# Infinitely Many Periodic Solutions of Duffing Equations with Singularities via Time Map

Authors: Tiantian Ma; Zaihong Wang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We study the periodic solutions of Duffing equations with singularitiesx′′+g(x)=p(t). By using Poincaré-Birkhoff twist theorem, we prove that the given equation possesses infinitely many positive periodic solutions provided that g satisfies the singular condition and the time map related to autonomous system x′′+g(x)=0 tends to zero.


---

## Body

## 1. Introduction

In this paper, we are concerned with the periodic solutions of singular Duffing equations:(1)x′′+g(x)=p(t),
where g:(0,+∞)→R is locally Lipschitz continuous and has a singularity at the origin and p:R→R is continuous and periodic, whose least period is 2π.The periodic problem of equations with singularities has been widely studied lately because of their background in applied sciences [1–15]. For example, the oscillation problem of a spherical thick shell made of an elastic material can also be modeled by this kind of equations [1].The opening work on the existence of periodic solutions of ordinary differential equations with singularities was done by Lazer and Solimini [2], in which the equations
(2)x′′-1xν=p(t)
were studied. It was proved in [2] that if ν≥1, then (2) has at least one positive 2π-periodic solution if and only if
(3)∫02π‍p(t)dt<0.
Meanwhile, if 0<ν<1, then they constructed a periodic function p(t) with negative mean value such that (2) does not have any 2π-periodic solution.It is well known that time map plays an important role in studying the existence and multiplicity of periodic solutions of Duffing equations without singularities. In case wheng has a singularity, we can also use time map to deal with the periodic solutions of (1) (see [4] and the related references therein).Assume thatg satisfies
(h1)lim⁡x→0+g(x)=-∞,
and the primitive function G of g satisfies
(h2)lim⁡x→0+G(x)=+∞,(G(x)=∫1x‍g(s)ds).
Moreover, the following condition holds:
(h3)lim⁡x→+∞g(x)=+∞.Let us define(4)τ(c)=∫1c‍dsG(c)-G(s).
The map τ is usually called time map, which is continuous for c large enough. We shall deal with the multiplicity of periodic solutions of (1) by means of asymptotic property of the time map τ. Assume that the time map satisfies
(τ)lim⁡c→+∞τ(c)=0.
It is easy to check that if g satisfies superlinear condition
