Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 38, 1-14; http://www.math.u-szeged.hu/ejqtde/

Green’s function and positive solutions of a singular nth-order three-point boundary value problem on time scales

Dapeng Xie^{a,}^{∗}, Yang Liu^{a}, Chuanzhi Bai^{b}

a Department of Mathematics, Hefei Teachers College, Hefei, Anhui 230061, P R China b Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsu 223300, P R China

Abstract

In this paper, we investigate the existence of positive solutions for a class of singular nth-order three-point boundary value problem. The associated Green’s function for the boundary value problem is given at first, and some useful properties of the Green’s function are obtained. The main tool is fixed-point index theory. The results obtained in this paper essentially improve and generalize some well-known results.

Mathematics Subject Classification(2000): 34B16; 34B18; 34B27. Keywords: Green’s function; Positive solution; Three-point boundary value problem; Cone; Singular.

1. Introduction

In recent years, the existence of positive solutions for higher-order boundary value problems has been studied by many authors using various methods (see [1-4, 7-14] and the references therein). For example, in paper [3], by using the Krasnosel’skii fixed point theorem, Eloe and Ahmad established the existence of at least one positive solution for the following nth-order three-point boundary value problem

\{ \begin{matrix}u^{(n)}(t) + a(t)f(u(t)) = 0, t ∈ (0, 1), \\ u(0) = 0, u′(0) = 0, ⋯, u^{(n−2)}(0) = 0, u(1) = αu(η), \end{matrix} (1.1)

where 0 < η < 1, 0 < αη^{n−1} < 1, a : [0, 1] → [0, ∞) is continuous and nonsingular, f ∈ C([0, ∞), [0, ∞)) is either superlinear or sublinear.

In this paper, we study the existence of positive solutions for a singular nth-order three-point boundary value problem as follows

\{ \begin{matrix}u^{(n)}(t) + h(t)f(t, u(t)) = 0, t ∈ [a, b], \\ u(a) = αu(η), u′(a) = 0, ⋯, u^{(n−2)}(a) = 0, u(b) = βu(η), \end{matrix} (1.2)

∗E-mail address: xiedapeng9@yahoo.com.cn

where a < η < b, 0 ≤ α < 1, 0 < β(η − a)^{n−1} < (1 − α)(b − a)^{n−1} + α(η − a)^{n−1}, f ∈ C([a, b] × [0, ∞), [0, ∞)) and h ∈ C([a, b], [0, +∞)) may be singular at t = a and t = b.

Up to now, no paper has appeared in the literature which discusses the existence of positive solutions for the problem (1.2). This paper attempts to fill this gap in the literature. In order to obtain our result, we give at first the associated Green’s function for the problem (1.2), which is the base for further discussion. Our results extend and improve the results of Eloe and Ahmad [3] (α = a = 0, b = 1 and f(t, u) = f(u)). Our results are obtained under certain suitable weaker conditions than that in [3]. It is also noted that our method here is different from that of Eloe and Ahmad [3].

2. Expression and properties of Green’s function

Lemma 2.1. If y ∈ C[a, b], then the problem

\{ \begin{matrix}u^{(n)}(t) + y(t) = 0, a ≤ t ≤ b, \\ u(a) = 0, u′(a) = 0, ⋯, u^{(n−2)}(a) = 0, u(b) = 0, \end{matrix} (2.1)

has a unique solution

w(t) = ∫_{a}^{b} H(t, s)y(s)ds, (2.2)

where

H(t, s) = \{ \begin{matrix}\frac{(b − s)^{n−1}(t − a)^{n−1} − (b − a)^{n−1}(t − s)^{n−1}}{(b − a)^{n−1}(n − 1)!} , & a ≤ s ≤ t ≤ b, \\ \frac{(b − s)^{n−1}(t − a)^{n−1}}{(b − a)^{n−1}(n − 1)!} , & a ≤ t < s ≤ b. \end{matrix} (2.3)

Proof. In fact, if w(t) is a solution of the problem (2.1), then we may suppose that

w(t) = − ∫_{a}^{t} \frac{(t − s)^{n−1}}{(n − 1)!} y(s)ds + A(t − a)^{n−1} + ∑_{i=1}^{n−2} A_{i}(t − a)^{i} + B.

Since w^{(i)}(a) = 0 for i = 0, 1, 2, ⋯, n − 2, we get B = A_{i} = 0 for i = 1, 2, ⋯, n − 2. By w(b) = 0, we have

A = \frac{1}{(b − a)^{n−1}} ∫_{a}^{b} \frac{(b − s)^{n−1}}{(n − 1)!} y(s)ds.

