Positive solutions for a higher-order four-point boundary value problem with a p -Laplacian

Computers and Mathematics with Applications 58 (2009) 1103–1112

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Positive solutions for a higher-order four-point boundary value problem with a p-LaplacianI Junfang Zhao ∗ , Weigao Ge Department of Mathematics, Beijing Institute of Technology, Beijing 100081, PR China

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Article history: Received 10 September 2007 Received in revised form 18 January 2009 Accepted 8 April 2009 Keywords: Multi-point boundary value problem Higher order Positive solution Fixed point theorem Cone

abstract In this paper, we study the existence of positive solutions of the following nth-order fourpoint boundary value problem with one dimensional p-Laplacian

 (φp (u(n−1) ))0 + h(t )f (t , u(t ), u0 (t ), . . . , u(n−2) (t )) = 0,    (i) u (0) = 0, (n−1)  (0) − α u(n−2) (ξ ) = 0, u  (n−1) u (1) + β u(n−2) (η) = 0,

0 < t < 1, n ≥ 3, 0 ≤ i ≤ n − 3, n ≥ 3, n ≥ 3,

1 , 0 < ξ < η < 1. By using a fixed where φp (s) = |s|p−2 s, p > 1, 0 < α ≤ ξ1 , 0 < β ≤ 1−η point theorem in cones, the existence of a positive solution and multiple positive solutions is obtained. Also, at the end of this paper, we present a generalization of this problem. © 2009 Elsevier Ltd. All rights reserved.

1. Introduction In this paper, we study the existence of positive solutions for the following nth-order four-point boundary value problem with one dimensional p-Laplacian

 (φp (u(n−1) ))0 + h(t )f (t , u(t ), u0 (t ), . . . , u(n−2) (t )) = 0,     (i) u (0) = 0, BVP BC u(n−1) (0) − α u(n−2) (ξ ) = 0,     (n−1) u (1) + β u(n−2) (η) = 0,

0 0 n n

< t < 1, n ≥ 3, ≤ i ≤ n − 3, ≥ 3, ≥ 3,

(1.1)

1 where φp (s) = |s|p−2 s, p > 1, 0 < α ≤ ξ1 , 0 < β ≤ 1−η , 0 < ξ < η < 1. Throughout this paper, we assume that

(H1 ) h ∈ L([0, 1], [0, +∞)), h(t ) is not identically zero on any subinterval of (0, 1). (H2 ) f ∈ C ([0, 1] × [0, +∞)n−1 , [0, +∞)), f 6≡ 0. The existence of positive solutions for higher-order boundary value problems (BVPs for short) has been studied by many authors for its practical significance. In fact, particular cases of the continuous BVP occur in various physical phenomena [1– 13], especially thermal self-ignition of a chemically active mixture of gases in a vessel, catalysis theory, chemically reacting systems, as well as adiabatic tubular reactor processes. And the deformations of an elastic beam in equilibrium state, whose

I Supported by NNSF of China (10671012) and SRFDP of China (20050007011).

∗

Corresponding author. E-mail addresses: [email protected] (J. Zhao), [email protected] (W. Ge).

0898-1221/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2009.04.022

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J. Zhao, W. Ge / Computers and Mathematics with Applications 58 (2009) 1103–1112

two ends are simply supported, can be described by the a fourth-order ordinary differential equation with some boundary conditions. Motivated by some of such investigations, in [14], Agarwal and Wong studied the following higher-order BVP:

 (n) u + f (t , u(t ), u0 (t ), . . . , u(n−2) (t )) = 0,    (i) u (0) = 0, α u(n−2) (0) − β u(n−1) (0) = 0,    (n−2) γu (1) + δ u(n−1) (1) = 0,

0 0 n n

< t < 1, n ≥ 3 ≤ i ≤ n − 3, ≥ 3, ≥ 3,

(1.2)

where α, β, γ , δ > 0. They established many theorems that guarantee the existence of multiple positive solutions. For other related works, we refer the readers to [15–21]. Multi-point boundary value problems arise in a variety of different areas of applied mathematics and physics. For example, the vibrations of a guy wire of a uniform cross section and composed of N parts of different densities can be set up as a multi-point boundary value problem (see [22]). Many problems in the theory of elastic stability can be handled by the method of multi-point problems (see [23]). Small size bridges are often designed with two supported points, which leads to a standard two-point boundary value condition. And large size bridges are sometimes contrived with multi-point supports, which corresponds to a multi-point boundary value condition [24]. The study of multi-point BVPs for linear second-order ordinary differential equations was initiated by Il’in and Moiseev [25]. Since then, multi-point BVPs have received a great deal of attention. We refer the readers to [26–28] and the references therein. Equations of the p-Laplacian form occur in the study of non-Newtonian fluid theory and the turbulent flow of a gas in a porous medium. Since 1980s, many authors have studied such kinds of boundary value problems. Recently, in [29], the authors considered

