Electronic Journal of Qualitative Theory of Differential Equations 2008, No. 18, 1-10; http://www.math.u-szeged.hu/ejqtde/

Two positive solutions for a nonlinear four-point boundary value problem with a p-Laplacian operator ^{∗}

Ruixi Liang^{1†} Jun Peng^{2} Jianhua Shen^{1,3}

1 Department of Mathematics, Hunan Normal University Changsha, Hunan 410081, China 2 Department of Mathematics, Central South University Changsha, Hunan 410083, China 3 Department of Mathematics, College of Huaihua Huaihua, Hunan 418008, China

Abstract: In this paper, we study the existence of positive solutions for a nonlinear fourpoint boundary value problem with a p-Laplacian operator. By using a three functionals fixed point theorem in a cone, the existence of double positive solutions for the nonlinear four-point boundary value problem with a p-Laplacian operator is obtained. This is different than previous results. Key words: p-Laplacian operator; Positive solution; Fixed point theorem; Four-point boundary value problem

1. Introduction

In this paper we are interested in the existence of positive solutions for the following nonlinear four-point boundary value problem with a p-Laplacian operator:

(ϕ_{p}(u′))′ + e(t)f(u(t)) = 0, 0 < t < 1,  (1.1)
μϕ_{p}(u(0)) − ωϕ_{p}(u′(ξ)) = 0, ρϕ_{p}(u(1)) + τϕ_{p}(u′(η)) = 0.  (1.2)

where ϕ_{p}(s) is a p-Laplacian operator, i.e., ϕ_{p}(s) = |s|^{p−2}s, p > 1, ϕ_{q} = (ϕ_{p})^{−1}, \frac{1}{q} + \frac{1}{p} = 1, μ > 0, ω ≥ 0, ρ > 0, τ ≥ 0, ξ, η ∈ (0, 1) is prescribed and ξ < η, e : (0, 1) → [0, ∞), f : [0, +∞) → [0, +∞).

In recent years, because of the wide mathematical and physical background [1,2,12], the existence of positive solutions for nonlinear boundary value problems with p-Laplacian has received wide attention. There exists a very large number of papers devoted to the existence of solutions of the p-Laplacian operators with two or three-point boundary conditions, for example,

u(0) = 0, u(1) = 0,
u(0) − B_{0}(u′(0)) = 0, u(1) + B_{1}(u′(1)) = 0,

∗Supported by the NNSF of China (No. 10571050), the Key Project of Chinese Ministry of Education and Innovation Foundation of Hunan Provincial Graduate Student

†The corresponding author. Email: liangruixi123@yahoo.com.cn

u(0) − B_{0}(u′(0)) = 0, u′(1) = 0,
u′(0) = 0, u(1) + B_{1}(u′(1)) = 0,

and

u(0) = 0, u(1) = u(η),
u(0) − B_{0}(u′(η)) = 0, u(1) + B_{1}(u′(1)) = 0,
u(0) − B_{0}(u′(0)) = 0, u(1) + B_{1}(u′(η)) = 0,
au(r) − bp(r)u′(r) = 0, cu(R) + dp(R)u′(R) = 0.

For further knowledge, see [3-11,13]. The methods and techniques employed in these papers involve the use of Leray-Shauder degree theory [4], the upper and lower solution method [5], fixed point theorem in a cone [3,6-8,10,11,13], and the quadrature method [9]. However, there are several papers dealing with the existence of positive solutions for four-point boundary value problem [13-15,18].

Motivated by results in [14], this paper is concerned with the existence of two positive solutions of the boundary value problem (1.1)-(1.2). Our tool in this paper will be a new double fixed point theorem in a cone [11,16,17,19] . The result obtained in this paper is essentially different from the previous results in [14].

In the rest of the paper, we make the following assumptions:

(H1) f ∈ C([0, +∞), [0, +∞));

(H2) e(t) ∈ C((0, 1), [0, +∞)), and 0 < ∫_{0}^{1} e(t)dt < ∞. Moreover, e(t) does not vanish identically on any subinterval of (0, 1).

