Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 43, 1-14; http://www.math.u-szeged.hu/ejqtde/

Positive Solutions for Singular m-Point Boundary Value Problems with Sign Changing Nonlinearities

Depending on x′ ^{∗}

Ya Ma, Baoqiang Yan ^{†}

Department of Mathematics, Shandong Normal University, Jinan, 250014, P.R. China

Abstract

Using the theory of fixed point theorem in cone, this paper presents the existence of positive solutions for the singular m-point boundary value problem

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

where 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, α_{i} ∈ [0, 1), i = 1, 2, ⋯, m − 2 , with

0 < ∑_{i=1}^{m−2} α_{i} < 1 and f may change sign and may be singular at x = 0 and x′ = 0. Keywords: m-point boundary value problem; Singularity; Positive solutions; Fixed point theorem Mathematics subject classification: 34B15, 34B10

1. Introduction

The study of multi-point BVP (boundary value problem) for linear second-order ordinary differential equations was initiated by Il’in and Moiseev [3-4]. Since then, many authors studied more general nonlinear multi-point BVP, for examples [2, 5-8], and references therein. In [7], Gupta, Ntouyas, and Tsamatos considered the existence of a C^{1}[0, 1] solution for the m-point boundary value problem

\{ \begin{matrix}x′′(t) = f(t, x(t), x′(t)) + e(t), 0 < t < 1, \\ x′(0) = 0, x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), \end{matrix}

∗The project is supported by the fund of National Natural Science (10871120), the fund of Shandong Education Committee (J07WH08) and the fund of Shandong Natural Science (Y2008A06)

†Corresponding author: yanbqcn@yahoo.com.cn

where ξ_{i} ∈ (0, 1), i = 1, 2, ⋯, m − 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, a_{i} ∈ R,

i = 1, 2, ⋯, m − 2, have the same sign, ∑_{i=1}^{m−2} a_{i} ≠ 1, e ∈ L^{1}[0, 1], f : [0, 1] × R^{2} → R

is a function satisfying Carathéodory’s conditions and a growth condition of the form | f(t, u, v) |≤ p_{1}(t)|u| + q_{1}(t)|v| + r_{1}(t) with p_{1}, q_{1}, r_{1} ∈ L^{1}[0, 1]. Recently, using Leray-Schauder continuation theorem, R.Ma and Donal O’Regan proved the existence of positive solutions of C^{1}[0, 1) solutions for the above BVP, where f : [0, 1] × R^{2} → R satisfies the Carathéodory’s conditions (see [8]).

Motivated by the works of [7,8], in this paper, we discuss the equation

\{ \begin{matrix}x′′(t) + a(t)f(t, x(t), x′(t)) = 0, 0 < t < 1, \\ x′(0) = 0, x(1) = ∑_{i=1}^{m−2} α_{i}x(ξ_{i}), \end{matrix}  (1.1)

where 0 < ξ_{i} < 1, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, α_{i} ∈ [0, 1) with 0 < ∑_{i=1}^{m−2} α_{i} < 1 and f

may change sign and may be singular at x = 0 and x′ = 0.

Our main features are as follows. Firstly, the nonlinearity af possesses singularity, that is, a(t)f(t, x, x′) may be singular at t = 0, t = 1, x = 0 and x′ = 0; also the degree of singularity in x and x′ may be arbitrary(i. e., if f contains \frac{1}{x^{α}} and \frac{1}{(−x′)^{γ}}, α and γ may be big enough). Secondly, f is allowed to change sign. Finally, we discuss the maximal and minimal solutions for equations (1.1). Some ideas come from [11-12].

2. Preliminaries

Now we list the following conditions for convenience .

(H_{1}) β, a, k ∈ C((0, 1), R_{+}), F ∈ C(R_{+}, R_{+}), G ∈ C(R_{−}, R_{+}), ak ∈ L[0, 1]; (H_{2}) F is bounded on any interval [z, +∞), z > 0;

(H_{3}) ∫_{−∞}^{−1} \frac{1}{G(y)}dy = +∞;

and the following conditions are satisfied (P_{1}) f ∈ C((0, 1) × R_{+} × R_{−}, R);

(P_{2}) 0 < ∑_{i=1}^{m−2} α_{i} < 1, 0 < ξ_{i} < 1 and |f(t, x, y)| ≤ k(t)F (x)G(y);

(P_{3}) There exists δ > 0 such that f(t, x, y) ≥ β(t), y ∈ (−δ, 0); where R_{+} = (0, +∞), R_{−} = (−∞, 0), R = (−∞, +∞).

Lemma 2.1^{[1]} Let E be a Banach space, K a cone of E, and B_{R} = {x ∈ E : ∥x∥ < R}, where 0 < r < R. Suppose that F: K ∩ \overline{B_{R}\B_{r}} = K_{R,r} → K is a completely continuous operator and the following conditions are satisfied

(1) ∥F (x)∥ ≥ ||x|| for any x ∈ K with ∥x∥ = r.

(2) If x ≠ λF (x) for any x ∈ K with ∥x∥ = R and 0 < λ < 1. Then F has a fixed point in K_{R,r}.

Let C[0, 1] = {x : [0, 1] → R|x(t) is continuous on [0, 1]} with norm ∥y∥ = max_{t∈[0,1]} |y(t)|. Then C[0, 1] is a Banach space.

