Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 94, pp. 1–22. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu PERIODIC SOLUTIONS FOR CONFORMABLE TYPE NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS YUANLIN DING, JINRONG WANG Abstract. In this article we study a type of conformable non-instantaneous impulsive equation with periodic effects. We find a Cauchy matrix that can provide solutions of linear and nonlinear problems and prove some of their properties. Also we study the existence of periodic solution of different types of conformable non-instantaneous impulsive differential equation. Some examples also are given to illustrate our theoretical results. 1. Introduction Hernández and O’Regan [11] introduced the concept of non-instantaneous im- pulsive equations. After that many results about this equations have been ob- tained, including periodic differential equations with periodic non-instantaneous impulses; see for example [1, 4, 6, 7, 10, 12, 15, 17, 18, 20, 23, 24, 27]. In [9] an effective framework is given for obtaining periodic solutions of first-order pe- riodic non-instantaneous impulsive problems. Fečkan et al. [19] studied the peri- odic solutions of second order non-instantaneous impulsive problems. Wang et al. [13, 14, 22, 25, 28, 29, 30] established many results about periodic solutions with non-instantaneous impulses. Abdeljawad [2] introduced the concept of conformable derivatives and studied its basic theory. Recently, many articles about conformable derivatives have appeared, see [3, 5, 8, 16, 21, 26]. In this article, we study the conformable homogeneous linear non-instantaneous impulsive differential equation Dσk β z(ι) = Pz(ι), ι ∈ (σk, ιk+1], k = 0, 1, 2, . . . , z(ι+k ) = Qz(ι−k ), k = 1, 2, . . . , z(ι) = Qz(ι−k ), ι ∈ (ιk, σk], k = 1, 2, . . . , z(σ+ k ) = z(σ−k ), k = 1, 2, . . . , z(a) = za ∈ Rn, (1.1) 2010 Mathematics Subject Classification. 34A37, 34C25. Key words and phrases. Periodic solution; conformable derivative; Cauchy matrix; non-instantaneous impulsive differential equations. ©2021. This work is licensed under a CC BY 4.0 license. Submitted August 18, 2021. Published November 30, 2021. 1 2 Y. DING, J. WANG EJDE-2021/94 the conformable non-homogeneous linear non-instantaneous impulsive differential equation Dσk β z(ι) = Pz(ι) + h(ι), ι ∈ (σk, ιk+1], k = 0, 1, 2, . . . z(ι+k ) = Qz(ι−k ) + dk, k = 1, 2, . . . , z(ι) = Qz(ι−k ) + dk, ι ∈ (ιk, σk], k = 1, 2, . . . , z(σ+ k ) = z(σ−k ), k = 1, 2, . . . , z(a) = za ∈ Rn, (1.2) and the conformable nonlinear non-instantaneous impulsive differential equation Dσk β z(ι) = Pz(ι) + h(t, z(ι)), ι ∈ (σk, ιk+1], k = 0, 1, 2, . . . z(ι+k ) = Qz(ι−k ) + dk, k = 1, 2, . . . , z(ι) = Qz(ι−k ) + dk, ι ∈ (ιk, σk], k = 1, 2, . . . , z(σ+ k ) = z(σ−k ), k = 1, 2, . . . , z(a) = za ∈ Rn, (1.3) where P and Q are n× n constant matrices with PQ = QP , 0 < β < 1. ιk and σk satisfy a = σ0 < ι1 < σ1 < · · · < ιk < σk < ιk+1 . . . , k = 1, 2, . . . , dk ∈ Rn. Let E be the unit matrix, I = ⋃∞ k=0(σk, ιk+1] and J = ⋃∞ k=1(ιk, σk] and h(·) ∈ C(I,Rn), h(·, ·) ∈ C(I× Rn,Rn). We use the assumption (A1) ιk and σk satisfy ιk+q = ιk + T , σk+q = σk + T for n(a, T ) = q in which n(a, T ) denotes the number of impulsive points existing in (a, T ). dk are constant vectors with dk+q = dk + T . We set I = [a,+∞) and PC(I,Rn) := {y : I → Rn : y ∈ C ( (ιk, ιk+1],Rn ) , k = 0, 1, . . . . There exists z(ι−k ) and z(ι+k ), k = 1, 2, . . . with z(ι−k ) = z(ιk)}, where C ( (ιk, ιk+1],Rn ) denotes the space of all continuous functions from (ιk, ιk+1] into Rn. We denote a vector θ = (θ1, . . . , θn)> ∈ Rn with its norm ‖θ‖ = ∑n i=1 |θi| and a matrix κ : Rn → Rn with its matrix norm ‖κ‖ = max‖y‖=1 ‖κy‖. This article is organized as follows: In Section 2, we introduce some basic theory and give the solution of (1.1) and (1.2) for non-instantaneous impulsive Cauchy matrix W (·, ·). Also we give some properties of W (·, ·). Section 3 concerns the existence of T -periodic solutions of (1.2) with two types of conditions. In Section 4, two lemmas prove the existence of T -periodic solutions of (1.3). 