Define

f_{0} = lim_{u→0^{+}} \frac{f(u)}{u^{p−1}} , f_{∞} = lim_{u→∞} \frac{f(u)}{u^{p−1}} .

2. Some background definitions

In this section we provide some background material from the theory of cones in Banach space, and we state a two fixed point theorem due to Avery and Henderson [19].

If P ⊂ E is a cone, we denote the order induced by P on E by ≤. That is

x ≤ y if and only if y − x ∈ P.

Definition 2.1 Given a cone P in a real Banach spaces E, a functional ψ : P → R is said to be increasing on P, provided ψ(x) ≤ ψ(y), for all x, y ∈ P with x ≤ y.

Definition 2.2 Given a nonnegative continuous functional γ on a cone P of a real Banach space E(i.e., γ : P → [0, +∞) continuous), we define, for each d > 0, the set

P (γ, d) = {x ∈ P |γ(x) < d}.

In order to obtain multiple positive solutions of (1.1)-(1.2), the following fixed point theorem of Avery and Henderson will be fundamental.

Theorem 2.1 [19] Let P be a cone in a real Banach space E. Let α and γ be increasing, nonnegative continuous functional on P, and let θ be a nonnegative continuous functional on P with θ(0) = 0 such that, for some c > 0 and M > 0,

γ(x) ≤ θ(x) ≤ α(x), and ∥x∥ ≤ Mγ(x)

for all x ∈ \overline{P (γ, c)}. Suppose there exist a completely continuous operator Φ : \overline{P (γ, c)} → P and 0 < a < b < c such that

θ(λx) ≤ λθ(x) for 0 ≤ λ ≤ 1 and x ∈ ∂P (θ, b),

and (i) γ(Φx) < c, for all x ∈ ∂P (γ, c), (ii) θ(Φx) > b for all x ∈ ∂P (θ, b), (iii) P (α, a) ≠ ∅ and α(Φx) < a, for x ∈ ∂P (α, a). Then Φ has at least two fixed points x_{1} and x_{2} belonging to \overline{P (γ, c)} satisfying

a < α(x_{1}) with θ(x_{1}) < b,

and

b < θ(x_{2}) with γ(x_{2}) < c.

3. Existence of two positive solutions of (1.1)-(1.2)

In this section, by defining an appropriate Banach space and cones, we impose growth conditions on f which allow us to apply the above two fixed point theorem in establishing the existence of double positive solutions of (1.1)-(1.2). Firstly, we mention without proof several fundamental results.

Lemma 3.1 [Lemma 2.1, 14]. If condition (H2) holds, then there exists a constant δ ∈ (0, \frac{1}{2}) that satisfies

0 < ∫_{δ}^{1−δ} e(t)dt < ∞.

Furthermore, the function:

y_{1}(t) = ∫_{δ}^{t} ϕ_{q}( ∫_{s}^{t} e(r)dr)ds + ∫_{t}^{1−δ} ϕ_{q}( ∫_{t}^{s} e(r)dr)ds, t ∈ [δ, 1 − δ],

is a positive continuous function on [δ, 1 − δ]. Therefore y_{1}(t) has a minimum on [δ, 1 − δ], so it follows that there exists L_{1} > 0 such that

min_{t∈[δ,1−δ]} y_{1}(t) = L_{1}.

If E = C[0, 1], then E is a Banach space with the norm ∥u∥ = sup_{t∈[0,1]} |u(t)|. We note that, from the nonnegativity of e and f, a solution of (1.1)-(1.2) is nonnegative and concave on [0, 1]. Define

P = {u ∈ E : u(t) ≥ 0, u(t) is concave function, t ∈ [0, 1]}.

Lemma 3.2 [Lemma 2.2, 14]. Let u ∈ P and δ be as Lemma 3.1, then

u(t) ≥ δ∥u∥, t ∈ [δ, 1 − δ].

