A class of second-order nonlocal indefinite impulsive differential systems
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1 Jiao and Zhang Boundary Value Problems 28) 28:63 R E S E A R C H Open Access A class of second-order nonlocal indefinite impulsive differential systems Lishuai Jiao and Xuemei Zhang * * Correspondence: zxm74@sina.com Department of Mathematics and Physics, North China Electric Power University, Beijing, People s Republic of China Abstract We consider the second-order nonlocal impulsive differential system x = at)xy ωt)fx), < t <,t t k, y = bt)x, <t <,t t k, x t=tk = I k xt k )), k =,2,..., n, y t=tk = J k yt k )), k =,2,..., n, x) = ht)xt) dt, x ) =, y) = gt)yt) dt, y ) =, where the weight functions at), bt), and ωt) change sign on [,], and gt) and ht) on [, ]. By constructing a cone K K 2, which is the Cartesian product of two cones in space PC[, ], and applying the well-known fixed point theorem of cone expansion and compression in K K 2, we obtain conditions for the existence and multiplicity of positive solutions of a nonlocal indefinite impulsive differential system. An example is given to illustrate the main results. Keywords: Indefinite weights; Nonlocal impulsive differential system; Positive solutions; Existence and multiplicity; Fixed point technology Introduction It is generally accepted that the theory and applications of differential equations with impulsive effects are an important area of investigation, since it is far richer than the corresponding theory of differential equations without impulsive effects. Various population models, biological system models, ecology models, biotechnology models, pharmacokinetics models, and optimal control models, which are characterized by the fact that per sudden changing of their state, can be expressed by impulsive differential equations. For an introduction of general theory of impulsive differential equations, we refer the reader to the references []and[2], whereas the applications of impulsive differential equations can be found in [3 5]. Some classical methods have been widely used to study impulsive differential equations: the theory of critical point theory and variational methods [6 8], fixed point theorems in cones [9 25], and bifurcation theory [26, 27]. In particular, we would like to mention some results of Lin and Jiang [28] and Feng and Xie [29]. Lin and Jiang [28] The Authors) 28. This article is distributed under the terms of the Creative Commons Attribution 4. International License which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original authors) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
2 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 2 of 34 considered the following Dirichlet boundary value problem with impulse effects: u t)=ft, ut)), t J, t t k, u t tk = I k ut k )), k =,2,...,m, u) = u) =,.) and by means of the fixed point index theory in cones the authors obtained some sufficient conditions for the existence of multiple positive solutions for problem.). Recently, using fixed point theorems in a cone, Feng and Xie [29] studied the existence of positive solutions for the following problem: u t)=ft, ut)), t J, t t k, u t tk = I k ut k )), k =,2,...,n, u) = m=2 i= a iu i ), u) = m=2 i= b iu i )..2) In comparison with numerous results on the impulsive differential equations, it is less known about the impulsive differential systems, even for the nonlocal impulsive differential systems. Moreover, we see that increasing attention has been paid to the study of boundary value problems with integral boundary conditions; for example, see Liu, Sun, Zhang, and Wu [3], Zhang, Feng, and Ge [3], Zhang andge [32], Haoet al. [33 35], Yan, Zuo, and Hao [36], Zhang et al. [37, 38], Sun, Liu, and Wu [39], Lin and Zhao [4], and Ahmad, Alsaedi, and Alghamdi [4]. This problem contains two-, three-, and multipoint boundary value problems as particular cases; for instance, see Karakostas and Tsamatos [42], Feng and Ge [43], Jiang, Liu, and Wu [44], Lan [45], Zhang et al. [46 5], Feng, Du, and Ge [5], Ahmad and Alsaedi [52], Mao and Zhao [53], Liu, Hao, and Wu [54], and the references therein. Specifically, Boucherif [55] exploited the fixed point theorem in cones to study the following problem: u t)=ft, ut)), < t <, u) cu ) = g t)ut) dt, u) du ) = g t)ut) dt..3) The author obtained several excellent results on the existence of positive solutions to problem.3). Feng, Ji, and Ge [56] began to study the following boundary value problem with integral boundary conditions in abstract spaces: u t)ft, ut)) = θ, <t <, u) = gt)ut) dt, u) = θ..4) Applying the fixed point theory in a cone for strict set contraction operators, the authors investigated the existence, nonexistence, and multiplicity of positive solutions for problem.4).
3 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 3 of 34 Recently, Kong [57] considered the existence and uniqueness of positive solutions for the second-order singular boundary value problem: u t)λfut)) =, t, ), u) = us) das), u) = us) dbs)..5) The author examined the uniqueness of the solution and its dependence on the parameter λ for problem.5) by using the mixed monotone operator theory. Simultaneously, an indefinite problem has attracted the attention of Ma and Han [58], López-Gómez and Tellini [59], Boscaggin and Zanolin [6], Feltrin and Zanolin [6], Boscaggin et al. [62, 63], Sovrano and Zanolin [64], Bravo and Torres [65], Wang and An [66], and Yao [67]. Ma and Han [58] considered the following boundary value problem: u λat)f u)=, <t <, u) = u) =,.6) where a C[, ] may change sign, and λ is a parameter. They proved the existence, multiplicity, and stability of positive solutions forproblem.6) by applying bifurcation techniques. Aapplying the shooting method, Sovrano and Zanolin [6] presented a multiplicity result for positive solutions for the Neumann problem u at)f u)=, <t <, ut)>, t [, T], u ) = u T)=,.7) where the weight function a C[, ] has indefinite sign. Recently, Wang and An [66] dealt with the existence and multiplicity of positive solutions for the second-order differential system u = at)ϕu ht)f u), < t <, ϕ = bt)u, <t <, u) = u) =, ϕ) = ϕ) =,.8) where at), bt), and gt) are allowed to change sign on [, ]. For the latest results of indefinite problems, please refer to Jiao and Zhang [68], Feltrin and Sovrano [69], and Zhang [7]. For all we know, in the literature there are no papers on multiple positive solutions for analogous indefinite impulsive differential systems with nonlocal boundary value conditions. More precisely, the study of at), bt), and ωt) changing sign on [, ] is still open
