Hindawi Publishing Corporation Advances in Difference Equations Volume 2010, Article ID 185701, 14 doc

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Hindawi Publishing Corporation Advances in Difference Equations Volume 2010, Article ID 185701, 14 doc

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Hindawi Publishing Corporation Advances in Difference Equations Volume 2010, Article ID 185701, 14 pages doi:10.1155/2010/185701 Research Article Positive Solutions for Impulsive Equations of Third Order in Banach Space Jingjing Cai Department of Mathematics, Tongji University, Shanghai 200092, China Correspondence should be addressed to Jingjing Cai, cjjing1983@163.com Received 4 September 2010; Accepted 30 November 2010 Academic Editor: John Graef Copyright q 2010 Jingjing Cai. This is an o pen access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Using the fixed-point theorem, this paper is devoted t o study the multiple and single positive solutions of third-order boundary value problems for impulsive differential equations in ordered Banach spaces. The arguments are based on a specially constructed cone. At last, an example is given to illustrate the main results. 1. Introduction The purpose of this paper is to establish the existence of positive solutions for the following third-order three-point boundary value problems BVP, for short in Banach space E −x   t   λf 1  t, x  t  ,y  t   ,t∈  0, 1  \ { t 1 ,t 2 , ,t m } , −y   t   μf 2  t, x  t  ,y  t   ,t∈  0, 1  \ { t 1 ,t 2 , ,t m } , Δx   t k   −I 1,k  x  t k  , Δy   t k   −I 2,k  y  t k   ,k 1, 2, ,m, x  0   x   0   θ, x   1  − αx   η   θ, y  0   y   0   θ, y   1  − αy   η   θ, 1.1 where f i ∈ C0, 1 × P × P, P, I i,k ∈ CP, P, i  1, 2, k  1, 2, ,m. Δx  t k x  t  k  − x  t − k , Δy  t k y  t  k  − y  t − k ,μ>0, λ>0. θ is the zero element of E. Recently, third-order boundary value problems cf. 1–9 have attracted many authors attention due to their wide range of applications in applied mathematics, physics, and engineering, especially in the bridge issue. To our knowledge, most papers in literature 2AdvancesinDifference Equations concern mainly about the existence of positive solutions for the cases in which the spaces are real and the equations have no parameters. And many authors consider nonlinear term have same linearity. In this paper, we consider the existence of solutions when the nonlinear terms have different properties, the space is abstract and the equations have two different parameters. In 3, Guo et al. studied the following nonlinear three-point boundary value problem: u   t   a  t  f  u  t   0, u  0   u   0   0,u   1   αu   η  , 1.2 where a ∈ C0, 1, 0, ∞, f ∈ C0, ∞, 0, ∞. The authors obtained at least one positive solutions of BVP 1.2 by using fixed-point theorem when f is sublinear or suplinear. In 8, Yao and Feng used the upper and lower solutions method proved some existence results for the following third-order two-point boundary value problem u   t   f  t, u  t   0, 0 ≤ t ≤ 1, u  0   u   0   u   1   0. 