Existence results for nonlinear boundary-value problems with integral boundary conditions in Banach spaces

Nonlinear Analysis 69 (2008) 3310–3321 www.elsevier.com/locate/na

Existence results for nonlinear boundary-value problems with integral boundary conditions in Banach spacesI Xuemei Zhang a,c , Meiqiang Feng b,c,∗ , Weigao Ge c a Department of Mathematics and Physics, North China Electric Power University, Beijing, 102206, PR China b Department of Fundamental Sciences, Beijing Information Technology Institute, Beijing, 100101, PR China c Department of Mathematics, Beijing Institute of Technology, Beijing, 100081, PR China

Received 21 February 2007; accepted 14 September 2007

Abstract This paper investigates the multiplicity of positive solutions for a class of nonlinear boundary-value problems of fourth-order differential equations with integral boundary conditions in ordered Banach spaces. The arguments are based upon a specially constructed cone and the fixed point index theory in cone for strict set contraction operator. The nonexistence of positive solution is also studied. Meanwhile, an example is worked out to demonstrate the main results. c 2007 Elsevier Ltd. All rights reserved.

MSC: 34B15 Keywords: Boundary-value problem; Positive solution; Fixed point index theory; Existence; Measure of noncompactness

1. Introduction The theory of boundary-value problems with integral boundary conditions for ordinary differential equations arises in different areas of applied mathematics and physics. For example, heat conduction, chemical engineering, underground water flow, thermo-elasticity, and plasma physics can be reduced to the nonlocal problems with integral boundary conditions. For boundary-value problems with integral boundary conditions and comments on their importance, we refer the reader to the papers by Gallardo [1], Karakostas and Tsamatos [2], Lomtatidze and Malaguti [3] and the references therein. For more information about the general theory of integral equations and their relation with boundary-value problems we refer to the book of Corduneanu [4] and Agarwal and O’Regan [5]. Moreover, boundary-value problems with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multipoint and nonlocal boundary-value problems as special cases. The existence and multiplicity of positive solutions for such problems have received a great deal of attention. To identify a few, we refer the reader to [2,6–12,16–27] and references therein. But the corresponding theory for multipoint boundary-value problems for ordinary differential equations in Banach spaces, especially in the case that I Supported by NNSF of China (10671012) and SRFDP of China (20050007011). ∗ Corresponding author at: Department of Fundamental Sciences, Beijing Information Technology Institute, Beijing, 100101, PR China.

E-mail address: [email protected] (M. Feng). c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.09.020

X. Zhang et al. / Nonlinear Analysis 69 (2008) 3310–3321

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the boundary-value problem with integral boundary conditions in Banach spaces, has not been investigated till now. Now, in this paper, we shall use fixed point index theory in cone for strict set contraction operators to investigate the existence of positive solutions for a class of nonlinear boundary-value problems with integral boundary conditions of fourth-order differential equations in a Banach space. Let E be a real Banach space with norm k · k and P ⊂ E be a cone of E. The purpose of this paper is to investigate the existence of positive solutions of the following fourth-order boundary-value problem with integral boundary conditions  (4) x (t) − λ f (t, x(t)) = θ, 0 < t < 1,   Z 1    x(0) = x(1) = g(s)x(s)ds, (1.1) 0Z  1    x 00 (0) = x 00 (1) = h(s)x 00 (s)ds, 0

