International Journal of Analysis and Applications ISSN 2291-8639 Volume 15, Number 2 (2017), 198-210 DOI: 10.28924/2291-8639-15-2017-198 COUNTABLY INFINITELY MANY POSITIVE SOLUTIONS FOR EVEN ORDER BOUNDARY VALUE PROBLEMS WITH STURM-LIOUVILLE TYPE INTEGRAL BOUNDARY CONDITIONS ON TIME SCALES K. R. PRASAD∗ AND MD. KHUDDUSH Abstract. In this paper, we establish the existence of countably infinitely many positive solutions for a certain even order two-point boundary value problem with integral boundary conditions on time scales by using Hölder’s inequality and Krasnoselskii’s fixed point theorem for operators on a cone. 1. Introduction The study of dynamic equations on time scales unifies existing results in differential and finite dif- ference equations, and provides powerful new tools for exploring connections between the traditionally separated fields. For details refer to the books by Bohner and Peterson [6], [7], Lakshmikantham et al. [23] and the papers [1], [3], [19]. The boundary value problems with integral boundary conditions occur in the study of nonlocal phenomena in many different areas of applied mathematics, physics and engineering, in particular, in heat conduction, chemical engineering, underground waterflow, thermo-elasticity, plasma physics, [2], [10], [11], [21], [22], [27], [33], [36] and reference therin. Recently, authors established the existence of positive solutions to boundary value problems with integral boundary conditions on time scales; for details, see [9], [12], [13], [18], [26], [28], [32], [34] and reference therein. However, to the best of our knowledge, little work has been done on the existence of positive solutions for higher order boundary value problems with integral boundary conditions on time scales. We would like to mention some results of Karasa and Tokmak [20], Li and Wang [25], Li and Sun [24], Cetin and Topal [8], and Sreedhar et al [31] which motivate us to consider the problem (1.6)-(1.7). In 2013, Karasa and Tokmak [20] established the existence of a positive solution of the following third order boundary value problem with integral boundary conditions,( φ(−u∆∆(t)) )∆ + q(t)f(t,u(t),u∆(t)) = 0, t ∈ [0, 1]T , au(0) − bu∆(0) = ∫ 1 0 g1(s)u(s)∆s, cu(1) + du ∆(1) = ∫ 1 0 g2(s)u(s)∆s, u ∆∆(1) = 0,   (1.1) by using four functionals fixed point theorem. In the same year, Li and Wang [25] studied the existence of a positive solution of the following nonlinear third order boundary value problem with integral boundary conditions,( φ(−u∆∆(t)) )∆ + q(t)f(t,u(t),u∆(t)) = 0, t ∈ [0, 1]T , au(0) − bu∆(0) = ∫ 1 0 g1(s,u(s))∆s, cu(1) + du ∆(1) = ∫ 1 0 g2(s)u(s)∆s, u ∆∆(1) = 0,   (1.2) by applying a generalization of the Leggett-Williams fixed point theorem. In [24], Li and Sun studied the following boundary value problem on time scales, x∆(t) + p(t)xσ(t) = f ( t,xσ(t) ) , t ∈ (0,T)T, x(0) −βxσ(T) = α ∫ σ(T) 0 xσ(s)∆g(s),   (1.3) Received 14th August, 2017; accepted 18th October, 2017; published 1st November, 2017. 2010 Mathematics Subject Classification. 