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Multiple monotone positive solutions for higher order differential equations with integral boundary conditions

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Abstract

This paper investigates the higher order differential equations with nonlocal boundary conditions

{ u ( n ) ( t ) + f ( t , u ( t ) , u ( t ) , , u ( n 2 ) ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = = u ( n 3 ) ( 0 ) = 0 , u ( n 2 ) ( 0 ) = 0 1 u ( n 2 ) ( s ) d A ( s ) , u ( n 2 ) ( 1 ) = 0 1 u ( n 2 ) ( s ) d B ( s ) .

The existence results of multiple monotone positive solutions are obtained by means of fixed point index theory for operators in a cone.

MSC:34B10, 34B18.

1 Introduction

In this paper, we are concerned with the existence of multiple monotone positive solutions for the higher order differential equation

u ( n ) (t)+f ( t , u ( t ) , u ( t ) , , u ( n 2 ) ( t ) ) =0,t(0,1),
(1.1)

subject to the following integral boundary conditions:

{ u ( 0 ) = u ( 0 ) = = u ( n 3 ) ( 0 ) = 0 , u ( n 2 ) ( 0 ) = 0 1 u ( n 2 ) ( s ) d A ( s ) , u ( n 2 ) ( 1 ) = 0 1 u ( n 2 ) ( s ) d B ( s ) ,
(1.2)

where n3, f:[0,1]× ( R + ) n 1 R + is continuous in which R + =[0,+), A and B are right continuous on [0,1), left continuous at t=1, and nondecreasing on [0,1], with A(0)=B(0)=0; 0 1 v(s)dA(s) and 0 1 v(s)dB(s) denote the Riemann-Stieltjes integrals of v with respect to A and B, respectively.

Boundary value problems (BVPs for short) for nonlinear differential equations arise in many areas of applied mathematics and physics. Many authors have discussed the existence of positive solutions for second order or higher order differential equations with boundary conditions defined at a finite number of points, for instance, [112]. In [2], Graef and Yang considered the following n th-order multi-point BVP:

{ u ( n ) ( t ) + λ g ( t ) f ( u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = i = 1 m a i u ( n 2 ) ( ξ i ) u ( n 2 ) ( 1 ) = 0 ,

where n3, λ>0 is a parameter, g and f are continuous functions, 1 2 ξ 1 < ξ 2 << ξ m <1, a i >0 for 1im and i = 1 m a i =1. The authors obtained the existence and nonexistence results of positive solutions by using Krasnosel’skii’s fixed point theorem in cones. In [5], we studied the following second order m-point nonhomogeneous BVP:

{ u ( t ) + a ( t ) f ( t , u ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) i = 1 m 2 k i u ( ξ i ) = b ,

where b>0, k i >0 (i=1,2,,m2), 0< ξ 1 < ξ 2 << ξ m 2 <1, i = 1 m 2 k i ξ i <1. The authors obtained the existence, nonexistence and multiplicity of positive solutions by using the Krasnosel’skii-Guo fixed point theorem, the upper-lower solutions method and topological degree theory.

Boundary value problems with integral boundary conditions for ordinary differential equations represent a very interesting and important class of problems and arise in the study of various physical, biological and chemical processes [1315], such as heat conduction, chemical engineering, underground water flow, thermo-elasticity, and plasma physics. They include two, three, multi-point and nonlocal BVPs as special cases. The existence and multiplicity of positive solutions for such problems have received a great deal of attention, see [1632] and the references therein. In [17], Feng, Ji and Ge considered the existence and multiplicity of positive solutions for a class of nonlinear boundary value problems of second order differential equations with integral boundary conditions in ordered Banach spaces

{ x ( t ) + f ( t , x ( t ) ) = θ , t ( 0 , 1 ) , x ( 0 ) = 0 1 g ( t ) x ( t ) d t , x ( 1 ) = θ , or x ( 0 ) = θ , x ( 1 ) = 0 1 g ( t ) x ( t ) d t .

The arguments are based upon a specially constructed cone and fixed point theory in a cone for strict set contraction operators.

Motivated by the works mentioned above, in this paper, we consider the existence of multiple monotone positive solutions for BVP (1.1) and (1.2). In comparison with previous works, our paper has several new features. Firstly, we consider higher order boundary value problems, and we allow the nonlinearity f to contain derivatives of the unknown function u(t) up to n2 order. Secondly, we discuss the boundary value problem with integral boundary conditions, i.e., BVP (1.1) and (1.2), which includes two-point, three-point, multi-point and nonlocal boundary value problems as special cases. Thirdly, we consider the existence of multiple monotone positive solutions. To our knowledge, few papers have considered the monotone positive solutions for a higher order differential equation with integral boundary conditions. We shall emphasize here that with these new features our work improves and generalizes the results of [2] and some other known results to some degree. In this work we shall also utilize the following fixed point theorem in cones.

