# On the Existence of Solutions for Impulsive Duffing Dynamic Equations on Time Scales with Dirichlet Boundary Conditions

Authors: Yongkun Li; Tianwei Zhang
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

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.


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## Body

## 1. Introduction

Consider the following Duffing dynamic equations on time scales with impulsive effects(1.1)uΔΔ(t)+CuΔ(σ(t))-r(t)u(σ(t))+f(σ(t),u(σ(t)))=h(t),a.e.t∈[0,σ(T)]𝕋κ2,ΔuΔ(tj)=uΔ(tj+)-uΔ(tj-)=Ij(u(tj)),j=1,2,…,p,u(0)=0=u(σ(T)),
where T>0, C is a regressive constant, t0=0<t1<t2<⋯<tp<tp+1=σ(T), r∈L∞[0,σ(T)]𝕋, h∈L2[0,σ(T)]𝕋, f:[0,σ(σ(T))]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) (1.2)u′′(t)+g(u(t))=h(t),u′′(t)+Cu′(t)+g(t,u(t))=h(t).
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 (1.3)uΔΔ(t)+f(σ(t),u(σ(t)))=0,a.e.t∈[0,σ(T)]𝕋κ2,u(0)=0=u(σ(T)),u′′(t)+r(t)u(t)=f(t,u(t)),a.e.t∈[0,T],Δu′(tj)=u′(tj+)-u′(tj-)=Ij(tj),j=1,2,…,p,u(0)=0=u(T),
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.Definition 2.1 (see [13]).
A time scale𝕋 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
(2.1)σ(t):=inf{s∈𝕋:s>t},ρ(t):=sup{s∈𝕋:s<t},μ(t):=σ(t)-t,ν(t):=t-ρ(t).
The point t∈𝕋 is called left dense, left scattered, right dense, or right scattered if ρ(t)=t, ρ(t)<t, σ(t)=t, or σ(t)>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]).
Forf:𝕋→ℝ and t∈𝕋κ, then the delta derivative of f at the point t is defined to be the number fΔ(t) (provided it exists) with the property that for each ϵ>0, there is a neighborhood U of t such that
(2.2)|f(σ(t))-f(s)-fΔ(t)(σ(t)-s)|≤ϵ|σ(t)-s|,∀s∈U.Definition 2.3 (see [13]).
A functionf 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 Crd(𝕋).Definition 2.4 (see [13]).
Assume that𝕋 is an arbitrary time scale. We say that a function p:𝕋→ℝ is regressive provided
(2.3)1+μ(t)p(t)≠0,∀t∈𝕋κ.Definition 2.5 (see [13]).
Assume that𝕋 is an arbitrary time scale and a function p:𝕋→ℝ is regressive, then we define the exponential function on 𝕋 by
(2.4)ep(t,s)=exp(∫stξμ(τ)(p(τ))Δτ),fors,t∈𝕋,
in which
(2.5)ξh(z)=1hLog(1+zh)forh>0,ξh(z)=ξ0(z)=z for h=0, here Log is the principal logarithm function.A functionf:𝕋→ℝ is called rd continuous provided it is continuous at each right-dense point and its left-sided limit exists (finite) at each left-dense point in 𝕋. We write f∈Crd(𝕋)=Crd(𝕋,ℝ). If f is differentiable with fΔ∈Crd(𝕋κ), we write f∈Crd1(𝕋). If fΔ is differentiable on 𝕋κ2=(𝕋κ)κ with fΔ2=(fΔ)Δ∈Crd(𝕋κ2), we write f∈Crd2(𝕋). 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]).
Letf,g:𝕋→ℝ be two functions and a,b∈𝕋. Then one has the following.  (1)
Letf and g be Riemann delta integrable functions on [a,b], and α,β∈ℝ. Then αf+βg is Riemann delta integrable and(2.6)∫ab(αf(s)+βg(s))Δs=α∫abf(s)Δs+β∫abg(s)Δs.(2)
Fundamental Theorem of Calculus. Letf be a continuous function on [a,b]𝕋 such that f is delta differentiable on [a,b). If fΔ is Riemann delta integrable from a to b, then(2.7)∫abfΔ(s)Δs=f(b)-f(a).(3)
Integration by Parts. Letf and g be continuous functions on [a,b] that are differentiable on [a,b). If fΔ and gΔ are Riemann delta integrable from a to b, then(2.8)∫abf(s)gΔ(s)Δs=[f(s)g(s)]|ab-∫abfΔ(s)g(σ(s))Δs.(4)
Iff is Riemann delta integrable on [a,b], then so is |f|, and(2.9)|∫abfΔ(s)Δs|=∫ab|fΔ(s)|Δs.The construction of theΔ-measure on 𝕋 and the following concepts can be found in [35].  (1)
For eacht0∈𝕋∖{max𝕋}, the single-point set t0 is Δ-measurable, and its Δ-measure is given by(2.10)μΔ({t0})=σ(t0)-t0=μ(t0).(2)
Ifa,b∈𝕋 and a≤b, then(2.11)μΔ([a,b))=b-a,μΔ((a,b))=b-σ(a).(3)
Ifa,b∈𝕋∖{max𝕋} and a≤b, then(2.12)μΔ((a,b])=σ(b)-σ(a),μΔ([a,b])=σ(b)-a.The Lebesgue integral associated with the measureμΔ 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 ∫Ef(s)Δs. 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 real-valued 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
(2.13)(R)∫abf(s)Δs=(L)∫[a,b)f(s)Δs,
where R and L indicate the Riemann delta integral and Lebesgue delta integral, respectively.Assume thatp∈ℝ and p>1. Let LLSp[a,b] be the set of(2.14){f:[a,b]→ℝ:∫ab|f(s)|pΔs<+∞}.
