# Existence of Solution for Impulsive Differential Equations with Nonlinear Derivative Dependence via Variational Methods

Authors: Lizhao Yan; Jian Liu; Zhiguo Luo
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We use variational methods and iterative methods to investigate the solutions of impulsive differential equations with nonlinear derivative dependence. The conditions for the existence of solutions are established. The main results are also demonstrated with examples.


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## Body

## 1. Introduction

Many dynamical systems have an impulsive dynamical behavior due to abrupt changes at certain instants during the evolution process. The mathematical description of these phenomena leads to impulsive differential equations. Recent development in this field has been motivated by many applied problems, such as control theory, population dynamics, and medicine [1–9].We consider the following nonlinear Dirichlet boundary value problems for impulsive differential equations:(1)-(|u′(t)|p-2u′(t))′+g(t)|u(t)|p-2u(t)=f(t,u(t),u′(t)),t≠tj,a.e.t∈[0,T],Δu′(tj)=|u′(tj+)|p-2u′(tj+)-|u′(tj-)|p-2u′(tj-)=Ij(u(tj)),j=1,2,…,l,u(0)=u(T)=0,
where p≥2, 0=t0<t1<t2<⋯<tl<tl+1=T, g∈L∞[0,T], f:[0,T]×R×R→R is continuous, and Ij:R→R, j=1,2,…,l, are continuous.The characteristic of (1) is the presence of the first order derivative in the nonlinearity term. Most of the results concerning the existence of solutions of these equations are obtained using upper and lower solutions methods, coincidence degree theory, and fixed point theorems [10–14]. However, to the best of our knowledge, there are few papers concerned with the existence of solutions for impulsive boundary value problems like problem (1) by using variational methods. Motivated by [15, 16], in this paper we will fill the gap in this area.When there is no derivative in the nonlinearity term, problem (1) has been extensively studied by [17–23], using variational methods. We know, contrary to these equations, (1) is not variational and the well-developed critical point theory is of no avail for, at least, a direct attack to problem (1). The technique used in this paper consists of, associating with problem (1), a family of the following Dirichlet boundary value problems with no dependence on the derivative of the solution. Namely, for each w∈W01,p, we consider the problem
(2)-(|u′(t)|p-2u′(t))′+g(t)|u(t)|p-2u(t)=f(t,u(t),w′(t)),t≠tj,a.e.t∈[0,T],Δu′(tj)=|u′(tj+)|p-2u′(tj+)-|u′(tj-)|p-2u′(tj-)=Ij(u(tj)),j=1,2,…,l,u(0)=u(T)=0.
Now problem (2) is variational and we can treat it by variational methods.In this paper, we need the following conditions.(f1)
f:[0,T]×R×R→R is measurable in t∈[0,T] for every (x,ζ)∈R×R and continuous in (x,ζ)∈R×R for a.e. t∈[0,T].(f2)
f(t,x,ζ)=o(|x|p-1) as x→0 uniformly for t∈[0,T] and ζ∈R and f(t,x,0)≠0 for t∈[0,T] and x∈R.(f3)
There exist constantsC>0 and r∈(p,+∞) such that
(3)&|f(t,x,ζ)|≤C(1+|x|r-1),∀t∈[0,T],x∈R,ζ∈R.(f4)
There exist constantsμ>p and x0>0 such that
(4)0<μF(t,x,ζ)≤xf(t,x,ζ),∀t∈[0,T],|x|≥x0,ζ∈R,whereF(t,x,ζ)=∫0x‍f(t,s,ζ)ds.(f5)
There exist constantsa,b>0 such that
(5)F(t,x,ζ)≥a|x|μ-b,∀t∈[0,T],x∈R,ζ∈R.(f6)
The functionf satisfies the following conditions:
(6)|f(t,x,ζ)-f(t,x′,ζ)|≤L1|x-x′|p-1,∀t∈[0,T],x,x′∈[-ρ1,ρ1],ζ∈R,|f(t,x,ζ)-f(t,x,ζ′)|≤L2|ζ-ζ′|p-1,∀t∈[0,T],x∈[-ρ1,ρ1],ζ,ζ′∈R,whereρ1 is a constant.(I1)
Ij(j=1,2,…,l) are odd and nondecreasing, and there exist constants aj>0, bj>0, and rj∈[0,p-1),j=1,2,…,l such that
(7)|Ij(u)|≤aj+bj|u|rj,forj=1,2,…,l.(I2)
There exists0<θ≤p such that
(8)θ∫0u‍Ij(s)ds≥uIj(u)>0,∀u>0.(I3)
|Ij(u)-Ij(v)|≤αj|u-v|p-1, for all u,v∈[-ρ1,ρ1], j=1,2,…,l.The paper is organized as follows: Section2 is the preliminaries of this paper, Section 3 is devoted to show the solvability of problem (2), and Section 4 will show the solvability of problem (1).


