Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 27, 1-15; http://www.math.u-szeged.hu/ejqtde/

Multiplicity of Positive Solutions for a Fourth-Order Quasilinear Singular Differential Equation^{∗}

Zhichang Guo

Department of Mathematics, Jilin University, 130012 Changchun, P.R.China

Jingxue Yin

School of Mathematical Sciences, South China Normal University, 510631 Guangzhou, P.R.China

Yuanyuan Ke ^{†}

Department of Mathematics, Renmin University of China, 100872 Beijing, P.R.China

Abstract

This paper is concerned with the multiplicity of positive solutions of boundary value problem for the fourth-order quasilinear singular differential equation

(|u′′|^{p−2}u′′)′′ = λg(t)f(u), 0 < t < 1,

where p > 1, λ > 0. We apply the fixed point index theory and the upper and lower solutions method to investigate the multiplicity of positive solutions. We have found a threshold λ^{∗} < +∞, such that if 0 < λ ≤ λ^{∗}, then the problem admits at least one positive solution; while if λ > λ^{∗}, then the problem has no positive solution. In particular, there exist at least two positive solutions for 0 < λ < λ^{∗}.

Key words: Multiplicity; Positive solutions; Fourth-order; Quasilinear; Singular.

1 Introduction

In this paper, we consider the multiplicity of positive solutions for the following fourthorder quasilinear singular differential equation

(|u′′|^{p−2}u′′)′′ = λg(t)f(u), 0 < t < 1,  (1.1)

subject to the boundary value conditions

u(0) = u(1) = u′′(0) = u′′(1) = 0,  (1.2)

where p > 1 and λ is a positive parameter.

∗This work is partially supported by the National Science Foundation of China, and partially supported by a Specific Foundation for Ph.D Specialities of Educational Department of China.

†Corresponding author. Tel.: +86-10-8250-0694 Email address: ke yy@163.com

In the past few years, some fourth-order nonlinear equations have been proposed for image (signal) processing (i.e., edge detection, image denoising, etc.). And a number of authors hoped that these methods might perform better than some second-order equations [15–19]. Indeed, there are good reasons to consider fourth-order nonlinear equations. First, fourth-order differential damps oscillations at high frequencies (e.g., noise) much faster than second-order differential. Second, there is the possibility of having schemes that include effects of curvature, i.e., the second derivatives of the image (signal), in the dynamics. The equation (1.1) can be regarded as the analogue of the Euler-Lagrange equations from the variation problem in [15]. Similarly, the solution of the equation (1.1) can be regarded as the steady-state case of the fourth-order anisotropic diffusion equation in [16–19].

Equations of the form (1.1), especially the special case p = 2, have been the subject of intensive study during the last thirty years, see for example [1,2,5,9–14]. In particular, in a recent paper [5], under some structure conditions which permit some singularities for g(t), the authors discussed the special case p = 2 subject to the boundary value conditions (1.2), and revealed the relation between the existence of positive solutions and the parameter λ. While for the case p > 1 with g(t) ≡ 1 and f(s) being of power type, in [6], P. Dráek, M. Ôani considered the corresponding initial value problem, and obtain the local existence and uniqueness of solutions, see also [7] for some extension of the results.

In this paper, we mainly discuss the boundary value problem for the fourth-order quasilinear differential equation (1.1), namely, the problem (1.1), (1.2). Throughout this paper, we assume the following basic conditions:

(H1) f ∈ C([0, +∞), (0, +∞)) and is nondecreasing on [0, +∞). Furthermore, there exist \overline{δ} > 0, m > p − 1 such that f(u) > \overline{δ}u^{m}, u ∈ [0, +∞);

(H2) g ∈ C((0, 1), (0, +∞)) and g(t) ≢ 0 on any subinterval of (0, 1).

The main purpose of this paper is to investigate the existence, nonexistence and multiplicity of positive solutions of the problem (1.1), (1.2). Different from the known works, the equation we consider is quasilinear, which might have degeneracy or singularities. In fact, if p > 2, then the equation is degenerate at the points where u′′ = 0; while if 1 < p < 2, then the equation has singularity at the points where u′′ = 0. It should be noticed that there is no Green’s function compared with the special case p = 2.

The main result of this paper is the following

Theorem 1.1 Let (H1) and (H2) be satisfied. If

∫_{0}^{1} s(1 − s)g(s)ds < +∞,  (1.3)

then there exists a threshold 0 < λ^{∗} < +∞ such that the problem (1.1), (1.2) has no positive solution for λ > λ^{∗}, has at least one positive solution for λ = λ^{∗}, and has at least two positive solutions for 0 < λ < λ^{∗}.

Our method can also be applied to the discussion of the equation subject to another boundary value condition

u(0) = u′(1) = u′′(0) = (|u′′(t)|^{p−2}u′′(t))′|_{t=1} = 0.  (1.4)

The corresponding extension of the result is as follows.

Theorem 1.2 Let (H1) and (H2) be satisfied. If

∫_{0}^{1} sg(s)ds < +∞,  (1.5)

then there exists a threshold 0 < λ^{∗} < +∞ such that the problem (1.1), (1.4) has no positive solution for λ > λ^{∗}, has at least one positive solution for λ = λ^{∗}, and has at least two positive solutions for 0 < λ < λ^{∗}.