Lemma 3.3 [Lemma 2.3, 14]. Suppose that conditions (H1), (H2) hold. Then u(t) ∈ E ∩C^{2}(0, 1) is a solution of boundary value problem (1.1)-(1.2) if and only if u(t) ∈ E is a solution of the following integral equation:

u(t) = \{ \begin{matrix}ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{σ} e(r)f(u(r))dr) + ∫_{0}^{t} ϕ_{q}( ∫_{s}^{σ} e(r)f(u(r))dr)ds, 0 ≤ t ≤ σ, \\ ϕ_{q}(\frac{τ}{ρ} ∫_{σ}^{η} e(r)f(u(r))dr) + ∫_{t}^{1} ϕ_{q}( ∫_{σ}^{s} e(r)f(u(r))dr)ds, σ ≤ t ≤ 1, \end{matrix}

where σ ∈ [ξ, η] ⊂ (0, 1) and u′(σ) = 0.

By means of the well known Guo-Krasnoselskii fixed point theorem in a cone, Su et al. [14] established the existence of at least one positive solution for (1.1)-(1.2) under some superlinear and sublinear assumptions imposed on the nonlinearity of f, which can be listed as

(i) f_{0} = 0 and f_{∞} = +∞ (superlinear), or

(ii) f_{0} = +∞ and f_{∞} = 0 (sublinear).

Using the same theorem, the authors also proved the existence of two positive solutions of (1.1)-(1.2) when f satisfies

(iii) f_{0} = f_{∞} = 0, or

(iv) f_{0} = f_{∞} = +∞.

When f_{0}, f_{∞} ∉ {0, +∞}, set

θ^{∗} = \frac{2}{L_{1}} , θ_{∗} = \frac{1}{(1 + ϕ_{q}(\frac{ω}{μ}))ϕ_{q}( ∫_{0}^{1} e(r)dr)},

and in the following, always assume δ be as in Lemma 3.1, the existence of double positive solutions of boundary value problem (1.1)-(1.2) can be list as follows:

Theorem 3.1 [Theorem 4.3, 14]. Suppose that conditions (H1),(H2) hold. Also assume that f satisfies

(A1) f_{0} = λ_{1} ∈ [0, (\frac{θ_{∗}}{4} )^{p−1});

(A2) f_{∞} = λ_{2} ∈ [0, (\frac{θ_{∗}}{4} )^{p−1});

(A3) f(u) ≤ (MR)^{p−1}, 0 ≤ u ≤ R, where M ∈ (0, θ_{∗}). Then the boundary value problem (1.1)-(1.2) has at least two positive solutions u_{1}, u_{2} such that

0 < ∥u_{1}∥ < R < ∥u_{2}∥.

Theorem 3.2 [Theorem 4.4, 14]. Suppose that conditions (H1),(H2) hold. Also assume that f satisfies

(A4) f_{0} = λ_{1} ∈ [(\frac{2θ^{∗}}{δ} )^{p−1}, ∞); (A5) f_{∞} = λ_{2} ∈ [(\frac{2θ^{∗}}{δ} )^{p−1}, ∞); (A6) f(u) ≥ (mr)^{p−1}, δr ≤ u ≤ r,

where m ∈ (θ^{∗}, ∞). Then the boundary value problem (1.1)-(1.2) has at least two positive solutions u_{1}, u_{2} such that

0 < ∥u_{1}∥ < r < ∥u_{2}∥.

When we see such a fact, we cannot but ask “Whether or not we can obtain a similar conclusion if neither f_{0} ∈ [(\frac{2θ^{∗}}{δ})^{p−1}, ∞) nor f_{0} ∈ [0, (\frac{θ_{∗}}{4} )^{p−1}).” Motivated by the above mentioned results, in this paper, we attempt to establish simple criteria for the existence of at least two positive solutions of (1.1)-(1.2). Our result is based on Theorem 2.1 and gives a positive answer to the question stated above.

