In this paper, we study a class of second order impulsive vibration equation containing fractional derivatives. By using monotone iterative technique, some new results on the multiplicity for solutions of the equations under nonlinear boundary conditions are obtained, and the properties of the solutions are discussed. Finally, the practicability of our results is discussed through a concrete example.
Citation: Xue Wang, Xiping Liu, Mei Jia. A class of impulsive vibration equation with fractional derivatives[J]. AIMS Mathematics, 2021, 6(2): 1965-1990. doi: 10.3934/math.2021120
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In this paper, we study a class of second order impulsive vibration equation containing fractional derivatives. By using monotone iterative technique, some new results on the multiplicity for solutions of the equations under nonlinear boundary conditions are obtained, and the properties of the solutions are discussed. Finally, the practicability of our results is discussed through a concrete example.
As we all know, the vibration equation is one of the important research topics in mechanics, physics and other disciplines. In the recent decades, researchers have been paying more and more attentions to the fractional differential equations due to its wide applications on mechanics, physical science, biological sciences and engineering disciplines, etc., see [3,6,7,12,13,17,18,19,20,25] and the references therein. The development of fractional differential equations provides some new theoretical bases for the study of vibration problems. In [22], the vibration equations with fractional derivatives are used to describe the vibration behavior of viscoelastic polymers and good results are obtained. The theoretical study of vibration equations with fractional derivative has also been widely concerned. These studies include the mechanical properties, dynamic characteristics of the system and the correlation functions with various influences for the vibration equations with fractional derivatives, and so on see[15,16,21,22]. In addition, mutation often occurs in vibration system. These abrupt changes can be simulated by the impulsive vibration equation. And this simulation are effective in describing the behavior of real system. There have been a large number of references for the study of fractional impulsive differential equations, see [1,4,5,9,11,23,24,26].
In this paper, we study a class of second order impulsive vibration equation containing fractional derivatives under the nonlinear boundary conditions
{x″(t)−λcDα0+x(t)=f(t,x(t),x′(t)),t∈J′,Δx(t)|t=tk=Ik(x(tk)),k=1,2,⋯,m,Δx′(t)|t=tk=Qk(x(tk)),k=1,2,⋯,m,g0(x(0),x(1))=0,g1(x′(0),x′(1))=0, | (1.1) |
where 0<α<1, λ>0 and cDα0+ is the Caputo derivative, 0=t0<t1<⋯<tm<tm+1=1. Set J=[0,1],J′=J∖{t1,t2,⋯,tm},J0=[0,t1],Jk=(tk,tk+1],k=1,2,⋯,m. Δx(tk)=x(t+k)−x(t−k),Δx′(tk)=x′(t+k)−x′(t−k). x(t−k),x(t+k) denote the left limit and the right limit of x(t) at t=tk. x′(t−k),x′(t+k) denote the left limit and the right limit of x′(t) at t=tk. Let x(tk)=x(t−k). f∈C(J×R2,R),Ik,Qk∈C(R,R),k=1,2,⋯, m. g0,g1∈C(R2,R) are given nonlinear functions. By using monotone iterative technique, some new results on multiplicity of boundary value problems are obtained, and the properties of the solutions are discussed. Finally, an example is given out to illustrate the applicability of our main results.
In this section, we present some basic definitions and lemmas, which will be used to prove our main results.
Definition 2.1. (See[8], P67) Let α,β>0. The function Eα,β is defined by
Eα,β(z)=∞∑j=0zjΓ(jα+β), |
whenever the series converge is called the two parameters Mittag-Leffler function with parameters α and β.
Lemma 2.1. (See[8], P68) Consider the two parameters Mittag-Leffler function Eα,β for some α,β>0. The power series defining Eα,β is convergent for all z∈C. In other words, Eα,β is an entire function.
Lemma 2.2. Let α,β>0,k=0,1,2,⋯,z∈R. Then
E(k)α,β(z)=∞∑j=0zjΓ(k+j+1)Γ(j+1)Γ(α(k+j)+β). |
Proof. By Definition 2.1 and Lemma 2.1, we get
E(k)α,β(z)=dkdzkEα,β(z)=∞∑j=0dkdzk(zjΓ(jα+β))=∞∑j=kΓ(j+1)zj−kΓ(j−k+1)Γ(αj+β)=∞∑j=0zjΓ(k+j+1)Γ(j+1)Γ(α(k+j)+β). |
Lemma 2.3. (See[14], P314) Let 0<α<β,n−1<α≤n,l−1<β≤l(n,l∈N,n≤l,λ∈R). Then
cDβ0+x(t)−λcDα0+x(t)=0(t>0) |
has its linearly independent solutions given by
xj(t)=tjEβ−α,j+1(λtβ−α)−λtβ−α+jEβ−α,β−α+j+1(λtβ−α)(j=0,1,⋯,n−1), | (2.1) |
xj(t)=tjEβ−α,j+1(λtβ−α)(j=n,⋯,l−1). | (2.2) |
Lemma 2.4. (See[14], P324) Let l−1<β≤l(l∈N),0<α<β be such that β−l+1≥α, λ∈R, and h(t) be a given real function defined on R+. The general solution to the nonhomogeneous linear differential equation
cDβ0+x(t)−λcDα0+x(t)=h(t)(t>0) |
is given by
x(t)=∫t0(t−s)β−1Eβ−α,β(λ(t−s)β−α)h(s)ds+l−1∑j=0cjxj(t), |
where xj(t) are given by (2.1) and (2.2), cj are arbitrary real constants (j=0,1,⋯,l−1).
Lemma 2.5. Let L[0,1] denote the space of Lebesgue integrable functions on [0, 1], h∈L[0,1] and 0<α<1, then the Cauchy problem of the second order vibration equation with fractional derivative
{x″(t)−λcDα0+x(t)=h(t),t∈J,x(ξ)=x0,x′(ξ)=x1,ξ∈J |
has a unique solution, which is given by
x(t)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds−∫ξ0(ξ−s)E2−α,2(λ(ξ−s)2−α)h(s)ds+x0+tE2−α,2(λt2−α)−ξE2−α,2(λξ2−α)E2−α,1(λξ2−α)(x1−∫ξ0E2−α,1(λ(ξ−s)2−α)h(s)ds), | (2.3) |
and x(t) is derivable while its derivative is given by
x′(t)=∫t0E2−α,1(λ(t−s)2−α)h(s)ds+E2−α,1(λt2−α)E2−α,1(λξ2−α)(x1−∫ξ0E2−α,1(λ(ξ−s)2−α)h(s)ds). | (2.4) |
Proof. In view of Lemma 2.4, for β=2,0<α<1, the general solution of the equation
x″(t)−λcDα0+x(t)=h(t) |
is given by
x(t)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds+c1tE2−α,2(λt2−α)+c0, |
where c0,c1∈R, and
x′(t)=∫t0E2−α,1(λ(t−s)2−α)h(s)ds+c1E2−α,1(λt2−α). |
Therefore,
x(ξ)=∫ξ0(ξ−s)E2−α,2(λ(ξ−s)2−α)h(s)ds+c1ξE2−α,2(λξ2−α)+c0=x0,x′(ξ)=∫ξ0E2−α,1(λ(ξ−s)2−α)h(s)ds+c1E2−α,1(λξ2−α)=x1. |
We get
c1=1E2−α,1(λξ2−α)(x1−∫ξ0E2−α,1(λ(ξ−s)2−α)h(s)ds),c0=x0−∫ξ0(ξ−s)E2−α,2(λ(ξ−s)2−α)h(s)ds−c1ξE2−α,2(λξ2−α). |
So
x(t)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds−∫ξ0(ξ−s)E2−α,2(λ(ξ−s)2−α)h(s)ds+x0+c1(tE2−α,2(λt2−α)−ξE2−α,2(λξ2−α))=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds−∫ξ0(ξ−s)E2−α,2(λ(ξ−s)2−α)h(s)ds+x0+tE2−α,2(λt2−α)−ξE2−α,2(λξ2−α)E2−α,1(λξ2−α)(x1−∫ξ0E2−α,1(λ(ξ−s)2−α)h(s)ds), |
and
x′(t)=∫t0E2−α,1(λ(t−s)2−α)h(s)ds+E2−α,1(λt2−α)E2−α,1(λξ2−α)(x1−∫ξ0E2−α,1(λ(ξ−s)2−α)h(s)ds). |
The proof is completed.