34B18, 34N05. Key words and phrases. Green’s function; boundary value problem; positive solution; cone. c©2017 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 198 COUNTABLE MANY POSITIVE SOL. FOR EOBVPS WITH IBCS ON TIME SCALES 199 where xσ = x ◦ σ, and by using fixed point index theory the existence of infinitely many positive solutions for (1.3) are obtained. In [8], Cetin and Topal investigated the existence of solutions for integral boundary value problems on time scales, [p(t)x∆(t)]∇ + q(t)x(t) = f(t,x(t)), t ∈ [a,b]T , αx(ρ(a)) −βx[∆](ρ(a)) = ∫ b ρ(a) h1(x(s))∇s,γx(b) + δx[∆](b) = ∫ b ρ(a) h2(x(s))∇s,   (1.4) by using Schauder fixed point theorem in a cone and by the method of upper and lower solutions. In 2017, Sreedhar et al [31] considered the 2nth order boundary value problem with integral boundary conditions on time scales, (−1)nu∆ 2n (t) = f ( t,u(t) ) , t ∈ (0, 1)T, u∆ 2i (0) = u∆ 2i (1) = ∫ 1 0 ai+1(x)u ∆2i(x)∆x, 0 ≤ i ≤ n− 1,   (1.5) where n act as positive. By using Avery-Henderson fixed point theorem, the authors established the existence of even number of positive solutions for (1.5). Motivated by the work mentioned above, in this paper we investigate the existence of infinitely many positive solutions for the even order boundary value problem on time scales given by (−1)nu(∆∇) n (t) = ω(t)f ( u(t) ) , t ∈ [0, 1]T, (1.6) satisfying the Sturm-Liouville type integral boundary conditions αi+1u (∆∇)i(0) −βi+1u(∆∇) i∆(0) = ∫ 1 0 ai+1(s)u (∆∇)i(s)∇s, 0 ≤ i ≤ n− 1, γi+1u (∆∇)i(1) + δi+1u (∆∇)i∆(1) = ∫ 1 0 bi+1(s)u (∆∇)i(s)∇s, 0 ≤ i ≤ n− 1,   (1.7) where n ≥ 1,T is a time scale, f ∈ C ( [0, +∞), [0, +∞) ) ,ω(t) ∈ Lp∇[0, 1] for some p ≥ 1 and has countably many singularities in (0, 1 2 )T. We show that the boundary value problem (1.6)-(1.7) has countably infinitely many positive solutions by imposing suitable conditions on ω and f. The key tool in our approach is the Hölder’s inequality and Krasnoselskii’s fixed point theorem for operators on a cone. 2. Preliminaries In this section, we provide some definitions and lemmas which are useful for our later discussions; for details, see [2], [4], [5], [6], [16], [30], [35]. Definition 2.1. A time scale T is a nonempty closed subset of the real numbers R. T has the topology that it inherits from the real numbers with the standard topology. It follows that the jump operators σ,ρ : T → T, σ(t) = inf{r ∈ T : r > t}, ρ(t) = sup{r ∈ T : r < t} (supplemented by inf ∅ := sup T and sup∅ := inf T) are well defined. The point t ∈ T is left-dense, left-scattered, right-dense, right-scattered if ρ(t) = t, ρ(t) < t, σ(t) = t, σ(t) < t, respectively. By an interval time scale, we mean the intersection of a real interval with a given time scale. i.e., [a,b]T = [a,b] ∩T other intervals can be defined similarly. Definition 2.2. Let µ∆ and µ∇ be the Lebesgue ∆− measure and the Lebesgue ∇−measure on T, respectively. If A ⊂ T satisfies µ∆(A) = µ∇(A), then we call A is measurable on T, denoted µ(A) and this value is called the Lebesgue measure of