Open Access

Solvability of n th-order Lipschitz equations with nonlinear three-point boundary conditions

Boundary Value Problems20142014:239

https://doi.org/10.1186/s13661-014-0239-7

Received: 26 June 2014

Accepted: 3 November 2014

Published: 22 November 2014

Abstract

In this paper, we investigate the solvability of n th-order Lipschitz equations y ( n ) = f ( x , y , y , , y ( n 1 ) ) , x 1 x x 3 , with nonlinear three-point boundary conditions of the form k ( y ( x 2 ) , y ( x 2 ) , , y ( n 1 ) ( x 2 ) ; y ( x 1 ) , y ( x 1 ) , , y ( n 1 ) ( x 1 ) ) = 0 , g i ( y ( i ) ( x 2 ) , y ( i + 1 ) ( x 2 ) , , y ( n 1 ) ( x 2 ) ) = 0 , i = 0 , 1 , , n 3 , h ( y ( x 2 ) , y ( x 2 ) , , y ( n 1 ) ( x 2 ) ; y ( x 3 ) , y ( x 3 ) , , y ( n 1 ) ( x 3 ) ) = 0 , where n 3 , x 1 < x 2 < x 3 . By using the matching technique together with set-valued function theory, the existence and uniqueness of solutions for the problems are obtained. Meanwhile, as an application of our results, an example is given.

MSC: 34B10, 34B15.

Keywords

nth-order Lipschitz equationnonlinear three-point boundary value problemmatching methodexistenceuniqueness

1 Introduction

As is well known, the differential equations with right hand sides satisfying the Lipschitz conditions (Lipschitz equations for short) are important, and thus their solvability has attracted much attention from many researchers. Among a substantial number of works dealing with higher order Lipschitz equations with three-point boundary conditions, we mention [1]–[14] and references therein. Most of these results are obtained via applying control theory methods (Pontryagin maximum principle), matching methods, and topological degree methods etc. To the best of our knowledge, most of the three-point boundary conditions in the above mentioned references are limited to simple boundary conditions.

In 1973, Barr and Sherman [2] showed by the matching technique that the third-order three-point boundary value problem with α = β = 0 has a unique solution, under the following four conditions:
  1. (A)

    f ( x , y , y , y ) is continuous on [ x 1 , x 3 ] × R 3 ;

     
  2. (B)
    f ( x , y , y , y ) satisfies the monotonicity conditions, i.e., y 1 y 2 , z 1 < z 2 implies
    f ( x , y 1 , z 1 , w ) < f ( x , y 2 , z 2 , w ) on  ( x 1 , x 2 ] ,
     
and y 1 y 2 , z 1 < z 2 implies
f ( x , y 1 , z 1 , w ) < f ( x , y 2 , z 2 , w ) on  [ x 2 , x 3 ) ;
  1. (C)
    for any ( x , y 1 , z 1 , w 1 ) , ( x , y 2 , z 2 , w 2 ) [ x 1 , x 3 ] × R 3 ,
    | f ( x , y 1 , z 1 , w 1 ) f ( x , y 2 , z 2 , w 2 ) | L 0 | y 1 y 2 | + L 1 | z 1 z 2 | + L 2 | w 1 w 2 | ,
     

where L 0 , L 1 , and L 2 are nonnegative constants;

(D1) for each i = 1 , 2 ,
3 27 L 0 h i 3 + 1 3 L 1 h i 2 + L 2 h i < 1 ,

where h i = x i + 1 x i , i = 1 , 2 .

In 1978, Moorti and Garner [12] by using the matching technique showed that BVP () with α , β { 0 , 1 } and α + β 0 has a unique solution, under the conditions (A), (B), (C), and

(D2) for each i = 1 , 2 ,
1 3 L 0 h i 3 + 1 2 L 1 h i 2 + L 2 h i < 1 .

Since then, many authors improved the condition (D i ), i = 1 , 2 . For example, in [4], Das and Lalli proved that BVP () with α = β = 0 has a unique solution, under the conditions of (A), (B), (C), and

(D3) for each i = 1 , 2 ,
1 60 L 0 h i 3 + 1 6 L 1 h i 2 + 2 3 L 2 h i < 1 .

In [1], Agarwal showed that BVP () with α = β = 0 has a unique solution, under the conditions of (A), (B), (C), and

(D4) for each i = 1 , 2 ,
3 160 L 0 h i 3 + 33 320 L 1 h i 2 + 3 8 L 2 h i < 1 .

In [14], Piao and Shi generalized the above results. They not only generalized the simple three boundary conditions to the nonlinear boundary conditions, but also they weakened the monotonicity condition (B) and removed the restriction (D i ) on the length of the interval.

Recently, Pei and Chang [13] generalized the results of Piao and Shi [14].

The purpose of this paper is to study the solvability of n th-order Lipschitz equations with more general nonlinear three-point boundary conditions of the form ( n 3 )
y ( n ) = f ( x , y , y , , y ( n 1 ) ) , x 1 x x 3 ,
(1.1)
{ k ( y ( x 2 ) , y ( x 2 ) , , y ( n 1 ) ( x 2 ) ; y ( x 1 ) , y ( x 1 ) , , y ( n 1 ) ( x 1 ) ) = 0 , g i ( y ( i ) ( x 2 ) , y ( i + 1 ) ( x 2 ) , , y ( n 1 ) ( x 2 ) ) = 0 , i = 0 , 1 , , n 3 , h ( y ( x 2 ) , y ( x 2 ) , , y ( n 1 ) ( x 2 ) ; y ( x 3 ) , y ( x 3 ) , , y ( n 1 ) ( x 3 ) ) = 0 ,
(1.2)

where < x 1 < x 2 < x 3 < + .

The paper is organized as follows. In Section 2, as a preliminary, we state some useful results as regards the solvability for the n th-order Lipschitz equation with the nonlinear two-point boundary conditions and a lemma of the differential inequality for n th-order differential equations. In Section 3, by using the matching technique together with set-valued function theory and nested interval theorem, we establish the existence and uniqueness theorems of solutions for BVP (1.1), (1.2). Our results improve and generalize widely the results of [1], [2], [4], [12]–[14].

We remark that the matching technique used in this paper is different from the classical one. In fact, by using the classical matching technique to obtain a matching solution of a three-point boundary value problem, it needs usually four two-point boundary value problems and among them two two-point boundary value problems need to have unique solutions, the other two two-point boundary value problems need to have at most one solution. However, our matching technique needs only two two-point boundary value problems and each of them needs to have at least one solution. For more about the three-point boundary value problems, we refer the readers to the references [15]–[19], with matching techniques, and to [20]–[35], with other techniques.

