Volume 2011, Article ID 456426,19pages doi:10.1155/2011/456426
Research Article
Positive Solutions of nth-Order Nonlinear
Impulsive Differential Equation with Nonlocal Boundary Conditions
Meiqiang Feng,
1Xuemei Zhang,
2and Xiaozhong Yang
21School of Science, Beijing Information Science & Technology University, Beijing 100192, China
2Department of Mathematics and Physics, North China Electric Power University, Beijing 102206, China
Correspondence should be addressed to Meiqiang Feng,[email protected] Received 25 March 2010; Accepted 9 May 2010
Academic Editor: Feliz Manuel Minh ´os
Copyrightq2011 Meiqiang Feng et al. 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.
This paper is devoted to study the existence, nonexistence, and multiplicity of positive solutions for thenth-order nonlocal boundary value problem with impulse effects. The arguments are based upon fixed point theorems in a cone. An example is worked out to demonstrate the main results.
1. Introduction
The theory of impulsive differential equations describes processes which experience a sudden change of their state at certain moments. Processes with such a character arise naturally and often, especially in phenomena studied in physics, chemical technology, population dynamics, biotechnology, and economics. For an introduction of the basic theory of impulsive differential equations, see Lakshmikantham et al.1; for an overview of existing results and of recent research areas of impulsive differential equations, see Benchohra et al.2. The theory of impulsive differential equations has become an important area of investigation in the recent years and is much richer than the corresponding theory of differential equations;
see, for instance,3–14and their references.
At the same time, a class of boundary value problems with integral boundary conditions arise naturally in thermal conduction problems 15, semiconductor problems 16, hydrodynamic problems 17. Such problems include two, three, and multipoint boundary value problems as special cases and attract much attention; see, for instance, 7,8,11,18–44and references cited therein. In particular, we would like to mention some results of Eloe and Ahmad19 and Pang et al. 22. In 19, by applying the fixed point
theorem in cones due to the work of Krasnosel’kii and Guo, Eloe and Ahmad established the existence of positive solutions of the followingnth boundary value problem:
xnt atft, xt 0, t∈0,1, x0 x0 · · ·xn−20 0,
x1 αx
η .
1.1
In22, Pang et al. considered the expression and properties of Green’s function for thenth-orderm-point boundary value problem
xnt atfxt 0, 0< t <1, x0 x0 · · ·xn−20 0,
x1 m−2
i1
βixξi,
1.2
where 0 < ξ1 < ξ2 < · · · < ξm−2 <1, βi >0,m−2
i1 βiξm−1i < 1. Furthermore, they obtained the existence of positive solutions by means of fixed point index theory.
Recently, Yang and Wei 23 and the author of 24 improved and generalized the results of Pang et al.22by using different methods, respectively.
On the other hand, it is well known that fixed point theorem of cone expansion and compression of norm type has been applied to various boundary value problems to show the existence of positive solutions; for example, see7,8,11,19,23,24. However, there are few papers investigating the existence of positive solutions ofnth impulsive differential equations by using the fixed point theorem of cone expansion and compression. The objective of the present paper is to fill this gap. Being directly inspired by19,22, using of the fixed point theorem of cone expansion and compression, this paper is devoted to study a class of nonlocal BVPs fornth-order impulsive differential equations with fixed moments.
Consider the following nth-order impulsive differential equations with integral boundary conditions:
xnt ft, xt 0, t∈J, t /tk,
−Δxn−1|ttk Ikxtk, k1,2, . . . , m, x0 x0 · · ·xn−20 0, x1
1
0
htxtdt.
1.3
Here J 0,1, f ∈ CJ ×R, R, Ik ∈ CR, R,andR 0,∞, tkk 1,2, . . . , m wheremis fixed positive integerare fixed points with 0 < t1 < t2 < · · · < tk < · · · < tm <
1,Δxn−1|ttk xn−1tk−xn−1t−k,wherexn−1tkandxn−1t−krepresent the right-hand limit and left-hand limit ofxn−1tatttk, respectively,h∈L10,1is nonnegative.