Lemma 1.1 ([33, 34])

Let K be a cone in a Banach space E. Let D be an open bounded subset of E with D K =DK and D ¯ K K. Assume that A: D ¯ K K is a compact operator such that uAu for u D K . Then the following results hold.

  1. (1)

    If Auu, u D K , then i K (A, D K )=1.

  2. (2)

    If there exists eK{0} such that uAu+λe for all u D K and λ>0, then i K (A, D K )=0.

  3. (3)

    Let U be open in E such that U ¯ D K . If i K (A, D K )=1 and i K (A, U K )=0, then A has a fixed point in D K U ¯ K . The same result holds if i K (A, D K )=0 and i K (A, U K )=1.

2 Preliminary lemmas

Let E={u C n 2 [0,1]: u ( i ) (0)=0,0in3}, then E is a Banach space with the norm u= sup t [ 0 , 1 ] | u ( n 2 ) (t)| for each uE.

We make the following assumptions:

( H 1 ) f:[0,1]× ( R + ) n 1 R + is continuous.

( H 2 ) k 1 >0, k 4 >0, k>0, where k= k 1 k 4 k 2 k 3 ,

k 1 = 1 0 1 ( 1 s ) d A ( s ) , k 2 = 0 1 s d A ( s ) , k 3 = 0 1 ( 1 s ) d B ( s ) , k 4 = 1 0 1 s d B ( s ) .

Lemma 2.1 Assume that ( H 2 ) holds. Then, for any yC[0,1], the BVP

{ u ( t ) = y ( t ) , t ( 0 , 1 ) , u ( 0 ) = 0 1 u ( t ) d A ( t ) , u ( 1 ) = 0 1 u ( t ) d B ( t )
(2.1)

has a unique solution u that can be expressed in the form

u(t)= 0 1 H(t,s)y(s)ds,t[0,1],
(2.2)

where

H ( t , s ) = G ( t , s ) + t k 3 + ( 1 t ) k 4 k G A ( s ) + t k 1 + ( 1 t ) k 2 k G B ( s ) , G A ( s ) = 0 1 G ( τ , s ) d A ( τ ) , G B ( s ) = 0 1 G ( τ , s ) d B ( τ ) , G ( t , s ) = { t ( 1 s ) , 0 t s 1 , s ( 1 t ) , 0 s t 1 .
(2.3)

Proof Firstly, we prove that if u is a solution of BVP (2.1), then it will take the form of (2.2). Now, integrating differential equation (2.1) from 0 to t twice, we have

u(t)=u(0)+ u (0)t 0 t (ts)y(s)ds.
(2.4)

Letting t=1 in (2.4), we get

u (0)=u(1)u(0)+ 0 1 (1s)y(s)ds.
(2.5)

Substituting the boundary conditions of (2.1) and (2.5) into (2.4) yields

u ( t ) = u ( 0 ) + [ u ( 1 ) u ( 0 ) + 0 1 ( 1 s ) y ( s ) d s ] t 0 t ( t s ) y ( s ) d s = ( 1 t ) 0 1 u ( t ) d A ( t ) + t 0 1 u ( t ) d B ( t ) + 0 1 G ( t , s ) y ( s ) d s ,
(2.6)

and, consequently,

0 1 u ( t ) d A ( t ) = 0 1 ( 1 t ) d A ( t ) 0 1 u ( t ) d A ( t ) + 0 1 t d A ( t ) 0 1 u ( t ) d B ( t ) + 0 1 0 1 G ( t , s ) y ( s ) d s d A ( t ) , 0 1 u ( t ) d B ( t ) = 0 1 ( 1 t ) d B ( t ) 0 1 u ( t ) d A ( t ) + 0 1 t d B ( t ) 0 1 u ( t ) d B ( t ) + 0 1 0 1 G ( t , s ) y ( s ) d s d B ( t ) .