Lemma 2.2 Let (H_{1})-(P_{3}) hold. For each given natural number n > 0, there exists y_{n} ∈ C[0, 1] with y_{n}(t) ≤ −\frac{1}{n} such that

y_{n}(t) = −\frac{1}{n} − ∫_{0}^{t} a(s)f(s, (Ay_{n})(s) + \frac{1}{n}, y_{n}(s))ds,  t ∈ [0, 1],  (2.1)

where

(Ay)(t) = \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y(τ)dτ,  t ∈ [0, 1].

Proof. For y ∈ P = {y ∈ C[0, 1] : y(t) ≤ 0, t ∈ [0, 1]}, define a operator as follows

(T_{n}y)(t) = −\frac{1}{n} + min\{0, − ∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds\}, t ∈ [0, 1],

(2.2)

where n > 0 is a natural number. For y ∈ P, we have

(Ay)(t) = \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y(τ)dτ

≥ \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{1} −y(τ)dτ

≥ \frac{∑_{i=1}^{m−2} α_{i}}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{m−2}} −y(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}}

≥ \frac{∑_{i=1}^{m−2} α_{i}}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{ξ_{m−2}}^{1} −y(τ)dτ

≥ 0, t ∈ [0, 1]. Let

c(y(t)) = − ∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds, t ∈ [0, 1],

c(y_{k}(t)) = − ∫_{0}^{t} a(s)f(s, (Ay_{k})(s) + \frac{1}{n}, min\{y_{k}(s), −\frac{1}{n}\})ds, t ∈ [0, 1].

By the equality min\{c, 0\} = \frac{c − |c|}{2} , it is easy to know

(T_{n}y)(t) = −\frac{1}{n} + \frac{c(y(t)) − |c(y(t)|}{2}, t ∈ [0, 1].

Let y_{k}, y ∈ P with lim_{k→+∞} ∥y_{k} − y∥ = 0. Then, there exists a constant h > 0, such that ∥y_{k}∥ ≤ h and ∥y∥ ≤ h. Thus, | min\{y_{k}(s), −\frac{1}{n}\} − min\{y(s), −\frac{1}{n}\}| → 0, uniformly for s ∈ [0, 1] as k → +∞. Therefore, |(Ay_{k})(s)+\frac{1}{n}−((Ay)(s)+\frac{1}{n})| → 0 for all s ∈ [0, 1] as k → +∞. (P_{1}) implies that \{a(s)f(s, (Ay_{k})(s) + \frac{1}{n}, min\{y_{k}(s), −\frac{1}{n}\})\} → \{a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})\}, for s ∈ (0, 1) as k → +∞. By the Lebesgue dominated convergence

EJQTDE, 2009 No. 43, p. 3 theorem (the dominating function a(s)k(s)F [\frac{1}{n}, +∞)G[−h − \frac{1}{n}, −\frac{1}{n}]), we have ∥cy_{k} − cy∥ → 0, which yields that

∥T_{n}y_{k} − T_{n}y∥ = ∥\frac{c(y_{k}) − c(y) − |c(y_{k})| + |c(y)|}{2}∥

≤ ∥\frac{c(y_{k}) − c(y) + |c(y_{k}) − c(y)|}{2}∥

≤ ∥c(y_{k}) − c(y)∥ → 0, as k → +∞.

Consequently, T_{n} is a continuous operator.

Let C be a bounded set in P, i.e., there exists h_{1} > 0 such that ∥y∥ ≤ h_{1}, for any y ∈ C. For any t_{1}, t_{2} ∈ [0, 1], t_{1} < t_{2}, y ∈ C,

|(T_{n}y)(t_{2}) − (T_{n}y)(t_{1})|

= |\frac{− ∫_{t_{1}}^{t_{2}} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\}ds}{2}

+\frac{| ∫_{0}^{t_{2}} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n})ds| − | ∫_{0}^{t_{1}} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n})ds|}{2}|

≤ |\frac{− ∫_{t_{1}}^{t_{2}} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\}ds}{2}|

+\frac{| ∫_{t_{1}}^{t_{2}} a(s)f(s, (Ay)(s), min\{y(s), −\frac{1}{n}\}ds|}{2}

≤ | ∫_{t_{1}}^{t_{2}} a(s)k(s)ds| sup F [\frac{1}{n}, +∞) sup G[−h_{1} − \frac{1}{n}, −\frac{1}{n}].

According to the absolute continuity of the Lebesgue integral, for any ϵ > 0, there exists δ > 0 such that | ∫_{t_{1}}^{t_{2}} a(s)k(s)ds| < ϵ, |t_{2} − t_{1}| < δ. Therefore, {T_{n}y, y ∈ C} is equicontinuous.

|(T_{n}y)(t)| = | − \frac{1}{n} + min\{0, − ∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds\}|

≤ |\frac{1}{n}| + | ∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds|

≤ 1 + ∫_{0}^{t} a(s)|f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\}|)ds

≤ 1 + ∫_{0}^{1} a(s)k(s)ds sup F [\frac{1}{n}, +∞)G[−h − \frac{1}{n}, \frac{1}{n}], t ∈ [0, 1].

Therefore {T_{n}y, y ∈ C} is bounded.

Hence T_{n} is a completely continuous operator.

By (H_{3}), choose a sufficiently large R_{n} > 1 to fit ∫_{−R_{n}}^{−1} \frac{dy}{G(y)} > ∫_{0}^{1} a(s)k(s)ds sup F [\frac{1}{n}, +∞).