For the case ofh≡0, problem1.3reduces to the problem studied by Samo˘ılenko and Perestyuk in4. By using the fixed point index theory in cones, the authors obtained some
sufficient conditions for the existence of at least one or two positive solutions to the two-point BVPs.
Motivated by the work above, in this paper we will extend the results of4,19,22–
24to problem1.3. On the other hand, it is also interesting and important to discuss the existence of positive solutions for problem1.3when Ik/0k 1,2, . . . , m,, n ≥ 2, and h /≡0. Many difficulties occur when we deal with them; for example, the construction of cone and operator. So we need to introduce some new tools and methods to investigate the existence of positive solutions for problem1.3. Our argument is based on fixed point theory in cones45.
To obtain positive solutions of 1.3, the following fixed point theorem in cones is fundamental which can be found in45, page 93.
Lemma 1.1. Let Ω1 and Ω2 be two bounded open sets in Banach spaceE, such that 0 ∈ Ω1 and Ω1 ⊂ Ω2. LetP be a cone inEand let operatorA :P∩Ω2\Ω1 → P be completely continuous.
Suppose that one of the following two conditions is satisfied:
iAx/≥x,∀x∈P∩∂Ω1;Ax/≤x,∀x∈P∩∂Ω2; iiAx/≤x,∀x∈P∩∂Ω1;Ax/≥x,∀x∈P∩∂Ω2. Then,Ahas at least one fixed point inP∩Ω2\Ω1.
2. Preliminaries
In order to define the solution of problem1.3, we will consider the following space.
LetJJ\ {t1, t2, . . . , tn}, and P Cn−10,1
x∈C0,1:xn−1|tk,tk1∈Ctk, tk1, xn−1
t−k
xn−1tk,∃ xn−1 tk
, k1,2, . . . , m.
2.1
ThenP Cn−10,1is a real Banach space with norm xpcn−1max
x∞, x
∞, x
∞, . . . , xn−1
∞
, 2.2
wherexn−1∞supt∈J|xn−1t|, n1,2, . . . .
A functionx ∈ P Cn−10,1∩CnJis called a solution of problem1.3if it satisfies 1.3.
To establish the existence of multiple positive solutions in P Cn−10,1 ∩ CnJ of problem1.3, let us list the following assumptions:
H1f ∈CJ×R, R, Ik∈CR, R; H2μ∈0,1, whereμ1
0httn−1dt.
Lemma 2.1. Assume thatH1 and H2 hold. Then x ∈ P Cn−10,1∩CnJ is a solution of problem1.3if and only ifxis a solution of the following impulsive integral equation:
xt 1
0
Ht, sfs, xsdsm
k1
Ht, tkIkxtk, 2.3
where
Ht, s G1t, s G2t, s, 2.4
G1t, s 1 n−1!
⎧⎨
⎩
tn−11−sn−1−t−sn−1, 0≤s≤t≤1,
tn−11−sn−1, 0≤t≤s≤1, 2.5
G2t, s tn−1 1−1
0httn−1dt 1
0
htG1t, sdt. 2.6
Proof. First suppose thatx ∈P Cn−10,1∩CnJis a solution of problem1.3. It is easy to see by integration of1.3that
xn−1t xn−10− t
0
fs, xsds
0<tk<t
xn−1 tk
−xn−1tk
xn−10− t
0
fs, xsds−
0<tk<t
Ikxtk.
2.7
Integrating again and by boundary conditions, we can get
xn−2t xn−10t− t
0
t−sfs, xsds−
0<tk<t
Ikxtkt−tk. 2.8
Similarly, we get
xt − 1
n−1!
t
0
t−sn−1fs, xsdsxn−10 tn−1 n−1!−
tk<t
Ikxtkt−tkn−1
n−1! . 2.9
Lettingt1 in2.9, we find
xn−10 n−1!x1 1
0
1−sn−1fs, xsds
tk<1
Ikxtk1−tkn−1.