Throughout this paper, we make the following assumptions:

( H ¯ 1 ) f ( x , y 0 , y 1 , , y n 1 ) is continuous on [ x 1 , x 3 ] × R n ;

( H ¯ 2 ) If x [ x 2 , x 3 ] and y i y ¯ i , i = 0 , 1 , , n 2 , then
f ( x , y 0 , y 1 , , y n 2 , y n 1 ) f ( x , y ¯ 0 , y ¯ 1 , , y ¯ n 2 , y n 1 ) .
Also if x [ x 1 , x 2 ] and ( 1 ) n + i y i ( 1 ) n + i y ¯ i , i = 0 , 1 , , n 2 , then
f ( x , y 0 , y 1 , , y n 2 , y n 1 ) f ( x , y ¯ 0 , y ¯ 1 , , y ¯ n 2 , y n 1 ) ;
( H ¯ 3 ) For any ( x , y 0 , y 1 , , y n 1 ) , ( x , y ¯ 0 , y ¯ 1 , , y ¯ n 1 ) [ x 1 , x 3 ] × R n ,
| f ( x , y 0 , y 1 , , y n 1 ) f ( x , y ¯ 0 , y ¯ 1 , , y ¯ n 1 ) | i = 0 n 1 L i | y i y ¯ i | ,

where L i , i = 0 , 1 , , n 1 , are nonnegative Lipschitz constants;

( H ¯ 4 ) g i ( y i , y i + 1 , , y n 1 ) , i = 0 , 1 , , n 3 , are continuously differentiable on R n i , g i y i δ > 0 , g i y j 0 , i = 0 , 1 , , n 3 , j = i + 1 , i + 2 , , n 1 , on R n i , and for any bounded set D i R n i 1 , i = 0 , 1 , , n 3 , the functions g i y j , j = i + 1 , i + 2 , , n 1 , are bounded on R × D i ;

( H ¯ 5 ) The functions h ( y 0 , y 1 , , y n 1 ; z 0 , z 1 , , z n 1 ) , k ( y 0 , y 1 , , y n 1 ; z 0 , z 1 , , z n 1 ) are continuously differentiable on R 2 n , and for each i = 0 , 1 , , n 1 , h y i 0 , h z i 0 , ( 1 ) n + i k y i 0 , ( 1 ) n + i k z i 0 on R 2 n ;

( H ¯ 6 ) i = 0 n 2 h z i δ > 0 , i = 0 n 1 ( 1 ) n + i k z i δ > 0 on R 2 n ;

( H ¯ 6 ) i = 0 n 1 h z i δ > 0 , i = 0 n 2 ( 1 ) n + i k z i δ > 0 on R 2 n ;

( H ¯ 7 ) h y n 1 + i = 0 n 1 h z i δ > 0 , h y n 2 + i = 0 n 2 h z i δ > 0 , k y n 1 + i = 0 n 1 ( 1 ) n + i k z i δ > 0 on R 2 n ;

( H ¯ 7 ) h y n 1 + i = 0 n 1 h z i δ > 0 , k y n 2 + i = 0 n 2 ( 1 ) n + i k z i δ > 0 , k y n 1 + i = 0 n 1 ( 1 ) n + i k z i δ > 0 on R 2 n .

In the above conditions, δ denotes a constant.

2 Preliminary results

In this section, we introduce some lemmas which will be useful in the proof of our main results.

Consider the following nonlinear two-point boundary value problems for the n th-order differential equation ( n 3 ):
y ( n ) = f ( x , y , y , , y ( n 1 ) ) , a x b ,
(2.1)
with nonlinear two-point boundary conditions
{ g i ( y ( i ) ( a ) , y ( i + 1 ) ( a ) , , y ( n 1 ) ( a ) ) = 0 , i = 0 , 1 , , n 2 , h ( y ( a ) , y ( a ) , , y ( n 1 ) ( a ) ; y ( b ) , y ( b ) , , y ( n 1 ) ( b ) ) = 0 ,
(2.2)

where < a < b < + .

Let us list the following conditions for convenience.

(H1) f ( x , y 0 , y 1 , , y n 1 ) is continuous on [ a , b ] × R n ;

(H2) for any ( x , y 0 , , y n 2 , y n 1 ) , ( x , y ¯ 0 , , y ¯ n 2 , y n 1 ) [ a , b ] × R n , if y i y ¯ i , i = 0 , 1 , , n 2 , then
f ( x , y 0 , , y n 2 , y n 1 ) f ( x , y ¯ 0 , , y ¯ n 2 , y n 1 ) ;
(H3) for any ( x , y 0 , y 1 , , y n 1 ) , ( x , y ¯ 0 , y ¯ 1 , , y ¯ n 1 ) [ a , b ] × R n ,
| f ( x , y 0 , y 1 , , y n 1 ) f ( x , y ¯ 0 , y ¯ 1 , , y ¯ n 1 ) | i = 0 n 1 L i | y i y ¯ i | ,

where L i , i = 0 , 1 , , n 1 , are nonnegative constants;

( H 3 ) for any ( x , y 0 , , y n 2 , y n 1 ) , ( x , y 0 , , y n 2 , y ¯ n 1 ) [ a , b ] × R n ,
| f ( x , y 0 , , y n 2 , y n 1 ) f ( x , y 0 , , y n 2 , y ¯ n 1 ) | L n 1 | y n 1 y ¯ n 1 | ,

where L n 1 is a nonnegative constant;

(H4) g i ( y i , y i + 1 , , y n 1 ) , i = 0 , 1 , , n 2 , are continuously differentiable on R n i and h ( y 0 , y 1 , , y n 1 ; z 0 , z 1 , , z n 1 ) is continuously differentiable on R 2 n ;

(H5) g i y i δ > 0 , i = 0 , 1 , , n 2 on R n i , g i y j 0 , i = 0 , 1 , , n 2 , j = i + 1 , i + 2 , , n 1 on R n i ;

( H 5 ) g i y i δ > 0 , i = 0 , 1 , , n 3 on R n i , g n 2 y n 2 0 on R 2 , g i y j 0 , i = 0 , 1 , , n 3 , j = i + 1 , i + 2 , , n 1 on R n i , g n 2 y n 1 δ on R 2 ;

(H6) h y i 0 , i = 0 , 1 , , n 1 on R 2 n ;

(H7) h z i 0 , i = 0 , 1 , , n 1 , i = 0 n 1 h z i δ > 0 on R 2 n ;

( H 7 ) h z i 0 , i = 0 , 1 , , n 1 , i = 0 n 2 h z i δ > 0 on R 2 n ;

(H8) h y i 0 , h z i 0 , i = 0 , 1 , , n 1 , h y n 1 + i = 0 n 1 h z i δ > 0 on R 2 n ;

( H 8 ) h y i 0 , h z i 0 , i = 0 , 1 , , n 1 , h y n 2 + i = 0 n 2 h z i δ > 0 on R 2 n .

