Nonlinear Analysis 70 (2009) 220–230www.elsevier.com/locate/na

Successive iteration and positive solutions for a third-ordermultipoint generalized right-focal boundary value problem

with p-LaplacianI

Bo Suna,∗, Junfang Zhaoa, Pinghua Yangb, WeiGao Gea

a Department of Mathematics, Beijing Institute of Technology, Beijing 100081, PR Chinab Department of Mathematics, Mechanical Engineering College, Shijiazhuang Hebei 050003, PR China

Received 20 March 2007; accepted 29 November 2007

Abstract

In this paper, we study the existence of positive solutions for the one-dimensional p-Laplacian differential equation,

(φp(u′′))′(t) = q(t) f (t, u(t), u′(t), u′′(t)), 0 ≤ t ≤ 1,

subject to the following multipoint generalized right-focal boundary condition,

u(0) =m∑

i=1

αi u(ξi ), u′(η) = 0, u′′(1) =n∑

i=1

βi u′′(θi ),

by applying a monotone iterative method. We not only obtain the existence of positive solutions for the problem, but also establishiterative schemes for approximating the solutions.c© 2007 Elsevier Ltd. All rights reserved.

Keywords: Successive iteration; Positive solutions; Third-order multipoint boundary value problem; p-Laplacian

1. Introduction

In this paper, we will consider the positive solutions to the following third-order multipoint generalized right-focalboundary value problem with p-Laplacian,

(φp(u′′))′(t) = q(t) f (t, u(t), u′(t), u′′(t)), 0 ≤ t ≤ 1, (1.1)

u(0) =m∑

i=1

αi u(ξi ), u′(η) = 0, u′′(1) =n∑

i=1

βi u′′(θi ), (1.2)

I This work is sponsored by the National Natural Science Foundation of China (10671012) and the Doctoral Program Foundation of EducationMinistry of China (20050007011).∗ Corresponding author.

E-mail addresses: sunbo19830328@163.com (B. Sun), gew@bit.edu.cn (W. Ge).

0362-546X/$ - see front matter c© 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2007.11.048

B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230 221

where φp(s) = |s|p−2s with 1 < p ≤ 2, and the following conditions hold,

(H1): 0 < ξ1 < ξ2 < · · · < ξm < η, 12 ≤ η ≤ 1, and 0 < θi < 1, i = 1, 2, . . . , n;

(H2): 0 ≤ αi < 1 (i = 1, 2, . . . ,m), 0 ≤ βi < 1 (i = 1, 2, . . . , n) satisfy 0 ≤∑m

i=1 αi ,∑n

i=1 βi < 1;(H3): f (t, x, y, z) ∈ C([0, 1] × [0,+∞) × R × (+∞, 0] → [0,+∞)), q(t) is a nonnegative continuous function

defined on (0, 1), and q(t) 6≡ 0 on any subinterval of (0, 1). In addition,∫ 1

0 q(t)dt < +∞.

The existence and multiplicity of positive solutions for second-order linear and nonlinear multipoint boundaryvalue problems have been widely studied by many authors, one may see [1–7] and the references therein.

In the past ten years or so, many authors also studied some third-order boundary value problems. Third-orderequations arise in a variety of different areas of applied mathematics and physics. They have been studied by differenttype of techniques. For example, in [10], Erbe and Wang applied Krasnoselskii’s work to study the existence of atleast one positive solution to a third-order boundary value problem. In [11], Chu and Zhou studied the existence ofpositive solutions for nonlinear third-order periodic boundary value problem. They obtained the existence of positivesolutions via a combination of Krasnoselskii’s fixed point theorem and a nonlinear alternative of Leray–Schaudertype. In [12–14], Anderson and Davis got the existence of multiple positive solutions for some generalized right-focalboundary value problems. And other methods have been used to study third-order boundary value problems, one maysee [12–21] and the references therein.

We recall the existing literature on studying the nonlinear third-order boundary value problems. Firstly, most ofthem studied the following third-order differential equations,

u′′′(t)− f (u(t)) = 0, t ∈ (0,+∞), (1.3)

u′′′(t)− f (t, u(t)) = 0, t ∈ [a, b], (1.4)

u′′′(t)− λa(t) f (t, u(t)) = 0, t ∈ (0, 1). (1.5)

We know that the corresponding Green’s function of equations like (1.3)–(1.5) exists, so most of the authors reducedthem to first- and/or second-order equation and used Green’s function and comparison principles to study the third-order boundary value problems. But, in (1.1), when p 6= 2, the p-Laplacian operator is nonlinear and we cannotconstruct a Green’s function. Then, the methods we mentioned above are not available to (1.1). And there were fewpapers studying the third-order boundary value problems with p-Laplacian as we know.

