DOI 10.1007/s10986-016-9311-6

Lithuanian Mathematical Journal, Vol. 56, No. 2, April, 2016, pp. 143–163

On higher-order nonlinear boundary value problems with nonlocal multipoint integral boundary conditions Bashir Ahmad, Ahmed Alsaedi, and Nada Al-Malki Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia (e-mail: [email protected]; [email protected]; [email protected]) Received April 13, 2015; revised August 13, 2015

Abstract. In this paper, we study the existence of solutions for nonlinear nth-order ordinary differential equations and inclusions with nonlocal multipoint integral boundary conditions. Fixed point theorems due to Schaefer and Banach are employed to prove the existence results for the single-valued case, whereas the existence of solutions for the multivalued problem is established by means of a nonlinear alternative for Kakutani maps and Covitz–Nadler fixed point theorem. The obtained results are well explained by examples. We extend our discussion to some new problems. MSC: 34B10, 34B15, 34A60 Keywords: differential equations, inclusions, nth-order, integral boundary conditions, nonlocal, fixed point

1 Introduction In this paper, we discuss some existence and uniqueness results for a boundary value problem of nth-order ordinary differential equations with nonlocal multipoint integral boundary conditions. Precisely, we consider the following problem: ⎧   (n) (t) = f t, u(t) , u t ∈ [0, 1], ⎪ ⎪ ⎨ ξ u(0) = δ 0 u(s) ds, u (0) = 0, u (0) = 0, . . . , u(n−2) (0) = 0, ⎪ ⎪  βi ⎩ 0 < ξ < β1 < β2 < · · · < βm < 1, αu(1) + βu (1) = m i=1 γi 0 u(s) ds,

(1.1)

where f : [0, 1] × R → R is a given continuous function, and α, β , γi , δ, ξ , βi (i = 1, 2, . . . , m) are appropriately chosen real constants. Nonlinear boundary value problems constitute an important field of research due to their continuous development and extensive applications in diverse disciplines such as one-dimensional hyperbolic conservation laws [12], nano boundary layer fluid flows [38], mathematical physics [7], and so on. The primary interest in the study of nonlinear boundary value problems is to examine the effect of nonlinearity on the solutions of the given problem. The available literature on the topic ranges from theoretical aspects of existence and c 2016 Springer Science+Business Media New York 0363-1672/16/5602-0143 

143

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uniqueness of solutions to analytic and numerical methods of solution for boundary value problems. Classical boundary conditions cannot take into account some peculiarities of physical, chemical, or other processes happening inside the domain. This gives rise to the consideration of nonlocal conditions that connect the boundary values of the unknown function to its values at some interior positions of the domain. For details and development of nonlocal nonlinear boundary value problems, we refer the reader to the papers [1, 3, 8, 9, 10, 15, 18, 19, 20, 21, 23, 24, 25, 28, 36, 39, 40, 42, 43, 44, 45, 49] and the references therein. Computational fluid dynamics (CFD) methods possess the potential to augment the data obtained from in vitro methods by providing a complete characterization of hemodynamic conditions (blood velocity and pressure as functions of space and time) under precisely controlled conditions. It has been observed that CFD studies of blood flows are directly related to the boundary data. Taylor et al. [41] assumed a very long circular vessel geometry upstream the inlet section so that the analytic solution of Womersley [46] can be prescribed. However, it is not always justified to assume a circular cross-section. To overcome this problem, an effective approach is to consider integral boundary conditions [37]. The validity of this approach is approved by computing both steady and pulsated channel flows for Womersley numbers up to 15. For applications of integral boundary conditions to thermal conduction, semiconductor, and hydrodynamic problems, we respectively refer the reader to the works [13, 14, 30]. As a matter of fact, integral boundary conditions have extensive applications in applied sciences such as blood flow problems, chemical engineering, thermoelasticity, underground water flow, population dynamics, and so on. Further details of boundary value problems involving integral boundary conditions can be found in a series of papers [2, 4, 5, 6, 11, 26, 29, 31, 32, 35, 47, 48, 50] and the references therein. ξ The main objective of this paper is to analyze the effect of the nonlocal integral condition u(0) = δ 0 u(s) ds on the solution of problem (1.1). We recall that problem (1.1) with u(0) = 0 was considered in [5,6]. Thus, the work presented in [5, 6]. Besides, we study the case  ξ in this paper generalizes and extends the results obtained ξ u(0) = δ 0 u (s) ds replacing the boundary condition u(0) = δ 0 u(s) ds in problem (1.1). Furthermore, we outline a uniqueness result for an nth-order integro-differential equation. The paper is organized as follows. In Section 2, we prove a fundamental lemma that is essential for the sequel. In Section 3, we present two existence results for the single-valued problem (1.1); the first one deals with the existence of solutions and relies on Schaefer’s fixed point theorem, and the second one shows the uniqueness of solutions and is based on Banach’s principle for contractive maps. Section 4 contains the existence results for the multivalued case of problem (1.1). In Section 5, we consider some new problems companion to (1.1) and outline a strategy for proving the existence results for these problems.

2 Preliminary result The following lemma plays a key role in defining the solution for problem (1.1). Lemma 1. For any function y ∈ C([0, 1], R), the linear problem ⎧ u(n) (t) = y(t), t ∈ [0, 1], ⎪ ⎪ ⎨ ξ u(0) = δ 0 u(s) ds, u (0) = 0, u (0) = 0, . . . , u(n−2) (0) = 0, ⎪ ⎪  βi ⎩ 0 < ξ < β1 < β2 < · · · < βm < 1, αu(1) + βu (1) = m i=1 γi 0 u(s) ds,

(2.1)

is equivalent to the integral equation

t u(t) = 0

 m βi (βi − s)n (t − s)n−1 y(s) ds + τ1 (t) y(s) ds γi (n − 1)! n! i=1

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n y(s) ds + τ2 (t) y(s) ds, (n − 1)! n! 0

(2.2)

On higher-order nonlinear boundary value problems

where

145

  1 δξ n n−1 τ1 (t) = +t (1 − δξ) , ρ n    m m δ γi βin n−1 τ2 (t) = α + β(n − 1) − −t α− γi βi , ρ n i=1 i=1     m m n γi βi δξ n = 0. ρ = α + (n − 1)β − (1 − δξ) + α − γi βi n n i=1

(2.3) (2.4)

(2.5)

i=1

Proof. It is well known that the solution of the differential equation in (2.1) can be written as

t u(t) = 0

(t − s)n−1 y(s) ds + c0 + c1 t + c2 t2 + · · · + cn−2 tn−2 + cn−1 tn−1 , (n − 1)!

