Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 38, 1-15; http://www.math.u-szeged.hu/ejqtde/

Multiple positive solutions for second order

impulsive boundary value problems in Banach spaces^{∗}

Zhi-Wei Lv ^{a}, Jin Liang ^{b}^{†}and Ti-Jun Xiao ^{c}

a Department of Mathematics, University of Science and Technology of China Hefei, Anhui 230026, People’s Republic of China sdlllzw@mail.ustc.edu.cn b Department of Mathematics, Shanghai Jiao Tong University Shanghai 200240, People’s Republic of China jinliang@sjtu.edu.cn c Shanghai Key Laboratory for Contemporary Applied Mathematics School of Mathematical Sciences, Fudan University Shanghai 200433, People’s Republic of China tjxiao@fudan.edu.cn

Abstract By means of the fixed point index theory of strict set contraction operators, we establish new existence theorems on multiple positive solutions to a boundary value problem for second-order impulsive integro-differential equations with integral boundary conditions in a Banach space. Moreover, an application is given to illustrate the main result. Keywords Fixed point index; Impulsive differential equation; Positive solution; Measure of noncompactness.

1 Introduction.

Impulsive differential equations can be used to describe a lot of natural phenomena such as the dynamics of populations subject to abrupt changes (harvesting, diseases, etc.), which cannot be described using classical differential equations. That is why in recent years they have attracted much attention of investigators (cf., e.g., [2, 3, 7, 8, 9]). Meanwhile, the boundary value problem with integral boundary conditions has been the subject of investigations along the line with impulsive differential equations because of their wide applicability in various fields (cf., e.g., [1, 2, 6, 10]).

∗This work was supported partially by the NSF of China (10771202), the Research Fund for Shanghai Key Laboratory for Contemporary Applied Mathematics (08DZ2271900) and the Specialized Research Fund for the Doctoral Program of Higher Education of China (2007035805).

†Corresponding author. E-mail: jinliang@sjtu.edu.cn

In [3], D. Guo discussed the following second-order impulsive differential equations

\{ \begin{matrix}−x′′ = f(t, x), t ≠ t_{k} , k = 1, 2, ⋯, m, \\ Δx|_{t=t_{k}} = I_{k}(x(t_{k})), k = 1, 2, ⋯, m, \\ ax(0) − bx′(0) = θ, cx(1) + dx′(1) = θ, \end{matrix}

where f ∈ C[J × P, P], J = [0, 1], P is a cone in real Banach space E, θ denotes the zero element of E. I_{k} ∈ C[P, P], 0 < t_{1} < ⋯ < t_{k} < ⋯ < t_{m} < 1. a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0 and ac + ad + bc > 0.

In [1], A. Boucherif investigated the existence of positive solutions to the following boundary value problem

\{ \begin{matrix}y′′(t) = f(t, y(t)), 0 < t < 1, \\ y(0) − ay′(0) = ∫_{0}^{1} g_{0}(s)y(s)ds, \\ y(1) − by′(1) = ∫_{0}^{1} g_{1}(s)y(s)ds, \end{matrix}

where f : [0, 1] × R → R is continuous, g_{0}, g_{1} : [0, 1] → [0, +∞) are continuous and positive, a and b are nonnegative real parameters.

In [2], M. Feng, B. Du and W. Ge studied the existence of multiple positive solutions for a class of second-order impulsive differential equations with p-Laplacian and integral boundary conditions

\{ \begin{matrix}−(ϕ_{p}(u′(t)))′ = f(t, u(t)), t ≠ t_{k}, t ∈ (0, 1), \\ −Δu|_{t=t_{k}} = I_{k}(u(t_{k})), k = 1, 2, ⋯, n, \end{matrix}

subject to the following boundary condition: u′(0) = 0, u(1) = ∫_{0}^{1} g(t)u(t)dt, where ϕ_{p}(s) is a p-Laplacian operator, 0 < t_{1} < ⋯ < t_{k} < ⋯ < t_{n} < 1, f ∈ C([0, 1] × [0, +∞), [0, +∞)), I_{k} ∈ C([0, +∞), [0, +∞)).

In this paper, we are concerned with the existence of multiple positive solutions of the following second-order impulsive differential equations with integral boundary conditions in real Banach space E

\{ \begin{matrix}x′′ = f(t, x, x′, T x, Sx), t ∈ J , t ≠ t_{k}, \\ Δx|_{t=t_{k}} = −I_{k}(x(t_{k}), x′(t_{k})), k = 1, 2, ⋯, m, \\ Δx′|_{t=t_{k}} = \overline{I}_{k}(x(t_{k}), x′(t_{k})), k = 1, 2, ⋯, m, \\ x(0) − ax′(0) = θ, \\ x(1) − bx′(1) = ∫_{0}^{1} g(s)x(s)ds, \end{matrix}  (1.1)

where a + 1 > b > 1, J = [0, 1], J′ = J\{t_{1}, ⋯ t_{m}}, 0 < t_{1} < ⋯ < t_{k} < ⋯ < t_{m} < 1, θ denotes the zero element of Banach space E, T and S are the linear operators defined as follows

(T x)(t) = ∫_{0}^{t} k(t, s)x(s)ds, (Sx)(t) = ∫_{0}^{1} h(t, s)x(s)ds,

in which k ∈ C[𝔻, R_{+}], h ∈ C[𝔻_{0}, R_{+}], 𝔻 = {(t, s) ∈ J × J : t ≥ s}, 𝔻_{0} = {(t, s) ∈ J × J : 0 ≤ t, s ≤ 1}, R_{+} = [0, +∞), Δx|_{t=t_{k}} denotes the jump of x(t) at t = t_{k}, i.e., Δx|_{t=t_{k}} = x(t_{k}^{+})−x(t_{k}^{−}), where x(t_{k}^{+}), x(t_{k}^{−}) represent the right and left limits of x(t) at t = t_{k}, respectively. By means of the fixed point index theory of strict set contraction operators, we establish new existence theorems on multiple positive solutions to (1.1). Moreover, an application is given to illustrate the main result.

