Volume 2008, Article ID 257680,19pages doi:10.1155/2008/257680
Research Article
Twin Positive Solutions of a Nonlinear
m -Point Boundary Value Problem for Third-Order p -Laplacian Dynamic Equations on Time Scales
Wei Han1and Guang Zhang2
1Department of Mathematics, North University of China, Taiyuan 030051, China
2Department of Mathematics, School of Science, Tianjin University of Commerce, Tianjin 300134, China
Correspondence should be addressed to Wei Han,qd [email protected] Received 20 April 2008; Accepted 20 June 2008
Recommended by Binggen Zhang
Several existence theorems of twin positive solutions are established for a nonlinear m-point boundary value problem of third-order p-Laplacian dynamic equations on time scales by using a fixed point theorem. We present two theorems and four corollaries which generalize the results of related literature. As an application, an example to demonstrate our results is given. The obtained conditions are different from some known results.
Copyrightq2008 W. Han and G. Zhang. 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
A time scaleT is a nonempty closed subset ofR. We make the blanket assumption that 0 andT are points inT. By an interval0, T, we always mean the intersection of the real interval0, T with the given time scale, that is,0, T∩T.
In this paper, we will be concerned with the existence of positive solutions of the p- Laplacian dynamic equations on time scales:
φpuΔ∇∇atft, ut 0, t∈0, T, 1.1
φpuΔ∇0 m−2
i1
aiφpuΔ∇ξi, uΔ0 0, uT m−2
i1
biuξi, 1.2
whereφpsisp-Laplacian operator; that is,φps |s|p−2s, p >1, φ−1p φq, 1/p1/q1, 0<
ξ1<· · ·< ξm−2< ρT, and
H1ai, bi∈0,∞, i1,2, . . . ,satisfy 0<m−2
i1 ai<1 andm−2
i1 bi<1;
H2at∈Cld0, T ,0,∞and there existst0∈ξm−2, Tsuch thatat0>0;
H3f ∈C0, T ×0,∞,0,∞.
We point out that theΔ-derivative and the∇-derivative in1.2and theCldspace inH2are defined inSection 2.
Recently, there has been much attention paid to the existence of positive solutions for third-order nonlinear boundary value problems of differential equations. For example, see 1–10 and the listed references. Anderson2 considered the following third-order nonlinear problem:
xt ft, xt, t1≤t≤t3,
xt1 xt2 0, γxt3 δxt3 0. 1.3
He used the Krasnoselskii and the Leggett and Williams fixed-point theorems to prove the existence of solutions to the nonlinear problem1.3. Li6 considered the existence of single and multiple positive solutions to the nonlinear singular third-order two-point boundary value problem:
ut λatfut 0, 0< t <1,
u0 u0 u1 0. 1.4
Under various assumptions onaandf, they established intervals of the parameterλwhich yield the existence of at least two and infinitely many positive solutions of the boundary value problem by using Krasnoselski’s fixed-point theorem of cone expansion-compression type. Liu et al.7 discussed the existence of at least one or two nondecreasing positive solutions for the following singular nonlinear third-order differential equations:
xt λαtft, xt 0, a < t < b,
xa xa xb 0. 1.5
Green’s function and the fixed-point theorem of cone expansion-compression type are utilized in their paper. In8 , Sun considered the following nonlinear singular third-order three-point boundary value problem:
ut−λatFt, ut 0, 0< t <1,
u0 uη u1 0. 1.6
He obtained various results on the existence of single and multiple positive solutions to the boundary value problem1.6by using a fixed-point theorem of cone expansion-compression type due to Krasnosel’skii. In10 , Zhou and Ma studied the existence and iteration of positive solutions for the following third-order generalized right-focal boundary value problem with p-Laplacian operator:
φput qtft, ut, 0≤t≤1, u0 m
i1αiuξi, uη 0, u1 n
i1βiuθi. 1.7
They established a corresponding iterative scheme for1.7by using the monotone iterative technique.
