Symmetric positive solutions for p -Laplacian fourth-order differential equations with integral boundary conditions

Journal of Computational and Applied Mathematics 222 (2008) 561–573 www.elsevier.com/locate/cam

Symmetric positive solutions for p-Laplacian fourth-order differential equations with integral boundary conditionsI Xuemei Zhang a,c,∗ , Meiqiang Feng b,c , Weigao Ge c a Department of Mathematics and Physics, North China Electric Power University, Beijing, 102206, PR China b Department of Fundamental Sciences, Beijing Information Technology Institute, Beijing, 100101, PR China c Department of Applied Mathematics, Beijing Institute of Technology, Beijing, 100081, PR China

Received 26 May 2007; received in revised form 28 November 2007

Abstract This paper investigates the existence and multiplicity of symmetric positive solutions for a class of p-Laplacian fourth-order differential equations with integral boundary conditions. The arguments are based upon a specially constructed cone and the fixed point theory for cones. The nonexistence of a positive solution is also studied. c 2007 Elsevier B.V. All rights reserved.

MSC: 34B15 Keywords: Boundary value problem; Symmetric positive solution; Fixed point theorem; Existence

1. Introduction Boundary value problems for ordinary differential equations arise in different areas of applied mathematics and physics and so on; the existence and multiplicity of positive solutions for such problems have become an important area of investigation in recent years. To identify a few, we refer the reader to [1–8,10,14,16–34] and references therein. In [3], Bai and Wang considered the following fourth-order two-point boundary value problem:  (4) x (t) − λ f (t, x(t)) = 0, 0 < t < 1, x(0) = x(1) = x 00 (0) = x 00 (1) = 0. The authors studied the existence, uniqueness and multiplicity of positive solutions by using the fixed point theorem and degree theory. In [4], Graef, Qian and Yang considered the following fourth-order three-point boundary value problem:  (4) x (t) = λg(t) f (x(t)), 0 < t < 1, x(0) = x 0 (1) = x 00 (0) = x 00 ( p) − x 00 (1) = 0, I Supported by NNSF of China (10671012) and SRFDP of China (20050007011) and the Science and Technology Innovation Foundation for postgraduate students of Beijing Institute of Technology. ∗ Corresponding author at: Department of Mathematics and Physics, North China Electric Power University, Beijing, 102206, PR China. E-mail address: [email protected] (X. Zhang).

c 2007 Elsevier B.V. All rights reserved. 0377-0427/$ - see front matter doi:10.1016/j.cam.2007.12.002

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X. Zhang et al. / Journal of Computational and Applied Mathematics 222 (2008) 561–573

where p ∈ (0, 1) is a constant. The authors obtained the existence and nonexistence of positive solutions by using the Krasnoselskii fixed point theorem. In [5], Zhang and Liu considered the following fourth-order four-point boundary value problem:  (φ p (x 00 (t)))00 = w(t) f (t, x(t)), t ∈ [0, 1], x(0) = 0, x(1) = ax(ξ ),  00 x (0) = 0, x 00 (1) = bx 00 (η), where 0 < ξ, η < 1, 0 ≤ a < b < 1 and f ∈ C((0, 1) × (0, +∞), [0, +∞)). The authors gave sufficient conditions for the existence of one positive solution by using the upper and lower solution method, fixed point theorems and the properties of the Green’s function. We notice that a type of symmetric problem has received much attention. For example, in [6] Yao considered the following two-point BVP:  00 x (t) + w(t) f (x(t)) = 0, t ∈ (0, 1), αx(0) − βx 0 (0) = 0, αx(1) + βx 0 (1) = 0. The author obtained the existence of n symmetric positive solutions and established a corresponding iterative scheme by using a monotone iterative technique. In [7] Avery and Henderson studied the existence of multiple symmetric positive solutions for the following twopoint BVP:  00 x (t) + f (x(t)) = 0, t ∈ [0, 1], x(0) = 0 = x(1). The main tool is a fixed point theorem due to Avery [8] which is a generalization of the Leggett–Williams fixed point theorem. At the same time, a class of boundary value problems with integral boundary conditions appeared in heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics. Such problems include two-point, three-point, multi-point and nonlocal boundary value problems as special cases and have attracted the attention of Gallardo [9], Karakostas and Tsamatos [10], Lomtatidze and Malaguti [11] (and see the references therein). For more information about the general theory of integral equations and their relation to boundary value problems we refer the reader to the book of Corduneanu [12] and Agarwal and O’Regan [13]. Motivated by the works mentioned above, in this paper, we study the existence and nonexistence of symmetric positive solutions of the following fourth-order boundary value problem with integral boundary conditions:  (φ p (x 00 (t)))00 = w(t) f (t, x(t)), 0 < t < 1,   Z 1    x(0) = x(1) = g(s)x(s)ds, (1.1) 0 Z 1     φ p (x 00 (0)) = φ p (x 00 (1)) = h(s)φ p (x 00 (s))ds, 0