2.10
Substitutingx1 1
0htxtdtand2.10into2.9, we obtain
xt − 1
n−1!
t
0
t−sn−1fs, xsds tn−1 n−1!
n−1!
1
0
htxtdt
1
0
1−sn−1fs, xsds
tk<1
Ikxtk1−tkn−1
−
tk<t
Ikxtkt−tkn−1 n−1!
1
0
G1t, sfs, xsdsm
k1
G1t, tkIkxtk tn−1 1
0
htxtdt.
2.11
Multiplying2.11withhtand integrating it, we have 1
0
htxtdt
1
0
ht 1
0
G1t, sfs, xsds dt 1
0
htm
k1
G1t, tkIkxtkdt
1
0
httn−1dt 1
0
htxtdt,
2.12
that is,
1
0
htxtdt 1
1−1
0httn−1dt 1
0
ht 1
0
G1t, sfs, xsds dt
1
0
htm
k1
G1t, tkIkxtkdt
.
2.13
Then we have
xt 1
0
G1t, sfs, xsdsm
k1
G1t, tkIkxtk
tn−1 1−1
0httn−1dt 1
0
ht 1
0
G1t, sfs, xsds dt 1
0
htm
k1
G1t, tkIkxtkdt
. 2.14
Then, the proof of sufficient is complete.
Conversely, ifx is a solution of2.3, direct differentiation of2.3implies that, for t /tk,
xt 1
n−2!
t
0
tn−21−sn−1−t−sn−2
fs, xsds
1 n−2!
1
t
tn−21−sn−1fs, xsds
− 1 n−2!
tk<t
tn−21−tkn−1−t−tkn−2
Ikxtk
1 n−2!
tk≥t
tn−21−tkn−1Ikxtk
n−1tn−2 1−1
0httn−1dt
1 0
ht 1
0
G1t, sfs, xsds dt
1
0
htm
k1
G1t, tkIkxtkdt
, ...
xn−1t 1
0
1−sn−1fs, xsds− t
0
fs, xsds
tk<1
1−tkn−1Ikxtk−
tk<t
Ikxtk
n−1!
1−1
0httn−1dt
1 0
ht 1
0
G1t, sfs, xsds dt
1
0
htm
k1
G1t, tkIkxtkdt
.
2.15
Evidently,
Δxn−1|ttk −Ikxtk, k1,2, . . . , m, 2.16
xnt −ft, xt. 2.17
Sox∈CnJandΔxn−1|ttk −Ikxtk,k1,2, . . . , m, and it is easy to verify that x0 x0 · · ·xn−20 0, x1 1
0htxtdt, and the lemma is proved.
Similar to the proof of that from22, we can prove thatHt, s, G1t, s, andG2t, s have the following properties.
Proposition 2.2. The functionG1t, sdefined by2.5satisfyongG1t, s≥0 is continuous for all t, s∈0,1, G1t, s>0,∀t, s∈0,1.
Proposition 2.3. There existsγ >0 such that
t∈tminm,1G1t, s≥γG1τs, s, ∀s∈0,1, 2.18
whereτsis defined in2.20.
Proposition 2.4. Ifμ∈0,1, then one has
iG2t, s≥0 is continuous for allt, s∈0,1, G2t, s>0,∀t, s∈0,1;
iiG2t, s≤1/1−μ1
0htG1t, sdt,∀t∈0,1, s∈0,1.
Proof. From the properties of G1t, s and the definition of G2t, s, we can prove that the results ofProposition 2.4hold.
