# Subharmonic Solutions of Nonautonomous Second Order Differential Equations with Singular Nonlinearities

Authors: N. Daoudi-Merzagui; F. Derrab; A. Boucherif
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We discuss the existence of subharmonic solutions for nonautonomous second order differential equations with singular nonlinearities. Simple sufficient conditions are provided enable us to obtain infinitely many distinct subharmonic solutions. Our approach is based on a variational method, in particular the saddle point theorem.


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## Body

## 1. Introduction and Main Result

In this paper we discuss the problem of the existence of infinitely many subharmonic solutions for nonautonomous second order differential equations with singular nonlinearities of the form(1.1)u′′(t)+f(t,u(t))=e(t),
where f:ℝ2→ℝ is continuous, is T-periodic, in its first argument with T>0, and presents a singularity with respect to its second argument. Here by a subharmonic solution we mean a kT-periodic solution for any integer k if T>0 is the minimal period. When the solution is not T-periodic we call it a true subharmonic. It was pointed out in [1] that singular differential equations of the form (1.1) appear in the description of many phenomena in the applied sciences, such as the Brillouin focusing system and nonlinear elasticity. Several authors have investigated the problem of existence of periodic solutions for second order differential equations with singular nonlinearities (see [2–4] and the references therein). Topological and variational methods are the two main techniques that have been developed for the study of (1.1). We refer the interested reader to the paper [1] for details and references on the topological methods. In this work we shall rely on the saddle point theorem, see [5, 6], to prove our main result. We use the truncation techniques introduced in [7] to modify our problem to one without singularities. We assume that the nonlinearity f is monotone with respect to its first variable t. When f is increasing, our result generalizes the result in [8]. We can obtain the same result by considering the monotonicity of the potential function instead of the field f. For more results on the subject and different techniques one can consult the papers [9–12]. We should point out some related recent articles, for instance [13, 14].Throughout this paper we shall use the following notations.  LetI=[0,T].Lp(I) is the classical Lebesgue space of functions u:I→ℝ such that |u(·)|p is integrable, and for u∈Lp(I) we define its norm by(1.2)‖u‖Lp=(∫0T|u(t)|pdt)1/p.Let∥u∥∞=sup{|u(t)|;t∈[0,kT]}. For T>0 and k∈ℕ we let HkT1={u∈W1,2([0,kT],ℝ);u(0)=u(kT)} and for u∈HkT1 we define its norm by(1.3)‖u‖HkT1=(‖u‖L2+‖u′‖L2)1/2.HkT1 endowed with the norm ∥·∥HkT1 is a reflexive Banach space. Also HkT1=H+⊕H-, orthogonal decomposition, where H+ is the subspace of constant functions in HkT1 and H- denotes the subspace of functions in HkT1 with mean value zero; so that u∈HkT1 can be written as u=u¯+ũ with u¯∈H+ and ũ∈H-.We shall assume thate:ℝ→ℝ is a locally integrable T-periodic function. We denote the mean value of e by e¯,thatis,e¯=1/T∫0Te(t)dt. It follows that e¯, ∥e∥L1 and ∥e∥L2 are bounded. Moreover, since e is T-periodic, we have ∫0kTe(t)dt=∑j=0k-1∫0Te(t+jT)dt=∑j=0k-1∫0Te(t)dt=k∫0Te(t)dt=kTe¯.LetF(t,u)=∫1uf(t,s)ds be an antiderivative of f defined for all u∈ℝ and for all t∈I.We introduce the following assumptions on the nonlinearity.(H1)
f:ℝ×(0,+∞)→ℝ is continuous, f(t+T,u)=f(t,u) for all (t,u)∈ℝ×(0,+∞), and such that
(i)
t→f(t,u) is monotone for each fixed u in (0,+∞),(ii)
limu→0+f(t,u)=-∞, uniformly in t∈I,(iii)
f(0,u)=f(T,u)>e¯ for all u>1.(H2)
(i)
limu→+∞2F(t,u)/u2=0, uniformly in t∈I.(ii)
limu→+∞∫0T[F(t,u)-e¯u]dt=+∞.Theorem 1.1.
Assume thate is a locally integrable T-periodic function. If (H1) and (H2) are satisfied then (1.1) has a sequence (uk)k≥1 of kT-periodic solutions whose amplitudes and minimal periods tend to infinity. In particular, if T is the minimal period of e and of f with respect tot, (1.1) admits solutions with minimal periods kT for every sufficiently Large Integer k.


## 2. Proof of Theorem1.1

The proof of this result will be based on several auxiliary results.