Solving the above two equations for 0 1 u(t)dA(t) and 0 1 u(t)dB(t), we have

( k 1 k 2 k 3 k 4 ) ( 0 1 u ( t ) d A ( t ) 0 1 u ( t ) d B ( t ) ) = ( 0 1 0 1 G ( t , s ) y ( s ) d s d A ( t ) 0 1 0 1 G ( t , s ) y ( s ) d s d B ( t ) ) ,

and so

( 0 1 u ( t ) d A ( t ) 0 1 u ( t ) d B ( t ) ) = k 1 ( k 4 k 2 k 3 k 1 ) ( 0 1 0 1 G ( t , s ) y ( s ) d s d A ( t ) 0 1 0 1 G ( t , s ) y ( s ) d s d B ( t ) ) .
(2.7)

Hence, (2.2) follows from (2.6) and (2.7).

Next we prove that the u given by (2.2) satisfies the differential equation and boundary conditions of (2.1). From (2.2), we have

u ( t ) = 0 t s ( 1 t ) y ( s ) d s + t 1 t ( 1 s ) y ( s ) d s + t k 3 + ( 1 t ) k 4 k 0 1 [ 0 1 G ( τ , s ) d A ( τ ) ] y ( s ) d s + t k 1 + ( 1 t ) k 2 k 0 1 [ 0 1 G ( τ , s ) d B ( τ ) ] y ( s ) d s .
(2.8)

Direct differentiation of (2.8) gives u (t)=y(t). Also, from (2.2) we have

0 1 u ( t ) d A ( t ) = 0 1 0 1 G ( t , s ) y ( s ) d s d A ( t ) + k 1 [ k 3 k 2 + k 4 ( 1 k 1 ) ] 0 1 [ 0 1 G ( τ , s ) d A ( τ ) ] y ( s ) d s + k 1 [ k 1 k 2 + k 2 ( 1 k 1 ) ] 0 1 [ 0 1 G ( τ , s ) d B ( τ ) ] y ( s ) d s = k 1 k 4 0 1 0 1 G ( τ , s ) y ( s ) d A ( τ ) d s + k 1 k 2 0 1 0 1 G ( τ , s ) y ( s ) d B ( τ ) d s ,

and, similarly,

0 1 u(t)dB(t)= k 1 k 3 0 1 0 1 G(τ,s)y(s)dA(τ)ds+ k 1 k 1 0 1 0 1 G(τ,s)y(s)dB(τ)ds.

Therefore, by solving the above two equations with the double integrals as unknowns, we have

0 1 0 1 G(τ,s)y(s)dA(τ)ds= k 1 0 1 u(t)dA(t) k 2 0 1 u(t)dB(t)
(2.9)

and

0 1 0 1 G(τ,s)y(s)dB(τ)ds= k 3 0 1 u(t)dA(t)+ k 4 0 1 u(t)dB(t).
(2.10)

Hence (2.6) follows from (2.2), (2.9) and (2.10), and thus u(0)= 0 1 u(t)dA(t), u(1)= 0 1 u(t)dB(t). This completes the proof. □

Defining

G 3 (t,s)= 0 t H(v,s)dv, G i (t,s)= 0 t G i 1 (v,s)dv,i4,

then G n (t,s) is the Green function of BVP (1.1) and (1.2). Moreover, solving BVP (1.1) and (1.2) is equivalent to finding a solution of the following integral equation:

u(t)= 0 1 G n (t,s)f ( s , u ( s ) , u ( s ) , , u ( n 2 ) ( s ) ) ds,t[0,1].

Remark 2.1 If ( H 2 ) holds, then for any t,s[0,1], it is easy to testify that

G(t,s)0,H(t,s)0, G n (t,s)0,n3.
(2.11)

Lemma 2.2 Let δ(0, 1 2 ), then for any t[δ,1δ], η,s[0,1], we have

H(t,s)δH(η,s).

Proof It is easy to show that G(t,s)δG(η,s), t[δ,1δ], η,s[0,1]. For t[δ,1δ], η,s[0,1], we have

H ( t , s ) = G ( t , s ) + t k 3 + ( 1 t ) k 4 k 0 1 G ( τ , s ) d A ( τ ) + t k 1 + ( 1 t ) k 2 k 0 1 G ( τ , s ) d B ( τ ) δ G ( η , s ) + δ ( k 3 + k 4 ) k 0 1 G ( τ , s ) d A ( τ ) + δ ( k 1 + k 2 ) k 0 1 G ( τ , s ) d B ( τ ) δ [ G ( η , s ) + η k 3 + ( 1 η ) k 4 k 0 1 G ( τ , s ) d A ( τ ) + η k 1 + ( 1 η ) k 2 k 0 1 G ( τ , s ) d B ( τ ) ] = δ H ( η , s ) .