(5)lim⁡x→+∞g(x)x=+∞,
then condition (τ) is satisfied. However, the converse is not true. In fact, we can find functions g, which do not satisfy (5). But the corresponding time maps satisfy the condition (τ). For example, let us define
(6)g(x)={3-1x,0<x≤1,x+(x-1)3+(x-1)3sin(x-1)4,x≥1.
Obviously, conditions (hi)(i=1,2,3) hold and condition (5) does not hold. Next, we will show that condition (τ) is satisfied. In case when x≥1, we have
(7)G(x)=∫1x‍(s+(s-1)3+(s-1)3sin(s-1)4)ds=12x2+14(x-1)4-14cos⁡(x-1)4-14.
Therefore, we have
(8)lim⁡c→+∞τ(c)=lim⁡c→+∞∫1c‍(ds)hhhhhhhhhh×(12(c2-s2)+14((c-1)4-(s-1)4)hhhhhhhhhhhhh+14(cos⁡⁡(s-1)4-cos⁡⁡(c-1)4))-1/2=lim⁡c→+∞1c∫1/c1(dt)‍hhhhhhhhhhhh×(12c2(1-t2)+14c4((c-1)4-(ct-1)4)hhhhhhhhhhhhhhh+14c4(cos⁡(ct-1)4-cos⁡(c-1)4))-1/2.
Since
(9)lim⁡c→+∞∫1/c1‍(dt)×(12c2(1-t2)+14c4((c-1)4-(ct-1)4)+14c4(cos⁡⁡(ct-1)4-cos⁡(c-1)4))-1/2=∫01‍2dt1-t4,
we get
(10)lim⁡c→+∞τ(c)=0.When the conditions(h1), (h2), and (5) hold, it was proved in [6] that (1) has infinitely many periodic solutions. In the present paper, we will deal with the multiplicity of periodic solutions of (1) under the conditions (h1), (h2), (h3), and (τ). Obviously, the conditions (h3) and (τ) generalize the condition (5). Since (5) does not hold, the estimating method in [6] is invalid. By taking some new estimating skills, we obtain the following results.Theorem 1.
Assume that conditions(hi)(i=1,2,3) and (τ) hold. Then (1) has infinitely many positive harmonic solutions {xj(t)} satisfying
(11)lim⁡j→∞(min⁡0≤t≤2π(xj(t)+|xj′(t)|))=0,lim⁡j→∞(max⁡0≤t≤2π(xj(t)+|xj′(t)|))=+∞.Theorem 2.
Assume that conditions(hi)(i=1,2,3) and (τ) hold. Then for any integer m≥2, (1) has infinitely many positive m-order subharmonic solutions {xj(t)} satisfying
(12)lim⁡j→∞(min⁡0≤t≤2mπ(xj(t)+|xj′(t)|))=0,lim⁡j→∞(max⁡0≤t≤2mπ(xj(t)+|xj′(t)|))=+∞.Remark 3.
In the following, for convenience and brevity, we move the singular point0 to the point -1. In fact, we can take a transformation x=u+1 to achieve this aim. We will consider singular equations as follows:
(1′)x′′+g(x)=p(t),
where g:(-1,+∞)→R is continuous and has a singularity at x=-1. We now assume that the following conditions hold:
(h1′)lim⁡x→-1+g(x)=-∞,(h2′)lim⁡x→-1+G(x)=+∞.
Next, we will deal with the existence and multiplicity of periodic solutions of (1′) under conditions (h1′), (h2′), (h3), and (τ).


