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Hindawi Publishing Corporation
Boundary Value Problems
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,
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
tft, ut,u
t, ,u
n−1
t et, t ∈ 0, 1, with one of the following boundary condi-
tions: u
i
00, i 1, 2, , n − 2, u
n−1
0u
n−1
ξ, u
n−2
1
m−2
j1
β
j
u
n−2
η
j
,andu
i
00,
i 1, 2, , n − 1, u
n−2
1
m−2
j1
β
j
u
n−2
η
j
, where f : 0, 1 × R
n
→R −∞, ∞ is a continuous
function, et ∈ L
1
0, 1 β
j
∈ R 1 ≤ j ≤ m − 2,m≥ 4,0<η
1
<η
2
< ··· <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 problem BVP for second- or third-order or-
dinary differential equation has been extensively studied, and a series of better results is ob-
tained in 1–10. But the multi-point boundary value problems for higher order are seldom
seen 11, 12.
In this paper, we consider the following higher-order differential equation:
u
n
tf
t, ut,u
t, ,u
n−1
t
et,t∈ 0, 1, 1.1
with one of the following boundary conditions:
u
i
00,i 1, 2, ,n− 2,u
n−1
0u
n−1
ξ,u
n−2
1
m−2
j1
β
j
u
n−2
η
j
, 1.2
u
i
00,i 1, 2, ,n− 1,u
n−2
1
m−2
j1
β
j
u
n−2
η
j
, 1.3
2 Boundary Value Problems
where f : 0, 1 ×
R
n
→ R −∞, ∞ is a continuous function, et ∈ L
1
0, 1, m ≥ 4,n≥ 2are
two integers, β
j
∈ R, η
j
∈ 0, 1j 1, 2, ,m− 2 are constants satisfying 0 <η
1
<η
2
< ··· <
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 case 2, 9, 10, 12.
The purpose of this paper is to study the existence of solutions for BVP 1.1, 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 by 8. 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 :domL ⊂ 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
:domL ∩ 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
that dom L ∩ Ω
/
∅,themapN : Y → Z will be called L-compact on Ω if QNΩ is bounded
and K
P
I − QN : Ω → Y is compact.
The theorem we use is of 13, 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,whereQ : 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.
We use the classical space C
n−1
0, 1,forx ∈ C
n−1
0, 1, we use the norm x
max{x
∞
, x
∞
, ,x
n−1
∞
}, the norm x
i
∞
max
t∈0,1
|x
i
t|,i 0, 1, ,n − 1, and
denote the norm in Z L
1
0, 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 a bsolutely
continuous on 0, 1 with x
n
∈ L
1
0, 1
.
1.4
Throughout this paper, we assume that the β
j
’s have not the same sign, or there exist
j
1
,j
2
∈{1, 2, ,m− 2} such that signβ
j
1
· β
j
2
−1.
2. Main results
In this section, we will firstly prove existence results for BVP 1.1, 1.2.Todothis,welet
Y C
n−1
0, 1, Z L
1
0, 1 and let L be the linear operator L :domL ⊂ Y → Z with
Y. Gao and M . P ei 3
dom L
u ∈ W
n,1
0, 1 : u
i
00,i 1, 2, ,n− 2,
u
n−1
0u
n−1
ξ,u
n−2
1
m−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 BVP 1.1, 1.2 can be written by Lu Nu.
Lemma 2.1. If
m−2
j1
β
j
1,
m−2
j1
β
j
η
j
/
1, then L :domL ⊂ 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
s
1
ds
1
,
2.3
and linear operator K
P
:ImL → dom L ∩ Ker P can be written by
K
P
v
t
n−1
n − 1!
m−2
j1
β
j
η
j
− 1
m−2
j1
β
j
1
η
j
s
1
0
v
s
1
ds
1
ds
2
t
0
s
n
0
···
s
2
0
v
s
1
ds
1
···ds
n
,
2.4
with
K
P
v≤Δ
1
v
1
, ∀v ∈ Im L,
2.5
where
Δ
1
1
1
m−2
j1
β
j
η
j
− 1
m−2
j1
β
j
1 − η
j
.
2.6
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
s
1
ds
1
0
. 2.7
Since the equation
u
n
v 2.8
4 Boundary Value Problems
has solution ut which satisfies
u
i
00,i 1, 2, ,n− 2,
u
n−1
0u
n−1
ξ,
u
n−2
1
m−2
j1
β
j
u
n−2
η
j
,
2.9
if and only if
ξ
0
v
s
1
ds
1
0. 2.10
In fact, if 2.8 has solution ut satisfying 2.9,then
ξ
0
v
s
1
ds
1
ξ
0
u
n
s
1
ds
1
u
n−1
ξ − u
n−1
00.