A. Let P denote a proposition with respect to t ∈ T. (i) If there exists E1 ⊂ A with µ∆(E1) = 0 such that P holds on A\E1, then P is said to hold ∆–a.e. on A. (ii) If there exists E2 ⊂ A with µ∇(E2) = 0 such that P holds on A\E2, then P is said to hold ∇–a.e. on A. 200 PRASAD AND KHUDDUSH Definition 2.3. Let E ⊂ T be a ∇–measurable set and p ∈ R̄ ≡ R∪{−∞, +∞} be such that p ≥ 1 and let f : E → R̄ be ∇–measurable function. We say that f belongs to Lp∇(E) provided that either∫ E |f|p(s)∇s < ∞ if p ∈ R, or there exists a constant M ∈ R such that |f| ≤ M, ∇−a.e. on E if p = +∞. Lemma 2.1. Let E ⊂ T be a ∇–measurable set. If f : T → R is a ∇–integrable on E, then∫ E f(s)∇s = ∫ E f(s)ds + ∑ i∈IE ( ti −ρ(ti) ) f(ti), where IE := {i ∈ I : ti ∈ E} and {ti}i∈I,I ⊂ N, is the set of all left-scattered points of T. For convenience, we introduce the following notation throughout the paper: For τ ∈ (0, 1 2 )T, φi(t) := γi + δi −γit, ψi(t) := βi + αit, di := γiβi + αiδi + αiγi, Gi(t,s) := 1 di { φi(s)ψi(t), t ≤ s, φi(t)ψi(s), s ≤ t, A := ∫ 1 0 [∫ 1 0 Gi(s,r)ai(s)∇s ] ω(r)∇r, B := ∫ 1 0 [∫ 1 0 Gi(s,r)bi(s)∇s ] ω(r)∇r, ui := 1 di ∫ 1 0 ai(t)φi(t)∇t, u∗i := 1 di ∫ 1 0 ai(t)ψi(t)∇t, vi := 1 di ∫ 1 0 bi(t)ψi(t)∇t, v∗i := 1 di ∫ 1 0 bi(t)φi(t)∇t, θi(t) := (1 −vi)φi(t) + v∗i ψi(t) di ( (1 −ui)(1 −vi) −u∗iv ∗ i ), ζi(t) := (1 −ui)ψi(t) + u∗iφi(t) di ( (1 −ui)(1 −vi) −u∗iv ∗ i ), θ∗i := max t∈J0 θi(t), θ ∗∗ i := max t∈[τ, 1−τ]T θi(t), ζ ∗ i := max t∈J0 ζi(t), ζ ∗∗ i := max t∈[τ, 1−τ]T ζi(t), a∗i := ∫ 1 0 ai(t)∇t, ai(τ) := ∫ 1−τ τ ai(t)∇t, b∗i := ∫ 1 0 bi(t)∇t, bi(τ) := ∫ 1−τ τ bi(t)∇t, gi := ∫ 1 0 Gi(t,t)∇t, gi(τ) := ∫ 1−τ τ Gi(t,t)∇t, and J0 := [0, 1]T. We make the following assumptions throughout the paper: (H1) there exists a sequence {tk}∞k=1 (k ∈ N), t1 < 12, limk→∞tk = t ∗ ≥ 0 and lim t→tk ω(t) = +∞ for all k = 1, 2, 3, · · · . (H2) ω ∈ Lp∇(J0) for some 1 ≤ p ≤ +∞ and there exists m > 0 such that ω(t) ≥ m for all [t∗, 1 − t∗]T, (H3) f : J0 × [0, +∞) → [0, +∞) is continuous, (H4) ai,bi ∈ L1∇(J0) for all 1 ≤ i ≤ n, are nonnegative and (1 − ui)(1 − vi) − v ∗ i u ∗ i > 0 for all 1 ≤ i ≤ n, on J0, (H5) αi,βi,γi,δi ≥ 0 such that di := γiβi + αiδi + αiγi > 0 for each 1 ≤ i ≤ n. The rest of the paper is organized in the following fashion. In Section 2, we provide some definitions and lemmas that provide us with some useful information concerning the behavior of solution of the boundary value problem (1.6)-(1.7). In Section 3, we construct the Green’s function for the homogeneous problem corresponding to (1.6)-(1.7), estimate bounds for the Green’s function, and some lemmas which are needed in establishing our main results are provided. In Section 4, we establish a criteria for the existence of countable number of positive solutions for the boundary value problem (1.6)-(1.7) by applying Krasnoselskii’s fixed point theorem in cones. Finally, we provide an example of a family of functions ω(t) that satisfy conditions (H1) − (H3). COUNTABLE MANY POSITIVE SOL. FOR EOBVPS WITH IBCS ON TIME SCALES 201 3. Green’s Function