1 1 1 where φ p (t) = |t| p−2 t, p > 1, φq = φ −1 p , p + q = 1, w ∈ L [0, 1] is nonnegative, symmetric on the interval [0, 1] (i.e., w(1 − t) = w(t) for t ∈ [0, 1]), f : [0, 1] × [0, +∞) → [0, +∞) is continuous, f (1 − t, x) = f (t, x) for all (t, x) ∈ [0, 1] × [0, +∞), and g, h ∈ L 1 [0, 1] are nonnegative, symmetric on [0, 1]. The main features of this paper are as follows. First, comparing with [3–5], we discuss the boundary value problem with integral boundary conditions, i.e., problem (1.1) including fourth-order two-point, three-point, multi-point and nonlocal boundary value problems as special cases. Secondly, comparing with [3–5], we consider the existence of symmetric positive solutions. To our knowledge, no paper has considered the fourth-order p-Laplacian equation with integral boundary conditions. Hence we improve and generalize the results of [3–5] to some degree, and so it is interesting and important to study the existence and nonexistence of positive solutions for problem (1.1). The arguments are based upon a specially constructed cone and the fixed point theory for cones. In this paper, a symmetric positive solution x of (1.1) means a solution x of (1.1) satisfying x > 0, t ∈ (0, 1) and x(t) = x(1 − t), t ∈ [0, 1].

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To obtain positive solutions of (1.1), the following fixed point theorem for cones is fundamental; it can be found in [15], p. 94. Lemma 1.1 (Fixed Point Theorem of Cone Expansion and Compression of Norm Type). Let P be a cone of real Banach space E, and Ω1 and Ω2 be two bounded open sets in E such that 0 ∈ Ω1 and Ω¯ 1 ⊂ Ω2 . Let operator A : P ∩ (Ω¯ 2 \ Ω1 ) → P be completely continuous. Suppose that one of the two conditions (i) kAxk ≤ kxk, ∀x ∈ P ∩ ∂Ω1 and kAxk ≥ kxk, ∀x ∈ P ∩ ∂Ω2 and (ii) kAxk ≥ kxk, ∀x ∈ P ∩ ∂Ω1 , and kAxk ≤ kxk, ∀x ∈ P ∩ ∂Ω2 is satisfied. Then A has at least one fixed point in P ∩ (Ω¯ 2 \ Ω1 ). The organization of this paper is as follows. In Section 2, we provide some necessary background. In particular, we state some properties of the Green’s function associated with problem (1.1). In Section 3, we discuss the existence of at least one symmetric positive solution for problem (1.1). In Section 4, we will prove the existence of two or n symmetric positive solutions, where n is an arbitrary natural number. In Section 5, we study the nonexistence of a positive solution for problem (1.1). Finally, in Section 6, one example is also included to illustrate the main results. 2. Preliminaries The basic space used in this paper is E = C[0, 1]. It is well known that E is a real Banach space with the norm k · k defined by kxk = max |x(t)|. 0≤t≤1

Let K be a cone of E, and K r = {x ∈ K : kxk ≤ r },

∂ K r = {x ∈ K : kxk = r },

K¯ r,R = {x ∈ K : r ≤ kxk ≤ R},

where 0 < r < R. We make the following assumptions: (H1 ) w ∈ L 1 [0, 1] is nonnegative, symmetric on [0, 1] and w(t) 6≡ 0 on any subinterval of [0, 1]; (H2 ) f : [0, 1] × [0, +∞) → [0, +∞) is continuous and f (·, x) is symmetric on [0, 1] for all x ≥ 0; (H3 ) g, h ∈ L 1 [0, 1] are nonnegative, symmetric on [0, 1], and µ ∈ (0, 1), ν ∈ (0, 1), where Z 1 Z 1 h(s)ds. g(s)ds, ν= µ= 0

0

We recall that the function x is said to be concave on [0, 1] if x(τ t2 + (1 − τ )t1 ) ≥ τ x(t2 ) + (1 − τ )x(t1 ),

t1 , t2 , τ ∈ [0, 1],

and the function x is said to be symmetric on [0, 1] if x(t) = x(1 − t), t ∈ [0, 1]. In our main results, we will make use of the following lemmas. Lemma 2.1. Assume that (H3 ) holds. Then for any y ∈ E, the boundary value problem  00 t ∈ (0, 1), −x (t) = −φq (y(t)), Z 1

x(0) = x(1) =

g(s)x(s)ds,

has a unique solution x and x can be expressed in the form Z 1 x(t) = − H (t, s)φq (y(s))ds, 0

(2.1)

0

(2.2)

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where Z 1 1 H (t, s) = G(t, s) + G(s, τ )g(τ )dτ, 1−µ 0  t (1 − s), 0 ≤ t ≤ s ≤ 1, G(t, s) = s(1 − t), 0 ≤ s ≤ t ≤ 1.