2.11
On the other hand, if 2.10 holds, setting
utc
0
ct
n−1
t
0
s
n
0
···
s
2
0
v
s
1
ds
1
···ds
n
,
2.12
where c
0
is an arbitrary constant, c
m−2
j1
β
j
1
η
j
s
2
0
vs
1
ds
1
ds
2
/n − 1!
m−2
j1
β
j
η
j
− 1,thenut
is a solution of 2.8, and satisfies 2.9. Hence 2.7 holds.
For v ∈ Z, taking the projector
Qv
1
ξ
ξ
0
v
s
1
ds
1
.
2.13
Let v
1
v − Qv.By
ξ
0
v
1
s
1
ds
1
0, then v
1
∈ Im L, hence Z Im L R. Since Im L ∩ R {0},
we have Z Im L ⊕
R,thus
dim Ker L dim
R co dim Im L 1. 2.14
Hence L is a Fredholm operator of index zero.
Taking P : Y → Y as follows:
Pu u0,
2.15
then the generalized inverse K
P
:ImL → dom L ∩ Ker P of L can be written by
K
P
v
t
n−1
n − 1!
m−2
j1
β
j
η
j
− 1
m−2
j1
β
j
1
η
j
s
2
0
v
s
1
ds
1
ds
2
t
0
s
n
0
···
s
2
0
v
s
1
ds
1
···ds
n
.
2.16
Y. Gao and M . P ei 5
In fact, for v ∈ Im L,wehave
LK
P
v K
P
v
n
tvt, 2.17
and for all u ∈ dom L ∩ Ker P,wehave
K
P
L
u
t
n−1
n − 1!
m−2
j1
β
j
η
j
− 1
m−2
j1
β
j
1
η
j
s
2
0
u
n
s
1
ds
1
ds
2
t
0
s
n
0
···
s
2
0
u
n
s
1
ds
1
···ds
n
ut − u0.
2.18
In view of u ∈ dom L ∩ Ker P, Pu u00, thus
K
P
L
utut, 2.19
this shows that K
P
L|
dom L∩ Ker P
−1
.
Again since for i 0, 1, ,n− 1, we have
K
P
v
i
t
t
n−1−i
n − 1 − i!
m−2
j1
β
j
η
j
− 1
m−2
j1
β
j
1
η
j
s
2
0
v
s
1
ds
1
ds
2
t
0
s
n−i
0
···
s
2
0
v
s
1
ds
1
···ds
n−i
,
2.20
consequently, for i 0, 1, ,n− 1, we have
K
P
v
i
t
≤
1
m−2
j1
β
j
η
j
− 1
m−2
j1
β
j
1 − η
j
1
v
1
Δ
1
v
1
,
2.21
where Δ
1
1/|
m−2
j1
β
j
η
j
− 1|
m−2
j1
|β
j
|1 − η
j
1. Thus
K
P
v
i
∞
≤ Δ
1
v
1
,i 0, 1, ,n− 1, 2.22
then K
P
v≤Δ
1
v
1
. This completes the proof of Lemma 2.1.
Theorem 2.2. Let f : 0, 1 × R
n
→ R be a continuous function. Assume that there exists n
1
∈
{1, 2, ,m− 3} m ≥ 4 such that β
j
> 0 j 1, 2, ,n
1
,β
j
< 0 j n
1
1,n
1
2, ,m− 2.
Furthermore, the following conditions are satisfied:
A
1
m−2
j1
β
j
1,
m−2
j1
β
j
η
j
/
1;
A
2
there exist functions a
0
,a
1
, ,a
n−1
,b,r ∈ L
1
0, 1, constant σ ∈ 0, 1, and some j ∈
{0, 1, ,n− 1} such that for all u
0
,u
1
, ,u
n−1
∈ R
n
,t∈ 0, 1,
f
t, u
0
, ,u
n−1
≤
n−1
i0
a
i
t
u
i
bt
u
j
σ
rt; 2.23
j
6 Boundary Value Problems
A
3
there exists M>0 such that for u
1
,u
2
, ,u
n−1
∈ R
n−1
,if|u| >M,then
f
t, u, u
1
, ,u
n−1
≥ α|u|−
n−1
i1
α
i
u
i
− γ, t ∈ 0, 1, 2.24
where α>0,α
i
≥ 0,i 1, 2, ,n− 1,γ≥ 0;
A
4
there exists M
∗
> 0 such that for any d ∈ R,if|d| >M
∗
, then either
d · ft, d, 0, ,0 ≤ 0 2.25
or
d · ft, d, 0, ,0 ≥ 0. 2.26
Then, for every e ∈ L
1
0, 1,BVP1.1, 1.2 has at least one solution in C
n−1
0, 1 provided that
n−1
i0
a
i
1
< 1/Δ
2
,whereΔ
2
Δ
1
1 1/α
n−1
i1
α
i
, Δ
1
as in Lemma 2.1.
