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Volume 2008, Article ID 723828, 14 pagesdoi:10.1155/2008/723828 Research Article Solvability for Two Classes of Higher-Order Multi-Point Boundary Value Problems at Resonance Yunzhu Gao a

Volume 2008, Article ID 723828, 14 pages

doi:10.1155/2008/723828

Research Article

Solvability for Two Classes of Higher-Order

Multi-Point Boundary Value Problems at Resonance

Yunzhu Gao and Minghe Pei

Department of Mathematics, Beihua University, Jilin City 132013, China

Correspondence should be addressed to Minghe Pei, peiminghe@ynu.ac.kr

Received 13 July 2007; Revised 14 December 2007; Accepted 29 January 2008

Recommended by Ivan Kiguradze

Using the theory of coincidence degree, we establish existence results of positive solutions for higher-order multi-point boundary value problems at resonance for ordinary differential equation

u n t  ft, ut, ut, , u n−1 t  et, t ∈ 0, 1, with one of the following boundary condi-tions: u i 0  0, i  1, 2, , n − 2, u n−1 0  u n−1 ξ, u n−21 m−2

j1 β j u n−2 η j , and u i0  0,

i  1, 2, , n − 1, u n−21 m−2

j1 β j u n−2 η j , where f : 0, 1 ×Rn→R  −∞, ∞ is a continuous

function, et ∈ L10, 1 β j ∈R 1 ≤ j ≤ m − 2, m ≥ 4, 0 < η1< η2< · · · < η m−2 < 1, 0 < ξ < 1, all the

β −s −j have not the same sign We also give some examples to demonstrate our results.

Copyright q 2008 Y Gao and M Pei This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1 Introduction

In recent years, the multi-point boundary value problemBVP for second- or third-order or-dinary differential equation has been extensively studied, and a series of better results is

seen11,12

In this paper, we consider the following higher-order differential equation:

u n t  ft, ut, ut, , u n−1 t et, t ∈ 0, 1, 1.1 with one of the following boundary conditions:

u i 0  0, i  1, 2, , n − 2, u n−1 0  u n−1 ξ, u n−21 m−2

j1

β j u n−2

η j



u i 0  0, i  1, 2, , n − 1, u n−21 m−2

j1

β j u n−2

η j



where f : 0, 1 × R n → R  −∞, ∞ is a continuous function, et ∈ L10, 1, m ≥ 4, n ≥ 2 are two integers, β j ∈ R, η j ∈ 0, 1 j  1, 2, , m − 2 are constants satisfying 0 < η1< η2< · · · <

η m−2 < 1.

For certain boundary condition case such that the linear operator Lu  u n, defined in

a suitable Banach space, is invertible, this is the so-called nonresonance case, otherwise, the so-called resonance case2,9,10,12

The purpose of this paper is to study the existence of solutions for BVP1.1, 1.2 and BVP 1.1, 1.3 at resonance case, and establish some existence theorems under nonlinear

growth restriction of f The boundary value problems 1.1, 1.2 and 1.1, 1.3 with n  2

have been studied by8 Our results generalize the corresponding result in 8 Our method

is based upon the coincidence degree theory of Mawhin 13,14 Finally, we also give some examples to demonstrate our results

Now, we will briefly recall some notations and an abstract existence result

Let Y, Z be real Banach spaces, let L : dom L ⊂ Y → Z be a Fredholm map of index zero, and let P : Y → Y, Q : Z → Z be continuous projectors such that Im P  Ker L, Ker Q  Im L, and Y  Ker L ⊕ Ker P, Z  Im L ⊕ Im Q It follows that L| dom L∩Ker P : dom L ∩ Ker P → Im L is invertible We denote the inverse of that map by K P IfΩ is an open-bounded subset of Y such

The theorem we use is of13, Theorem 2.4 or of 14, Theorem IV.13

Theorem 1.1 see13,14 Let L be a Fredholm operator of index zero and let N be L-compact on Ω Assume that the following conditions are satisfied:

i Lx / λNx for every x, λ ∈ dom L \ Ker L ∩ ∂Ω × 0, 1;

ii Nx/∈Im L for every x ∈ Ker L ∩ ∂Ω;

iii degQN| Ker L , Ω ∩ Ker L, 0 /  0, where Q : Z → Z is a projection as above with Im L  Ker Q.

