# On Stability of Periodic Solutions of Lienard Type Equations

Authors: Zijian Yin; Hongbin Chen
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

We use the Floquet theory to analyze the stability of periodic solutions of Lienard type equations under the asymptotic linear growth of restoring force in this paper. We find that the existence and the stability of periodic solutions are determined primarily by asymptotic behavior of damping term. For special type of Lienard equation, the uniqueness and stability of periodic solutions are obtained. Furthermore, the sharp rate of exponential decay of the stable periodic solutions is determined under suitable conditions imposed on restoring force.


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## Body

## 1. Introduction

This paper is devoted to the stability of large periodic solutions of the following Lienard type equation:(1)x′′t+ddtft,xt+gxt=ht,where d/dtft,xt=ftt,xt+fxt,xtx′t,g(x) is a continuous function and ht is a T-periodic function. The existence and multiplicity of periodic solutions of (1) or more general types of nonlinear second-order differential equations (2)x′′t+ddtft,xt+gt,xt=hthave been investigated extensively by many authors. For details, we refer the reader to [1–4]. One can mention, for example, the papers by Fonda and Habets [5] or more recent papers by Qian [6] and the literature therein. In these papers, the asymptotic behavior of restoring force and damping term are controlled by the inequalities (3)at≤liminfx→∞gt,xx≤limsupx→∞⁡gt,xx≤bt,ct≤liminfx→∞ft,xx≤limsupx→∞⁡ft,xx≤dt.These tend to keep away the combinational effects of the quotients gt,x/x and ft,x/x from the spectrum of the linear operator Lx=-x′′ as x→∞. However, the sign condition on damping term on solvability of (1) seems to be ignored. The aim of the paper is to show that sign condition plays dominate role on the existence of periodic solutions of (1) under the condition that g is asymptotically linear. Resonance phenomenon may appear when limx→∞⁡ft,x/x≡0. In this case, the Landesman-Lazer type solvability condition should be imposed. In this paper, the following hypotheses on f and g are imposed: (4)limx→∞⁡gxx=c≠0,limx→∞ft,xx=at≫0or(5)limx→∞⁡gxx=c≠2nπT2,limx→∞⁡ft,xx≡0,where the second limit converges uniformly for t∈0,T.When the hypotheses(6)limx→∞⁡ft,xx≡0,c=2nπT2,are satisfied, which can be treated as the resonance conditions, (1) does not exist with T-periodic solution in general, even if fxt,x does not change sign for all x∈R. A counterexample will be given at the end of Section 2. In this case, an additional condition should be imposed to guarantee the existence.The related results about the stability of periodic solutions were less extensively studied. In [7], Lazer and Mckenna established stability results by converting the equation to a fixed point problem. Recently, more complete results concerning the stability and the sharpness of the rate of decay of periodic solutions were obtained by Chen and Li in [8, 9].The following notations will be used throughout the rest of the paper:(1)
LTp: T-periodic function u∈Lp0,T with up for 1≤p≤∞;(2)
CTk: T-periodic function u∈Ck0,T, k≥0 with Ck-norm;(3)
αt≫βt: if αt≥βt on 0,T and αt>βt on some positive measure subset.The main results of this paper are the following.Theorem 1.
Assumeg(x)∈C(R×R), f(t,x)∈C1(R×R), and T-periodic in t, satisfying (4) or (5). Then (1) has at least one T-periodic solution.Let us introduce the following symbols:(7)α±t=liminfx→±∞⁡qt,x,β±t=limsupx→±∞⁡qt,x,at=limx→+∞⁡ft,x,bt=limx→-∞⁡ft,x,where at,bt∈CT1.Theorem 2.
Assume thatqt,x,ft,x is bounded, where gt,x=2nπ/T2+qt,x and (6). Then (2) has a T-periodic solution provided that either (8)∫0Thzdt+∫z>0a′zdt+∫z<0b′zdt<∫z>0α+zdt+∫z<0β-zdt,or (9)∫0Thzdt+∫z>0a′zdt+∫z<0b′zdt<∫z<0α-zdt+∫z>0β+zdt,for any zt∈sin⁡2nπ/Tt,cos⁡2nπ/Tt.Remark 3.
Whenat and bt are continuous, the conclusion of Theorem 2 still holds if the second and third term in (8) or (9) are replaced by -∫z>0az′dt+∫z<0bz′dt.In particular, by Theorem2, we see that the following well known Landesman-Lazer type solvability conditions hold, if both at and bt are all constants.Theorem 4.
Under the same conditions of Theorem2, where at and bt are constants, then (1) has a T-periodic solution if either (10)∫0Thzdt<∫z>0α+zdt+∫z<0β-zdt,or(11)∫0Thzdt<∫z<0α-zdt+∫z>0β+zdt.Whengx is a linear function, more refined results can be obtained. Concerning uniqueness and stability as well as the rate of decay to the unique periodic solution, we have the following.Theorem 5.
Assume thatgx=cx where c>0 and ft,x∈C1R×R, such thatH1fx′t,xt>0,limn→∞⁡ft,xx=at≫0.
Then(1)
(1) has a unique T-periodic solution;(2)
the uniqueT-periodic solution is locally exponential asymptotically stable;(3)
the uniqueT-periodic solution is globally asymptotically stable.
Furthermore, if the conditions,H212ftx′′t,xt+12fxx′′t,xtx′t-14fx′2t,xt+c≫n2π2T2,H312ftx′′t,xt+12fxx′′t,xtx′t-14fx′2t,xt+c≪n+12π2T2,hold for some n≥1, then every solution to (1) that is near the unique T-periodic solution decays uniformly at the exponential rate of 1/2∫0Tfx′(t,x0(t))dt.