Proposition 2.5. Ifμ∈0,1, the functionHt, sdefined by2.4satisfies iHt, s≥0 is continuous for allt, s∈0,1, Ht, s>0,∀t, s∈0,1;
iiHt, s≤Hs≤H0for eacht, s∈0,1, and
t∈tminm,1Ht, s≥γ∗Hs, ∀s∈0,1, 2.19
whereγ∗min{γ, tn−1m }, and
Hs G1τs, s G21, s, τs s
1−1−s11/n−2, H0max
s∈J Hs, 2.20
γis defined inProposition 2.3.
Proof. iFrom Propositions2.2and2.4, we obtain thatHt, s≥0 is continuous for allt, s∈ 0,1, andHt, s>0,∀t, s∈0,1.
iiFromiiofProposition 2.2andiiofProposition 2.4, we haveHt, s≤Hsfor eacht, s∈0,1.
Now, we show that2.19holds.
In fact, fromProposition 2.3, we have
t∈tminm,1Ht, s≥γG1τs, s tn−1m 1−μ
1
0
htG1t, sdt
≥γ∗
G1τs, s 1 1−μ
1
0
htG1t, sdt
γ∗Hs, ∀s∈0,1.
2.21
Then the proof ofProposition 2.5is completed.
Remark 2.6. From the definition ofγ∗, it is clear that 0< γ∗<1.
Lemma 2.7. Assume thatH1andH2hold. Then, the solutionxof problem1.3satisfiesxt≥ 0,∀t∈J.
Proof. It is an immediate subsequence of the facts thatHt, s≥0 on0,1×0,1.
Remark 2.8. FromiiofProposition 2.5, one can find that
γ∗Hs≤Ht, s≤Hs, t∈tm,1, s∈J. 2.22
For the sake of applyingLemma 1.1, we construct a coneKinP Cn−10,1by
K
x∈P Cn−10,1:x≥0, xt≥γ∗xs, t∈tm,1, s∈J
. 2.23
DefineT :K → Kby
Txt 1
0
Ht, sfs, xsdsm
k1
Ht, tkIkxtk. 2.24
Lemma 2.9. Assume thatH1andH2hold. Then,TK ⊂ K, andT : K → K is completely continuous.
Proof. FromProposition 2.5and2.24, we have
t∈tminm,1Txt min
t∈tm,1
1
0
Ht, sfs, xsdsm
k1
Ht, tkIkxtk
≥ 1
0
t∈tminm,1Ht, sfs, xsdsm
k1
t∈tminm,1Ht, tkIkxtk
≥γ∗ 1
0
Hsfs, xsdsm
k1
HtkIkxtk
≥γ∗ 1
0
t∈0,1maxHt, sfs, xsdsm
k1
t∈0,1maxHt, tkIkxtk
≥γ∗max
t∈0,1
1
0
Ht, sfs, xsdsm
k1
Ht, tkIkxtk
γ∗Tx, ∀x∈K.
2.25
Thus,TK⊂K.
Next, by similar arguments to those in8one can prove thatT:K → Kis completely continuous. So it is omitted, and the lemma is proved.
3. Main Results
Write
fβlim sup
x→β
maxt∈J
ft, x
x , fβlim inf
x→β min
t∈J
ft, x x , Iβk lim inf
x→β
Ikx
x , Iβk lim sup
x→β
Ikx x ,
3.1
whereβdenotes 0or∞.
In this section, we applyLemma 1.1to establish the existence of positive solutions for BVP1.3.
Theorem 3.1. Assume thatH1andH2hold. In addition, lettingf andIksatisfy the following conditions:
H3f0 0 andI0k 0, k1,2, . . . , m;
H4f∞∞orI∞k ∞, k1,2, . . . , m, BVP1.3has at least one positive solution.
Proof. ConsideringH3, there existsη >0 such that
ft, x≤εx, Ikx≤εkx, k1,2, . . . , m, ∀0≤x≤η, t∈J, 3.2
whereε, εk>0 satisfy
max{H0,1G0}
εm
k1
εk
<1; 3.3
here
G0max
G10, G20, . . . , Gn−10 , G10 max
t,s∈J,t /sG2tt, s max
t,s∈J,t /s
n−1tn−2 1−μ
1
0
htG1t, sdt,
G20 max
t,s∈J,t /sG2tt, s max
t,s∈J,t /s
n−1n−2tn−3 1−μ
1
0
htG1t, sdt, ...