Let us first recall some basic information on cone (see more from [4, 5]). Let E be a real Banach space and P be a cone in E which defined a partial ordering in E by x ≤ y if and only if y − x ∈ P. P is said to be normal if there exists a positive constant N such that θ ≤ x ≤ y implies ∥x∥ ≤ N∥y∥. P is called solid if its interior P^{∘} is nonempty. If x ≤ y and x ≠ y, we write x < y. If P is solid and y − x ∈P^{∘}, we write x ≪ y.

Let P C[J, E]={x : x is a map from J into E such that x(t) is continuous at t ≠ t_{k}, left continuous at t = t_{k} and x(t_{k}^{+}) exists for k = 1, 2, 3, ⋯, m} and

P C^{1}[J, E] := {x ∈ P C[J, E] : x′(t) is continuous at t ≠ t_{k},
and x′(t_{k}^{+}), x′(t_{k}^{−}) exist for k = 1, 2, 3, ⋯, m}.

Clearly, P C[J, E] is a Banach space with the norm ∥x∥_{P C} = sup_{t∈J} ∥x(t)∥ and P C^{1}[J, E] is a Banach space with the norm ∥x∥_{P C}_{1} = max{∥x∥_{P C}, ∥x′∥_{P C}}.

By a positive solution of BVP (1.1), we mean a map x ∈ P C^{1}[J, E] ∩ C^{2}[J′, E] such that x(t) ≥ θ , x′(t) ≥ θ , x(t) ≢ θ for t ∈ J and x(t) satisfies (1.1).

Let α, α_{P C^{1}} be the Kuratowski measure of noncompactness in E and P C^{1}[J, E], respectively (see [4, 5], for further understanding). Moreover, we set J_{1} = [0, t_{1}], J_{k} = (t_{k−1}, t_{k}] (k = 2, 3, ⋯, m), and for u_{i} ∈ P, i = 1, 2, 3, 4,

f^{∞} = lim sup_{∑_{i=1}^{4} ∥u_{i}∥→∞} max_{t∈J} _{\frac{∥f(t,u_{1},u_{2},u_{3},u_{4})∥}{∑_{i=1}^{4} ∥u_{i}∥}}, f^{0} = lim sup_{∑_{i=1}^{4} ∥u_{i}∥→0} max_{t∈J} _{\frac{∥f(t,u_{1},u_{2},u_{3},u_{4})∥}{∑_{i=1}^{4} ∥u_{i}∥}},

I^{∞}(k) = lim sup_{∥u_{1}∥+∥u_{2}∥→∞} \frac{∥I_{k}(u_{1},u_{2})∥}{∥u_{1}∥+∥u_{2}∥}, I^{0}(k) = lim sup_{∥u_{1}∥+∥u_{2}∥→0} \frac{∥I_{k}(u_{1},u_{2})∥}{∥u_{1}∥+∥u_{2}∥}.

Similarly, we denote \overline{I}^{∞}(k), \overline{I}^{0}(k).

The following lemmas are basic, which can be found in [5].

Lemma 1.1 If W ⊂ P C^{1}[J, E] is bounded and the elements of W′ are equicontinuous on each J_{k} (k = 1, 2, ⋯, m). Then α_{P C^{1}}(W) = max { sup_{t∈J} α(W (t)), sup_{t∈J} α(W′(t))}.

Lemma 1.2 Let K be a cone in real Banach space E and Ω be a nonempty bounded open convex subset of K. Suppose that A : \overline{Ω} → K is a strict set contraction and A(\overline{Ω}) ⊂ Ω, when \overline{Ω} denotes the closure of Ω in K. Then the fixed-point index i(A, Ω, K) = 1.

2 Main results

(H_{1}) f ∈ C[J × P × P × P × P, P], and for any r > 0, f is uniformly continuous on J × P_{r}^{4}, I_{k}, \overline{I}_{k} ∈ C[P × P, P] (k = 1, 2, ⋯, m) are bounded on P_{r} × P_{r}, where P_{r} = \{x ∈ P : ∥x∥ ≤ r\}. (H_{2}) g ∈ L^{1}[0, 1] is nonnegative, and u ∈ [0, a + 1 − b), where u = ∫_{0}^{1}(a + s)g(s)ds. (H_{3}) There exist nonnegative constants c_{i}, d_{k}, \overline{d}_{k}, i = 1, 2, 3, 4, k = 1, 2 such that

α(f(t, B_{1}, B_{2}, B_{3}, B_{4})) ≤ ∑_{i=1}^{4} c_{i}α(B_{i}), ∀ t ∈ J , B_{i} ⊂ P_{r} (i = 1, 2, 3, 4),  (2.1)

