Volume 2011, Article ID 867615, 17 pagesdoi:10.1155/2011/867615 Research Article Multiple Positive Solutions for Second-Order p-Laplacian Dynamic Equations with Integral Boundary Conditi
Trang 1Volume 2011, Article ID 867615, 17 pages
doi:10.1155/2011/867615
Research Article
Multiple Positive Solutions for Second-Order
p-Laplacian Dynamic Equations with Integral
Boundary Conditions
Yongkun Li and Tianwei Zhang
Department of Mathematics, Yunnan University, Kunming, Yunnan 650091, China
Correspondence should be addressed to Yongkun Li,yklie@ynu.edu.cn
Received 13 July 2010; Revised 21 November 2010; Accepted 25 November 2010
Academic Editor: Gennaro Infante
Copyrightq 2011 Y Li and T 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
We are concerned with the following second-order p-Laplacian dynamic equations on time scales
ϕ p xΔt∇ λft, xt, xΔt 0, t ∈ 0, TT, with integral boundary conditions xΔ0 0,
αx T−βx0 T
0g sxs∇s By using Legget-Williams fixed point theorem, some criteria for the
existence of at least three positive solutions are established An example is presented to illustrate the main result
1 Introduction
Boundary value problems with p-Laplacian have received a lot of attention in recent years They often occur in the study of the n-dimensional p-Laplacian equation, non-Newtonian
fluid theory, and the turbulent flow of gas in porous medium1 7 Many works have been carried out to discuss the existence of solutions or positive solutions and multiple solutions for the local or nonlocal boundary value problems
On the other hand, the study of dynamic equations on time scales goes back to its founder Stefan Hilger 8 and is a new area of still fairly theoretical exploration in mathematics Motivating the subject is the notion that dynamic equations on time scales can build bridges between continuous and discrete equations Further, the study of time scales has led to several important applications, for example, in the study of insect population models, neural networks, heat transfer, and epidemic models, we refer to8 10 In addition, the study of BVPs on time scales has received a lot of attention in the literature, with the pioneering existence results to be found in11–16
However, existence results are not available for dynamic equations on time scales with integral boundary conditions Motivated by above, we aim at studying the second-order
Trang 2p-Laplacian dynamic equations on time scales in the form of
ϕ p
xΔt∇ λft, x t, xΔt 0, t ∈ 0, TT 1.1 with integral boundary condition
xΔ0 0, αx T − βx0
T
0
g sxs∇s, 1.2
where λ is positive parameter, ϕ p s |s| p−2s for p > 1 with ϕ−1p ϕ q and 1/p 1/q 1,
Δ is the delta derivative, ∇ is the nabla derivative, T is a time scale which is a nonempty closed subset ofR with the topology and ordering inherited from R, 0 and T are points in T,
an interval0, TT: 0, T ∩ T, f ∈ C0, TT× R2, 0, ∞ with ft, 0, 0 / 0 for all t ∈ 0, TT,
g ∈ Cld0, TT, 0, ∞, α, β > 0 with α − g0> β, and where g0T
0 g s∇s.
The main purpose of this paper is to establish some sufficient conditions for the existence of at least three positive solutions for BVPs1.1-1.2 by using Legget-Williams fixed point theorem This paper is organized as follows InSection 2, some useful lemmas are established In Section 3, by using Legget-Williams fixed point theorem, we establish sufficient conditions for the existence of at least three positive solutions for BVPs 1.1-1.2
An illustrative example is given inSection 4
2 Preliminaries
In this section, we will first recall some basic definitions and lemmas which are used in what follows
Definition 2.1see 8 A time scale T is an arbitrary nonempty closed subset of the real set R with the topology and ordering inherited fromR The forward and backward jump operators
σ, ρ : T → T and the graininess μ, ν : T → Rare defined, respectively, by
σ t : inf{s ∈ T : s > t}, ρ t : sup{s ∈ T : s < t}, μ t : σt − t, ν t : t − ρt.
