Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 40, 1-13; http://www.math.u-szeged.hu/ejqtde/

Positive solutions for a multi-point eigenvalue

problem involving the one dimensional

p-Laplacian ^{∗}^{†}
Youyu Wang^{1} ^{‡} Weigao Ge^{2} Sui Sun Cheng^{3}

1. Department of Mathematics, Tianjin University of Finance and Economics, Tianjin 300222, P. R. China

2. Department of Mathematics, Beijing Institute of Technology, Beijing 100081, P. R. China 3. Department of Mathematics, Tsing Hua University, Hsinchu, Taiwan 30043.

Abstract A multi-point boundary value problem involving the one dimensional p-Laplacian and depending on a parameter is studied in this paper and existence of positive solutions is established by means of a fixed point theorem for operators defined on Banach spaces with cones. Keywords Positive solutions; Boundary value problems; One-dimensional p-Laplacian

1. Introduction

In this paper we study the existence of positive solutions to the boundary value problem (BVP) for the one-dimensional p-Laplacian

(ϕ_{p}(u′))′ + λq(t)f(t, u) = 0, t ∈ (0, 1),  (1.1)
u′(0) = ∑_{i=1}^{n−2} α_{i}u′(ξ_{i}),  u(1) = ∑_{i=1}^{n−2} β_{i}u(ξ_{i}),  (1.2)

where ϕ_{p}(s) = |s|^{p−2}s, p > 1, ξ_{i} ∈ (0, 1) with 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{n−2} < 1 and α_{i}, β_{i}, f satisfy

(H_{1}) α_{i}, β_{i} ∈ [0, ∞) satisfy 0 < ∑_{i=1}^{n−2} α_{i} < 1, and ∑_{i=1}^{n−2} β_{i} < 1;

(H_{2}) f ∈ C([0, 1] × [0, ∞), [0, ∞));

EJQTDE, 2009 No. 40, p. 1

∗This work is sponsored by the National Natural Science Foundation of China (No. 10671012).

†2000 Mathematics Subject Classification 34B10, 34B15 ‡Email: wang_{−}youyu@163.com

(H_{3}) q(t) ∈ C(0, 1) ∩ L^{1}[0, 1] with q > 0 on (0, 1). The number λ is regarded as a parameter and is to be determined among positive numbers.

The study of multipoint boundary value problems for linear second-order ordinary differential equations was initiated by Il’in and Moiseev [1,2]. Since then there has been much current attention focused on the study of nonlinear multipoint boundary value problems, see [3,4,5,7]. The methods include the Leray-Schauder continuation theorem, nonlinear alternatives of Leray-Schauder, coincidence degree theory, and fixed point theorem in cones. For example,

In [6], R. Ma and N. Castaneda studied the following BVP

\{ \begin{matrix}x′′(t) + a(t)f(x(t)) = 0, 0 ≤ t ≤ 1, \\ x′(0) = ∑_{i=1}^{m−2} α_{i}x′(ξ_{i}),  x(1) = ∑_{i=1}^{m−2} β_{i}x(ξ_{i}), \end{matrix}

where 0 < ξ_{1} < ⋯ < ξ_{m−2} < 1, α_{i}, β_{i} ≥ 0 with 0 < ∑_{i=1}^{m} α_{i} < 1 and ∑_{i=1}^{m} β_{i} < 1. They showed the existence of at least one positive solution if f is either superlinear or sublinear by applying the fixed point theorem in cones.

The authors in [7] considered the multi-point BVP for one dimensional p-Laplacian

(ϕ_{p}(u′))′ + f(t, u) = 0, t ∈ (0, 1),

ϕ_{p}(u′(0)) = ∑_{i=1}^{n−2} α_{i}ϕ_{p}(u′(ξ_{i})), u(1) = ∑_{i=1}^{n−2} β_{i}u(ξ_{i}).

Using a fixed point theorem in a cone, we provided sufficient conditions for the existence of multiple positive solutions to the above BVP.

In paper [8], we investigated the following more general multi-point BVPs

(ϕ_{p}(u′))′ + q(t)f(t, u, u′) = 0, t ∈ (0, 1),

u(0) = ∑_{i=1}^{n−2} α_{i}u(ξ_{i}), u′(1) = ∑_{i=1}^{n−2} β_{i}u′(ξ_{i}),

u′(0) = ∑_{i=1}^{n−2} α_{i}u′(ξ_{i}), u(1) = ∑_{i=1}^{n−2} β_{i}u(ξ_{i}).

