Acta Appl Math (2010) 110: 353–365DOI 10.1007/s10440-008-9415-z

Existence of Three Positive Solutions of Three-Orderwith m-Point Impulsive Boundary Value Problems

Sihua Liang · Jihui Zhang

Received: 15 September 2008 / Accepted: 15 December 2008 / Published online: 20 December 2008© Springer Science+Business Media B.V. 2008

Abstract This paper deals with the existence of three positive solutions for the followingimpulsive boundary value problem

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(ϕ(−u′′)(t))′ + a(t)f (u(t)) = 0, t �= tk, 0 < t < 1,

�u|t=tk = Ik(u(tk)), k = 1,2, . . . ,N,

u(0) = ∑m−2i=1 αiu(ξi),

u′(1) = 0, u′′(0) = 0,

where ϕ : R → R is the increasing homeomorphism (see Definition 1.1) and positive homo-morphism and ϕ(0) = 0. Using the five functionals fixed point theorem, we provide suffi-cient conditions for the existence of three positive solutions for the above boundary valueproblems.

Keywords Impulsive boundary value problems · The five functionals fixed-point theorem ·Positive solutions · Cone

Mathematics Subject Classification (2000) 34B18 · 34B40

S. Liang (�) · J. ZhangInstitute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal University,210097 Jiangsu, People’s Republic of Chinae-mail: liangsihua@163.com

J. Zhange-mail: jihuiz@jlonline.com

S. LiangCollege of Mathematics, Changchun Normal University, Changchun 130032, Jilin,People’s Republic of China

354 S. Liang, J. Zhang

1 Introduction

Mathematics is successfully applied in numerous technical problems, biology, economics,control theory, etc. In general, taking into account different properties and effects intrinsic ofa considered system, it gives us the possibility to study processes and phenomena by meansof mathematical simulation more exactly. For instance, many areas of industry deal withtechnical systems being subjected to shocks, impacts or impulses. Thus, in cases when anexternal perturbation for the system is of a short-term (shock, impact or impulse) nature andits duration could be disregarded while formulating the corresponding mathematical modelwe have to study dynamical system with discontinuous trajectories. A classical example ofthat problem is the model of a clock mechanism. Impulsive differential equations describeprocesses which experience a sudden change of their state at certain moments. The theoryof impulsive differential equations has become important in recent years in mathematicalmodels of real processes rising in phenomena studied in physics, chemical technology, pop-ulation dynamics, biotechnology and economics; see [2]. There has been a significant de-velopment in impulsive theory especially in the area of impulsive differential equations withfixed moments [4].

In this paper, we introduce a new operator, which improves and generates a p-Laplaceoperator for some p > 1, and we study the existence of three positive solutions for thefollowing three-order with m-point nonlinear impulsive boundary value problem of the form

{(ϕ(−u′′)(t))′ + a(t)f (u(t)) = 0, t �= tk, 0 < t < 1,

�u|t=tk = Ik(u(tk)), k = 1,2, . . . ,N,(1)

with the following boundary value condition:

{u(0) = ∑m−2

i=1 αiu(ξi),

u′(1) = 0, u′′(0) = 0,(2)

where ϕ : R → R is an increasing homeomorphism (see Definition 1.1) and positivehomomorphism and ϕ(0) = 0. Here ξi ∈ [0,1] with 0 ≤ ξ1 < ξ2 < · · · < ξm ≤ 1, f ∈C([0,+∞), [0,+∞)), tk (k = 1,2, . . . ,N , where N is a fixed positive integer) arefixed points with 0 < t1 < t2 < · · · < tk < · · · < tN < 1 and ξi �= tk , i = 1,2, . . . ,m, k =1,2, . . . ,N , �u|t=tk denotes the jump of u(t) at t = tk , i.e.,

�u|t=tk = u(t+k ) − u(t−k ),

where u(t+k ) and u(t−k ) represent the right-hand limit and left-hand limit of u(t) at t = tk ,respectively.

