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Hindawi Publishing Corporation
Advances in Difference Equations
Volume 2009, Article ID 209707, 18 pages
doi:10.1155/2009/209707
Research Article
Existence Results for Higher-Order
Boundary Value Problems on Time Scales
Jian Liu1 and Yanbin Sang2
1
2
School of Mathematics and Statistics, Shandong Economics University, Jinan Shandong 250014, China
Department of Mathematics, North University of China, Taiyuan Shanxi 030051, China
Correspondence should be addressed to Jian Liu, kkword@163.com
Received 22 March 2009; Revised 5 June 2009; Accepted 16 June 2009
Recommended by Victoria Otero-Espinar
By using the fixed-point index theorem, we consider the existence of positive solutions for the
following nonlinear higher-order four-point singular boundary value problem on time scales
n
n−2
i
n−2
n−1
uΔ tgtfut, uΔ t, . . . , uΔ t 0, 0 < t < T ; uΔ 0 0, 0 ≤ i ≤ n−3; αuΔ 0−βuΔ ξ 0,
Δn−2
Δn−1
T δu
η 0, n ≥ 3, where α > 0, β ≥ 0, γ > 0, δ ≥ 0, ξ, η ∈ 0, T , ξ < η, and
n ≥ 3; γu
g : 0, T → 0, ∞ is rd-continuous.
Copyright q 2009 J. Liu and Y. Sang. 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
Time scales and time-scale notation are introduced well in the fundamental texts by Bohner
and Peterson 1, 2, respectively, as important corollaries. In, the recent years, many authors
have paid much attention to the study of boundary value problems on time scales see, e.g.,
3–17. In particular, we would like to mention some results of Anderson et al. 3, 5, 6, 14,
16, DaCunha et al. 4, and Agarwal and O’Regan 7, which motivate us to consider our
problem.
In 3, Anderson and Karaca discussed the dynamic equation on time scales
2n
−1n yΔ t f t, yσ t 0,
αi1 y
Δ2i
t ∈ a, b,
2i1
2i
2i 2i
η βi1 yΔ a yΔ a, γi1 yΔ η yΔ σb,
1.1
and the eigenvalue problem
2n
−1n yΔ t λf t, yσ t 0,
t ∈ a, b,
1.2
2
Advances in Difference Equations
with the same boundary conditions where λ is a positive parameter. They obtained some
results for the existence of positive solutions by using the Krasnoselskii, the Schauder, and
the Avery-Henderson fixed-point theorem.
In 4, by using the Gatica-Oliker-Waltman fixed-point theorem, DaCunha, Davis, and
Singh proved the existence of a positive solution for the three-point boundary value problem
on a time scale T given by
yΔΔ t f x, y 0,
x ∈ 0, 1T ,
y p y σ 2 1 ,
y0 0,
1.3
where p ∈ 0, 1 ∩ T is fixed, and fx, y is singular at y 0 and possibly at x 0, y ∞.
Anderson et al. 5 gave a detailed presentation for the following higher-order selfadjoint boundary value problem on time scales:
Lyt n
∇n−i−1 Δ
∇n−1 Δ
n−i−1
n−1
−1n−i pi yΔ ∇
t −1n p0 yΔ ∇
t · · ·
i0
− pn−3 yΔ
2
∇
∇ 2 Δ
t pn−2 yΔ∇
∇Δ
t − pn−1 yΔ
∇
1.4
t pn tyt,
and got many excellent results.
In related papers, Sun 11 considered the following third-order two-point boundary
value problem on time scales:
uΔΔΔ t f t, ut, uΔΔ t 0,
ua A,
u σ b B,
t ∈ a, σb,
uΔΔ a C,
1.5
where a, b ∈ T and a < b. Some existence criteria of solution and positive solution are
established by using the Leray-Schauder fixed point theorem.