Therefore, the problem (2.1) has a unique solution

w(t) = − ∫_{a}^{t} \frac{(t − s)^{n−1}}{(n − 1)!} y(s)ds + ∫_{a}^{b} \frac{(b − s)^{n−1}(t − a)^{n−1}}{(b − a)^{n−1}(n − 1)!} y(s)ds = ∫_{a}^{b} H(t, s)y(s)ds,

where H(t, s) is defined by (2.3).

Lemma 2.2. H(t, s) has the following properties (i) 0 ≤ H(t, s) ≤ k(s), ∀t, s ∈ [a, b], where

k(s) = \frac{(s − a)(b − s)^{n−1}}{(b − a)(n − 2)!} ;

(ii) H(t, s) ≥ ϕ(t)k(s), ∀t, s ∈ [a, b], where

ϕ(t) = \{ \begin{matrix}\frac{(t − a)^{n−1}}{(n − 1)(b − a)^{n−1}} , & a ≤ t ≤ \frac{a+b}{2}, \\ \frac{(b − t)(t − a)^{n−2}}{(n − 1)(b − a)^{n−1}} , & \frac{a+b}{2} ≤ t ≤ b. \end{matrix}

Proof. It is obvious that H(t, s) is nonnegative. Moreover,

H(t, s) = \{ \begin{matrix}\frac{(b − s)^{n−1}(t − a)^{n−1} − (b − a)^{n−1}(t − s)^{n−1}}{(b − a)^{n−1}(n − 1)!} , & a ≤ s ≤ t ≤ b, \\ \frac{(b − s)^{n−1}(t − a)^{n−1}}{(b − a)^{n−1}(n − 1)!} , & a ≤ t < s ≤ b, \end{matrix}

= \frac{1}{(b − a)^{n−1}(n − 1)!}\{ \begin{matrix}(s − a)(b − t)\{[(t − a)(b − s)]^{n−2} \\ +[(t − a)(b − s)]^{n−3}(b − a)(t − s) + ⋯ \\ +(t − a)(b − s)[(b − a)(t − s)]^{n−3} \\ +[(b − a)(t − s)]^{n−2}\}, & a ≤ s ≤ t ≤ b, \\ (b − s)^{n−1}(t − a)^{n−1}, & a ≤ t < s ≤ b, \end{matrix}

≤ \frac{1}{(b − a)^{n−1}(n − 1)!}\{ \begin{matrix}(n − 1)(s − a)(b − s)[(b − a)(b − s)]^{n−2}, & a ≤ s ≤ t ≤ b, \\ (b − s)^{n−1}(s − a)^{n−1}, & a ≤ t < s ≤ b, \end{matrix}

≤ \frac{(s − a)(b − s)^{n−1}}{(b − a)(n − 2)!} = k(s), t, s ∈ [a, b].

Thus, (i) holds.

If s = a or s = b, we easily see that (ii) holds. If s ∈ (a, b) and t ∈ [a, b], we have

\frac{H(t, s)}{k(s)} = \{ \begin{matrix}\frac{(b − s)^{n−1}(t − a)^{n−1} − (b − a)^{n−1}(t − s)^{n−1}}{(n − 1)(s − a)(b − s)^{n−1}(b − a)^{n−2}} , & a ≤ s ≤ t ≤ b, \\ \frac{(t − a)^{n−1}}{(n − 1)(s − a)(b − a)^{n−2}} , & a ≤ t < s ≤ b, \end{matrix}

= \{ \begin{matrix}\frac{1}{(n − 1)(s − a)(b − s)^{n−1}(b − a)^{n−2}} (s − a)(b − t)\{[(t − a)(b − s)]^{n−2} \\ +[(t − a)(b − s)]^{n−3}(b − a)(t − s) + ⋯ \\ +(t − a)(b − s)[(b − a)(t − s)]^{n−3} + [(b − a)(t − s)]^{n−2}\}, & a ≤ s ≤ t ≤ b, \\ \frac{(t − a)^{n−1}}{(n − 1)(s − a)(b − a)^{n−2}} , & a ≤ t < s ≤ b, \end{matrix}

≥ \{ \begin{matrix}\frac{(s − a)(b − t)(t − a)^{n−2}(b − s)^{n−2}}{(n − 1)(s − a)(b − s)^{n−1}(b − a)^{n−2}} , & a ≤ s ≤ t ≤ b, \\ \frac{(t − a)^{n−1}}{(n − 1)(s − a)(b − a)^{n−2}} , & a ≤ t < s ≤ b, \end{matrix}

≥ \{ \begin{matrix}\frac{(b − t)(t − a)^{n−2}}{(n − 1)(b − a)^{n−1}} , & a ≤ s ≤ t ≤ b, \\ \frac{(t − a)^{n−1}}{(n − 1)(b − a)^{n−1}} , & a ≤ t < s ≤ b, \end{matrix}

since,

(b − t)(t − a)^{n−2} ≥ (t − a)^{n−1}, for t ∈ [a, \frac{a+b}{2}],

and

(t − a)^{n−1} ≥ (b − t)(t − a)^{n−2}, for t ∈ [\frac{a+b}{2}, b],

then

H(t, s) ≥ ϕ(t)k(s), for s ∈ (a, b) and t ∈ [a, b].