## 2. Preliminaries

Firstly, we recall some facts which will be used in the proof of our main result. It has been shown, for instance, in [16] that the set of all eigenvalues of the following problem
(9)(|u′(t)|p-2u′(t))′+λ|u(t)|p-2u(t)=0,t∈[0,T],u(0)=u(T)=0
is given by the sequence of positive numbers
(10)λk=(p-1)(kπpT)p,k=1,2,…,
where
(11)πp=2∫01‍ds(1-sp)1/p=2πpsin(π/p).
Each eigenvalue λk is simple with the associated eigenfunction
(12)φk(x)=sinp(kπpxT),for0≤x≤T.
We recall the well-known characterization of λ1 as the best constant in the Poincaré inequality
(13)∫0T‍|u′(t)|pdt≥λ1∫0T‍|u(t)|pdt,∀u∈W01,p(0,T).LetW01,p(0,T) be the Sobolev space endowed with the norm
(14)∥u∥=(∫0T‍|u′(t)|pdt+∫0T‍g(t)|u(t)|pdt)1/p.
Throughout the paper, it will be assumed that essinft∈[0,T]g(t)=m>-λ1. We also consider the norm
(15)∥u∥W01,p=(∫0T‍|u′(t)|pdt)1/p,∥u∥p=(∫0T|u(t)|p‍dt)1/p.We need the following Lemmas.Lemma 1 (see [24, Lemma 2.4]).
Ifessinft∈[0,T]g(t)=m>-λ1, then the norms ∥·∥ and ∥·∥W01,p are equivalent.Lemma 2 (see [24, Lemma 2.5]).
There existsC2>0 such that if u∈W01,p, then
(16)∥u∥∞≤(Tq+1)1/q(∫0T‍|u′(s)|pds)1/p,(17)∥u∥∞≤C2∥u∥.Lemma 3 (see [24, Lemma 2.6]).
Letu∈W01,p(0,T); then there exists C1∈(0,1) such that
(18)∥u∥p≤1λ1C1p∥u∥.Foru∈W1,p(0,T), we have that u and u′ are both absolutely continuous. Hence Δu′(t)=u′(t+)-u′(t-)=0 for any t∈[0,T]. If u∈W01,p(0,T), then u is absolutely continuous. In this case, Δu′(t)=u′(t+)-u′(t-)=0 may not hold for some t∈[0,T]. It leads to the impulsive effects. As a consequence, we need to introduce a different concept of solution.Definition 4.