Gn−10 max
t,s∈J,t /sGn−12t t, s max
t,s∈J,t /s
n−1!
1−μ 1
0
htG1t, sdt.
3.4
Now, for 0< r < η, we prove that
Tx/≥x, x∈K, xpcn−1r. 3.5
In fact, if there existsx1 ∈K,x1pcn−1rsuch thatTx1≥x1. Noticing3.2, then we have
0≤x1t≤ 1
0
Ht, sfs, x1sdsm
k1
Ht, tkIkx1tk
≤εr 1
0
Hsdsr
m k1
Htkεk
≤rH0
εm
k1
εk
< rx1pcn−1, x1t≤
1
0
Htt, sfs, x1sdsm
k1
Htt, tkIkx1tk
≤ 1
0
G1tt, sG2tt, sfs, x1sds
m
k1
G1tt, tkG2tt, tkIkx1tk
≤ 1
0
1G10
fs, x1sdsm
k1
1G10
Ikx1tk
≤r
1G10 εm
k1
εk
< rx1pcn−1, x1t≤
1
0
Htt, sfs, x1sdsm
k1
Htt, tkIkx1tk
≤ 1
0
G1tt, sG2tt, sfs, x1sds
m
k1
G1tt, tkG2tt, tkIkx1tk
≤ 1
0
1G20
fs, x1sdsm
k1
1G20
Ikx1tk
≤r
1G20 εm
k1
εk
< rx1pcn−1, ...
xn−11 t≤ 1
0
Htn−1t, sfs, x1sdsm
k1
Htn−1t, tkIkx1tk
≤ 1
0
Gn−11t t, sGn−12t t, s
fs, x1sds m
k1
Gn−11t t, tkGn−12t t, tk
Ikx1tk
≤ 1
0
1Gn0
fs, x1sdsm
k1
1Gn0
Ikx1tk
≤r
1Gn0 εm
k1
εk
< rx1pcn−1, 3.6
where
G1tt, s 1 n−2!
⎧⎨
⎩
tn−21−sn−1−t−sn−2 if 0≤s≤t≤1, tn−21−sn−1 if 0≤t≤s≤1,
G1tt, s 1 n−3!
⎧⎨
⎩
tn−31−sn−1−t−sn−3 if 0≤s≤t≤1, tn−31−sn−1 if 0≤t≤s≤1,
...
Gn−11t t, s
⎧⎨
⎩
1−sn−1−1 if 0≤s≤t≤1, 1−sn−1 if 0≤t≤s≤1,
t,s∈J,t /maxs
GN1t t, s1, N1,2, . . . , n−1.
3.7
Therefore,x1pcn−1<x1pcn−1, which is a contraction. Hence,3.2holds.
Next, turning toH4. Case1.f∞∞. There existsτ >0 such that
ft, x≥Mx, t∈J, x≥τ, 3.8
whereM >γ∗H01−tm−1. Choose
R >max r, τ
γ∗−1
. 3.9
We show that
Tx/≤x, x∈K, xpcn−1 R. 3.10
In fact, if there existsx0∈K,x0pcn−1Rsuch thatTx0≤x0, then
x0t≥γ∗x0s, t∈tm,1, s∈J. 3.11
This and3.9imply that
t∈tminm,1x0t≥γ∗x0pcn−1 γ∗R > τ. 3.12
So, we have
t∈J⇒x0t≥Tx0t≥ min
t∈tm,1
1
tm
Ht, sfs, x0sds≥γ∗H0M 1
tm
x0sds, 3.13
that is,
1
tm
x0tdt≥γ∗H0M1−tm 1
tm
x0sds. 3.14
It is easy to see that
1
tm
x0sds >0. 3.15
In fact, if1
tmx0sds0, thenx0t 0, fort∈tm,1. Sincex0 ∈K, x0s 0,∀s∈J.
