Existence of solutions of three-point boundary value problems in Banach spaces

Mathematical and Computer Modelling 49 (2009) 780–788

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Existence of solutions of three-point boundary value problems in Banach spaces Haibo Chen a,∗ , Peiluan Li a,b a

Department of Mathematics, Central South University, Changsha 410075, PR China

b

Department of Mathematics and Physics, Henan University of Science and Technology, Luoyang 471003, PR China

article

a b s t r a c t

info

Article history: Received 16 December 2007 Received in revised form 15 April 2008 Accepted 7 May 2008

In this paper, we study the second-order three-point boundary value problem u00 (t ) + f (t , u(t ), u0 (t )) = θ , u (0) = θ, 0

0 < t < 1,

u(1) = α u(η)

in a Banach space E, where θ is the zero element of E , 0 < α < 1, 0 < η < 1. By using the Sadovskii fixed point theorem, we get the existence of at least one (positive) solution. As an application, we give an example to demonstrate our results. © 2008 Elsevier Ltd. All rights reserved.

Keywords: Boundary value problem Fixed point theorem Banach spaces Existence of (positive) solution

1. Introductions Over the past twenty years, the theory of ordinary differential equations in Banach spaces has become a new important branch (see, for e.g. [1–3] and references therein). On the other hand, the multi-point boundary value problems for scalar ordinary differential equations have been studied extensively (see, for instance, [4–7] and references therein). However, to the author’s knowledge, few results can be found in the literature concerning the multi-point boundary value problems in Banach spaces. In paper [8], Liu has discussed the existence of at least one solution of a two-point boundary value problem for second-order nonlinear ordinary differential equations in a Banach space. Being directly inspired by [8], in the present paper, by using the Sadovskii fixed point theorem, the authors consider the following three-point BVP : u00 (t ) + f (t , u(t ), u0 (t )) = θ , u (0) = θ , 0

0 < t < 1,

u(1) = α u(η)

(1.1) (1.2)

in a Banach space E, where θ is the zero element of E, I = [0, 1], 0 < α < 1, 0 < η < 1, f ∈ C [I × E × E , E ]. Our purpose here is to get some existence results for (positive) solutions. In scalar space, Liu [7] has solved successfully the existence of positive solutions of the following boundary value problem: u00 (t ) + a(t )f (u(t )) = 0, u (0) = 0, 0

t ∈ (0, 1),

u(1) = α u(η)

under the following assumptions: (A1) f ∈ C ([0, +∞), [0, +∞)); (A2) a(t ) ∈ C ([0, 1], [0, +∞)) and there exists x0 ∈ [η, 1] such that a(x0 ) > 0.

∗

Corresponding author. E-mail addresses: [email protected] (H. Chen), [email protected] (P. Li).

0895-7177/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2008.05.003

(1.3) (1.4)

H. Chen, P. Li / Mathematical and Computer Modelling 49 (2009) 780–788

781

But in Banach space, this BVP has not been discussed. In the present paper, by using the Sadovskii fixed point theorem, we obtain some existence results of solutions and positive solutions of the BVP (1.1)–(1.2). Let the real Banach space E with norm k·k be partially ordered by a cone P of E, i.e., x ≤ y if and only if y − x ∈ P. P ∗ denotes the dual cone of P. A cone is called normal if θ ≤ x ≤ y implies kxk ≤ N kyk (x, y ∈ P , N is a constant). N is called the normal constant. Let I = [0, 1]; Obviously, (C [I , E ], k·kc ) is a Banach space with kxkc = maxt ∈I kx(t )k. At the end of this section we state some lemmas which will be used in Sections 2 and 3. Lemma 1.1 ([2]). If H ⊂ C [I , E ] is bounded and equicontinuous, then α(H (t )) is continuous on I and

αc (H ) = max α(H (t ))