Thus, (ii) holds. The proof is completed.

Theorem 2.3. Suppose that Δ =: (1−α)(b−a)^{n−1} +(α−β)(η −a)^{n−1} ≠ 0, then for any y ∈ C[a, b], the problem

\{ \begin{matrix}u^{(n)}(t) + y(t) = 0, a ≤ t ≤ b, \\ u(a) = αu(η), u′(a) = 0, ⋯, u^{(n−2)}(a) = 0, u(b) = βu(η), \end{matrix} (2.4)

has a solution

u(t) = ∫_{a}^{b} G(t, s)y(s)ds,

where

G(t, s) = H(t, s) + \frac{1}{Δ}\{α[(b − a)^{n−1} − (t − a)^{n−1}] + β(t − a)^{n−1}\}H(η, s), (2.5)

here H(t, s) is given by (2.3).

Proof. The three-point boundary value problem (2.4) can be obtained from replacing u(a) = 0 by u(a) = αu(η) and u(b) = 0 by u(b) = βu(η) in (2.1). Thus, we suppose the solution of the three-point boundary value problem (2.4) can be expressed by

u(t) = w(t) + ∑_{j=0}^{n−1} γ_{j}(t − a)^{j}w(η), (2.6)

where w(t) is given as (2.2), γ_{j} (j = 0, 1, 2, ⋯, n − 1) are constants that will be determined. By calculating, we obtain

u^{(i)}(t) = w^{(i)}(t) + ∑_{j=i}^{n−1} \frac{j!}{(j − i)!}γ_{j}(t − a)^{(j−i)}w(η). (2.7)

Since u^{(i)}(a) = 0 for i = 1, 2, ⋯, n − 2 and (2.7), we obtain

γ_{j} = 0, for j = 1, 2, ⋯, n − 2. (2.8) In view of (2.6) and (2.8), we have

u(t) = w(t) + [γ_{0} + γ_{n−1}(t − a)^{n−1}]w(η). (2.9) Then, by u(a) = αu(η) and u(b) = βu(η) (note (2.9)), we get

γ_{0}w(η) = αw(η) + α[γ_{0} + γ_{n−1}(η − a)^{n−1}]w(η), and

[γ_{0} + γ_{n−1}(b − a)^{n−1}]w(η) = βw(η) + β[γ_{0} + γ_{n−1}(η − a)^{n−1}]w(η).

From this

\{ \begin{matrix}γ_{0} = α[1 + γ_{0} + γ_{n−1}(η − a)^{n−1}], \\ γ_{0} + γ_{n−1}(b − a)^{n−1} = β[1 + γ_{0} + γ_{n−1}(η − a)^{n−1}]. \end{matrix} (2.10)

By calculating, we obtain by (2.10) that

\{ \begin{matrix}γ_{0} = \frac{α}{1 − α} + \frac{α(β − α)(η − a)^{n−1}}{(1 − α) · Δ} , \\ γ_{n−1} = \frac{β − α}{Δ}. \end{matrix} (2.11)

Hence, by (2.9) and (2.11), we obtain

u(t) = w(t) + [\frac{α}{1 − α} + \frac{α(β − α)(η − a)^{n−1}}{(1 − α) · Δ} + \frac{(β − α)(t − a)^{n−1}}{Δ} ]w(η)

= w(t) + \frac{1}{Δ}\{β(t − a)^{n−1} + α[(b − a)^{n−1} − (t − a)^{n−1}]\}w(η).

Using this and (2.2), we see that

u(t) = ∫_{a}^{b} H(t, s)y(s)ds + \frac{1}{Δ}\{β(t − a)^{n−1} + α[(b − a)^{n−1} − (t − a)^{n−1}]\} ∫_{a}^{b} H(η, s)y(s)ds.

Thus, the Green’s function G(t, s) for the BVP (2.4) is described by (2.5).