4 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 4 of 34 for the second-order nonlocal impulsive differential system x = at)xy ωt)f x), < t <,t t k, y = bt)x, <t <, x t=tk = I k xt k )), k =,2,...,n, y t=tk = J k yt k )), k =,2,...,n, x) = ht)xt) dt, x ) =, y) = gt)yt) dt, y ) =,.9) where at), ωt), bt) change sign on [, ], t k k =,2,...,n; wheren is a fixed positive integer) are fixed points such that < t < t 2 < < t k < < t n <, x t=tk denotes the jump of xt) att = t k,thatis, x t=tk = xt k ) xt k ), where xt k )andxt k )representthe right- and left-hand limits of xt) att = t k,respectively; y t=tk has a similar meaning for yt). In addition, a, ω, b, f, I k,andj k k =,2,...,n)satisfyH ) a, ω, b : [, ], ) are continuous, and there exists a constant, ) such that at), ωt), bt), t [, ], at)), ωt), bt), t [,]. Moreover, at), ωt), bt) do not vanish identically on any subinterval of [, ]. H 2 ) f :[, ) [, ) is continuous. H 3 ) I k :[, ) [, ) is continuous. H 4 ) J k :[, ) [, ) is continuous. H 5 ) h, g L [, ] are nonnegative, and ν, ν [, ),where ν = gs) ds, ν = hs) ds..) Let J = [, ] and J = J\{t, t 2,...,t n }. The basic space used in this paper is PC[, ] = {u u : [, ] R is continuous at t t k, ut k )=ut k), and ut k ) exists, k =,2,...,n}. Then PC[, ] is a real Banach space with the norm u PC = max ut). t J For convenience, consider PC [, ] = {x : x is continuous at t t k, xt k )=xt k), and xt k ) exists, k =,2,...,n}, which is a real Banach space with norm x PC = max xt), t J and PC 2 [, ] = {y : y is continuous at t t k, xt k )=xt k), and yt k ) exists, k =,2,...,n}, which is a real Banach space with norm y PC2 = max yt). t J
5 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 5 of 34 Clearly, PC [, ] PC 2 [, ] is also a real Banach space with norm x, y) = max { x PC, y PC2 }. By a positive solution of system.9) we mean a pair of functions x, y)withx C 2 J ) PC [, ] and y C 2 J ) PC 2 [, ] such that x, y)satisfiessystem.9)andx, y, t J, x, y. Remark. The technique to deal with the impulsive term is completely different from that of [6 27]. Remark.2 When we consider nonlocal differential systems with indefinite weights, another difficulty is to prove T : K K 2 K K 2 ; for details, see Lemma 2.3. Remark.3 It is not difficult to see that Proposition 2.3 of [67] plays key roles in the proofs of main results of [66] and[67]. However, it is invalid for nonlocal problems; for details, see Corollary 4.. Remark.4 In comparison with other related indefinite problems [58 66], the main featuresofthispaperareasfollows. i) I k, J k k =,2,...,n)areintroduced. ii) Nonlocal boundary conditions are introduced. iii) K K 2 is the Cartesian product of two cones in the space PC[, ]. We define a ± t), ω ± t), and b ± t)as a t)=max { at), }, a t)= min { at), }, ω t)=max { ωt), }, ω t)= min { ωt), }, b t)=max { bt), }, b t)= min { bt), }, so that at)=a t) a t), ωt)=ω t) ω t), bt)=b t) b t), t [, ]. Inspired by the references mentioned, in this paper, we investigate the existence and multiplicity of positive solutions for system.9). By constructing a cone K K 2,whichis the Cartesian product of two cones in the space PC[, ], and using the well-known fixed point theorem of cone expansion and compression, we obtain conditions for the existence and multiplicity of positive solutions of system.9). We remark that this is probably the first time that the existence and multiplicity of positive solutions of impulsive differential systems with indefinite weight and integral boundary conditions have been studied. The rest of this paper is organized as follows. In Sect. 2,wegivesomepreliminaryresults. Section 3 is devoted to state and prove the main results. Finally, an example is given in Sect. 4.
6 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 6 of 34 2 Preliminaries In this section, we give some preliminary results for the convenience of later use and reference. It is clear that system.9) is equivalent to the following two boundary value problems: y = bt)x, <t <,t t k, y t=tk = J k yt k )), k =,2,...,n, 2.) y) = gt)yt) dt, y ) =, and x = at)xy ωt)f x), < t <,t t k, x t=tk = I k xt k )), k =,2,...,n, x) = ht)xt) dt, x ) =. 2.2) Lemma 2. Assume that H ), H 2 ), H 4 ), and H 5 ) hold. Then problem 2.) has a unique solution y, which can be expressed in the form where yt)= Ht, s)bs)xs) ds Ht, s)=gt, s) ν H s t, s)=g s t, s) ν t, t s, Gt, s)= s, s t, G s, t s, t, s)=, s t. k)j k ytk ) ), 2.3) Gτ, s)gτ) dτ, 2.4) G s τ, s)gτ) dτ, 2.5) 2.6) 2.7) Proof First, suppose that u is a solution of problem 2.). It is easy to see by integration of problem 2.)that y t) y ) = t Integrating again, we get yt)=y) y )t bs)xs) ds. 2.8) t Letting t =in2.8), we find y ) = t s)bs)xs) ds J k ytk ) ). 2.9) t k <t bs)xs) ds. 2.)
7 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 7 of 34 Substituting the boundary condition y) = gt)yt) dt and 2.)into2.9), we obtain where yt)= = gs)ys) ds t gs)ys) ds bs)xs) ds t Gt, s)bs)xs) ds t s)bs)xs) ds J k ytk ) ) t k <t G s t, s)j k ytk ) ), gs)ys) ds = Therefore we have = [ gs) gτ)yτ) dτ G τ s, τ)j k ytk ) )] ds gs) ds gτ)yτ) dτ G τ s, τ)j k ytk ) )] ds. gs)ys) ds = ν [ gs) Gs, τ)bτ)xτ) dτ Gs, τ)bτ)xτ) dτ [ gs) Gs, τ)bτ)xτ) dτ G τ s, τ)j k ytk ) )] ds and Let yt)= ν = ν [ gs) Gs, τ)bτ)xτ) dτ Gt, s)bs)xs) ds ν G τ s, τ)j k ytk ) )] ds G s t, s)j k ytk ) ) [ ] Gτ, s)gτ) dτ bs)ys) ds [ G s τ, s)gτ)j k ytk ) )] dτ Gt, s)bs)xs) ds Ht, s)=gt, s) ν H s t, s)=g s t, s) ν G s t, s)j k ytk ) ). Gτ, s)gτ) dτ, G s τ, s)gτ) dτ.