Definition 1.1 A projection ϕ : R → R is called an increasing homeomorphism and positivehomomorphism if the following conditions are satisfied:

(1) If x ≤ y, then ϕ(x) ≤ ϕ(y), for all x, y ∈ R;(2) ϕ is a continuous bijection and its inverse is also continuous;(3) ϕ(xy) = ϕ(x)ϕ(y), for all x, y ∈ R+ = [0,+∞).

In the above definition, condition (3) can be replaced by the following stronger condition:

(4) ϕ(xy) = ϕ(x)ϕ(y), for all x, y ∈ R, where R = (−∞,+∞).

Existence of Three Positive Solutions of Three-Order with m-Point 355

Remark 1.1 If conditions (1), (2) and (4) hold, then ϕ is homogeneous and generates ap-Laplace operator, i.e., ϕ(x) = |x|p−2x, for some p > 1.

Recently, the existence and multiplicity of positive solutions for linear and nonlinearsecond-order impulsive differential equations have been studied extensively. To identify afew, we refer the reader to see [1, 5, 7, 8, 12]. For the special case, the existence and mul-tiplicity of positive solutions for the p-Laplacian operator, i.e., Ik = 0, k = 1,2, . . . ,N andϕ(x) = |x|p−2x, for some p > 1, have received wide attention, see [3, 6, 9, 14, 15] and ref-erences therein. We know that the oddness of a p-Laplacian operator key to the proof. How-ever in this paper we define a new operator which improves and generates a p-Laplacianoperator for some p > 1 and ϕ is not necessarity odd. Moreover for increasing homeomor-phism and positive homomorphism operator and research has proceeded very slowly, see[10, 11]. Especially the existence of three positive solutions for three-order with m-pointimpulsive boundary value problems still remains unknown.

In [11], Liu and Zhang studied the existence of positive solutions of quasi-linear differ-ential equation

(ϕ(x ′))′ + a(t)f (x(t)) = 0, t ∈ (0,1),

x(0) − βx ′(0) = 0, x(1) + δx ′(1) = 0,

subject to linear mixed boundary value conditions by a simple application of a fixed pointindex theorem in cones, where ϕ : R → R is an increasing homeomorphism and positivehomomorphism and ϕ(0) = 0.

Recently, Avery and Henderson [3] studied the three point boundary problem

(g(u′))′ + a(t)f (u) = 0, 0 < t < 1,

u(0) = 0, u(v) = u(1),

where v ∈ (0,1) is a constant, the authors proved that the existence of the at least threepositive pseudo-symmetric solutions by applying the five functionals fixed point theorem.

In [5], the authors studied the following boundary value problems with one dimensionalp-Laplacian with impulse effects

⎧⎪⎨

⎪⎩

(ϕ(x ′))′ + a(t)f (t, x(t)) = 0, 0 < t < 1,

−�x|t=tk = Ik(x(tk)), k = 1,2, . . . ,N,

x ′(0) = 0, x(1) = ∫ 10 g(t)x(t)dt,

where ϕ(x) = |x|p−2x, for some p > 1 is a p-Laplacian operator. They proved that theexistence of one or two positive solutions by using the classical fixed-point index theoremfor compact map.

In [13], Sang and Su studied the existence of positive solutions of the following third-order differential equations.

(φ(u′′))′(t) = a(t)f (t, u), 0 < t < 1,

φ(u′′(0)) = ∑m−2i=1 aiφ(u′′(ξi)) u′(0) = 0, u(1) = ∑m−2

i=1 biu(ξi),

where φ : R → R is an increasing homeomorphism and homomorphism and φ(0) = 0, f

may change sign.

356 S. Liang, J. Zhang

Remark 1.2 It is well known that a p-Laplacian operator is odd. However, the operatorwhich we defined above is not necessary odd, see Example 5.1. We emphasize that theresults of the paper [3, 5, 13, 15] are not replaced by ϕ which we defined above.