(2.6)

where ci (i = 0, 1, . . . n − 1) are arbitrary real constants. Using the boundary conditions u (0) = 0, u (0) = 0, . . . , u(n−2) (0) = 0 in (2.6), we find that c1 = c2 = · · · = cn−2 = 0, and consequently, (2.6) takes the form

t u(t) = 0

(t − s)n−1 y(s) ds + c0 + cn−1 tn−1 . (n − 1)!

(2.7)

Next, using the remaining boundary conditions

ξ u(0) = δ

u(s) ds

and αu(1) + βu (1) =

m

βi γi

i=1

0

u(s) ds 0

in (2.7), we get A1 c0 − A2 cn−1 = A3

and E1 c0 + E2 cn−1 = E3 ,

(2.8)

where δξ n A2 = , n

A1 = 1 − δξ, E1 = α −

m

ξ A3 = δ 0

E2 = α + (n − 1)β −

γi βi ,

i=1

E3 =

m i=1

βi γi 0

(βi − s)n y(s) ds − β n!

(ξ − s)n y(s) ds, n! m γi β n i

i=1

1 0

(1 − s)n−2 y(s) ds − α (n − 2)!

1 0

n

,

(1 − s)n−1 y(s) ds. (n − 1)!

Solving system (2.8) for c0 and cn−1 , we obtain c0 =

1 (E2 A3 + E3 A2 ), ρ

cn−1 =

1 (A1 E3 − E1 A3 ), ρ

where ρ is given by (2.5). Substituting the values of c0 , cn−1 into (2.7) and using (2.3) and (2.4), we obtain the  solution (2.2). This completes the proof.  Lith. Math. J., 56(2):143–163, 2016.

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In view of Lemma 1, we define the operator G : C([0, 1], R) → C([0, 1], R) by

t (Gu)(t) = 0

 m βi (βi − s)n    (t − s)n−1  γi f s, u(s) ds + τ1 (t) f s, u(s) ds (n − 1)! n! i=1

1 − 0

0

 (1 − s)n−2 [β(n − 1) + α(1 − s)]  f s, u(s) ds (n − 1)!

ξ

 (ξ − s)n  f s, u(s) ds, n!

+ τ2 (t) 0

(2.9)

where τ1 (t) and τ2 (t) are respectively given by (2.3) and (2.4). Observe that problem (1.1) has solutions only if the operator equation Gu = u has fixed points.

3

Existence and uniqueness results

Let C([0, 1], R) denote the Banach space of all continuous functions from [0, 1] into R endowed with the norm u = supt∈[0,1] |u(t)|. For convenience, we set  m

γi β n+1 1 (βn + α) ξ n+1 i Q= + b1 + + b2 , (3.1) n! (n + 1)! n! (n + 1)! i=1

where b1 = maxt∈[0,1] |τ1 (t)| and b2 = maxt∈[0,1] |τ2 (t)|. Now we are in a position to present the main results of this paper. Our first result is based on Schaefer’s fixed point theorem. Theorem 1. Assume that (A1) the function f : [0, 1] × R → R is continuous; (A2) there exists a positive constant M such that |f (t, u)|  M for all t ∈ [0, 1] and u ∈ R. Then problem (1.1) has at least one solution on [0, 1]. Proof. In order to show that the operator G defined by (2.9) has a fixed point, we use Schaefer’s fixed point theorem. The proof will be given in several steps. Step 1: G is continuous. Let {uk } be a sequence such that uk → u in C([0, 1], R). Then for each t ∈ [0, 1], we have   G(uk )(t) − G(u)(t) 

t 0

   (t − s)n−1   f s, uk (s) − f s, u(s)  ds (n − 1)!

 m

βi      (βi − s)n   γi f s, uk (s) − f s, u(s)  ds + τ1 (t) n! i=1

1 + 0

0

   (1 − s)n−2 [β(n − 1) + α(1 − s)]   f s, uk (s) − f s, u(s)  ds (n − 1)!

On higher-order nonlinear boundary value problems

  + τ2 (t)

ξ 0

 t

   (ξ − s)n   f s, uk (s) − f s, u(s)  ds n!    (t − s)n−1   f s, uk (s) − f s, u(s)  ds (n − 1)!

 sup t∈[0,1]

147

0

 m

βi      (βi − s)n   γi f s, uk (s) − f s, u(s)  ds + τ1 (t) n! i=1

1 + 0

0

   (1 − s)n−2 [β(n − 1) + α(1 − s)]   f s, uk (s) − f s, u(s)  ds (n − 1)!

  + τ2 (t)

ξ 0

   (ξ − s)n   f s, uk (s) − f s, u(s)  ds n!

      f ·, uk (·) − f ·, u(·)  sup

t∈[0,1]

1 + 0

 t 0



 m

βi   (t − s)n−1 (βi − s)n ds + τ1 (t) ds γi (n − 1)! n! i=1

  (1 − s)n−2 [β(n − 1) + α(1 − s)] ds + τ2 (t) (n − 1)!

      Qf ·, uk (·) − f ·, u(·) ,

ξ 0

(ξ − s)n ds n!

where Q is given by (3.1). Using condition (A1), we get that        G(uk ) − G(u)  Qf ·, uk (·) − f ·, u(·)  → 0

0



as k → ∞.

