NONAUTONOMOUS DYNAMICAL SYSTEMS
MARTIN SCHECHTERReceived 13 March 2006; Revised 10 May 2006; Accepted 15 May 2006
We study the existence of periodic solutions for second-order nonautonomous dynamical systems. We give four sets of hypotheses which guarantee the existence of solutions. We were able to weaken the hypotheses considerably from those used previously for such systems. We employ a new saddle point theorem using linking methods.
Copyright © 2006 Martin Schechter. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
We consider the following problem. One wishes to solve
−x(t)= ∇xVt,x(t), (1.1)
where
x(t)=x1(t),...,xn(t)
(1.2)
is a map fromI=[0,T] toRnsuch that each componentx
j(t) is a periodic function in H1with periodT, and the functionV(t,x)=V(t,x
1,...,xn) is continuous fromRn+1to
Rwith
∇xV(t,x)=
∂V ∂x1,...,
∂V ∂xn
∈CRn+1,Rn. (1.3)
Here H1 represents the Hilbert space of periodic functions in L2(I) with generalized derivatives inL2(I). The scalar product is given by
(u,v)H1=(u,v) + (u,v). (1.4)
For eachx∈Rn, the functionV(t,x) is periodic intwith periodT.
Hindawi Publishing Corporation Boundary Value Problems
We will study this problem under the following assumptions: (1)
V(t,x)≥0, t∈I,x∈Rn; (1.5)
(2) there are constantsm >0,α≤6m2/T2such that
V(t,x)≤α, |x| ≤m,t∈I,x∈Rn; (1.6)
(3) there is a constantμ >2 such that Hμ(t,x)
|x|2 ≤W(t)∈L 1
(I), |x| ≥C,t∈I,x∈Rn, (1.7)
lim sup
|x|→∞
Hμ(t,x)
|x|2 ≤0, (1.8)
where
Hμ(t,x)=μV(t,x)− ∇xV(t,x)·x; (1.9)
(4) there is a subsete⊂Iof positive measure such that
lim inf
|x|→∞
V(t,x)
|x|2 >0, t∈e. (1.10)
We have the following theorem.
Theorem1.1. Under the above hypotheses, the system (1.1) has a solution. As a variant ofTheorem 1.1, we have the following one.
Theorem1.2. The conclusion inTheorem 1.1is the same if Hypothesis (2) is replaced by (2) there is a constantq >2such that
V(t,x)≤C|x|q+ 1, t∈I,x∈Rn, (1.11)
and there are constantsm >0,α <2π2/T2such that
V(t,x)≤α|x|2, |x| ≤m,t∈I,x∈Rn. (1.12)
We also have the following theorem.
Theorem1.3. The conclusions of Theorems1.1and1.2hold if Hypothesis (3) is replaced by (3) there is a constantμ <2such that
Hμ(t,x)
|x|2 ≥ −W(t)∈L
1(I), |x| ≥C,t∈I,x∈Rn,
lim inf
|x|→∞
Hμ(t,x) |x|2 ≥0.
And we have the following theorem.
Theorem1.4. The conclusion ofTheorem 1.1holds if Hypothesis (1) is replaced by (1)
0≤V(t,x)≤C|x|2+ 1, t∈I,x∈Rn (1.14)
and Hypothesis (3) by (3) the function given by
H(t,x)=2V(t,x)− ∇xV(t,x)·x (1.15)
satisfies
H(t,x)≤W(t)∈L1(I), |x| ≥C,t∈I,x∈Rn,
H(t,x)−→ −∞, |x| −→ ∞,t∈I,x∈Rn. (1.16)
The periodic nonautonomous problem
x(t)= ∇xVt,x(t) (1.17)
has an extensive history in the case of singular systems (cf., e.g., Ambrosetti-Coti Zelati [1]). The first to consider it for potentials satisfying (1.3) were Berger and Schechter [3]. We proved the existence of solutions to (1.17) under the condition that
V(t,x)−→ ∞ as |x| −→ ∞ (1.18)
uniformly for a.e.t∈I. Subsequently, Willem [16], Mawhin [6], Mawhin and Willem [8], Tang [11,12], Tang and Wu [13–15], Wu and Tang [17] and others proved existence under various conditions (cf. the references given in these publications).
The periodic problem (1.1) was studied by Mawhin and Willem [7,8], Long [5], Tang and Wu [13–15] and others (cf. the refernces quoted in them). Ben-Naoum et al. [2] and Nirenberg (cf. Ekeland and Ghoussoub [4]) proved the existence of nonconstant solutions.