Seeing such a fact, we cannot but ask “whether or not we can obtain three positive solu-tions of the three-order with m-point impulsive boundary value problem (1) and (2)?”

Motivated by papers [5, 11, 13], we prove in this work the existence of three positivesolutions to (1) and (2) which have specific scope. The arguments developed in this paperare five functionals fixed-point theorem [3] and our main result extends the study madein [3], So the goal of the present paper is to improve and generate p-Laplacian operatorand establish some criteria for the existence of three positive solutions by means of fivefunctionals fixed-point theorem.

We shall assume that the following conditions are satisfied.

(C1) f ∈ C([0,+∞), [0,+∞)), and αi satisfy 0 <∑m−2

i=1 αi < 1;(C2) Ik ∈ C([0,+∞), [0,+∞)), k = 1,2, . . . ,N ;(C3) a(t) is a nonnegative measurable function defined in (0,1) and a(t) does not identi-

cally vanish on any subinterval of (0,1) and

0 <

∫ 1

0a(t)dt < +∞.

The plan of the paper is as follows. In Sect. 2 for the convenience of the reader we givesome background and definitions. In Sect. 3 we present some lemmas in order to proveour main results. Section 4 is developed to presenting and proving our main results. Someexamples are presented in Sect. 5 to demonstrate the application of our main results.

2 Preliminaries

In this section, we provide background definitions from the cone theory in Banach spaces.

Definition 2.1 Let (E, ‖.‖) be a real Banach space. A nonempty, closed, convex set P ⊂ E

is said to be a cone provided the following are satisfied:

(a) If y ∈ P and λ ≥ 0, then λy ∈ P ;(b) If y ∈ P and −y ∈ P , then y = 0.

If P ⊂ E is a cone, we denote the order induced by P on E by ≤, that is, x ≤ y if andonly if y − x ∈ P .

Definition 2.2 A map α is said to be a nonnegative, continuous, concave functional on acone P of a real Banach space E, if

α : P → [0,∞)

is continuous, and

α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y)

for all x, y ∈ P and t ∈ [0,1].

Existence of Three Positive Solutions of Three-Order with m-Point 357

Definition 2.3 An operator is called completely continuous if it is continuous and mapsbounded sets into precompact sets.

Let γ , β , θ be nonnegative continuous convex functions on P , α, ψ be a nonnegativecontinuous concave functional on P . Then for nonnegative positive numbers h,a, b, c andd , we define the following convex sets:

P (γ, c) = {x ∈ P : γ (x) < c},P (γ,α, a, c) = {x ∈ P : a ≤ α(x), γ (x) ≤ c},Q(γ,β, d, c) = {x ∈ P : β(x) ≤ d, γ (x) ≤ c},P (γ, θ,α, a, b, c) = {x ∈ P : a ≤ α(x), θ(x) ≤ b, γ (x) ≤ c} and

Q(γ,β,ψ,h, d, c) = {x ∈ P : h ≤ ψ(x), β(x) ≤ d, γ (x) ≤ c}.We introduce the five functionals fixed-point theorem due to Avery and Henderson [3].

Theorem 2.1 Let P be a cone in a real Banach space E. Suppose there exist positive num-bers c and M , nonnegative, continuous, concave functionals α, ψ on P , and nonnegative,continuous, convex functionals γ , β and θ on P satisfying:

α(x) ≤ β(x), ‖x‖ ≤ Mγ(x) (3)

for all x ∈ P (γ, c). Suppose

T : P (γ, c) �→ P (γ, c) (4)

is completely continuous and there exist nonnegative numbers h,a, k, b with 0 < a < b suchthat

(S1) {x ∈ P (γ, θ,α, b, k, c) | α(x) > b} �= ∅ and α(T x) > b for x ∈ P (γ, θ,α, b, k, c);(S2) {x ∈ Q(γ,β,ψ,h, a, c) | β(x) < a} �= ∅ and β(T x) < a for x ∈ Q(γ,β,ψ,h, a, c);(S3) α(T x) > b for x ∈ P (γ,α, b, c) with θ(T x) > k;(S4) β(T x) < a for x ∈ P (γ,β, a, c) with ψ(T x) < h.