In this paper, we consider the existence of positive solutions for the following higherorder four-point singular boundary value problem BVP on time scales
n
n−2
uΔ t gtf ut, uΔ t, . . . , uΔ t 0,
uΔ 0 0,
i
n ≥ 3,
η 0,
n ≥ 3,
γuΔ T δuΔ
n−2
1.6
0 ≤ i ≤ n − 3,
αuΔ 0 − βuΔ ξ 0,
n−2
0 < t < T,
n−1
n−1
1.7
where α > 0, β ≥ 0, γ > 0, δ ≥ 0, ξ, η ∈ 0, T , ξ < η, and g : 0, T → 0, ∞ is rd-continuous.
In the rest of the paper, we make the following assumptions:
H1 f ∈ C0, ∞n−1 , 0, ∞
T
H2 0 < 0 gtΔt < ∞.
Advances in Difference Equations
3
In this paper, by constructing one integral equation which is equivalent to the BVP
1.6 and 1.7, we study the existence of positive solutions. Our main tool of this paper is the
following fixed-point index theorem.
Theorem 1.1 18. Suppose E is a real Banach space, K ⊂ E is a cone, let Ωr {u ∈ K : u
≤ r}.
x, ∀ x ∈ ∂Ωr . Then
Let operator T : Ωr → K be completely continuous and satisfy T x /
i if T x
≤ x
, ∀ x ∈ ∂Ωr , then iT, Ωr , K 1
ii if T x
≥ x
, ∀ x ∈ ∂Ωr , then iT, Ωr , K 0.
The outline of the paper is as follows. In Section 2, for the convenience of the reader
we give some definitions and theorems which can be found in the references, and we present
some lemmas in order to prove our main results. Section 3 is developed in order to present
and prove our main results. In Section 4 we present some examples to illustrate our results.
2. Preliminaries and Lemmas
For convenience, we list the following definitions which can be found in 1, 2, 9, 14, 17. A
time scale T is a nonempty closed subset of real numbers R. For t < sup T and r > inf T, define
the forward jump operator σ and backward jump operator ρ, respectively, by
σt inf{τ ∈ T : τ > t} ∈ T,
ρr sup{τ ∈ T : τ < r} ∈ T,
2.1
for all t, r ∈ T. If σt > t, t is said to be right scattered, and if ρr < r, r is said to be left
scattered; if σt t, t is said to be right dense, and if ρr r, r is said to be left dense. If T
has a right scattered minimum m, define Tκ T − {m}; otherwise set Tκ T. If T has a left
scattered maximum M, define Tκ T − {M}; otherwise set Tκ T. In this general time-scale
setting, Δ represents the delta or Hilger derivative 13, Definition 1.10,
zσt − zs
zσ t − zs
lim
,
s→t
s→t
σt − s
σt − s
zΔ t : lim
2.2
where σt is the forward jump operator, μt : σt − t is the forward graininess function,
and z ◦ σ is abbreviated as zσ . In particular, if T R, then σt t and xΔ x , while if T hZ
for any h > 0, then σt t h and
xΔ t xt h − xt
.
h
2.3
A function f : T → R is right-dense continuous provided that it is continuous at each rightdense point t ∈ T a point where σt t and has a left-sided limit at each left-dense point
t ∈ T. The set of right-dense continuous functions on T is denoted by Crd T. It can be shown
4
Advances in Difference Equations
that any right-dense continuous function f has an antiderivative a function Φ : T → R with
the property ΦΔ t ft for all t ∈ T. Then the Cauchy delta integral of f is defined by
t1
ftΔt Φt1 − Φt0 ,
2.4
t0
where Φ is an antiderivative of f on T. For example, if T Z, then
t1
ftΔt t0
t
1 −1
ft,
2.5
ftdt.
2.6
tt0
and if T R, then
t1
ftΔt t0
t1
t0
Throughout we assume that t0 < t1 are points in T, and define the time-scale interval t0 , t1 T {t ∈ T : t0 ≤ t ≤ t1 }. In this paper, we also need the the following theorem which can be found
in 1.
Theorem 2.1. If f ∈ Crd and t ∈ Tk , then
σt
fτΔτ σt − tft.
2.7
t
In this paper, let
i
Δn−2
E u ∈ Crd
0, T : uΔ 0 0, 0 ≤ i ≤ n − 3 .