2. Preliminaries Definition 2.1 (see [2, Definition 2.1]). The conformable derivative with lower index a of a function x : [a,∞)→ R is defined as Da βx(ι) = lim ε→0 x(ι+ ε(ι− a)1−β)− x(ι) ε , ι > a, 0 < β < 1, Da βx(a) = lim ι→a+ Da βx(ι). Remark 2.2 ([2]). If Da βx(ιa) exists and is finite, we say that x is β-differentiable at ιa. If x ∈ C1((a,∞],R), then Da βx(ι) = (ι−a)1−βx′(ι). The conformable derivative Da βx(ι) exists if and only if x is differentiable at ι and Da βx(ι) = (ι− a)1−βx′(ι) for ι > a. EJDE-2021/94 NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS 3 Definition 2.3 (see [2, Notation]). The conformable integral with lower index a of a function x : [a,∞)→ R is written as Jaβx(ι) = ∫ ι a x(σ)dβ(σ, a) = ∫ ι a (σ − a)β−1x(σ)dσ, ι ≥ a, 0 < β < 1, if a = 0, then we write dβ(σ, a) as dβ(σ). Lemma 2.4. The solution z(·, ·, ·) ∈ PC([a,∞) × [a,∞),Rn) of (1.1) with the initial condition z(σ) = zσ has the form z(ι) := z(ι, σ, zσ) = W (ι, σ)zσ, ι ≥ a, (2.1) in which W (ι, σ) = Qn(a,ι)−n(a,σ)e P β [( (ι−σn(a,ι)) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] , (2.2) where j+ := max{0, j}, j ∈ R. When n(a, ι) = n(a, σ), we have ∑n(a,ι)−1 k=n(a,σ) = 0. In particular with σ = a, z(ι, a, za) = Qn(a,ι)e P β [( (ι−σn(a,ι)) β )+ + ∑n(a,ι)−1 k=n(a,σ) ((ιk+1−σk)β ] za = W (ι, a)za. (2.3) Proof. There are several cases to be considered. Case 1:: n(a, ι) and n(a, σ) satisfy n(a, ι) = n(a, σ). (i) For any t, σ ∈ (σk, ιk+1], k = 0, 1, 2, . . . , when ι ∈ (a, ι1], we have z(ι) = e P β (ι−σ0) β za. When ι ∈ (ι1, σ1], we have z(ι) = Qz(ι−1 ) = Qe P β (ι1−σ0) β za. Then, for ι, σ ∈ (σ1, ι2], we have z(ι) = e P β (ι−σ1) β z(σ1) = e P β (ι−σ1) β Qe P β (ι1−σ0) β za, z(σ) = e P β (σ−σ1) β z(σ1) = e P β (σ−σ1) β Qe P β (ι1−σ0) β za, so W (ι, σ) = e P β ( (ι−σ1) β−(σ−σ1) β ) . In summary for any ι, σ ∈ (σk, ιk+1], it holds W (ι, σ) = e P β ( (ι−σk)β−(σ−σk)β ) . (ii) For any ι, σ ∈ (ιk, σk], k = 1, 2, . . . , we have z(ι) = Qz(ι−k ), k = 1, 2, . . . , so z(ι) = z(σ). (iii) For any ι ∈ (σn(a,ι), ιn(a,ι)+1) and any σ ∈ (ιn(a,σ), σn(a,σ)], we have z(ι) = e P β (ι−σn(a,ι)) β z(σ+ n(a,ι)) = e P β (ι−σn(a,ι)) β z(σ−n(a,ι)) = e P β (ι−σn(a,ι)) β z(σ), so W (ι, σ) = e p β (ι−σn(a,ι)) β . Case 2: n(a, ι) and n(a, σ) satisfy n(a, ι) = n(a, σ) + 1. 4 Y. DING, J. WANG EJDE-2021/94 (i) For any ι ∈ (ιn(a,ι), σn(a,ι)] and any σ ∈ (σn(a,σ), ιn(a,σ)+1], we have z(ι) = Qz(ι−n(a,ι)) and z(ι) = Qz(ι−n(a,ι)) = Qe P β ( (ιn(a,ι)−σn(a,σ)) β−(σ−σn(a,σ)) β ) z(σ), so W (ι, σ) = Qe P β ( (ιn(a,ι)−σn(a,σ)) β−(σ−σn(a,σ)) β ) . (ii) For any ι ∈ (σn(a,ι), ιn(a,ι)+1] and any σ ∈ (σn(a,σ), ιn(a,σ)+1], we have z(ι) = e P β (ι−σn(a,ι)) β z(σ+ n(a,ι)) = e P β (ι−σn(a,ι)) β z(σ−n(a,ι)) = e P β (ι−σn(a,ι)) β Qz(ι−n(a,ι)) = e P β (ι−σn(a,ι)) β Qe P β ( (ιn(a,ι)−σn(a,σ)) β−(σ−σn(a,σ)) β ) z(σ), so W (ι, σ) = Qe P β ( (ι−σn(a,ι)) β−(σ−σn(a,σ)) β+(ιn(a,ι)−σn(a,σ)) β ) . (iii) For any ι ∈ (ιn(a,ι), σn(a,ι)] and any σ ∈ (ιn(a,σ), σn(a,σ)],there is z(ι) = Qz(ι−n(a,ι)) = Qe P β (ιn(a,ι)−σn(a,σ)) β z(σ+ n(a,σ)) = Qe P β (ιn(a,ι)−σn(a,σ)) β z(σ−n(a,σ)) = Qe P β (ιn(a,ι)−σn(a,σ)) β z(σ), so W (ι, σ) = Qe P β (ιn(a,ι)−σn(a,σ)) β . (iv) For any ι ∈ (σn(a,ι), ιn(a,ι)+1] and any σ ∈ (ιn(a,σ), σn(a,σ)], we have z(ι) = e P β (ι−σn(a,ι)) β z(σ+ n(a,ι)) = e P β (ι−σn(a,ι)) β z(σ−n(a,ι)) = e P β (ι−σn(a,ι)) β Qz(ι−n(a,ι)) = e P β (ι−σn(a,ι)) β Qe P β (ιn(a,ι)−σn(a,σ)) β z(σ+ n(a,ι)) = e P β (ι−σn(a,ι)) β ) Qe P β (ιn(a,ι)−σn(a,σ)) β z(σ), so W (ι, σ) = Qe P β ( (ι−σn(a,ι)) β+(ιn(a,ι)−σn(a,σ)) β ) . Case 3: n(a, ι) and n(a, σ) satisfy n(a, ι) = n(a, σ) + 