A function(19)u∈{|u′|p-2u∈W01,p(0,T):|u′|p-2u′∈W1,∞(0,T)∖{t1,t2,…,tl}|u′|p-2}
is said to be a classical solution of problem (1) if u satisfies the equation a.e. on [0,T]∖{t1,t2,…,tl} and the limits u′(tj+), u′(tj-) and j=1,2,…,l, exist and satisfy the impulsive condition
(20)Δu′(tj)=|u′(tj+)|p-2u′(tj+)-|u′(tj-)|p-2u′(tj-)=Ij(u(tj))
and the boundary condition u(0)=u(T)=0.We have the following fact. Takev∈W01,p(0,T) and multiply the equation in problem (1) by v and integrate from 0 to T:
(21)-∫0T‍(|u′(t)|p-2u′(t))′v(t)dt+∫0T‍g(t)|u(t)|p-2u(t)v(t)dt=∫0T‍f(t,u(t),u′(t))v(t)dt.
The first term is now
(22)-∫0T‍(|u′(t)|p-2u′(t))′v(t)dt=-∑j=0l‍∫tjtj+1‍(|u′(t)|p-2u′(t))′v(t)dt,∫tjtj+1‍(|u′(t)|p-2u′(t))′v(t)dt=|u′(tj+1-)|p-2u′(tj+1-)v(tj+1)-|u′(tj+)|p-2u′(tj+)v(tj)-∫tjtj+1‍|u′(t)|p-2u′(t)v′(t)dt.
Hence
(23)-∫0T‍(|u′(t)|p-2u′(t))′v(t)dt=∑j=1l‍Δu′(tj)v(tj)-|u′(T)|p-2u′(T)v(T)+|u′(0)|p-2u′(0)v(0)+∫0T‍|u′(t)|p-2u′(t)v′(t)dt=∑j=1l‍Ij(u(tj))v(tj)+∫0T‍|u′(t)|p-2u′(t)v′(t)dt.Definition 5.
We say that a functionu∈W01,p(0,T) is a weak solution of problem (1) if the identity
(24)∫0T‍|u′(t)|p-2u′(t)v′(t)dt+∫0T‍g(t)|u(t)|p-2u(t)v(t)dt+∑j=1l‍Ij(u(tj))v(tj)=∫0T‍f(t,u(t),u′(t))v(t)dt
holds for any v∈W01,p(0,T).Proposition 6.
Under the hypotheses(f1) and (f2), the functional φw:W01,p(0,T)→R defined by
(25)φw(u)=1p∥u∥p-∫0T‍F(t,u(t),w′(t))dt+∑j=1l‍∫0u(tj)‍Ij(s)ds
is continuous and differentiable and
(26)φw′(u)v=∫0T‍|u′(t)|p-2u′(t)v′(t)dt+∫0T‍g(t)|u(t)|p-2u(t)v(t)dt+∑j=1l‍Ij(u(tj))v(tj)-∫0T‍f(t,u(t),w′(t))v(t)dt
for any v∈W01,p(0,T). Moreover, the critical point of φw is a classical solutions of problem (2).Proof.
Using the assumptions(f1), we can obtain the continuity and differentiability of φw and that φw′:W01,p(0,T)→(W01,p(0,T))* is defined by
(27)φw′(u)v=∫0T‍|u′(t)|p-2u′(t)v′(t)dt+∫0T‍g(t)|u(t)|p-2u(t)v(t)dt+∑j=1l‍Ij(u(tj))v(tj)-∫0T‍f(t,u(t),w′(t))v(t)dt
for any v∈W01,p(0,T). It follows that the critical point of φw is the weak solution of (2). Moreover, it is a classical solution of problem (2).
Evidently,u(0)=u(T)=0 since u∈W01,p(0,T). By the definition of weak solution, we have
(28)∫0T‍|u′(t)|p-2u′(t)v′(t)dt+∫0T‍g(t)|u(t)|p-2u(t)v(t)dt+∑j=1l‍Ij(u(tj))v(tj)-∫0T‍f(t,u(t),w′(t))v(t)dt=0.
Choose v∈W01,p(0,T) with v(t)=0 for every t∈[0,tj]⋃‍[tj+1,T]; then
(29)∫tjtj+1‍|u′(t)|p-2u′(t)v′(t)dt+∫tjtj+1‍g(t)|u(t)|p-2u(t)v(t)dt=∫tjtj+1‍f(t,u(t),w′(t))v(t)dt.