2.8
Then E is a Banach space with the norm u
maxt∈0,T |uΔ t|. Define a cone K by
n−2
n−2
n
K u ∈ E : uΔ t ≥ 0, uΔ t ≤ 0, t ∈ 0, T .
2.9
Obviously, K is a cone in E. Set Kr {u ∈ K : u
≤ r}. If uΔΔ ≤ 0 on 0, T , then we say u is
concave on 0, T . We can get the following.
Lemma 2.2. Suppose condition H2 holds. Then there exists a constant θ ∈ 0, T/2 satisfies
T −θ
0<
θ
gtΔt < ∞.
2.10
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5
Furthermore, the function
At t t
θ
gs1 Δs1 Δs T −θ s
s
t
gs1 Δs1 Δs,
t ∈ θ, T − θ
2.11
t
is a positive continuous function on θ, T − θ, therefore At has minimum on θ, T − θ. Then there
exists L > 0 such that At ≥ L, t ∈ θ, T − θ.
Lemma 2.3. Let u ∈ K and θ ∈ 0, T/2 in Lemma 2.2. Then
uΔ t ≥ θ
u
,
n−2
t ∈ θ, T − θ.
2.12
Proof. Suppose τ inf{ξ ∈ 0, T : supt∈0,T uΔ t uΔ ξ}.
We will discuss it from three perspectives.
n−2
n−2
i τ ∈ 0, θ. It follows from the concavity of uΔ t that
n−2
uΔ T − uΔ τ
t − τ,
T −τ
n−2
uΔ t ≥ uΔ τ n−2
n−2
n−2
t ∈ θ, T − θ,
2.13
then
Δn−2
u
t ≥ min
u
t∈θ,T −θ
uΔ T − uΔ τ
τ t − τ
T −τ
n−2
Δn−2
uΔ T − uΔ τ
T − θ − τ
T −τ
n−2
uΔ τ n−2
n−2
n−2
2.14
T − θ − τ Δn−2
θ
n−2
u T uΔ τ ≥ θuτ,
T −τ
T −τ
which means uΔ t ≥ θ
u
, t ∈ θ, T − θ.
n−2
ii τ ∈ θ, T − θ. If t ∈ θ, τ, we have
uΔ τ − uΔ 0
t − τ,
τ
n−2
uΔ t ≥ uΔ τ n−2
n−2
n−2
t ∈ θ, τ,
2.15
then
Δn−2
u
t ≥ min
t∈θ,T −θ
uΔ τ − uΔ 0
τ t − τ
τ
n−2
Δn−2
u
n−2
θ n−2
τ − θ Δn−2
n−2
uΔ τ u 0 ≥ θuΔ τ,
τ
τ
2.16
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Advances in Difference Equations
If t ∈ τ, T − θ, we have
uΔ T − uΔ τ
t − τ,
T −τ
n−2
uΔ t ≥ uΔ τ n−2
n−2
n−2
t ∈ τ, T − θ,
2.17
then
Δn−2
u
t ≥ min
u
t∈θ,T −θ
uΔ T − uΔ τ
τ t − τ
T −τ
n−2
Δn−2
n−2
2.18
T − θ − τ Δn−2
θ
n−2
uΔ τ u T T −τ
T −τ
≥ θuΔ τ,
n−2
and this means uΔ t ≥ θ
u
, t ∈ θ, T − θ.
n−2
iii τ ∈ T − θ, T . Similarly, we have
uΔ τ − uΔ 0
τ t − τ,
τ
n−2
u
Δn−2
Δn−2
t ≥ u
n−2
t ∈ θ, T − θ,
2.19
then
Δn−2
u
t ≥ min
t∈θ,T −θ
uΔ τ − uΔ 0
uτ t − τ
τ
n−2
n−2
τ − θ Δn−2
θ Δn−2
u τ u 0
τ
τ
2.20
≥ θuΔ τ,
n−2
which means uΔ t ≥ θ
u
, t ∈ θ, T − θ.
n−2
From the above, we know uΔ t ≥ θ
u
, t ∈ θ, T − θ. The proof is complete.
n−2
Lemma 2.4. Suppose that conditions H1 , H2 hold, then ut is a solution of boundary value
problem 1.6, 1.7 if and only if ut ∈ E is a solution of the following integral equation:
ut t s1
0
0
···
sn−3
0
wsn−2 Δsn−2 Δsn−3 · · · Δs1 ,
2.21
Advances in Difference Equations
7
where
⎧ δ
β
⎪
⎪
Δ
Δn−2
⎪
gsf
us,
u
.