On the other hand, the existence of positive solutions for third-order nonlinear boundary value problems of difference equations is also extensively studied by a number of authorssee 1,3,5,9 and the listed references. The present work is motivated by a recent paper4 . In 4 , Henderson and Yin considered the existence of solutions for a third-order boundary value problem on a time-scale equation of the form
uΔ3ft, u, uΔ, uΔΔ, t∈T, 1.8 which is uniform for the third-order difference equation and the third-order differential equation.
2. Preliminaries and lemmas
For convenience, we list the following definitions which can be found in4,11–15 .
Definition 2.1. Let T be a time scale. Fort < supT and r > infT, define the forward jump operatorσand the backward jump operatorρ, respectively, by
σt inf{τ∈T|τ > t} ∈T,
ρr sup{τ∈T|τ < r} ∈T 2.1
for allt, r ∈ T. Ifσt > t,tis said to be right-scattered, and ifρr < r, r is said to be left- scattered; ifσt t,tis said to be right-dense, and ifρr r, r is said to be left-dense. If T has a right-scattered minimumm, defineTk T− {m}; otherwise setTk T. If T has a left-scattered maximumM, defineTkT− {M}; otherwise setTkT.
Definition 2.2. Forf : T→Randt ∈ Tk, the delta derivative off at the pointtis defined to be the numberfΔt provided that it exists, with the property that for each >0 there is a neighborhoodUoftsuch that
|fσt−fs−fΔtσt−s| ≤|σt−s| 2.2 for alls∈U.
Forf:T→Randt∈Tk, the nabla derivative offattis denoted byf∇t provided that it exists, with the property that for each >0 there is a neighborhoodUoftsuch that
|fρt−fs−f∇tρt−s| ≤|ρt−s| 2.3 for alls∈U.
Definition 2.3. A functionfis left-dense continuousi.e.,ld-continuousiff is continuous at each left-dense point inT, and its right-sided limit exists at each right-dense point in T.
Definition 2.4. IfφΔt ft, then one defines the delta integral by b
a
ftΔtφb−φa. 2.4
IfF∇t ft, then one defines the nabla integral by b
a
ft∇tFb−Fa. 2.5
To prove the main results in this paper, we will employ several lemmas. These lemmas are based on the linear BVP
φpuΔ∇∇ht 0, t∈0, T, 2.6 φpuΔ∇0 m−2
i1
aiφpuΔ∇ξi, uΔ0 0, uT m−2
i1
biuξi. 2.7
Lemma 2.5. Ifm−2
i1 ai/1 andm−2
i1 bi/1, then forh∈Cld0, T the BVP2.6-2.7has the unique solution
ut −
t
0
t−sφq
s 0
hτ∇τ−A
∇sC, 2.8
where
A− m−2
i1 ai
ξi
0hτ∇τ
1−m−2
i1 ai
,
C T
0T−sφqs
0hτ∇τ−A∇s−m−2
i1 bi
ξi
0ξi−sφqs
0hτ∇τ−A∇s
1−m−2
i1 bi
.