### 2.1. Modification of the Problem

Define the truncation functionfr:ℝ2→ℝ,0<r≤1, by(2.1)fr(t,u)={f(t,u),u≥r,f(t,r),u<r.
Note that condition (H1) implies that fr is continuous with respect to (t,u)∈I×ℝ and T-periodic with respect to its first variable t.Lemma 2.1.
Assume (H1) and (H2) (ii) are satisfied. Then there existsd>1 such that for every u∈(0,1/d)∪(d,+∞)(2.2)(f(t,u)-e¯)(u-1)>0,uniformlyint∈I.Proof.
First, it follows from (H1)(ii) that for anyA>0, there is δA>0such that for every u∈(0,δA) we have f(t,u)<-A, uniformly in t∈I.
In particular forA>2|e¯|+1, there is δA>0 such that for every u∈(0,δA) it holds(2.3)f(t,u)-e¯<-|e¯|-1<0.
Choose d1>1 such that 1/d1<δA. Then for every u∈(0,1/d1), we have f(t,u)-e¯<0.
Therefore, condition (H1)(ii) implies that there existsd1>1 such that for every u∈(0,1/d1) it holds (f(t,u)-e¯)(u-1)>0, uniformly in t∈I.
Next, condition (H2)(ii) implies that for anyB>1, sufficiently large, there is χB>0, large enough, such that for every u>χB we have(2.4)∫0T[F(t,u)-e¯u]dt>B.
Hence, there exists d2, sufficiently large such that d2>max(1,χB) and for every u>d2 it holds
(2.5)∫0T[F(t,u)-e¯u]dt>B.
We show that for u>d2 we have f(t,u)-e¯>0, uniformly in t∈I. Assume, on the contrary, that there exists y>d2 for which f(t0,y)-e¯≤0, for some t0∈(0,T). By the continuity of f on I and ((H1)(iii)), there exists σ∈(0,T) such that f(σ,u)=maxt∈If(t,u). If t0≤σ then 0<f(0,y)-e¯<f(t0,y)-e¯≤0, and this is a contradiction. Similarly, if t0≥σ then 0<f(0,y)-e¯=f(T,y)-e¯<f(t0,y)-e¯≤0, and we again arrive at a contradiction.
Hence we deduce that if (H1) and (H2)(ii) hold, then there existsd>max(d1,d2) such that for any u∈(0,1/d)∪(d,+∞)(f(t,u)-e¯)(u-1)>0, uniformly in t∈I. This completes the proof of Lemma 2.1.Lemma 2.2.
For every positive integerk, there exist rk and Rk with 0<rk<1/d<d<Rk such that for any μ∈(0,rk] each kT-periodic solution u of
(2.6)u′′(t)+fμ(t,u(t))=e(t),
satisfies
(2.7)rk≤u(t)≤Rk,∀t∈R.
In particular, any kT-periodic solution of (2.6), with μ=rk is a solution of (1.1).Proof.
This is essentially Proposition  2.1 in [8]. We shall use some ideas from [8] (see also [15]). Fix k∈ℕ, and suppose, on the contrary, that for each integer n, there exist μn∈(0,1/n) and a kT-periodic solution un satisfying
(2.8)un′′(t)+fμn(t,un(t))=e(t)
and {un(t);t∈ℝ}⊈[1/n,n].Claim 1.
Letd be as in Lemma 2.1 and let un be as above. Then for every n there exists τn∈[0,kT] such that un(τn)∈[1/d,d].
Indeed, it follows from (2.8) that(2.9)∫0kTfμn(t,un(t))dt=kTe¯.
Now, if un(t)>d for all t∈[0,kT], then Lemma 2.1 implies that fμn(t,un(t))-e¯>0, which in turn yields ∫0kTfμn(t,un(t))dt>kTe¯.
This contradicts (2.9). On the other hand, if un(t)<1/d for all t∈[0,kT], then fμn(t,un(t))-e¯<0, so that ∫0kTfμn(t,un(t))dt<kTe¯. This is again a contradiction to (2.9).Claim 2.
There existsR>0 such that Mn=maxt∈[0,kT]un(t)≤R for each integer n. To prove the claim notice, there exists tn1∈[0,kT] such that un(tn1)=Mn. If un(tn1)=Mn∈[1/d,d] then un(t)≤R for any R>d. So, assume that there exists a subsequence of (un)n, which we label the same, for which Mn→+∞ when n→+∞. So that Mn>d for n large enough. Since un(τn)<d, there exists an interval [αn,βn], containing tn1, such that
(2.10)βn-αn≤kT,un(αn)=d=un(βn),d≤un(t)≤un(tn1),forallt∈[αn,βn].
Equation (2.8) can be written as
(2.11)un′(t)=vn(t)+∫αnt[e(s)-e¯]ds,vn′(t)=-fμn(t,un(t))+e¯.
Since for allt∈(αn,βn),un(t)>d and μn<1/n, then the second equation in (2.11) is equivalent to(2.12)vn′(t)=-f(t,un(t))+e¯.
Lemma2.1 implies that f(t,un(t))-e¯>0 for all t∈[αn,βn]. Then vn′(t)<0 for all t∈[αn,βn] and hence vn is decreasing on [αn,βn]. The first equation in (2.11) implies(2.13)un′(t)≤vn(αn)+∫αnt[e(s)-e¯]ds∀t∈[αn,βn].
This yields
(2.14)un′(t)≤vn(αn)+‖e‖L1+kT|e¯|∀t∈[αn,βn].
Integrating the above inequality over [αn,tn1]⊂[αn,βn] we obtain
(2.15)un(tn1)-un(αn)=Mn-d≤kT[vn(αn)+‖e‖L1+kT|e¯|].
Equation (2.15) leads to
(2.16)lim⁡n→+∞(1-dMn)≤kTlim⁡n→+∞(vn(αn)+‖e‖L1+kT|e¯|Mn).
Hence
(2.17)1≤kTlim⁡n→+∞vn(αn)Mn.
It is clear from (2.17) that vn(αn)→+∞ when n→+∞. So that, for n large enough, we have
(2.18)vn(αn)≥‖e‖L1+kT|e¯|.
Since Mn=un(tn1), we have un′(tn1)=0. It follows from the first equation in (2.11) that
(2.19)|vn(tn1)|=|∫αntn1[e(s)-e¯]ds|≤‖e‖L1+kT|e¯|.
We see from (2.18) and (2.19) that for n large enough
(2.20)vn(αn)≥‖e‖L1+kT|e¯|≥vn(tn1).
Since vn(·) is continuous on [αn,βn], then, for n large enough, there exists at least one tn*∈(αn,tn1) such that vn(tn*)=∥e∥L1+kT|e¯|. We denote by tn2, the first such tn*. Then
(2.21)vn(tn2)=‖e‖L1+kT|e¯|.
We distinguish two cases.Case 1.
f(·,s) is increasing for each fixed s∈(0,+∞).
Consider the functionB1 defined by(2.22)B1(t)=F(tn2,un(t))-e¯un(t)+12(vn(t)-‖e‖L1-kT|e¯|)2.
Then
(2.23)B1′(t)=(f(tn2,un(t))-e¯)un′(t)+(vn(t)-‖e‖L1-kT|e¯|)vn′(t).
Since vn′(t)=-f(t,un(t))+e¯, it follows from the first equation in (2.11)
(2.24)B1′(t)=(f(tn2,un(t))-e¯)(vn(t)+∫αnt[e(s)-e¯]ds)+(vn(t)-‖e‖L1-kT|e¯|)(-f(t,un(t))+e¯)=(f(tn2,un(t))-e¯)(vn(t)+∫αnt[e(s)-e¯]ds+‖e‖L1+kT|e¯|-‖e‖L1-kT|e¯|)+(vn(t)-‖e‖L1-kT|e¯|)(-f(t,un(t))+e¯).
Hence
(2.25)B1′(t)=(f(tn2,un(t))-e¯)(∫αnt[e(s)-e¯]ds+‖e‖L1+kT|e¯|)+(f(tn2,un(t))-f(t,un(t)))(vn(t)-‖e‖L1-kT|e¯|)⋅
Since, for all t∈[αn,βn](2.26)|∫αnt[e(s)-e¯]ds|≤‖e‖L1+kT|e¯|,
it follows that for all t∈[αn,βn](2.27)∫αnt[e(s)-e¯]ds+‖e‖L1+kT|e¯|≥0.
Also, Lemma 2.1 implies that f(tn2,un(t))-e¯>0 for all t∈[αn,βn]. Furthermore, the monotonicity of f implies that f(tn2,un(t))-f(t,un(t))≥0 for all t∈[αn,tn2]. Since vn(·) is decreasing on [αn,tn2] and vn(tn2)=∥e∥L1+kT|e¯|, it follows that
(2.28)vn(t)-‖e‖L1-kT|e¯|≥vn(tn2)-‖e‖L1-kT|e¯|=0
for all t∈[αn,tn2]. Now, Lemma 2.1 combined with (2.27), (2.28), and the monotonicity of f with respect to its first variable shows that
(2.29)B1′(t)≥0,∀t∈[αn,tn2].
Thus, the function B1 is increasing on [αn,tn2]. Since un(αn)=d,
(2.30)B1(αn)=F(tn2,d)-e¯d+12(vn(αn)-‖e‖L1-kT|e¯|)2≤B1(tn2)=F(tn2,un(tn2))-e¯un(tn2)+12(vn(tn2)-‖e‖L1-kT|e¯|)2.
Since vn(tn2)-∥e∥L1-kT|e¯|=0, it follows that
(2.31)B1(αn)=F(tn2,d)-e¯d≤F(tn2,un(tn2))-e¯un(tn2).
Notice that
(2.32)F(tn2,un(tn2))-e¯un(tn2)=∫1d(f(tn2,s)-e¯)ds+∫dun(tn2)(f(tn2,s)-e¯)ds-e¯.
Also, f(tn2,s)-e¯>0 if s∈(d,un(tn2)]⊂[d,un(tn1)] (see Lemma 2.1). It follows that
(2.33)∫dun(tn2)[f(tn2,s)-e¯]ds≤∫dun(tn1)[f(tn2,s)-e¯]ds.
Hence
(2.34)B1(αn)≤∫1d[f(tn2,s)-e¯]ds+∫dun(tn1)[f(tn2,s)-e¯]ds-e¯=F(tn2,un(tn1))-e¯un(tn1).
Set
(2.35)ρn=12(vn(αn)-‖e‖L1-kT|e¯|Mn)2.
Then
(2.36)lim⁡n→+∞ρn=0.
Indeed, since Mn=un(tn1), we have
(2.37)B1(αn)=F(tn2,d)-e¯d+ρnMn2≤F(tn2,un(tn1))-e¯un(tn1)=F(tn2,Mn)-e¯Mn.
From (2.37) we deduce
(2.38)lim⁡n→+∞F(tn2,d)(Mn)2+lim⁡n→+∞ρn≤lim⁡n→+∞F(tn2,Mn)(Mn)2.
Notice that F(tn2,d)=∫1df(tn2,s)ds is bounded, so that
(2.39)lim⁡n→+∞F(tn2,d)(Mn)2=0.
Also, (H2)(i) implies that
(2.40)lim⁡n→+∞F(tn2,Mn)(Mn)2=0.
Therefore (2.36) holds; that is, limn→+∞ρn=0. Since ∥e∥L1 and |e¯| are bounded, it follows that
(2.41)lim⁡n→+∞vn(αn)Mn=0.
It is clear that (2.41) contradicts (2.17).Case 2.
f(·,s) is decreasing for each fixed s∈(0,+∞).
In this case we consider the functionB2 defined by(2.42)B2(t)=F(αn,un(t))-e¯un(t)+12(vn(t)-‖e‖L1-kT|e¯|)2.
Then
(2.43)B2′(t)=(f(αn,un(t))-e¯)(∫αnt[e(s)-e¯]ds+‖e‖L1+kT|e¯|)+(f(αn,un(t))-f(t,un(t)))(vn(t)-‖e‖L1-kT|e¯|)≥0∀t∈[αn,tn2].
Repeating the same reasoning as in Case 1, we arrive at a contradiction.
Therefore, we deduce that there must existR>0 such that Mn=maxt∈[0,kT]un(t)≤R for each n.
Next, we prove that there isr>0 such that un(t)≥r for every t∈[0,kT]. Assuming that this is not true we will obtain a contradiction.
Consider the following sets(2.44)I1/n={t∈[0,kT];un(t)<1n},I1/n,1/d={t∈[0,kT];1n≤un(t)<1d},I1/d,R={t∈[0,kT];1d≤un(t)≤R}.
It is clear that for n≥R, un(t)≤R≤n. Also, we cannot have un(t)≥1/n for every t∈[0,kT], for otherwise we would have
(2.45)1n≤un(t)≤n∀t∈[0,kT],
which contradicts the assumption {un(t);t∈ℝ}⊈[(1/n),n]. Hence, for n≥R there exists tn3∈[0,kT] such that un(tn3)<1/n. This shows that I1/n≠∅. The continuity of un implies that I1/n is open, and so meas(I1/n)≠0.
Define(2.46)Ψn=∫0kT[fμn(t,un(t))-e¯]dt,n∈N.
It follows from (2.9) that Ψn=0. On the other hand
(2.47)Ψn=∫I1/n[fμn(t,un(t))-e¯]dt+∫I1/n,1/d[fμn(t,un(t))-e¯]dt+∫I1/d,R[fμn(t,un(t))-e¯]dt.
(i)Assume we are integrating positively on all subintervals of[0,kT].
Ift∈I1/n,1/d, then un(t)∈[1/n,1/d)⊂(0,1/d). So that, by Lemma 2.1,(2.48)∫I1/n,1/d[fμn(t,un(t))-e¯]dt<0.
For t∈I1/d,R we have un(t)∈[1/d,R]. This means that un(t) is bounded uniformly in t∈I1/d,R. Since fμn is continuous it is bounded on I1/d,R.
Let(2.49)c=max⁡{|fμn(t,x)|;t∈[0,kT],1d≤x≤R}=max⁡I1/d,R{|fμn(t,x)|}.
Then
(2.50)|∫I1/d,R[fμn(t,un(t))-e¯]dt|≤∫I1/d,R[|fμn(t,un(t))|+|e¯|]dt≤kT(c+|e¯|)
It follows from (2.47), (2.48), and (2.50) that
(2.51)Ψn<∫I1/n[fμn(t,un(t))-e¯]dt+kT(c+|e¯|).Claim 3.
(2.52)lim⁡n→∞⁡∫I1/n[fμn(t,un(t))-e¯]dt=-∞.Proof.
Recall thatμn∈(0,1/n) and un(t)<1/n for each t∈I1/n. Then, if un(t)<μn we have fμn(t,un(t))=f(t,μn), and if un(t)∈[μn,1/n), we have fμn(t,un(t))=f(t,un(t)). In both cases condition (H1)(ii) and the continuity of f imply that limn→∞(fμn(t,un(t))-e¯)=-∞ for every t∈I1/n.
Sincee¯ is bounded, then (2.51) implies that(2.53)lim⁡n→+∞Ψn=-∞,
which is a contradiction with (2.9).
(ii) If we integrate negatively on all subintervals of[0,kT] we will obtain limn→+∞Ψn=+∞, which, again, contradicts (2.9). Thus, the proof of Lemma 2.2 is complete.Remark 2.3.
Lemma2.2 shows that any kT-periodic solution u of (2.6), with μ=rk is a solution of (1.1), since it satisfies u(t)≥rk for all t∈ℝ and frk(t,u(t))=f(t,u(t)).
In the remainder of the paper we shall deal with (2.6), with μ=rk instead of (1.1). Let Frk(t,u)=∫1ufrk(t,s)ds be a primitive of frk defined for all t∈I and u∈ℝ.Lemma 2.4.
If (H1) and (H2) hold, thenfrk and Frk satisfy the following conditions.(L1)
frk is defined and continuous in (t,u)∈I×ℝ and T-periodic with respect to t∈I.(L2)
liminf|u|→+∞2Frk(t,u)/u2=0, uniformly in t∈I.(L3)
∃d>1 such that foru∈(-∞,1/d)∪(d,+∞) it holds (frk(t,u)-e¯)(u-1)>0 uniformly in t∈I.(L4)
lim|u|→+∞∫0T[Frk(t,u)-e¯u]dt=+∞.


