# Weak Solutions for ap-Laplacian Antiperiodic Boundary  Value Problem with Impulsive Effects

Authors: Keyu Zhang; Jiafa Xu; Wei Dong
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

By virtue of variational method and critical point theory, we will investigate the existence of weak solutions for ap-Laplacian impulsive differential equation with antiperiodic boundary conditions.


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## Body

## 1. Introduction

In this work, we will study the existence of weak solutions for thep-Laplacian impulsive differential equation with antiperiodic boundary conditions
(1)-(|u′|p-2u′)′=f(x,u),inΩ,Δ|u′(xk)|p-2u′(xk)=Ik(u(xk)),k=1,2,…,m,u(0)=-u(T),u′(0)=-u′(T),
where p>1, T>0, 0=x0<x1<x2<⋯<xm<xm+1=T, Ω:=[0,T]∖{x1,…,xm}, f∈C([0,T]×ℝ,ℝ), and Ik∈C(ℝ,ℝ),  k=1,2,…,m. Moreover, Δ|u′(xk)|p-2u′(xk)=|u′(xk+)|p-2u′(xk+)-|u′(xk-)|p-2u′(xk-), where u′(xk+) and u′(xk-) denote the right and left limits, respectively, of u′(x) at x=xk, for k=1,2,…,m.Impulsive differential equations have become more important in recent years in some mathematical models of real processes and phenomena studied in physics, chemical technology, population dynamics, biotechnology, and economics. We mention, for instance, the books [1–3] dealing with impulsive differential equations.Recently, many researchers pay their attention to impulsive differential equations by variational method and critical point theory, to the best our knowledge; we refer the reader to [4–8] and references cited therein. Meanwhile, some people begin to study p-Laplacian differential equations with impulsive effects; for example, see [9–14].Chen and Tang [9] adopted the least action principle and the saddle point theorem to obtain some existence theorems for a second-order p-Laplacian system with or without impulsive effects under weak sublinear growth conditions. In [10], they also considered that a second-order impulsive differential equation with Dirichlet problems has one or infinitely many solutions under more relaxed assumptions on their nonlinearity f, which satisfies a kind of new superquadratic and subquadratic conditions.The problem of finding infinitely many large energy solutions is a very classical problem; there is an extensive literature concerning the existence of infinitely many large energy solutions of a plethora of problems via the symmetric mountain pass theorem and fountain theorem; for instance, see [15–22].Motivated by the works cited above, in this paper, we shall discuss the problem (1). Firstly, we adopt Browder theorem, introduced in [23], to prove that the problem (1) has only one weak solution. Secondly, we shall utilize Fountain theorem under Cerami condition (C), which is introduced in [24],  to prove that the problem (1) has infinitely many weak solutions. The results obtained here improve some existing results in the literature.


## 2. Variational Structure

We first introduce the Banach spaceX:={u∈W1,p([0,T]):u(0)=-u(T)}, endowed with the norm
(2)∥u∥=(∫0T‍|u′(x)|pdx)1/p,foru∈X.
Then we see that (X,∥·∥) is a reflexive Banach space. We denote by ∥u∥p the norm in Lp([0,T]) and ∥u∥∞ in C[0,T].In what follows, we shall convert the problem (1) into an integral equation. For all v∈X, multiply (1) by v and integrate between 0 and T to obtain
(3)∫0T‍-(|u′(x)|p-2u′(x))′v(x)dx=∫0T‍f(x,u(x))v(x)dx.
Noting the impulsive effects, for the left integral of (3), we have
(4)∫0T‍-(|u′(x)|p-2u′(x))′v(x)dx=∑j=0m‍∫xj+xj+1-‍-(|u′(x)|p-2u′(x))′v(x)dx=-∑j=0m‍|u′(x)|p-2u′(x)v(x)|xj+xj+1-+∑j=0m‍∫xj+xj+1-‍|u′(x)|p-2u′(x)v′(x)dx=∑j=1m‍[|u′(xj+)|p-2u′(xj+)-|u′(xj-)|p-2u′(xj-)]v(xj)-|u′(T)|p-2u′(T)v(T)+|u′(0)|p-2u′(0)v(0)+∫0T‍|u′(x)|p-2u′(x)v′(x)dx=∑j=1m‍Ij(u(xj))v(xj)+∫0T‍|u′(x)|p-2u′(x)v′(x)dx.