Then the equation Lx  Nx has at least one solution in dom L ∩ Ω.

max{x∞, x∞, , x n−1∞}, the norm x i∞  maxt∈0,1 |x i t|, i  0, 1, , n − 1, and denote the norm in Z  L10, 1 by · 1 We also use the Sobolev space

W n,1 0, 1 x : 0, 1 −→ R | x, x, , x n−1 that are absolutely

Throughout this paper, we assume that the β j’s have not the same sign, or there exist

j1, j2∈ {1, 2, , m − 2} such that signβ j1 · β j2  −1

2 Main results

In this section, we will firstly prove existence results for BVP 1.1, 1.2 To do this, we let

Y  C n−1 0, 1, Z  L10, 1 and let L be the linear operator L : dom L ⊂ Y → Z with

dom L 



u ∈ W n,1 0, 1 : u i 0  0, i  1, 2, , n − 2,

j1

β j u n−2

η j



,

2.1

and Lu  u n , u ∈ dom L We also define N : Y → Z by setting

Nu  f

t, ut, , u n−1 t et, t ∈ 0, 1. 2.2 Then BVP1.1, 1.2 can be written by Lu  Nu.

Lemma 2.1 Ifm−2

j1 β j  1, m−2

j1 β j η j /  1, then L : dom L ⊂ Y → Z is a Fredholm operator of index zero Furthermore, the linear continuous projector operator Q : Z → Z can be defined by

Qv  1 ξ

ξ

0

v

s1



and linear operator K P : Im L → dom L ∩ Ker P can be written by

n − 1!m−2

j1 β j η j− 1

m−2

j1

β j

1

η j

s1

0

v

s1



ds1ds2 t

0

s n

0

· · · s2 0

v

s1



ds1· · · ds n , 2.4

with

where

Δ1 1  m−21

j1 β j η j− 1 m−2j1 β j 1− η j

Proof It is clear that Ker L  {u ∈ dom L : u  d, d ∈ R} We now show that

Im L 

v ∈ Z :

ξ

0

v

s1



Since the equation

has solution ut which satisfies

u i 0  0, i  1, 2, , n − 2,

u n−1 0  u n−1 ξ,

u n−21 m−2

j1

β j u n−2

η j



,

2.9

if and only if

ξ

0

v

s1



In fact, if2.8 has solution ut satisfying 2.9, then

ξ

0

v

s1



ds1 ξ

0

u n

s1



On the other hand, if2.10 holds, setting

ut  c0 ct n−1 t

0

s n

0

· · · s2 0

v

s1



where c0is an arbitrary constant, c m−2

j1 β j

1

η j

s2

0 vs1ds1ds2/n − 1!m−2

j1 β j η j − 1, then ut

is a solution of2.8, and satisfies 2.9 Hence 2.7 holds

For v ∈ Z, taking the projector

Qv  1 ξ

ξ

0

v

s1



Let v1  v − Qv Byξ

0v1s1ds1  0, then v1 ∈ Im L, hence Z  Im L  R Since Im L ∩ R  {0},

we have Z  Im L ⊕ R, thus

Hence L is a Fredholm operator of index zero.

Taking P : Y → Y as follows:

then the generalized inverse K P : Im L → dom L ∩ Ker P of L can be written by

n − 1!m−2

j1 β j η j− 1

m−2

j1

β j

1

η j

s2

0

v

s1



ds1ds2 t

0

s n

0

· · · s2 0

v

s1



ds1· · · ds n 2.16

In fact, for v ∈ Im L, we have

and for all u ∈ dom L ∩ Ker P , we have



K P L

n − 1!m−2

j1 β j η j− 1

m−2

j1

β j

1

η j

s2

0

u n

s1



ds1ds2 t

0

s n

0

· · · s2 0

u n

s1



ds1· · · ds n

 ut − u0.