Proof. Set
Ω
1
u ∈ dom L \ Ker L : Lu λNu, λ ∈ 0, 1
. 2.27
Then for u ∈ Ω
1
,Lu λNu,thusλ
/
0,Nu∈ Im L Ker Q. Hence
ξ
0
f
t, ut, ,u
n−1
t
et
dt 0. 2.28
Thus, there exists t
0
∈ 0,ξ such that
f
t
0
,u
t
0
,u
t
0
, ,u
n−1
t
0
−
1
ξ
ξ
0
etdt.
2.29
This yields
f
t
0
,u
t
0
,u
t
0
, ,u
n−1
t
0
≤
1
ξ
e
1
. 2.30
If for some t
1
∈ 0, 1, |ut
1
|≤M,thenwehave
u0
u
t
1
−
t
1
0
u
tdt
≤ M u
∞
.
2.31
Otherwise, if |ut| >Mfor any t ∈ 0, 1,from2.30 and A
3
,weobtain
u
t
0
≤
1
α
n−1
i1
α
i
u
i
t
0
1
α
γ
1
ξ
e
1
≤
1
α
n−1
i1
α
i
u
i
∞
1
α
γ
1
ξ
e
1
.
2.32
Y. Gao and M . P ei 7
Thus
u0
u
t
0
−
t
0
0
u
tdt
≤
u
t
0
u
∞
≤
1
α
n−1
i1
α
i
u
i
∞
1
α
γ
1
ξ
e
1
u
∞
.
2.33
Again, since u
i
00,i 1, 2, ,n− 2, then for all t ∈ 0, 1,wehave
u
i
t
u
i
0
t
0
u
i1
tdt
≤
u
i1
∞
. 2.34
Thus
u
i
∞
≤
u
i1
∞
,i 1, 2, ,n− 2. 2.35
Therefore, we have
u
i
∞
≤
u
n−1
∞
,i 1, 2, ,n− 2. 2.36
Hence
Pu
u0
≤
1
1
α
n−1
i1
α
i
u
n−1
∞
1
α
γ
1
ξ
e
1
M. 2.37
According to the conditions β
j
> 0 j 1, 2, ,n
1
,β
j
< 0 j n
1
1,n
1
2, ,m− 2,and
u
n−2
1
m−2
j1
β
j
u
n−2
η
j
,wehave
u
n−2
1 −
m−2
jn
1
1
β
j
u
n−2
η
j
n
1
j1
β
j
u
n−2
η
j
. 2.38
Again, since there exist t
2
∈ η
n
1
1
, 1,t
3
∈ η
1
,η
n
1
such that
u
n−2
t
2
1
1 −
m−2
jn
1
1
β
j
u
n−2
1 −
m−2
jn
1
1
β
j
u
n−2
η
j
, 2.39
u
n−2
t
3
1
1 −
n
1
j1
β
j
n
1
j1
β
j
u
n−2
η
j
, 2.40
thus, in view of
m−2
j1
β
j
1, from 2.38–2.40,weget
u
n−2
t
2
u
n−2
t
3
. 2.41
8 Boundary Value Problems
Since η
n
1
<η
n
1
1
,thent
2
/
t
3
,sofrom2.41, there exists t
∗
∈ t
2
,t
3
such that u
n−1
t
∗
0.
Hence, in view of u
n−1
tu
n−1
t
∗
t
t
∗
u
n
tdt,wehave
u
n−1
∞
≤
u
n
1
Lu
1
≤Nu
1
.
2.42
Therefore, from 2.37 and 2.42, one has
Pu≤
1
1
α
n−1
i1
α
i
Nu
1
1
α
γ
1
ξ
e
1
M. 2.43
Again, for u ∈ Ω
1
,u∈ dom L \ Ker L,thenI − Pu ∈ dom L \ Ker L, LPu 0. Thus from
Lemma 2.1,wehave
I − Pu
K
p
LI − Pu
≤ Δ
1
LI − Pu
1
Δ
1
Lu
1
≤ Δ
1
Nu
1
.