(2.3)

Proof. The proof follows by routine calculations. From (2.3) and the definition of G(t, s), we can prove that H (t, s), G(t, s) have the following properties.



Proposition 2.1. If (H3 ) holds, then we have H (t, s) > 0,

G(t, s) > 0,

for t, s ∈ (0, 1),

(2.4)

H (t, s) ≥ 0,

G(t, s) ≥ 0,

for t, s ∈ [0, 1].

(2.5)

Proposition 2.2. For t, s ∈ [0, 1], we have e(t)e(s) ≤ G(t, s) ≤ G(t, t) = t (1 − t) = e(t) ≤ e¯ = max e(t) = t∈[0,1]

G(1 − t, 1 − s) = G(t, s).

1 , 4

(2.6) (2.7)

Proposition 2.3. If (H3 ) holds, then for t, s ∈ [0, 1], we have ρe(s) ≤ H (t, s) ≤ γ s(1 − s) = γ e(s) ≤

1 γ, 4

(2.8)

where and 1 γ = , 1−µ

R1 ρ=

0

e(τ )g(τ )dτ . 1−µ

Proof. By (2.3) and (2.6), we have Z 1 1 H (t, s) = G(t, s) + G(s, τ )g(τ )dτ 1−µ 0 Z 1 1 G(s, τ )g(τ )dτ ≥ 1−µ 0 R1 e(τ )g(τ )dτ ≥ 0 s(1 − s) 1−µ = ρe(s), t ∈ [0, 1]. On the other hand, noticing G(t, s) ≤ s(1 − s), we obtain Z 1 1 H (t, s) = G(t, s) + G(s, τ )g(τ )dτ 1−µ 0 Z 1 1 ≤ s(1 − s) + s(1 − s)g(τ )dτ 1−µ 0 Z 1 1 ≤ s(1 − s)[1 + g(τ )dτ ] 1−µ 0 1 ≤ s(1 − s) 1−µ = γ e(s), t ∈ [0, 1]. 

(2.9)

(2.10)

(2.11)

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Proposition 2.4. If (H3 ) holds, then for all t, s ∈ [0, 1] we have H (1 − t, 1 − s) = H (t, s).

(2.12)

Proof. By (2.7), we have H (1 − t, 1 − s) = G(1 − t, 1 − s) +

1 1−µ

Z

1

G(1 − s, τ )g(τ )dτ

0

Z 0 1 G(1 − s, 1 − τ )g(1 − τ )d(1 − τ ) 1−µ 1 Z 1 1 G(s, τ )g(τ )dτ = G(t, s) + 1−µ 0 = H (t, s).  = G(t, s) +

Lemma 2.2. If (H3 ) holds, then for any y ∈ E, BVP  00 f (t, x(t)), t ∈ (0, 1), −y (t) = −w(t) Z 1  y(0) = y(1) = h(s)y(s)ds,

(2.13)

(2.14)

0

has a unique solution Z 1 y(t) = − H1 (t, s)w(s) f (s, x(s))ds,

(2.15)

0

where H1 (t, s) = G(t, s) +

1 1−ν

Z

1

G(s, v)h(v)dv.

(2.16)

0

Remark 1. From (2.16), we can prove that the properties of H1 (t, s) are similar to those of H (t, s). Suppose that x is a solution of problem (1.1). Then from Lemma 2.1, we have Z 1 x(t) = − H (t, s)φq (y(s))ds.

(2.17)

0

From Lemma 2.2, we have Z 1 Z x(t) = H (t, s)φq 0

1

! H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds.

0

For applying Lemma 1.1, we construct a cone in E via   K = x ∈ E : x ≥ 0, x(t) is symmetric, concave on [0, 1], and min x(t) ≥ δkxk , t∈[0,1]

where R1

q−1

δ=

ρρ1

q−1 γ γ1

,

ρ1 =

0

e(τ )h(τ )dτ 1 , γ1 = . 1−ν 1−ν

It is easy to see that K is a closed convex cone of E. Remark 2. By the definition of ρ, ρ1 , γ and γ1 , we have 0 < δ < 1.