2.18

In view of u ∈ dom L ∩ Ker P , P u  u0  0, thus



K P L

this shows that K P  L| dom L∩ Ker P−1.

Again since for i  0, 1, , n − 1, we have



K P vi

t t n−1−i

n − 1 − i!m−2

j1 β j η j− 1

m−2

j1

β j

1

η j

s2

0

v

s1



ds1ds2 t

0

s n−i

0

· · · s2 0

v

s1



ds1· · · ds n−i ,

2.20

consequently, for i  0, 1, , n − 1, we have

K P vi t ≤ 1

m−2 j1 β j η j− 1 m−2j1 β j 1− η j

 1



whereΔ1 1/|m−2

j1 β j η j − 1|m−2

j1 |β j |1 − η j  1 Thus

K P vi∞ ≤ Δ1v1, i  0, 1, , n − 1, 2.22 thenK P v ≤ Δ1v1 This completes the proof ofLemma 2.1

Theorem 2.2 Let f : 0, 1 × R n → R be a continuous function Assume that there exists n1 ∈

{1, 2, , m − 3} m ≥ 4 such that β j > 0 j  1, 2, , n1, β j < 0 j  n1 1, n1 2, , m − 2 Furthermore, the following conditions are satisfied:

A1m−2

j1 β j  1, m−2

j1 β j η j / 1;

A2 there exist functions a0, a1, , a n−1 , b, r ∈ L10, 1, constant σ ∈ 0, 1, and some j ∈ {0, 1, , n − 1} such that for all u0, u1, , u n−1 ∈ Rn , t ∈ 0, 1,

f

t, u0, , u n−1 ≤n−1

i0

a i t u i  bt u j σ  rt; 2.23j

A3 there exists M > 0 such that for u1, u2, , u n−1 ∈ Rn−1 , if |u| > M, then

f

t, u, u1, , u n−1  ≥ α|u| −n−1

i1

where α > 0, α i ≥ 0, i  1, 2, , n − 1, γ ≥ 0;

A4 there exists M∗> 0 such that for any d ∈ R, if |d| > M∗, then either

or

Then, for every e ∈ L10, 1, BVP 1.1, 1.2 has at least one solution in C n−1 0, 1 provided that

n−1

i0 a i1< 1/Δ2, whereΔ2 Δ1 1  1/αn−1

i1 α i , Δ1as in Lemma 2.1 Proof Set

Ω1u ∈ dom L \ Ker L : Lu  λNu, λ ∈ 0, 1

Then for u ∈ Ω1, Lu  λNu, thus λ /  0, Nu ∈ Im L  Ker Q Hence

ξ

0



f

t, ut, , u n−1 t etdt  0. 2.28

Thus, there exists t0∈ 0, ξ such that

f

t0, u

t0



, u

t0



, , u n−1

t0



 −1

ξ

ξ

0

This yields

f

t0, u

t0



, u

t0



, , u n−1

t0 ≤ 1

If for some t1∈ 0, 1, |ut1| ≤ M, then we have

u0  u

t1



− t1 0

utdt

Otherwise, if|ut| > M for any t ∈ 0, 1, from 2.30 and A3, we obtain

u

t0 ≤ 1

α

n−1



i1

α i u i

t0  1

α



γ  1

ξ e1



α

n−1



i1

α iu i∞ 1

α



γ  1

ξ e1



.

2.32

u0  u

t0



− t0 0

utdt

≤ u

t0  u ∞

α

n−1



i1

α iu i∞ 1

α



γ  1

ξ e1



 u∞.