The main tool is a fixed point theorem due to Avery and Peterson[9], we provided sufficient conditions for the existence of multiple positive solutions.

In view of the common concern about multi-point boundary value problems as exhibited in [6,7,8] and their references, it is of interests to continue the investigation and study the problem (1.1) and (1.2).

EJQTDE, 2009 No. 40, p. 2

Motivated by the works of [7] and [8], the aim of this paper is to show the existence of positive solutions u to BVP(1.1) and (1.2). For this purpose, we consider the Banach space E = C[0, 1] with the maximum norm ∥x∥ = max_{0≤t≤1} |x(t)|. By a positive solution of (1.1) and (1.2), we means a function u ∈ C^{1}[0, 1], (ϕ_{p}(u′))′(t) ∈ C(0, 1) ∩ L[0, 1] which is positive on [0, 1] and satisfies the differential equation (1.1) and the boundary conditions (1.2).

Our main results will depend on the following Guo-Krasnoselskii fixed-point theorem . Theorem A[10][11]. Let E be a Banach space and let K ⊂ E be a cone in E. Assume Ω_{1}, Ω_{2}

are open subsets of E with 0 ∈ Ω_{1}, \overline{Ω}_{1} ⊂ Ω_{2}, and let

T : K ∩ (\overline{Ω}_{2} \ Ω_{1}) → K

be a completely continuous operator such that either (i) ∥T x∥ ≤ ∥x∥, x ∈ K ∩ ∂Ω_{1} and ∥T x∥ ≥ ∥x∥, x ∈ K ∩ ∂Ω_{2}, or (ii) ∥T x∥ ≥ ∥x∥, x ∈ K ∩ ∂Ω_{1} and ∥T x∥ ≤ ∥x∥, x ∈ K ∩ ∂Ω_{2}. Then T has a fixed point in K ∩ (\overline{Ω}_{2} \ Ω_{1}). In section 3, we shall present some sufficient conditions with λ belonging to an open interval to

ensure the existence of positive solutions to problems (1.1) and (1.2). To the author’s knowledge, no one has studied the existence of positive solutions for problems (1.1) and (1.2) using the Guo-Krasnoselskii fixed-point theorem.

2. The preliminary lemmas Let

C^{+}[0, 1] = {ω ∈ C[0, 1] : ω(t) ≥ 0, t ∈ [0, 1]}.

Lemma 2.1 Let (H_{1}) − (H_{3}) hold. Then for x ∈ C^{+}[0, 1], the problem

(ϕ_{p}(u′))′ + λq(t)f(t, x(t)) = 0, t ∈ (0, 1),  (2.1)
u′(0) = ∑_{i=1}^{n−2} α_{i}u′(ξ_{i}),  u(1) = ∑_{i=1}^{n−2} β_{i}u(ξ_{i}),  (2.2)

has a unique solution

u(t) = B_{x} − ∫_{t}^{1} ϕ_{p}^{−1} (A_{x} − λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds,  (2.3)

where A_{x}, B_{x} satisfy

ϕ_{p}^{−1}(A_{x}) = ∑_{i=1}^{n−2} α_{i}ϕ_{p}^{−1} (A_{x} − λ ∫_{0}^{ξ_{i}} q(s)f(s, x(s))ds) ,  (2.4)

B_{x} = −\frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (A_{x} − λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds.

Lemma 2.2 Suppose β_{i}, r_{i} ≥ 0 (i = 1, 2, ⋯, n−2) and ∑_{i=1}^{n−2} β_{i} < 1, denote r = min_{1≤i≤n−2} r_{i}, R = max_{1≤i≤n−2} r_{i}, then there exists a unique

x ∈ [\frac{ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i})}{1 − ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i})}r, \frac{ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i})}{1 − ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i})}R]

such that

ϕ_{p}(x) = ∑_{i=1}^{n−2} β_{i}ϕ_{p}(x + r_{i}).

Proof. Define

H(x) = ϕ_{p}(x) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(x + r_{i}).

Then, H(x) ∈ C((−∞, +∞), R).