By Lemma 3.6 in [36], LLSp[a,b] is a complete linear space with the norm ∥·∥p defined by(2.15)∥f∥p=(∫ab|f(s)|pΔs)1/p.LetCTS[a,b] denote the linear space of all continuous functions f:[a,b]→ℝ with the maximum norm ∥f∥C=maxt∈[a,b]|f(t)|.Lemma 2.8 (Hölder inequality [37]).
Letf,g∈Crd([a,b]),p>1 and q  the conjugate number of p. Then
(2.16)∫ab|f(s)g(s)|Δs≤(∫ab|f(s)|pΔs)1/p·(∫ab|g(s)|qΔs)1/q.At the end of this section, we recall some notation and known results from critical point theory.Let𝕏 be a real normed space and let φ be a functional from 𝕏 to ℝ.  (1)
φ is called weakly continuous if(2.17)xk⇀x⇒φ(xk)→φ(x).(2)
φ is called lower semicontinuous if(2.18)xk→x⇒limk→∞infφ(xk)≥φ(x).(3)
φ is called weakly lower semicontinuous if(2.19)xk⇀x⇒limk→∞infφ(xk)≥φ(x).Let𝕏 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 {φ(xn)} is bounded and φ′(xn)→0 as n→∞ possesses a convergent subsequence in 𝕏.LetBr 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]).
If𝕏 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]).
Let𝕏 be a real Banach space, φ∈C1(𝕏,ℝ) be weakly lower(upper) semicontinuous, and
(2.20)lim∥x∥→∞φ(x)=+∞(lim∥x∥→∞φ(x)=-∞).
Then, there exists x0∈𝕏 such that
(2.21)φ(x0)=infx∈𝕏φ(x)(φ(x0)=supx∈𝕏φ(x)).Lemma 2.11 (Mountain Pass Theorem [40]).
Let𝕏 be a real Hilbert space and let φ∈C1(𝕏,ℝ) satisfy P.S. condition, and φ satisfies (J1)
There are constantsρ,α>0 such that φ(x)≥α for all x∈∂Bρ, where Bρ={x∈𝕏:∥x∥<ρ}.(J2)
φ(0)≤0 and there exists x0∈¯Bρ such that φ(x0)≤0.Then φ possesses a critical value c≥α. Moreover, c can be characterized as
(2.22)c=infg∈Γmaxs∈[0,1]φ(g(s)),
where
(2.23)Γ={g∈C([0,1],𝕏)∣g(0)=0,g(1)=x0}.


## 3. Variational Structure

In this section, we will establish the corresponding variational framework for problem (1.1).Letℙ be the function space of the form(3.1)ℙ={u∣u∈Crd1[0,σ(T)]𝕋κ,u∈CTS[0,σ(T)]𝕋,u(0)=0=u(σ(T))}.ℙ can be equipped with the inner product(3.2)(u,v)ℙ=∫0σ(T)uΔ(t)vΔ(t)Δt,∀u,v∈ℙ,
inducing the norm(3.3)∥u∥ℙ=(u,u)ℙ=(∫0σ(T)|uΔ(t)|2Δt)1/2,∀u∈ℙ.Let{uk}∈ℙ be a Cauchy sequence in ℙ, that is, ∥ukΔ-ulΔ∥2=∥uk-ul∥ℙ→0 as k,l→∞. Since LLS2[0,σ(T)]𝕋 is a complete space, there exists a v∈LLS2[0,σ(T)]𝕋 such that ∥ukΔ-v∥2→0 as k→∞. Define a function u by(3.4)u(t)=∫0tv(s)Δs.
The function v is called a weak derivative of u and is also denoted by uΔ.Remark 3.1.
Ifu has a weak derivative, then this weak derivative is unique in LLS2[0,σ(T)]𝕋, that is, if v1,v2∈LLS2[0,σ(T)]𝕋 are both weak derivatives of u, then ∥v1-v2∥2=0. If u is (delta) differential on [0,σ(T)]𝕋, then its weak derivative is its (delta) derivative.From (3.4), it is clear that u(0)=0 and(3.5)|u(σ(T))|=|∫0σ(T)v(s)Δs|=|∫0σ(T)(v(s)-ukΔ(s))Δs|≤∫0σ(T)|v(s)-ukΔ(s)|Δs≤(∫0σ(T)12Δs)1/2(∫0σ(T)|v(s)-ukΔ(s)|2Δs)1/2=σ(T)∥v-ukΔ∥2→0,ask→∞.
Thus, u(σ(T))=0.Define the spaceℍ as(3.6)ℍ={u(t)=∫0tv(s)Δs∣∃{uk}∈ℙsatisfying∥ukΔ-v∥2→0ask→∞}.