## 2. Basic Lemmas

In this section, we will perform some phase-plane analyses for(1′) when conditions (h1′), (h2′), and (h3) hold. Consider the equivalent system of (1′):
(13)x′=y,y′=-g(x)+p(t).
Let (x(t),y(t))=(x(t,x0,y0),y(t,x0,y0)) be the solution of (13) through the initial point:
(14)x(0,x0,y0)=x0,y(0,x0,y0)=y0.Lemma 4.
Assume that conditions(h2′) and (h3) hold. Then every solution (x(t),y(t)) of system (13) exists uniquely on the whole t-axis.Proof.
Define a potential function(15)V(x,y)=12y2+G(x).
Set
(16)V(t)=12y2(t)+G(x(t)).
Then we have
(17)V′(t)=p(t)y(t)≤M|y(t)|,
where M=max⁡{|p(t)|:t∈R}. From (h2′) and (h3) we know that there exists a constant M′>0 such that
(18)G(x)+M′≥0,x∈(-1,+∞).
From (17) and (18) we get
(19)V′(t)≤M|y(t)|+G(x(t))+M′≤V(t)+M′′,
where M′′=M′+M2/2. Then, for any finite T>0, we have
(20)V(t)≤V(0)eT+M′′(eT-1),t∈[0,T).
Therefore, (x(t),y(t)) is bounded for t∈[0,T). Furthermore, (x(t),y(t)) exists on the interval [0,+∞). Similarly, we can prove that (x(t),y(t)) exists on the interval (-∞,0]. The uniqueness of the solution (x(t),y(t)) follows directly from the local Lipschitzian condition on g.On the basis of Lemma4, we can define the Poincaré map P:(-1,+∞)×R→R2 as follows:
(21)P:(x0,y0)⟶(x1,y1)=(x(2π,x0,y0),y(2π,x0,y0)).
We know that fixed points of the Poincaré map P correspond to 2π-periodic solutions of (13).To show the position of orbit(x(t),y(t)) of (13), we introduce a function ζ:(-1,+∞)×R→R+,
(22)ζ(x,y)=x2+y2+1(1+x)2.Lemma 5.
There exists a constantc0>0 such that, for any c≥c0, Γc:ζ(x,y)=c is a closed star-shaped curve around the origin.Proof.
Consider autonomous system:(23)x′=y,y′=-x+1(1+x)3.
Obviously, Γc is one orbit of the autonomous system above. From the expression of ζ we know that there exists c1>0 such that, for c≥c1, Γc is a closed curve around the origin. Applying the polar coordinate transformation x=ρcos⁡ϑ, y=ρsinϑ to this system, we get
(24)ϑ′(t)=-1+cos⁡ϑρ(1+ρcos⁡ϑ)3.
In the case when -1<ρcos⁡ϑ≤0, we have ϑ′(t)≤-1. In the case when cos⁡ϑ≥0 and ρ≥2, we have cos⁡ϑ/(ρ(1+ρcos⁡ϑ)3)≤1/2, which implies ϑ′(t)≤-1/2. Therefore, there exists c2>0 such that, for ζ(ρcos⁡ϑ,ρsinϑ)≥c2, ϑ(t) is decreasing strictly. Take c0=max⁡{c1,c2}. Then for c≥c0, Γc is a closed star-shaped curve around the origin.Lemma 6 (see [1]).
Assume that conditions(h1′), (h2′), and (h3) hold. Then, for any T>0 and ϱ>0, there exists ϱ0>0 sufficiently large such that, for ζ(x0,y0)≥ϱ02,
(25)ζ(x(t),y(t))≥ϱ2,t∈[0,T],
where (x(t),y(t)) is the solution of (13) through the initial point (x0,y0).From Lemma6 we know that if ζ(x0,y0) is large enough, then x2(t)+y2(t)>0, t∈[0,T]. Therefore, we can take the polar coordinate transformation
(26)x=rcos⁡θ,y=rsinθ.
Under this transformation, system (13) becomes