and Bounds In this section, we construct the Green’s function for the homogeneous problem corresponding to (1.6)-(1.7) and estimate bounds for the Green’s function. Lemma 3.1. Let (H4), (H5) hold. Then for any h(t) ∈ C(J0), the boundary value problem, −u∆∇(t) = h(t), t ∈ J0, (3.1) αiu(0) −βiu∆(0) = ∫ 1 0 ai(s)u(s)∇s, 1 ≤ i ≤ n, (3.2) γiu(1) + δiu ∆(1) = ∫ 1 0 bi(s)u(s)∇s, 1 ≤ i ≤ n, (3.3) has a unique solution u(t) = ∫ 1 0 Ki(t,s)h(s)∇s, for 1 ≤ i ≤ n, (3.4) where, Ki(t,s) = Gi(t,s) + θi(t) ∫ 1 0 Gi(r,s)aj(r)∇r + ζi(t) ∫ 1 0 Gi(r,s)bi(r)∇r, (3.5) for 1 ≤ i ≤ n, Proof. Suppose that u is a solution of (3.1), then, we have u(t) = − ∫ t 0 ∫ s 0 h(r)∇r∆s + At + B = − ∫ t 0 (t−s)h(s)∇s + At + B where A = lim t→0+ u∆(t) and B = u(0). Using the boundary conditions (3.1), (3.2) we can determined A and B as A = 1 di ∫ 1 0 [αibi(s) −γiai(s)]u(s)∇s + αi di ∫ 1 0 [γi(1 −s) + δi]h(s)∇s B = 1 di ∫ 1 0 [(γi + δi)ai(s) + βibi(s)]u(s)∇s + βi di ∫ 1 0 [γi(1 −s) + δi]h(s)∇s Thus, we have u(t) = 1 di [∫ t 0 (γi + δi −γit)(βi + αis)h(s)∇s + ∫ 1 t (γi + δi −γis)(βi + αit)h(s)∇s ] + 1 di [∫ 1 0 [(γi + δi −γit)ai(s) + (βi + αit)bi(s)]u(s)∇s ] = 1 di [∫ t 0 φi(t)ψi(s)h(s)∇s + ∫ 1 t φi(s)ψi(t)h(s)∇s ] + 1 di ∫ 1 0 [φi(t)ai(s) + ψi(t)bi(s)]u(s)∇s, from which, we obtain u(t) = ∫ 1 0 Gi(t,s)h(s)∇s + 1 di φi(t) ∫ 1 0 ai(s)u(s)∇s + 1 di ψi(t) ∫ 1 0 bi(s)u(s)∇s. (3.6) After certain computations we can determined,∫ 1 0 ai(s)u(s)∇s = (1 −vi)Ai + u∗iBi (1 −ui)(1 −vi) −v∗i u ∗ i . (3.7) ∫ 1 0 bi(s)u(s)∇s = v∗i Ai + (1 −ui)Bi (1 −ui)(1 −vi) −v∗i u ∗ i . (3.8) 202 PRASAD AND KHUDDUSH From (3.7) and (3.8), (3.6) can be written as u(t) = ∫ 1 0 Gi(t,s)h(s)∇s + 1 di φi(t) (1 −vi)Ai + u∗iBi (1 −ui)(1 −vi) −v∗i u ∗ i + 1 di ψi(t) v∗i Ai + (1 −ui)Bi (1 −ui)(1 −vi) −v∗i u ∗ i = ∫ 1 0 Gi(t,s)h(s)∇s + (1 −vi)φi(t) + v∗i ψi(t) di ( (1 −ui)(1 −vi) −v∗i u ∗ i )Ai + (1 −ui)ψi(t) + u∗iφi(t) di ( (1 −ui)(1 −vi) −v∗i u ∗ i )Bi = ∫ 1 0 Gi(t,s)h(s)∇s + θi(t)Ai + ζi(t)Bi = ∫ 1 0 Gi(t,s)h(s)∇s + θi(t) ∫ 1 0 (∫ 1 0 Gi(r,s)ai(r)∇r ) h(s)∇s + ζi(t) ∫ 1 0 (∫ 1 0 Gi(r,s)bi(r)∇r ) h(s)∇s = ∫ 1 0 [ Gi(t,s) + θi(t) ∫ 1 0 Gi(r,s)ai(r)∇r + ζi(t) ∫ 1 0 Gi(r,s)bi(r)∇r ] h(s)∇s = ∫ 1 0 Ki(t,s)h(s)∇s, where Ki(t,s) is defined in (3.5). The proof is complete. � Lemma 3.2. Assume that (H4), (H5) hold and for τ ∈ (0, 1 2 )T define ηi(τ) = min { αiτ + βi αi + βi , γiτ + δi γi + δi } < 1. Then Gi(t,s) for 1 ≤ i ≤ n, satisfies the following properties: (i) 0 ≤ Gi(t,s) ≤ Gi(s,s) for all t,s ∈ J0, (ii) 0 ≤ ηi(τ)Gi(s,s) ≤ Gi(t,s) for all t ∈ [τ, 1 − τ]T and s ∈ J0, Lemma 3.3. Assume that (H4), (H5) holds. Then Ki(t,s) for 1 ≤ i ≤ n, have the following proper- ties: (i) 0 ≤ Ki(t,s) ≤ ( 1 + θ∗i a ∗ i + ζ ∗ i b ∗ i ) Gi(s,s) for all t,s ∈ J0, (ii) 0 ≤ ηi(τ) ( 1 + θ∗∗i ai(τ) + ζ ∗∗ i bi(τ) ) Gi(s,s) ≤ Ki(t,s) for all t ∈ [τ, 1 − τ]T and s ∈ J0 Lemma 3.4. Assume that (H4), (H5) hold and Kj(t,s) for 1 ≤ j ≤ n, is given in (3.5). Let H1(t,s) = K1(t,s) and recursively define Hj(t,s) = ∫ 1 0 Hj−1(t,r)Kj(r,s)∇r, for 2 ≤ j ≤ n. (3.9) Then Hn(t,s) is the Green’s function for the