Existence of Positive Solutions for Boundary Value Problems of Second-order Functional Differential Equations

Daqing Jiang Northeast Normal University Changchun 130024, China

Peixuan Weng ^{1} South China Normal University Guangzhou 510631, China

Abstract

We use a fixed point index theorem in cones to study the existence of positive solutions for boundary value problems of second-order functional differential equations of the form

\{ \begin{matrix}y′′(x) + r(x)f(y(w(x))) = 0, & 0 < x < 1, \\ αy(x) − βy′(x) = ξ(x), & a ≤ x ≤ 0, \\ γy(x) + δy′(x) = η(x), & 1 ≤ x ≤ b; \end{matrix}

where w(x) is a continuous function defined on [0, 1] and r(x) is allowed to have singularities on [0, 1]. The result here is the generalization of a corresponding result for ordinary differential equations.

Keywords: functional differential equation, boundary value problem, positive solution, superlinear and sublinear

1991 AMS classification: 34K10, 34B15

1 Introduction

Boundary value problems (abbr. as BVP) associated with second order diffferential equations have a long history and many different methods and techniques have been used and developed in order to obtain various qualitative properties of the solutions (see [1-5, 8, 11, 15, 17]). In recent years, accompanied by the development of theory of functional differential equations (abbr. as FDE), many authors have paid attention to BVP of second order FDE such as

[p(t)x′(t)]′ = f(t, x_{t}, x(t)),

or

x′′(t) = f(t, x(t), x(σ(t)), x′(t), x′(τ(t)))

(for example see [6-7, 10, 12-14, 16]). As pointed out by the authors of [6], the background for these problems lies in many areas of physics, applied mathematics and variational problems of control theory.

1Supported by NSF of China and Guangdong Province

In this paper, we investigate the existence of positive solutions for BVP of secondorder FDE with the form:

\{ \begin{matrix}y′′(x) + r(x)f(y(w(x))) = 0, & 0 < x < 1, \\ αy(x) − βy′(x) = ξ(x), & a ≤ x ≤ 0, \\ γy(x) + δy′(x) = η(x), & 1 ≤ x ≤ b. \end{matrix}  (1.1)

Here we assume that (P_{1}) w(x) is a continuous function defined on [0, 1] satisfying

c = inf{w(x); 0 ≤ x ≤ 1} < 1, d = sup{w(x); 0 ≤ x ≤ 1} > 0.

Thus E := {x ∈ [0, 1]; 0 ≤ w(x) ≤ 1} is a compact set and mes E > 0;

(P_{2}) ξ(x) and η(x) are continuous functions defined on [a, 0] and [1, b] respectively, where a := min{0, c}, b := max{1, d}; furthermore, ξ(0) = η(1) = 0; ξ(x) ≥ 0 as β = 0 ; ∫_{x}^{0} e^{− \frac{α}{β} s}ξ(s)ds ≥ 0 as β > 0; η(x) ≥ 0 as δ = 0; ∫_{1}^{x} e^{\frac{γ}{δ} s}η(s)ds ≥ 0 as δ > 0.

For the case of w(x) ≡ x, BVP (1.1) is related to two point BVP of ODE for which Erbe and Wang^{[4]} have got the following theorem:

Theorem A.^{[4]} Assume that w(x) ≡ x, ξ(x) ≡ 0, η(x) ≡ 0 and (A_{1}) f ∈ C([0, +∞), [0, +∞)); (A_{2}) r ∈ C([0, 1], [0, +∞)) and r(x) ≢ 0 for any subinterval of [0, 1]; (A_{3}) α, β, γ, δ ≥ 0, ρ := γβ + αγ + αδ > 0.

Then any one of the following is a sufficient condition for the existence of at least one positive solution of BVP (1.1):

(1) lim_{v↓0} \frac{f(v)}{v} = +∞ and lim_{v↑+∞} \frac{f(v)}{v} = 0 (sublinear case);

(2) lim_{v↓0} \frac{f(v)}{v} = 0 and lim_{v↑+∞} \frac{f(v)}{v} = +∞ (superlinear case).

Motivated by [4], in this paper we shall extend the results of [4] to BVP (1.1). Firstly, we have the following hypotheses:

(H_{1}) α, β, γ, δ ≥ 0, ρ := γβ + αγ + αδ > 0; (H_{2}) r(x) is a measurable function defined on [0, 1], and

0 < ∫_{E} h(x)r(x)dx ≤ ∫_{0}^{1} h(x)r(x)dx < +∞,

0 < ∫_{E} ϕ_{1}(x)r(x)dx ≤ ∫_{0}^{1} ϕ_{1}(x)r(x)dx < +∞,

where E is defined as in (P_{1}); h(x) : [0, 1] → [0, 1] is defined by

h(x) = \{ \begin{matrix}1, & δβ > 0, \\ x, & β = 0, δ > 0, \\ 1 − x, & β > 0, δ = 0, \\ x(1 − x), & β = δ = 0 \end{matrix}

and ϕ_{1}(x) (ϕ_{1}(x) > 0, x ∈ (0, 1)) is the eigenfunction related to the smallest eigenvalue λ_{1} (λ_{1} > 0) of the eigenvalue problem

−ϕ′′ = λϕ,  αϕ(0) − βϕ′(0) = 0,  γϕ(1) + δϕ′(1) = 0;

(H_{3}) f(y) is a nonnegative continuous function defined on [0, +∞).