Secondly, there are few papers which cover the computational methods of boundary value problems. Most of thepapers studied the existence of the positive solutions of various boundary value problems, then, it a question comesto us, that “How can we find the solutions when they are known to exist?” Motivated by this question, more recently,Ma, Du and Ge [7] and our papers [8,9] proved the existence of positive solutions for some second-order p-Laplacianboundary value problems via monotone iterative technique.

So, motivated as we mentioned above, we will investigate the iteration and existence of positive solutions forthe third-order multipoint generalized right-focal boundary value problem with p-Laplacian (1.1), (1.2). We will notrequire the existence of lower and upper solution. And by applying monotone iterative techniques, we will constructsome successive iterative schemes whose starting points are simple quadratic functions or some known constantfunctions.

2. Preliminaries

In this section, we give the preliminaries and some definitions.

Definition 2.1. Let E be a real Banach space. A nonempty closed set P ⊂ E is said to be a cone provided that,

♦ au + bv ∈ P for all u, v ∈ P and all a ≥ 0, b ≥ 0, and♦ u,−u ∈ P implies u = 0.

Definition 2.2. The map α is said to be concave on [0, 1], if

α(tu + (1− t)v) ≥ tα(u)+ (1− t)α(v)

for all u, v ∈ [0, 1] and t ∈ [0, 1].

222 B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230

Let the Banach space E = C2[0, 1] be endowed with the norm,

‖u‖ := maxmax0≤t≤1

|u(t)|, max0≤t≤1

|u′(t)|, max0≤t≤1

|u′′(t)|.

We denote

E+ = C2+[0, 1] = u ∈ E |u(t) ≥ 0, t ∈ [0, 1],

and define the cone P ⊂ E by

P = u ∈ E |u(t) ≥ 0, u is concave and nondecreasing on [0, η] nonincreasing on [η, 1].

Throughout, it is assumed that (H1), (H2) and (H3) hold.For ∀x ∈ C2

+[0, 1], suppose u is a solution of the problem (1.1), (1.2), then we have,

u′′(t) = φ−1p

(Ax −

∫ 1

tq(s) f (s, x(s), x ′(s), x ′′(s))ds

),

u′(t) =

−

∫ η

tφ−1

p

[Ax −

∫ 1

sq(r) f (r, x(r), x ′(r), x ′′(r))dr

]ds, 0 ≤ t ≤ η,∫ t

η

φ−1p

[Ax −

∫ 1

sq(r) f (r, x(r), x ′(r), x ′′(r))dr

]ds, η ≤ t ≤ 1,

and

u(t) = −1

1−m∑

i=1αi

m∑i=1

αi

∫ ξi

0

[∫ η

tφ−1

p

(Ax −

∫ 1

sq(r) f (r, x(r), x ′(r), x ′′(r))dr

)ds

]dt

−

∫ t

0

[∫ η

sφ−1

p

(Ax −

∫ 1

τ

q(r) f (r, x(r), x ′(r), x ′′(r))dr

)dτ

]ds,

where Ax satisfies the boundary conditions, i.e.,

φ−1p (Ax ) =

n∑i=1

βiφ−1p

(Ax −

∫ 1

θi

q(s) f (s, x(s), x ′(s), x ′′(s))ds

). (2.1)

Lemma 2.1. For ∀x ∈ C2+[0, 1], there exists a unique Ax with

Ax ∈

−φp

(n∑

i=1βi

)1− φp

(n∑

i=1βi

) ∫ 1

0q(s) f (s, x(s), x ′(s), x ′′(s))ds, 0

satisfying (2.1).

Proof. For ∀x ∈ C2+[0, 1], we define,

Hx (c) = φ(c)−n∑

i=1

βiφ−1p

(c −

∫ 1

θi

q(s) f (s, x(s), x ′(s), x ′′(s))ds

),

then Hx (c) ∈ C((−∞,+∞), R) and Hx (0) ≥ 0. In what follows, we can prove that Hc(0) = 0 has a unique solutionon (−∞,+∞), which means that there exists a unique Ax ∈ (−∞,+∞) satisfying (2.1). The proof process can begiven in a similar way as that of Lemma 2.2 in [7]. So, we omit it here.