By Theorem 2.3, we obtain the following corollary. Corollary 2.4. Suppose that Δ =: (1−α)(b−a)^{n−1} +(α−β)(η−a)^{n−1} ≠ 0, then for any y ∈ C[a, b], the problem (2.4) has a unique solution

u(t) = ∫_{a}^{b} G(t, s)y(s)ds, where G(t, s) is given as in (2.5). Proof. We need only prove the uniqueness. Suppose that u_{1}(t) is also a solution of the problem (2.4). Let

x(t) = u_{1}(t) − u(t), t ∈ [a, b]. Obviously,

x^{(n)}(t) = u_{1}^{(n)}(t) − u^{(n)}(t) = 0, t ∈ [a, b], (2.12) x(a) = u_{1}(a) − u(a) = αu_{1}(η) − αu(η) = αx(η), (2.13) x^{(i)}(a) = u_{1}^{(i)}(a) − u^{(i)}(a) = 0, i = 1, 2, ⋯, n − 2, (2.14) x(b) = u_{1}(b) − u(b) = βu_{1}(η) − βu(η) = βx(η). (2.15)

In view of (2.12), we have

x(t) = c_{0} + c_{1}(t − a) + c_{2}(t − a)^{2} + ⋯ + c_{n−1}(t − a)^{n−1}, (2.16) where c_{j} (j = 0, 1, 2, ⋯, n − 1) are undetermined constants. By calculating, one has

x^{(i)}(t) = ∑_{j=i}^{n−1} \frac{j!}{(j − i)!}c_{j}(t − a)^{(j−i)}, i = 1, 2, ⋯, n − 1. (2.17)

From (2.14) and (2.17), we get

c_{1} = c_{2} = ⋯ = c_{n−2} = 0, (2.18) By (2.13), (2.15), (2.16) and (2.18), one has

c_{0} = α[c_{0} + c_{n−1}(η − a)^{n−1}], (2.19) and

c_{0} + c_{n−1}(b − a)^{n−1} = β[c_{0} + c_{n−1}(η − a)^{n−1}]. (2.20) Thus, by (2.19), (2.20) and Δ ≠ 0, we can easily know that c_{0} = c_{n−1} = 0. So x(t) = 0, t ∈ [a, b], which implies that the solution of the problem (2.4) is unique. Theorem 2.5. G(t, s) has the following properties

(i) G(t, s) ≥ 0, ∀t, s ∈ [a, b]; (ii) G(t, s) ≤ M_{1}k(s), ∀t, s ∈ [a, b], where k(s) as in Lemma 2.2, and

M_{1} = max \{\frac{1}{Δ}[(b − a)^{n−1} + α(η − a)^{n−1}], \frac{1}{Δ}\{(b − a)^{n−1} + β[(b − a)^{n−1} − (η − a)^{n−1}]\}\};

(iii) min_{t∈[τ, \frac{a+b}{2} ]} G(t, s) ≥ M_{2}k(s), ∀t, s ∈ [a, b], where τ ∈ (a, \frac{a+b}{2}) and

M_{2} = \frac{(τ − a)^{n−1}}{(n − 1)(b − a)^{n−1}} + \frac{1}{Δ}[β(τ − a)^{n−1} + \frac{α(2^{n−1} − 1)(b − a)^{n−1}}{2^{n−1}} ]ϕ(η),

here ϕ(t) and k(s) as in Lemma 2.2. Proof. It is clear that (i) holds. Next we will divide the proof of (ii) into two cases. Case (1) If a ≤ t ≤ η, then, by Lemma 2.2 (i), we have

G(t, s) = H(t, s) + \frac{1}{Δ}\{β(t − a)^{n−1} + α[(b − a)^{n−1} − (t − a)^{n−1}]\}H(η, s)

≤ k(s) + \frac{1}{Δ}[β(η − a)^{n−1} + α(b − a)^{n−1}]k(s)

= \frac{1}{Δ}[(b − a)^{n−1} + α(η − a)^{n−1}]k(s) ≤ M_{1}k(s).

Case (2) If η ≤ t ≤ b, similarly, we obtain

G(t, s) = H(t, s) + \frac{1}{Δ}\{β(t − a)^{n−1} + α[(b − a)^{n−1} − (t − a)^{n−1}]\}H(η, s)

≤ k(s) + \frac{1}{Δ}\{β(b − a)^{n−1} + α[(b − a)^{n−1} − (η − a)^{n−1}]\}k(s)

= \frac{1}{Δ}\{(b − a)^{n−1} + β[(b − a)^{n−1} − (η − a)^{n−1}]\}k(s) ≤ M_{1}k(s).

By the inequality above, we know that (ii) holds.

