Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 42, 1-16 ; http://www.math.u-szeged.hu/ejqtde/

Positive solutions of singular four-point boundary

value problem with p-Laplacian ^{∗}

Chunmei Miao^{1,2}^{†} Huihui Pang^{3} Weigao Ge^{2}

1 College of Science, Changchun University, Changchun 130022, PR China. 2 Department of Mathematics, Beijing Institute of Technology, Beijing 100081, PR China.

3 College of Science, China Agricultural University, Beijing 100083, PR China. Email: miaochunmei@yahoo.com.cn, phh2000@163.com , gew@bit.edu.cn

Abstract

In this paper, we deal with the following singular four-point boundary value problem with p-Laplacian

\{ \begin{matrix}(ϕ_{p}(u′(t)))′ + q(t)f(t, u(t)) = 0, t ∈ (0, 1), \\ u(0) − αu′(ξ) = 0, u(1) + βu′(η) = 0, \end{matrix}

where f(t, u) may be singular at u = 0 and q(t) may be singular at t = 0 or 1. By imposing some suitable conditions on the nonlinear term f, existence results of at least two positive solutions are obtained. The proof is based upon theory of Leray-Schauder degree and Krasnosel’skii’s fixed point theorem.

Keywords. Singular, Four-point boundary value problem, Multiple positive solutions, p-Laplacian, Leray-Schauder degree, Fixed point theorem

2000 Mathematics Subject Classification. 34B10, 34B16, 34B18

1 Introduction

Krasnosel’skii M A, Positive solutions of operator equations Singular boundary value problems (BVPs) arise in a variety of problems in applied mathematics and physics such as gas dynamics, nuclear physics, chemical reactions, studies of atomic structures, and atomic calculations [1]. They also arise in the study of positive radial solutions of a nonlinear elliptic equations. Such problems have been studied extensively in recent years, see, for instance, [2-8] and references therein. At the very beginning, most literature in this area concentrated on singular two-point boundary value problems. More recently, several authors begin to pay attention to singular multi-point boundary value problems [9-17]. In the existing literatures, the following multi-point boundary conditions

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

†Corresponding author.

\begin{matrix}u(0) = 0, u(1) = βu(η); & u(0) = αu(ξ), u(1) = 0; \\ u(0) = 0, u(1) = ∑_{i=1}^{m−2} β_{i}u(η_{i}); & u(0) = ∑_{i=1}^{m−2} α_{i}u(ξ_{i}), u(1) = 0; \\ u′(0) = 0, u(1) = ∑_{i=1}^{m−2} β_{i}u(η_{i}); & u(0) = ∑_{i=1}^{m−2} α_{i}u(ξ_{i}), u′(1) = 0; \\ u′(0) = 0, u(1) = u(η); & u(0) = ∑_{i=1}^{m−2} α_{i}u(ξ_{i}), u′(1) = ∑_{i=1}^{m−2} β_{i}u′(η_{i}); \\ u(0) = αu(ξ), u(1) = βu(η); & u(0) = ∑_{i=1}^{m−2} α_{i}u(ξ_{i}), u(1) = ∑_{i=1}^{m−2} β_{i}u(η_{i}), \end{matrix}

where α, β, α_{i}, β_{i} > 0, 0 < ξ, η, ξ_{i}, η_{i} < 1(i = 1, 2, ⋯, m − 1), have been studied.

However, to our knowledge, there are few papers investigating the singular four-point boundary value problem. The aim of the present paper is to fill this gap.

In this paper, we establish sufficient conditions which guarantee the existence theory for single and multiple positive solutions to the following singular four-point BVP

\{ \begin{matrix}(ϕ_{p}(u′(t)))′ + q(t)f(t, u(t)) = 0, t ∈ (0, 1), \\ u(0) − αu′(ξ) = 0, u(1) + βu′(η) = 0, \end{matrix}  (1.1)

where ϕ_{p}(s) = |s|^{p−2}s, p > 1, (ϕ_{p})^{−1} = ϕ_{q}, \frac{1}{p} + \frac{1}{q} = 1, α > 0, β > 0, 0 < ξ < η < 1. f(t, u) may be singular at u = 0 and q(t) may be singular at t = 0 or 1.

Existence results for one solution are obtained by using the existence principle guaranteed by the property of Leray-Schauder degree, and for two solutions by using a fixed point theorem in cones. In order to obtain the positivity of solution, u(0) ≥ 0, u(1) ≥ 0 is required. While note the boundary conditions u(0) = αu′(ξ), u(1) = −βu′(η), so we need to ensure u′(ξ) ≥ 0, u′(η) ≤ 0. Hence it is vital that the maximum of this solution must be achieved between ξ and η. To this end, we need to establish some suitable conditions on the nonlinear term f (see (H_{4})), in course of that, we overcome more difficulties since the singularity of the nonlinear term f.

Throughout, we always suppose the following conditions are satisfied:

(H_{1a}) q ∈ ℂ(0, 1) ∩ 𝕃^{1}[0, 1] with q(t) ≥ 0, q(t) ≢ 0 on any subinterval of [0, 1] and nondecreasing on (0, 1);

(H_{1b}) q ∈ ℂ(0, 1) ∩ 𝕃^{1}[0, 1] with q(t) ≥ 0, q(t) ≢ 0 on any subinterval of [0, 1] and nonincreasing on (0, 1);

(H_{2}) f : [0, 1] × (0, +∞) → (0, +∞) is a continuous function;

(H_{3}) 0 < f(t, u) ≤ f_{1}(u) + f_{2}(u) on [0, 1] × (0, +∞) with f_{1} > 0 continuous, nonincreasing on (0, +∞) and ∫_{0}^{L} f_{1}(u)du < +∞ for any fixed L > 0; f_{2} ≥ 0 is continuous on [0, +∞); \frac{f_{2}}{f_{1}} nondecreasing on (0, +∞);