2.1
The point t ∈ T is called left-dense, left-scattered, right-dense, or right-scattered if ρt t,
ρ t < t, and σt t or σt > t, respectively Points that are right-dense and left-dense
at the same time are called dense IfT has a left-scattered maximum m1, definedTκ T −
{m1}; otherwise, set Tκ T If T has a right-scattered minimum m2, definedTκ T − {m2}; otherwise, setTκ T
Definition 2.2see 8 For f : T → R and t ∈ T κ , then the delta derivative of f at the point
t is defined to be the number fΔ
there is a neighborhood U of t such that
fσt − fs − fΔtσt − s 2.2
Trang 3For f : T → R and t ∈ T κ , then the nabla derivative of f at the point t is defined to
be the number f∇
neighborhood U of t such that
fρ t− fs − f∇tρ t − s ρ t − s ∀s ∈ U. 2.3
Definition 2.3see 8 A function f is rd-continuous provided it is continuous at each
right-dense point in T and has a left-sided limit at each left-dense point in T The set of
rd-continuous functions f will be denoted by CrdT A function g is left-dense continuous i.e.,
ld-continuous if g is continuous at each left-dense point in T and its right-sided limit exists
finite at each right-dense point in T The set of left-dense continuous functions g will be denoted by CldT
Definition 2.4see 8 If FΔt ft, then we define the delta integral by
b
a
f sΔs Fb − Fa. 2.4
If G∇t gt, then we define the nabla integral by
b
a
g s∇s Gb − Ga. 2.5
Lemma 2.5 see 8 If f ∈ CrdT and t ∈ T κ , then
σ t
t
f sΔs μtft. 2.6
If g ∈ CldT and t ∈ T κ , then
t
ρ t g s∇s νtgt. 2.7 Let the Banach space
B C1
ld0, TT
x : 0, TT−→ R | x is Δ-differentiable on 0, TT, and xΔis ld-continuous on0, TT
2.8
be endowed with the norm x max{ x 0, xΔ 0}, where
x 0 sup
t ∈0,T |xt|, xΔ
0 sup
t ∈0,T
xΔt
2.9
Trang 4and choose a coneP ⊂ B defined by
P
⎧
⎪
⎪
x ∈ B : xt ≥ 0, xΔt ≤ 0, xΔ∇t ≤ 0 ∀t ∈ 0, TT,
αx T − βx0
T
0
g sxs∇s
⎫
⎪
⎪. 2.10
Lemma 2.6 If x ∈ P, then xt ≥ β/α − g0 x 0for all t ∈ 0, TT.
Proof If x ∈ P, then xΔ≤ 0 It follows that
x T min
t ∈0,TT
x t, x 0 x0 max
t ∈0,TT
x t. 2.11
With αxT − βx0 T
0 g sxs∇s and xΔ≤ 0, one obtains
αx T βx0
T
0
g sxs∇s ≥ βx0
T
0
g s∇sxT βx0 g0x T. 2.12 Therefore,
x T ≥ β
α − g0
x0 β
α − g0 x 0. 2.13 From2.11–2.13, we can get that
x t ≥ min
t ∈0,TTx t xT ≥ β
α − g0
x0 β
α − g0 x 0. 2.14
SoLemma 2.6is proved
Lemma 2.7 x ∈ B is a solution of BVPs 1.1-1.2 if and only if x ∈ B is a solution of the following
integral equation:
x t
T
0
Θβ V sϕ q
s
0
λf
r, x r, xΔr∇r
Δs
T
t
ϕ q
s
0
λf
r, x r, xΔr∇r
Δs,
2.15
where
Θ 1
α − β −T
0 g s∇s
1
α − β − g0
,
V t
t
0
g s∇s ∀t ∈ 0, TT.