For any s[0,1], we define H(s)= max t [ 0 , 1 ] H(t,s). From Lemma 2.2, we know that

δH(s)H(t,s)H(s),t[δ,1δ],s[0,1].
(2.12)

 □

Lemma 2.3 Assume that ( H 2 ) holds. If u C n [0,1] satisfies the boundary conditions (1.2) and

u ( n ) (t)0,t[0,1],

then

u(t)0and u (t)0fort[0,1].
(2.13)

Proof Let m(t)= u ( n 2 ) (t), t[0,1], then we have

{ m ( t ) 0 , t [ 0 , 1 ] , m ( 0 ) = 0 1 m ( t ) d A ( t ) , m ( 1 ) = 0 1 m ( t ) d B ( t ) .

For t[0,1], m (t)0 implies that

m(t)=m [ ( 1 t ) 0 + t 1 ] (1t)m(0)+tm(1).

Thus

m ( 0 ) = 0 1 m ( t ) d A ( t ) m ( 0 ) 0 1 ( 1 t ) d A ( t ) + m ( 1 ) 0 1 t d A ( t ) = ( 1 k 1 ) m ( 0 ) + k 2 m ( 1 ) ,

i.e.,

m(0) k 2 k 1 m(1).

On the other hand,

m ( 1 ) = 0 1 m ( t ) d B ( t ) m ( 0 ) 0 1 ( 1 t ) d B ( t ) + m ( 1 ) 0 1 t d B ( t ) = k 3 m ( 0 ) + ( 1 k 4 ) m ( 1 ) ,

and so

m(1) k 3 k 4 m(0) k 2 k 3 k 1 k 4 m(1),

i.e., km(1)0, therefore m(1)0, and so m(0)0.

Now, m(0)0, m(1)0 and m(t) is concave downward, so we have

m(t)= u ( n 2 ) (t)0,t[0,1].
(2.14)

From (2.14) and u(0)= u (0)== u ( n 3 ) (0)=0, we obtain (2.13). This completes the proof of Lemma 2.3. □

Remark 2.2 From Lemma 2.3, if u is a positive solution of BVP (1.1) and (1.2), then u is nondecreasing on [0,1], i.e., u is a monotone positive solution of BVP (1.1) and (1.2).

Let

K= { u E : u ( n 2 ) ( t ) 0 , t [ 0 , 1 ] , u ( n 2 ) ( t ) δ u , t [ δ , 1 δ ] } .

Obviously, K is a cone in E. For any ρ>0, let K ρ ={uK:u<ρ}, K ρ ={uK:u=ρ} and K ¯ ρ ={uK:uρ}. Define an operator T:KE as follows:

(Tu)(t)= 0 1 G n (t,s)f ( s , u ( s ) , u ( s ) , , u ( n 2 ) ( s ) ) ds,t[0,1].
(2.15)

Then u is a solution of BVP (1.1) and (1.2) if and only if u solves the operator equation u=Tu.

Lemma 2.4 Suppose that ( H 1 ) and ( H 2 ) hold, then T:KK is completely continuous.

Proof For all uK, t[0,1], by ( H 1 ), (2.11), (2.12) and (2.15), we have

( T u ) ( n 2 ) ( t ) = 0 1 H ( t , s ) f ( s , u ( s ) , u ( s ) , , u ( n 2 ) ( s ) ) d s 0 , ( T u ) ( n 2 ) ( t ) = 0 1 H ( t , s ) f ( s , u ( s ) , u ( s ) , , u ( n 2 ) ( s ) ) d s 0 1 H ( s ) f ( s , u ( s ) , u ( s ) , , u ( n 2 ) ( s ) ) d s ,

and

Tu 0 1 H(s)f ( s , u ( s ) , u ( s ) , , u ( n 2 ) ( s ) ) ds.

Thus, further from the first inequality of (2.12), we have

( T u ) ( n 2 ) ( t ) = 0 1 H ( t , s ) f ( s , u ( s ) , u ( s ) , , u ( n 2 ) ( s ) ) d s δ 0 1 H ( s ) f ( s , u ( s ) , u ( s ) , , u ( n 2 ) ( s ) ) d s δ T u , t [ δ , 1 δ ] .

Hence, TuK and T(K)K.