(2.18)

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Define an operator T : E → E by "Z # Z 1 1 (T x)(t) = H (t, s)φq H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds. 0

(2.19)

0

From (2.19), we know that x ∈ E is a solution of problem (1.1) if and only if x is a fixed point of operator T , and we can obtain the following Lemma 2.3. Lemma 2.3. Assume that (H1 )–(H3 ) hold. If x ∈ E is a solution of the following integral equation: "Z # Z 1 1 x(t) = (T x)(t) = H (t, s)φq H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds, 0

(2.20)

0

then x ∈ C[0, 1] ∩ C 4 (0, 1), and x is a solution of BVP (1.1). Lemma 2.4. Suppose that (H1 )–(H3 ) hold. Then T (K ) ⊂ K and T : K → K is completely continuous. Proof. For all x ∈ K , we have by (2.19) "Z 1

(T x) = −φq 00

#

H1 (t, s)w(s) f (s, x(s))ds ≤ 0,

(2.21)

0

which implies that T x is concave on [0, 1]. On the other hand, by (2.19) we obtain "Z Z 1 (T x)(0) = (T x)(1) = H (0, s)φq 0

1

# H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds ≥ 0.

0

It follows that (T x)(t) ≥ 0 for t ∈ [0, 1]. Noticing that w(t) is symmetric on (0, 1), x(t) is symmetric on [0, 1], and f (t, .) is symmetric on [0, 1], we have "Z # Z 1 1 (T x)(1 − t) = H (1 − t, s)φq H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds 0

Z

0 0

=

H (1 − t, 1 − s)φq

1

Z

1

H (t, s)φq

"Z

0

# H1 (1 − s, τ )w(τ ) f (τ, x(τ ))dτ d(1 − s)

1

# H1 (1 − s, τ )w(τ ) f (τ, x(τ ))dτ ds

0 1

=

H (t, s)φq

"Z

0

Z

1

0

= Z

"Z

0

# H1 (1 − s, 1 − τ )w(1 − τ ) f (1 − τ, x(1 − τ ))d(1 − τ ) ds

1 1

=

H (t, s)φq

"Z

0

1

# H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds

0

= (T x)(t), i.e., (T x)(1 − t) = (T x)(t), t ∈ [0, 1]. Therefore, (T x)(t) is symmetric on [0, 1]. On the other hand, by (2.11) we obtain "Z # Z 1 1 (T x)(t) = H (t, s)φq H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds 0

≤

q−1 γ γ1

0

"Z

1

Z

1

e(s)φq 0

0

# e(τ )w(τ ) f (τ, x(τ ))dτ ds.

X. Zhang et al. / Journal of Computational and Applied Mathematics 222 (2008) 561–573

Similarly, by (2.10) we obtain "Z Z 1 (T x)(t) = H (t, s)φq 0

≥

=

1

567

# H1 (s, τ )w(τ ) f (τ, x(τ ))dτ

0

q−1 ρρ1

"Z

1

Z

#

1

e(τ )w(τ ) f (τ, x(τ ))dτ ds

e(s)φq 0

q−1 δγ γ1

0

"Z

1

Z

1

e(s)φq 0

≥ δkT xk,

# e(τ )w(τ ) f (τ, x(τ ))dτ ds

0

t ∈ [0, 1].

So, T x ∈ K and then T (K ) ⊂ K . Next by standard methods and the Ascoli–Arzela theorem one can prove that T : K → K is completely continuous. So this is omitted.  3. Existence of one symmetric positive solution In this section, we impose growth conditions on f which allow us to apply Lemma 1.1 to establish the existence of one symmetric positive solution of BVP (1.1), and we begin by introducing some notation: f (t, x) , φ p (x)

f β = lim sup max

t∈[0,1]

x→β

where β denotes 0 or ∞, and "Z Z 1 q−1 σ = γ γ1 e(s)φq 0

1

f β = lim inf min

x→β t∈[0,1]

f (t, x) , φ p (x)

# e(τ )w(τ )dτ ds.

0

Theorem 3.1. Assume that (H1 )–(H3 ) hold. In addition, we suppose that one of the following conditions is satisfied: (C1 ) There exist two constants r, R with 0 < r ≤ δ R such that f (t, x) ≤ φ p (σ −1r ) for all (t, x) ∈ [0, 1] × [δr, r ], 1 and f (t, x) ≥ φ p ( δσ R) for all (t, x) ∈ [0, 1] × [δ R, R]. 1 (C2 ) f 0 > φ p ( σ δ 2 ) and f ∞ < φ p (σ −1 ) (in particular, f 0 = ∞ and f ∞ = 0). (C3 ) There exist two constants r2 , R2 with 0 < r2 ≤ R2 such that f (t, .) is nondecreasing on [0, R2 ] for all t ∈ [0, 1], 1 and f (t, δr2 ) ≥ φ p ( δσ r2 ), and f (t, R2 ) ≤ φ p (σ −1 R2 ) for all t ∈ [0, 1]. Then problem (1.1) has at least one symmetric positive solution. Proof. Let T be the cone preserving, completely continuous operator that was defined by (2.19). Case (1); let us consider (C1 ). For x ∈ K , from (2.18) we obtain that mint∈[0,1] x(t) ≥ δkxk. Therefore, for x ∈ ∂ K r , we have x ∈ [δr, r ], which implies f (t, x(t)) ≤ φ p (σ −1r ). Thus for t ∈ [0, 1] we have "Z # Z 1 1 (T x)(t) = H (t, s)φq H1 (s, τ )w(τ ) f (τ, x(τ ))dτ 0