Under the assumptions (H_{1}) − (H_{3}), we allow that r(x) ≡ 0 on some subset of E, and r(x) has some kind of singularities on [0, 1]. For example, if β = δ = 0, then ϕ_{1}(x) = sin πx and

r(x) = x^{−m}(1 − x)^{−n}, 0 < m < 2, 0 < n < 2,

satisfies (H_{2}).

To the best of the authors’ knowledge, there has not been much work done about the positive solutions for the singular boundary value problems with deviating arguments, although they have importance in applications.

2 Main Theorem

First, we give the following definitions of solution and positive solution of BVP (1.1). Definition. y(x) is said to be a solution of BVP (1.1) if it satisfies the following:

1. y(x) is nonnegative and continuous on [a, b];

2. y(x) = y(a; x) as x ∈ [a, 0], where y(a; x) : [a, 0] → [0, +∞) is defined by

y(a; x) = \{ \begin{matrix}e^{\frac{α}{β} x} (\frac{1}{β} ∫_{x}^{0} e^{− \frac{α}{β} s}ξ(s)ds + y(0)) , & β > 0, \\ \frac{1}{α}ξ(x), & β = 0; \end{matrix}  (2.1)

3. y(x) = y(b; x) as x ∈ [1, b], where y(b; x) : [1, b] → [0, +∞) is defined by

y(b; x) = \{ \begin{matrix}e^{− \frac{γ}{δ} x} (\frac{1}{δ} ∫_{1}^{x} e^{\frac{γ}{δ} s}η(s)ds + e^{\frac{γ}{δ}} y(1)) , & δ > 0, \\ \frac{1}{γ}η(x), & δ = 0; \end{matrix}  (2.2)

4. while δβ > 0, y′(x) exists and is absolutely continuous on [0, 1]; while β = 0, δ > 0, y′(x) exists and is locally absolutely continuous on (0, 1]; while β > 0, δ = 0, y′(x) exists and is locally absolutely continuous on [0, 1); while β = δ = 0, y′(x) exists and is locally absolutely continuous on (0, 1);

5. y′′(x) = −r(x)f(y(w(x))) for x ∈ (0, 1) almost everywhere.

Furthermore, a solution y(x) of (1.1) is called a positive solution if y(x) > 0 for x ∈ (0, 1).

Suppose that y(x) is a solution of BVP (1.1), then it could be expressed as

y(x) = \{ \begin{matrix}y(a; x), & a ≤ x ≤ 0, \\ ∫_{0}^{1} G(x, t)r(t)f(y(w(t)))dt, & 0 ≤ x ≤ 1, \\ y(b; x), & 1 ≤ x ≤ b, \end{matrix}  (2.3)

and Green’s function

G(x, t) := \{ \begin{matrix}\frac{1}{ρ}(δ + γ − γx)(β + αt), & 0 ≤ t ≤ x ≤ 1, \\ \frac{1}{ρ}(δ + γ − γt)(β + αx), & 0 ≤ x ≤ t ≤ 1, \end{matrix}

where ρ is given in (H_{1}). It is obvious that 0 < G(x, t) ≤ G(t, t) for (x, t) ∈ (0, 1) × (0, 1).

By an elementary calculation, one can find constants λ and B such that

λBh(x)h(t) ≤ G(x, t) ≤ Bh(t), (x, t) ∈ [0, 1] × [0, 1],  (2.4)

where h(x) is decided by (H_{2}).

By using (2.3) and (2.4), we know that for every solution y(x) of BVP (1.1), one has

\{ \begin{matrix}∥y∥_{[0,1]} ≤ B ∫_{0}^{1} h(t)r(t)f(y(w(t)))dt, \\ y(x) ≥ λ∥y∥_{[0,1]}h(x),  x ∈ [0, 1], \end{matrix}  (2.5)

where ∥y∥_{[0,1]} := sup{|y(x)|; 0 ≤ x ≤ 1}.

Choose σ ∈ (0, \frac{1}{4}) such that

∫_{E_{σ}} h(s)r(s)ds > 0, ∫_{E_{σ}} ϕ_{1}(s)r(s)ds > 0,  (2.6)

where E_{σ} := {x ∈ E; σ ≤ w(x) ≤ 1 − σ}. In this paper, we always assume that σ satisfies (2.6). Then we have from (2.5) that

y(x) ≥ λ∥y∥_{[0,1]}h(x) ≥ λσ_{o}∥y∥_{[0,1]}, x ∈ [σ, 1 − σ],

here

σ_{o} = \{ \begin{matrix}1, & δβ > 0, \\ σ, & β > 0, δ = 0 or β = 0, δ > 0, \\ σ(1 − σ), & δ = β = 0. \end{matrix}  (2.7)

The following theorem is our main result.

Theorem 1. If (P_{1}), (P_{2}), (H_{1})−(H_{3}) are satisfied, then any one of the following is a sufficient condition for the existence of at least one positive solution of BVP(1.1):

(H_{4}) lim inf_{v→0+} \frac{f(v)}{v} > kλ_{1}, lim sup_{v→+∞} \frac{f(v)}{v} < qλ_{1};

(H_{5}) lim inf_{v→+∞} \frac{f(v)}{v} > kλ_{1}, lim sup_{v→0+} \frac{f(v)}{v} < qλ_{1}, ξ(x) ≡ 0, η(x) ≡ 0;

where k > 0 is large enough such that

kλσ_{o} ∫_{E_{σ}} r(x)ϕ_{1}(x)dx ≥ ∫_{0}^{1} ϕ_{1}(x)dx,

and q > 0 is small enough such that

q ∫_{0}^{1} r(x)ϕ_{1}(x)dx ≤ λσ_{o} ∫_{σ}^{1−σ} ϕ_{1}(x)dx

(λ and σ_{o} are defined as in (2.4) and (2.7) respectively).