In the above conditions, δ denotes a constant.

Now we recall the results [36] of the existence and uniqueness of solutions for BVP (2.1), (2.2) and a lemma for a differential inequality for differential equation (2.1) of the n th order.

Lemma 2.1

(See [36], Theorem 3.1])

Assume that (H1), (H2), (H3), (H4), (H5), and (H8) hold. Then BVP (2.1), (2.2) has at least one solution.

Lemma 2.2

(See [36], Theorem 3.2])

Assume that (H1), (H2), (H3), (H4), ( H 5 ), and ( H 8 ) hold. Then BVP (2.1), (2.2) has at least one solution.

Lemma 2.3

(See [36], Theorem 3.3])

Assume that (H1), (H2), (H3), (H4), (H5), (H6), and (H7) hold. Then BVP (2.1), (2.2) has exactly one solution.

Lemma 2.4

(See [36], Theorem 3.4])

Assume that (H1), (H2), (H3), (H4), ( H 5 ), (H6), and ( H 7 ) hold. Then BVP (2.1), (2.2) has exactly one solution.

Lemma 2.5

(See [36], Lemma 2.4])

Assume that (H1), (H2), and ( H 3 ) hold. Let ϕ 1 ( x ) , ϕ 2 ( x ) be solutions of the differential equation (2.1) on some interval [ a 1 , b 1 ) [ a , b ] satisfying
ϕ 1 ( i ) ( a 1 ) ϕ 2 ( i ) ( a 1 ) , i = 0 , 1 , , n 1 ,
and
ϕ 1 ( n 2 ) ( a 1 ) + ϕ 1 ( n 1 ) ( a 1 ) < ϕ 2 ( n 2 ) ( a 1 ) + ϕ 2 ( n 1 ) ( a 1 ) .

Then ϕ 1 ( n 1 ) ( x ) ϕ 2 ( n 1 ) ( x ) for x [ a 1 , b 1 ) .

3 Main results

In order to obtain the existence and uniqueness of solutions for BVP (1.1), (1.2) by using the matching technique, we need first to discuss the existence and uniqueness of solutions for the n th-order Lipschitz equation (1.1) with one of the following sets of two-point boundary conditions:
{ g i ( y ( i ) ( x 2 ) , y ( i + 1 ) ( x 2 ) , , y ( n 1 ) ( x 2 ) ) = 0 , i = 0 , 1 , , n 3 , y ( n 2 ) ( x 2 ) = μ , h ( y ( x 2 ) , y ( x 2 ) , , y ( n 1 ) ( x 2 ) ; y ( x 3 ) , y ( x 3 ) , , y ( n 1 ) ( x 3 ) ) = 0 ,
(3.1)
{ k ( y ( x 2 ) , y ( x 2 ) , , y ( n 1 ) ( x 2 ) ; y ( x 1 ) , y ( x 1 ) , , y ( n 1 ) ( x 1 ) ) = 0 , g i ( y ( i ) ( x 2 ) , y ( i + 1 ) ( x 2 ) , , y ( n 1 ) ( x 2 ) ) = 0 , i = 0 , 1 , , n 3 , y ( n 2 ) ( x 2 ) = μ ,
(3.2)
{ g i ( y ( i ) ( x 2 ) , y ( i + 1 ) ( x 2 ) , , y ( n 1 ) ( x 2 ) ) = 0 , i = 0 , 1 , , n 3 , y ( n 1 ) ( x 2 ) = μ , h ( y ( x 2 ) , y ( x 2 ) , , y ( n 1 ) ( x 2 ) ; y ( x 3 ) , y ( x 3 ) , , y ( n 1 ) ( x 3 ) ) = 0 ,
(3.3)

where μ R = ( , + ) .

Let x = t and y ( x ) = ( 1 ) n z ( t ) . Then BVP (1.1), (3.2) becomes an equivalent boundary value problem:
where
F ( t , y 0 , y 1 , , y n 1 ) = f ( t , ( 1 ) n y 0 , ( 1 ) n + 1 y 1 , , ( 1 ) 2 n 1 y n 1 ) , G i ( y i , y i + 1 , , y n 1 ) = g i ( ( 1 ) n + i y i , ( 1 ) n + i + 1 y i + 1 , , ( 1 ) 2 n 1 y n 1 ) , H ( y 0 , y 1 , , y n 1 ; z 0 , z 1 , , z n 1 ) = k ( ( 1 ) n y 0 , ( 1 ) n + 1 y 1 , , ( 1 ) 2 n 1 y n 1 ; ( 1 ) n z 0 , ( 1 ) n + 1 z 1 , , ( 1 ) 2 n 1 z n 1 ) .

This shows that BVP (1.1), (3.2) on the interval [ x 1 , x 2 ] can be transformed to the same type as BVP (1.1), (3.1) on the interval [ x 2 , x 1 ] .

Lemma 3.1

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 7 ) hold. Then each of BVP (1.1), (3.1), BVP (1.1), (3.2), and BVP (1.1), (3.3) has at least one solution.

Proof

It is easy to check that conditions ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 7 ) imply conditions (H1), (H2), (H3), (H4), (H5), and (H8) for BVP (1.1), (3.1) as well as conditions (H1), (H2), (H3), (H4), ( H 5 ), and ( H 8 ) for BVP (1.1), (3.3), respectively. Hence by Lemma 2.1 and 2.2, each of BVP (1.1), (3.1) and BVP (1.1), (3.3) has at least one solution.

Similarly, by Lemma 2.1 BVP (1.1′), (3.2′) has at least one solution. Hence BVP (1.1), (3.2) has at least one solution. □

Lemma 3.2

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 6 ) hold. Then each of BVP (1.1), (3.1), BVP (1.1), (3.2), and BVP (1.1), (3.3) has exactly one solution.

Proof

Similarly to the proof of Lemma 3.1 by Lemma 2.3 and 2.4, the lemma follows. □

In order to prove our main results, we introduce some concepts as follows.

Definition 3.1

A set-valued function T : R 2 R is said to be upper semi-continuous at μ 0 R if for any open set U with T ( μ 0 ) U , there exists a neighborhood V of μ 0 such that μ V T ( μ ) U .

Definition 3.2

Let I 1 and I 2 be subsets of .
  1. (1)

    If for any t 1 I 1 and t 2 I 2 , t 1 t 2 holds, then we denote I 1 I 2 and say that I 1 is not greater than I 2 .