α

t ∈I

Z

x(t )dt : x ∈ H



Z

α(H (t ))dt

≤

I

I

where I = [A, B], H (t ) = {x(t ) : x ∈ H }, t ∈ I , αc (·), α(·) are the Kuratovski noncompactness measures of H (in C [I , E ]) and H (t ) (in E) (for details, see [3,10]). Lemma 1.2 ([9, Sadovskii]). Let D be a bounded, closed and convex subset of Banach space E. If operator A : D → D is condensing, then A has a fixed point in D. The paper is organized as follows. The preliminary lemmas are in Section 2. The main results are given in Section 3. Finally, in Section 4, we give one example to illustrate our results. 2. The preliminary lemmas In order to discuss the BVP (1.1)–(1.2), the preliminary lemmas are given in this Section. Now, we denote

 kx(t )k < +∞ t ∈I 1 + t  

kx(t )k DC 1 [I , E ] = x ∈ C 1 [I , E ] : sup < +∞ and sup x0 (t ) < +∞ t ∈I 1 + t t ∈I  Q = x ∈ DC 1 [I , E ] : x(t ) ≥ θ , θ is the zero element of E . FC [I , E ] =



x ∈ C [I , E ] : sup

Evidently, C 1 [I , E ] ⊂ C [I , E ], DC 1 [I , E ] ⊂ FC [I , E ], Q is a cone in DC 1 [I , E ]. It is easy to see that FC [I , E ] is a Banach space with norm kx(t )k kxkF = sup t ∈I 1 + t and DC 1 [I , E ] is a Banach space with norm

 kxkD = max kxkF , x0 C



where x0 = supt ∈I x0 (t ) . C

The basic space used in this paper is DC 1 [I , E ]. A map x ∈ C 2 [I , E ] ∩ DC 1 [I , E ] is called a solution of the BVP (1.1)–(1.2) if it satisfies Eqs. (1.1)–(1.2). A map x is called a positive solution of the BVP (1.1)–(1.2) if it satisfies Eqs. (1.1)–(1.2) and x ∈ Q ∩ C 2 [I , E ]. For a bounded subset V of Banach space E, let α(V ) be the Kuratovski noncompactness measure of V . In this paper, the Kuratovski measures of noncompactness of bounded set in E , C [I , E ], FC [I , E ] and DC 1 [I , E ] are denoted by αE (·), αC (·), αF (·), αD (·), respectively. Ra Let x(t ) : I → E be continuous, the abstract generalized integral 0 x(t )dt (a ∈ I ) can be defined similarly just as in the

Ra

scalar spaces and 0 x(t )dt ∈ E. The convergence and divergence of the abstract generalized integral are similar to that of the integral in scalar space. For convenience, let us list some conditions.

(H1 ) There exist nonnegative functions a, b, c ∈ C [0, 1] such that kf (t , x, y)k ≤ a(t ) kxk + b(t ) kyk + c (t ) ∀t ∈ I , x, y ∈ E . (H2 ) For any r > 0, [a, b] ∈ I , f (t , x, y) is uniformly continuous on [a, b] × BE (θ , r ) × BE (θ , r ), where θ is the zero element of E , BE (θ , r ) = {x ∈ E : kxk ≤ r }. (H3 ) There exist l1 , l2 ∈ L[0, +∞) such that αE (f (t , D1 , D2 )) ≤ l1 (t )αE (D1 ) + l2 (t )αE (D2 ) ∀t ∈ I , bounded sets D1 , D2 ∈ E and

 1+

1+α 1−α

1

Z

[(1 + t )l1 (t ) + l2 (t )]dt < 1. 0

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H. Chen, P. Li / Mathematical and Computer Modelling 49 (2009) 780–788

Lemma 2.1. Let α 6= 1 and (H1 ) be satisfied, the problem u00 (t ) + f (t , u(t ), u0 (t )) = θ , u0 (0) = θ ,

0 < t < 1,

u(1) = α u(η)

has a unique solution

α (t − s)f (s, u(s), u (s))ds − u(t ) = − 1 − α 0 Z 1 1 + (1 − s)f (s, u(s), u0 (s))ds. 1−α 0 Z

t

0

η

Z

(η − s)f (s, u(s), u0 (s))ds

0

Proof. The proof of this lemma is easy, so we omit it.

(2.1)



For u ∈ DC 1 [I , E ], we define an operator A by

(Au)(t ) = −

t

Z

(t − s)f (s, u(s), u0 (s))ds −

0

+

1−α

η

Z

(η − s)f (s, u(s), u0 (s))ds

0

1

Z

1

α 1−α

(1 − s)f (s, u(s), u0 (s))ds.