 (φp (u(n−1) ))0 + g (t )f (t , u(t ), u0 (t ), . . . , u(n−2) (t )) = 0,     (i) u (0) = 0, BVP BC α u(n−2) (0) − β u(n−1) (ξ ) = 0,     (n−2) γu (1) + δ u(n−1) (η) = 0,

0 < t < 1, 0 ≤ i ≤ n − 3, n ≥ 3, n ≥ 3,

(1.3)

where φp (s) = |s|p−2 s, p > 1, 0 < ξ < η < 1, α > 0, β ≥ 0, γ > 0, δ ≥ 0. Unfortunately, there is something wrong with Lemma 2.3 in this paper, so the result needs to be reconsidered. When n = 2, in [30], the authors corrected the result in [29] to some extend (the counterpart of Lemma 2.3 was corrected). However, to the best knowledge of the authors, there is no paper concerned with the higher-order multi-point BVPs like (1.1) up until now. So, motivated by the works mentioned above, in this paper, we introduce sufficient conditions on the nonlinearity f which allow us to make use of the Guo–Krasnosel’skii fixed point theorem, and in turn to get the existence of at least one or two positive solutions for the BVP (1.1). We point out that when ξ converges to 0, and η converges to 1, the BC in (1.1) becomes the BC in (1.2). So our result can be seen as a generalization of the BVP (1.2). Moreover, our results cannot be deduced trivially from any of the earlier published results. And at the end of this paper, we give a new generalization of the BVP (1.1), which is more generous. In contrast to [14] (1.2), we claim that when α > 0, β > 0, the solution of BVP (1.1) is not always positive. In fact, when p = 2, consider the following third-order four-point BVP:

u000 + 2 = 0,   u(0) = 0,      1 = 0, u00 (0) − 8u0   4    u00 (1) + 4u0 1 = 0.

0 < t < 1, (1.4)

2

3 3 After a series of calculations, we obtain that u0 (t ) = − 13 t 3 + 11 t 2 − 40 t is a solution of BVP (1.4). Obviously, u( 40 ) < 0. So 20 in this paper, we need to give some restrictions on α and β , so that the solutions of BVP (1.1) are nonnegative. This is done in Section 2.

2. Preliminaries In this section, for convenience, we give some background material, which is important in the proof of our main results. Definition 2.1. Let E be a real Banach Space. A nonempty closed convex set P ⊂ E is called a cone if it satisfies the following two conditions: (i) au ∈ P for all a ≥ 0, (ii) u, −u ∈ P implies u = 0. Every cone P ⊂ E induces an ordering in E given by x ≥ y, if and only if y − x ∈ P .

J. Zhao, W. Ge / Computers and Mathematics with Applications 58 (2009) 1103–1112

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Definition 2.2. A map α is said to be a nonnegative concave continuous functional on a cone P ∈ X provided that α : P → [0, ∞) is continuous and

α(λx + (1 − λ)y) > λα(x) + (1 − λ)α(y) for all x, y ∈ P and 0 6 λ 6 1. Let E = C n−2 [0, 1] be endowed with the maximum norm kuk = max0≤i≤n−2 {kuki }, where kuki = maxt ∈[0,1] |u(i) (t )|. In what follows, we denote by φq the inverse to φp , where q > 1 such that 1p + 1q = 1. Consider the following BVP,

 (φp (u(n−1) ))0 + ω(t ) = 0,    (i) u (0) = 0, u(n−1) (0) − α u(n−2) (ξ ) = 0,    (n−1) u (1) + β u(n−2) (η) = 0,

0 0 n n

< t < 1, n ≥ 3, ≤ i ≤ n − 3, ≥ 3, ≥ 3.

(2.1)

Lemma 2.1. Suppose that ω ∈ L1 [0, 1], ω(t ) > 0, t ∈ [0, 1] and w(t ) 6≡ 0 on any subinterval of [0, 1]. Further if the BVP (2.1) has a solution, then the solution is nonnegative. Proof. Firstly, we assume that u(t ) is a solution of BVP (2.1). If we denote v(t ) = u(n−2) (t ), we have that u(t ) can be expressed as u( t ) =

Z tZ

s1

sn−3

Z ···

0

0

v(sn−2 )dsn−2 dsn−3 · · · ds1 .

