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
j1βjun−2ηj, wheref:0,1×Rn→R −∞,∞is a continuous function,et∈L10,1βj ∈R1≤j ≤m−2, m≥4, 0< η1< η2<· · · < ηm−2<1, 0< ξ <1, all the β−s−jhave 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 in1–10. 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 f
t, ut, ut, . . . , un−1t
et, 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
, 1.3
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 1.1,1.3at resonance case, and establish some existence theorems under nonlinear growth restriction off. The boundary value problems1.1,1.2and1.1,1.3withn 2 have been studied by8. Our results generalize the corresponding result in8. Our method is based upon the coincidence degree theory of Mawhin13,14. Finally, we also give some examples to demonstrate our results.
Now, we will briefly recall some notations and an abstract existence result.
LetY, Zbe 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,KerQImL, andY KerL⊕KerP, ZImL⊕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 letNbeL-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, whereQ : Z →Zis a projection as above with ImL KerQ.
Then the equationLxNxhas at least one solution in domL∩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 norm xi∞ 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−1that are absolutely continuous on0,1withxn∈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
In this section, we will firstly prove existence results for BVP 1.1,1.2. To do this, we let Y Cn−10,1,ZL10,1and letLbe the linear operatorL: domL⊂Y →Zwith
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
Nuf
t, ut, . . . , un−1t
et, t∈0,1. 2.2
Then BVP1.1,1.2can be written byLuNu.
Lemma 2.1. Ifm−2
j1βj1, m−2
j1 βjηj/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
v s1
ds1, 2.3
and linear operatorKP: ImL→domL∩KerPcan be written by
KPv tn−1
n−1!m−2
j1 βjηj−1
m−2
j1
βj
1
ηj
s1
0
v s1
ds1ds2 t
0
sn
0
· · · s2
0
v s1
ds1· · ·dsn, 2.4
with
KPv ≤Δ1v1, ∀v∈ImL, 2.5
where
Δ11 1 m−2j1βjηj−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
v s1
ds10
. 2.7
Since the equation
unv 2.8
has solutionutwhich satisfies
ui0 0, i1,2, . . . , n−2, un−10 un−1ξ,
un−21 m−2
j1βjun−2 ηj
,
2.9
if and only if
ξ
0
v s1
ds10. 2.10
In fact, if2.8has solutionutsatisfying2.9, then ξ
0
v s1
ds1 ξ
0
un s1
ds1un−1ξ−un−10 0. 2.11
On the other hand, if2.10holds, setting ut c0ctn−1
t
0
sn
0
· · · s2
0 v s1
ds1· · ·dsn, 2.12
wherec0is an arbitrary constant,cm−2
j1βj
1
ηj
s2
0 vs1ds1ds2/n−1!m−2
j1βjηj−1, thenut is a solution of2.8, and satisfies2.9. Hence2.7holds.
Forv∈Z, taking the projector
Qv 1 ξ
ξ
0v s1
ds1. 2.13
Letv1v−Qv. Byξ
0v1s1ds10, 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 →Yas follows:
P uu0, 2.15
then the generalized inverseKP : ImL→domL∩KerPofLcan be written by
KPv tn−1
n−1!m−2
j1βjηj−1
m−2
j1
βj
1
ηj
s2
0
v s1
ds1ds2 t
0
sn
0
· · · s2
0
v s1
ds1· · ·dsn. 2.16
In fact, forv∈ImL, we have
LKPv KPvnt vt, 2.17 and for allu∈domL∩KerP, we have
KPL
u tn−1
n−1!m−2
j1βjηj−1
m−2
j1
βj
1
ηj
s2
0
un s1
ds1ds2 t
0
sn
0
· · · s2
0
un s1
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 KPvi
t tn−1−i
n−1−i!m−2
j1 βjηj−1
m−2
j1βj
1
ηj
s2
0
v s1
ds1ds2 t
0
sn−i
0
· · · s2
0
v s1
ds1· · ·dsn−i,
2.20
consequently, fori0,1, . . . , n−1, we have KPvit≤
1
m−2j1 βjηj−1m−2
j1
βj1−ηj
1
v1 Δ1v1, 2.21
whereΔ1 1/|m−2
j1βjηj−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 existsn1 ∈ {1,2, . . . , m−3}m≥4such thatβj >0 j1,2, . . . , n1, βj <0 j n11, n12, . . . , m−2.