Let

\underline{x} = \frac{ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i})}{1 − ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i})} · r, \overline{x} = \frac{ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i})}{1 − ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i})} · R,

then we have

H(\underline{x}) = ϕ_{p}(\underline{x}) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(\underline{x} + r_{i})
≤ ϕ_{p}(\underline{x}) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(\underline{x} + r)
= ϕ_{p}(\underline{x}) − ϕ_{p} [ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i}) (\underline{x} + r)]
= ϕ_{p}(\underline{x}) − ϕ_{p}(\underline{x}) = 0,

and

H(\overline{x}) = ϕ_{p}(\overline{x}) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(\overline{x} + r_{i})
≥ ϕ_{p}(\overline{x}) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(\overline{x} + R)
= ϕ_{p}(\overline{x}) − ϕ_{p} [ϕ_{p}^{−1} (∑_{i=1}^{n−2} β_{i}) (\overline{x} + R)]
= ϕ_{p}(\overline{x}) − ϕ_{p}(\overline{x}) = 0.

EJQTDE, 2009 No. 40, p. 4 The zero point theorem guarantees that there exists an x_{0} ∈ [\underline{x}, \overline{x}] such that H(x_{0}) = 0.

If there exist two constants x_{1}, x_{2} ∈ [\underline{x}, \overline{x}] satisfying H(x_{1}) = H(x_{2}) = 0, then

Case 1. x_{1} = 0.

i.e., H(0) = 0, ∑_{i=1}^{n−2} β_{i}ϕ_{p}(r_{i}) = 0. So, β_{i}ϕ_{p}(r_{i}) = 0, (i = 1, 2, ⋯, n − 2), then, r_{i}ϕ_{p}^{−1}(β_{i}) =

0, (i = 1, 2, ⋯, n − 2). Therefore,

H(x) = ϕ_{p}(x) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(x + r_{i})
= ϕ_{p}(x) − ∑_{i=1}^{n−2} ϕ_{p} (ϕ_{p}^{−1}(β_{i})x + ϕ_{p}^{−1}(β_{i})r_{i})
= ϕ_{p}(x) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(x) = (1 − ∑_{i=1}^{n−2} β_{i}) ϕ_{p}(x).

Obviously, there exists a unique x = 0 satisfying H(x) = 0. So, x_{1} = x_{2} = 0.

Case 2. x_{1} ≠ 0.

(i) If x_{1} ∈ (−∞, 0), then

H(x_{1}) = ϕ_{p}(x_{1}) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(x_{1} + r_{i})
≤ ϕ_{p}(x_{1}) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(x_{1})
= (1 − ∑_{i=1}^{n−2} β_{i}) ϕ_{p}(x_{1}) < 0.

So, when, x_{1} ∈ (−∞, 0), H(x_{1}) ≠ 0.

(ii) If x_{1} ∈ (0, +∞), then

H(x_{1}) = ϕ_{p}(x_{1}) − ∑_{i=1}^{n−2} β_{i}ϕ_{p}(x_{1} + r_{i})
= ϕ_{p}(x_{1}) [1 − ∑_{i=1}^{n−2} β_{i}ϕ_{p} (1 + \frac{r_{i}}{x_{1}} )]
= ϕ_{p}(x_{1})H̃(x_{1}),

where

H̃(x) = 1 − ∑_{i=1}^{n−2} β_{i}ϕ_{p} (1 + \frac{r_{i}}{x} ) .

As H(x_{1}) = 0 and x_{1} ≠ 0, so there must exist i_{0} ∈ {1, 2, ⋯, n − 2}, such that β_{i_{0}}ϕ_{p}(r_{i_{0}}) > 0. Thus, we get H̃(x) is strictly increasing on (0, +∞).

If H(x_{1}) = H(x_{2}) = 0, then H̃(x_{1}) = H̃(x_{2}) = 0. So, x_{1} = x_{2} since H̃(x) is strictly increasing on (0, +∞).

EJQTDE, 2009 No. 40, p. 5

Therefore, H(x) = 0 has a unique solution on (0, +∞). Combining case 1, case 2, we obtain H(x) = 0 has a unique solution on (−∞, +∞) and x ∈ [\underline{x}, \overline{x}].