### 2.2. Existence ofkT-Periodic Solutions for  (2.6), μ=rk

Using a variational method we shall show that  (2.6), with μ=rk has infinitely many kT-periodic solutions. In fact, we have the following.Lemma 2.5.
Assume thate is locally integrable T-periodic function and the conditions (L1), (L2), (L3), (L4) hold. Then (2.6), with μ=rk admits a kT-periodic solution uk.Proof.
We shall rely on a variational method and more precisely on the saddle point theorem.
Define for eachk≥1 the action functional Jk:HkT1→ℝ by(2.54)Jk(u)=∫0kT[12[u′(t)]2-Frk(t,u(t))+e(t)u(t)]dt.Jk is well defined on HkT1, weakly lower semicontinuous and continuously differentiable on HkT1. Furthermore,
(2.55)〈Jk′(u),v〉=∫0kT[u′(t)v′(t)-frk(t,u(t))v(t)+e(t)v(t)]dt,∀u,v∈HkT1.
The critical points of Jk are precisely the weak solutions of  (2.6), with μ=rk.
First, we show that the functionalJk satisfies the Palais-Smale condition.
For this, letk≥1 be fixed and let (un)n∈ℕ be a sequence in HkT1 such that (Jk(un))n∈ℕ is bounded and limn→+∞Jk′(un)=0. Then (un)n∈ℕ has a convergent subsequence.
Suppose, on the contrary, thatlimn→+∞∥un∥HkT1=+∞. Condition (L2) implies that for any ε>0, small enough, there is Cε≥0 such that(2.56)Frk(t,u)≤εu2+Cε,∀u∈R.
Writing un(t)=un¯+uñ(t) for t∈[0,kT], we obtain
(2.57)-∫0kTFrk(t,un(t))dt≥-εkT|un¯|2-ε∫0kT|uñ(t)|2dt-2εun¯∫0kTuñ(t)dt-CεkT.
Since uñ∈H-, we have ∫0kTuñ(t)dt=kTuñ___=0, so that
(2.58)-∫0kTFrk(t,un(t))dt≥-ε‖uñ‖L22-εkT|un¯|2-CεkT.
Now, Hölder’s inequality gives
(2.59)∫0kTe(t)un(t)dt=un¯∫0kTe(t)dt+∫0kTe(t)uñ(t)dt≥-|un¯|kT|e¯|-‖e‖L2‖uñ‖L2.
Since un′(t)=uñ′(t) for t∈[0,kT], it follows from (2.54), (2.58), and (2.59)
(2.60)Jk(un)≥12‖uñ′‖L22-ε‖uñ‖L22-εkT|un¯|2-CεkT-kT|e-||un¯|-‖e‖L2‖uñ‖L2.
Wirtinger’s inequality
(2.61)‖uñ′‖L22≥(4π24π2+(kT)2)‖uñ‖HkT12
combined with the inequality ∥uñ∥L2≤∥uñ∥HkT1 give
(2.62)Jk(un)+(εkT|un¯|2+kT|e¯||un¯|)+CεkT≥‖uñ‖HkT1([2π24π2+(kT)2-ε]‖uñ‖HkT1-‖e‖L2).
This leads to
(2.63)lim⁡n→+∞|un¯|=+∞.
Indeed, if (2.63) does not hold then there would exist a subsequence of (un¯)n∈ℕ, still denoted the same, which is bounded. Since (Jk(un))n∈ℕ and e¯ are bounded and ε is chosen arbitrarily small, then (2.62) implies that ∥uñ∥HkT1 is bounded. It follows from the inequality(2.64)‖un‖HkT1≤kT|un¯|+‖uñ‖HkT1
that ∥un∥HkT1 is bounded, but this contradicts our assumption limn→+∞∥un∥HkT1=+∞. Therefore, (2.63) holds.
Using Wirtinger’s inequality(2.65)‖uñ‖L2≤kT2π‖uñ′‖L2
in (2.60), we get
(2.66)‖uñ′‖L2|un¯|([12-ε(kT2π)2]‖uñ′‖L2|un¯|-(kT2π)‖e‖L2|un¯|)≤Jk(un)|un¯|2+εkT+kT|e¯||un¯|+CεkT|un¯|2.
It follows from (2.63) that
(2.67)lim⁡n→+∞‖uñ′‖L2|un¯|=0.
Using Sobolev’s inequality we obtain
(2.68)lim⁡n→+∞‖uñ‖∞|un¯|≤kT12lim⁡n→+∞‖uñ′‖L2|un¯|=0.
The identity un(t)=un¯(1+(uñ(t)/un¯)) for all t∈[0,kT] and (2.63) imply that
(2.69)lim⁡n→+∞min⁡t∈[0,kT]|un(t)|=+∞.
Assume that limn→+∞mint∈[0,kT]un(t)=+∞ (the other case can be treated similarly). Then for n large enough, un(t)>d uniformly in t∈[0,kT]. By (L3) we have for all t∈[0,kT](2.70)frk(t,un(t))-e¯>0.
Consequently, for n large enough
(2.71)∫0kT|frk(t,un(t))-e¯|dt=∫0kT[frk(t,un(t))-e¯]dt=|∫0kT[frk(t,un(t))-e¯]dt|.
Since limn→+∞Jk′(un)=0, then for all v∈HkT1 and for n large enough
(2.72)|∫0kT[un′(t)v′(t)-frk(t,un(t))v(t)+e(t)v(t)]dt|≤εn‖v‖HkT1,
where εn>0 for every n, and limn→+∞εn=0. In particular, if we take v(t)=-1 in the above inequality we obtain for every n∈ℕ(2.73)|∫0kT[frk(t,un(t))-e¯]dt|≤εnkT,
which infer
(2.74)∫0kT|frk(t,un(t))|dt≤∫0kT|frk(t,un(t))-e¯|dt+∫0kT|e¯|dt=|∫0kT[frk(t,un(t))-e¯]dt|+∫0kT|e¯|dt≤εnkT+kT|e¯|.
Now, taking v=uñ in (2.72) we obtain
(2.75)εn‖uñ‖HkT1≥‖uñ′‖L22-|∫0kT[frk(t,un(t))-e(t)]uñ(t)dt|.
Obviously, we have for n large enough
(2.76)|∫0kT[frk(t,un(t))-e(t)]uñ(t)dt|≤sup⁡t∈[0,kT]|uñ(t)|(∫0kT|frk(t,un(t))|dt+∫0kT|e(t)|dt)≤‖uñ‖∞(εnkT+kT|e¯|+‖e‖L1).
Thus, for n large enough, (2.75) implies that
(2.77)εn‖uñ‖HkT1≥‖uñ′‖L22-‖uñ‖∞(εnkT+kT|e¯|+‖e‖L1).
Sobolev’s inequality
(2.78)‖uñ‖∞≤kT12‖uñ′‖L2≤kT12‖uñ‖HkT1
and Wirtinger’s inequality combined with (2.77) give, for n large enough,
(2.79)εn‖uñ‖HkT1≥(4π24π2+(kT)2)‖uñ‖HkT12-kT12‖uñ‖HkT1(εnkT+kT|e¯|+‖e‖L1).
So, for n large enough, we deduce that
(2.80)‖uñ‖HkT1≤l:=(4π2+(kT)24π2)(1+kT12[kT+kT|e¯|+‖e‖L1]).
Hence (uñ)n is bounded in HkT1. Consequently ∥un′∥L2=∥uñ′∥L2≤∥uñ∥HkT1≤ℓ. Since (Jk(un))n∈ℕ is bounded, it follows that
(2.81)∫0kT[Frk(t,un(t))-e(t)un(t)]dt
is bounded. Hölder’s inequality gives
(2.82)|∫0kTe(t)uñ(t)dt|≤‖e‖L2‖uñ‖L2≤‖e‖L2‖uñ‖HkT1≤l‖e‖L2.
Since
(2.83)∫0kTFrk(t,un(t))dt-un-∫0kTe(t)dt=∫0kT[Frk(t,un(t))-e(t)un(t)]dt+∫0kTe(t)uñ(t)dt,
it follows that
(2.84)∫0kTFrk(t,un(t))dt-un-∫0kTe(t)dt,
is bounded. But,
(2.85)un-∫0kTe(t)dt=kTe¯un-=kTe¯1kT∫0kTun(t)dt=∫0kTe¯un(t)dt.
Consequently, there exists C>0 such that
(2.86)∫0kT[Frk(t,un(t))-e¯un(t)]dt≤C.
On the other hand, extending Frk by T-periodicity we obtain
(2.87)∫0kT[Frk(t,un(t))-e¯un(t)]dt=∑k-1j=0∫0T[Frk(t+jT,un(t+jT))-e¯un(t+jT)]dt=∑k-1j=0∫0T[Frk(t,un(t+jT))-e¯un(t+jT)]dt.
Setting xn=un(t+jT) for t∈[0,T], we get
(2.88)∫0kT[Frk(t,un(t))-e¯un(t)]dt=∑k-1j=0∫0T[Frk(t,xn)-e¯xn]dt=k∫0T[Frk(t,xn)-e¯xn]dt.
From (2.75) |xn|=|un(t+jT)|→+∞ when n→+∞ uniformly in t∈[0,kT]. By (L4) we have
(2.89)lim⁡n→+∞∫0kT[Frk(t,un(t))-e¯un(t)]dt=klim⁡|x|→+∞∫0T[Frk(t,x)-e¯x]dt=+∞.
This is a clear contradiction to (2.86). Therefore, (un)n∈ℕ is bounded in HkT1, and so it has a convergent subsequence. This shows that Jk satisfies the Palais-Smale condition. Next, we show that Jk has a geometry of a Saddle. For, let u∈H- then we have u=uñ and u¯=0, so that
(2.90)Jk(uñ)=12‖uñ′‖L22-∫0kT[Frk(t,uñ(t))-e(t)uñ(t)]dt.
Proceeding as before, we get an inequality similar to (2.72) by replacing un by uñ and u¯nby 0,
(2.91)Jk(uñ)≥‖uñ‖HkT1([2π24π2+(kT)2-ε]‖uñ‖HkT1-‖e‖L2)-CεkT.
Since ε is chosen arbitrary small, we obtain
(2.92)lim⁡‖uñ‖HkT1→+∞Jk(uñ)=+∞,
which shows that Jk is coercive. Hence, Jk admits a bounded minimizing sequence. Furthermore, Jk is weakly lower semicontinuous on HkT1, then
(2.93)inf⁡H-Jk>-∞.
For s, a constant function, we have ∥s∥HkT1=|s|kT→+∞ if and only if |s|→+∞. Then
(2.94)Jk(s)=-∫0kT[Frk(t,s)-e(t)s]dt.
Extending Frk by T-periodicity we obtain
(2.95)Jk(s)=-∫0kT[Frk(t,s)-e¯s]dt=-∑k-1i=0∫0T[Frk(t+iT,s)-e¯s]dts=-k∫0T[Frk(t,s)-e¯s]dt.
Condition (L4) implies that
(2.96)lim⁡|s|→+∞[-Jk(s)]=klim⁡|s|→+∞∫0T[Frk(t,s)-e¯s]dt=+∞.
Hence, for each k≥1,(-Jk) is coercive on the space of constant functions. Then for each k≥1, there exists ηk>0, large enough, such that
(2.97)Jk(ηk)→-∞,andJk(-ηk)→-∞.
Thus,
(2.98)max⁡(Jk(-ηk),Jk(ηk))→-∞whenηk→+∞.
Therefore,
(2.99)max⁡(Jk(-ηk),Jk(ηk))<inf⁡H-Jk.
Let Iηk be the open interval of ℝ centered at 0 and with radius ηk. Since H+=ℝ it is clear that
(2.100)∂Iηk∩H+={-ηk,ηk}.
Therefore, we have
(2.101)max⁡∂Iηk∩H+Jk<inf⁡H-Jk,
with H+⊕H-=HkT1. Thus Jk has a geometry of Saddle.
Finally, all conditions of the Saddle point theorem are satisfied. Then for eachk≥1,Jk admits a critical point βk, which is characterized by(2.102)Jk(βk)=inf⁡ψ∈Γkmax⁡s∈[-ηk,ηk]Jk(ψ(s)),
where
(2.103)Γk={ψ∈C([-ηk,ηk],HkT1);ψ(-ηk)=-ηk,ψ(ηk)=ηk}.
Thus for each k≥1,βk is a weak kT-periodic solution of  (2.6) with μ=rk If, furthermore, e is assumed continuous, then βk is a classical solution of  (2.6) with μ=rk.
This completes the proof of Lemma2.5.Remark 2.6.
As a consequence of Lemmas2.2, 2.5 and the Remark 2.3, we conclude that if (H1) and (H2) are satisfied, then (1.1) admits a sequence (uk)k≥1 of kT-periodic solutions.