This paper is organized as follows. As preliminaries, we state some necessary lemmas in Section 2. Subsequently, in Section 3, we apply the fixed point index theory and the upper and lower solutions method to the proof of Theorem 1.1. In the last section, we are concerned with the proof of Theorem 1.2. Owing to the similarity with the proof of Theorem 1.1, we only give a sketch and omit the details of the proof of Theorem 1.2.

2 Preliminaries

In this section, we first present necessary definitions and introduce some auxiliary lemmas, including those from the fixed point index theory and the theory based on the upper and lower solutions method. Subsequently, we construct some key integral operators, which is closely related to the problem (1.1), (1.2). As an important preparation to the proof of our main result, we devote the remaining part of this section to the proof of the complete continuity and monotonicity of the operator. We also give the a priori estimates on the positive solutions of the problem (1.1), (1.2).

First, because of the possible degeneracy and singularities, the exact meaning of solutions should be clarified.

Definition 2.1 A function u(t) is said to be a positive solution of the problem (1.1), (1.2), if u(t) > 0, for t ∈ (0, 1), u(t) ∈ C^{2}[0, 1] ∩ C^{4}(0, 1), |u′′|^{p−2}u′′ ∈ C^{2}(0, 1), and u(t) satisfies (1.1) and (1.2).

The following two lemmas from the fixed point index theory and the theory based on upper and lower solutions method will be used to obtain the multiplicity of positive solutions, see [4].

Lemma 2.1 Let D be a cone in a real Banach space X, Ω be a bounded open subset of X with θ ∈ Ω, and A : D ∩ \overline{Ω} → D is a completely continuous operator. If

Ax = μx, x ∈ D ∩ ∂Ω ⇒ μ < 1,

then i(A, D ∩ Ω, D) = 1; While if

(1) inf_{x∈D∩∂Ω} ∥Ax∥ > 0;

(2) Ax = μx, x ∈ D ∩ ∂Ω ⇒ μ ∉ (0, 1],

then i(A, D ∩ Ω, D) = 0.

Lemma 2.2 Suppose X is a partially ordered Banach space, D is a normal cone of X, and A : D → X is an increasing and completely continuous operator. If there exist x_{0}, y_{0} ∈ D such that x_{0} ≤ y_{0}, ⟨x_{0}, y_{0}⟩ ⊂ D, and x_{0}, y_{0} are a lower solution and an upper solution of the equation

x − Ax = 0,

respectively, then the above equation has a minimal solution x^{∗} and a maximal solution y^{∗} in the ordered interval ⟨x_{0}, y_{0}⟩, and x^{∗} ≤ y^{∗}.

The following technical lemma, see [8], will be used to characterize some properties of a related operator.

Lemma 2.3 If y ∈ C^{2}(0, 1) ∩ C[0, 1] with y(t) ≥ 0 for 0 ≤ t ≤ 1, and y′′(t) ≤ 0 for 0 < t < 1, then

y(t) ≥ \frac{ξ}{4} max_{0≤s≤1} |y(s)|, \frac{ξ}{4} ≤ t ≤ 1 − \frac{ξ}{4},

where 0 ≤ ξ ≤ 1.

We also need the following technical lemma on the property of the function f, see [3].

Lemma 2.4 Suppose that f : [0, +∞) → (0, +∞) is continuous. For s > 0 and M > 0, there exists \overline{s} > s and h_{0} > 0 such that

sf(u + h) < \overline{s}f(u), u ∈ [0, M], h ∈ (0, h_{0}).

We are now in the position to make preparation to prove our main result. We use the notation φ_{p}(s) = |s|^{p−2}s, for p > 1. Let q = \frac{p}{p−1}. Then φ_{q} is the inverse function of φ_{p}. Let E = C[0, 1] with the norm ∥u∥ = max_{0≤t≤1} |u(t)| and

P = {u ∈ E; u(t) ≥ 0, t ∈ [0, 1]}.

It is clear that P is a normal cone of E. Define a set

S={λ > 0; such that the problem (1.1), (1.2) has at least one positive solution }.

To show the existence, it is necessary to construct an appropriate operator and solve the corresponding operator equation. For this purpose, we first notice that u(t) is a solution of the problem (1.1), (1.2), if and only if u(t) is a solution of the following problem

v′′ = λg(t)f(u),  (2.1)
u′′ = φ_{q}(v),  (2.2)
v(0) = v(1) = 0,  (2.3)
u(0) = u(1) = 0.  (2.4)

Because of (2.1) and (2.3), v(t) can be expressed by

v(t) = −λ ∫_{0}^{1} k(t, s)g(s)f(u(s))ds,

where k(t, s) is the Green function of the equation (2.1) with the boundary value (2.3), and

k(t, s) = \{ \begin{matrix}t(1 − s), & t ≤ s, \\ s(1 − t), & t ≥ s. \end{matrix}

Then u(t) can be expressed by

u(t) = λ^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u(s))ds) dτ.  (2.5)

It is easy to prove that the Green function k(t, s) has the following properties.