2.9
Proof. iLetube a solution, then we will show that2.8holds. By taking the nabla integral of problem2.6on0, t, we have
φpuΔ∇t − t
0
hτ∇τA 2.10
then
uΔ∇t φq
− t
0
hτ∇τA
−φq
t
0
hτ∇τ−A
. 2.11
By taking the nabla integral of2.11on0, t, we can get uΔt −
t
0
φq
s
0
hτ∇τ−A
∇sB. 2.12
By taking the delta integral of2.12on0, t, we can get
ut −
t
0
t−sφq
s
0
hτ∇τ−A
∇sBtC. 2.13
Similarly, lett0 on2.10, then we haveφpuΔ∇0 A; lettξion2.10, then we have
φpuΔ∇ξi − ξi
0
hτ∇τA. 2.14
Lett0 on2.12, then we have
uΔ0 B. 2.15
LettT on2.13, then we have
uT −
T
0
T−sφq
s 0
hτ∇τ−A
∇sBTC. 2.16
Similarly, lettξion2.13, then we have
uξi − ξi
0
ξi−sφq
s 0
hτ∇τ−A
∇sBξiC. 2.17
By the boundary condition2.7, we can get
B0, 2.18
Am−2
i1ai
− ξi
0
hτ∇τA
. 2.19
Solving2.19, we get
A− m−2
i1 ai
ξi
0hτ∇τ
1−m−2
i1 ai
. 2.20
By the boundary condition2.7, we can obtain
− T
0
T−sφq
s
0
hτ∇τ−A
∇sCm−2
i1
bi
− ξi
0
ξi−sφq
s
0
hτ∇τ−A
∇sC . 2.21
Substituting2.20in the above expression, one has
C T
0T−sφqs
0hτ∇τ−A∇s−m−2
i1 bi
ξi
0ξi−sφqs
0hτ∇τ−A∇s
1−m−2
i1 bi
. 2.22
iiWe show that the functionugiven in2.8is a solution.
Letube as in2.8. By12, Theorem 2.10iii and taking the delta derivative of2.8, we have
uΔt − t
0
φq
s
0
hτ∇τ−A
∇s; 2.23
moreover, we get
uΔ∇t −φq
t
0
hτ∇τ−A
,
φpuΔ∇ − t
0
hτ∇τ−A
.
2.24
Taking the nabla derivative of this expression yields φpuΔ∇∇ −ht. Also, routine calculation verifies thatusatisfies the boundary value conditions in2.7so thatugiven in 2.8is a solution of2.6and2.7. The proof is complete.
Lemma 2.6. AssumeH1holds. Forh∈Cld0, T andh≥0, the unique solutionuof2.6and2.7 satisfies
ut≥0 for t∈0, T . 2.25
Proof. Let
ϕ0s φq
s 0
hτ∇τ−A
. 2.26
Since
s
0
hτ∇τ−A
s
0
hτ∇τ
m−2
i1 ai
ξi
0hτ∇τ
1−m−2
i1 ai
≥0, 2.27
thenϕ0s≥0.
According toLemma 2.5, we get
u0 C
T
0T−sϕ0s∇s−m−2
i1 bi
ξi
0ξi−sϕ0s∇s 1−m−2
i1 bi
≥ T
0T−sϕ0s∇s−m−2
i1 bi
ξi
0T−sϕ0s∇s 1−m−2
i1 bi
T
0T−sϕ0s∇s−m−2
i1 biT
0T−sϕ0s∇s−T
ξiT−sϕ0s∇s 1−m−2
i1 bi
T
0
T−sϕ0s∇s m−2
i1 bi
T
ξiT−sϕ0s∇s 1−m−2
i1 bi
≥0,
uT −
T
0
T−sϕ0s∇sC
− T
0
T−sϕ0s∇s T
0T−sϕ0s∇s−m−2
i1 bi
ξi
0ξi−sϕ0s∇s 1−m−2
i1 bi
≥ − T
0
T−sϕ0s∇s T
0T−sϕ0s∇s−m−2
i1 bi
ξi
0T−sϕ0s∇s 1−m−2
i1 bi
m−2
i1 bi
T
ξiT−sϕ0s∇s 1−m−2
i1 bi
≥0.
2.28
Ift∈0, T, we have
ut −
t
0
t−sϕ0s∇s 1 1−m−2
i1 bi
T
0
T−sϕ0s∇s−m−2
i1
bi
ξi
0
ξi−sϕ0s∇s
≥ − T
0
T−sϕ0s∇s 1 1−m−2
i1 bi
T
0
T−sϕ0s∇s−m−2
i1
bi
ξi
0
T−sϕ0s∇s
1 1−m−2
i1 bi
−
1−m−2
i1
bi
T
0
T−sϕ0s∇s T
0
T−sϕ0s∇s−m−2
i1
bi
ξi
0
T−sϕ0s∇s
1 1−m−2
i1 bi m−2
i1
bi
T
ξi
T−sϕ0s∇s≥0.