Hence,x0pcn−1 xn−10 ∞ x0∞ 0, which contracts x0pcn−1 R. So,3.15holds.
Therefore,M≤γ∗H01−tm−1, this is also a contraction. Hence,3.10holds.
Case2.I∞k ∞, k1,2, . . . , m. There existsτ1>0 such that
Ikx≥Mkx, x≥τ1, 3.16
whereMk > γ∗H0−1, k 1,2, . . . , m. If we defineM∗ min{Mk : k 1,2, . . . , m}, then M∗>γ∗H0−1. Choose
R >max r, τ1
γ∗−1
. 3.17
We prove that3.10holds.
In fact, if there existsx00 ∈K,x00pcn−1 Rsuch thatTx00≤x00, then
x00t≥γ∗x00s, t∈tm,1, s∈J. 3.18
This and3.17imply that
t∈tminm,1x00t≥γ∗x00pcn−1γ∗R > τ1. 3.19
So, we have
t∈J⇒x00t≥Tx00t≥ min
t∈tm,1
m k1
Ht, tkIkx00tk
≥γ∗H0
m k1
Mkx00tk
≥γ∗H0M∗ m k1
x00tk.
3.20
From3.20, we obtain that
x00t1≥γ∗H0M∗ m k1
x00tk,
x00t2≥γ∗H0M∗ m k1
x00tk, ...
x00tk≥γ∗H0M∗ m k1
x00tk.
3.21
So, we have
m k1
x00tk≥mγ∗H0M∗ m k1
x00tk. 3.22
From the definition ofM∗, we can find that m
k1
x00tk> m m k1
x00tk, x00∈K,x00pcn−1R. 3.23
Similar to the proof in case1, we can show thatm
k1x00tk> 0. Then, from3.23, we havem <1, which is a contraction. Hence,3.10holds.
ApplyingiofLemma 1.1to3.2and3.10yields thatT has a fixed pointx∈Kr,R {x: r ≤ xpcn−1 ≤ R}. Thus, it follows that BVP1.3has at least one positive solution, and the theorem is proved.
Theorem 3.2. Assume thatH1andH2hold. In addition, lettingf andIksatisfy the following conditions:
H5f∞0 andI∞k 0, k1,2, . . . , m;
H6f0∞orI0k ∞, k1,2, . . . , m, BVP1.3has at least one positive solution.
Proof. Considering H5, there exists r > 0 such that ft, x ≤ εr, Ikx ≤ εkr,andk 1,2, . . . , m, forx≥r, t∈J, whereε, εk>0 satisfy max{H0,1G0}εm
k1εk<1.
Similar to the proof of3.2, we can show that
Tx/≥x, x∈K, xpc1r. 3.24
Next, turning toH6. Under conditionH6, similar to the proof of3.10, we can also show that
Tx/≤x, x∈K, xpc1R. 3.25
Applyingi of Lemma 1.1to 3.24 and 3.25 yields that T has a fixed point x ∈ Kr,R {x :r ≤ xpcn−1 ≤ R}. Thus, it follows that BVP1.3has one positive solution, and the theorem is proved.