(27)drdt=rsinθcos⁡θ-g(rcos⁡θ)sinθ+p(t)sinθ,dθdt=-sin2θ-1rg(rcos⁡θ)cos⁡θ+1rp(t)cos⁡θ.
Let (r(t),θ(t))=(r(t,r0,θ0),θ(t,r0,θ0)) be the solution of (27) satisfying condition
(28)r(0)=r0,θ(0)=θ0
with x0=r0cos⁡θ0, y0=r0sinθ0.Then we can rewrite the Poincaré mapP as follows:
(29)P:(r0,θ0)⟶(r1,θ1)=(r(2π,r0,θ0),θ(2π,r0,θ0)),
with x0=r0cos⁡θ0>-1, y0=r0sinθ0.Lemma 7.
Assume that conditions(h1′), (h2′), and (h3) hold. Then, for any T>0, there exist ρ0>0 and ω>0 such that, for ζ(x0,y0)≥ρ02,
(30)θ′(t)≤-ω,t∈[0,T].Proof.
From(h3) we know that there exist constants α>0 and c>0 such that
(31)g(x)-p(t)x≥α,x∈(c,+∞),t∈R.
Moreover, we know from (h1′) that there exist β>0 and -1<d<0 such that
(32)g(x)-p(t)x≥β,x∈(-1,d),t∈R.
If x(t)>c, t∈[0,T], then
(33)θ′(t)≤-sin2θ-αcos⁡2θ≤-min⁡(1,α).
If -1<x(t)<d<0, t∈[0,T], then
(34)θ′(t)≤-sin2θ-βcos⁡2θ≤-min⁡(1,β).
On the other hand, we know from Lemma 6 that there exists ρ0>0 large enough such that if ζ(x0,y0)≥ρ02 and x(t)∈[d,c], t∈[0,T], then
(35)|g(x(t))-p(t)|r(t)≤13,|sinθ(t)|≥22.
Hence,
(36)θ′(t)≤-sin2θ(t)+|g(x(t))-p(t)|r(t)|cos⁡θ(t)|≤-16.
Consequently, the conclusion of Lemma 7 holds.Lemma 8.
Assume that conditions(h1′), (h2′), (h3), and (τ) hold. Let A≥0 be a given constant. Then we have
(37)lim⁡c→+∞∫0c‍dsG(c)-G(s)-A(c-s)=0,lim⁡c→-1+∫c0‍dsG(c)-G(s)+A(c-s)=0.Proof.
We now prove the first estimation. From condition(h3) we know that there exists a constant η>1 such that, for η≤s≤c,
(38)G(c)-G(s)≥2A(c-s).
Then, for η≤s≤c, we have
(39)G(c)-G(s)-A(c-s)≥12[G(c)-G(s)].
Write
(40)∫0c‍dsG(c)-G(s)-A(c-s)=I1+I2,
where
(41)I1=∫0η‍dsG(c)-G(s)-A(c-s),I2=∫ηc‍dsG(c)-G(s)-A(c-s).
From condition (h3) we can derive easily that lim⁡c→+∞I1=0. From (39) we get
(42)I2≤2∫ηc‍dsG(c)-G(s)≤2∫1c‍dsG(c)-G(s).
According to condition (τ), we have that lim⁡c→+∞I2=0. Hence, we get
(43)lim⁡c→+∞∫0c‍dsG(c)-G(s)-A(c-s)=0.
Next, we prove the second estimation. Let 0<ɛ<1 be a sufficiently small constant. In the case when -1<c<-1+ɛ, we write
(44)∫c0‍dsG(c)-G(s)+A(c-s)=J1+J2,
where
(45)J1=∫c-1+ɛ‍dsG(c)-G(s)+A(c-s),J2=∫-1+ɛ0‍dsG(c)-G(s)+A(c-s).
If s∈(c,-1+ɛ)⊂(-1,-1+ɛ), then we have
(46)G(c)-G(s)=g(ζ)(c-s),ζ∈(c,s)⊂(-1,-1+ɛ).
Set
(47)ξ(ɛ)=sup⁡{g(x):x∈(-1,-1+ɛ)}.
Obviously, g(ζ)≤ξ(ɛ). From condition (h1′) we know
(48)lim⁡ɛ→0+ξ(ɛ)=-∞.
According to (46), we get that, for s∈(c,-1+ɛ)⊂(-1,-1+ɛ),
(49)G(c)-G(s)=g(ζ)(c-s)≥ξ(ɛ)(c-s).
Hence,
(50)J1≤∫c-1+ɛ‍dsξ(ɛ)(c-s)+A(c-s)=2-1+ɛ-c-ξ(ɛ)-A,
which, together with (48), means that lim⁡c→-1+J1=0. On the other hand, we can infer easily from (h1′) that lim⁡c→-1+J2=0. Consequently, we have
(51)lim⁡c→-1+∫c0‍dsG(c)-G(s)+A(c-s)=0.
Thus, the proof is completed.Lemma 9.
Assume that conditions(h1′), (h2′), (h3), and (τ) hold. Let m be a given positive integer. Then, for any given positive integer n, there is a constant Rn>0 such that, for ζ(x0,y0)≥Rn2,