Let PC1(J)={x:J→R|x,x′∈C(J′,R),x(t+k),x(t−k),x′(t+k),x′(t−k)exist,andx(tk)=x(t−k),k=1,2,⋯,m} and endowed with the normal ‖x‖=max{supt∈[0,1]|x(t)|,supt∈[0,1]|x′(t)|}. Then PC1(J) is a Banach space.
For x∈PC1(J), by the Lagrange mean value theorem, there exists ξk∈[tk−ε,tk] such that
x(tk)−x(tk−ε)=x′(ξk)ε, |
and
x′−(tk)=limε→0+x(tk)−x(tk−ε)ε=limε→0+x′(ξk)εε=x′(t−k),k=1,2,⋯,m. |
Thus, for x∈PC1(J), we denote
x′(tk)=x′−(tk)=x′(t−k),k=1,2,⋯,m. | (2.5) |
Let P={x∈PC1(J)|x(t)≥0,x′(t)≥0,t∈J}. It is obvious that P⊂PC1(J) is a normal solid cone. We denote x≺_y∈PC1(J) if and only if x(t)≤y(t) and x′(t)≤y′(t) on t∈[0,1], i.e. y−x∈P. We denote x≺y if x≺_y∈PC1(J) and x≠y, and x≺≺y if y−x∈˚P.
Lemma 2.6. (See[10], P220, [2], P666) Let E be a Banach space, and P⊂E be a normal solid cone. Suppose that there exist α1,β1,α2,β2∈E with α1≺β1≺α2≺β2 and A:[α1,β2]→E is a completely continuous strongly increasing operator such that
α1≺_Aα1,Aβ1≺β1,α2≺Aα2,Aβ2≺_β2. |
Then the operator A has at least three distinct fixed points x1,x2,x3 on [α1,β2] such that
α1≺_x1≺≺β1,α2≺≺x2≺_β2,α2⊀_x3⊀_β1. |
In this section, we obtain the solution of the linear impulsive vibration equation and discuss the properties of its kernel function.
Lemma 3.1. For any pk,qk∈R (k=1,2,⋯,m), mi,ni∈R (i=1,2) and h∈L[0,1], the following boundary value problem of the second order impulsive vibration equation with fractional derivative
{x″(t)−λcDα0+x(t)=h(t),t∈J′,Δx(t)|t=tk=pk,k=1,2,⋯,m,Δx′(t)|t=tk=qk,k=1,2,⋯,m,m1x(0)+m2x(1)=γ0,n1x′(0)+n2x′(1)=γ1 | (3.1) |
has a unique solution, which is given by
x(t)=∫10G(t,s)h(s)ds+φ(t)+(tE2−α,2(λt2−α)n1+n2E2−α,1(λ)−m2E2−α,2(λ)(m1+m2)(n1+n2E2−α,1(λ)))γ1+γ0m1+m2,t∈J, | (3.2) |
where
G(t,s)={(t−s)E2−α,2(λ(t−s)2−α)−n2tE2−α,2(λt2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ)+m2n2E2−α,2(λ)E2−α,1(λ(1−s)2−α)(m1+m2)(n1+n2E2−α,1(λ))−m2(1−s)E2−α,2(λ(1−s)2−α)m1+m2,0≤s≤t≤1,m2n2E2−α,2(λ)E2−α,1(λ(1−s)2−α)(m1+m2)(n1+n2E2−α,1(λ))−n2tE2−α,2(λt2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ)−m2(1−s)E2−α,2(λ(1−s)2−α)m1+m2,0≤t<s≤1, | (3.3) |
φ(t)=∑0<ti<tpi−m2m1+m2m∑i=1pi+∑0<ti<ttE2−α,2(λt2−α)−tiE2−α,2(λt2−αi)E2−α,1(λt2−αi)qi+m∑i=1(m2n2E2−α,2(λ)E2−α,1(λ)(m1+m2)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi)−m2(E2−α,2(λ)−tiE2−α,2(λt2−αi))(m1+m2)E2−α,1(λt2−αi)−n2E2−α,1(λ)tE2−α,2(λt2−α)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi))qi,t∈J. | (3.4) |
Furthermore,
x′(t)=∫10G′t(t,s)h(s)ds+φ′(t)+E2−α,1(λt2−α)n1+n2E2−α,1(λ)γ1,t∈J, | (3.5) |
G′t(t,s)={E2−α,1(λ(t−s)2−α)−n2E2−α,1(λt2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ),0≤s≤t≤1,−n2E2−α,1(λt2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ),0≤t<s≤1, | (3.6) |
and
φ′(t)=∑0<ti<tE2−α,1(λt2−α)E2−α,1(λt2−αi)qi−m∑i=1n2E2−α,1(λ)E2−α,1(λt2−α)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi)qi,t∈J. | (3.7) |
Proof. For t∈[0,t1], let ξ=0,x(0)=c0,x′(0)=c1, by Lemma 2.5, Cauchy problem
{x″(t)−λcDα0+x(t)=h(t),x(0)=c0,x′(0)=c1 |
has a unique solution
x(t)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds+c1tE2−α,2(λt2−α)+c0, |
and
x′(t)=∫t0E2−α,1(λ(t−s)2−α)h(s)ds+c1E2−α,1(λt2−α),c0,c1∈R. |
So
x(t−1)=∫t10(t1−s)E2−α,2(λ(t1−s)2−α)h(s)ds+c1t1E2−α,2(λt2−α1)+c0,x′(t−1)=∫t10E2−α,1(λ(t1−s)2−α)h(s)ds+c1E2−α,1(λt2−α1). |
For t∈(t1,t2], let ξ=t1,x(t+1)=x(t−1)+p1,x′(t+1)=x′(t−1)+q1. By Lemma 2.5, we can obtain that
x(t)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds−∫t10(t1−s)E2−α,2(λ(t1−s)2−α)h(s)ds+x(t+1)+tE2−α,2(λt2−α)−t1E2−α,2(λt2−α1)E2−α,1(λt2−α1)(x′(t+1)−∫t10E2−α,1(λ(t1−s)2−α)h(s)ds)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds−∫t10(t1−s)E2−α,2(λ(t1−s)2−α)h(s)ds+x(t−1)+p1+tE2−α,2(λt2−α)−t1E2−α,2(λt2−α1)E2−α,1(λt2−α1)(x′(t−1)+q1−∫t10E2−α,1(λ(t1−s)2−α)h(s)ds)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds−∫t10(t1−s)E2−α,2(λ(t1−s)2−α)h(s)ds+p1+∫t10(t1−s)E2−α,2(λ(t1−s)2−α)h(s)ds+c1t1E2−α,2(λt2−α1)+c0+tE2−α,2(λt2−α)−t1E2−α,2(λt2−α1)E2−α,1(λt2−α1)(∫t10E2−α,1(λ(t1−s)2−α)h(s)ds+c1E2−α,1(λt2−α1)+q1−∫t10E2−α,1(λ(t1−s)2−α)h(s)ds)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds+tE2−α,2(λt2−α)c1+c0+p1+tE2−α,2(λt2−α)−t1E2−α,2(λt2−α1)E2−α,1(λt2−α1)q1, |
and
x′(t)=∫t0E2−α,1(λ(t−s)2−α)h(s)ds+E2−α,1(λt2−α)E2−α,1(λt2−α1)(x′(t−1)+q1−∫t10E2−α,1(λ(t1−s)2−α)h(s)ds)=∫t0E2−α,1(λ(t−s)2−α)h(s)ds+E2−α,1(λt2−α)c1+E2−α,1(λt2−α)E2−α,1(λt2−α1)q1. |