For any x ∈ C2+[0, 1], let Ax be the unique constant satisfying Eq. (2.1) corresponding to x , then we have the

following lemma.

B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230 223

Lemma 2.2. Ax : C2+[0, 1] → R has the following properties,

(a) Ax : C2+[0, 1] → R is continuous about x;

(b) If f (t, x, y, z) is nondecreasing about x, nondecreasing about |y| and nonincreasing about z on [0, 1] ×[0,+∞)× R × (+∞, 0], then Ax is nonincreasing on P.

Proof. The proof can be given in a similar way as that of Lemma 2.3 in [7]. So, we omit it here.

3. Main results

For notational convenience, we denote,

A = max

m∑

i=1αiξi

(η −

ξi2

)+

η2

2

(1−

m∑i=1

αi

)(

1−m∑

i=1αi

)φ−1

p

(1− φp

(n∑

i=1βi

)) , 1

φ−1p

(1− φp

(n∑

i=1βi

))φ−1p

(∫ 1

0q(s)ds

).

We will prove the following existence results.

Theorem 3.1. Assume that (H1), (H2), (H3) hold, and there exists a > 0, such that,

(S1): f (t, x1, y1, z1) ≤ f (t, x2, y2, z2) for any 0 ≤ t ≤ 1, 0 ≤ x1 ≤ x2 ≤ a, 0 ≤ |y1| ≤ |y2| ≤ a−a ≤ z2 ≤ z1 ≤ 0;(S2): max0≤t≤1 f (t, a, a,−a) ≤ φp(

aA );

(S3): f (t, 0, 0, 0) 6≡ 0 for 0 ≤ t ≤ 1.

Then the boundary value problem (1.1), (1.2) has two positive, nondecreasing on [0, η] nonincreasing on [η, 1]and concave solutions w∗ and v∗, such that,

0 < w∗ ≤ a, 0 ≤ |(w∗)′| ≤ a, −a ≤ (w∗)′′ ≤ 0,

and limn→∞

wn = limn→∞

T nw0 = w∗, lim

n→∞(wn)

′= lim

n→∞(T nw0)

′= (w∗)′,

limn→∞

(wn)′′= lim

n→∞(T nw0)

′′= (w∗)′′,

where w0(t) = a + at

(η −

t

2

), 0 ≤ t ≤ 1,

and

0 < v∗ ≤ a, 0 ≤ |(v∗)′| ≤ a, −a ≤ (v∗)′′ ≤ 0,

and limn→∞

vn = limn→∞

T nv0 = v∗, lim

n→∞(vn)

′= lim

n→∞(T nv0)

′= (v∗)′,

limn→∞

(vn)′′= lim

n→∞(T nv0)

′′= (v∗)′′,

where v0(t) = 0, 0 ≤ t ≤ 1,

where

(T u)(t) = −1

1−m∑

i=1αi

m∑i=1

αi

∫ ξi

0

[∫ η

tφ−1

p

(Au −

∫ 1

sq(r) f (r, u(r), u′(r), u′′(r))dr

)ds

]dt

−

∫ t

0

[∫ η

sφ−1

p

(Au −

∫ 1

τ

q(r) f (r, u(r), u′(r), u′′(r))dr

)dτ

]ds. (3.1)

The iterative schemes in Theorem 3.1 are w0(t) = a + at (η − t2 ), wn+1 = Twn = T nw0, n = 0, 1, 2 . . . and

v0(t) = 0, vn+1 = T vn = T nv0, n = 0, 1, 2 . . . . They start off with a known simple quadratic function and the zerofunction respectively.

Proof. We define an operator T : P → E by (3.1), then from the definition of T , we deduce that for each u ∈ P ,there is T u ∈ C2

[0, 1] which satisfies (1.2). And,

224 B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230

(T u)′′(t) = φ−1p (Au −

∫ 1

tq(s) f (s, u(s), u′(s), u′′(s))ds), (3.2)

(T u)′(t) =

−

∫ η

tφ−1

p

[Au −

∫ 1

sq(r) f (r, u(r), u′(r), u′′(r))dr

]ds, 0 ≤ t ≤ η,∫ t

η

φ−1p

[Au −

∫ 1

sq(r) f (r, u(r), u′(r), u′′(r))dr

]ds, η ≤ t ≤ 1.