where λ is a positive parameter, f ∈ C([0, 1] × P, P), θ is the zero element of E, and g, h ∈ L 1 [0, 1]. The main features of this paper are as follows. First, we discuss the existence results in abstract space E, not E = R as in [10]. Secondly, comparing with [10], the number of solutions is multiple. Finally, we consider the existence results in the case λ > 0, not λ = 1 as in [10]. To our knowledge, no paper has considered fourth-order nonlocal boundary-value problem with integral boundary conditions in abstract spaces. This paper fills this gap in the literature. The purpose of this paper is to improve and generalize the results in the above mentioned references. We will show that the number of positive solutions of BVP (1.1) is determined by the parameter λ. The arguments are based upon a specially constructed cone and the fixed point index theory in cone for strict set contraction operator. The organization of this paper is as follows. We shall introduce some lemmas and notations in the rest of this section. In Section 2, we provide some necessary background. In particular, we state some properties of the Green’s function associated with BVP (1.1). In Section 3, the main results will be stated and proved. Finally, one example is also included to illustrate the main results. Basic facts about ordered Banach space E can be found in [13–15]. Here we just recall a few of them. The cone P in E induces a partial order on E, i.e., 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 kxk ≤ N kyk. Without loss of generality, suppose, in the present paper, the normal constant N = 1. We consider problem (1.1) in C(J, E), in which J = [0, 1]. Evidently, (C(J, E), k · k) is a Banach space with norm kxkc = maxt∈J kx(t)k for x ∈ C(J, E). In the following, x ∈ C(J, E) is called a solution of (1.1) if it satisfies (1.1). x is a positive solution of (1.1) if, in addition, x(t) > θ for t ∈ (0, 1). R1 Let x(t) : (0, 1] → E be continuous, the abstract generalized integral 0 x(t)dt is called convergent if the R1 limit limε→0+ ε x(t)dt exists. The convergency or divergency of other kinds of generalized integrals can be defined similarly. For a bounded set V in Banach space E, we denote α(V ) the Kuratowski measure of noncompactness (see [13–15], for further understanding). The operator A : D → E(D ⊂ E) is said to be a k-set contraction if A : D → E is continuous and bounded and there is a constant k ≥ 0 such that α(A(S)) ≤ kα(S) for any bounded S ⊂ D; a k-set contraction with k < 1 is called a strict set contraction. In the following, denote the Kuratowski’s measure of noncompactness by α(·). For the application in the sequel, we first state the following definition and lemmas which can be found in [13]. Definition 1.1. Let P be a cone of real Banach space E. If P ∗ = {Ψ ∈ E ∗ |Ψ (x) ≥ 0, ∀x ∈ P}, then P ∗ is dual cone of cone P. Lemma 1.1. Let K be a cone in real Banach space E and Ω be a nonempty bounded open convex subset of K . Suppose that A : Ω¯ −→ K is a strict set contraction and A(Ω¯ ) ⊂ Ω , when Ω¯ denotes the closure of Ω in K . Then the fixed point index i(A, Ω , K ) = 1. Lemma 1.2. Let K be a cone in real Banach space E, K r = {x ∈ K : kxk < r } (r > 0) and K¯ r = {x ∈ K : kxk ≤ r }. Suppose that A : K¯ r −→ K is a strict set contraction. If kAxk ≥ kxk and Ax 6= x for x ∈ ∂ Pr , then the fixed

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point index i(A, K r , K ) = 0. 2. Preliminaries To establish the existence of multiple positive solutions in C(J, P) of (1.1), let us list the following assumptions, which will stand throughout this paper: (H1 ) f ∈ C(J × P, P), and for any l > 0, f is uniformly continuous on J × (P ∩ Tl ). Further suppose that µ ∈ (0, 1), ν ∈ (0, 1) and there exist nonnegative constants ηl with 1 ληl < 1 8(1 − µ)(1 − ν) such that α( f (t, S)) ≤ ηl α(S),

t ∈ J, S ⊂ P ∩ Tl , R1 R1 where Tl = {x ∈ E : kxk ≤ l}, µ = 0 g(s)ds, ν = 0 h(s)ds; (H2 ) There exists a function m ∈ C(J, [0, ∞)) which does not vanish identically on any subset of positive measure such that Ψ ( f (t, x)) ≥ m(t)kxk for all t ∈ J, x ∈ P, where Ψ ∈ P ∗ and kΨ k = 1; (H3 ) There exist a function n ∈ C(Jδ , [0, ∞)) which does not vanish identically on any subset of positive measure and a positive constant d such that Ψ ( f (t, x)) ≥ n(t) for all t ∈ Jδ , x ∈ P, kxk ≥ d, where Jδ = [δ, 1 − δ], δ ∈ (0, 12 ). In our main results, we will make use of the following lemmas. R1 Lemma 2.1. Assume that σ := 0 φ(s)ds 6= 1. Then for any ϕ ∈ C(J, P), the following boundary-value problem  00 −x (t) = ϕ(t),Z 0 < t < 1, 1 (2.1) x(0) = x(1) = φ(s)x(s)ds, 0

has a unique solution Z 1 x(t) = H (t, s)ϕ(s)ds, 0

where H (t, s) = G(t, s) +  G(t, s) =

t (1 − s), s(1 − t),

1 1−σ

1

Z

G(s, τ )φ(τ )dτ,

(2.2)

0

0 ≤ t ≤ s ≤ 1, 0 ≤ s ≤ t ≤ 1.