(iii) By Lemma 2.2 (ii), we have

min_{t∈[τ, \frac{a+b}{2} ]} G(t, s) = min_{t∈[τ, \frac{a+b}{2} ]}\{H(t, s) + \frac{1}{Δ}\{β(t − a)^{n−1} + α[(b − a)^{n−1} − (t − a)^{n−1}]\}H(η, s)\}

≥ min_{t∈[τ, \frac{a+b}{2} ]}\{ϕ(t)k(s) + \frac{1}{Δ}[β(τ − a)^{n−1} + \frac{α(2^{n−1} − 1)(b − a)^{n−1}}{2^{n−1}} ]ϕ(η)k(s)\}

= \{\frac{(τ − a)^{n−1}}{(n − 1)(b − a)^{n−1}} + \frac{1}{Δ}[β(τ − a)^{n−1} + \frac{α(2^{n−1} − 1)(b − a)^{n−1}}{2^{n−1}} ]ϕ(η)\}k(s)

= M_{2}k(s).

Thus, (iii) holds.

3. Main results Throughout this paper, we assume the following conditions hold.

(H_{1}) h : [a, b] → [0, +∞) is continuous, and

0 < ∫_{a}^{b}(b − s)^{n−1}h(s)ds < +∞. (H_{2}) f : [a, b] × [0, +∞) → [0, +∞) is continuous.

Let C[a, b] be Banach space with the norm ∥u∥ = max_{a≤t≤b} |u(t)|. Define the cones K and P by

K = {u ∈ C[a, b] : u(t) ≥ 0, t ∈ [a, b]}, and

P = \{u ∈ C[a, b] : u(t) ≥ 0, t ∈ [a, b], and min_{t∈[τ, \frac{a+b}{2} ]} u(t) ≥ γ∥u∥\},

where τ ∈ (a, \frac{a+b}{2}), γ := \frac{M_{2}}{M_{1}} (here M_{1} and M_{2} are defined as in Theorem 2.5). Obviously, K and P are cones of C[a, b]. Define the operators A_{1}, A_{2} and T by

(A_{1}u)(t) = ∫_{a}^{b} G(t, s)h(s)u(s)ds, ∀t ∈ [a, b], (3.1)

(A_{2}u)(t) = ∫τ^{\frac{a+b}{2}} G(t, s)h(s)u(s)ds, ∀t ∈ [a, b], (3.2)

and

(T u)(t) = ∫_{a}^{b} G(t, s)h(s)f(s, u(s))ds, ∀t ∈ [a, b], (3.3) It is clear that the problem (1.2) has a positive solution u = u(t) if and only if u is a fixed point of T. Lemma 3.1. Suppose (H_{0}) − (H_{2}) hold. Then

(i) The operator A_{1} : C[a, b] → C[a, b] is completely continuous and satisfies A_{1}(K) ⊂ K. (ii) The operator A_{2} : C[a, b] → C[a, b] is completely continuous and satisfies A_{2}(K) ⊂ K. (iii) The operator T : P → C[a, b] is completely continuous and satisfies T (P) ⊂ P. Proof. It is obvious that (i) and (ii) hold. By Theorem 2.3, we know that T (P) ⊂ P. Next, we will prove that the operator T is completely continuous.

For m ≥ 2, define h_{m} by

h_{m}(t) = \{ \begin{matrix}inf_{a≤s< \frac{am+b}{m+1}} h(s), & a ≤ t < \frac{am+b}{m+1}, \\ h(t), & \frac{am+b}{m+1} ≤ t ≤ \frac{a+bm}{m+1}, \\ inf_{\frac{a+bm}{m+1} <s≤b} h(s), & \frac{a+bm}{m+1} < t ≤ b, \end{matrix}

and define the operator T_{m} as follows

(T_{m}u)(t) = ∫_{a}^{b} G(t, s)h_{m}(s)f(s, u(s))ds, t ∈ [a, b]. It is easy to show that the operator T_{m} is compact on P for all m ≥ 2 by using Arzela-Ascoli theorem. In addition, the continuity of G(t, s)h_{m}(s) on [a, b] × [a, b] implies the continuity of T_{m} : P → P.

Therefore, T_{m} is a completely continuous operator. It follows from (H_{1}) that

0 < ∫_{a}^{b} k(s)h(s) < \frac{1}{(n − 1)!} ∫_{a}^{b}(b − s)^{n−1}h(s)ds < ∞.

Using this and by the absolute continuity of the integral, we have

lim_{m→∞} ∫_{e(m)} k(s)h(s)ds = 0,

where k(s) is given as in Lemma 2.2, e(m) = [a, \frac{am+b}{m+1}] ∪ [\frac{a+bm}{m+1}, b]. For each R > 0, set Ω_{R} = {u ∈ P : ∥u∥ ≤ R} and M = max_{t∈[a,b],u∈[0,R]} f(t, u). Then fix R > 0 and u ∈ Ω_{R}, we have from Theorem 2.5 and (H_{2}) that

|(T_{m}u)(t) − (T u)(t)| = |∫_{a}^{b} G(t, s)(h_{m}(s) − h(s))f(s, u(s))ds|

≤ M_{1}M ∫a^{\frac{am+b}{m+1}} k(s)|h(s) − h_{m}(s)|ds + M_{1}M ∫ b_{\frac{a+bm}{m+1}} k(s)|h(s) − h_{m}(s)|ds

≤ M_{1}M ∫_{e(m)} k(s)h(s)ds → 0, (m → ∞).