homogeneous boundary value problem (−1)nu(∆∇) n (t) = 0, t ∈ J0, αi+1u (∆∇)i(0) −βi+1u(∆∇) i∆(0) = ∫ 1 0 ai+1(s)u (∆∇)i(s)∇s, 0 ≤ i ≤ n− 1, γi+1u (∆∇)i(1) + δi+1u (∆∇)i∆(1) = ∫ 1 0 bi+1(s)u (∆∇)i(s)∇s, 0 ≤ i ≤ n− 1. COUNTABLE MANY POSITIVE SOL. FOR EOBVPS WITH IBCS ON TIME SCALES 203 Lemma 3.5. Assume that (H4), (H5) hold. Define g = n−1∏ j=1 gj, φ = n∏ j=1 ( 1 + θ∗ja ∗ j + ζ ∗ j b ∗ j ) , ητ = n∏ j=1 ηj(τ) ( 1 + θ∗∗j aj(τ) + ζ ∗∗ j bj(τ) ) , gτ = n−1∏ j=1 gj(τ), then the Green’s function Hn(t,s) satisfies the following inequalities: (i) 0 ≤ Hn(t,s) ≤ gφGn(s,s), for all t,s ∈ J0 and (ii) Hn(t,s) ≥ ητgτGn(s,s), for all t ∈ [τ, 1 − τ]T and s ∈ J0, Proof. It is clear that Green’s function Hn(t,s) ≥ 0, for all t,s ∈ J0. Now we prove the inequality by induction on n and denote the statement by p(n). From (3.5) we have H1(t,s) = G1(t,s) + θ1(t) ∫ 1 0 G1(r,s)a1(r)∇r + ζ1(t) ∫ 1 0 G1(r,s)b1(r)∇r ≤ G1(s,s) + θ1(t) ∫ 1 0 G1(s,s)a1(r)∇r + ζ1(t) ∫ 1 0 G1(s,s)b1(r)∇r ≤ ( 1 + θ1(t) ∫ 1 0 a1(r)∇r + ζ1(r) ∫ 1 0 b1(r)∇r ) G1(s,s) ≤ ( 1 + θ∗1a ∗ 1 + ζ ∗ 1 b ∗ 1 ) G1(s,s) and for t ∈ [τ, 1 − τ]T, H1(t,s) = G1(t,s) + θ1(t) ∫ 1 0 G1(r,s)a1(r)∇r + ζ1(t) ∫ 1 0 G1(r,s)b1(r)∇r ≥ G1(t,s) + θ1(t) ∫ 1−τ τ G1(r,s)a1(r)∇r + ζ1(t) ∫ 1−τ τ G1(r,s)b1(r)∇r ≥ η1(τ)G1(s,s) + θ1(t)η1(τ)G1(s,s) ∫ 1−τ τ a1(r)∇r + ζ1(t)η1(τ)G1(s,s) ∫ 1−τ τ b1(r)∇r ≥ ( 1 + θ1(t) ∫ 1−τ τ a1(r)∇r + ζ1(r) ∫ 1−τ τ b1(r)∇r ) η1(τ)G1(s,s) ≥ ( 1 + θ∗∗1 a1(τ) + ζ ∗∗ 1 b1(τ) ) η1(τ)G1(s,s) Hence, p(1) is true. From (3.9), we have Hm+1(t,s) = ∫ 1 0 Hm(t,r)Km+1(r,s)∇r ≤ ∫ 1 0 Gm(r,r)gφ ( 1 + θ∗m+1a ∗ m+1 + ζ ∗ m+1b ∗ m+1 ) Gm+1(s,s)∇r ≤ (∫ 1 0 Gm(r,r)∇r ) gφ ( 1 + θ∗m+1a ∗ m+1 + ζ ∗ m+1b ∗ 1 ) Gm+1(s,s) ≤ ( m∏ i=1 gi )m+1∏ i=1 ( 1 + θ∗i a ∗ i + ζ ∗ i b ∗ i ) Gm+1(s,s) 204 PRASAD AND KHUDDUSH and for t ∈ [τ, 1 − τ]T, Hm+1(t,s) = ∫ 1 0 Hm(t,r)Km+1(r,s)∇r ≥ ∫ 1−τ τ ητgτGm(r,r)ηm+1(τ) ( 1 + θ∗∗m+1am+1(τ) + ζ ∗∗ m+1bm+1(τ) ) Gm+1(s,s)∇r ≥ gτ (∫ 1−τ τ Gm(r,r)∇r ) ητηm+1(τ) ( 1 + θ∗∗m+1am+1(τ) + ζ ∗∗ m+1bm+1(τ) ) Gm+1(s,s) ≥ ( m∏ i=1 gi(τ) )m+1∏ i=1 ηm+1(τ) ( 1 + θ∗∗i ai(τ) + ζ ∗∗ i bi(τ) ) Gm+1(s,s) So, p(m + 1) holds. Thus p(n) is true by induction � Let X denotes the Banach space Cld(J0,R) with norm ‖u‖ = max t∈J0 |u(t)|. For τ ∈ (0, 1 2 )T, define the cone Pτ ⊂ X by Pτ = { u ∈ X : u(t) ≥ 0 and min t∈[τ, 1−τ]T u(t) ≥ ξτ‖u(t)‖ } , where ξτ = ητgτ gφ . For any u ∈ Pτ, define an operator T : Pτ → X by (Tu)(t) = ∫ 1 0 Hn(t,s)ω(s)f ( u(s) ) ∇s. (3.10) Lemma 3.6. Assume that (H1)-(H3) hold. Then T(Pτ ) ⊂ Pτ and T : Pτ → Pτ is completely continuous for each τ ∈ (0, 1 2 )T. Proof. Fix τ ∈ (0, 1 2 ). Since ω(s)f(u(s)) ≥ 0 for all s ∈ J0,u ∈ Pτ and since Hn(t,s) ≥ 0 for all t,s ∈ J0, then T(u(t)) ≥ 0 for all t ∈ J0,u ∈ Pτ. On the other hand, by Lemma 3.5 we obtain (Tu)(t) = ∫ 1 0 Hn(t,s)ω(s)f ( u(s) ) ∇s ≤ gφ ∫ 1 0 Gn(s,s)ω(s)f ( u(s) ) ∇s, min t∈[τ,1−τ]T u(t) = min t∈[τ,1−τ]T ∫ 1 0 Hn(t,s)ω(s)f ( u(s) ) ∇s ≥ ∫ 1 0 min t∈[τ,1−τ]T Hn(t,s)ω(s)f ( u(s) ) ∇s = ητgτ ∫ 1 0 Gn(s,s)ω(s)f ( u(s) ) ∇s ≥ ξτTu(t) for all t ∈ J0. Thus min t∈[τ,1−τ]T u(t) ≥ ξτ‖Tu‖. So, Tu ∈ Pτ and