8 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 8 of 34 Then yt)= Ht, s)bs)xs) ds k)j k ytk ) ). The proof of sufficiency is complete. Conversely, let u be a solution of 2.). Direct differentiation of 2.4) and2.5) implies, for t t k, y t)= Evidently, bs)xs) t bs)xs). y = bt)x, y t=tk = J k ytk ) ), k =,2,...,n, y) = gt)yt) dt, y ) =. The lemma is proved. Proposition 2. Let Gt, s), G s t, s), Ht, s), and H s t, s) be given as in Lemma 2.. If ν [, ), then we get Gt, s)>, Ht, s)>, t, s, ), 2.) Gt, t)gs, s) Gt, s) Gs, s)=s, t, s J, 2.2) ρgt, t)gs, s) Ht, s) Hs, s)=γ Gs, s) γ, t, s J, 2.3) G s t, s), H s t, s) γ, t, s J, 2.4) where γ = ν, ρ = τgτ) dτ. 2.5) ν Proof By the definition of Gt, s) andht, s), relations 2.) and2.2) aresimpletoprove. Next, we consider 2.3). In fact, from.)and2.2) weget Ht, s) Gs, s) Gs, s)gτ) dτ ν = Gs, s) ) gτ) dτ ν = Gs, s) ν γ
9 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 9 of 34 and Ht, s) Gt, t)gs, s) Gs, s)gτ, τ)gτ) dτ ν = Gs, s) Gt, t) ) Gτ, τ)gτ) dτ ν Gt, t) ) Gs, s) Gt, t) Gτ, τ)gτ) dτ ν Gt, t) ) = Gs, s)gt, t) τgτ) dτ ν = ρgt, t)gs, s). This shows that 2.3)holds. Similarly, by the the definition of G s t, s)andh s t, s), we can prove that 2.4)holds. Remark 2. From 2.5) we can prove that H s t, s) ν gτ) dτ, t, s [, ]. Proof It follows from 2.5)and2.7)that H s ν t, s) = G s τ, s)gτ) dτ, t s, ν G s τ, s)gτ) dτ, s t ν = [ s G s τ, s)gτ) dτ s G s τ, s)gτ) dτ], t s, ν [ s G s τ, s)gτ) dτ s G s τ, s)gτ) dτ], s t ν s gτ) dτ, t s, = ν s gτ) dτ, s t gτ) dτ, t, s [, ]. ν Lemma 2.2 Assume that H ) H 3 ) and H 5 ) hold. Then problem 2.2) has a unique solution x given by where xt)= H t, s)as)xs)ys) ds H t, s)ωs)f xs) ) ds k)i k xtk ) ), 2.6) H t, s)=gt, s) ν H s t, s)=g s t, s) ν Gτ, s)hτ) dτ, 2.7) G s τ, s)hτ) dτ. 2.8)
10 Jiao and Zhang Boundary Value Problems 28) 28:63 Page of 34 Proof The proof of Lemma 2.2 issimilartothatoflemma2.. Proposition 2.2 Let H and H s begivenasinlemma2.. If ν [, ), then we get ρ Gt, t)gs, s) H t, s) H s, s)=γ Gs, s) γ, t, s J, 2.9) G s t, s), H s t, s) γ, t, s J, 2.2) where γ = ν, ρ = Remark 2.2 From 2.8) we can prove that τhτ) dτ. 2.2) ν H s t, s) ν hτ) dτ, t, s [, ]. Remark 2.3 Let x, y) be a solution of system.9). Then from Lemma 2. and Lemma 2.2 we have xt)= yt)= H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)as)xs) H τ s, t k)j k ytk ) )) ds Ht, s)bs)xs) ds k)j k ytk ) ), H t, s)ωs)f xs) ) ds k)i k xtk ) ), 2.22) where H τ s, τ)=g τ s, τ) ν G τ, τ)g) d. To obtain the existence and multiplicity of a positive solution of system.9), we make the following hypotheses: H 6 ) There exists a constant σ satisfying <σ < such that σ Ht, s)b s) ds σ Ht, s)b s) ds; H 7 ) There exists a constant satisfying < < such that ρ H t, s)gs, s)a s) ds γ H t, s)a s) ds; H 8 ) There exists a constant μ satisfying <μ such that f ω) μϕω), ω [, ), where ϕω)=max{f ρ): ρ ω};
11 Jiao and Zhang Boundary Value Problems 28) 28:63 Page of 34 H 9 ) Thereexistconstants<α < with α and k, k 2, l, l 2, m, m 2 >such that k x α f x) k 2 x α, l x α I k x) l 2 x α, m y α J k y) m 2 y α, x, y [, ); H ) Thereexists<σ 3 < satisfying σ 3 2 < t < σ 3 such that σ α 3 μ2 k σ 3 H t, s)ω s) ds k 2 α H t, s)ω s) ds. Obviously, ϕ :[, ) [, ) is nondecreasing. Moreover, if f is nondecreasing, then f = ϕ and μ =. We denote } PC {x [, ] = PC [, ] : min xt) andx) = ht)xt) dt, x ) =, t K = { x PC [, ] : x is concave on [, ], and convex on [,]}, } PC 2 {y [, ] = PC 2 [, ] : min yt) andy) = gt)yt) dt, y ) =, t K 2 = { y PC 2 [, ] : y is concave on [, ], and convex on [,]}. If x K,thenitiseasytoseethat x PC = max t xt). Similarly, we have y PC2 = max t yt).also,forapositivenumberr,wedefine Ω r = { x, y) K K 2, x, y) < r }, and then we get Ω r = { x, y) K K 2, x, y) = r }. For any x, y) K K 2, define the mappings T : K PC [, ], T 2 : K 2 PC 2 [, ], and T : K K 2 PC [, ] PC 2 [, ] as follows: T x)t)= T 2 y)t)= H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)ωs)f xs) ) ds H t, s)as)xs) H τ s, t k)j k ytk ) )) ds k)i k xtk ) ), Ht, s)bs)xs) ds k)j k ytk ) ), 2.23) Tx, y) ) t)= T x)t), T 2 y)t) ).