(H_{4}) There exists R > 0 such that

∫_{0}^{ξ} q(t)N(t)dt ≤ Γ ∫_{ξ}^{η} q(t)n(t)dt, ∫_{η}^{1} q(t)N(t)dt ≤ Γ ∫_{ξ}^{η} q(t)n(t)dt,

where Γ = (min\{\frac{α}{1−η} + \frac{ξ}{1−η} , \frac{β}{ξ} + \frac{1−η}{ξ} \})^{p−1}, n(t) = inf_{u∈(0,R]}\{f(t, u), t ∈ [ξ, η]\}, N(t) =

sup_{u∈(0,R]}{f(t, u), t ∈ [0, ξ] ∪ [η, 1]};

(H_{5}) For each constant H > 0, there exists a function ψ_{H} continuous on [0, 1] and positive on (0, 1) such that f(t, u) ≥ ψ_{H}(t) on (0, 1) × (0, H] ;

(H_{6}) ∫_{0}^{ξ} f_{1}(k_{1}s)q(s)ds + ∫_{η}^{1} f_{1}(k_{2}(1 − s))q(s)ds < +∞ for any k_{1} > 0, k_{2} > 0.

2 Preliminaries

Consider the Banach space X = ℂ[0, 1] with the maximum norm ||u|| = max_{t∈[0,1]} |u(t)|. By a positive solution u(t) to BVP(1.1) we mean that u(t) satisfies (1.1), u ∈ ℂ^{1}[0, 1], (ϕ_{p}(u′))′ ∈ ℂ(0, 1) ∩ 𝕃^{1}[0, 1] and u(t) > 0 on (0, 1).

We suppose F : [0, 1] × ℝ → (0, +∞) is continuous, q ∈ ℂ(0, 1) ∩ 𝕃^{1}[0, 1] with q(t) ≥ 0 on (0, 1) and q(t) ≢ 0 on any subinterval of [0, 1]. For any x ∈ X, we consider the following BVP

\{ \begin{matrix}(ϕ_{p}(u′(t)))′ + q(t)F (t, x(t)) = 0, t ∈ (0, 1), \\ u(0) − αu′(ξ) = a, u(1) + βu′(η) = a, \end{matrix}  (2.1)

where a is a fixed positive constant. Then we have

Lemma 2.1. ([18]) For any x ∈ X, BVP (2.1) has a unique solution u(t) which can be expressed as

u(t) = \{ \begin{matrix}αϕ_{q}(∫_{ξ}^{σ} q(s)F (s, x(s))ds) + ∫_{0}^{t} ϕ_{q}(∫_{s}^{σ} q(τ)F (τ, x(τ))dτ)ds + a, 0 ≤ t ≤ σ, \\ βϕ_{q}(∫_{σ}^{η} q(s)F (s, x(s))ds) + ∫_{t}^{1} ϕ_{q}(∫_{σ}^{s} q(τ)F (τ, x(τ))dτ)ds + a, σ ≤ t ≤ 1, \end{matrix}

(2.2) where σ is the unique solution of the equation υ_{1}(t) − υ_{2}(t) = 0, 0 < t < 1, in which

\begin{matrix}υ_{1}(t) = αϕ_{q}(∫_{ξ}^{σ} q(s)F (s, x(s))ds) + ∫_{0}^{t} ϕ_{q}(∫_{s}^{σ} q(τ)F (τ, x(τ))dτ)ds, \\ υ_{2}(t) = βϕ_{q}(∫_{σ}^{η} q(s)F (s, x(s))ds) + ∫_{t}^{1} ϕ_{q}(∫_{σ}^{s} q(τ)F (τ, x(τ))dτ)ds. \end{matrix}  (2.3)

Lemma 2.2. For any x ∈ X, assume that

∫_{0}^{ξ} q(t)F (t, x(t))dt ≤ Γ ∫_{ξ}^{η} q(t)F (t, x(t))dt  (2.4)

and

∫_{η}^{1} q(t)F (t, x(t))dt ≤ Γ ∫_{ξ}^{η} q(t)F (t, x(t))dt,  (2.5)

where Γ = (min\{\frac{α}{1−η} + \frac{ξ}{1−η} , \frac{β}{ξ} + \frac{1−η}{ξ} \})^{p−1}. If u(t) is a solution of BVP (2.1), then (i) u(t) is concave on [0, 1]; (ii) There exists σ ∈ [ξ, η] such that u′(σ) = 0, u(σ) = ||u||; u(t) ≥ a for t ∈ [0, 1], u(t)

is nondecreasing on [0, ξ] and nonincreasing on [η, 1]; (iii) u(t) ≥ ω(t)||u|| for t ∈ [0, 1], where ω(t) = min\{\frac{1}{η}t, \frac{1}{1−ξ} (1 − t)\}.

Proof. Suppose u(t) is a solution to BVP(2.1), then

(i) (ϕ_{p}(u′(t)))′ = −q(t)F (t, x(t)) ≤ 0, so ϕ_{p}(u′(t)) is nonincreasing on [0, 1]. Therefore, u′(t) is nonincreasing on [0, 1] which implies the concavity of u(t).

(ii) From Lemma 2.1, we know that there exists σ ∈ (0, 1) such that u′(σ) = 0. Now we show that σ ∈ [ξ, η]. If not, then there exists σ ∈ (0, ξ) such that u′(σ) = 0. By Lemma 2.1 and (2.4), we have

u(σ) = αϕ_{q}(∫_{ξ}^{σ} q(s)F (s, x(s))ds) + ∫_{0}^{σ} ϕ_{q}(∫_{s}^{σ} q(τ)F (τ, x(τ))dτ)ds
< ∫_{0}^{ξ} ϕ_{q}(∫_{0}^{ξ} q(τ)F (τ, x(τ))dτ)ds
≤ ξϕ_{q}(Γ ∫_{ξ}^{η} q(τ)F (τ, x(τ))dτ)
≤ βϕ_{q}(∫_{ξ}^{η} q(τ)F (τ, x(τ))dτ) + (1 − η)ϕ_{q}(∫_{ξ}^{η} q(s)F (s, x(s))ds)
≤ βϕ_{q}(∫_{σ}^{η} q(s)F (s, x(s))ds) + ∫_{σ}^{1} ϕ_{q}(∫_{σ}^{s} q(τ)F (τ, x(τ))dτ)ds
= u(σ)

which is a contradiction. So, σ ∉ (0, ξ). Similarly, σ ∉ (η, 1). The concavity of u(t) guarantees that u(t) is nondecreasing on [0, ξ] and nonincreasing on [η, 1]. By the boundary conditions, we have u(0) = αu′(ξ) + a ≥ a, u(1) = −βu′(η) + a ≥ a. Therefore, for any t ∈ [0, 1], u(t) ≥ 0 since u(t) is concave on [0,1].