Furthermore, the following conditions are satisfied:
A1m−2
j1βj1, m−2
j1 βjηj/1;
A2there exist functions a0, a1, . . . , an−1, b, r ∈ L10,1, constantσ ∈ 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
aituibtujσrt; 2.23j
A3there existsM >0 such 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∗>0 such 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, BVP1.1,1.2 has at least one solution inCn−10,1provided that n−1
i0ai1<1/Δ2, whereΔ2 Δ11 1/αn−1
i1αi,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
f
t, ut, . . . , un−1t et
dt0. 2.28
Thus, there existst0∈0, ξsuch that f
t0, u t0
, u t0
, . . . , un−1 t0
−1 ξ
ξ
0
etdt. 2.29
This yields
f t0, u
t0
, u t0
, . . . , un−1
t0≤ 1
ξe1. 2.30
If for somet1∈0,1,|ut1| ≤M, then we have u0
u 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 t01
α
γ1 ξe1
≤ 1 α
n−1
i1
αiui∞ 1 α
γ1
ξe1
.
2.32
Thus
u0 u
t0
− t0
0
utdt
≤u
t0u∞
≤ 1 α
n−1
i1
αiui∞1 α
γ1
ξe1
u∞.
2.33
Again, sinceui0 0, i1,2, . . . , n−2, then for allt∈0,1, we have uit
ui0 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≤
11 α
n−1 i1
αi
un−1∞1 α
γ1
ξe1
M. 2.37
According to the conditionsβj >0j 1,2, . . . , n1, βj <0j n11, n12, . . . , m−2, and un−21 m−2
j1βjun−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−2 t2
1 1−m−2
jn11βj
un−21− m−2
jn11βjun−2 ηj
, 2.39
un−2 t3
1 1−n1
j1βj n1
j1
βjun−2 ηj
, 2.40
thus, in view ofm−2
j1βj1, from2.38–2.40, we get un−2
t2
un−2 t3
. 2.41
Sinceηn1 < ηn11, thent2/t3, so from2.41, there existst∗ ∈t2, t3such thatun−1t∗ 0.
Hence, in view ofun−1t un−1t∗ t
t∗untdt, we have un−1∞≤un
1Lu1≤ Nu1. 2.42
Therefore, from2.37and2.42, one has
P u ≤
1 1 α
n−1 i1
αi
Nu11 α
γ1
ξe1
M. 2.43
Again, foru ∈ Ω1, u ∈ domL\KerL, thenI−Pu ∈ domL\KerL, LP u 0. Thus from Lemma 2.1, we have
I−PuKpLI−Pu
≤Δ1LI−Pu
1
Δ1Lu1
≤Δ1Nu1.
2.44
From2.43and2.44, we get
u ≤ P uI−Pu
≤
1 Δ11 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∞≤ Δ2
1−Δ2a0
1a1
1
n−1
i2
ai1ui∞b1un−1σ∞c
. 2.48
In general, fork2,3, . . . , n−2, we have uk∞≤ Δ2
1−Δ2k
i0ai1 n−1
ik1
ai
1ui∞b1un−1σ∞c
, 2.49k
un−1∞≤ Δ2b1
1−Δ2n−1
i0ai
1
un−1σ∞ Δ2c 1−Δ2n−1
i0ai
1
. 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
umax
u∞,u∞, . . . ,un−1∞
≤max
M0, 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 et
dt0. 2.53
Thus, there existst4∈0, ξsuch that
f
t4, d,0, . . . ,0 −1
ξ ξ
0
etdt. 2.54
This yields
f
t4, d,0, . . . ,0≤1
ξe1. 2.55
Since either|d| ≤ Mor|d| > M, if|d| > M, then in view ofA3and2.55, we have|d| ≤ 1/αγ 1/ξe1. Therefore, it follows that
|d| ≤max
M,1 α
γ1
ξe1
. 2.56
HenceΩ2is bounded.
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−λnλn·QN
dn
dn . 2.58
Since λn ∈ 0,1, then {λn} has a convergent subsequence, and we writefor simplicity of notationλn→λ0n→ ∞.