It is clear that ℙ⊂ℍ, and we define the inner product (·,·)ℍ in ℍ as(3.7)(u,v)ℍ=∫0σ(T)uΔ(t)vΔ(t)Δt,∀u,v∈ℍ,
inducing the norm(3.8)∥u∥ℍ=(u,u)ℍ=(∫0σ(T)|uΔ(t)|2Δt)1/2,∀u∈ℍ,
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
(3.9)∫0σ(T)|uσ(t)|2Δt≤σ(T)2∫0σ(T)|uΔ(t)|2Δt,∀u∈ℍ
holds, that is, ∥uσ∥2≤σ(T)∥u∥ℍ.Also, we consider the inner product(3.10)(u,v)=∫0σ(T)eC(t,0)r(t)uσ(t)vσ(t)Δt+∫0σ(T)eC(t,0)uΔ(t)vΔ(t)Δt,∀u,v∈ℍ,
inducing the norm(3.11)∥u∥=(∫0σ(T)eC(t,0)r(t)|uσ(t)|2Δt+∫0σ(T)eC(t,0)|uΔ(t)|2Δt)1/2,∀u∈ℍ.Throughout this paper, it will be assumed that essinft∈[0,σ(T)]𝕋r(t)=r0>-λ0, where λ0 is the smallest eigenvalue of the problem (3.12)uΔΔ(t)+λuσ(t)=0,t∈[0,σ(T)]𝕋κ2,u(0)=0=u(σ(T)).
From [41], λ0 may be defined as(3.13)λ0=minu≠0u(0)=0=u(σ(T))-∫0σ(T)uΔΔ(t)uσ(t)Δt∥uσ∥22.
Applying the integration by parts in Lemma 2.6, we have(3.14)λ0=minu≠0u(0)=0=u(σ(T))∥u∥ℍ2∥uσ∥22>0,
which implies that(3.15)∥uσ∥22≤1λ0∥u∥ℍ2,∀u∈ℍ.For convenience, we introduce the following notations:(3.16)el=mint∈[0,σ(T)]𝕋eC(t,0),eL=maxt∈[0,σ(T)]𝕋eC(t,0),h0=∫0σ(T)|h(t)|Δt.Lemma 3.3.
Suppose that(H1)
r0eL+λ0el>0,forallr0∈(-λ0,0).Then one has
(3.17)ω∥u∥ℍ2≤∥u∥2≤eL(∥r∥∞σ2(T)+1)∥u∥ℍ2,
which implies that the norm ∥·∥ and the norm ∥·∥ℍ are equivalent, where
(3.18)ω={r0eLλ0+el,-λ0<r0<0,el,r0≥0.Proof
Case 1 (r0≥0).
For everyu∈ℍ, we have
(3.19)∥u∥2=∫0σ(T)eC(t,0)r(t)|uσ(t)|2Δt+∫0σ(T)eC(t,0)|uΔ(t)|2Δt≥∫0σ(T)eC(t,0)|uΔ(t)|2Δt≥el∫0σ(T)|uΔ(t)|2Δt=el∥u∥ℍ2.
On the other hand, by Lemma 2.6, one has
(3.20)∥u∥2=∫0σ(T)eC(t,0)r(t)|uσ(t)|2Δt+∫0σ(T)eC(t,0)|uΔ(t)|2Δt≤eL(∫0σ(T)|r(t)||uσ(t)|2Δt+∫0σ(T)|uΔ(t)|2Δt)≤eL(∥r∥∞σ2(T)+1)∫0σ(T)|uΔ(t)|2Δt=eL(∥r∥∞σ2(T)+1)∥u∥ℍ2.
Therefore, the norm ∥·∥ and the norm ∥·∥ℍ are equivalent.
Case 2 (-λ0<r0<0).
For everyu∈ℍ, we have from (3.15) that
(3.21)∥u∥2=∫0σ(T)eC(t,0)r(t)|uσ(t)|2Δt+∫0σ(T)eC(t,0)|uΔ(t)|2Δt≥eLr0∫0σ(T)|uσ(t)|2Δt+el∫0σ(T)|uΔ(t)|2Δt≥r0eLλ0∫0σ(T)|uΔ(t)|2Δt+el∫0σ(T)|uΔ(t)|2Δt=(r0eLλ0+el)∫0σ(T)|uΔ(t)|2Δt=(r0eLλ0+el)∥u∥ℍ2.
Similar to Case 1, it is easy to obtain the desired results. This completes the proof.Together with (3.15) and Lemma 3.3, it is not difficult to obtain the following.Lemma 3.4 (Poincaré inequality).
For allu∈ℍ, one has
(3.22)∫0σ(T)|uσ(t)|2Δt≤1λ0ω(∫0σ(T)eC(t,0)r(t)|uσ(t)|2Δt+∫0σ(T)eC(t,0)|uΔ(t)|2Δt),
that is, ∥uσ∥2≤(1/λ0ω)∥u∥.Lemma 3.5.
∥u∥∞≤a0∥u∥ for all u∈ℍ, where a0=σ(T)/ω.Proof.