2.44
From 2.43 and 2.44,weget
u≤Pu
I − Pu
≤
1 Δ
1
1
α
n−1
i1
α
i
Nu
1
c
1
Δ
2
Nu
1
c
1
,
2.45
where c
1
M 1/αγ 1/ξe
1
.
If 2.23
j
n−1
holds, then from 2.45,weget
u≤Δ
2
n−1
i0
a
i
1
u
i
∞
b
1
u
n−1
σ
∞
c
, 2.46
where c r
1
e
1
c
1
/Δ
2
.Inviewof2.46,weobtain
u
∞
≤u≤
Δ
2
1 − Δ
2
a
0
1
n−1
i1
a
i
1
u
i
∞
b
1
u
n−1
σ
∞
c
. 2.47
Again, u
∞
≤u,from2.46 and 2.47, one has
u
∞
≤
Δ
2
1 − Δ
2
a
0
1
a
1
1
n−1
i2
a
i
1
u
i
∞
b
1
u
n−1
σ
∞
c
. 2.48
Y. Gao and M . P ei 9
In general, for k 2, 3, ,n− 2, we have
u
k
∞
≤
Δ
2
1 − Δ
2
k
i0
a
i
1
n−1
ik1
a
i
1
u
i
∞
b
1
u
n−1
σ
∞
c
, 2.49
k
u
n−1
∞
≤
Δ
2
b
1
1 − Δ
2
n−1
i0
a
i
1
u
n−1
σ
∞
Δ
2
c
1 − Δ
2
n−1
i0
a
i
1
. 2.50
Since σ ∈ 0, 1,thenfrom2.50, there exists M
n−1
> 0 such that u
n−1
∞
≤ M
n−1
.Thusfrom
2.49
k
, there exist M
k
> 0,k 0, 1, ,n− 2, such that u
k
∞
≤ M
k
,k 0, 1, ,n− 2. Hence
u max
u
∞
,
u
∞
, ,
u
n−1
∞
≤ max
M
0
,M
1
, ,M
n−1
.
2.51
Therefore, Ω
1
is bounded.
If 2.23
j
, j ∈{0, 1, ,n− 2} holds, similar to 2.23
j
n−1
argument, we can prove that Ω
1
is bounded too.
Set
Ω
2
{u ∈ Ker L : Nu ∈ Im L}. 2.52
Then for u ∈ Ω
2
, u ∈ Ker L {u ∈ dom L : u d,d ∈ R},andQNu 0, one has
ξ
0
ft, d, 0, ,0et
dt 0. 2.53
Thus, there exists t
4
∈ 0,ξ such that
f
t
4
,d,0, ,0
−
1
ξ
ξ
0
etdt. 2.54
This yields
f
t
4
,d,0, ,0
≤
1
ξ
e
1
. 2.55
Since either |d|≤M or |d| >M,if|d| >M,theninviewofA
3
and 2.55,wehave|d|≤
1/αγ 1/ξe
1
. Therefore, it follows that
|d|≤max
M,
1
α
γ
1
ξ
e
1
. 2.56
Hence Ω
2
is bounded.
10 Boundary Value Problems
Now, according to condition A
4
, we have the following two cases.
Case 1. For any d ∈
R,if|d| >M
∗
,thend · ft, d, 0, ,0 ≤ 0,t∈ 0, 1. In this case, we set
Ω
3
u ∈ Ker L : −1 − λJu λQNu 0,λ∈ 0, 1
, 2.57
where J :KerL → Im Q is the linear isomorphism given by Jdd, d ∈
R.
In the following, we will show that Ω
3
is bounded. Suppose u
n
td
n
∈ Ω
3
and |d
n
|→
∞ n →∞, then there exists λ
n
∈ 0, 1,forsufficiently large n, such that
1 − λ
n
λ
n
·
QN
d
n
d
n
. 2.58
Since λ
n
∈ 0, 1,then{λ
n
} has a convergent subsequence, and we write for simplicity of
notation λ
n
→ λ
0
n →∞.
If 2.23
j
j
, j ∈{1, 2, ,n− 1} holds, then
QN
d
n
d
n
1
d
n
1
ξ
ξ
0
f
t, d
n
, 0, ,0
et
dt
≤
1
d
n
1
ξ
a
0
1
d
n
r
1
e
1
1
ξ
a
0
1
1
ξ
r
1
e
1
d
n
.