For n > \frac{1}{δ}, we prove that

y(t) ≠ λ(T_{n}y)(t) = \frac{−λ}{n}+λ min\{0, − ∫_{0}^{t} a(s)f(s, (Ay)(s)+\frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds\}, t ∈ [0, 1],

(2.3) for any y ∈ P with ||y|| = R_{n} and 0 < λ < 1.

In fact, if there exists y ∈ P with ∥y∥ = R_{n} and 0 < λ < 1 such that

y(t) = \frac{−λ}{n} + λ min\{0, − ∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds\}, t ∈ [0, 1]. (2.4)

y(0) = \frac{−λ}{n} . Since n > \frac{1}{δ} , we have −δ < y(0) < 0, which implies there exists δ_{0} > 0 such that y(t) > −δ, t ∈ (0, δ_{0}). (P_{3}) implies

∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds > 0, t ∈ [0, 1].

Let t^{∗} = sup\{s ∈ [0, 1]| ∫_{0}^{t} a(τ)f(τ, (Ay)(τ) + \frac{1}{n}, min\{y(τ), −\frac{1}{n}\})dτ > 0, 0 ≤ t ≤ s\}.

We show that t^{∗} = 1. If t^{∗} < 1, we have

\{ \begin{matrix}∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds > 0, t ∈ (0, t^{∗}), \\ ∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds = 0, t = t^{∗}, \end{matrix}

y(t) = \frac{−λ}{n} − λ ∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds, t ∈ (0, t^{∗}],  (2.5)

y(t^{∗}) = \frac{−λ}{n} > −δ. (2.6)

(2.6) and (P_{3}) imply there exists r > 0 such that f(t, x, y) ≥ β(t), t ∈ (t^{∗} − r, t^{∗}). So

\begin{aligned}y(t^{∗}) & = \frac{−λ}{n} − λ ∫_{0}^{t^{∗}} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds \\ & ≤ \frac{−λ}{n} − λ ∫_{0}^{t^{∗}−r} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds − λ ∫_{t^{∗}−r}^{t^{∗}} a(s)β(s)ds, \end{aligned}

∫_{0}^{t^{∗}−r} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds + ∫_{t^{∗}−r}^{t^{∗}} a(s)β(s)ds < 0,

which is a contradiction. Then, t^{∗} = 1. Hence,

y(t) = \frac{−λ}{n} − λ ∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, min\{y(s), −\frac{1}{n}\})ds, t ∈ [0, 1].  (2.7)

Since ∥y∥ = R_{n} > 1 and y ∈ P, there exists a t_{0} ∈ (0, 1) with y(t_{0}) = −R_{n} < −1 and a t_{1} ∈ (0, 1) such that y(t) < −1 < −\frac{1}{n}, t ∈ (t_{0}, t_{1}], which together with (2.7) implies that

y(t) = \frac{−λ}{n} − λ ∫_{0}^{t} a(s)f(s, (Ay)(s) + \frac{1}{n}, y(s))ds, t ∈ (t_{0}, t_{1}].  (2.8)

Differentiating (2.8) and using (H_{2}), we obtain

−y′(t) = λa(t)f(t, (Ay)(t) + \frac{1}{n}, y(t)) ≤ a(t)F ((Ay)(t) + \frac{1}{n})G(y(t)), t ∈ (t_{0}, t_{1}].

And then

\frac{−y′(t)}{G(y(t))} ≤ a(t)k(t) sup F [(Ay)(t) + \frac{1}{n}, +∞) ≤ a(t)k(t) sup F [\frac{1}{n}, +∞), t ∈ (t_{0}, t_{1}). (2.9)

Integrating for (2.9) from t_{0} to t_{1}, we have

∫_{y(t_{0})}^{y(t_{1})} \frac{dy}{G(y)} ≤ ∫_{t_{0}}^{t_{1}} a(s)k(s)ds sup F [\frac{1}{n}, +∞), t ∈ (t_{0}, t_{1}).  (2.10)

Then

∫_{−R_{n}}^{−1} \frac{dy}{G(y)} ≤ ∫_{−R_{n}}^{y(t_{1})} \frac{dy}{G(y)} ≤ ∫_{t_{0}}^{t_{1}} a(s)k(s)ds sup F [\frac{1}{n}, +∞) ≤ ∫_{0}^{1} a(s)k(s)ds sup F [\frac{1}{n}, +∞),

which contradicts

∫_{−R_{n}}^{−1} \frac{dy}{G(y)} > ∫_{0}^{1} a(s)k(s)ds sup F [\frac{1}{n}, +∞).

Hence(2.3) holds. Then put r = \frac{1}{n}, Lemma 2.1 leads to the desired result. This completes the proof.

Lemma 2.3^{[10]} Let {x_{n}(t)} be an infinite sequence of bounded variation function on [a, b] and {x_{n}(t_{0})}(t_{0} ∈ [a, b]) and {V (x_{n})} be bounded(V (x) denotes the total variation of x). Then there exists a subsequence {x_{n_{k}}(t)} of {x_{n}(t)}, i ≠ j, n_{i} ≠ n_{j}, such that {x_{n_{k}}(t)} converges everywhere to some bounded variation function x(t) on [a, b].