For everyu∈ℍ, we have
(3.23)|u(t)|=|u(t)-u(0)|=|∫0tuΔ(s)Δs|≤∫0t|uΔ(s)|Δs≤t1/2(∫0t|uΔ(s)|2Δs)1/2≤t1/2∥u∥ℍ.
Therefore,
(3.24)∥u∥∞≤σ(T)∥u∥ℍ≤σ(T)ω∥u∥.
This completes the proof.Foru∈ℍ⋂Crd2[0,σ(T)]𝕋κ2, we have that u and uΔ are both absolutely continuous. Hence, ΔuΔ(t)=uΔ(t+)-uΔ(t-)=0 for any t∈[0,σ(T)]𝕋κ.Ifu∈ℍ, then u is absolutely continuous and uΔ∈LTS2[0,σ(T)]𝕋. In this case, ΔuΔ(t)=uΔ(t+)-uΔ(t-)=0 may not hold for some t∈[0,σ(T)]𝕋κ. It leads to the impulsive effects.Takev∈ℍ and multiply the two sides of the equality(3.25)uΔΔ(t)+CuΔ(σ(t))-r(t)u(σ(t))+f(σ(t),u(σ(t)))=h(t)
by eC(t,0),vσ(t) and integrate from 0 to σ(T); we have(3.26)∫0σ(T)(eC(t,0)uΔ(t))Δvσ(t)Δt=∫0σ(T)eC(t,0)h(t)vσ(t)Δt-∫0σ(T)eC(t,0)r(t)u(σ(t))vσ(t)Δt-∫0σ(T)eC(t,0)f(σ(t),u(σ(t)))vσ(t)Δt.
Furthermore,(3.27)∫0σ(T)(eC(t,0)uΔ(t))Δvσ(t)Δt=∑j=0j=p∫tjtj+1(eC(t,0)uΔ(t))Δvσ(t)Δt=∑j=0j=p(eC(tj+1-,0)uΔ(tj+1-)v(tj+1-)-eC(tj+,0)uΔ(tj+)v(tj+)-∫tjtj+1eC(t,0)uΔ(t)vΔ(t)Δt)=-∑j=1j=peC(tj,0)ΔuΔ(tj)v(tj)+eC(σ(T),0)uΔ(σ(T))v(σ(T))-eC(0,0)uΔ(0)v(0)-∫0σ(T)eC(t,0)uΔ(t)vΔ(t)Δt=-∑j=1j=peC(tj,0)Ij(u(tj))v(tj)-∫0σ(T)eC(t,0)uΔ(t)vΔ(t)Δt.
Combining (3.26), we have(3.28)∫0σ(T)eC(t,0)uΔ(t)vΔ(t)Δt+∫0σ(T)eC(t,0)r(t)u(σ(t))vσ(t)Δt+∑j=1j=peC(tj,0)Ij(u(tj))v(tj)=∫0σ(T)eC(t,0)f(σ(t),u(σ(t)))vσ(t)Δt-∫0σ(T)eC(t,0)h(t)vσ(t)Δt.
Considering the above, we introduce the following concept solution for problem (1.1).Definition 3.6.
We say that a functionu∈ℍ is a weak solution of problem (1.1) if the identity
(3.29)∫0σ(T)eC(t,0)uΔ(t)vΔ(t)Δt+∫0σ(T)eC(t,0)r(t)u(σ(t))vσ(t)Δt+∑j=1peC(tj,0)Ij(u(tj))v(tj)=∫0σ(T)eC(t,0)f(σ(t),u(σ(t)))vσ(t)Δt-∫0σ(T)eC(t,0)h(t)vσ(t)Δt
holds for any v∈ℍ.Consider the functionalφ:ℍ→ℝ defined by(3.30)φ(u)=12∥u∥2+∫0σ(T)eC(t,0)[h(t)uσ(t)-F(σ(t),uσ(t))]Δt+∑j=1peC(tj,0)∫0u(tj)Ij(s)ds,
where F(t,u)=∫0uf(t,s)ds. Using the continuity of f and Ij, j=1,2,…,p, one has that φ∈C1(ℍ,ℝ). For any v∈ℍ, we have(3.31)φ′(u)v=∫0σ(T)eC(t,0)uΔ(t)vΔ(t)Δt+∫0σ(T)eC(t,0)r(t)uσ(t)vσ(t)Δt-∫0σ(T)eC(t,0)f(σ(t),uσ(t))vσ(t)Δt+∫0σ(T)eC(t,0)h(t)vσ(t)Δt+∑j=1peC(tj,0)Ij(u(tj))v(tj).
Therefore, the solutions of problem (1.1) are the corresponding critical points of φ.Lemma 3.7 (see [33]).
The functionalsψ1,ψ2,ψ3:ℍ→ℝ defined by
(3.32)ψ1(u)=∫0σ(T)F(σ(t),uσ(t))Δt,ψ2(u)=∫0σ(T)h(t)uσ(t)Δt,ψ3(u)=∑j=1p∫0u(tj)Ij(s)ds
are weakly continuous on ℍ.


## 4. Main Results

Theorem 4.1.