For t∈(tk,tk+1], let ξ=tk,x(t+k)=x(t−k)+pk,x′(t+k)=x′(t−k)+qk,k=2,3,⋯,m. In the same way, we have
x(t)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds+tE2−α,2(λt2−α)c1+c0+∑0<ti<tpi+∑0<ti<ttE2−α,2(λt2−α)−tiE2−α,2(λt2−αi)E2−α,1(λt2−αi)qi, |
and
x′(t)=∫t0E2−α,1(λ(t−s)2−α)h(s)ds+E2−α,1(λt2−α)c1+∑0<ti<tE2−α,1(λt2−α)E2−α,1(λt2−αi)qi. |
Hence,
x(1)=∫10(1−s)E2−α,2(λ(1−s)2−α)h(s)ds+E2−α,2(λ)c1+c0+m∑i=1pi+m∑i=1E2−α,2(λ)−tiE2−α,2(λt2−αi)E2−α,1(λt2−αi)qi,x′(1)=∫10E2−α,1(λ(1−s)2−α)h(s)ds+E2−α,1(λ)c1+m∑i=1E2−α,1(λ)E2−α,1(λt2−αi)qi. |
By the boundary conditions m1x(0)+m2x(1)=γ0,n1x′(0)+n2x′(1)=γ1, we can get that
{−(m1+m2)c0−m2E2−α,2(λ)c1=m2∫10(1−s)E2−α,2(λ(1−s)2−α)h(s)ds+m2m∑i=1pi+m2m∑i=1E2−α,2(λ)−tiE2−α,2(λt2−αi)E2−α,1(λt2−αi)qi−γ0,−(n1+n2E2−α,1(λ))c1=n2∫10E2−α,1(λ(1−s)2−α)h(s)ds+m∑i=1n2E2−α,1(λ)E2−α,1(λt2−αi)qi−γ1. |
So
c0=∫10(−m2(1−s)E2−α,2(λ(1−s)2−α)m1+m2+m2n2E2−α,2(λ)E2−α,1(λ(1−s)2−α)(m1+m2)(n1+n2E2−α,1(λ)))h(s)ds−m2m1+m2m∑i=1pi+m∑i=1(m2n2E2−α,2(λ)E2−α,1(λ)(m1+m2)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi)−m2(E2−α,2(λ)−tiE2−α,2(λt2−αi))(m1+m2)E2−α,1(λt2−αi))qi−m2E2−α,2(λ)γ1(m1+m2)(n1+n2E2−α,1(λ))+γ0m1+m2, |
c1=−n2n1+n2E2−α,1(λ)(∫10E2−α,1(λ(1−s)2−α)h(s)ds+m∑i=1E2−α,1(λ)E2−α,1(λt2−αi)qi)+γ1n1+n2E2−α,1(λ). |
Therefore,
x(t)=∫t0(t−s)E2−α,2(λ(t−s)2−α)h(s)ds+∫10(m2n2E2−α,2(λ)E2−α,1(λ(1−s)2−α)(m1+m2)(n1+n2E2−α,1(λ))−n2tE2−α,2(λt2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ)−m2(1−s)E2−α,2(λ(1−s)2−α)m1+m2)h(s)ds+∑0<ti<tpi−m2m1+m2m∑i=1pi+∑0<ti<ttE2−α,2(λt2−α)−tiE2−α,2(λt2−αi)E2−α,1(λt2−αi)qi+m∑i=1(m2n2E2−α,2(λ)E2−α,1(λ)(m1+m2)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi)−m2(E2−α,2(λ)−tiE2−α,2(λt2−αi))(m1+m2)E2−α,1(λt2−αi)−n2E2−α,1(λ)tE2−α,2(λt2−α)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi))qi+γ0m1+m2+(tE2−α,2(λt2−α)n1+n2E2−α,1(λ)−m2E2−α,2(λ)(m1+m2)(n1+n2E2−α,1(λ)))γ1=∫10G(t,s)h(s)ds+φ(t)+(tE2−α,2(λt2−α)n1+n2E2−α,1(λ)−m2E2−α,2(λ)(m1+m2)(n1+n2E2−α,1(λ)))γ1+γ0m1+m2,t∈[0,1]. |
And
x′(t)=∫t0E2−α,1(λ(t−s)2−α)h(s)ds−∫10n2E2−α,1(λt2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ)h(s)ds+∑0<ti<tE2−α,1(λt2−α)E2−α,1(λt2−αi)qi−m∑i=1n2E2−α,1(λ)E2−α,1(λt2−α)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi)qi+E2−α,1(λt2−α)n1+n2E2−α,1(λ)γ1=∫10G′t(t,s)h(s)ds+φ′(t)+E2−α,1(λt2−α)n1+n2E2−α,1(λ)γ1,t∈[0,1]. |
Therefore, boundary value problem (3.1) has a unique solution x=x(t) which is given by (3.2), and G(t,s),φ(t) are given by (3.3) and (3.4), respectively. Furthermore, x′(t) is also established.
For convenience, we give out the following hypothesis:
(H1) The constants mi,ni∈R(i=1,2) satisfy m2(m1+m2)<0 and n2(n1+n2E2−α,1(λ))<0.
Lemma 3.2. Suppose that (H1) holds. Then functions G and φ defined by (3.3) and (3.4) satisfy the following properties:
(1) G(t,s) is continuous for t,s∈[0,1].
(2) G(t,s)>0 for t,s∈[0,1] and maxt∈[0,1]G(t,s)=G(1,s),mint∈[0,1]G(t,s)=G(0,s).
(3) G′t(t,s)>0 for t,s∈[0,1] and maxt∈[0,1]G′t(t,s)=G′t(1,s),mint∈[0,1]G′t(t,s)=G′t(0,s).
(4) If pk≥0,qk≥0,k=1,2,⋯,m, then φ(t)≥0,φ′(t)≥0, for t∈Jk.
Proof. (1) By the definition of G(t,s),G∈C([0,1]×[0,1]) is obvious.
(2) By (H1), for 0≤s≤t≤1,
G′t(t,s)=∂G(t,s)∂t=E2−α,1(λ(t−s)2−α)−n2E2−α,1(λt2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ)>0. |
Then G(s,s)≤G(t,s)≤G(1,s), for s∈[0,1] and t∈[s,1]. And
G(s,s)=m2n2E2−α,2(λ)E2−α,1(λ(1−s)2−α)(m1+m2)(n1+n2E2−α,1(λ))−n2sE2−α,2(λs2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ)−m2(1−s)E2−α,2(λ(1−s)2−α)m1+m2>0. |
Hence, G(t,s)>0 for 0≤s≤t≤1 and G(t,s)≤G(1,s) for t∈[s,1].
For 0≤t<s≤1,
∂G(t,s)∂t=−n2E2−α,1(λt2−α)E2−α,1(λ(1−s)2−α)n1−n2E2−α,1(λ)>0, |
we can get that G(0,s)≤G(t,s)<G(s,s), for s∈[0,1] and t∈[0,s). And
G(0,s)=m2n2E2−α,2(λ)E2−α,1(λ(1−s)2−α)(m1+m2)(n1+n2E2−α,1(λ))−m2(1−s)E2−α,2(λ(1−s)2−α)m1+m2>0. |
Hence, G(t,s)>0 for 0≤t<s≤1 and G(t,s)<G(s,s) for t∈[0,s).
Therefore, G(t,s)>0 for any t,s∈[0,1]. And G(t,s) is monotone increasing with respect to t∈[0,1], so maxt∈[0,1]G(t,s)=G(1,s),mint∈[0,1]G(t,s)=G(0,s).
(3) Since
(E2−α,1(λt2−α))′=(∞∑k=0(λt2−α)kΓ((2−α)k+1))′=∞∑k=1(2−α)kλkt(2−α)k−1Γ((2−α)k+1)≥0,t∈[0,1]. |
By (H1), for 0≤s≤t≤1,
∂2G(t,s)∂t2=(E2−α,1(λ(t−s)2−α))′−n2E2−α,1(λ(1−s)2−α)(E2−α,1(λt2−α))′n1+n2E2−α,1(λ)≥0. |
Then G′t(s,s)≤G′t(t,s)≤G′t(1,s), for any s∈[0,1],t∈[s,1]. Because
G′t(s,s)=1−n2E2−α,1(λs2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ)>0, |
we can get that G′t(t,s)>0 for 0≤s≤t≤1.