α(I_{k}(B_{1}, B_{2})) ≤ d_{1}α(B_{1}) + d_{2}α(B_{2}), B_{1}, B_{2} ⊂ P_{r} ,  (2.2)
α(\overline{I}_{k}(B_{1}, B_{2})) ≤ \overline{d}_{1}α(B_{1}) + \overline{d_{2}}α(B_{2}), B_{1}, B_{2} ⊂ P_{r} ,  (2.3)

and

l = max{l_{1}, l_{2}} < 1,

where

l_{1} = 2m_{2}(c_{1} + c_{2} + k^{∗}c_{3} + h^{∗}c_{4}) + m_{2}m(\overline{d}_{1} + \overline{d}_{2}) + \overline{m}_{2}m(d_{1} + d_{2}),
l_{2} = 2m_{4}(c_{1} + c_{2} + k^{∗}c_{3} + h^{∗}c_{4}) + m_{4}m(\overline{d}_{1} + \overline{d}_{2}) + \overline{m}_{4}m(d_{1} + d_{2}),

in which

k^{∗} = max{k(t, s), t, s ∈ 𝔻}, h^{∗} = max{h(t, s), t, s ∈ 𝔻_{0}}.

(H_{4}) f^{∞} = f^{0} = 0, I^{∞}(k) = I^{0}(k) = 0, \overline{I}^{∞}(k) = \overline{I}^{0}(k) = 0.

Lemma 2.1 Let (H_{1}) and (H_{2}) hold. Then x ∈ P C^{1}[J, E] ∩ C^{2}[J′, E] is a solution to (1.1) if and only if x ∈ P C^{1}[J, E]∩C^{2}[J′, E] is a solution to the following impulsive integral equation:

\begin{aligned}x(t) & = & ∫_{0}^{1} H_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds + ∑_{k=1}^{m} H_{1}(t, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})) \\ & & + ∑_{k=1}^{m} H_{2}(t, t_{k})I_{k}(x(t_{k}), x′(t_{k})), \end{aligned}

(2.4)

where

H_{1}(t, s) = G_{1}(t, s) + \frac{a + t}{a + 1 − b − u} ∫_{0}^{1} G_{1}(τ, s)g(τ)dτ,

H_{2}(t, s) = G_{2}(t, s) + \frac{a + t}{a + 1 − b − u} ∫_{0}^{1} G_{2}(τ, s)g(τ)dτ,

G_{1}(t, s) = \{ \begin{matrix}\frac{1}{a+1−b}(a + t)(b + s − 1), t ≤ s, \\ \frac{1}{a+1−b}(a + s)(b + t − 1), s ≤ t, \end{matrix}

G_{2}(t, s) = \{ \begin{matrix}\frac{a+t}{a+1−b}, t ≤ s, \\ \frac{b+t−1}{a+1−b}, s ≤ t. \end{matrix}

Proof. “⟹”.

Suppose that x ∈ P C^{1}[J, E] ∩ C^{2}[J′, E] is a solution to problem (1.1). From (1.1), we get

x′(t) = x′(0) + ∫_{0}^{t} f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds + ∑_{0<t_{k}<t} \overline{I}_{k}(x(t_{k}), x′(t_{k})),

and

\begin{aligned}x(t) & = & x(0) + tx′(0) + ∫_{0}^{t}(t − s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds \\ & & + ∑_{0<t_{k}<t}(t − t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})) − ∑_{0<t_{k}<t} I_{k}(x(t_{k}), x′(t_{k})). \end{aligned}  (2.5)

In particular,

x′(1) = x′(0) + ∫_{0}^{1} f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds + ∑_{0<t_{k}<1} \overline{I}_{k}(x(t_{k}), x′(t_{k})),

and

\begin{aligned}x(1) & = & x(0) + x′(0) + ∫_{0}^{1}(1 − s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds \\ & & + ∑_{0<t_{k}<1}(1 − t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})) − ∑_{0<t_{k}<1} I_{k}(x(t_{k}), x′(t_{k})). \end{aligned}

From this and the boundary conditions in (1.1), and by induction, we obtain

x(0) = ax′(0),

and

x′(0) = \frac{1}{a + 1 − b}( ∫_{0}^{1}(b + s − 1)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∑_{0<t_{k}<1}(b + t_{k} − 1)\overline{I}_{k}(x(t_{k}), x′(t_{k})) + ∑_{0<t_{k}<1} I_{k}(x(t_{k}), x′(t_{k}))

+ ∫_{0}^{1} g(s)x(s)ds).

This, together with (2.5), implies

x(t) = \frac{a + t}{a + 1 − b}( ∫_{0}^{1}(b + s − 1)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∑_{0<t_{k}<1}(b + t_{k} − 1)\overline{I}_{k}(x(t_{k}), x′(t_{k})) + ∑_{0<t_{k}<1} I_{k}(x(t_{k}), x′(t_{k}))

+ ∫_{0}^{1} g(s)x(s)ds) + ∫_{0}^{t}(t − s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∑_{0<t_{k}<t}(t − t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})) − ∑_{0<t_{k}<t} I_{k}(x(t_{k}), x′(t_{k}))

= \frac{1}{a + 1 − b} ∫_{0}^{t}(a + s)(b + t − 1)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+\frac{1}{a + 1 − b} ∫_{t}^{1}(a + t)(b + s − 1)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+\frac{1}{a + 1 − b} ∑_{0<t_{k}<t}(a + t_{k})(b + t − 1)\overline{I}_{k}(x(t_{k}), x′(t_{k}))

+\frac{1}{a + 1 − b} ∑_{t≤t_{k}<1}(a + t)(b + t_{k} − 1)\overline{I}_{k}(x(t_{k}), x′(t_{k}))

+\frac{1}{a + 1 − b} ∑_{0<t_{k}<t}(b + t − 1)I_{k}(x(t_{k}), x′(t_{k}))

+\frac{1}{a + 1 − b} ∑_{t≤t_{k}<1}(a + t)I_{k}(x(t_{k}), x′(t_{k})) + \frac{a + t}{a + 1 − b} ∫_{0}^{1} g(s)x(s)ds.