.
.
,
u
Δs
s,
s
⎪
⎪
α ξ
⎪
⎪
⎪
⎪
t δ
⎪
⎪
⎪
⎪
Δ
Δn−2
⎪
ΔrΔs,
grf
ur,
u
.
.
.
,
u
r,
r
⎪
⎨
0 s
wt ⎪
⎪
δ η
n−2
⎪
⎪
gsf us, uΔ s, . . . , uΔ s Δs
⎪
⎪
⎪
γ δ
⎪
⎪
⎪
⎪
1 s
⎪
⎪
n−2
⎪
⎪
⎩ grf ur, uΔ r, . . . , uΔ r ΔrΔs,
t
0 ≤ t ≤ δ,
2.22
δ ≤ t ≤ T.
δ
Proof. Necessity. By the equation of the boundary condition, we see that uΔ ξ ≥ 0, uΔ η ≤
n−1
0, then there exists a constant δ ∈ ξ, η ⊂ 0, T such that uΔ δ 0. Firstly, by delta
integrating the equation of the problems 1.6 on δ, t, we have
n−1
uΔ t uΔ δ −
n−1
n−1
t
n−1
n−2
gsf us, uΔ s, . . . , uΔ s Δs,
2.23
n−2
grf ur, uΔ r, . . . , uΔ r Δr Δs.
2.24
δ
thus
Δn−2
u
Δn−2
t u
δ −
t s
δ
δ
By uΔ δ 0 and the boundary condition 1.7, let t η on 2.23, we have
n−1
u
Δn−1
η −
η
n−2
gsf us, uΔ s, . . . , uΔ s Δs.
2.25
δ
By the equation of the boundary condition 1.7, we get
uΔ T −
n−2
δ Δn−1 η ,
u
γ
2.26
then
Δn−2
u
δ
T γ
η
n−2
gsf us, uΔ s, . . . , uΔ sΔs .
2.27
δ
Secondly, by 2.24 and let t T on 2.24, we have
Δn−2
u
δ
δ γ
η
n−2
gsf us, uΔ s, . . . , uΔ s Δs
δ
T s
δ
δ
grf ur, uΔ r, . . . , uΔ r Δr Δs.
n−2
2.28
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Advances in Difference Equations
Then
uΔ t n−2
δ
γ
η
n−2
gsf us, uΔ s, . . . , uΔ s Δs
δ
T s
t
2.29
grf ur, uΔ r, . . . , uΔ r Δr Δs.
n−2
δ
Then by delta integrating 2.29 for n − 2 times on 0, T , we have
t s1
sn−3 η
δ
n−2
ut ···
gsf us, uΔ s, . . . , uΔ s Δs Δsn−2 · · · Δs2 Δs1
γ δ
0 0
0
t s1 sn−3 T s
Δ
Δn−2
···
grf ur, u r, . . . , u r Δr Δs Δsn−2 · · · Δs2 Δs1 .
0
0
sn−2
0
δ
2.30
Similarly, for t ∈ 0, δ, by delta integrating the equation of problems 1.6 on 0, δ, we have
ut t s1
0
0
sn−3 δ
δ
Δ
Δn−2
···
gsf us, u s, . . . , u s Δs Δsn−2 · · · Δs2 Δs1
γ ξ
0
t s1
0
0
···
sn−3 sn−2 s
0
n−2
grf ur, uΔ r, . . . , uΔ r Δr Δs Δsn−2 · · · Δs2 Δs1 .
δ
0
2.31
Therefore, for any t ∈ 0, T , ut can be expressed as the equation
ut t s1
0
···
0
sn−3
wsn−2 Δsn−2 Δsn−3 · · · Δs1 ,
2.32
wsn−2 Δsn−2 Δsn−3 · · · Δs1 ,
2.33
0
where wt is expressed as 2.22.