Lemma 2.3 Let k = \frac{ϕ_{p} (∑_{i=1}^{n−2} α_{i})}{1 − ϕ_{p} (∑_{i=1}^{n−2} α_{i})}, then there exists a unique real number A_{x} that

satisfies (2.4). Furthermore, A_{x} is contained in the interval [−λk ∫_{0}^{1} q(s)f(s, x(s))ds, 0]. Proof The equation is equivalent to

ϕ_{p}^{−1}(−A_{x}) = ∑_{i=1}^{n−2} α_{i}ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{ξ_{i}} q(s)f(s, x(s))ds) .

By Lemma 2.2, we can easily obtain

−A_{x} ∈ [λk ∫_{0}^{ξ_{1}} q(s)f(s, x(s))ds, λk ∫_{0}^{ξ_{n−2}} q(s)f(s, x(s))ds]
⊂ [0, λk ∫_{0}^{1} q(s)f(s, x(s))ds] .

So the conclusion is obvious.

Lemma 2.4 Let (H_{1}) − (H_{3}) hold. If x ∈ C^{+}[0, 1] and λ > 0, then the unique solution of problem (2.1)-(2.2) satisfies u(t) ≥ 0 for t ∈ [0, 1].

Proof: According to Lemmas 2.1 and 2.3, we first have

u(1) = B_{x}
= −\frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (A_{x} − λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
= \frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≥ \frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≥ 0,

and

u(0) = B_{x} − ∫_{0}^{1} ϕ_{p}^{−1} (A_{x} − λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
= u(1) + ∫_{0}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≥ u(1) + ∫_{0}^{1} ϕ_{p}^{−1} (λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≥ 0.

EJQTDE, 2009 No. 40, p. 6 If t ∈ (0, 1), we have

u(t) = B_{x} − ∫_{t}^{1} ϕ_{p}^{−1} (A_{x} − λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
= u(1) + ∫_{t}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≥ u(1) + ∫_{t}^{1} ϕ_{p}^{−1} (λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≥ 0.

So u(t) ≥ 0, t ∈ [0, 1].

Lemma 2.5 Let (H_{1}) − (H_{3}) hold. If x ∈ C^{+}[0, 1] and λ > 0, then the unique solution of problem (2.1)-(2.2) satisfies

min_{t∈[0,1]} u(t) ≥ γ∥u∥,

where

γ = \frac{∑_{i=1}^{n−2} β_{i}(1 − ξ_{i})}{1 − ∑_{i=1}^{n−2} β_{i}ξ_{i}}.

Proof: Clearly u′(t) = ϕ_{p}^{−1} (A_{x} − λ ∫_{0}^{t} q(s)f(s, x(s))ds) = −ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{t} q(s)f(s, x(s))ds) ≤ 0. This implies that

∥u∥ = u(0) and min_{t∈[0,1]} u(t) = u(1).

Furthermore, it is easy to see that u′(t_{2}) ≤ u′(t_{1}) for any t_{1}, t_{2} ∈ [0, 1] with t_{1} ≤ t_{2}. Hence u′(t) is a decreasing function on [0,1]. This means that the graph of u′(t) is concave down on (0,1). For each i ∈ {1, 2, ⋯, n − 2} we have

\frac{u(ξ_{i}) − u(1)}{1 − ξ_{i}} ≥ \frac{u(0) − u(1)}{1},

i.e.,

u(ξ_{i}) − ξ_{i}u(1) ≥ (1 − ξ_{i})u(0)

so that

∑_{i=1}^{n−2} β_{i}u(ξ_{i}) − ∑_{i=1}^{n−2} β_{i}ξ_{i}u(1) ≥ ∑_{i=1}^{n−2} β_{i}(1 − ξ_{i})u(0)

and, by means of the boundary condition u(1) = ∑_{i=1}^{n−2} β_{i}u(ξ_{i}), we have

u(1) ≥ \frac{∑_{i=1}^{n−2} β_{i}(1 − ξ_{i})}{1 − ∑_{i=1}^{n−2} β_{i}ξ_{i}}u(0).

This completes the proof.