Corollary. Using the following (H_{6}) or (H_{7}) instead of (H_{4}) or (H_{5}), the conclusion of Theorem 1 is true.

(H_{6}) lim_{v↓0} \frac{f(v)}{v} = +∞, lim_{v↑+∞} \frac{f(v)}{v} = 0 (sublinear);

(H_{7}) lim_{v↓0} \frac{f(v)}{v} = 0, lim_{v↑+∞} \frac{f(v)}{v} = +∞ (superlinear), ξ(x) ≡ 0, η(x) ≡ 0.

It is obvious that our corollary is an extension of Theorem A, and Theorem 1 is an improvement of Theorem A even for the case w(x) = x. We remark here that only if δβ > 0 and r(x) is continuous on [0, 1], every positive solution of BVP (1.1) belongs to C^{1}[a, b] ⋂ C^{2}[0, 1].

3 Proof of Theorem

In this section, we shall show the conclusion of Theorem 1 only for the situation β > 0, δ = 0. The arguments for the other three cases are similar. First we give a lemma which will be used later (see [5] or [11]).

Lemma 1. Assume that X is a Banach space, and K ⊂ X is a cone in X. Let K_{p} = {u ∈ K; ∥u∥ ≤ p}. Furthermore, assume that Φ : K → K is a compact map, and Φu ≠ u for u ∈ ∂K_{p} = {u ∈ K; ∥u∥ = p}. Then one has the following conclusions:

1. if ∥u∥ ≤ ∥Φu∥ for u ∈ ∂K_{p}, then i(Φ, K_{p}, K) = 0;

2. if ∥u∥ ≥ ∥Φu∥ for u ∈ ∂K_{p}, then i(Φ, K_{p}, K) = 1.

If u_{o}(x) is a solution of BVP (1.1) for f ≡ 0, then it can be expressed as

u_{o}(x) = \{ \begin{matrix}\frac{1}{β} e^{\frac{α}{β} x} ∫_{x}^{0} e^{− \frac{α}{β} s}ξ(s)ds, & a ≤ x ≤ 0, \\ 0, & 0 ≤ x ≤ 1, \\ \frac{1}{γ}η(x), & 1 ≤ x ≤ b. \end{matrix}  (3.1)

If y(x) is a solution of BVP(1.1), let u(x) = y(x) − u_{o}(x), noting that u(x) ≡ y(x) as 0 ≤ x ≤ 1, then by using (3.1), we have

u(x) = \{ \begin{matrix}e^{\frac{α}{β} x}u(0), & a ≤ x ≤ 0, \\ ∫_{0}^{1} G(x, t)r(t)f(u(w(t)) + u_{o}(w(t)))dt, & 0 ≤ x ≤ 1, \\ 0, & 1 ≤ x ≤ b. \end{matrix}  (3.2)

Let K be a cone in the Banach space X = C[a, b] which is defined as

K = {u ∈ C[a, b]; u(x) ≥ λσ_{o}∥u∥, x ∈ [σ, 1 − σ]},

where ∥u∥ := sup{|u(x)|; a ≤ x ≤ b} (noting that ∥u∥_{[0,1]} is defined as in (2.5)). Define an operator Φ : K → K by

(Φu)(x) = \{ \begin{matrix}e^{\frac{α}{β} x} ∫_{0}^{1} G(0, t)r(t)f(u(w(t)) + u_{o}(w(t)))dt, & a ≤ x ≤ 0, \\ ∫_{0}^{1} G(x, t)r(t)f(u(w(t)) + u_{o}(w(t)))dt, & 0 ≤ x ≤ 1, \\ 0, & 1 ≤ x ≤ b. \end{matrix}  (3.3)

Then we have the following four lemmas.

Lemma 2. Φ(K) ⊂ K.

Proof. It is obvious that 0 ≤ (Φu)(x) ≤ (Φu)(0) as a ≤ x ≤ 0, then one has ∥Φu∥ = ∥Φu∥_{[0,1]}. We have from (2.4) and (3.3) that (noting that h(x) = 1 − x)

∥Φu∥_{[0,1]} ≤ B ∫_{0}^{1}(1 − t)r(t)f(u(w(t)) + u_{o}(w(t)))dt,

thus we have

(Φu)(x) ≥ λB ∫_{0}^{1}(1 − x)(1 − t)r(t)f(u(w(t)) + u_{o}(w(t)))dt,
≥ λ(1 − x)∥Φu∥_{[0,1]},
= λ(1 − x)∥Φu∥, x ∈ [0, 1].

Thus (Φu)(x) ≥ λσ_{o}∥Φu∥, σ ≤ x ≤ 1 − σ, i.e. Φ(K) ⊂ K.

Lemma 3. Φ : K → K is completely continuous.

Proof. We can obtain the continuity of Φ from the continuity of f. In fact, if u_{n}, u ∈ K and ∥u_{n}−u∥ → 0 as n → ∞, then we have from (3.3) that for a ≤ x ≤ b,

\begin{aligned}& |(Φu_{n})(x) − (Φu)(x)| \\ ≤ & max_{0≤x≤1} |f(u_{n}(w(x)) + u_{o}(w(x))) − f(u(w(x)) + u_{o}(w(x)))| B ∫_{0}^{1}(1 − t)r(t)dt, \end{aligned}

which implies that ∥Φu_{n} − Φu∥ → 0 as n → ∞.