2. (i) For any ι ∈ (ιn(a,ι), σn(a,ι)] and any σ ∈ (σn(a,σ), ιn(a,σ)+1], we have z(ι) = Qz(ι−n(a,ι)) = Qe p β (ιn(a,ι)−σn(a,σ)+1) β z(σ+ n(a,σ)+1) = Qe P β (ιn(a,ι)−σn(a,σ)+1) β z(σ−n(a,σ)+1) = Qe P β (ιn(a,ι)−σn(a,σ)+1) β Qz(ι−n(a,σ)+1) EJDE-2021/94 NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS 5 = Qe P β (ιn(a,ι)−σn(a,σ)+1) β Qe P β ( (ιn(a,σ)+1−σn(a,σ)) β−(σ−σn(a,σ)) β ) z(σ), so that W (ι, σ) = Q2e P β ( −(σ−σn(a,σ)) β+(ιn(a,ι)−σn(a,σ)+1) β+(ιn(a,σ)+1−σn(a,σ)) β ) . (ii) For any ι ∈ (σn(a,ι), ιn(a,ι)+1] and any σ ∈ (σn(a,σ), ιn(a,σ)+1], we have z(ι) = e P β (ι−σn(a,ι)) β z(σ+ n(a,ι)) = e P β (ι−σn(a,ι)) β z(σ−n(a,ι)) = e P β (ι−σn(a,ι)) β Qz(ι−n(a,ι)) = e P β (ι−σn(a,ι)) β Qe P β (ιn(a,ι)−σn(a,σ)+1) β z(σ+ n(a,ι)−1) = e P β (ι−σn(a,ι)) β Qe P β (ιn(a,ι)−σn(a,σ)+1) β z(σ−n(a,σ)+1) = e P β (ι−σn(a,ι)) β Qe P β (ιn(a,ι)−σn(a,σ)+1) β ) Qz(ι−n(a,σ)+1) = e P β (ι−σn(a,ι)) β Qe P β (ιn(a,ι)−σn(a,σ)+1) β Qe P β ( (ιn(a,σ)+1−σn(a,σ)) β−(σ−σn(a,σ)) β ) z(σ), so W (ι, σ) = Q2e P β ( (ι−σn(a,ι)) β+(ιn(a,ι)−σn(a,σ)+1) β+(ιn(a,σ)+1−σn(a,σ)) β−(σ−σn(a,σ)) β ) . (iii) For any ι ∈ (ιn(a,ι), σn(a,ι)] and any σ ∈ (ιn(a,σ), σn(a,σ)], we have z(ι) = Qz(ι−n(a,ι)) = Qe P β (ιn(a,ι)−σn(a,σ)+1) β z(σ+ n(a,σ)+1) = Qe P β (ιn(a,ι)−σn(a,σ)+1) β z(σ−n(a,σ)+1) = Qe P β (ιn(a,ι)−σn(a,σ)+1) β ) Qz(ι−n(a,σ)+1) = Qe P β (ιn(a,ι)−σn(a,σ)+1) β Qe P β (ιn(a,σ)+1−σn(a,σ)) β z(σ), so W (ι, σ) = Q2e P β ( (ιn(a,ι)−σn(a,σ)+1) β+(ιn(a,σ)+1−σn(a,σ)) β ) . Case 4: General n(a, ι) and n(a, σ). (i) For any ι ∈ (σn(a,ι), ιn(a,ι)+1] and any σ ∈ (ιn(a,σ), σn(a,σ)], we have W (ι, σ) = Qn(a,ι)−n(a,σ)e P β [ (ι−σn(a,ι)) β+ ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] . (ii) For any ι ∈ (ιn(a,ι), σn(a,ι)] and any σ ∈ (σn(a,σ), ιn(a,σ)+1], we have W (ι, σ) = Qn(a,ι)−n(a,σ)e P β [ −(σ−σn(a,σ)) β+ ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β )] . (iii) For any ι ∈ (σn(a,ι), ιn(a,ι)+1] and any σ ∈ (σn(a,σ), ιn(a,σ)+1], we have W (ι, σ) = Qn(a,ι)−n(a,σ)e P β [ (ι−σn(a,ι)) β−(σ−σn(a,σ)) β+ ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] . (iv) For any ι ∈ (ιn(a,ι), σn(a,ι)] and any σ ∈ (ιn(a,σ), σn(a,σ)], we have W (ι, σ) = Qn(a,ι)−n(a,σ)e P β ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β . 6 Y. DING, J. WANG EJDE-2021/94 Thus W (ι, σ) can be written in the form W (ι, σ) =  e P β ( (ι−σk)β−(σ−σk)β ) , if ι, σ ∈ (σk, ιk+1], k = 0, 1, 2, . . . ; E, if ι, σ ∈ (ιk, σk], k = 1, 2, . . . , n(a, ι); Qn(a,ι)−n(a,σ)e P β [ −(σ−σn(a,σ)) β+ ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] , if σ ∈ (σn(a,σ), ιn(a,σ)+1], ι ∈ (ιn(a,ι), σn(a,ι)]; Qn(a,ι)−n(a,σ)e P β [ (ι−σn(a,ι)) β+ ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] , if σ ∈ (ιn(a,σ), σn(a,σ)], ι ∈ (σn(a,ι), ιn(a,ι)+1]; Qn(a,ι)−n(a,σ)e P β [ ((ι−σn(a,ι)) β−(σ−σn(a,σ)) β+ ∑n(a,ι)−1 k=n(a,σ) ((ιk+1−σk)β ] , if σ ∈ (σn(a,σ), ιn(a,σ)+1], ι ∈ (σn(a,ι), ιn(a,ι)+1]; Qn(a,ι)−n(a,σ)e P β ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β , if σ ∈ (ιn(a,σ), σn(a,σ)], ι ∈ (ιn(a,ι), σn(a,ι)], so W (ι, σ) can be written as W (ι, σ) = Qn(a,ι)−n(a,σ)e P β [( (ι−σn(a,ι)) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] . (2.4) In particular, if σ = a, we have W (ι, a) = Qn(a,ι)e P β [( (ι−σn(a,ι)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] . The proof is complete. � Lemma 2.5. For ι, σ ∈ I, η ∈ I, if (A1) holds, then W (ι, σ) = W (ι, η)W (η, σ), σ ≤ η ≤ ι. Proof. From ι, σ ∈ I and η ∈ I, we obtain W (ι, σ) = Qn(a,ι)−n(a,σ)e P β [( (ι−σn(a,ι)) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] = Qn(a,ι)−n(a,η)e P β [( (ι−σn(a,ι)) β )+ − ( (η−σn(a,η)) β )+ + ∑n(a,ι)−1 k=n(a,η) (ιk+1−σk)β ] ×Qn(a,η)−n(a,σ)e P β [( (η−σn(a,η)) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,η)−1 k=n(a,σ) (ιk+1−σk)β ] = W (ι, η)W (η, σ). The proof is complete. � Lemma 2.6. If (A1) holds, then W (·+ T, ·+ T ) = W (·, ·). Proof. It is clear that n(a, ι+T ) = n(a, T )+n(T, ι+T ) = q+n(T, ι+T ) = q+n(a, ι), so σn(a,ι+T ) = σn(a,ι)+q = σn(a,ι) + T . According (2.4), for a ≤ σ < ι ≤ T , we obtain W (ι+ T, σ + T ) = Qn(a,ι+T )−n(a,σ+T )e { P β [( (ι+ T − σn(a,ι+T )) β )+ EJDE-2021/94 NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS 7 − ( (σ + T − σn(a,σ+T )) β )+ + n(a,ι+T )−1∑ k=n(a,σ+T ) (ιk+1 − σk)β ]} = Qn(a,T )+n(T,ι+T )−(n(a,T )+n(T,σ+T ))e { P β [( (ι+ T − σn(a,ι+T )) β )+ − ( (σ + T − σn(a,σ+T )) β )+ + n(a,ι)+q−1∑ k=n(a,σ)+q (ιk+1 − σk)β ]} = Qn(a,ι)−n(a,σ)e P β [( (ι−σn(a,ι)) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] . The proof is complete. � Lemma 2.7. Suppose that (A1) holds, and for every t, σ ∈ I, η ∈ I with σ ≤ η ≤ ι, then W (ι, a)W (T, σ) = W (ι, σ)W (T, a). Proof. Calculating each side of the equality yields W (ι, a)W (T, σ) = W (ι+ T, T )W (T, σ) = W (ι+ T, σ), W (ι, σ)W (T, a) = W (ι, σ)W (ι+ T, ι) = W (ι+ T, ι)W (ι, σ) = W (ι+ T, σ). So W (ι, a)W (T, σ) = W (ι, σ)W (T, a). and he proof is complete. � Lemma 2.8. A solution z ∈ PC(I,Rn) of (1.2) with z(a) = za ∈ Rn has the form z(ι, a, za) = W (ι, a)za + n(a,ι)−1∑ k=0 ∫ ιk+1 σk W (ι, σ)h(σ)(σ − σk)β−1dσ + ∫ ι σn(a,ι) W (ι, σ)h(σ)(σ − σn(a,ι))β−1dσ + n(a,ι)∑ k=1 W (ι, σk)dk. (2.5) Set χ(ι) = { (ι− σk)β−1, ι ∈ (σk, ιk+1], k = 0, 1, 2, . . . , 0, ι ∈ (ιk, σk], k = 1, 2, . . . . So (2.5) can be rewritten as z(ι, a, za) = W (ι, a)za + ∫ ι a χ(ι)W (ι, σ)h(σ)dσ + n(a,ι)∑ k=1 W (ι, σk)dk. Proof. When ι ∈ [σ0, ι1], it holds Dσ0 β z(ι) = Pz(ι), and the solution of the above equation is z(ι) = W (ι, a)za. When za = za(ι), we obtain Dσ0 β z(ι) = Dσ0 β W (ι, a)za(ι) +W (ι, a)Dσ0 β za(ι) = Pz(ι) +W (ι, a)(ι− σ0)1−βz′a(ι) = Pz(ι) + c(ι). Then y′a(ι) = W−1(ι, a)c(ι)(ι− σ0)β−1, za(ι) = ∫ ι σ0 W−1(σ, a)h(σ)(σ − σ0)β−1dσ + za. 8 Y. DING, J. WANG EJDE-2021/94 By comparing both side of the equation, we obtain z(ι) = W (ι, a) [ ∫ ι σ0 W−1(σ, a)h(σ)(σ − σ0)β−1dσ + za ] = W (ι, a)za + ∫ ι σ0 W (ι, σ)h(σ)(σ − σ0)β−1dσ. When ι ∈ (ι1, σ1], we have z(ι) = Qz(ι−1 ) + d1 = NW (ι1, a)za +N ∫ ι1 σ0 W (ι1, σ)h(σ)(σ − σ0)β−1dσ + d1. When ι ∈ (σ1, ι2], we have z(ι) = W (ι, σ1)z(σ1) + ∫ ι σ1 W (ι, σ)h(σ)(σ − σ1)β−1dσ = W (ι, σ1)QW (ι1, a)za +W (ι, σ1)Q ∫ ι1 σ0 W (ι1, σ)h(σ)(σ − σ0)β−1dσ +W (ι, σ1)d1 + ∫ ι σ1 W (ι, σ)h(σ)(σ − σ1)β−1dσ = W (ι, a)za + ∫ ι1 σ0 W (ι, σ)h(σ)(σ − σ0)β−1dσ + ∫ ι σ1 W (ι, σ)h(σ)(σ − σ1)β−1dσ +W (ι, σ1)d1. (2.6) For a positive integer n and ι ∈ (σn(a,ι), ιn(a,ι)+1], it holds z(ι) = W (ι, a)za + n(a,ι)−1∑ k=0 ∫ ιk+1 σk W (ι, σ)h(σ)(σ − σk)β−1dσ + ∫ ι σn(a,ι) W (ι, σ)h(σ)(σ − σn(a,ι))β−1dσ + n(a,ι)∑ k=1 W (ι, σk)dk. When ι ∈ (ιn(a,ι)+1, σn(a,ι)+1], we have z(ι) = Qz(ι−n(a,ι)+1) + dn(a,ι)+1 = Q [ W (ιn(a,ι)+1, σ0)za + n(a,ι)−1∑ k=0 ∫ ιk+1 σk W (ιn(a,ι)+1, σ)h(σ)(σ − σk)β−1dσ + ∫ ιn(a,ι)+1 σn(a,ι) W (ιn(a,ι)+1, σ)h(σ)(σ − σn(a,ι))β−1dσ + n(a,ι)∑ k=1 W (ιn(a,ι)+1, σk)dk ] + dn(a,ι)+1. When ι ∈ (σn(a,ι)+1, σn(a,ι)+2], we have z(ι) = W (ι, σn(a,ι)+1)z(σn(a,ι)+1) + ∫ ι σn(a,ι)+1 W (ι, σ)h(σ)(σ − σn(a,ι)+1)β−1dσ = W (ι, σn(a,ι)+1)QW (ιn(a,ι)+1, a)za EJDE-2021/94 NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS 9 +W (ι, σn(a,ι)+1)Q n(a,ι)−1∑ k=0 ∫ ιk+1 σk W (ιn(a,ι)+1, σ)h(σ)(σ − σk)β−1dσ +W (ι, σn(a,ι)+1)Q ∫ ιn(a,ι)+1 σn(a,ι) W (ιn(a,ι)+1, σ)h(σ)(σ − σn(a,ι))β−1dσ +W (ι, σn(a,ι)+1)Q n(a,ι)∑ k=1 W (ιn(a,ι)+1, σk)dk +W (ι, σn(a,ι)+1)dn(a,ι)+1 + ∫ ι σn(a,ι)+1 W (ι, σ)h(σ)(σ − σn(a,ι)+1)β−1dσ = W (ι, a)za + n(a,ι)∑ k=0 ∫ ιk+1 σk W (ι, σ)h(σ)(σ − σk)β−1dσ + ∫ ι σn(a,ι)+1 W (ι, σ)h(σ)(σ − σn(a,ι)+1)β−1dσ + n(a,ι)+1∑ k=1 W (ι, σk)dk. Using induction, we can show that the solution of (1.2) with z(a) = za ∈ Rn has the form z(ι) = W (ι, a)za + n(a,ι)−1∑ k=0 ∫ ιk+1 σk W (ι, σ)h(σ)(σ − σk)β−1dσ + ∫ ι σn(a,ι) W (ι, σ)h(σ)(σ − σn(a,ι))β−1dσ + n(a,ι)∑ k=1 W (ι, σk)dk = W (ι, a)za + ∫ ι a χ(ι)W (ι, σ)h(σ)dσ + n(a,ι)∑ k=1 W (ι, σk)dk. (2.7) The proof is complete. � A function z(·, a, za) ∈ PC([a,∞),Rn) is T -periodic if z(ι, a, za) = z(ι+T, a, za), ι ≥ a. We define PCT (I,Rn) = {z ∈ PC(I,Rn) : z(ι) = z(ι+ T ), ι ≥ a}. Theorem 2.9. If (A1) holds, then (1.1) has a solution z ∈ PCT (I,Rn) if an only if (E −W (T, a))za = 0. Proof. Lemma 2.4 gives z(ι, a, za) = W (ι, a)za, so that z(ι+ T, a, za) = z(ι, a, za)⇐⇒W (ι+ T, a)za = W (ι, a)za ⇐⇒W (ι+ T, T )W (T, a)za = W (ι, a)za ⇐⇒W (ι, a)W (T, a)za = W (ι, a)za ⇐⇒ (E −W (T, a))za = 0. The proof is complete. � Example 2.10. Consider (1.1) and let β = 1/2, σ0 = 0, sk = k, ιk = k − 1 2 , k = 1, 2, . . . , T = 1, q = 1. We set P = ( 2 1 0 1 ) , Q = ( 1 1 0 1 ) , z0 = ( 1 1 ) , 10 Y. DING, J. WANG EJDE-2021/94 so that ePtβ = ( e4t e4t − e2t 0 e2t ) . And we can obtain W (ι, 0) = Qn(0,ι)e P β [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ] = e4 [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ] ( 1 n(0, ι) 0 1 ) × 1 1− e−2 [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ] 0 e−2 [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ]  = e4 [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ] × 1 1 + (n(0, ι)− 1)e−2 [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ] 0 e−2 [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ]  . Then z(ι, 0, z0) = W (ι, 0)z0 = e4 [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ] × 2 + (n(0, ι)− 1)e−2 [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ] e−2 [( (ι−σn(0,ι)) β )+ + ∑n(0,ι)−1 k=0 (ιk+1−σk)β ]  . Then W (1, 0) = ( e2 3/2 e2 3/2 0 e2 1/2 ) , (E −W (1, 0))za 6= 0, so (1.1) only has the trivial 1-periodic solution. 3. Nonhomogeneous linear non-instantaneous impulsive problem In this section, we study the existence of T -periodic solution of (1.2) in this section and consider the following assumptions: (A2) det(E −W (T, a)) 6= 0; (A3) det(E −W (T, a)) = 0; (A4) there are constants u ∈ R and J ≥ 1 such that ‖ exp{Aι}‖ ≤ Jeut, ι ≥ a; (A5) for any ι ∈ I, it holds h(ι+ T ) = h(ι). Lemma 3.1. Assume (A1), (A2), (A5). Then the solution z ∈ PC([a, T ],Rn) of (1.2) with z(T ) = z(a) has the form z(ι, a, za) = ∫ T a χ(ι)φ(ι, σ)h(σ)dσ + q∑ k=1 φ(ι, σk)dk, where φ(ι, σ) = {( W (T, a)(E −W (T, a))−1 + E ) W (ι, σ), a < s < t, W (ι, a)(E −W (T, a))−1W (T, σ), ι ≤ σ ≤ T. (3.1) EJDE-2021/94 NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS 11 Proof. From Lemma (2.4), we obtain z(T, a, za) = W (T, a)za + q−1∑ k=0 ∫ ιk+1 σk W (T, σ)h(σ)(σ − σk)β−1dσ + ∫ T σq W (T, σ)h(σ)(σ − σq)β−1dσ + q∑ k=1 W (T, σk)dk = za, so that za = (E −W (T, a))−1 ( q−1∑ k=0 ∫ ιk+1 σk W (T, σ)h(σ)(σ − σk)β−1dσ + ∫ T σq W (T, σ)h(σ)(σ − σq)β−1dσ + q∑ k=1 W (T, σk)dk ) Then, the solution of (1.2) has the form z(ι, a, za) = W (ι, a)(E −W (T, a))−1 ( q−1∑ k=0 ∫ ιk+1 σk W (T, σ)h(σ)(σ − σk)β−1dσ + ∫ T σq W (T, σ)h(σ)(σ − σn(a,ι))β−1dσ + q∑ k=1 W (T, σk)dk ) + n(a,ι)−1∑ k=0 ∫ ιk+1 σk W (ι, σ)h(σ)(σ − σk)β−1dσ + ∫ ι σn(a,ι) W (ι, σ)h(σ)(σ − σn(a,ι))β−1dσ + n(a,ι)∑ k=1 W (ι, σk)dk = ( q−1∑ k=0 ∫ ιk+1 σk W (ι, a)(E −W (T, a))−1W (T, σ)h(σ)(σ − σk)β−1dσ + ∫ T σq W (ι, a)(E −W (T, a))−1W (T, σ)h(σ)(σ − σq)β−1dσ + q∑ k=1 W (ι, a)(E −W (T, a))−1W (T, σk)dk ) + n(a,ι)−1∑ k=0 ∫ ιk+1 σk W (ι, σ)h(σ)(σ − σk)β−1dσ + ∫ ι σn(a,ι) W (ι, σ)h(σ)(σ − σn(a,ι))β−1dσ + n(a,ι)∑ k=1 W (ι, σk)dk = n(a,ι)−1∑ k=0 ∫ ιk+1 σk ( W (T, a)(E −W (T, a))−1 + E ) W (ι, σ)h(σ)(σ − σk)β−1dσ + ∫ ι σn(a,ι) ( W (T, a)(E −W (T, a))−1 + E ) W (ι, σ)h(σ)(σ − σn(a,ι))β−1dσ 12 Y. DING, J. WANG EJDE-2021/94 + ∫ ιn(a,ι)+1 ι W (ι, a)(E −W (T, a))−1W (T, σ)h(σ)(σ − σn(a,ι))β−1dσ + q−1∑ k=n(a,ι)+1 ∫ ιk+1 σk W (ι, a)(E −W (T, a))−1W (T, σ)h(σ)(σ − σk)β−1dσ + ∫ T σq W (ι, a)(E −W (T, a))−1W (T, σ)h(σ)(σ − σq)β−1dσ + n(a,ι)∑ k=1 ( W (T, a)(E −W (T, a))−1 + E ) W (ι, σk)dk + q∑ k=n(a,ι)+1 W (ι, a)(E −W (T, a))−1W (T, σk)dk = ∫ T a χ(ι)φ(ι, σ)h(σ)dσ + q∑ k=1 φ(ι, σk)dk. The proof