(2.2)

0

From (2.2) and (H1 ), we have

 Z 1

(Au)(t )



≤ 1+ 1+α

f (s, u(s), u0 (s)) ds

1+t 1−α 0  Z 1

  α+1 ≤ 1+ a(s) ku(s)k + b(s) u0 (s) + c (s) ds 1−α 0 Z 1   

ku(s)k α+1 (1 + s)a(s) + b(s) u0 (s) + c (s) ds ≤ 1+ 1−α 1+s 0   Z 1  Z 1 α+1 ≤ 1+ [(1 + s)a(s) + b(s)]ds · kukD + c (s)ds . 1−α 0 0

(2.3)

Then also by (H1 ), we know a(t ), b(t ), c (t ) ∈ C [0, 1], u(t ) ∈ DC 1 [I , E ], so a(t ), b(t ), c (t ) are bounded in I and kukD < +∞. Together with (2.3), we get that

(Au)(t )

(2.4)

1 + t < +∞. From (2.2), we also have

 Z t 0 Z t (Au)0 (t ) = − t f (s, u(s), u0 (s))ds − sf (s, u(s), u0 (s))ds 0 0 Z t =− f (s, u(s), u0 (s))ds. 0

So



(Au)0 (t ) ≤

1

Z



f (s, u(s), u0 (s))ds 0

Z ≤

1

[(1 + s)a(s) + b(s)]ds · kukD +

0

1

Z

c (s)ds



0

≤ +∞.

(2.5)

By (2.4) and (2.5) we know (Au)(t ) is well defined and (Au)(t ) ∈ DC [I , E ]. 1

Lemma 2.2. Let 0 < α < 1. If f ∈ C [I × E × E , E ] and f ≥ θ , then the unique solution u of the BVP (1.1)–(1.2) satisfies u(t ) ≥ θ , i.e. u ∈ Q .

H. Chen, P. Li / Mathematical and Computer Modelling 49 (2009) 780–788

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Proof. For any φ ∈ P ∗ , setting p(t ) = φ(u(t )), t ∈ I, from the fact p00 (t ) = −ϕ(f ) ≤ 0, we know that p(t ) is concave down on (0, 1) and p0 (t ) monotone decreasing. Thus p0 (t ) < p0 (0) = φ(u0 (0)) = 0 and p(t ) is a monotone decreasing function, this is p(t ) ≥ p(1)(t ∈ [0, 1]). So if u(1) ≥ θ , p(1) = φ(u(1)) ≥ 0, then p(t ) = φ(u(t )) ≥ 0 for t ∈ [0, 1]. That is φ(u(t )) ≥ 0, which implies u(t ) ≥ θ , t ∈ I because φ ∈ P ∗ is arbitrary. If u(1) < θ , then p(1) = φ(u(1)) < 0, p(η) = φ(u(η)) = φ( α1 u(1)) = α1 p(1) < 0, then p(1) = φ(u(1)) = φ(α u(η)) = α p(η) > p(η), this contradicts the fact that p(t ) is a monotone decreasing function. So u(1) ≥ θ and u(t ) ≥ θ , t ∈ I, i.e. u(t ) ∈ Q , t ∈ I .  Lemma 2.3. Suppose (H1 ) and (H2 ) are satisfied. Then A : DC 1 [I , E ] → DC 1 [I , E ] is continuous and bounded. Proof. First, by (2.4) and (2.5) we get (Au)(t ) ∈ DC 1 [I , E ] for any u ∈ DC 1 [I , E ], so A is bounded. Next we prove that A is continuous on DC 1 [I , E ]. Let {un }, {u} ⊂ DC 1 [I , E ] and kun − ukD → 0(n → +∞). Hence {un } is a bounded subset of DC 1 [I , E ].Thus, there exists R > 0 such that kun kD ≤ R for n ≥ 1. Taking the limit, we have kukD ≤ R. On the other hand, by (2.2) we have

 Z 1

(Aun )(t ) (Au)(t )

α+1

f (s, un (s), u0 (s)) − f (s, u(s), u0 (s)) ds.

