Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2010, Article ID 152460, 27 pages
doi:10.1155/2010/152460
Research Article
On the Existence of Solutions for Impulsive
Duffing Dynamic Equations on Time Scales with
Dirichlet Boundary Conditions
Yongkun Li and Tianwei Zhang
Department of Mathematics, Yunnan University, Yunnan, Kunming 650091, China
Correspondence should be addressed to Yongkun Li, yklie@ynu.edu.cn
Received 4 June 2010; Accepted 16 August 2010
Academic Editor: Douglas Robert Anderson
Copyright q 2010 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.
By using critical point theory, some new sufficient conditions for the existence of solutions of
impulsive Duffing dynamic equations on time scales with Dirichlet boundary conditions are
obtained. Some examples are also given to illustrate our results.
1. Introduction
Consider the following Duffing dynamic equations on time scales with impulsive effects
2
uΔΔ t CuΔ σt − rtuσt fσt, uσt ht, a.e. t ∈ 0, σT κT ,
ΔuΔ tj uΔ tj − uΔ t−j Ij u tj , j 1, 2, . . . , p,
1.1
u0 0 uσT ,
where T > 0, C is a regressive constant, t0 0 < t1 < t2 < · · · < tp < tp1 σT , r ∈
L∞ 0, σT T , h ∈ L2 0, σT T , f : 0, σσT T × R → R is continuous, and Ij : × R → R,
j 1, 2, . . . , p are continuous.
Obviously, system 1.1 covers Duffing equations when T R
u t gut ht,
u t Cu t gt, ut ht.
1.2
2
Abstract and Applied Analysis
The Duffing equation has been used to model the nonlinear dynamics of special types
of mechanical and electrical systems. This differential equation has been named after
the studies of Duffing in 1918 1
, has a cubic nonlinearity, and describes an oscillator.
The main applications have been in electronics, but it can also have applications in
mechanics and in biology. For example, the brain is full of oscillators at micro- and
macrolevel 2
. There are applications in neurology, ecology, secure communications, chaotic
synchronization, and so on. Due to the rich behaviour of these equations, the most
general forced forms of the Duffing equation 1.2 have been studied by many researchers
3–12
.
The study of dynamic equations on time scales goes back to its founder Stefan Hilger
13
, 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 14–16
.
Impulsive effects exist widely in many evolution processes in which their states
are changed abruptly at certain moments of time. The theory of impulsive differential
systems has been developed by numerous mathematicians see 17–24
. Applications
of impulsive differential equations with or without delays occur in biology, medicine,
mechanics, engineering, chaos theory, and so on see 25–32
.
In addition, system 1.1 also includes
uΔΔ t fσt, uσt 0,
2
a. e. t ∈ 0, σT κT ,
u0 0 uσT ,
u t rtut ft, ut, a.e. t ∈ 0, T ,
Δu tj u tj − u t−j Ij tj , j 1, 2, . . . , p,
1.3
u0 0 uT ,
which were studied by papers 33, 34
, and some existence results were obtained by using
some critical point theorems.
Our purpose in this paper is to study the variational structure of problem 1.1 in an
appropriate space of functions and the existence of solutions for problem 1.1 by means of
some critical point theorems. The organization of this paper is as follows. In Section 2, we
make some preparations. In Section 3, we will study the variational structure of problem
1.1 and give some important lemmas which will be used in later section. In Section 4, by
applying some critical point theorems, we establish sufficient conditions for the existence of
solutions to problem 1.1. Some illustrative examples are also given in Section 4.
2. Preliminaries
In this section, we will first recall some basic definitions and lemmas which are used in what
follows.
Abstract and Applied Analysis
3
Definition 2.1 see 13
. A time scale T is an arbitrary nonempty closed subset of the real set R
with the topology and ordering inherited from R. The forward and backward jump operators
σ, ρ : T → T, and the graininess μ, ν : T → R are defined, respectively, by
σt : inf{s ∈ T : s > t},
ρt : sup{s ∈ T : s < t},
μt : σt − t,
νt : t − ρt.
2.1
The point t ∈ T is called left dense, left scattered, right dense, or right scattered if ρt t,
ρt < t, σt t, or σt > t, respectively. Points that are right dense and left dense at the same
time are called dense. If T has a left-scattered maximum m1 , defined Tκ T−{m1 }; otherwise,
set Tκ T. If T has a right-scattered minimum m2 , defined Tκ T − {m2 }; otherwise, set
Tκ T.
Definition 2.2 see 13
. For f : T → R and t ∈ Tκ , then the delta derivative of f at the point
t is defined to be the number f Δ t provided it exists with the property that for each > 0,
there is a neighborhood U of t such that
fσt − fs − f Δ tσt − s ≤ |σt − s|,
∀s ∈ U.
2.2
Definition 2.3 see 13
. 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 rdcontinuous functions f will be denoted by Crd T.
Definition 2.4 see 13
. Assume that T is an arbitrary time scale. We say that a function
p : T → R is regressive provided
1 μtpt /
0,
2.3
∀t ∈ Tκ .
Definition 2.5 see 13
. Assume that T is an arbitrary time scale and a function p : T → R is
regressive, then we define the exponential function on T by
t
ξμτ pτ Δτ ,
ep t, s exp
for s, t ∈ T,
2.4
s
in which
ξh z 1
Log1 zh
h
2.5
for h > 0,
ξh z ξ0 z z for h 0, here Log is the principal logarithm function.