Lemma 2.5 For all t, s ∈ [0, 1], we have

k(t, s) > 0, for (t, s) ∈ (0, 1) × (0, 1); k(t, s) ≤ k(s, s) = s(1 − s), for t, s ∈ [0, 1];

0 ≤ k(t, s) ≤ \frac{1}{4}, for t, s ∈ [0, 1].

Lemma 2.6 For all t ∈ [θ, 1 − θ], we have

k(t, s) ≥ θk(s, s), θ ∈ (0, \frac{1}{2}), s ∈ [0, 1].

Proof. In fact, for all t ∈ [θ, 1 − θ], we have

\frac{k(t, s)}{k(s, s)} = \{ \begin{matrix}\frac{t}{s}, & 1 ≥ s ≥ t ≥ 0, \\ \frac{1 − t}{1 − s}, & 0 ≤ s ≤ t ≤ 1, \end{matrix}

≥ \{ \begin{matrix}t ≥ θ, & s ≥ t, \\ 1 − t ≥ θ, & s ≤ t. \end{matrix}

Therefore, For all t ∈ [θ, 1 − θ], we have

k(t, s) ≥ θk(s, s), θ ∈ (0, \frac{1}{2}), s ∈ [0, 1].

The proof is complete. □ Next, we consider the following approximate problem

(|u′′|^{p−2}u′′)′′ = λg(t)f(u), 0 < t < 1,  (2.6)
u(0) = u(1) = h ≥ 0,  u′′(0) = u′′(1) = 0.  (2.7)

Define a cone K ⊂ P as follows

K = \{u ∈ E; u(t) ≥ 0, t ∈ [0, 1] and min_{1/8≤t≤7/8} u(t) ≥ \frac{1}{8}∥u∥\} .

And we define an integral operator T_{λ}^{h} on K by

T_{λ}^{h}u(t) = h + λ^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u(s))ds) dτ.  (2.8)

Now, we discuss the properties of the operator T_{λ}^{h}. First, by the definition of T_{λ}^{h} and the assumption (H1), we have Lemma 2.7 Let (H1) and (H2) be satisfied. Then the operator T_{λ}^{h} defined by (2.8) is non-decreasing, that is, the inequality u_{1} ≤ u_{2} implies T_{λ}^{h}u_{1} ≤ T_{λ}^{h}u_{2}, where “ ≤ ” is the partial order defined on K.

The complete continuity of T_{λ}^{h} can be obtained by the following lemma.

Lemma 2.8 Let (H1), (H2) and (1.3) be satisfied. Then the operator T_{λ}^{h} defined by (2.8) is completely continuous, and T_{λ}^{h}K ⊂ K.

Proof. Firstly, we testify the complete continuity of T_{λ}^{h}. Let {u_{n}} ⊂ K, u ∈ K with ∥u_{n} − u∥ → 0 as n → +∞. Then we have

∥(T_{λ}^{h}u_{n})(t) − (T_{λ}^{h}u)(t)∥

≤ λ^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ) |φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u_{n}(s))ds) − φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u(s))ds)| dτ

≤ λ^{\frac{1}{p−1}} sup_{t∈[0,1]} |φ_{q} (∫_{0}^{1} k(t, s)g(s)f(u_{n}(s))ds) − φ_{q} (∫_{0}^{1} k(t, s)g(s)f(u(s))ds)| .

Since ∥u_{n} − u∥ → 0 as n → ∞ and u_{n}, u ∈ K, {u_{n}(t)} is bounded uniformly. Then there exists a constant M_{0} > 0, such that |u(t)| ≤ M_{0}, |u_{n}(t)| ≤ M_{0}, for any t ∈ [0, 1], n = 1, 2, · · ·. Due to the continuity of f(s), f(u_{n}) is bounded uniformly in [0, M_{0}]. Moreover, because of the continuity of φ_{q}(s), (1.3) and (H2), by the Lebesgue dominated convergence theorem, we have

∥T_{λ}^{h}u_{n} − T_{λ}^{h}u∥ → 0 (n → ∞).

Therefore, we see that T_{λ}^{h} is continuous. And the compactness of the operator T_{λ}^{h} is easily obtained from the Arzela-Ascoli theorem.

Next, we testify that T_{λ}^{h}K ⊂ K. For each u ∈ K, it is easy to check that (T_{λ}^{h}u)′′(t) ≤ 0 for 0 < t < 1 and (T_{λ}^{h}u)(t) ≥ 0 for 0 ≤ t ≤ 1. Then according to Lemma 2.3, the following inequality holds

(T_{λ}^{h}u)(t) ≥ \frac{1}{8}∥T_{λ}^{h}u∥, \frac{1}{8} ≤ t ≤ \frac{7}{8},

which implies that (T_{λ}^{h}u)(t) ∈ K. Hence we testify that T_{λ}^{h}K ⊂ K. The proof is complete. □

By the definition of T_{λ}^{h}, namely (2.8), and a direct computation, we obtain the following lemma.

Lemma 2.9 Let (H1) and (H2) be satisfied. Then the problem (1.1), (1.2) has a positive solution u if and only if u is a fixed point of T_{λ}^{0}. And the problem (2.6), (2.7) has a positive solution u if and only if u is a fixed point of T_{λ}^{h}.

