Volume 2008, Article ID 723828,14pages 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,[email protected]
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
unt ft, ut, ut, . . . , un−1t et, t ∈0,1, with one of the following boundary
condi-tions:ui0 0,i1,2, . . .,n−2,un−10 un−1ξ,un−21 m−2
j1βjun−2ηj, andui0 0,
i1,2, . . .,n−1,un−21 m−2
j1 βjun−2ηj, wheref:0,1×Rn→R −∞,∞is a continuous function,et∈L10,1β
j ∈R1≤j ≤m−2, m≥4, 0< η1< η2<· · · < ηm−2 <1, 0< ξ <1, all the
β−−sj have not the same sign. We also give some examples to demonstrate our results.
Copyrightq2008 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 problemBVPfor 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 seen11,12.
In this paper, we consider the following higher-order differential equation:
unt ft, ut, ut, . . . , un−1tet, t∈0,1, 1.1
with one of the following boundary conditions:
ui0 0, i1,2, . . . , n−2, un−10 un−1ξ, un−21
m−2
j1
βjun−2
ηj
, 1.2
ui0 0, i1,2, . . . , n−1, un−21
m−2
j1
βjun−2
ηj
wheref :0,1×Rn→R −∞,∞is a continuous function,et∈L10,1,m≥4, n≥2 are two integers,βj ∈R,ηj ∈0,1 j 1,2, . . . , m−2are constants satisfying 0< η1< η2< · · ·< ηm−2<1.
For certain boundary condition case such that the linear operatorLu un, 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 BVP1.1,1.2and BVP 1.1, 1.3 at resonance case, and establish some existence theorems under nonlinear growth restriction off. The boundary value problems 1.1, 1.2and1.1, 1.3withn 2 have been studied by8. Our results generalize the corresponding result in8. 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.
LetY, Z be real Banach spaces, letL: domL⊂Y →Zbe a Fredholm map of index zero, and letP :Y →Y, Q : Z →Zbe continuous projectors such that ImP KerL, KerQ ImL, andY KerL⊕KerP, Z ImL⊕ImQ. It follows thatL|domL∩KerP : domL∩KerP → ImLis
invertible. We denote the inverse of that map byKP. IfΩis an open-bounded subset ofY such
that domL∩Ω/∅, the mapN :Y → Zwill be calledL-compact onΩifQNΩis bounded andKPI−QN :Ω→Y is compact.
The theorem we use is of13, Theorem 2.4or of14, Theorem IV.13.
Theorem 1.1see13,14. LetLbe a Fredholm operator of index zero and letN beL-compact onΩ. Assume that the following conditions are satisfied:
iLx /λNxfor everyx, λ∈domL\KerL∩∂Ω×0,1;
iiNx/∈ImLfor everyx∈KerL∩∂Ω;
iiidegQN|KerL,Ω∩KerL,0/0, where Q : Z → Z is a projection as above with ImL
KerQ.
Then the equationLxNxhas at least one solution indomL∩Ω.
We use the classical space Cn−10,1, for x ∈ Cn−10,1, we use the norm x max{x∞,x∞, . . . ,xn−1∞}, the normxi∞ maxt∈0,1|xit|, i 0,1, . . . , n−1, and denote the norm inZL10,1by · 1. We also use the Sobolev space
Wn,10,1 x:0,1−→R|x, x, . . . , xn−1 that are absolutely
continuous on 0,1with xn∈L10,1. 1.4
Throughout this paper, we assume that theβj’s have not the same sign, or there exist
j1, j2∈ {1,2, . . . , m−2}such that signβj1 · βj2 −1.
2. Main results
domL
u∈Wn,10,1:ui0 0, i1,2, . . . , n−2,
un−10 un−1ξ, un−21 m−
2
j1
βjun−2
ηj
,
2.1
andLuun, u∈domL. We also defineN :Y →Zby setting
Nuft, ut, . . . , un−1tet, t∈0,1. 2.2
Then BVP1.1,1.2can be written byLuNu.