(2.2)

0

In order to prove that u(t ) is nonnegative, it is sufficient to verify that v(t ) ≥ 0. So in what follows, we aim to prove that v(t ) = u(n−2) (t ) is nonnegative. From (2.1), it is clear that (φp (v 0 (t )))0 = −ω(t ) ≤ 0 on t ∈ (0, 1), thus φp (v 0 (t )) is non-increasing on t ∈ (0, 1). Then, according to the definition of φp , we can see that v 0 (t ) is non-increasing on t ∈ (0, 1), which implies that v 00 (t ) ≤ 0 for t ∈ (0, 1). Therefore v(t ) is concave on t ∈ [0, 1]. Consequently, v(t ) achieves its minimum at 0 or 1 and, without loss of generality, we assume that v(t ) achieves its minimum at 0. Then v(0) ≤ v(ξ ). On the other hand,

v(ξ ) = v(0) +

ξ

Z

v 0 (s)ds ≤ v(0) +

0

ξ

Z

v 0 (0)ds

0

= v(0) + v 0 (0)ξ = v(0) + αv(ξ )ξ , 1 i.e. v(0) ≥ v(ξ )(1 − αξ ). Considering 0 < α ≤ ξ1 , it is immediate that v(0) ≥ 0. When 0 < β ≤ 1−η , essentially the same reasoning as above establishes that v(1) ≥ 0. To sum up, we have that v(t ) ≥ 0, v(t ) 6≡ 0, t ∈ [0, 1]. That is u(t ) ≥ 0, u(t ) 6≡ 0, t ∈ [0, 1]. The proof is complete. 

Lemma 2.2. Suppose that ω(t ) satisfies all the conditions in Lemma 2.1, then BVP (2.1) has a unique solution u ∈ E ∩ C (n−1) (0, 1) which can be expressed as (2.2). Moreover, there exists unique solution σ of V1 (t ) − V2 (t ) = 0, where V1 (t ) = V2 (t ) =

1

α 1

β

φq

t

Z

 Z t Z t  ω(s)ds + φq ω(τ )dτ ds, ξ

0

φq

and then v(t ) = u

Z

1



ω(s)ds +

t

s

η

Z t

φq

Z

s

 ω(τ )dτ ds,

t

(n−2)

(t ) can be expressed as Z σ  Z t Z σ   1   ω(s)ds + φq ω(τ )dτ ds,  φq α Z0  Z ξ η Zs s  v(t ) = 1 1    φq ω(s)ds + φq ω(τ )dτ ds, β σ t σ

t ∈ [0, σ ], (2.3) t ∈ [σ , 1].

Proof. Let v(t ) = u(n−2) (t ), then it can be seen easily that BVP (2.1) has a positive solution if the following BVP has a positive solution.



(φp (v 0 (t )))0 + ω(t ) = 0, v 0 (0) − αv(ξ ) = 0,

0 < t < 1, v 0 (1) + βv(η) = 0.

(2.4)

To see this, it is enough to prove that BVP (2.4) has a positive solution that can be expressed as (2.3). And in order to prove this, we need to verify the following two facts:

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J. Zhao, W. Ge / Computers and Mathematics with Applications 58 (2009) 1103–1112

(i) : V1 (t ) and V2 (t ) must intersect at one point σ ∈ (0, 1). (ii) : There exists σ ∈ (0, 1) such that v 0 (σ ) = 0, where v(t ) is the solution of BVP (2.4). They can be proved easily, so we omit them here.



Remark 2.1. We can see that when n = 2, BVP (1.1) becomes



(φp (u0 ))0 + h(t )f (t , u(t )) = 0, u0 (0) − α u(ξ ) = 0,

0 < t < 1, u0 (1) + β u(η) = 0,

(2.5)

then v(t ) = u(t ), thus BVP (2.5) can be seen as a special case of BVP (1.1). Since we aim to research higher-order BVP, we just let n ≥ 3. But the results of this paper can also be applied to n = 2. Define cone P ⊂ E as u(i) (0) = 0, (0 ≤ i ≤ n − 3), u(n−2) (t ) ≥ 0, u(n−2) (t ) is concave on u∈E: . t ∈ [0, 1], there exists one point σ ∈ (0, 1) such that u(n−1) (σ ) = 0

 P =



Define T : P → E as:

(Tu)(t ) =

Z tZ

s1

sn−3

Z ···

0

0

v(sn−2 )dsn−2 dsn−3 · · · ds1 ,

(2.6)

0

where

Z σ  Z t Z σ   1   h(s)f (s, u(s), . . . , u(n−2) (s))ds + φq h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds, t ∈ [0, σ ],  φq α Z0  Z ξ η Zs s  v(t ) = 1 1 ( n−2)   φq h(s)f (s, u(s), . . . , u (s))ds + φq h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds, t ∈ [σ , 1]. β σ t σ Lemma 2.3 ([31]). For u ∈ P and θ ∈ (0, 21 ), the following inequality holds u(n−2) (t ) ≥ θ max u(n−2) (s), s∈[0,1]

t ∈ [θ , 1 − θ ].