In order to apply the fixed point index lemmas, we need the a priori estimates on positive solutions of the problem (1.1), (1.2).

Lemma 2.10 Let (H1), (H2) and (1.3) be satisfied. And suppose that λ ∈ S, S_{1} = (λ, +∞) ∩ S ≢ ∅. Then there exists a constant R(λ) > 0, such that ∥u_{λ′}∥ ≤ R(λ), where λ′ ∈ S_{1}, and the function u_{λ′} ∈ K is a positive solution of the problem (1.1), (1.2) with λ replaced by λ′.

Proof. For any fixed λ′ ∈ S, let u_{λ′} be a positive solution of the problem (1.1), (1.2). Then, by Lemma 2.9, we have

u_{λ′}(t) = T_{λ′}^{0}u_{λ′}(t) = (λ′)^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u_{λ′}(s))ds) dτ.

Let

R(λ) = max \{(λ^{\frac{1}{p−1}} (\frac{1}{8})^{\frac{m+p}{p−1}} ∫_{1/8}^{7/8} k(τ, τ)φ_{q} (∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ)^{−(p−1)/(m−p+1)} , 1\} .

We conclude that ∥u_{λ′}∥ ≤ R(λ). Indeed, if ∥u_{λ′}∥ < 1, the result is easily obtained; while if ∥u_{λ′}∥ ≥ 1, by (H1), Lemma 2.3 and Lemma 2.6, we have

∥u_{λ′}∥ ≥ (λ′)^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u_{λ′}(s))ds) dτ

≥ λ^{\frac{1}{p−1}} ∫_{1/8}^{7/8} k(t, τ)φ_{q} ((\frac{1}{8})^{m+1} ∥u_{λ′}∥^{m} ∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ

≥ λ^{\frac{1}{p−1}} (\frac{1}{8})^{(m+p)/(p−1)} ∥u_{λ′}∥^{m/(p−1)} ∫_{1/8}^{7/8} k(τ, τ)φ_{q} (∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ.

Consequently,

∥u_{λ′}∥^{m/(p−1)−1} ≤ (λ^{\frac{1}{p−1}} (\frac{1}{8})^{(m+p)/(p−1)} ∫_{1/8}^{7/8} k(τ, τ)φ_{q} (∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ)^{−1} ,

∥u_{λ′}∥ ≤ (λ^{\frac{1}{p−1}} (\frac{1}{8})^{(m+p)/(p−1)} ∫_{1/8}^{7/8} k(τ, τ)φ_{q} (∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ)^{−(p−1)/(m−p+1)} .

Therefore, ∥u_{λ′}∥ ≤ R(λ), which completes the proof of the lemma. □

3 Proofs of the Main Result

In this section, we give the proof of the main result, that is, Theorem 1.1. The proof will be divided into two parts. Firstly, by the upper and lower solutions method, we investigate the basic existence of positive solutions of the problem (1.1), (1.2). Exactly, we will determine the threshold λ^{∗} of the parameter λ, such that the problem is solvable if and only if 0 < λ ≤ λ^{∗}. Furthermore, by the fixed point index theory, we establish the multiplicity of positive solutions for 0 < λ < λ^{∗}.

We first present and prove the basic existence result of positive solutions of the problem (1.1), (1.2).

Proposition 3.1 Let (H1), (H2) and (1.3) be satisfied. Then there exists a threshold λ^{∗} with 0 < λ^{∗} < +∞, such that the problem (1.1), (1.2) admits at least one positive solution for any λ ∈ (0, λ^{∗}], and has no positive solution for any λ > λ^{∗}.

Proof. Let S be the set defined in the previous section, namely

S = {λ > 0; such that the problem (1.1), (1.2) has at least one positive solution}.

We first show that S ≠ ∅. Let β(t) be a solution of the boundary value problem

\begin{matrix}(|u′′|^{p−2}u′′)′′ = g(t), 0 < t < 1, \\ u(0) = u(1) = u′′(0) = u′′(1) = 0. \end{matrix}  (3.1)

Then, by (2.5) we have

β(t) = ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)ds) dτ.

Let β_{0} = max_{t∈[0,1]} β(t). Then by (H1) and (2.8), we have

T_{λ}^{0}β(t) ≤ T_{λ}^{0}β_{0}

≤ λ^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(β_{0})ds) dτ.

≤ λ^{\frac{1}{p−1}} f(β_{0})^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)ds) dτ.

≤ β(t), ∀ 0 < λ < \frac{1}{f(β_{0})},

which implies that β(t) is an upper solution of T_{λ}^{0}. On the other hand, let α(t) ≡ 0, t ∈ [0, 1]. Then α(t) is a lower solution of T_{λ}^{0}, and α(t) ≤ β(t), t ∈ (0, 1). Hence, by Lemma 2.2, Lemma 2.7 and Lemma 2.8, T_{λ}^{0} has a fixed point u_{λ} ∈ [α, β], for 0 < λ < \frac{1}{f(β_{0})}. So u_{λ} is a positive solution of the problem (1.1), (1.2). Therefore, for any 0 < λ < \frac{1}{f(β_{0})}, we have λ ∈ S, which implies that S ≠ ∅.