2.29 Sout≥0, t∈0, T .
Lemma 2.7. AssumeH1holds. Ifh∈Cld0, T andh≥0, then the unique solutionuof 2.6and 2.7satisfies
t∈0,T inf ut≥γu, 2.30
where
γ m−2
i1 biT−ξi T−m−2
i1 biξi
, u max
t∈0,T |ut|. 2.31
Proof. It is easy to check thatuΔt −t
0ϕ0s∇s≤0; this implies that uu0, min
t∈0,T ut uT. 2.32
It is easy to see that uΔt2 ≤ uΔt1for any t1, t2 ∈ 0, T with t1 ≤ t2. Hence,uΔt is a decreasing function on0, T . This means that the graph ofutis concave down on0, T.
For eachi∈ {1,2, . . . , m−2}, we have
uT−u0
T−0 ≥ uT−uξi
T−ξi , 2.33
that is,
Tuξi−ξiuT≥T−ξiu0, 2.34
so that
T
m−2
i1
biuξi−m−2
i1
biξiuT≥m−2
i1
biT−ξiu0. 2.35
With the boundary conditionuT m−2
i1 biuξi, we have uT≥
m−2
i1 biT−ξi T−m−2
i1 biξi
u0. 2.36
This completes the proof.
Let the norm onCld0, T be the maximum norm. Then, theCld0, T is a Banach space.
It is easy to see that BVP1.1-1.2has a solutionuutif and only ifuis a fixed point of the operator
Aut − t
0
t−sφq
s 0
aτfτ, uτ∇τ−A
∇sC, 2.37 where
A− m−2
i1 ai
ξi
0aτfτ, uτ∇τ 1−m−2
i1 ai
,
C T
0T−sφqs
0aτfτ, uτ∇τ−A∇s− m−2
i1 bi
ξi
0ξi−sφqs
0aτfτ, uτ∇τ−A∇s 1−m−2
i1 bi
. 2.38 Denote
K
u|u∈Cld0, T , ut≥0, inf
t∈0,T ut≥γu
, 2.39
whereγ is the same as inLemma 2.7. It is obvious thatKis a cone inCld0, T . ByLemma 2.7, AK⊂K. So by applying Arzela-Ascoli theorem on time scales16 , we can obtain thatAK is relatively compact. In view of Lebesgue’s dominated convergence theorem on time scales 13 , it is easy to prove thatAis continuous. Hence,A:K→Kis completely continuous.
Lemma 2.8. A:K→Kis completely continuous.
Proof. First, we show thatAmaps bounded set into bounded set.
Assumec >0 is a constant andu∈Kc{u∈K:u ≤c}. Note that the continuity off guarantees that there isc>0 such thatft, ut≤φpcfort∈0, T . So
Au max
t∈0,T |Aut| ≤C
≤ T
0T−sφqs
0aτfτ, uτ∇τ−A∇s 1−m−2
i1 bi
≤cT
0T−sφqs
0aτ∇τm−2
i1 ai
ξi
0aτ∇τ/1−m−2
i1 ai∇s 1−m−2
i1 bi
.
2.40
That is,AKcis uniformly bounded.
In addition, notice that for anyt1, t2∈0, T , we have
|Aut1−Aut2|
t1
0
t2−t1φq
s
0
aτfτ, uτ∇τ−A
∇s t2
t1
t2−sφq
s
0
aτfτ, uτ∇τ−A
∇s
≤c|t1−t2| T
0
φq
s
0
aτ∇τ m−2
i1 ai
ξi
0aτ∇τ
1−m−2
i1 ai
∇s Tmax
s∈0,T φq
s 0
aτ∇τ m−2
i1 ai
ξi
0aτ∇τ
1−m−2
i1 ai
.
2.41
So, by applying Arzela-Ascoli theorem on time scales16 , we obtain thatAKc is relatively compact.