A function f : T → R is called rd continuous provided it is continuous at each rightdense point and its left-sided limit exists finite at each left-dense point in T. We write
1
T. If
f ∈ Crd T Crd T, R. If f is differentiable with f Δ ∈ Crd Tκ , we write f ∈ Crd
Δ
f Δ is differentiable on Tκ Tκ κ with f Δ f Δ 2
2
2
2
∈ Crd Tκ , we write f ∈ Crd
T.
4
Abstract and Applied Analysis
Similar to the classical Riemann’s definition of integrability, the concept of the Riemann delta
integral on time scales is given in 35
. We know that many familiar functions, including
monotone continuous, piecewise continuous, right-dense continuous functions, are Riemann
delta integrable. In the following lemma, we present some properties of the integral that will
be needed later.
Lemma 2.6 see 35
. Let f, g : T → R be two functions and a, b ∈ T. Then one has the following.
1 Let f and g be Riemann delta integrable functions on a, b
, and α, β ∈ R. Then αf βg
is Riemann delta integrable and
b
αfs βgs Δs α
a
b
fsΔs β
b
a
gsΔs.
2.6
a
2 Fundamental Theorem of Calculus. Let f be a continuous function on a, b
T such that f
is delta differentiable on a, b. If f Δ is Riemann delta integrable from a to b, then
b
f Δ sΔs fb − fa.
2.7
a
3 Integration by Parts. Let f and g be continuous functions on a, b
that are differentiable
on a, b. If f Δ and g Δ are Riemann delta integrable from a to b, then
b
a
b
fsg Δ sΔs fsgs a −
b
f Δ sgσsΔs.
2.8
a
4 If f is Riemann delta integrable on a, b
, then so is |f|, and
b
b
Δ
f sΔs f Δ sΔs.
a
a
35
.
2.9
The construction of the Δ-measure on T and the following concepts can be found in
1 For each t0 ∈ T\{max T}, the single-point set t0 is Δ-measurable, and its Δ-measure
is given by
μΔ {t0 } σt0 − t0 μt0 .
2.10
Abstract and Applied Analysis
5
2 If a, b ∈ T and a ≤ b, then
μΔ a, b b − a,
μΔ a, b b − σa.
2.11
3 If a, b ∈ T \ {max T} and a ≤ b, then
μΔ a, b
σb − σa,
μΔ a, b
σb − a.
2.12
The Lebesgue integral associated with the measure μΔ on T is called the Lebesgue
delta integral. For a measurable
set E ⊂ T and a function f : E → R, the corresponding
integral of f on E is denoted by E fsΔs. All theorems of the general Lebesgue integration
theory hold also for the Lebesgue delta integral on T. Moreover, comparing the Lebesgue
delta integral with the Riemann delta integral, we have the following.
Lemma 2.7 see 35
. Let a, b
be a closed bounded interval in T and let f be a bounded realvalued function defined on a, b
. If f is Riemann delta integrable from a to b, then f is Lebesgue
delta integrable on a, b and
R
b
fsΔs L
fsΔs,
a,b
a
2.13
where R and L indicate the Riemann delta integral and Lebesgue delta integral, respectively.
p
Assume that p ∈ R and p > 1. Let LLS a, b
be the set of
f : a, b
−→ R :
b
fsp Δs < ∞ .
2.14
a
p
By Lemma 3.6 in 36
, LLS a, b
is a complete linear space with the norm · p defined by
f p
b
1/p
fsp Δs
.
2.15
a
Let CT S a, b
denote the linear space of all continuous functions f : a, b
→ R with
the maximum norm fC maxt∈a,b
|ft|.
Lemma 2.8 Hölder inequality 37
. Let f, g ∈ Crd a, b
, p > 1 and q the conjugate number of
p. Then
b
a
fsgsΔs ≤
b
a
1/p b
1/q
q
gs Δs
·
.
fsp Δs
a
2.16
6
Abstract and Applied Analysis
At the end of this section, we recall some notation and known results from critical
point theory.
Let X be a real normed space and let ϕ be a functional from X to R.
1 ϕ is called weakly continuous if
xk x ⇒ ϕxk −→ ϕx.
2.17
2 ϕ is called lower semicontinuous if
xk −→ x ⇒ lim inf ϕxk ≥ ϕx.
2.18
k→∞
3 ϕ is called weakly lower semicontinuous if
xk x ⇒ lim inf ϕxk ≥ ϕx.
2.19
k→∞
Let X be a real Hilbert space, ϕ ∈ C1 X, R, which means that ϕ is a continuously
Fréchet-differentiable functional defined on X. ϕ is said to satisfy the Palais-Smale condition
P.S. condition if any sequence {xn } ⊂ X for which {ϕxn } is bounded and ϕ xn → 0 as
n → ∞ possesses a convergent subsequence in X.
Let Br be the open ball in X with radius r and centered at 0 and let ∂Br denote its
boundary. The following lemma is taken from 38, 39
and will play an important role in the
proof of our main results.
Lemma 2.9 see 38
. If X is a real normed space and ϕ : X → R is lower semicontinuous and
convex, then ϕ is weakly lower semicontinuous.
Lemma 2.10 see 39
. Let X be a real Banach space, ϕ ∈ C1 X, R be weakly lower(upper)
semicontinuous, and
lim ϕx ∞
x → ∞
lim ϕx −∞ .
x → ∞
2.20
Then, there exists x0 ∈ X such that
ϕx0 inf ϕx
x∈X
ϕx0 sup ϕx .