(52)θ(2mπ)-θ0<-2nπ.Proof.
From Lemmas6 and 7 we know that, for any sufficiently large ϱ>0, there is a constant ϱ0>ϱ such that, for ζ(x0,y0)≥ϱ02 and t∈[0,2mπ],
(53)ζ(x(t),y(t))≥ϱ2,θ′(t)<0.
Let (x(t),y(t)) be a solution of (13) satisfying ζ(x0,y0)≥ϱ02. Then the solution (x(t),y(t)) will move clockwise during the time period [0,2mπ]. Without loss of generality, we assume that (x0,y0) lies in the first quadrant. Then there exist t0=0<t1<t2<t3<t4<t5 such that
(54)x(t1)>0,y(t1)=0;x(t)>0,y(t)>0,t∈(t0,t1),x(t2)=0,y(t2)<0;x(t)>0,y(t)<0,t∈(t1,t2),x(t3)<0,y(t3)=0;x(t)<0,y(t)<0,t∈(t2,t3),x(t4)=0,y(t4)>0;x(t)<0,y(t)>0,t∈(t3,t4),x(t5)>0,y(t5)=0;x(t)>0,y(t)>0,t∈(t4,t5).
Next, we will estimate ti-ti-1 (i=1,2,3,4,5), respectively. We first estimate t1-t0. If t∈[t0,t1], then y(t)≥0. Let us define an auxiliary function
(55)w(t)=12y2(t)+G(x(t))-Mx(t),
where M=max⁡{|p(t)|:t∈[0,2π]}. Then we have that, for t∈[t0,t1],
(56)w′(t)=y(t)y′(t)+g(x(t))x′(t)-Mx′(t)=y(t)(p(t)-M)≤0,
which implies that w(t) is decreasing on the interval [t0,t1]. Therefore, we get that, for t∈[t0,t1],
(57)12y2(t)+G(x(t))-Mx(t)≥G(x(t1))-Mx(t1),
which means
(58)x′(t)≥2(G(x(t1))-G(x(t)))-2M(x(t1)-x(t)).
Hence, we obtain
(59)t1-t0≤∫x(t0)x(t1)‍dx2(G(x(t1))-G(x))-2M(x(t1)-x)≤∫0x(t1)‍dx2(G(x(t1))-G(x))-2M(x(t1)-x).
Similarly, we can obtain
(60)t5-t4≤∫0x(t5)‍dx2(G(x(t5))-G(x))-2M(x(t5)-x).
We next estimate t2-t1. If t∈[t1,t2], then y(t)≤0. Therefore, we have
(61)w′(t)=y(t)(p(t)-M)≥0,t∈[t1,t2],
which implies that w(t) is increasing on the interval [t1,t2]. Furthermore, we have that, for t∈[t1,t2],
(62)12y2(t)+G(x(t))-Mx(t)≥G(x(t1))-Mx(t1),
which yields
(63)-x′(t)≥2(G(x(t1))-G(x(t)))-2M(x(t1)-x(t)).
Consequently, we get
(64)t2-t1≤∫0x(t1)‍dx2(G(x(t1))-G(x))-2M(x(t1)-x).
We now estimate t3-t2. If t∈[t2,t3], then y(t)≤0. Define
(65)w~(t)=12y2(t)+G(x(t))+Mx(t).
Then we have that, for t∈[t2,t3],
(66)w~(t)=y(t)(p(t)+M)≤0,
which implies that w~(t) is decreasing on the interval [t2,t3]. Therefore, we get that, for t∈[t2,t3],
(67)12y2(t)+G(x(t))+Mx(t)≥G(x(t3))+Mx(t3),
which implies
(68)-x′(t)≥2(G(x(t3))-G(x(t)))+2M(x(t3)-x(t)).
Hence,
(69)t3-t2≤∫x(t3)0‍dx2(G(x(t3))-G(x))+2M(x(t3)-x).
Similarly, we get
(70)t4-t3≤∫x(t3)0‍dx2(G(x(t3))-G(x))+2M(x(t3)-x).
According to Lemma 6, if we take ϱ0>>1, then we have x(t1)>>1, 0<1+x(t3)<<1 and x(t5)>>1. From Lemma 8 and (59)–(70) we know that, for any sufficiently small ɛ>0, there exists ϱ0>>1 such that
(71)ti-ti-1<ɛ,(i=1,2,3,4,5).
It follows that
(72)t5-t0<5ɛ.
Therefore, the motion (x(t),y(t)) rotates clockwise a turn in a period less than 5ɛ. Consequently, (x(t),y(t)) can rotate a sufficiently large number of turns during the period 2mπ provided that ζ(x0,y0)≥ϱ02 (ϱ0>>1) is satisfied.
The proof is thus completed.