(3.3)

Since (T u)′′(t) ≤ 0, T u is concave. And since (T u)′(t) ≥ 0 on [0, η], and (T u)′(t) ≤ 0 on [η, 1], we can obtain that(T u)(t) is nondecreasing on [0, η], and nonincreasing on [η, 1]. Moreover, (T u)(0) ≥ 0, then,

(T u)(1) = −1

1−m∑

i=1αi

m∑i=1

αi

∫ ξi

0

[∫ η

tφ−1

p

(Au −

∫ 1

sq(r) f (r, u(r), u′(r), u′′(r))dr

)ds

]dt

−

∫ 1

0

[∫ η

sφ−1

p

(Au −

∫ 1

τ

q(r) f (r, u(r), u′(r), u′′(r))dr

)dτ

]ds

≥ −

∫ 1

0

[∫ η

sφ−1

p

(Au −

∫ 1

τ

q(r) f (r, u(r), u′(r), u′′(r))dr

)dτ

]ds

=

∫ η

0

[∫ η

sφ−1

p

(−Au +

∫ 1

τ

q(r) f (r, u(r), u′(r), u′′(r))dr

)dτ

]ds

−

∫ 1

η

[∫ s

η

φ−1p

(−Au +

∫ 1

τ

q(r) f (r, u(r), u′(r), u′′(r))dr

)dτ

]ds

≥

∫ η

0

[∫ η

sφ−1

p

(−Au +

∫ 1

η

q(r) f (r, u(r), u′(r), u′′(r))dr

)dτ

]ds

−

∫ 1

η

[∫ s

η

φ−1p

(−Au +

∫ 1

η

q(r) f (r, u(r), u′(r), u′′(r))dr

)dτ

]ds

= φ−1p

(−Au +

∫ 1

η

q(r) f (r, u(r), u′(r), u′′(r))dr

)(η −

12

)≥ 0,

so, by the concavity of T , then, (T u)(t) ≥ 0, 0 ≤ t ≤ 1. Hence, T : P → P .In what follows, we will prove that T : P → P is completely continuous. The continuity of T is obvious. Now,

we prove that T is compact. Let Ω ⊂ P be a bounded set. It is easy to prove that T (Ω) is bounded and equi-continuous. Then the Arzela–Ascoli theorem guarantees that TΩ is relatively compact, which means T is compact.Then, T : P → P is completely continuous, and each fixed point of T in P is a solution of (1.1), (1.2).

By (S1) and Lemma 2.2 we have, for any ui ∈ P (i = 1, 2) with u1 ≤ u2, |u′1| ≤ |u′

2| and u′′1 ≥ u′′2 . LetAui (i = 1, 2) be two constants decided in Eq. (2.1) corresponding to ui ∈ P (i = 1, 2), then we know thatAu1 ≥ Au2 . From the definition of T , we can easily get T u1 ≤ T u2.

We denote,

Pa = u ∈ P | ‖u‖ ≤ a.

Then, in what follows, we first prove that T : Pa → Pa . If u ∈ Pa , then ‖u‖ ≤ a, we have,

0 ≤ u(t) ≤ u(η) = max0≤t≤1

|u(t)| ≤ ‖u‖ ≤ a,

0 ≤ max0≤t≤1

|u′(t)| ≤ ‖u‖ ≤ a,

−a ≤ −‖u‖ ≤ − max0≤t≤1

|u′′(t)| = u′′(0) ≤ u′′(t) ≤ 0.

B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230 225

So by (S1), (S2) we have,

0 ≤ f (t, u(t), u′(t), u′′(t)) ≤ f (t, a, a,−a) ≤ max0≤t≤1

f (t, a, a,−a) ≤ φp

( a

A

), for 0 ≤ t ≤ 1.

In fact,

‖T u‖ = maxmax0≤t≤1

|(T u)(t)|, max0≤t≤1

|(T u)′(t)|, max0≤t≤1

|(T u)′′(t)|

= max(T u)(η), (T u)′(0),−(T u)′(1),−(T u)′′(0).