Thus,

\begin{aligned}x(t) & = & ∫_{0}^{1} G_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds + ∑_{k=1}^{m} G_{1}(t, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})) \\ & & + ∑_{k=1}^{m} G_{2}(t, t_{k})I_{k}(x(t_{k}), x′(t_{k})) + \frac{a + t}{a + 1 − b} ∫_{0}^{1} g(s)x(s)ds. \end{aligned}

On the other hand,

∫_{0}^{1} g(t)x(t)dt = ∫_{0}^{1} g(t)( ∫_{0}^{1} G_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∑_{k=1}^{m} G_{1}(t, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k}))

+ ∑_{k=1}^{m} G_{2}(t, t_{k})I_{k}(x(t_{k}), x′(t_{k})) + \frac{a + t}{a + 1 − b} ∫_{0}^{1} g(s)x(s)ds)dt,

= ∫_{0}^{1} ∫_{0}^{1} g(t)G_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))dsdt

+ ∫_{0}^{1} g(t)( ∑_{k=1}^{m} G_{1}(t, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})))dt

+ ∫_{0}^{1} g(t)( ∑_{k=1}^{m} G_{2}(t, t_{k})I_{k}(x(t_{k}), x′(t_{k})))dt + ∫_{0}^{1} \frac{a + t}{a + 1 − b}g(t)dt ∫_{0}^{1} g(t)x(t)dt,

and also, EJQTDE, 2010 No. 38, p. 6

∫_{0}^{1} g(s)x(s)ds = \frac{1}{1 − ∫_{0}^{1} \frac{a+s}{a+1−b}g(s)ds}( ∫_{0}^{1}(∫_{0}^{1} G_{1}(τ, s)g(τ)dτ)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∫_{0}^{1} g(τ)(∑_{k=1}^{m} G_{1}(τ, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})))dτ

+ ∫_{0}^{1} g(τ)(∑_{k=1}^{m} G_{2}(τ, t_{k})I_{k}(x(t_{k}), x′(t_{k})))dτ).

Therefore, we have

x(t) = ∫_{0}^{1} G_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∑_{k=1}^{m} G_{1}(t, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})) + ∑_{k=1}^{m} G_{2}(t, t_{k})I_{k}(x(t_{k}), x′(t_{k}))

+\frac{a + t}{a + 1 − b − ∫_{0}^{1}(a + s)g(s)ds}( ∫_{0}^{1}(∫_{0}^{1} G_{1}(τ, s)g(τ)dτ)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∫_{0}^{1} g(τ)(∑_{k=1}^{m} G_{1}(τ, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})))dτ + ∫_{0}^{1} g(τ)(∑_{k=1}^{m} G_{2}(τ, t_{k})I_{k}(x(t_{k}), x′(t_{k})))dτ)

= ∫_{0}^{1} H_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds + ∑_{k=1}^{m} H_{1}(t, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k}))

+ ∑_{k=1}^{m} H_{2}(t, t_{k})I_{k}(x(t_{k}), x′(t_{k})).

“⟸”

If x ∈ P C^{1}[J, E] ∩ C^{2}[J′, E] is a solution of Eq. (2.4), then a direct differentiation of (2.4) yields, for t ≠ t_{k}

x′(t) = ∫_{0}^{t} \frac{a + s}{a + 1 − b}f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∫_{t}^{1} \frac{b + s − 1}{a + 1 − b}f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∑_{0<t_{k}<t} \frac{a + t_{k}}{a + 1 − b}\overline{I}_{k}(x(t_{k}), x′(t_{k})) + ∑_{t≤t_{k}<1} \frac{b + t_{k} − 1}{a + 1 − b}\overline{I}_{k}(x(t_{k}), x′(t_{k}))

+\frac{1}{a + 1 − b} ∑_{k=1}^{m} I_{k}(x(t_{k}), x′(t_{k})) + \frac{1}{a + 1 − b − ∫_{0}^{1}(a + s)g(s)ds}

( ∫_{0}^{1}(∫_{0}^{1} G_{1}(τ, s)g(τ)dτ)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds

+ ∫_{0}^{1} g(τ)(∑_{k=1}^{m} G_{1}(τ, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})))dτ

+ ∫_{0}^{1} g(τ)(∑_{k=1}^{m} G_{2}(τ, t_{k})I_{k}(x(t_{k}), x′(t_{k})))dτ).

Thus,

\begin{aligned}x′(t) & = & ∫_{0}^{1} H′_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds+ \\ & & ∑_{k=1}^{m} H′_{1}(t, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})) + ∑_{k=1}^{m} H′_{2}(t, t_{k})I_{k}(x(t_{k}), x′(t_{k})), \end{aligned}  (2.6)

where

H′_{1}(t, s) = G′_{1}(t, s) + \frac{1}{a + 1 − b − u} ∫_{0}^{1} G_{1}(τ, s)g(τ)dτ,

H′_{2}(t, s) = \frac{1}{a + 1 − b} + \frac{1}{a + 1 − b − u} ∫_{0}^{1} G_{2}(τ, s)g(τ)dτ,

G′_{1}(t, s) = \{ \begin{matrix}\frac{b+s−1}{a+1−b}, t ≤ s, \\ \frac{a+s}{a+1−b}, s ≤ t. \end{matrix}

Differentiating (2.6), we see

x′′(t) = f(t, x(t), x′(t), (T x)(t), (Sx)(t)).