For 0≤t<s≤1,
∂G2(t,s)∂t2=−n2E2−α,1(λ(1−s)2−α)(E2−α,1(λt2−α))′n1−n2E2−α,1(λ)≥0, |
we can get that G′t(0,s)≤G′t(t,s)≤G′t(s,s) for any s∈[0,1] and t∈[0,s). Since
G′t(0,s)=−n2E2−α,1(λ(1−s)2−α)n1−n2E2−α,1(λ)>0, |
we have G′t(t,s)>0 for 0≤t<s≤1.
Therefore, G′t(t,s)>0 for any t,s∈[0,1] and maxt∈[0,1]G′t(t,s)=G′t(1,s),mint∈[0,1]G′t(t,s)=G′t(0,s).
(4) If pk,qk≥0, k=1,2,⋯,m, by (H1), (3.4) and (3.7), we can easily get that
φ(t)≥0,φ′(t)≥0,t∈Jk. |
The proof is completed.
In this section, we will establish the existence results of the solutions for the boundary value problem (1.1).
For any u∈PC1(J), we consider the following boundary value problem
{x″(t)−λcDα0+x(t)=f(t,u(t),u′(t)),t∈J′,Δx(t)|t=tk=Ik(u(tk)),k=1,2,⋯,m,Δx′(t)|t=tk=Qk(u(tk)),k=1,2,⋯,m,m1x(0)+m2x(1)=g0(u(0),u(1))+m1u(0)+m2u(1):=γu,0,n1x′(0)+n2x′(1)=g1(u′(0),u′(1))+n1u′(0)+n2u′(1):=γu,1. | (4.1) |
By Lemma 3.1, we can get that boundary value (4.1) is equivalent to the following integral equation
x(t)=∫10G(t,s)f(s,u(s),u′(s))ds+φu(t)+(tE2−α,2(λt2−α)n1+n2E2−α,1(λ)−m2E2−α,2(λ)(m1+m2)(n1+n2E2−α,1(λ)))γu,1+γu,0m1+m2,t∈J, |
and
x′(t)=∫10G′t(t,s)f(s,u(s),u′(s))ds+φ′u(t)+E2−α,1(λt2−α)n1+n2E2−α,1(λ)γu,1,t∈J, |
where
φu(t)=∑0<ti<tIi(u(ti))−m2m1+m2m∑i=1Ii(u(ti))+∑0<ti<ttE2−α,2(λt2−α)−tiE2−α,2(λt2−αi)E2−α,1(λt2−αi)Qi(u(ti))+m∑i=1(m2n2E2−α,2(λ)E2−α,1(λ)(m1+m2)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi)−m2(E2−α,2(λ)−tiE2−α,2(λt2−αi))(m1+m2)E2−α,1(λt2−αi)−n2E2−α,1(λ)tE2−α,2(λt2−α)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi))Qi(u(ti)),t∈J | (4.2) |
and
φ′u(t)=∑0<ti<tE2−α,1(λt2−α)E2−α,1(λt2−αi)Qi(u(ti))−m∑i=1n2E2−α,1(λ)E2−α,1(λt2−α)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi)Qi(u(ti)),t∈J. | (4.3) |
We define an operator T:PC1(J)→PC1(J) by
Tu(t)=∫10G(t,s)f(s,u(s),u′(s))ds+φu(t)+(tE2−α,2(λt2−α)n1+n2E2−α,1(λ)−m2E2−α,2(λ)(m1+m2)(n1+n2E2−α,1(λ)))γu,1+γu,0m1+m2,t∈J. |
By Lemma 3.1 and (3.5),
(Tu)′(t)=∫10G′t(t,s)f(s,u(s),u′(s))ds+φ′u(t)+E2−α,1(λt2−α)n1+n2E2−α,1(λ)γu,1. |
We can easily get that the following Lemma 4.1 holds.
Lemma 4.1. The function x=x(t) is the solution of boundary value problem (1.1) if and only if x is a fixed point of the operator T in PC1(J).
Lemma 4.2. If (H1) holds, then T:PC1(J)→PC1(J) is completely continuous.
Proof. Step 1: T is a continuous operator.
Suppose that {un}⊂PC1(J) and there exists u∈PC1(J) such that ‖un−u‖→0(n→∞). Then there exists a constant M>0 such that ‖un‖≤M,‖u‖≤M.
Since f∈C(J×R2,R),Ik,Qk∈C(R,R),k=1,2,⋯,m, g0,g1∈C(R2,R) and γun,0,γun,1,γu,0,γu,1,∈R, then
limn→∞|f(t,un(t),u′n(t))−f(t,u(t),u′(t))|=0,t∈J, |
limn→∞|Ik(un(tk))−Ik(u(tk))|=0,limn→∞|Qk(un(tk))−Qk(u(tk))|=0,k=1,2,⋯,m, |
limn→∞|γun,0−γu,0|=0,limn→∞|γun,1−γu,1|=0. |
By (H1), Lemma 3.2 and Lebesgue dominated convergence theorem, for any t∈J, we have
|Tun(t)−Tu(t)|=|∫10G(t,s)(f(s,un(s),u′n(s))−f(s,u(s),u′(s)))ds+(φun(t)−φu(t))+(tE2−α,2(λt2−α)n1+n2E2−α,1(λ)−m2E2−α,2(λ)(m1+m2)(n1+n2E2−α,1(λ)))(γun,1−γu,1)+γun,0−γu,0m1+m2|≤∫10G(1,s)|f(s,un(s),u′n(s))−f(s,u(s),u′(s))|ds+|m1+m2|+|m2||m1+m2|m∑i=1|Ii(un(ti))−Ii(u(ti))|+(|m1|+2|m2|)(|n1|+2|n2|E2−α,1(λ))E2−α,2(λ)|m1+m2||n1+n2E2−α,1(λ)|m∑i=1|Qi(un(ti))−Qi(u(ti))|+(|m1|+2|m2|)E2−α,2(λ)|m1+m2||n1+n2E2−α,1(λ)||γun,1−γu,1|+|γun,0−γu,0||m1+m2|→0(n→∞), |
and
|(Tun)′(t)−(Tu)′(t)|=|∫10G′t(t,s)(f(s,un(s),u′n(s))−f(s,u(s),u′(s)))ds+(φ′un(t)−φ′u(t))+E2−α,1(λt2−α)n1+n2E2−α,1(λ)(γun,1−γu,1)|≤∫10G′t(1,s)|f(s,un(s),u′n(s))−f(s,u(s),u′(s))|ds+m∑i=1(|n1|+2|n2|E2−α,1(λ))E2−α,1(λ)|n1+n2E2−α,1(λ)||Qi(un(ti))−Qi(u(ti))|+E2−α,1(λ)|n1+n2E2−α,1(λ)||γun,1−γu,1|→0(n→∞). |
So ||Tun−Tu||→0 as n→∞, which means T is continuous.
Step 2: T is relatively compact.
Let Ω⊂PC1(J) be a bounded set. By the continuity of the functions f,Ik,Qk(k=1,2,⋯,m), g0 and g1, there exists a constant L>0, for any u∈Ω and t∈J,
|f(t,u(t),u′(t))|≤L,|γu,0|≤L,|γu,1|≤L, |
|Ik(u(t))|≤L,|Qk(u(t))|≤L,k=1,2,⋯,m. |
Then
|φu(t)|≤m(|m1|+2|m2|)(|n1|(1+E2−α,2(λ))+|n2|E2−α,1(λ)(1+2E2−α,2(λ)))|m1+m2||n1+n2E2−α,1(λ)|L, |
|φ′u(t)|≤m(|n1|+2|n2|E2−α,1(λ))E2−α,1(λ)|n1+n2E2−α,1(λ)|L. |
By Lemma 3.2, for any t∈J,
|Tu(t)|=|∫10G(t,s)f(s,u(s),u′(s))ds+φu(t)+γu,0m1+m2+(tE2−α,2(λt2−α)n1+n2E2−α,1(λ)−m2E2−α,2(λ)(m1+m2)(n1+n2E2−α,1(λ)))γu,1|≤(∫10G(1,s)ds+(|m1|+2|m2|)E2−α,2(λ)+|n1|+|n2|E2−α,1(λ)|m1+m2||n1+n2E2−α,1(λ)|+m(|m1|+2|m2|)(|n1|(1+E2−α,2(λ))+|n2|E2−α,1(λ)(1+2E2−α,2(λ)))|m1+m2||n1+n2E2−α,1(λ)|)L, |
and
|(Tu)′(t)|=|∫10G′t(t,s)f(s,u(s),u′(s))ds+φ′u(t)+E2−α,1(λt2−α)n1+n2E2−α,1(λ)γu,1|≤(∫10G′t(1,s)ds+m(|n1|+2|n2|E2−α,1(λ))E2−α,1(λ)+E2−α,1(λ)|n1+n2E2−α,1(λ)|)L. |
Therefore, the operator T(Ω) is uniformly bounded.