Then T has at least three fixed points x1, x2, x3 ∈ P (γ, c) such that

β(x1) < a, b < α(x2), and a < β(x3) with α(x3) < b.

3 Some Lemmas

Let J ′ = [0,1] \ {t1, t2, . . . , tN }. The basic space used in this paper is E = {u | u : [0,1] →R is continuous at t �= tk, u(t) left continuous at t = tk, and u(t+k ) exists, k = 1,2, . . . ,N}.Thus E is a Banach space when endowed with the norm ‖u‖ = supt∈[0,1] |u(t)|. By a solutionof (1) and (2), we mean a function u ∈ E ∩ C2(J ′) which satisfies (1) and (2). Define coneK ⊂ E by

K = {u ∈ E : u(t) is a nondecreasing and nonnegative concave function on [0,1]} .

Obviously K is a cone in E.We can easily get the following lemmas which are useful in the proof of our main results.

358 S. Liang, J. Zhang

Lemma 3.1 For any∑m−2

i=1 αi �= 1, then for p ∈ C[0,1], the problem

(ϕ(−u′′(t)))′ = p(t), 0 < t < 1, (5)

u(0) =m−2∑

i=1

αiu(ξi), u′(1) = 0, u′′(0) = 0 (6)

has a unique solution

u(t) = t

∫ 1

t

ϕ−1

(∫ s

0p(τ)dτ

)

ds +∫ t

0sϕ−1

(∫ s

0p(τ)dτ

)

ds

+∑

tk<t

Ik(u(tk)) +∑m−2

i=1 αi

1 − ∑m−2i=1 ai

∑

tk<ξi

Ik(u(tk))

+∑m−2

i=1 αiξi

∫ 1ξi

ϕ−1(∫ s

0 p(τ)dτ)ds + ∑m−2i=1 αi

∫ ξi0 sϕ−1(

∫ s

0 p(τ)dτ)ds

1 − ∑m−2i=1 αi

. (7)

Proof Necessity. By taking the integral of (5) on [0, t], we have

ϕ(−u′′(t)) − ϕ(−u′′(0)) =∫ t

0p(τ)dτ. (8)

By the boundary value condition and together with ϕ(0) = 0, we have

u′′(t) = −ϕ−1

(∫ t

0p(τ)dτ

)

. (9)

By taking the integral of (9) on [t,1], we can get

u′(1) − u′(t) = −∫ 1

t

ϕ−1

(∫ s

0p(τ)dτ

)

ds. (10)

By the boundary value condition u′(1) = 0, we can get

u′(t) =∫ 1

t

ϕ−1

(∫ s

0p(τ)dτ

)

ds. (11)

If t1 < t ≤ t2, then integrate (11) from 0 to t1, we have

u(t1) − u(0) =∫ t1

0

∫ 1

s

ϕ−1

(∫ τ

0p(η)dη

)

dτds. (12)

Integrate (11) from t1 to t , we have

u(t) − u(t+1 ) =∫ t

t1

∫ 1

s

ϕ−1

(∫ τ

0p(η)dη

)

dτds. (13)

So, (12) and (13) imply

u(t) − u(0) =∫ t

0

∫ 1

s

ϕ−1

(∫ τ

0p(η)dη

)

dτds + I1(u(t1)), t1 < t < t2. (14)

Existence of Three Positive Solutions of Three-Order with m-Point 359

Repeating the above procession, for t ∈ [0,1], we have

u(t) = u(0) +∫ t

0

∫ 1

s

ϕ−1

(∫ τ

0p(η)dη

)

dτds +∑

tk<t

Ik(u(tk)). (15)