This establishes the continuity of G . Step 2: G maps bounded sets into bounded sets in C([0, 1], R). Indeed, it suffices to show that for any η ∗ > 0, there exists a positive constant L such that for each u ∈ Bη∗ = {u ∈ C([0, 1], R): u  η ∗ }, we have G(u)  L. Using condition (A2), for each t ∈ [0, 1], we have   G(u)(t) 

t 0

 m

βi     (t − s)n−1   (βi − s)n   f s, u(s)  ds + τ1 (t) |γi | f s, u(s)  ds (n − 1)! n! i=1

1 + 0

0

 (1 − s)n−2 [β(n − 1) + α(1 − s)]   f s, u(s)  ds (n − 1)!

  + τ2 (t)

ξ 0

 (ξ − s)n   f s, u(s)  ds n!

 QM = L, where Q is given by (3.1). Taking the norm for t ∈ [0, 1], we get G(u)  L. Lith. Math. J., 56(2):143–163, 2016.

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Step 3: G maps bounded sets into equicontinuous sets of C([0, 1], R). Let t1 , t2 ∈ (0, 1), t1 < t2 , and Bη∗ be a bounded set in C([0, 1], R) as in Step 2. Then, for u ∈ Bη∗ , we obtain   G(u)(t2 ) − G(u)(t1 )

 t1   

t2   n−1 n−1 n−1     (t − s) (t − s) (t − s)   2 1 2  − f s, u(s) ds + f s, u(s) ds   (n − 1)! (n − 1)! (n − 1)! 0

t1

 

βi m 1   n−1   (βi − s)n    +  (1 − δξ) t2 − tn−1 |γ | f s, u(s)  ds i 1  ρ n! i=1

1 + 0

0

 (1 − s)n−2 [|β|(n − 1) + |α|(1 − s)]   f s, u(s)  ds (n − 1)!

    ξ m 1   (ξ − s)n       n−1 n−1 f s, u(s)  ds +  δ t2 − t1 α− γi βi  ρ   n! i=1

0

  t1  

t2  (t2 − s)n−1 (t1 − s)n−1 (t2 − s)n−1   M  − ds + ds   (n − 1)! (n − 1)! (n − 1)! 0

t1

  m

βi 1  n−1  (βi − s)n n−1  ds +  (1 − δξ) t2 − t1 |γ | i  ρ n! i=1

1 + 0

0

(1 − s)n−2 [|β|(n − 1) + |α|(1 − s)] ds (n − 1)!

   ξ  m 1   (ξ − s)n    +  δ tn−1 − tn−1 α− γi βi  ds 1 ρ 2  n! i=1

0

 m

γi β n+1      M |1 − δξ| (|β|n + |α|) M i tn−1 − tn−1  + 2(t2 − t1 )n + tn2 − tn1  +  2 1 n! |ρ| (n + 1)! n! i=1    m  ξ n+1      n−1 . + M δ tn−1 − t α − γ β i i  1  |ρ|(n + 1)!  2 i=1

Clearly, the right-hand side of this inequality does not depend on u and tends to zero as t1 → t2 . Thus, in view of Steps 1 to 3, by the Arzelá–Ascoli theorem we conclude that G : C([0, 1], R) → C([0, 1], R) is completely continuous. Step 4: A priori bounds. Now we show that the set ε = {u ∈ C([0, 1], R): u = λG(u) for some 0 < λ < 1} is bounded. Let u ∈ ε. Then u = λG(u) for some λ ∈ (0, 1). Thus, for each t ∈ [0, 1], we get  t u(t) = λ 0

 m βi (βi − s)n    (t − s)n−1  f s, u(s) ds + τ1 (t) f s, u(s) ds γi (n − 1)! n! i=1

0

On higher-order nonlinear boundary value problems

1 − 0

 (1 − s)n−2 [β(n − 1) + α(1 − s)]  f s, u(s) ds (n − 1)!

ξ

+ τ2 (t) 0

149

  (ξ − s)n  f s, u(s) ds . n!

By condition (A2) it is easy to show that u  G(u)  M Q, where Q is given by (3.1). This shows that the set ε is bounded. Hence, by Schaefer’s fixed point theorem the operator G has a fixed point, which is a solution of problem (1.1). This completes the proof.   Our next result is based on Banach’s fixed point theorem. Theorem 2. Assume that f : [0, 1] × R → R is a continuous function satisfying Lipschitz condition: |f (t, u) − f (t, v)|  L|u − v| for all u, v ∈ R, t ∈ [0, 1]. Then the boundary value problem (1.1) has a unique solution if LQ < 1, where Q is given by (3.1). Proof. Let us fix supt∈[0,1] |f (t, 0)| = M , and choose R  M Q/(1 − LQ), where Q is given by (3.1). Let us show that GBR ⊂ BR , where BR = {u ∈ C([0, 1], R): u  R}. For u ∈ BR and t ∈ [0, 1], we have  t

  (Gu)(t)  sup t∈[0,1]

 (t − s)n−1   f s, u(s)  ds (n − 1)!

0

 m

βi    (βi − s)n   γi f s, u(s)  ds + τ1 (t) n! i=1

1 + 0

0

 (1 − s)n−2 [β(n − 1) + α(1 − s)]   f s, u(s)  ds (n − 1)!

  + τ2 (t)

ξ 0

 t



    (t − s)n−1   f s, u(s) − f (s, 0) + f (s, 0) ds (n − 1)!

 sup t∈[0,1]

 (ξ − s)n   f s, u(s)  ds n!

0

 m

βi       (βi − s)n   γi f s, u(s) − f (s, 0) + f (s, 0) ds + τ1 (t) n! i=1

1 + 0

0

    (1 − s)n−2 [β(n − 1) + α(1 − s)]   f s, u(s) − f (s, 0) + f (s, 0) ds (n − 1)!