Because G(t,s) is continuous on [0,1]×[0,1], then it is uniformly continuous on [0,1]×[0,1]. Thus, for any ε>0, there exists a constant δ1>0 such that for any s∈[0,1], t′1,t′2∈Jk,k=0,1,⋯,m, whenever |t′1−t′2|<δ1, we can get that
|G(t′1,s)−G(t′2,s)|<ε2L. |
Denote
G0(t,s)=−n2tE2−α,2(λt2−α)E2−α,1(λ(1−s)2−α)n1+n2E2−α,1(λ). |
By (3.6),
G′t(t,s)={E2−α,1(λ(t−s)2−α)+G0(t,s),0≤s≤t≤1,G0(t,s),0≤t<s≤1. |
Similarly, for the ε>0, there exists a constant δ2>0 such that
|G0(t′1,s)−G0(t′2,s)|<ε3L |
whenever |t′1−t′2|<δ2 and t′1,t′2∈Jk,k=0,1,⋯,m.
By the uniformly continuity of functions tE2−α,2(λt2−α) and E2−α,1(λt2−α) on t∈[0,1], we can show that for the ε>0, there exists a constant δ3>0 such that
|t′1E2−α,2(λt′12−α)−t′2E2−α,2(λt′22−α)|<|n1+n2E2−α,1(λ)|ε2(m(|n1|+2|n2|E2−α,1(λ))+1)L, |
and
|E2−α,1(λt′2−α1)−E2−α,1(λt′22−α)|<|n1+n2E2−α,1(λ)|ε3(m(|n1|+2|n2|E2−α,1(λ))+1)L, |
whenever |t′1−t′2|<δ3 and t′1,t′2∈Jk,k=0,1,⋯,m. Hence, by (4.2) and (4.3), we have
|φu(t′1)−φu(t′2)|≤m∑i=1|t′1E2−α,2(λt′12−α)−t′2E2−α,2(λt′22−α)||Qi(u(ti))|+m∑i=1|n2|E2−α,1(λ)|t′1E2−α,2(λt′12−α)−t′2E2−α,2(λt′22−α)||n1+n2E2−α,1(λ)||Qi(u(ti))|≤mL(|n1|+2|n2|E2−α,1(λ))|n1+n2E2−α,1(λ)||t′1E2−α,2(λt′12−α)−t′2E2−α,2(λt′22−α)|, |
and
|φ′u(t′1)−φ′u(t′2)|≤m∑i=1|E2−α,1(λt′12−α)−E2−α,1(λt′22−α)||Qi(u(ti))|+m∑i=1|n2|E2−α,1(λ)|E2−α,1(λt′12−α)−E2−α,1(λt′22−α)||n1+n2E2−α,1(λ)||Qi(u(ti))|≤mL(|n1|+2|n2|E2−α,1(λ))|n1+n2E2−α,1(λ)||E2−α,1(λt′12−α)−E2−α,1(λt′22−α)|. |
By the uniformly continuity of E2−α,1(λ(t−s)2−α) on D={(t,s)|0≤s≤t≤1}, for the ε>0, there exists a constant 0<δ4<ε6LE2−α,1(λ), for (t′1,s),(t′2,s)∈D, |t′1−t′2|<δ4 and t′1,t′2∈Jk,k=0,1,⋯,m, we have
|E2−α,1(λ(t′1−s)2−α)−E2−α,1(λ(t′2−s)2−α)|<ε6L. |
Then
|∫t′10E2−α,1(λ(t′1−s)2−α)f(s,u(s),u′(s))ds−∫t′20E2−α,1(λ(t′2−s)2−α)f(s,u(s),u′(s))ds|≤|∫t′10E2−α,1(λ(t′1−s)2−α)f(s,u(s),u′(s))ds−∫t′10E2−α,1(λ(t′2−s)2−α)f(s,u(s),u′(s))ds−∫t′2t′1E2−α,1(λ(t′2−s)2−α)f(s,u(s),u′(s))ds|≤L∫t′10|(E2−α,1(λ(t′1−s)2−α)−E2−α,1(λ(t′2−s)2−α))|ds+L|∫t′2t′1E2−α,1(λ(t′2−s)2−α)ds|≤L|(E2−α,1(λ(t′1−s)2−α)−E2−α,1(λ(t′2−s)2−α))|+L|E2−α,1(λ)||t′2−t′1|<ε3. |
We take δ=min{δ1,δ2,δ3,δ4}. Therefore, for any ε>0, there exists a constant δ>0 such that for t′1,t′2∈Jk,k=0,1,⋯,m whenever |t′1−t′2|<δ and any u∈Ω, we can get that
|Tu(t′1)−Tu(t′2)|≤∫10|G(t′1,s)−G(t′2,s)||f(s,u(s),u′(s))|ds+|φu(t′1)−φu(t′2)|+|t′1E2−α,2(λt′2−α1)−t′2E2−α,2(λt′2−α2)||n1+n2E2−α,1(λ)|γu,1≤L∫10|G(t′1,s)−G(t′2,s)|ds+|φu(t′1)−φu(t′2)|+L|n1+n2E2−α,1(λ)||t′1E2−α,2(λt′12−α)−t′2E2−α,2(λt′22−α)|<ε2+(m(|n1|+2|n2|E2−α,1(λ))+1)L|n1+n2E2−α(λ)||t′1E2−α,2(λt′12−α)−t′2E2−α,2(λt′22−α)|<ε2+ε2=ε, |
and
|(Tu)′(t′1)−(Tu)′(t′2)|=|∫t′10E2−α,1(λ(t′1−s)2−α)f(s,u(s),u′(s))ds−∫t′20E2−α,1(λ(t′2−s)2−α)f(s,u(s),u′(s))ds+∫10(G0(t′1,s)−G0(t′1,s))f(s,u(s),u′(s))ds+φ′u(t′1)−φ′u(t′2)+γu,1n1+n2E2−α,1(λ)(E2−α,1(λt′2−α1)−E2−α,1(λt′22−α))|≤|∫t′10E2−α,1(λ(t′1−s)2−α)f(s,u(s),u′(s))ds−∫t′20E2−α,1(λ(t′2−s)2−α)f(s,u(s),u′(s))ds|+L|G0(t′1,s)−G0(t′2,s)|+(m(|n1|+2|n2|E2−α,1(λ))+1)L|n1+n2E2−α(λ)||E2−α,1(λt′12−α)−E2−α,1(λt′22−α)|<ε3+ε3+ε3=ε. |
Thus, the operator T(Ω) is equicontinuous on every interval Jk.
According to the Arzela-Ascoli theorem, T(Ω) is relatively compact.
Therefore, T:PC1(J)→PC1(J) is completely continuous.
In the following, we give out some hypotheses.
(H2) f(t,x1,y1)≤f(t,x2,y2) if t∈J, x1≤x2, y1≤y2∈R. Furthermore, f(t,x1,y1)<f(t,x2,y2) if x1<x2 and y1≤y2.
And for x1≤x2∈R, Ik(x1)≤Ik(x2),Qk(x1)≤Qk(x2),k=1,2,⋯,m.