(2.3)

Proof. First suppose that x ∈ C[J, E] is a solution of problem (2.1). It is easy to see by integration of (2.1) that Z t x 0 (t) = x 0 (0) − ϕ(s)ds. 0

Integrate again, we can get x(t) = x(0) + x (0)t − 0

t

Z

(t − s)ϕ(s)ds.

(2.4)

0

Letting t = 1 in (2.4), we find Z 1 x 0 (0) = (1 − s)ϕ(s)ds. 0

(2.5)

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R1 Substituting x(0) = 0 φ(s)x(s)ds and (2.5) into (2.4), we obtain Z t Z 1 Z 1 (t − s)ϕ(s)ds (1 − s)ϕ(s)ds − φ(s)x(s)ds + t x(t) = 0

0

0 1

Z

1

Z

φ(s)x(s)ds,

G(t, s)ϕ(s)ds +

=

(2.6)

0

0

where 1

Z

φ(s)x(s)ds =

1

Z

φ(s)

"Z

1

Z

G(s, τ )ϕ(τ )dτ +

=

φ(s)

"Z

1

Z

1

# G(s, τ )ϕ(τ )dτ ds +

φ(s)x(s)ds ds 1

Z

φ(s)ds

1

Z

φ(s)x(s)ds,

(2.7)

0

0

0

0

#

0

0

0

0

1

and so, 1

Z

φ(s)x(s)ds =

0

1−

R1 0

1

Z

1 φ(s)ds

φ(s)

"Z

0

1

# G(s, τ )ϕ(τ )dτ ds.

(2.8)

0

Substituting (2.8) into (2.6), we obtain "Z # Z 1 Z 1 1 1 x(t) = G(t, s)ϕ(s)ds + φ(s) G(s, τ )ϕ(τ )dτ ds R1 0 0 1 − 0 φ(s)ds 0 Z 1 = H (t, s)ϕ(s)ds,

(2.9)

0

where H (t, s) is defined by (2.2). R1 Conversely, suppose x(t) = 0 H (t, s)ϕ(s)ds, then Z

t

x(t) =

1

Z s(1 − t)ϕ(s)ds +

0

t (1 − s)ϕ(s)ds +

t

1 1−

R1 0

φ(s)ds

1

Z

φ(s)

0

"Z

1

# G(s, τ )ϕ(τ )dτ ds.

(2.10)

0

Direct differentiation of (2.10) implies Z t Z 1 x 0 (t) = − sϕ(s)ds + t (1 − t)ϕ(t) + (1 − s)ϕ(s)ds − t (1 − t)ϕ(t) 0 1

Z =

t

(1 − s)ϕ(s)ds −

t

t

Z

(2.11)

sϕ(s)ds, 0

and x 00 (t) = −tϕ(t) − (1 − t)ϕ(t) = −ϕ(t), R1 and it is easy to verify that x(0) = x(1) = 0 φ(s)x(s)ds. The proof is complete.



From (2.2) and (2.3), we can prove that H (t, s), G(t, s) have the following properties. Proposition 2.1. If σ ∈ (0, 1), then for t, s ∈ (0, 1), we have H (t, s) > 0, G(t, s) > 0. Proposition 2.2. If σ ∈ (0, 1), then for t, s ∈ J , we have 0 ≤ H (t, s),

0 ≤ G(t, s) ≤ G(s, s) ≤

1 . 4

(2.12)

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Proposition 2.3. For all t ∈ Jδ , s, u ∈ [0, 1] we have G(t, s) ≥ δG(u, s).