  + τ2 (t)

ξ 0

Lith. Math. J., 56(2):143–163, 2016.

    (ξ − s)n   f s, u(s) − f (s, 0) + f (s, 0) ds n!



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B. Ahmad, A. Alsaedi, and N. Al-Malki

 t  (LR + M ) sup t∈[0,1]

1 + 0

0

 m

βi   (t − s)n−1 (βi − s)n ds + τ1 (t) ds γi (n − 1)! n! i=1

  (1 − s)n−2 [β(n − 1) + α(1 − s)] ds + τ2 (t) (n − 1)!

0 ξ

0

(ξ − s)n ds n!



 Q(LR + M )  R, and thus Gu  R, that is, Gu ∈ BR . Consequently, GBR ⊂ BR . Now, for u, v ∈ C([0, 1], R) and t ∈ [0, 1], we obtain   (Gu)(t) − (Gv)(t)  t

   (t − s)n−1   f s, u(s) − f s, v(s)  ds (n − 1)!

 sup t∈[0,1]

0

 m

βi      (βi − s)n   γi f s, u(s) − f s, v(s)  ds + τ1 (t) n! i=1

1 + 0

0

   (1 − s)n−2 [β(n − 1) + α(1 − s)]   f s, u(s) − f s, v(s)  ds (n − 1)!

  + τ2 (t)

ξ 0

   (ξ − s)n   f s, u(s) − f s, v(s)  ds n!  t

 Lu − v sup t∈[0,1]

1 + 0

0



 m

βi   (t − s)n−1 (βi − s)n ds + τ1 (t) γi ds (n − 1)! n! i=1

  (1 − s)n−2 [β(n − 1) + α(1 − s)] ds + τ2 (t) (n − 1)!

0

ξ 0

 (ξ − s)n ds . n!

Taking the maximum for t ∈ [0, 1], we get   (Gu) − (Gv)  LQu − v, where Q is given by (3.1). Since LQ < 1, the operator G is a contraction. Hence, it follows by Banach’s  contraction mapping principle that problem (1.1) has a unique solution. This completes the proof.  Example 1. Consider the boundary value problem ⎧    (t) = f t, u(t) , u t ∈ [0, 1], ⎪ ⎪ ⎨ ξ u(0) = δ 0 u(s) ds, u (0) = 0, ⎪ ⎪ β ⎩ αu(1) + βu (1) = 4i=1 γi 0 i u(s) ds,

(3.2)

On higher-order nonlinear boundary value problems

151

where n = 3, α = 1, β = 1, δ = 1/2, m = 4, ξ = 1/4, β1 = 1/3, β2 = 1/2, β3 = 2/3, β4 = 3/4, γ1 = 3/4, γ2 = 2/3, γ3 = 1/2, γ4 = 1/3, and f (t, u) = √

  1 arctan u(t) + u + e−t . t + 16

Observe that |f (t, u) − f (t, v)|  |u − v|/2, which implies that L = 1/2. Using the given values, we find that 1/ρ  0.398735 (ρ is given by (2.5)) and  Q=

 m

 γi β n+1 1 βn + α ξ n+1 i + b1 + + b2  0.403871. n! (n + 1)! n! (n + 1)! i=1

Clearly, LQ  0.2019355 < 1. Thus, all the conditions of Theorem 2 are satisfied. Hence, by Theorem 2 problem (3.2) has a unique solution.

4

Multivalued case

In this section, we consider the multivalued case of problem (1.1) given by ⎧ u(n) (t) ∈ F (t, u(t)), t ∈ [0, 1], ⎪ ⎪ ⎨ ξ u(0) = δ 0 u(s) ds, u (0) = 0, u (0) = 0, . . . , u(n−2) (0) = 0, ⎪ ⎪  βi ⎩ αu(1) + βu (1) = m 0 < ξ < β1 < β2 < · · · < βm < 1, i=1 γi 0 u(s) ds,

(4.1)

where F : [0, 1] × R → P(R), and P(R) is the family of all nonempty subsets of R. Before proceeding to the main results of problem (4.1), we outline some background material [17, 27]. For a normed space (X , ·), we define Pcl (X ) = {Y ∈ P(X ): Y is closed}, P b (X ) = {Y ∈ P(X ): Y is bounded}, Pcp (X ) = {Y ∈ P(X ): Y is compact}, and Pcp,c (X ) = {Y ∈ P(X ): Y is compact and convex}. A multivalued map G : X → P(X ) is convex-valued (closed-valued) if G(a) is convex (closed) for all a ∈ X . The map G is bounded on bounded sets if G(B) = ∪x∈B G(x) is bounded in X for all B ∈ P b (X ) (i.e., supx∈B {sup{|y|: y ∈ G(x)}} < ∞). The map G is called upper semicontinuous (u.s.c.) on A if for each a0 ∈ X , the set G(a0 ) is a nonempty closed subset of X and if for each open set N of X containing G(a0 ), there exists an open neighborhood N0 of a0 such that G(N0 ) ⊆ N . The map G is said to be completely continuous if G(B) is relatively compact for every B ∈ P b (X ). The map G has a fixed point if there is a ∈ A such that a ∈ G(a). We denote by Fix G the set of fixed points of the multivalued operator G. A multivalued map G : J → Pcl (R) is said to be measurable if for every b ∈ R, the function t → d(b, G(t)) = inf{|b−c|: c ∈ G(t)} is measurable. We define the graph of G to be the set Gr G = {(x, y) ∈ X × Y, y ∈ G(x)} and recall the following relationship between closed graphs and upper-semicontinuity: If G : X → Pcl (X ) is u.s.c., then Gr G is a closed subset of X × Y , that is, for every sequence {xn }n∈N ⊂ X and {yn }n∈N ⊂ X , if xn → x∗ , yn → y∗ as n → ∞ and yn ∈ G(xn ), then y∗ ∈ G(x∗ ). Conversely, if G is completely continuous and has a closed graph, then it is upper semicontinuous. D EFINITION 1. A multivalued map F : [0, 1] × R → P(R) is said to be Carathéodory if (i) t → F (t, x) is measurable for each x ∈ R; (ii) x → F (t, x) is upper semicontinuous for almost all t ∈ [0, 1]; Further, a Carathéodory function F is called L1 -Carathéodory if Lith. Math. J., 56(2):143–163, 2016.