### 2.3. Existence of Distinct Subharmonic Solutions for  (2.6) with  μ=rk

Note thatJk=Jm on HkT1∩HmT1 for each k and m. This justifies the following definition (see [10, Definition 2.1 page 653]).Definition 2.7.
The level ofu∈∪kHkT1 is defined by Jm(u) when u∈HmT1.
Every functionalJk admits at least a critical level which is given by βk=minHkT1Jk.Note that nondistinct subharmonic solutions have the same level. Then we deduce that in order to find the multiplicity of distinct subharmonic solutions, we have to search the multiple critical levels. The sequence(βk)k≥1 is not always increasing, which means that there exists m∈ℕ such that if k∈mℕ then βk≤βm.Ifβk<β1 for an integer k, then βk is not a level for the T-periodic functions and every global minimum of Jk is in fact a subharmonic solution of (2.6).In our caseJk(uk) is not necessary a global minimum for Jk and then even if the condition above is verified, it is still insufficient to deduce the existence of true subharmonic solutions. This is why we prove also that the amplitudes and the minimal periods tend to infinity.Lemma 2.8.
The minimal periods of the solutionsuk of (2.6), with μ=rk tend to infinity.Proof.
Letuk be a weak solution of (2.6) with μ=rk  Then uk is a critical point of Jk. We show that
(2.104)lim⁡k→+∞1kJk(uk)=-∞.
Let ηk≥k and let φk∈Γk be defined for all s∈[-ηk,ηk] and for all t∈[0,kT] by
(2.105)(φk(s))(t)=s+(ηk-|s|).
We have (φk(ηk))(t)=ηk and (φk(-ηk))(t)=-ηk for all t∈[0,kT]. (φk(s))(·) is constant with respect to t for all s∈[-ηk,ηk] and φk(s)∈HkT1. Let sk∈[-ηk,ηk] be such that
(2.106)Jk(φk(sk))=max⁡s∈[-ηk,ηk]Jk(φk(s)).
We have
(2.107)Jk(uk)≤Jk(φk(sk)).
Since (φk(sk))/(t)=0 for all t∈[0,kT], (2.54) implies that
(2.108)Jk(φk(sk))=-∫0kT[Frk(t,(φk(sk))(t))-e(t)(φk(sk))(t)]dt.
Extending Frk by T-periodicity, we obtain for k≥2(2.109)1kJk(uk)≤1kJk(φk(sk))=-1k∑k-1j=0∫0kT[Frk(t+jT,(φk(sk))(t+jT))-e¯(φk(sk))(t+jT)]dts=-∫0T[Frk(t,(φk(sk))(t))-e¯(φk(sk))(t)]dt.
We have limk→+∞|(φk(sk))(t)|=+∞ for all t∈[0,kT]. Apply (L4) with u=(φk(sk))(t) to obtain
(2.110)lim⁡k→+∞1kJk(uk)≤-lim⁡|u|→+∞∫0T[Frk(t,u)-e¯u]dt=-∞.
Hence (2.104) holds.
Now, assume by contradiction that we can extract from the sequence(uk)k≥1, of solutions of (2.6) with μ=rk a subsequence whose minimal periods are bounded. Then for this subsequence we can find a common period k0T. The sequence (un)n≥1 of the critical points of Jk0 satisfies(2.111)Jk0(un)=1nJn(un).
Assuming limn→+∞∥un∥∞=+∞ and proceeding as before we arrive at the conclusion (Jk0(un))n is bounded and this contradicts (2.104). This completes the proof of Lemma 2.8.Lemma 2.9.
The amplitudesAk:=(max[0,kT]uk-min[0,kT]uk) of the the solutions uk of equation  (2.6) with μ=rk tend to infinity.Proof.
We have to show thatlimk→+∞Ak=+∞. First, we have
(2.112)lim⁡k→+∞‖uk‖∞=+∞.
Otherwise, we can extract from (uk)k≥1 a subsequence converging to some u* with period n0T, for some n0>0. But, this would contradict Lemma 2.8. Next, we must prove that limk→+∞∥ũk∥∞=+∞. Assume, on the contrary, that ∥ũk∥∞ is bounded. If we suppose further that (|ũk|)k≥1 is bounded, then ∥uk∥∞ would be bounded and this contradicts (2.112). Hence there exists a subsequence of (uk)k≥1, which we label the same, such that limk→+∞|u̅k|=+∞. This implies limk→+∞min[0,kT]|uk(t)|=+∞. Then, for k sufficiently large and each j=0,1,…,k-1, uk(t+jT)>d, uniformly in t∈(0,T). It follows from (L3) that for k large enough frk(t,uk(t+jT))-e¯>0, uniformly in t∈(0,T).  (2.6) with μ=rk and the T-periodicity of frk with respect to t give
(2.113)∫0T[1k∑k-1j=0frk(t,uk(t+jT))-e¯]dt=0.
Hence we can use Fatou’s Lemma to obtain
(2.114)0=∫0Tliminf⁡k→+∞[1k∑k-1i=0frk(t,uk(t+iT))-e¯]dt=∫0Tliminf⁡x→+∞[frk(t,x)-e¯]dt>0.
This is a contradiction. Hence limk→+∞∥ũk∥∞=+∞, and the proof of Lemma 2.9 is complete.From the above auxiliary results we deduce that equation  (2.6) with μ=rk, and consequently (1.1), admits a sequence (uk)k≥1 of distinct kT-periodic solutions whose amplitudes and minimal periods tend to infinity. Thus, the proof of Theorem 1.1 is complete.