Next by standard methods and the Ascoli-Arzela theorem, one can prove that T:KK is completely continuous. So this is omitted. □

Let

Ω ρ = { u K : min δ t 1 δ u ( n 2 ) ( t ) < δ ρ } = { u E : u ( n 2 ) ( t ) 0 , t [ 0 , 1 ] , δ u min δ t 1 δ u ( n 2 ) ( t ) < δ ρ } .

Proceeding as for the proof of Lemma 2.5 in [33], we have the following.

Lemma 2.5 Ω ρ has the following properties:

  1. (a)

    Ω ρ is open relative to K;

  2. (b)

    K δ ρ Ω ρ K ρ ;

  3. (c)

    u Ω ρ if and only if min δ t 1 δ u ( n 2 ) (t)=δρ;

  4. (d)

    if u Ω ρ , then δρ u ( n 2 ) (t)ρ for δt1δ.

Now for convenience we introduce the following notations:

f δ ρ , ρ = min { min δ t 1 δ f ( t , u 0 , , u n 2 ) ρ : u 0 , u 1 , , u n 3 [ 0 , ρ ] , u n 2 [ δ ρ , ρ ] } , f 0 , ρ = max { max 0 t 1 f ( t , u 0 , u 1 , , u n 2 ) ρ : ( u 0 , u 1 , , u n 2 ) [ 0 , ρ ] n 1 } , f 0 = lim u 0 , u 1 , , u n 2 0 max 0 t 1 f ( t , u 0 , u 1 , , u n 2 ) u n 2 , f = lim u 0 + u 1 + + u n 2 + max 0 t 1 f ( t , u 0 , u 1 , , u n 2 ) u 0 + u 1 + + u n 2 , f 0 = lim u 0 , u 1 , , u n 2 0 min δ t 1 δ f ( t , u 0 , u 1 , , u n 2 ) u n 2 , f = lim u 0 + u 1 + + u n 2 + min δ t 1 δ f ( t , u 0 , u 1 , , u n 2 ) u 0 + u 1 + + u n 2 , m = ( max 0 t 1 0 1 H ( t , s ) d s ) 1 , M = ( min δ t 1 δ δ 1 δ H ( t , s ) d s ) 1 .

To prove our main results, we need the following lemmas.

Lemma 2.6 Assume that ( H 1 ), ( H 2 ) hold and f satisfies

f 0 , ρ manduTuforu K ρ ,
(2.16)

then i K (T, K ρ )=1.

Proof For u K ρ , we have 0 u ( n 2 ) (t)ρ and 0 u ( i ) (t)= 0 t u ( i + 1 ) (s)ds max 0 t 1 u ( i + 1 ) (t)u=ρ, t[0,1], i=0,1,,n3. Then by (2.16) we have, for t[0,1],

( T u ) ( n 2 ) ( t ) = 0 1 H ( t , s ) f ( s , u ( s ) , u ( s ) , , u ( n 2 ) ( s ) ) d s m ρ 0 1 H ( t , s ) d s ρ = u .

This implies that Tuu for u K ρ . By the point (1) in Lemma 1.1, we have i K (T, K ρ )=1. □

Lemma 2.7 Assume that ( H 1 ), ( H 2 ) hold and f satisfies

f δ ρ , ρ δManduTuforu Ω ρ ,
(2.17)

then i K (T, K ρ )=0.

Proof Let e(t)= t n 2 ( n 2 ) ! , t[0,1], then eK with e=1. Next we prove that

uTu+λe,u Ω ρ ,λ>0.

In fact, if not, then there exist u 0 Ω ρ and λ 0 >0 such that u 0 =T u 0 + λ 0 e. By (2.17) and the point (d) in Lemma 2.5, we have, for t[δ,1δ],

u 0 ( n 2 ) ( t ) = ( T u 0 ) ( n 2 ) ( t ) + λ 0 e ( n 2 ) ( t ) = 0 1 H ( t , s ) f ( s , u 0 ( s ) , u 0 ( s ) , , u 0 ( n 2 ) ( s ) ) d s + λ 0 δ 1 δ H ( t , s ) f ( s , u 0 ( s ) , u 0 ( s ) , , u 0 ( n 2 ) ( s ) ) d s + λ 0 δ M ρ δ 1 δ H ( t , s ) d s + λ 0 δ ρ + λ 0 > δ ρ .

This implies that min t [ δ , 1 δ ] u 0 ( n 2 ) (t)>δρ, and so by the point (c) in Lemma 2.5, this is a contradiction. It follows from the point (2) of Lemma 1.1 that i K (T, K ρ )=0. □

3 Main results

In the following, we shall give the main results on the existence of multiple positive solutions of BVP (1.1) and (1.2).