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(iii) for each α > 0, there exists ϕα ∈ L1 ([0, 1], R+ ) such that     F (t, x) = sup |v|: v ∈ F (t, x)  ϕα (t) for all x  α and for a.e. t ∈ [0, 1]. For each y ∈ C([0, 1], R), define the set of selections of F by       SF,y := v ∈ L1 [0, 1], R : v(t) ∈ F t, y(t) for a.e. t ∈ [0, 1] . We need the following known results to establish the first existence result for problem (4.1). Lemma 2 [Nonlinear alternative for Kakutani maps]. (See [22].) Let E be a Banach space, C a closed convex subset of E , U an open subset of C , and 0 ∈ U . Suppose that F : U → Pc,cv (C) is an upper semicontinuous compact map; here Pc,cv (C) denotes the family of nonempty compact convex subsets of C . Then either (i) F has a fixed point in U , or (ii) there is u ∈ ∂U and λ ∈ (0, 1) with u ∈ λF (u). Lemma 3. (See [34].) Let X be a Banach space. Let F : [0, 1] × R → Pcp,c (R) be an L1 -Carathéodory multivalued map, and let Θ be a linear continuous mapping from L1 ([0, 1], R) to C([0, 1], R). Then the operator      Θ ◦ SF : C [0, 1], R → Pcp,c C [0, 1], R ,

x → (Θ ◦ SF )(x) = Θ(SF,x ),

is a closed graph operator in C([0, 1], R) × C([0, 1], R). Now we present the first existence result for the multivalued problem (4.1). Theorem 3. Assume that (H1) F : [0, 1] × R → P(R) is Carathéodory and has compact and convex values; (H2) there exists a continuous nondecreasing function ψ : [0, ∞) → (0, ∞) and a function p ∈ C([0,1], R+) such that F (t, u)p := sup{|v|: v ∈ F (t, u)}  p(t)ψ(u) for all (t, u) ∈ [0, 1] × R; (H3) there exists a constant M > 0 such that M/(ψ(M )pQ) > 1, where Q is given by (3.1). Then problem (4.1) has at least one solution on [0, 1]. Proof. Define the operator ΩF : C([0, 1], R) → P(C([0, 1], R)) by 

t

ΩF (u) = h ∈ C([0, 1], R):

 + τ1 (t)

0 m i=1

ξ + τ2 (t) 0

for ν ∈ SF ,u .

h(t) =

βi γi 0

(t − s)n−1 ν(s) ds (n − 1)!

(βi − s)n ν(s) ds − n!

(ξ − s)n ν(s) ds, n!



1 0

(1 − s)n−2 [β(n − 1) + α(1 − s)] ν(s) ds (n − 1)!

On higher-order nonlinear boundary value problems

153

We will show that ΩF satisfies the assumptions of the nonlinear alternative of Leray–Schauder type. The proof consists of several steps. As a first step, we show that ΩF is convex for each u ∈ C([0, 1], R). To this end, let h1 , h2 ∈ ΩF (u). Then there exist ν1 , ν2 ∈ SF,u such that for each t ∈ [0, 1], i = 1, 2, we have  m βi (βi − s)n (t − s)n−1 νi (s) ds + τ1 (t) γi νi (s) ds (n − 1)! n!

t hi (t) =

i=1

0

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n νi (s) ds + τ2 (t) νi (s) ds. (n − 1)! n! 0

Let 0  ω  1. Then, for each t ∈ [0, 1], we have 

 ωh1 + (1 − ω)h2 (t)

t = 0

 (t − s)n−1  ων1 (s) + (1 − ω)ν2 (s) ds (n − 1)! 

+ τ1 (t)

m i=1

1

βi γi 0

 (βi − s)n  ων1 (s) + (1 − ω)ν2 (s) ds n!

 (1 − s)n−2 [β(n − 1) + α(1 − s)]  ων1 (s) + (1 − ω)ν2 (s) ds (n − 1)!

− 0

ξ + τ2 (t) 0

 (ξ − s)n  ων1 (s) + (1 − ω)ν2 (s) ds. n!

Since SF,u is convex (F has convex values), it follows that ωh1 + (1 − ω)h2 ∈ ΩF (u). Next, we show that ΩF maps bounded sets into bounded sets in C([0, 1], R). For a positive number ρ∗ , let ∗ Bρ = {u ∈ C([0, 1], R): u  ρ∗ } be a bounded set in C([0, 1], R). Then, for all h ∈ ΩF (u) and u ∈ Bρ∗ , there exists ν ∈ SF,u such that

t h(t) = 0

 m βi (βi − s)n (t − s)n−1 ν(s) ds + τ1 (t) ν(s) ds γi (n − 1)! n! i=1

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n ν(s) ds + τ2 (t) ν(s) ds. (n − 1)! n! 0

Then, for t ∈ [0, 1], we have   h(t) 

t 0

Lith. Math. J., 56(2):143–163, 2016.

 m

βi     (t − s)n−1  (βi − s)n  ν(s) ds + τ1 (t) γi ν(s) ds (n − 1)! n! i=1

0

154

B. Ahmad, A. Alsaedi, and N. Al-Malki

1 + 0

ξ    (ξ − s)n   (1 − s)n−2 [β(n − 1) + α(1 − s)]  ν(s) ds ν(s) ds + τ2 (t) (n − 1)! n!  t

   ψ x p

0

1 + 0

0



m (t − s)n−1 ds + τ1 (t) γi (n − 1)! i=1



βi 0

(βi − s)n ds n!