## 2.1. Modification of the Problem

Define the truncation functionfr:ℝ2→ℝ,0<r≤1, by(2.1)fr(t,u)={f(t,u),u≥r,f(t,r),u<r.
Note that condition (H1) implies that fr is continuous with respect to (t,u)∈I×ℝ and T-periodic with respect to its first variable t.Lemma 2.1.
Assume (H1) and (H2) (ii) are satisfied. Then there existsd>1 such that for every u∈(0,1/d)∪(d,+∞)(2.2)(f(t,u)-e¯)(u-1)>0,uniformlyint∈I.Proof.
First, it follows from (H1)(ii) that for anyA>0, there is δA>0such that for every u∈(0,δA) we have f(t,u)<-A, uniformly in t∈I.
In particular forA>2|e¯|+1, there is δA>0 such that for every u∈(0,δA) it holds(2.3)f(t,u)-e¯<-|e¯|-1<0.
Choose d1>1 such that 1/d1<δA. Then for every u∈(0,1/d1), we have f(t,u)-e¯<0.
Therefore, condition (H1)(ii) implies that there existsd1>1 such that for every u∈(0,1/d1) it holds (f(t,u)-e¯)(u-1)>0, uniformly in t∈I.
Next, condition (H2)(ii) implies that for anyB>1, sufficiently large, there is χB>0, large enough, such that for every u>χB we have(2.4)∫0T[F(t,u)-e¯u]dt>B.
Hence, there exists d2, sufficiently large such that d2>max(1,χB) and for every u>d2 it holds
(2.5)∫0T[F(t,u)-e¯u]dt>B.
We show that for u>d2 we have f(t,u)-e¯>0, uniformly in t∈I. Assume, on the contrary, that there exists y>d2 for which f(t0,y)-e¯≤0, for some t0∈(0,T). By the continuity of f on I and ((H1)(iii)), there exists σ∈(0,T) such that f(σ,u)=maxt∈If(t,u). If t0≤σ then 0<f(0,y)-e¯<f(t0,y)-e¯≤0, and this is a contradiction. Similarly, if t0≥σ then 0<f(0,y)-e¯=f(T,y)-e¯<f(t0,y)-e¯≤0, and we again arrive at a contradiction.
Hence we deduce that if (H1) and (H2)(ii) hold, then there existsd>max(d1,d2) such that for any u∈(0,1/d)∪(d,+∞)(f(t,u)-e¯)(u-1)>0, uniformly in t∈I. This completes the proof of Lemma 2.1.Lemma 2.2.
For every positive integerk, there exist rk and Rk with 0<rk<1/d<d<Rk such that for any μ∈(0,rk] each kT-periodic solution u of
(2.6)u′′(t)+fμ(t,u(t))=e(t),
satisfies
(2.7)rk≤u(t)≤Rk,∀t∈R.
In particular, any kT-periodic solution of (2.6), with μ=rk is a solution of (1.1).Proof.
This is essentially Proposition  2.1 in [8]. We shall use some ideas from [8] (see also [15]). Fix k∈ℕ, and suppose, on the contrary, that for each integer n, there exist μn∈(0,1/n) and a kT-periodic solution un satisfying
(2.8)un′′(t)+fμn(t,un(t))=e(t)
and {un(t);t∈ℝ}⊈[1/n,n].Claim 1.
Letd be as in Lemma 2.1 and let un be as above. Then for every n there exists τn∈[0,kT] such that un(τn)∈[1/d,d].
Indeed, it follows from (2.8) that(2.9)∫0kTfμn(t,un(t))dt=kTe¯.
Now, if un(t)>d for all t∈[0,kT], then Lemma 2.1 implies that fμn(t,un(t))-e¯>0, which in turn yields ∫0kTfμn(t,un(t))dt>kTe¯.
This contradicts (2.9). On the other hand, if un(t)<1/d for all t∈[0,kT], then fμn(t,un(t))-e¯<0, so that ∫0kTfμn(t,un(t))dt<kTe¯. This is again a contradiction to (2.9).Claim 2.
There existsR>0 such that Mn=maxt∈[0,kT]un(t)≤R for each integer n. To prove the claim notice, there exists tn1∈[0,kT] such that un(tn1)=Mn. If un(tn1)=Mn∈[1/d,d] then un(t)≤R for any R>d. So, assume that there exists a subsequence of (un)n, which we label the same, for which Mn→+∞ when n→+∞. So that Mn>d for n large enough. Since un(τn)<d, there exists an interval [αn,βn], containing tn1, such that
(2.10)βn-αn≤kT,un(αn)=d=un(βn),d≤un(t)≤un(tn1),forallt∈[αn,βn].
Equation (2.8) can be written as
(2.11)un′(t)=vn(t)+∫αnt[e(s)-e¯]ds,vn′(t)=-fμn(t,un(t))+e¯.
Since for allt∈(αn,βn),un(t)>d and μn<1/n, then the second equation in (2.11) is equivalent to(2.12)vn′(t)=-f(t,un(t))+e¯.
Lemma2.1 implies that f(t,un(t))-e¯>0 for all t∈[αn,βn]. Then vn′(t)<0 for all t∈[αn,βn] and hence vn is decreasing on [αn,βn]. The first equation in (2.11) implies(2.13)un′(t)≤vn(αn)+∫αnt[e(s)-e¯]ds∀t∈[αn,βn].
This yields
(2.14)un′(t)≤vn(αn)+‖e‖L1+kT|e¯|∀t∈[αn,βn].
Integrating the above inequality over [αn,tn1]⊂[αn,βn] we obtain
(2.15)un(tn1)-un(αn)=Mn-d≤kT[vn(αn)+‖e‖L1+kT|e¯|].
Equation (2.15) leads to
(2.16)lim⁡n→+∞(1-dMn)≤kTlim⁡n→+∞(vn(αn)+‖e‖L1+kT|e¯|Mn).
Hence
(2.17)1≤kTlim⁡n→+∞vn(αn)Mn.
It is clear from (2.17) that vn(αn)→+∞ when n→+∞. So that, for n large enough, we have
(2.18)vn(αn)≥‖e‖L1+kT|e¯|.
Since Mn=un(tn1), we have un′(tn1)=0. It follows from the first equation in (2.11) that
(2.19)|vn(tn1)|=|∫αntn1[e(s)-e¯]ds|≤‖e‖L1+kT|e¯|.
We see from (2.18) and (2.19) that for n large enough
(2.20)vn(αn)≥‖e‖L1+kT|e¯|≥vn(tn1).
Since vn(·) is continuous on [αn,βn], then, for n large enough, there exists at least one tn*∈(αn,tn1) such that vn(tn*)=∥e∥L1+kT|e¯|. We denote by tn2, the first such tn*. Then
(2.21)vn(tn2)=‖e‖L1+kT|e¯|.
We distinguish two cases.Case 1.
f(·,s) is increasing for each fixed s∈(0,+∞).
Consider the functionB1 defined by(2.22)B1(t)=F(tn2,un(t))-e¯un(t)+12(vn(t)-‖e‖L1-kT|e¯|)2.
Then
(2.23)B1′(t)=(f(tn2,un(t))-e¯)un′(t)+(vn(t)-‖e‖L1-kT|e¯|)vn′(t).
Since vn′(t)=-f(t,un(t))+e¯, it follows from the first equation in (2.11)
(2.24)B1′(t)=(f(tn2,un(t))-e¯)(vn(t)+∫αnt[e(s)-e¯]ds)+(vn(t)-‖e‖L1-kT|e¯|)(-f(t,un(t))+e¯)=(f(tn2,un(t))-e¯)(vn(t)+∫αnt[e(s)-e¯]ds+‖e‖L1+kT|e¯|-‖e‖L1-kT|e¯|)+(vn(t)-‖e‖L1-kT|e¯|)(-f(t,un(t))+e¯).