12 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 2 of 34 Remark 2.4 It follows from Lemmas and Remark 2.3 that x, y) is a solution of system.9)ifandonlyifx, y) is a fixed point of operator T. Lemma 2.3 Assume that H ) H ) hold. Then TK K 2 ) K K 2, and T : K K 2 K K 2 is completely continuous. Proof For any x, y) K K 2,weprovethatTx, y) K K 2,thatis,T x K and T 2 y K 2.Inviewof2.2), we know that T x) t)= where zs) ds t zs) ds, zs)=as)xy ωs)f xs) ), and then we have T x) ) =. From 2.4)and2.2)weget and T x)) = = ν H, s)hs, τ)as)bτ)xs)xτ) dτ ds H, s)ωs)f xs) ) ds H, s)as)xs) H τ s, t k)j k ytk ) )) ds H s, t k)i k xtk ) ) ν ν ν as)xs) as)xs) Gτ, s)hτ) dτ ds Hs, τ)bτ)xτ) dτ Gτ, s)hτ) dτ H τ s, t k)j k ytk ) )) ds ωs)f xs) ) Gτ, s)hτ) dτ ds G s τ, s)hτ) dτi k xtk ) ) ht)t x)t) dt = ht) [ = H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)as)xs) H τ s, t k)j k ytk ) )) ds H t, s)ωs)f xs) ) ds ht) dt ht) dt H t, s)as)xs) ds k)i k xtk ) )] dt Hs, τ)bτ)xτ) dτ H t, s)as)xs) H τ s, t k)j k ytk ) )) ds
13 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 3 of 34 ht) dt = ν ν ν ν =T x)). H t, s)ωs)f xs) ) ds H s t, t k)ht) dti k xtk ) ) as)xs) as)xs) Gτ, s)hτ) dτ ds Hs, τ)bτ)xτ) dτ Gτ, s)hτ) dτ H τ s, t k)j k ytk ) )) ds ωs)f xs) ) Gτ, s)hτ) dτ ds G s τ, s)hτ) dτi k xtk ) ) Similarly, we have T 2 y) ) =, T 2 y)) = gt)t 2y)t) dt. Define the function q : [, ] [, ] as follows: if x) =,then { t qt)=min, t }, t J; if x) >,then { } t qt)=min,, t J. Since σ <, max t qt)=andmin σi t qt)= σ i, i =,2,3. Let x K.Thenx is concave on [, ] andconvexon[, ]. Noticing that x) = ht)xt) dt and x ) =, we get xt) x)qt), t [, ]; xt) x)qt), t [,]. Firstofall,foranyx K, we show that Ht, s)bs)xs) ds σ Indeed, for x K,weobtain Ht, s)b s)xs) ds, t J. Ht, s)bs)xs) ds σ = Ht, s)b s)xs) ds σ Ht, s)b s)xs) ds σ Ht, s)b s)qs)x) ds Ht, s)b s)xs) ds Ht, s)b s)qs)x) ds
14 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 4 of 34 [ x) min qs) Ht, s)b s) ds max qs) s [σ,] σ s [,] [ ] σ = x) Ht, s)b s) ds Ht, s)b s) ds. σ ] Ht, s)b s) ds Then by H 6 )wehave Ht, s)bs)xs) ds σ Ht, s)b s)xs) ds. Secondly, for any x K,weprovethat H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)hs, τ)a s)bτ)xs)xτ) dτ ds. For t J,since Ht, s)bs)xs) ds, we have = H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)hs, τ)a s)bτ)xs)xτ) dτ ds H t, s)hs, τ)a s)bτ)xs)xτ) dτ ds H t, s)hs, τ)a s)bτ)xs)xτ) dτ ds H t, s)a s)qs)x) H t, s)a s)qs)x) H t, s)a s) min qs)x) s [,] x) x) x) x) Hs, τ)bτ)xτ) dτ ds Hs, τ)bτ)xτ) dτ ds H t, s)a s) max qs)x) s [,] H t, s)a s) H t, s)a s) H t, s)a s) H t, s)a s) Hs, τ)bτ)xτ) dτ ds Hs, τ)bτ)xτ) dτ ds Hs, τ)bτ)xτ) dτ ds Hs, τ)bτ)xτ) dτ ds ρgs, s)gτ, τ)bτ)xτ) dτ ds γ Gτ, τ)bτ)xτ) dτ ds
15 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 5 of 34 = x) γ [ σ2 Gτ, τ)bτ)xτ) dτ ρ ] H t, s)a s) ds, and then it follows from H 7 )that H t, s)gs, s)a s) ds H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)hs, τ)a s)bτ)xs)xτ) dτ ds. Similarly, for any y K 2,since n H τ s, t k)i k yt k )), we get H t, s)as)xs) H τ s, t k)j k ytk ) )) ds H t, s)as)xs) H τ s, t k)j k ytk ) )) ds. Thirdly, for any x K,weprovethat H t, s)ωs)f xs) ) ds Since ϕ is nondecreasing, we also obtain H t, s)ω s)f xs) ) ds, t J. ϕ xt) ) ϕ qt)x) ), ϕ xt) ) ϕ qt)x) ), t [, ], t [,]. Therefore, for any x K, it follows from H 8 ) H )that H t, s)ωs)f xs) ) ds = H t, s)ω s)f xs) ) ds σ 3 μ μ μ μ H t, s)ω s)f xs) ) ds σ 3 H t, s)ω s)ϕ xs) ) ds σ 3 H t, s)ω s)ϕ x)qs) ) ds σ 3 H t, s)ω s)f x)qs) ) ds μ H t, s)ω s)f xs) ) ds H t, s)ω s)ϕ xs) ) ds H t, s)ω s)k x α )q α s) ds σ 3 μ [ x α ) min σ 3 s qα s)μk H t, s)ω s)ϕ x)qs) ) ds H t, s)ω s)f x)qs) ) ds H t, s)ω s)k 2 x α )q α s) ds σ 3 H t, s)ω s) ds max s qα s) μ k 2 ] H t, s)ω s) ds