2.21
x∈X
Lemma 2.11 Mountain Pass Theorem 40
. Let X be a real Hilbert space and let ϕ ∈ C1 X, R
satisfy P.S. condition, and ϕ satisfies
J1 There are constants ρ, α > 0 such that ϕx ≥ α for all x ∈ ∂Bρ , where Bρ {x ∈ X :
x < ρ}.
J2 ϕ0 ≤ 0 and there exists x0 ∈ Bρ such that ϕx0 ≤ 0.
Abstract and Applied Analysis
7
Then ϕ possesses a critical value c ≥ α. Moreover, c can be characterized as
c inf max ϕ gs ,
2.22
Γ g ∈ C0, 1
, X | g0 0, g1 x0 .
2.23
g∈Γ s∈0,1
where
3. Variational Structure
In this section, we will establish the corresponding variational framework for problem 1.1.
Let P be the function space of the form
1
P u | u ∈ Crd
0, σT κT , u ∈ CT S 0, σT T , u0 0 uσT .
3.1
P can be equipped with the inner product
u, vP σT uΔ tvΔ tΔt,
∀u, v ∈ P,
3.2
0
inducing the norm
uP u, uP σT 1/2
Δ 2
u t Δt
,
∀u ∈ P.
3.3
0
− uΔ
uk − ul P → 0 as
Let {uk } ∈ P be a Cauchy sequence in P, that is, uΔ
k
l 2
2
k, l → ∞. Since LLS 0, σT T is a complete space, there exists a v ∈ L2LS 0, σT T such that
− v2 → 0 as k → ∞. Define a function u by
uΔ
k
ut t
vsΔs.
3.4
0
The function v is called a weak derivative of u and is also denoted by uΔ .
Remark 3.1. If u has a weak derivative, then this weak derivative is unique in L2LS 0, σT T ,
that is, if v1 , v2 ∈ L2LS 0, σT T are both weak derivatives of u, then v1 − v2 2 0. If u is
delta differential on 0, σT T , then its weak derivative is its delta derivative.
8
Abstract and Applied Analysis
From 3.4, it is clear that u0 0 and
σT vsΔs
|uσT | 0
σT vs − uΔ
k s Δs
0
σT ≤
s
Δs
vs − uΔ
k
3.5
0
≤
σT 1/2 σT 1 Δs
2
0
0
1/2
2
s
Δs
vs − uΔ
k
σT v − uΔ
k −→ 0,
2
as k −→ ∞.
Thus, uσT 0.
Define the space H as
H
ut t
0
Δ
vsΔs | ∃{uk } ∈ P satisfying uk − v −→ 0 as k −→ ∞ .
2
3.6
It is clear that P ⊂ H, and we define the inner product ·, ·H in H as
u, vH σT uΔ tvΔ tΔt,
3.7
∀u, v ∈ H,
0
inducing the norm
σT 1/2
Δ 2
,
uH u, uH u t Δt
∀u ∈ H,
3.8
0
where uΔ represents the weak derivative of u. Moreover, we have the following result.
Lemma 3.2 see 33
. H is a complete space in the norm · H and inequality
σT |uσ t|2 Δt ≤ σT 2
σT Δ 2
u t Δt,
0
3.9
∀u ∈ H
0
holds, that is, uσ 2 ≤ σT uH .
Also, we consider the inner product
u, v σT 0
eC t, 0rtu tv tΔt σ
σ
σT 0
eC t, 0uΔ tvΔ tΔt,
∀u, v ∈ H,
3.10
Abstract and Applied Analysis
9
inducing the norm
u σT 2
eC t, 0rt|u t| Δt σ
σT 0
1/2
2
eC t, 0uΔ t Δt
,
∀u ∈ H.
3.11
0
Throughout this paper, it will be assumed that ess inft∈0,σT T rt r0 > −λ0 , where λ0
is the smallest eigenvalue of the problem
2
uΔΔ t λuσ t 0, t ∈ 0, σT κT ,
u0 0 uσT .
3.12
From 41
, λ0 may be defined as
σT λ0 min
u/
0
u00uσT −
0
uΔΔ tuσ tΔt
uσ 22
3.13
.
Applying the integration by parts in Lemma 2.6, we have
λ0 min
u2H
> 0,
3.14
∀u ∈ H.
3.15
u/
0
uσ 22
u00uσT which implies that
uσ 22 ≤
1
u2H ,
λ0
For convenience, we introduce the following notations:
el min eC t, 0,
t∈0,σT T
eL max eC t, 0,
t∈0,σT T
h0 σT |ht|Δt.
3.16
0
Lemma 3.3. Suppose that
H1 r0 eL λ0 el > 0, for all r0 ∈ −λ0 , 0.
Then one has
ωu2H ≤ u2 ≤ eL r∞ σ 2 T 1 u2H ,
3.17
which implies that the norm · and the norm · H are equivalent, where
⎧
L
⎪
⎨ r0 e el ,
λ0
ω
⎪
⎩e l ,
−λ0 < r0 < 0,
r0 ≥ 0.
3.18
10
Abstract and Applied Analysis
Proof
Case 1 r0 ≥ 0. For every u ∈ H, we have
u2 σT eC t, 0rt|uσ t|2 Δt σT 0
≥
2
eC t, 0uΔ t Δt
0
σT 2
eC t, 0uΔ t Δt
3.19
0
σT Δ 2
≥ el
u t Δt
0
e u2H .
l
On the other hand, by Lemma 2.6, one has
u2 σT 0
≤ eL
eC t, 0rt|uσ t|2 Δt σT σT 2
eC t, 0uΔ t Δt
0
σT Δ 2
2
σ
|rt||u t| Δt u t Δt
0
0
σT Δ 2
≤ eL r∞ σ 2 T 1
u t Δt
eL
3.20
0
r∞ σ 2 T 1 u2H .