Lemma 2.4. Suppose that conditions (H1 ) and (H2 ) hold, the solution u(t ) of BVP (2.1) satisfies u(t ) ≤ u0 (t ) ≤ · · · ≤ u(n−3) (t ),

u(n−3) (t ) ≤ kukn−2 ,

t ∈ [0, 1],

and further, for θ ∈ (0, ), the following inequality holds 1 2

u(n−2) (t ) ≥ θ u(n−3) (t ),

t ∈ [θ , 1 − θ ].

Proof. Suppose that u(t ) is the solution of BVP (2.1), in view of Lemma 2.2, we have

 Z t    ω(s)ds , φ q σ Z  u(n−1) (t ) = σ   ω(s)ds , −φq

0 ≤ t ≤ σ, (2.7)

σ ≤ t ≤ 1.

t

From (2.7) we have u u(i) (t ) =

t

Z

(n−2)

(t ) is concave on [0, 1], thus u(i) (t ) ≥ 0, i = 0, 1, . . . , n − 2, t ∈ [0, 1]. Then we obtain

u(i+1) (s)ds ≤ tu(i+1) (t ) ≤ u(i+1) (t ),

i = 0, 1, . . . , n − 4, t ∈ [0, 1],

0

which means that u(t ) ≤ u0 (t ) ≤ · · · ≤ u(n−3) (t ), t ∈ [0, 1]. And further u(n−3) (t ) =

t

Z

u(n−2) (s)ds ≤ t kukn−2 ≤ kukn−2 ,

0

by Lemma 2.3 it is immediate that θ u(n−3) (t ) ≤ u(n−2) (t ), t ∈ [θ , 1 − θ ]. The proof is complete. Remark 2.2. It can be seen easily from Lemma 2.4 that for u ∈ P , kuk = kukn−2 . Lemma 2.5. T : P → P is completely continuous.



J. Zhao, W. Ge / Computers and Mathematics with Applications 58 (2009) 1103–1112

Proof. The proof is standard, we omit it here, for more similar details, we refer the readers to [32].

1107



For convenience, we provide here Guo–Krasnosel’skii fixed point theorem. Theorem 2.6 ([33]). Let E be a Banach space and P ⊂ E be a cone. Suppose Ω1 , Ω2 ⊂ E open and bounded, 0 ∈ Ω1 ⊂ Ω 1 ⊂ Ω2 ⊂ Ω 2 . Assume that T : (Ω 2 \ Ω1 ) ∩ P → P is completely continuous. If one of the following conditions holds. (i) kTxk 6 kxk, ∀x ∈ ∂ Ω1 ∩ P, kTxk > kxk, ∀x ∈ ∂ Ω2 ∩ P; (ii) kTxk 6 kxk, ∀x ∈ ∂ Ω2 ∩ P, kTxk > kxk, ∀x ∈ ∂ Ω1 ∩ P. Then T has a fixed point in (Ω 2 \ Ω1 ) ∩ P. 3. Existence of positive solutions Choose θ > 0 small enough such that ξ , η ∈ (θ , 1 − θ ). In what follows, for convenience, we introduce some notations. Set

Z

ξ

L = min θ

λ1 =

1 L

,

φq

s

Z θ

ds,

1−θ

Z

φq

η

n

max α + 1 − ξ , β + η 1

1−θ

Z

h(τ )dτ





ds ,

s

1

λ2 =

m ∈ (λ1 , ∞),

h(τ )dτ



1

o  R  , 1 φq 0 h(s)ds

M ∈ (0, λ2 ).