(2.13)

In fact, for t ∈ [δ, 1 − δ], we have Case 1. If max{t, u} ≤ s, then G(t, s) t (1 − s) t = = ≥ t ≥ δ. G(u, s) u(1 − s) u Case 2. If s ≤ min{t, u}, then s(1 − t) 1−t G(t, s) = = ≥ 1 − t ≥ δ. G(u, s) s(1 − u) 1−u Case 3. If t ≤ s ≤ u, then G(t, s) t (1 − s) t = ≥ ≥ t ≥ δ. G(u, s) s(1 − u) s Case 4. If u ≤ s ≤ t, then s(1 − t) 1−t G(t, s) = ≥ ≥ 1 − t ≥ δ. G(u, s) u(1 − s) 1−s Therefore, for t ∈ Jδ , s, u ∈ [0, 1] we have G(t, s) ≥ δ. G(u, s) Lemma 2.2. Suppose that (H1 ) holds. We define an operator T by Z 1 Υ (t, s) f (s, x(s))ds, (T x)(t) = λ

(2.14)

0

where Υ (t, s) =

1

Z

Υ1 (t, τ )Υ2 (τ, s)dτ,

(2.15)

0

Υ1 (t, τ ) = G(t, τ ) + Υ2 (τ, s) = G(τ, s) +

1 1−µ 1 1−ν

Z

1

0 Z 1

G(τ, v)g(v)dv,

(2.16)

G(s, v)h(v)dv.

(2.17)

0

Then x is a nonnegative solution of (1.1) if and only if x is a fixed point of T . Proof. (1) Necessity. Let y = x 00 . Then, (1.1) implies  00 x(t)), 0 < t < 1, −y (t) = −λ f (t, Z 1  y(0) = y(1) = h(s)y(s)ds.

(2.18)

0

By Lemma 2.1, we have Z 1 y(t) = −λ Υ2 (t, s) f (s, x(s))ds. 0

(2.19)

X. Zhang et al. / Nonlinear Analysis 69 (2008) 3310–3321

Similarly, from  00 −x (t) = −y(t), Z

0 < t < 1, 1

(2.20)

g(s)x(s)ds,

x(0) = x(1) =

3315

0

we obtain 1

Z x(t) = −

Υ1 (t, s)y(s)ds.

(2.21)

0

Substituting (2.19) into (2.21), we have # "Z Z 1 1 Υ2 (τ, s) f (s, x(s))ds dτ Υ1 (t, τ ) x(t) = λ 0

0

=λ

1

Z 0

=λ

! Υ1 (t, τ )Υ2 (τ, s)dτ

f (s, x(s))ds

0 1

Z

1

Z

Υ (t, s) f (s, x(s))ds.

0

The necessity is proved. (2) Sufficiency. Let Z 1 y(τ ) = −λ Υ2 (τ, s) f (s, x(s))ds.

(2.22)

0

R1 R1 Then, by x(t) = λ 0 Υ1 (t, τ )dτ 0 Υ2 (τ, s) f (s, x(s))ds, we have Z 1 x(t) = − Υ1 (t, τ )y(τ )dτ.

(2.23)

0

Lemma 2.1 and (2.22) imply that y(τ ) is a solution of the following BVP  00 −y (τ ) = −λ f Z(τ, x(τ )), 0 < τ < 1, 1  y(0) = y(1) = h(τ )y(τ )dτ.

(2.24)

0

Similarly, Lemma 2.1 and (2.23) imply that x(t) is a solution of BVP  00 0 < t < 1, −x (t) = −y(t), Z 1 x(0) = x(1) = g(t)x(t)dt.

(2.25)

0

R1 0

From (2.24) and (2.25), we have x (4) (t) = λ f (t, x(t)), ∀0 < t < 1. Obviously, x(0) = x(1) = R1 g(t)x(t)dt, x 00 (0) = x 00 (1) = 0 h(t)x 00 (t)dt, and x ∈ C 4 [J, E]. The sufficiency is proved.  From (2.15), we can obtain that Υ (t, s) has the following properties.