≤

=

q−1 γ γ1

0

Z

"Z

1

1

e(s)φq 0

q−1 γ γ1 σ −1r

# e(τ )w(τ )φ p (σ

−1

r )dτ ds

0

Z

"Z

1

e(s)φq 0

1

# e(τ )w(τ )dτ ds

0

= r = kxk, i.e., x ∈ ∂ K r implies kT xk ≤ kxk.

(3.1)

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X. Zhang et al. / Journal of Computational and Applied Mathematics 222 (2008) 561–573

On the other hand, for x ∈ ∂ K R , we have x ∈ [δ R, R], which implies f (t, x(t)) ≥ φ p we have "Z # Z 1 1 (T x)(t) = H (t, s)φq H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds 0

≥

=



1 σδ

 R . Thus for t ∈ [0, 1]

0

q−1 ρρ1

"Z

1

Z

1

e(s)φq 0

e(τ )w(τ )φ p



0

Z

q−1 1 δγ γ1 σδ

"Z

1

R

1

e(s)φq 0

 # 1 R dτ ds σδ #

e(τ )w(τ )dτ ds

0

= R = kxk, i.e., x ∈ ∂ K R implies (3.2)

kT xk ≥ kxk.

holds. Considering f 0 > φ p ( σ1δ 2 ), there exists r1 > 0 such that f (t, x) ≥ [0, r1 ], where ε1 > 0 satisfies σ δ 2 ( f 0 −ε1 )q−1 ≥ 1. Then, for t ∈ [0, 1], x ∈ ∂ K r1 ,

Case (2); the condition (C2 ) ( f 0 −ε1 )φ p (x), for t ∈ [0, 1], x ∈ we have "Z Z 1 (T x)(t) = H (t, s)φq 0

≥

≥

=

1

# H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds

0

q−1 ρρ1

Z

q−1 ρρ1

Z

"Z

1

1

e(s)φq 0

# e(τ )w(τ )( f 0 − ε1 )φ p (x(τ ))dτ ds

0

"Z

1

1

e(s)φq 0

# e(τ )w(τ )( f 0 − ε1 )φ p (δkxk)dτ ds

0

q−1 δγ γ1 ( f 0

− ε1 )

q−1

δkxk

"Z

1

Z

1

e(s)φq 0

# e(τ )w(τ )dτ ds

0

≥ kxk, i.e., x ∈ ∂ K r1 implies (3.3)

kT xk ≥ kxk. f∞

Next, turning to

< φp

(σ −1 ),

f (t, x) ≤ ( f ∞ + ε2 )φ p (x),

there exists R¯ 1 > 0 such that for t ∈ [0, 1], x ∈ ( R¯ 1 , ∞),

where ε2 > 0 satisfies σ ( f ∞ + ε2 )q−1 ≤ 1. Set M=

max

0≤x≤ R¯ 1 ,t∈[0,1]

f (t, x).

Then f (t, x) ≤ M + ( f ∞ + ε2 )φ p (x). Choose R1 > max{r1 , R¯ 1 , M q−1 σ (1 − σ ( f ∞ + ε2 )q−1 )−1 }. Thus for x ∈ K R1 , we have "Z # Z 1 1 (T x)(t) = H (t, s)φq H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds 0

≤

q−1 γ γ1

0

"Z

1

Z

1

e(s)φq 0

# e(τ )w(τ )(M + ( f ∞ + ε2 )φ p (x(τ )))dτ ds

0

≤ M q−1 σ + ( f ∞ + ε2 )q−1 kxkσ

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569

< R1 − σ R1 ( f ∞ + ε2 )q−1 + ( f ∞ + ε2 )q−1 kxkσ = R1 , i.e., x ∈ ∂ K R1 implies kT xk < kxk.

