Boundedness of solutions for a class of impact oscillators with time-denpendent polynomial potentials

0.17 MB·English

by  Daxiong Piao

Boundedness of solutions for a class of impact oscillators with
3
time-denpendent polynomial potentials
1
0
2
Daxiong Piao∗, Xiang Sun†
n
a School of Mathematical Sciences, Ocean University of China
J Qingdao 266100, P.R. China

8
2

]
S Abstract
D
In this paper, we consider the boundedness of solutions for a class of impact oscillators
.
h with time dependent polynomial potentials,
t
a
2n
m
 x¨+x2n+1+ pi(t)xi =0, for x(t)>0,
[
 x(t)≥0, Xi=0
1
48v where n∈N+, pi(t+1)x˙=(t+0pi)(t=) −anx˙d(tp−0i)(,t)∈C5(R/Z). if x(t0)=0,
4
6 Keywords: Impactor oscillators;Boundedness of solutions; Canonicaltransformation;Time-
. dependent polynomial potentials; Moser’s small twist theorem.
1
0
3
1
:
v
i
X 1 Introduction and main result
r
a
InRef. [1],DieckerhoffandZehnderprovedtheboundednessofsolutionsforatime-dependent
nonlinear differential equation:
2n
x¨+x2n+1+ p (t)xi = 0, n ≥ 1, (1.1)
i
Xi=0
with p (t+1) = p (t) and p (t) ∈ C∞ and asked whether or not the boundedness phenomenon
i i i
is related to the smoothness of p (t). Then Laederich and Levi [2] relaxed the smoothness re-
i
quirement of p (t) to C5+ε with ε > 0, Yuan [3, 4] relaxed the smoothness requirement of p (t).
i i
to C2, etc.
∗Supported bythe NSFof ShangdongProvince(No.ZR2012AM018), E-mail: [email protected]
†E-mail: [email protected]
1
Boundednessfor impact oscillators 2
Motivatedby[1,2,3,4],wearegoingtoconsidertheproblemoftheboundednessforEq.(1.1)
with impacts:
2n
x¨+x2n+1+ p (t)xi = 0, for x(t) > 0,
 i
 Xi=0 (1.2)

 x(t) ≥ 0,
x˙(t+) = −x˙(t−), if x(t ) = 0,
 0 0 0


where n ∈N+, p (t+1) = p (t) and p (t)∈ C5(R/Z).
i i i
The system is included in the system of impact oscillators given by
x¨+V (x,t) = 0, for x(t) > 0,
x
 x(t)≥ 0, (1.3)
 x˙(t+)= −x˙(t−), if x(t ) =0,
0 0 0

which serve as models of dynamical systems with discontinuities [5]. From the viewpoint of
mechanics, this equation models the motion of particle attached to a nonlinear spring and
bouncingelastically against the fixed barrier. Thesystem of this form also relate to the research
of the Fermi accelerator [6], dual billiards [7] and celestial mechanics [8].
The nonsmoothness caused by the impact limit the applications of many powerful mathe-
matical tools. However there are also many interesting papers on the impact oscillators, see
[9]-[16] and their references.
To overcome the problem of nonsmoothness, Qian and Torres [9]and Qian and Sun[10]used
success mappingin their papersandproved theexistence of invariant tori by avariant version of
Moser’s small twist theorem by Ortega [17]. In Zharnitsky [11], Wang [12, 13], they exchanged
the position of angle and time variables of the Hamiltonian function which only C0 in angle
variable, and finally obtained the boundedness of solutions by Moser’s small twist theorem.
In [12], Z.Wang and Y.Wang studied the following impact oscillator:
x¨+x2n+1 = p(t), for x(t) >0,
 x(t) ≥ 0, (1.4)
 x˙(t+) = −x˙(t−), if x(t )= 0,
0 0 0