By (3.1)–(3.3) and Lemma 2.1, we have,

(T u)(η) = −1

1−m∑

i=1αi

m∑i=1

αi

∫ ξi

0

[∫ η

tφ−1

p

(Au −

∫ 1

sq(r) f (r, u(r), u′(r), u′′(r))dr

)ds

]dt

−

∫ η

0

[∫ η

sφ−1

p

(Au −

∫ 1

τ

q(r) f (r, u(r), u′(r), u′′(r))dr

)dτ

]ds

≤1

1−m∑

i=1αi

m∑i=1

αi

∫ ξi

0

∫ η

tφ−1

p

1

1− φp(n∑

i=1βi )

∫ 1

0q(r) f (r, u(r), u′(r), u′′(r))dr

ds

dt

+

∫ η

0

∫ η

sφ−1

p

1

1− φp

(n∑

i=1βi

) ∫ 1

0q(r) f (r, u(r), u′(r), u′′(r))dr

dτ

ds

≤

m∑i=1

αiξi

(η −

ξi2

)+

η2

2

(1−

m∑i=1

αi

)(

1−m∑

i=1αi

)φ−1

p

(1− φp

(n∑

i=1βi

)) a

Aφ−1

p

(∫ 1

0q(s)ds

)≤ a,

(T u)′(0) = −∫ η

0φ−1

p

[Au −

∫ 1

sq(r) f (r, u(r), u′(r), u′′(r))dr

]ds

≤

∫ η

0φ−1

p

1

1− φp

(n∑

i=1βi

) ∫ 1

0q(r) f (r, u(r), u′(r), u′′(r))dr

ds

≤η

φ−1p

(1− φp

(n∑

i=1βi

)) a

Aφ−1

p

(∫ 1

0q(s)ds

)≤ a,

−(T u)′(1) =∫ 1

η

φ−1p

[−Au +

∫ 1

sq(r) f (r, u(r), u′(r), u′′(r))dr

]ds

≤

∫ 1

η

φ−1p

1

1− φp

(n∑

i=1βi

) ∫ 1

0q(r) f (r, u(r), u′(r), u′′(r))dr

ds

≤1− η

φ−1p

(1− φp

(n∑

i=1βi

)) a

Aφ−1

p

(∫ 1

0q(s)ds

)≤ a,

226 B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230

and

−(T u)′′(0) = φ−1p

(−Au +

∫ 1

0q(s) f (s, u(s), u′(s), u′′(s))ds

)

≤ φ−1p

1

1− φp

(n∑

i=1βi

) ∫ 1

0q(s) f (s, u(s), u′(s), u′′(s))ds

≤

1

φ−1p

(1− φp

(n∑

i=1βi

)) a

Aφ−1

p

(∫ 1

0q(s)ds

)≤ a.

Thus, we obtain that,

‖T u‖ = max(T u)(η), (T u)′(0),−(T u)′(1),−(T u)′′(0) ≤ a.

Hence, we assert that T : Pa → Pa .

Let w0(t) = a+ at (η− t2 ), 0 ≤ t ≤ 1, then w0(t) ∈ Pa . Let w1 = Tw0, w2 = T 2w0, then w1 ∈ Pa and w2 ∈ Pa .

We denote wn+1 = Twn = T nw0, n = 0, 1, 2 . . . . Since T : Pa → Pa , we have wn ∈ T Pa ⊆ Pa, n = 1, 2, . . . .Since T is completely continuous, we assert that wn

∞

n=1 is a sequentially compact set.

Since

w1(t) = Tw0(t)

= −1

1−m∑

i=1αi

m∑i=1

αi

∫ ξi

0

[∫ η

tφ−1

p

(Aw0 −

∫ 1

sq(r) f (r, w0(r), w

′

0(r), w′′

0(r))dr

)ds

]dt

−

∫ t

0

[∫ η

sφ−1

p

(Aw0 −

∫ 1

τ

q(r) f (r, w0(r), w′

0(r), w′′

0(r))dr

)dτ

]ds,

≤1

1−m∑

i=1αi

m∑i=1

αi

∫ ξi

0

∫ η

tφ−1

p

1

1− φp

(n∑

i=1βi

) ∫ 1

0q(r) f (r, w0(r), w

′

0(r), w′′

0(r))dr

ds

dt

+

∫ t

0

∫ η

sφ−1

p

1

1− φp

(n∑

i=1βi

) ∫ 1

0q(r) f (r, w0(r), w

′

0(r), w′′

0(r))dr

dτ

ds

≤

m∑i=1

αiξi

(η −

ξi2

)(

1−m∑

i=1αi

)φ−1

p

(1− φp

(n∑

i=1βi

)) a

Aφ−1

p

(∫ 1

0q(s)ds

)