Assume that (H1) holds. Suppose further the following.(H2)
There exista,b>0 and θ∈[0,1) such that(4.1)|f(t,u)|≤a+b|u|θ,∀(t,u)∈[0,σ(T)]𝕋×ℝ.(H3)
There existaj,bj>0 and θj∈[0,1)(j=1,2,…,p) such that(4.2)|Ij(u)|≤aj+bj|u|θj,∀u∈ℝ(j=1,2,…,p).
Then problem (1.1) has at least one weak solution.Proof.
According to Lemma3.7, ψ1(u),ψ2(u),ψ3(u) are weakly continuous on ℍ. On the other hand, it is clear that ∥u∥2 is continuous and convex on ℍ. By Lemma 2.9,∥u∥2 is weakly lower semicontinuous. So φ(u) is weakly lower semicontinuous on ℍ.
From  (H2)-(H3), we have(4.3)φ(u)=12∥u∥2+∫0σ(T)eC(t,0)[h(t)uσ(t)-F(σ(t),uσ(t))]Δt+∑j=1peC(tj,0)∫0u(tj)Ij(s)ds≥12∥u∥2-eL∫0σ(T)|h(t)uσ(t)|Δt-eL∫0σ(T)(a|uσ(t)|+b|uσ(t)|θ+1)Δt-eL∑j=1p∫0u(tj)(aj+bj|s|θj)ds≥12∥u∥2-eLh0∥u∥∞-eLσ(T)(a∥u∥∞+b∥u∥∞θ+1)-eL∑j=1p(aj∥u∥∞+bj∥u∥∞θj+1)≥12∥u∥2-eLh0a0∥u∥-eLσ(T)(aa0∥u∥+ba0θ+1∥u∥θ+1)-eL∑j=1p(aja0∥u∥+bja0θj+1∥u∥θj+1)
for all u∈ℍ. This implies that lim∥u∥→∞φ(u)=∞, 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(4.4)uΔΔ(t)+2uΔ(t)-3u(t)+|u(t)|=110esint,a.e.t∈[0,1]𝕋,ΔuΔ(12)=uΔ(12+)-uΔ(12-)=u(12)3,u(0)=0=u(σ(1)).
Then, problem (4.4) has at least one solution.Proof.
The result is easy to obtain from Theorem4.1 and we should omit it. This completes the proof.Theorem 4.3.
Assume that (H1) holds. Suppose further the following.(H4)
There exista,b,c>0 and θ∈[0,1) such that(4.5)|f(t,u)|≤a|u|+b|u|θ+c,∀(t,u)∈[0,σ(T)]𝕋×ℝ.(H5)
There existaj,bj,cj>0 and θj∈[0,1)(j=1,2,…,p) such that(4.6)|Ij(u)|≤aj|u|+bj|u|θj+cj,∀u∈ℝ(j=1,2,…,p).(H6)
1/2-aσ(T)eLa02-∑j=1pajeLa02>0. Then problem (1.1) has at least one weak solution.Proof.
According to Theorem4.1, we know that φ(u) is weakly lower semicontinuous on ℍ.
From  (H4)–(H6), we have(4.7)φ(u)=12∥u∥2+∫0σ(T)eC(t,0)[h(t)uσ(t)-F(σ(t),uσ(t))]Δt+∑j=1peC(tj,0)∫0u(tj)Ij(s)ds≥12∥u∥2-eL∫0σ(T)|h(t)uσ(t)|Δt-eL∫0σ(T)(a|uσ(t)|2+b|uσ(t)|θ+1+c|uσ(t)|)Δt-eL∑j=1p∫0u(tj)(aj|s|+bj|s|θj+cj)ds≥12∥u∥2-eLh0∥u∥∞-eLσ(T)(a∥u∥∞2+b∥u∥∞θ+1+c∥u∥∞)-eL∑j=1p(aj∥u∥∞2+bj∥u∥∞θj+1+cj∥u∥∞)≥12∥u∥2-eLh0a0∥u∥-eLσ(T)(aa02∥u∥2+ba0θ+1∥u∥θ+1+ca0∥u∥)-eL∑j=1p(aja02∥u∥2+bja0θj+1∥u∥θj+1+cja0∥u∥)=(12-aσ(T)eLa02-∑j=1pajeLa02)∥u∥2-eLh0a0∥u∥-eLσ(T)(ba0θ+1∥u∥θ+1+ca0∥u∥)-eL∑j=1p(bja0θj+1∥u∥θj+1+cja0∥u∥)
for all u∈ℍ. This implies that lim∥u∥→∞φ(u)=∞, 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:(4.8)uΔΔ(t)+2uΔ(t)-3u(t)+15σ3(1)ν0u(t)+|u(t)|=sint,a.e.t∈[0,1]𝕋,ΔuΔ(12)=uΔ(12+)-uΔ(12-)=15σ2(1)ν0u(12),u(0)=0=u(σ(1)),
where
(4.9)ν0=exp(∫0σ(1)log(1+3μ(s))1/μ(s)Δs).
Then problem (4.8) has at least one solution.Proof.
By an easy calculation, we obtain(4.10)eL=exp(∫0σ(1)log(1+3μ(s))1/μ(s)Δs),el=1.
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.Lemma 4.5.