is complete. � Then we consider det(E −W (T, a)) = 0. We assume that Q is invertible and consider the adjoint system of (1.1) with the form Dσk β x(ι) = −P>x(ι), ι ∈ (σk, ιk+1], k = 0, 1, 2, . . . , x(ι+k ) = [Q>]−1x(ι−k ), k = 1, 2, . . . , x(ι) = [Q>]−1x(ι−k ), ι ∈ (ιk, σk], k = 1, 2, . . . , x(σ+ k ) = x(σ−k ), k = 1, 2, . . . , x(a) = xa ∈ Rn. (3.2) Theorem 3.2. Let y and x be the solution of (1.1) and (3.2), respectively. Then 〈z(ι), x(ι)〉 = c for ι ≥ a, where c is a constant. Proof. Let ι ∈ (σk, ιk+1], k = 0, 1, . . . . Then Dσk β 〈z(ι), x(ι)〉 = 〈Dσk β z(ι), x(ι)〉+ 〈z(ι),Dσk β x(ι)〉 = 〈Pz(ι), x(ι)〉+ 〈z(ι),−P>x(ι)〉 = 〈z(ι), P>x(ι)〉+ 〈z(ι),−P>x(ι)〉 = 0. Let ι ∈ (ιk, σk], k = 1, 2, . . . . Then 〈z(ι), x(ι)〉 = 〈Qz(ι−k ), [Q>]−1x(ι−k )〉 = 〈z(ι−k ), Q>[Q>]−1x(ι−k )〉 = 〈z(ι−k ), x(ι−k )〉. Let t = ιk, k = 1, 2, . . . . Then 〈z(ι+k ), x(ι+k )〉 = 〈Qz(ι−k ), [Q>]−1x(ι−k )〉 = 〈z(ι−k ), Q>[Q>]−1x(ι−k )〉 = 〈z(ι−k ), x(ι−k )〉. Therefore 〈z(ι), x(ι)〉 = c. The proof is complete. � EJDE-2021/94 NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS 13 Lemma 3.3. Suppose that (A1), (A3) hold and rank(E −W (T, a)) = n − l with 1 ≤ l ≤ n. Then the adjoint system (3.2) has l linearly independent T -periodic solutions. Proof. By (A3) and rank(E −W (T, a)) = n − l, Equation (1.1) has linearly inde- pendent solutions. And the solution of (3.2) is x(ι) = W>(ι, a)xa, where W>(ι, a) = [(Q>)−1]n(a,ι)e −P>β [( (ι−σn(a,ι)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] . Now x(ι) = [ e P> β [( (ι−σn(a,ι)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ] (Q>)n(a,ι) ]−1 xa. Then x(T ) = xa ⇐⇒ [ e P> β [( (T−σn(a,ι)) β )+ + ∑n(a,T )−1 k=n(a,σ) (ιk+1−σk)β ] (Q>)n(a,T ) ]−1 xa = xa ⇐⇒ [ E − e P> β [( (T−σn(a,ι)) β )+ + ∑n(a,T )−1 k=n(a,σ) (ιk+1−σk)β ] (Q>)n(a,T ) ] xa = 0 ⇐⇒ xa ∈ ker [ E −Qn(a,T )e P β [( (T−σn(a,ι)) β )+ + ∑n(a,T )−1 k=n(a,σ) (ιk+1−σk)β ]]> = ker(E −W (T, a))>. So that dim ker(E−W (T, a))> = n−rank(E−W (T, a))> = n−rank(E−W (T, a)) = n− l. The proof is complete. � Theorem 3.4. Suppose that (A1), (A3) hold. Then(1.2) has a T -periodic solution if and only if 〈xa, φq〉 = 0, for every initial value xa of a T -periodic solution of (3.2), in which φq = ∫ T a χ(ι)W (T, σ)h(σ)dσ + n(a,T )∑ k=1 W (T, σk)dk. Proof. By Lemma 2.4, we obtain z(T ) = W (T, a)za + φq = za ⇐⇒ φq = (E −W (T, a))za ⇐⇒ φq ∈ Im(E −W (T, a)) ⇐⇒ φq ∈ [ker(E −W (T, a))>]⊥ ⇐⇒ φq ∈ [ker(E −W (T, a)>)]⊥. The proof is complete. � Example 3.5. Consider (1.2) with β = 1/2, σ0 = 0, σk = k, ιk = k − 1/2, k ∈ N , T = 1, q = 1, and h(ι) = { ((ι− k)1/2, 0)>, ι ∈ (k, k + 1 2 ], k = 0, 1, . . . ((−t− k + 1)1/2, 0)>, ι ∈ (k + 1 2 , k + 1], k = 0, 1, . . . . Put P = ( 1 2 0 1 ) , Q = ( 1 1 0 1 ) , di = ( 0 0 ) , 14 Y. DING, J. WANG EJDE-2021/94 so that ePt/β = ( e2t 4te2t 0 e2t ) . It is easy to obtain W (ι, σ) = Qn(0,ι)−n(0,σ)e P β [( (ι−σn(0,ι)) β )+ − ( (σ−σn(0,σ)) β )+ + ∑n(0,ι)−1 k=n(0,σ) (ιk+1−σk)β ] = e 2 [( (ι−σn(0,ι)) β )+ − ( (σ−σn(0,σ)) β )+ + ∑n(0,ι)−1 k=n(0,σ) (ιk+1−σk)β ] ( 1 n(0, ι)− n(0, σ) 0 1 ) × ( 1 4 [( (ι− σn(0,ι))β )+ − ((σ − σn(0,σ))β)+ + ∑n(0,ι)−1 k=n(0,σ)(ιk+1 − σk)β ] 0 1 ) = e 2 [( (ι−σn(0,ι)) β )+ − ( (σ−σn(0,σ)) β )+ + ∑n(0,ι)−1 k=n(0,σ) (ιk+1−σk)β ] ( 1 B̃ 0 1 ) . where B̃ = 4 [( (ι− σn(0,ι))β )+ − ((σ − σn(0,σ))β)+ + n(0,ι)−1∑ k=n(0,σ) (ιk+1 − σk)β ] + n(0, ι)− n(0, σ) . Then φ1 = ∫ 1 0 χ(σ)W (1, σ)h(σ)dσ + n(0,1)∑ k=1 W (1, σk)dk = ∫ 1/2 0 W (1, σ)h(σ)s− 1 2 dσ + (0, 0)> = ( e2 1/2 2 − 1 2 − 21/2 2 , 0)>. From W (1, 0) = ( e2 1/2 (1 + 25/2)e2 1/2 0 e2 1/2 ) and (E −W (1, 0))−1 =  1 1−e21/2 − (1+25/2)e2 1/2 1−e21/2 0 1 1−e21/2  , we have z0 = (E −W (1, 0))−1φ1 = (e21/2 − 1− 21/2 2(1− e21/2) , 0 )> . Therefore, z(ι, 0, z0) = W (ι, 0)z0 + ∫ ι 0 χ(σ)W (ι, σ)h(σ)dσ + n(0,ι)∑ k=1 W (1, σk)dk = e2(ι−σn(0,ι)) 1/2+21/2n(0,ι) (e21/2 − 1− 21/2 2(1− e21/2) , 0 )> EJDE-2021/94 NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS 15 + n(0,ι)−1∑ k=0 ∫ ιk+1 σk ( e 2 [( (ι−σn(0,ι)) 1/2 )+ − ( (σ−σn(0,σ)) 1/2 )+ + ∑n(0,ι)−1 k=n(0,σ) (ιk+1−σk)1/2 ] , 0 )> dσ + ∫ ι σn(0,ι) ( e 2 [( (ι−σn(0,ι)) 1/2 )+ − ( (σ−σn(0,σ)) 1/2 )+ + ∑n(0,ι)−1 k=n(0,σ) (ιk+1−σk)1/2 ] , 0 )> dσ = e2(ι−σn(0,ι)) 1/2 ( e2 1/2n(0,ι) e 21/2 − 1− 21/2 2(1− e21/2) + n(0,ι)−1∑ k=0 ∫ k+ 1 2 k e−2 ( (σ−σn(0,σ)) 1/2 )+ +(n(0,ι)−k)21/2 + ∫ ι σn(0,ι) e−2 ( (σ−σn(0,σ)) 1/2 )+ dσ, 0 )> = e2(ι−σn(0,ι)) 1/2 ( e2 1/2n(0,ι) e 21/2 − 1− 21/2 2(1− e21/2) + ( − 21/2 2 e−2 1/2 − e−2 1/2 2 + 1 2 )e21/2(1− e2 1 2 n(0,ι) ) 1− e21/2 − (ι− σn(0,ι))1/2e−2(ι−σn(0,ι)) 1/2 − 1 2 e−2(ι−σn(0,ι)) 1/2 + 1 2 , 0 )> = ( e2(ι−σn(0,ι)) 1/2 e2 1/2 2(1− e21/2) − (ι− σn(0,ι))1/2 − 1 2 + 1 2 e2(ι−σn(0,ι)) 1/2 , 0 )> . Then z(ι+ 1, 0, z0) = ( e2(ι+1−σn(0,t+1)) 1/2 e2 1/2 2(1− e21/2) − (ι+ 1− σn(0,t+1)) 1/2 − 1 2 + 1 2 e2(ι+1−σn(0,ι)+1) 1/2 , 0 )> = z(ι, 0, z0), so there is a 1-periodic solution. 4. Nonlinear non-instantaneous impulsive problems In this section, we study the nonlinear non-instantaneous impulsive problem (1.3). We use the following assumptions: (A6) For all ι ∈ I and x ∈ Rn, we have h(ι+ T, x) = h(ι, x); (A7) there is a constant Lh > 0 such that ‖h(ι, x1)−h(ι, x2)‖ ≤ Lh‖x1−x2‖ for all ι ∈ I and x1, x2 ∈ Rn; (A8) there are constant A,B ≥ 0 such that ‖h(ι, x)‖ ≤ A‖x‖ + B for any ι ∈ I and x ∈ Rn. Two important lemmas are given first. Lemma 4.1. When ι ∈ [a, T ] and Lemma 3.1 holds, we obtain q∑ i=1 ‖φ(ι, σi)di‖ 16 Y. DING, J. WANG EJDE-2021/94 ≤ Fu :=  euqT β max{‖Q‖q, 1}max{‖W (T, a)‖, 1} ×J [ ‖(E −W (T, a))−1‖+ 1 ]∑q i=1 ‖di‖, if u > 0, max{‖Q‖q, 1}max{‖W (T, a)‖, 1} ×J [ ‖(E −W (T, a))−1‖+ 1 ]∑q i=1 ‖di‖, if u ≤ 0. Proof. Using (3.1) we have q∑ i=1 ‖φ(ι, σi)di‖ ≤ q∑ i=1 ‖φ(ι, σi)‖‖di‖ = ∑ a<σi 0, then q∑ i=1 ‖φ(ι, σi)‖‖di‖ ≤ max{‖Q‖q, 1}max{‖W (T, a)‖, 1}J [ ∑ a 0, T β max{‖Q‖q, 1}J(‖(E −W (T, a))−1‖+ 1), u ≤ 0. Proof. Using (3.1), we have∫ T a ‖χ(σ)φ(ι, σ)‖dσ ≤ ∫ T a ‖φ(ι, σ)T β−1‖dσ ≤ T β−1 [ ∫ ι a ‖(E −W (T, a))−1‖‖W (ι, σ)‖+ ‖W (ι, σ)‖dσ + ∫ T ι ‖W (ι, a)‖‖(E −W (T, a))−1‖‖W (T, σ)‖dσ ] = T β−1 [ ∫ ι a ‖(E −W (T, a))−1‖ × ∥∥∥Qn(a,ι)−n(a,σ)ePβ [((ι−σn(a,ι)) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ]∥∥∥ + ∥∥∥Qn(a,ι)−n(a,σ)ePβ [((ι−σn(a,ι)) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ]∥∥∥dσ + ∫ T ι ∥∥∥Qn(a,ι)ePβ [((ι−σn(a,ι)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ]∥∥∥‖(E −W (T, a))−1‖ × ∥∥∥Qn(a,T )−n(a,σ)e P β [( (T−σn(a,T )) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,T )−1 k=n(a,σ) (ιk+1−σk)β ]∥∥∥dσ] 18 Y. DING, J. WANG EJDE-2021/94 ≤ T β−1 max{‖Q‖q, 1}J ∫ ι a ‖(E −W (T, a))−1‖ × ∥∥∥eu[((ι−σn(a,ι)) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ]∥∥∥ + ∥∥∥eu[((ι−σn(a,ι)) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ]∥∥∥dσ + ∫ T ι ∥∥∥eu[((ι−σn(a,ι)) β )+ + ∑n(a,ι)−1 k=n(a,σ) (ιk+1−σk)β ]∥∥∥‖(E −W (T, a))−1‖ × ∥∥∥eu[((T−σn(a,T )) β )+ − ( (σ−σn(a,σ)) β )+ + ∑n(a,T )−1 k=n(a,σ) (ιk+1−σk)β ]∥∥∥dσ. If u > 0, then∫ T a ‖χ(ι)φ(ι, σ)‖dσ ≤ T β−1 max{‖Q‖q, 1}J [ ‖(E −W (T, a))−1‖ ∫ T a euqT β dσ + ∫ ι a euqT β dσ ] ≤ T β max{‖Q‖q, 1}JeuqT β (‖(E −W (T, a))−1‖+ 1). If u ≤ 0, then∫ T a ‖χ(ι)φ(ι, σ)‖dσ ≤ T β−1 max{‖Q‖q, 1}J [ ‖(E −W (T, a))−1‖ ∫ T a euqT β dσ + ∫ ι a euqT β dσ ] ≤ T β max{‖Q‖q, 1}J(‖(E −W (T, a))−1‖+ 1). The proof is complete. � Theorem 4.3. Suppose that (A1), (A2), (A4), (A6), (A7) hold. If 0 < LhKu < 1, equation (1.3) has a unique T -periodic solution z ∈ PCT (I,Rn) satisfying ‖y‖ ≤ Lh‖z(a)‖‖Ku + ‖ha‖Ku + Fu 1− LhKu , where ‖ha‖ = maxι∈[a,T ] |h(ι, a)|. Proof. For each z ∈ PCT , it holds z(ι+ T ) + z(ι). By (A6), h(ι+ T, z(ι+ T )) = h(ι+ T, z(T )) = h(ι, z), ι ∈ R, so h(·, (·)) ∈ PCT . According to Lemma 3.1, we consider the equation z(ι, a, za) = ∫ T a χ(σ)φ(ι, σ)h(σ, z(σ))dσ + q∑ k=1 φ(ι, σk)dk. By using the operator H : PC([a, T ],Rn)→ PC([a, T ],Rn), we have