Open Access

Symmetric positive solutions of higher-order boundary value problems

Boundary Value Problems20142014:78

DOI: 10.1186/1687-2770-2014-78

Received: 31 October 2013

Accepted: 20 March 2014

Published: 3 April 2014

Abstract

We study the higher-order boundary value problems. The existence of symmetric positive solutions of the problem is discussed. Our results extend some recent work in the literature. The analysis of this paper mainly relies on the monotone iterative technique.

MSC:34B15, 34B18.

Keywords

higher-order boundary value problems symmetric positive solutions existence monotone iterative technique

1 Introduction

We study the boundary value problem (BVP)
{ ( 1 ) n u ( 2 n ) ( t ) = f ( t , u ) , t ( 0 , 1 ) , α i u ( 2 i 2 ) ( j ) + ( 1 ) j + 1 β i u ( 2 i 1 ) ( j ) = 0 , i = 1 , , n , j = 0 , 1 ,
(1.1)

where n 1 is an integer, f : ( 0 , 1 ) × ( 0 , ) [ 0 , ) is continuous, α i , β i are nonnegative constants, α i 2 + 2 α i β i > 0 , i = 1 , , n . f ( t , u ) may be singular at u = 0 , t = 0 (and/or t = 1 ). If a function u : [ 0 , 1 ] R is continuous and satisfies u ( t ) = u ( 1 t ) for t [ 0 , 1 ] , then we say that u ( t ) is symmetric on [ 0 , 1 ] . By a symmetric positive solution of BVP (1.1) we mean a symmetric function u C 2 n [ 0 , 1 ] such that u ( t ) > 0 for t ( 0 , 1 ) and u ( t ) satisfies (1.1).

In recent years, many authors have studied BVP (1.1), they only considered that f is nondecreasing or nonincreasing in u, or the boundary condition depends only on derivatives of even orders; see [18] and references cited therein. To the best of the author’s knowledge, there is no such results involving (1.1). In this note, we intend to fill in such gaps in the literature.

The organization of this paper is as follows. After this introduction, in Section 2, we state the assumptions and some preliminary lemmas. By applying the monotone iterative technique, we discuss the existence of symmetric positive solutions for (1.1) and obtain the main results in Section 3.

2 Preliminaries

For convenience, in this paper we let E = C ( 2 n ) [ 0 , 1 ] , Δ i = α i 2 + 2 α i β i ,
G i ( t , s ) = { 1 Δ i ( β i + α i s ) ( β i + α i ( 1 t ) ) , 0 s < t 1 , 1 Δ i ( β i + α i t ) ( β i + α i ( 1 s ) ) , 0 t s 1 ,
(2.1)
e i ( t ) = G i ( t , t ) = 1 Δ i ( β i + α i t ) ( β i + α i ( 1 t ) ) , t [ 0 , 1 ] ,
(2.2)
where i = 1 , , n , and define
P = { u E : u ( 0 ) 0 , u ( t ) > 0  for  t ( 0 , 1 ) , u ( t ) = u ( 1 t )  and there P = exists constant  l u ( 0 , 1 )  satisfying  l u e 1 ( t ) u ( t ) l u 1 e 1 ( t )  for  t [ 0 , 1 ] } .

Remark 2.1 The set P is not a cone as it is not closed.

Throughout this paper, we assume the following:

(H1) α i , β i are nonnegative constants, α i 2 + 2 α i β i > 0 , i = 1 , , n . f : ( 0 , 1 ) × ( 0 , ) [ 0 , ) is continuous and symmetric in t, i.e., f satisfies
f ( 1 t , u ) = f ( t , u ) , t ( 0 , 1 ) .
(2.3)
(H2) For ( t , u ) ( 0 , 1 ) × ( 0 , ) , f ( t , u ) is nondecreasing in u and there exists a constant λ ( 0 , 1 ) such that if σ ( 0 , 1 ] , then
σ λ f ( t , u ) f ( t , σ u ) .
(2.4)
(H2′) For ( t , u ) ( 0 , 1 ) × ( 0 , ) , f ( t , u ) is nonincreasing in u and there exists a constant λ ( 0 , 1 ) such that if σ ( 0 , 1 ] , then
σ λ f ( t , u ) f ( t , σ u ) .
(2.5)

(H3) 0 < 0 1 f ( t , e 1 ( t ) ) d t < .