Lemma 2.4^{[9]}(Zorn) If X is a partially ordered set in which every chain has an upper bound, then X has a maximal element.

3. Main results

Theorem 3.1 Let (H_{1})-(P_{3}) hold. Then the m-point boundary value problem (1.1) has at least one positive solution.

Proof. Put M_{n} = min{y_{n}(t) : t ∈ [0, ξ_{m−2}]}, (H_{1}) implies γ = sup{M_{n}} < 0. In fact, if γ = 0, there exists n_{k} > N > 0 such that M_{n_{k}} → 0 and −δ < y_{n_{k}} < 0. (H_{1}) implies

y_{n_{k}}(t) = −\frac{1}{n} − ∫_{0}^{t} a(s)f(s, (Ay_{n_{k}})(s) + \frac{1}{n}, y_{n_{k}}(s))ds

< −\frac{1}{n} − ∫_{0}^{t} a(s)β(s)ds

< − ∫_{0}^{t} a(s)β(s)ds, t ∈ [0, ξ_{m−2}].

Then y_{n_{k}}(ξ_{m−2}) < − ∫_{0}^{ξ_{m−2}} a(s)β(s)ds, which contradicts to M_{n_{k}} → 0.

Set τ = max\{γ, −δ, − ∫_{0}^{ξ_{m−2}} a(s)β(s)\}. In the remainder of the proof, assume n > −\frac{1}{τ} .

1). First, we prove there exists a t_{n} ∈ (0, ξ_{m−2}] with y_{n}(t_{n}) = τ. In fact, since y_{n}(0) = −\frac{1}{n} > τ, there exists δ_{0} > 0 such that y_{n}(t) > τ, t ∈ (0, δ_{0}). Let t_{n} = sup\{t|s ∈ [0, t], y_{n}(s) > τ} .Then y_{n}(t_{n}) = τ. If t_{n} > ξ_{m−2}, we have y_{n}(t) > τ > −δ, t ∈ [0, ξ_{m−2}] . (H_{1}) shows that

y_{n}(t) = −\frac{1}{n} − ∫_{0}^{t} a(s)f(s, (Ay_{n})(s) + \frac{1}{n}, y_{n}(s))ds

≤ −\frac{1}{n} − ∫_{0}^{t} a(s)β(s)ds

≤ − ∫_{0}^{t} a(s)β(s)ds, t ∈ [0, ξ_{m−2}].

Then τ < y_{n}(ξ_{m−2}) ≤ − ∫_{0}^{ξ_{m−2}} a(s)β(s)ds < τ, which is a contradiction.

Second, we prove

y_{n}(t) ≤ τ, t ∈ [t_{n}, 1].  (3.1)

In fact, if there exists a t ∈ (t_{n}, 1] such that y_{n}(t) > τ, and we choose t′, t′′ ∈ [t_{n}, 1], t′ < t′′

to fit y_{n}(t′) = τ, τ < y_{n}(t) < −\frac{1}{n}, t ∈ (t′, t′′], from (2.1)

0 < ∫_{t′}^{t′′} a(s)f(s, (Ay_{n})(s) + \frac{1}{n}, y_{n}(s))ds = y_{n}(t′) − y_{n}(t′′) < 0.

This contradiction implies that (3.1) holds. Then

\{ \begin{matrix}y_{n}(t) ≤ − ∫_{0}^{t} a(s)β(s)ds, t ∈ [0, t_{n}], \\ y_{n}(t) ≤ τ,  t ∈ [t_{n}, 1]. \end{matrix}

Let W (t) = max\{− ∫_{0}^{t} a(s)β(s)ds, τ\}, t ∈ (0, 1). Obviously, W (t) is bounded on [\frac{1}{3k} , 1 −

\frac{1}{3k}] and y_{n}(t) ≤ W (t), t ∈ [0, 1].

2). \{y_{n}(t)\} is equicontinuous on [\frac{1}{3k} , 1− \frac{1}{3k}](k ≥ 1 is a natural number) and uniformly bounded on [0, 1].

Notice that

\begin{aligned}(Ay_{n})(t) + \frac{1}{n} & = \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{n}(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y_{n}(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y_{n}(τ)dτ + \frac{1}{n} \\ & > \frac{∑_{i=1}^{m−2} α_{i}}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{ξ}^{1} −y_{n}(τ)dτ ≥ \frac{∑_{i=1}^{m−2} α_{i}}{1 − ∑_{i=1}^{m−2} α_{i}} (−τ)(1 − ξ) = Θ, t ∈ [0, 1]. \end{aligned}

We know from (2.9)

∫_{y_{n}(t)}^{− \frac{1}{n}} \frac{dy_{n}}{G(y_{n})} ≤ ∫_{0}^{t} a(s)k(s)ds sup F [Θ, +∞), t ∈ [0, 1].  (3.2)