EJQTDE, 2009 No. 42, p. 4

(iii) Since u(t) is nondecreasing on [0, σ] and nonincreasing on [σ, 1] for σ ∈ [ξ, η], we have

\begin{matrix}\frac{u(t)}{t} ≥ \frac{u(σ)}{σ} ≥ \frac{1}{η}||u||, & t ∈ [0, σ], \\ \frac{u(t)}{1 − t} ≥ \frac{u(σ)}{1 − σ} ≥ \frac{1}{1 − ξ} ||u||, & t ∈ [σ, 1]. \end{matrix}

Let ω(t) = min\{\frac{1}{η}t, \frac{1}{1−ξ} (1 − t)\}, it follows that

u(t) ≥ ω(t)||u||, t ∈ [0, 1].

The proof is completed.

We shall consider the following boundary value problem

\{ \begin{matrix}(ϕ_{p}(u′(t)))′ + q(t)F (t, u(t)) = 0, t ∈ (0, 1), \\ u(0) − αu′(ξ) = a, u(1) + βu′(η) = a. \end{matrix}  (2.6)

Define an operator T : X → X by

(T u)(t) = \{ \begin{matrix}αϕ_{q}(∫_{ξ}^{σ} q(s)F (s, u(s))ds) + ∫_{0}^{t} ϕ_{q}(∫_{s}^{σ} q(τ)F (τ, u(τ))dτ)ds + a, 0 ≤ t ≤ σ, \\ βϕ_{q}(∫_{σ}^{η} q(s)F (s, u(s))ds) + ∫_{t}^{1} ϕ_{q}(∫_{σ}^{s} q(τ)F (τ, u(τ))dτ)ds + a, σ ≤ t ≤ 1. \end{matrix}

(2.7)

We have the following result: Lemma 2.3. ([18], Lemma 3.1) T : X → X is completely continuous.

Now we state an existence principle which plays an important role in our proof of existence results for one solution. Lemma 2.4. (Existence principle) Assume that there exists a constant M > a independent of λ, such that for λ ∈ (0, 1), ||u|| ≠ M, where u(t) satisfies

\{ \begin{matrix}(ϕ_{p}(u′(t)))′ + λq(t)F (t, u(t)) = 0, t ∈ (0, 1), \\ u(0) − αu′(ξ) = a, u(1) + βu′(η) = a. \end{matrix}  (2.7)_{λ}

Then (2.7)_{1} has at least one solution u(t) with ||u|| ≤ M. Proof. For any λ ∈ [0, 1], define an operator

N_{λ}u(t) = \{ \begin{matrix}λαϕ_{q}(∫_{ξ}^{σ} q(s)F (s, u(s))ds) + ∫_{0}^{t} ϕ_{q}(∫_{s}^{σ} q(τ)λF (τ, u(τ))dτ)ds + a, 0 ≤ t ≤ σ, \\ λβϕ_{q}(∫_{σ}^{η} q(s)F (s, u(s))ds) + ∫_{t}^{1} ϕ_{q}(∫_{σ}^{s} q(τ)λF (τ, u(τ))dτ)ds + a, σ ≤ t ≤ 1. \end{matrix}

By Lemma 2.3, N_{λ} : X → X is completely continuous. It can be verified that a solution of BVP (2.7)_{λ} is equivalent to a fixed point of N_{λ} in X. Let Ω = {u ∈ X : ||u|| < M}, then Ω is an open set in X. If there exists u ∈ ∂Ω such that N_{1}u = u, then u(t) is a solution of (2.7)_{1} with ||u|| ≤ M. Thus the conclusion is true. Otherwise, for any u ∈ ∂Ω, N_{1}(u) ≠ u. If λ = 0, for u ∈ ∂Ω, (I − N_{0})u(t) = u(t) − N_{0}u(t) = u(t) − a ≢ 0 since ||u|| = M > a. For λ ∈ (0, 1), if there is a solution u(t) to BVP (2.7)_{λ}, by the assumption, one gets ||u|| ≠ M, which is a contradiction to u ∈ ∂Ω.

In a word, for any u ∈ ∂Ω and λ ∈ [0, 1], N_{λ}u ≠ u. Homotopy invariance of Leray-Schauder degree deduce that

Deg{I − N_{1}, Ω, 0} = Deg{I − N_{0}, Ω, 0} = 1.

Hence, N_{1} has a fixed point u in Ω. And BVP (2.7)_{1} has a solution u(t) with ||u|| ≤ M. The proof is completed.

To obtain two positive solutions of BVP (1.1), we need the following well-known fixed point theorem of compression and expansion of cones [19].

Theorem 2.5. (Krasnosel’skii [19, p.148]) Let X be a Banach space and P (⊂ X) be a cone. Assume that are Ω_{1}, Ω_{2} are open subsets of X with 0 ∈ Ω_{1}, \overline{Ω}_{1} ⊂ Ω_{2}, and let T : \overline{Ω}_{2}\Ω_{1} ∩ P → P be a continuous and compact operator such that either

(i)∥T x∥ ⩽ ∥x∥, ∀x ∈ ∂Ω_{1} ∩ P and ∥T x∥ ⩾ ∥x∥, ∀x ∈ ∂Ω_{2} ∩ P; or

(ii)∥T x∥ ⩽ ∥x∥, ∀x ∈ ∂Ω_{2} ∩ P and ∥T x∥ ⩾ ∥x∥, ∀x ∈ ∂Ω_{1} ∩ P. Then T has a fixed point theorem in (\overline{Ω}_{2} \ Ω_{1}) ∩ P.