EJQTDE, 2009 No. 40, p. 7 Now we define

K = {ω|ω ∈ C^{+}[0, 1], min_{0≤t≤1} ω(t) ≥ γ∥ω∥},

where γ is defined in Lemma 2.5. For any λ > 0, define operator T_{λ} : C^{+}[0, 1] → K by

(T_{λ}x)(t) = −\frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (A_{x} − λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds

− ∫_{t}^{1} ϕ_{p}^{−1} (A_{x} − λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds. (2.5)

By Lemmas 2.1 and 2.3, we know T_{λ}x is well defined. Furthermore, we have the following result. Lemma 2.6[Lemma 2.4, 7] T : K → K is completely continuous. Let

min f_{∞} := lim_{x→∞} inf min_{t∈[0,1]} \frac{f(t, x)}{ϕ_{p}(x)}, max f_{0} := lim_{x→0^{+}} sup max_{t∈[0,1]} \frac{f(t, x)}{ϕ_{p}(x)},

min f_{0} := lim_{x→0^{+}} inf min_{t∈[0,1]} \frac{f(t, x)}{ϕ_{p}(x)}, max f_{∞} := lim_{x→∞} sup max_{t∈[0,1]} \frac{f(t, x)}{ϕ_{p}(x)}.

3. Main results We now give our results on the existence of positive solutions of BVP(1.1) and (1.2). Theorem 3.1 Let (H_{1}) − (H_{3}) hold and min f_{∞} > 0, max f_{0} < ∞. If

\frac{1}{γ^{p−1} min f_{∞} · N^{p−1}} < λ < \frac{1}{(k + 1) max f_{0} · M^{p−1}} .  (3.1)

where

M = \frac{1 − ∑_{i=1}^{n−2} β_{i}ξ_{i}}{1 − ∑_{i=1}^{n−2} β_{i}} · ϕ_{p}^{−1} (∫_{0}^{1} q(s)ds) ,
N = \frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (∫_{0}^{s} q(τ)dτ) ds + ∫_{0}^{1} ϕ_{p}^{−1} (∫_{0}^{s} q(τ)dτ) ds,

then the problem (1.1)(1.2) has at least one positive solution.

Proof. Define the operator T_{λ} as (2.5). Under the condition (3.1), there exists an ε > 0 such that

\frac{1}{γ^{p−1}(min f_{∞} − ε) · N^{p−1}} ≤ λ ≤ \frac{1}{(k + 1)(max f_{0} + ε) · M^{p−1}} .  (3.2)

(i). Since max f_{0} < ∞, there exists an H_{1} > 0 such that for x : 0 ≤ x ≤ H_{1},

f(t, x) ≤ (max f_{0} + ε)ϕ_{p}(x).  (3.3)

EJQTDE, 2009 No. 40, p. 8 Let Ω_{1} = {x ∈ E : ∥x∥ < H_{1}}, then for x ∈ K ∩ ∂Ω_{1}, we have

−A_{x} + λ ∫_{0}^{s} q(s)f(s, x(s))ds ≤ λk ∫_{0}^{1} q(s)f(s, x(s))ds + λ ∫_{0}^{s} q(s)f(s, x(s))ds
≤ λ(k + 1) ∫_{0}^{1} q(s)f(s, x(s))ds
≤ λ(k + 1)(max f_{0} + ε) ∫_{0}^{1} q(s)ϕ_{p}(x(s))ds
≤ λ(k + 1)(max f_{0} + ε)ϕ_{p}(∥x∥) ∫_{0}^{1} q(s)ds.

So,

ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(s)f(s, x(s))ds) ≤ [λ(k + 1)(max f_{0} + ε)]^{q−1} ϕ_{p}^{−1} (∫_{0}^{1} q(s)ds) ∥x∥.

Therefore,

∥T_{λ}x∥ = (T_{λ}x)(0) = \frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
+ ∫_{0}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≤ \frac{1 − ∑_{i=1}^{n−2} β_{i}ξ_{i}}{1 − ∑_{i=1}^{n−2} β_{i}} [λ(k + 1)(max f_{0} + ε)]^{q−1} ϕ_{p}^{−1} (∫_{0}^{1} q(s)ds) ∥x∥
= M[λ(k + 1)(max f_{0} + ε)]^{q−1}∥x∥
≤ ∥x∥.

Thus ∥T_{λ}x∥ ≤ ∥x∥.