Lemma 2.1. Ifmj−12βj 1, jm−12βjηj /1,thenL: domL⊂Y →Zis a Fredholm operator of index
zero. Furthermore, the linear continuous projector operatorQ:Z→Zcan be defined by
Qv 1
ξ ξ
0 vs1
ds1, 2.3
and linear operatorKP : ImL→ domL∩KerP can be written by
KPv t
n−1
n−1!mj−12βjηj−1
m
−2
j1 βj
1
ηj s1
0 vs1
ds1ds2
t
0 sn
0 · · ·
s2
0 vs1
ds1· · ·dsn, 2.4
with
KPv ≤Δ1v1, ∀v ∈ImL, 2.5
where
Δ11 m−21
j1βjηj−1 m−2
j1
βj1−ηj
. 2.6
Proof. It is clear that KerL{u∈domL:ud, d∈R}. We now show that
ImL
v ∈Z:
ξ
0 vs1
ds10
. 2.7
Since the equation
has solutionutwhich satisfies
ui0 0, i1,2, . . . , n−2,
un−10 un−1ξ,
un−21
m−2
j1
βjun−2
ηj
,
2.9
if and only if
ξ
0 vs1
ds10. 2.10
In fact, if2.8has solutionutsatisfying2.9, then
ξ
0 vs1
ds1
ξ
0 uns
1
ds1un−1ξ−un−10 0. 2.11
On the other hand, if2.10holds, setting
ut c0ctn−1
t
0 sn
0 · · ·
s2
0 vs1
ds1· · ·dsn, 2.12
wherec0is an arbitrary constant,cmj−12βj
1
ηj s2
0 vs1ds1ds2/n−1! m−2
j1 βjηj−1, thenut
is a solution of2.8, and satisfies2.9. Hence2.7holds. Forv∈Z, taking the projector
Qv 1
ξ ξ
0 vs1
ds1. 2.13
Letv1 v−Qv. By ξ
0v1s1ds1 0, thenv1 ∈ImL, henceZImLR. Since ImL∩R{0}, we haveZImL⊕R, thus
dim KerLdimRco dim ImL1. 2.14
HenceLis a Fredholm operator of index zero. TakingP :Y →Y as follows:
P uu0, 2.15
then the generalized inverseKP : ImL→domL∩KerPofLcan be written by
KPv t
n−1
n−1!mj−12βjηj−1
m
−2
j1 βj
1
ηj s2
0 vs1
ds1ds2
t
0 sn
0 · · ·
s2
0 vs1
In fact, forv∈ImL, we have
LKPv KPvnt vt, 2.17
and for allu∈domL∩KerP, we have
KPL
u tn
−1
n−1!mj−12βjηj−1
m
−2
j1 βj
1
ηj s2
0
uns1
ds1ds2
t
0 sn
0 · · ·
s2
0
uns1
ds1· · ·dsn
ut−u0.
2.18
In view ofu∈domL∩KerP,P uu0 0, thus
KPL
ut ut, 2.19
this shows thatKP L|domL∩KerP−1.
Again since fori0,1, . . . , n−1,we have
KPv
i
t t
n−1−i
n−1−i!jm−12βjηj−1
m− 2
j1 βj
1
ηj s2
0 vs1
ds1ds2
t
0 sn−i
0 · · ·
s2
0 vs1
ds1· · ·dsn−i,
2.20
consequently, fori0,1, . . . , n−1, we have
KPvi
t≤
1 m−2
j1βjηj−1 m−2
j1
βj1−ηj
1
v1 Δ1v1, 2.21
whereΔ1 1/| m−2
j1 βjηj −1|
m−2
j1|βj|1−ηj 1. Thus
KPvi
∞ ≤Δ1v1, i0,1, . . . , n−1, 2.22
thenKPv ≤Δ1v1. This completes the proof ofLemma 2.1.
Theorem 2.2. Let f : 0,1×Rn → R be a continuous function. Assume that there exists n1 ∈ {1,2, . . . , m−3}m ≥4such thatβj >0 j 1,2, . . . , n1, βj <0 j n11, n12, . . . , m−2.
Furthermore, the following conditions are satisfied:
A1mj−12βj 1, mj−12βjηj /1;
A2there exist functions a0, a1, . . . , an−1, b, r ∈ L10,1, constant σ ∈ 0,1, and some j ∈ {0,1, . . . , n−1}such that for allu0, u1, . . . , un−1∈Rn, t∈0,1,
f
t, u0, . . . , un−1≤
n−1
i0
A3there existsM >0such that foru1, u2, . . . , un−1∈Rn−1, if|u|> M, then
f
t, u, u1, . . . , un−1≥α|u| −
n−1
i1
αiui−γ, t∈0,1, 2.24
whereα >0, αi≥0, i1,2, . . . , n−1, γ ≥0;
A4there existsM∗>0such that for anyd∈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 everye ∈ L10,1, BVP 1.1, 1.2has at least one solution in Cn−10,1provided that n−1
i0ai1<1/Δ2, whereΔ2 Δ11 1/αni−11αi, Δ1as inLemma 2.1.