16 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 6 of 34 [ σ x α α ) 3 μk α H t, s)ω s) ds σ 3 μ k 2, ] H t, s)ω s) ds which shows that H t, s)ωs)f xs) ) ds Thus, for x, y) K K 2, T x)t)= T 2 y)t)=, H t, s)ω s)f xs) ) ds. H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)as)xs) H τ s, t k)j k ytk ) )) ds H t, s)a s)xs) H t, s)ω s)f xt k ) ) ds Hs, τ)bτ)xτ) dτ ds H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds Ht, s)bs)xs) ds σ. Ht, s)b s)xs) ds H t, s)ωs)f xt k ) ) ds k)i k xtk ) ) k)j k ytk ) ) k)j k ytk ) ) k)i k xtk ) ) Moreover, by direct calculation we derive T x) t)= a t)xt)yt) ω t)f x), t [, ], T 2 x) t)=a t)xt)yt)ω t)f x), t [,], T y) t)= b t)xt), t [, ], T 2 y) t)=b t)xt), t [,], which shows that T x and T 2 y are concave on [, ] andconvexon[, ]. It follows that T x K and T 2 y K 2.ThusTK K 2 ) K K 2. Finally, by standard methods and the Arzelà Ascoli theorem we can prove that T is completely continuous. Remark 2.5 In [66]and[67], it is not difficult to see that the function qt)playsanimportant role in the proof of completely continuous operator. If x) = x) =, then we can
17 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 7 of 34 define qt) =min{ t, t }.However,ifx) = ht)xt) dt and x ) =, then the definition of qt) is invalid. This shows that when x) = ht)xt) dt and x ) =, we require a special technique to give a fine definition of qt). In fact, a fine definition of qt) is very difficult to give when x) = ht)xt) dt and x ) =. This is probably the main reason why there is almost no paper studying the existence of positive solutions for the class of second-order nonlocal differential systems with indefinite weights and even for second-order nonlocal impulsive differential systems with indefinite weights. Remark 2.6 The idea of the proof of Lemma 2.3 comes from Theorem 3. of [67]. The following lemma is very crucial in our argument. Lemma 2.4 Theorem of [7], Fixedpoint theoremof coneexpansion and compression of norm type) Let Ω and Ω 2 be two bounded open sets in a real Banach space E such that Ω and Ω Ω 2. Let an operator T : K Ω 2 \Ω ) K be completely continuous, where K is a cone in E. Suppose that one of the following two conditions is satisfied: a) Tx x, x K Ω, and Tx x, x K Ω 2, and b) Tx x, x K Ω, and Tx x, x K Ω 2, is satisfied. Then T has at least one fixed point in K Ω 2 \Ω ). 3 Main results In this part, applying Lemma 2.4, we obtain the following three existence theorems. Theorem 3. Assume that H ) H ) hold. If α >,then system.9) admits at least one positive solution. Proof On one hand, considering the case α >,byh 9 )weget f x) lim x x lim k 2 x α I k x) l 2 x α =, lim lim x x x x x x =, J k y) m 2 y α lim lim =. y y y y Furthermore, there exist r, r >suchthat f x) ε x, I k x) ε 2 x, k =,2,...,n, x r, J k y) ε 3 y, k =,2,...,n, y r, where ε, ε 2, ε 3 satisfy 4γ ε ω s) ds <, { ε 3 < min 4nγ r a s) ds, nγ 4nε 2 <, ν γ )} b s) ds.
18 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 8 of 34 Let A = γ A = γ n Hs, τ)a s)b τ) dτ ds, a s) ds, 3.) and choose r = min{4a),4ε 3 A ), r, r }. Then for any x, y) K K 2 ) Ω r,wehave x, y) = r,andby2.2)weget T x)t)= = H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)ωs)f xs) ) ds H t, s)as)xs) H τ s, t k)j k ytk ) )) ds k)i k xtk ) ) H t, s)hs, τ)a s)bτ)xs)xτ) dτ ds H t, s)hs, τ)a s)bτ)xs)xτ) dτ ds H t, s)ω s)f xs) ) ds H t, s)ω s)f xs) ) ds H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds k)i k xtk ) ) = H t, s)hs, τ)a s)bτ)xs)xτ) dτ ds H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds H t, s)hs, τ)a s)b τ)xs)xτ) dτ ds H t, s)ω s)f xs) ) ds k)i k xtk ) ) H t, s)hs, τ)a s)b τ)xs)xτ) dτ ds H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds H t, s)hs, τ)a s)b τ)xs)xτ) dτ ds H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds γ H t, s)ω s)f xs) ) ds k)i k xtk ) ) H t, s)ω s)f xs) ) ds k)i k xtk ) ) Hs, τ)a s)b τ)xs)xτ) dτ ds γ ω s)ε xds
19 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 9 of 34 T 2 y)t)= γ ν A x 2 PC γ ε nε 2 x PC ν Ar 2 γ ε r ε 3 y a s)xs) ds ε 2 x ν ω s) ds x PC γ ν nε 3 a s) ds x PC y PC2 < 4 r 4 r 4 r 4 r ω s) ds A ε 3 r 2 ν nε 2 r = r, 3.2) Ht, s)bs)xs) ds Ht, s)b s)xs) ds γ γ γ k)j k ytk ) ) b s)xs) ds ν k)j k ytk ) ) J k ytk ) ) b s) ds x PC ν nε 3 y PC2 b s) dsr γ nε 3 r < r. 3.3) Consequently, Tx, y) < x, y), x, y) K K 2 ) Ω r. 3.4) On the other hand, since α >, it follows from H 9 )that f x) lim x x lim k x α =, x x I k x) l x α lim lim x x x x =, J k y) m y α lim lim =, y y y y which shows that there exist R, R >suchthat f x) ε 4 x, I k x) ε 5 x, k =,2,...,n, x R, J k y) ε 6 y, k =,2,...,n, y R,