Therefore, the norm · and the norm · H are equivalent.
Case 2 −λ0 < r0 < 0. For every u ∈ H, we have from 3.15 that
u2 σT eC t, 0rt|uσ t|2 Δt 0
≥ eL r0
σT 2
eC t, 0uΔ t Δt
0
σT 0
σT Δ 2
2
σ
l
|u t| Δt e
u t Δt
0
σT r0 eL σT Δ 2
Δ 2
≥
u t Δt el
u t Δt
λ0 0
0
σT r0 eL
Δ 2
el
u t Δt
λ0
0
r0 eL
l
e u2H .
λ0
Similar to Case 1, it is easy to obtain the desired results. This completes the proof.
3.21
Abstract and Applied Analysis
11
Together with 3.15 and Lemma 3.3, it is not difficult to obtain the following.
Lemma 3.4 Poincaré inequality. For all u ∈ H, one has
σT 0
1
|u t| Δt ≤
λ0 ω
σ
2
σT 2
eC t, 0rt|u t| Δt σ
σT 0
Δ 2
eC t, 0u t Δt ,
3.22
0
that is, uσ 2 ≤ 1/ λ0 ωu.
Lemma 3.5. u∞ ≤ a0 u for all u ∈ H, where a0 σT /ω.
Proof. For every u ∈ H, we have
|ut| |ut − u0|
t
Δ
u sΔs
0
t
≤ uΔ sΔs
3.23
0
≤t
1/2
t
1/2
Δ 2
u s Δs
0
≤ t1/2 uH .
Therefore,
u∞ ≤
σT uH ≤
σT u.
ω
3.24
This completes the proof.
2
2
For u ∈ H Crd
0, σT κT , we have that u and uΔ are both absolutely continuous.
Hence, ΔuΔ t uΔ t − uΔ t− 0 for any t ∈ 0, σT κT .
If u ∈ H, then u is absolutely continuous and uΔ ∈ L2T S 0, σT T . In this case, ΔuΔ t Δ u t − uΔ t− 0 may not hold for some t ∈ 0, σT κT . It leads to the impulsive effects.
Take v ∈ H and multiply the two sides of the equality
uΔΔ t CuΔ σt − rtuσt fσt, uσt ht
3.25
12
Abstract and Applied Analysis
by eC t, 0, vσ t and integrate from 0 to σT ; we have
σT Δ
eC t, 0u t
Δ
v tΔt σ
0
σT eC t, 0htvσ tΔt
0
−
σT eC t, 0rtuσtvσ tΔt
3.26
0
−
σT eC t, 0fσt, uσtvσ tΔt.
0
Furthermore,
σT Δ
eC t, 0uΔ t vσ tΔt
0
jp tj1 Δ
eC t, 0uΔ t vσ tΔt
tj
j0
jp
eC t−j1 , 0
Δ
u
t−j1
tj1
v t−j1 − eC tj , 0 uΔ tj v tj −
eC t, 0uΔ tvΔ tΔt
tj
j0
−
jp
eC tj , 0 ΔuΔ tj v tj eC σT , 0uΔ σT vσT − eC 0, 0uΔ 0v0
j1
σT −
eC t, 0uΔ tvΔ tΔt
0
−
σT jp
e C tj , 0 I j u tj v t j −
eC t, 0uΔ tvΔ tΔt.
0
j1
3.27
Combining 3.26, we have
σT eC t, 0uΔ tvΔ tΔt 0
σT eC t, 0rtuσtvσ tΔt 0
σT 0
eC t, 0fσt, uσtvσ tΔt −
jp
e C tj , 0 I j u tj v t j
j1
σT eC t, 0htvσ tΔt.
0
Considering the above, we introduce the following concept solution for problem 1.1.
3.28
Abstract and Applied Analysis
13
Definition 3.6. We say that a function u ∈ H is a weak solution of problem 1.1 if the identity
σT eC t, 0uΔ tvΔ tΔt σT 0
eC t, 0rtuσtvσ tΔt 0
σT p
e C tj , 0 I j u tj v t j
j1
eC t, 0fσt, uσtvσ tΔt −
0
σT eC t, 0htvσ tΔt
0
3.29
holds for any v ∈ H.
Consider the functional ϕ : H → R defined by
ϕu σT 1
u2 2
eC t, 0htuσ t − Fσt, uσ t
Δt 0
p
e C tj , 0
utj Ij s ds,
0
j1
3.30
u
where Ft, u 0 ft, sds. Using the continuity of f and Ij , j 1, 2, . . . , p, one has that
ϕ ∈ C1 H, R. For any v ∈ H, we have
ϕ uv σT σT eC t, 0uΔ tvΔ tΔt 0
−
0
σT eC t, 0fσt, uσ tvσ tΔt 0
eC t, 0rtuσ tvσ tΔt
σT eC t, 0htvσ tΔt
3.31
0
p
e C tj , 0 I j u tj v t j .
j1
Therefore, the solutions of problem 1.1 are the corresponding critical points of ϕ.