1.3 Inspired by the above work, the aim of this paper is to establish some simple criteria for the existence of nontrivial solutions for BVP 1.1 under some weaker conditions. The new features of this paper mainly include the following aspects. Firstly, we consider the system 1.1 in abstract space while 3, 8 talk about equations in real space E  R.Secondly,we obtained the positive solutions when the two parameters have different ranges. Thirdly, f 1 and f 2 in system 1.1 may have different properties. Fourthly, f i i  1, 2 in system 1.1 not only contains x, y but also t, which is much more complicated. Finally, the main technique used here is the fixed-point theory and a special cone is constructed to study the existence of nontrivial solutions. We recall some basic facts about ordered Banach spaces E. The cone P in E induces a partial order on E,thatis,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, without loss of generality, suppose, in present paper, the normal constant N  1. α· denotes the measure of noncompactness cf. 10. Some preliminaries and a number of lemmas to the derivation of the main results are given in Section 2, then the proofs of the theorems are given in Section 3, followed by an example, in Section 4, to demonstrate the validity of our main results. 2. Preliminaries and Lemmas In this paper we will consider the Banach space E, ·,denoteJ 0 , 1 and PC 2 J, E{x | x  ∈ CJ, E, x  is continuous at t /  t k and x  is left continuous at t  t k , the right limit x  t  k  exists, k  1, 2, ,m}.Foranyx ∈ PC 2 J, E we define x 1  sup t∈J xt and x, y 2  x 1  y 1 for x, y ∈ PC 2 J, E × PC 2 J, E. For convenience, let us list the following assumption. Advances in Difference Equations 3 A f i ∈ C0, 1 × P × P, P, I i,k ∈ CP, P, i  1, 2, k  1, 2, ,m.Foranyt ∈ 0, 1 and r>0, ft, P r ,P r {ft, u, v : u, v ∈ P r } is relatively compact in E,where P r  {x ∈ P |x≤r}. Lemma 2.1. Assume that αη /  1, then for any y ∈ C0, 1, the following boundary value problem: −u   t   y  t  ,t∈  0, 1  \ { t 1 ,t 2 , ,t m } , Δu   t k   −I k  u  t k  ,k 1, 2, ,m, u  0   u   0   θ, u   1  − αu   η   θ 2.1 has a unique solution u  t    1 0 G  t, s  y  s  ds  m  k1 G  t, t k  I k  u  t k  , 2.2 where G  t, s   1 2  1 − αη  ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩  2ts − s 2  1 − αη   t 2 s  α − 1  ,s≤ min  η, t  , t 2  1 − αη   t 2 s  α − 1  ,t≤ s ≤ η,  2ts − s 2  1 − αη   t 2  αη − s  ,η≤ s ≤ t, t 2  1 − s  , max  η, t  ≤ s. 2.3 Proof. The proof is similar to Lemma 2.2 in 3,weomitit. Lemma 2.2 see 3. Assume that 0 <η<1 and 1 <α<1/η.Then0 ≤ Gt, s ≤ gs for any t, s ∈ 0, 1 × 0, 1,wheregs1  α/1 − αηs1 − s, s ∈ 0, 1. Lemma 2.3 see 3. Let 0 <η<1 and 1 <α<1/η, then for any t, s ∈ η/α, η × 0 , 1, Gt, s ≥ σgs,where 0 <σ η 2 2α 2  1  α  min { α − 1, 1 } < 1. 2.4 Inthepaper,wedefineconeK as follows: K   x ∈ PC 2  J, E  | x  t  ≥ θ, x  t  ≥ σx  s  ,t∈  η α ,η  ,s∈  0, 1   . 2.5 Lemma 2.4 see 10. Let E be a Banach space and K ⊂ E be a cone. Suppose Ω 1 and Ω 2 ∈ E are bounded open sets, θ ∈ Ω 1 , Ω 1 ⊂ Ω 2 , A : K ∩ Ω 2 \ Ω 1  → K is completely continuous such that 4AdvancesinDifference Equations either i Au≤u for any u ∈ K ∩ ∂Ω 1 and Au≥u for any u ∈ K ∩ ∂Ω 2 or ii Au≥u for any u ∈ K ∩ ∂Ω 1 and Au≤u for any u ∈ K ∩ ∂Ω 2 . Then A has a fixed-point in K ∩  Ω 2 \ Ω 1 . Lemma 2.5. The vector x, y ∈ PC 2 J, E × PC 2 J, E is a solution of differential systems 1.1 if and only if x, y ∈ PC 2 J, E is the solution of the following integral systems: x  t   λ  1 0 G  t, s  f 1  s, x  s  ,y  s   ds  m  k1 G  t, t k  I 1,k  x  t k  , y  t   μ  1 0 G  t, s  f 2  s, x  s  ,y  s   ds  m  k1 G  t, t k  I 2,k  y  t k   . 2.6 Define operators T 1 : K → K, T 2 : K → K and T : K × K → K × K as follows: T 1  x, y   λ  1 0 G  t, s  f 1  s, x  s  ,y  s   ds  m  k1 G  t, t k  I 1,k  x  t k  , T 2  x, y   μ  1 0 G  t, s  f 2  s, x  s  ,y  s   ds  m  k1 G  t, t k  I 2,k  y  t k   , T  x, y   t    T 1  x, y  ,T 2  x, y   t  . 2.7 As we know, BVP 1.1 has a positive solution x, y ifandonlyifx, y ∈ K × K is the fixed-point of T. Lemma 2.6. T : K × K → K × K is completely continuous. Proof. By condition A we get T 1 x, yt ≥ θ, T 2 x, yt ≥ θ,forallx, y ∈ K.Forany t ∈ η/α, η,wehave T 1  x, y   t    1 0 G  t, s  f 1  s, x  s  ,y  s   ds  m  k1 G  t, t k  I 1,k  x  t k  ≥ σ  1 0 g  s  f 1  s, x  s  ,y  s   ds  σ m  k1 g  t k  I 1,k  x  t k  ≥ σ  1 0 G  u, s  f 1  s, x  s  ,y  s   ds  σ m  k1 G  u, t k  I 1,k  x  t k   σT 1  x, y   u  ,u∈  0, 1  . 2.8 Advances in Difference Equations 5 Similarly T 2  x, y   t  ≥ σT 2  x, y   u  ,u∈  0, 1  . 2.9 So T : K × K → K × K. Next, we prove T : K × K → K × K is completely continuous. We first prove that T 1 is continuous. Let x n ,y n  ∈ Kn  1, 2,  and x 0 ,y 0  ∈ K such that x n ,y n  − x 0 ,y 0  2 → 0 n →∞.Letr  sup n x n ,y n  2 ,then   x 0 ,y 0    2 ≤ r,  x 0  1 ≤ r,   y 0   1 ≤ r,  x n  1 ≤ r,   y n   1 ≤ r. 2.10 By A,weobtain f i  t, x n  t  ,y n  t   −→ f i  t, x 0  t  ,y 0  t   ,  n −→ ∞  , for any t ∈  0, 1  ,i 1, 2, I 1,k  x n  t k  −→ I 1,k  x 0  t k  ,  n −→ ∞  ,k 1, 2, ,m, I 2,k  y n  t k   −→ I 2,k  y 0  t k   ,  n −→ ∞  ,k 1, 2, ,m. 2.11 Hence   T 1  x n ,y n   t  − T 1  x 0 ,y 0   t           1 0 G  t, s  f 1  s, x n  s  ,y n  s   ds  m  k1 G  t, t k  I 1,k  x n  t k  −  1 0 G  t, s  f 1  s, x 0  s  ,y 0  s   ds − m  k1 G  t, t k  I 1,k  x 0  t k       ≤  1 0 G  t, s    f 1  s, x n  s  ,y n  s   − f 1  s, x 0  s  ,y 0  s     ds  m  k1 G  t, t k  I 1,k  x n  t k  − I 1,k  x 0  t k  ≤  1 0 g  s    f 1  s, x n  s  ,y n  s   − f 1  s, x 0  s  ,y 0  s     ds  m  k1 g  t k  I 1,k  x n  t k  − I 1,k  x 0  t k  . 2.12 Since   T 1 x n ,y n  − T 1  x 0 ,y 0    1  sup t∈0,1   T 1  x n ,y n   t  − T 1  x 0 ,y 0   t    . 2.13 6AdvancesinDifference Equations By 2.11–2.13 and Lebesgue-dominated convergence theorem   T 1 x n ,y n  − T 1 x 0 ,y 0    1 −→ 0  n −→ ∞  . 2.14 So T 1 is continuous. Similarly, T 2 is continuous. It follows that T is continuous. Next we prove T is compact. Let V  {x n ,y n }⊂K × K be bounded, V 1  {x n } and V 2  {y n }.Letx n ,y n  2 ≤ r for some r>0, then x n  1 ≤ r, y n  1 ≤ r. It is easy to see that {T 1 x n ,y n t} is equicontinuous. By condition A we have α  T 1 V  t   α   1 0 G  t, s  f 1  s, x n  s  ,y n  s   ds  m  k1 G  t, t k  I 1,k  x n  t k  : x n ∈ V 1 ,y n ∈ V 2  ≤ 2  1 0 α  G  t, s  f 1  s, V 1  s  ,V 2  s    m  k1 α  G  t, t k  I 1,k  V 1  t k   0 2.15 which implies that αT 1 V 0. So, αTV0, it follows that T is compact. The lemma is proved. In this paper, denote f β i  lim sup  x    y  → β max t∈0,1   f  t, x, y     x     y   ,f i,β  lim inf  x    y  → β min t∈η/α,η   f  t, x, y     x     y   ,  ψf i  β  lim sup  x    y  → β max t∈0,1 ψ  f  t, x, y   x     y   ,  ψf i  β  lim inf  x    y  → β min t∈η/α,η ψ  f  t, x, y   x     y   . I i,β  k   lim inf  x  → β  I i,k  x   x  ,I β i  k   lim sup  x  → β  I i,k  x   x  ,k 1, 2, ,m. 2.16 where β  0orβ ∞, ψ ∈ P ∗  {ψ ∈ E ∗ : ψx ≥ θ, ∀x ∈ P} and ψ  1. P ∗ is a dual cone of P. We list the assumptions: H 1 ψf 1  0 >m 1 , ψf 2  ∞ >m 2 ,wherem 1 ,m 2 ∈ 0, ∞; H 2  f 0 i <m 3 , ψf 1  ∞ >m 4 ,I i,0 k0, i  1, 2, where m 3 ,m 4 > 0andm 3  m 4 ; H 3 ψf 1  0 >m 5 , f ∞ i <m 6 , I ∞ i k0, i  1, 2, where m 5 ,m 6 > 0andm 6  m 5 . Advances in Difference Equations 7 For convenience, denote a 1  1 4  m 3  1 0 gsds  −1 ,α 2   m 4 σ  η α/η Gη, sds  −1 , a 3   m 5 σ  η η/α Gη, sds  −1 ,α 4  1 4  m 6  1 0 gsds  −1 . 