## 2. Preliminaries

In this section we shall recall some basic results about topological methods.Consider the periodic boundary value problem(12)x′=Ft,x,x0=xT,where F:0,T×Rn→Rn is a continuous function and T-periodic in t. In order to use a homotopic method to compute the degree, we assume that h:0,T×Rn×0,1→Rn is a continuous function such that (13)ht,x,1=Ft,x,ht,x,0=Gx,where Gx is continuous. The following continuation theorem is due to Mawhin [10].Lemma 6.
LetΩ⊂CT be an open bounded set such that the following conditions are satisfied. (1)
There is nox∈∂Ω such that x′t=ht,x,λ,∀λ∈0,1.(2)
deg⁡G,Ω∩Rn,0≠0.
Then (12) has at least one solution.Next, we consider the homogeneous periodic equation(14)x′′t+αtxt′+cxt=0,where c≠0 is constant and αt∈LT∞.The following Lemma is crucial to the argument for existence of periodic solutions.Lemma 7.
Assume thatαt∈LT∞ satisfies either αt≫0 or αt≪0. Then (14) does not admit any nontrivial T-periodic solutions.Proof.
Suppose on the contrary that there exists a nontrivialT-periodic solution xt. Multiplying both sides of (14) by x′t and αtxt, respectively, integrating by parts, and applying the boundary condition, we get (15)∫0Tx′′tαtxtdt=0,∫0Tx′′tαtxtdt+c∫0Tαtx2tdt=0;then(16)c∫0Tαtx2tdt=0.
This implies we havext≡0 on positive measure subset of 0,T. It follows from Rolle’s theorem that the derivative x′t has a zero, between two zeros of xt. Let t0 be an accumulation point of zero of xt, such that t0 exists. Otherwise, the zeros of xt are isolated; hence the set consisting of zeros of xt is a zero measure set. Evidently, at such point xt0=x′(t0)=0. According to a theorem concerning the uniqueness of initial value problem, we have xt≡0 on 0,T.Let us give a counterexample, which demonstrates that (1) does not possess any periodic solutions under condition (6).Consider the following equation:(17)x′′+11+x2x′+n2x=acos⁡nt,where ft,x=arctan⁡x and T=2π satisfies condition (6) but (17) does not admit any 2π-periodic solutions for a>nπ.Indeed, ifxt is a 2π-periodic solution of (17), multiplying both sides of (17) by cos⁡nt and integrating over a period, we obtain (18)aπ=∫02πacos2⁡ntdt=∫02πcos⁡nt1+x2x′dt=n∫02πsin⁡ntarctan⁡xdt,which means that (19)aπ<2πn×1×π2=π2n.Thus, the equation does not have any2π-periodic solutions for a>nπ. Moreover, according to Massera’s theorem [11], we obtain that any solutions of (17) are unbounded for a>nπ.For convenience, we begin with a definition.Definition 8.