(ii). Next, since min f_{∞} > 0, there exists an \overline{H}_{2} > 0 such that for x > \overline{H}_{2},

f(t, x) ≥ (min f_{∞} − ε)ϕ_{p}(x).  (3.4)

Take H_{2} = max\{\overline{H}_{2}, 2H_{1}\} and Ω_{2} = \{x ∈ E : ∥x∥ < H_{2}\}. Then for x ∈ K ∩ ∂Ω_{2}, we have

−A_{x} + λ ∫_{0}^{s} q(τ)f(s, x(τ))dτ ≥ λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ
≥ λ(min f_{∞} − ε) ∫_{0}^{s} q(τ)ϕ_{p}(x(τ))dτ
≥ λϕ_{p}(γ)(min f_{∞} − ε)ϕ_{p}(∥x∥) ∫_{0}^{s} q(τ)dτ.

So,

ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(s)f(s, x(s))ds) ≥ [λ(min f_{∞} − ε)]^{q−1} ϕ_{p}^{−1} (∫_{0}^{s} q(τ)dτ) γ∥x∥.

Therefore,

∥T_{λ}x∥ = (T_{λ}x)(0) = \frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
+ ∫_{0}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≥ [λ(min f_{∞} − ε)]^{q−1} γ∥x∥\frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (∫_{0}^{s} q(τ)dτ) ds
+ [λ(min f_{∞} − ε)]^{q−1} γ∥x∥ ∫_{0}^{1} ϕ_{p}^{−1} (∫_{0}^{s} q(τ)dτ) ds
= N [λ(min f_{∞} − ε)]^{q−1} γ∥x∥
≥ ∥x∥.

So ∥T_{λ}x∥ ≥ ∥x∥.

Therefore, by the first part of Theorem A, T_{λ} has a fixed point x^{∗} ∈ K ∩ (\overline{Ω}_{2} \ Ω_{1}) such that H_{1} ≤ ∥x^{∗}∥ ≤ H_{2}. It is easily checked that x^{∗}(t) is a positive solution of problems (1.1) and (1.2). The proof is complete.

Theorem 3.2. Let (H_{1}) − (H_{3}) hold and min f_{0} > 0, max f_{∞} < ∞. If

\frac{1}{γ^{p−1} min f_{0} · N^{p−1}} < λ < \frac{1}{(k + 1) max f_{∞} · M^{p−1}} .  (3.5)

then the problem (1.1)(1.2) has at least one positive solution.

Proof. Under the condition (3.5), there exists an ε > 0 such that

\frac{1}{γ^{p−1}(min f_{0} − ε) · N^{p−1}} ≤ λ ≤ \frac{1}{(k + 1)(max f_{∞} + ε) · M^{p−1}} .  (3.6)

(i). Since min f_{0} > 0, there exists an H_{1} > 0 such that for x : 0 ≤ x ≤ H_{1},

f(t, x) ≥ (min f_{0} − ε)ϕ_{p}(x).  (3.7)

Let Ω_{1} = {x ∈ E : ∥x∥ < H_{1}}, then for x ∈ K ∩ ∂Ω_{1}, we have

−A_{x} + λ ∫_{0}^{s} q(τ)f(s, x(τ))dτ ≥ λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ
≥ λ(min f_{0} − ε) ∫_{0}^{s} q(τ)ϕ_{p}(x(τ))dτ
≥ λϕ_{p}(γ)(min f_{0} − ε)ϕ_{p}(∥x∥) ∫_{0}^{s} q(τ)dτ.

So,

ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(s)f(s, x(s))ds) ≥ [λ(min f_{0} − ε)]^{q−1} ϕ_{p}^{−1} (∫_{0}^{s} q(τ)dτ) γ∥x∥.

Therefore,

∥T_{λ}x∥ = (T_{λ}x)(0) = \frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
+ ∫_{0}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≥ [λ(min f_{0} − ε)]^{q−1} γ∥x∥\frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (∫_{0}^{s} q(τ)dτ) ds
+ [λ(min f_{0} − ε)]^{q−1} γ∥x∥ ∫_{0}^{1} ϕ_{p}^{−1} (∫_{0}^{s} q(τ)dτ) ds
= N [λ(min f_{0} − ε)]^{q−1} γ∥x∥
≥ ∥x∥.

So ∥T_{λ}x∥ ≥ ∥x∥.