where n ∈ N+, p(t) is a C5 periodic function with 1, and the term x2n+1 models a hard spring.
They proved the Lagrangian stability for system (1.4) by exchanging the roles of angle and time
variables and using Moser’s small twist theorem.
In this paper we will extend (1.4) to the case of time-periodic polynomial potentials with
constantleadingcoefficientas(1.2)andusesimilarmethodtoprovetheboundednessofsolutions
for (1.2). More exactly, we obtain the following conclusion:
Theorem 1 Every solution of (1.2) is bounded, i.e., x(t) exits for t ∈ R and
sup(|x(t)|+|x˙(t)|) < +∞.
Boundednessfor impact oscillators 3
The idea of proving the boundedness of solutions of (1.2) is as follows. First of all, we
describe (1.2) by a Hamiltonian function H (ρ,φ,t) (see (2.6)) in action-angle variables defined
3
on the whole space R+ × S1 × S1 by means of transformation theory. Due to the existence
of impact, H (ρ,φ,t) is only continuous in φ, but is C5 smooth in ρ and t. Exchanging the
3
roles of the variables φ and t outside of a large disk D = {(ρ,φ),ρ < r} in (ρ,φ)-plane, (2.9) is
r
transformedintoaperturbationofanintegrableHamiltoniansystemH (I,θ,τ)(see(3.3))which
4
is sufficiently smooth in I and θ. ThePoincar´e mappingwith respect to the new time τ is closed
to a so-called twist mapping in (R+ ×S1)\D , and satisfies Moser’s invariant curve theorem
r
after a scaling transformation. Then Moser’s theorem guarantees the existence of arbitrarily
large invariant curves diffeomorphic to υ = const. over υ = 0 in (υ,θ)-plane. Go back to the
equivalent system (2.1), every such curve is a base of a time-periodic and flow-invariant cylinder
intheextendedphasespace(x,y,t) ∈R+×R×R,whichconfinesthesolutionsintheinteriorand
which leads to a bound of these solutions if the uniqueness of initial value problem holds. Note
that, from the proof below, system (2.1) is equivalent to a smooth system (2.9) in (R+×S1)\D
r
which has uniqueness. Hence system (2.1) has uniqueness.
Remark 1.1 The time-dependence of the stiffness coefficient p (t) of the spring can be produced
i
by periodic changes of the temperature or other physical variables. Following Laederich and Levi
[2] and Z.Wang and Y. Wang [12], we assume p (t) ∈C5.
i
Remark 1.2 Compared with (1.4), the potentials of (1.1) become more complicated which will
make more difficulties in the proof of Theorem 1.
Remark 1.3 The main difficulty in this paper is also the nonsmoothness caused by the impact.
Similar to [12], we will also exchange the roles of angle and time variables of the Hamiltonian
function H (ρ,φ,t) (see below), and then obtain the boundedenes of solutions by Moser’s small
3
twist theorem.
The rest of this paper is organized as follows. In section 2, we will make some action-angle
transformations to transform the system into an equivalent Hamiltonian H (I,θ,τ) (see below)
4
which is defined in the whole plane R+ ×S1 ×S1 and C5 in I,θ, but only C0 in τ. In section
3, we do more canonical transformations such that the Poincar´e mapping of the new system is
closed to the twist mapping and then use Moser’s theorem to complete the proof of Theorem 1.
Finally, we will give the proofs of some technical lemmas in section 4 and draw a conclusion in
section 5.
Throughout this paper, we denote A < .B if A < C B, and A > .B if A > C B , where
1 2
C ,C are positive constants.
1 2
2 Action-angle variables
Inthis section, wewill make someaction-angle transformations totransform thesystem into
anequivalentHamiltonianH (I,θ,τ)(seebelow)whichisdefinedinthewholeplaneR+×S1×S1
4
and C5 in I,θ, but only C0 in τ.
Boundednessfor impact oscillators 4
Without impact, Eq.(1.2) is just Eq.(1.1) which is of second order and equivalent to the one
order system
x˙ = y,
 2n (2.1)
 y˙ = −x2n+1− p (t)xi.
 i
Xi=0