Hence, the completely continuous operator T_{m} converges uniformly to T as m → ∞ on any bounded subset of P, and T : P → P is completely continuous.

By Virtue of Krein-Rutmann theorems, It is easy to see that the following lemma holds.

Lemma 3.2 [6]. Suppose A is a completely continuous operator and A(K) ⊂ K. If there exists ψ ∈ C[a, b]\(−K) and a constant c > 0 such that cKψ ≥ ψ, then the spectral radius r(A) ≠ 0 and A has a positive eigenfunction φ corresponding to its first eigenvalue λ = (r(A))^{−1}, that is, φ = λAφ.

Lemma 3.3. Suppose that (H_{0}) and (H_{1}) hold. Then the spectral radius r(A_{1}) ≠ 0 and A_{1} has a positive eigenfunction corresponding to its first eigenvalue λ_{1} = (r(A_{1}))^{−1}.

Proof. It follows from (2.5) that

G(t, t) ≥ H(t, t) = \frac{(b − t)^{n−1}(t − a)^{n−1}}{(b − a)^{n−1}(n − 1)!} > 0, ∀t ∈ (a, b).

From this and (H_{1}), we know that there exists t_{0} ∈ (a, b), such that G(t_{0}, t_{0})h(t_{0}) > 0, then there is [α, β] such that t_{0} ∈ (α, β) and G(t, s)h(s) > 0, ∀t, s ∈ [α, β]. We take ψ ∈ C[a, b] such that ψ(t) ≥ 0, ∀t ∈ [a, b], ψ(t_{0}) > 0 and ψ(t) = 0, ∀t ∉ [α, β]. Then for all t ∈ [a, b], we have

(A_{1}ψ)(t) = ∫_{a}^{b} G(t, s)h(s)ψ(s)ds = ∫_{α}^{β} G(t, s)h(s)ψ(s)ds > 0.

Then there exists a constant c > 0 such that c(A_{1}ψ) ≥ ψ, ∀t ∈ [a, b]. By Lemma 3.2, we see that the spectral radius r(A_{1}) ≠ 0 and A_{1} has a positive eigenfunction corresponding to its first eigenvalue λ_{1} = (r(A_{1}))^{−1}.

To establish the existence of positive solutions of the problem (1.2), we will employ the following lemmas. Lemma 3.4 [5]. Let E be Banach space, P be a cone in E, and Ω be a bounded open set in E. Suppose that T : P ∩ \overline{Ω} → P is a completely continuous operator. If there exists u_{0} ∈ P \\{θ\} such that

u − T u ≠ μu_{0}, ∀u ∈ P ∩ ∂Ω, μ ≥ 0.

Then the fixed point index i(T, P ∩ \overline{Ω}, P) = 0.

Lemma 3.5 [5]. Let E be Banach space, P be a cone in E, and Ω be a bounded open set in E. Suppose that T : P ∩ \overline{Ω} → P is a completely continuous operator. If

T u ≠ μu, ∀u ∈ P ∩ ∂Ω, μ ≥ 1.

Then the fixed point index i(T, P ∩ Ω, P) = 1.

For convenience, we introduce the following notations

f_{0} = lim inf_{u→0^{+}}min_{a≤t≤b}(f(t, u)/u), f^{0} = lim sup_{u→0^{+}}max_{a≤t≤b}(f(t, u)/u), f_{∞} = lim inf_{u→∞}min_{a≤t≤b}(f(t, u)/u), f^{∞} = lim sup_{u→+∞}max_{a≤t≤b}(f(t, u)/u).

Theorem 3.6. Let λ_{1} be the first eigenvalue of A_{1} defined as in (3.1). Suppose the previous hypotheses (H_{0}) − (H_{2}) hold, in addition we assume f_{0} > λ_{1} and f^{∞} < λ_{1}. Then problem (1.2) has at least one positive solution.