If2.23jj,j ∈ {1,2, . . . , n−1}holds, then QN
dn
dn
1 dn
1 ξ
ξ
0
f
t, dn,0, . . . ,0 et
dt
≤ 1 dn1
ξa01dnr1e1
1 ξa0
11 ξ
r1e1
dn .
2.59
If2.23j0holds, then QN
dn
dn
1 dn
1 ξ
ξ
0
f
t, dn,0, . . . ,0 et
dt
≤ 1 dn1
ξa0
1dnb1dnσr1e1
1 ξa0
11 ξ · b1
dn1−σ 1
ξ·r1e1
dn .
2.60
Since|dn| → ∞, then from2.59or2.60, we know{|QNdn/dn|}is bounded. From2.58, we haveλn→λ0/0. Hence fornsufficiently large,λn/0, and we have
1−λn
λn 1 ξ
ξ 0
f
t, 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
f
t, 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 f
t, dn,0, . . . ,0
dn ≤ −α
2 <0. 2.63
Hence, according to Fatou lemma, we obtain
n→∞lim ξ
0
f
t, dn,0, . . . ,0
dn dt 1
dn
ξ
0
etdt
≤ lim
n→∞
ξ
0
f
t, dn,0, . . . ,0
dn dt
≤ ξ
0 n→∞lim
f
t, 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 ofYsuch that3
i1Ωi⊂Ω. By using the Ascoli-Arzela theorem, we can prove thatKPI−QN :Y → Y is compact, thusNisL-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,
deg
QN|KerL,Ω∩KerL,0 deg
H·,0,Ω∩KerL,0 deg
H·,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
j1βjun−2 ηj
. 2.67
Lemma 2.3. Ifm−2
j1 βj1,m−2
j1 βjη2j/1 , thenL: domL⊂Y→Zis a Fredholm operator of index zero. Furthermore, the linear continuous projectorQ:Z→Zcan be defined by
Qv 2
1−m−2
j1 βjη2j
m−2
j1βj
1
ηj
s2
0 v s1
ds1ds2, 2.68
and linear operatorKPImL→domL∩KerPcan be written as
KPv t
0
sn
0
· · · s2
0 v s1
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 vs1ds1ds20}. Thus, by using the same method as the proof ofLemma 2.1, we can proveLemma 2.3, and we omit it.
Theorem 2.4. Letf : 0,1×Rn → Rbe a continuous function. Assume that conditionA2of Theorem 2.2and the following conditions are satisfied:
A5m−2
j1βj1,m−2
j1βjηj2/1;
A6there existsM >0, such that foru∈domL, if|ut|> Mfor allt∈0,1, then
m−2
j1
βj
1
ηj
s2
0
f s1, u
s1
, . . . , un−1 s1
e s1
ds1ds2/0; 2.71
A7there existsM∗>0 such that for anyd∈R, if|d|> M∗, then either
d·m−2
j1
βj
1
ηj
s2
0
f
s1, d,0, . . . ,0 e
s1
ds1ds2<0, 2.72
or else
d·m−2
j1
βj
1
ηj
s2
0
f
s1, d,0, . . . ,0 e
s1
ds1ds2>0. 2.73
Then, for everye ∈ L10,1, BVP1.1,1.3 has at least one solution inCn−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 f
t, ut, ut, ut
et, t∈0,1, u0 uξ, u0 0, u1 6u
1 6
−3u 1
3
−2u 1
2
, 2.74
whereft, u, v, w 1/22u 1/44v 1/44wsinw1/5, e∈L10,1, ξ∈0,1.
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 1
3j1βjηj−13
j1
βj1−ηj
5,
Δ2 Δ11 1 α
2 i1
αi5117, a0
1a1
1a2
1 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 f
t, ut, ut, ut
et, 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 j1βj
1
ηj
s2
0
f s1, u
s1
, u s1
, u s1
e s1
ds1ds2/0; 2.77
ivfor anyd∈R,one has
d 3 j1
βj
1
ηj
s2
0
f
s1, d,0,0
ds1ds2 5
192d2>0. 2.78
Furthermore
a0
1a1
1a2
11 81
81 8 3
8 < 1
2. 2.79
Hence fromTheorem 2.4, for everye∈L10,1,BVP2.76has at least one solutionu∈C20,1.
Acknowledgment
The authors thank the referee for valuable suggestions which led to improvement of the origi- nal manuscript.
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