Clearly,

Δx|_{t=t_{k}} = −I_{k}(x(t_{k}), x′(t_{k})), Δx′|_{t=t_{k}} = \overline{I}_{k}(x(t_{k}), x′(t_{k})),
x(0) − ax′(0) = θ, x(1) − bx′(1) = ∫_{0}^{1} g(s)x(s)ds.

The proof is then complete.□ The following “Facts” are clearly known. Fact I. For t, s ∈ [0, 1], we have

\frac{a(b − 1)}{a + 1 − b} ≤ G_{1}(t, s) ≤ \frac{(a + 1)b}{a + 1 − b},

\frac{b − 1}{a + 1 − b} ≤ G_{2}(t, s) ≤ \frac{a + 1}{a + 1 − b},

\frac{b − 1}{a + 1 − b} ≤ G′_{1}(t, s) ≤ \frac{a + 1}{a + 1 − b}.

Fact II. For t, s ∈ [0, 1], there exist positive constants m_{i} , \overline{m}_{i} (i = 1, 2, 3, 4) such that

m_{1} = \frac{a(b − 1)}{a + 1 − b} + \frac{a^{2}(b − 1)u_{1}}{u_{2}} ≤ H_{1}(t, s) ≤ \frac{(a + 1)b}{a + 1 − b} + \frac{(a + 1)^{2}bu_{1}}{u_{2}} = m_{2} ,

\overline{m}_{1} = \frac{b − 1}{a + 1 − b} + \frac{a(b − 1)u_{1}}{u_{2}} ≤ H_{2}(t, s) ≤ \frac{a + 1}{a + 1 − b} + \frac{(a + 1)^{2}u_{1}}{u_{2}} = \overline{m}_{2} ,

m_{3} = \frac{b − 1}{a + 1 − b} + \frac{a(b − 1)u_{1}}{u_{2}} ≤ H′_{1}(t, s) ≤ \frac{a + 1}{a + 1 − b} + \frac{(a + 1)bu_{1}}{u_{2}} = m_{4} ,

EJQTDE, 2010 No. 38, p. 8

\overline{m}_{3} = \frac{1}{a + 1 − b} + \frac{(b − 1)u_{1}}{u_{2}} ≤ H′_{2}(t, s) ≤ \frac{1}{a + 1 − b} + \frac{(a + 1)u_{1}}{u_{2}} = \overline{m}_{4} ,

where

u_{1} = ∫_{0}^{1} g(s)ds, u_{2} = (a + 1 − b − u)(a + 1 − b).

We shall reduce BVP (1.1) to an impulsive integral equation in E. To this end, we first consider operator A defined by

\begin{aligned}(Ax)(t) & = & ∫_{0}^{1} H_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds+ \\ & & ∑_{k=1}^{m} H_{1}(t, t_{k})\overline{I}_{k}(x(t_{k}), x′(t_{k})) + ∑_{k=1}^{m} H_{2}(t, t_{k})I_{k}(x(t_{k}), x′(t_{k})). \end{aligned}  (2.7)

In what follows, we write

Q = {x ∈ P C^{1}[J, E] : x(t) ≥ θ, x′(t) ≥ θ, t ∈ J}, B_{r} = {x ∈ P C^{1}[J, E] : ∥x∥_{P C^{1}} ≤ r}.

Obviously, Q is a cone in space P C^{1}[J, E].

Lemma 2.2 Let (H_{1}) − (H_{3}) hold. Then for any r > 0, A : Q ∩ B_{r} → Q is a strict set contraction.

Proof. By (H_{1}) and (H_{2}), we know that A : Q ∩ B_{r} → Q is continuous and bounded. Let C ⊂ Q ∩ B_{r}. From (2.6) and (2.7), it follows that the elements of (AC)′ are equicontinuous on each J_{k} (k = 1, ⋯, m). Lemma 1.1 shows us that

α_{P C^{1}}(AC) = max { sup_{t∈J} α((AC)(t)), sup_{t∈J} α((AC)′(t))}.

By (2.7), we obtain

α((AC)(t)) ≤ α(\overline{co}\{H_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s)) : s ∈ [0, t], t ∈ J, x ∈ C\})

+ ∑_{k=1}^{m} α(H_{1}(t, t_{k})\overline{I}_{k}(C(t_{k}), C′(t_{k}))) + ∑_{k=1}^{m} α(H_{2}(t, t_{k})I_{k}(C(t_{k}), C′(t_{k})))

≤ m_{2}α(f(s, C(s), C′(s), (T C)(s), (SC)(s)), s ∈ J)

+m_{2} ∑_{k=1}^{m} α(\overline{I}_{k}(C(t_{k}), C′(t_{k}))) + \overline{m}_{2} ∑_{k=1}^{m} α(I_{k}(C(t_{k}), C′(t_{k})))

≤ m_{2}(c_{1}α(C(J)) + c_{2}α(C′(J)) + c_{3}α((T C)(J)) + c_{4}α((SC)(J)))

+m_{2} ∑_{k=1}^{m} (\overline{d}_{1}α(C(t_{k})) + \overline{d}_{2}α(C′(t_{k}))) + \overline{m}_{2} ∑_{k=1}^{m} (d_{1}α(C(t_{k})) + d_{2}α(C′(t_{k}))).