Proposition 2.4. If (H1 ) holds, then for t, s ∈ J we have 0 ≤ Υ (t, s) ≤

1 . 16(1 − µ)(1 − ν)

(2.26)

Proposition 2.5. If (H1 ) holds, then for t ∈ Jδ , s, u ∈ [0, 1] we have Υ (t, s) ≥ δΥ (u, s).

(2.27)

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In fact, from (2.13) and (2.15), we have Z 1 1 Υ1 (t, τ ) ≥ δG(u, τ ) + G(τ, v)g(v)dv = δΥ1 (u, τ ), δ 1−µ 0

t ∈ Jδ , τ, u ∈ J.

This and (2.15) imply that (2.27) holds. To obtain a positive solution, we construct a cone K by K = {x ∈ Q : x(t) ≥ δx(s), t ∈ Jδ , s ∈ [0, 1]},

(2.28)

where Q = {x ∈ C(J, E) : x(t) ≥ θ, t ∈ J }. It is easy to see that K is a cone of C(J, E) and K r,R = {x ∈ K : r ≤ kxk ≤ R} ⊂ K , K ⊂ Q. In the following, let Bl = {x ∈ C(J, E) : kxkc ≤ l},

l > 0.

Lemma 2.3. Suppose that (H1 ) holds. Then for each l > 0, T is strict set contraction on Q ∩ Bl , i.e., there exists a constant 0 ≤ kl < 1 such that α(T (S)) ≤ kl α(S) for any S ⊂ Q ∩ Bl . Proof. For any l > 0, suppose S ⊂ Q ∩ Bl . Clearly, T is continuous and bounded from Q ∩ Bl into Q. 1 On the other hand, it is clear that 0 ≤ Υ (t, s) ≤ 16(1−µ)(1−ν) and using a similar method as in the proof of Lemma 2 in [12], we can get that α(T (S)) ≤ 2λ

1 ηl α(S). 16(1 − µ)(1 − ν)

Therefore α(T (S)) ≤ kl α(S),

S ⊂ Q ∩ Bl ,

1 ηl , 0 ≤ kl < 1. The proof is complete. where kl = λ 8(1−µ)(1−ν)



Lemma 2.4. Suppose that (H1 ) holds. Then T (K ) ⊂ K and T : K r,R → K is strict set contraction. Proof. From (2.27) and (2.14), we obtain Z 1 min(T x)(t) = λ min Υ (t, s) f (s, x(s))ds t∈Jδ

t∈Jδ

≥ λδ

0 1

Z

Υ (u, s) f (s, x(s))ds

0

≥ δ(T x)(u),

u ∈ J.

Therefore T (x) ∈ K , i.e., T (K ) ⊂ K . Also we have T (K r,R ) ⊂ K by K r,R ⊂ K . Hence we have T : K r,R → K . Next by Lemma 2.3, one can prove that T : K r,R → K is a strict set contraction. So it is omitted.  3. Main results Write f β = lim sup max kxk→β t∈J

k f (t, x)k , kxk

(Ψ f )β = lim inf min kxk→β t∈J

Ψ ( f (t, x)) , kxk

where β denotes 0 or ∞, Ψ ∈ P ∗ , and kΨ k = 1. In this section, we apply Lemmas 1.1 and 1.2 to establish the existence of positive solutions for problem (1.1).

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Theorem 3.1. Assume (H1 ), (H2 ), P is normal and for any x ∈ P, (Ψ f )0 = (Ψ f )∞ = ∞. Then, for λ sufficiently small, problem (1.1) has at least two positive solutions, whereas for λ sufficiently large, problem (1.1) has no positive solution. Proof. Let T be cone preserving, strict set contraction that was defined by (2.14). For L > 0, let Z 1 1 F(L) = max k f (s, x(s))kds. 16(1 − µ)(1 − υ) x∈K ,kxkc =L 0 Since (Ψ f )∞ = ∞, there exists r1 > 0 such that F(r1 ) > 0. Let λ1 = x ∈ K , kxkc = r1 , we have Z 1 1 k f (s, x(s))kds k(T x)(t)k ≤ λ 16(1 − µ)(1 − υ) 0 Z 1 1 < λ1 k f (s, x(s))kds 16(1 − µ)(1 − υ) 0 ≤ λ1 F(r1 ) = kxkc ,

r1 F(r1 ) .