Theorem 3.1. Suppose that (H1 ) and (H2 ) hold, further assume that there exist r , ρ, R ∈ R such that either 0 < r < ρ < R or 0 < ρ < θ r < r < R and the following conditions hold

M m

ρ <

(S1 ) f (t , u0 , . . . , un−2 ) ≥ φp (mr ), for all t ∈ [θ , 1 − θ], θ r ≤ un−2 ≤ r , 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 ; (S2 ) f (t , u0 , . . . , un−2 ) ≤ φp (M ρ), for all t ∈ [0, 1], 0 ≤ un−2 ≤ ρ, 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . Then boundary value problem (1.1) has at least one positive solution u ∈ Ω R ∩ P such that kuk lies between r and ρ , where ΩR = {x ∈ E | kxk < R}. Proof. Let E , P be defined as above. Notice that T : Ω R ∩ P → P is completely continuous. It can be easily seen that the fixed point of T is the solution of BVP (1.1). So we intend to prove that T has at least one fixed point in Ω R ∩ P. Without loss of generality, we suppose that 0 < r < M ρ < ρ < R. Denote m

Ωa = {x ∈ E | kxk < a},

where a = r , ρ.

For t ∈ [θ , 1 − θ ] and u ∈ ∂ Ωr ∩ P , 0 < θ r = θ kuk ≤ u(n−2) (t ) ≤ kuk = r, in view of (S1 ) and Lemma 2.3, we will discuss three cases. Case (a): σ ∈ [θ , 1 − θ ].

kTuk = |(Tu)(n−2) (σ )| Z σ  Z σ Z σ  1 (n−2) (n−2) = φq h(s)f (s, u(s), . . . , u (s))ds + φq h(τ )f (τ , u(τ ), . . . , u (τ ))dτ ds α 0 ξ s  Z θ Z θ ≥ φq h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds ξ

s

ξ

Z = θ

φq

s

Z θ

h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ

 ds

≥ mrL > r = kuk. Using the same reasoning, we have

kTuk = |(Tu)(n−2) (σ )| Z 1  Z η Z s  1 = φq h(s)f (s, u(s), . . . , u(n−2) (s))ds + φq h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds β σ σ σ Z s  Z η ≥ φq h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds 1−θ

1−θ

1−θ

Z = η

φq

1−θ

Z s

≥ mrL > r = kuk.

h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ

 ds

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J. Zhao, W. Ge / Computers and Mathematics with Applications 58 (2009) 1103–1112

(b): σ ∈ (0, θ ).

kTuk = |(Tu)(n−2) (σ )|   Z η Z s Z 1 1 h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds φq = φq h(s)f (s, u(s), . . . , u(n−2) (s))ds + β σ σ σ   Z η Z s Z 1 Z 1 h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds φq h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds + φq ≥ 1

φq

≥ η

θ

1

Z η

θ

1−θ

Z

φq

≥ θ

s

Z

φq

θ

h(τ )f (τ , u(τ ), . . . , u

θ

1

Z =

h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ

s

Z

φq

≥

1

Z

σ

σ

σ

η

Z

(n−2)

s

h(τ )f (τ , u(τ ), . . . , u

θ

ds + θ

(τ ))dτ ds +

(n−2)

η

Z



h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ

Z



η

Z θ

φq φq

s

Z θ

h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ

s

Z

h(τ )f (τ , u(τ ), . . . , u

θ

(n−2)

 ds



(τ ))dτ ds

 ds



(τ ))dτ ds

≥ mrL > r = kuk. It remains to consider case (c): σ ∈ (1 − θ , 1). In this case,

kTuk = |(Tu)(n−2) (σ )| Z σ  Z σ Z σ  1 (n−2) (n−2) = φq h(s)f (s, u(s), . . . , u (s))ds + φq h(τ )f (τ , u(τ ), . . . , u (τ ))dτ ds α 0 ξ s Z  Z  Z ξ Z σ σ σ (n−2) (n−2) ≥ φq h(τ )f (τ , u(τ ), . . . , u (τ ))dτ ds + φq h(τ )f (τ , u(τ ), . . . , u (τ ))dτ ds 0

≥

φq φq

1−θ

Z

0

φq

=

θ

h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ

s 1−θ

φq

ξ



h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds

s 1−θ

ξ

1−θ

Z

φq

1−θ

Z

φq

1−θ

Z

s



h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ

ds

 ds

s

≥ mrL > r = kuk. To sum up, we have kTuk ≥ kuk for any u ∈ ∂ Ωr ∩ P . On the other hand, for u ∈ ∂ Ωρ ∩ P, we have u(t ) ≤ kuk = ρ , in view of (S2 ) we have