## 3. Infinity of Harmonic Solutions

To prove Theorem1, we first prove the following proposition.Proposition 10.
Assume that conditions(h1′), (h2′), (h3), and (τ) hold. Then (1′) has infinitely many harmonic solutions {xj(t)} satisfying
(73)lim⁡j→∞(min⁡0≤t≤2π(1+xj(t)+|xj′(t)|))=0,lim⁡j→∞(max⁡0≤t≤2π(1+xj(t)+|xj′(t)|))=+∞.Proof.
From Lemma6 we know that there exist a1>c0 (c0 is given in Lemma 5) and ω1>0 such that, for ζ(x0,y0)≥a12, ζ(x(t),y(t))≥2 and θ′(t)<-ω1, t∈[0,2π]. For ζ(x0,y0)≥a12, we consider
(74)Φ(r0,θ0)=θ(2π,r0,θ0)-θ0,
with x0=r0cos⁡θ0, y0=r0sinθ0. Obviously, there exists an integer k≥1 such that
(75)θ(2π,r0,θ0)-θ0>-2kπ,forζ(r0cos⁡θ0,r0sinθ0)=a12.
On the other hand, it follows from Lemma 9 that there exists a2>a1 such that
(76)θ(2π,r0,θ0)-θ0<-2kπ,forζ(r0cos⁡θ0,r0sinθ0)=a22.
Meanwhile, there exists an integer k′>k such that
(77)θ(2π,r0,θ0)-θ0>-2k′π,forζ(r0cos⁡θ0,r0sinθ0)=a22.
From (75) and (76) we know that the area-preserving homeomorphism P is twisting on the annulus A1={(x,y)∈(-1,+∞)×R:a1≤ζ(x,y)≤a2}. Obviously, we have r(2π)>0 provided that ζ(x0,y0)≥a12. Hence, O∈P(D), where D is an open region with boundary ζ(x,y)=a12. Finally, we know from Lemma 5 that both Γa1:ζ(x,y)=a12 and Γa2:ζ(x,y)=a22 are closed star-shaped curves with respect to the origin O. Thus, we have proved that all conditions of the generalized Poincaré-Birkhoff theorem [16, 17] are satisfied. Consequently, the Poincaré map P has at least two fixed points (r1i,θ1i)(i=1,2) in annulus A1 and then (13) has two 2π periodic solutions (x1i(t),y1i(t))=(x(t,x1i,y1i),y(t,x1i,y1i)) (x1i=r1icos⁡θ1i, y1i=r1isinθ1i). Therefore, x1i(t) are 2π periodic solutions of (1′). On the other hand, since the period of any periodic solution of (1′) must be multiple of the period of p(t), then 2π is the minimal period of x1i(t). Therefore, x1i(t) are harmonic solutions of (1′).
Similarly, we can find a sequence(78)(a1<a2<)a3<⋯<aj<aj+1<⋯,lim⁡j→∞aj=+∞,
such that the area-preserving homeomorphism P is twisting on the annuli
(79)Aj={(x,y)∈(-1,+∞)×R:aj≤ζ(x,y)≤aj+1},j=2,3,….
Therefore, the Poincaré map P has at least two fixed points (rji,θji)(i=1,2) in each Aj, (j=2,3,…). Consequently, (13) has two 2π periodic solutions (xji(t),yji(t))=(x(t,xji,yji), y(t,xji,yji)) (xji=rjicos⁡θji,yji=rjisinθji) and then xji(t) are 2π periodic solutions of (1′). Similarly, we know that xji(t) are harmonic solutions of (1′). Since lim⁡j→∞aj=+∞, we have
(80)min⁡{x+1+|y|:ζ(x,y)=aj}⟶0,j⟶∞,max⁡{x+1+|y|:ζ(x,y)=aj}⟶+∞,j⟶∞.
Furthermore, we know from Lemma 6 that, for i=1,2,
(81)lim⁡j→∞(min⁡0≤t≤2π(1+xji(t)+|xji′(t)|))=0,lim⁡j→∞(max⁡0≤t≤2π(1+xji(t)+|xji′(t)|))=+∞.
Thus we have proved Proposition 10.Proof of Theorem1.
Consider the equivalent equation of (1):
(82)u′′+g~(u)=p(t),
where g~(u)=g(1+u). Obviously, g~ satisfies conditions (h1′), (h2′), and (h3). To use Proposition 10, we only need to prove that condition (τ) holds for function G~(u)(=∫1u‍g~(s)ds). Set
(83)τ~(c)=∫1c‍dsG~(c)-G~(s).
Then we have
(84)τ~(c)=∫11+c‍dsG(1+c)-G(s)-∫12‍dsG(1+c)-G(s).
From conditions (τ) and (h3) we get lim⁡c→+∞τ~(c)=0. Therefore, all conditions of Proposition 10 are satisfied. Accordingly, (82) has infinitely many harmonic solutions {uj(t)} satisfying
(85)lim⁡j→∞(min⁡0≤t≤2π(1+uj(t)+|uj′(t)|))=0,lim⁡j→∞(max⁡0≤t≤2π(1+uj(t)+|uj′(t)|))=+∞.
Recalling that (82) is obtained by taking a parallel transformation x=1+u to (1), we know that the conclusion of Theorem 1 holds.Remark 11.
In [16], the Poincaré-Birkhoff theorem was proved in case that the inner closed curve of the annulus is star shaped. From [17] we know that there is a need for both boundaries of the annulus to be star shaped in the Poincaré-Birkhoff theorem.