Letpt be a T-periodic solution of x′t=fx,t, where fx,t∈C1R×R. Then p is stable if for each ε>0 there exists a δ>0, such that if ψt is any solution of x′t=fx,t for which (20)ψt0-pt0<δat some t0, then (21)ψt-pt<εfor any t≥t0.And next, we shall recall a principle of linearized stability for periodic systems.Letx0 be a T-periodic solution of (12); then we associate the T-periodic solution x0 with the linearized equation (22)y′=Fx′t,x0y.LetMt be the fundamental matrix of (22) and μ1 and μ2 the eigenvalues of the matrix MT. Then xt,x0 is exponential asymptotically stable if and only if μi<1, i=1,2. Otherwise, if there exists an eigenvalue of MT with modulus greater than one, then xt,x0 is unstable.In order to show that every solution of the nonlinear equation (1) locally decays at the rate of 1/2∫0Tfx′t,x0tdt to the unique T-periodic solution, we need the following C1 version of the Hartman-Grobman theorem [12].Lemma 9.
LetU be an open neighborhood of 0 and f:U⊂Rn→Rn be a C1 function such that fx′0:Rn→Rn is a contraction mapping. Then f is C1 conjugate equivalent to fx′0.Consider the following boundary value problem:(23)x′′t+qtxt=0,xξ=xξ+T=0,sgn⁡x′ξ=sgn⁡x′ξ+T;we have the following lemma which is given by the author in [9]Lemma 10.
Letqt∈L∞0,T such that (24)2nπ2T2≪qt≪2n+1π2T2for some n∈N. Then (23) does not admit any nontrivial solution.Proof.
Consider the eigenvalue problem(25)x′′t+λxt=0,xξ=xξ+T=0;we have λn=nπ2/T2, the nth eigenvalue of the above equation. If x is a nontrivial solution of (23), then x is an eigenfunction associated with eigenvalue μkqt=0 for some k of the eigenvalue problem(26)x′′t+qtxt+μxt=0,xξ=xξ+T=0.
Sinceλ2n≪qt≪λ2n+1, it follows from the comparison principle that (27)λk-λ2n+1=μkλ2n+1<μkqt=0<μkλ2n=λk-λ2n.Thus 2n<k<2n+1,k=2n+1. On the other hand, it follows from Sturm theory that the eigenfunction corresponding to the eigenvalue μ2n+1 of (26) has exact 2n zeros in ξ,ξ+T, which indicates that sgn⁡x′ξ=-sgn⁡x′ξ+T. Together with the boundary condition of (23), we have x′ξ=x′ξ+T=0. This is a contradiction.Now we consider the homogeneous periodic equation(28)y′′t+ddtfx′t,x0tyt+cyt=0.We have the following.Lemma 11.
Assume that there exists an integern∈N such that H412ftx′′t,xt+12fxx′′t,xtx′t-14fx′2t,xt+c≫2nπ2T2,H512ftx′′t,xt+12fxx′′t,xtx′t-14fx′2t,xt+c≪2n+1π2T2.Then (28) does not admit any positive Floquet multipliers. In particular, (28) does not admit any nontrivial T-periodic solutions.Proof.
Suppose that there is a nontrivialT-periodic solution of (28) such that yt+T=αyt with α>0.