     
  2. (2)

    If for any t 1 I 1 and t 2 I 2 , t 1 < t 2 holds, then we denote I 1 < I 2 and say that I 1 is less than I 2 .

     

Definition 3.3

  1. (1)
    Define a set-valued function T 1 : R 2 R by
    T 1 ( μ ) = S μ for any  μ R ,
     
where S μ = { y ( n 1 ) ( x 2 , μ ) : y ( x , μ )  are solutions of BVP (1.1), (3.1) } ;
  1. (2)
    Define a set-valued function T 2 : R 2 R by
    T 2 ( μ ) = J μ for any  μ R ,
     

where J μ = { y ( n 1 ) ( x 2 , μ ) : y ( x , μ )  are solutions of BVP (1.1), (3.2) } .

Lemma 3.3

  1. (1)
    Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ) and ( H ¯ 7 ) hold. If μ 1 < μ 2 , then
    T 1 ( μ 1 ) T 1 ( μ 2 ) , T 2 ( μ 1 ) T 2 ( μ 2 ) .
     
  1. (2)
    Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ) and ( H ¯ 6 ) hold. If μ 1 < μ 2 , then
    T 1 ( μ 1 ) > T 1 ( μ 2 ) .
     

Proof

  1. (1)
    Let us show first the inequality with respect to T 1 . To do this, we take any y 1 ( n 1 ) ( x 2 , μ 1 ) S μ 1 , y 2 ( n 1 ) ( x 2 , μ 2 ) S μ 2 . Suppose that y 1 ( n 1 ) ( x 2 , μ 1 ) y 2 ( n 1 ) ( x 2 , μ 2 ) is false, i.e., y 1 ( n 1 ) ( x 2 , μ 1 ) < y 2 ( n 1 ) ( x 2 , μ 2 ) . Then, for each i = 0 , 1 , , n 3 , from (3.1) we have by the mean value theorem
    0 = g i ( y 2 ( i ) ( x 2 , μ 2 ) , , y 2 ( n 1 ) ( x 2 , μ 2 ) ) g i ( y 1 ( i ) ( x 2 , μ 1 ) , , y 1 ( n 1 ) ( x 2 , μ 1 ) ) = g i y i ( y 2 ( i ) ( x 2 , μ 2 ) y 1 ( i ) ( x 2 , μ 1 ) ) + j = i + 1 n 1 g i y j ( y 2 ( j ) ( x 2 , μ 2 ) y 1 ( j ) ( x 2 , μ 1 ) ) ,
     
and y 1 ( n 2 ) ( x 2 , μ 1 ) = μ 1 < μ 2 = y 2 ( n 2 ) ( x 2 , μ 2 ) . By ( H ¯ 4 ) we can inductively show that, for each i = n 3 , , 1 , 0 , y 1 ( i ) ( x 2 , μ 1 ) y 2 ( i ) ( x 2 , μ 2 ) . Consequently by Lemma 2.5 we have y 1 ( n 1 ) ( x , μ 1 ) y 2 ( n 1 ) ( x , μ 2 ) for x 2 x x 3 . Furthermore one can inductively get for each i = n 2 , , 1 , 0 the result y 1 ( i ) ( x , μ 1 ) < y 2 ( i ) ( x , μ 2 ) for x 2 < x x 3 . Now by ( H ¯ 5 ) and ( H ¯ 7 ) we get
h ( y 2 ( x 2 , μ 2 ) , , y 2 ( n 1 ) ( x 2 , μ 2 ) ; y 2 ( x 3 , μ 2 ) , , y 2 ( n 1 ) ( x 3 , μ 2 ) ) h ( y 1 ( x 2 , μ 1 ) , , y 1 ( n 1 ) ( x 2 , μ 1 ) ; y 1 ( x 3 , μ 1 ) , , y 1 ( n 1 ) ( x 3 , μ 1 ) ) = i = 0 n 1 h y i ( y 2 ( i ) ( x 2 , μ 2 ) y 1 ( i ) ( x 2 , μ 1 ) ) + i = 0 n 1 h z i ( y 2 ( i ) ( x 3 , μ 2 ) y 1 ( i ) ( x 3 , μ 1 ) ) > 0 .
This is a contradiction to (3.1). Thus we conclude that
y 1 ( n 1 ) ( x 2 , μ 1 ) y 2 ( n 1 ) ( x 2 , μ 2 ) ,

i.e., T 1 ( μ 1 ) T 1 ( μ 2 ) for μ 1 < μ 2 .

By similar arguments, we can show the inequality for T 2 .
  1. (2)
    Since ( H ¯ 5 ) and ( H ¯ 6 ) imply ( H ¯ 7 ), for any y 1 ( n 1 ) ( x 2 , μ 1 ) S μ 1 and y 2 ( n 1 ) ( x 2 , μ 2 ) S μ 2 , we have by (1), y 1 ( n 1 ) ( x 2 , μ 1 ) y 2 ( n 1 ) ( x 2 , μ 2 ) . Suppose y 1 ( n 1 ) ( x 2 , μ 1 ) = y 2 ( n 1 ) ( x 2 , μ 2 ) . Then both y 1 ( x , μ 1 ) and y 2 ( x , μ 2 ) are solutions of BVP (1.1), (3.3) with μ = y 1 ( n 1 ) ( x 2 , μ 1 ) = y 2 ( n 1 ) ( x 2 , μ 2 ) . By Lemma 3.2 of the uniqueness, we conclude y 1 ( x , μ 1 ) = y 2 ( x , μ 2 ) for x 2 x x 3 , which implies
    μ 1 = y 1 ( n 2 ) ( x 2 , μ 1 ) = y 2 ( n 2 ) ( x 2 , μ 2 ) = μ 2 .
     

This is a contradiction. Thus y 1 ( n 1 ) ( x 2 , μ 1 ) > y 2 ( n 1 ) ( x 2 , μ 2 ) , i.e., T 1 ( μ 1 ) > T 1 ( μ 2 ) for μ 1 < μ 2 . □

Lemma 3.4

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ) and ( H ¯ 7 ) (or ( H ¯ 6 )) hold. Then, for any μ R , both S μ and J μ are compact and connected subsets of .

Proof

If ( H ¯ i ), i = 1 , 2 , 3 , 4 , 5 , 6 hold, then by Lemma 3.2, each of BVP (1.1), (3.1) and BVP (1.1), (3.2) has exactly one solution. Consequently both S μ and J μ are single-point sets. Hence the theorem holds.