Now (H_{3}) and (3.2) show that ω(t) = inf{y_{n}(t)} > −∞ is bounded on [0, 1]. On the other hand, it follows from (2.1) and (3.1) that

|y′_{n}(t)| ≤ k(t)a(t) sup F [Θ, +∞) sup G[ω_{k}, max\{τ, W (\frac{1}{3k})\}], (n ≥ k),  (3.3)

where ω_{k} = inf\{ω(t), t ∈ [\frac{1}{3k} , 1− \frac{1}{3k}]\}. Thus (3.3) and the absolute continuity of Lebesgue

integral show that \{y_{n}(t)\} is equicontinuous on [\frac{1}{3k} , 1 − \frac{1}{3k}]. Now the Arzela-Ascoli theorem guarantees that there exists a subsequence of {y_{n}^{(k)}(t)}, which converges uniformly on [\frac{1}{3k} , 1 − \frac{1}{3k}]. When k = 1, there exists a subsequence \{y_{n}^{(1)}(t)\} of \{y_{n}(t)\}, which

converges uniformly on [\frac{1}{3}, \frac{2}{3}].When k = 2, there exists a subsequence \{y_{n}^{(2)}(t)\} of \{y_{n}^{(1)}(t)\},

which converges uniformly on [\frac{1}{6}, \frac{5}{6}]. In general, there exists a subsequence \{y_{n}^{(k+1)}(t)\}

of \{y_{n}^{(k)}(t)\}, which converges uniformly on [\frac{1}{3(k + 1)}, 1 − \frac{1}{3(k + 1)}]. Then the diagonal

sequence {y_{k}^{(k)}(t)} converges pointwise in (0, 1) and it is easy to verify that {y_{k}^{(k)}(t)} converges uniformly on any interval [c, d] ⊆ (0, 1). Without loss of generality, let {y_{k}^{(k)}(t)} be itself of {y_{n}(t)} in the rest. Put y(t) = lim_{n→∞} y_{n}(t), t ∈ (0, 1). Then y(t) is continuous on (0, 1) and since y_{n}(t) ≤ W (t) < 0, we have y(t) ≤ 0, t ∈ (0, 1).

3) Now (3.2) shows

sup{max{−y_{n}(t), t ∈ [0, 1]}} < +∞.

We have

lim_{t→0^{+}} sup{∫_{0}^{t} −y_{n}(s)ds} = 0, lim_{t→1^{−}} sup{∫_{t}^{1} −y_{n}(s)ds} = 0, t ∈ [0, 1],  (3.4)

and

\begin{aligned}(Ay_{n})(t) & = \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{n}(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y_{n}(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y_{n}(τ)dτ \\ & < \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{n}(τ)dτ \\ & < +∞, t ∈ [0, 1]. \end{aligned} (3.5)

Since (3.4) and (3.5) hold, the Fatou theorem of the Lebesgue integral implies (Ay)(t) < +∞, for any fixed t ∈ (0, 1).

4) y(t) satisfies the following equation

y(t) = − ∫_{0}^{t} a(s)f(s, (Ay)(s), y(s))ds, t ∈ (0, 1).  (3.6)

Since y_{n}(t) converges uniformly on [a, b] ⊂ (0, 1), (3.4) implies that (Ay_{n})(s) converges to (Ay)(s) for any s ∈ (0, 1). For fixed t ∈ (0, 1) and any d , 0 < d < t, we have

y_{n}(t) − y_{n}(d) = − ∫_{d}^{t} a(s)f(s, (Ay_{n})(s) + \frac{1}{n}, y_{n}(s))ds.  (3.7)

for all n > k. Since y_{n}(s) ≤ max\{τ, W (d)\} , (Ay_{n})(s) + \frac{1}{n} ≥ Θ , s ∈ [d, t] , \{(Ay_{n})(s)\} and {y_{n}(s)} are bounded and equicontinuous on [d, t]

y(t) − y(d) = − ∫_{d}^{t} a(s)f(s, (Ay)(s), y(s))ds.  (3.8)

Putting t = d in (3.2), we have

∫_{y_{n}(d)}^{− \frac{1}{n}} \frac{dy_{n}}{G(y_{n})} ≤ ∫_{0}^{d} a(s)k(s)ds sup F [Θ, +∞).  (3.9)

Letting n → ∞ and d → 0^{+}, we obtain

y(0^{+}) = lim_{d→0^{+}} y(d) = 0.

Letting d → 0^{+} in (3.8), we have

y(t) = − ∫_{0}^{t} a(s)f(s, (Ay)(s), y(s))ds, t ∈ (0, 1),  (3.10)

and

(Ay)(1) = ∑_{i=1}^{m−2} α_{i}(Ay)(ξ_{i}).

Hence x(t) = (Ay)(t) is a positive solution of (1.1). □

Theorem 3.2 Suppose that (H_{1})-(P_{3}) hold. Then the set of positive solutions of (1.1) is compact in C^{1}[0, 1].

Proof Let M = {y ∈ C[0, 1]: (Ay)(t) is a positive solution of equation (1.1) }. We show that

(1) M is not empty;

(2) M is relatively compact(bounded, equicontinuous);

(3) M is closed.

Obviously, Theorem 3.1 implies M is not empty.

First, we show that M ⊂ C[0, 1] is relatively compact. For any y ∈ M, differentiating (3.10) and using (H_{2}), we obtain

−y′(t) = a(t)f(t, (Ay)(t), y(t))
≤ a(t)|f(t, (Ay)(t), y(t))|
≤ a(t)k(t)F [Θ, +∞)G(y(t)), t ∈ (0, 1),
\begin{aligned}\frac{−y′(t)}{G(y(t))} & ≤ a(t)k(t) sup F [(Ay)(t), +∞) \\ & ≤ a(t)k(t) sup F [Θ, +∞), t ∈ [0, 1]. \end{aligned}  (3.11)

Integrating for (3.11) from 0 to t, we have

∫_{y(t)}^{0} \frac{dy}{G(y)} ≤ ∫_{0}^{1} a(s)k(s)ds sup F [Θ, +∞), t ∈ [0, 1].  (3.12)

Now (H_{3}) and (3.12) show that for any y ∈ M, there exists K > 0 such that |y(t)| < K,∀ t ∈ [0, 1]. Then M is bounded.