Substituting u(0) = ∑m−2i=1 αiu(ξi) and integration by parts, we obtain

u(t) = t

∫ 1

t

ϕ−1

(∫ s

0p(τ)dτ

)

ds +∫ t

0sϕ−1

(∫ s

0p(τ)dτ

)

ds

+m−2∑

i=1

αiu(ξi) +∑

tk<t

Ik(u(tk)). (16)

Letting t = ξi on (16), it follows that

u(ξi) = ξi

∫ 1

ξi

ϕ−1

(∫ s

0p(τ)dτ

)

ds +∫ ξi

0sϕ−1

(∫ s

0p(τ)dτ

)

ds

+m−2∑

i=1

αiu(ξi) +∑

tk<ξi

Ik(u(tk))

and

m−2∑

i=1

αiu(ξi) =m−2∑

i=1

αiξi

∫ 1

ξi

ϕ−1

(∫ s

0p(τ)dτ

)

ds +m−2∑

i=1

αi

∫ ξi

0sϕ−1

(∫ s

0p(τ)dτ

)

ds

+m−2∑

i=1

αi

m−2∑

i=1

αiu(ξi) +m−2∑

i=1

αi

∑

tk<ξi

Ik(u(tk)).

Therefore

m−2∑

i=1

αiu(ξi) =∑m−2

i=1 αiξi

∫ 1ξi

ϕ−1(∫ s

0 p(τ)dτ)ds + ∑m−2

i=1 αi

∫ ξi0 sϕ−1

(∫ s

0 p(τ)dτ)ds

1 − ∑m−2i=1 αi

+∑m−2

i=1 αi

1 − ∑m−2i=1 ai

∑

tk<ξi

Ik(u(tk)).

By (15), we have

u(t) = t

∫ 1

t

ϕ−1

(∫ s

0p(τ)dτ

)

ds +∫ t

0sϕ−1

(∫ s

0p(τ)dτ

)

ds

+∑

tk<t

Ik(u(tk)) +∑m−2

i=1 αi

1 − ∑m−2i=1 ai

∑

tk<ξi

Ik(u(tk))

+∑m−2

i=1 αiξi

∫ 1ξi

ϕ−1(∫ s

0 p(τ)dτ)ds + ∑m−2

i=1 αi

∫ ξi0 sϕ−1

(∫ s

0 p(τ)dτ)ds

1 − ∑m−2i=1 αi

.

360 S. Liang, J. Zhang

Sufficiency: Let u be as in (7). Taking the derivative of (7), it implies that

u′(t) =∫ 1

t

ϕ−1

(∫ s

0p(τ)dτ

)

ds.

Furthermore, we get

u′′(t) = −ϕ−1

(∫ t

0p(τ)dτ

)

and

ϕ(−u′′(t)) =∫ t

0p(τ)dτ,

taking the derivative of this expression yields (ϕ(−u′′(t)))′ = p(t). Routine calculation ver-ifies that u satisfies the boundary value conditions (6), so that u given in (5) is a solutionof (5) and (6).

It is easy to see that problem (ϕ(−u′′(t)))′ = 0, u(0) = ∑m−2i=1 αiu(ξi), u′(∞) = 0,

u′′(0) = 0, has only the trivial solution if∑m−2

i=1 αi �= 1. Thus u in (5) is the unique solu-tion of (1) and (2). The proof is complete. �

Lemma 3.2 For any p(t) ∈ C[0,1] and p(t) ≥ 0, then the unique solution u(t) of (1)and (2) satisfies

u(t) ≥ 0, for t ∈ [0,1].

Lemma 3.3 Let u ∈ K , then u(t) ≥ t‖u‖, for t ∈ J ′.

Proof Since u(t) is concave and nondecreasing, together with u′(1) = 0, then ‖u‖ = u(1)

and

u(t) − u(0)

t − 0≥ u(1) − u(0)

1 − 0, t ∈ J ′.

Thus, u(t) ≥ tu(1) + (1 − t)u(0) ≥ tu(1) = t‖u‖ for t ∈ J ′. �

Now, fix l ∈ J ′ such that

0 < ξ1 < l < 1

and∫ 1

l

a(τ )dτ.