Proof. In view of f_{0} > λ_{1}, there exists R_{1} > 0, such that

f(t, u) ≥ λ_{1}u, for all t ∈ [a, b], u ∈ [0, R_{1}]. (3.4)

Let Ω_{R_{1}} = {u ∈ C[a, b] : ∥u∥ < R_{1}}, for u ∈ P ∩ ∂Ω_{R_{1}}, we have by (3.4) that

(T u)(t) = ∫_{a}^{b} G(t, s)h(s)f(s, u(s))ds ≥ λ_{1} ∫_{a}^{b} G(t, s)h(s)u(s)ds = λ_{1}(A_{1}u)(t), ∀t ∈ [a, b]. (3.5)

Let u^{∗} be the positive eigenfunction of A_{1} corresponding to λ_{1}, thus u^{∗} = λ_{1}A_{1}u^{∗}. We may suppose that T has no fixed point on P ∩ ∂Ω_{R_{1}}, otherwise, the proof is finished. In the following we will show that

u − T u ≠ μu^{∗}, ∀u ∈ P ∩ ∂Ω_{R_{1}}, μ ≥ 0. (3.6)

If (3.6) is not true, then there is u_{0} ∈ P ∩ ∂Ω_{R_{1}} and μ_{0} ≥ 0 such that u_{0} − T u_{0} = μ_{0}u^{∗}. It is clear that μ_{0} > 0 and

u_{0} = T u_{0} + μ_{0}u^{∗} ≥ μ_{0}u^{∗}.

Set

μ^{∗} = sup{μ : u_{0} ≥ μu^{∗}}. (3.7) Obviously, μ^{∗} ≥ μ_{0} > 0. It follows from A_{1}(K) ⊂ K that

λ_{1}A_{1}u_{0} ≥ μ^{∗}λ_{1}A_{1}u^{∗} = μ^{∗}u^{∗}. Using this and (3.5), we have

u_{0} = T u_{0} + μ_{0}u^{∗} ≥ λ_{1}A_{1}u_{0} + μ_{0}u^{∗} ≥ μ^{∗}u^{∗} + μ_{0}u^{∗}, which contradicts (3.7). Thus, (3.6) holds. By Lemma 3.4, we have

i(T, P ∩ Ω_{R_{1}}, P) = 0. (3.8) On the other hand, it follows from f^{∞} < λ_{1} that there are 0 < ρ < 1 and R_{2} > R_{1} such that

f(t, u) ≤ ρλ_{1}u, ∀t ∈ [a, b], u ≥ R_{2}. (3.9) Put

B = \{u ∈ K : u = σT u, 0 ≤ σ ≤ 1\} , \overline{u}(t) = min\{u(t), R_{2}\} and w(t) = \{t ∈ [a, b] : u(t) > R_{2}\}. Now we will show that B is bounded. For all u in B, we have by (3.9) and Theorem 2.5 (ii) that

u(t) = σ(T u)(t) ≤ ∫_{a}^{b} G(t, s)h(s)f(s, u(s))ds

= ∫_{w(t)} G(t, s)h(s)f(s, u(s))ds + ∫_{[a,b]\w(t)} G(t, s)h(s)f(s, u(s))ds
≤ ρλ_{1} ∫_{a}^{b} G(t, s)h(s)u(s)ds + M_{1} ∫_{a}^{b} k(s)h(s)f(s, \overline{u}(s))ds
= ρλ_{1}(A_{1}u)(t) + M = (\overline{A}_{1}u)(t) + M.

where \overline{A}_{1}u = ρλ_{1}A_{1}u, u ∈ C[a, b] and M = M_{1} sup_{u∈P ∩\overline{Ω}_{R_{2}}} ∫_{a}^{b} k(s)h(s)f(s, \overline{u}(s))ds. Thus, ((I − \overline{A}_{1})u)(t) ≤ M, t ∈ [a, b]. Since λ_{1} is the first eigenvalue of A_{1} and 0 < ρ < 1, then the first eigenvalue of \overline{A}_{1}, (r(\overline{A}_{1}))^{−1} > 1. Thus, the inverse operator (I − \overline{A}_{1})^{−1} exists and

(I − \overline{A}_{1})^{−1} = I + \overline{A}_{1} + \overline{A}_{1}^{2} + ⋯ + \overline{A}_{1}^{n} + · · ·. In view of \overline{A}_{1}(P) ⊂ P, we get that (I − \overline{A}_{1})^{−1}(P) ⊂ P. Then we have u(t) ≤ (I − \overline{A}_{1})^{−1}M, t ∈ [a, b] and B is bounded.

Take R_{3} > max{R_{2}, sup B}, let Ω_{R_{3}} = {u ∈ C[a, b] : ∥u∥ < R_{3}}. Then by Lemma 3.5, one has

EJQTDE, 2009 No. 38, p. 11

i(T, P ∩ Ω_{R_{3}}, P) = i(θ, P ∩ Ω_{R_{3}}, P) = 1. (3.10)

It follows from (3.8) and (3.10) that

i(T, (P ∩ Ω_{R_{3}}) \ (P ∩ \overline{Ω}_{R_{1}}), P) = i(θ, P ∩ Ω_{R_{3}}, P) − i(θ, P ∩ Ω_{R_{1}}, P) = 1.