Suppose that A ⊂ K is a bounded set, and there exists a constant M_{1} > 0 such that ∥u∥ ≤ M_{1} for u ∈ A. Let ∥u_{o}∥ = M_{2}, then ∥u + u_{o}∥ ≤ M_{1} + M_{2} = M for u ∈ A. We have from (3.3) that

∥Φu∥ ≤ B max_{0≤v≤M} f(v) ∫_{0}^{1}(1 − t)r(t)dt.  (3.4)

Thus Φ(A) is bounded in K. Now it is easy to see that Φu ∈ C^{1}[a, 1) ∩ C[a, b], and

\begin{aligned}(Φu)′(x) = & \frac{−1}{β+α} ∫_{0}^{x}(β + αt)r(t)f(u(w(t)) + u_{o}(w(t)))dt \\ & +\frac{α}{β+α} ∫_{x}^{1}(1 − t)r(t)f(u(w(t)) + u_{o}(w(t)))dt, 0 ≤ x < 1, \\ (Φu)′(x) = & \frac{α}{β}e^{\frac{α}{β} x} ∫_{0}^{1} G(0, t)r(t)f(u(w(t)) + u_{o}(w(t)))dt, a ≤ x ≤ 0. \end{aligned}  (3.5)

For u ∈ A, 0 ≤ x ≤ 1, we have

(Φu)(x) ≤ F (x) := max_{0≤v≤M} f(v) ∫_{0}^{1} G(x, t)r(t)dt, 0 ≤ x ≤ 1.  (3.6)

Noting the facts that F (1) = 0 and the continuity of F (x) on [0, 1], we have from (3.6) that for any ϵ > 0, one can find a δ_{1} > 0 (independent with u) such that, 0 < δ_{1} < \frac{1}{4} and

(Φu)(x) < \frac{ϵ}{2}, 1 − 2δ_{1} < x < 1.  (3.7)

On the other hand, for x ∈ [0, 1 − δ_{1}] one has

\begin{aligned}|(Φu)′(x)| & ≤ max_{0≤v≤M} f(v) \{∫_{0}^{1−δ_{1}} r(t)dt + ∫_{0}^{1}(1 − t)r(t)dt\} \\ & ≤ \frac{1+δ_{1}}{δ_{1}} max_{0≤v≤M} f(v) ∫_{0}^{1}(1 − t)r(t)dt = L_{1}. \end{aligned}

For x ∈ [a, 0], one has from (3.5) and (3.4) that

|(Φu)′(x)| ≤ \frac{α}{β}|(Φu)(x)| ≤ \frac{α}{β}B max_{0≤v≤M} f(v) ∫_{0}^{1} r(t)(1 − t)dt = L_{2}.

Let δ_{2} = \frac{ϵ}{max\{L_{1},L_{2}\}}, then for x_{1}, x_{2} ∈ [a, 1 − δ_{1}], |x_{1} − x_{2}| < δ_{2}, we have

|(Φu)(x_{1}) − (Φu)(x_{2})| ≤ max{L_{1}, L_{2}}|x_{1} − x_{2}| < ϵ.  (3.8)

Define δ_{o} = min{δ_{1}, δ_{2}}, then by using (3.7) − (3.8) and the fact that (Φu)(x) ≡ 0 for x ∈ [1, b] , we obtain that

|(Φu)(x_{1}) − (Φu)(x_{2})| < ϵ,

for x_{1}, x_{2} ∈ [a, b], |x_{1} −x_{2}| < δ_{o}, which implies that Φ(A) is equicontinuous. In view of the Arzela-Ascoli lemma, we know that Φ̄(A) is a compact set; thus Φ : K → K is completely continuous.

Lemma 4. (H_{6}) implies that there exists r_{o}, R_{o} : 0 < r_{o} < R_{o} such that

i(Φ, K_{r}, K) = 0, 0 < r ≤ r_{o}; i(Φ, K_{R}, K) = 1, R ≥ R_{o}.

Proof. By using the first equality of (H_{6}) we can choose r_{o} > 0 such that

f(v) ≥ Mv, 0 ≤ v ≤ r_{o},

where M satisfies λ^{2}T Bσ_{o}M > 2 and

T = ∫_{E_{σ}}(1 − s)r(s)ds.

If u ∈ ∂K_{r} (0 < r ≤ r_{o}), one has

u(x) ≥ λσ_{o}∥u∥ = λσ_{o}r,  x ∈ [σ, 1 − σ].  (3.9)

Then we obtain ( noting that u_{o}(x) ≡ 0 as x ∈ [0, 1])

(Φu)(\frac{1}{2}) = ∫_{0}^{1} G(\frac{1}{2}, s)r(s)f(u(w(s)) + u_{o}(w(s)))ds
≥ \frac{1}{2}λB ∫_{E_{σ}}(1 − s)r(s)f(u(w(s)))ds
≥ \frac{1}{2}λ^{2}Bσ_{o}rT M
> r = ∥u∥.

This leads to

∥Φu∥ > ∥u∥, ∀u ∈ ∂K_{r}.

Thus we have from Lemma 1 i(Φ, K_{r}, K) = 0, for 0 < r ≤ r_{o}.