Hence
(2.25)B1′(t)=(f(tn2,un(t))-e¯)(∫αnt[e(s)-e¯]ds+‖e‖L1+kT|e¯|)+(f(tn2,un(t))-f(t,un(t)))(vn(t)-‖e‖L1-kT|e¯|)⋅
Since, for all t∈[αn,βn](2.26)|∫αnt[e(s)-e¯]ds|≤‖e‖L1+kT|e¯|,
it follows that for all t∈[αn,βn](2.27)∫αnt[e(s)-e¯]ds+‖e‖L1+kT|e¯|≥0.
Also, Lemma 2.1 implies that f(tn2,un(t))-e¯>0 for all t∈[αn,βn]. Furthermore, the monotonicity of f implies that f(tn2,un(t))-f(t,un(t))≥0 for all t∈[αn,tn2]. Since vn(·) is decreasing on [αn,tn2] and vn(tn2)=∥e∥L1+kT|e¯|, it follows that
(2.28)vn(t)-‖e‖L1-kT|e¯|≥vn(tn2)-‖e‖L1-kT|e¯|=0
for all t∈[αn,tn2]. Now, Lemma 2.1 combined with (2.27), (2.28), and the monotonicity of f with respect to its first variable shows that
(2.29)B1′(t)≥0,∀t∈[αn,tn2].
Thus, the function B1 is increasing on [αn,tn2]. Since un(αn)=d,
(2.30)B1(αn)=F(tn2,d)-e¯d+12(vn(αn)-‖e‖L1-kT|e¯|)2≤B1(tn2)=F(tn2,un(tn2))-e¯un(tn2)+12(vn(tn2)-‖e‖L1-kT|e¯|)2.
Since vn(tn2)-∥e∥L1-kT|e¯|=0, it follows that
(2.31)B1(αn)=F(tn2,d)-e¯d≤F(tn2,un(tn2))-e¯un(tn2).
Notice that
(2.32)F(tn2,un(tn2))-e¯un(tn2)=∫1d(f(tn2,s)-e¯)ds+∫dun(tn2)(f(tn2,s)-e¯)ds-e¯.
Also, f(tn2,s)-e¯>0 if s∈(d,un(tn2)]⊂[d,un(tn1)] (see Lemma 2.1). It follows that
(2.33)∫dun(tn2)[f(tn2,s)-e¯]ds≤∫dun(tn1)[f(tn2,s)-e¯]ds.
Hence
(2.34)B1(αn)≤∫1d[f(tn2,s)-e¯]ds+∫dun(tn1)[f(tn2,s)-e¯]ds-e¯=F(tn2,un(tn1))-e¯un(tn1).
Set
(2.35)ρn=12(vn(αn)-‖e‖L1-kT|e¯|Mn)2.
Then
(2.36)lim⁡n→+∞ρn=0.
Indeed, since Mn=un(tn1), we have
(2.37)B1(αn)=F(tn2,d)-e¯d+ρnMn2≤F(tn2,un(tn1))-e¯un(tn1)=F(tn2,Mn)-e¯Mn.
From (2.37) we deduce
(2.38)lim⁡n→+∞F(tn2,d)(Mn)2+lim⁡n→+∞ρn≤lim⁡n→+∞F(tn2,Mn)(Mn)2.
Notice that F(tn2,d)=∫1df(tn2,s)ds is bounded, so that
(2.39)lim⁡n→+∞F(tn2,d)(Mn)2=0.
Also, (H2)(i) implies that
(2.40)lim⁡n→+∞F(tn2,Mn)(Mn)2=0.
Therefore (2.36) holds; that is, limn→+∞ρn=0. Since ∥e∥L1 and |e¯| are bounded, it follows that
(2.41)lim⁡n→+∞vn(αn)Mn=0.
It is clear that (2.41) contradicts (2.17).Case 2.
f(·,s) is decreasing for each fixed s∈(0,+∞).
In this case we consider the functionB2 defined by(2.42)B2(t)=F(αn,un(t))-e¯un(t)+12(vn(t)-‖e‖L1-kT|e¯|)2.
Then
(2.43)B2′(t)=(f(αn,un(t))-e¯)(∫αnt[e(s)-e¯]ds+‖e‖L1+kT|e¯|)+(f(αn,un(t))-f(t,un(t)))(vn(t)-‖e‖L1-kT|e¯|)≥0∀t∈[αn,tn2].
Repeating the same reasoning as in Case 1, we arrive at a contradiction.
Therefore, we deduce that there must existR>0 such that Mn=maxt∈[0,kT]un(t)≤R for each n.
Next, we prove that there isr>0 such that un(t)≥r for every t∈[0,kT]. Assuming that this is not true we will obtain a contradiction.
Consider the following sets(2.44)I1/n={t∈[0,kT];un(t)<1n},I1/n,1/d={t∈[0,kT];1n≤un(t)<1d},I1/d,R={t∈[0,kT];1d≤un(t)≤R}.
It is clear that for n≥R, un(t)≤R≤n. Also, we cannot have un(t)≥1/n for every t∈[0,kT], for otherwise we would have
(2.45)1n≤un(t)≤n∀t∈[0,kT],
which contradicts the assumption {un(t);t∈ℝ}⊈[(1/n),n]. Hence, for n≥R there exists tn3∈[0,kT] such that un(tn3)<1/n. This shows that I1/n≠∅. The continuity of un implies that I1/n is open, and so meas(I1/n)≠0.
Define(2.46)Ψn=∫0kT[fμn(t,un(t))-e¯]dt,n∈N.
It follows from (2.9) that Ψn=0. On the other hand
(2.47)Ψn=∫I1/n[fμn(t,un(t))-e¯]dt+∫I1/n,1/d[fμn(t,un(t))-e¯]dt+∫I1/d,R[fμn(t,un(t))-e¯]dt.
(i)Assume we are integrating positively on all subintervals of[0,kT].
Ift∈I1/n,1/d, then un(t)∈[1/n,1/d)⊂(0,1/d). So that, by Lemma 2.1,(2.48)∫I1/n,1/d[fμn(t,un(t))-e¯]dt<0.
For t∈I1/d,R we have un(t)∈[1/d,R]. This means that un(t) is bounded uniformly in t∈I1/d,R. Since fμn is continuous it is bounded on I1/d,R.
Let(2.49)c=max⁡{|fμn(t,x)|;t∈[0,kT],1d≤x≤R}=max⁡I1/d,R{|fμn(t,x)|}.
Then
(2.50)|∫I1/d,R[fμn(t,un(t))-e¯]dt|≤∫I1/d,R[|fμn(t,un(t))|+|e¯|]dt≤kT(c+|e¯|)
It follows from (2.47), (2.48), and (2.50) that
(2.51)Ψn<∫I1/n[fμn(t,un(t))-e¯]dt+kT(c+|e¯|).Claim 3.
(2.52)lim⁡n→∞⁡∫I1/n[fμn(t,un(t))-e¯]dt=-∞.Proof.
Recall thatμn∈(0,1/n) and un(t)<1/n for each t∈I1/n. Then, if un(t)<μn we have fμn(t,un(t))=f(t,μn), and if un(t)∈[μn,1/n), we have fμn(t,un(t))=f(t,un(t)). In both cases condition (H1)(ii) and the continuity of f imply that limn→∞(fμn(t,un(t))-e¯)=-∞ for every t∈I1/n.
Sincee¯ is bounded, then (2.51) implies that(2.53)lim⁡n→+∞Ψn=-∞,
which is a contradiction with (2.9).
(ii) If we integrate negatively on all subintervals of[0,kT] we will obtain limn→+∞Ψn=+∞, which, again, contradicts (2.9). Thus, the proof of Lemma 2.2 is complete.Remark 2.3.
Lemma2.2 shows that any kT-periodic solution u of (2.6), with μ=rk is a solution of (1.1), since it satisfies u(t)≥rk for all t∈ℝ and frk(t,u(t))=f(t,u(t)).
In the remainder of the paper we shall deal with (2.6), with μ=rk instead of (1.1). Let Frk(t,u)=∫1ufrk(t,s)ds be a primitive of frk defined for all t∈I and u∈ℝ.Lemma 2.4.
If (H1) and (H2) hold, thenfrk and Frk satisfy the following conditions.(L1)
frk is defined and continuous in (t,u)∈I×ℝ and T-periodic with respect to t∈I.(L2)
liminf|u|→+∞2Frk(t,u)/u2=0, uniformly in t∈I.(L3)
∃d>1 such that foru∈(-∞,1/d)∪(d,+∞) it holds (frk(t,u)-e¯)(u-1)>0 uniformly in t∈I.(L4)
lim|u|→+∞∫0T[Frk(t,u)-e¯u]dt=+∞.