(H3.1) If m1+m2>0,m2<0,n1+n2E2−α,1(λ)>0,n2<0, then for y1≤y2, z1≤z2∈R,
g0(z2,y2)−g0(z1,y1)≥−m1(z2−z1)−m2(y2−y1), |
and
g1(z2,y2)−g1(z1,y1)≥−n1(z2−z1)−n2(y2−y1). |
(H3.2) If m1+m2<0,m2>0,n1+n2E2−α,1(λ)>0,n2<0, then for y1≤y2, z1≤z2∈R,
g0(z2,y2)−g0(z1,y1)≤−m1(z2−z1)−m2(y2−y1), |
and
g1(z2,y2)−g1(z1,y1)≥−n1(z2−z1)−n2(y2−y1). |
(H3.3) If m1+m2>0,m2<0,n1+n2E2−α,1(λ)<0,n2>0, then for y1≤y2, z1≤z2∈R,
g0(z2,y2)−g0(z1,y1)≥−m1(z2−z1)−m2(y2−y1), |
and
g1(z2,y2)−g1(z1,y1)≤−n1(z2−z1)−n2(y2−y1). |
(H3.4) If m1+m2<0,m2>0,n1+n2E2−α,1(λ)<0,n2>0, then for y1≤y2, z1≤z2∈R,
g0(z2,y2)−g0(z1,y1)≤−m1(z2−z1)−m2(y2−y1), |
and
g1(z2,y2)−g1(z1,y1)≤−n1(z2−z1)−n2(y2−y1). |
Remark. It is easy to see that if one of (H3.1)–(H3.4) holds, then (H1) holds.
Lemma 4.3. For mi,ni∈R(i=1,2), if one of the following conditions holds, then x(t)≥0 and x′(t)≥0 for t∈J.
(1) m1+m2>0,m2<0,n1+n2E2−α,1(λ)>0,n2<0 and x∈PC1(J) satisfies
{x″(t)−λcDα0+x(t)≥0,t∈J′,Δx(t)|t=tk≥0,k=1,2,⋯,m,Δx′(t)|t=tk≥0,k=1,2,⋯,m,m1x(0)+m2x(1)≥0,n1x′(0)+n2x′(1)≥0. |
(2) m1+m2<0,m2>0,n1+n2E2−α,1(λ)>0,n2<0 and x∈PC1(J) satisfies
{x″(t)−λcDα0+x(t)≥0,t∈J′,Δx(t)|t=tk≥0,k=1,2,⋯,m,Δx′(t)|t=tk≥0,k=1,2,⋯,m,m1x(0)+m2x(1)≤0,n1x′(0)+n2x′(1)≥0. |
(3) m1+m2>0,m2<0,n1+n2E2−α,1(λ)<0,n2>0 and x∈PC1(J) satisfies
{x″(t)−λcDα0+x(t)≥0,t∈J′,Δx(t)|t=tk≥0,k=1,2,⋯,m,Δx′(t)|t=tk≥0,k=1,2,⋯,m,m1x(0)+m2x(1)≥0,n1x′(0)+n2x′(1)≤0. |
(4) m1+m2<0,m2>0,n1+n2E2−α,1(λ)<0,n2>0 and x∈PC1(J) satisfies
{x″(t)−λcDα0+x(t)≥0,t∈J′,Δx(t)|t=tk≥0,k=1,2,⋯,m,Δx′(t)|t=tk≥0,k=1,2,⋯,m,m1x(0)+m2x(1)≤0,n1x′(0)+n2x′(1)≤0. |
Proof. (1) Denote x″(t)−λcDα0+x(t)=h(t),Δx(t)|t=tk=pk,Δx′(t)|t=tk=qk,k=1,2,⋯,m,m1x(0)+m2x(1)=γ0,n1x′(0)+n2x′(1)=γ1. Then h(t)≥0,pk≥0,qk≥0,k=1,2,⋯,m.
By Lemma 3.1, the following boundary value problem
{x″(t)−λcDα0+x(t)=h(t),t∈J′,Δx(t)|t=tk=pk,k=1,2,⋯,m,Δx′(t)|t=tk=qk,k=1,2,⋯,m,m1x(0)+m2x(1)=γ0,n1x′(0)+n2x′(1)=γ1, |
has a unique solution
x(t)=∫t0G(t,s)h(s)ds+φ(t)+(tE2−α,2(λt2−α)n1+n2E2−α,1(λ)−m2E2−α,2(λ)(m1+m2)(n1+n2E2−α,1(λ)))γ1+γ0m1+m2,t∈J, |
and
x′(t)=∫10G′t(t,s)h(s)ds+φ′(t)+E2−α,1(λt2−α)n1+n2E2−α,1(λ)γ1,t∈J. |
It's easy to see that x(t)≥0 and x′(t)≥0 for t∈J.
Similarly, (2)–(4) are easy to be proved.
Lemma 4.4. Suppose (H2) and one of (H3.1)-(H3.4) holds, then T is a strongly increasing operator.
Proof. Here, we will only prove the conclusion when (H3.1) holds, and other situations are similar.
For any u1,u2∈PC1(J), and u1≺u2 which implies that u1(t)≤u2(t),u′1(t)≤u′2(t) and u1(t)≢u2(t) for t∈J. By (H2), for any t∈J,
f(t,u2(t),u′2(t))−f(t,u1(t),u′1(t))≥0,Ik(u2(tk))−Ik(u1(t2))≥0,Qk(u2(tk))−Qk(u2(tk))≥0. |
Since u1(t)≢u2(t), there exists an interval [a,b]⊂J such that u1(t)<u2(t) for t∈[a,b]. It follows from (H2)
f(t,u2(t),u′2(t))−f(t,u1(t),u′1(t))>0,t∈[a,b]. | (4.4) |
By (H3.1), we can get that
γu2,0−γu1,0=g0(u2(0),u2(1))−g0(u1(0),u1(1))+(m1u2(0)+m2u2(1))−(m1u1(0)+m2u1(1))≥0,γu2,1−γu1,1=g1(u2(0),u2(1))−g1(u1(0),u1(1))+(n1u2(0)+n2u2(1))−(n1u1(0)n2u1(1))≥0. |
By (4.2), (4.3), (H3.1) and (H2), we can get that for t∈J,
φu2(t)−φu1(t)=∑0<ti<t(Ii(u2(ti))−Ii(u1(ti)))−m2m1+m2m∑i=1(Ii(u2(ti))−Ii(u1(ti)))+∑0<ti<ttE2−α,2(λt2−α)−tiE2−α,2(λt2−αi)E2−α,1(λt2−αi)(Qi(u2(ti))−Qi(u1(ti)))+m∑i=1(−m2(E2−α,2(λ)−tiE2−α,1(λt2−αi))(m1+m2)E2−α,1(λt2−αi)−n2E2−α,1(λ)(tE2−α,2(λt2−α))(n1+n2E2−α,1(λ))E2−α,1(λt2−αi)+m2n2E2−α,2(λ)E2−α,1(λ)(m1+m2)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi))(Qi(u2(ti))−Qi(u1(ti)))≥∑0<ti<t(Ii(u2(ti))−Ii(u1(ti)))−m2m1+m2m∑i=1(Ii(u2(ti))−Ii(u1(ti)))+m∑i=1m2n2E2−α,2(λ)E2−α,1(λ)(m1+m2)(n1+n2E2−α,1(λ))E2−α,1(λ)(Qi(u2(ti))−Qi(u1(ti)))≥0, |
and
φ′u2(t)−φ′u1(t)=∑0<ti<tE2−α,1(λt2−α)E2−α,1(λt2−αi)(Qi(u2(ti))−Qi(u1(ti)))−m∑i=1n2E2−α,1(λ)E2−α,1(λt2−α)(n1+n2E2−α,1(λ))E2−α,1(λt2−αi)(Qi(u2(ti))−Qi(u1(ti)))≥∑0<ti<t1E2−α,1(λ)(Qi(u2(ti))−Qi(u1(ti)))−m∑i=1n2E2−α,1(λ)(n1+n2E2−α,1(λ))E2−α,1(λ)(Qi(u2(ti))−Qi(u1(ti)))≥0. |
By Lemma 3.2, (H2) and (4.4), for any u1,u2∈PC1(J) and u1≺u2, as t∈J, we have
Tu2(t)−Tu1(t)=∫10G(t,s)(f(s,u2(s),u′2(s))−f(s,u1(s),u′1(s)))ds+(φu2(t)−φu1(t))+(tE2−α,2(λt2−α)n1+n2E2−α,1(λ)−m2E2−α,2(λ)(m1+m2)(n1+n2E2−α,1(λ)))(γu2,1−γu1,1)+γu2,0−γu1,0m1+m2≥∫baG(0,s)(f(s,u2(s),u′2(s))−f(s,u1(s),u′1(s)))ds>0, |
and
(Tu2)′(t)−(Tu1)′(t)=∫10G′t(t,s)(f(s,u2(s),u′2(s))−f(s,u1(s),u′1(s)))ds+(φ′u2(t)−φ′u1(t))+E2−α,1(λt2−α)n1+n2E2−α,1(λ)(γu2,1−γu1,1)≥∫baG′t(0,s)(f(s,u2(s),u′2(s))−f(s,u1(s),u′1(s)))ds>0. |
Thus,
Tu2−Tu1∈˚P |
which implies that T is a strongly increasing operator.