Set

y_{2}(t) := ϕ_{q}( ∫_{δ}^{t} e(r)dr) + ϕ_{q}( ∫_{t}^{1−δ} e(r)dr), δ ≤ t ≤ 1 − δ.

For notational convenience, we introduce the following constants:

L_{2} = min_{δ≤t≤1−δ} y_{2}(t),

and

L_{3} = δϕ_{q}( ∫_{0}^{1} e(r)dr) + max \{ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{η} e(r)dr), ϕ_{q}(\frac{τ}{ρ} ∫_{ξ}^{η} e(r)dr)\},

Q = ϕ_{q}( ∫_{0}^{1} e(r)dr) + max \{ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{η} e(r)dr), ϕ_{q}(\frac{τ}{ρ} ∫_{ξ}^{η} e(r)dr)\}.

Finally, we define the nonnegative, increasing continuous functions γ, θ and α by

γ(u) = min_{t∈[δ,1−δ]} u(t),

θ(u) = \frac{1}{2}[u(δ) + u(1 − δ)], α(u) = max_{0≤t≤1} u(t).

We observe here that, for every u ∈ P,

γ(u) ≤ θ(u) ≤ α(u).

It follows from Lemma 3.2 that, for each u ∈ P, one has γ(u) ≥ δ∥u∥, so ∥u∥ ≤ \frac{1}{δ}γ(u), for all u ∈ P. We also note that θ(λu) = λθ(u), for 0 ≤ λ ≤ 1, and u ∈ ∂P (θ, b).

The main result of this paper is as follows:

Theorem 3.3 Assume that (H_{1}) and (H_{2}) hold, and suppose that there exist positive constants 0 < a < b < c such that 0 < a < δb < \frac{δ^{2}L_{2}}{2L_{3}} c, and f satisfies the following conditions

(D1) f(v) < ϕ_{p}(\frac{a}{Q}), if 0 ≤ v ≤ a;

(D2) f(v) > ϕ_{p}(\frac{2b}{δL_{2}} ), if δb ≤ v ≤ \frac{b}{δ};

(D3) f(v) < ϕ_{p}(\frac{c}{L_{3}} ), if 0 ≤ v ≤ \frac{c}{δ}; Then, the boundary value problem (1.1) and (1.2) has at least two positive solutions u_{1} and u_{2} such that

a < max_{t∈[0,1]} u_{1}(t), with \frac{1}{2}[u_{1}(δ) + u_{1}(1 − δ)] < b;

and

b < \frac{1}{2}[u_{2}(δ) + u_{2}(1 − δ)], with min_{t∈[δ,1−δ]} u_{2}(t) < c.

Proof. We define the operator: Φ : P → P,

(Φu)(t) := \{ \begin{matrix}ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{σ} e(r)f(u(r))dr) + ∫_{0}^{t} ϕ_{q}( ∫_{s}^{σ} e(r)f(u(r))dr)ds, 0 ≤ t ≤ σ, \\ ϕ_{q}(\frac{τ}{ρ} ∫_{σ}^{η} e(r)f(u(r))dr) + ∫_{t}^{1} ϕ_{q}( ∫_{σ}^{s} e(r)f(u(r))dr)ds, σ ≤ t ≤ 1, \end{matrix}

for each u ∈ P, where σ ∈ [ξ, η] ⊂ (0, 1). It is shown in Lemma 3.3 that the operator Φ : P → P is well defined with ∥Φu∥ = Φu(σ). In particular, if u ∈ P (γ, c), we also have Φu ∈ P, moreover, a standard argument shows that Φ : P → P is completely continuous (see [Lemma 2.4, 14]) and each fixed point of Φ in P is a solution of (1.1)-(1.2).

We now show that the conditions of Theorem 2.1 are satisfied.