Lemma 3.7 see 33
. The functionals ψ1 , ψ2 , ψ3 : H → R defined by
ψ1 u σT Fσt, uσ tΔt,
0
ψ2 u σT 0
htuσ tΔt,
ψ3 u p utj j1
Ij sds
0
3.32
are weakly continuous on H.
14
Abstract and Applied Analysis
4. Main Results
Theorem 4.1. Assume that (H1) holds. Suppose further the following.
H2 There exist a, b > 0 and θ ∈ 0, 1 such that
ft, u ≤ a b|u|θ ,
4.1
∀t, u ∈ 0, σT T × R.
H3 There exist aj , bj > 0 and θj ∈ 0, 1 j 1, 2, . . . , p such that
Ij u ≤ aj bj |u|θj ,
∀u ∈ R j 1, 2, . . . , p .
4.2
Then problem 1.1 has at least one weak solution.
Proof. According to Lemma 3.7, ψ1 u, ψ2 u, ψ3 u are weakly continuous on H. On the other
hand, it is clear that u2 is continuous and convex on H. By Lemma 2.9, u2 is weakly lower
semicontinuous. So ϕu is weakly lower semicontinuous on H.
From H2-H3, we have
1
ϕu u2 2
σT −e
σ
0
1
≥ u2 − eL
2
L
eC t, 0htu t − Fσt, u t
Δt σ
p
eC
tj , 0
utj j1
σT |htu t|Δt − e
σ
L
0
Ij sds
0
σT a|uσ t| b|uσ t|θ1 Δt
0
p utj aj bj |s|θj ds
j1
0
≥
p 1
θj 1
L
−
e
aj u∞ bj u∞
u2 − eL h0 u∞ − eL σT au∞ buθ1
∞
2
j1
≥
1
θ1
u
u2 − eL h0 a0 u − eL σT aa0 u baθ1
0
2
− eL
p θj 1
aj a0 u bj a0
uθj 1
j1
4.3
Abstract and Applied Analysis
15
for all u ∈ H. This implies that limu → ∞ ϕu ∞, and ϕ is coercive. By Lemma 2.10, ϕ has a
minimum point on H, which is a critical point of ϕ. Therefore, problem 1.1 has at least one
weak solution. This completes the proof.
Example 4.2. Consider the following impulsive Duffing equations on time scales
uΔΔ t 2uΔ t − 3ut 1
sin t,
|ut| 10e
a.e. t ∈ 0, 1
T ,
− 1
1
Δ 1
Δ 1
u
,
Δu
−u
3u
2
2
2
2
4.4
Δ
u0 0 uσ1.
Then, problem 4.4 has at least one solution.
Proof. The result is easy to obtain from Theorem 4.1 and we should omit it. This completes
the proof.
Theorem 4.3. Assume that (H1) holds. Suppose further the following.
H4 There exist a, b, c > 0 and θ ∈ 0, 1 such that
ft, u ≤ a|u| b|u|θ c,
4.5
∀t, u ∈ 0, σT T × R.
H5 There exist aj , bj , cj > 0 and θj ∈ 0, 1 j 1, 2, . . . , p such that
Ij u ≤ aj |u| bj |u|θj cj ,
H6 1/2 − aσT eL a20 −
p
j1
∀u ∈ R j 1, 2, . . . , p .
4.6
aj eL a20 > 0.
Then problem 1.1 has at least one weak solution.
Proof. According to Theorem 4.1, we know that ϕu is weakly lower semicontinuous on H.
From H4−H6, we have
ϕu 1
u2 2
σT 0
eC t, 0htuσ t − Fσt, uσ t
Δt p
j1
e C tj , 0
utj Ij sds
0
σT σT 1
2
L
σ
L
a|uσ t|2 b|uσ t|θ1 c|uσ t| Δt
≥ u − e
|htu t|Δt − e
2
0
0
p
utj − eL
aj |s| bj |s|θj cj ds
j1
0
16
Abstract and Applied Analysis
≥
1
cu
u2 − eL h0 u∞ − eL σT au2∞ buθ1
∞
∞
2
− eL
p θj 1
aj u2∞ bj u∞
cj u∞
j1
≥
1
θ1
ca
u
u
u2 − eL h0 a0 u − eL σT aa20 u2 baθ1
0
0
2
− eL
p θj 1
aj a20 u2 bj a0
uθj 1 cj a0 u
j1
⎞
p
1
⎝ − aσT eL a20 − aj eL a20 ⎠u2 − eL h0 a0 u
2
j1
⎛
p θj 1
θ1
L
−
e
− eL σT baθ1
bj a0 uθj 1 cj a0 u
ca
u
u
0
0
j1
4.7
for all u ∈ H. This implies that limu → ∞ ϕu ∞, and ϕ is coercive. By Lemma 2.10, ϕ has a
minimum point on H, which is a critical point of ϕ. Therefore, problem 1.1 has at least one
weak solution. This completes the proof.
Example 4.4. Consider the following impulsive Duffing equations on time scales:
1
uΔΔ t 2uΔ t − 3ut ΔuΔ
5σ 3 1ν0
Δ 1
1
u
2
ut − uΔ
2
|ut| sin t,
1−
2
a.e. t ∈ 0, 1
T ,
1
1
,
u
2
2
5σ 1ν0
4.8
u0 0 uσ1,
where
ν0 exp
σ1
log 1 3μs
1/μs
Δs .
4.9
0
Then problem 4.8 has at least one solution.
