:Y→Y, Q:Z→Z be continuous projectors such that
KerL=ImP,ImL=KerQ,Y=KerL⊕KerP,Z=ImL⊕ImQ. |
It follows that
L|domL∩KerP:domL∩KerP→ImL |
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(I−Q)N:¯Ω→Y 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)Lx≠λNx for each (x,λ)∈[(domL∖KerL)∩∂Ω]×(0,1);
(2)Nx∉ImL for each x∈KerL∩∂Ω;
(3)deg(JQN|KerL,Ω∩KerL,0)≠0, where Q:Z→Z is a continuous projection as above with ImL=KerQ and J:ImQ→KerL 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‖∞=max0≤t≤1|u(t)| and X={u(t):u(t),u′(t),u″(t)∈E} with the norm ‖u‖X=‖u‖∞+‖u′‖∞+‖u″(t)‖∞. It is clear that (E,‖⋅‖∞) and (X,‖⋅‖X) are Banach spaces.
Define
L:domL→E,u↦(Dγ0+(Dβ0+(Dα0+u)))(t), | (3.1) |
N:X→E,u↦f(t,u(t),u′(t),u″(t)), | (3.2) |
where
domL={u∈X: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={u∈X:u=c,c∈R}, | (3.3) |
ImL={x∈E: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,c0∈R. 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,c1∈R. 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,c∈R. Hence, we obtain (3.3).
Next, we prove (3.4) hold. Let x∈ImL, so there exists u∈domL 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:domL⊂Y→Z 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=u−Pu+Pu that Y=KerP+KerL. For u∈KerL∩KerP, then u=c,c∈R. Furthermore, by the definition of KerP, we have c=0. Thus,
Y=KerL⊕KerP. |
The linear operator Q can be defined as
Qx(t)=Γ(1+α+β+γ)ξα+β+γIα+β+γ0+x(ξ). |
Obviously, Qx≅R. 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=x−Qx+Qx that E=ImL+ImQ. Moreover, by KerQ=ImL and Q2=Q, we get ImL∩ImQ={0}. Hence,
E=ImL⊕ImQ. |
Now, IndL=dimKerL−codimImL=0, and so L is a Fredholm mapping of index zero.
For every u∈X, we have
‖Pu‖X=|u(0)|. | (3.5) |
Furthermore, the operator KP:ImL→domL∩KerP 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 u∈domL∩KerP, 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=(LdomL∩KerP)−1.
For x∈ImL and α+β+γ>2, we have
‖KPx‖X=‖Iα+β+γ0+x‖X=‖Iα+β+γ0+x‖∞+‖(Iα+β+γ0+x)′‖∞+‖(Iα+β+γ0+x)″‖∞=[1Γ(1+α+β+γ)+1Γ(α+β+γ)+1Γ(α+β+γ−1)]‖x‖∞=[1+(α+β+γ)2Γ(1+α+β+γ)]‖x‖∞=p‖x‖∞. | (3.8) |
Again for u∈Ω1, u∈dom(L)∖Ker(L), then (I−P)u∈domL∩KerP and LPu=0, thus from (3.8), we have
‖(I−P)u‖X=‖KPL(I−P)u‖X=‖KPLu‖X=p‖Nu‖∞. |
With the similar proof showed in [13], we have the following lemma.