## 2.2. Existence ofkT-Periodic Solutions for  (2.6), μ=rk

Using a variational method we shall show that  (2.6), with μ=rk has infinitely many kT-periodic solutions. In fact, we have the following.Lemma 2.5.
Assume thate is locally integrable T-periodic function and the conditions (L1), (L2), (L3), (L4) hold. Then (2.6), with μ=rk admits a kT-periodic solution uk.Proof.
We shall rely on a variational method and more precisely on the saddle point theorem.
Define for eachk≥1 the action functional Jk:HkT1→ℝ by(2.54)Jk(u)=∫0kT[12[u′(t)]2-Frk(t,u(t))+e(t)u(t)]dt.Jk is well defined on HkT1, weakly lower semicontinuous and continuously differentiable on HkT1. Furthermore,
(2.55)〈Jk′(u),v〉=∫0kT[u′(t)v′(t)-frk(t,u(t))v(t)+e(t)v(t)]dt,∀u,v∈HkT1.
The critical points of Jk are precisely the weak solutions of  (2.6), with μ=rk.
First, we show that the functionalJk satisfies the Palais-Smale condition.
For this, letk≥1 be fixed and let (un)n∈ℕ be a sequence in HkT1 such that (Jk(un))n∈ℕ is bounded and limn→+∞Jk′(un)=0. Then (un)n∈ℕ has a convergent subsequence.
Suppose, on the contrary, thatlimn→+∞∥un∥HkT1=+∞. Condition (L2) implies that for any ε>0, small enough, there is Cε≥0 such that(2.56)Frk(t,u)≤εu2+Cε,∀u∈R.
Writing un(t)=un¯+uñ(t) for t∈[0,kT], we obtain
(2.57)-∫0kTFrk(t,un(t))dt≥-εkT|un¯|2-ε∫0kT|uñ(t)|2dt-2εun¯∫0kTuñ(t)dt-CεkT.
Since uñ∈H-, we have ∫0kTuñ(t)dt=kTuñ___=0, so that
(2.58)-∫0kTFrk(t,un(t))dt≥-ε‖uñ‖L22-εkT|un¯|2-CεkT.
Now, Hölder’s inequality gives
(2.59)∫0kTe(t)un(t)dt=un¯∫0kTe(t)dt+∫0kTe(t)uñ(t)dt≥-|un¯|kT|e¯|-‖e‖L2‖uñ‖L2.
Since un′(t)=uñ′(t) for t∈[0,kT], it follows from (2.54), (2.58), and (2.59)
(2.60)Jk(un)≥12‖uñ′‖L22-ε‖uñ‖L22-εkT|un¯|2-CεkT-kT|e-||un¯|-‖e‖L2‖uñ‖L2.
Wirtinger’s inequality
(2.61)‖uñ′‖L22≥(4π24π2+(kT)2)‖uñ‖HkT12
combined with the inequality ∥uñ∥L2≤∥uñ∥HkT1 give
(2.62)Jk(un)+(εkT|un¯|2+kT|e¯||un¯|)+CεkT≥‖uñ‖HkT1([2π24π2+(kT)2-ε]‖uñ‖HkT1-‖e‖L2).
This leads to
(2.63)lim⁡n→+∞|un¯|=+∞.
Indeed, if (2.63) does not hold then there would exist a subsequence of (un¯)n∈ℕ, still denoted the same, which is bounded. Since (Jk(un))n∈ℕ and e¯ are bounded and ε is chosen arbitrarily small, then (2.62) implies that ∥uñ∥HkT1 is bounded. It follows from the inequality(2.64)‖un‖HkT1≤kT|un¯|+‖uñ‖HkT1
that ∥un∥HkT1 is bounded, but this contradicts our assumption limn→+∞∥un∥HkT1=+∞. Therefore, (2.63) holds.
Using Wirtinger’s inequality(2.65)‖uñ‖L2≤kT2π‖uñ′‖L2
in (2.60), we get
(2.66)‖uñ′‖L2|un¯|([12-ε(kT2π)2]‖uñ′‖L2|un¯|-(kT2π)‖e‖L2|un¯|)≤Jk(un)|un¯|2+εkT+kT|e¯||un¯|+CεkT|un¯|2.
It follows from (2.63) that
(2.67)lim⁡n→+∞‖uñ′‖L2|un¯|=0.
Using Sobolev’s inequality we obtain
(2.68)lim⁡n→+∞‖uñ‖∞|un¯|≤kT12lim⁡n→+∞‖uñ′‖L2|un¯|=0.
The identity un(t)=un¯(1+(uñ(t)/un¯)) for all t∈[0,kT] and (2.63) imply that
(2.69)lim⁡n→+∞min⁡t∈[0,kT]|un(t)|=+∞.
Assume that limn→+∞mint∈[0,kT]un(t)=+∞ (the other case can be treated similarly). Then for n large enough, un(t)>d uniformly in t∈[0,kT]. By (L3) we have for all t∈[0,kT](2.70)frk(t,un(t))-e¯>0.
Consequently, for n large enough
(2.71)∫0kT|frk(t,un(t))-e¯|dt=∫0kT[frk(t,un(t))-e¯]dt=|∫0kT[frk(t,un(t))-e¯]dt|.
Since limn→+∞Jk′(un)=0, then for all v∈HkT1 and for n large enough
(2.72)|∫0kT[un′(t)v′(t)-frk(t,un(t))v(t)+e(t)v(t)]dt|≤εn‖v‖HkT1,
where εn>0 for every n, and limn→+∞εn=0. In particular, if we take v(t)=-1 in the above inequality we obtain for every n∈ℕ(2.73)|∫0kT[frk(t,un(t))-e¯]dt|≤εnkT,
which infer
(2.74)∫0kT|frk(t,un(t))|dt≤∫0kT|frk(t,un(t))-e¯|dt+∫0kT|e¯|dt=|∫0kT[frk(t,un(t))-e¯]dt|+∫0kT|e¯|dt≤εnkT+kT|e¯|.
Now, taking v=uñ in (2.72) we obtain
(2.75)εn‖uñ‖HkT1≥‖uñ′‖L22-|∫0kT[frk(t,un(t))-e(t)]uñ(t)dt|.
Obviously, we have for n large enough
(2.76)|∫0kT[frk(t,un(t))-e(t)]uñ(t)dt|≤sup⁡t∈[0,kT]|uñ(t)|(∫0kT|frk(t,un(t))|dt+∫0kT|e(t)|dt)≤‖uñ‖∞(εnkT+kT|e¯|+‖e‖L1).
Thus, for n large enough, (2.75) implies that
(2.77)εn‖uñ‖HkT1≥‖uñ′‖L22-‖uñ‖∞(εnkT+kT|e¯|+‖e‖L1).
Sobolev’s inequality
(2.78)‖uñ‖∞≤kT12‖uñ′‖L2≤kT12‖uñ‖HkT1
and Wirtinger’s inequality combined with (2.77) give, for n large enough,
(2.79)εn‖uñ‖HkT1≥(4π24π2+(kT)2)‖uñ‖HkT12-kT12‖uñ‖HkT1(εnkT+kT|e¯|+‖e‖L1).
So, for n large enough, we deduce that
(2.80)‖uñ‖HkT1≤l:=(4π2+(kT)24π2)(1+kT12[kT+kT|e¯|+‖e‖L1]).
Hence (uñ)n is bounded in HkT1. Consequently ∥un′∥L2=∥uñ′∥L2≤∥uñ∥HkT1≤ℓ. Since (Jk(un))n∈ℕ is bounded, it follows that
(2.81)∫0kT[Frk(t,un(t))-e(t)un(t)]dt
is bounded. Hölder’s inequality gives
(2.82)|∫0kTe(t)uñ(t)dt|≤‖e‖L2‖uñ‖L2≤‖e‖L2‖uñ‖HkT1≤l‖e‖L2.
Since
(2.83)∫0kTFrk(t,un(t))dt-un-∫0kTe(t)dt=∫0kT[Frk(t,un(t))-e(t)un(t)]dt+∫0kTe(t)uñ(t)dt,
it follows that
(2.84)∫0kTFrk(t,un(t))dt-un-∫0kTe(t)dt,
is bounded. But,
(2.85)un-∫0kTe(t)dt=kTe¯un-=kTe¯1kT∫0kTun(t)dt=∫0kTe¯un(t)dt.
Consequently, there exists C>0 such that
(2.86)∫0kT[Frk(t,un(t))-e¯un(t)]dt≤C.
On the other hand, extending Frk by T-periodicity we obtain
(2.87)∫0kT[Frk(t,un(t))-e¯un(t)]dt=∑k-1j=0∫0T[Frk(t+jT,un(t+jT))-e¯un(t+jT)]dt=∑k-1j=0∫0T[Frk(t,un(t+jT))-e¯un(t+jT)]dt.
Setting xn=un(t+jT) for t∈[0,T], we get
(2.88)∫0kT[Frk(t,un(t))-e¯un(t)]dt=∑k-1j=0∫0T[Frk(t,xn)-e¯xn]dt=k∫0T[Frk(t,xn)-e¯xn]dt.
From (2.75) |xn|=|un(t+jT)|→+∞ when n→+∞ uniformly in t∈[0,kT]. By (L4) we have
(2.89)lim⁡n→+∞∫0kT[Frk(t,un(t))-e¯un(t)]dt=klim⁡|x|→+∞∫0T[Frk(t,x)-e¯x]dt=+∞.
This is a clear contradiction to (2.86). Therefore, (un)n∈ℕ is bounded in HkT1, and so it has a convergent subsequence. This shows that Jk satisfies the Palais-Smale condition. Next, we show that Jk has a geometry of a Saddle. For, let u∈H- then we have u=uñ and u¯=0, so that
(2.90)Jk(uñ)=12‖uñ′‖L22-∫0kT[Frk(t,uñ(t))-e(t)uñ(t)]dt.
Proceeding as before, we get an inequality similar to (2.72) by replacing un by uñ and u¯nby 0,
(2.91)Jk(uñ)≥‖uñ‖HkT1([2π24π2+(kT)2-ε]‖uñ‖HkT1-‖e‖L2)-CεkT.
Since ε is chosen arbitrary small, we obtain
(2.92)lim⁡‖uñ‖HkT1→+∞Jk(uñ)=+∞,
which shows that Jk is coercive. Hence, Jk admits a bounded minimizing sequence. Furthermore, Jk is weakly lower semicontinuous on HkT1, then
(2.93)inf⁡H-Jk>-∞.
For s, a constant function, we have ∥s∥HkT1=|s|kT→+∞ if and only if |s|→+∞. Then
(2.94)Jk(s)=-∫0kT[Frk(t,s)-e(t)s]dt.
Extending Frk by T-periodicity we obtain
(2.95)Jk(s)=-∫0kT[Frk(t,s)-e¯s]dt=-∑k-1i=0∫0T[Frk(t+iT,s)-e¯s]dts=-k∫0T[Frk(t,s)-e¯s]dt.
Condition (L4) implies that
(2.96)lim⁡|s|→+∞[-Jk(s)]=klim⁡|s|→+∞∫0T[Frk(t,s)-e¯s]dt=+∞.
Hence, for each k≥1,(-Jk) is coercive on the space of constant functions. Then for each k≥1, there exists ηk>0, large enough, such that
(2.97)Jk(ηk)→-∞,andJk(-ηk)→-∞.
Thus,
(2.98)max⁡(Jk(-ηk),Jk(ηk))→-∞whenηk→+∞.
Therefore,
(2.99)max⁡(Jk(-ηk),Jk(ηk))<inf⁡H-Jk.
Let Iηk be the open interval of ℝ centered at 0 and with radius ηk. Since H+=ℝ it is clear that
(2.100)∂Iηk∩H+={-ηk,ηk}.
Therefore, we have
(2.101)max⁡∂Iηk∩H+Jk<inf⁡H-Jk,
with H+⊕H-=HkT1. Thus Jk has a geometry of Saddle.
Finally, all conditions of the Saddle point theorem are satisfied. Then for eachk≥1,Jk admits a critical point βk, which is characterized by(2.102)Jk(βk)=inf⁡ψ∈Γkmax⁡s∈[-ηk,ηk]Jk(ψ(s)),
where
(2.103)Γk={ψ∈C([-ηk,ηk],HkT1);ψ(-ηk)=-ηk,ψ(ηk)=ηk}.
Thus for each k≥1,βk is a weak kT-periodic solution of  (2.6) with μ=rk If, furthermore, e is assumed continuous, then βk is a classical solution of  (2.6) with μ=rk.
This completes the proof of Lemma2.5.Remark 2.6.
As a consequence of Lemmas2.2, 2.5 and the Remark 2.3, we conclude that if (H1) and (H2) are satisfied, then (1.1) admits a sequence (uk)k≥1 of kT-periodic solutions.