Finally, we prove that A : Kc→K is continuous. Suppose that {un}∞n1 ⊂ Kc and unt converges tou∗tuniformly on 0, T .Hence, {Aunt}∞n1 is uniformly bounded and equicontinuous on0, T . The Arzela-Ascoli theorem on time scales 16 tells us that there exists uniformly convergent subsequence in{Aunt}∞n1.Let{Aunmt}∞m1be a subsequence which converges tovtuniformly on0, T .In addition,
0≤Aunt≤cT
0T−sφqs
0aτ∇τm−2
i1 ai
ξi
0aτ∇τ/1−m−2
i1 ai∇s 1−m−2
i1 bi
. 2.42
Observe the expression of{Aunmt},and then lettingm→∞,we obtain
vt −
t
0
t−sφq
s
0
aτfτ, u∗τ∇τ−A∗
∇sC∗, 2.43
where
A∗− m−2
i1 ai
ξi
0aτfτ, u∗τ∇τ 1−m−2
i1 ai
,
C∗ 1
1−m−2
i1 bi
T 0
T−sφq
s 0
aτfτ, u∗τ∇τ−A∗
∇s
−m−2
i1
bi
ξi
0
ξi−sφq
s 0
aτfτ, u∗τ∇τ−A∗
∇s .
2.44
Here, we have used the Lebesgue dominated convergence theorem on time scales13 . From the definition ofA, we know thatvt Au∗ton0, T .This shows that each subsequence of{Aunt}∞n1uniformly converges toAu∗t.Therefore, the sequence{Aunt}∞n1uniformly converges toAu∗t.This means thatAis continuous atu∗ ∈ Kc. So,Ais continuous onKc
sinceu∗is arbitrary. Thus,Ais completely continuous. This proof is complete.
Lemma 2.9. Let
ϕs φq
s 0
aτfτ, uτ∇τ−A
. 2.45
Forξi i1, . . . , m−2, ξi
0
ξi−sϕs∇s≤ ξi
T T
0
T−sϕs∇s. 2.46
Proof. Since s
0
aτfτ, uτ∇τ−A s
0
aτfτ, uτ∇τ m−2
i1 ai
ξi
0aτfτ, uτ∇τ 1−m−2
i1 ai
≥0,
2.47
thenϕs≥0. For all t∈0, T , we have t
0t−sϕs∇s t
∇ tt
0ϕs∇s−t
0t−sϕs∇s
tρt ≥0. 2.48
In fact, letψt tt
0ϕs∇s−t
0t−sϕs∇s; taking the nabla derivative of this expression, we have
ψ∇t t
0
ϕs∇stϕt− t
0
ϕs∇stϕt≥0. 2.49
Hence,ψtis a nondecreasing function on0, T . That is,
ψt≥0. 2.50
For all t∈0, T ,
t
0t−sϕs∇s
t ≤
T
0T−sϕs∇s
T . 2.51
By2.51, forξi i1, . . . , m−2, we have ξi
0
ξi−sϕs∇s≤ ξi
T T
0
T−sϕs∇s. 2.52
Lemma 2.10see17 . LetEbe a Banach space, and letK ⊂Ebe a cone. AssumeΩ1,Ω2are open bounded subsets ofEwith 0∈Ω1,Ω1⊂Ω2,and let
F:K∩Ω2\Ω1→Kbe a completely continuous operator such that iFu ≤ u, u∈K∩∂Ω1,andFu ≥ u, u∈K∩∂Ω2, or iiFu ≥ u, u∈K∩∂Ω1, andFu ≤ u, u∈K∩∂Ω2. Then,Fhas a fixed point inK∩Ω2\Ω1.
Now, we introduce the following notations. Let A0
1 1−m−2
i1 bi
T
0
T−sφq
s
0
aτ∇τ
m−2
i1 ai
ξi
0aτ∇τ
1−m−2
i1 ai
∇s −1
,
B0
Tm−2
i1 bi−m−2
i1 biξi
T1−m−2
i1 bi T
0
T−sφq
s 0
aτ∇τ
m−2
i1 ai
ξi
0aτ∇τ
1−m−2
i1 ai
∇s −1
.