This implies that
(30)-(|u′(t)|p-2u′(t))′+g(t)|u(t)|p-2u(t)=f(t,u(t),w′(t)),a.e.t∈(tj,tj+1).
Hence, |u|p-2u′∈W1,∞(tj,tj+1) for every j=1,2,…,l. The impulsive condition in (2) is satisfied. This completes the proof.We will obtain the critical points ofφw by using the Mountain Pass Theorem. Therefore, we state this theorem precisely.Lemma 7 (see [25]).
LetX be a real Banach space and I∈C′(X,R) satisfy (PS)-condition. Suppose that I satisfies the following conditions:
(
i
)
I(0)=0;
(
ii
)
there exists constants ρ,α>0 such that I|∂Bρ(0)≥α;
(
iii
)
there exists e∈X∖B-ρ(0) such that I(e)≤0.
ThenI possesses a critical value c≥α given by
(31)c=infg∈Γmaxs∈[0,1]I(g(s)),
where Bρ(0) is an open ball in X of radius ρ centered at 0 and
(32)Γ={g∈C([0,1],X):g(0)=0,g(1)=e}.


## 3. The Solvability of (2)

Theorem 8.
Suppose that(f1)–(f5) and (I1) hold; then there exist positive constants d1 and d2 such that, for each w∈W01,p(0,T), problem (2) has one solution uw such that d1≤∥uw∥≤d2.Proof.
(I) We show thatφw satisfies the (PS)-condition.
Assume that{un}n∈N⊂W01,p(0,T) is a sequence such that {φw(un)}n∈N is bounded and φw′(un)→0 as n→+∞. We will prove that the sequence {un}n∈N is bounded. Obviously, there exists a constant C3>0 such that
(33)|φw(un)|≤C3,|φw′(un)|≤C3forn∈N.
We set
(34)m0=max{a1,a2,…,al},M0=max{b1,b2,…,bl}.
From (17), (25), (26), and (33), (f4), and (I1), we have
(35)μφw(un)-φw′(un)un=(μp-1)∥un∥p-∫0T‍(μF(t,un,w′)-f(t,un,w′)un)dt+∑j=1l‍[μ∫0un(tj)‍Ij(s)ds-Ij(un(tj))un(tj)]≥(μp-1)∥un∥p-∑j=1l‍[(aj+bj|un(tj)|rj)un(tj)]-∑j=1l‍(aj+bj|s|rj)ds-C4≥(μp-1)∥un∥p-(μ+1)(C2lm0∥un∥+M0∑j=1l‍C2rj+1∥un∥rj+1)-C4.
By p>rj+1, we obtain that {un} is bounded in W01,p(0,T).
SinceW01,p(0,T) is a reflexive Banach space, passing to a subsequence if necessary, we can assume that
(36)un⇀uinW01,p(0,T),un⟶uinLp(0,T),un⟶uuniformlyinC[0,T].
Hence
(37)∫0T‍(f(t,un,w′)-f(t,u,w′)un)(un(t)-u(t))dt⟶0,(38)∑j=1l‍[Ij(un(tj))-Ij(u(tj))](un(tj)-u(tj))⟶0.
Notice that
(39)〈φw′(un)-φw′(u),un-u〉=∫0T‍(|un′(t)|p-2un′(t)-|u′(t)|p-2u′(t))×(un′(t)-u′(t))dt+∫0T‍(g(t)|un(t)|p-2un(t)-|u(t)|p-2u(t))×(un(t)-u(t))dt+∑j=1l‍(Ij(un(tj))-Ij(u(tj))(un(tj)-u(tj)))+∫0T‍(f(t,un,w′)-f(t,u,w′))(un(t)-u(t))dt.