Proof. By an easy calculation, we obtain
e exp
L
σ1
1/μs
log 1 3μs
Δs ,
el 1.
4.10
0
It is easy to verify that all the conditions of Theorem 4.3 are satisfied. From Theorem 4.3,
problem 4.8 has at least one solution. This completes the proof.
Abstract and Applied Analysis
17
Lemma 4.5. Assume that (H1) and (H3) hold. Suppose further the following.
H7 There exist constants μ > 2 and R > 0, such that
0 < μFt, u ≤ uft, u,
where Ft, u u
0
∀|u| ≥ R, t ∈ 0, σT T ,
4.11
ft, sds.
Then ϕ satisfies the P.S. condition in H.
Proof. Let {un } be a sequence in H satisfying that {ϕun } is bounded and ϕ un → 0 as
n → ∞. Let Ωn {t ∈ 0, σT T | |un t| ≥ R} for all n ∈ N. By the continuity of f and
H7, there exists a constant c1 > 0 such that
βϕun − ϕ un , un σT β
2
− 1 un eC t, 0 uσn fσt, uσn − βFσt, uσn Δt
2
0
p
p
un tj β e C tj , 0
Ij sds − eC tj , 0 Ij un tj un tj
0
j1
σT 0
≥
j1
β − 1 eC t, 0htuσn Δt
β
− 1 un 2 eC t, 0 uσn fσt, uσn − βFσt, uσn Δt
2
Ωn
4.12
p θj 1
aj a0 u bj a0 uθj 1
− β 1 eL
j1
− β − 1 eL h0 a0 un − c1
≥
p β
θj 1
aj a0 u bj a0 uθj 1
− 1 un 2 − β 1 eL
2
j1
− β − 1 eL h0 a0 un − c1 ,
which implies that {un } is bounded in H. Taking into account {un } ⊂ H ⊂ CT S 0, σT T
and employing Lemma 3.5, we know that {un } is relatively compact in CT S 0, σT T . Thus
{un } possesses a convergent subsequence {unk } in CT S 0, σT T , that is, uni − unj ∞ → 0 as
i, j → ∞. since un ∈ H, we know that uσn T 0 for all n ∈ N. From the above inequality,
18
Abstract and Applied Analysis
it is easy to see that
sup uni t − unj t −→ 0,
as i, j −→ ∞.
t∈0,σT κT
4.13
From the continuity of f and Ij , j 1, 2, . . . , p, we have
f σt, uσni t − f σt, uσnj t −→ 0,
Ij uni t − Ij unj t −→ 0
4.14
hold uniformly for t ∈ 0, σT κT as i, j → ∞. Set
f σt, uσni t − f σt, uσnj t ∞
Ij uni t − Ij unj t ∞
sup f σt, uσni t − f σt, uσnj t ,
t∈0,σT κT
4.15
sup Ij uni t − Ij unj t .
t∈0,σT κT
Clearly,
f σt, uσni t − f σt, uσnj t −→ 0,
∞
Ij uni t − Ij unj t −→ 0,
∞
4.16
as i, j → ∞. We rewrite 3.31 as
ϕ u, v u, v −
σT eC t, 0fσt, uσtvσ tΔt
0
σT p
eC t, 0htv tΔt eC tj , 0 Ij u tj v tj
4.17
σ
0
j1
for any u, v ∈ H. Therefore,
p
% $ v tj
uni − unj , v ϕ uni − ϕ unj , v − eC tj , 0 Ij uni tj − Ij unj tj
j1
σT 0
$
%
eC t, 0 fσt, uni σt − f σt, unj σt vσ tΔt.
4.18
Abstract and Applied Analysis
19
Thus, for any v ∈ H, it follows that
p uni − unj , v ≤ ϕ uni − ϕ unj v eL Ij uni tj − Ij unj tj vtj ∞
∞
j1
σT e fσt, uni σt − f σt, unj σt |vσ t|Δt
L
∞
0
p ≤ ϕ uni − ϕ unj v eL a0 Ij uni tj − Ij unj tj v
∞
j1
eL σT fσt, uni σt − f σt, unj σt vσ 2
∞
≤ ϕ uni − ϕ unj v eL a0
Ij uni tj − Ij unj tj v
p ∞
j1
σT e fσt, uni σt − f σt, unj σt v.
∞
λ0 ω
L
4.19
Hence,
uni − unj sup uni − unj , v v≤1
p ≤ ϕ uni − ϕ unj eL a0 Ij uni tj − Ij unj tj ∞
j1
4.20
σT e fσt, uni σt − f σt, unj σt .
∞
λ0 ω
L
Observing that ϕ un → 0 as n → ∞, it is easy to see that
ϕ uni − ϕ unj −→ 0,
as i, j −→ ∞.
4.21
Together with 4.16−4.21, we get
uni − unj −→ 0,
as i, j −→ ∞.
4.22
In other words, {un } possesses a convergent subsequence {unk } in H. The P.S. condition is
now satisfied. This completes the proof.
20
Abstract and Applied Analysis
Theorem 4.6. Assume that (H1), (H3), and (H7) hold. Suppose further the following.
H8 There exists constant ν ≥ 2 such that
0 < νFt, u ≤ uft, u,
∀u ∈ −1, 0
&
0, 1
, t ∈ 0, σT T ,
4.23
p
p
H9 a20 /2 − eL MσT − eL σT h2 − eL j1 aj − eL j1 bj > 0, where
M : sup{Ft, u : t ∈ 0, σT T , |u| 1}.