We can easy to get the following lemma from the concavity of u(t) on [0,1].

Lemma 3.4 Let u ∈ K , then ξ1u(l) ≤ lu(ξ1).

Define T : K → E by

(T u)(t) = t

∫ 1

t

ϕ−1

(∫ s

0a(τ)f (u(τ))dτ

)

ds +∫ t

0sϕ−1

(∫ s

0a(τ)f (u(τ))dτ

)

ds

Existence of Three Positive Solutions of Three-Order with m-Point 361

+∑

tk<t

Ik(u(tk)) +∑m−2

i=1 αi

1 − ∑m−2i=1 ai

∑

tk<ξi

Ik(u(tk))

+∑m−2

i=1 αiξi

∫ 1ξi

ϕ−1(∫ s

0 a(τ)f (u(τ))dτ)ds

1 − ∑m−2i=1 αi

+∑m−2

i=1 αi

∫ ξi0 sϕ−1(

∫ s

0 a(τ)f (u(τ))dτ)ds

1 − ∑m−2i=1 αi

. (17)

Obviously, From Lemma 3.2, we have (T u)(t) ≥ 0, for t ∈ (0,1) and we also have(T u)′(t)=∫ +∞t

ϕ−1(∫ s

0 a(τ)f (u(τ))dτ)ds ≥ 0, furthermore (T u)′′(t) = −ϕ−1(∫ t

0 a(s)f (u(s))ds ≤ 0.This shows (T K) ⊂ K .

From (13) and the Arzela-Ascoli Theorem, it is easy to obtain the following result.

Lemma 3.5 Let conditions (C1)–(C3) hold. Then T : K → K is completely continuous.

We define the nonnegative, continuous, concave functionals α, ψ and nonnegative, con-tinuous, convex functionals β , θ , γ on the cone K by

γ (u) = θ(u) = max0≤t≤ξ1

u(t) = u(ξ1), α(u) = minl≤t≤1

u(t) = u(l),

β(u) = max0≤t≤l

u(t) = u(l), ψ(u) = minξ1≤t≤1

u(t) = u(ξ1).

It is clear that α(u) = β(u) for all u ∈ K . By Lemma 3.3 we have γ (u) = u(ξ1) ≥ ξ1‖u‖.Hence ‖u‖ ≤ 1

ξ1γ (u) for all u ∈ K .

For notational convenience we denote by

M ={

1

1 − ∑m−2i=1 αi

∫ 1

0ϕ−1

(∫ s

0a(τ))dτ

)

ds + N

1 − ∑m−2i=1 αi

}

,

m = l

∫ 1

l

ϕ−1

(∫ s

l

a(τ )dτ

)

ds.

4 Main Results

The main results of this paper are the following.

Theorem 4.1 Suppose that conditions (C1)–(C3) hold. Let 0 <ξ1la < a < lb < lξ1c, Mb <

mc and suppose that f and Ik satisfy the following conditions:

(C4) f (v) < ϕ( cM

), for all 0 ≤ v ≤ cξ1

;

(C5) f (v) > ϕ( bm), for all b ≤ v ≤ b

ξ21

;

(C6) f (v) < ϕ( aM

), for all 0 ≤ v ≤ cξ1

;(C7) sup{Ik(u(tk)), k = 1,2, . . . ,N} ≤ a

M.

Then the boundary value problem (1) and (2) has at least three positive solutions u1, u2

and u3 such that

β(u1) < a, b < α(u2), and a < β(u3) with α(u3) < b.

362 S. Liang, J. Zhang

Proof We define the completely continuous operator T by (17). Let u ∈ P (γ, c), thenγ (u) = max0≤t≤ξ1 u(t) = u(ξ1) ≤ c. This implies that 0 ≤ u(t) ≤ c for t ∈ [0, ξ1]. If werecall that ‖u‖ ≤ 1

ξ1γ (u) ≤ c

ξ1, thus, we have

0 ≤ u(t) ≤ c

ξ1, t ∈ [0,1].