Then for λ ∈ (0, λ1 ) and

i.e., x ∈ K , kxkc = r1 , imply kT xkc < r1 . Thus, Lemma 1.1 implies i(T, K r1 , K ) = 1. Considering (Ψ f )∞ = ∞, there exists r¯2 > 0 such that Ψ ( f (t, x(t))) ≥ ε1 kxk, for x ∈ P, t ∈ J, kxk ≥ r¯2 , R 1−δ where ε1 > 0 satisfies ε1 λδ δ Υ ( 21 , s)ds > 1. Let r2 = max{2r1 , r¯δ2 }. Then, for t ∈ Jδ , x ∈ K , kxkc = r2 , we have by (2.28), kx(t)k ≥ δkxkc ≥ r¯2 , and

      Z 1 

1

(T x) 1 ≥ Ψ (T x) 1 , s Ψ ( f (s, x(s)))ds =λ Υ

2 2 2 0  Z 1−δ  1 ≥λ Υ , s Ψ ( f (s, x(s)))ds 2 δ  Z 1−δ  1 , s kx(s)kds ≥ λε1 Υ 2 δ  Z 1−δ  1 ≥ λε1 δkxkc Υ , s ds 2 δ > kxkc , i.e., x ∈ K , kxkc = r2 , imply kT xkc > r2 . Thus, Lemma 1.2 implies i(T, K r2 , K ) = 0. Similarly, since(Ψ f )0 = ∞ there exists r¯3 with 0 < r¯3 < r1 such that Ψ ( f (t, x)) ≥ ε2 kxk for t ∈ J, x ∈ R 1−δ P, kxk ≤ r¯3 , where ε2 > 0 satisfies ε2 λγ δ Υ ( 21 , s)ds > 1. Let 0 < r3 < r¯3 . Then for x ∈ K , kxkc = r3 , we have kT xkc > r3 . Thus, Lemma 1.2 implies i(T, K r3 , K ) = 0, too. Applying the additivity of the fixed point index, we have i(T, K r2 / K¯ r1 , K ) = −1,

i(T, K r1 / K¯ r3 , K ) = 1.

Thus, T has two fixed points in K r2 / K¯ r1 and K r1 / K¯ r3 , respectively, which are the desired positive solutions for problem (1.1). Our next result corresponds to the case when (1.1) has no positive solutions for λ sufficiently large. we note by (H2 ), there exists a function m ∈ C(J, [0, ∞)) which does not vanish identically on any subset of positive measure such that Ψ ( f (t, x)) ≥ m(t)kxk for all t ∈ J, kxk ≥ 0. Let x ∈ C[J, E] be a positive solution for problem (1.1), one has x ∈ K . Now choose λ large enough so that Z 1−δ λδ Υ (t, s)m(s)ds > 1. δ

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Thus, we have kxkc ≥ kx(t)k ≥ Ψ (x(t)) = λ

Z

≥λ

Z

1

Υ (t, s)Ψ ( f (s, x(s)))ds

0 1−δ δ

≥ λδkxkc > kxkc ,

Υ (t, s)m(s)kx(s)kds Z δ

1−δ

Υ (t, s)m(s)ds

which is a contradiction and the proof is completed.



Theorem 3.2. Assume (H1 ), (H3 ), P is normal and for any x ∈ P, f 0 = f ∞ = 0. Then, for λ sufficiently large, problem (1.1) has at least two positive solutions, whereas for λ sufficiently small, problem (1.1) has no positive solution. Proof. By (H3 ), there exists a number r4 = dδ > 0 such that   Z 1−δ  Z 1−δ  1 1 ∗ , s Ψ ( f (s, x(s)))ds ≥ , s n(s)ds > 0. F (r4 ) = min Υ Υ x∈K ,kxkc =r4 δ 2 2 δ Let λ2 = F ∗r(r4 4 ) . Then for λ ∈ (λ2 , ∞) and x ∈ K , kxkc = r4 , we have