Now let ( H ¯ i ), i = 1 , 2 , 3 , 4 , 5 , 7 hold. First, we prove that S μ is an interval. To do this, let us take any y 1 ( n 1 ) ( x 2 , μ ) , y 2 ( n 1 ) ( x 2 , μ ) S μ with y 1 ( n 1 ) ( x 2 , μ ) < y 2 ( n 1 ) ( x 2 , μ ) . We need to show that if y 1 ( n 1 ) ( x 2 , μ ) < y 0 ( n 1 ) < y 2 ( n 1 ) ( x 2 , μ ) , then y 0 ( n 1 ) S μ . By ( H ¯ 4 ), it is easy to see inductively that y 1 ( i ) ( x 2 , μ ) y 2 ( i ) ( x 2 , μ ) , i = n 3 , , 1 , 0 , and for any fixed y 0 ( n 1 ) ( y 1 ( n 1 ) ( x 2 , μ ) , y 2 ( n 1 ) ( x 2 , μ ) ) there exist unique y 0 ( i ) [ y 1 ( i ) ( x 2 , μ ) , y 2 ( i ) ( x 2 , μ ) ] , i = n 3 , , 1 , 0 , such that
g i ( y 0 ( i ) , y 0 ( i + 1 ) , , y 0 ( n 3 ) , μ , y 0 ( n 1 ) ) = 0 , i = 0 , 1 , , n 3 .
Now let y 0 ( x ) be the unique solution of (1.1) which satisfies the initial conditions y 0 ( i ) ( x 2 ) = y 0 ( i ) , i = 0 , 1 , , n 1 , where y 0 ( n 2 ) = μ . Then by Lemma 2.5, y 1 ( n 1 ) ( x , μ ) y 0 ( n 1 ) ( x ) for x 2 x x 3 . Furthermore we have y 1 ( i ) ( x , μ ) y 0 ( i ) ( x ) for x 2 x x 3 , i = 0 , 1 , , n 2 . Similarly we can show that y 0 ( i ) ( x ) y 2 ( i ) ( x , μ ) for x 2 x x 3 , i = 0 , 1 , , n 1 . Hence by ( H ¯ 5 ), we have
h ( y 0 ( x 2 ) , y 0 ( x 2 ) , , y 0 ( n 1 ) ( x 2 ) ; y 0 ( x 3 ) , y 0 ( x 3 ) , , y 0 ( n 1 ) ( x 3 ) ) h ( y 1 ( x 2 , μ ) , y 1 ( x 2 , μ ) , , y 1 ( n 1 ) ( x 2 , μ ) ; y 1 ( x 3 , μ ) , y 1 ( x 3 , μ ) , , y 1 ( n 1 ) ( x 3 , μ ) ) = 0
and
h ( y 0 ( x 2 ) , y 0 ( x 2 ) , , y 0 ( n 1 ) ( x 2 ) ; y 0 ( x 3 ) , y 0 ( x 3 ) , , y 0 ( n 1 ) ( x 3 ) ) h ( y 2 ( x 2 , μ ) , y 2 ( x 2 , μ ) , , y 2 ( n 1 ) ( x 2 , μ ) ; y 2 ( x 3 , μ ) , y 2 ( x 3 , μ ) , , y 2 ( n 1 ) ( x 3 , μ ) ) = 0 .
Thus
h ( y 0 ( x 2 ) , y 0 ( x 2 ) , , y 0 ( n 1 ) ( x 2 ) ; y 0 ( x 3 ) , y 0 ( x 3 ) , , y 0 ( n 1 ) ( x 3 ) ) = 0 .

Hence y 0 ( x ) satisfies the boundary condition (3.1), which implies that y 0 ( x ) is the solution of BVP (1.1), (3.1), and then y 0 ( n 1 ) = y 0 ( n 1 ) ( x 2 ) S μ .

Next, we show that S μ is closed. To do this, for any sequence { y m ( n 1 ) } m = 1 in S μ with y m ( n 1 ) y 0 ( n 1 ) as m , we need to show y 0 ( n 1 ) S μ . By the definition of S μ , corresponding to { y m ( n 1 ) } m = 1 there exists a sequence { y m ( x , μ ) } m = 1 of solutions of BVP (1.1), (3.1) such that y m ( n 1 ) = y m ( n 1 ) ( x 2 , μ ) . By ( H ¯ 4 ), it is easy to see that, for each y m ( n 1 ) , there exist y m ( i ) , i = 0 , 1 , , n 3 , such that
g i ( y m ( i ) , y m ( i + 1 ) , , y m ( n 3 ) , μ , y m ( n 1 ) ) = 0 , i = 0 , 1 , , n 3 .
Furthermore we have, by ( H ¯ 4 ),
y m ( i ) = y m ( i ) ( x 2 , μ ) , i = 0 , 1 , , n 3 , m = 1 , 2 , .
Now let us show that the sequences { y m ( i ) } m = 1 , i = 0 , 1 , , n 3 , are convergent. In fact, when i = n 3 , for any positive integers m , p N we have
0 = g n 3 ( y m ( n 3 ) , μ , y m ( n 1 ) ) g n 3 ( y m + p ( n 3 ) , μ , y m + p ( n 1 ) ) = g n 3 y n 3 ( y m ( n 3 ) y m + p ( n 3 ) ) + g n 3 y n 1 ( y m ( n 1 ) y m + p ( n 1 ) ) .
Consequently by ( H ¯ 4 ), we get
| y m ( n 3 ) y m + p ( n 3 ) | δ 1 | g n 3 y n 1 | | y m ( n 1 ) y m + p ( n 1 ) | .

Since { y m ( n 1 ) } m = 1 is a Cauchy sequence, so is the sequence { y m ( n 3 ) } m = 1 . Hence { y m ( n 3 ) } m = 1 converges to a number y 0 ( n 3 ) . Similarly we can show inductively that, for each i = n 4 , , 1 , 0 , the sequence { y m ( i ) } m = 1 converges to a number y 0 ( i ) .

We note that y m ( n 2 ) = y 0 ( n 2 ) = μ , m = 1 , 2 ,  . Then by Kamke’s standard convergence theorem [37], there exists a solution y = y ˆ ( x ) of (1.1) defined on [ x 2 , x 3 ] satisfying initial conditions y ˆ ( i ) ( x 2 ) = y 0 ( i ) , i = 0 , 1 , , n 1 , and there exists a subsequence { y m j ( x , μ ) } j = 1 of { y m ( x , μ ) } m = 1 such that, for each i = 0 , 1 , , n 1 , the sequence { y m j ( i ) ( x , μ ) } j = 1 uniformly converges to y ˆ ( i ) ( x ) on [ x 2 , x 3 ] . It is easy to see that y = y ˆ ( x ) is the solution of BVP (1.1), (3.1). Hence y 0 ( n 1 ) = y ˆ ( n 1 ) ( x 2 ) S μ .