(3.4)

Case (3); the condition (C3 ) is satisfied. For x ∈ K , from (2.18) we obtain that mint∈[0,1] x(t) ≥ δkxk. Therefore, for x ∈ ∂ K r2 , we have x(t) ≥ δkxk = δr2 for t ∈ [0, 1]; putting this together with (C3 ), we have "Z # Z 1 1 (T x)(t) = H (t, s)φq H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds 0

Z

0 1

≥

H (t, s)φq

"Z

0

≥

=

#

1

H1 (s, τ )w(τ ) f (τ, δr2 )dτ ds

0

q−1 ρρ1

"Z

1

Z

1

e(s)φ p

q−1 δγ γ1

0

0

Z

"Z

1

e(τ )w(τ )dτ ds 1

e(s)φ p 0

#

1 r2 σδ

# e(τ )w(τ )dτ ds

0

1 r2 σδ

= r2 = kxk, i.e., x ∈ ∂ K r2 implies (3.5)

kT xk ≥ kxk. On the other hand, for x ∈ ∂ K R2 , we have x(t) ≤ R2 for t ∈ [0, 1]; putting this together with (C3 ), we have # "Z Z 1 1 H1 (s, τ )w(τ ) f (τ, x(τ ))dτ ds H (t, s)φq (T x)(t) = 0

0

Z

1

≤

H (t, s)φq

0

"Z

1

# H1 (s, τ )w(τ ) f (τ, R2 )dτ ds

0 q−1

≤ γ γ1

Z

1

0

e(τ )dτ

1

Z

e(s)w(s)dsσ −1 R2

0

= R2 = kxk, i.e., x ∈ ∂ K R2 implies kT xk ≤ kxk.

(3.6)

Applying Lemma 1.1 to (3.1) and (3.2), or (3.3) and (3.4), or (3.5) and (3.6) yields that T has a fixed point x ∗ ∈ K¯ r,R or x ∗ ∈ K¯ ri ,Ri (i = 1, 2) with x ∗ (t) ≥ δkx ∗ k > 0, t ∈ [0, 1]. Thus it follows that problem (1.1) has a symmetric positive solution x ∗ , and the theorem is proved.  Remark 3. Under the conditions of (C1 )–(C3 ), we consider the existence of one symmetric positive solution of problem (1.1). So, we deal with three cases in theorem (Theorem 3.1). By the same method as for Theorem 3.1, we can also obtain the following result. Theorem 3.2. Assume that (H1 )–(H3 ) hold. In addition, we suppose that one of the following two conditions is satisfied: (C4 ) There exist two constants r, R with 0 < r ≤ δ R such that f (t, x) ≥ φ p ( σ1δ r ) for all (t, x) ∈ J × [δr, r ], and f (t, x) ≤ φ p (σ −1 R) for all (t, x) ∈ J × [δ R, R]. (C5 ) f 0 < φ p (σ −1 ) and f ∞ > φ p ( σ1δ 2 ) (in particular, f 0 = 0 and f ∞ = ∞). Then problem (1.1) has at least one symmetric positive solution.

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4. The existence of multiple positive solutions Now we discuss the multiplicity of symmetric positive solutions for problem (1.1). We obtain the following existence results. Theorem 4.1. Assume that (H1 )–(H3 ) hold, as do the following two conditions: (C6 ) f 0 > φ p ( σ1δ 2 ) and f ∞ > φ p ( σ1δ 2 ) (in particular, f 0 = f ∞ = ∞); (C7 ) There exists b > 0 such that maxt∈[0,1],x∈∂ K b f (t, x) < φ p (σ −1 b). Then problem (1.1) has at least two positive solutions x ∗ (t), x ∗∗ (t), which satisfy 0 < kx ∗∗ k < b < kx ∗ k.

(4.1)

Proof. We consider the condition (C6 ). Choose r, R with 0 < r < b < R. If f 0 > φ p ( σ1δ 2 ), then by the proof of (3.3), we have kT xk ≥ kxk,

for x ∈ ∂ K r .

(4.2)

If f ∞ > φ p ( σ1δ 2 ), then like in the proof of (3.3), we have kT xk ≥ kxk,

for x ∈ ∂ K R .

(4.3)

On the other hand, by (C7 ), for x ∈ ∂ K b we have "Z # Z 1 1 q−1 kT xk ≤ γ γ1 e(s)φq e(τ )w(τ ) f (τ, x(τ ))dτ ds 0

< =

q−1 γ γ1

Z

q−1 γ γ1

Z

0

"Z

1

1

e(s)φq 0

# e(τ )w(τ )φ p (σ

−1

b)dτ ds

0

"Z

1

e(s)φq 0

1

# e(τ )w(τ )dτ dsσ −1 b

0

(4.4)

= b. By (4.4), we have k(T x)k < b = kxk.