then T(Pτ ) ⊂ Pτ. Next, by stan- dard methods and the Arzela-Ascoli theorem, one can easily prove that the operator T is completely continuous. The proof is complete. � 4. Main Results In this section, we establish the existence of countably infinitely many positive solutions for the boundary value problem (1.6)-(1.7) by applying Krasnoselskii’s fixed point theorem in cones. Theorem 4.1. [14] Let B be a Banach space and let P ⊂B be a cone in B. Assume that Ω1, Ω2 are open with 0 ∈ Ω1, Ω̄1 ⊂ Ω2, and let T : P ∩ (Ω̄2\Ω1) → P be a completely continuous operator such that either (i) ‖Tu‖≤‖u‖,u ∈ P ∩∂Ω1, and ‖Tu‖≥‖u‖,u ∈ P ∩∂Ω2, or COUNTABLE MANY POSITIVE SOL. FOR EOBVPS WITH IBCS ON TIME SCALES 205 (ii) ‖Tu‖≥‖u‖,u ∈ P ∩∂Ω1, and ‖Tu‖≤‖u‖,u ∈ P ∩∂Ω2. Then T has a fixed point in P ∩ (Ω̄2\Ω1). Theorem 4.2. [4, 29] Let f ∈ Lp∇(J) with p > 1,g ∈ L q ∇(J) with q > 1, and 1 p + 1 q = 1. Then fg ∈ L1∇(J) and ‖fg‖L1∇ ≤‖f‖Lp∇‖g‖Lq∇. where ‖f‖Lp∇ :=   [∫ J |f|p(s)∇s ]1 p , p ∈ R, inf { M ∈ R/ |f| ≤ M ∇−a.e., onJ } , p = ∞, and J = (a,b]. Moreover, if f ∈ L1∇(J) and g ∈ L ∞ ∇ (J). Then fg ∈ L 1 ∇(J) and ‖fg‖L1∇ ≤‖f‖L1∇‖g‖L∞∇ . We consider the following three cases for ω ∈ Lp∇(J0) : p > 1,p = 1,p = ∞. Case p > 1 is treated in the following theorem. Theorem 4.3. Assume that (H1)−(H5) hold, let {τk}∞k=1 be such that tk+1 < τk < tk, k = 1, 2, 3, · · · . Let {Rk}∞k=1 and {rk} ∞ k=1 be such that Rk+1 < ξτkrk < Crk < Rk, k ∈ N, where C = max { 1 ητ1gτ1m ∫ 1−τ1 τ1 Gn(s,s)∇s , 1 } . Assume that f satisfies (A1) f(u) ≤ M1Rk for all t ∈ J0, 0 ≤ u ≤ Rk, where M1 < 1 gφ‖Gn(s,s)‖Lq∇‖ω‖Lp∇ , (A2) f(u) ≥ Crk for all t ∈ [τk, 1 − τk]T, ξτrk ≤ u ≤ rk. Then the boundary value problem (1.6)-(1.7) has countably infinitely many positive solutions {uk}∞k=1. Furthermore, rk ≤‖uk‖≤ Rk for each k ∈ N. Proof. Consider the sequences {Ω1,k}∞k=1 and {Ω2,k} ∞ k=1 of open subsets of E defined by Ω1,k = {u ∈B : ‖u‖ < Rk}, Ω2,k = {u ∈B : ‖u‖ < rk}. Let {τk}∞k=1 be as in the hypothesis and note that t ∗ < tk+1 < τk < tk < 1 2 , for all k ∈ N. For each k ∈ N, define the cone Pτk by Pτk = { u ∈ X : u(t) ≥ 0 and min t∈[τk, 1−τk]T u(t) ≥ ξτk‖u(t)‖ } . Let u ∈ Pτk ∩∂Ω1,k. Then, u(s) ≤ Rk = ‖u‖ for all s ∈ J0. By (A1), ‖Tu‖ = max t∈J0 ∫ 1 0 Hn(t,s)ω(s)f ( u(s) ) ∇s ≤ gφ ∫ 1 0 Gn(s,s)ω(s)∇sM1Rk ≤ gφ‖Gn(s,s)‖Lq∇‖ω‖Lp∇M1Rk ≤ Rk. Since ‖u‖ = Rk for all u ∈ Pτk ∩∂Ω1,k, then ‖Tu‖≤‖u‖. (4.1) Let s ∈ [τk, 1 − τk]T. Then, rk = ‖u‖≥ u(s) ≥ min s∈[τk, 1−τk]T u(s) ≥ ξτ‖u‖≥ ξτkrk. 206 PRASAD AND KHUDDUSH By (A2), ‖Tu‖ = max t∈J0 ∫ 1 0 Hn(t,s)ω(s)f ( u(s) ) ∇s ≥ max t∈J0 ∫ 1−τk τk Hn(t,s)ω(s)f ( u(s) ) ∇s ≥ max t∈J0 ∫ 1−τk τk Hn(t,s)ω(s)∇sCrk ≥ Crkm max t∈[τ1, 1−τ1]T ∫ 1−τ1 τ1 Hn(t,s)∇s ≥ ητ1gτ1mCrk max t∈[τ1, 1−τ1]T ∫ 1−τ1 τ1 Gn(s,s)∇s ≥ rk = ‖u‖. Thus, if u ∈ Pτ ∩∂Ω2,k, then ‖Tu‖≥‖u‖. (4.2) It is obvious that 0 ∈ Ω2,k ⊂ Ω̄2,k ⊂ Ω1,k. By (4.1),(4.2), it follows from Theorem 4.1 that the operator T has a fixed point uk ∈ Pτk ∩ ( Ω̄1,k\Ω2,k ) such that rk ≤‖uk‖≤ Rk. Since k ∈ N was arbitrary, the proof is complete. � Now we deal with the case p = 1. Theorem 