Theorem 3.1 Suppose that ( H 1 ) and ( H 2 ) are satisfied. In addition, assume that one of the following conditions holds.

( H 3 ) There exist ρ 1 , ρ 2 , ρ 3 (0,+) with ρ 1 <δ ρ 2 and ρ 2 < ρ 3 such that

f 0 , ρ 1 m, f δ ρ 2 , ρ 2 δM,uTuforu Ω ρ 2 and f 0 , ρ 3 m.

( H 4 ) There exist ρ 1 , ρ 2 , ρ 3 (0,+) with ρ 1 < ρ 2 <δ ρ 3 such that

f δ ρ 1 , ρ 1 δM, f 0 , ρ 2 m,uTuforu K ρ 2 and f δ ρ 3 , ρ 3 δM.

Then BVP (1.1) and (1.2) has two nondecreasing positive solutions u 1 , u 2 in K. Moreover, if in ( H 3 ) f 0 , ρ 1 m is replaced by f 0 , ρ 1 <m, then BVP (1.1) and (1.2) has a third nondecreasing positive solution u 3 K ρ 1 .

Proof Assume that ( H 3 ) holds. We show that either T has a fixed point u 1 K ρ 1 or in Ω ρ 2 K ¯ ρ 1 . If uTu for u K ρ 1 K ρ 3 , by Lemmas 2.6 and 2.7, we have i K (T, K ρ 1 )=1, i K (T, Ω ρ 2 )=0, and i K (T, K ρ 3 )=1. By Lemma 2.5(b), we have K ¯ ρ 1 K δ ρ 2 Ω ρ 2 since ρ 1 <δ ρ 2 . It follows from Lemma 1.1(3) that T has a fixed point u 1 Ω ρ 2 K ¯ ρ 1 . Similarly, T has a fixed point u 2 K ρ 3 Ω ¯ ρ 2 . The proof is similar when ( H 4 ) holds. □

Corollary 3.1 If there exists ρ>0 such that one of the following conditions holds:

( H 5 ) 0 f 0 <m, f δ ρ , ρ δM, uTu for u Ω ρ , 0 f < m n 1 ,

( H 6 ) M< f 0 , f 0 , ρ m, uTu for u K ρ , M< f ,

then BVP (1.1) and (1.2) has at least two nondecreasing positive solutions in K.

Proof We show that ( H 5 ) implies ( H 3 ). It is easy to verify that 0 f 0 <m implies that there exists ρ 1 (0,δρ) such that f 0 , ρ 1 <m. Let k( f , m n 1 ), by f < m n 1 , there exists r>ρ such that f(t, u 0 , u 1 ,, u n 2 )k( u 0 + u 1 ++ u n 2 ) for t[0,1], u 0 + u 1 ++ u n 2 [r,+). Let

M = max { max t [ 0 , 1 ] f ( t , u 0 , u 1 , , u n 2 ) : u 0 + u 1 + + u n 2 [ 0 , r ] } , ρ 3 > max { M m ( n 1 ) k , r } ,

then for ( u 0 , u 1 ,, u n 2 ) [ 0 , ρ 3 ] n 1 , we have

max t [ 0 , 1 ] f(t, u 0 , u 1 ,, u n 2 ) M +k( u 0 + u 1 ++ u n 2 ) M +k(n1) ρ 3 <m ρ 3 .

This implies that f 0 , ρ 3 m and ( H 3 ) holds. Similarly, ( H 6 ) implies ( H 4 ). This completes the proof. □

Remark 3.1 We establish the multiplicity of monotone positive solutions for a higher order differential equation with integral boundary conditions, and we allow the nonlinearity f to contain derivatives of the unknown function u(t) up to n2 order, so our work improves and generalizes the results of [2] to some degree.

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Acknowledgements

Research supported by the National Natural Science Foundation of China (11371221, 11201260), the Specialized Research Fund for the Doctoral Program of Higher Education of China (20123705120004, 20123705110001), a Project of Shandong Province Higher Educational Science and Technology Program (J11LA06) and Foundation of Qufu Normal University (BSQD20100103).

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Correspondence to Xinan Hao.

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The authors declare that they have no competing interests.

Authors’ contributions

XH wrote the first manuscript and LL corrected and improved the final version. Both authors read and approved the final manuscript.

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Keywords

  • monotone positive solutions
  • multiplicity
  • higher order differential equations
  • nonlocal boundary conditions