On the other hand, the second equality of (H_{6}) leads to: for ∀ϵ > 0, there is a R′ > r_{o} + ∥u_{o}∥ such that

f(v) ≤ ϵv,  v > R′,  (3.10)

where ϵ satisfies

ϵB(1 + ∥u_{o}∥) ∫_{0}^{1}(1 − s)r(s)ds < \frac{1}{2}.  (3.11)

Choose

R_{o} > 1 + 2B max{f(v); 0 ≤ v ≤ R′ + ∥u_{o}∥} ∫_{0}^{1}(1 − s)r(s)ds.  (3.12)

Thus, if u ∈ ∂K_{R} and R ≥ R_{o}, then we have from (3.10)-(3.12) that

(Φu)(x) ≤ B ∫_{0}^{1}(1 − s)r(s)f(u(w(s)) + u_{o}(w(s)))ds
= B ∫_{u(w(s))>R′}(1 − s)r(s)f(u(w(s)) + u_{o}(w(s)))ds
+B ∫_{0≤u(w(s))≤R′}(1 − s)r(s)f(u(w(s)) + u_{o}(w(s)))ds
≤ ϵB(∥u∥ + ∥u_{o}∥) ∫_{0}^{1}(1 − s)r(s)ds
+B max{f(v); 0 ≤ v ≤ R′ + ∥u_{o}∥} ∫_{0}^{1}(1 − s)r(s)ds
< \frac{1}{2}∥u∥ + \frac{1}{2} + B max\{f(v); 0 ≤ v ≤ R′ + ∥u_{o}∥\} ∫_{0}^{1}(1 − s)r(s)ds
< \frac{1}{2}∥u∥ + \frac{1}{2}R, 0 ≤ x ≤ 1.

That is

∥Φu∥ = ∥Φu∥_{[0,1]} < ∥u∥, ∀u ∈ ∂K_{R}, .

Thus i(Φ, K_{R}, K) = 1 for R ≥ R_{o}.

Lemma 5. (H_{7}) implies that there exists r_{o}, R_{o} : 0 < r_{o} < R_{o} such that

i(Φ, K_{r}, K) = 1, 0 < r ≤ r_{o}; i(Φ, K_{R}, K) = 0, R ≥ R_{o}.

Proof. Since ξ(x) ≡ 0, η(x) ≡ 0, we have u_{o}(x) ≡ 0. By the first equality of (H_{7}), one can choose a r_{o} > 0 such that

f(v) ≤ ϵv,  0 ≤ v ≤ r_{o},  (3.13)

where ϵ > 0 satisfies

0 < ϵB ∫_{0}^{1}(1 − s)r(s)ds < \frac{1}{2}.  (3.14)

If u ∈ ∂K_{r}, 0 < r ≤ r_{o}, then we have from (3.3), (3.13) and (3.14) that

0 ≤ (Φu)(x) ≤ B ∫_{0}^{1}(1 − s)r(s)f(u(w(s)))ds
≤ ϵB ∫_{0}^{1}(1 − s)r(s)u(w(s))ds
≤ ϵB∥u∥ ∫_{0}^{1}(1 − s)r(s)ds
< ∥u∥, 0 ≤ x ≤ 1,

That is

∥Φu∥ = ∥Φu∥_{[0,1]} < ∥u∥, ∀u ∈ ∂K_{r}.

So we have the conclusion that i(Φ, K_{r}, K) = 1, 0 < r ≤ r_{o}.

On the other hand, the second equality of (H_{7}) implies that ∀M > 0, there is an R_{o} > r_{o} such that

f(v) ≥ Mv, v > λσ_{o}R_{o}; (3.15)

here we choose M > 0 such that λ^{2}T Bσ_{o}M > 2. For u ∈ ∂K_{R}, R ≥ R_{o}, we have from the definition of K_{R} that

u(x) ≥ λσ_{o}∥u∥ = λσ_{o}R,  x ∈ [σ, 1 − σ].  (3.16)

Thus we have from (3.3),(3.15)-(3.16) that

(Φu)(\frac{1}{2}) = ∫_{0}^{1} G(\frac{1}{2}, s)r(s)f(u(w(s)))ds
≥ \frac{1}{2}λB ∫_{E_{σ}}(1 − s)r(s)f(u(w(s)))ds
≥ \frac{1}{2}λ^{2}Bσ_{o}RM ∫_{E_{σ}}(1 − s)r(s)ds
= \frac{1}{2}λ^{2}Bσ_{o}RT M
> R = ∥u∥,

which leads to

∥Φu∥ > ∥u∥ ∀u ∈ ∂K_{R}.

Thus i(Φ, K_{R}, K) = 0 for R ≥ R_{o}.

Now by using the above lemmas, we can show the conclusions of Theorem 1. Proof of Theorem 1. For 0 < m < 1 < n, define f_{1}(u) = u^{m}, f_{2}(u) = u^{n}, u ≥

0, so that f_{1}(u) satisfies (H_{6}) and f_{2}(u) satisfies (H_{7}). Define Φ_{i} : K → K (i = 1, 2) as follows:

(Φ_{i}u)(x) = \{ \begin{matrix}e^{\frac{α}{β} x} ∫_{0}^{1} G(0, t)r(t)f_{i}(u(w(t)) + u_{o}(w(t)))dt, & a ≤ x ≤ 0, \\ ∫_{0}^{1} G(x, t)r(t)f_{i}(u(w(t)) + u_{o}(w(t)))dt, & 0 ≤ x ≤ 1, \\ 0, & 1 ≤ x ≤ b. \end{matrix}  (3.17)

Then Φ_{i}u (i = 1, 2) are completely continuous operators. One has from Lemma 4-5 that

i(Φ_{1}, K_{r}, K) = 0, 0 < r ≤ r_{o};  i(Φ_{1}, K_{R}, K) = 1, R ≥ R_{o},  (3.18)

and

i(Φ_{2}, K_{r}, K) = 1, 0 < r ≤ r_{o};  i(Φ_{2}, K_{R}, K) = 0, R ≥ R_{o},  (3.19)