Example 2.1 Consider the equation
f ( t , u ) = | 2 t 1 | u 1 4 , ( t , u ) ( 0 , 1 ) × ( 0 , ) .

It is easy to see that the function f satisfies assumptions (H1) and (H2). In fact, if σ ( 0 , 1 ] , there exists constant λ with 1 4 λ < 1 such that f ( t , σ u ) σ λ f ( t , u ) .

Remark 2.2 It is easy to see that (H2) implies that if σ [ 1 , ) , then
f ( t , σ u ) σ λ f ( t , u ) ,
(2.6)
and (H2′) implies that if σ [ 1 , ) , then
f ( t , σ u ) σ λ f ( t , u ) .
(2.7)

Now, we present several lemmas that will be used in the proof of our results. By routine calculations we have the following results.

Lemma 2.1 Let v be integrable on ( 0 , 1 ) , then the BVP
{ ( 1 ) n u ( 2 n ) ( t ) = v ( t ) , t ( 0 , 1 ) , α i u ( 2 i 2 ) ( j ) + ( 1 ) j + 1 β i u ( 2 i 1 ) ( j ) = 0 , i = 1 , , n , j = 0 , 1
has a unique solution
u ( t ) = 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) v ( s ) d s d s 1 d s n 1 ,

where G i ( t , s ) are defined by (2.1).

Lemma 2.2 For any t , s [ 0 , 1 ] , we have
ε i e i ( t ) e i ( s ) G i ( t , s ) e i ( t ) ,
(2.8)
G i ( 1 t , 1 s ) = G i ( t , s ) ,
(2.9)

where ε i = Δ i ( α i + β i ) 2 , i = 1 , , n .

3 Main results

Define the operator T : E E by
T u ( t ) = 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , u ( s ) ) d s d s 1 d s n 1 ,
(3.1)

where G i ( t , s ) are defined by (2.1). It is clear that u is a solution of (1.1) if and only if u is a fixed point of T.

Theorem 3.1 Assume (H1)-(H3) hold. Then BVP (1.1) has at least one symmetric positive solution.

Proof

Claim 3.1 T : P P is completely continuous and nondecreasing.

In fact, for u P , it is obvious that T u E , T u ( t ) > 0 for t ( 0 , 1 ) and T u ( 0 ) 0 . (2.3), (2.9) and a change of variables imply
T u ( 1 t ) = 0 1 0 1 G 1 ( 1 t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , u ( s ) ) d s d s 1 d s n 1 = 0 1 0 1 G 1 ( 1 t , 1 s n 1 ) G 2 ( 1 s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , u ( s ) ) d s d s n 1 = = 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , u ( s ) ) d s d s 1 d s n 1 = T u ( t ) , t [ 0 , 1 ] .
(3.2)
For any u P , from (2.4), (2.6), (2.8), and (H3), we have
T u ( t ) e 1 ( t ) 0 1 0 1 G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , l u 1 e 1 ( s ) ) d s d s 1 d s n 1 l u λ e 1 ( t ) 0 1 0 1 G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , e 1 ( s ) ) d s d s 1 d s n 1 l T u 1 e 1 ( t ) ,
(3.3)
T u ( t ) ε 1 e 1 ( t ) 0 1 0 1 e 1 ( s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , l u e 1 ( s ) ) d s d s 1 d s n 1 T u ( t ) l u λ ε 1 e 1 ( t ) 0 1 0 1 e 1 ( s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , e 1 ( s ) ) d s d s 1 d s n 1 T u ( t ) l T u e 1 ( t )
(3.4)
for t [ 0 , 1 ] , where l T u satisfies
0 < l T u < min { 1 , l u λ 0 1 0 1 G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , e 1 ( s ) ) d s d s 1 d s n 1 , l u λ ε 1 0 1 0 1 e 1 ( s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , e 1 ( s ) ) d s d s 1 d s n 1 } .

Thus, it follows from (3.3) and (3.4) that T u ( t ) P , and so T : P P . Next by a standard method and the Ascoli-Arzela theorem one can prove that T : P P is completely continuous, we omit it here. From (H2), it is easy to see that T is nondecreasing in u. Hence, Claim 3.1 holds.