2.33

Again, since u i 0  0, i  1, 2, , n − 2, then for all t ∈ 0, 1, we have

u i t  u i0  t

0

u i1 tdt

Thus

Therefore, we have

Hence

Pu  u0 ≤1 1

α

n−1



i1

α i



u n−1∞ 1

α



γ  1

ξ e1



According to the conditions β j > 0 j  1, 2, , n1, β j < 0 j  n1 1, n1 2, , m − 2, and

j1 β j u n−2 η j, we have

u n−21 − m−2

jn1 1

β j u n−2

η j

n1

j1

β j u n−2

η j

Again, since there exist t2∈ η n1 1, 1, t3∈ η1, η n1 such that

u n−2

t2



1−m−2 jn1 1β j



u n−21 − m−2

jn1 1β j u n−2

η j



u n−2

t3



1−n1

j1 β j

n1



j1

β j u n−2

η j



thus, in view ofm−2

j1 β j  1, from 2.38–2.40, we get

u n−2

t2



 u n−2

t3



Since η n1 < η n1 1, then t2 /  t3, so from2.41, there exists t∗ ∈ t2, t3 such that u n−1 t∗  0.

Hence, in view of u n−1 t  u n−1 t∗ t

t∗u n tdt, we have

u n−1∞≤u n

Therefore, from2.37 and 2.42, one has

Pu ≤



1 1

α

n−1



i1

α i



Nu1 1

α



γ  1

ξ e1



I − Pu  K p LI − P u

≤ Δ1LI − P u

1

 Δ1Lu1

≤ Δ1Nu1.

2.44

From2.43 and 2.44, we get

u ≤ Pu  I − Pu

≤



1 Δ1 1

α

n−1



i1

α i



Nu1 c1

 Δ2Nu1 c1,

2.45

where c1 M  1/αγ  1/ξe1

If2.23jn−1holds, then from2.45, we get

u ≤ Δ2



n−1



i0

a i

1u i∞ b1u n−1σ



where c  r1 e1 c1/Δ2 In view of2.46, we obtain

1− Δ2a0

1

n−1



i1

a i

1u i∞ b1u n−1σ



Again,u∞≤ u, from 2.46 and 2.47, one has

1− Δ2a0

1a1

1





n−1



i2

a i

1u i∞ b1u n−1σ



In general, for k  2, 3, , n − 2, we have

1− Δ2k

i0a i

1

n−1



ik1

a i

1u i∞ b1u n−1σ



1− Δ2n−1

i0a i

1

u n−1σ

1− Δ2n−1

i0a i

1

Since σ ∈ 0, 1, then from 2.50, there exists M n−1 > 0 such that u n−1∞≤ M n−1 Thus from

2.49k , there exist M k > 0, k  0, 1, , n − 2, such that u k∞≤ M k , k  0, 1, , n − 2 Hence

u  maxu∞,u∞, ,u n−1∞

≤ maxM0, M1, , M n−1



Therefore,Ω1is bounded

If2.23j , j ∈ {0, 1, , n − 2} holds, similar to 2.23jn−1argument, we can prove thatΩ1

is bounded too

Set

Then for u ∈ Ω2, u ∈ Ker L  {u ∈ dom L : u  d, d ∈ R}, and QNu  0, one has

ξ

0



ft, d, 0, , 0  et

Thus, there exists t4∈ 0, ξ such that

f

t4, d, 0, , 0

 −1

ξ

ξ

0

This yields

f

t4, d, 0, , 0 ≤ 1

Since either|d| ≤ M or |d| > M, if |d| > M, then in view of A3 and 2.55, we have |d| ≤

1/αγ  1/ξe1 Therefore, it follows that

|d| ≤ max

M, 1 α



γ  1

Now, according to conditionA4, we have the following two cases.

Case 1 For any d ∈ R, if |d| > M∗, then d · ft, d, 0, , 0 ≤ 0, t ∈ 0, 1 In this case, we set

Ω3 u ∈ Ker L : −1 − λJu  λQNu  0, λ ∈ 0, 1

where J : Ker L → Im Q is the linear isomorphism given by Jd  d, d ∈ R.

In the following, we will show thatΩ3is bounded Suppose u n t  d n∈ Ω3and|d n| →

∞ n → ∞, then there exists λ n ∈ 0, 1, for sufficiently large n, such that

1− λ n  λ n·QN



d n



Since λ n ∈ 0, 1, then {λ n} has a convergent subsequence, and we write for simplicity of notation λn → λ0n → ∞.