4.4. Assume that (H1)−(H5) hold, let {τk}∞k=1 be such that tk+1 < τk < tk, k = 1, 2, 3, · · · . Let {Rk}∞k=1 and {rk} ∞ k=1 be such that Rk+1 < ξτkrk < Crk < Rk, k ∈ N, Assume that f satisfies (B1) f(u) ≤ M2Rk for all t ∈ J0, 0 ≤ u ≤ Rk, where M2 < min { 1 gφ‖Gn(s,s)‖L∞∇‖ω‖L1∇ , C } and (A2). Then the boundary value problem (1.6)-(1.7) has countably infinitely many positive solutions {uk}∞k=1. Furthermore, rk ≤‖uk‖≤ Rk for each k ∈ N. Proof. For a fixed k, let Ω1,k be as in the proof of Theorem 4.3 and let u ∈ Pτk ∩∂Ω2,k. Again u(s) ≤ Ak = ‖u‖, for all s ∈ J0. By (B1) and Theorem 4.3, ‖Tu‖ = max t∈J0 ∫ 1 0 Hn(t,s)ω(s)f ( u(s) ) ∇s ≤ gφ ∫ 1 0 Gn(s,s)ω(s)∇sM2Rk ≤ gφ‖Gn(s,s)‖L∞∇‖ω‖L1∇M2Rk ≤ Rk. Thus, ‖Tu‖≤‖u‖, for u ∈ Pτk∩∂Ω1,k. Now define Ω2,k = {u ∈B : ‖u‖ < rk}. Let u ∈ Pτk∩∂Ω2,k and let s ∈ [τk, 1−τk]T. Then, the argument leading to (4.2) carries over to the present case and completes the proof. � Finally we consider the case of p = ∞. COUNTABLE MANY POSITIVE SOL. FOR EOBVPS WITH IBCS ON TIME SCALES 207 Theorem 4.5. Assume that (H1) − (H5) hold. Let {Rk}∞k=1 and {rk} ∞ k=1 be such that Rk+1 < ξτrk < Crk < Rk, k ∈ N, Assume that f satisfies (E1) f(u) ≤ M3Rk for all t ∈ J0, 0 ≤ u ≤ Rk, where M3 < min { 1 gφ‖Gn(s,s)‖L1∇‖ω‖L∞∇ , C } and (A2). Then the boundary value problem (1.6)-(1.7) has countably infinitely many positive solutions {uk}∞k=1. Furthermore, rk ≤‖uk‖≤ Rk for each k ∈ N. Proof. By (E1), ‖Tu‖ = max t∈J0 ∫ 1 0 Hn(t,s)ω(s)f ( u(s) ) ∇s ≤ gφ ∫ 1 0 Gn(s,s)ω(s)∇sM3Rk ≤ gφ‖Gn(s,s)‖L1∇‖ω‖L∞∇ M3Rk ≤ Rk. This shows that if u ∈ Pτk ∩∂Ω1,k, where Ω1,k = {u ∈B : ‖u‖ < Rk}, Then, ‖Tu‖≤‖u‖. Define Ω2,k = {u ∈B : ‖u‖ < rk} and let u ∈ Pτk ∩∂Ω2,k. Then, the argument employed in the proof of Theorem 4.3 applies directly to yield ‖Tu‖≥‖u‖. By the Theorem 4.1, completes the proof. � 5. Example In this section, we provide an example of a family of functions ω(t) that satisfy conditions (H1), (H2) corresponding to the cases p = 1 and p = 2. Let T = [0, 1 2 ] ∪{3 5 , 3 4 , 4 5 }∪ [ 5 6 , 1] and consider the family of functions ω(t,�) : T → (0, +∞] given by ω(t,�) =   ∞∑ k=1 χ[νk, νk−1] |t− tk|� if 0 ≤ t ≤ 1 2 , 1 |t− 1 2 |� if 1 2 < t < 5 6 , 1 |t− 4 5 |� if 5 6 ≤ t ≤ 1, where t0 = 5 16 , tk = t0 − k−1∑ i=0 1 (i + 2)4 , k = 1, 2, 3, · · · , ν0 = 1, νk = 1 2 (tk + tk+1), k = 1, 2, 3, · · · . At first, it is easily seen that ω(t,�) ≥ ω(1,�) = 1|1−4 5 |� = 5 �, t1 = 1 4 < 1 2 , tk − tk+1 = 1(k+2)4 , k = 1, 2, 3, · · · , and note that ∑∞ k=1 1 k4 = π 4 90 . t∗ = lim k→∞ tk = 5 16 − ∞∑ i=0 1 (i + 2)4 = 5 16 − ( π4 90 − 1 ) = 21 16 − π4 90 > 1 5 . 208 PRASAD AND KHUDDUSH We claim that if � = 1 2 , then ω(t,�) ∈ L1∇[0, 1]. Note that ∑∞ k=1 1 k2 = π 2 6 , we have∫ 1 0 ω(t,�)∇t = ∫ 1 2 0 ω(t,�)∇t + ∫ 1 1 2 ω(t,�)∇t = ∫ 1 2 0 ∞∑ k=1 χ[νk, νk−1] |t− tk|� ∇t + ∫ 1 5 6 1 |t− 4 5 |� ∇t + [( 3 5 − 1 2 ) ω ( 3 5 ,� ) + ( 3 4 − 3 5 ) ω ( 3 4 ,� ) + ( 4 5 − 3 4 ) ω ( 4 5 ,� )( 5 6 − 4 5 ) ω ( 5 6 ,� )] = ∞∑ k=1 ∫ νk−1 νk 1 |t− tk|� ∇t + ∫ 1 5 6 1( t− 4 5 )�∇t + [ 1 10 × 10� + 3 20 × 4� + 1 20 × ( 10 3 )� + 1 30 × 