Claim 3.2 Let 0 < δ < 1 be fixed number satisfying
0 < δ l T e 1 1 / ( 1 λ ) ,
(3.5)
where λ is defined in (H2) in which σ = δ , and assume
u 0 = δ e 1 ( t ) , v 0 = δ 1 e 1 ( t ) ,
(3.6)
u n = T u n 1 , v n = T v n 1 , n = 1 , 2 , .
(3.7)
Then
u 0 u 1 u n v n v 1 v 0 ,
(3.8)
and there exists u P such that
u n ( t ) u ( t ) , v n ( t ) u ( t ) , uniformly on [ 0 , 1 ] .
(3.9)

In fact, 0 < l T e 1 < 1 since T e 1 P . So, from (3.5) and noting that 0 < λ < 1 , 0 < δ < 1 . From (3.6), we have u 0 , v 0 P and u 0 v 0 .

On the other hand, from (2.4) and (2.6), we have
u 1 = T u 0 ( t ) = 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , δ e 1 ( s ) ) d s d s 1 d s n 1 δ λ 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , e 1 ( s ) ) d s d s 1 d s n 1 = δ λ T e 1 δ λ l T e 1 e 1 ( t ) δ λ δ 1 λ e 1 ( t ) = u 0 , v 1 = T v 0 ( t ) = 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , δ 1 e 1 ( s ) ) d s d s 1 d s n 1 δ λ 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , e 1 ( s ) ) d s d s 1 d s n 1 = δ λ T e 1 δ λ l T e 1 1 e 1 ( t ) δ λ δ ( 1 λ ) e 1 ( t ) = v 0 .

Since u 0 v 0 and T is nondecreasing, by induction, (3.8) holds.

Let ς = δ 2 , then 0 < ς < 1 . It follows from
T ( c u ) c λ T u , if  0 < c < 1 , u P
that, for any natural number n,
u n = T u n 1 = T n u 0 = T n ( δ e 1 ( t ) ) = T n ( ς δ 1 e 1 ( t ) ) ς λ n T n ( δ 1 e 1 ( t ) ) = ς λ n v n .
Thus, for all natural numbers n and p, we have
0 u n + p u n v n u n ( 1 ς λ n ) v n ( 1 ς λ n ) δ 1 e 1 ( t ) ,

which implies that there exists u P such that (3.9) holds, and Claim 3.2 holds.

Letting n in (3.7), we obtain u ( t ) = T u ( t ) , which is a symmetric positive solution of BVP (1.1), and this completes the proof of the theorem.  □

Theorem 3.2 Assume (H1), (H2′) and (H3) hold. Then BVP (1.1) has at least one symmetric positive solution.

Proof

Claim 3.3 T : P P is completely continuous and nonincreasing.

The proof of Claim 3.3 is similar to the proof of Claim 3.1, so this is omitted.

Claim 3.4 Let 0 < ξ < 1 be fixed number, η > 1 be sufficiently large constant satisfying
( ξ λ + 1 ) η l T e 1 ,
(3.10)
where λ is defined in (H2′) in which σ = ξ η , and assume
u 0 = ξ η e 1 ( t ) , u n = T u n 1 , n = 1 , 2 , .
(3.11)
Then
u 0 u 2 u 2 n u 2 n + 1 u 3 u 1 ,
(3.12)
and there exists u P such that
u 2 n ( t ) u ( t ) , u 2 n + 1 ( t ) u ( t ) , uniformly on [ 0 , 1 ] .
(3.13)
In fact, 0 < ξ η < 1 since 0 < ξ < 1 and η > 1 . So from (3.11),
u 0 ( t ) e 1 ( t ) .
(3.14)
From (2.5), (3.10), (3.14), and noting that T is nonincreasing in u, we have
u 1 = T u 0 ( t ) T e 1 ( t ) l T e 1 e 1 ( t ) ( ξ λ + 1 ) η e 1 ( t ) = ( ξ λ ) η ξ η e 1 ( t ) u 0 ,
(3.15)
u 1 = T u 0 ( t ) u 1 = 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , ξ η e 1 ( s ) ) d s d s 1 d s n 1 u 1 ( ξ η ) λ 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , e 1 ( s ) ) d s d s 1 d s n 1 u 1 = ξ η λ T e 1 ξ η λ l T e 1 1 e 1 ( t ) ξ η e 1 ( t ) .
(3.16)
Therefore,
u 2 = T u 1 ( t ) T u 0 ( t ) = u 1 ,
(3.17)
u 2 = T u 1 ( t ) T ( ξ η e 1 ( t ) ) = 0 1 0 1 G 1 ( t , s n 1 ) G 2 ( s n 1 , s n 2 ) G n ( s 1 , s ) f ( s , ξ η e 1 ( s ) ) d s d s 1 d s n 1 ( ξ η ) λ T e 1 ( t ) ξ η λ l T e 1 e 1 ( t ) ξ η e 1 ( t ) = u 0 .
(3.18)