(2.1) is a system defined in the whole phase plane XOY, which has a Hamiltonian function
1 1 2n p (t)xi+1
H(x,y,t) = y2+ x2n+2+ i (2.2)
2 2n+2 i+1
Xi=0
with the symplectic form dx∧dy.
In order to make an action-angle transformation, we consider the system
x˙ = y,
(2.3)
(cid:26) y˙ = −x2n+1.
with the Hamiltonian H (x,y,t) = 1y2+ 1 x2n+2.
0 2 2n+2
Clearly, H > 0 on R2 except at the only equilibrium point (x,y) = (0,0) where H = 0.
0 0
Let(C(t),S(t))isthesolutionofEq.(2.3)satisfyingtheinitialcondition:(C(0),S(0)) = (1,0).
Let T > 0 be its minimal period.
0
From Eq.(2.3), we can find that C(t) and S(t) satisfy:
(1) C(t), S(t) ∈ C∞(R), and C(t+T )= C(t), S(t+T ) = S(t) with C(0) = 1, S(0) = 0;
0 0
(2) C˙(t) = S(t) and S˙(t) = −C(t)2n+1;
(3) (n+1)S(t)2 +C(t)2n+1 = 1;
(4) C(−t)= C(t) and S(−t) = −S(t).
Now we can make coordinate transformation by the mapping Ψ : R+×S1 → R2/{0}, where
1
(x,y) = Ψ (λ,ϑ) defined by the formula
1
x = (aλ)αC(ϑT ), y = (aλ)αS(ϑT ),
0 0
where α = 1 , β = 1 − α, a = T0 are constants. We can claim that Ψ is a symplectic
n+2 α 1
diffeomorphism from R+×S1 onto R2/{0}. Indeed, the Jacobian determinant of Ψ is 1, so Ψ
1 1
is measure preserving. Moreover, since (C(t),S(t)) is a solution of Eq.(2.3) and has T as its
0
minimal period, which concludes that Ψ is one to one and onto. This proves the claim.
1
Under Ψ , Hamiltonian H is transformed into
1
a2β 2n p (t)(aλ)α(i+1)[C(ϑT )]i+1
H (λ,ϑ,t) = λ2β + i 0 , (2.4)
1
2n+2 i+1
Xi=0
which is C5 in λ, t and C0 in ϑ.
Boundednessfor impact oscillators 5
Considering the impact case, the phase space is only a half plane (x ≥ 0) of the original
phaseplane XOY. Since impact, the smoothness with variable ϑ of H is disappeared. We shall
1
study another Hamiltonian H (ρ,φ,t) as defined below. Let C(ϑT ), we will find two solutions
3 0
ϑ = ϑ ,ϑ in interval [0,1). By the symmetry of H , we can easily find ϑ = 0,ϑ = 1.
1 2 0 1 2 2
Under Ψ , (λ,0(mod 1)) is mapped into the positive Y-axis, (λ, 1(mod 1)) is mapped into
1 2
the negative Y-axis.
Define a sympletic transformation Ψ :(φ,ρ) 7→ (ϑ,λ), given by ϑ = 1, λ = 2ρ. And H is
2 2 1
transformed into
2n p (t)(2aρ)α(i+1)[C(1φT )]i+1
H (ρ,φ,t) = dρ2β + i 2 0 , (2.5)
2
i+1
Xi=0
(2a)2β
where d= .
2n+2
To get a system which is equivalent to (1.2), we could define a new Hamiltonian function by
2n p (t)(2aρ)α(i+1){C(1(φ−[φ])T )}i+1
H (ρ,φ,t) = H (ρ,φ−[φ],t) = dρ2β + i 2 0 , (2.6)
3 2
i+1
Xi=0
where[φ]denotesthelargestinteger lessthanorequaltoφ. AndthecorrespondingHamiltonian
system is
ρ˙ = ∂H3, when φ(t) ∈ (k,k+1),k ∈ Z,
∂φ
 φ˙ = −∂H3, when φ(t) ∈ (k,k+1),k ∈ Z, (2.7)
∂ρ