## 2.3. Existence of Distinct Subharmonic Solutions for  (2.6) with  μ=rk

Note thatJk=Jm on HkT1∩HmT1 for each k and m. This justifies the following definition (see [10, Definition 2.1 page 653]).Definition 2.7.
The level ofu∈∪kHkT1 is defined by Jm(u) when u∈HmT1.
Every functionalJk admits at least a critical level which is given by βk=minHkT1Jk.Note that nondistinct subharmonic solutions have the same level. Then we deduce that in order to find the multiplicity of distinct subharmonic solutions, we have to search the multiple critical levels. The sequence(βk)k≥1 is not always increasing, which means that there exists m∈ℕ such that if k∈mℕ then βk≤βm.Ifβk<β1 for an integer k, then βk is not a level for the T-periodic functions and every global minimum of Jk is in fact a subharmonic solution of (2.6).In our caseJk(uk) is not necessary a global minimum for Jk and then even if the condition above is verified, it is still insufficient to deduce the existence of true subharmonic solutions. This is why we prove also that the amplitudes and the minimal periods tend to infinity.Lemma 2.8.
The minimal periods of the solutionsuk of (2.6), with μ=rk tend to infinity.Proof.
Letuk be a weak solution of (2.6) with μ=rk  Then uk is a critical point of Jk. We show that
(2.104)lim⁡k→+∞1kJk(uk)=-∞.
Let ηk≥k and let φk∈Γk be defined for all s∈[-ηk,ηk] and for all t∈[0,kT] by
(2.105)(φk(s))(t)=s+(ηk-|s|).
We have (φk(ηk))(t)=ηk and (φk(-ηk))(t)=-ηk for all t∈[0,kT]. (φk(s))(·) is constant with respect to t for all s∈[-ηk,ηk] and φk(s)∈HkT1. Let sk∈[-ηk,ηk] be such that
(2.106)Jk(φk(sk))=max⁡s∈[-ηk,ηk]Jk(φk(s)).
We have
(2.107)Jk(uk)≤Jk(φk(sk)).
Since (φk(sk))/(t)=0 for all t∈[0,kT], (2.54) implies that
(2.108)Jk(φk(sk))=-∫0kT[Frk(t,(φk(sk))(t))-e(t)(φk(sk))(t)]dt.
Extending Frk by T-periodicity, we obtain for k≥2(2.109)1kJk(uk)≤1kJk(φk(sk))=-1k∑k-1j=0∫0kT[Frk(t+jT,(φk(sk))(t+jT))-e¯(φk(sk))(t+jT)]dts=-∫0T[Frk(t,(φk(sk))(t))-e¯(φk(sk))(t)]dt.
We have limk→+∞|(φk(sk))(t)|=+∞ for all t∈[0,kT]. Apply (L4) with u=(φk(sk))(t) to obtain
(2.110)lim⁡k→+∞1kJk(uk)≤-lim⁡|u|→+∞∫0T[Frk(t,u)-e¯u]dt=-∞.
Hence (2.104) holds.
Now, assume by contradiction that we can extract from the sequence(uk)k≥1, of solutions of (2.6) with μ=rk a subsequence whose minimal periods are bounded. Then for this subsequence we can find a common period k0T. The sequence (un)n≥1 of the critical points of Jk0 satisfies(2.111)Jk0(un)=1nJn(un).
Assuming limn→+∞∥un∥∞=+∞ and proceeding as before we arrive at the conclusion (Jk0(un))n is bounded and this contradicts (2.104). This completes the proof of Lemma 2.8.Lemma 2.9.
The amplitudesAk:=(max[0,kT]uk-min[0,kT]uk) of the the solutions uk of equation  (2.6) with μ=rk tend to infinity.Proof.
We have to show thatlimk→+∞Ak=+∞. First, we have
(2.112)lim⁡k→+∞‖uk‖∞=+∞.
Otherwise, we can extract from (uk)k≥1 a subsequence converging to some u* with period n0T, for some n0>0. But, this would contradict Lemma 2.8. Next, we must prove that limk→+∞∥ũk∥∞=+∞. Assume, on the contrary, that ∥ũk∥∞ is bounded. If we suppose further that (|ũk|)k≥1 is bounded, then ∥uk∥∞ would be bounded and this contradicts (2.112). Hence there exists a subsequence of (uk)k≥1, which we label the same, such that limk→+∞|u̅k|=+∞. This implies limk→+∞min[0,kT]|uk(t)|=+∞. Then, for k sufficiently large and each j=0,1,…,k-1, uk(t+jT)>d, uniformly in t∈(0,T). It follows from (L3) that for k large enough frk(t,uk(t+jT))-e¯>0, uniformly in t∈(0,T).  (2.6) with μ=rk and the T-periodicity of frk with respect to t give
(2.113)∫0T[1k∑k-1j=0frk(t,uk(t+jT))-e¯]dt=0.
Hence we can use Fatou’s Lemma to obtain
(2.114)0=∫0Tliminf⁡k→+∞[1k∑k-1i=0frk(t,uk(t+iT))-e¯]dt=∫0Tliminf⁡x→+∞[frk(t,x)-e¯]dt>0.
This is a contradiction. Hence limk→+∞∥ũk∥∞=+∞, and the proof of Lemma 2.9 is complete.From the above auxiliary results we deduce that equation  (2.6) with μ=rk, and consequently (1.1), admits a sequence (uk)k≥1 of distinct kT-periodic solutions whose amplitudes and minimal periods tend to infinity. Thus, the proof of Theorem 1.1 is complete.



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*Source: 903281-2012-03-29.xml*