kTuk = |(Tu)(n−2) (σ )| Z σ  Z σ Z σ  1 = φq f (s, u(s), . . . , u(n−2) (s))ds + φq f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds α 0 ξ s Z σ  Z σ Z σ  1 ≤ φq f (s, u(s), . . . , u(n−2) (s))ds + φq f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds α 0 ξ 0 Z 1  Z 1 Z 1  1 ≤ φq f (s, u(s), . . . , u(n−2) (s))ds + φq f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds α 0 ξ 0   Z 1  1 = + 1 − ξ φq h(s)f (s, u(s), . . . , u(n−2) (s))ds α 0   Z 1  1 ≤ Mρ + 1 − ξ φq h(s)ds < ρ = kuk. α 0 Similarly,

kTuk = |(Tu)(n−2) (σ )|



h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds

s 1−θ

Z

θ

Z

0

≥

 Z h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds +

s 1−θ

Z

 Z h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ ds +

0

ξ

≥

1−θ

Z

0

Z

ξ

0

ξ

Z

J. Zhao, W. Ge / Computers and Mathematics with Applications 58 (2009) 1103–1112

= ≤ ≤

1

β 1

β 1

β 

=

φq φq

1

Z σ

1

Z

h(s)f (s, u(s), . . . , u(n−2) (s))ds



h(s)f (s, u(s), . . . , u(n−2) (s))ds



+ σ

1

Z

h(s)f (s, u(s), . . . , u(n−2) (s))ds) +

σ η

Z

β

≤ Mρ

φq φq

φq

1

 Z + η φq

s

Z σ

1

Z σ 1

Z

0

0

1

η

Z +

0

φq (

η

Z

1109

h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ



h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ



h(τ )f (τ , u(τ ), . . . , u(n−2) (τ ))dτ

ds ds

 ds

0

h(s)f (s, u(s), . . . , u(n−2) (s))ds

0



1

β

1

 Z + η φq

h(s)ds



< ρ = kuk,

0

which means that for any u ∈ ∂ Ωρ ∩ P, there holds kTuk ≤ kuk. Thus, by using Theorem 2.6, we obtain that T has at least one fixed point u ∈ (Ω ρ \ Ωr ) ∩ P with r < kuk < ρ , which implies that boundary value problem (1.1) has at least one positive solution u ∈ P. The proof is complete.  Theorem 3.2. Assume in addition to (H1 ) and (H2 ) that the following inequalities hold: f (t ,u0 ,...,un−2 ) φp (un−2 ) f (t ,u ,...,u ) lim infun−2 →∞ min0≤t ≤1 φ 0(u n)−2 . p n−2

(S3 ) f 0 < φp (M ), where f 0 = lim supun−2 →0+ max0≤t ≤1 (S4 ) f∞ > φp ( θ ), where f∞ = m

;

Then BVP (1.1) has at least one positive solution u ∈ P. Proof. From (S3 ) and the definition of f 0 , we have: for ε > 0 small enough, there exists ρ > 0 such that f (t , u0 , . . . , un−2 ) ≤ (φp (M ) + ε)φp (un−2 ) ≤ (φp (M ) + ε)φp (ρ), t ∈ [0, 1], un−2 ∈ [0, ρ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . Obviously, condition (S2 ) holds. Similarly, from (S4 ), there exists r > 0 with θ r > ρ such that



f (t , u0 , . . . , un−2 ) ≥ φp

m θ

 − ε φp (un−2 ),

t ∈ [0, 1], un−2 ∈ [θ r , +∞], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . Hence, by Lemma 2.3, we obtain



f (t , u0 , . . . , un−2 ) ≥ φp

m θ

 − ε φp (θ r ),

t ∈ [θ , 1 − θ ], un−2 ∈ [θ r , r ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 , thus (S1 ) holds. Therefore, it follows immediate from Theorem 3.1 that there exists at least one positive solution with its norm lying between r and ρ . The proof is complete.  Theorem 3.3. (H1 ) and (H2 ) hold, further we assume that f (t ,u0 ,...,un−2 ) ; φp (un−2 ) f (t ,u ,...,u ) lim infun−2 →0+ min0≤t ≤1 φ 0(u n)−2 . p n−2

(S5 ) f ∞ < φp (M ), where f ∞ = lim supun−2 →∞ max0≤t ≤1 (S6 ) f0 > φp ( mθ ), where f0 =

Then BVP (1.1) has at least one positive solution u ∈ P . Proof. For any ε > 0 small enough, considering (S5 ), there exists ρ > 0 such that f (t , u0 , . . . , un−2 ) ≤ (φp (M ) + ε)φp (un−2 ), t ∈ [0, 1], un−2 ∈ [ρ, ∞], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . Here, there are two cases to be considered, namely, where f is bounded, and f is unbounded. Case I. f is bounded, then there exists N > 0 such that f (t , u0 , . . . , un−2 ) ≤ N , t ∈ [0, 1], un−2 ∈ [0, ∞), 0 ≤ u0 ≤ · · · ≤ θ un−3 ≤ un−2 . Choose ρ > ρ such that φp (M ρ) ≥ N . Hence, f (t , u0 , . . . , un−2 ) ≤ N ≤ (φp (M ) + ε)φp (ρ), t ∈ [0, 1], un−2 ∈ [0, ρ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 .