 φ(t ) = k, ρ(t ) = lim ρ(t), when lim φ(t) = k,k ∈ Z.
0 0
t→t0 t→t0

Obviously, His periodic in φ with 1, and C5 in ρ,φ when φ ∈/ Z, but only continuous in φ
3
when φ ∈ Z. In fact, H is right continuous when φ∈ Z.
3
Remark 2.1 The systems mentioned above are equivalent, see [12] and [13], we omit the proof.
3 The proof of Theorem 1
Now weareconcernedwiththeHamiltonian system(2.7)withHamiltonian functionH given
3
by (2.6). To cope with the nonsmoothness in φ and use Moser’s small twist theorem to prove
the Lagrangian stability, we will exchange the positions of variables (ρ,φ) and (H ,t) below.
3
This trick has been used in [18, 19, 20].
From (2.6), we have that
lim H =∞
3
ρ→∞
and
∂H 2n α(i+1)p (t)(2a)α(i+1)ρα(i+1)−1{C(1(φ−[φ])T )}i+1
3 = 2dβρ2β−1 + i 2 0 > 0,
∂ρ i+1
Xi=0
Boundednessfor impact oscillators 6
if ρ≥ 1. By the implicit theorem, we know that there is a function R = R(H ,t,φ) such that
3
ρ(H3,t,φ) = d−21βH321β −R(H3,t,φ). (3.1)
Then we make another transformation Ψ : (ρ,φ,t) 7→ (I,θ,τ) given by
3
I = H (ρ,φ,t), θ =t, τ = φ. (3.2)
3
This transformation also leads to a Hamiltonian system with new Hamiltonian function
− 1 1
H4(I,,θ,τ) = d 2βI2β −R(I,θ,τ), (3.3)
which is C5 in θ, C∞ in I and C0 in τ.
And the system is given by
I˙ = −∂H4, τ ∈/ Z,
∂θ
 θ˙ = ∂H4, τ ∈/ Z, (3.4)
 ∂I
 θ(k) = lim θ(τ), I(k) = lim I(τ), k ∈ Z.
τ→k τ→k

Then we have 
Lemma 3.1 For I large enough, the following estimate about R hold ture
|DjDjR|< .I12−1, for j +k ≤ 5. (3.5)
I θ
To be convenient for readers, the proof is given in section 4.
In order to use the Moser’s small twist theorem, we introduce a new variable υ and a small
positive ε by the formula I = υ, where υ ∈ [1,2], ε > 0 and ε → 0, when I → ∞. Then the
ε
system (3.4) is equivalent to the following system
υ˙ = ε∂R, τ ∈/ Z,
∂θ
 θ˙ = 1 d−21βε1−21βυ21β−1−ε∂R, τ ∈/ Z, (3.6)
 2β ∂υ
 θ(k) = lim θ(τ), υ(k) = lim υ(τ), k ∈ Z.
τ→k τ→k