+1

φ−1p

(1− φp

(n∑

i=1βi

)) a

Aφ−1

p

(∫ 1

0q(s)ds

)∫ t

0

[∫ η

sdτ]

ds

≤ a + at

(η −

t

2

)= w0(t), 0 ≤ t ≤ 1,

B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230 227

|w′1(t)| = |(Tw0)′(t)|

=

∫ η

tφ−1

p

[−Aw0 +

∫ 1

sq(r) f (r, w0(r), w

′

0(r), w′′

0(r))dr

]ds, 0 ≤ t ≤ η,

∫ t

η

φ−1p

[−Aw0 +

∫ 1

sq(r) f (r, w0(r), w

′

0(r), w′′

0(r))dr

]ds, η ≤ t ≤ 1,

≤

a(η − t), 0 ≤ t ≤ η,a(t − η), η ≤ t ≤ 1,

= |a(η − t)| = |w′0(t)|, 0 ≤ t ≤ 1,

and

w′′1(t) = (Tw0)′′(t)

= φ−1p

(Aw0 −

∫ 1

tq(s) f (s, w0(s), w

′

0(s))ds

)

≥ φ−1p

−φp

(n∑

i=1βi

)1− φp

(n∑

i=1βi

) ∫ 1

0q(s) f (s, w0(s), w

′

0(s))ds −∫ 1

0q(s) f (s, w0(s), w

′

0(s))ds

≥ φ−1p

− 1

1− φp

(n∑

i=1βi

) ∫ 1

0q(s) f (s, w0(s), w

′

0(s))ds

≥ −

a

A

φ−1p

(∫ 10 q(s)ds

)φ−1

p

(1− φp

(n∑

i=1βi

)) ≥ −a = w′′0(t), 0 ≤ t ≤ 1.

So,

w2(t) = Tw1(t) ≤ Tw0(t) = w1(t), 0 ≤ t ≤ 1,

|w′2(t)| = |(Tw1)′(t)| ≤ |(Tw0)

′(t)| = |w′1(t)|, 0 ≤ t ≤ 1,

w′′2(t) = (Tw1)′′(t) ≥ (Tw0)

′′(t) = w′′1(t), 0 ≤ t ≤ 1.

Hence, by the induction, we have,

wn+1 ≤ wn, |w′n+1(t)| ≤ |w′n(t)|, w′′n+1(t) ≥ w

′′n(t), 0 ≤ t ≤ 1, n = 0, 1, 2 . . . .

Thus, there exists w∗ ∈ Pa such that wn → w∗. Applying the continuity of T and wn+1 = Twn , we get Tw∗ = w∗.Let v0(t) = 0, 0 ≤ t ≤ 1, then v0(t) ∈ Pa . Let v1 = T v0, v2 = T 2v0, then v1 ∈ Pa and v2 ∈ Pa . We denote

vn+1 = T vn = T nv0, n = 0, 1, 2 . . . . Since T : Pa → Pa , we have vn ∈ T Pa ⊆ Pa, n = 1, 2, . . . . Since T iscompletely continuous, we assert that vn

∞

n=1 is a sequentially compact set.Since v1 = T v0 ∈ Pa , we have,

v1(t) = T v0(t) = (T 0)(t) ≥ 0, 0 ≤ t ≤ 1,

|v′1(t)| = |T v′

0(t)| = |(T 0)′(t)| ≥ 0, 0 ≤ t ≤ 1,

v′′1 (t) = (T v0)′′(t) = (T 0)′′(t) ≤ 0, 0 ≤ t ≤ 1.

So,

v2(t) = T v1(t) ≥ (T 0)(t) = v1(t), 0 ≤ t ≤ 1,

|v′2(t)| = |T v′

1(t)| ≥ |(T 0)′(t)| = |v′1(t)|, 0 ≤ t ≤ 1,

v′′2 (t) = (T v1)′′(t) ≤ (T 0)′′(t) = v′′1 (t), 0 ≤ t ≤ 1.