Substituting (4) into (3), we easily find that
(5)∫0T‍|u′(x)|p-2u′(x)v′(x)dx+∑j=1m‍Ij(u(xj))v(xj)-∫0T‍f(x,u(x))v(x)dx=0.Definition 1.
If for eachv∈X there is a u∈X such that (5) holds, then u∈X is called a weak solution for (1).
From (5), we can obtain the weak solutions for (1) that coincide with critical points of the energy functional
(6)J(u)=1p∫0T‍|u′(x)|pdx+∑j=1m‍∫0u(xj)‍Ij(x)dx-∫0T‍F(x,u(x))dx,
where F(x,u)=∫0u‍f(x,t)dt. Clearly, J is class of C1 and its derivative is
(7)(J′(u),v)=∫0T‍|u′(x)|p-2u′(x)v′(x)dx+∑j=1m‍Ij(u(xj))v(xj)-∫0T‍f(x,u(x))v(x)dx,∀u,v∈X.Lemma 2.
ConsiderX↪↪C[0,T] and X↪↪Lp([0,T]).Proof.
Foru∈X, by mean value theorem, u(x)=u(0)+∫0x‍u′(s)ds and u(x)=u(T)-∫xT‍u′(s)ds. It follows from Hölder inequality that
(8)u(x)=12[u(0)+∫0x‍u′(s)ds+u(T)-∫xT‍u′(s)ds]≤12∫0T‍|u′(s)|ds≤12T(p-1)/p∥u∥.
Consequently, we have
(9)∥u∥∞≤12T(p-1)/p∥u∥,foru∈X.
Furthermore, for u∈X, we have t,s∈[0,T],
(10)|u(t)-u(s)|=|∫st‍u′(τ)dτ|≤∫0T‍|u′(τ)|dτ≤T(p-1)/p∥u∥.
Hence every bounded set in X is equicontinuous in C[0,T]. By Arzela-Ascoli theorem, we claim that X↪↪C[0,T]. By the same manner, we can prove that X↪↪Lp([0,T]). Furthermore, we can obtain by (9)
(11)∥u∥p≤T2∥u∥,foru∈X.
This completes the proof.Definition 3 (see [23, page 303]).
LetX be a reflexive real Banach space and X* its dual. The operator ℒ:X→X* is said to be demicontinuous if ℒ maps strongly convergent sequences in X to weakly convergent sequences in X*.Lemma 4 (Browder theorem, see [23, Theorem 5.3.22]).
LetX be a reflexive real Banach space. Moreover, let ℒ:X→X* be an operator satisfying the following conditions:(i)
ℒ is bounded and demicontinuous;(ii)
ℒ is coercive, that is, lim∥x∥→∞(ℒ(x),x)/∥x∥=+∞;(iii)
ℒ is monotone on the space X; that is, for all x,y∈X, one has
(12)(ℒ(x)-ℒ(y),x-y)≥0.
Then the equationℒ(x)=f* has at least one solution x∈X for every f*∈X*. If, moreover, the inequality (12) is strict for all x,y∈X, x≠y, then the equation ℒ(x)=f* has precisely one solution x∈X for all f*∈X*.In what follows, we shall introduce Fountain theorem under Cerami condition (C). We first give the definition of Cerami condition (C).Definition 5 (see [24, Definition 1.1]).
Assume thatX is a Banach space with norm ∥·∥; we say that J∈C1(X,ℝ) satisfies Cerami condition (C), if for all d∈ℝ,(i)
any bounded sequence{un}⊂X satisfying J(un)→d, J′(un)→0 possesses a convergent subsequence;(ii)
there existδ,ξ,ρ>0 such that for any u∈J-1([d-δ,d+δ]) with ∥u∥≥ξ, ∥J′(u)∥·∥u∥≥ρ.
AsX is a reflexive Banach space, there exist (see [25, Section 17]) {en}n=1∞⊂X and {fn}n=1∞⊂X* such that fn(em)=δn,m, X=span¯{en:n=1,2,…} and X*=span¯W*{fn:n=1,2,…}. For j,k∈ℕ, denote Xj:=span{ej}, Yk:=⊕j=1kXj, and Zk:=⊕j=k∞Xj¯. Clearly, X=⊕j∈ℕXj¯ with dimXj<∞ for all j∈ℕ. Denote Sρ:={u∈X:∥u∥=ρ}.Lemma 6 (see [24, Proposition 1.2]).