Integrate the system (3.6) from τ = 0 to τ = 1, we obtain the Poincar´e mapping P of the
form
υ1 = υ0+f1(υ0,θ0), θ1 = θ0+ 21βd−21βε1−21βυ021β−1+f2(υ0,θ0), (3.7)
where f (υ ,θ ) = 1ε∂Rdτ, f (υ ,θ )= − 1ε∂Rdτ, which satisfy:
1 0 0 0 ∂θ 2 0 0 0 ∂υ
Lemma 3.2 If ε sRmall enough, then R
|Dυj0Dθk0f1(υ0,θ0)| < .ε12, for j +k ≤ 4; (3.8)
and
|Dυj0Dθk0f2(υ0,θ0)|< .ε12, for j+k ≤ 4. (3.9)
Boundednessfor impact oscillators 7
Proof. Note I = υ/ε, we have DjR(υ/ε,θ,τ) = ε−jDjR(I,θ,τ), and ε → 0, when I → ∞.
υ I
Thus, with (3.5)-(3.7), it is readily to get the results, see [1]. We omit it.
Lemma 3.3 The mapping P possesses the intersection property in the annulus [1,2]×S1, i.e.,
if Γ is an embedded circle in [1,2]×S1 homotopic to a circle υ = const., then P(Γ)∩Γ 6= ∅.
The proof can be found in [1].
Proof of Theorem 1. Until now, we have verified that the mapping P satisfies all the condi-
tions of Moser’s small twist theorem [21, 22]. Hence for any ω ∈ ( 1 d−21βε1−21β, 1 d−21β21−21β)
2β 2β
satisfying
p
|ω− |≥ .ε|q|−5/2, (3.10)
q
there is an invariant curve Γ of P surrounding υ = 1 if ε is sufficiently small. This means that
0
the Poincar´e mapping of the system (3.4) has an invariant curve Γ surroundingI = 1 and in the
ε
annulus 1 < I < 2. Then transform the system (3.4) back to the system (2.7), we can find that
ε ε e
there exit arbitrary large invariant tori in (ρ,φ,t) space, and this implies that all the solutions
are bounded. Finally, according to the equivalent of the system (1.1) and (2.7), we can imply
the boundedness of all the solutions of Eq.(1.1). The proof of Theorem 1 is completed.
4 The proof of Lemma 3.1
Proof. From (3.1) and (3.2), we have
− 1 1
ρ= d 2βI2β −R (4.1)
where α= 1 , β =1−α = n+1, a = T0 and d = (2a)2β.
n+2 n+2 α 2n+2
When j +k = 0, substitute ρ in (2.6), we have
I = d(d−21βI21β −R)2β + 2n pi(θ)(2a)α(i+1)(d−21βI21β −R)α(i+1){C(21(τ −[τ])T0)}i+1, (4.2)
i+1
Xi=0
and
I − 2n pi(θ)(2a)α(i+1)(d−21βI21β −R)α(i+1){C(12(τ −[τ])T0)}i+1 = d(d−21βI21β −R)2β. (4.3)
i+1
Xi=0
Using the Taylor expansion of f (x) = (1 +x)σ = 1+ σx +x2 1f′′(sx)(1 −s)ds, where
σ 0 σ
f′′(sx)= σ(σ−1)(1+sx)σ−2, and with the constants α, β, d, we haRve
σ
d(d−21βI21β −R)2β = I(1−d21βI−21βR)21β = I −2βd21βI1−21βR+IR1, (4.4)
Boundednessfor impact oscillators 8
where
1
R1 = (d21βI−21βR)2 f2′′β(sd21βI−21βR)(1−s)ds, (4.5)
Z
0
and
I − 2n pi(θ)(2a)α(i+1)(d−21βI21β −R)α(i+1){C(21(τ −[τ])T0)}i+1 =
i+1
Xi=0 (4.6)
2n
α(i+1) 1 − 1
I − µi(θ,τ)I 2β (1−α(i+1)d2βI 2βR+R2),
Xi=0
where
µ (θ,τ) = pi(θ)(2a)α(i+1)d−α(2i+β1){C(21(τ −[τ])T0)}i+1, (4.7)
i
i+1
1
R2 = (d21βI−21βR)2Z fα′′(i+1)(sd21βI−21βR)(1−s)ds. (4.8)
0
From (4.3)-(4.8), we have
R(I,θ,τ) = g (I,θ,τ)+g (I,θ,τ)+g (I,θ,τ)+g (I,θ,τ), (4.9)
1 2 3 4
where
2n
g1(I,θ,τ) = (2βd21β)−1µi(θ,τ)Iα(2i+β1)−1, (4.10)
Xi=0
2n
g2(I,θ,τ) = − (2β)−1µi(θ,τ)Iα(i+2β1)+1−1R, (4.11)