20 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 2 of 34 where ε 4, ε 5, ε 6 satisfy where σ σ3 2 3ρ 2 ε 4 min δt) G σ s, s)ω s) ds >, 32 σ t σ ε 5 min ν ν δt) σ 32 t σ 3 gτ) dτε 6 hτ) dτ >, min δt)>, σ 32 t σ 3 { t δt)=min, t }, t [, ]. 3.5) If x, y) K K 2,thenx, y are two nonnegative concave functions on [, ]. So we get xt) δt) x PC, yt) δt) y PC2, t [, ], t [, ]. 3.6) It follows that min σ i 2 t σ i xt) θ i x PC, min σ i 2 t σ i yt) θ i y PC2, i =,2,3,where Let θ i = { σi min δt)=min σ i 2 t σ i 2, σ } i >. 3.7) B = ρ 2 θ θ 2 σ 2 G s, s)hs, τ)a s)b τ) dτ ds >, 3.8) R > max{3b), R θ 3, r}, R 2 > max{ R θ 3, r}, andr = max{r, R 2 }. Then for any x, y) K K 2 ) Ω R,wehave R = x, y) = max { x PC, y PC2 } = max{r, R 2 }, min xt) min δt) x σ 32 σ PC θ 3 R > R, t σ 32 3 t σ 3 min yt) min δt) y σ 32 σ PC2 θ 3 R 2 > R, t σ 32 3 t σ 3 and T x PC { = max H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)ωs)f xs) ) ds t J H t, s)as)xs) H τ s, t k)j k ytk ) )) ds k)i k xtk ) )}
21 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 2 of 34 { max H t, s)a s)xs) Hs, τ)bτ)xτ) dτ ds H t, s)ω s)f xs) ) ds t J σ2 H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds k)i k xtk ) )} min H σ t, s)a s)xs) 22 t min 2 2 σ t 22 min ρ 2 2 σ t 32 ρ 2 ρ 2 ρ σ Hs, τ)b τ)xτ) dτ ds H t, s)as)xs) H τ s, t k)j k ytk ) ) ) ds σ 32 <t k <σ 3 H t, s)ω s)f xs) ) ds min σ 3 t t 2 <t k<t H s t, t k)i k xtk ) ) σ G s, s)a s)δs) x PC Hs, τ)b τ)δτ) x PC dτ ds G s, s)ω s)ε 3 xs) ds σ 32 ν hτ) dτ σ 32 <t k <σ 3 ε 4 xt k ) σ G s, s)a s)δs) x PC Hs, τ)b τ)δτ) x PC dτ ds G s, s)ω s)ε 3 xs) ds σ 32 ν B x 2 PC ρ 2 ε 3 min σ 32 t σ 3 δt) ν hτ) dτε 4 B x 2 PC ρ 2 ε 3 min σ 32 t σ 3 δt) ν hτ) dτε 4 σ 32 min σ 32 t σ 3 δt) x PC σ 32 min σ 32 t σ 3 δt) x PC G s, s)ω s) ds x PC G s, s)ω s) ds x PC hτ) dτε 4 xt ) σ 32 <t <σ 3 > 3 x PC 3 x PC 3 x PC = x PC, 3.9) { T 2 y PC2 = max Ht, s)bs)xs) ds k)j k ytk ) )} t J { max Ht, s)b s)xs) ds t J min 2 t ν k)j k ytk ) )} Ht, s)b s)xs) ds min σ 3 t gτ) dτε 6 yt ) σ 32 <t <σ 3 t 2 <t k<t k)j k ytk ) )
22 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 22 of 34 ν gτ) dτε 6 min σ 32 t σ 3 δt) y PC2 = y PC2. 3.) Consequently, Tx, y) > x, y), x, y) K K 2 ) Ω R. 3.) Therefore, applying Lemma 2.4 to 3.4)and3.), we can show that T has at least one fixed point x, y) K K 2 ) Ω R \ Ω r ). The proof of Theorem 3. is completed. The following theorem deals with the multiplicity of system.9). For convenience, we introduce the following notations: D =3γ ω s) ds, Λ =3nγ, Γ = ν ) γ b s) ds, n D = ρ G s, s)ω s) ds, Λ =2γ hτ) dτ, σ 32 Γ = ν n gτ) dτ). Theorem 3.2 Assume that H ) H ) hold. Suppose that <α <and there exist constants d and r satisfying <d < min{4a),4a Γ ), r} such that min f x)<d d, x,y) K K 2 ) Ω d min I k x)<λ d, 3.2) x,y) K K 2 ) Ω d min x,y) K K 2 ) Ω d J k y)<γ d, where A and A are defined in 3.). Then system.9) admits at least two positive solutions. Proof If < α <,thenbyh 9 )weknowthat f x) k i) lim x lim x α I x x =, lim k x) x x x lim y J k y) y ii) lim x f x) x m lim y α y = ; y k lim 2 x α x x =, lim x I k x) x lim x l x α x =, lim x l 2 x α x =, J lim k y) m y lim 2 y α y y =. y From i) it follows that there exists a sufficiently small positive constant r such that f x) ε 7 x, I k x) ε 8 x, k =,2,...,n, x r, J k y) ε 9 y, k =,2,...,n, y r,
23 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 23 of 34 where ε 7, ε 8, ε 9 satisfy ρ ε 7 min δt) σ 32 t σ 3 σ 32 G s, s)ω s) ds >, 2ε 8 hτ) dτ ν gτ) dτε 9 ν min δt)>, σ 32 t σ 3 min δt)>, σ 32 t σ 3 with δt)definedin3.5). Therefore, for any x, y) K K 2 ) Ω r,wegetfrom3.6)that T x PC = max t J { H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)ωs)f xs) ) ds H t, s)as)xs) H τ s, t k)j k ytk ) )) ds { max H t, s)a s)xs) t J H t, s)ω s)f xs) ) ds k)i k xtk ) )} Hs, τ)bτ)xτ) dτ ds H t, s)as)xs) H τ s, t k)j k ytk ) )) ds min H σ t, s)a s)xs) 22 t min ρ 2 ρ σ t 32 σ H t, s)ω s)f xs) ) ds min G s, s)ω s)f xs) ) ds σ 32 ν G s, s)ω s)ε 7 xs) ds σ 32 ν ρ 2 ε 7 min σ 32 t σ 3 δt) ε 8 ν hτ) dτ σ 32 k)i k xtk ) )} Hs, τ)b τ)xτ) dτ ds t <t k < G s, s)ω s) ds x PC min σ 32 t σ 3 δt) x PC H s t, t k)i k xtk ) ) hτ) dτ hτ) dτ <t k < I k xtk ) ) σ 32 <t <σ 3 ε 8 xt ) > 2 x PC 2 x PC = x PC, 3.3)