2.59
If 2.23
j
0
holds, then
QN
d
n
d
n
1
d
n
1
ξ
ξ
0
f
t, d
n
, 0, ,0
et
dt
≤
1
d
n
1
ξ
a
0
1
d
n
b
1
d
n
σ
r
1
e
1
1
ξ
a
0
1
1
ξ
·
b
1
d
n
1−σ
1
ξ
·
r
1
e
1
d
n
.
2.60
Since |d
n
|→∞,thenfrom2.59 or 2.60,weknow{|QNd
n
/d
n
|} is bounded. From 2.58,
we have λ
n
→ λ
0
/
0. Hence for n sufficiently large, λ
n
/
0, and we have
1 − λ
n
λ
n
1
ξ
ξ
0
f
t, d
n
, 0, ,0
d
n
dt
1
d
n
ξ
0
etdt
.
2.61
In view of |d
n
|→∞, we can assume that |d
n
| > max{M, M
∗
},thusforn sufficiently large, from
A
3
,weget
f
t, d
n
, 0, ,0
d
n
≥ α −
γ
d
n
≥
α
2
> 0. 2.62
Again since d
n
· ft, d
n
, 0, ,0 ≤ 0,t∈ 0, 1,from2.62, one has
f
t, d
n
, 0, ,0
d
n
≤−
α
2
< 0.
2.63
Y. Gao and M . P ei 11
Hence, according to Fatou lemma, we obtain
lim
n→∞
ξ
0
f
t, d
n
, 0, ,0
d
n
dt
1
d
n
ξ
0
etdt
≤ lim
n→∞
ξ
0
f
t, d
n
, 0, ,0
d
n
dt
≤
ξ
0
lim
n→∞
f
t, d
n
, 0, ,0
d
n
dt
≤−
α
2
ξ<0,
2.64
which contradicts with 1 − λ
n
/λ
n
≥ 0. Thus Ω
3
is bounded.
Case 2. For any d ∈
R,if|d| >M
∗
,thend · ft, d, 0, ,0 ≥ 0,t∈ 0, 1. In this case, we set
Ω
3
u ∈ Ker L : 1 − λJu λQNu 0,λ∈ 0, 1
, 2.65
where J as in above. Similar to the above argument, we can also show that Ω
3
is bounded.
In the following, we will prove that all the conditions of Theorem 1.1 are satisfied. Set Ω
to be an open-bounded subset of Y such that
3
i1
Ω
i
⊂ Ω. By using the Ascoli-Arzela theorem,
we can prove that K
P
I − QN : Y → Y is compact, thus N is L-compact on Ω. Then by the
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 ∈ KerL ∩ ∂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
QN|
Ker L
, Ω ∩ Ker L, 0
deg
H·, 0, Ω ∩ Ker L, 0
deg
H·, 1, Ω ∩ Ker L, 0
deg
± J, Ω ∩ Ker L, 0
/
0.
2.66
Then by Theorem 1.1, Lu Nu has at least one solution in dom L ∩
Ω,sothatBVP1.1, 1.2
has solution in C
n−1
0, 1. The proof is finished.
Now, we will consider existence results for BVP 1.1, 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
00,i 1, 2, ,n− 1,u
n−2
1
m−2
j1
β
j
u
n−2
η
j
.
2.67
Lemma 2.3. If
m−2
j1
β
j
1,
m−2
j1
β
j
η
2
j
/
1 ,thenL :domL ⊂ Y → Z is a Fredholm operator of index
zero. Furthermore, the linear continuous projector Q : Z → Z can be defined by
Qv
2
1 −
m−2
j1
β
j
η
2
j
m−2
j1
β
j
1
η
j
s
2
0
v
s
1
ds
1
ds
2
,
2.68
12 Boundary Value Problems
and linear operator K
P
Im L → dom L ∩ Ker P can be written as
K
P
v
t
0
s
n
0
···
s
2
0
v
s
1
ds
1
···ds
n
,
2.69
with
K
P
v
≤v
1
, ∀v ∈ Im L. 2.70
Notice that the Ker L {u ∈ dom L : u d, d ∈
R,t∈ 0, 1} and Im L {v ∈ Z :
m−2
j1
β
j
1
η
j
s
2
0
vs
1
ds
1
ds
2
0}. Thus, by using the same method as the proof of Lemma 2.1,we
can prove Lemma 2.3, and we omit it.