Then assumption (C4) implies

f (u(t)) < ϕ

(c

M

)

, 0 ≤ t ≤ 1.

Therefore, (C4) and (C7) imply

γ (T u) = max0≤t≤ξ1

(T u)(t) = (T u)(ξ1)

= ξ1

∫ 1

ξ1

ϕ−1

(∫ s

0a(τ)f (u(τ))dτ

)

ds +∫ ξ1

0sϕ−1

(∫ s

0a(τ)f (u(τ))dτ

)

ds

+∑

tk<ξ1

Ik(u(tk)) +∑m−2

i=1 αi

1 − ∑m−2i=1 ai

∑

tk<ξi

Ik(u(tk))

+∑m−2

i=1 αiξi

∫ 1ξi

ϕ−1(∫ s

0 a(τ)f (u(τ))dτ)ds

1 − ∑m−2i=1 αi

+∑m−2

i=1 αi

∫ ξi0 sϕ−1

(∫ s

0 a(τ)f (u(τ))dτ)ds

1 − ∑m−2i=1 αi

<c

M

∫ 1

0ϕ−1

(∫ s

0a(τ)dτ

)

ds + Na

M+

∑m−2i=1 αi

1 − ∑m−2i=1 ai

Na

M

+∑m−2

i=1 αi

∫ 10 ϕ−1

(∫ s

0 a(τ)dτ)ds

1 − ∑m−2i=1 αi

c

M

= c

M

{1

1 − ∑m−2i=1 αi

∫ 1

0ϕ−1

(∫ s

0a(τ))dτ

)

ds + N

1 − ∑m−2i=1 αi

}

= c.

Therefore T u ∈ P (γ, c).Now we show that the conditions (S1)–(S4) of Theorem 2.1 are satisfied.Firstly let u(t) ≡ b

ξ1and k = b

ξ1, it follows that

α(u) = u(l) = b

ξ1> b, θ(u) = u(ξ1) = b

ξ1= k,

γ (u) = u(ξ1) = b

ξ1< c,

which shows that {u ∈ P (γ, θ,α, b, k, c)|α(u) > b} �= ∅ and for u ∈ P (γ, θ,α, b, bξ1

, c), wehave α(u) = minl≤t≤1 u(t) = u(l), this means u(t) > b for t ∈ [l,1]. Observing that ‖u‖ ≤

Existence of Three Positive Solutions of Three-Order with m-Point 363

1ξ1

γ (u) = 1ξ1

θ(u) ≤ b

ξ21

. So we have

b ≤ u(t) ≤ b

ξ 21

, t ∈ [l,1].

Then assumption (C5) implies

f (u(t)) > ϕ

(b

m

)

, l ≤ t ≤ 1.

Therefore

α(T u) = maxl≤t≤1

(T u)(t) = (T u)(l)

≥ l

∫ 1

l

ϕ−1

(∫ s

0a(τ)f (u(τ))dτ

)

ds +∫ l

0sϕ−1

(∫ s

0a(τ)f (u(τ))dτ

)

ds

>b

ml

∫ 1

l

ϕ−1

(∫ s

l

a(τ )dτ

)

ds = b.

So the condition (S1) of Theorem 2.1 is satisfied.Secondly we show that (S2) of Theorem 2.1 is fulfilled. We take u(t) ≡ ξ1

la, h = ξ1

la,

then

γ (u) ≡ ξ1

la < c, ψ(u) = ξ1

la = h, β(u) = ξ1

la < a.

From this we know that {u ∈ Q(γ,β,ψ,h, a, c)|β(u) < a} �= ∅. If we take u ∈ Q(γ,β,ψ,ξ1la, a, c), together with the definition of β , γ and Lemma 3.3, we have ‖u‖ ≤ a

l. Thus, we

have

0 ≤ u(t) ≤ a

l, t ∈ [0,1].