(1 − s)n−2 [β(n − 1) + α(1 − s)] ds + τ2 (t) (n − 1)!

ξ



  Qψ u p. As a consequence, we get

0

(ξ − s)n ds n!



  h  Qψ ρ∗  p.

Now we show that ΩF maps bounded sets into equicontinuous sets of C([0, 1], R). Let t1 , t2 ∈ (0, 1) with t1 < t2 and u ∈ Bρ∗ . For each h ∈ ΩF (u), we obtain   h(t2 ) − h(t1 )

 t2 

t1  (t − s)n−1  (t1 − s)n−1   2  ν(s) ds − ν(s) ds   (n − 1)! (n − 1)! 0

0

 

βi 1  m  n−1  (βi − s)n n−1   +  (1 − δξ) t2 − t1 γ ν(s) ds i  ρ n! i=1

1 + 0

0

(1 − s)n−2 [β(n − 1) + α(1 − s)] ν(s) ds (n − 1)!

  ξ  m  (ξ − s)n  1  n−1   +  δ t2 − tn−1 α− γi βi  ν(s) ds 1  ρ n! i=1

0

  t1  

t2   n−1 (t2 − s)n−1 (t1 − s)n−1 (t − s)   2 − p(s)ψ(ρ∗ ) ds  p(s)ψ(ρ∗ ) ds +   (n − 1)! (n − 1)! (n − 1)! 0

t1

  m

βi 1  n−1  (βi − s)n n−1  +  (1 − δξ) t2 − t1 γi p(s)ψ(ρ∗ ) ds  ρ n! i=1

1 + 0

0

(1 − s)n−2 [β(n − 1) + α(1 − s)] p(s)ψ(ρ∗ ) ds (n − 1)!

   ξ m 1   (ξ − s)n    n−1 p(s)ψ(ρ∗ ) ds. +  δ tn−1 − t α − γ β i i  1 ρ 2  n! i=1

0

On higher-order nonlinear boundary value problems

155

Obviously, the right-hand side of this inequality does not depend on u ∈ Bρ∗ and tends to zero as t2 − t1 → 0. Since ΩF satisfies all three assumptions, it follows by the Arzelá–Ascoli theorem that ΩF : C([0, 1], R) → P(C([0, 1]), R)) is completely continuous. In our next step, we show that ΩF has a closed graph. Let uk → u∗ , hk ∈ ΩF (uk ), and hk → h∗ . Then we need to show that h∗ ∈ ΩF (u∗ ). Associated with hk ∈ ΩF (uk ), there exist νk ∈ SF ,uk such that for each t ∈ [0, 1], we have  m βi (βi − s)n (t − s)n−1 νk (s) ds + τ1 (t) γi νk (s) ds (n − 1)! n!

t hk (t) =

i=1

0

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n νk (s) ds + τ2 (t) νk (s) ds. (n − 1)! n! 0

Then it suffices to show that there exists ν∗ ∈ SF ,u∗ such that for each t ∈ [0, 1],  m βi (βi − s)n (t − s)n−1 ν∗ (s) ds + τ1 (t) ν∗ (s) ds γi (n − 1)! n!

t h∗ (t) =

i=1

0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n ν∗ (s) ds + τ2 (t) ν∗ (s) ds. (n − 1)! n!

1 − 0

0

Let us consider the linear operator Θ : L1 ([0, 1], R) → C([0, 1], R) given by

t ν → Θ(ν)(t) =

 m βi (βi − s)n (t − s)n−1 ν(s) ds + τ1 (t) γi ν(s) ds (n − 1)! n! i=1

0

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n ν(s) ds + τ2 (t) ν(s) ds. (n − 1)! n! 0

Observe that   hk (t) − h∗ (t)  t  m βi  (t − s)n−1  (βi − s)n     = νk (u) − ν∗ (u) ds + τ1 (t) γi νk (u) − ν∗ (u) ds  (n − 1)! n! i=1

0

1 − 0

0

 (1 − s)n−2 [β(n − 1) + α(1 − s)]  νk (u) − ν∗ (u) ds (n − 1)!

ξ

+ τ2 (t) 0

   (ξ − s)n   νk (u) − ν∗ (u) ds → 0  n!

Lith. Math. J., 56(2):143–163, 2016.

as k → ∞.

156

B. Ahmad, A. Alsaedi, and N. Al-Malki

Thus, it follows by Lemma 3 that Θ ◦ SF is a closed graph operator. Further, we have hk (t) ∈ Θ(SF ,uk ). Since uk → u∗ , therefore, we have

t h∗ (t) = 0

 m βi (βi − s)n (t − s)n−1 ν∗ (s) ds + τ1 (t) γi ν∗ (s) ds (n − 1)! n! i=1

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n ν∗ (s) ds + τ2 (t) ν∗ (s) ds (n − 1)! n! 0

for some ν∗ ∈ SF ,u∗ . Finally, we show there exists an open set U ⊆ C([0, 1], R) with u ∈ / ΩF (u) for all σ ∈ (0, 1) and u ∈ ∂U . Let σ ∈ (0, 1) and u ∈ σΩF (u). Then there exists ν ∈ L1 ([0, 1], R) with ν ∈ SF ,u such that, for t ∈ [0, 1], we have

t u(t) = 0

 m βi (βi − s)n (t − s)n−1 ν(s) ds + τ1 (t) γi ν(s) ds (n − 1)! n! i=1

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n ν(s) ds + τ2 (t) ν(s) ds. (n − 1)! n! 0

As in the second step, we can easily obtain   u  Qψ u p, which means that

u  1. Qψ(u)p

In view of (H3), there exists M such that u = M . Let     U = u ∈ C [0, 1], R : u < M . ¯ → P(C([0, 1], R)) is upper semicontinuous and completely continuous. By Note that the operator ΩF : U the choice of U , there is no u ∈ ∂U such that u ∈ σΩF for some σ ∈ (0, 1). Consequently, by the nonlinear ¯ , which is a solution alternative of Leray–Schauder type (Lemma 2) we deduce that ΩF has a fixed point u ∈ U  of problem (4.1). This completes the proof. 