Assume that (H1) and (H3) hold. Suppose further the following.(H7)
There exist constantsμ>2 and R>0, such that
(4.11)0<μF(t,u)≤uf(t,u),∀|u|≥R,t∈[0,σ(T)]𝕋,
where F(t,u)=∫0uf(t,s)ds.Then φ satisfies the P.S. condition in ℍ.Proof.
Let{un} be a sequence in ℍ satisfying that {φ(un)} is bounded and φ′(un)→0 as n→+∞. Let Ωn={t∈[0,σ(T)]𝕋∣|un(t)|≥R} for all n∈ℕ. By the continuity of f and (H7), there exists a constant c1>0 such that
(4.12)βφ(un)-(φ′(un),un)=(β2-1)∥un∥2+∫0σ(T)eC(t,0)[unσf(σ(t),unσ)-βF(σ(t),unσ)]Δt+β∑j=1peC(tj,0)∫0un(tj)Ij(s)ds-∑j=1peC(tj,0)Ij(un(tj))un(tj)+∫0σ(T)(β-1)eC(t,0)h(t)unσΔt≥(β2-1)∥un∥2+∫ΩneC(t,0)[unσf(σ(t),unσ)-βF(σ(t),unσ)]Δt-(β+1)eL∑j=1p(aja0∥u∥+bja0θj+1∥u∥θj+1)-(β-1)eLh0a0∥un∥-c1≥(β2-1)∥un∥2-(β+1)eL∑j=1p(aja0∥u∥+bja0θj+1∥u∥θj+1)-(β-1)eLh0a0∥un∥-c1,
which implies that {un} is bounded in ℍ. Taking into account {un}⊂ℍ⊂CTS[0,σ(T)]𝕋 and employing Lemma 3.5, we know that {un} is relatively compact in CTS[0,σ(T)]𝕋. Thus {un} possesses a convergent subsequence {unk} in CTS[0,σ(T)]𝕋, that is, ∥uni-unj∥∞→0 as i,j→+∞. since un∈ℍ, we know that unσ(T)=0 for all n∈ℕ. From the above inequality, it is easy to see that
(4.13)supt∈[0,σ(T)]𝕋κ|uni(t)-unj(t)|→0,asi,j→+∞.
From the continuity of f and Ij, j=1,2,…,p, we have
(4.14)|f(σ(t),uniσ(t))-f(σ(t),unjσ(t))|→0,|Ij(uni(t))-Ij(unj(t))|→0
hold uniformly for t∈[0,σ(T)]𝕋κ as i,j→+∞. Set
(4.15)∥f(σ(t),uniσ(t))-f(σ(t),unjσ(t))∥∞=supt∈[0,σ(T)]𝕋κ|f(σ(t),uniσ(t))-f(σ(t),unjσ(t)),∥Ij(uni(t))-Ij(unj(t))∥∞=supt∈[0,σ(T)]𝕋κ|Ij(uni(t))-Ij(unj(t))|.
Clearly,
(4.16)∥f(σ(t),uniσ(t))-f(σ(t),unjσ(t))∥∞→0,∥Ij(uni(t))-Ij(unj(t))∥∞→0,
as i,j→+∞. We rewrite (3.31) as
(4.17)(φ′(u),v)=(u,v)-∫0σ(T)eC(t,0)f(σ(t),u(σ(t)))vσ(t)Δt+∫0σ(T)eC(t,0)h(t)vσ(t)Δt+∑j=1peC(tj,0)Ij(u(tj))v(tj)
for any u,v∈ℍ. Therefore,
(4.18)(uni-unj,v)=(φ′(uni)-φ′(unj),v)-∑j=1peC(tj,0)[Ij(uni(tj))-Ij(unj(tj))]v(tj)+∫0σ(T)eC(t,0)[f(σ(t),uni(σ(t)))-f(σ(t),unj(σ(t)))]vσ(t)Δt.
Thus, for any v∈ℍ, it follows that
(4.19)|(uni-unj,v)|≤∥φ′(uni)-φ′(unj)∥∥v∥+eL∑j=1p∥Ij(uni(tj))-Ij(unj(tj))∥∞∥v(tj)∥∞+eL∥f(σ(t),uni(σ(t)))-f(σ(t),unj(σ(t)))∥∞∫0σ(T)|vσ(t)|Δt≤∥φ′(uni)-φ′(unj)∥∥v∥+eLa0∑j=1p∥Ij(uni(tj))-Ij(unj(tj))∥∞∥v∥+eLσ(T)∥f(σ(t),uni(σ(t)))-f(σ(t),unj(σ(t)))∥∞∥vσ∥2≤∥φ′(uni)-φ′(unj)∥∥v∥+eLa0∑j=1p∥Ij(uni(tj))-Ij(unj(tj))∥∞∥v∥+eLσ(T)λ0ω∥f(σ(t),uni(σ(t)))-f(σ(t),unj(σ(t)))∥∞∥v∥.
Hence,
(4.20)∥uni-unj∥=sup∥v∥≤1|(uni-unj,v)|≤∥φ′(uni)-φ′(unj)∥+eLa0∑j=1p∥Ij(uni(tj))-Ij(unj(tj))∥∞+eLσ(T)λ0ω∥f(σ(t),uni(σ(t)))-f(σ(t),unj(σ(t)))∥∞.