For any y ∈ M, we obtain from (3.11)

−y′(t) = a(t)f(t, (Ay)(t), y(t))
≤ a(t)|f(t, (Ay)(t), y(t))|
≤ a(t)k(t)F [Θ, +∞)G(y(t)), t ∈ (0, 1),

and

y′(t) = −a(t)f(t, (Ay)(t), y(t))
≤ a(t)|f(t, (Ay)(t), y(t))|
≤ a(t)k(t)F [Θ, +∞)G(y(t)), t ∈ (0, 1),

which yields

\frac{−y′(t)}{G(y(t)) + 1} ≤ a(t)k(t) sup F [Θ, +∞), t ∈ (0, 1),  (3.13)

and

\frac{y′(t)}{G(y(t)) + 1} ≤ a(t)k(t) sup F [Θ, +∞), t ∈ (0, 1).  (3.14)

Notice that the rights are always positive in (3.13) and (3.14). Let I(y(t)) = ∫_{0}^{y(t)} \frac{dy}{G(y) + 1}. For any t_{1}, t_{2} ∈ [0, 1], integrating for (3.13) and (3.14) from t_{1} to t_{2} ,we obtain

|I(y(t_{1})) − I(y(t_{2}))| ≤ ∫_{t_{1}}^{t_{2}} a(t)k(t)F [Θ, +∞)dt.  (3.15)

Since I^{−1} is uniformly continuous on [I(−K), 0], for any \overline{ϵ} > 0, there is a ϵ′ > 0 such that

|I^{−1}(s_{1}) − I^{−1}(s_{2})| < \overline{ϵ}, ∀|s_{1} − s_{2}| < ϵ′, s_{1}, s_{2} ∈ [I(−K), 0].  (3.16)

And (3.15) guarantees that for ϵ′ > 0, there is a δ′ > 0 such that

|I(y(t_{1})) − I(y(t_{2}))| < ϵ′, ∀|t_{1} − t_{2}| < δ′, t_{1}, t_{2} ∈ [0, 1].  (3.17)

Now (3.16) and (3.17) yield that

|y(t_{1}) − y(t_{2})| = |I^{−1}(I(y(t_{1})) − I^{−1}(I(y(t_{2}))| < \overline{ϵ}, t_{1}, t_{2} ∈ [0, 1],  (3.18)

which means that M is equicontinuous. So M is relatively compact.

Second, we show that M is closed. Suppose that {y_{n}} ⊆ M and lim_{n→+∞} max_{t∈[0,1]} |y_{n}(t) − y_{0}(t)| = 0. Obviously y_{0} ∈ C[0, 1] and lim_{n→+∞}(Ay_{n})(t) = (Ay_{0})(t), t ∈ [0, 1]. Moreover,

(Ay_{n})(t) = \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{n}(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y_{n}(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y_{n}(τ)dτ

< \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{n}(τ)dτ

< \frac{K}{1 − ∑_{i=1}^{m−2} α_{i}} , t ∈ [0, 1].

(3.19)

For y_{n} ∈ M, from (3.10) we obtain

y_{n}(t) = − ∫_{0}^{t} a(s)f(s, (Ay_{n})(s), y_{n}(s))ds, t ∈ (0, 1).  (3.20)

For fixed t ∈ (0, 1) , there exists 0 < d < t such that

y_{n}(t) − y_{n}(d) = − ∫_{d}^{t} a(s)f(s, (Ay_{n})(s), y_{n}(s))ds.  (3.21)

Since y_{n}(s) ≤ max{τ, W (d)}, (Ay_{n})(s) ≥ Θ, s ∈ [d, t], the Lebesgue Dominated Convergence Theorem yields that

y_{0}(t) − y_{0}(d) = − ∫_{d}^{t} a(s)f(s, (Ay_{0})(s), y_{0}(s))ds, t ∈ (0, 1).  (3.22)

From (3.10), we have

\begin{aligned}−y′_{n}(t) & = a(t)f(t, (Ay_{n})(s), y_{n}(s)) \\ & ≤ a(t)k(t)F [Θ, +∞)G(y_{n}(t)), t ∈ (0, 1), \end{aligned}

which yields

\frac{−y′_{n}(t)}{G(y_{n}(t))} ≤ a(t)k(t) sup F [Θ, +∞), t ∈ (0, 1).

Integrating from 0 to d

∫_{y_{n}(d)}^{0} \frac{dy_{n}}{G(y_{n})} ≤ ∫_{0}^{d} a(s)k(s)ds sup F [Θ, +∞).  (3.23)

Letting n → ∞ and d → 0^{+}, we obtain

y_{0}(0^{+}) = lim_{d→0^{+}} y_{0}(d) = 0.

Letting d → 0^{+} in (3.22), we have

y_{0}(t) = − ∫_{0}^{t} a(s)f(s, (Ay_{0})(s), y_{0}(s))ds, t ∈ (0, 1),  (3.24)

and

(Ay_{0})(1) = ∑_{i=1}^{m−2} α_{i}(Ay_{0})(ξ_{i}).