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Case II. suppose f is unbounded, then there exists ρ > u0n−2 > ρ and f (t , u0 , . . . , un−2 ) ≤ f (t 0 , u00 , . . . , u0n−2 ) ≤ (φp (M ) + ε)φp (u0n−2 ) ≤ (φp (M ) + ε)φp (ρ), t ∈ [0, 1], un−2 ∈ [ρ, ρ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . What is more, f (t , u0 , . . . , un−2 ) ≤ (φp (M ) + ε)φp (un−2 ) ≤ (φp (M ) + ε)φp (ρ), t ∈ [0, 1], un−2 ∈ [0, ρ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . In both case I and case II, we have f (t , u0 , . . . , un−2 ) ≤ (φp (M ) + ε)φp (ρ), t ∈ [0, 1], u ∈ [0, ρ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 , which implies that (S2 ) holds. Similarly, by condition (S6 ), for ε small enough, there exists r > 0 with r <

M m

ρ such that

 m  f (t , u0 , . . . , un−2 ) ≥ φp − ε φp (un−2 ), θ t ∈ [0, 1], un−2 ∈ [0, r ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . Therefore,



f (t , u0 , . . . , un−2 ) ≥ φp

m θ

 − ε φp (un−2 ),

t ∈ [θ , 1 − θ ], un−2 ∈ [θ r , r ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . Obviously, condition (S1 ) holds. It follows directly from Theorem 3.1 that boundary value problem (1.1) has at least one positive solution u ∈ P. The proof is complete.  Remark 3.1. In fact, we can see that both Theorems 3.2 and 3.3 are corollaries of Theorem 3.1. And by the above three Theorems, we can obtain the following results. Corollary 3.4. Assume that (S1 ) holds, further, (S3 ) or (S5 ) holds, then boundary value problem (1.1) has at least one positive solution u ∈ P . Corollary 3.5. Assume that (S2 ) holds, further, (S4 ) or (S6 ) holds, then boundary value problem (1.1) has at least one positive solution u ∈ P . In what follows, we give sufficient conditions for the existence of at least two positive solutions of BVP (1.1). Theorem 3.6. (S4 ) and (S6 ) hold. Further assume that there exists ρ > 0 such that (S2 ) holds and, u 6= Tu, u ∈ ∂ Ωρ ∩ P. Then BVP (1.1) has at least two positive solutions u1 , u2 ∈ P such that ku1 k < ρ < ku2 k. Proof. In view of Theorem 3.1, we know from (S2 ) that for u ∈ ∂ Ωρ ∩ P, there holds kTuk ≤ kuk. From Theorem 3.2, (S4 ) implies that there exists R > ρ such that kTuk ≥ kuk for u ∈ ∂ ΩR ∩ P . By Theorem 3.3, (S6 ) implies that there exists r ∈ (0, M ρ) such that kTuk ≥ kuk for u ∈ ∂ Ωr ∩ P. Moreover, u 6= Tu, u ∈ ∂ Ωρ ∩ P. Therefore, from Theorem 2.6 we obtain m that T has at least two fixed points u1 ∈ (Ωρ \ Ωr ) and u2 ∈ (ΩR \ Ωρ ). Thus BVP (1.1) has at least two positive solutions u1 , u2 ∈ P such that ku1 k < ρ < ku2 k.  Remark 3.2. u 6= Tu, u ∈ ∂ Ωρ ∩ P in Theorem 3.6 can be satisfied if (S2 ) was changed to

(S ∗ 2 ) f (t , u0 , . . . , un−2 ) < φp (M ρ), for all t ∈ [0, 1], 0 ≤ un−2 ≤ ρ, 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . The proof of the following two theorems are similar to that of Theorem 3.6, so we omit them here. Theorem 3.7. (S3 ) and (S5 ) hold. Further assume that there exists r > 0 such that (S1 ) holds and, u 6= Tu, u ∈ ∂ Ωr ∩ P. Then BVP (1.1) has at least two positive solutions u1 , u2 ∈ P such that u1 6= u2 , ku1 k < r < ku2 k. Theorem 3.8. Assume that there exists r > 0 such that one of the following conditions holds

(S7 ) f (t , u0 , . . . , un−2 ) > φp (mr ), for all t ∈ [θ , 1 − θ], un−2 ∈ [θ r , r ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 and f 0 = 0, f ∞ = 0; (S8 ) f (t , u0 , . . . , un−2 ) < φp (Mr ), for all t ∈ [0, 1], un−2 ∈ [0, r ], 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 and f0 = +∞, f∞ = ∞. Then BVP (1.1) has at least two positive solutions u1 , u2 ∈ P such that ku1 k < r < ku2 k.