We will prove Theorems1.1–1.4in the next section. We use a linking method of critical point theory (cf. [9,10]). These methods allow us to improve the previous results.
2. Proofs of the theorems
We now give the proof ofTheorem 1.1.
Proof. LetXbe the set of vector functionsx(t) given by (1.2) and described above. It is a Hilbert space with norm satisfying
x2
X= n
j=1 xj2
We also write
x2=n
j=1 xj2
, (2.2)
where · is theL2(I) norm. Let
N=x(t)∈X:xj(t)≡ constant, 1≤j≤n (2.3)
andM=N⊥. The dimension ofN isn, andX=M⊕N. Proof of the following lemma can be found in [7].
Lemma2.1. Ifx∈M,then
x2
∞≤12Tx2, x ≤2πT x. (2.4)
We define
G(x)= x2−2
IV
t,x(t)dt, x∈X. (2.5)
For eachx∈X writex=v+w, wherev∈N,w∈M. For convenience, we will use the following equivalent norm forX:
x2
X= w2+v2. (2.6)
Ifx∈Mand
x2=ρ2=12 Tm
2, (2.7)
thenLemma 2.1implies thatx∞≤m, and we have by Hypothesis (2) thatV(t,x)≤α.
Hence,
G(x)≥ x2−2
|x(t)|<mαdt≥ρ
2−
2αT≥0. (2.8)
We also note that Hypothesis (1) implies
G(v)≤0, v∈N. (2.9)
Take
A=∂Bρ∩M, ρ2=12 Tm
2,
B=N, (2.10)
where
Bσ=
By [9, Theorem 1.1],AlinksB. (For background material on linking theory, cf. [10].) Moreover, by (2.8) and (2.9), we have
sup A
[−G]≤0≤inf
B [−G]. (2.12)
Hence, we may apply [9, Theorem 1.1] to conclude that there is a sequence{x(k)} ⊂X such that
Gx(k)=x(k)2−2
IV
t,x(k)(t)dt−→c≤0, (2.13)
Gx(k),z
2 =
x(k),z−
I∇xV
t,x(k)(t)·z(t)dt−→0, z∈X, (2.14)
Gx(k),x(k)
2 =
x(k)2−
I∇xV
t,x(k)(t)·x(k)(t)dt−→0. (2.15)
If
ρk=x(k)X≤C, (2.16)
then there is a renamed subsequence such thatx(k)converges to a limitx∈Xweakly in Xand uniformly onI. From (2.14) we see that
G(x),z
2 =(x
,z)−
I∇xV
t,x(t)·z(t)dt=0, z∈X, (2.17)
from which we conclude easily thatxis a solution of (1.1). If
ρk=x(k)X−→ ∞, (2.18)
letx(k)=x(k)/ρ
k. Then,x(k)X=1. Letx(k)=w(k)+v(k), wherew(k)∈Mandv(k)∈N. There is a renamed subsequence such that[x(k)] →randx(k) →τ, wherer2+τ2=1. From (2.13) and (2.15) we obtain
x(k)2−2
IV
t,x(k)(t)dt ρ2
k
−→0,
x(k)2−
I∇xV
t,x(k)(t)·x(k)(t)dt ρ2k −→0.