4.24
Then problem 1.1 has at least one weak solution.
Proof. It suffices to show that ϕ possesses a nonzero critical point. Now, we need to verify
that all assumptions of the Mountain Pass Theorem hold. The P.S. condition follows from
Lemma 4.5. Next, we will check conditions J1 and J2 of Lemma 2.11. Assume that 0 <
u∞ ≤ 1. In view of H8, for ξ ∈ 0, 1
, we get that
d
d $ −1 ν %
F t, ξ u ξ dξ
dξ
(
' ξ−1 u
ft, sds ξ
ν
νF t, ξ−1 u − ξ−1 uf t, ξ−1 u ξν−1 ≤ 0.
0
4.25
Hence,
u
Ft, u ≤ F t,
|u|ν ,
|u|
for 0 < |u| ≤ 1, t ∈ 0, σT T ,
4.26
which implies that
σT 0
eC t, 0|Fσt, uσ t|Δt ≤
σT 0
≤ eL M
uσ t σ ν
eC t, 0F σt, σ
|u t| Δt
|u t| σT |uσ t|2 Δt
0
≤ eL M
σ 2 T u2 .
ω
4.27
Abstract and Applied Analysis
21
At the same time, we can obtain that
σT σT eC t, 0|htuσ t|Δt ≤ eL h2 uσ 2 ≤ eL √ h2 u,
ω
0
p
p utj utj Ij sds ≤ eL
aj bj |s|θj ds
e C tj , 0
0
j1
0
j1
≤e
L
p aj u∞ θj 1
bj u∞
4.28
j1
≤e
L
p θj 1
aj a0 u bj a0
uθj 1 .
j1
Together 4.27-4.28 and the expression of functional ϕ, it follows that
ϕu ≥
√
1
u2 − eL MσT a20 u2 − eL ωa20 h2 u
2
− eL
p θj 1
aj a0 u bj a0
uθj 1
j1
p
1
θj 1
− eL MσT a20 u2 − eL bj a0 uθj 1
2
j1
⎛
− ⎝e
√
L
⎞
p
ωa20 h2 eL aj a0 ⎠u
j1
⎛
⎞
p
1
⎝ − eL MσT a20 − eL σT a20 h2 − eL aj a20 ⎠u2
2
j1
⎛
⎝e
L
σT a20 h2
4.29
⎞
p
2⎠
e
aj a0 u2
L
j1
⎛
− ⎝e
√
L
⎞
p
ωa20 h2 eL aj a0 ⎠u
j1
− eL
p
θj 1
bj a0
uθj 1 .
j1
Setting
1
u ρ :
a0
ω
.
σT 4.30
22
Abstract and Applied Analysis
Then we have
⎛
⎞
⎞
p
p
√
⎝eL σT a2 h eL aj a2 ⎠ρ2 − ⎝eL ωa2 h eL aj a0 ⎠ρ 0,
2
2
0
0
0
⎛
j1
4.31
j1
which implies from H9 that
⎛
⎞
p
p
1
θj 1
L
2
L
2
L
2
ϕu ≥ ⎝ − e MσT a0 − e σT a0 h2 − e
aj a0 ⎠ρ2 − eL bj a0 ρθj 1
2
j1
j1
p
p
a20
− eL MσT − eL σT h2 − eL aj − eL bj
2
j1
j1
4.32
α > 0.
By Lemma 3.5, if u ρ, then 0 < u∞ ≤ 1. Set Bρ {u ∈ H | u < ρ 1/a0 }, then
ϕu ≥ α > 0 for all u ∈ ∂Bρ . Condition J1 is satisfied.
Secondly, we verify the condition J2 of Lemma 2.11. In view of assumption H7, we
have
μ ft, u
≤
,
u Ft, u
μ ft, u
≥
,
u Ft, u
u ≥ R,
4.33
u ≤ −R.
4.34
Integrating 4.33 and 4.34 for u on R, u
and u, −R
, respectively, we get
μ ln
u
Ft, u
≤ ln
,
R
Ft, u
u ≥ R,
4.35
Ft, −R
R
≥ ln
,
μ ln
−u
Ft, u
u ≤ −R,
that is,
Ft, u ≥ Ft, R
u μ
Ft, u ≥ Ft, −R
R
−u
R
u ≥ R,
,
μ
,
u ≤ −R.
4.36
4.37
Combining 4.36 and 4.37, one has
Ft, u ≥ α1 |u|μ ,
|u| ≥ R,
4.38
Abstract and Applied Analysis
23
where
)
α1 R−μ min
min Ft, R,
t∈0,σT T
*
min Ft, −R > 0.
4.39
t∈0,σT T
On the other hand, by the continuity of Ft, u, Ft, u is bounded on 0, σT T × −R, R
,
there exists K > 0 such that
Ft, u ≥ −K ≥ α1 |u|μ − α1 Rμ − K,
|u| ≤ R.
4.40
Combining 4.38 and 4.40, we have
Ft, u ≥ α1 |u|μ − α2 ,
∀t, u ∈ 0, σT T × R,
4.41
where α2 α1 Rμ K.
According to the Hölder inequality, we get
σT 2
1−2/μ
|u t| Δt ≤ σT σ
σT 0
2/μ
μ
|u t| Δt
σ
4.42
.
0
Thus,
σT μ
2−μ/2
|u t| Δt ≥ σT σ
σT 0
μ/2
σT 2−μ/2 uσ 2 .