3 Existence of positive solutions

First we give some notations.

Denote

G(c) = ∫_{0}^{c}[f_{1}(u) + f_{2}(u)]du, I(c) = ∫_{0}^{c} ϕ_{q}(t)dt = (\frac{p−1}{p})c^{\frac{p}{p−1}} for c > 0.

Clearly, G(c) is increasing in c, I^{−1}(c) exists and I^{−1}(c) = (\frac{p}{p−1})^{\frac{p−1}{p}} c^{\frac{p−1}{p}}. Since p > 1, (\frac{p}{p−1})^{\frac{p−1}{p}} > 1, we have I^{−1}(uυ) ≤ I^{−1}(u)I^{−1}(υ) for any u > 0, υ > 0. For c < 0, it is easy to see that I(−c) = I(c).

We now give the main results for BVP (1.1) in this paper.

Theorem 3.1. Suppose (H_{1a})-(H_{6}) hold. Furthermore, we assume that (H_{7}) there exists r > 0 such that

\frac{r}{M_{β,η}(r)} > 1,

where

M_{β,η}(r) = ϕ_{q}(I^{−1}(G(r)))[βϕ_{q}(I^{−1}(q(η))) + ∫_{0}^{1} ϕ_{q}(I^{−1}(q(t)))dt].

Then BVP (1.1) has a positive solution u(t) with ||u|| ≤ r.

Proof. From (H_{7}), we choose r > 0 and 0 < ε < r such that

\frac{r}{ε + M_{β,η}(r)} > 1.  (3.1)

Let n_{0} ∈ \{1, 2, 3, ⋯ \} satisfying that \frac{1}{n_{0}} ≤ ε. Set ℕ_{0} = \{n_{0}, n_{0} + 1, n_{0} + 2, ⋯ \}. In what follows, we show that the following BVP

\{ \begin{matrix}(ϕ_{p}(u′(t)))′ + q(t)f(t, u(t)) = 0, t ∈ (0, 1), \\ u(0) − αu′(ξ) = \frac{1}{m}, u(1) + βu′(η) = \frac{1}{m} \end{matrix}  (3.2)

has a solution for each m ∈ ℕ_{0}.

In order to obtain a solution of BVP (3.2) for each m ∈ ℕ_{0}, we consider the following BVP

\{ \begin{matrix}(ϕ_{p}(u′(t)))′ + q(t)f^{∗}(t, u(t)) = 0, t ∈ (0, 1), \\ u(0) − αu′(ξ) = \frac{1}{m}, u(1) + βu′(η) = \frac{1}{m}, \end{matrix}  (3.2)_{m}

where

f^{∗}(t, u) = \{ \begin{matrix}f(t, u), & u ≥ \frac{1}{m}, \\ f(t, \frac{1}{m}), & u ≤ \frac{1}{m}. \end{matrix}

Clearly, f^{∗} ∈ ℂ([0, 1] × ℝ, (0, +∞)).

To obtain a solution of BVP (3.2)_{m} for each m ∈ ℕ_{0}, by applying Lemma 2.4, we consider the family of BVPs

\{ \begin{matrix}(ϕ_{p}(u′(t)))′ + λq(t)f^{∗}(t, u(t)) = 0, t ∈ (0, 1), \\ u(0) − αu′(ξ) = \frac{1}{m}, u(1) + βu′(η) = \frac{1}{m}, \end{matrix}  (3.2)_{m}^{λ}

where λ ∈ [0, 1]. Let u(t) be a solution of (3.2)_{m}^{λ}. From (H_{4}) and Lemma 2.2, we observe that u(t) is concave, u(t) ≥ \frac{1}{m} on [0,1] and there exists σ̂ ∈ [ξ, η] such that u(σ̂) = ||u||, u′(σ̂) = 0, u′(t) ≥ 0, t ∈ [0, σ̂] and u′(t) ≤ 0, t ∈ [σ̂, 1].

For t ∈ [σ̂, 1] and λ ∈ (0, 1), in view of (H_{3}), we have

0 ≤ −(ϕ_{p}(u′(t)))′ = λq(t)f^{∗}(t, u(t)) = λq(t)f(t, u(t)) ≤ q(t)[f_{1}(u(t)) + f_{2}(u(t))]. (3.3)

EJQTDE, 2009 No. 42, p. 7 Multiplying (3.3) by −u′(t), for t ∈ [σ̂, 1], it follows that

(ϕ_{p}(u′(t)))′ϕ_{q}(ϕ_{p}(u′(t))) ≤ q(t)[f_{1}(u(t)) + f_{2}(u(t))](−u′(t)).  (3.4)

Integrating (3.4) from σ̂ to t (t ≥ σ̂), by (H_{1}) and the fact that G(c) is increasing on c, we obtain

∫_{0}^{ϕ_{p}(u′(t))} ϕ_{q}(z)dz ≤ q(t) ∫_{u(t)}^{u(σ̂)}[f_{1}(z) + f_{2}(z)]dz
= q(t)[G(u(σ̂)) − G(u(t))]
≤ q(t)G(u(σ̂)),

this implies

I(−ϕ_{p}(u′(t))) = I(ϕ_{p}(u′(t))) ≤ q(t)G(u(σ̂)),
0 ≤ −u′(t) ≤ ϕ_{q}(I^{−1}(q(t)))ϕ_{q}(I^{−1}(G(u(σ̂)))),  (3.5)

therefore,

0 ≤ −u′(η) ≤ ϕ_{q}(I^{−1}(q(η)))ϕ_{q}(I^{−1}(G(u(σ̂)))).