By

α(C(J)) ≤ 2α_{P C}_{1}(C), α(C′(J)) ≤ 2α_{P C}_{1}(C),  (2.8)

α(C(t_{k})) ≤ α_{P C}_{1}(C), α(C′(t_{k})) ≤ α_{P C}_{1}(C),  (2.9)

we have

α((AC)(t)) ≤ l_{1}α_{P C}_{1}(C).

In the same way, by virtue of (2.6)-(2.9) and (H_{3}), we get

α((AC)′(t)) ≤ l_{2}α_{P C}_{1}(C).

Thus,

α_{P C}_{1}(AC) ≤ lα_{P C}_{1}(C).

Since l < 1, we assert that A : Q ∩ B_{r} → Q is a strict set contraction.□

Theorem 2.1. Let (H_{1}) − (H_{4}) hold, P be normal and solid. Let there exist v ≫ θ, 0 < t_{∗} < t^{∗} < 1 and σ ∈ C[I, R_{+}] (I = [t_{∗}, t^{∗}]) such that I ⊂ J_{k} for some k, and

f(t, u_{1}, u_{2}, u_{3}, u_{4}) ≥ σ(t)v (∀ t ∈ I),

u_{1} ≥ v, u_{i} ≥ θ (i = 2, 3, 4), \overline{m} ∫_{t_{∗}}^{t^{∗}} σ(s)ds > 1,

where \overline{m} = min\{m_{1}, m_{3}\}. Then (1.1) has at least two positive solutions x_{1}, x_{2} ∈ Q ∩ C^{2}[J′, E] satisfying x_{1}(t) ≫ v and x′_{1}(t) ≫ v for t ∈ I.

Proof. By Lemma 2.2, A : Q ∩ B_{r} → Q is a strict set contraction. Write

ϵ = \frac{1}{6(2 + k^{∗} + h^{∗})m^{(1)}} , ϵ_{1} = \frac{1}{12mm^{(1)}} , ϵ_{2} = \frac{1}{12mm^{(1)}} ,  (2.10)

where

m^{(1)} = max\{m_{2}, \overline{m}_{2}, m_{4}, \overline{m}_{4}\}.

By (H_{1}) and (H_{4}), we know that there exist M_{1} > 0, M_{2} > 0 and M_{3} > 0 such that

∥f(t, u_{1}, u_{2}, u_{3}, u_{4})∥ ≤ ε ∑_{i=1}^{4} ∥u_{i}∥ + M_{1}, ∀ t ∈ J, u_{i} ∈ P,  (2.11)
∥I_{k}(u_{1}, u_{2})∥ ≤ ϵ_{1}(∥u_{1}∥ + ∥u_{2}∥) + M_{2}, ∀ u_{1}, u_{2} ∈ P,  (2.12)
∥\overline{I}_{k}(u_{1}, u_{2})∥ ≤ ϵ_{2}(∥u_{1}∥ + ∥u_{2}∥) + M_{3}, ∀ u_{1}, u_{2} ∈ P.  (2.13)

Now, in view of (2.7), (2.10)-(2.13), we get

∥(Ax)(t)∥ ≤ m_{2} ∫_{0}^{1} ∥f(s, x(s), x′(s), (T x)(s), (Sx)(s))∥ds

+m_{2} ∑_{k=1}^{m} ∥\overline{I}_{k}(x(t_{k}), x′(t_{k}))∥ + \overline{m}_{2} ∑_{k=1}^{m} ∥I_{k}(x(t_{k}), x′(t_{k}))∥

≤ m_{2} ∫_{0}^{1} (ε(∥x(s)∥ + ∥x′(s)∥ + ∥(T x)(s)∥ + ∥(Sx)(s)∥) + M_{1})ds

+m_{2} ∑_{k=1}^{m} (ϵ_{2}(∥x(t_{k})∥ + ∥x′(t_{k})∥) + M_{3}) + \overline{m}_{2} ∑_{k=1}^{m} (ϵ_{1}(∥x(t_{k})∥ + ∥x′(t_{k})∥) + M_{2})

≤ m_{2}(ε(2 + k^{∗} + h^{∗})∥x∥_{P C^{1}} + M_{1}) + m_{2}m(2ϵ_{2}∥x∥_{P C^{1}} + M_{3})

+\overline{m}_{2}m(2ϵ_{1}∥x∥_{P C^{1}} + M_{2})

= ((2 + k^{∗} + h^{∗})m_{2}ε + 2m_{2}mε_{2} + 2\overline{m}_{2}mε_{1})∥x∥_{P C^{1}}

+m_{2}M_{1} + m_{2}mM_{3} + \overline{m}_{2}mM_{2}

≤ \frac{1}{2}∥x∥_{P C^{1}} + \overline{M}_{1} ,
(2.14)

where

\overline{M}_{1} = m_{2}M_{1} + m_{2}mM_{3} + \overline{m}_{2}mM_{2}.

Similarly, from (2.6), (2.7), (2.10)-(2.13), we have

∥(Ax)′(t)∥ ≤ \frac{1}{2}∥x∥_{P C^{1}} + \overline{M}_{2} ,  (2.15)

where

\overline{M}_{2} = m_{4}M_{1} + m_{4}mM_{3} + \overline{m}_{4}mM_{2}.