## 4. Infinity of Subharmonic Solutions

To prove Theorem2, we first prove the following proposition.Proposition 12.
Assume that conditions(h1′), (h2′), (h3), and (τ) hold. Then, for any given integer m≥2, (1′) has infinitely many m-order subharmonic solutions {xj(t)} satisfying
(86)lim⁡j→∞(min⁡0≤t≤2mπ(1+xj(t)+|xj′(t)|))=0,lim⁡j→∞(max⁡0≤t≤2mπ(1+xj(t)+|xj′(t)|))=+∞.Proof.
Letm≥2 be a given integer. From Lemmas 6 and 9 we know that there exists b1>c0 (c0 is given in Lemma 5) and ω1′>0 such that, for ζ(x0,y0)≥b12,
(87)θ′(t)<-ω1′,ζ(x(t),y(t))≥2,t∈[0,2mπ],(88)θ(2π,r0,θ0)-θ0<-2π.
For ζ(x0,y0)≥b12, we consider
(89)Ψ(r0,θ0)=θ(2mπ,r0,θ0)-θ0,
with x0=r0cos⁡θ0, y0=r0sinθ0. Obviously, there exists a positive prime integer q such that
(90)θ(2mπ,r0,θ0)-θ0>-2qπ,forζ(r0cos⁡θ0,r0sinθ0)=b12.
On the other hand, it follows from Lemma 8 that there exists b2>b1 such that
(91)θ(2mπ,r0,θ0)-θ0<-2qπ,forζ(r0cos⁡θ0,r0sinθ0)=b22.
From (90) and (91) we know that the map Pm is twisting on the annulus B1={(x,y)∈(-1,+∞)×R:b1≤ζ(x,y)≤b2}. Using the generalized Poincaré-Birkhoff twist theorem, we know that Pm has at least two fixed points (rm1i,θm1i)(i=1,2) in B1, which satisfy
(92)θ(2mπ,rm1i,θm1i)-θm1i=-2qπ,(i=1,2).
It follows that (13) has two 2mπ-periodic solutions (xm1i(t),ym1i(t)) and then (1′) has two 2mπ-periodic solutions xm1i(t).
Next, we will prove that2mπ is the minimal period of xm1i(t). Assume by contradiction that 2lπ(1≤l≤m-1) is the minimal period of xm1i(t). Then we have m=nl (n≥2). Since (xm1i(t),ym1i(t)) is 2lπ periodic, we know from (88) that there exists a positive integer ki≥2 such that
(93)θ(2lπ,rm1i,θm1i)-θm1i=-2kiπ.
Furthermore,
(94)θ(2mπ,rm1i,θm1i)-θm1i=-2nkiπ.
Hence, we have nki=q. Since q is a prime integer, we get a contradiction. This proves that 2mπ is the minimal period of xm1i(t). Consequently, xm1i(t) are m-order subharmonic solutions of (1′).
In a similar manner, we can find a sequence(95)(b1<b2<)b3<⋯<bj<bj+1<⋯,lim⁡j→∞bj=+∞,
such that the area-preserving homeomorphism Pm is twisting on the annuli
(96)Bj={(x,y)∈(-1,+∞)×R:bj≤ζ(x,y)≤bj+1}.
Therefore, the Poincaré map Pm has at least two fixed points (rmji,θmji)(i=1,2) in each Bj, (j=2,3,…). Consequently, (13) has two 2mπ periodic solutions (xmji(t),ymji(t)) and then xmji(t) are 2mπ periodic solutions of  (1′). Similarly, we know that xmji(t) are m-order subharmonic solutions of  (1′). Since lim⁡j→∞bj=+∞, we have
(97)min⁡{x+1+|y|:ζ(x,y)=bj}⟶0,j⟶∞,max⁡{x+1+|y|:ζ(x,y)=bj}⟶+∞,j⟶∞.
Furthermore, we know from Lemma 6 that, for i=1,2,
(98)lim⁡j→∞(min⁡0≤t≤2mπ(1+xmji(t)+|xmji′(t)|))=0,lim⁡j→∞(max⁡0≤t≤2mπ(1+xmji(t)+|xmji′(t)|))=+∞.
Thus we have proved Proposition 12.Proof of Theorem2.
Using Proposition12 and the same method as proving Theorem 1, we can prove Theorem 2.



---

*Source: 398512-2014-04-16.xml*