Hz(ι, a, za) = ∫ T a χ(σ)φ(ι, σ)h(σ, z(σ))dσ + q∑ k=1 φ(ι, σk)dk. (4.1) EJDE-2021/94 NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS 19 If each y, z ∈ PC([a, T ],Rn), we obtain ‖Hz(ι)−Hz(ι)‖ ≤ ∥∥∥∫ T a χ(σ)φ(ι, σ)h(σ, z(σ))− ∫ T a χ(σ)φ(ι, σ)h(σ, z(σ)) ∥∥∥dσ ≤ ∫ T a ‖χ(σ)φ(ι, σ)‖‖h(σ, z(σ))− h(σ, z(σ))‖dσ ≤ Lh‖y − z‖ ∫ T a χ(σ)φ(ι, σ)dσ ≤ LhKu‖y − z‖ From 0 < LhKu < 1, we know that H is a contraction mapping and H has a unique fixed point. Then we obtain ‖y‖ = ‖Hy‖ ≤ ∫ T a ‖χ(σ)φ(ι, σ)‖‖h(σ, z(σ))‖dσ + q∑ k=1 ‖φ(ι, σk)‖dk‖ ≤ ∫ T a ‖χ(σ)φ(ι, σ)‖‖h(σ, z(σ))− h(σ, z(a)) + h(σ, z(a))‖dσ + q∑ k=1 ‖φ(ι, σk)‖dk‖ ≤ Lh‖y − z(a)‖ ∫ T a ‖χ(σ)φ(ι, σ)‖dσ + ∫ T a ‖χ(σ)φ(ι, σ)‖‖h(σ, z(a))‖dσ + q∑ k=1 ‖φ(ι, σk)‖dk‖ ≤ Lh‖y − z(a)‖Ku + ‖ha‖Ku + Fu ≤ Lh(‖y‖+ ‖z(a)‖)‖Ku + ‖ha‖Ku + Fu, so ‖y‖ ≤ Lh‖z(a)‖‖Ku + ‖ha‖Ku + Fu 1− LhKu . The proof is complete. � Theorem 4.4. Suppose that (A1), (A2), (A4), (A6), (A8) hold. If 0 < AKu < 1, then (1.3) has a unique T -periodic solution z ∈ PCT (I,Rn). Proof. We use the operator H in (4.1) defined on Cτ := {z ∈ PC([a, T ],Rn)|‖y‖ ≤ τ, τ ≥ BKu+Fu 1−AKu }. For any a ≤ ι ≤ T , z ∈ Cτ , by lemma 4.1 and lemma 4.2, we have ‖Hz(ι)‖ ≤ ∫ T a ‖χ(σ)φ(ι, σ)‖‖h(σ, z(σ))‖dσ + q∑ k=1 ‖φ(ι, σk)dk‖ ≤ A ∫ T a ‖χ(σ)φ(ι, σ)‖‖z(σ)‖dσ +B ∫ T a ‖χ(σ)φ(ι, σ)‖dσ + q∑ k=1 ‖φ(ι, σk)dk‖ ≤ AKu‖y‖+BKuFu = τ, 20 Y. DING, J. WANG EJDE-2021/94 so ‖Hy‖ ≤ τ and H(Cτ ) ⊂ Cτ . We can show that H is continuous and H(Cτ ) is pre-compact. By Schauder’s fixed-point Theorem, (1.3) has at least one T -periodic solution z ∈ PCT (I,Rn). � Example 4.5. We consider (1.3), with z(ι) = ( z1(ι) z2(ι) ) , P = ( 1 −2 3 −4 ) , B = ( 1 0 0 1 ) , dk = ( 1 0 ) , h(ι, z(ι)) = sin t cos z(ι), ιk = 2k − 1 4 π, σk = k 2 π, k = 1, 2, . . . , σ0 = 0. Let T = π. Then PQ = QP , ιk+2 = 2k + 3 4 π = 2k − 1 4 π + π = ιk + π, σk+2 = k + 2 2 π = k 2 π + π = σk + π, dk+2 = dk for k = 0, 1, 2, . . . . Then we obtain q = 2, so (A1) holds. We obtain ePt/β = ( 3e−2t − 2e−4t −2e−2t + 2e−4t 3e−2t − 3e−4t −2e−2t + 3e−4t ) and W (T, 0) = W (π, 0) = Qn(0,π)e P β [( (π−σn(0,π)) β )+ + ∑n(0,π)−1 k=0 (ιk+1−σk)β ] = Q2e P β [ (ι1−σ0) β+(ι2−σ1) β ] = Q2e P β π 1/2 = ( 3e−2π 1/2 − 2e−4π 1/2 −2e−2π 1/2 + 2e−4π 1/2 3e−2π 1/2 − 3e−4π 1/2 −2e−2π 1/2 + 3e−4π 1/2 ) , ‖W (π, a)‖ = 0.169. So det(E −W (π, 0)) 6= 0 and (A2) holds. Then (E −W (π, 0))−1 = 1 (1− e−2π1/2)(1− e−4π1/2) × ( 1 + 2e−2π 1/2 − 3e−4π 1/2 −2e−2π 1/2 + 2e−4π 1/2 3e−2π 1/2 − 3e−4π 1/2 1− 3e−2π 1/2 + 2e−4π 1/2 ) = ( 3 1−e−2π1/2 + −2 1−e−4π1/2 −2 1−e−2π1/2 + 2 1−e−4π1/2 3 1−e−2π1/2 + −3 1−e−4π1/2 −2 1−e−2π1/2 + 3 1−e−4π1/2 ) , and ‖(E −W (π, 0))−1‖ = max {∣∣∣ 3 1− e−2π1/2 + −2 1− e−4π1/2 ∣∣∣+ ∣∣∣ 3 1− e−2π1/2 + −3 1− e−4π1/2 ∣∣∣,∣∣∣ −2 1− e−2π1/2 + 2 1− e−4π1/2 ∣∣∣+ ∣∣∣ −2 1− e−2π1/2 + 3 1− e−4π1/2 ∣∣∣} = 1.9617. Next, h(ι + π, z) = b sin(ι + π) cos(z) = bh(ι, z) and (A6) holds. By |h(t, x) − h(t, z)| ≤ |b|| cosx − cos y| ≤ |b||x − y|, it follows that Lh = |b| and (A7) holds. Because σ(Pβ ) = {−2,−4}, (A5) satisfies with u = −2. Now J = sup ι≥0 e2t‖e P β ι‖ EJDE-2021/94 NON-INSTANTANEOUS IMPULSIVE DIFFERENTIAL EQUATIONS 21 = sup ι≥0 max{|3− 2e−2t|+ |3− 3e−2t|, | − 2 + 2e−2t|+ | − 2 + 3e−2t|} = 6, and Fu = 35.5404, Ku = 31.4969. If Lh = |b| < 0.0281, then 0 < LhKu < 1 and the conditions of Theorem 4.3 hold. 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Yuanlin Ding Department of Mathematics, School of Mathematics and Statistics, Guizhou University, Guiyang, Guizhou 550025, China Email address: yldingmath@126.com Jinrong Wang (corresponding author) Department of Mathematics, School of Mathematics and Statistics, Guizhou University, Guiyang, Guizhou 550025, China Email address: jrwang@gzu.edu.cn 1. Introduction 2. Preliminaries 3. Nonhomogeneous linear non-instantaneous impulsive problem 4. Nonlinear non-instantaneous impulsive problems Acknowledgments References