Hence, T has at least one fixed point on (P ∩ Ω_{R_{3}}) \ (P ∩ \overline{Ω}_{R_{1}}). This implies that the problem (1.2) has at least one positive solution.

Theorem 3.7. Let λ_{1} be the first eigenvalue of A_{1} defined as in (3.1), λ_{2} be the first eigenvalue of A_{2} defined as in (3.2). Suppose the previous hypotheses (H_{0}) − (H_{2}) hold, in addition we assume f^{0} < λ_{1} and f_{∞} > λ_{2}. Then problem (1.2) has at least one positive solution.

Proof. From f^{0} < λ_{1}, we know that there exists R_{4} > 0, such that

f(t, u) ≤ λ_{1}u, ∀t ∈ [a, b], u ∈ [0, R_{4}]. (3.11)

Let Ω_{R_{4}} = {u ∈ C[a, b] : ∥u∥ < R_{4}}, for u ∈ P ∩ ∂Ω_{R_{4}}, we have by (3.11) that (T u)(t) ≤ λ_{1}(A_{1}u)(t). Now we show that

i(T, P ∩ Ω_{R_{4}}, P) = 1. (3.12)

We may suppose that T has no fixed point on P ∩ ∂Ω_{R_{4}}, otherwise, the proof is finished. In the following we will show that

T u ≠ ϱu, ∀u ∈ P ∩ ∂Ω_{R_{4}}, ϱ ≥ 1. (3.13)

If otherwise, there exist u_{0} ∈ P ∩ ∂Ω_{R_{4}} and ϱ_{0} ≥ 1 such that T u_{0} = ϱ_{0}u_{0}. So ϱ_{0} > 1 and ϱ_{0}u_{0} = T u_{0} ≤ λ_{1}A_{1}u_{0}. From induction, we get that ϱ_{0}^{j}u_{0} ≤ λ_{1}^{j}A_{1}^{j}u_{0}, ∀j ∈ N. Therefore, we have by Gelfand’s formula that

r(A_{1}) = lim_{j→∞} ^{j} \sqrt{∥A_{1}^{j}∥} ≥ lim_{j→∞} ^{j} \sqrt{\frac{∥A_{1}^{j}u_{0}∥}{∥u_{0}∥}} ≥ lim_{j→∞} ^{j} \sqrt{\frac{ϱ_{0}^{j}∥u_{0}∥}{λ_{1}^{j}∥u_{0}∥}} = \frac{ϱ_{0}}{λ_{1}} > \frac{1}{λ_{1}} ,

which is a contradiction with r(A_{1}) = \frac{1}{λ_{1}} , and so (3.13) holds. It follows from Lemma 3.5 that (3.12) holds.

On the other hand, by f_{∞} > λ_{2}, there exists R_{5} > R_{4} > 0 such that

f(t, u) ≥ λ_{2}u, ∀t ∈ [a, b], u ≥ γR_{5}. (3.14)

Let Ω_{R_{5}} = \{u ∈ C[a, b] : ∥u∥ < R_{5}\}, then, for all u ∈ P ∩ ∂Ω_{R_{5}}, we have that min_{t∈[τ, \frac{a+b}{2} ]} u(t) ≥ γ∥u∥ =

γR_{5}. Using this and (3.14), one has

(T u)(t) ≥ λ_{2} ∫τ^{\frac{a+b}{2}} G(t, s)h(s)u(s)ds = λ_{2}(A_{2}u)(t), ∀t ∈ [a, b].

Hence, by the same way as in Theorem 3.6, we obtain

i(T, P ∩ Ω_{R_{5}}, P) = 0. (3.15)

According to (3.12) and (3.15), we have

i(T, (P ∩ Ω_{R_{5}}) \ (P ∩ \overline{Ω}_{R_{4}}), P) = i(θ, P ∩ Ω_{R_{5}}, P) − i(θ, P ∩ Ω_{R_{4}}, P) = −1.

Consequently, T has at least one fixed point on (P ∩ Ω_{R_{5}}) \ (P ∩ \overline{Ω}_{R_{4}}). Hence, the problem (1.2) has

at least one positive solution.

Corollary 3.8. Assume that (H_{0}) − (H_{2}) hold. If either

(i) f^{0} = 0, f_{∞} = ∞, or

(ii) f_{0} = ∞, f^{∞} = 0.

Then problem (1.2) has at least one positive solution.

Acknowledgements

The authors are very grateful to the referees for careful reading of the original manuscript and

for valuable suggestions on improving this paper. Project supported by the National Natural Science

Foundation of China (10771212) and the Natural Science Foundation of Jiangsu Education Office

(06KJB110010).

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(Received October 26, 2008)