2.53
Forl >0, Ωl{u∈K:u< l}, and∂Ωl{u∈K:ul},
αl sup{Au: u∈∂Ωl}, βl inf{Au: u∈∂Ωl}, 2.54 byLemma 2.6, whereαandβare well defined.
3. Main results
Theorem 3.1. AssumeH1,H2, andH3hold, and assume that the following conditions hold:
A1pi∈C0,∞,0,∞, i1,2, and
l→0lim p1l
lp−1 < Ap−10 , lim
l→∞
p2l
lp−1 < Ap−10 ; 3.1
A2ki∈L10, T ,0,∞, i1,2;
A3there exist 0< c1≤c2and 0≤λ2< p−1< λ1such that
ft, l≤p1l k1tlλ1, t, l∈0, T ×0, c1 ,
ft, l≤p2l k2tlλ2, t, l∈0, T ×c2,∞; 3.2
A4there existsb >0 such that
min{ft, l:t, l∈0, T ×γb, b } ≥bB0p−1. 3.3
Then, problem1.1-1.2has at least two positive solutionsu∗1, u∗2satisfying 0<u∗1< b <u∗2. Theorem 3.2. AssumeH1,H2, andH3hold, and assume that the following conditions hold:
B1pi∈C0,∞,0,∞, i3,4, and
l→0lim p3l
lp−1 >
B0
γ p−1
, lim
l→∞
p4l lp−1 >
B0
γ p−1
; 3.4
B2ki∈L10, T ,0,∞, i3,4;
B3there exist 0< c3≤c4and 0≤λ4< p−1< λ3such that
ft, l≥p3l−k3tlλ3, t, l∈0, T ×0, c3 ,
ft, l≥p4l−k4tlλ4, t, l∈0, T ×c4,∞; 3.5
B4there existsa >0 such that
max{ft, l:t, l∈0, T ×0, a } ≤aA0p−1. 3.6
Then, problem1.1-1.2has at least two positive solutionsu∗3, u∗4satisfying 0<u∗3< a <u∗4. Proof ofTheorem 3.1. Let
1
2min
Ap−10 −lim
l→0
p1l
lp−1 , Ap−10 −lim
l→∞
p2l
lp−1 , 3.7
then there exist 0< a1≤c1andc2≤a2<∞such that
p1l≤Ap−10 −lp−1, 0≤l≤a1,
p2l≤Ap−10 −lp−1, a2≤l≤∞. 3.8
If 0≤l≤a1, u∈∂Ωl, then 0≤ut≤l, 0≤t≤T. By conditionA3, we have ft, ut≤p1ut k1tuλ1t
≤Ap−10 −up−1t k1tuλ1t
≤Ap−10 −up−1k1tuλ1 Ap−10 −lp−1k1tlλ1
3.9
so thats
0
aτfτ, uτ∇τ−A
s
0
aτfτ, uτ∇τ m−2
i1 ai
ξi
0aτfτ, uτ∇τ 1−m−2
i1 ai
≤ s
0
aτAp−10 −lp−1k1τlλ1 ∇τ m−2
i1 ai
ξi
0aτAp−10 −lp−1k1τlλ1 ∇τ 1−m−2
i1 ai
. 3.10 Therefore,
Au ≤C 1 1−m−2
i1 bi
T
0
T−sϕs∇s−m−2
i1bi
ξi
0
ξi−sϕs∇s
≤ 1
1−m−2
i1 bi
T
0
T−sϕs∇s
≤ 1
1−m−2
i1 bi
× T
0
T−sφq
s
0
aτAp−10 −lp−1k1τlλ1 ∇τ
m−2
i1 ai
ξi
0aτAp−10 −lp−1k1τlλ1 ∇τ 1−m−2
i1 ai
∇s.