2.16
Trang 5Proof First assume x∈ B is a solution of BVPs 1.1-1.2; then we have
ϕ p
xΔt ϕ p
xΔ0−
t
0
λf
s, x s, xΔs∇s −
t
0
λf
s, x s, xΔs∇s. 2.17
That is,
xΔt −ϕ q
t
0
λf
s, x s, xΔs∇s
−Ht. 2.18
Integrating2.18 from t to T, it follows that
x t xT
T
t
H sΔs. 2.19
Together with2.19 and αxT − βx0 T
0 g sxs∇s, we obtain
αx T − β
x T
T
0
H sΔs
T
0
g s
x T
T
s
H rΔr
∇s. 2.20
Thus,
α − β −
T
0
g s∇s
x T β
T
0
H sΔs
T
0
g s
T
s
H rΔr
∇s
β
T
0
H sΔs
T
0
T
s
V s − V rHrΔr
∇
∇s
β
T
0
H sΔs −
T
0
V 0 − V sHsΔs
β
T
0
H sΔs
T
0
V sHsΔs,
2.21
namely,
x T βΘ
T
0
H sΔs Θ
T
0
V sHsΔs. 2.22
Trang 6Substituting2.22 into 2.19, we obtain
x t βΘ
T
0
H sΔs Θ
T
0
V sHsΔs
T
t
H sΔs
T
0
Θβ V sϕ q
s
0
λf
r, x r, xΔr∇r
Δs
T
t
ϕ q
s
0
λf
r, x r, xΔr∇r
Δs.
2.23
The proof of sufficiency is complete
Conversely, assume x∈ B is a solution of the following integral equation:
x t
T
0
Θβ V sϕ q
s
0
λf
r, x r, xΔr∇r
Δs
T
t
ϕ q
s
0
λf
r, x r, xΔr∇r
Δs
T
0
Θβ V sH sΔs
T
t
H sΔs.
2.24
It follows that
xΔt −ϕ q
t
0
λf
s, x s, xΔs∇s
−Ht,
ϕ p
xΔt∇ λft, x t, xΔt 0.
2.25
So xΔ0 0 Furthermore, we have
αx T − βx0 α
T
0
Θβ V sH sΔs − β
T
0
Θβ V sH sΔs − β
T
0
H sΔs
α − β T
0
Θβ V sH sΔs − β
T
0
H sΔs,
T
0
g sxs∇s
T
0
g s
T
0
Θβ V rH rΔr
T
s
H rΔr
∇s
T
0
g s∇s
T
0
Θβ V sH sΔs
T
0
T
s
g sHrΔr∇s
Trang 7T
0
g s∇s
T
0
Θβ V sH sΔs
T
0
T
s
V s − V rHrΔr
∇
∇s
T
0
g s∇s
T
0
Θβ V sH sΔs
T
0
V sHsΔs,
2.26 which imply that
αx T − βx0 −
T
0
g sxs∇s α − β T
0
Θβ V sH sΔs
− β
T
0
H sΔs −
T
0
g s∇s
T
0
Θβ V sH sΔs
−
T
0
V sHsΔs
0.
2.27
The proof ofLemma 2.7is complete
Define the operatorΨ : P → B by
Ψxt
T
0
Θβ V sϕ q
s
0
λf
r, x r, xΔr∇r
Δs
T
t
ϕ q
s
0
λf
r, x r, xΔr∇r
Δs
2.28
for all t ∈ 0, TT Obviously,Ψxt ≥ 0 for all t ∈ 0, TT
Lemma 2.8 If x ∈ P, then Ψx ∈ P.
Proof It is easily obtained from the second part of the proof in Lemma 2.7 The proof is complete
Lemma 2.9 Ψ : P → P is complete continuous.
Proof First, we show that Ψ maps bounded set into itself Assume c is a positive constant and
x∈ Pc {x ∈ P : x ≤ c} Note that the continuity of ft, x, xΔ guarantees that there is a
Trang 8C > 0 such that f t, x, xΔ ≤ ϕ p C for all t ∈ 0, TT So we get fromΨΔx≤ 0 and ΨΔ∇x≤ 0 that
Ψx 0 Ψx0
T
0
Θβ V sϕ q
s
0
λf
r, x r, xΔr∇r
Δs
T
0
ϕ q
s
0
λf
r, x r, xΔr∇r
Δs
≤ Cλ q−1T q−1T
0
Θβ V sΔs Cλ q−1T q ,
2.29
ΨΔx
0ΨΔx T
ϕ q
T
0
λf
r, x r, xΔr∇r
≤ Cλ q−1T q−1.