Observing that φ′(un)→0 as n→+∞, it is easy to see that
(4.21)∥φ′(uni)-φ′(unj)∥→0,asi,j→+∞.
Together with (4.16)–(4.21), we get
(4.22)∥uni-unj∥→0,asi,j→+∞.
In other words, {un} possesses a convergent subsequence {unk} in ℍ. The P.S. condition is now satisfied. This completes the proof.Theorem 4.6.
Assume that (H1), (H3), and (H7) hold. Suppose further the following.(H8)
There exists constantν≥2 such that(4.23)0<νF(t,u)≤uf(t,u),∀u∈[-1,0)⋃(0,1],t∈[0,σ(T)]𝕋,(H9)
a02/2-eLMσ(T)-eLσ(T)∥h∥2-eL∑j=1paj-eL∑j=1pbj>0, where(4.24)M:=sup{F(t,u):t∈[0,σ(T)]𝕋,|u|=1}.
Then problem (1.1) has at least one weak solution.Proof.
It suffices to show thatφ 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
(4.25)ddξ[F(t,ξ-1u)ξν]=ddξ[∫0ξ-1uf(t,s)dsξν]=(νF(t,ξ-1u)-ξ-1uf(t,ξ-1u))ξν-1≤0.
Hence,
(4.26)F(t,u)≤F(t,u|u|)|u|ν,for0<|u|≤1,t∈[0,σ(T)]𝕋,
which implies that
(4.27)∫0σ(T)eC(t,0)|F(σ(t),uσ(t))|Δt≤∫0σ(T)eC(t,0)|F(σ(t),uσ(t)|uσ(t)|)||uσ(t)|νΔt≤eLM∫0σ(T)|uσ(t)|2Δt≤eLMσ2(T)ω∥u∥2.
At the same time, we can obtain that
(4.28)∫0σ(T)eC(t,0)|h(t)uσ(t)|Δt≤eL∥h∥2∥uσ∥2≤eLσ(T)ω∥h∥2∥u∥,∑j=1peC(tj,0)∫0u(tj)|Ij(s)|ds≤eL∑j=1p∫0u(tj)(aj+bj|s|θj)ds≤eL∑j=1p(aj∥u∥∞+bj∥u∥∞θj+1)≤eL∑j=1p(aja0∥u∥+bja0θj+1∥u∥θj+1).
Together (4.27)-(4.28) and the expression of functional φ, it follows that
(4.29)φ(u)≥12∥u∥2-eLMσ(T)a02∥u∥2-eLωa02∥h∥2∥u∥-eL∑j=1p(aja0∥u∥+bja0θj+1∥u∥θj+1)=(12-eLMσ(T)a02)∥u∥2-eL∑j=1pbja0θj+1∥u∥θj+1-(eLωa02∥h∥2+eL∑j=1paja0)∥u∥=(12-eLMσ(T)a02-eLσ(T)a02∥h∥2-eL∑j=1paja02)∥u∥2+(eLσ(T)a02∥h∥2+eL∑j=1paja02)∥u∥2-(eLωa02∥h∥2+eL∑j=1paja0)∥u∥-eL∑j=1pbja0θj+1∥u∥θj+1.
Setting
(4.30)∥u∥=ρ:=1a0=ωσ(T).
Then we have
(4.31)(eLσ(T)a02∥h∥2+eL∑j=1paja02)ρ2-(eLωa02∥h∥2+eL∑j=1paja0)ρ=0,
which implies from (H9) that
(4.32)φ(u)≥(12-eLMσ(T)a02-eLσ(T)a02∥h∥2-eL∑j=1paja02)ρ2-eL∑j=1pbja0θj+1ρθj+1=a022-eLMσ(T)-eLσ(T)∥h∥2-eL∑j=1paj-eL∑j=1pbj=α>0.
By Lemma 3.5, if ∥u∥=ρ, then 0<∥u∥∞≤1. Set Bρ={u∈ℍ∣∥u∥<ρ=1/a0}, then φ(u)≥α>0 for all u∈∂Bρ. Condition (J1) is satisfied.
Secondly, we verify the condition  (J2) of Lemma2.11. In view of assumption  (H7), we have(4.33)μu≤f(t,u)F(t,u),u≥R,(4.34)μu≥f(t,u)F(t,u),u≤-R.
Integrating (4.33) and (4.34) for u on [R,u] and [u,-R], respectively, we get
(4.35)μlnuR≤lnF(t,u)F(t,u),u≥R,μlnR-u≥lnF(t,-R)F(t,u),u≤-R,
that is,
(4.36)F(t,u)≥F(t,R)(uR)μ,u≥R,(4.37)F(t,u)≥F(t,-R)(-uR)μ,u≤-R.
Combining (4.36) and (4.37), one has
(4.38)F(t,u)≥α1|u|μ,|u|≥R,
where
(4.39)α1=R-μmin{mint∈[0,σ(T)]𝕋F(t,R),mint∈[0,σ(T)]𝕋F(t,-R)}>0.
On the other hand, by the continuity of F(t,u), F(t,u) is bounded on [0,σ(T)]𝕋×[-R,R], there exists K>0 such that
(4.40)F(t,u)≥-K≥α1|u|μ-α1Rμ-K,|u|≤R.