Define H_{i}(s, u) = (1 − s)Φu + sΦ_{i}u (i = 1, 2) so that for any s ∈ [0, 1], H_{i} is a completely continuous operator. Furthermore, for any ω > 0 and i = 1, 2, we have

|H_{i}(s_{1}, u) − H_{i}(s_{2}, u)| ≤ |s_{1} − s_{2}|[∥Φ_{i}u∥ + ∥Φu∥]

as s_{1}, s_{2} ∈ [0, 1], u ∈ K_{ω}. Note that ∥Φ_{i}u∥+∥Φu∥ is uniformly bounded in K_{ω}. Thus H_{i}(s, u) is continuous on u ∈ K_{ω} uniformly for s ∈ [0, 1]. According to Lemma 7.2.3 in [18], we conclude that H_{i}(s, u) is a completely continuous operater on [0, 1] × K_{ω}.

Suppose that (H_{4}) holds. By using the first inequality of (H_{4}) and the definition of f_{1}, one can find ϵ > 0 and r_{1} : 0 < r_{1} ≤ r_{o} such that

\{ \begin{matrix}f(u) ≥ (kλ_{1} + ϵ)u, & 0 ≤ u ≤ r_{1}, \\ f_{1}(u) ≥ (kλ_{1} + ϵ)u, & 0 ≤ u ≤ r_{1}. \end{matrix}  (3.20)

In what follows, we shall show that H_{1}(s, u) ≠ u for u ∈ ∂K_{r_{1}} and s ∈ [0, 1]. If this is not true, then there exist s_{1} : 0 ≤ s_{1} ≤ 1 and u_{1} ∈ ∂K_{r_{1}} such that H_{1}(s_{1}, u_{1}) = u_{1}. Note that u_{1}(x) satisfies

\begin{aligned}−u′′_{1}(x) = & (1 − s_{1})r(x)f(u_{1}(w(x)) + u_{o}(w(x))) \\ & +s_{1}r(x)f_{1}(u_{1}(w(x)) + u_{o}(w(x))),  0 < x < 1; \end{aligned}  (3.21)

and

\{ \begin{matrix}αu_{1}(x) − βu′_{1}(x) = 0, & a ≤ x ≤ 0, \\ u_{1}(x) = 0, & 1 ≤ x ≤ b. \end{matrix}  (3.22)

Multiplying both sides of (3.21) by ϕ_{1}(x) and then integrating it from 0 to 1, after two times of integrating by parts, we get from (3.22) that

\begin{aligned}λ_{1} ∫_{0}^{1} u_{1}(x)ϕ_{1}(x)dx & = ∫_{0}^{1} ϕ_{1}(x)[(1 − s_{1})r(x)f(u_{1}(w(x)) + u_{o}(w(x))) \\ & +s_{1}r(x)f_{1}(u_{1}(w(x)) + u_{o}(w(x)))]dx. \end{aligned}  (3.23)

Noting that u_{o}(w(x)) ≡ 0 as x ∈ E_{σ}, we obtain from (3.20) and (H_{4}) that

λ_{1} ∫_{0}^{1} u_{1}(x)ϕ_{1}(x)dx ≥ ∫_{E_{σ}} ϕ_{1}(x)r(x)[(1 − s_{1})f(u_{1}(w(x)))) + s_{1}f_{1}(u_{1}(w(x)))]dx
≥ ∫_{E_{σ}} ϕ_{1}(x)r(x)[(1 − s_{1})(kλ_{1} + ϵ)u_{1}(w(x))
+s_{1}(kλ_{1} + ϵ)u_{1}(w(x))]dx
≥ (λ_{1} + \frac{ϵ}{k})kλσ_{o}∥u_{1}∥ ∫_{E_{σ}} ϕ_{1}(x)r(x)dx
≥ (λ_{1} + \frac{ϵ}{k})∥u_{1}∥ ∫_{0}^{1} ϕ_{1}(x)dx.
(3.24)

We also have

λ_{1} ∫_{0}^{1} u_{1}(x)ϕ_{1}(x)dx ≤ λ_{1}∥u_{1}∥ ∫_{0}^{1} ϕ_{1}(x)dx,  (3.25)

which together with (3.24) leads to

λ_{1} ≥ λ_{1} + \frac{ϵ}{k} .

This is impossible. Thus H_{1}(s, u) ≠ u for u ∈ ∂K_{r_{1}} and s ∈ [0, 1]. In view of the homotopic invariant property of topological degree (see [9] or [18]) and (3.18) we know that

\begin{aligned}i(Φ, K_{r_{1}}, K) & = i(H_{1}(0, ·), K_{r_{1}}, K) \\ & = i(H_{1}(1, ·), K_{r_{1}}, K) = i(Φ_{1}, K_{r_{1}}, K) = 0. \end{aligned}  (3.26)

On the other hand, according to the second inequality of (H_{4}), there exist ϵ > 0 and R′ > R_{o} such that

f(u) ≤ (qλ_{1} − ϵ)u, u ≥ R′.