The proof is completed.
Theorem 4.5. If (H2) and (H3.1) hold, there exist zi,yi∈PC1(J),i=1,2, with z1≺y1≺z2≺y2 such that
{z″i(t)−λcDα0+zi(t)≤f(t,zi(t),z′i(t)),t∈J′,Δzi(t)|t=tk≤Ik(zi(tk)),k=1,2,⋯,m,Δz′i(t)|t=tk≤Qk(zi(tk)),k=1,2,⋯,m,g0(zi(0),zi(1))≥0,g1(z′i(0),z′i(1))≥0, | (4.5) |
and
{y″i(t)−λcDα0+yi(t)≥f(t,yi(t),y′i(t)),t∈J′,Δyi(t)|t=tk≥Ik(yi(tk)),k=1,2,⋯,m,Δy′i(t)|t=tk≥Qk(yi(tk)),k=1,2,⋯,m,g0(yi(0),yi(1))≤0,g1(y′i(0),y′i(1))≤0, | (4.6) |
where i=1,2, z2 and y1 are not the solutions of boundary value problem (1.1). Then boundary value problem (1.1) has at least three distinct solutions x1,x2,x3∈[z1,y2] and satisfies
z1(t)≤x1(t)<y1(t),z2(t)<x2(t)≤y2(t),z2(t)≰x2(t)≰y1(t),t∈J. |
Proof. By Lemma 4.2 and Lemma 4.4, we can get that T:[z1,y2]→PC1(J) is a completely continuous strongly increasing operator.
By the definition of operator T, we can show that
{(Tz1)″(t)−λcDα0+(Tz1)(t)=f(t,z1(t),z′1(t)),t∈J′,Δ(Tz1)(t)|t=tk=Ik(z1(tk)),k=1,2,⋯,m,Δ(Tz1)′(t)|t=tk=Qk(z1(tk)),k=1,2,⋯,m,m1(Tz1)(0)+m2(Tz1)(1)=g0(z1(0),z1(1))+m1z1(0)+m2z1(1),n1(Tz1)′(0)+n2(Tz1)′(1)=g1(z′1(0),z′1(1))+n1z′1(0)+n2z′1(1). |
Let x=Tz1−z1. By (4.5),
x″(t)−λcDα0+x(t)=((Tz1)″(t)−z″1(t))−λ(cDα0+(Tz1)(t)−cDα0+z1(t))=((Tz1)″(t)−λcDα0+(Tz1)(t))−(z″1(t)−λcDα0+z1(t))=f(t,z1(t),z′1(t))−(z″1(t)−λcDα0+z1(t))≥0,t∈J′. |
Δx(t)|t=tk=ΔTz1(t)|t=tk−Δz1(t)|t=tk=Ik(z1(tk))−Δz1(t)|t=tk≥0.Δx′(t)|t=tk=Δ(Tz1)′(t)|t=tk−Δz′1(t)|t=tk=Qk(z1(tk))−Δz′1(t)|t=tk≥0. |
m1x(0)+m2x(1)=m1((Tz1)(0)−z1(0))+m2((Tz1)(1)−z1(1))=m1((Tz1)(0)+m2(Tz1)(1))−(m1z1(0)+m2z1(1))=g0(z1(0),z1(1))+(m1z1(0)+m2z1(1))−(m1z1(0)+m2z1(1))=g0(z1(0),z1(1))≥0. |
Similarly, we can get that
n1x′(0)+n2x′(1)≥0. |
By (1) in Lemma 4.3, we have
x(t)=(Tz1)(t)−z1(t)≥0,x′(t)=(Tz1)′(t)−z′1(t)≥0,t∈J. |
Therefore, z1≺_Tz1.
It is similar that we can obtain z2≺_Tz2. Because z2 is not the solution of boundary value problem (1.1), then Tz2≠z2. Thus, z2≺Tz2.
By using the same method, we can easily get Ty1≺y1,Ty2≺_y2.
By using Lemma 2.6, we can get that the operator T has at least three distinct fixed points x1,x2,x3∈[z1,y2], and satisfies
z1≺_x1≺≺y1,z2≺≺x2≺_y2,z2⊀_x3⊀_y1. |
Therefore, by Lemma 4.1, we can obtain that boundary value problem (1.1) has at least three distinct solutions x1,x2,x3∈[z1,y2], and
z1(t)≤x1(t)<y1(t),z2(t)<x2(t)≤y2(t),z2(t)≰x2(t)≰y1(t),t∈J. |
The proof is completed.
In the similar way, the following three theorems can be established.
Theorem 4.6. If (H2) and (H3.2) hold, there exist zi,yi∈PC1(J),i=1,2, with z1≺y1≺z2≺y2 such that
{z″i(t)−λcDα0+zi(t)≤f(t,zi(t),z′i(t)),t∈J′,Δzi(t)|t=tk≤Ik(zi(tk)),k=1,2,⋯,m,Δz′i(t)|t=tk≤Qk(zi(tk)),k=1,2,⋯,m,g0(zi(0),zi(1))≤0,g1(z′i(0),z′i(1))≥0, |
and
{y″i(t)−λcDα0+yi(t)≥f(t,yi(t),y′i(t)),t∈J′,Δyi(t)|t=tk≥Ik(yi(tk)),k=1,2,⋯,m,Δy′i(t)|t=tk≥Qk(yi(tk)),k=1,2,⋯,m,g0(yi(0),yi(1))≥0,g1(y′i(0),y′i(1))≤0, |
where i=1,2, z2 and y1 are not the solutions of boundary value problem (1.1). Then the boundary value problem (1.1) has at least three distinct solutions x1,x2,x3∈[z1,y2] and satisfies
z1(t)≤x1(t)<y1(t),z2(t)<x2(t)≤y2(t),z2(t)≰x2(t)≰y1(t),t∈J. |
Theorem 4.7. If (H2) and (H3.3) hold, there exist zi,yi∈PC1(J),i=1,2, with z1≺y1≺z2≺y2 such that
{z″i(t)−λcDα0+zi(t)≤f(t,zi(t),z′i(t)),t∈J′,Δzi(t)|t=tk≤Ik(zi(tk)),k=1,2,⋯,m,Δz′i(t)|t=tk≤Qk(zi(tk)),k=1,2,⋯,m,g0(zi(0),zi(1))≥0,g1(z′i(0),z′i(1))≤0, |
and
{y″i(t)−λcDα0+yi(t)≥f(t,yi(t),y′i(t)),t∈J′,Δyi(t)|t=tk≥Ik(yi(tk)),k=1,2,⋯,m,Δy′i(t)|t=tk≥Qk(yi(tk)),k=1,2,⋯,m,g0(yi(0),yi(1))≤0,g1(y′i(0),y′i(1))≥0, |
where i=1,2, z2 and y1 are not the solutions of boundary value problem (1.1). Then the boundary value problem (1.1) has at least three distinct solutions x1,x2,x3∈[z1,y2] and satisfies
z1(t)≤x1(t)<y1(t),z2(t)<x2(t)≤y2(t),z2(t)≰x2(t)≰y1(t),t∈J. |
Theorem 4.8. If (H2) and (H3.4) hold, there exist zi,yi∈PC1(J),i=1,2, with z1≺y1≺z2≺y2 such that
{z″i(t)−λcDα0+zi(t)≤f(t,zi(t),z′i(t)),t∈J′,Δzi(t)|t=tk≤Ik(zi(tk)),k=1,2,⋯,m,Δz′i(t)|t=tk≤Qk(zi(tk)),k=1,2,⋯,m,g0(zi(0),zi(1))≤0,g1(z′i(0),z′i(1))≤0, |
and
{y″i(t)−λcDα0+yi(t)≥f(t,yi(t),y′i(t)),t∈J′,Δyi(t)|t=tk≥Ik(yi(tk)),k=1,2,⋯,m,Δy′i(t)|t=tk≥Qk(yi(tk)),k=1,2,⋯,m,g0(yi(0),yi(1))≥0,g1(y′i(0),y′i(1))≥0, |
where i=1,2, z2 and y1 are not the solutions of boundary value problem (1.1). Then the boundary value problem (1.1) has at least three distinct solutions x1,x2,x3∈[z1,y2] and satisfies
z1(t)≤x1(t)<y1(t),z2(t)<x2(t)≤y2(t),z2(t)≰x2(t)≰y1(t),t∈J. |
In this section, we discuss the applicability of our main results.