Now we show the existence of solutions for problem (4.1) by means of the Covitz–Nadler fixed point theorem. Lemma 4 [Covitz–Nadler lemma]. (See [16].) Let (X, d) be a complete metric space. If N : X → Pcl (X) is a contraction, then Fix N = ∅. Remark. Let (X, d) be the metric space induced from the normed space (X; ·). Consider Hd : P(X) × P(X) → R ∪ {∞} given by   Hd (A, B) = max sup d(a, B), sup d(A, b) , a∈A

b∈B

On higher-order nonlinear boundary value problems

157

where d(A, b) = inf a∈A d(a; b) and d(a, B) = inf b∈B d(a; b). Then (Pb,cl (X), Hd ) is a metric space, and (Pcl (X), Hd ) is a generalized metric space (see [33]). Theorem 4. Assume that the following conditions hold: (H4) F : [0, 1] × R → Pcp (R) is such that F (·, u) : [0, 1] → Pcp (R) is measurable for each u ∈ R; (H5) Hd (F (t, u), F (t, u¯))  m(t)|u − u ¯| for almost all t ∈ [0, 1] and u, u¯ ∈ R with m ∈ C([0, 1], R+ ) and d(0, F (t, 0))  m(t) for almost all t ∈ [0, 1]. Then the boundary value problem (4.1) has at least one solution on [0, 1] if mQ < 1, where Q is given by (3.1). Proof. We transform problem (4.1) into a fixed point problem. Consider the set-valued map ΩF : C([0, 1], R) → P(C([0, 1], R)) defined at the beginning of the proof of Theorem 3. It is clear that the fixed points of ΩF are solutions of problem (4.1). Note that, by assumption (H4), since the set-valued map F (·, u) is measurable, it admits a measurable selection f : [0, 1] → R. Moreover, from assumption (H5) we have     f (t)  m(t) + m(t)u(t), that is, f (·) ∈ C([0, 1], R). Therefore, the set SF,u is nonempty, and consequently ΩF (u) = ∅ for any u ∈ C([0, 1], R). Now let us show that the operator ΩF satisfies the assumptions of Lemma 4. To show that ΩF (u) ∈ Pcl (C(0, 1], R) for each u ∈ C([0, 1], R), let {uk }k0 ∈ ΩF (u) be such that uk → u (k → ∞) in C([0, 1], R). Then u ∈ C([0, 1], R), and there exist νk ∈ SF,uk such that, for each t ∈ [0, 1], we have

t uk (t) = 0

 m βi (βi − s)n (t − s)n−1 νk (s) ds + τ1 (t) γi νk (s) ds (n − 1)! n! i=1

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n νk (s) ds + τ2 (t) νk (s) ds. (n − 1)! n! 0

Since F takes compact values, we may pass to a subsequence (if necessary) to obtain that νk converges to ν in C([0, 1], R). Thus, ν ∈ SF,u , and for each t ∈ [0, 1],

t uk (t) → u(t) = 0

 m βi (βi − s)n (t − s)n−1 ν(s) ds + τ1 (t) ν(s) ds γi (n − 1)! n! i=1

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n ν(s) ds + τ2 (t) ν(s) ds. (n − 1)! n! 0

Thus, u ∈ ΩF , and hence ΩF (u) is closed. Next, let us show that ΩF is a contraction on C([0, 1], R), that is, there exists γ < 1 such that     Hd Ω(u), Ω(¯ u)  γu − u ¯ for all u, u ¯ ∈ C [0, 1], R . Lith. Math. J., 56(2):143–163, 2016.

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B. Ahmad, A. Alsaedi, and N. Al-Malki

Let u, u ¯ ∈ C([0, 1], R) and h1 ∈ ΩF (u). Then there exists ν1 (t) ∈ F (t, u(t)) such that, for each t ∈ [0, 1],

t h1 (t) = 0

 m βi (βi − s)n (t − s)n−1 γi ν1 (s) ds + τ1 (t) ν1 (s) ds (n − 1)! n! i=1

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n ν1 (s) ds + τ2 (t) ν1 (s) ds. (n − 1)! n! 0

By (H5) we have     Hd F (t, u), F (t, u¯)  m(t)u(t) − u ¯(t). So, there exists w ∈ F (t, u¯(t)) such that     ν1 (t) − w  m(t)u(t) − u ¯(t),

t ∈ [0, 1].

Define U : [0, 1] → P(R) by      U (t) = w ∈ R: ν1 (t) − w  m(t)u(t) − u ¯(t) . Since the multivalued operator U (t) ∩ F (t, u ¯(t)) is measurable, there exists a function ν2 (t) that is a measurable selection for U . So ν2 (t) ∈ F (t, u¯(t)), and for each t ∈ [0, 1], we have |ν1 (t)−ν2 (t)|  m(t)|u(t)− u ¯(t)|. For each t ∈ [0, 1], let us define

t h2 (t) = 0

 m βi (βi − s)n (t − s)n−1 ν2 (s) ds + τ1 (t) γi ν2 (s) ds (n − 1)! n! i=1

1 − 0

0

ξ (1 − s)n−2 [β(n − 1) + α(1 − s)] (ξ − s)n ν2 (s) ds + τ2 (t) ν2 (s) ds. (n − 1)! n! 0

Thus,   h1 (t) − h2 (t)

t  0

 m

βi     (t − s)n−1  (βi − s)n     ν1 (s) − ν2 (s) ds + τ1 (t) γi ν1 (s) − ν2 (s) ds (n − 1)! n! i=1

1 + 0

0

 (1 − s)n−2 [β(n − 1) + α(1 − s)]  ν1 (s) − ν2 (s) ds (n − 1)!