Theorem 2.4. Let f : 0, 1 ×
R
n
→ R be a continuous function. Assume that condition A
2
of
Theorem 2.2 and the following conditions are satisfied:
A
5
m−2
j1
β
j
1,
m−2
j1
β
j
η
2
j
/
1;
A
6
there exists M>0,suchthatforu ∈ dom L,if|ut| >Mfor all t ∈ 0, 1,then
m−2
j1
β
j
1
η
j
s
2
0
f
s
1
,u
s
1
, ,u
n−1
s
1
e
s
1
ds
1
ds
2
/
0; 2.71
A
7
there exists M
∗
> 0 such that for any d ∈ R,if|d| >M
∗
, then either
d ·
m−2
j1
β
j
1
η
j
s
2
0
f
s
1
,d,0, ,0
e
s
1
ds
1
ds
2
< 0, 2.72
or else
d ·
m−2
j1
β
j
1
η
j
s
2
0
f
s
1
,d,0, ,0
e
s
1
ds
1
ds
2
> 0. 2.73
Then, for every e ∈ L
1
0, 1,BVP1.1, 1.3 has at least one solution in C
n−1
0, 1 provided that
n−1
i0
a
i
1
< 1/2.
The proof of Theorem 2.4 is similar to the proof of Theorem 2.2, and we omit it.
Next we give two examples to demonstrate the applications of the main results.
Example 2.5. Consider the boundary value problems
u
tf
t, ut,u
t,u
t
et,t∈ 0, 1,
u
0u
ξ,u
00,u
16u
1
6
− 3u
1
3
− 2u
1
2
,
2.74
where ft, u, v, w1/22u 1/44v 1/44w sinw
1/5
,e ∈ L
1
0, 1,ξ∈ 0, 1.
Y. Gao and M . P ei 13
Since β
1
6,β
2
−3,β
3
−2,η
1
1/6,η
2
1/3,η
3
1/2, then
i β
1
β
2
β
3
1,β
1
η
1
β
2
η
2
β
3
η
3
/
1;
ii |ft, u, v, w|≤1/22|u| 1/44|v| 1/44|w| |w|
1/5
;
iii |ft, u, v, w|≥1/22|u|−1/44|v|−1/44|w|−1;
iv for any d ∈
R, d · ft, d, 0, 01/22d
2
≥ 0.
Furthermore
Δ
1
1
1
3
j1
β
j
η
j
− 1
3
j1
β
j
1 − η
j
5,
Δ
2
Δ
1
1
1
α
2
i1
α
i
5 1 1 7,
a
0
1
a
1
1
a
2
1
1
22
1
44
1
44
1
11
<
1
7
.
2.75
Hence from Theorem 2.2, for every e ∈ L
1
0, 1,BVP2.74 has at least one solution u ∈ C
2
0, 1.
Example 2.6. Consider the boundary value problems
u
tf
t, ut,u
t,u
t
et,t∈ 0, 1,
u
00,u
00,u
1−4u
1
4
3u
1
3
2u
1
2
,
2.76
where ft, u, v, w1/8u 1/8v 1/8w sin
2
usinw
1/3
,e∈ L
1
0, 1,ξ∈ 0, 1.
Since β
1
−4,β
2
3,β
3
2,η
1
1/4,η
2
1/3,η
3
1/2, then
i β
1
β
2
β
3
1,β
1
η
2
1
β
2
η
2
2
β
3
η
2
3
/
1;
ii |ft, u, v, w|≤1/8|u| 1/8|v| 1/8|w| |w|
1/3
;
iii let M 8, then for |ut| >M, one has
3
j1
β
j
1
η
j
s
2
0
f
s
1
,u
s
1
,u
s
1
,u
s
1
e
s
1
ds
1
ds
2
/
0; 2.77
iv for any d ∈
R, one has
d
3
j1
β
j
1
η
j
s
2
0
f
s
1
,d,0, 0
ds
1
ds
2
5
192
d
2
> 0.
2.78
Furthermore
a
0
1
a
1
1
a
2
1
1
8
1
8
1
8
3
8
<
1
2
.
2.79
Hence from Theorem 2.4, for every e ∈ L
1
0, 1, BVP 2.76 has at least one solution u ∈ C
2
0, 1.
14 Boundary Value Problems
Acknowledgment
The authors thank the referee for valuable suggestions which led to improvement of the origi-
nal manuscript.
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