Integrating (3.5) from σ̂ to 1, by the boundary condition of (3.2)_{m}^{λ}, we have

\begin{aligned}u(σ̂) & ≤ \frac{1}{m} + ϕ_{q}(I^{−1}(G(u(σ̂))))[βϕ_{q}(I^{−1}(q(η))) + ∫_{0}^{1} ϕ_{q}(I^{−1}(q(t)))dt] \\ & ≤ ε + ϕ_{q}(I^{−1}(G(u(σ̂))))[βϕ_{q}(I^{−1}(q(η))) + ∫_{0}^{1} ϕ_{q}(I^{−1}(q(t)))dt]. \end{aligned}

Hence,

\frac{u(σ̂)}{ε + ϕ_{q}(I^{−1}(G(u(σ̂))))[βϕ_{q}(I^{−1}(q(η))) + ∫_{0}^{1} ϕ_{q}(I^{−1}(q(t)))dt]} ≤ 1,

i.e.,

\frac{u(σ̂)}{ε + M_{β,η}(u(σ̂))} ≤ 1.  (3.6)

Due to (3.1) and (3.6), we have u(σ̂) = ||u|| ≠ r. Further, Lemma 2.4 implies that (3.2)_{m} has at least a solution u^{m} ∈ ℂ^{1}[0, 1] and (ϕ_{p}((u^{m})′))′ ∈ ℂ(0, 1) ∩ 𝕃^{1}[0, 1] with ||u^{m}|| ≤ r (independent of m) for any fixed m. From Lemma 2.2, we note that u^{m}(t) ≥ \frac{1}{m} > 0. So f^{∗}(t, u^{m}(t)) = f(t, u^{m}(t)). Therefore, u^{m}(t) is a solution to BVP (3.2).

Using Arzelà-Ascoli theorem, we shall show that BVP (1.1) has at least a positive solution u(t) which satisfies lim_{m→∞} u^{m}(t) = u(t) for t ∈ (0, 1). Note that

0 < \frac{1}{m} ≤ u^{m}(t) ≤ r, t ∈ [0, 1].

(H_{5}) implies that there is a continuous function ψ_{r} : (0, 1) → (0, +∞)(independent of m) satisfying

f(t, u^{m}(t)) ≥ ψ_{r}(t), t ∈ (0, 1),

EJQTDE, 2009 No. 42, p. 8 hence,

−(ϕ_{p}(u^{m}(t))′)′ ≥ ψ_{r}(t)q(t), t ∈ (0, 1).  (3.7)

For any m ∈ ℕ_{0}, by Lemma 2.2, there exists t^{m} ∈ [ξ, η] such that u^{m}(t^{m}) = ||u^{m}||, (u^{m})′(t^{m}) = 0, (u^{m})′(t) ≥ 0 for t ∈ [0, t^{m}] and (u^{m})′(t) ≤ 0 for t ∈ [t^{m}, 1].

If t ∈ [0, ξ], integrating (3.7) from t to t^{m}, we obtain

ϕ_{p}((u^{m})′(t)) ≥ ∫_{t}^{t^{m}} q(s)ψ_{r}(s)ds.  (3.8)

Integrating (3.8) from 0 to t, we have

\begin{aligned}u^{m}(t) & ≥ \frac{1}{m} + ∫_{0}^{t} ϕ_{q}(∫_{s}^{t^{m}} q(τ)ψ_{r}(τ)dτ)ds \\ & ≥ ∫_{0}^{t} ϕ_{q}(∫_{s}^{ξ} q(τ)ψ_{r}(τ)dτ)ds, \end{aligned}

further,

u^{m}(ξ) ≥ ∫_{0}^{ξ} ϕ_{q}(∫_{s}^{ξ} q(τ)ψ_{r}(τ)dτ)ds = θ_{1} > 0.

Since u^{m}(t) is concave in [0, ξ], we have

\frac{u^{m}(t)}{t} ≥ \frac{u^{m}(ξ)}{ξ} ⇒ u^{m}(t) ≥ \frac{u^{m}(ξ)}{ξ}t ≥ \frac{θ_{1}}{ξ} t.  (3.9)

If t ∈ [η, 1], integrating (3.7) from t^{m} to t, we obtain

−ϕ_{p}((u^{m})′(t)) ≥ ∫_{t^{m}}^{t} q(s)ψ_{r}(s)ds.  (3.10)

Integrating (3.10) from t to 1, we have

\begin{aligned}u^{m}(t) & ≥ \frac{1}{m} + ∫_{t}^{1} ϕ_{q}(∫_{t^{m}}^{s} q(τ)ψ_{r}(τ)dτ)ds \\ & ≥ ∫_{t}^{1} ϕ_{q}(∫_{t^{m}}^{s} q(τ)ψ_{r}(τ)dτ)ds, \end{aligned}

therefore,

u^{m}(η) ≥ ∫_{η}^{1} ϕ_{q}(∫_{η}^{s} q(τ)ψ_{r}(τ)dτ)ds = θ_{2} > 0.

Since u^{m}(t) is concave in [η, 1], we get

\frac{u^{m}(t)}{1 − t} ≥ \frac{u^{m}(η)}{1 − η} ⇒ u^{m}(t) ≥ \frac{u^{m}(η)}{1 − η}(1 − t) ≥ \frac{θ_{2}}{1 − η} (1 − t).  (3.11)

If t ∈ [ξ, η], in view of concavity of u^{m}(t), we have

u^{m}(t) ≥ min{u^{m}(ξ), u^{m}(η)} ≥ θ,  (3.12)

where θ = min{θ_{1}, θ_{2}}.

EJQTDE, 2009 No. 42, p. 9 Let

δ(t) = \{ \begin{matrix}\frac{θ}{ξ} t, & t ∈ [0, ξ], \\ θ, & t ∈ [ξ, η], \\ \frac{θ}{1 − η} (1 − t), & t ∈ [η, 1], \end{matrix}  (3.13)

where θ = min{θ_{1}, θ_{2}}.

By (3.9), (3.11), (3.12) and (3.13), for any m ∈ ℕ_{0}, we have

u^{m}(t) ≥ δ(t), t ∈ [0, 1].