Next, we show that if λ_{1} ∈ S, then (0, λ_{1}) ⊂ S. In fact, if u_{λ_{1}} be a positive solution of the problem (1.1), (1.2), then, by Lemma 2.9, we have

u_{λ_{1}}(t) = T_{λ_{1}}^{0}u_{λ_{1}}(t), t ∈ [0, 1].

Therefore, for any λ ∈ (0, λ_{1}), by (2.8), we have

T_{λ}^{0}u_{λ_{1}}(t) = λ^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u_{λ_{1}}(s))ds) dτ
≤ λ_{1}^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u_{λ_{1}}(s))ds) dτ
= T_{λ_{1}}^{0}u_{λ_{1}}(t)
= u_{λ_{1}}(t),

which implies that u_{λ_{1}} is an upper solution of T_{λ}^{0}. Taking this into account, noticing the fact that the function α(t) ≡ 0 is a lower solution of T_{λ}^{0}, and using Lemma 2.2, Lemma 2.7 and Lemma 2.8, we see that the problem (1.1), (1.2) has a positive solution, and therefore λ ∈ S, which implies that (0, λ_{1}) ⊂ S.

Now, we conclude that sup S < +∞. If this were not true, then we would have ℕ ⊂ S, where ℕ denotes the set of natural numbers. Therefore, for any n ∈ ℕ, by Lemma 2.8, there exists u_{n} ∈ K satisfying

u_{n} = T_{n}^{0}u_{n} = n^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u_{n}(s))ds) dτ.

If ∥u_{n}∥ ≥ 1, then we have

∥u_{n}∥ ≥ n^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u_{n}(s))ds) dτ

≥ n^{\frac{1}{p−1}} ∫_{1/8}^{7/8} k(t, τ)φ_{q} ((\frac{1}{8})^{m+1} ∥u_{n}∥^{m} ∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ

≥ n^{\frac{1}{p−1}} (\frac{1}{8})^{(m+p)/(p−1)} ∥u_{n}∥^{m/(p−1)} ∫_{1/8}^{7/8} k(τ, τ)φ_{q} (∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ.

Consequently,

1 ≥ n^{\frac{1}{p−1}} (\frac{1}{8})^{(m+p)/(p−1)} ∫_{1/8}^{7/8} k(τ, τ)φ_{q} (∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ.  (3.2)

If ∥u_{n}∥ ≤ 1, then

1 ≥ ∥u_{n}∥ ≥ n^{\frac{1}{p−1}} (\frac{1}{8})^{p/(p−1)} ∫_{1/8}^{7/8} k(τ, τ)φ_{q} (∫_{1/8}^{7/8} k(s, s)g(s)f(0)ds) dτ > 0. (3.3)

Letting n → +∞ in (3.2) and (3.3), we get a contradiction. Therefore, sup S < +∞. We are now in a position to determine the threshold λ^{∗}. We conclude that

λ^{∗} = sup S.

It remains to show that λ^{∗} ∈ S. Let \{λ_{n}\} be an increasing sequence in [\frac{λ^{∗}}{2} , λ^{∗}) with λ_{n} → λ^{∗}(n → +∞), and let u_{n} be the corresponding solutions of the problem (1.1), (1.2) with λ replaced by λ_{n}. By Lemma 2.10, there exists R(\frac{λ^{∗}}{2} ) > 0 such that

∥u_{n}∥ ≤ R (\frac{λ^{∗}}{2} ) , n = 1, 2, · · · .

Therefore, {u_{n}} is an equicontinuous and uniformly bounded subset of C[0, 1]. By the Arzela-Ascoli theorem, {u_{n}} has a convergent subsequence, and denoted also by itself, with u_{n} → u^{∗} as n → +∞. Since u_{n} = T_{λ_{n}}^{0}u_{n}, due to the continuity of f(s), we see that f(u_{n}) is bounded uniformly in [0, R(\frac{λ^{∗}}{2} )], which together with the continuity of φ_{q}(s) and (H2), by the Lebesgue dominated convergence theorem, implies that u^{∗} = T_{λ^{∗}}^{0}u^{∗}. Hence, by Lemma 2.9, u^{∗} is a positive solution of the problem (1.1), (1.2) with λ replaced by λ^{∗}. Summing up, we have completed the proof of the proposition. □

To complete the proof of our main result, that is Theorem 1.1, it remains to show the multiplicity of solutions for 0 < λ < λ^{∗}. To do this, we need also the following lemma.

Lemma 3.1 Let (H1), (H2) and (1.3) be satisfied. Then there exists h_{0} ∈ (0, 1), such that for any h̄ ∈ (0, h_{0}) and λ ∈ (0, λ^{∗}), the problem (2.6), (2.7) admits at least one positive solution.

Proof. Let α(t) ≡ h̄ for t ∈ [0, 1]. It is obvious that, for any fixed λ ∈ (0, λ^{∗}), α(t) is a lower solution of the operator T h̄ ; λ. On the other hand, by Lemma 2.10, there exists R(λ) > 0 such that ∥u_{λ′}∥ ≤ R(λ), where λ′ ∈ [λ, λ^{∗}] and u_{λ′} is a positive solution of the problem (1.1), (1.2) with λ replaced by λ′. By Lemma 2.4, there exist \overline{λ} ∈ (λ, λ^{∗}) and h_{0} ∈ (0, 1) satisfying

λf(u + h̄) < \overline{λ}f(u), u ∈ [0, R(λ)], h̄ ∈ (0, h_{0}).