From [26, Lemma 4.2], we see that there exists Cp>0, for any x,y∈R:
(40)(|x|p-2x-|y|p-2y)(x-y)≥Cp|x-y|p,p≥2.
Combining this inequality with (39), we have
(41)Cp(∫0T‍(|un′(t)-u′(t)|p+g(t)|un(t)-u(t)|p)dt)≤∥φw′(un)-φw′(u)∥∥un-u∥+∑j=1l(Ij(un(tj))-Ij(u(tj))(un(tj)-u(tj)))‍+∫0T(f(t,un,w′)-f(t,u,w′))(un(t)-u(t))dt‍.
It follows from (37)–(39) that un→u in W01,p(0,T). Hence, φw satisfies (PS)-condition.
(II) We verify assumption (ii) of Lemma7.
By (f2), there exists δ∈(0,1) such that
(42)|f(t,x,ζ)|≤λ1C12|x|p-1fort∈[0,T],|x|≤δ,ζ∈R.
Since F(t,0,ζ)=0, it follows that
(43)|F(t,x,ζ)|≤λ1C12p|x|pfort∈[0,T],|x|≤δ,ζ∈R.
Using (I1), we have
(44)∑j=1l‍∫0u(tj)‍Ij(t)dt≥0.
Hence, from (18), (25), (35), (f2), and (I1), for ∥u∥≤δ/C2, we have
(45)φw(u)=1p∥u∥p-∫0T‍F(t,u(t),w′(t))dt+∑j=1l‍∫0u(tj)‍Ij(t)dt≥1p∥u∥p-λ1C12p∫0T‍|u(t)|pdt≥1p∥u∥p-12p∥u∥p=12p∥u∥p.
Set α=(1/2p)(δ/C2)p, ρ=δ/C2. Equation (45) shows that ∥u∥=ρ implies that φw(u)≥α; that is, φw satisfies assumption (ii) of Lemma 7.
(III) We verify assumption (iii) of Lemma7.
By(I1) and (f5), we know that for s>1(46)φw(su)=1p∥su∥p-∫0T‍F(t,su(t),w′(t))dt+∑j=1l‍∫0su(tj)‍Ij(t)dt≤1p∥u∥psp+lC2∥u∥∑j=1l‍(aj+bj|s|rj)C2rj∥u∥rj-a|s|μ∫0T‍|u(t)|μdt+bT.
Take v0∈W01,p(0,T) such that ∥v0∥=1. Since μ>p, a>0, and 0≤rj<p-1, (46) implies that there exists ξ1>1 such that ∥e∥>ρ, and φw(e)<0 if we set e=ξ1v0. By Lemma 7, φw possesses a critical value c′≥α>0 given by
(47)c′=infg∈Γmaxs∈[0,1]φw(g(s)),
where Γ={g∈C([0,1],W01,p(0,T)):g(0)=0,g(1)=e}. Obviously, φw(0)=0, so according to Lemma 7, there exists uw≠0 and uw∈X, such that
(48)φw(uw)=c′,φw(uw)=0.
(IV) We prove thatd1≤∥uw∥≤d2.
Sinceuw is the solution of problem (2), we have
(49)∫0T‍|uw′|pdt+∫0T‍g(t)|uw|pdt+∑j=1l‍Ij(uw(tj))uw(tj)=∫0T‍f(t,uw,w′)uwdt.
So
(50)∥uw∥p+∑j=1l‍Ij(uw(tj))uw(tj)=∫0T‍f(t,uw,w′)uwdt.
It follows from (f2) and (f3) that, given ϵ>0, there exists a positive constant Cϵ, independent of w, such that
(51)|f(t,x,ζ)|≤ϵ|x|p-1+Cϵ|x|r-1.
Hence, using (I1), we have
(52)∥uw∥p≤∥uw∥p+∑j=1l‍Ij(uw(tj))uw(tj)=∫0T‍f(t,uw,w′)uwdt≤ϵ∫0T‍|uw|pdt+Cϵ∫0T‍|uw|rdt.