≤ 1 − 1 + n

1+t

1+t 1+t 1−α 0 It follows from (H2 ) that ∀ε > 0, there exists N > 0 such that  −1 

f (s, un (s), u0 (s)) − f (s, u(s), u0 (s)) ≤ 1 + α + 1 ε, for n ≥ N , ∀t ∈ I . n 1−α Therefore, ∀ε > 0, for any t ∈ [0, 1] and n ≥ N, by (2.6) and (2.7) we know

(Aun )(t ) (Au)(t ) 1

1 + t − 1 + t ≤ 1 + t ε ≤ ε.

(2.6)

(2.7)

(2.8)

Similarly, ∀ε > 0, for any t ∈ [0, 1] and n ≥ N, we have



(Aun )0 (t ) − (Au)0 (t ) ≤ ε.

(2.9)

So, A is continuous from DC [I , E ] to DC [I , E ]. 1

1



Lemma 2.4. Let (H1 ) be satisfied, V be a bounded subset of DC 1 [I , E ], then Proof. In order to show

(AV )(t ) , (AV )0 (t ) 1+t

(AV )(t ) , 1+t

(AV )0 (t ) are equicontinuous on [0, 1].

are equicontinuous on [0, 1], we only need to prove the following conclusions:

(1) ∀ε > 0, there exists a δ1 > 0, such that for any u ∈ V , t1 , t2 ∈ [0, 1], |t1 − t2 | < δ2 ,



(Au)(t1 ) (Au)(t2 )

< ε. −

1+t 1 + t2 1 (2) ∀ε > 0, there exists a δ2 > 0, such that for any u ∈ V , t1 , t2 ∈ [0, 1], |t1 − t2 | < δ2 ,

(Au)0 (t1 ) − (Au)0 (t2 ) < ε.

(2.10)

(2.11)

(AV )(t )

Firstly we claimed that 1+t is equicontinuous. For u ∈ V , t1 , t2 ∈ I, we assume t1 < t2 , then by (2.2),we get



Z t2 Z t1

(Au)(t2 ) (Au)(t1 ) t2

t1 0 0

f (s, u(s), u (s))ds − f (s, u(s), u (s))ds

1 + t − 1 + t ≤ 1 + t

1 + t1 0 2 1 2 0

Z t2 Z t1

1

1 + sf (s, u(s), u0 (s))ds − sf (s, u(s), u0 (s))ds

1 + t

1 + t1 0 2 0 Z 1  

1+α 1 1

f (s, u(s), u0 (s)) ds + − 1−α 1 + t2 1 + t1 0 Z t1 Z t2



f (s, u(s), u0 (s))ds + t2

f (s, u(s), u0 (s))ds ≤ |t2 − t1 | 0

t1 t1

Z + |t2 − t1 |



f (s, u(s), u0 (s))ds +

0

 ≤ 2+  ≤ 4+

1+α



1−α 1+α 1−α

Z

t2



f (s, u(s), u0 (s))ds +

t1

(t2 − t1 )

1

Z



f (s, u(s), u0 (s)) ds + 2 0



(t2 − t1 ) max f (s, u(s), u0 (s)) . 0≤s≤1

Z

t2



1+α 1−α



Z |t2 − t1 |

1



f (s, u(s), u0 (s)) ds

0



f (s, u(s), u0 (s))ds

t1

(2.12)

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H. Chen, P. Li / Mathematical and Computer Modelling 49 (2009) 780–788

In fact, from (H1 ), we know: (1) f (s, u(s), u0 (s))ds ≤ [(1 + s)a(s) + b(s)] kukD + c (s). (2) Nonnegative functions a(t ), b(t ), c (t ) ∈ C [0, 1], so they are bounded in [0, 1].