Proof. Set
Ω1
u∈domL\KerL:LuλNu, λ∈0,1. 2.27
Then foru∈Ω1, LuλNu, thusλ /0, Nu∈ImLKerQ. Hence ξ
0
ft, ut, . . . , un−1tetdt0. 2.28
Thus, there existst0∈0, ξsuch that
ft0, u
t0
, ut0
, . . . , un−1t 0
−1
ξ ξ
0
etdt. 2.29
This yields
f t0, u
t0
, ut0
, . . . , un−1t0≤ 1
ξe1. 2.30
If for somet1∈0,1,|ut1| ≤M, then we have
u0u t1
−
t1
0
utdt≤Mu∞. 2.31
Otherwise, if|ut|> Mfor anyt∈0,1, from2.30andA3, we obtain
u
t0≤ 1
α
n−1
i1 αiui
t0 1
α
γ 1
ξe1
≤ 1
α
n−1
i1
αiui∞ 1
α
γ 1
ξe1
.
Thus
u0u t0
−
t0
0
utdt
≤ut0u∞
≤ 1
α
n−1
i1
αiui∞ 1
α
γ 1
ξe1
u∞.
2.33
Again, sinceui0 0, i1,2, . . . , n−2, then for allt∈0,1, we have
uitui0 t
0
ui1tdt≤ui1∞. 2.34
Thus
ui
∞≤ui1∞, i1,2, . . . , n−2. 2.35
Therefore, we have
ui
∞≤un−1∞, i1,2, . . . , n−2. 2.36
Hence
P uu0≤
1 1
α
n−1
i1 αi
un−1
∞
1 α
γ 1
ξe1
M. 2.37
According to the conditionsβj > 0 j 1,2, . . . , n1, βj < 0 j n11, n12, . . . , m−2, and
un−21 m−2
j1 βjun−2ηj, we have
un−21−
m−2
jn11
βjun−2
ηj
n1
j1
βjun−2
ηj
. 2.38
Again, since there existt2∈ηn11,1, t3∈η1, ηn1such that
un−2t2
1
1−mj−n211βj
un−21−
m−2
jn11
βjun−2
ηj
, 2.39
un−2t3
1
1−n1
j1βj
n1
j1
βjun−2
ηj
, 2.40
thus, in view ofmj−12βj 1, from2.38–2.40, we get
un−2t2
un−2t3
Since ηn1 < ηn11, thent2 /t3, so from2.41, there exists t∗ ∈ t2, t3such thatun−
1t∗ 0.
Hence, in view ofun−1t un−1t∗ t
t∗untdt, we have
un−1
∞≤un1Lu1≤ Nu1. 2.42
Therefore, from2.37and2.42, one has
P u ≤
1 1
α
n−1
i1 αi
Nu1 1
α
γ 1
ξe1
M. 2.43
Again, for u ∈ Ω1, u ∈ domL\KerL, thenI−Pu ∈ domL\KerL, LP u 0. Thus from
Lemma 2.1, we have
I−PuKpLI−Pu
≤Δ1LI−Pu1
Δ1Lu1
≤Δ1Nu1.
2.44
From2.43and2.44, we get
u ≤ P uI−Pu
≤
1 Δ1
1 α
n−1
i1 αi
Nu1c1
Δ2Nu1c1,
2.45
wherec1M 1/αγ 1/ξe1.
If2.23jn−1holds, then from2.45, we get
u ≤Δ2
n−1
i0 ai
1ui∞b1un− 1σ
∞c
, 2.46
wherecr1e1c1/Δ2. In view of2.46, we obtain
u∞≤ u ≤ Δ2
1−Δ2a01 n−1
i1 ai
1ui∞b1un− 1σ
∞c
. 2.47
Again,u∞≤ u, from2.46and2.47, one has
u
∞≤ 1−Δ Δ2
2a01a11
n−1
i2 ai
1ui∞b1un− 1σ
∞c
In general, fork2,3, . . . , n−2, we have
uk ∞ ≤
Δ2 1−Δ2ki0ai1
n−1
ik1 ai
1ui∞b1un− 1σ
∞c
, 2.49k
un−1
∞≤
Δ2b1 1−Δ2ni−01ai1
un−1σ
∞
Δ2c 1−Δ2ni−01ai1
. 2.50
Sinceσ ∈0,1, then from2.50, there existsMn−1 >0 such thatun−1∞≤Mn−1. Thus from
2.49k, there existMk >0, k0,1, . . . , n−2, such thatuk∞≤Mk, k0,1, . . . , n−2.Hence
umaxu∞,u∞, . . . ,un−1∞
≤maxM0, M1, . . . , Mn−1
. 2.51
Therefore,Ω1is bounded.