By Sobolev embedding theorem, we obtain
(53)∥uw∥p≤ϵ·1λ1C1∥uw∥p+Cϵ·1(λ1C1)r/p∥uw∥r.
Set ϵ/λ1C1=ϵ′ and Cϵ/(λ1C1)r/p=Cϵ′; then
(54)(1-ϵ′)∥uw∥p≤Cϵ′∥uw∥r,
which implies
(55)∥uw∥≥d1.
From the inf max characterization of uw in (III), we obtain
(56)φw(uw)≤maxφw(sv0)
with v0 chosen in (III). We estimate φw(sv0) using (f5) and (I1):
(57)φw(sv0)≤1psp-a|s|μ∫0T‍|v0|μdt+bT-ajTφw(sv0)≤-bj|s|rj∫0T‍|v0|rjdt=:h(s),
whose maximum is achieved at some s0>0 and the value h(s0) can be taken as d2. Clearly it is independent of w. Which implies
(58)∥uw∥≤d2.
This completes the proof.Lemma 9.
Suppose that(f1)–(f3) hold and (f4), (f5), and (I1) hold only for positive x; then there exist positive constants d3, and d4 such that, for each w∈W01,p(0,T), problem (2) has a positive solution uw such that d3≤∥uw∥≤d4.Proof.
Set(59)f¯(t,u,ζ)={f(t,u,ζ),ifu≥0,0,ifu<0,I¯j(u)={Ij(u),ifu≥0,0,ifu<0.
Consider the function
(60)-(|u′(t)|p-2u′(t))′+g(t)|u(t)|p-2u(t)=f¯(t,u(t),u′(t)),t≠tj,a.e.t∈[0,T],Δu′(tj)=I¯j(u(tj)),j=1,2,…,l,u(0)=u(T)=0.
Obviously, f¯ satisfies (f1)–(f5), and I¯j satisfies (I1), so using Theorem 8, we obtained a solution uw of (60). Multiplying the equation by uw- and integrating by parks, we conclude that uw-≡0. So uw is positive.Theorem 10.
Suppose that(f1)–(f5) and (I2) hold; then there exist positive constants d5 and d6 such that, for each w∈W01,p(0,T), problem (2) has one solution uw such that d5≤∥uw∥≤d6.Proof.
(I) We show thatφw satisfies the (PS)-condition.
Assume that{un}n∈N⊂W01,p(0,T) is a sequence such that {φw(un)}n∈N is bounded and φw′(un)→0 as n→+∞. We will prove that the sequence {un}n∈N is bounded.
It follows from(f3), (f4), and (I2) that
(61)μφw(un)-φw′(un)un=(μp-1)∥un∥p-∫0T‍(μF(t,un,w′)-f(t,un,w′)un)dt+∑j=1l‍[μ∫0un(tj)‍Ij(s)ds-Ij(un(tj))un(tj)]≥(μp-1)∥un∥p-∫|un|≥x0‍(μF(t,un,w′)-f(t,un,w′)un)dt-∫|un|≤x0‍(μF(t,un,w′)-f(t,un,w′)un)dt≥(μp-1)∥un∥p-C6.
Hence, {un} is bounded in W01,p(0,T). The following proof of (PS)-condition is similar to that in (I) of Theorem 8.
(II) We verify assumption (ii) of Lemma7.
Using(I2), we have
(62)∑j=1l‍∫0u(tj)‍Ij(t)dt≥0.
The proof is similar to that in (II) of Theorem 8.
(III) We verify assumption (iii) of Lemma7.
Takev0∈W01,p(0,T) such that ∥v0∥=1. By (I2) and (f5), using Sobolev embedding theorem, we have
(63)φw(sv0)=1psp-∫0T‍F(t,sv0(t),w′(t))+∑j=1l‍∫0sv0(tj)‍Ij(t)dt≤1psp+ls∥u∥∞θ-C7|s|μ∫0T‍|v0(t)|μdt-C8T≤1psp+C9sθ-C10|s|μ-C8T.