μ
2
|u t| Δt
σ
0
4.43
Together 4.41 and 4.43, we know that
σT 0
eC t, 0Fσt, uσ tΔt ≥
σT eC t, 0α1 |uσ t|μ Δt −
0
σT α2 eC t, 0Δt
0
≥ α1 el σT 2−μ/2 uσ 2 − α2 eL σT μ
μ
≥ α3 uσ 2 − α4 ,
where α3 α1 el σT 2−μ/2 , and α4 α2 eL σT .
4.44
24
Abstract and Applied Analysis
Obviously, ϕ0 0. Let v0 ∈ H satisfy v0 1. For any ς > 0, it follows from 4.44
and 4.28 that
ϕςv0 σT ς2
eC t, 0F σt, ςv0σ t Δt
v0 2 −
2
0
σT p
ςv0 tj σ
eC t, 0htv0 tΔt eC tj , 0
Ij sds
ς
0
j1
0
μ
ς2
σT ≤ v0 2 − α3 ςμ v0σ 2 ςeL √ h2 v0 2
ω
eL
p θj 1
ςaj a0 v0 bj ςθj 1 a0
4.45
v0 θj 1 α4 .
j1
Since μ > 2, we see that ϕςv0 → −∞ as ς → ∞. we can choose sufficiently large ς0 such
that u0 ς0 v0 ∈ H satisfying u0 ∈∂Bρ and ϕu0 < 0. Condition J2 is satisfied.
According to Lemma 2.11, ϕ possesses a critical value c ≥ α > 0 given by
c inf max ϕ gs ,
4.46
Γ g ∈ C0, 1
, E | g0 0, g1 u0 .
4.47
g∈Γ s∈0,1
where
Therefore, problem 1.1 has at least one weak solution. This completes the proof.
Example 4.7. Let T R. Consider the following impulsive Duffing equations with Dirichlet
boundary condition
1
1
sin t, a.e. t ∈ 0, 1
,
u t u t − ut ft, ut 2 10e
−
1
1
1
1
1
5
Δu
u
,
u
− u
2
2
2
10e
2
u0 0 u1,
4.48
where
⎧
⎪
⎨
u
,
10et1
ft, u 3
⎪
⎩ u ,
10et1
∀t, u ∈ 0, 1
× −1, 1
,
&
∀t, u ∈ 0, 1
× −∞, −1 1, ∞ .
Then problem 4.48 has at least one solution.
4.49
Abstract and Applied Analysis
25
Proof. By an easy calculation, eL e, el 1, h2 ≤ 1/10e, a0 1, a1 0, b1 1/10e, and
M 1/20e. It is easy to verify that all the conditions of Theorem 4.6 are satisfied. From
Theorem 4.6, problem 4.48 has at least one solution. This completes the proof.
Corollary 4.8. Assume that (H1), (H3), (H7), and (H8) hold. Suppose further the following.
H10 Ij t ≥ 0 for all t ∈ 0, σT T , j 1, 2, . . . , p.
H11 a20 /2 − eL MσT − eL σT h2 > 0, where
M : sup{Ft, u : t ∈ 0, σT T , |u| 1}.
4.50
Then problem 1.1 has at least one weak solution.
Corollary 4.9. Assume that (H1), (H3), (H7), (H8), and (H10) hold. Suppose further the following.
H12 ht ≡ 0 for all t ∈ 0, σT T .
H13 a20 /2 > eL MσT , where M : sup{Ft, u : t ∈ 0, σT T , |u| 1}.
Then problem 1.1 has at least one weak solution.
Corollary 4.10. Assume that (H1), (H3), (H7), (H10), and (H12) hold. Suppose further the
following.
H14 supt∈0,σT T lims → 0 ft, s/s 0.
Then problem 1.1 has at least one weak solution.
Proof. From the proof of Theorem 4.6, it is easy to see that ϕ ∈ C1 E, R and the condition J2
of Lemma 2.11 is satisfied. We only need to verify the condition J1 in Lemma 2.11.
In view of H14, for λ0 ω/2eL > 0, there is a constant δ > 0 such that
ft, s ≤ |s|,
∀0 ≤ |s| ≤ δ, t ∈ 0, σT T .
4.51
As a result, we have
|Fσt, uσ t| uσ t
fσt, sds ≤
0
|uσ t|
σ 2
|u t|
2
4.52
∀0 ≤ |s| ≤ δ.
4.53
|s|ds 0
which implies from Lemma 3.4 that
σT |Fσt, uσ t|Δt ≤
0
σ 2
u 2 ≤
u2 ,
2
2λ0 ω
From the above inequality, H10 and H12, it follows that
ϕu ≥
1
eL
1
u2 −
u2 u2 .
2
2λ0 ω
4
4.54
26
Abstract and Applied Analysis
Take ρ δ/a0 , α δ2 /4a20 . By Lemma 3.5, if u ρ, then 0 < u∞ ≤ δ. Set Bρ {u ∈ H |
u < ρ δ/a0 } , then ϕu ≥ α > 0 for all u ∈ ∂Bρ . Condition J1 is satisfied. Therefore,
problem 1.1 has at least one weak solution. This completes the proof.
Remark 4.11. Let C 0, rt ≡ 0 and let Ij t ≡ 0 for all t ∈ 0, σT T , j 1, 2, . . . , p, then
Corollary 4.10 reduces to Theorem 4.2 in 33
.
Acknowledgment
This work is supported by the National Natural Sciences Foundation of People’s Republic of
China under Grant 10971183.
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