It follows from (2.14) and (2.15) that

∥Ax∥_{P C^{1}} ≤ \frac{1}{2}∥x∥_{P C^{1}} + \overline{M} ,  (2.16)

where

\overline{M} = max\{\overline{M}_{1}, \overline{M}_{2}\}.

On the other hand, the condition (H_{4}) implies that there exist \overline{l}_{1} > 0, \overline{l}_{2} > 0 and \overline{l}_{3} > 0 such that

∥f(t, u_{1}, u_{2}, u_{3}, u_{4})∥ ≤ ε ∑_{i=1}^{4} ∥u_{i}∥, ∀ t ∈ J, u_{i} ∈ P, ∑_{i=1}^{4} ∥u_{i}∥ ≤ \overline{l}_{1},  (2.17)
∥I_{k}(u_{1}, u_{2})∥ ≤ ϵ_{1}(∥u_{1}∥ + ∥u_{2}∥), ∀ u_{1}, u_{2} ∈ P, ∥u_{1}∥ + ∥u_{2}∥ ≤ \overline{l}_{2},  (2.18)
∥\overline{I}_{k}(u_{1}, u_{2})∥ ≤ ϵ_{2}(∥u_{1}∥ + ∥u_{2}∥), ∀ u_{1}, u_{2} ∈ P, ∥u_{1}∥ + ∥u_{2}∥ ≤ \overline{l}_{3},  (2.19)

where ε, ε_{1}, ε_{2} defined by (2.10).

Let r_{1} = min\{\overline{l}_{1}, \overline{l}_{2}, \overline{l}_{3}\}. Then by (2.6),(2.7),(2.17)-(2.19), we deduce that for x ∈ Q, ∥x∥_{P C^{1}} ≤ \frac{r_{1}}{2+k^{∗}+h^{∗}},

∥Ax∥_{P C^{1}} ≤ \frac{1}{2}∥x∥_{P C^{1}}.  (2.20)

Fix R > max\{2\overline{M}, 4∥v∥\}. Let ∪_{1} = \{x ∈ Q, ∥x∥_{P C^{1}} < R\}. By (2.16), we have

∥Ax∥_{P C^{1}} ≤ \frac{1}{2}∥x∥_{P C^{1}} + \overline{M} < \frac{1}{2}∥x∥_{P C^{1}} + \frac{1}{2}R ≤ R, ∀ x ∈ \overline{∪}_{1},

which gives

A(\overline{∪}_{1}) ⊂ ∪_{1}.  (2.21)

Choose 0 < r < min\{∥v∥ , \frac{r_{1}}{2+k^{∗}+h^{∗}} \}, and let ∪_{2} = \{x ∈ Q, ∥x∥_{P C^{1}} < r\}. Then by (2.20), we get

∥Ax∥_{P C^{1}} ≤ \frac{1}{2}∥x∥_{P C^{1}} < r,

which implies

A(\overline{∪}_{2}) ⊂ ∪_{2}.  (2.22)

Let ∪_{3} = {x ∈ Q : ∥x∥_{P C^{1}} < R, x(t) ≫ v, x′(t) ≫ v, ∀ t ∈ [t_{∗}, t^{∗}]}. Then it is easy to check that ∪_{3} is open in Q. Set w(t) = 2v + 2tv. Then w ∈ Q and w(t) ≫ v, w′(t) ≫ v, for t ∈ [t_{∗}, t^{∗}]. Hence w ∈ ∪_{3}, and so, ∪_{3} ≠ ∅. By (2.21), we know that ∥Ax∥_{P C^{1}} < R, ∀ x ∈ \overline{∪}_{3}. On the other hand, for x ∈ \overline{∪}_{3}, we have

\begin{aligned}(Ax)(t) & ≥ & ∫_{t_{∗}}^{t^{∗}} H_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds \\ & ≥ & m_{1} ∫_{t_{∗}}^{t^{∗}} f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds \\ & ≥ & m_{1} ∫_{t_{∗}}^{t^{∗}} σ(s)vds ≫ v, \end{aligned}  (2.23)

\begin{aligned}(Ax)′(t) & ≥ & ∫_{t_{∗}}^{t^{∗}} H′_{1}(t, s)f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds \\ & ≥ & m_{3} ∫_{t_{∗}}^{t^{∗}} f(s, x(s), x′(s), (T x)(s), (Sx)(s))ds \\ & ≥ & m_{3} ∫_{t_{∗}}^{t^{∗}} σ(s)vds ≫ v. \end{aligned}  (2.24)

Therefore,

A(\overline{∪}_{3}) ⊂ ∪_{3}.  (2.25)

Since ∪_{1},∪_{2},∪_{3} are nonempty bounded open convex sets of Q, by (2.21), (2.22), (2.25) and Lemma 1.2, we see

i(A, ∪_{i}, Q) = 1, i = 1, 2, 3.  (2.26)

Clearly,

∪_{2} ⊂ ∪_{1}, ∪_{3} ⊂ ∪_{1}, ∪_{2} ∩ ∪_{3} = ∅.  (2.27)

It follows from (2.26) and (2.27) that

i(A, ∪_{1}\(\overline{∪}_{2} ∪ \overline{∪}_{3}), Q) = i(A, ∪_{1}, Q) − i(A, ∪_{2}, Q) − i(A, ∪_{3}, Q) = −1.  (2.28)