At the same time, it follows from (H_{3}) and (H_{6}) that

\begin{aligned}0 ≤ −(ϕ_{p}(u^{m})′(t))′ & = q(t)f(t, u^{m}(t)) ≤ q(t)[f_{1}(u^{m}(t)) + f_{2}(u^{m}(t))] \\ & ≤ q(t)f_{1}(δ(t)) + max_{0≤τ≤r} f_{2}(τ)q(t). \end{aligned}

Thus,

|ϕ_{p}((u^{m})′(t))| ≤ ∫_{0}^{1} q(s)f_{1}(δ(s))ds + max_{0≤τ≤r} f_{2}(τ) ∫_{0}^{1} q(s)ds

and

|(u^{m})′(t)| ≤ ϕ_{q}(∫_{0}^{1} q(s)f_{1}(δ(s))ds + max_{0≤τ≤r} f_{2}(τ) ∫_{0}^{1} q(s)ds) < +∞.  (3.14)

Therefore, {u^{m}(t)}_{m∈ℕ_{0}} is equi-continuous on [0,1]. Furthermore, from the fact that

0 < u^{m}(t) ≤ r, t ∈ [0, 1],

we have {u^{m}(t)}_{m∈ℕ_{0}} is uniformly bounded on [0,1].

The Arzelà-Ascoli theorem guarantees that there is a subsequence ℕ^{∗} ⊂ ℕ_{0}, a function u ∈ ℂ^{1}[0, 1] satisfying u^{m}(t) → u(t) uniformly on [0, 1] and t^{m} → σ as m → +∞ in ℕ^{∗}. From the definition of u^{m}(t), we have

u^{m}(t) = \{ \begin{matrix}αϕ_{q}(∫_{ξ}^{t^{m}} q(s)f(s, u^{m}(s))ds) + ∫_{0}^{t} ϕ_{q}(∫_{s}^{t^{m}} q(τ)f(τ, u^{m}(τ))dτ)ds + \frac{1}{m}, 0 ≤ t ≤ t^{m}, \\ βϕ_{q}(∫_{t^{m}}^{η} q(s)f(s, u^{m}(s))ds) + ∫_{t}^{1} ϕ_{q}(∫_{t^{m}}^{s} q(τ)f(τ, u^{m}(τ))dτ)ds + \frac{1}{m}, t^{m} ≤ t ≤ 1. \end{matrix}

(3.15)

Let m → +∞ in ℕ^{∗} in (3.15), by the continuity of f and Lebesgue’s dominated convergence theorem, we get

u(t) = \{ \begin{matrix}αϕ_{q}(∫_{ξ}^{σ} q(s)f(s, u(s))ds) + ∫_{0}^{t} ϕ_{q}(∫_{s}^{σ} q(τ)f(τ, u(τ))dτ)ds, 0 ≤ t ≤ σ, \\ βϕ_{q}(∫_{σ}^{η} q(s)f(s, u(s))ds) + ∫_{t}^{1} ϕ_{q}(∫_{σ}^{s} q(τ)f(τ, u(τ))dτ)ds, σ ≤ t ≤ 1, \end{matrix}

hence,

\{ \begin{matrix}(ϕ_{p}(u′(t)))′ + q(t)f(t, u(t)) = 0, t ∈ (0, 1), \\ u(0) − αu′(ξ) = 0, u(1) + βu′(η) = 0. \end{matrix}

From δ(t) ≤ u^{m}(t) ≤ r, t ∈ [0, 1], we have δ(t) ≤ u(t) ≤ r, t ∈ [0, 1]. So u(t) > 0, t ∈ (0, 1) and −(ϕ_{p}(u′(t)))′ = q(t)f(t, u) ≤ q(t)f_{1}(δ(t)) + max_{0≤τ≤r} f_{2}(τ)q(t) ∈ 𝕃^{1}[0, 1]. Therefore, (ϕ_{p}(u′(t)))′ ∈ 𝕃^{1}[0, 1] which means u(t) is a positive solution to BVP (1.1). The proof is completed.

Theorem 3.2. Suppose (H_{1b})-(H_{6}) hold. Furthermore, we assume that (H_{8}) there exists r > 0 such that

\frac{r}{M_{α,ξ}(r)} > 1,

where

M_{α,ξ}(r) = ϕ_{q}(I^{−1}(G(r)))[αϕ_{q}(I^{−1}(q(ξ))) + ∫_{0}^{1} ϕ_{q}(I^{−1}(q(t)))dt].

Then BVP (1.1) has a positive solution u(t) with ||u|| ≤ r.

Theorem 3.3. Suppose (H_{1a})-(H_{7}) hold. Furthermore, we assume that

(H_{9}) there exists a fixed constant δ ∈ (0, \frac{1}{2}) with [ξ, η] ⊂ [δ, 1 − δ], and μ ∈ ℂ[δ, 1 − δ] with μ > 0 on [δ, 1 − δ] such that

q(t)f(t, u) ≥ μ(t)[f_{1}(u) + f_{2}(u)] on [δ, 1 − δ] × (0, +∞);

(H_{10}) there exists R > r such that

\frac{2Rϕ_{q}(f_{1}(ω_{1}(δ)R))}{ϕ_{q}[f_{1}(R)f_{1}(ω_{1}(δ)R) + f_{1}(R)f_{2}(ω_{1}(δ)R)]} ≤ b_{0},

where b_{0} = min{α, β} min{1, 2^{2−q}}ϕ_{q}(∫_{ξ}^{η} μ(s)ds),

ω_{1}(δ) = min\{\frac{δ}{η} , \frac{δ}{1 − ξ} \}.

Then BVP (1.1) has a positive solution u_{1}(t) with r < ||u_{1}|| ≤ R.

Proof. Let

K = {u ∈ X : u(t) ≥ 0, u(t) is concave, u(t) ≥ ω(t)||u||, t ∈ [0, 1]}.