Finally, we show that S μ is bounded. To do this, we take μ 1 , μ 2 R with μ 1 < μ < μ 2 . Then from Lemma 3.3, we have
S μ 2 S μ S μ 1 .

This implies the boundedness of S μ .

By a similar argument for BVP (1.1′), (3.2′), we can show that J μ is also a compact and connected subset of . □

Lemma 3.5

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 7 ) hold. Then there exist sequences { y m ( x , μ m ) } m = 1 and { y m ( x , ν m ) } m = 1 of solutions of BVP (1.1), (3.1) with μ = μ m and of BVP (1.1), (3.1) with μ = ν m , respectively, for which
lim m y m ( n 1 ) ( x 2 , μ m ) = , lim m y m ( n 1 ) ( x 2 , ν m ) = .

Proof

Let us take a sequence { y m ( n 1 ) } m = 1 with lim m y m ( n 1 ) = . Then, by Lemma 3.1, BVP (1.1), (3.3) with μ = y m ( n 1 ) has a solution, denoted by y m ( x ) . It is easy to see that y m ( x ) is the solution of BVP (1.1), (3.1) with μ = y m ( n 2 ) ( x 2 ) . Let μ m = y m ( n 2 ) ( x 2 ) and let y m ( x , μ m ) = y m ( x ) . Then y m ( n 1 ) ( x 2 , μ m ) S μ m and
lim m y m ( n 1 ) ( x 2 , μ m ) = lim m y m ( n 1 ) = .
Similarly one can show that there exists y m ( n 1 ) ( x 2 , ν m ) S ν m , for which
lim m y m ( n 1 ) ( x 2 , ν m ) = .

 □

Lemma 3.6

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ) and ( H ¯ 7 ) hold. Then
  1. (1)

    for any μ 0 R and ε > 0 , there exists ρ > 0 such that if | μ μ 0 | < ρ , then, for any y ( n 1 ) ( x 2 , μ ) S μ , there exists y ( n 1 ) ( x 2 , μ 0 ) S μ 0 satisfying | y ( n 1 ) ( x 2 , μ ) y ( n 1 ) ( x 2 , μ 0 ) | < ε ;

     
  2. (2)

    for any μ 0 R and ε > 0 , there exists ρ > 0 such that if | μ μ 0 | < ρ , then, for any z ( n 1 ) ( x 2 , μ ) J μ , there exists z ( n 1 ) ( x 2 , μ 0 ) J μ 0 satisfying | z ( n 1 ) ( x 2 , μ ) z ( n 1 ) ( x 2 , μ 0 ) | < ε .

     

Proof

Let us prove only (1), since (2) can be shown similarly.

Suppose the conclusion (1) is false. Then there exist μ 0 R and ε 0 > 0 such that, for each ρ = 1 m , m = 1 , 2 ,  , there exist μ m ( μ 0 1 m , μ 0 + 1 m ) and y ( n 1 ) ( x 2 , μ m ) S μ m such that, for any y ( n 1 ) ( x 2 , μ 0 ) S μ 0 ,
| y ( n 1 ) ( x 2 , μ m ) y ( n 1 ) ( x 2 , μ 0 ) | ε 0 .
Since μ 0 1 m < μ m < μ 0 + 1 m , m = 1 , 2 ,  , we have by Lemma 3.3
T 1 ( μ 0 + 1 ) T 1 ( μ 0 + 1 m ) { y ( n 1 ) ( x 2 , μ m ) } T 1 ( μ 0 1 m ) T 1 ( μ 0 1 ) .
Thus { y ( n 1 ) ( x 2 , μ m ) } m = 1 is bounded. Without loss of generality, we may assume that y ( n 1 ) ( x 2 , μ m ) y 0 ( n 1 ) as m . For any positive integers m , p N , we have, for each i = 0 , 1 , , n 3 ,
0 = g i ( y ( i ) ( x 2 , μ m ) , , y ( n 1 ) ( x 2 , μ m ) ) g i ( y ( i ) ( x 2 , μ m + p ) , , y ( n 1 ) ( x 2 , μ m + p ) ) = g i y i ( y ( i ) ( x 2 , μ m ) y ( i ) ( x 2 , μ m + p ) ) + j = i + 1 n 1 g i y j ( y ( j ) ( x 2 , μ m ) y ( j ) ( x 2 , μ m + p ) ) .
Hence, for each i = 0 , 1 , , n 3 , by ( H ¯ 4 ) we have
| y ( i ) ( x 2 , μ m ) y ( i ) ( x 2 , μ m + p ) | δ 1 j = i + 1 n 1 | g i y j | | y ( j ) ( x 2 , μ m ) y ( j ) ( x 2 , μ m + p ) | .
Since { y ( n 1 ) ( x 2 , μ m ) } m = 1 and { y ( n 2 ) ( x 2 , μ m ) } m = 1 = { μ m } m = 1 are convergent, { y ( n 3 ) ( x 2 , μ m ) } m = 1 is a Cauchy sequence, and thus { y ( n 3 ) ( x 2 , μ m ) } m = 1 is convergent. Similarly one can show inductively that, for each i = n 4 , , 1 , 0 , { y ( i ) ( x 2 , μ m ) } m = 1 is also convergent. Set lim m y ( i ) ( x 2 , μ m ) = y 0 ( i ) , i = 0 , 1 , , n 1 , where y 0 ( n 2 ) = μ 0 . Then by Kamke’s convergence theorem, there exists a solution y = y ˆ ( x ) of (1.1) defined on [ x 2 , x 3 ] satisfying the initial conditions y ˆ ( i ) ( x 2 ) = y 0 ( i ) , i = 0 , 1 , , n 1 and there exists a subsequence { y ( x , μ m j ) } j = 1 of { y ( x , μ m ) } m = 1 such that, for each i = 0 , 1 , , n 1 , the sequence { y ( i ) ( x , μ m j ) } j = 1 uniformly converges to y ˆ ( i ) ( x ) on [ x 2 , x 3 ] . It is easy to see that y ˆ ( x ) is the solution of BVP (1.1), (3.1) with μ = μ 0 . Consequently y 0 ( n 1 ) = y ˆ ( n 1 ) ( x 2 ) S μ 0 , and hence
| y ( n 1 ) ( x 2 , μ m ) y 0 ( n 1 ) | ε 0 ,

which is a contradiction to lim m y ( n 1 ) ( x 2 , μ m ) = y 0 ( n 1 ) . Thus (1) holds. □

Lemma 3.7

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 7 ) hold. Then both T 1 and T 2 are upper semi-continuous on .