Sufficiency. Suppose that
ut t s1
0
0
···
sn−3
0
then by 2.22, we have
⎧ δ
⎪
⎪
Δ
Δn−2
⎪
Δs ≥ 0,
gsf
us,
u
.
.
.
,
u
s,
s
⎪
⎨
n−1
t
uΔ t t
⎪
⎪
n−2
⎪
⎪
⎩− gsf us, uΔ s, . . . , uΔ s Δs ≤ 0,
δ
0 ≤ t ≤ δ,
2.34
δ ≤ t ≤ T,
Advances in Difference Equations
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So,
n
n−2
uΔ t gtf ut, uΔ t, . . . , uΔ t 0,
0 < t < T,
2.35
which imply that 1.6 holds. Furthermore, by letting t 0 and t T on 2.22 and 2.34, we
can obtain the boundary value equations of 1.7. The proof is complete.
Δ
Now, we define a mapping T : K → Crd
0, T given by
n−1
T ut t s1
0
···
0
sn−3
wsn−2 Δsn−2 Δsn−3 · · · Δs1 ,
2.36
0
where wt is given by 2.22.
Lemma 2.5. Suppose that conditions H1 , H2 hold, the solution ut of problem 1.6, 1.7
satisfies
ut ≤ T uΔ t ≤ · · · ≤ T n−3 uΔ t,
n−3
t ∈ 0, T ,
2.37
and for θ ∈ 0, T/2 in Lemma 2.2, one has
uΔ t ≤
n−3
T Δn−2
u t,
θ
t ∈ θ, T − θ.
2.38
Proof. If ut is the solution of 1.6, 1.7, then uΔ t is a concave function, and ui t ≥ 0, i 0, 1, . . . , n − 2, t ∈ 0, T , thus we have
n−1
uΔ t i
t
uΔ sΔs ≤ tuΔ t ≤ T uΔ t,
i1
i1
i1
i 0, 1, . . . , n − 4,
2.39
0
that is,
ut ≤ T uΔ t ≤ · · · ≤ T n−3 uΔ t,
n−3
t ∈ 0, T .
2.40
By Lemma 2.3, for t ∈ θ, T − θ, we have
uΔ t ≥ θ
u
,
n−2
then uΔ t n−3
t
0
2.41
uΔ sΔs ≤ tuΔ t ≤ T u
≤ T/θuΔ t.The proof is complete.
n−2
n−2
n−2
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Advances in Difference Equations
Lemma 2.6. T : K → K is completely continuous.
Proof. Because
⎧
δ
⎪
⎪
Δ
Δn−2
⎪
Δs ≥ 0,
gsf
us,
u
.
.
.
,
u
s,
s
⎨
n−1
t
T uΔ t wΔ t t
⎪
n−2
⎪
⎪
⎩− gsf us, uΔ s, . . . , uΔ s Δs ≤ 0,
0 ≤ t ≤ δ,
2.42
δ≤t≤T
δ
is continuous, decreasing on 0, T , and satisfies T uΔ δ 0. Then, T u ∈ K for each u ∈ K
n−2
n−2
and T uΔ δ maxt∈0,T T uΔ t. This shows that T K ⊂ K. Furthermore, it is easy to
check that T : K → K is completely continuous by Arzela-ascoli Theorem.
For convenience, we set
n−1
θ∗ 2
,
L
1
,
1
grΔr
1 β/α
0
θ∗ 2.43
where L is the constant from Lemma 2.2. By Lemma 2.5, we can also set
fu1 , u2 , . . . , un−1 ,
un−1 → 0 0≤u1 ≤T u2 ≤···≤T n−2 un−2 ≤T/θun−1
un−1
f0 lim
max
fu1 , u2 , . . . , un−1 min
.
f∞ lim
un−1 → ∞ 0≤u1 ≤T u2 ≤···≤T n−2 un−2 ≤T/θun−1
un−1
2.44
3. The Existence of Positive Solution
Theorem 3.1. Suppose that conditions (H1 ), (H2 ) hold. Assume that f also satisfies
A1 fu1 , u2 , . . . , un−1 ≥ mr, for θr ≤ un−1 ≤ r, 0 ≤ u1 ≤ T u2 ≤ · · · ≤ T n−2 un−2 ≤
T/θun−1 ,
A2 fu1 , u2 , . . . , un−1 ≤ MR, for 0 ≤ un−1 ≤ R, 0 ≤ u1 ≤ T u2 ≤ · · · ≤ T n−2 un−2 ≤
T/θun−1 ,
where m ∈ θ∗ , ∞, M ∈ 0, θ∗ .