3� ] = ∞∑ k=1 [∫ tk νk 1 (tk − t)� ∇t + ∫ νk−1 tk 1 (t− tk)� ∇t ] + 1 1 − � [ 1 51−� − 1 301−� ] + [ 1 10 × 10� + 3 20 × 4� + 1 20 × ( 10 3 )� + 1 30 × 3� ] = ∞∑ k=1 [∫ tk tk+tk+1 2 1 (tk − t)� ∇t + ∫ tk−1+tk 2 tk 1 (t− tk)� ∇t ] + 1 1 − � [ 1 51−� − 1 301−� ] + [ 10�−1 + 3 5 × 4�−1 + 1 20 × ( 10 3 )� + 1 10 × 3�−1 ] = 1 1 − � ∞∑ k=1 [( tk − tk+1 2 )1−� + ( tk−1 − tk 2 )1−� + 1 1 − � [ 1 51−� − 1 301−� ] + [ 10�−1 + 3 5 × 4�−1 + 1 20 × ( 10 3 )� + 1 10 × 3�−1 ] = 1 21−�(1 − �) ∞∑ k=1 [ 1 (k + 2)4(1−�) + 1 (k + 1)4(1−�) ] + 1 1 − � [ 1 51−� − 1 301−� ] + [ 10�−1 + 3 5 × 4�−1 + 1 20 × ( 10 3 )� + 1 10 × 3�−1 ] = √ 2 ∞∑ k=1 [ 1 (k + 1)2 + 1 (k + 1)2 ] + 1 15 (6 √ 5 − √ 30) + 1 60 [6( √ 10 + 3) + √ 3( √ 10 + 2)] = √ 2 ( π2 3 − 9 4 ) + 1 60 [24 √ 5 − 3 √ 30 + 6 √ 10 + 2 √ 3 + 18], which implies that ω(t,�) ∈ L1∇[0, 1]. Next, we claim that if � = 1 4 , then ω(t,�) ∈ L2∇[0, 1]. In this case, we need the cauchy product, ∞∑ k=1 ak · ∞∑ k=1 bk = ∞∑ k=1 ck, (5.1) where ck = k∑ n=1 anbk−n+1. (5.2) Note that ∫ 1 0 ω2(t,�)∇t = ∫ 1 2 0 [ ∞∑ k=1 χ[νk, νk−1] |t− tk|� ]2 ∇t + ∫ 1 1 2 ω2(t,�)∇t, (5.3) COUNTABLE MANY POSITIVE SOL. FOR EOBVPS WITH IBCS ON TIME SCALES 209 we use (5.1) and (5.2) and the fact that, if X ∩Y = ∅, then χ[X] ·χ[Y ] = 0 to simplify the integrand,[ ∞∑ k=1 χ[νk, νk−1] |t− tk|� ]2 = ∞∑ k=1 k∑ n=1 χ[νn, νk−1] |t− tn|� χ[νk−n+1, νk−n] |t− tk−n+1|� = ∞∑ k=1 χ[νk, νk−1] |t− tk|2� a.e., and so (5.3) may be written as∫ 1 0 ω2(t,�)∇t = ∞∑ k=1 ∫ 1 2 0 χ[νk, νk−1] |t− tk|2� ∇t + ∫ 1 1 2 ω2(t,�)∇t = ∞∑ k=1 ∫ νk−1 νk 1 |t− tk|2� ∇t + ∫ 1 5 6 ω2(t,�)∇t + + [( 3 5 − 1 2 ) ω2 ( 3 5 ,� ) + ( 3 4 − 3 5 ) ω2 ( 3 4 ,� ) + ( 4 5 − 3 4 ) ω2 ( 4 5 ,� )( 5 6 − 4 5 ) ω2 ( 5 6 ,� )] = ∞∑ k=1 [∫ tk νk 1 (tk − t)2� ∇t + ∫ νk−1 tk 1 (t− tk)2� ∇t ] + ∫ 1 5 6 1( t− 4 5 )2�∇t + [ 1 10 × 102� + 3 20 × 42� + 1 20 × ( 10 3 )2� + 1 30 × 32� ] = ∞∑ k=1 [∫ tk tk+tk+1 2 1 (tk − t)2� ∇t + ∫ tk−1+tk 2 tk 1 (t− tk)2� ∇t ] + 1 1 − 2� [ 1 51−2� − 1 301−2� ] + [ 102�−1 + 3 5 × 42�−1 + 1 20 × ( 10 3 )2� + 1 10 × 32�−1 ] = 1 1 − 2� ∞∑ k=1 [( tk − tk+1 2 )1−2� + ( tk−1 − tk 2 )1−2� + 1 1 − 2� [ 1 51−2� − 1 301−2� ] + [ 102�−1 + 3 5 × 42�−1 + 1 20 × ( 10 3 )2� + 1 10 × 32�−1 ] = 1 21−2�(1 − 2�) ∞∑ k=1 [ 1 (k + 2)4(1−2�) + 1 (k + 1)4(1−2�) ] + 1 1 − 2� [ 1 51−2� − 1 301−2� ] + [ 102�−1 + 3 5 × 42�−1 + 1 20 × ( 10 3 )2� + 1 10 × 32�−1 ] = √ 2 ∞∑ k=1 [ 1 (k + 1)2 + 1 (k + 1)2 ] + 1 15 (6 √ 5 − 5 √ 6) + 1 60 [6( √ 10 + 3) + √ 3( √ 10 + 2)] = √ 2 ( π2 3 − 9 4 ) + 1 60 [24 √ 5 − 20 √ 6 + 6( √ 10 + 3) + √ 3( √ 10 + 2)], which implies ω(t,�) ∈ L2∇[0, 1]. 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Ge, Existence of solutions of boundary value problems with integral boundary conditions for second order impulsive integro-differential equations in Banach spaces, Comput. Appl. Math., 233(2010), 1915- 1926. Department of Applied Mathematics, Andhra University, Visakhapatnam, 530 003, India ∗Corresponding author: rajendra92@rediffmail.com 1. Introduction 2. Preliminaries 3. Green's Function and Bounds 4. Main Results 5. Example References