Obviously, K is a cone of X. Since r < R, denote open subsets Ω_{1} and Ω_{2} of X:

Ω_{1} = {u ∈ X : ||u|| < r}, Ω_{2} = {u ∈ X : ||u|| < R}.

EJQTDE, 2009 No. 42, p. 11 Let A : K ∩ (Ω̄_{2} \ Ω_{1}) → X be defined by

(Au)(t) = \{ \begin{matrix}αϕ_{q}(∫_{ξ}^{σ} q(s)f(s, u(s))ds) + ∫_{0}^{t} ϕ_{q}(∫_{s}^{σ} q(τ)f(τ, u(τ))dτ)ds, 0 ≤ t ≤ σ, \\ βϕ_{q}(∫_{σ}^{η} q(s)f(s, u(s))ds) + ∫_{t}^{1} ϕ_{q}(∫_{σ}^{s} q(τ)f(τ, u(τ))dτ)ds, σ ≤ t ≤ 1. \end{matrix}

(3.16)

It is easy to verify that the fixed points of operator A are the positive solutions of BVP (1.1). So it suffices to show that A has at least one fixed point.

First we prove A : K ∩ (Ω̄_{2} \ Ω_{1}) → K. If u ∈ K ∩ (Ω̄_{2} \ Ω_{1}), differentiating (3.16) for t, we obtain

(Au)′(t) = \{ \begin{matrix}ϕ_{q}(∫_{t}^{σ} q(τ)f(τ, u(τ))dτ), 0 ≤ t ≤ σ, \\ − ϕ_{q}(∫_{σ}^{t} q(τ)f(τ, u(τ))dτ), σ ≤ t ≤ 1 \end{matrix}

and

(Au)′′(t) = \{ \begin{matrix}− (q − 1)(∫_{t}^{σ} q(τ)f(τ, u(τ))dτ)^{q−2}q(t)f(t, u(t)), 0 < t ≤ σ, \\ − (q − 1)(∫_{σ}^{t} q(τ)f(τ, u(τ))dτ)^{q−2}q(t)f(t, u(t)), σ ≤ t < 1. \end{matrix}  (3.17)

In view of (3.16) and (3.17), note that (Au)′′(t) ≤ 0 for any t ∈ (0, 1), and Au(0) ≥ 0, Au(1) ≥ 0. Therefore, Au(t) is concave and Au(t) ≥ 0 on [0, 1]. By Lemma 2.2, we have Au(t) ≥ ω(t)||Au||. Consequently, Au ∈ K, i.e., A : K ∩ (Ω̄_{2} \ Ω_{1}) → K.

By Lemma 2.3, we have that A : K ∩ (Ω̄_{2} \ Ω_{1}) → K is completely continuous.

Now we shall show that

||Au|| < ||u|| for u ∈ K ∩ ∂Ω_{1}.  (3.18)

Let u ∈ K ∩ ∂Ω_{1}, then ||u|| = r. As in the proof of (3,7), we have

||Au|| = (Au)(σ)
≤ ϕ_{q}(I^{−1}(G(r)))[βϕ_{q}(I^{−1}(q(η))) + ∫_{0}^{1} ϕ_{q}(I^{−1}(q(t)))dt]
= M_{β,η}(r)
< r = ||u||.

Consequently, ||Au|| < ||u||. So (3.18) holds.

Furthermore, we give that

||Au|| ≥ ||u|| for u ∈ K ∩ ∂Ω_{2}.  (3.19)

Let u ∈ K ∩ ∂Ω_{2}, so ||u|| = R and u(t) ≥ ω(t)R for t ∈ [0, 1]. In particular, u(t) ∈ [ω_{1}(δ)R, R] for t ∈ [δ, 1 − δ], where ω_{1}(δ) = min\{\frac{δ}{η} , \frac{δ}{1−ξ} \} (δ is the same as in (H_{9})). By (H_{3}), (H_{7}) and (H_{8}), we obtain

||Au|| =(Au)(σ)
=\frac{1}{2}[αϕ_{q}(∫_{ξ}^{σ} q(s)f(s, u(s))ds) + ∫_{0}^{σ} ϕ_{q}(∫_{s}^{σ} q(τ)f(τ, u(τ))dτ)ds
+ βϕ_{q}(∫_{σ}^{η} q(s)f(s, u(s))ds) + ∫_{σ}^{1} ϕ_{q}(∫_{σ}^{s} q(τ)f(τ, u(τ))dτ)ds]
≥\frac{1}{2}[αϕ_{q}(∫_{ξ}^{σ} q(s)f(s, u(s))ds) + ∫_{δ}^{σ} ϕ_{q}(∫_{s}^{σ} q(τ)f(τ, u(τ))dτ)ds
+ βϕ_{q}(∫_{σ}^{η} q(s)f(s, u(s))ds) + ∫_{σ}^{1−δ} ϕ_{q}(∫_{σ}^{s} q(τ)f(τ, u(τ))dτ)ds]
≥\frac{1}{2}ϕ_{q}(f_{1}(R))ϕ_{q}[1 + \frac{f_{2}(ω_{1}(δ)R)}{f_{1}(ω_{1}(δ)R)}][αϕ_{q}(∫_{ξ}^{σ} μ(s)ds) + ∫_{δ}^{σ} ϕ_{q}(∫_{s}^{σ} μ(τ)dτ)ds
+ βϕ_{q}(∫_{σ}^{η} μ(s)ds) + ∫_{σ}^{1−δ} ϕ_{q}(∫_{σ}^{s} μ(τ)dτ)ds]
≥\frac{1}{2}ϕ_{q}(f_{1}(R))ϕ_{q}[1 + \frac{f_{2}(ω_{1}(δ)R)}{f_{1}(ω_{1}(δ)R)}] min\{α, β\} min\{1, 2^{2−q}\}ϕ_{q}(∫_{ξ}^{η} μ(s)ds)
≥R = ||u||.