If2.23jj , j ∈ {1, 2, , n − 1} holds, then

QNd n



d n

 d1

n 1ξ 0ξf

t, d n , 0, , 0

 etdt

≤ d1n 1ξa0

1 d n  r1 e1



 1

ξa0

ξ

r1 e1

d n .

2.59

If2.23j0holds, then

QNd n



d n

 d1

n 1ξ 0ξf

t, d n , 0, , 0

 etdt

≤ d1n 1ξa0

1 d n  b1 d n σ  r1 e1



ξa0

ξ · b1

d n 1−σ 1

ξ · r1 e1

d n .

2.60

Since|d n| → ∞, then from 2.59 or 2.60, we know {|QNd n /d n|} is bounded From 2.58,

we have λ n → λ0/  0 Hence for n sufficiently large, λ n / 0, and we have

1− λ n

ξ

ξ

0

f

t, d n , 0, , 0

d n

ξ

0

etdt



In view of|d n | → ∞, we can assume that |d n | > max{M, M∗}, thus for n sufficiently large, from

A3, we get

ft, d n , 0, , 0

d n

≥ α − d γ

Again since d n · ft, d n , 0, , 0 ≤ 0, t ∈ 0, 1, from 2.62, one has

f

t, d n , 0, , 0

Hence, according to Fatou lemma, we obtain

lim

n→∞

ξ

0

f

t, d n , 0, , 0

d n

ξ

0

etdt



≤ lim

n→∞

ξ

0

f

t, d n , 0, , 0

0

lim

n→∞

f

t, d n , 0, , 0

≤ −α

2ξ < 0,

2.64

which contradicts with1 − λ n /λ n≥ 0 Thus Ω3is bounded

Case 2 For any d ∈ R, if |d| > M∗, then d · ft, d, 0, , 0 ≥ 0, t ∈ 0, 1 In this case, we set

Ω3 u ∈ Ker L : 1 − λJu  λQNu  0, λ ∈ 0, 1

where J as in above Similar to the above argument, we can also show that Ω3is bounded

In the following, we will prove that all the conditions ofTheorem 1.1are satisfied SetΩ

to be an open-bounded subset of Y such that3

i1Ωi⊂ Ω By using the Ascoli-Arzela theorem,

above argument, we have the following

i Lu / λNu for every u, λ ∈ dom L \ Ker L ∩ ∂Ω × 0, 1.

ii Nu/∈Im L for u ∈ Ker L ∩ ∂Ω.

ii Hu, λ  ±λJu1−λQNu According to the above argument, we know Hu, λ / 0 for every u ∈ Ker L ∩ ∂Ω Thus, by the homotopy property of degree,

deg

 degH·, 0, Ω ∩ Ker L, 0

 degH·, 1, Ω ∩ Ker L, 0

 deg± J, Ω ∩ Ker L, 0/  0.

2.66

Then byTheorem 1.1, Lu  Nu has at least one solution in dom L ∩Ω, so that BVP 1.1, 1.2

has solution in C n−1 0, 1 The proof is finished.

Now, we will consider existence results for BVP1.1, 1.3 In the following, the

map-ping N and linear operator L are the same as above, and let

dom L 



u ∈ W n,1 0, 1 : u i 0  0, i  1, 2, , n − 1, u n−21 m−2

j1

β j u n−2

η j



Lemma 2.3 Ifm−2

j1 β j  1, m−2

j1 β j η2

j /  1 , then L : dom L ⊂ Y → Z is a Fredholm operator of index zero Furthermore, the linear continuous projector Q : Z → Z can be defined by

1−m−2 j1 β j η2

j

m−2

j1

β j

1

η j

s2

0

v

s1





Since the equation

has solution ut which satisfies

u i... 2.16

In fact, for v ∈ Im L, we have

and for all u ∈ dom L ∩ Ker P , we... 2.23j