(2.19)
Thus,
2IVt,x(k)(t)dt ρ2k −→r
2, (2.20)
I∇xV
t,x(k)(t)·x(k)(t)dt ρ2
k
Hence, by (1.9),
IHμ
t,x(k)(t)dt ρ2
k
−→
μ 2−1
r2. (2.22)
Note that
x(k)(t)≤Cx(k)X=C. (2.23)
If
x(k)
(t)−→ ∞, (2.24)
then by (1.8)
lim supHμ
t,x(k)(t) ρ2
k
=lim supHμ
t,x(k)(t) x(k)(t)2 x
(k)(t)2≤0. (2.25)
If
x(k)(t)≤C, (2.26)
then
Hμt,x(k)(t) ρ2
k
−→0. (2.27)
Hence,
lim sup
IHμt,x(k)(t)dt ρ2
k
≤0. (2.28)
Hence by (2.22)
μ 2−1
r2≤0. (2.29)
Ifr=0, this contradicts the fact thatμ >2. Ifr=0, then w(k)→0 uniformly inI by Lemma 2.1. Moreover,T|v(k)|2= v(k)2→1. Hence, there is a renamed subsequence such thatv(k)→vinN with|v|2=1/T. Hence,x(k)→vuniformly inI. Consequently,
|x(k)| → ∞uniformly inI. Thus, by Hypothesis (4),
lim inf
IV
t,x(k)(t)dt ρ2
k
≥
elim inf
Vt,x(k)(t) x(k)(t)2 x
(k)(t)2dt >0. (2.30)
The proof ofTheorem 1.2is similar to that ofTheorem 1.1with the exception of the inequality (2.8) resulting from Hypothesis (2). In its place we reason as follows: ifx∈M, we have by Hypothesis (2),
G(x)≥ x2−2
|x|<mα
x(t)2 dt−2C
|x(t)|>m
x(t)q+ 1 dt
≥ x2−2αx2−2C1 +m−q
|x(t)|>m
x(t)q dt
≥ x21−2αT2 4π2
−C
|x(t)|>m
x(t)q dt ≥ 1− αT2 2π2
x2
X−C
Ix q Xdt ≥ 1− αT2 2π2
x2
X−CxqX
= 1− αT2 2π2
−CxqX−2
x2
X
(2.31)
byLemma 2.1. Hence, we have the following lemma. Lemma2.2.
G(x)≥εx2
X, xX≤ρ,x∈M (2.32)
forρ >0sufficiently small, whereε <1−[αT2/2π2]. The remainder of the proof is essentially the same.
In provingTheorem 1.3we follow the proof of Theorem 1.1until we reach (2.20). Then we reason as follows. If
x(k)(t)−→ ∞, (2.33)
then
lim infHμ
t,x(k)(t)
ρ2k =lim inf Hμ
t,x(k)(t) x(k)(t)2 x
(k)(t)2≥0. (2.34)
If
x(k)
(t)≤C, (2.35)
then by Hypothesis (3),
Hμt,x(k)(t) ρ2
k
−→0. (2.36)
Hence,
lim inf
IHμt,x(k)(t)dt ρ2
k
Thus by (2.22)
μ 2−1
r2≥0. (2.38)
Ifr=0, this contradicts the fact thatμ <2. Ifr=0, then w(k)→0 uniformly inI by Lemma 2.1. Moreover,T|v(k)|2= v(k)2→1. Hence, there is a renamed subsequence such thatv(k)→vinN with|v|2=1/T. Hence,x(k)→vuniformly inI. Consequently,
|x(k)| → ∞uniformly inI. Thus, by Hypothesis (4),
lim inf
IV
t,x(k)(t)dt ρ2k ≥
elim inf
Vt,x(k)(t) x(k)(t)2 x
(k)(t)2dt >0. (2.39)
This contradicts (2.20). Hence theρkare bounded, and the proof is complete.
In proving Theorem 1.4, we follow the proof ofTheorem 1.1until (2.20). Assume first thatr >0. Note that (2.13) and (2.15) imply that
IH
t,x(k)(t)dt−→ −c. (2.40)
On the other hand, by Hypothesis (1), we have
0←−x(k)2−2
I
Vt,x(k)(t)dt ρ2
k
≥x(k)2−2C
Ix
(k)(t)2+ρ−2
k
dt
−→r2−2C
Ix(t)
2 dt.
(2.41)
Hence,x(t)≡0. LetΩ0⊂Ibe the set on whichx(t) =0.The measure ofΩ0is positive. Moreover,|x(k)(t)| → ∞ask→ ∞fort∈Ω
0. Thus,
IH
t,x(k)(t)dt≤
Ω0
Ht,x(k)(t)dt+
I\Ω0
W(t)dt−→ −∞ (2.42)
by Hypothesis (3). But this contradicts (2.40). Ifr=0, thenw(k)→0 uniformly inI byLemma 2.1. Moreover,T|v(k)|2= v(k)2→1. Thus, there is a renamed subsequence such thatv(k)→vinNwith|v|2=1/T. Hence,x(k)(t)→vuniformly inI. Consequently,
|x(k)(t)| → ∞uniformly inI. Thus, by Hypothesis (4),
lim inf
IV
t,x(k)(t)dt ρ2k ≥
elim inf
Vt,x(k)(t) x(k)(t)2 x
(k)
(t)2dt >0. (2.43)
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Martin Schechter: Department of Mathematics, University of California, Irvine, CA 92697-3875, USA