Proof

For any μ 0 R , let us show T 1 is upper semi-continuous at μ = μ 0 .

From Lemma 3.4, T 1 ( μ 0 ) is a compact and connected subset of . Hence without loss of generality, we may assume that
T 1 ( μ 0 ) = [ y 1 ( n 1 ) ( x 2 , μ 0 ) , y 2 ( n 1 ) ( x 2 , μ 0 ) ] .
Take any open set U with T 1 ( μ 0 ) U . Then there exists ε > 0 such that
( y 1 ( n 1 ) ( x 2 , μ 0 ) ε , y 2 ( n 1 ) ( x 2 , μ 0 ) + ε ) U .
Thus from Lemma 3.6, there exists ρ > 0 such that if | μ μ 0 | < ρ , then, for any y ( n 1 ) ( x 2 , μ ) S μ , there exists y ( n 1 ) ( x 2 , μ 0 ) S μ 0 = T 1 ( μ 0 ) for which
| y ( n 1 ) ( x 2 , μ ) y ( n 1 ) ( x 2 , μ 0 ) | < ε ,

and so S μ U . Hence T 1 is upper semi-continuous at μ = μ 0 .

The upper semi-continuity of T 2 on can be shown similarly. □

Theorem 3.1

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 7 ) hold. Then BVP (1.1), (1.2) has at least one solution.

Proof

We consider two cases as follows.

Case 1. Suppose there exists μ 0 R such that S μ 0 J μ 0 . Then BVP (1.1), (3.1) with μ = μ 0 and BVP (1.1), (3.2) with μ = μ 0 have solutions y ( x , μ 0 ) and z ( x , μ 0 ) , respectively, such that y ( n 1 ) ( x 2 , μ 0 ) = z ( n 1 ) ( x 2 , μ 0 ) . Since y ( n 2 ) ( x 2 , μ 0 ) = μ 0 = z ( n 2 ) ( x 2 , μ 0 ) , by ( H ¯ 4 ) it is easy to see that y ( i ) ( x 2 , μ 0 ) = z ( i ) ( x 2 , μ 0 ) , i = 0 , 1 , , n 3 . Hence, if we let
u ( x ) : = { y ( x , μ 0 ) , x [ x 2 , x 3 ] , z ( x , μ 0 ) , x [ x 1 , x 2 ] ,

then u ( x ) is a solution of BVP (1.1), (1.2).

Case 2. Suppose for any μ R , S μ J μ = . Then by Lemma 3.3 and 3.5, there exist μ 1 and μ 2 with μ 1 < μ 2 , such that
S μ 1 > J μ 1 , S μ 2 < J μ 2 , S μ 1 > S μ 2 .
In fact, let us take any μ 0 R and z ( n 1 ) ( x 2 , μ 0 ) J μ 0 . Then by Lemma 3.5, there exists some y 1 ( n 1 ) ( x 2 , ν 1 ) S ν 1 such that y 1 ( n 1 ) ( x 2 , ν 1 ) > z ( n 1 ) ( x 2 , μ 0 ) . Take μ 1 with μ 1 < min { ν 1 , μ 0 } . Then by Lemma 3.3, we have
S μ 1 { y 1 ( n 1 ) ( x 2 , ν 1 ) } > { z ( n 1 ) ( x 2 , μ 0 ) } J μ 1 .
Also by Lemma 3.5, there exists some y 2 ( n 1 ) ( x 2 , ν 2 ) S ν 2 such that
y 2 ( n 1 ) ( x 2 , ν 2 ) < min { z ( n 1 ) ( x 2 , μ 1 ) , y 1 ( n 1 ) ( x 2 , ν 1 ) } .
Again take μ 2 > max { μ 1 , ν 2 } . Then by Lemma 3.3, we have
S μ 2 { y 2 ( n 1 ) ( x 2 , ν 2 ) } < { z ( n 1 ) ( x 2 , μ 1 ) } J μ 2
and
S μ 1 { y 1 ( n 1 ) ( x 2 , ν 1 ) } > { y 2 ( n 1 ) ( x 2 , ν 2 ) } S μ 2 .
Now we apply a bisection argument as follows. Set a 0 = μ 1 , b 0 = μ 2 . Then we have two cases, i.e.,
S a 0 + b 0 2 > J a 0 + b 0 2 or S a 0 + b 0 2 < J a 0 + b 0 2 .
If S a 0 + b 0 2 > J a 0 + b 0 2 , set a 1 = a 0 + b 0 2 and b 1 = b 0 . If S a 0 + b 0 2 < J a 0 + b 0 2 , set a 1 = a 0 and b 1 = a 0 + b 0 2 . In summary, there exist a 1 , b 1 [ a 0 , b 0 ] such that
a 1 < b 1 , b 1 a 1 = 1 2 ( b 0 a 0 ) , S a 1 > J a 1 , S b 1 < J b 1 .
By continuing this bisection process, we can get sequences { a m } m = 1 and { b m } m = 1 with a m , b m [ a m 1 , b m 1 ] [ a 0 , b 0 ] , m = 1 , 2 ,  , such that
a m < b m , b m a m = 1 2 m ( b 0 a 0 ) , S a m > J a m , S b m < J b m .

Hence by the nested interval theorem, there uniquely exists ξ R such that ξ m = 1 [ a m , b m ] , actually a m , b m squeeze to the common limit ξ.

Suppose S ξ > J ξ . Then since both S ξ and J ξ are compact and connected subsets of and S ξ J ξ = , there exist two open interval U S and U J such that U S S ξ , U J J ξ and U S U J = . Consequently U S > U J . Since both T 1 and T 2 are upper semi-continuous on by Lemma 3.7, there exists ρ > 0 such that if | μ ξ | < ρ then T 1 ( μ ) = S μ U S and T 2 ( μ ) = J μ U J , and thus S μ > J μ . On the other hand since b m ξ as m , there exists m 0 N such that | b m 0 ξ | < ρ , consequently S b m 0 > J b m 0 , which is a contradiction.

If S ξ < J ξ , then we can similarly obtain a contradiction. Hence the case 2 cannot occur. This completes the proof of the theorem. □

Theorem 3.2

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 6 ) hold. Then BVP (1.1), (1.2) has exactly one solution.

Proof

Since ( H ¯ 5 ) and ( H ¯ 6 ) imply ( H ¯ 7 ), by Theorem 3.1, BVP (1.1), (1.2) has at least one solution.