If C = max_{0≤u≤R′} |f(u) − (qλ_{1} − ϵ)u| + 1, then we deduce that

f(u) ≤ (qλ_{1} − ϵ)u + C, ∀u ≥ 0.  (3.27)

Define H(s, u) = sΦu, s ∈ [0, 1]. We shall show that there exists a R_{1} ≥ R′ such that

H(s, u) ≠ u, ∀s ∈ [0, 1], u ∈ K, ∥u∥ ≥ R_{1}.  (3.28)

If ∃s_{1} ∈ [0, 1], u_{1} ∈ K such that H(s_{1}, u_{1}) = u_{1}, then it is similar to the argument of (3.24)-(3.25) that

\begin{aligned}& λ_{1} ∫_{0}^{1} u_{1}(x)ϕ_{1}(x)dx \\ = & s_{1} ∫_{0}^{1} r(x)ϕ_{1}(x)f(u_{1}(w(x)) + u_{o}(w(x)))dx \\ ≤ & q(λ_{1} − \frac{ϵ}{q})∥u_{1} + u_{o}∥ ∫_{0}^{1} r(x)ϕ_{1}(x)dx + C ∫_{0}^{1} r(x)ϕ_{1}(x)dx \\ ≤ & q(λ_{1} − \frac{ϵ}{q})∥u_{1}∥ ∫_{0}^{1} r(x)ϕ_{1}(x)dx + C_{1} ∫_{0}^{1} r(x)ϕ_{1}(x)dx, \end{aligned}  (3.29)

and

\begin{aligned}λ_{1} ∫_{0}^{1} u_{1}(x)ϕ_{1}(x)dx & ≥ λ_{1}λσ_{o}∥u_{1}∥ ∫_{σ}^{1−σ} ϕ_{1}(x)dx \\ & ≥ λ_{1}q∥u_{1}∥ ∫_{0}^{1} r(x)ϕ_{1}(x)dx, \end{aligned}  (3.30)

where C_{1} = q(λ_{1} − \frac{ϵ}{q})∥u_{o}∥ + C. Combining (3.29) with (3.30), we have

∥u_{1}∥ ≤ \frac{C_{1}}{ϵ} = R̄.

Define R_{1} = max{R′, R̄}, then (3.28) is true. By the homotopic invariant property of topological degree, one has

\begin{aligned}i(Φ, K_{R_{1}}, K) & = i(H(1, ·), K_{R_{1}}, K) \\ & = i(H(0, ·), K_{R_{1}}, K) = i(θ, K_{R_{1}}, K) = 1, \end{aligned}  (3.31)

where θ is zero operator. In view of (3.26),(3.31), we obtain

i(Φ, K_{R_{1}}\K_{r_{1}}, K) = 1.

Thus Φ has a fixed point in K_{R_{1}}\K_{r_{1}}.

Now assume that (H_{5}) is true. The first inequality and the definition of f_{2} lead to: ∃ϵ > 0 and R′ > R_{o} such that

\{ \begin{matrix}f(u) ≥ (kλ_{1} + ϵ)u, & u > R′, \\ f_{2}(u) ≥ (kλ_{1} + ϵ)u, & u > R′. \end{matrix}

Let

C = max_{0≤u≤R′} |f(u) − (kλ_{1} + ϵ)u| + max_{0≤u≤R′} |f_{2}(u) − (kλ_{1} + ϵ)u| + 1,

then we have

f(u) ≥ (kλ_{1} + ϵ)u − C, f_{2}(u) ≥ (kλ_{1} + ϵ)u − C, ∀u ≥ 0.  (3.32)

We want to show that ∃R_{1} ≥ R′ such

H_{2}(s, u) ≠ u, ∀s ∈ [0, 1], u ∈ K, ∥u∥ ≥ R_{1}.  (3.33)

In fact, if there are s_{1} ∈ [0, 1], u_{1} ∈ K such that H_{2}(s_{1}, u_{1}) = u_{1}, then using (3.32), it is analogous to the argument of (3.24)-(3.25) that

λ_{1} ∫_{0}^{1} u_{1}(x)ϕ_{1}(x)dx ≥ ∫_{E_{σ}} ϕ_{1}(x)r(x){(1 − s_{1})[(kλ_{1} + ϵ)u_{1}(w(x)) − C]
+s_{1}[(kλ_{1} + ϵ)u_{1}(w(x)) − C]}dx

≥ (λ_{1} + \frac{ϵ}{k})kλσ_{o}∥u_{1}∥ ∫_{E_{σ}} ϕ_{1}(x)r(x)dx − ∫_{E_{σ}} Cϕ_{1}(x)r(x)dx;

(3.34)
\begin{aligned}λ_{1} ∫_{0}^{1} u_{1}(x)ϕ_{1}(x)dx & ≤ λ_{1}∥u_{1}∥ ∫_{0}^{1} ϕ_{1}(x)dx \\ & ≤ λ_{1}kλσ_{o}∥u_{1}∥ ∫_{E_{σ}} r(x)ϕ_{1}(x)dx. \end{aligned}  (3.35)

(3.34)-(3.35) lead to ∥u_{1}∥ ≤ \frac{C}{λσ_{o}ϵ} = R̄. Let R_{1} = max\{R′, R̄\}. We obtain (3.33) and then we have

i(Φ, K_{R_{1}}, K) = i(Φ_{2}, K_{R_{1}}, K) = 0.  (3.36)

On the other hand, noting that ξ(x) ≡ 0, μ(x) ≡ 0, one has u_{o}(x) ≡ 0 for x ∈ [a, b]. Define H(s, u) as above. By the second inequality of (H_{5}), there exist ϵ > 0 and r_{1} : 0 < r_{1} ≤ r_{o} such that

f(u) ≤ (qλ_{1} − ϵ)u, 0 ≤ u ≤ r_{1}.  (3.37)

We could also show that

H(s, u) ≠ u, ∀s ∈ [0, 1], u ∈ ∂K_{r_{1}}.