Finally, (2.26) and (2.28) yield that A has two fixed point x_{1} ∈ ∪_{3} and x_{2} ∈ ∪_{1}\(\overline{∪}_{2} ∪ \overline{∪}_{3}). It is easy to see that

x_{1}(t) ≫ v x′_{1}(t) ≫ v, for every t ∈ [t_{∗}, t^{∗}],

and ∥x_{2}∥_{P C^{1}} > r. Hence x_{1}(t) ≢ θ and x_{2}(t) ≢ θ. The proof is then complete.□

3 An Example

Example 3.1. Consider the following boundary value problem for scalar second-order impulsive integro-differential equation

\{ \begin{matrix}x′′(t) = 32(x(t) + 2x′(t) + 3 ∫_{0}^{t} e^{−s}x(s)ds + 4 ∫_{0}^{1} e^{−2s}x(s)ds)^{2} \\ (1 + x(t) + x′(t) + ∫_{0}^{t} e^{−s}x(s)ds + ∫_{0}^{1} e^{−2s}x(s)ds)^{−2}, t ∈ J, t ≠ t_{1}, \\ Δx|_{t_{1}= \frac{1}{2}} = −\frac{1}{100} \frac{(x(\frac{1}{2}))^{2} + (x′(\frac{1}{2}))^{2}}{1 + (x(\frac{1}{2}))^{2} + (x′(\frac{1}{2}))^{2}} , \\ Δx′|_{t_{1}= \frac{1}{2}} = \frac{1}{200} \frac{(x(\frac{1}{2}))^{2} + (x′(\frac{1}{2}))^{2}}{1 + (x(\frac{1}{2}) + x′(\frac{1}{2}))^{2}} , \\ x(0) − 3x′(0) = 0, \\ x(1) − 2x′(1) = ∫_{0}^{1} \frac{1}{10}x(s)ds. \end{matrix}  (3.1)

Conclusion. Problem (3.1) has at least two positive solutions x_{1}(t) and x_{2}(t) such that x_{1}(t) > 1, x′_{1}(t) > 1 for t ∈ [\frac{1}{4}, \frac{1}{2}].

Proof. Let E = R^{1} and P = R_{+}. Then P is a normal and solid cone in E and problem (3.1) can be regarded as a BVP in the form of (1.1) in E. In this case,

k(t, s) = e^{−s}, h(t, s) = e^{−2s}, a = 3, b = 2, m = 1,

t_{1} = \frac{1}{2}, g(s) = \frac{1}{10}, t_{∗} = \frac{1}{4}, t^{∗} = \frac{1}{2}, v = 1,

and

f(t, u_{1}, u_{2}, u_{3}, u_{4}) = 32(\frac{u_{1} + 2u_{2} + 3u_{3} + 4u_{4}}{1 + u_{1} + u_{2} + u_{3} + u_{4}} )^{2}, ∀ t ∈ J, u_{i} ≥ 0, i = 1, 2, 3, 4,  (3.2)

I_{1}(u_{1}, u_{2}) = \frac{1}{100} \frac{u_{1}^{2} + u_{2}^{2}}{1 + u_{1}^{2} + u_{2}^{2}} ,  (3.3)
\overline{I}_{1}(u_{1}, u_{2}) = \frac{1}{200} \frac{u_{1}^{2} + u_{2}^{2}}{1 + (u_{1} + u_{2})^{2}} .  (3.4)

Clearly,

f ∈ C[J × P × P × P × P, P],
I_{1} ∈ C[P × P, P], \overline{I}_{1} ∈ C[P × P, P];

for any r > 0, f is bounded and uniformly continuous on J × P_{r} × P_{r} × P_{r} × P_{r}, I_{1} and \overline{I}_{1} are bounded on P_{r} × P_{r}. So (H_{1}) is satisfied.

u = ∫_{0}^{1}(a + s)g(s)ds = ∫_{0}^{1}(3 + s)\frac{1}{10}ds = \frac{7}{20}, u ∈ [0, a + 1 − b) = [0, 2).

This means that (H_{2}) is satisfied.

As in Example 3.2.1 in [5], we can prove that (2.1) is satisfied for c_{i} = 0 (i = 1, 2, 3, 4). By (3.3) and (3.4), we know that (2.2) and (2.3) are satisfied for

d_{1} = d_{2} = \frac{1}{50}, \overline{d}_{1} = \overline{d}_{2} = \frac{1}{100}.

By “Fact II”, we have

m_{1} = \frac{39}{22}, m_{2} = \frac{164}{33}, \overline{m}_{2} = \frac{82}{33}, m_{3} = \frac{13}{22}, m_{4} = \frac{74}{33}, \overline{m}_{4} = \frac{41}{66}.

So

l_{1} < \frac{11}{50}, l_{2} < \frac{1}{10}

and l < 1. Hence, (H_{3}) is satisfied.

Moreover, (3.2)-(3.4) implies that (H_{4}) holds.

On the other hand,

f(t, u_{1}, u_{2}, u_{3}, u_{4}) ≥ 32(\frac{u_{1} + u_{2} + u_{3} + u_{4}}{1 + u_{1} + u_{2} + u_{3} + u_{4}} )^{2} ≥ 32 × \frac{1}{4} = 8 = σ(t),

\overline{m} = min\{m_{1}, m_{3}\} = \frac{13}{22}, \overline{m} ∫_{t_{∗}}^{t^{∗}} σ(s)ds = \frac{13}{22} × \frac{1}{4} × 8 > 1.

Thus, our conclusion follows from Theorem 2.1.□

4 Acknowledgements

The authors are grateful to the referees for very valuable comments and suggestions.

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(Received December 29, 2009)