Then, the boundary value problem 1.6, 1.7 has a solution u such that u
lies between r
and R.
Theorem 3.2. Suppose that conditions (H1 ), (H2 ) hold. Assume that f also satisfies
A3 f0 ϕ ∈ 0, θ∗ /4
A4 f∞ λ ∈ 2θ∗ /θ, ∞.
Then, the boundary value problem 1.6, 1.7 has a solution u such that u
lies between r and R.
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11
Theorem 3.3. Suppose that conditions (H1 ), (H2 ) hold. Assume that f also satisfies
A5 f∞ λ ∈ 0, θ∗ /4
A6 f0 ϕ ∈ 2θ∗ /θ, ∞.
Then, the boundary value problem 1.6, 1.7 has a solution u such that u
lies between r and R.
Proof of Theorem 3.1. Without loss of generality, we suppose that r < R. For any u ∈ K, by
Lemma 2.3, we have
uΔ t ≥ θ
u
,
n−2
t ∈ θ, T − θ.
3.1
Ω2 {u ∈ K : u
< R}.
3.2
We define two open subsets Ω1 and Ω2 of E:
Ω1 {u ∈ K : u
< r},
For any u ∈ ∂Ω1 , by 3.1 we have
r u
≥ uΔ t ≥ θ
u
θr,
n−2
t ∈ θ, T − θ.
3.3
For t ∈ θ, T − θ and u ∈ ∂Ω1 , we will discuss it from three perspectives.
i If δ ∈ θ, T − θ, thus for u ∈ ∂Ω1 , by A1 and Lemma 2.4, we have
2
T u
2T uΔ δ
n−2
≥
δ δ
0
Δn−1
grf ur, u r, . . . , u
T s
r Δr Δs
θ
n−1
grf ur, uΔ r, . . . , uΔ r Δr Δs
δ
δ δ
Δ
s
δ
≥
grf ur, uΔ r, . . . , uΔ r Δr Δs
n−1
s
T −θ s
δ
n−1
grf ur, uΔ r, . . . , uΔ r Δr Δs
δ
≥ mrAδ ≥ mrL > 2r 2
u
.
3.4
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ii If δ ∈ T − θ, T , thus for u ∈ ∂Ω1 , by A1 and Lemma 2.4, we have
T u
T uΔ δ
n−2
β
≥
α
δ
n−1
gsf us, uΔ s, . . . , uΔ s Δs
ξ
δ δ
0
≥
n−1
grf us, uΔ s, . . . , uΔ s ΔrΔs
3.5
s
T −θ T −θ
Δ
grf ur, u r, . . . , u
θ
Δn−1
r Δr Δs
s
≥ mrAT − θ ≥ mrL > 2r > r u
.
iii If δ ∈ 0, θ, thus for u ∈ ∂Ω1 , by A1 and Lemma 2.4, we have
T u
T uΔ δ
n−2
≥
δ
γ
η
δ
1 s
δ
≥
n−1
gsf us, uΔ s, . . . , uΔ s Δs
n−1
grf ur, uΔ r, . . . , uΔ rΔr Δs
T −θ s
θ
3.6
δ
n−1
grf ur, uΔ r, . . . , uΔ r Δr Δs
θ
≥ mrAθ ≥ mrL > 2r > r u
.
Therefore, no matter under which condition, we all have
T u
≥ u
,
∀u ∈ ∂Ω1 .
3.7
Then by Theorem 2.1, we have
iT, Ω1 , K 0.