Then, we let a(t ) ≤ M1 , b(t ) ≤ M2 , c (t ) ≤ M3 , for any t ∈ I , M = max{M1, M2 , M3 }. Since V is bounded, there exists M 0 > 0, which satisfies kukD ≤ M 0 , ∀u ∈ V . Together with (2.12), we get

 

(Au)(t2 ) (Au)(t1 )

≤ 4 + 1 + α (t2 − t1 )((2 + t )M · M 0 + M ) −

1+t 1 + t1 1−α 2   1+α ≤ 4+ (t2 − t1 )(3M · M 0 + M ). 1−α Let δ = {(4 + 11+α )(3M · M 0 + M )}−1 · 2ε . −α So for any u ∈ V , t1 , t2 ∈ I, t1 < t2 , when |t1 − t2 | < δ , we have



(Au)(t2 ) (Au)(t1 ) ε

≤ < ε. −

1+t 1 + t1 2 2 (AV )(t )

If t1 ≥ t2 , we can also get (2.10) similarly. So 1+t is equicontinuous on [0, 1]. Similarly, we can prove (2.11) and (AV )0 (t ) is equicontinuous on [0, 1]. To sum up, the proof of Lemma 2.4 is complete.  Lemma 2.5. Let (H1 ) be satisfied, V be a bounded subset of DC 1 [I , E ], then



αD (AV ) = max sup αE t ∈I



  (AV )(t ) 0 , sup αE ((AV ) (t )) . 1+t t ∈I

Proof. By Lemma 2.3, we know AV is bounded subset of DC 1 [I , E ]. So



d =: max sup αE



t ∈I

  (AV )(t ) , sup αE ((AV )0 (t )) < +∞. 1+t t ∈I

Firstly, we prove that αD (AV ) ≤ d. (AV )(t ) From Lemma 2.4, we know 1+t , (AV )0 (t ) are equicontinuous on [0, 1]. By Lemma 1.1, we know

   (AV )(t ) ≤ d, αF (AV ) = max αE t ∈I 1+t  αc ((AV )0 ) = max αE ((AV )0 (t )) ≤ d, t ∈I

Therefore, there exist V1 , V2 , . . . , Vn ⊂ V and W1 , . . . , Wm ⊂ V such that V =

n [

Vi =

m [

i =1

Wj

j =1

satisfying AV =

n [

AVi ,

diamF (AVi ) < d + ε, i = 1, . . . , n,

(2.13)

i =1

(AV )0 =

m [ (AWj )0 ,

diamc ((AWj )0 ) < d + ε, j = 1, . . . , m,

(2.14)

j =1

where diamF (·), diamc (·) denote the diameters of the bounded subsets of FC 1 [I , E ] and C [I , E ], respectively. At the same time, for any Au1 , Au2 ∈ AVi , by (2.13) we obtain



(Au1 )(t ) (Au2 )(t )

≤ d + ε. −

1+t 1+t

(2.15)

Similarly, for Au1 , Au2 ∈ AWj , we can get



(Au1 )0 (t ) − (Au2 )0 (t ) ≤ d + ε.

(2.16)

H. Chen, P. Li / Mathematical and Computer Modelling 49 (2009) 780–788

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Let Yij =: {Au ∈ AV : Au ∈ AVi , (Au)0 ∈ (AWj )0 }, i = 1, 2, . . . , n, j = 1, 2, . . . , m, By using (2.13)–(2.16) we can get diamF (Yij ) ≤ d + ε,

diamC (Yij0 ) ≤ d + ε,

this means diamD (Yij ) ≤ d + ε , i = 1, 2, . . . , n, j = 1, 2, . . . , m. Then it follows from

[

AV =

Yij

i=1,...,n, j=1,...,m,

that αD (AV ) ≤ d. On the other hand, for any ε > 0, there exist Ui ⊂ V , i = 1, . . . , k, such that k [

AV =

AUi

diamD (AUi ) ≤ αD (AV ) + ε.

and

i=1

Hence, for ∀t ∈ I , ∀u1 , u2 ∈ Ui , i = 1, . . . , k, we have

k(Au1 )(t ) − (Au2 )(t )k ≤ kAu1 − Au2 kF ≤ kAu1 − Au2 kD ≤ αD (AV ) + ε. 1+t Sk From the fact (AV )(t ) = i=1 (AUi )(t ) and (2.17), we have   (AV )(t ) ≤ αD (AV ) + ε ∀t ∈ I , αE 1+t

(2.17)

so sup αE



t ∈I

(AV )(t ) 1+t



≤ αD (AV ) + ε.

Since ε > 0 is arbitrary, we obtain sup αE



t ∈I

(AV )(t ) 1+t



≤ αD (AV ).

Similarly, it follows that sup αE ((AV )0 ) ≤ αD (AV ). t ∈I

Consequently, the proof of Lemma 2.5 is complete.