3.11
It follows that αl
l ≤ 1
1−m−2
i1 bi
× T
0
T−sφq
s 0
aτAp−10 −k1τlλ1−p1 ∇τ
m−2
i1 ai
ξi
0aτAp−10 −k1τlλ1−p1 ∇τ 1−m−2
i1 ai
∇s.
3.12
Noticingλ1−p1>0, we have liml→0
αl
l ≤ 1
1−m−2
i1 bi
T
0
T−sφq
s
0
aτAp−10 −∇τ m−2
i1 ai
ξi
0aτAp−10 −∇τ 1−m−2
i1 ai
∇s
Ap−10 −1/p−1 1−m−2
i1 bi
T
0
T−sφq
s 0
aτ∇τ
m−2
i1 ai
ξi
0aτ∇τ
1−m−2
i1 ai
∇s
Ap−10 −1/p−1A−10 1−A−p−10 1/p−1<1.
3.13 Therefore, there exist 0< a1< a1such thatαa1< a1. It implies thatAu<u, u∈∂Ωa1.
Ifa2 ≤ l <∞andu∈∂Ωl, then 0≤ut≤ l. Similar to the above argument, noticing thatλ2−p1 <0, we can get liml→∞αl/l<1. Therefore, there exist 0< a2 < a2such that αa2< a2. It implies thatAu<u, u∈∂Ωa2.
On the other hand, sincef :0, T ×0,∞→0,∞is continuous, by conditionA4, there exista1< b1< b < b2< a2such that
min{ft, l:t, l∈0, T ×γbi, bi } ≥biB0p−1, i1,2. 3.14 Ifu∈∂Ωb1, thenγb1≤ut≤b1, 0≤t≤T. ApplyingLemma 2.9, it follows that
Aumax
0≤t≤T|Aut|
≥ − T
0
T−sϕs∇s 1
1−m−2
i1 bi
T
0
T−sϕs∇s−m−2
i1
bi
ξi
0
ξi−sϕs∇s
m−2
i1 bi
1−m−2
i1 bi
T
0
T−sϕs∇s− m−2
i1 bi
ξi
0ξi−sϕs∇s 1−m−2
i1 bi
≥
m−2
i1 bi
1−m−2
i1 bi
T
0
T−sϕs∇s−
m−2
i1 biξi
T1−m−2
i1 bi T
0
T−sϕs∇s
Tm−2
i1 bi−m−2
i1 biξi
T1−m−2
i1 bi T
0
T−sϕs∇s
≥ Tm−2
i1 bi−m−2
i1 biξi
T1−m−2
i1 bi T
0
T−sφq
s 0
aτb1B0p−1∇τ m−2
i1 ai
ξi
0aτb1B0p−1∇τ 1−m−2
i1 ai
∇s Tm−2
i1 bi−m−2
i1 biξi
T1−m−2
i1 bi b1B0
T
0
T−sφq
s 0
aτ∇τ
m−2
i1 ai
ξi
0aτ∇τ
1−m−2
i1 ai
∇s b1B0B0−1b1u.
3.15
In the same way, we can prove that ifu∈∂Ωb2, thenAu ≥ u.
Now, we consider the operatorAonΩb1\Ωa1andΩa2\Ωb2, respectively. ByLemma 2.10, we assert that the operatorAhas two fixed pointsu∗1, u∗2 ∈ K such thata1 ≤ u∗1 ≤ b1 and b2≤ u∗2 ≤a2. Therefore,u∗i, i1,2, are positive solutions of problem1.1-1.2.
Proof ofTheorem 3.2. Let
1
2min
l→0lim p3l
lp−1 − B0
γ p−1
, lim
l→∞
p4l lp−1 −
B0
γ p−1
, 3.16
then there exist 0< b3≤c3andc4≤b4<∞such that
p3l≥ B0
γ p−1
lp−1, 0≤l≤b3,
p4l≥ B0
γ p−1
lp−1, b4≤l≤∞.