Then x_{0}(t) = (Ay_{0})(t) is a positive solution of (1.1). So y_{0} ∈ M and M is a closed set. Hence {Ay, y ⊆ M} ∈ C^{1}[0, 1] is compact. Theorem 3.3 Suppose (H_{1})-(P_{3}) hold. Then (1.1) has a minimal positive solution

and a maximal positive solution in C^{1}[0, 1].

Proof. Let Ω = {x(t) : x(t) is a C^{1}[0, 1] positive solution of (1.1)}. Theorem 3.1 implies that is nonempty. Define a partially ordered ≤ in Ω : x ≤ y iff x(t) ≤ y(t) for any t ∈ [0, 1]. We prove only that any chain in < Ω, ≤> has a lower bound in Ω. The rest is obtained from Zorn’s lemma. Let {x_{α}(t)} be a chain in < Ω, ≤>. Since C[0, 1] is a separable Banach space, there exists countable set at most {x_{n}(t)}, which is dense in {x_{α}(t)}. Without loss of generality, we may assume that {x_{n}(t)} ⊆ {x_{α}(t)}. Put z_{n}(t) = min{x_{1}(t), x_{2}(t), ⋯, x_{n}(t)}. Since {x_{α}(t)} is a chain, z_{n}(t) ∈ Ω for any n (in fact, z_{n}(t) equals one of x_{n}(t)) and z_{n+1}(t) ≤ z_{n}(t) for any n. Put z(t) = lim_{m→+∞} z_{n}(t). We prove that z(t) ∈ Ω.

By Lemma 2.2, there exists y_{n}(t) (e.g., y_{n}(t) may be z′_{n}(t)), which is a solution of

(T y)(t) = − ∫_{0}^{t} a(s)f(s, (Ay)(s), y(s))ds t ∈ [0, 1],

such that

z_{n}(t) = \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{n}(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y_{n}(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y_{n}(τ)dτ.

(3.2) imply that {∥y_{n}∥} is bounded. From Lemma 2.3, there exists a subsequence {y_{n_{k}}(t)} of {y_{n}(t)}, i ≠ j, n_{i} ≠ n_{j}, which converges everywhere on [0, 1]. Without loss of generality, let {y_{n_{k}}(t)} be itself of {y_{n}(t)}. Put y_{0}(t) = lim_{m→+∞} y_{n}(t), t ∈ [0, 1]. Use y_{n}(t), y_{0}(t), and 0 in place of y_{n}(t), y(t), and 1/n in Theorem 3.1, respectively. A similar argument to show Theorem 3.1 yields that y_{0}(t) is a solution of

y(t) = − ∫_{0}^{t} a(s)f(s, (Ay)(s), y_{n}(s))ds, t ∈ [0, 1].

The boundedness of {∥y_{n}∥} leads to

z(t) = lim_{m→+∞} z_{n}(t)

= lim_{m→+∞}[\frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{n}(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y_{n}(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y_{n}(τ)dτ]

= \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{0}(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y_{0}(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y_{0}(τ)dτ.

Hence z ∈ Ω. By Lemma 2.2, for any x ∈ {x_{α}}, there exists {x_{n_{k}}} ⊆ {x_{n}} such that ∥x_{n_{k}} − x∥ → 0. Notice that x_{n_{k}}(t) ≥ z_{n_{k}}(t) ≥ z(t), t ∈ [0, 1]. Letting k → +∞, we have x(t) ≥ z(t), t ∈ [0, 1]; i.e.,{x_{α}} has lower boundedness in Ω. Zorn’s lemma shows that (1.1) has a minimal C^{1}[0, 1] positive solution. By a similar proof, we can get the a maximal C^{1}[0, 1] positive solution. The proof is complete.

Theorem 3.4 Suppose that (H_{1})-(P_{3}) hold , f(t, x, z) is decreasing in x for all (t, z) ∈ [0, 1] × R_{−}, a(0)f(0, x, z) ≠ 0 and lim_{t→0} f(t, x, y) ≠ +∞. Then (1.1) has an unique positive solution in C^{1}[0, 1].

Proof. Assume that x_{1} and x_{2} are two positive different solutions to (1.1), i.e., there exists t_{0} ∈ (0, 1] such that x_{1}(t_{0}) ≠ x_{2}(t_{0}). Without loss of generality, assume that x_{1}(t_{0}) > x_{2}(t_{0}). Let φ(t) = x_{1}(t) − x_{2}(t) for all t ∈ [0, 1]. Obviously, φ ∈ C[0, 1] ∩ C^{1}(0, 1] with φ(t_{0}) > 0.

Let t_{∗} = inf{0 < t < t_{0}|φ(s) > 0 for all s ∈ t ∈ [t, t_{0}]} and t^{∗} = sup{t_{0} < t < 1|φ(s) > 0 for all s ∈ t ∈ [t_{0}, t]}. It is easy to see that φ(t) > 0 for all t ∈ (t_{∗}, t^{∗}) and φ has maximum in [t_{∗}, t^{∗}]. Let t′ satisfying that φ(t′) = max_{t∈[t_{∗},t^{∗}]} φ(t). There are three cases:

(1) t′ ∈ (t_{∗}, t^{∗}); (2) t′ = t^{∗} = 1;(3) t′ = 0.