Theorem 3.3. Assume thatH1,H2,H3,andH5hold. In addition, lettingfandIk satisfy the following condition:
H7there is aς >0 such thatγ∗ς≤x≤ςandt∈Jimplies
ft, x≥τς, Ikx≥τkς, k1,2, . . . , 3.26
whereτ, τk≥0 satisfyτm
k1τk>0, τ1
tmH1/2, sdsm
k1τkH1/2, tk>1, BVP 1.3has at least two positive solutionsx∗andx∗∗with 0<x∗pcn−1 < ς <x∗∗pcn−1. Proof. We chooseρ, ξwith 0 < ρ < ς < ξ. IfH3holds, similar to the proof of3.2, we can prove that
Tx/≥x, x∈K, xpc1ρ. 3.27
IfH5holds, similar to the proof of3.24, we have
Tx/≥x, x∈K, xpcn−1 ξ. 3.28
Finally, we show that
Tx/≤x, x∈K, xpcn−1 ς. 3.29
In fact, if there existsx2∈Kwithx2pcn−1ς,then by2.23, we have
x2t≥γ∗x2pcn−1γ∗ς, 3.30
and it follows fromH7that
x2t≥ 1
tm
H 1
2, s
fs, x2sdsm
k1
H 1
2, tk
Ikx2tk
≥ς
τ 1
tm
H 1
2, s
dsm
k1
τkH 1
2, tk
> ςx2pcn−1,
3.31
that is,x2pcn−1>x2pcn−1, which is a contraction. Hence,3.29holds.
ApplyingLemma 1.1to3.27, 3.28, and 3.29yields that T has two fixed points x∗, x∗∗ withx∗ ∈ Kρ,ς, x∗∗ ∈ Kς,ξ. Thus it follows that BVP1.3has two positive solutions x∗, x∗∗with 0<x∗pcn−1< ς <x∗∗pcn−1. The proof is complete.
Our last results corresponds to the case when problem1.3has no positive solution.
Write
Δ H01m. 3.32
Theorem 3.4. AssumeH1,H2, ft, x<Δ−1x, t∈J, x >0, andIkx<Δ−1x,∀x >0, then problem1.3has no positive solution.
Proof. Assume to the contrary that problem1.3has a positive solution, that is,Thas a fixed pointy. Theny∈K, y >0 fort∈0,1, and
y
∞≤ 1
0
Hsf
s, ys
dsm
k1
HtkIk
ytk
<
1
0
HsΔ−1ysdsm
k1
HtkΔ−1 y
∞
≤H0Δ−1 y
∞m
k1
H0Δ−1 y
∞
H0Δ−11m y
∞
y
∞,
3.33
which is a contradiction, and this completes the proof.
To illustrate how our main results can be used in practice we present an example.
Example 3.5. Consider the following boundary value problem:
−x4t 3
t51x5tanhx, t∈J, t /1 2,
−Δx3|t11/2x3 1
2
,
x0 x0 x0 0, x1
1
0
txtdt.
3.34
Conclusion. BVP3.34has at least one positive solution.
Proof. BVP3.34can be regarded as a BVP of the form1.3, where
ht t, μ
1
0
t·t3dt 1
5, t1 1
2, ft, x 3
t51x5tanhx, I1x x3,
G1t, s 1 6
⎧⎨
⎩
t31−s3−t−s3, 0≤s≤t≤1, t31−s3, 0≤t≤s≤1, G2t, s 1
24t3 3
4s−2s23 2s3−1
4s5
.
3.35
It is not difficult to see that conditionsH1andH2hold. In addition,
f0lim sup
x→0
maxt∈J
ft, x
x 0, I0k lim sup
x→0
Ikx
x 0,
f∞lim inf
x→ ∞ min
t∈J
ft, x
x ∞.
3.36
Then, conditions H3 and H4 of Theorem 3.1 hold. Hence, by Theorem 3.1, the conclusion follows, and the proof is complete.
Acknowledgment
This work is supported by the National Natural Science Foundation of China10771065, the Natural Sciences Foundation of Heibei Province A2007001027, the Funding Project for Academic Human Resources Development in Institutions of Higher Learning Under the Jurisdiction of Beijing Municipality PHR201008430, the Scientific Research Common Program of Beijing Municipal Commission of EducationKM201010772018 and Beijing Municipal Education Commission71D0911003. The authors thank the referee for his/her careful reading of the paper and useful suggestions.
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