To fulfill property (i) of Theorem 2.2, we choose u ∈ ∂P (γ, c), thus γ(u) = min_{t∈[δ,1−δ]} u(t) = c. Recalling that ∥u∥ ≤ \frac{1}{δ}γ(u) = \frac{c}{δ}, we have

0 ≤ u(t) ≤ ∥u∥ ≤ \frac{1}{δ}γ(u) = \frac{c}{δ} , 0 ≤ t ≤ 1.

Then assumption (D3) of Theorem 3.2 implies

f(u(t)) < ϕ_{p}(\frac{c}{L_{3}} ), 0 ≤ t ≤ 1.

(i) If σ ∈ (0, δ), we have

γ(Φu) = min_{t∈[δ,1−δ]}(Φu)(t) = (Φu)(1 − δ)

= ϕ_{q}(\frac{τ}{ρ} ∫_{σ}^{η} e(r)f(u(r))dr) + ∫_{1−δ}^{1} ϕ_{q}( ∫_{σ}^{s} e(r)f(u(r))dr)ds

≤ ϕ_{q}(\frac{τ}{ρ} ∫_{ξ}^{η} e(r)f(u(r))dr) + ∫_{1−δ}^{1} ϕ_{q}( ∫_{0}^{1} e(r)f(u(r))dr)ds

≤ [ϕ_{q}(\frac{τ}{ρ} ∫_{ξ}^{η} e(r)dr) + δϕ_{q}( ∫_{0}^{1} e(r)dr)] · \frac{c}{L_{3}} < c.

(ii) If σ ∈ [δ, 1 − δ], we have

γ(Φu) = min_{t∈[δ,1−δ]}(Φu)(t) = min{(Φu)(δ), (Φu)(1 − δ)}

= min \{ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{σ} e(r)f(u(r))dr) + ∫_{0}^{δ} ϕ_{q}( ∫_{s}^{σ} e(r)f(u(r))dr)ds,

ϕ_{q}(\frac{τ}{ρ} ∫_{σ}^{η} e(r)f(u(r))dr) + ∫_{1−δ}^{1} ϕ_{q}( ∫_{σ}^{s} e(r)f(u(r))dr)ds\}

≤ max \{ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{η} e(r)f(u(r))dr) + ∫_{0}^{δ} ϕ_{q}( ∫_{0}^{1} e(r)f(u(r))dr)ds,

ϕ_{q}(\frac{τ}{ρ} ∫_{ξ}^{η} e(r)f(u(r))dr) + ∫_{1−δ}^{1} ϕ_{q}( ∫_{0}^{1} e(r)f(u(r))dr)ds\}

< [ max \{ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{η} e(r)dr), ϕ_{q}(\frac{τ}{ρ} ∫_{ξ}^{η} e(r)dr)\} + δϕ_{q}( ∫_{0}^{1} e(r)dr)] · \frac{c}{L_{3}}

= c.

(iii) If σ ∈ (1 − δ, 1), we have

γ(Φu) = min_{t∈[δ,1−δ]}(Φu)(t) = Φu(δ)

= ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{σ} e(r)f(u(r))dr) + ∫_{0}^{δ} ϕ_{q}( ∫_{s}^{σ} e(r)f(u(r))dr)ds

≤ ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{η} e(r)f(u(r))dr) + ∫_{0}^{δ} ϕ_{q}( ∫_{0}^{1} e(r)f(u(r))dr)ds

≤ [ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{η} e(r)dr) + δϕ_{q}( ∫_{0}^{1} e(r)dr)] · \frac{c}{L_{3}}

< c.

Therefore, condition (i) of Theorem 2.2 is satisfied. We next address (ii) of Theorem 2.2. For this, we choose u ∈ ∂P (θ, b) so that θ(u) =

\frac{1}{2}[u(δ) + u(1 − δ)] = b. Noting that

∥u∥ ≤ (1/δ)γ(u) ≤ (1/δ)θ(u) = b/δ,

we have

δb < δ∥u∥ ≤ u(t) ≤ \frac{b}{δ} , for t ∈ [δ, 1 − δ].