Since μ>p≥θ, (63) implies that there exists ξ2∈R∖{0} such that ∥e∥>ρ and φw(e)<0 if we set e=ξ2v0. By Lemma 7, φw possess a critical value c′′≥α>0, given by
(64)c′′=infg∈Γmaxs∈[0,1]φw(g(s)),
where Γ={g∈C([0,1],W01,p(0,T)):g(0)=0,g(1)=e}. Obviously, φw(0)=0, so according to Lemma 7, there exists uw≠0 and uw∈X, such that
(65)φw(uw)=c′′,φw(uw)=0.
(IV) We prove thatd5≤∥uw∥≤d6.
Like (IV) of Theorem8, we can obtain that there exist d5>0 such that
(66)∥uw∥≥d5.
It follows from uw satisfying (65) that
(67)(μp-1)∥uw∥p≤μφw(uw)-φw′(uw)uw+C11=μφw(uw)+C11=μinfg∈Γmaxs∈[0,1]φw(g(s))+C11≤μmaxs≥0φw(sv0)+C11≤μ(1p|s|p+∑j=1l‍∫0sv0(tj)‍Ij(t)dt-a|s|μ∫0T‍|v0|μdt-bT∑j=1l‍∫0sv0(tj)‍)+C11≤μ(∫0T‍|v0|μdt1p|s|p+|s|θl∥v0∥∞θ-a|s|μ∫0T‍|v0|μdt+bT)+C11.
Let
(68)r(t)=1p|t|p+tθl∥v0∥∞θ-atμ∫0T‍|v0|μdt,t≥0.
Since μ>p≥θ, then r(t) can achieve its maximum at some t, which implies there exist d6>0 such that
(69)∥uw∥≤d6.
This completes the proof.Lemma 11.
Suppose that(f1)–(f3) hold and (f4), (f5), and (I2) hold only for positive x, then there exist positive constants d7, and d8 such that, for each w∈W01,p(0,T), problem (2) has a positive solution uw such that d7≤∥uw∥≤d8.Proof.
The proof is similar to that of Lemma9, so we omit it.Remark 12.
By the same method, we can discuss the negative solution of (2).


## 4. The Solvability of (1)

Theorem 13.
Assume that(f1)–(f6) and (I1) and (I3) hold; then problem (1) has one nontrivial solution provided
(70)k=L2λ1-1/pC1-1-(1/p)Cp-L1λ1-1C1-1-C2p∑j=1lαj
satisfying 0<k<1.Proof.
We construct a sequence{un}∈W01,p(0,T) as solutions of the following problem:
(1.1n)-(|un′(t)|p-2un′(t))′+g(t)|un(t)|p-2un(t)=f(t,un(t),un-1′(t)),t≠tj,a.e.t∈[0,T],Δun′(tj)=Ij(un(tj)),j=1,2,…,l,un(0)=un(T)=0
obtained in Theorem 8, starting with an arbitrary u0∈W01,p(0,T). It follows from the Sobolev embedding theorem that ∥un∥∞≤ρ1. Using (1.1n) and (1.1n+1), we obtain
(71)∫0T‍|un′(t)|p-2un′(t)(un+1′-un′)dt+∫0T‍g(t)|un(t)|p-2un(t)(un+1-un)dt+∑j=1l‍Ij(un(tj))(un+1(tj)-un(tj))=∫0T‍f(t,un,un-1′)(un+1-un)dt,∫0T‍|un+1′(t)|p-2un+1′(t)(un+1′-un′)dt+∫0T‍g(t)|un+1(t)|p-2un+1(t)(un+1-un)dt+∑j=1l‍Ij(un+1(tj))(un+1(tj)-un(tj))=∫0T‍f(t,un+1,un′)(un+1-un)dt.