Then assumption (C6) implies

f (u(t)) < ϕ

(a

M

)

, 0 ≤ t ≤ 1.

Therefore, by (C6) and (C7), we have

β(T u) = max0≤t≤l

(T u)(t) = (T u)(l)

= l

∫ 1

l

ϕ−1

(∫ s

0a(τ)f (u(τ))dτ

)

ds +∫ l

0sϕ−1

(∫ s

0a(τ)f (u(τ))dτ

)

ds

+∑

tk<l

Ik(u(tk)) +∑m−2

i=1 αi

1 − ∑m−2i=1 ai

∑

tk<ξi

Ik(u(tk))

+∑m−2

i=1 αiξi

∫ 1ξi

ϕ−1(∫ s

0 a(τ)f (u(τ))dτ)ds

1 − ∑m−2i=1 αi

+∑m−2

i=1 αi

∫ ξi0 sϕ−1

(∫ s

0 a(τ)f (u(τ))dτ)ds

1 − ∑m−2i=1 αi

364 S. Liang, J. Zhang

<a

M

∫ 1

0ϕ−1

(∫ s

0a(τ)dτ

)

ds + Na

M+

∑m−2i=1 αi

1 − ∑m−2i=1 ai

Na

M

+∑m−2

i=1 αi

∫ 10 ϕ−1

(∫ s

0 a(τ)dτ)ds

1 − ∑m−2i=1 αi

a

M

= a

M

{1

1 − ∑m−2i=1 αi

∫ 1

0ϕ−1

(∫ s

0a(τ))dτ

)

ds + N

1 − ∑m−2i=1 αi

}

= a.

Thirdly we show that (S3) of Theorem 2.1 is satisfied. If u ∈ P (γ,α, b, c) and θ(T (u)) =(T u)(ξ1) > b

ξ1, then

α(T u) = (T u)(l) ≥ (T u)(ξ1) > ξ1(T u)(ξ1) > b.

Finally if u ∈ Q(γ,β, a, c) and ψ(T u) = (T u)(ξ1) <ξ1la. From Lemma 3.4 we have

β(T u) = (T u)(l) ≤ l

ξ1(T u)(ξ1) <

l

ξ1× ξ1

la = a.

Which shows that the condition (S4) of Theorem 2.1 is satisfied.Thus all the conditions in Theorem 2.1 are satisfied. So the boundary value problem (1)

and (2) has at least three solutions u1, u2, u3 and

β(u1) < a, b < α(u2), and a < β(u3) with α(u3) < b. �

5 Examples and Remark

Example 5.1 As an example we mention the boundary value problem⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(ϕ(−u′′)(t))′ + a(t)f (t, u(t)) = 0, t �= tk, 0 < t < 1,

�u|t=tk = Ik(u(tk)), k = 1,2, . . . ,N,

u(0) = ∑m−2i=1 αiu(ξi),

u′(1) = 0, u′′(0) = 0,

(18)

where

ϕ(u) =⎧⎨

⎩

u5

1+u3 , u ≤ 0,

u6 u > 0.

Here a(t) ∈ ((0, 1), [0,∞)) and f satisfies the conditions of Theorem 4.1. It is clear thatϕ : R → R is an increasing homeomorphism and positive homomorphism and ϕ(0) = 0.

Remark 5.1 Because p-Laplacian operators are odd, they do not apply to our example.Hence we generalize boundary value problems p-Laplacian operators and the results [1, 5,9, 13–15] do not apply to Example 5.1.

Acknowledgements The authors of this paper wish to thank the referee for the valuable suggestions re-garding the original manuscript. This work was supported by NSFC (10871096), Foundation of Major Projectof Science and Technology of Chinese Education Ministry, SRFDP of Higher Education, the project of Nan-jing Normal University (181200000214) and Project of Graduate Education Innovation of Jiangsu Province.

Existence of Three Positive Solutions of Three-Order with m-Point 365

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