Now we need to show the uniqueness. By Theorem 3.1, BVP (1.1), (1.2) has a solution u ( x ) , for which we denote
u ( x ) : = { y ( x , μ 0 ) , x [ x 2 , x 3 ] , z ( x , μ 0 ) , x [ x 1 , x 2 ] .

Let v ( x ) be any solution of BVP (1.1), (1.2), and let z 1 ( x ) = v ( x ) for x 1 x x 2 , y 1 ( x ) = v ( x ) for x 2 x x 3 and v ( n 2 ) ( x 2 ) = μ . Then y 1 ( x ) and z 1 ( x ) are the solutions of BVP (1.1), (3.1) with μ = μ and BVP (1.1), (3.2) with μ = μ , respectively.

If μ > μ 0 , then by Lemma 3.3 we have
y ( n 1 ) ( x 2 , μ 0 ) > y 1 ( n 1 ) ( x 2 ) = z 1 ( n 1 ) ( x 2 ) z ( n 1 ) ( x 2 , μ 0 ) ,

which is a contradiction.

If μ < μ 0 , then by Lemma 3.3 we have
z ( n 1 ) ( x 2 , μ 0 ) z 1 ( n 1 ) ( x 2 ) = y 1 ( n 1 ) ( x 2 ) > y ( n 1 ) ( x 2 , μ 0 ) ,

which is also a contradiction. Hence μ = μ 0 . Consequently by Lemma 3.2, we get z 1 ( x ) = z ( x , μ 0 ) for x 1 x x 2 and y 1 ( x ) = y ( x , μ 0 ) for x 2 x x 3 . Thus u ( x ) v ( x ) on [ x 1 , x 3 ] . This completes the proof of the theorem. □

Remark 3.1

Theorem 3.2 includes the results of [1], [2], [4], [12]–[14] as particular cases.

It is easy to see that the linear boundary conditions in the next corollary satisfy ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 6 ).

Corollary 3.1

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), and ( H ¯ 3 ) hold. Suppose further that a i a i + 1 0 , i = 0 , 1 , , n 2 , i = 0 n 1 | a i | > 0 ; b i i b i j 0 , i = 0 , 1 , , n 3 , j = i + 1 , i + 2 , , n 1 , | b i i | > 0 , i = 0 , 1 , , n 3 ; c i c i + 1 0 , i = 0 , 1 , , n 2 , i = 0 n 2 | c i | > 0 . Then, for any λ i R , i = 0 , 1 , , n 1 , the three-point boundary value problem of (1.1) with linear boundary conditions
{ i = 0 n 1 a i y ( i ) ( x 1 ) = λ 0 , j = i n 1 b i j y ( j ) ( x 2 ) = λ i + 1 , i = 0 , 1 , , n 3 , i = 0 n 1 c i y ( i ) ( x 3 ) = λ n 1

has exactly one solution.

By using the transformations x = t and y ( x ) = ( 1 ) n z ( t ) , from Theorem 3.1 we can easily obtain the following.

Theorem 3.3

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 7 ) hold. Then BVP (1.1), (1.2) has at least one solution.

Similarly to the proof of Theorem 3.2, from Theorem 3.3 we can get the following.

Theorem 3.4

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), ( H ¯ 3 ), ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 6 ) hold. Then BVP (1.1), (1.2) has exactly one solution.

It is easy to see that the linear boundary conditions in the next corollary satisfy ( H ¯ 4 ), ( H ¯ 5 ), and ( H ¯ 6 ).

Corollary 3.2

Suppose that ( H ¯ 1 ), ( H ¯ 2 ), and ( H ¯ 3 ) hold. Suppose further that a i a i + 1 0 , i = 0 , 1 , , n 2 , i = 0 n 2 | a i | > 0 ; b i i b i j 0 , i = 0 , 1 , , n 3 , j = i + 1 , i + 2 , , n 1 , | b i i | > 0 , i = 0 , 1 , , n 3 ; c i c i + 1 0 , i = 0 , 1 , , n 2 , i = 0 n 1 | c i | > 0 . Then, for any λ i R , i = 0 , 1 , , n 1 , the three-point boundary value problems of (1.1) with linear boundary conditions
{ i = 0 n 1 a i y ( i ) ( x 1 ) = λ 0 , j = i n 1 b i j y ( j ) ( x 2 ) = λ i + 1 , i = 0 , 1 , , n 3 , i = 0 n 1 c i y ( i ) ( x 3 ) = λ n 1

has exactly one solution.

Finally, as an application, we give an example to demonstrate our results.

Example 3.1

Consider a third-order three-point boundary value problem
y = ( sin x ) y 3 1 + y 2 + ( cos x ) arctan y + | y | + 1 , π 2 x π 2 ,
(3.4)
{ a 0 y ( π 2 ) + a 1 y ( π 2 ) + a 2 y ( π 2 ) = λ 0 , b 0 y ( 0 ) + b 1 y ( 0 ) + b 2 y ( 0 ) = λ 1 , c 0 y ( π 2 ) + c 1 y ( π 2 ) + c 2 y ( π 2 ) = λ 2 ,
(3.5)

where a i , b i , c i , λ i R , i = 0 , 1 , 2 , are constants.

Let
f ( x , y , z , w ) = ( sin x ) y 3 1 + y 2 + ( cos x ) arctan z + | w | + 1 on  [ π 2 , π 2 ] × R 3 .

Then it is easy to check that the assumptions ( H ¯ 1 ), ( H ¯ 2 ), and ( H ¯ 3 ) are satisfied. Hence from either Corollary 3.1 or Corollary 3.2, BVP (3.4), (3.5) has exactly one solution under either of the following conditions:

(i): a 0 a 1 0 , a 1 a 2 0 , | a 0 | + | a 1 | + | a 1 | > 0 ;

(ii): b 0 b 1 0 , b 0 b 2 0 , b 0 0 ;

(iii): c 0 c 1 0 , c 1 c 2 0 , | c 0 | + | c 1 | > 0 ,

or the following conditions:

(i′): a 0 a 1 0 , a 1 a 2 0 , | a 0 | + | a 1 | > 0 ;

(ii′): b 0 b 1 0 , b 0 b 2 0 , b 0 0 ;

(iii′): c 0 c 1 0 , c 1 c 2 0 , | c 0 | + | c 1 | + | c 2 | > 0 .

We note that the results of [1], [2], [4], [12]–[14] cannot guarantee that the above third-order three-point boundary value problem has a unique solution, unless b 1 = 0 .

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Beihua University, JiLin, People’s Republic of China
(2)
Department of Mathematics, Yeungnam University, Kyongsan, Korea

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© Pei and Chang; licensee Springer. 2014

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