If2.23j,j ∈ {0,1, . . . , n−2}holds, similar to2.23jn−1argument, we can prove thatΩ1
is bounded too. Set
Ω2{u∈KerL:Nu∈ImL}. 2.52
Then foru∈Ω2,u∈KerL{u∈domL:ud, d∈R}, andQNu0, one has
ξ
0
ft, d,0, . . . ,0 etdt0. 2.53
Thus, there existst4∈0, ξsuch that
ft4, d,0, . . . ,0
−1
ξ ξ
0
etdt. 2.54
This yields
f
t4, d,0, . . . ,0≤ 1
ξe1. 2.55
Since either|d| ≤ M or |d| > M, if |d| > M, then in view of A3and2.55, we have|d| ≤
1/αγ 1/ξe1. Therefore, it follows that
|d| ≤max
M, 1
α
γ 1
ξe1
. 2.56
Now, according to conditionA4, we have the following two cases.
Case 1. For anyd∈R, if|d|> M∗, thend·ft, d,0, . . . ,0≤0, t∈0,1. In this case, we set
Ω3
u∈KerL:−1−λJuλQNu0, λ∈0,1, 2.57
whereJ : KerL→ ImQis the linear isomorphism given byJd d, d∈R.
In the following, we will show thatΩ3is bounded. Supposeunt dn∈Ω3and|dn| →
∞n→ ∞, then there existsλn∈0,1, for sufficiently largen, such that
1−λn λn·
QNdn
dn .
2.58
Since λn ∈ 0,1, then {λn} has a convergent subsequence, and we write for simplicity of
notationλn→λ0n→ ∞.
If2.23jj,j ∈ {1,2, . . . , n−1}holds, then
QNdn
dn
1
dn
1ξξ
0
ft, dn,0, . . . ,0
etdt
≤ 1
dn
1
ξa01dnr1e1
1
ξa01
1 ξ
r1e1
dn .
2.59
If2.23j0holds, then
QNdn
dn
1
dn
1ξξ
0
ft, dn,0, . . . ,0
etdt
≤ 1
dn
1
ξa01dnb1dn
σ
r1e1
1
ξa01
1 ξ ·
b1 dn1−σ
1 ξ ·
r1e1
dn .
2.60
Since|dn| → ∞, then from2.59or 2.60, we know{|QNdn/dn|}is bounded. From 2.58,
we haveλn→λ0/0. Hence fornsufficiently large,λn/0, and we have
1−λn
λn
1 ξ
ξ
0
ft, dn,0, . . . ,0
dn dt
1 dn
ξ
0 etdt
. 2.61
In view of|dn| → ∞, we can assume that|dn|>max{M, M∗}, thus fornsufficiently large, from
A3, we get
ft, dn,0, . . . ,0
dn
≥α− γ
dn
≥ α
2 >0. 2.62
Again sincedn·ft, dn,0, . . . ,0≤0, t∈0,1, from2.62, one has
ft, dn,0, . . . ,0
dn ≤ −
α
Hence, according to Fatou lemma, we obtain
lim
n→∞
ξ
0
ft, dn,0, . . . ,0
dn dt
1 dn
ξ
0 etdt
≤ lim
n→∞
ξ
0
ft, dn,0, . . . ,0
dn dt
≤ ξ
0 lim
n→∞
ft, dn,0, . . . ,0
dn dt
≤ −α
2ξ <0,
2.64
which contradicts with1−λn/λn≥0. ThusΩ3is bounded.
Case 2. For anyd∈R, if|d|> M∗, thend·ft, d,0, . . . ,0≥0, t∈0,1. In this case, we set
Ω3
u∈KerL:1−λJuλQNu0, λ∈0,1, 2.65
whereJas in above. Similar to the above argument, we can also show thatΩ3is bounded.
In the following, we will prove that all the conditions ofTheorem 1.1are satisfied. SetΩ to be an open-bounded subset ofY such that3i1Ωi⊂Ω. By using the Ascoli-Arzela theorem,
we can prove that KPI−QN : Y → Y is compact, thusN isL-compact onΩ. Then by the
above argument, we have the following.
iLu /λNufor everyu, λ∈domL\KerL∩∂Ω×0,1.
iiNu/∈ImLforu∈KerL∩∂Ω.
iiHu, λ ±λJu1−λQNu. According to the above argument, we knowHu, λ/0 for everyu∈KerL∩∂Ω. Thus, by the homotopy property of degree,
degQN|KerL,Ω∩KerL,0
degH·,0,Ω∩KerL,0
degH·,1,Ω∩KerL,0
deg±J,Ω∩KerL,0/0.