Consider the following the boundary value problem
{x″(t)−1100cD120+x(t)=2t3πarctan(x(t)+x′(t)),t∈(0,12)∪(12,1),Δx(t)|t=12=12x(12),Δx′(t)|t=12=1209x(12),−cos(2πx(0))+0.02x(1)−0.52=0,−6πcos(πx′(0))+πx′(1)−6π=0, | (5.1) |
where α=12,λ=1100,f(t,x,y)=2t3πarctan(x+y),I1(x)=12x,Q1(x)=1209x.
g0(x,y)=−cos(2πx)+0.02y−0.52,g1(x,y)=−6πcos(πx)+πy−6π. |
Obviously, f∈C([0,1]×R2,R),I1,Q1∈C(R,R),g0,g1∈C(R2,R) are nonlinear functions. And f,I1,Q1 satisfy (H2).
Let m1=2π,m2=−0.02,n1=6π,n2=−π. Since
g0(x2,y2)−g0(x1,y1)=−cos(2πx2)+cos(2πx1)+0.02(y2−y1)≥−2π(x2−x1)+0.02(y2−y1),g1(x2,y2)−g1(x1,y1)=−6πcos(πx2)+6πcos(πx1)+π(y2−y1)≥−6π(x2−x1)+π(y2−y1), |
then g0,g1 satisfy (H3.1).
For t∈[0,1], we take
z1(t)={32+t,t∈[0,12],2+t,t∈(12,1],y1(t)={4+32t+18t2+14t4,t∈[0,12],9+32t+16t2+14t4,t∈(12,1], |
z2(t)={232+3t,t∈[0,12],12+3t,t∈(12,1],y2(t)={16+72t+25t2+16t4,t∈[0,12],33+72t+34t2+16t4,t∈(12,1]. |
We can easily get that
z″1(t)−1100cD120+z1(t)−2t3πarctan(z1(t)+z′1(t))={−√t50√π−2t3arctan(52+t)π<0,t∈(0,12),−√t50√π−2t3arctan(3+t)π<0,t∈(12,1), |
y″1(t)−1100cD120+y1(t)−2t3πarctan(y1(t)+y′1(t))={14+3t2−√t(315+35t+96t3)10500√π−2t3arctan(112+7t4+t28+t3+t44)π>0,t∈[0,12],84000√π√t(1+9t2)−7560t−840t2−2304t4−35√2(√t(2t−1)+4√t3(2t−1))252000√π√t−2t3arctan(212+11t6+t26+t3+t44)π>0,t∈(12,1], |
z″2(t)−1100cD120+z2(t)−2t3πarctan(z2(t)+z′2(t))={−3√t50√π−2t3arctan(292+3t)π<0,t∈(0,12),−3√t50√π−2t3arctan(15+3t)π<0,t∈(12,1), |
and
y″2(t)−1100cD120+y2(t)−2t3πarctan(y2(t)+y′2(t))={45+2t2−√t(735+112t+64t3)10500√π−2t3arctan(392+43t10+2t25+2t33+t46)π>0,t∈[0,12],21000√π√t(3+4t2)−2940t−448t2−256t4−49√2(√t(2t−1)+4√t3(2t−1))42000√π√t−2t3arctan(732+5t+3t24+2t33+t46)π>0,t∈(12,1]. |
So
{z″1(t)−1100cD120+z1(t)<2t3πarctan(z1(t)+z′1(t)),t∈(0,12)∪(12,1),Δz1(t)|t=12=z1(12+)−z1(12−)=12<12z1(12)=1,Δz′1(t)|t=12=z′1(12+)−z′1(12−)=0<1209z1(12)=2209,−cos(2πz1(0))+0.02z1(1)−0.52>0,−6πcos(πz′1(0))+πz′1(1)−6π>0, |
{y″1(t)−1100cD120+y1(t)>2t3πarctan(y1(t)+y′1(t)),t∈(0,12)∪(12,1),Δy1(t)|t=12=y1(12+)−y1(12−)=48196>12y1(12)=307128,Δy′1(t)|t=12=y′1(12+)−y′1(12−)=124>1209y1(12)=30713376,−cos(2πy1(0))+0.02y1(1)−0.52<0,−6πcos(πy′1(0))+πy′1(1)−6π<0, |
{z″2(t)−1100cD120+z2(t)<2t3πarctan(z2(t)+z′2(t)),t∈(0,12)∪(12,1),Δz2(t)|t=12=z2(12+)−z2(12−)=12<12z2(12)=132,Δz′2(t)|t=12=z′2(12+)−z′2(12−)=0<1209z2(12)=13209,−cos(2πz2(0))+0.02z2(1)−0.52>0,−6πcos(πz′2(0))+πz′2(1)−6π>0 |
and
{y″2(t)−1100cD120+y2(t)>2t3πarctan(y2(t)+y′2(t)),t∈(0,12)∪(12,1),Δy2(t)|t=12=y2(12+)−y2(12−)=136780>12y2(12)=8573960,Δy′2(t)|t=12=y′2(12+)−y′2(12−)=720>1209y2(12)=8573100320,−cos(2πy2(0))+0.02y2(1)−0.52<0,−6πcos(πy′2(0))+πy′2(1)−6π<0. |
We can easily get that z1(t),z2(t) satisfy (4.5) and y1(t),y2(t) satisfy (4.6) with z1≺y1≺z2≺y2.
The conditions of Theorem 4.5 are all satisfied. So by Theorem 4.5, the boundary value problem (5.1) has at least three distinct solutions x1,x2,x3∈[z1,y2], and moreover,
z1(t)≤x1(t)<y1(t),z2(t)<x2(t)≤y2(t),z2(t)≰x3(t)≰y1(t),t∈[0,1]. |
In this work, we investigate the existence of solutions for a class of second order impulsive vibration equation with fractional derivatives. Some sufficient conditions for existence of the multiplicity solutions are established by applying monotone iterative technique. Finally, a concrete example is given to illustrate the wide range of potential applications of our main results.
Further extensions of this paper are to study the motion state of the vibrator in the system described by boundary value problem (1.1) and the existence of solutions to the boundary value problems with other boundary conditions. Moreover, fractional differential equation models such as the rheological model of the fractional derivative and singular systems model of fractional differential equations have real world applications. So, we also can consider using the boundary value problem of impulsive differential equations to simulate the abrupt changes in the systems described by these models.
The authors would like to thank college mathematics characteristic pilot team of University of Shanghai for science and technology for its support to this project.
The authors declare that they have no conflicts of interest.
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