  + τ2 (t)

ξ 0

 (ξ − s)n  ν1 (s) − ν2 (s) ds, n!

which implies that h1 − h2   Qmu − u ¯.

On higher-order nonlinear boundary value problems

159

Similarly, interchanging the roles of u and u, we obtain   Hd ΩF (u), ΩF (¯ u)  γu − u ¯  Qmu − u ¯. Since Qm < 1, ΩF is a contraction. Thus, it follows by Lemma 4 that ΩF has a fixed point u, which is a solution of problem (4.1).   Example 2. Consider the problem ⎧   u (t) ∈ F t, u(t) , t ∈ [0, 1], ⎪ ⎪ ⎨ ξ u(0) = δ 0 u(s) ds, u (0) = 0, ⎪ ⎪ β ⎩ αu(1) + βu (1) = 4i=1 γi 0 i u(s) ds

(4.2)

with parameters as in Example 1; hence, 1/ρ  0.398735 and Q  0.403871 (from Example 1). (a) For illustration of Theorem 3, we choose 

 et |u|7 2t|u|3 1 F (t, u) = + 2, + . 2(|u|7 + 1) (|u|3 + 6) 2 For f ∈ F , we have



et |u|7 2t|u|3 1 |f |  max + 2, + 2(|u|7 + 1) (|x|3 + 6) 2

Thus,





 et  +2 , 2

u ∈ R, t ∈ [0, 1].

 t        F (t, u) := sup |y|: y ∈ F (t, u)  e + 2 = p(t)ψ u , P 2

u ∈ R,

with p(t) = [et /2 + 2], ψ(u) = 1. Further, using condition (H3), we find that M > M1 , where M1 = 1.356659. Clearly, all the conditions of Theorem 3 are satisfied. So there exists at least one solution of problem (4.2) (b) To illustrate Theorem 4, consider the multivalued map F : [0, 1] × R → P(R) given by   3 sin u 1 + . F (t, u) = 0, 2 (t + 4) 2

(4.3)

Then we have   sup |v|: v ∈ F (t, u) 

1 3 + (t2 + 4) 2

  and Hd F (t, u), F (t, u¯)  m(t)|u − u ¯|,

where m(t) = 3/(t2 + 4) and mQ < 1. Thus, by Theorem 4, problem (4.2) with F (t, u) given by (4.3) has at least one solution on [0, 1].

5

Some more problems

In this section, we discuss two problems that can be termed as companions to problem (1.1). Lith. Math. J., 56(2):143–163, 2016.

160

B. Ahmad, A. Alsaedi, and N. Al-Malki

ξ ξ (I) In problem (1.1), we replace the condition u(0) = δ 0 u(s) ds by u(0) = δ 0 u (s) ds. In this case, the operator GI : C([0, 1], R) → C([0, 1], R) analogous to that given by (2.9) is

t (GI u)(t) = 0

 m βi (βi − s)n    (t − s)n−1  f s, u(s) ds + τ1 (t) γi f s, u(s) ds (n − 1)! n! i=1

1 − 0

0

 (1 − s)n−2 [β(n − 1) + α(1 − s)]  f s, u(s) ds (n − 1)!

ξ

+ τ2 (t) 0

and

 QI =

 (ξ − s)n−1  f s, u(s) ds, (n − 1)!

(5.1)

 m

 (n+1) n 1  γi βi [βn + α] ξ b2 + b1 + + , n! (n + 1)! n! n!

(5.2)

i=1

where  b1 = maxt∈[0,1] |τ1 (t)| and b2 = maxt∈[0,1] |τ2 (t)|. Using the operator GI and the constant QI defined by (5.1) and (5.2), respectively, we can obtain the existence results for this problem similar to those established in Sections 3 and 4. (II) Consider the nth-order integro-differential boundary value problem ⎧   u(n) (t) = f t, u(t), (φu)(t) , t ∈ [0, 1], ⎪ ⎪ ⎨ ξ (5.3) u(0) = δ 0 u(s) ds, u (0) = 0, u (0) = 0, . . . , u(n−2) (0) = 0, ⎪ ⎪  ⎩ βi 0 < ξ < β1 , αu(1) + βu (1) = m i=1 γi 0 u(s) ds, where f : [0, 1] × R × R → R is a given continuous function, α, β , γi , δ, βi , ξ (i = 1, 2, . . . , m) are real t constants to be chosen appropriately, and for κ : [0, 1] × [0, 1] → [0, ∞), (φu)(t) = 0 κ(t, s)u(s) ds. Theorem 5 [Uniqueness result]. Assume that f : [0, 1] × R × R → R is a continuous function satisfying the Lipschitz condition: |f (t, u(t), (φu)(t)) − f (t, v(t), (φv)(t))|  L1 |u − v| + L2 |φu − φv| for all u, v ∈ R, t ∈ [0, 1], where L1 , L2 are Lipschitz constants. Then the boundary  t value problem (5.3) has a unique solution if Q(L1 + L2 κ0 ) < 1, where Q is given by (3.1), and supt∈[0,1] 0 |κ(t, s)| ds = κ0 . Proof. The proof is similar to that of Theorem 2, so we omit it.

 

The existence results for problem (5.3) analogous to those established in Sections 3 and 4 can be obtained in a similar manner. Acknowledgment. We thank the reviewers for their useful and constructive remarks that led to the improvement of the original manuscript.

References 1. A.R. Aftabizadeh, Existence and uniqueness theorems for fourth-order boundary value problems, J. Math. Anal. Appl., 116:415–426, 1986. 2. B. Ahmad and A. Alsaedi, Existence of approximate solutions of the forced Duffing equation with discontinuous type integral boundary conditions, Nonlinear Anal., Real World Appl., 10:358–367, 2009.

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