      Z 1 

1

(T x) 1 ≥ Ψ (T x) 1 = λ , s Ψ ( f (s, x(s)))ds Υ

2 2 2 0  Z 1−δ  1 > λ2 Υ , s Ψ ( f (s, x(s)))ds 2 δ ≥ λ2 F ∗ (r4 ) = kxkc , i.e., x ∈ K , kxkc = r4 , imply kT xkc > r4 . Thus, Lemma 1.2 implies i(T, K r4 , K ) = 0. Considering f 0 = 0, there exists r¯5 with 0 < r¯5 < r4 such that k f (t, x(t))k ≤ ε3 kxk, for t ∈ J, x ∈ P, kxk ≤ r¯5 , 1 where ε3 > 0 satisfies ε3 λ 16(1−µ)(1−υ) < 1. Let 0 < r5 < r¯5 . Then, for t ∈ Jδ , x ∈ K , kxkc = r5 , we have Z 1 1 k f (s, x(s))kds k(T x)(t)k ≤ λ 16(1 − µ)(1 − υ) 0 Z 1 1 kx(s)kds ≤ λε3 16(1 − µ)(1 − υ) 0 1 kxkc ≤ λε3 16(1 − µ)(1 − υ) < kxkc , i.e., x ∈ K , kxkc = r5 , imply kT xkc < r5 . Thus, Lemma 1.1 implies i(T, K r5 , K ) = 1. Next, turning to f ∞ = 0, there exists r¯6 with r¯6 > r4 such that k f (t, x)k ≤ ε4 kxk, for x ∈ P, kxk ≥ r¯6 , t ∈ J , 1 < 12 . where ε4 > 0 satisfies ε4 λ 16(1−µ)(1−υ) Let Z 1 1 k f (t, x(t))kdt. M= λ sup 16(1 − µ)(1 − υ) x∈K ,kxkc =¯r6 ,t∈J 0 It is not difficult to see that M < +∞. Choosing r6 > max{r4 , r¯6 , 2M}, then we get M < 21 r6 . Now, we choose x ∈ ∂ K r6 arbitrarily. Letting kx(t)k ¯ = min{kx(t)k, r¯6 }, then kx(t)k ¯ ≤ r¯6 . In addition, write 4(x) = {t ∈ J : kx(t)k > r¯6 }. Therefore, for t ∈ 4(x), we get r¯6 < kx(t)k ≤ kxkc = r6 . By the choice of r¯6 , for t ∈ 4(x), we have k f (t, x(t))k ≤ ε4r6 .

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Thus for x ∈ K , kxkc = r6 , we have Z 1 1 k f (s, x(s))kds 16(1 − µ)(1 − υ) 0 Z Z 1 1 k f (s, x(s))kds + λ k f (s, x(s))kds λ 16(1 − µ)(1 − υ) 4(x) 16(1 − µ)(1 − υ) J \4(x) Z Z 1 1 1 k f (s, x(s))kds + λ k f (s, x(s))kds ¯ λ 16(1 − µ)(1 − υ) 4(x) 16(1 − µ)(1 − υ) 0 1 λ ε4 r 6 + M 16(1 − µ)(1 − υ) 1 1 r6 + r6 2 2 r6 = kxkc ,

k(T x)(t)k ≤ λ = ≤ ≤ < =

i.e., x ∈ K , kxkc = r6 imply kT xkc < kxkc . Thus, Lemma 1.1 implies i(T, K r6 , K ) = 1. Applying the additivity of the fixed point index, we have i(T, K r6 / K¯ r4 , K ) = 1,

i(T, K r4 / K¯ r5 , K ) = −1.