24 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 24 of 34 T 2 y PC2 = max t J { Ht, s)bs)xs) ds k)j k ytk ) )} { max Ht, s)b s)xs) ds t J min 2 t ν ν k)j k ytk ) )} Ht, s)b s)xs) ds min σ 3 t gτ) dτε 9 yt ) σ 32 <t <σ 3 gτ) dτε 9 min σ 32 t σ 3 δt) y PC2 t 2 <t k<t k)j k ytk ) ) = y PC2. 3.4) Consequently, Tx, y) > x, y), x, y) K K 2 ) Ω r. 3.5) Next, let us turn to ii), which shows that there exist R, R > r such that f x) ε x, I k x) ε 2 x, x R, J k y) ε 3 y, y R, where ε, ε 2, ε 3 satisfy Let 5γ ε ω s) ds <, { ε 3 < max 6A R, γ n n 5nε 2 <, ν } b s) ds. { } η = max f x), { η2 = max Ik x) }, x [,R ] x [,R ] { η 3 = max Jk y) }, k =,2,...,n. y [,R ] Then f x) ε x η, I k x) ε 2 x η 2, J k y) ε 3 y η 3, x, y. 3.6) Let M = γ η ω s) ds, M = nη 2. ν
25 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 25 of 34 Choosing max{6m,6m, r} < R <6A), for any x, y) K K 2 ) Ω R, similarly to the proof of 3.2)and3.3), we get T x)t) H t, s)hs, τ)a s)b τ)xs)xτ) dτ ds H t, s)ω s)f xs) ) ds H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds k)i k xtk ) ) T 2 y)t)= γ γ ν n Hs, τ)a s)b τ) x 2 PC dτ ds γ ω s) ) ε x PC η ds AR 2 γ ε ) nε 2 ν a ) s) ds x PC ε3 y PC2 η 3 ν ω s) dsr M ε 3 A R 2 γ ν nη 3 R M < 6 R 6 R 6 R 6 R 6 R 6 R ) ε2 x PC η 2 a s) ds = R, 3.7) Ht, s)bs)xs) ds Ht, s)b s)xs) ds γ γ γ k)j k ytk ) ) b s)xs) ds ν k)j k ytk ) ) J k ytk ) ) b s) ds x PC ν nε 3 y PC2 b s) dsr γ nε 3 R = R, 3.8) which shows that Tx, y) < x, y), x, y) K K 2 ) Ω R. 3.9) Finally, since < d < min{4a),4a Γ ), r},forx, y) K K 2 ) Ω d, it follows from 3.2)that T x)t) H t, s)hs, τ)a s)b τ)xs)xτ) dτ ds H t, s)ω s)f xs) ) ds H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds k)i k xtk ) )
26 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 26 of 34 T 2 y)t)= γ γ ν n Hs, τ)a s)b τ) x 2 PC dτ ds γ ω s)f xs) ) ds a s)j k ytk ) ) ds x PC ν I k xtk ) ) Ad 2 γ ω s) dsd) d A Γ d 2 nλ) d ν < 4 d 4 d 4 d 4 d = d, 3.2) Ht, s)bs)xs) ds Ht, s)b s)xs) ds γ k)j k ytk ) ) b s)xs) ds ν k)j k ytk ) ) J k ytk ) ) γ b s) dsd ν nγ d = d, 3.2) which shows that Tx, y) < x, y), x, y) K K 2 ) Ω d. 3.22) Therefore, applying Lemma 2.4 to 3.5), 3.9), and 3.22), we can show that T has at least two fixed points x, y ) K K 2 ) Ω R \ Ω r ), x 2, y 2 ) K K 2 ) Ω r \ Ω d ). The proof of Theorem 3.2 is completed. Corollary 3. Assume that H ) H ) hold. If <α <,then system.9) admits at least one positive solution. Proof It follows from the proof of Theorem 3.2 that Corollary 3. holds. Corollary 3.2 Assume that H ) H ) hold. Suppose that α >and there exist two constants d and r satisfying <d < r = min{4a),4ε 3 A ), r, r } such that min f x)> D ) d, x,y) K K 2 ) Ω d min I k x)> Λ ) d, x,y) K K 2 ) Ω d min x,y) K K 2 ) Ω d J k y)>γ d,
27 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 27 of 34 where A, A, ε 3, r and r are defined in Theorem 3.. Then system.9) admits at least two positive solutions. Proof SimilarlytotheproofofTheorem3., we can obtain that 3.4)and3.)hold.Then, similarly to the proof 3.22), we get Tx, y) > x, y), x, y) K K 2 ) Ω d. 3.23) This finishes the proof of Corollary 3.2. Finally, in the case < α <, we consider the existence of three positive solutions for system.9). Theorem 3.3 Assume that H ) H ) hold and there exist four positive numbers η, η, η 2, and γ such that one of the following conditions is satisfied: H ) <α <, <η = max{η, η 2 } < min{r,4a),4a Γ ) } r max{6m,6m, r} < R <6A) < γ, and f x)<d) η, I k x)<λ) η, J k y)<γη, x [θ 3 η, η ], y [θ 3 η 2, η 2 ], f x)> D ) γ, Ik x)> Λ ) γ, J k y)>γ γ, x, y [, γ ], where r, R, A, M, M, D, D, γ, γ, θ 3, Λ, and Λ are defined in Theorems 3. and 3.2, respectively. Then system.9) admitsatleastthreepositivesolutions. Proof Since < α <, from the proof of Theorem 3.2 we know that Tx, y) > x, y), x, y) K K 2 ) Ω r, 3.24) Tx, y) < x, y), x, y) K K 2 ) Ω R. 3.25) By the first part of H ), for any x, y) K K 2 ) Ω η,weobtain x, y) = max { x PC, y PC2 } = η = max{η, η 2 }, θ 3 η θ 3 x PC θ 3 η 2 θ 3 y PC2 min xt) xt) η σ, 32 t σ 3 min yt) xt) η, 32 t σ 3 and similarly to the proof of 3.22), we get Tx, y) < x, y), x, y) K K 2 ) Ω η. 3.26)