2.17 3. Main Results Theorem 3.1. Assume that (A), (H 1 ) and the following condition H  hold, then BVP 1.1 has at least two positive solution while λ ∈ 0, 1/4M 1  1 0 gsds and μ ∈ 0 , 1/4M 2  1 0 gsds. H  : m 1 λσ  η η/α Gη, sds ≥ 1; m 2 μσ  η η/α Gη, sds ≥ 1;  2 i1  m k1 gt k M i,k < 1/2,where M i  max t∈0,1, 0≤uv≤1 f i t, u, v > 0, M i,k  max 0≤u≤1 {I i,k u}. Proof. Let Ω 1  {x, y ∈ K × K : x, y 2 < 1},thenforx, y ∈ ∂Ω 1 ,wehave   T 1  x, y   t    ≤      λ  1 0 g  s  f 1  x  s  ,y  s   ds            m  k1 g  t k  I 1,k  x  t k       ≤ λM 1  1 0 g  s  ds  m  k1 g  t k  M 1,k , 3.1 that is,   T 1 x, y   1 ≤ λM 1  1 0 g  s  ds  m  k1 g  t k  M 1,k , 3.2 Similarly   T 2 x, y   1 ≤ M 2 μ  1 0 g  s  ds  m  k1 g  t k  M 2,k . 3.3 So   Tx, y   2 ≤  λM 1  μM 2   1 0 g  s  ds  2  i1 m  k1 g  t k  M i,k < 1    x, y   2 . 3.4 Hence   Tx, y   2 <   x, y   2 , for any  x, y  ∈ ∂Ω 1 . 3.5 8AdvancesinDifference Equations Since ψf 1  0 >m 1 ,thereexistε 1 > 0and0<R 1 < 1suchthatψf 1 t, u, v ≥ m 1  ε 1 uv for 0 ≤uv≤R 1 and t ∈ η/α, η.LetΩ 2  {x, y ∈ K×K : x, y 2 <R 1 }. Then for any x, y ∈ ∂Ω 2 ,byH 1  and the definition of ψ,weobtain   T 1 x, y   1 ≥ ψ  T 1  x, y  η  ≥ λ  η η/α G  η, s  ψ  f 1  t, x  s  ,y  s   ds ≥  m 1  ε 1  λσ  η η/α G  η, s   x  1    y   1  ds  R 1  m 1  ε 1  λσ  η η/α G  η, s  ds. 3.6 By 3.6 and H    Tx, y   2 ≥   T 1 x, y   1 ≥ R 1  m 1  ε 1  λσ  η η/α G  η, s  ds >R 1    x, y   2 ,  x, y  ∈ ∂Ω 2 , 3.7 Similarly, by ψf 2  ∞ >m 2 ,thereexistε 2 > 0andR 2 > 1suchthatψf 2 t, u, v ≥ m 2  ε 2 u  v for t ∈ η/α, η and u, v ∈ P with 0 ≤u  v≤R 2 .LetΩ 3  {x, y ∈ K × K : x, y 2 <R 2 }.Thenforanyx, y ∈ ∂Ω 3 ,   T 2 x, y   1 ≥ R 2 μ  m 2  ε 2  σ  η η/α G  η, s  ds. 3.8 So we have by 3.8 and H    Tx, y   2 ≥   T 2 x, y   1 >R 2    x, y   2 , for any  x, y  ∈ ∂Ω 3 . 3.9 By 3.5, 3.7, 3.9 and Lemma 2.4 wegetthatBVP1.1 has at least two positive solutions with x 1 ,y 1  2 < 1 < x 2 ,y 2  2 . Corollary 3.2. Assume that (A) and the following condition h old, then the conclusion of Theorem 3.1 also holds.  ψf 1  0 >m 1 ,ψ  f 2  ∞ >m 2 , where m 1 ,m 2 ∈  0, ∞  . 3.10 Theorem 3.3. Assume that (A) and (H 2 )hold,thenBVP1.1 has at least one positive solution when λ ∈ a 2 ,a 1  and μ ∈ 0,a 1 . Proof. By Lemma 2.6,weseethatT : K × K → K × K is completely continuous. By H 2 , there exists r 1 > 0, ε 3 > 0, ε>0suchthatfori  1, 2,   f i  t, x  t  ,y  t     ≤  m 3 − ε 3    x  t     y  t     ,  I i,k  x  t k  ≤ ε  x  t k  , 3.11 Advances in Difference Equations 9 for any x, y ∈ K with 0 ≤x 1  y 1 ≤ r 1 ,wherem 3 − ε 3 > 0, ε>0suchthat ε m  k1 g  t k  ≤ 1 2 . 3.12 Let Ω 4  {x, y ∈ K × K : x, y 2 <r 1 }.Thenforanyx, y ∈ ∂Ω 4 ,weobtain   T 1  x, y   t          λ  1 0 G  t, s  f 1  s, x  s  ,y  s   ds  m  k1 G  t, t k  I 1,k  x  t k       ≤ λ       1 0 g  s  f 1  s, x  s  ,y  s   ds            m  k1 g  t k  I 1,k  x  t k       ≤ λ  m 3 − ε 3   1 0 g  s  ds   x  t     y  t      ε m  k1 g  t k  x  t k  ≤ λ  m 3 − ε 3   1 0 g  s  ds   x  1    y   1   ε m  k1 g  t k  x  1  λ  m 3 − ε 3  r 1  1 0 g  s  ds  ε m  k1 g  t k  x  1 ≤ 1 4 r 1  ε m  k1 g  t k  x  1 , 3.13 Similarly   T 2  x, y   t    ≤ μ  m 3 − ε 3  r 1  1 0 g  s  ds  ε m  k1 g  t k    y   1 ≤ 1 4 r 1  ε m  k1 g  t k    y   1 . 