But we omit the details. Thus we obtain

i(Φ, K_{r_{1}}, K) = i(θ, K_{r_{1}}, K) = 1.  (3.38)

In view of (3.36),(3.38), we obtain

i(Φ, K_{R_{1}}\K_{r_{1}}, K) = −1.

Thus Φ has a fixed point in K_{R_{1}}\K_{r_{1}}.

Suppose that u is the fixed point of Φ in K_{R_{1}}\K_{r_{1}}. Let y(x) = u(x) + u_{o}(x). Since y(x) = u(x) for x ∈ [0, 1] and 0 < r_{1} ≤ ∥u∥ = ∥u∥_{[0,1]} = ∥y∥_{[0,1]} ≤ R_{1}, we have from (2.5) that y(x) is the positive solution of BVP(1.1).

Thus we complete the proof.

Example. Let us introduce an example to illustrate the usage of our theorem. Consider the BVP:

\{ \begin{matrix}y′′(x) + r(x)y^{\frac{1}{2}} (x − \frac{1}{3}) = 0, & 0 < x < 1, \\ y(x) = − sin πx, & −\frac{1}{3} ≤ x ≤ 0, \\ y(1) = 0; \end{matrix}  (3.39)

where

r(x) = \{ \begin{matrix}π^{2} (sin πx) |sin π(x − \frac{1}{3})|^{− \frac{1}{2}} , & x ∈ [0, \frac{1}{3}) ∪ (\frac{1}{3}, 1], \\ 0, & x = \frac{1}{3}. \end{matrix}

Then w(x) = x − \frac{1}{3}, a = −\frac{1}{3}, b = 1, f(v) = v^{\frac{1}{2}} , α = γ = 1, β = δ = 0, E = [\frac{1}{3}, 1]. Since

\frac{f(v)}{v} = \frac{v^{\frac{1}{2}}}{v} = v^{− \frac{1}{2}}

we have lim_{v→+∞} \frac{f(v)}{v} = 0, lim_{v→0+} \frac{f(v)}{v} = +∞. Thus (P_{1}), (P_{2}), (H_{1}) − (H_{3}), (H_{6}) are

satisfied and (3.39) has at least one positive solution y(x). In fact,

y(x) = \{ \begin{matrix}− sin πx, & −\frac{1}{3} ≤ x ≤ 0, \\ sin πx, & 0 < x ≤ 1, \end{matrix}

is a positive solution of (3.39).

References

[1] C. Bandle and M. K. Kwong, Semilinear Elliptic Problems in Annular Domains, J Appl Math Phys, 40 (1989), 245-257.

[2] C. Bandle, C. V. Coffman and M. Marcus, Nonlinear Elliptic Problems in Annular Domains. J Differential Equations, 69 (1987), 322-345.

[3] C. V. Coffman and M. Marcus, Existence and Uniqueness Results for Semilinear Dirichlet Problems in Annuli. Arch Rational Mech Anal, 108 (1987), 293-307.

[4] L. H. Erbe and H. Y. Wang, On the Existence of Positive Solutions of Ordinary Differential Equations. Proc Amer Math Soc, 120 (1994), 743-748.

[5] L. H. Erbe and S. C. Hu, Multiple Positive Solution of Some Boundary Value Problems, J.Math.Anal.Appl., 184 (1994), 640-648.

[6] L. H. Erbe and Q. K. Kong, Boundary Value Problems for Singular Second-Order Functional Differential Equations. J Comput Appl Math, 53 (1994), 377-388.

[7] L. H. Erbe, Z. C. Wang and L. T. Li, Boundary Value Problems for Second Order Mixed Type Functional Differential Equations, Boundary Value Problems for Functional Differential Equations, World Scientific, 1995, 143-151.

[8] X. Garaizar, Existence of Positive Radial Solutions for Semilinear Elliptic Problems in the Annulus. J Differential Equations, 70 (1987), 69-72.

[9] D. J. Guo, Nonlinear Functional Analysis (in Chinese), Shandong Scientific Press (Jinan), 1985, 302-303.

[10] B. S. Lalli and B. G. Zhang, Boundary Value Problems for Second Order Functional Differential Equations. Ann. of Diff. Eqs, 8 (1992), 261-268.

[11] Z. L. Liu and F. Y. Li, Multiple Positive Solution of Nonlinear Two-point Boundary Value Problems, J.Math. Anal. Appl., 203 (1996), 610-625.

[12] J. W. Lee and D. O’Regan, Existence Results for Differential Delay Equations-I, J. Differential Equations, 102 (1993), 342-359.

[13] J. W. Lee and D. O’Regan, Existence Results for Differential Delay Equations-II, Nonlinear Analysis, 17 (1991), 683-702.

[14] S. K. Ntouyas, Y. G. Sficas and P. CH. Tsamatos, An Existence Principle for Boundary Value Problems for Second Order Functional Differential Equations, Nonlinear Analysis, 20 (1993), 215-222.

[15] H. Y. Wang, On the Existence of Positive Solutions for Semilinear Elliptic Equations in the Annulus. J Differential Equations, 109 (1994), 1-8.

[16] P. X. Weng, Boundary Value Problems for Second Order Mixed Type Functional Differential Equations, Appl. Math.-JCU, 12B (1997), 155-164.

[17] J. S. W. Wong, On the Generalized Emden-Fowler Equation. SIAM Rev, 17 (1975), 339-360.

[18] Z. Y. Li and Q. X. Ye, Theory of Reaction-Diffusive Equations (in Chinese), Scientific Press (Peijing), 1990.