Consequently, ||Au|| ≥ ||u||. So (3.19) holds. By Theorem 2.5, we can obtain that A has a fixed point u_{1} ∈ K ∩ (Ω̄_{2} \ Ω_{1}) with r < ||u_{1}|| ≤ R and u_{1}(t) ≥ ω(t)r for t ∈ [0, 1]. Thus u_{1}(t) > 0 for t ∈ (0, 1). The proof is completed.

Theorem 3.4. Suppose (H_{1a})-(H_{7}), (H_{9}) and (H_{10}) hold. Then BVP (1.1) has two positive solutions u(t), u_{1}(t) with ||u|| ≤ r < ||u_{1}|| ≤ R. Theorem 3.5. Suppose (H_{1b})-(H_{6}) and (H_{8})-(H_{10}) hold. Then BVP (1.1) has two positive solutions u(t), u_{1}(t) with ||u|| ≤ r < ||u_{1}|| ≤ R.

4 Example

In this section, we give an explicit example to illustrate our main result. Example 4.1. Consider four-point boundary value problem of second order differential equation

\{ \begin{matrix}u′′ + \frac{a}{\sqrt{1 − t}}(\frac{σ(t)}{u^{\frac{1}{4}}} + t) = 0, 0 < t < 1, \\ u(0) − u′(\frac{1}{4}) = 0, u(1) + u′(\frac{3}{4}) = 0, \end{matrix}  (4.1)

where σ(t) = max{0, (t − ξ)(η − t)}, a > 0 is a constant.

Conclusion: BVP (4.1) has at least one positive solution.

EJQTDE, 2009 No. 42, p. 13 Proof. Obviously, p = 2, α = 1, β = 1, ξ = \frac{1}{4}, η = \frac{3}{4} in BVP(4.1). Comparing to Theorem 3.1, we verify (H_{1a})-(H_{7}) as follows:

(H_{1a}) q(t) = \frac{a}{\sqrt{1−t}} ∈ ℂ(0, 1) ∩ 𝕃^{1}[0, 1], q(t) > 0 and nonincreasing on (0, 1); (H_{2}) f(t, u) = \frac{σ(t)}{u^{\frac{1}{4}}} + t is a continuous function on [0, 1] × (0, +∞); (H_{3}) 0 < f(t, u) = \frac{σ(t)}{u^{\frac{1}{4}}} + t ≤ _{\frac{1}{u^{\frac{1}{4}}}} + 1 + u^{4} = f_{1}(u) + f_{2}(u), where f_{1}(u) = _{\frac{1}{u^{\frac{1}{4}}}} > 0

continuous, nonincreasing on (0, +∞) and ∫_{0}^{L} f_{1}(u)du < +∞ for any fixed L > 0; f_{2}(u) = u^{4} + 1 > 0 is continuous on [0, +∞); \frac{f_{2}}{f_{1}} = u^{\frac{1}{4}} + u^{\frac{17}{4}} nondecreasing on (0, +∞);

(H_{4}) Let R = 1, so N(t) = sup_{u∈(0,1]} f(t, u) = t for t ∈ [0, \frac{1}{4}], n(t) = inf_{u∈(0,1]} f(t, u) = (t − \frac{1}{4})(\frac{3}{4} − t) + t for t ∈ [\frac{1}{4}, \frac{3}{4}], then

∫0^{\frac{1}{4}} q(t)N(t)dt = a ∫0^{\frac{1}{4}} \frac{t}{\sqrt{1 − t}}dt ≐ 0.034295227a,

∫ \frac{3}{4} \frac{1}{4} q(t)n(t)dt = a ∫ \frac{3}{4} \frac{1}{4} \frac{(t − \frac{1}{4})(\frac{3}{4} − t) + t}{\sqrt{1 − t}}dt ≐ 0.4124355660a,

Γ = \frac{13}{3}, Γa ∫ \frac{3}{4} \frac{1}{4} \frac{(t − \frac{1}{4})(\frac{3}{4} − t) + t}{\sqrt{1 − t}}dt ≐ 1.787220786a

so a ∫0^{\frac{1}{4}} \frac{t}{\sqrt{1−t}}dt ≤ Γa ∫ \frac{3}{4} \frac{1}{4} \frac{(t− \frac{1}{4} )( \frac{3}{4} −t)+t}{\sqrt{1−t}}dt;

(H_{5}) For each constant H > 0, there exists a function ψ_{H}(t) = t that is continuous on [0, 1] and positive on (0, 1) satisfying f(t, u) ≥ ψ_{H}(t) on (0, 1) × (0, H];

(H_{6}) Clearly, ∫0^{\frac{1}{4}} \frac{1}{\sqrt[4]{k_{1}s}\sqrt{1−s}}ds < +∞, ∫ 1_{\frac{3}{4}} \frac{1}{\sqrt[4]{k_{2}(1−s)}\sqrt{1−s}}ds < +∞, for any k_{1} > 0, k_{2} >

0; (H_{7}) When r > 0,

ϕ_{q}(r) = r, I^{−1}(r) = \sqrt{2r},

G(r) = ∫_{0}^{r}[f_{1}(s) + f_{2}(s)]ds = ∫_{0}^{r}[\frac{1}{s^{\frac{1}{4}}} + 1 + s^{4}]ds = \frac{4}{3}r^{\frac{3}{4}} + r + \frac{1}{5}r^{5},

M_{β,η}(r) = ϕ_{q}(I^{−1}(G(r)))[βϕ_{q}(I^{−1}(q(η))) + ∫_{0}^{1} ϕ_{q}(I^{−1}(q(t)))dt]

= \sqrt{\frac{8}{3}r^{\frac{3}{4}} + 2r + \frac{2}{5}r^{5}}(2 + \frac{4}{3}\sqrt{2})\sqrt{a},

let r = 0.1, a = 10^{−4}, so \frac{r}{M_{β,η}(r)} ≐ 3.134308209 > 1.

Therefore, By Theorem 3.1, we obtain that BVP (4.1) has at least one positive solution.

Acknowledgement. The authors thank the referee for his/her careful reading of the manuscript and valuable suggestions.

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(Received March 3, 2009)