LetX, Yk, Zk be defined as above. Assume that J∈C1(X,ℝ) satisfies condition (C), and J(-u)=J(u). For each k∈ℕ, there exist ρk>rk>0 such that(i)
bk:=infu∈Zk∩SrkJ(u)→+∞, k→∞,(ii)
ak:=maxu∈Yk∩SρkJ(u)≤0.
ThenJ has a sequence of critical points un, such that J(un)→+∞ as n→∞.


## 3. Main Results

Now, we list our assumptions onf and Ij, j=1,2,…,m.(H1)
f(x,u) is a decreased function about u, uniformly in x∈[0,T], and Ij(u)(j=1,2,…,m) are increased functions with u.(H2)
There existaj,bj>0 and γj∈[1,p) such that |Ij(u)|≤aj+bj|u|γj-1, for all u∈ℝ and j=1,2,…,m.(H3)
There existc1,c2>0 such that f(x,u)≤c1+c2|u|p-1, for all u∈ℝ,  x∈[0,T].(H4)
There is a positive constanta>0 such that lim|u|→∞(-pF(x,u)+f(x,u)u)/|u|≥a, uniformly in x∈[0,T].(H5)
p∫0u‍Ij(s)ds-Ij(u)u≥0,  ∫0u‍Ij(s)ds≥0,  for all u∈ℝ, j=1,2,…,m.(H6)
lim|u|→∞F(x,u)/|u|p=+∞, uniformly on x∈[0,T].(H7)
F(x,u) is an even function about u and Ij(u)(j=1,2,…,m) are odd functions about u, for all x∈[0,T].Theorem 7.
Let0<c2<(T/2)-p and (H1)–(H3) hold. Then (1) has precisely a weak solution.Proof.
Foru,v∈X, we define
(13)(L1(u),v):=∫0T‍|u′(x)|p-2u′(x)v′(x)dx,(L2(u),v):=∑j=1m‍Ij(u(xj))v(xj),(L3(u),v):=∫0T‍f(x,u(x))v(x)dx.
Let
(14)(ℒ(u),v)=(L1(u),v)+(L2(u),v)-(L3(u),v),∀u,v∈X.
Then, to find a weak solution of (1) is equivalent to finding a solution for the operator equation ℒ(u)=0. In what follows, we shall sketch the properties of operators Li(i=1,2,3). By Hölder inequality, we have
(15)∫0T‍|u′(x)|p-2u′(x)v′(x)dx≤(∫0T‍||u′(x)|p-2u′(x)|p/(p-1)dx)(p-1)/p×(∫0T‍|v′(x)|pdx)1/p=(∫0T‍|u′(x)|pdx)(p-1)/p(∫0T‍|v′(x)|pdx)1/p=∥u∥p-1∥v∥<∞,∀u,v∈X.
Consequently, L1 is bounded:
(16)|(L1(un)-L1(u),v)|=|∫0T‍[|un′(x)|p-2un′(x)-|u′(x)|p-2u′(x)]v′(x)dx|≤(∫0T‍||un′(x)|p-2un′(x)-|u′(x)|p-2u′(x)|p/(p-1)dx∫0T‍)(p-1)/p∥v∥.
The last integral tends to zero as ∥un-u∥→0 due to the continuity of the Nemytskii operator φ(s)=|s|p-2s. Hence, L1 is continuous. We assume that un⇀u in X; by Lemma 2, we know that there is a subsequence, still denoted by {un},  which strongly converges to u in C[0,T]. Therefore, utilizing the continuity of f, we have
(17)|(L3(un)-L3(u),v)|=|∫0T‍[f(x,un(x))-f(x,u(x))]v(x)dx|⟶0asn⟶∞.