Then (D2) yields

f(u(t)) > ϕ_{p}(\frac{2b}{δL_{2}} ), for t ∈ [δ, 1 − δ].

As Φu ∈ P :

(i) If σ ∈ (0, δ), we have

θ(Φu) = \frac{1}{2}(Φu(δ) + Φu(1 − δ)) ≥ Φu(1 − δ)

= ϕ_{q}(\frac{τ}{ρ} ∫_{σ}^{η} e(r)f(u(r))dr) + ∫_{1−δ}^{1} ϕ_{q}( ∫_{σ}^{s} e(r)f(u(r))dr)ds

≥ ∫_{1−δ}^{1} ϕ_{q}( ∫_{σ}^{1−δ} e(r)f(u(r))dr)ds

≥ ∫_{1−δ}^{1} ϕ_{q}( ∫_{δ}^{1−δ} e(r)f(u(r))dr)ds

= δϕ_{q}( ∫_{δ}^{1−δ} e(r)f(u(r))dr)

≥ δϕ_{q}( ∫_{δ}^{1−δ} e(r)dr) · \frac{2b}{δL_{2}} ≥ 2b > b.

(ii) If σ ∈ [δ, 1 − δ], we have

2θ(Φu) = [Φu(δ) + Φu(1 − δ)]

≥ ∫_{0}^{δ} ϕ_{q}( ∫_{s}^{σ} e(r)f(u(r))dr)ds + ∫_{1−δ}^{1} ϕ_{q}( ∫_{σ}^{s} e(r)f(u(r))dr)ds

≥ ∫_{0}^{δ} ϕ_{q}( ∫_{δ}^{σ} e(r)f(u(r))dr)ds + ∫_{1−δ}^{1} ϕ_{q}( ∫_{σ}^{1−δ} e(r)f(u(r))dr)ds

= δ[ϕ_{q}( ∫_{δ}^{σ} e(r)f(u(r))dr) + ϕ_{q}( ∫_{σ}^{1−δ} e(r)f(u(r))dr)]

≥ δ[ϕ_{q}( ∫_{δ}^{σ} e(r)dr) + ϕ_{q}( ∫_{σ}^{1−δ} e(r)dr)] · \frac{2b}{δL_{2}}

≥ 2b.

(iii) If σ ∈ (1 − δ, 1), we have θ(Φu) = \frac{1}{2}(Φu(δ) + Φu(1 − δ)) ≥ Φu(δ)

= ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{σ} e(r)f(u(r))dr) + ∫_{0}^{δ} ϕ_{q}( ∫_{s}^{σ} e(r)f(u(r))dr)ds

≥ ∫_{0}^{δ} ϕ_{q}( ∫_{δ}^{1−δ} e(r)f(u(r))dr)ds

> δϕ_{q}( ∫_{δ}^{1−δ} e(r)dr) · \frac{2b}{δL_{2}} ≥ 2b > b.

Hence, condition (ii) of Theorem 2.2 holds.

To fulfill property (iii) of Theorem 2.2, we note u_{∗}(t) ≡ a/2, 0 ≤ t ≤ 1, is a member of P (α, a) and α(u_{∗}) = a/2, so P (α, a) ≠ 0. Now, choose u ∈ ∂P (α, a), so that α(u) = max_{t∈[0,1]} u(t) = a and implies 0 ≤ u(t) ≤ a, 0 ≤ t ≤ 1. It follows from assumption (D1), f(u(t)) ≤ ϕ_{p}(a/Q), t ∈ [0, 1]. As before we obtain

α(Φu) = ∥Φu∥ = Φu(σ)

= ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{σ} e(r)f(u(r))dr) + ∫_{0}^{σ} ϕ_{q}( ∫_{s}^{σ} e(r)f(u(r))dr)ds

= ϕ_{q}(\frac{τ}{ρ} ∫_{σ}^{η} e(r)f(u(r))dr) + ∫_{σ}^{1} ϕ_{q}( ∫_{σ}^{s} e(r)f(u(r))dr)ds

≤ max \{ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{η} e(r)dr) + ϕ_{q}( ∫_{0}^{1} e(r)dr),

ϕ_{q}(\frac{τ}{ρ} ∫_{ξ}^{η} e(r)dr) + ϕ_{q}( ∫_{0}^{1} e(r)dr)\} · \frac{a}{Q}

≤ a.