Let u\overline{λ} be a positive solution of the problem (1.1), (1.2) with λ replaced by \overline{λ}, and \overline{u}_{λ}(t) = u\overline{λ} + h̄, h̄ ∈ (0, h_{0}). Then

\overline{u}_{λ}(t) = u\overline{λ} + h̄

= h̄ + ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)\overline{λ}g(s)f(u\overline{λ}(s))ds) dτ

≥ h̄ + λ^{\frac{1}{p−1}} ∫_{0}^{1} k(t, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u\overline{λ}(s) + h̄)ds) dτ

= T h̄ ; λ\overline{u}_{λ}(t),

which implies that \overline{u}_{λ}(t) is an upper solution of the operator T h̄ ; λ. Then, by Lemma 2.2, Lemma 2.7 and Lemma 2.8, the problem (2.6), (2.7) has a positive solution, which completes the proof. □

We can now establish the multiplicity of positive solutions by the fixed point index theory.

Proposition 3.2 Let (H1), (H2) and (1.3) be satisfied. If 0 < λ < λ^{∗}, then there exist at least two positive solutions of the problem (1.1), (1.2).

Proof. Just as mentioned above, the arguments are based on fixed point index theory. Exactly, we apply Lemma 2.1 to calculate the indexes of the corresponding operator in two different domains, and then complete the proof by the index theory.

Let K be the set defined in the previous section, namely

K = \{u ∈ E; u(t) ≥ 0, t ∈ [0, 1] and min_{1/8≤t≤7/8} u(t) ≥ \frac{1}{8}∥u∥\} .

To calculate the index of the operator T_{λ}^{0} on some subset of K, we need to check the validity of the conditions in Lemma 2.1. By Lemma 3.1, there exists h_{0} ∈ (0, 1), such that for any h̄ ∈ (0, h_{0}) and λ ∈ (0, λ^{∗}), the problem (2.6), (2.7) admits at least one positive solution. Let v_{λ}(t) be a positive solution of the problem (2.6), (2.7), and Ω = {u ∈ K; u(t) < v_{λ}(t), t ∈ [0, 1]}. It is clear that the set Ω ⊂ K is nonempty, bounded and open. If u ∈ ∂Ω, then there exists t_{0} ∈ [0, 1], such that u(t_{0}) = v_{λ}(t_{0}). Therefore, for any μ ≥ 1, h̄ ∈ (0, h_{0}) and u ∈ ∂Ω, we have

T_{λ}^{0}u(t_{0}) < h̄ + T_{λ}^{0}u(t_{0})

≤ h̄ + T_{λ}^{0}v_{λ}(t_{0})

= T h̄ ; λv_{λ}(t_{0}),

and by Lemma 2.9, we have

T h̄ ; λv_{λ}(t_{0}) = v_{λ}(t_{0}) = u(t_{0}) ≤ μu(t_{0}).

Hence, for any μ ≥ 1, we see that T_{λ}^{0}u ≠ μu, u ∈ ∂Ω. Therefore, by Lemma 2.1,

i(T_{λ}^{0}, Ω, K) = 1.  (3.4)

Now, we calculate the index of the operator T_{λ}^{0} on another relevant subset of K. For this purpose, we check the conditions of Lemma 2.1. Firstly, we check the validity of the condition (1) of the second part of Lemma 2.1. In fact, for any u ∈ K, we have

T_{λ}^{0}u(\frac{1}{2}) = λ^{\frac{1}{p−1}} ∫_{0}^{1} k(\frac{1}{2}, τ)φ_{q} (∫_{0}^{1} k(τ, s)g(s)f(u(s))ds) dτ

≥ λ^{\frac{1}{p−1}} ∫_{1/8}^{7/8} k(\frac{1}{2}, τ)φ_{q} ((\frac{1}{8})^{m+1} ∥u∥^{m} ∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ

≥ λ^{\frac{1}{p−1}} ∥u∥^{m/(p−1)} ∫_{1/8}^{7/8} k(\frac{1}{2}, τ)φ_{q} ((\frac{1}{8})^{m+1} ∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ

≥ λ^{\frac{1}{p−1}} ∥u∥^{(m−p+1)/(p−1)} ∫_{1/8}^{7/8} k(\frac{1}{2}, τ)φ_{q} ((\frac{1}{8})^{m+1} ∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ∥u∥. (3.5)

Choose \overline{R} > 0 such that

λ^{\frac{1}{p−1}} \overline{R}^{(m−p+1)/(p−1)} ∫_{1/8}^{7/8} k(\frac{1}{2}, τ)φ_{q} ((\frac{1}{8})^{m+1} ∫_{1/8}^{7/8} k(s, s)g(s)\overline{δ}ds) dτ > 1.  (3.6)

Therefore, for any R > \overline{R} > 0 and B_{R} ⊂ K, by (3.5) and (3.6) we have