(1) t′ ∈ (t_{∗}, t^{∗}). It is easy to see that φ′′(t′) ≤ 0 and φ′(t′) = 0. Then φ′′(t′) = x′′_{1}(t′) − x′′_{2}(t′)

= −a(t′)f(t′, x_{1}(t′), x′_{1}(t′)) + a(t′)f(t′, x_{2}(t′), x′_{2}(t′)) > 0,

a contradiction.

(2) t′ = t^{∗} = 1. Since t′ = t^{∗} = 1, we have ∑_{i=1}^{m−2} α_{i} max{φ(ξ_{i})} > ∑_{i=1}^{m−2} α_{i}φ(ξ_{i}) = φ(1), a

contradiction to 0 < ∑_{i=1}^{m−2} α_{i} < 1.

(3) t′ = 0. Since t′ = 0 and x_{1} and x_{2} are solutions, the proof of lemma 2.2 implies that there exist x_{n,1} and x_{n,2} such that

∥x_{n,1} − x_{1}∥ < \frac{φ(0)}{2}, ∥x_{n,2} − x_{2}∥ < \frac{φ(0)}{2}

where

x_{n,1}(t) = \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{n,1}(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y_{n,1}(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y_{n,1}(τ)dτ,  t ∈ [0, 1],

x_{n,2}(t) = \frac{1}{1 − ∑_{i=1}^{m−2} α_{i}} ∫_{0}^{1} −y_{n,2}(τ)dτ − \frac{∑_{i=1}^{m−2} α_{i} ∫_{0}^{ξ_{i}} −y_{n,2}(τ)dτ}{1 − ∑_{i=1}^{m−2} α_{i}} − ∫_{0}^{t} −y_{n,2}(τ)dτ,  t ∈ [0, 1],

and

y_{n,1}(t) = −\frac{1}{n} − ∫_{0}^{t} a(s)f(s, x_{n,1}(s) + \frac{1}{n}, y_{n,1}(s))ds, t ∈ [0, 1],

y_{n,2}(t) = −\frac{1}{n} − ∫_{0}^{t} a(s)f(s, x_{n,2}(s) + \frac{1}{n}, y_{n,2}(s))ds, t ∈ [0, 1],

y_{n,1}(t) ≤ −\frac{1}{n}, y_{n,2}(t) ≤ −\frac{1}{n} for all t ∈ [0, 1].

By a similar proof with above, there exists t_{1} ∈ (0, 1] such that x_{n,1}(t_{1}) ≠ x_{n,2}(t_{1}). Without loss of generality, assume that x_{n,1}(t_{1}) > x_{n,2}(t_{1}). Let φ_{n}(t) = x_{n,1}(t) − x_{n,2}(t) for all t ∈ [0, 1]. Obviously, φ_{n} ∈ C[0, 1] ∩ C^{1}(0, 1] with φ_{n}(t_{1}) > 0. Let t_{∗} = inf{0 < t < t_{1}|φ_{n}(s) > 0 for all s ∈ t ∈ [t, t_{1}]} and t^{∗} = sup{t_{1} < t < 1|φ_{n}(s) > 0 for all s ∈ t ∈ [t_{1}, t]}. It is easy to see that φ_{n}(t) > 0 for all t ∈ (t_{1∗}, t^{1∗}) and φ_{n} has maximum in [t_{1∗}, t^{1∗}]. Let t′′ satisfying that φ(t′′) = max_{t∈[t_{1∗},t^{1∗}]} φ(t). There are three cases: 1) t′′ ∈ (t_{1∗}, t^{1∗}); 2) t′′ = t^{∗} = 1; 3) t′′ = 0.

The proof of 1) and 2) are similar with (1) and (2).

3) t′′ = 0. We have φ_{n}(t) < φ_{n}(0), t ∈ (0, 1], φ′_{n}(0) = 0, φ′_{n}(t_{ξ}) < 0, t_{ξ} ∈ (0, 1). Then

\underline{lim}_{t_{ξ}→0+}φ′′_{n}(t) = \underline{lim}_{t_{ξ}→0+} \frac{φ′_{n}(t_{ξ}) − φ′_{n}(0)}{t_{ξ} − 0} ≤ 0.

On the other hand, since φ′′_{n}(0) = x′′_{n,1}(0) − x′′_{n,2}(0)

= −a(0)f(0, x_{n,1}(0) + \frac{1}{n}, x′_{n,1}(0)) + a(0)f(0, x_{n,2}(0) + \frac{1}{n}, x′_{n,2}(0)) > 0,

a contradiction. Then (1.1) has at most one solution. The proof is complete.

Example 3.1. In (1.1), let f(t, x, y) = k(t)[1+x^{−γ}+(−y)^{−σ}−(−y) ln(−y)], a(t) = t^{− \frac{1}{3}} , and

k(t) = t^{− \frac{1}{2}} , 0 < t < 1,

where γ > 0, σ < −2, and let F (x) = 1 + x^{−γ}, G(y) = 1 + (−y)^{−σ} − (−y) ln(−y). Then

f(t, x, y) ≤ k(t)F (x)G(y), δ = 1, β(t) = k(t),

and

∫_{−∞}^{−1} \frac{dy}{G(y)} = +∞.

By Theorem 3.1, (1.1) at least has a positive solution and Corollary 3.1 implies the set of solutions is compact.

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(Received September 19, 2008)