3.14 It follows that   Tx, y   2    T 1 x, y   1    T 2 x, y   1 ≤ r 1    x, y   2 , 3.15 which implies   Tx, y   2 ≤   x, y   2 , for any  x, y  ∈ ∂Ω 4 . 3.16 On the other hand, by ψf 1  ∞ >m 4 ,thereexistsR>0, ε 4 > 0suchthatψf 1 t, xt,yt ≥ m 4  ε 4 xt  yt for x 1  y 1 >Rand t ∈ η/α, η.LetR 1  max{2r 1 ,R/σ}, Ω 5  {x, y ∈ K × K : x, y 2 <R 1 }.Foranyx, y ∈ ∂Ω 5 ,wehave x  t  ≥ σx  s  ,y  t  ≥ σy  s  ,  x  t  ≥ σ  x  s  ,   y  t    ≥ σ   y  s    ,t∈  η α ,η  ,s∈  0, 1  . 3.17 10 Advances in Difference Equations By the definition of T 1 we get   T 1 x, y   1 ≥ ψ  T 1  x, y  η  ≥ λ  η η/α G  η, s  ψ  f 1  t, x  s  ,y  s   ds ≥ λ  m 4  ε 4   η η/α G  η, s   x  s     y  s     ds ≥ λ  m 4  ε 4  σ  η η/α G  η, s   x  u     y  u     ds. 3.18 So   T 1 x, y   1 ≥ λ  m 4  ε 4  σ  η η/α G  η, s   x  1    y   1  ds  R 1 λ  m 4  ε 4  σ  η η/α G  η, s  ds. 3.19 Hence   Tx, y   2 ≥   T 1 x, y   1 ≥ R 1 λ  m 4  ε 4  σ  η η/α G  η, s  ds ≥ R 1     x, y    2 . 3.20 Therefore   Tx, y   2 ≥   x, y   2 , ∀  x, y  ∈ ∂Ω 5 . 3.21 By 3.16, 3.21 and Lemma 2.4, it is easily seen that T has a fixed-point x ∗ ,y ∗  ∈ Ω 5 \ Ω 4 . Corollary 3.4. Let (A) and the following conditions hold, then BVP 1.1 has at least one positive solution while μ ∈ a 2 ,a 1  and λ ∈ 0,a 1 . f 0 i <m 3 ,  ψf 2  ∞ >m 4 ,I i,0  k   0,i 1, 2. 3.22 Theorem 3.5. Let (A) and (H 3 )hold,thenBVP1.1 has at least one positive solution while λ ∈ a 3 ,a 4  and μ ∈ 0,a 4 . [...]... fixed-point x∗ , y∗ ∈ Ω7 \ Ω6 Corollary 3.6 Assume that (A) and the following conditions hold, then BVP 1.1 has at least one positive solution while μ ∈ a3 , a4 and λ ∈ 0, a4 ψf2 0 > m5 , fi∞ < m6 , Ii∞ k 0, i 1, 2 3.32 Advances in Difference Equations 13 4 An Example In this section, we construct an example to demonstrate the application of our main results obtained in Section 3 Consider the following... 5 ⎪ ⎪ ⎪ ⎪2 ⎪t 1 − s , ⎩ x 0, y 0, where k 1, i 2 ,t , 5 2 t≤s≤ , 5 2 ≤ s ≤ t, 5 2 max , t ≤ s, 5 4.2 1, 2 In s≤ t2 s, t2 f t, x, y 4 −s , 5 4.3 α s 1 − s / 1 − αη 15s 1 − s 5, m2 3000 By computing, we get 1 4 −1 1 5g s ds 0 8 , 25 a2 4 3125 2/5 1/5 2 G , s ds 5 −1 4.4 14 Advances in Difference Equations Above all, the conditions of Theorem 3.3 are satisfied Then for any λ ∈ a2 , ∞ and μ ∈ 0, a1 , BVP... at resonance,” Nonlinear Analysis: Theory, Methods & Applications, vol 32, no 4, pp 493–499, 1998 6 Y Sun, “Positive solutions of singular third-order three-point boundary value problem,” Journal of Mathematical Analysis and Applications, vol 306, no 2, pp 589–603, 2005 7 J R L Webb, “Positive solutions of some three point boundary value problems via fixed point index theory,” Nonlinear Analysis: Theory,... 