∥T_{λ}^{0}u∥ > ∥u∥ > 0,  u ∈ ∂B_{R},  (3.7)

where B_{R} = {u ∈ K; ∥u∥ < R}. Hence the condition (1) of the second part of Lemma 2.1 is fulfilled. It remains to check the validity of the condition (2) of the second part of Lemma 2.1. In fact, if this condition were not satisfied, then there would exist a function u_{1} ∈ K ∩ ∂B_{R}, 0 < μ_{1} ≤ 1, such that T_{λ}^{0}u_{1} = μ_{1}u_{1}. Then, we have ∥T_{λ}^{0}u_{1}∥ ≤ ∥u_{1}∥, which conflicts with (3.7). Therefore by Lemma 2.1, we have

i(T_{λ}^{0}, B_{R}, K) = 0.  (3.8)

Finally, from (3.8), by the additivity of the fixed point index, we can now complete the proof of the proposition. In fact, we have

0 = i(T_{λ}^{0}, B_{R}, K) = i(T_{λ}^{0}, Ω, K) + i(T_{λ}^{0}, B_{R}\\overline{Ω}, K),

which, together with (3.4), implies that

i(T_{λ}^{0}, B_{R}\\overline{Ω}, K) = −1.  (3.9)

Consequently, from (3.4) and (3.9), by the properties of the fixed point index, there are a fixed point of T_{λ}^{0} in Ω and a fixed point of T_{λ}^{0} in B_{R}\\overline{Ω}, respectively. Therefore by Lemma 2.9, the problem (1.1), (1.2) has at least two positive solutions. The proof is complete.□

Combining Proposition 3.1 with Proposition 3.2, we have completed the proof of Theorem 1.1.

4 The Extension

Just as mentioned in the introduction, our approach for treating the problem (1.1), (1.2) can also be applied to the problem (1.1), (1.4). Since several details are quite similar, we only give a sketch for the proof of Theorem 1.2.

Similar to the proof of the Theorem 1.1, it is necessary to construct an appropriate operator and solve the corresponding operator equation. For this purpose, we first notice that u(t) is a solution of the problem (1.1), (1.4), if and only if u(t) is a solution of the following problem

v′′ = λg(t)f(u),  (4.1)
u′′ = φ_{q}(v),  (4.2)
v(0) = v′(1) = 0,  (4.3)
u(0) = u′(1) = 0,  (4.4)

where φ_{q} is the inverse function of φ_{p}. Because of (4.1) and (4.3), v(t) can be expressed by

v(t) = −λ ∫_{0}^{1} k̂(t, s)g(s)f(u(s))ds,  (4.5)

where k̂(t, s) is the Green function of the equation (4.1) with the boundary value (4.3), and

k̂(t, s) = \{ \begin{matrix}t, & t ≤ s, \\ s, & t ≥ s. \end{matrix}

From (4.2), (4.4), and (4.5), u(t) can be expressed by

u(t) = λ^{\frac{1}{p−1}} ∫_{0}^{1} k̂(t, τ)φ_{q} (∫_{0}^{1} k̂(τ, s)g(s)f(u(s))ds) dτ.

Next, we consider the following approximate problem

(|u′′|^{p−2}u′′)′′ = λg(t)f(u), 0 < t < 1,  (4.6)
u(0) = u′(1) = h ≥ 0, (|u′′(t)|^{p−2}u′′(t))|_{t=0} = (|u′′(t)|^{p−2}u′′(t))′|_{t=1} = 0.  (4.7)

Define an integral operator 𝒯_{λ}^{h} : E → E related to the problem (4.6), (4.7) by

(𝒯_{λ}^{h}u)(t) = (1 + t)h + λ^{\frac{1}{p−1}} ∫_{0}^{1} k̂(t, τ)φ_{q} (∫_{0}^{1} k̂(τ, s)g(s)f(u(s))ds) dτ.  (4.8)

A direct computation shows that Lemma 4.1 Let (H1) and (H2) be satisfied. Then the problem (1.1), (1.4) has a positive solution u if and only if u is a fixed point of 𝒯_{λ}^{0}. And the problem (4.6), (4.7) has a positive solution u if and only if u is a fixed point of 𝒯_{λ}^{h}.

Now, we discuss the properties of the operator 𝒯_{λ}^{h}. From (H1), we can easily obtain the monotonicity of the operator 𝒯_{λ}^{h}. While the complete continuity of 𝒯_{λ}^{h} can be obtained by the following lemma.

Lemma 4.2 Let (H1), (H2) and (1.5) be satisfied. Then the operator 𝒯_{λ}^{h} is completely continuous, and 𝒯_{λ}^{h}K ⊂ K.

Proof. Firstly, we testify the complete continuity of 𝒯_{λ}^{h}. The proof is parallel to that of the front part of Lemma 2.8, and so, we omit the details.