Lemma 3.3. KP(I−Q)N:Y→Y 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,c∈R, 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={u∈domL∖KerL:Lu=λNu,λ∈(0,1)}. |
For Lu=λNu∈ImL=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,u∈domL∖KerL, that is (Dγ0+(Dβ0+(Dα0+u)))(t)=λNu, we have
u(t)=λΓ(α+β+γ)∫t0(t−s)α+β+γ−1f(s,u,u′,u″)ds+c0. |
Substituting t=t0 into the above equation, we get
u(t0)=λΓ(α+β+γ)∫t00(t0−s)α+β+γ−1f(s,u,u′,u″)ds+c0. |
So, we have
u(t)−u(t0)=λΓ(α+β+γ)∫t0(t−s)α+β+γ−1f(s,u,u′,u″)ds−λΓ(α+β+γ)∫t00(t0−s)α+β+γ−1f(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(t0−s)α+β+γ−1f(s,u,u′,u″)ds. |
So
|u(0)|≤|u(t0)|+λΓ(α+β+γ)∫t00(t0−s)α+β+γ−1|f(s,u,u′,u″)|ds≤A+1Γ(α+β+γ)∫t00(t0−s)α+β+γ−1|f(s,u,u′,u″)|ds≤A+1Γ(α+β+γ)∫t00(t0−s)α+β+γ−1(|ψ|+|φ0||u|+|φ1||u′|+|φ2||u″|)ds≤A+1Γ(α+β+γ)⋅(‖ψ‖∞+‖u‖X⋅φ)⋅∫10(1−s)α+β+γ−1ds=A+1Γ(α+β+γ+1)(‖ψ‖∞+‖u‖X⋅φ). | (3.10) |
By (3.10) and (H1), we have
‖u‖X=‖Pu+(I−P)u‖X≤‖Pu‖Y+‖(I−P)u‖X≤|u(0)|+p‖Nu‖∞≤A+1Γ(α+β+γ+1)(‖ψ‖∞+‖u‖X⋅φ)+p|f(s,u,u′,u″)|∞≤A+1Γ(α+β+γ+1)(‖ψ‖∞+‖u‖X⋅φ)+p(‖ψ‖∞+‖u‖X⋅φ)≤A+1+pΓ(α+β+γ+1)Γ(α+β+γ+1)(‖ψ‖∞+‖u‖X⋅φ)=A+1+pΓ(α+β+γ+1)Γ(α+β+γ+1)‖ψ‖∞+‖u‖X⋅1+pΓ(α+β+γ+1)Γ(α+β+γ+1)φ=A+2+(α+β+γ)2Γ(α+β+γ+1)‖ψ‖∞+‖u‖X⋅2+(α+β+γ)2Γ(α+β+γ+1)φ |
According to (3.9), we can derive
‖u‖X≤A+2+(α+β+γ)2Γ(α+β+γ+1)‖ψ‖∞1−2+(α+β+γ)2Γ(α+β+γ+1)φ:=M | (3.11) |
Thus, we have Ω1 is bounded.
Let
Ω2={u∈KerL:Nu∈ImL}. |
For u∈KerL, then u(t)=c,c∈R. In view of Nu∈ImL=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={u∈KerL:λu+(1−λ)QNu=0,λ∈[0,1]}. |
For u∈KerL, then u(t)=c0. By the definition of the set Ω3, we have
λc0+(1−λ)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+(1−λ)c0⋅QN(c0)>0, |
which contradicts (3.13). Thus, Ω3 is bounded.
If the second part of (H3) holds, we can prove the set
Ω′3={u∈KerL:−λu+(1−λ)QNu=0,λ∈[0,1]} |
is bounded.
Let Ω={x∈X:‖x‖X<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 L−compact on ˉΩ. By (3.11) and (3.12), we get that the following two conditions of Theorem 2.5 are satisfied
(1). Lu≠λNu, for every u∈[(domL∖KerL)∩∂Ω]×(0,1);
(2). Nu∉ImL for every u∈KerL∩∂Ω.
Let H(u,λ)=±λIu+(1−λ)JQNu, where I is the identical operator. By the proof of the boundness of Ω3, we have that H(u,λ)≠0 for u∈KerL∩∂Ω. Therefore, via the homotopy property of degree, we obtain
deg(JQN|KerL,Ω∩KerL,0)=deg(H(⋅,0),Ω∩KerL,0)=deg(H(⋅,1),Ω∩KerL,0)=deg(I,Ω∩KerL,0)=1≠0. |
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)11.61≈0.19. |
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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