3.8
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13
On the other hand, for u ∈ ∂Ω2 , we have ut ≤ u
R; by A2 we know
T u
T uΔ δ
n−1
δ
β
≤
α
ξ
1 δ
0
≤
n−1
gsf us, uΔ s, . . . , uΔ s Δs
Δ
Δn−1
grf ur, u r, . . . , u
3.9
rΔr Δs
s
1
β
MR
1
grΔr ≤ R u
.
α
0
thus
T u
≤ u
,
3.10
∀u ∈ ∂Ω2 .
Then, by Theorem 2.1, we get
iT, Ω2 , K 1.
3.11
Therefore, by 3.8, 3.11, r < R, we have
i T, Ω2 \ Ω1 , K 1.
3.12
Then operator T has a fixed point u ∈ Ω1 \ Ω2 , and r ≤ u
≤ R. Then the proof of
Theorem 3.1 is complete .
Proof of Theorem 3.2. First, by f0 ϕ ∈ 0, θ∗ /4, for θ∗ /4 − ϕ, there exists an adequately
small positive number ρ, as 0 ≤ un−1 ≤ ρ, un−1 /
0, we have
fu1 , u2 , . . . , un−1 ≤ ϕ un−1 ≤
θ∗
θ∗
ρ ρ.
4
4
3.13
Then let R ρ, M θ∗ /4 ∈ 0, θ∗ , thus by 3.13
fu1 , u2 , . . . , un−1 ≤ MR,
0 ≤ un−1 ≤ R.
3.14
So condition A2 holds. Next, by condition A4 , f∞ λ ∈ 2θ∗ /θ, ∞, then for λ − 2θ∗ /θ, there exists an appropriately big positive number r / R, as un−1 ≥ θr, we have
fu1 , u2 , . . . , un−1 ≥ λ − un−1 ≥
2θ∗
θr 2θ∗ r.
θ
3.15
14
Advances in Difference Equations
Let m 2θ∗ > θ∗ , thus by 3.15, condition A1 holds. Therefore by Theorem 3.1 we know
that the results of Theorem 3.2 hold. The proof of Theorem 3.2 is complete.
Proof of Theorem 3.3. Firstly, by condition A6 , f0 ϕ ∈ 2θ∗ /θ, ∞, then for ϕ −
0, we
2θ∗ /θ, there exists an adequately small positive number r, as 0 ≤ un−1 ≤ r, un−1 /
have
2θ∗
un−1 ,
fu1 , u2 , . . . , un−1 ≥ ϕ − un−1 θ
3.16
thus when θr ≤ un−1 ≤ r, we have
fu1 , u2 , . . . , un−1 ≥
2θ∗
θr 2θ∗ r.
θ
3.17
Let m 2θ∗ > θ∗ , so by 3.17, condition A1 holds.
Secondly, by condition A5 , f∞ λ ∈ 0, θ∗ /4, then for θ∗ /4 − λ, there exists a
suitably big positive number ρ /
r, as un−1 ≥ ρ, we have
fu1 , u2 , . . . , un−1 ≤ λ un−1 ≤
θ∗
un−1 .
4
3.18
If f is unbounded, by the continuity of f on 0, T × 0, ∞n−1 , then there exist a constant
2 , . . . , u
n−1 ∈ 0, T × 0, ∞n−1 such that
R /
r ≥ ρ, and a point u1 , u
ρ≤u
n−1 ≤ R,
2 , . . . , u
n−1 ,
u1 , u
fu1 , u2 , . . . , un−1 ≤ f
0 ≤ un−1 ≤ R.
3.19
Thus, by ρ ≤ u0n−1 ≤ R, we know
2 , . . . , u
n−1 ≤
fu1 , u2 , . . . , un−1 ≤ f
u1 , u
θ∗
θ∗
u
n−1 ≤ R.
4
4
3.20
Choose M θ∗ /4 ∈ 0, θ∗ . Then, we have
fu1 , u2 , . . . , un−1 ≤ MR,
0 ≤ un−1 ≤ R.