Thus, T has two fixed points in K r6 / K¯ r4 and K r4 / K¯ r5 , respectively, which are the desired positive solutions for problem (1.1). Our next result corresponds to the case when (1.1) has no positive solutions for λ small enough. we note by f 0 = f ∞ = 0, there exists ε > 0 such that k f (t, x)k ≤ εkxk for all t ∈ J, x ∈ P, kxk ≥ 0. Let x ∈ C[J, E] be a positive solution for problem (1.1), then x ∈ K . Now choose λ small enough such that λε

1 < 1. 16(1 − µ)(1 − υ)

Thus, we have kxkc = ≤ ≤ <

1 max kx(t)k ≤ λ t∈J 16(1 − µ)(1 − υ) Z 1 1 λ ε kx(s)kds 16(1 − µ)(1 − υ) 0 1 λ εkxkc 16(1 − µ)(1 − υ) kxkc ,

1

Z

k f (s, x(s))kds

0

which is a contradiction and the proof is completed.



To illustrate how our main results can be used in practice we present an example. Example 3.1. Consider the following boundary-value problem of finite system of scalar differential equations ( p 1 3 xn(4) (t) = λ t 2 + 1xn3 tanh xn+1 , t ∈ J, (3.1) xn (0) = xn (1) = xn00 (0) = xn00 (1) = 0, where xn+m = xn (n = 1, 2 . . . , m). Conclusion. For λ sufficiently large, finite system (3.1) has at least two positive solutions, whereas for λ sufficiently small, finite system (3.1) has no positive solution. Proof. Let E = R m = {x = (x1 , x2 , . . . , xm ) : xn ∈ R, n = 1, 2, . . . , m} with the norm kxk = max1≤n≤m |xn |, and P = {x = (x1 , x2 , . . . , xm ) : xn ≥ 0, n = 1, 2, . . . , m}. Then P is a normal cone in E and system (3.1) can

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be translated into the form of (1.1) in E. In this situation, g(t) = h(t) ≡ 0, ∀t ∈ J, x = (x1 , x2 , . . . , xm ), f = ( f 1 , f 2 , . . . , f m ) and f n (n = 1, 2, . . . , m) is defined by p 1 3 f n (t, x) = t 2 + 1xn3 tanh xn+1 . On the other hand, we can get that P ∗ = P. If we choose Ψ = (1, 1, . . . , 1), then for any x ∈ P we have Ψ ( f (t, x)) =

m X

f n (t, x1 , x2 , . . . , xm ).

n=1

It is clear, (H1 ) is satisfied automatically when E is finite dimensional (here, E = R m ). Since α( f (t, D)) is identical to zero for any t ∈ J and D ⊂ P ∩ Tl , l > 0 (see Theorem 1.2.1 (i) in [13]). Now, we prove (H3 ) and f ∞ = f 0 = 0. Since g(t) = h(t) ≡ 0, we get Z 1 Υ (t, s) = G(t, ξ )G(ξ, s)dξ 0 2s − s 2 − t 2   t (1 − s) , 0 ≤ t ≤ s ≤ 1, 6 = 2 2   s(1 − t) 2t − t − s , 0 ≤ s ≤ t ≤ 1. 6 Noticing kxk ≤ Ψ (x) for x ∈ P, so we have  p  1 2d − 1 p 1 e 3 3 , Ψ ( f (t, x)) ≥ k f (t, x)k = t 2 + 1 max xn3 tanh xn+1 ≥ t 2 + 1d 3 2d 1≤n≤m e +1 for all t ∈ Jδ , x ∈ P, kxk ≥ d, where δ ∈ (0, 1), Jδ = [δ, 1 − δ], d > 0. Therefore the condition (H3 ) holds. In addition, f ∞ = lim sup max kxk→∞ t∈J

= lim sup max kxk→∞ t∈J

k f (t, x)k kxk  1  √ 3 2 3 t + 1 max xn tanh xn+1 1≤n≤m

max {xn }

1≤n≤m

= 0, and f 0 = lim sup max kxk→0 t∈J

= lim sup max kxk→0 t∈J

k f (t, x)k kxk  1  √ 3 2 t + 1 max xn3 tanh xn+1 1≤n≤m

max {xn }

1≤n≤m

= 0. Thus, our conclusion follows from Theorem 3.1, and the proof is complete.



Acknowledgement The authors thank the referee for his/her valuable comments and suggestions. References [1] J.M. Gallardo, Second order differential operators with integral boundary conditions and generation of semigroups, Rocky Mountain J. Math 30 (2000) 1265–1292.

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