228 B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230

By an induction argument similar to the one above we obtain,

vn+1 ≥ vn, |v′n+1(t)| ≥ |v′n(t)|, v′′n+1(t) ≤ v

′′n (t), 0 ≤ t ≤ 1, n = 0, 1, 2 . . . .

Hence, there exists v∗ ∈ Pa such that vn → v∗. Applying the continuity of T and vn+1 = T vn , we get T v∗ = v∗.If f (t, 0, 0) 6≡ 0, 0 ≤ t ≤ 1, then the zero function is not the solution of (1.1), (1.2). Thus, max0≤t≤1 |v

∗(t)| > 0,we have, v∗ ≥ mint, 1− tmax0≤t≤1 |v

∗(t)| > 0, 0 < t < 1.It is well known that each fixed point of T in P is a solution of (1.1), (1.2). Hence, we assert that w∗ and v∗ are

two positive, nondecreasing on [0, η] nonincreasing on [η, 1] and concave solutions of the problem (1.1), (1.2).The proof is completed.

The following corollaries follow easily.

Corollary 3.1. Assume that (H1), (H2), (H3), (S1), (S3) hold, and there exists a > 0, such that

(C3.1): lim`→+∞max0≤t≤1f (t,`,a,−a)`p−1 ≤

1Ap−1 , (particularly, lim`→+∞max0≤t≤1

f (t,`,a,−a)`p−1 = 0).

Then the boundary value problem (1.1), (1.2) has two positive, nondecreasing on [0, η] nonincreasing on [η, 1] andconcave solutions w∗ and v∗, such that the conclusions of Theorem 3.1 hold.

Corollary 3.2. Assume that (H1), (H2), (H3), (S3) hold, and there exists 0 < a1 < a2 < · · · < an , such that

(C3.2.1): f (t, x1, y1, z1) ≤ f (t, x2, y2, z2) for any 0 ≤ t ≤ 1, 0 ≤ x1 ≤ x2 ≤ ak , 0 ≤ |y1| ≤ |y2| ≤ ak ,−ak ≤ z2 ≤ z1 ≤ 0, k = 1, 2, . . . , n;

(C3.2.2): max0≤t≤1 f (t, ak, ak,−ak) ≤ φp(akA ), k = 1, 2, . . . , n.

Then the boundary value problem (1.1), (1.2) has 2n positive, nondecreasing on [0, η] nonincreasing on [η, 1] andconcave solutions w∗k and v∗k , such that

0 < w∗k ≤ ak, 0 < |w∗k | ≤ ak, −ak < (w∗k )′′≤ 0,

and limn→∞

wkn = limn→∞

T nwk0 = w∗

k , limn→∞

(wkn )′= lim

n→∞(T nwk0)

′= (w∗k )

′,

limn→∞

(wkn )′′= lim

n→∞(T nwk0)

′′= (w∗k )

′′,

where wk0(t) = ak + ak t (η −t

2), 0 ≤ t ≤ 1,

and

0 < v∗k ≤ ak, 0 < |v∗k | ≤ ak, −ak < (v∗k )′′≤ 0,

and limn→∞

vkn = limn→∞

T nvk0 = v∗

k , limn→∞

(vkn )′= lim

n→∞(T nvk0)

′= (v∗k )

′,

limn→∞

(vkn )′′= lim

n→∞(T nvk0)

′′= (v∗k )

′′,

where vk0(t) = 0, 0 ≤ t ≤ 1,

and (T u)(t) is defined the same as (3.1).

The iterative schemes in Corollary 3.2 are wk0(t) = ak + ak t (η − t2 ), wkn+1 = Twkn = T nwk0 , k = 1, 2 . . . n =

0, 1, 2 . . . and vk0(t) = 0, vkn+1 = T vkn = T nvk0 , k = 1, 2 . . . n = 0, 1, 2 . . . . They start off with known simplequadratic functions and zero functions respectively.

Corollary 3.3. Assume that (H1), (H2), (H3), (C3.2.1), (S3) hold, and there exists 0 < a1 < a2 < · · · < an , suchthat

(C3.3): lim`→+∞max0≤t≤1f (t,`,ak ,−ak )

`p−1 ≤1

Ap−1 , k = 1, 2, . . . , n, (particularly, lim`→+∞max0≤t≤1f (t,`,ak ,−ak )

`p−1 =

0, k = 1, 2, . . . , n).