Combining (4.38) and (4.40), we have
(4.41)F(t,u)≥α1|u|μ-α2,∀(t,u)∈[0,σ(T)]𝕋×ℝ,
where α2=α1Rμ+K.
According to the Hölder inequality, we get(4.42)∫0σ(T)|uσ(t)|2Δt≤σ(T)1-2/μ(∫0σ(T)|uσ(t)|μΔt)2/μ.
Thus,
(4.43)∫0σ(T)|uσ(t)|μΔt≥σ(T)(2-μ)/2(∫0σ(T)|uσ(t)|2Δt)μ/2=σ(T)(2-μ)/2∥uσ∥2μ.
Together (4.41) and (4.43), we know that
(4.44)∫0σ(T)eC(t,0)F(σ(t),uσ(t))Δt≥∫0σ(T)eC(t,0)α1|uσ(t)|μΔt-∫0σ(T)α2eC(t,0)Δt≥α1elσ(T)(2-μ)/2∥uσ∥2μ-α2eLσ(T)≥α3∥uσ∥2μ-α4,
where α3=α1elσ(T)(2-μ)/2, and α4=α2eLσ(T).
Obviously,φ(0)=0. Let v0∈ℍ satisfy ∥v0∥=1. For any ς>0, it follows from (4.44) and (4.28) that(4.45)φ(ςv0)=ς22∥v0∥2-∫0σ(T)eC(t,0)F(σ(t),ςv0σ(t))Δt+ς∫0σ(T)eC(t,0)h(t)v0σ(t)Δt+∑j=1peC(tj,0)∫0ςv0(tj)Ij(s)ds≤ς22∥v0∥2-α3ςμ∥v0σ∥2μ+ςeLσ(T)ω∥h∥2∥v0∥+eL∑j=1p(ςaja0∥v0∥+bjςθj+1a0θj+1∥v0∥θj+1)+α4.
Since μ>2, we see that φ(ςv0)→-∞ as ς→+∞. we can choose sufficiently large ς0 such that u0=ς0v0∈ℍ satisfying u0∈¯∂B¯ρ and φ(u0)<0. Condition (J2) is satisfied.
According to Lemma2.11, φ possesses a critical value c≥α>0 given by(4.46)c=infg∈Γmaxs∈[0,1]φ(g(s)),
where
(4.47)Γ={g∈C([0,1],𝔼)∣g(0)=0,g(1)=u0}.
Therefore, problem (1.1) has at least one weak solution. This completes the proof.Example 4.7.
Let𝕋=ℝ. Consider the following impulsive Duffing equations with Dirichlet boundary condition
(4.48)u′′(t)+12u′(t)-u(t)+f(t,u(t))=110esint,a.e.t∈[0,1],Δu′(12)=u′(12+)-u′(12-)=110eu(12)5,u(0)=0=u(1),
where
(4.49)f(t,u)={u10et+1,∀(t,u)∈([0,1]×[-1,1]),u310et+1,∀(t,u)∈([0,1]×(-∞,-1)⋃(1,+∞)).
Then problem (4.48) has at least one solution.Proof.
By an easy calculation,eL=e, el=1, ∥h∥2≤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,σ(T)]𝕋, j=1,2,…,p.(H11)
a02/2-eLMσ(T)-eLσ(T)∥h∥2>0, where(4.50)M:=sup{F(t,u):t∈[0,σ(T)]𝕋,|u|=1}.
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)
h(t)≡0 for all t∈[0,σ(T)]𝕋.(H13)
a02/2>eLMσ(T), where M:=sup{F(t,u):t∈[0,σ(T)]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,σ(T)]𝕋lims→0(f(t,s)/s)=0. Then problem (1.1) has at least one weak solution.Proof.
From the proof of Theorem4.6, 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), forϵ=λ0ω/2eL>0, there is a constant δ>0 such that(4.51)|f(t,s)|≤ϵ|s|,∀0≤|s|≤δ,t∈[0,σ(T)]𝕋.
As a result, we have
(4.52)|F(σ(t),uσ(t))|=∫0uσ(t)f(σ(t),s)ds≤∫0|uσ(t)|ϵ|s|ds=ϵ2|uσ(t)|2
which implies from Lemma 3.4 that
(4.53)∫0σ(T)|F(σ(t),uσ(t))|Δt≤ϵ2∥uσ∥22≤ϵ2λ0ω∥u∥2,∀0≤|s|≤δ.
From the above inequality, (H10) and (H12), it follows that
(4.54)φ(u)≥12∥u∥2-ϵeL2λ0ω∥u∥2=14∥u∥2.
Take ρ=δ/a0, α=δ2/4a02. By Lemma 3.5, if ∥u∥=ρ, then 0<∥u∥∞≤δ. Set Bρ={u∈ℍ∣∥u∥<ρ=δ/a0}, then φ(u)≥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.
LetC=0,  r(t)≡0 and let Ij(t)≡0 for all t∈[0,σ(T)]𝕋,  j=1,2,…,p, then Corollary 4.10 reduces to Theorem 4.2 in [33].



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*Source: 152460-2010-09-27.xml*