3. Main results The main result of this paper is as follows. Theorem 3.1. Let (H1 )–(H3 ) be satisfied. Then the BVP (1.1)–(1.2) has at least one solution belonging to DC 1 [I , E ]. Proof. We need to prove only the existence of a fixed point of operator A in DC 1 [I , E ]. Let R>

1

Z

c (t )dt ·

( 1+

0

1+α 1−α

−1

)−1

1

Z

[(1 + t )a(t ) + b(t )]dt

−

,

0

B =: BD (θ , R) = u ∈ DC 1 [I , E ]; kukD ≤ R .





In the following we proceed to show AB ⊂ B. In fact, for any u ∈ B, by (2.2) and (2.3) we know

 Z 1

(Au)(t )



≤ 1+ 1+α

f (s, u(s), u0 (s)) ds

1+t 1−α 0   Z 1  Z 1 α+1 ≤ 1+ [(1 + s)a(s) + b(s)]ds · kukD + c (s)ds 1−α 0 0 !)   (Z 1   −1 Z 1 α+1 1+α ≤ 1+ [(1 + s)a(s) + b(s)]ds · R + R 1+ − [(1 + s)a(s) + b(s)]ds 1−α 1−α 0 0 = R ∀t ∈ I .

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Similarly, we can get (Au)0 (t ) ≤ R. Thus, together with Lemma 2.3, AB ⊂ B follows.

Let Ω := COD (AB), i.e. Ω is the convex closure of AB in DC 1 [I , E ]. It is easy to see that Ω is a nonempty, bounded, convex and closed subset of B. (AB)(t ) By Lemma 2.4, we know that 1+t , (AB)0 (t ) are equicontinuous on [0, 1]. Together with the definition of Ω , we know (Ω )(t )

that 1+t , (Ω )0 (t ) are equicontinuous on [0, 1]. Now we are in a position to show A is a strict set contraction operator from Ω to Ω . First, since Ω ⊂ B and AB ⊂ Ω , we know A is an operator from Ω into Ω . Next from Lemma 2.3 we obtain A is a bounded and continuous operator from Ω to Ω . Finally, we prove that

αD (AV ) ≤ lαD (V ) for V ⊂ Ω , Z 1  1+α [l1 (s)(1 + s) + l2 (s)]ds < 1. where l = 1 + 1−α 0

(3.1)

In fact, by Lemma 2.5, we need to prove only that sup αE



t ∈I

(AV )(t ) 1+t



(AV )(t ) 1+t



≤ lαD (V ) and

sup αE ((AV )0 ) ≤ lαD (V ). t ∈I

Now we show sup αE t ∈I



≤ lαD (V ).

By (H2 ) and the definition of Ω , we know {f (s, u(s), u0 (s)) : u ∈ Ω } are equicontinuous on [0, 1]. Thus by virtue of Lemma 1.1 and condition (H3 ), we get

 Z t  
(AV )(t ) t −s ≤ αE f (s, u(s), u0 (s)) : u ∈ V ds 1+t 1+t 0 Z η Z 1 

α(η − s) (1 − s) + αE f (s, u(s)u0 (s)) : u ∈ V ds + αE f (s, u(s), u0 (s)) : u ∈ V ds 0 (1 + t )(1 − α) 0 (1 + t )(1 − α)  Z 1 
1+α ≤ 1+ αE f (s, u(s), u0 (s)) : u ∈ V ds 1−α 0  Z 1 1+α ≤ 1+ [l1 (s)αE (V (s)) + l2 (s)αE (V 0 (s))]ds 1−α 0 Z 1      V (s) 1+α (1 + s)l1 (s)αE + l2 (s)αE (V 0 (s)) ds ≤ 1+ 1−α 1+s 0 Z 1  1+α [(1 + s)l1 (s) + l2 (s)]dsαD (V ). ≤ 1+ 1−α 0

αE



Since t is arbitrary, we get sup αE t ∈I



(AV )(t ) 1+t



≤ lαD (V ).

(3.2)

Similarly, we can obtain sup αE ((AV )0 (t )) ≤ lαD (V ).