(4.5)

Applying Lemma 1.1 to (4.2), (4.3) and (4.5) yields that T ∈ K¯ r,b , and a fixed point x ∗ ∈ K¯ b,R . ∗ ∗∗ Thus it follows that BVP (1.1) has at least two positive solutions x and x . Noticing (4.5), we have kx ∗ k 6= b and kx ∗∗ k 6= b. Therefore (4.1) holds, and the proof is complete.  has a fixed point x ∗∗

Theorem 4.2. Assume (H1 )–(H3 ) hold, as do the following two conditions: (C8 ) f 0 < φ p (σ −1 ) and f ∞ < φ p (σ −1 ); (C9 ) there exists B > 0 such that mint∈[0,1],x∈∂ K B f (t, x) > φ p ( σ1δ B). Then problem (1.1) has at least two symmetric positive solutions x ∗ (t), x ∗∗ (t), which satisfy 0 < kx ∗∗ k < B < kx ∗ k. Theorem 4.3. Assume that (H1 )–(H3 ) hold. Suppose that there exist 2n positive numbers dk , Dk , k = 1, 2, . . . , n with d1 < δ D1 < D1 < d2 < δ D2 < D2 < · · · < dn < δ Dn < Dn such that (C10 ) f (t, x) ≤ φ p (σ −1 dk ) for (t, x) ∈ [0, 1] × [δdk , dk ] and f (t, x) ≥ φ p ( σ1δ Dk ) for (t, x) ∈ [0, 1] × [δ Dk , Dk ], k = 1, 2, . . . , n; or (C11 ) f (t, x) ≥ φ p ( σ1δ dk ) for (t, x) ∈ [0, 1] × [δdk , dk ] and f (t, x) ≤ φ p (σ −1 Dk ) for (t, x) ∈ [0, 1] × [δ Dk , Dk ], k = 1, 2, . . . , n. Then problem (1.1) has at least n symmetric positive solutions xk satisfying dk ≤ kxk k ≤ Dk , k = 1, 2, . . . , n.

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571

Theorem 4.4. Assume that (H1 )–(H3 ) hold. Suppose that there exist 2n positive numbers dk , Dk , k = 1, 2, . . . , n with d1 < D1 < d2 < D2 < · · · < dn < Dn such that (C12 ) f (t, ·) is nondecreasing on [0, Dn ] for all t ∈ [0, 1]; (C13 ) f (t, δdk ) ≥ φ p ( σ1δ d2 ), and f (t, Dk ) ≤ φ p (σ −1 Dk ), k = 1, 2, . . . , n. Then problem (1.1) has at least n symmetric positive solutions xk satisfying dk ≤ kxk k ≤ Dk , k = 1, 2, . . . , n. 5. The nonexistence of a positive solution Our last result corresponds to the case when problem (1.1) has no positive solution. Theorem 5.1. Assume (H1 )–(H3 ) and f (t, x) < φ p (σ −1 x), for all t ∈ [0, 1], x > 0; then problem (1.1) has no positive solution. Proof. Assume to the contrary that x(t) is a positive solution of the problem (1.1). Then x ∈ K , x(t) > 0 for t ∈ (0, 1), and kxk = max kx(t)k t∈[0,1]

≤ < =

q−1 γ γ1

Z

q−1 γ γ1

Z

q−1 γ γ1

Z

"Z

1

1

e(s)φq 0

# e(τ )w(τ ) f (τ, x(τ ))dτ ds

0

"Z

1

1

e(s)φq 0

# e(τ )w(τ )φ p (σ −1 x(τ ))dτ ds

0

"Z

1

1

e(s)φq 0

# e(τ )w(τ )dτ dsσ −1 kxk

0

= kxk, which is a contradiction, and this completes the proof.



Similarly, we have the following results. Theorem 5.2. Assume (H1 )–(H3 ) and f (t, x) > φ p ( σ1δ 2 x), for all x > 0; then problem (1.1) has no positive solution. 6. Example To illustrate how our main results can be used in practice we present an example. Let p = 3, g(t) = 12 , h(t) = 0 in (1.1). Now we consider the following boundary value problem:  (φ p (x 00 (t)))00 = w(t) f (t, x(t)), 0 < t < 1,   Z 1  1 x(0) = x(1) = x(s)ds,   0 2  00 00 φ p (x (0)) = φ p (x (1)) = 0,

(6.1)

where w(t) = 6, f (t, x) = (1 + sin πt)x 6 . Since p = 3, g(t) = 12 , h(t) = 0, then, by calculations we obtain that 1 q = 32 , µ = 12 , ν = 0, γ = 2, ρ = 61 , γ1 = 1, ρ1 = 16 , δ = 72 , σ = 13 , and Z 1 H (t, s) = G(t, s) + G(s, τ )dτ, H1 (t, s) = G(t, s), 0

where  t (1 − s), G(t, s) = s(1 − t),

0 ≤ t ≤ s ≤ 1, 0 ≤ s ≤ t ≤ 1.