3.21
If f is bounded, we suppose fu1 , u2 , . . . , un−1 ≤ M, un−1 ∈ 0, ∞, M ∈ R , there exists an
appropriately big positive number R > 4/θ∗ M, then choose M θ∗ /4 ∈ 0, θ∗ , we have
fu1 , u2 , . . . , un−1 ≤ M ≤
θ∗
R MR,
4
0 ≤ un−1 ≤ R.
3.22
Therefore, condition A2 holds. Thus, by Theorem 3.1, we know that the result of
Theorem 3.3 holds. The proof of Theorem 3.3 is complete.
Advances in Difference Equations
15
4. Application
In this section, in order to illustrate our results, we consider the following examples.
Example 4.1. Consider the following boundary value problem on the specific time scale T 0, 1/3 ∪ {1/2, 2/3, 1}:
ΔΔΔ
u
Δ
t tu
Δ
16/L 1e2u − 16/L
0,
u 5euΔ e2uΔ
t ∈ 0, 1T ,
4.1
u0 0,
uΔ 0 − uΔΔ
1
0,
4
uΔ 1 δuΔΔ
1
0,
2
1
,
4
θ
where
α γ 1,
β 1,
δ ≥ 0,
ξ
η
1
,
2
1
,
4
T 1,
4.2
and L is the constant defined in Lemma 2.2,
gt t,
Δ
f u, u
u
Δ
Δ
16/L 1e2u − 16/L
.
u 5euΔ e2uΔ
4.3
Then obviously
f u, uΔ
1
f0 ϕ lim max
,
Δ
Δ
Δ
6
u → 0 0≤u≤4u
u
f u, uΔ
16
f∞ λ lim min
1,
L
uΔ → ∞ 0≤u≤4uΔ
uΔ
4.4
By Theorem 2.1, we have
1
0
gtΔt 1/3
0
gtdt σ1/3
gtΔt 1/3
σ1/2
gtΔt σ2/3
1/2
2/3
gtΔt 5
,
12
4.5
so conditions H1 , H2 hold.
By simple calculations, we have
1
6
,
1
5
grΔr
1 β/α
0
θ∗ then θ∗ /4 3/10, that is, ϕ ∈ 0, θ∗ /4, so condition A3 holds.
4.6
16
Advances in Difference Equations
For θ 1/4, it is easy to see that
λ∈
2θ∗
, ∞ ,
θ
4.7
so condition A4 holds. Then by Theorem 3.2, BVP 4.1 has at least one positive solution.
Example 4.2. Consider the following boundary value problem on the specific time scale T 0, 1/3 ∪ 1/2, 1.
ΔΔΔ
u
Δ
t tu
Δ
1/4eu sin uΔ 16/L
0,
u euΔ
t ∈ 0, 1T ,
4.8
u0 0,
uΔ 0 − uΔΔ
1
1
0, uΔ 1 δuΔΔ
0,
4
2
where
α γ 1,
β 1,
δ ≥ 0,
ξ
1
,
4
η
1
,
2
θ
1
,
4
T 1,
4.9
and L is the constant from Lemma 2.2,
gt t,
Δ
f u, u
u
Δ
Δ
1/4eu sin uΔ 16/L
.
u euΔ
4.10
Then obviously
f u, uΔ
16 1
,
L 4
uΔ → 0 0≤u≤4uΔ
uΔ
f u, uΔ
1
f∞ λ lim min
,
Δ
Δ
Δ
4
u → ∞ 0≤u≤4u
u
f0 ϕ lim
max
4.11
By Theorem 2.1, we have
1
0
gtΔt 1/3
gtdt 0
σ1/3
1/3
gtΔt 1
gtdt 1/2
35
,
72
4.12
so conditions H1 , H2 hold. By simple calculations, we have
1
36
,
1
35
grdr
1 β/α
0
θ∗ then θ∗ /4 9/35, that is, λ ∈ 0, θ∗ /4, so condition A5 holds.
4.13
Advances in Difference Equations
17
For θ 1/4, it is easy to see that
ϕ∈
2θ∗
, ∞ ,
θ
4.14
then condition A6 holds. Thus by Theorem 3.3, BVP 4.8 has at least one positive solution.
Acknowledgment
The authors would like to thank the anonymous referee for his/her valuable suggestions,
which have greatly improved this paper.
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