28 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 28 of 34 Considering the second part of H ), for any x, y) K K 2 ) Ω γ,wehave T x PC = max t J { H t, s)hs, τ)as)bτ)xs)xτ) dτ ds H t, s)ωs)f xs) ) ds H t, s)as)xs) H τ s, t k)j k ytk ) )) ds { max H t, s)a s)xs) t J H t, s)ω s)f xs) ) ds k)i k xtk ) )} Hs, τ)bτ)xτ) dτ ds H t, s)a s)xs) H τ s, t k)j k ytk ) )) ds min H σ t, s)a s)xs) 22 t min 2 2 σ t 22 min 2 σ t 32 σ k)i k xtk ) )} Hs, τ)b τ)xτ) dτ ds H t, s)as)xs) H τ s, t k)j k ytk ) ) ) ds σ 32 <t k <σ 3 H t, s)ω s)f xs) ) ds min min H σ t, s)ω s)f xs) ) ds min 22 t ρ 2 ρ 2 T 2 y PC2 = max t J σ 32 G s, s)ω s)f x) ds σ 32 ν t <t k < t <t k < σ 32 G s, s)ω s) ds D ) γ ν hτ) dτ H s t, t k)i k xtk ) ) H s t, t k)i k xtk ) ) σ 32 <t <σ 3 I k x) hτ) dτ Λ ) γ > 2 γ 2 γ = γ, 3.27) { Ht, s)bs)xs) ds k)j k ytk ) )} { max Ht, s)b s)xs) ds t J min t [,] min t [,] n ν k)j k ytk ) )} Ht, s)b s)xs) ds min t [,] k)j k ytk ) ) k)j k ytk ) ) gτ) dτγ γ = γ. 3.28)
29 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 29 of 34 This shows that Tx, y) > x, y), x, y) K K 2 ) Ω γ. 3.29) Therefore, applying Lemma 2.4 to 3.24), 3.25), 3.26), and 3.29) respectively, we can show that T has at least three fixed points x i, y i )i =,2,3)satisfying x, y ) K K 2 ) Ω γ \ Ω R ), x 2, y 2 ) K K 2 ) Ω R \ Ω r ), x 3, y 3 ) K K 2 ) Ω r \ Ω η ). This gives the proof of Theorem Anexample Example 4. Let n =andt =. Consider the following system: x = at)xy ωt)x 3, <t <,t y = bt)x, <t <,t, x t= = I x )), y t= = J y )), x) = xt) dt, x ) =, y) = tyt) dt, y ) =,, 4.) where I x)= x3 2, J y)= y3, ht), gt)=t,and bt)= t), t [, 3 ], 8 t 3 ), t [ 3,], at)= 3 t), t [, 3 ], 8 t 3 ), t [ 3,], 92 3 gt)= t), t [, 3 ], 8 t 3 ), t [ 3,]. Conclusion 4. System 4.) admits at least one positive solution. For convenience, we give a corollary of Proposition 2.3 in [67]. Corollary 4. Consider the following system: x = kt)x α, <t <, x) = ht)xt) dt, x ) =, y = kt)y α, <t <, y) = gt)yt) dt, y ) =, 4.2) 4.3)
30 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 3 of 34 where α >,and kt) satisfies the changing-sign condition kt), t [, ], kt), t [,], and c x α f x)=x α c 2 x α, c y α f y)=y α c 2 y α, c, c 2 >. If there exists <σ < such that σ c σ α μ 2 k σ ) η c 2 α k η), η [, ], 4.4) then the following inequalities hold: σ α μ 2 σ σ α μ 2 σ Ht, s)k s) ds c 2 c α H t, s)k s) ds c 2 c α Ht, s)k s) ds, 4.5) H t, s)k s) ds. 4.6) Proof SimilarlytotheproofofProposition2.3in[67], we can prove that G t, σ ) η σ Gt, η), Hence it follows from 2.4)that η [, ]. H t, σ ) η = G t, σ ) η ν σ Gt, η) σ ν [ = σ Gt, η) ν = σ Ht, η), η [, ]. G τ, σ ) η gτ) dτ Gτ, η)gτ) dτ ] Gτ, η)gτ) dτ Next, letting s = σ η, η [, ], we get Ht, s)k s) ds = σ H t, σ )k σ η σ ) η dη; letting s = η, η [, ], we have Ht, s)k s) ds = Ht, η)k η) dη. Now, from assumption 4.4), for all t, η) [, ] [, ], we have σ c σ α μ 2 H t, σ )k η σ ) η c 2 α Ht, η)k η). 4.7)
31 Jiao and Zhang Boundary Value Problems 28) 28:63 Page 3 of 34 Finally, by integrating in η both sides of 4.7) fromto it follows that inequality 4.5)holds. Similarly, we can show that inequality 4.6)holds. Proof of Example 4. From the definitions of at), bt), and gt)weknowthat = 3. Step. We show that H 6 ) holds. For fixed c = c 2 =,σ =, μ =,andα =,4.4) is 6 equivalent to the inequality 48 b 3 ) ) [ 3 4 η b η, η, 2 ]. 3 Letting 3 4 τ = ϱ,thisinequalityisequivalentto ) [ 5 48 b ϱ) b 3 4ϱ, ϱ 4, ]. 3 By the definition of bt) the last inequality holds obviously. It is clear that by 4.5) H 6 )is reasonable. Step 2. We show H 7 ) holds. Similarly to Step, letting c =,c 2 = 36 5, = 6, μ =,and α =,by4.6)weget 3 H t, s)a s) ds It is easy to see by calculating that ν = gs) ds = sds= 2, γ = =2, ρ = ν H t, s)a s) ds. 4.8) τgτ) dτ = 5 ν 3. Furthermore, from inequality 4.8) it followsthat H t, s) a s) ds H t, s)a s) ds H t, s) min Gs, s)a s) ds 2 3 s [ 6 6, 3 ] H t, s)gs, s)a s) ds H t, s)a s) ds H t, s)a s) ds. So, H 7 )holds. Step 3. Similarly to Step, letting c = c 2 =,σ 3 = 6, μ =,andα =3,wegetthatH ) holds. Hence it followsfromtheorem3. that system 4.) admits at least one positive solution for α >.
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