(ii). Since max f_{∞} < ∞, there exists an \overline{H}_{2} > 0 such that for x > \overline{H}_{2},

f(t, x) ≤ (max f_{∞} + ε)ϕ_{p}(x).  (3.8)

There are two cases : Case 1, f is bounded, and Case 2, f is unbounded.

Case 1. Suppose that f is bounded, i.e., there exists N > 0 such that f(t, x) ≤ ϕ_{p}(N) for t ∈ [0, 1] and 0 ≤ x < ∞. Define

H_{2} = max \{2H_{1}, \frac{1 − ∑_{i=1}^{n−2} β_{i}ξ_{i}}{1 − ∑_{i=1}^{n−2} β_{i}} [λ(k + 1)]^{q−1} ϕ_{p}^{−1} (∫_{0}^{1} q(s)ds) N\} ,

and Ω_{2} = {x ∈ E : ∥x∥ < H_{2}}. Then for x ∈ K ∩ ∂Ω_{2}, we have

−A_{x} + λ ∫_{0}^{s} q(s)f(s, x(s))ds ≤ λk ∫_{0}^{1} q(s)f(s, x(s))ds + λ ∫_{0}^{s} q(s)f(s, x(s))ds
≤ λ(k + 1) ∫_{0}^{1} q(s)f(s, x(s))ds
≤ λ(k + 1)ϕ_{p}(N) ∫_{0}^{1} q(s)ds.

So,

ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(s)f(s, x(s))ds) ≤ [λ(k + 1)]^{q−1} ϕ_{p}^{−1} (∫_{0}^{1} q(s)ds) N.

Therefore,

∥T_{λ}x∥ = (T_{λ}x)(0) = \frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
+ ∫_{0}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≤ \frac{1 − ∑_{i=1}^{n−2} β_{i}ξ_{i}}{1 − ∑_{i=1}^{n−2} β_{i}} [λ(k + 1)]^{q−1} ϕ_{p}^{−1} (∫_{0}^{1} q(s)ds) N
≤ H_{2} = ∥x∥.

Case 2. We choose H_{2} > max\{2H_{1}, \overline{H}_{2}\} such that f(t, x) ≤ f(t, H_{2}) for t ∈ [0, 1] and 0 < x < H_{2}. Let Ω_{2} = {x ∈ E : ∥x∥ < H_{2}}. Then for x ∈ K ∩ ∂Ω_{2}, we have

−A_{x} + λ ∫_{0}^{s} q(s)f(s, x(s))ds ≤ λk ∫_{0}^{1} q(s)f(s, x(s))ds + λ ∫_{0}^{s} q(s)f(s, x(s))ds
≤ λ(k + 1) ∫_{0}^{1} q(s)f(s, x(s))ds
≤ λ(k + 1) ∫_{0}^{1} q(s)f(s, H_{2})ds
≤ λ(k + 1)(max f_{∞} + ε)ϕ_{p}(H_{2}) ∫_{0}^{1} q(s)ds.

So,

ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(s)f(s, x(s))ds) ≤ [λ(k + 1)(max f_{∞} + ε)]^{q−1} ϕ_{p}^{−1} (∫_{0}^{1} q(s)ds) H_{2}.

Therefore,

∥T_{λ}x∥ = (T_{λ}x)(0) = \frac{1}{1 − ∑_{i=1}^{n−2} β_{i}} ∑_{i=1}^{n−2} β_{i} ∫_{ξ_{i}}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
+ ∫_{0}^{1} ϕ_{p}^{−1} (−A_{x} + λ ∫_{0}^{s} q(τ)f(τ, x(τ))dτ) ds
≤ \frac{1 − ∑_{i=1}^{n−2} β_{i}ξ_{i}}{1 − ∑_{i=1}^{n−2} β_{i}} [λ(k + 1)(max f_{∞} + ε)]^{q−1} ϕ_{p}^{−1} (∫_{0}^{1} q(s)ds) H_{2}
= M [λ(k + 1)(max f_{∞} + ε)]^{q−1} H_{2}
≤ H_{2} = ∥x∥.

Therefore, by the second part of Theorem A, T_{λ} has a fixed point x^{∗} ∈ K ∩ (\overline{Ω}_{2} \ Ω_{1}) such that H_{1} ≤ ∥x^{∗}∥ ≤ H_{2}. It is easily checked that x^{∗}(t) is a positive solution of problems (1.1) and (1.2).

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(Received February 14, 2009)