2.30
That is,ΨPcis uniformly bounded In addition, notice that
|Ψxt1 − Ψxt2|
t1
t2
ϕ q
s
0
λf
r, x r, xΔr∇r
Δs
≤ Cλ q−1T q−1|t1− t2|,
2.31
which implies that
|Ψxt1 − Ψxt2| −→ 0 as t1− t2−→ 0,
ΨxΔt1p−1−ΨxΔt2p−1
ϕ p
ΨxΔt1− ϕ p
ΨxΔt2
t1
t2
λf
r, x r, xΔr∇r
≤ λϕ p C|t1− t2|,
2.32
which implies that
ΨxΔt1p−1−ΨxΔt2p−1
−→ 0 as t1− t2−→ 0. 2.33
That is,
ΨxΔt1 − ΨxΔt2 −→ 0 as t
1− t2−→ 0. 2.34
Trang 9SoΨx is equicontinuous for any x ∈ P c Using Arzela-Ascoli theorem on time scales17, we obtain thatΨ Pcis relatively compact In view of Lebesgue’s dominated convergence theorem
on time scales18, it is easy to prove that Ψ is continuous Hence, Ψ is complete continuous The proof of this lemma is complete
Let υ and ω be nonnegative continuous convex functionals on a pone P, ψ a
nonnegative continuous concave functional on P, and r, a, L positive numbers with r > a
we defined the following convex sets:
Pυ, r; ω, l {x ∈ P : υx < r, ωx < l}, Pυ, r; ω, l {x ∈ P : υx ≤ r, ωx ≤ l},
Pυ, r; ω, l; ψ, a
x ∈ P : υx < r, ωx < l, ψx > a,
Pυ, r; ω, l; ψ, a
x ∈ P : υx ≤ r, ωx ≤ l, ψx ≥ a
2.35
and introduce two assumptions with regard to the functionals υ, ω as follows:
H1 there exists M > 0 such that x ≤ M max{υx, ωx} for all x ∈ P;
H2 Pυ, r; ω, l / ∅ for any r > 0 and l > 0.
The following fixed point theorem duo to Bai and Ge is crucial in the arguments of our main result
Lemma 2.10 see 19 Let B be Banach space, P ⊂ B a cone, and r2≥ d > b > r1 > 0, l2≥ l1 > 0 Assume that υ and ω are nonnegative continuous convex functionals satisfying (H1) and (H2), ψ is
a nonnegative continuous concave functional on P such that ψx ≤ υx for all x ∈ Pυ, r2; ω, l2,
and Ψ : Pυ, r2; ω, l2 → Pυ, r2; ω, l2 is a complete continuous operator Suppose
C1 {x ∈ Pυ, d; ω, l2; ψ, b } / ∅, ψΨx > b for x ∈ Pυ, d; ω, l2; ψ, b;
C2 υΨx < r1, ω Ψx < l1for x ∈ Pυ, r1; ω, l1;
C3 ψΨx > b for x ∈ Pυ, r2; ω, l2; ψ, b with υΨx > d.
Then Ψ has at least three fixed points x1, x2, x3∈ Pυ, r2; ω, l2 with
x1∈ Pυ, r1; ω, l1,
x2 ∈ x∈ Pυ, r2; ω, l2; ψ, b
: ψx > b ,
x3∈ Pυ, r2; ω, l2 \Pυ, r2; ω, l2; ψ, b
∪ Pυ, r1; ω, l1.
2.36
3 Main Result
In this section, we will give sufficient conditions for the existence of at least three positive solutions to BVPs1.1-1.2
Trang 100 2 ≥ l1 > 0, and r2 > b >
r1 0 ∈ 0, TT, b/N ≤ min{r2/K, l2/L } and αb − g0b ≤ r2β such that the following conditions are satisfied.
H3 ft, u, v ≤ min{ϕ p r2/K , ϕ p l2/L } for all t, u, v ∈ 0, TT× 0, r2 × −l2, l2, where
K λ q−1T
0
Θβ V ss q−1Δs
T
0
s q−1Δs
, L λ q−1T q−1. 3.1
H4 ft, u, v < min{ϕ p r1/K , ϕ p l1/L } for all t, u, v ∈ 0, TT× 0, r1 × −l1, l1.