So, L3 is continuous. In the same methods, we see that L2 is also continuous. On the other hand, by (9) and (11), for u,v∈X, it follows from (H2) and (H3) that
(18)|(L2(u),v)|≤∑j=1m‍|Ij(u(xj))||v(xj)|≤∑j=1m‍(aj+bj|u(xj)|γj-1)|v(xj)|≤∑j=1m‍(aj+bj∥u∥∞γj-1)∥v∥∞≤m2T(p-1)/p(aj+bj(12T(p-1)/p)γj-1∥u∥γj-1)∥v∥<∞,|(L3(u),v)|=|∫0T‍f(x,u(x))v(x)dx|≤∫0T‍|f(x,u(x))||v(x)|dx≤∫0T‍(c1+c2|u(x)|p-1)|v(x)|dx≤c1T(p-1)/p∥v∥p+c2∥u∥pp-1∥v∥p≤c12T(2p-1)/p∥v∥+c2(T2)p∥u∥p-1∥v∥<∞.
Therefore, L2 and L3 are bounded. Up to now, we have proved that ℒ is bounded and continuous, so (i) of Lemma 4 holds. Finally, we shall show the monotonicity and coercivity of ℒ. Indeed, (H1) implies that for u,v∈X and u≠v,
(19)(ℒ(u)-ℒ(v),u-v)=∫0T‍[|u′(x)|p-2u′(x)-|v′(x)|p-2v′(x)]×(u′(x)-v′(x))dx+∑j=1m‍(Ij(u(xj))-Ij(v(xj)))(u(xj)-v(xj))-∫0T‍(f(x,u(x))-f(x,v(x)))(u(x)-v(x))dx≥∫0T‍[|u′(x)|p-2u′(x)-|v′(x)|p-2v′(x)]×(u′(x)-v′(x))dx≥∥u∥p-∥u∥p-1∥v∥-∥v∥p-1∥u∥+∥v∥p≥(∥u∥p-1-∥v∥p-1)(∥u∥-∥v∥)>0.
(H2) and (H3) enable us to get
(20)(ℒ(u),u)=∥u∥p+∑j=1m‍Ij(u(xj))u(xj)-∫0T‍f(x,u(x))u(x)dx≥∥u∥p-∑j=1m‍(aj+bj|u(xj)|γj-1)|u(xj)|-∫0T‍(c1+c2|u(x)|p-1)|u(x)|dx≥[1-c2(T2)p]∥u∥p-[maj2T(p-1)/p+c12T(2p-1)/p]∥u∥-mbj(12T(p-1)/p)γj∥u∥γj.
Therefore, by the span of c2∈(0,(T/2)-p), we arrive at lim∥u∥→∞(ℒ(u),u)/∥u∥=+∞. As a result, (ii) and (iii) of Lemma 4 hold. Hence, Lemma 4 implies that (1) has precisely a weak solution. This completes the proof.Lemma 8.
Let (H3)–(H5) hold. ThenJ satisfies Cerami condition (C).Proof.
For alld∈ℝ, we assume that {un}n=1∞⊂X is bounded and
(21)J(un)⟶d,J′(un)⟶0,n⟶∞.
Going, if necessary, to a subsequence, we can assume that un⇀u weakly in X, and then
(22)(J′(un)-J′(u),un-u)=∫0T‍(|un′|p-2un′-|u′|p-2u′)(un′-u′)dx+∑j=1m‍(Ij(un(xj))-Ij(u(xj)))(un(xj)-u(xj))-∫0T‍(f(x,un)-f(x,u))(un-u)dx.
By Lemma 2, X↪↪C[0,T] enables us to obtain that
(23)∑j=1m‍(Ij(un(xj))-Ij(u(xj)))(un(xj)-u(xj))⟶0,∫0T‍(f(x,un)-f(x,u))(un-u)dx⟶0asn⟶∞.
It follows from un⇀u weakly in X and (J′(un)-J′(u),un-u)→0 that
(24)∫0T‍(|un′|p-2un′-|u′|p-2u′)(un′-u′)dx⟶0asn⟶∞.
Noting (19), we have
(25)(∥un∥p-1-∥u∥p-1)(∥un∥-∥u∥)≤∫0T‍(|un′|p-2un′-|u′|p-2u′)(un′-u′)dx,
and thus ∥un-u∥→0 as n→∞. Hence, (i) of Definition 5 is satisfied. Next, we prove (ii) of Definition 5; if not, there exists a sequence {un}⊂X such that
(26)J(un)⟶d,∥J′(un)∥·∥un∥⟶0,n⟶∞,(27)∥un∥⟶∞,n⟶∞.