Thus, condition (iii) of Theorem 2.1 is also satisfied. Consequently, an application of Theorem 2.1 completes the proof. □

Finally, we present an example to explain our result.

Example. Consider the boundary value problem (1.1)-(1.2) with

p = \frac{3}{2}, μ = 2, ρ = ω = 1, ξ = \frac{1}{4}, η = \frac{1}{2}, τ = 1, δ = \frac{1}{4}, e(t) = t^{− \frac{1}{2}} ,

and

f(u) = \{ \begin{matrix}\frac{6\sqrt{2u}}{u + 1}, & 0 ≤ u ≤ 200, \\ \frac{40}{67} + \frac{1202}{335}(u − 200), & 200 ≤ u ≤ 250, \\ 180, & 250 < u, \end{matrix}

Then (1.1)-(1.2) has at least two positive solutions. Proof. In this example we have

L_{1} = min_{1/4≤x≤3/4} \{ ∫_{1/4}^{x} ϕ_{q}(∫_{s}^{x} t^{−1/2}dt)ds + ∫_{x}^{3/4} ϕ_{q}(∫_{x}^{s} t^{−1/2}dt)ds\} = \frac{3\sqrt{3} − 5}{9},

L_{2} = min_{1/4≤x≤3/4} (ϕ_{q}(∫_{1/4}^{x} t^{−1/2}dt) + ϕ_{q}(∫_{x}^{3/4} t^{−1/2}dt)) = 2 − \sqrt{3},

L_{3} = δϕ_{q}( ∫_{0}^{1} e(r)dr) + max \{ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{η} e(r)dr), ϕ_{q}(\frac{τ}{ρ} ∫_{ξ}^{η} e(r)dr)\} = 4 − 2\sqrt{2},

Q = ϕ_{q}( ∫_{0}^{1} e(r)dr) + max \{ϕ_{q}(\frac{ω}{μ} ∫_{ξ}^{η} e(r)dr), ϕ_{q}(\frac{τ}{ρ} ∫_{ξ}^{η} e(r)dr)\} = 7 − 2\sqrt{2}.

Let a = 80, b = 1000, c = 40000. Then we have

f(u) = \frac{6\sqrt{2u}}{u + 1} < ϕ_{p}(a/Q), for 0 ≤ u ≤ 80,

f(u) = 180 > ϕ_{p}((2b)/(δL_{2})), for 250 ≤ u ≤ 4000,

f(u) = 180 < ϕ_{p}(c/L_{3}), for 0 ≤ u ≤ 160000.

Therefore, by Theorem 3.3 we deduce that (1.1)-(1.2) has at least two positive solutions u_{1} and u_{2} satisfying

80 < max_{t∈[0,1]} u_{1}(t), with \frac{1}{2}[u_{1}(δ) + u_{1}(1 − δ)] < 1000;

and

1000 < \frac{1}{2}[u_{2}(δ) + u_{2}(1 − δ)], with min_{t∈[δ,1−δ]} u_{2}(t) < 40000.

Remark. We notice that in the above example, f_{0} = 6\sqrt{2} ≈ 8.48528, (\frac{θ_{∗}}{4} )^{p−1} = \frac{\sqrt{5}}{10} ≈ 0.223607

and (\frac{2θ^{∗}}{δ})^{p−1} = 6\sqrt{10 + 6\sqrt{3}} ≈ 27.0947. Therefore, Theorem 3.1 and Theorem 3.2 are not applicable to this example since conditions (A1) and (A4) fail.

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(Received January 16, 2008)