From (3.15), (3.17), (3.18), and noting that T 2 is nondecreasing, by induction, (3.12) holds.

On the other hand, from (2.5) and (2.7), for 0 < σ 1 ,
T 2 ( σ u ) T ( σ λ T u ) ( σ λ ) λ T 2 u = σ λ 2 T 2 u .
(3.19)
Then from (3.16) and (3.19), we have
u 2 n = T u 2 n 1 = T 2 n u 0 = T 2 n ( ξ η e 1 ( t ) ) = T 2 n 2 T 2 ( ξ 2 η ξ η e 1 ( t ) ) T 2 n 2 T 2 ( ξ 2 η u 1 ( t ) ) T 2 n 2 ( ( ξ 2 η ) λ 2 T 2 u 1 ( t ) ) ( ξ 2 η ) λ 2 n T 2 n u 1 ( t ) ) = ( ξ 2 η ) λ 2 n T 2 n + 1 u 0 ( t ) ) = ( ξ 2 η ) λ 2 n u 2 n + 1 ,
and thus
( ξ 2 η ) λ 2 n u 2 n + 1 ( t ) u 2 n ( t ) u 2 n + 1 ( t ) .
Therefore, for all natural numbers n and p, we have
0 u 2 n + 2 p ( t ) u 2 n ( t ) u 2 n + 1 ( t ) u 2 n ( t ) ( 1 ( ξ 2 η ) λ 2 n ) u 2 n + 1 ( t ) ( 1 ( ξ 2 η ) λ 2 n ) u 1 ( t ) ,
(3.20)
0 u 2 n + 1 ( t ) u 2 n + 2 p + 1 ( t ) u 2 n + 1 ( t ) u 2 n ( t ) ( 1 ( ξ 2 η ) λ 2 n ) u 1 ( t ) .
(3.21)

From (3.20) and (3.21), there exists u P such that (3.13) holds, and Claim 3.4 holds.

Letting n in (3.11), we obtain u ( t ) = T u ( t ) , which is a symmetric positive solution of BVP (1.1), and this completes the proof of the theorem.  □

Remark 3.1 [3, 5] only considered that f is nondecreasing or nonincreasing in u, and α i = 1 , β i = 0 in (1.1), so our results extend the work in the literature.

Example 3.1 Consider the BVP
{ u ( 6 ) ( t ) = f ( t , u ) , t ( 0 , 1 ) , α i u ( 2 i 2 ) ( j ) + ( 1 ) j + 1 β i u ( 2 i 1 ) ( j ) = 0 , i = 1 , 2 , 3 , j = 0 , 1 ,
(3.22)

where f ( t , u ) = u α sin π t for ( t , u ) ( 0 , 1 ) × ( 0 , ) , 0 < α < 1 , α i = β i = 1 , i = 1 , 2 , 3 .

It is easy to see that function f ( t , u ) satisfies (H1) and (H3). If σ ( 0 , 1 ] , there exists constant λ with 0 < α λ < 1 such that f ( t , σ u ) σ λ f ( t , u ) , so (H2) is also satisfied. Therefore, from Theorem 3.1, (3.22) has at least one symmetric positive solution.

Declarations

Acknowledgements

Research supported by the Scientific Research Fund of Hunan Provincial Education Department (13C319).

Authors’ Affiliations

(1)
Department of Mathematics, Hunan University of Science and Technology

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© Luo; licensee Springer. 2014

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