By (26), there exists a constant ε1>0 such that
(28)J(un)-1pJ′(un)un≤ε1.
On the other hand, (H4) implies that there is a M>0 such that -pF(x,u)+f(x,u)u≥a|u|, for all |u|>M and x∈[0,T]. Furthermore, -pF(x,u)+f(x,u)u is bounded for |u|≤M and x∈[0,T]. Therefore, there exists c>0 such that -F(x,u)+(1/p)f(x,u)u≥(a/p)|u|-c, for all u∈ℝ, x∈[0,T]. This, together with (H5), yields
(29)J(un)-1pJ′(un)un=∑j=1m‍∫0un(xj)‍Ij(x)dx-1pIj(un(xj))un(xj)+∫0T‍(-F(x,un)+1pf(x,un)un)dx≥∫0T‍(-F(x,un)+1pf(x,un)un)dx≥∫0T‍(ap|un|-c)dx,
which implies ∫0T‍|un|dx≤pa-1(Tc+ε1). Therefore, there is a ε2>0 such that ∥un∥∞≤ε2.
It follows from (H3) that there arec3,c4>0 such that
(30)F(x,u)≤c3|u|+c4|u|p,∀u∈ℝ,x∈[0,T].
By this and (H5), we can find
(31)J(un)=1p∫0T‍|un′|pdx+∑j=1m‍∫0un(xj)‍Ij(x)dx-∫0T‍F(x,un)dx≥1p∥un∥p-∫0T‍(c3|un|+c4|un|p)dx≥1p∥un∥p-c3T∥un∥∞-c4T∥un∥∞p≥1p∥un∥p-c3Tε2-c4Tε2p,
and thus J(un)→∞ if (27) holds, which contradicts J(un)→d in (26). This proves that J satisfies condition (C).Theorem 9.
Suppose that (H2)–(H7) hold, and then (1) has infinitely many weak solutions.Proof.
By (H7), we know thatJ is even. Denote βk=supu∈Zk∩S1∥u∥p, and by the compactness of the embedding X↪↪Lp([0,T]), we see that βk→0 as k→∞ (see [26, Lemma 3.8]). Noting (30), we have by (H5) and Hölder inequality, for any u∈Zk and ∥u∥=rk:=βk-1,
(32)J(u)≥1p∥u∥p-∫0T‍(c3|u|+c4|u|p)dx≥1p∥u∥p-c3T(p-1)/p∥u∥p-c4∥u∥pp≥1p∥u∥p-c3T(p-1)/pβk∥u∥-c4βkp∥u∥p≥βk-pp-c3T(p-1)/p-c4.
We easily have rk→∞ as k→∞ and J(u)≥βk-p/p-c3T(p-1)/p-c4→∞ as k→∞. Hence,
(33)bk:=infu∈Zk,∥u∥=rkJ(u)⟶∞ask⟶∞.
On the other hand, by (H6), we find that there areb,c>0 such that
(34)F(x,u)≥b|u|p-c,∀u∈ℝ,x∈[0,T].
Since all the norms of a finite dimensional normed space are equivalent, note that ∥·∥p is a norm of Yk, so there exists a ξ>0 such that
(35)∥u∥pp≥ξ∥u∥p,∀u∈Yk.
Noting (9), we have
(36)J(u)=1p∫0T‍|u′|pdx+∑j=1m‍∫0u(xj)‍Ij(x)dx-∫0T‍F(x,u)dx≤1p∥u∥p-∫0T‍(b|u|p-c)dx+∑j=1m‍∫0u(xj)‍(aj+bj|s|γj-1)dx≤1p∥u∥p-b∥u∥pp+cT+∑j=1m‍[aj|u(xj)|+bjγj|u(xj)|γj]≤1p∥u∥p-bξ∥u∥p+∑j=1m‍[aj∥u∥∞+bjγj∥u∥∞γj]+cT≤(1p-bξ)∥u∥p+∑j=1m‍[aj2T(p-1)/p∥u∥+bjγj(12T(p-1)/p)γj∥u∥γj]+cT.