(3.3)

t ∈I

Hence, by using Lemma 2.5 and (3.1)–(3.3), we obtain that A is a strict set contraction operator from Ω to Ω . Obviously, A is condensing too. It follows from Lemma 1.2 that A has at least one fixed point in Ω , that is, the BVP (1.1)–(1.2) has at least one solution in DC 1 [I , E ].  By Theorem 3.1 and Lemma 2.2, we can immediately get the following corollary about the existence of positive solution of BVP (1.1)–(1.2). Corollary 3.1. Let (H1 )–(H3 ) be satisfied, f ∈ C [I × E × E , E ] and f ≥ θ , then the BVP (1.1)–(1.2) has at least one positive solution u(t ) ≥ θ , i.e. u ∈ Q . Remark 3.1. If f (t , u, u0 ) = f (t , u) in the BVP (1.1)–(1.2), we may use a similar method to study BVP (1.1)–(1.2) in basic space FC [I , E ] to obtain the same result as Theorem 3.1 under the assumptions (H1 )–(H3 ) (here b(t ) ≡ 0, l2 (s) ≡ 0). Moreover, the proof may be easier since we need not estimate the derivative term.

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4. Example Now we consider an example to illustrate our results. Example 4.1. Consider the boundary value problem in E = l∞ = {x = x1 , . . . , xi , . . . , sup |xi | < +∞} with norm kxk = sup |xi | . i

t + xi

x00i =

+

4(t 2 + 2t + 1)

x0i (0) = 0,

xi (1) =

e− t

 2 sin x2i+1 + x0i+2 ,

18 1 2

i

q

  1

xi

2

t ∈ [0, 1],

(4.1)

.

(4.2)

Then the BVP (4.1)–(4.2) can be regarded as a BVP of form (1.1)–(1.2) in E. In this situation, I = [0, 1], x = (x1 , . . . , xi , . . .) ∈ E , f = (f1 , . . . , fi , . . .), fi = gi + hi , in which gi =

t + xi 4(t 2 + 2t + 1)

,

hi =

e−t 18

q  2 sin x2i+1 + x0i+2 ,

t ∈ I , x = (xi ) ∈ l∞ .

It is clear that f ∈ C [I × E × E , E ], now we verify that (H1 )–(H3 ) hold.



f (t , x, x0 ) = sup fi (t , x, x0 ) i

 ≤

1 4(t 2 + 2t + 1)

+

1

 kx k +

18et

0

x 18et

+

1 4(t 2 + 2t + 1)

.

Hence (H1 ) is satisfied for a( t ) =

1 4(t 2 + 2t + 1)

+

1 18et

,

b(t ) =

1 18et

,

c (t ) =

1 4(t 2 + 2t + 1)

.

In addition, it is easy to see that (H2 ) is satisfied. On the other hand, for any t ∈ I and bounded subsets D1 , D2 ⊂ E, by (4.1), we know

αE (g (t , D1 , D2 )) = αE



t + xi 4(

t2

+ 2t + 1)

 : xi ∈ D1

=

αE ({t + xi : xi ∈ D1 }) 4(t + 1)(t + 1)

αE (D1 ) = 4(t + 1)(t + 1)

(4.3)

and

kx k + x 0 0 0 ≤ hi (t , x, x ) ≤ ∀ t ∈ I , x, x0 ∈ E . t 18e

Similarly to the proof of [1, Example 2.12], we can obtain

αE (h(t , D1 , D2 )) = 0 ∀t ∈ I , bounded sets D1 , D2 ⊂ E .

(4.4)

Combining (4.3) with (4.4), we get

αE (f (t , D1 , D2 )) ≤

αE (D1 ) 4(t + 1)2

,

so, (H3 ) is satisfied for l1 (t ) =

1 4(t + 1)2

,

l2 (t ) ≡ 0

and

 l=

1+

1+α 1−α

1

Z

[(1 + t )l1 (t ) + l2 (t )]dt = 4 0

1

Z 0



1 4(1 + t )



dt = ln 2 < 1.

Hence, our conclusion follows from Theorem 3.1 Acknowledgements The authors would like to thank the referees for their kind suggestions and comments on this paper.

788

H. Chen, P. Li / Mathematical and Computer Modelling 49 (2009) 780–788

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