J. Zhao, W. Ge / Computers and Mathematics with Applications 58 (2009) 1103–1112

1111

Remark 3.3. With the same method, we can also consider the following 2m-point BVP

 (ϕp (u(n−1) ))0 + h(t )f (t , u(t ), u0 (t ), . . . , u(n−2) (t )) = 0,    (i)   u (0) = 0,      m −1 X     (n−1)  u ( 0 ) − αi u(n−2) (ξi ) = 0, BVP BC  i=1     m −1  X     (n−1)   (1) + βi u(n−2) (ηi ) = 0,  u

0 < t < 1, n ≥ 3, 0 ≤ i ≤ n − 3, n ≥ 3,

(3.1)

n ≥ 3,

i=1

where φp (s) = |s|p−2 s, p > 1, αi > 0, βi > 0, 0 < i=1 αi ξi ≤ 1, 0 < i=1 βi (1 − ηi ) ≤ 1, 0 = ξ0 < ξ1 < ξ2 < · · · < ξm−1 < η1 < η2 < · · · < ηm−1 < ηm = 1, i = 1, 2, . . . , m − 1. Similarly, n = 2 can also be seen as a special case of BVP (3.1).

Pm−1

Pm−1

4. Examples In this section, we will give examples to illustrate our main results. Consider the following fourth-order boundary value problem with p-Laplacian Example 4.1.

  00 5 sin t 5u (t )   (φp (x(3) (t )))0 + + = 0,

0 < t < 1,

(i)   x000 (0) = 0,00 x (0) − 2x (1/5) = 0,

i = 0, 1, x000 (1) + 4x00 (5/6) = 0,

5

20

(4.1)

where h(t ) = 1,

f (t , u0 , u1 , u2 ) =

And we can see that 1

ξ=

5

,

η=

5 6

,

sin t

+

5

α = 2,

 u 5 2

4

.

β = 4.

1 5 , θ = 10 . By direct calculation, we can get that λ1 = 10 125, λ2 = 10 , so we choose m = 12 800, M = 13 . Further, 13 1 let ρ = < θ r = 10 × 200 < r = 200. We can verify that hf satisfy (H1 ) and (H2 ) respectively, furthermore we have

Let p =

3 2 13 5

(i) f (t , u0 , . . . , un−2 ) ≥ φp (mr ) = 1.6 × 103 , for all t ∈ [θ , 1 − θ ], θ r ≤ un−2 ≤ r , 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 ; (ii) f (t , u0 , . . . , un−2 ) ≤ φp (M ρ) = 1, for all t ∈ [0, 1], 0 ≤ un−2 ≤ ρ, 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 . Then all the conditions of Theorem 3.1 are satisfied, thus problem (4.1) has at least one positive solution with its norm lying between ρ = 13 and r = 200. 5 Example 4.2.

  1   (u00 (t ))2 + (φp (x(3) (t )))0 + et 100

 x(i) (0) = 0,   000 x (0) − 3x00 (1/4) = 0,



1 1000

= 0,

0 < t < 1, i = 0, 1, x000 (1) + x00 (3/4) = 0,

(4.2)

where h ( t ) = et ,

f (t , u0 , u1 , u2 ) =

And we can see that 1

ξ=

4

,

η=

3 4

,

1 100

α = 3,

u22 +

1 1000

.

β = 1.

Let p = 2, θ = 51 . By direct calculation, we can get that λ2 ≈ 0.336, f0 = ∞, f∞ = ∞. Choose M = 0.2, r = 19. We can verify that h, f satisfy (H1 ) and (H2 ) respectively, furthermore we have f (t , u0 , . . . , un−2 ) < φp (Mr ) = 3.8, for all t ∈ [0, 1], 0 < un−2 < r , 0 ≤ u0 ≤ · · · ≤ un−4 ≤ θ un−3 ≤ un−2 ; Then the conditions in Theorem 3.8 are all satisfied, thus problem (4.2) has at least two positive solutions u1 , u2 ∈ P such that ku1 k ≤ 19 ≤ ku2 k.

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