3.17
If 0≤l≤b3, u∈∂Ωl, thenγl≤ut≤l, 0≤t≤T. ByLemma 2.9and conditionB3, we have Aumax
0≤t≤T|Aut|
≥ Tm−2
i1 bi−m−2
i1 biξi
T1−m−2
i1 bi T
0
T−sϕs∇s
≥ Tm−2
i1 bi−m−2
i1 biξi
T1−m−2
i1 bi
× T
0
T−sφq
s 0
aτp3uτ−k3τuλ3 ∇τ
m−2
i1 ai
ξi
0aτp3uτ−k3τuλ3 ∇τ 1−m−2
i1 ai
∇s
≥ Tm−2
i1 bi−m−2
i1 biξi
T1−m−2
i1 bi
× T
0
T−sφq
s 0
aτ
B0
γ p−1
γlp−1−k3τlλ3 ∇τ
m−2
i1 ai
ξi
0aτB0/γp−1γlp−1−k3τlλ3 ∇τ 1−m−2
i1 ai
∇s.
3.18
It follows that βl
l ≥ Tm−2
i1 bi−m−2
i1 biξi
T1−m−2
i1 bi
× T
0
T−sφq
s 0
aτB0p−1γp−1−k3τlλ3−p1 ∇τ
m−2
i1 ai
ξi
0aτBp−10 γp−1−k3τlλ3−p1 ∇τ 1−m−2
i1 ai
∇s.
3.19
Noticingλ3−p1>0, we get
l→0lim βl
l ≥ Tm−2
i1 bi−m−2
i1 biξi
T1−m−2
i1 bi
× T
0
T−sφq
s
0
aτBp−10 γp−1∇τ m−2
i1 ai
ξi
0aτBp−10 γp−1∇τ 1−m−2
i1 ai
∇s
B0p−1γp−11/p−1B−10 1γp−1B−p−10 1/p−1>1.
3.20 Therefore, there existsb3with 0 < b3 < asuch thatβb3> b3. It implies thatAu> ufor u∈∂Ωb3.
Ifb4 ≤ γl < ∞andu ∈∂Ωl, thenb4 ≤ γl ≤ ut≤ l, 0 ≤ t ≤ T. Similar to the above argument, noticing thatλ4−p1<0, we can get liml→∞βl/l>1.
Therefore, there existb4with 0< b4<∞such thatβb4> b4. It implies thatAu>u foru∈∂Ωb4.
By conditionB4, we can see that there existb3< a3< a < a4< b4such that
max{ft, l:t, l∈0, T ×0, ai } ≤aiA0p−1, i3,4. 3.21 Ifu∈∂Ωa3, then 0≤ut≤a3, 0≤t≤T, andft, ut≤a3A0p−1. It follows that
Au ≤ 1 1−m−2
i1 bi
T
0
T−sϕs∇s
≤ 1
1−m−2
i1 bi
a3A0
T
0
T−sφq
s
0
aτ∇τ
m−2
i1 ai
ξi
0aτ∇τ
1−m−2
i1 ai
∇s a3u.
3.22
Similarly, ifu∈∂Ωa4, thenAu ≤ u.
Now, we study the operatorAonΩa3\Ωb3andΩb4\Ωa4, respectively. ByLemma 2.10, we assert that the operatorAhas two fixed pointsu∗3, u∗4 ∈ K such thatb3 ≤ u∗3 ≤ a3 and a4≤ u∗4 ≤b4. Therefore,u∗i, i3,4, are positive solutions of problem1.1-1.2.
4. Further discussion
If the conditions of Theorems3.1 and3.2 are weakened, we will get the existence of single positive solution of problem1.1-1.2.
Corollary 4.1. AssumeH1,H2, andH3hold, and assume that the following conditions hold:
C1p1∈C0,∞,0,∞, and liml→0p1l/lp−1< Ap−10 ; C2k1∈L10, T ,0,∞;