Xi=0
2n
g3(I,θ,τ) = (2βd21β)−1µi(θ,τ)Iα(i+2β1)+1−1R2, (4.12)
Xi=0
g4(I,θ,τ) = (2βd21β)−1I21βR1. (4.13)
For I large enough, we can prove the following inequality hold true,
1
|R| < .I2, (4.14)
from (4.9)-(4.13).
Assume we have proved (3.5) for j +k ≤4 already, i.e., for I large enough,
|DIjDθkR|< .I12−j, for 0 ≤ j +k ≤ 4. (4.15)
We are going to prove |DIjDθkR|< .I21−j, for j +k =5. From (4.9), we have
DjDkR = DjDkg +DjDkg +DjDkg +DjDkg . (4.16)
I θ i θ 1 I θ 2 i θ 3 i θ 4
We will give the estimates of every term of the right hand of Eq.(4.15) as follows.
Boundednessfor impact oscillators 9
Lemma 4.1
|DIjDθkg1|< .I21−j, for j +k = 5. (4.17)
Proof. From (4.10), we have
2n
DIjDθkg1(I,θ,τ) = (2βd21β)−1DIjDθkµi(θ,τ)Iα(2i+β1)−1, (4.18)
Xi=0
Using Leibnitz’s rule, we can easily prove
|Dθkµi(θ,τ)Iα(2i+β1)−1| < .Iα(2i+β1)−1−j, forj +k = 5, (4.19)
from (4.7) and the definition of p (τ).
i
Finally, from (4.16), we have
2n
|DIjDθkg1(I,θ,τ)| = | (2βd21β)−1DIjDθkµi(θ,τ)Iα(2i+β1)−1| < .I21−j.
Xi=0
Lemma 4.2
|DIjDθkg2| < .I−2nn++22−j +I−2n1+2|DIjDθkR|, for j +k = 5. (4.20)
Proof. From (4.11), we have
2n
DijDθkg2(I,θ,τ) = − (2β)−1DijDθkµi(θ,τ)Iα(i+2β1)+1−1R, (4.21)
Xi=0
Using Leibnitz’s rule, we have
DIjDθkµi(θ,τ)Iα(2i+β1)−1R = DIj1Dθk1(µi(θ,τ)Iα(2i+β1)−1)DIj2Dθk2R
X
j1+j2=j,
k1+k2=k,
j+k=5 (4.22)
= DIj1Dθk1(µi(θ,τ)Iα(2i+β1)−1)DIj2Dθk2R+µi(θ,τ)Iα(2i+β1)−1DIjDθkR.
X
j1+j2=j,
k1+k2=k,
j+k≤4
Then we can prove that
|DIjDθkg2| < .I−2nn++22−j +I−2n1+2|DIjDθkR|, for j +k = 5,
from (4.7) and (4.15).
Boundednessfor impact oscillators 10
Lemma 4.3
|DIjDθkg3|< .I2nn−+12−j +I−2n2+2|DIjDθkR|, for j +k = 5. (4.23)
Proof. From (4.13), we have
2n
DIjDθkg3(I,θ,τ) = (2βd21β)−1DIjDθkµi(θ,τ)Iα(i+2β1)+1−1R2, (4.24)
Xi=0
Firstly, we give some claims before proving (4.23). Similar to [12] Lemma 3.3, we have
Lemma 4.4 Let F :R → R and f : R2 → R be sufficiently smooth, then for j +k ≥ 1
DjDk[F ◦f]= C (DsF(f))(Dj1Dk1f)...(DjsDksf) (4.25)
x y s,~j,~k f x y x y
X
1≤s≤j+k,
~j=(j1,...,js),|~j|=j,
~k=(k1,...,ks),|~k|=k
where |~j|= j +...+j , |~k| = k +...+k .
1 s 1 s
From (4.8), we have
1
DIjDθkR2 = DIjDθk(d21βI−21βR)2Z fα′′(i+1)(sd21βI−21βR)(1−s)ds. (4.26)
0
and
|DjDk 1f′′ (sd21βI−21βR)(1−s)ds| < .|DjDkf′′ (sd21βI−21βR)|
I θ 0 α(i+1) I θ α(i+1) (4.27)
< .|DIjDRθk(1+sd21βI−21βR)σ−2|≡ |DIjDθkFσ(I−21βR)|
1
Let Fσ(x) = (1+sd2βx)σ−2, then we have
DxsFσ(x) = ddxssFσ(x) = Cσs(1+sd21βx)σ−2−s (4.28)
where C are constants.
σs
With (4.25) and (4.28) and |I−21βR|< .1, provide I large enough, we obtain that
2
|DjDk[F(I−21βR)]|
I θ
= | Cs,~j,~k(DsF(I−21βR))(DIj1Dθk1(I−21βR))...(DIjsDθks(I−21βR))|
X
1≤s≤j+k,
~j=(j1,...,js),|~j|=j, (4.29)
~k=(k1,...,ks),|~k|=k
< . |(Dj1Dk1(I−21βR))|...|(DjsDks(I−21βR))|
I θ I θ
X
1≤s≤j+k,
~j=(j1,...,js),|~j|=j,
~k=(k1,...,ks),|~k|=k
For j +k ≤ 4, we have
|DjDk(I−21βR)| < .I−2n1+2−j. (4.30)
I θ