Note that we can choose a large enough b such that 1/p-bξ<0 by (H6) and p>γj by (H2), and then there exists positive constants dk such that
(37)J(u)≤0,foreachu∈Yk,∥u∥≥dk.
By this and (33), we can take ρk:=max{dk,rk+1}, and thus ak:=maxu∈Yk,∥u∥=ρkJ(u)≤0. Up until now, we have proved that the functional J satisfies all the conditions of Lemma 6, and then J has infinitely many critical points. Equivalently, (1) has infinitely many weak solutions. This completes the proof.Two Examples.(1) Let p=4 and x1=T/2. Consider the problem
(38)-(|u′|2u′)′=f(x,u),inΩ,Δ|u′(x1)|2u′(x1)=u(x1),u(0)=-u(T),u′(0)=-u′(T),
where f(x,u)=-η1-η2u3, η1≥0, η2∈(0,(T/2)-4). Clearly, (H1)–(H3) hold true. By Theorem 7, (38) has only a weak solution.(2) (H6) can be weaken that(39)lim|u|→∞F(x,u)|u|p>(pξ)-1,uniformlyonx∈[0,T],
where ξ is determined by (35). Indeed, by (39), we can obtain that there exist b>(pξ)-1 and c>0 such that (34) is satisfied. Furthermore, 1/p-bξ<0.Letp=4 and x1=T/2. Consider the problem
(40)-(|u′|2u′)′=f(x,u),inΩ,Δ|u′(x1)|2u′(x1)=u(x1)3,u(0)=-u(T),u′(0)=-u′(T),
where f(x,u)=-η+(ξ-1+4)u3 and Ij(u(xj))=u(x1)3, where η>0 and ξ is defined by (35).ForIj(u), we can easily have (H2) and (H7) hold. By computation, ∫0u‍Ij(s)ds=(3/4)u43≥0,  p∫0u‍Ij(s)ds-Ij(u)u=4×(3/4)u43-u43=2u43≥0, for allu∈ℝ, and hence we obtain that (H5) is satisfied.Forf(x,u) and F(x,u), we see f(x,u)=-η+(ξ-1+4)u3≤η+(ξ-1+4)|u|3, forallu∈ℝ,x∈[0,T]. Therefore, (H3) holds.It is obvious thatF(x,u)=∫0u‍f(x,s)ds=∫0u‍(-η+(ξ-1+4)s3)ds=-η|u|+((ξ-1+4)/4)|u|4, so (H7) holds; furthermore,
(41)lim|u|→∞F(x,u)|u|p=lim|u|→∞-η|u|+((ξ-1+4)/4)|u|4|u|4=ξ-1+44>(pξ)-1.
As a result of this, (39) (i.e., (H6)) is true.By(42)-pF(x,u)+f(x,u)u=-4[-η|u|+ξ-1+44|u|4]+(-η+(ξ-1+4)u3)u=4η|u|-ηu≥3η|u|,
we obtain that
(43)lim|u|→∞-pF(x,u)+f(x,u)u|u|≥lim|u|→∞3η|u||u|=3η,
uniformly on x∈[0,T]. Consequently, (H4) holds. Nowadays, we have proved that (H2)–(H7) hold, and then (40) has infinitely many weak solutions by Theorem 9.In [14], the condition (p2) is
(44)∃μ>p∃δ>0:∀|t|≥δ,0≤μF(x,t)≤tf(x,t).
As known to all, this condition is originally introduced in [27] and is still present in many works which is used to guarantee the boundedness of (P.S.) sequences of the corresponding functional.In the problem (40), note that there exists δ>0 such that |u|≥δ and
(45)μF(x,u)≤uf(x,u)⟹-ημ|u|+ξ-1+44μ|u|4≤-ηu+(ξ-1+4)|u|4.
This is impossible if μ>p=4 and δ is an adequately large positive number. On the other hand, if (H4) holds, then there is a δ sufficiently large such that
(46)-pF(x,u)+uf(x,u)≥a|u|≥0,forx∈[0,T],|u|≥δ.
Therefore, pF(x,u)≤uf(x,u),for all x∈[0,T]  and |u|≥δ, which is also weaker than the condition (p2) if F(x,u)≥0. This means that our results generalize the corresponding results in [14].



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*Source: 786548-2013-07-07.xml*