Hence,
(72)Cp(∫0T‍(|un+1′-un′|p+g(t)|un+1-un|p)dt)≤∫0T‍[f(t,un+1,un′)-f(t,un,un-1′)](un+1-un)dt+∑j=1l‍[Ij(un(tj))-Ij(un+1(tj))]×(un+1(tj)-un(tj))=∫0T‍[f(t,un+1,un′)-f(t,un,un′)]×(un+1-un)dt+∫0T‍[f(t,un,un′)-f(t,un,un-1′)](un+1-un)dt+∑j=1l‍[Ij(un(tj))-Ij(un+1(tj))]×(un+1(tj)-un(tj))≤L1∫0T‍|un+1-un|pdt+L2∫0T‍|un′-un-1′|p-1×|un+1-un|dt+∑j=1l‍αj|un+1(tj)-un(tj)|p.
Using (17) and (18) and by Hölder inequality, we have
(73)∥un+1-un∥p-1≤L2λ1-1/pC1-1/pCp-L1λ1-1C1-1-C2p∑j=1lαj[∫0T‍|un′-un-1′|pdt](p-1)/p=L2λ1-1/pC1-1/pCp-L1λ1-1C1-1-C2p∑j=1lαj∥un-un-1∥W01,pp-1≤L2λ1-1/pC1-1-(1/p)Cp-L1λ1-1C1-1-C2p∑j=1lαj∥un-un-1∥p-1;
that is,
(74)∥un+1-un∥≤k1/(p-1)∥un-un-1∥.
Since k is less than 1, then it follows that {un} strongly converges in W01,p(0,T), as it easily follows proving that {un} is a Cauchy sequence in W01,p(0,T). By Theorem 8, we know that u≠0. In this way we obtain a nontrivial solution of (1).Example 14.
Lett1∈(0,T) and p=2; consider the following nonlinear Dirichlet impulsive problem:
(75)-u′′(t)+(-1-t)u(t)=u3(t)(1+sinu′(t)),a.e.t∈[0,T],t≠t1,Δu′(t1)=u(t1)3,u(0)=u(T)=0.
Compared with (1), f(t,u,u′)=u3(1+sinu′) and I1(u)=u1/3. We can take μ=3 and θ=5/3. Then by simple computation, it is easy to verify that all conditions of Theorem 13 are satisfied. Hence, by Theorem 13, problem (75) has one nontrivial solution.Theorem 15.
Assume that(f1)–(f6) and (I2) and (I3) hold; then problem (1) has one nontrivial solution provided
(76)k=L2λ1-1/pC1-1-(1/p)Cp-L1λ1-1C1-1-C2p∑j=1lαj
satisfying 0<k<1.Proof.
We construct a sequence{un}∈W01,p(0,T) as solutions of the problem (1.1n) obtained in Theorem 10, starting with an arbitrary u0∈W01,p(0,T). The rest of proof is similar to that of Theorem 13, so we omit it.Example 16.
Lett1∈(0,T) and p=4; consider the following nonlinear Dirichlet impulsive problem:
(77)-(|u′(t)|2u′(t))′-t2|u(t)|2u(t)=u7(t)(2+cosu′(t)),a.e.t∈[0,T],t≠t1,Δu′(t1)=u2(t1),u(0)=u(T)=0.
Compared with (1), f(t,u,u′)=u7(2+cosu′) and I1(u)=u2. We can take μ=6 and r1=2. Then by simple computation, it is easy to verify that all conditions of Theorem 15 are satisfied. Hence, by Theorem 15, problem (77) has one nontrivial solution.Remark 17.
In order to discuss the positive solutions of (1), we can argue as in Theorem 8, by replacing f by f-. By the same method, we can discuss the negative solutions of (1).



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*Source: 908062-2013-10-02.xml*