Then the boundary value problem (1.1), (1.2) has 2n positive, nondecreasing on [0, η] nonincreasing on [η, 1] andconcave solutions w∗k and v∗k , such that the conclusions of Corollary 3.2 hold.

B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230 229

4. Example

Example 4.1. Let p = 32 and q(t) = 1 in (1.1), we consider the following boundary value problem,(

|u′′(t)|−12 u′′(t)

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

u(0) =19

u

(18

)+

29

u

(14

)+

29

u

(38

),

u′(1) = 0,

u′′(1) =116

u′′(

23

)+

116

u′′(

23

)+

116

u′′(

23

)+

116

u′′(

23

),

(4.1)

where

f (t, x, y, z) =18

t2+

116

x +1

16y −

116

z,

Choose a = 4, then we have A = 4.So, f (t, x, y) satisfies:

(1) f (t, x1, y1, z1) ≤ f (t, x2, y2, z2) for any 0 ≤ t ≤ 1, 0 ≤ x1 ≤ x2 ≤ 4, 0 ≤ y1 ≤ y2 ≤ 4, −4 ≤ z2 ≤ z1 ≤ 0;(2) max0≤t≤1 f (t, a, a,−a) = f (1, 4, 4,−4) < φ 3

2( a

A ) = 1;(3) f (t, 0, 0) 6≡ 0 for 0 ≤ t ≤ 1.

So by Theorem 3.1, the boundary value problem (4.1) has two positive, nondecreasing on [0, 1] and concave solutionsw∗ and v∗, such that

0 < w∗ ≤ 4, 0 ≤ (w∗)′ ≤ 4, −4 ≤ (w∗)′′ ≤ 0,

and limn→∞

wn = limn→∞

T nw0 = w∗, lim

n→∞(wn)

′= lim

n→∞(T nw0)

′= (w∗)′,

limn→∞

(wn)′′= lim

n→∞(T nw0)

′′= (w∗)′′,

where w0(t) = −2t2+ 4t + 4, 0 ≤ t ≤ 1,

and

0 < v∗ ≤ 4, 0 ≤ (v∗)′ ≤ 4, −4 ≤ (v∗)′′ ≤ 0,

and limn→∞

vn = limn→∞

T nv0 = v∗, lim

n→∞(vn)

′= lim

n→∞(T nv0)

′= (v∗)′,

limn→∞

(vn)′′= lim

n→∞(T nv0)

′′= (v∗)′′,

where v0(t) = 0, 0 ≤ t ≤ 1,

where T u(t) is as defined in (3.1).

For n = 0, 1, 2 . . . , the two iterative schemes are

w0(t) = −2t2+ 4t + 4, 0 ≤ t ≤ 1,

w1(t) = (Tw0)(t) =3

64t4−

14

t3+

12

t2−

2148

t −162 725

3145 728, 0 ≤ t ≤ 1,

· · · · · · · · ·

wn+1(t) = −94

3∑i=1

αi

∫ ξi

0

[∫ 1

tφ−1

p

(Awn −

∫ 1

s

(18

r2+

116wn(r)+

116w′n(r)−

116w′′n(r)

)dr

)ds

]dt

−

∫ t

0

[∫ 1

sφ−1

p

(Awn −

∫ 1

τ

(18

r2+

116wn(r)+

116w′n(r)−

116w′′n(r)

)dr

)dτ

]ds,

0 ≤ t ≤ 1.

v0(t) = 0,

230 B. Sun et al. / Nonlinear Analysis 70 (2009) 220–230

v1(t) = (T v0)(t) =1

1920t6−

2531 104

t4+

62552 488

t2−

82912099 520

t −169 447 9312024 908 480

, 0 ≤ t ≤ 1,

· · · · · · · · ·

vn+1(t) = −94

3∑i=1

αi

∫ ξi

0

[∫ 1

tφ−1

p

(Avn −

∫ 1

s

(18

r2+

116vn(r)+

116v′n(r)−

116v′′n (r)

)dr

)ds

]dt

−

∫ t

0

[∫ 1

sφ−1

p

(Avn −

∫ 1

τ

(18

r2+

116vn(r)+

116v′n(r)−

116v′′n (r)

)dr

)dτ

]ds, 0 ≤ t ≤ 1,

where Awn , Avn satisfy the boundary conditions, i.e., satisfy (2.1).They start off with a known simple quadratic function and the zero function respectively.

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