2.66
Then byTheorem 1.1,LuNuhas at least one solution in domL∩Ω, so that BVP1.1,1.2
has solution inCn−10,1. The proof is finished.
Now, we will consider existence results for BVP1.1, 1.3. In the following, the map-pingNand linear operatorLare the same as above, and let
domL
u∈Wn,10,1:ui0 0, i1,2, . . . , n−1, un−21
m−2
j1
βjun−2
ηj
. 2.67
Lemma 2.3. Ifmj−12βj 1,
m−2
j1βjη2j /1, thenL: domL⊂Y →Zis a Fredholm operator of index
zero. Furthermore, the linear continuous projectorQ:Z→Zcan be defined by
Qv 2
1−mj−12βjη2j m−2
j1 βj
1
ηj s2
0 vs1
and linear operatorKP ImL→domL∩KerP can be written as
KPv
t
0 sn
0 · · ·
s2
0 vs1
ds1· · ·dsn, 2.69
with
KPv≤ v1, ∀v∈ImL. 2.70
Notice that the KerL {u ∈ domL : u d, d ∈ R, t ∈ 0,1}and ImL {v ∈ Z : m−2
j1βj
1
ηj s2
0 vs1ds1ds2 0}. Thus, by using the same method as the proof ofLemma 2.1, we can proveLemma 2.3, and we omit it.
Theorem 2.4. Let f : 0,1×Rn → R be a continuous function. Assume that condition A
2 of
Theorem 2.2and the following conditions are satisfied:
A5mj−12βj 1, mj−12βjηj2/1;
A6there existsM >0, such that foru∈domL, if|ut|> Mfor allt∈0,1, then
m−2
j1 βj
1
ηj s2
0
fs1, u
s1
, . . . , un−1s1
es1
ds1ds2/0; 2.71
A7there existsM∗>0such that for anyd∈R, if|d|> M∗, then either
d·
m−2
j1 βj
1
ηj
s2
0
fs1, d,0, . . . ,0
es1
ds1ds2<0, 2.72
or else
d·
m−2
j1 βj
1
ηj s2
0
fs1, d,0, . . . ,0
es1
ds1ds2>0. 2.73
Then, for everye ∈ L10,1, BVP 1.1, 1.3has at least one solution in Cn−10,1provided that n−1
i0ai1<1/2.
The proof ofTheorem 2.4is similar to the proof ofTheorem 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
ut ft, ut, ut, utet, t∈0,1,
u0 uξ, u0 0, u1 6u
1 6
−3u
1 3
−2u
1 2
, 2.74
Sinceβ16, β2−3, β3−2, η11/6, η21/3, η31/2,then
iβ1β2β31, β1η1β2η2β3η3/1;
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;
ivfor anyd∈R,d·ft, d,0,0 1/22d2≥0.
Furthermore
Δ11 3 1
j1βjηj−1
3
j1
βj1−ηj
5,
Δ2 Δ11 1 α
2
i1
αi5117,
a0
1a11a21 1
22
1
44
1
44
1 11 <
1 7.
2.75
Hence fromTheorem 2.2, for everye∈L10,1, BVP2.74has at least one solutionu∈C20,1.
Example 2.6. Consider the boundary value problems
ut ft, ut, ut, utet, t∈0,1,
u0 0, u0 0, u1 −4u
1 4
3u
1 3
2u
1 2
, 2.76
whereft, u, v, w 1/8u 1/8v 1/8wsin2usinw1/3, e∈L10,1, ξ ∈0,1. Sinceβ1−4, β23, β32, η11/4, η21/3, η31/2,then
iβ1β2β31, β1η21β2η22β3η23/1;
ii|ft, u, v, w| ≤1/8|u| 1/8|v| 1/8|w||w|1/3;
iiiletM8, then for|ut|> M, one has
3
j1 βj
1
ηj s2
0
fs1, u
s1
, us1
, us1
es1
ds1ds2/0; 2.77
ivfor anyd∈R,one has
d 3
j1 βj
1
ηj s2
0 f
s1, d,0,0
ds1ds2
5 192d
2>0. 2.78
Furthermore
a0
1a11a21 1
8
1
8
1
8
3 8 <
1
2. 2.79
Acknowledgment
The authors thank the referee for valuable suggestions which led to improvement of the origi-nal manuscript.
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