H5 ft, u, v > ϕ p 0 T× b, αb − g0b /β × −l2, l2, where
N λ q−1
0q−1T
Θβ V sΔs. 3.2
Then BVPs1.1-1.2 have at least three positive solutions.
Proof By the definition of the operator Ψ and its properties, it suffices to show that the conditions ofLemma 2.10hold with respect to the operatorΨ
Let the nonnegative continuous convex functionals υ, ω and the nonnegative continuous concave functional ψ be defined on the coneP by
υ x max
t ∈0,TT|xt| x0, ω x max
t ∈0,TT
xΔt xΔT, ψ x min
t Tx t xT.
3.3
Then it is easy to see that x max{υx, ωx} and H1-H2 hold.
First of all, we show thatΨ : Pυ, r2; ω, l2 → Pυ, r2; ω, l2 In fact, if x ∈ Pυ, r2; ω, l2, then
υ x max
t ∈0,TT
|xt| ≤ r2, ω x max
t ∈0,TT
xΔt ≤ l
and assumptionH3 implies that
f
t, x t, xΔt≤ min
ϕ p r2
K
, ϕ p
l2 L
∀t ∈ 0, TT. 3.5
Trang 11On the other hand, for x ∈ P, there is Ψx ∈ P; thus
υ Ψx max
t ∈0,TT|Ψxt|
max
t ∈0,TT
T
0
Θβ V sϕ q
s
0
λf
r, x r, xΔr∇r
Δs
T
t
ϕ q
s
0
λf
r, x r, xΔr∇rΔs
T
0
Θβ V sϕ q
s
0
λf
r, x r, xΔr∇r
Δs
T
0
ϕ q
s
0
λf
r, x r, xΔr∇r
Δs
≤
T
0
Θβ V sϕ q
s
0
λϕ p r2
K
∇r
Δs
T
0
ϕ q
s
0
λϕ p r2
K
∇r
Δs
r2
K
T
0
Θβ V sϕ q
s
0
λ ∇r
Δs r2
K
T
0
ϕ q
s
0
λ ∇r
Δs
r2
K λ
q−1T
0
Θβ V ss q−1Δs
T
0
s q−1Δs
r2
K · K
r2,
ω Ψx max
t ∈0,TT
ΨxΔt
max
t ∈0,TT
−ϕ q
t
0
λf
r, x r, xΔr∇r
ϕ q
T
0
λf
r, x r, xΔr∇r
≤ ϕ q
T
0
λϕ p
l2 L
∇r
l2
L ϕ q
T
0
λ ∇r
Therefore,Ψ : Pυ, r2; ω, l2 → Pυ, r2; ω, l2
Trang 12In the same way, if x ∈ Pυ, r1; ω, l1, then assumption H4 implies
f
t, x t, xΔt< min
ϕ p r1
K
, ϕ p
l1 L
∀t ∈ 0, TT. 3.7
As in the argument above, we can get thatΨ : Pυ, r1; ω, l1 → Pυ, r1; ω, l1 Thus, condition
C2 ofLemma 2.10holds
To check conditionC1 inLemma 2.10 Let d αb − g0b /β We choose xt ≡ d > b for t ∈ 0, TT It is easy to see that
x t ≡ d ∈ Pυ, d; ω, l2; ψ, b
, ψ x d > b. 3.8 Consequently,
x∈ Pυ, d; ω, l2; ψ, b
: ψx > b / ∅. 3.9
Hence, for x ∈ Pυ, d; ω, l2; ψ, b, there are
b ≤ xt ≤ d, xΔt ≤ l
In view of assumptionH5, we have
f
t, x t, xΔt> ϕ p
b N
0 T. 3.11
It follows that
ψ Ψx min
t TΨxt
ΨxT
T
0
Θβ V sϕ q
s
0
λf
r, x r, xΔr∇r
Δs
≥
T
Θβ V sϕ q
s
0
λf
r, x r, xΔr∇r
Δs
≥
T
Θβ V sϕ q
0
λf
r, x r, xΔr∇r
Δs
≥
T
Θβ V sϕ q
0
λf
r, x r, xΔr∇r
Δs
>
T
Θβ V sϕ q
λϕ p
b N
∇r
Δs