2005 3 L.-J Guo, J.-P Sun, and Y.-H Zhao, “Existence of positive solutions for nonlinear third-order threepoint boundary value problems,” Nonlinear Analysis: Theory, Methods & Applications, vol 68, no 10, pp 3151–3158, 2008 4 C P Gupta, “Solvability of a three-point nonlinear boundary value problem for a second order ordinary differential equation,” Journal of Mathematical Analysis and Applications, vol... third-order two-point boundary value problem,” Applied Mathematics Letters, vol 15, no 2, pp 227–232, 2002 9 Q Yao, “The existence and multiplicity of positive solutions for a third-order three-point boundary value problem,” Acta Mathematicae Applicatae Sinica, vol 19, no 1, pp 117–122, 2003 10 D Guo, Nonlinear Functional Analysis, Shangdong Science and Technology Press, Jinan, China, 1985 .. .Advances in Difference Equations 11 v Proof Since ψf1 0 > m5 , we choose R3 > 0, ε5 > 0 such that ψ fi t, u, v ≥ m5 ε5 u for 0 ≤ u v ≤ R3 and t ∈ η/α, η Let Ω6 { x, y ∈ K × K : x, y 2 < R3 } Then for any x, y ∈ ∂Ω6 , T1 x,... partially supported by NSFC 10971155 and Innovation program of Shanghai Municipal Education Commission 09ZZ33 References 1 Z Du, W Ge, and X Lin, “Existence of solutions for a class of third-order nonlinear boundary value problems,” Journal of Mathematical Analysis and Applications, vol 294, no 1, pp 104–112, 2004 2 Y Feng and S Liu, “Solvability of a third-order two-point boundary value problem,” Applied... , y t ≤ m6 − ε6 for any x , 0 i 1, 2 , there exist M > 0, ε6 > 0, ε > 0 such , y t x t 1 2 y 1 x, y Ii,k x tk 2 ≥ M, ≤ ε x tk , t ∈ 0, 1 , 3.26 where ε satisfies m ε k 1 g tk ≤ 1 2 3.27 12 Advances in Difference Equations max{M, 2R3 } and Ω7 Let R4 x, y ∈ ∂Ω7 , we have { x, y | x, y ∈ K × K : 1 G t, s f1 s, x s , y s ds k 1 1 m g s f1 s, x s , y s ds g tk I1,k x tk 0 k 1 1 ≤ λ m6 − ε6 m y t g s ds x . Hindawi Publishing Corporation Advances in Difference Equations Volume 2010, Article ID 185701, 14 pages doi:10.1155/2010/185701 Research Article Positive Solutions for Impulsive Equations. to their wide range of applications in applied mathematics, physics, and engineering, especially in the bridge issue. To our knowledge, most papers in literature 2AdvancesinDifference Equations concern. 3.32 Advances in Difference Equations 13 4. An Example In this section, we construct an example to demonstrate the application of our main results obtained in Section 3. Consider the following

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