Now, we testify 𝒯_{λ}^{h}K ⊂ K. For each u ∈ K, it is easy to check that (𝒯_{λ}^{0})′′(t) ≤ 0 for 0 < t < 1 and (𝒯_{λ}^{0})(t) ≥ 0 for 0 < t < 1. Then according to Lemma 2.3, the following inequality holds

(𝒯_{λ}^{0}u)(t) ≥ \frac{1}{8}∥𝒯_{λ}^{0}u∥, \frac{1}{8} ≤ t ≤ \frac{7}{8},  (4.9)

which implies that (𝒯_{λ}^{0}u)(t) ∈ K. For any u ∈ K, by (4.8), we have

∥𝒯_{λ}^{h}u∥ ≤ ∥𝒯_{λ}^{0}u∥ + 2h.

On the other hand, by (4.9), we have

min_{1/8≤t≤7/8} 𝒯_{λ}^{h}u(t) ≥ min_{1/8≤t≤7/8} 𝒯_{λ}^{0} + min_{1/8≤t≤7/8}((1 + t)h)

≥ min_{1/8≤t≤7/8} 𝒯_{λ}^{0} + \frac{9}{8}h

≥ \frac{1}{8}∥𝒯_{λ}^{0}u∥ + \frac{9}{8}h

≥ \frac{1}{8}∥𝒯_{λ}^{h}u∥,

which implies that 𝒯_{λ}^{h}K ⊂ K. The proof is complete. □

Similar to Lemma 2.10, we obtain the following a priori estimates on positive solutions of the problem (1.1), (1.4).

Lemma 4.3 Let (H1), (H2) and (1.5) be satisfied. Suppose that λ ∈ S and S_{1} = (λ, +∞) ∩ S ≢ ∅, where S={λ > 0; such that the problem (1.1), (1.4) has at least one positive solution }. Then there exists R(λ) > 0, such that ∥u_{λ′}∥ ≤ R(λ), where λ′ ∈ S_{1}, and u_{λ′} ∈ K is a positive solution of the problem (1.1), (1.4) with λ replaced by λ′.

The Proof of Theorem 1.2. The proof is parallel to that of Theorem 1.1, and so, we omit the details. □

Acknowledgments

The authors would like to express their deep thanks to the referees for their suggestions of revising the manuscript.

References

[1] A. R. Aftabizadeh, Existence and uniqueness theorems for fourth-order boundary value problem, J. Math. Anal. Appl., 116(2)(1986), 415–426.

[2] C. P. Gupta, Existence and uniqueness theorems for the bending of an elastic beam equation, App. Anal., 26(4)(1988), 289–304.

[3] R. Dalmasso, Positive solutions of singular boundary value problems, Nonlinear Analysis, 27(6)(1996), 645–652.

[4] D. Guo, V. Lakshmikantham, X. Z. Liu, Nonlinear integral equations in abstract cones, Academic Press, New York, 1988.

[5] M. Q. Feng, W. G. Ge, Existence of positive solutions for singular eigenvalue problems, Electronic J. Differential Equations, 2006(105)(2006), 1–9.

[6] P. Dráek, M. Ôani, Global bifurcation result for the p-biharmonic operator, Electron. J. Differential Equations, 2001(48)(2001), 1–9.

[7] J. Benedikt, Uniqueness theorem for p-biharmonic equations, Electron. J. Differential Equations, 2002(53)(2002), 1–17.

[8] C. H. Jin, J. X. Yin, Positive solutions for the boundary value problems of one dimensional p-Laplacian with delay, J. Math. Anal. Appl., 330(2)(2007), 1238–1248.

[9] R. Ma, Multiple positive solutions for a semipositone fourth-order bounadry value problem, Hiroshima Math. J., 33 (2)(2003), 217–227.

[10] Z. Bai and H. Wang, On positive solutions of some nonlinear fourth-order beam equations, J. Math. Anal. Appl., 270(2)(2002), 357–368.

[11] J.R. Graef and B. Yang, On a nonlinear boundary value problem for fourth order equations, Appl. Anal., 72 (3–4) (1999), 439–448.

[12] R. Ma, Existence and uniqueness theorems for some fourth-order nonlinear boundary value problems, Int. J. Math. Math. Sci., 23(11) (2000), 783–788.

[13] Y. Yang, Fourth-order two-point boundary value problems, Proc. Amer. Math. Soc., 104 (1) (1988), 175–180.

[14] Q. Yao, Positive solutions for eigenvalue problems of fourth-order elastic beam equations, Appl. Math. Lett. 17(2) (2004), 237–243.

[15] T. Chan, A. Marquina, and P. Mulet, High-order total variation-based image restoration, SIAM J. Sci. Comput., 22 (2)(2000), 503–516.

[16] M. Lysaker, A. Lundervold, and X.-C. Tai, Noise removal using fourth-order partial differential equations with applications to medical magnetic resonance images in space and time, IEEE Trans. Image Process., 12(2003), 1579–1590.

[17] J. Tumblin and G. Turk, LCIS: A boundary hierarchy for detail-preserving contrast reduction, in Proceedings of the 26th Annual Conference on Computer Graphics and Interactive Techniques, ACM Press/Addison-Wesley, New York, 1999, 83–90.

[18] G. W. Wei, Generalized Perona-Malik equation for image processing, IEEE Signal Processing Letters, 6(7)(1999), 165–167.

[19] Y. L. You and M. Kaveh, Fourth-order partial differential equations for noise removal, IEEE Trans. Image Process., 9 (2000), 1723–1730.

(Received December 28, 2009)