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Clearly, the conditions (H1 )–(H3 ) hold. Now, we prove that the condition (C1 ) of Theorem 3.1 is satisfied. Choosing r = 21 , R = 72 × 217, we obtain that    2   1 3 1 −1 φ3 (σ r ) = φ3 3 × = , φp R = φ p (72 × 3 × 72 × 217) = (72 × 216 × 217)2 , 2 2 σδ  2 1 1 f (t, x) = (1 + sin π t)x 6 ≤ 2 × r 6 = 2 × = , for (t, x) ∈ [0, 1] × [δr, r ], 8 32  6 1 6 6 f (t, x) = (1 + sin π t)x ≥ (δ R) = × 72 × 217 = 2176 , for (t, x) ∈ [0, 1] × [δ R, R]. 72   Therefore, f (t, x) < φ p (σ −1r ) for all (t, x) ∈ J × [δr, r ], and f (t, x) > φ p σ1δ R for all (t, x) ∈ J × [δ R, R]. Hence, by Theorem 3.1, BVP (6.1) has at least one positive solution, and the proof is complete. Remark 4. This example implies that there are a large number of functions that satisfy the conditions of Theorem 3.1. In addition, the conditions of Theorem 3.1 are also easy to check. Acknowledgement The authors thank the referee for his/her careful reading of the manuscript and useful suggestions. References [1] C.P. Gupta, A generalized multi-point boundary value problem for second order ordinary differential equations, Appl. Math. Comput. 89 (1998) 133–146. [2] C.P. Gupta, S.K. Ntouyas, P.Ch. Tsamatos, Solvability of an m-point boundary value problem for second order ordinary differential equations, J. Math. Anal. Appl. 189 (1995) 575–584. [3] Z. Bai, H. Wang, On the positive solutions of some nonlinear fourth-order beam equations, J. Math. Anal. Appl. 270 (2002) 357–368. [4] J.R. Graef, C. Qian, B. Yang, A three point boundary value problem for nonlinear fourth order differential equations, J. Math. Anal. Appl. 287 (2003) 217–233. [5] Xinguang Zhang, Lishan Liu, Positive solutions of fourth-order four-point boundary value problems with p-Laplacian operator, J. Math. Anal. Appl. (2007), doi:10.1016/j.jmaa.2007.03.015. [6] Q. Yao, Existence and iteration of n symmetric positive solutions for a singular two-point boundary value problem, Comput. Math. Appl. 47 (2004) 1195–1200. [7] R.I. Avery, A.C. Henderson, Three symmetric positive solutions for a second-order boundary value problem, Appl. Math. Lett. 13 (2000) 1–7. [8] R.I. Avery, A generalization of the Leggett–Williams fixed-point theorem, Math. Sci. Res. Hot-line 2 (1998) 9–14. [9] J.M. Gallardo, Second order differential operators with integral boundary conditions and generation of semigroups, Rocky Mountain J. Math. 30 (2000) 1265–1292. [10] G.L. Karakostas, P.Ch. Tsamatos, Multiple positive solutions of some Fredholm integral equations arisen from nonlocal boundary-value problems, Electron. J. Differential Equations 30 (2002) 1–17. [11] A. Lomtatidze, L. Malaguti, On a nonlocal boundary-value problems for second order nonlinear singular differential equations, Georgian Math. J. 7 (2000) 133–154. [12] C. Corduneanu, Integral Equations and Applications, Cambridge University Press, Cambridge, 1991. [13] R.P. Agarwal, D. O’Regan, Infinite Interval Problems for Differential, Difference and Integral Equations, Kluwer Academic Publishers, Dordrecht, 2001. [14] R.P. Agarwal, D. O’Regan, V. Lakshmikantham, S. Leela, Existence of positive solutions for singular initial and boundary value problems via the classical upper and lower solution approach, Nonlinear Anal. 50 (2002) 215–222. [15] D.J. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, Inc., New York, 1988. [16] Y. Guo, W. Shan, W. Ge, Positive solutions for a second order m-point boundary value problem, J. Comput. Appl. Math. 151 (2003) 415–424. [17] Y. Guo, J. Tian, Positive solutions of m-point boundary value problems for higher order ordinary differential equations, Nonlinear Anal. 66 (2007) 1573–1586. [18] R. Ma, H. Wang, Positive solutions of nonlinear three-point boundary-value problems, J. Math. Anal. Appl. 279 (2003) 216–227. [19] R. Ma, H. Wang, On the existence of positive solutions of fourth-order ordinary differential equations, Appl. Anal. 59 (1995) 225–231. [20] Haiyan Wang, On the number of positive solutions of nonlinear systems, J. Math. Anal. Appl. 281 (2003) 287–306. [21] D.Q. Jiang, W.J. Gao, A. Wan, A monotone method for constructing extremal solutions to fourth-order periodic boundary value problems, Appl. Math. Comput. 132 (2002) 411–421. [22] B. Liu, Positive solutions of fourth-order two point boundary value problem, Appl. Math. Comput. 148 (2004) 407–420. [23] W. Feng, T.R.L. Weeb, Solvability of a m-point boundary value problem with Nonlinear growth, J. Math. Anal. Appl. 212 (1997) 467–480.

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