Set(29)yt=exp⁡-12∫0tfx′s,x0sdsut;then ut solves the following equation: (30)u′′t+c+12fxx′′t,x0tx0′t+12fxt′′t,x0t-14fx′2t,x0tut=0,with the Floquet multiplier β=αexp⁡1/2∫0Tfx′s,x0sds.
Claim. There exists some ξ∈0,T such that(31)yξ=0.
Assume, by way of contradiction, that (31) does not hold. Then ut≠0 for all t∈0,T. Dividing (30) by ut and integrating from 0 to T by parts, noticing that u′0/u0=u′T/uT, we obtain (32)∫0Tu′tut2dt+∫0Tc+12fxx′′t,x0tx0′t+12fxt′′t,x0t-14fx′2t,x0tdt=0,which contradicts condition H4. This implies that the claim (31) holds.
Sinceyξ+T=αyξ=0, we see that ut is a nontrivial solution of the Dirichlet boundary value problem(33)u′′t+qtut=0,uξ=uξ+T=0,sgn⁡u′ξ=sgn⁡u′ξ+T,where q(t)=[c+1/2fxx′′t,x0tx0′t+1/2fxt′′(t,x0(t))-1/4fx′2(t,x0(t))]. Multiplying (33) by u(t) and integrating from ξ and ξ+T, we have, by H5(34)∫ξξ+Tu′t2dt=∫ξξ+Tc+12fxx′′t,x0tx0′t+12fxt′′t,x0t-14fx′2t,x0tu2tdt<2n+12T2∫ξξ+Tu2tdt,which contradicts thepoincaré inequality.Similarly, consider the following Dirichlet boundary value problem:(35)x′′t+qtxt=0;xξ=xξ+T=0,sgn⁡x′ξ=-sgn⁡x′ξ+T,we can derive the following lemma.Lemma 12 (see [9]).
Letqt∈L∞0,T such that (36)2n-1π2T2≪qt≪2n+1π2T2for some n∈N. Then (35) does not admit any nontrivial solution.Lemma 13.
Suppose that there is an integern∈N such that H612ftx′′t,xt+12fxx′′t,xtx′t-14fx′2t,xt+c≫2n-1π2T2,H712ftx′′t,xt+12fxx′′t,xtx′t-14fx′2t,xt+c≪2n+1π2T2.Then (28) does not admit any negative Floquet multipliers. In particular, (28) does not admit any nontrivial subharmonic periodic solution of order 2.Combining Lemmas11 and 13, under the condition of Theorem 5, we can prove that (1) does not admit any real Floquet multiplier.


## 3. Proof of Main Results

Now we are ready to prove our main results.


### 3.1. Proof of Theorem1

Proof.
Without loss of generality, we may assume thatg0=0; otherwise, we can subtract g(0) from both sides of (1).
Consider the parametrized equation,(37)x′′t+λddtft,xt+1-λx′t+gxt=hλt,where hλt=λht+1-λ1/T∫0Thtdt.
First, we claim that there is anR>0 which is independent of λ∈0,1 such that x∞<R for any solution x of (37).
If there is not such anR, let xn be a sequence such that xn→∞ and λn∈0,1, and denote by zn the ratio xn/xn. Dividing (37) by xn, multiplying by φt∈CT2, and integrating from 0 to T gives (38)∫0Tznφ′′-φ′λnft,xnxn+1-λnzn+φgxnxndt=∫0Tφhλnxndt.
The condition of Theorem1 implies that λnft,xn+1-λnxn/xn is bounded. It is precompact in the weak∗ topology in L10,T. Thus there are subsequences such that ft,xn/xn→at and λn→λ. Passing to the limit in (38), we get (39)∫0Tzφ′′-wtzφ′+cφzdt=0,where wt=λat+1-λ.
If (4) holds, then wt=λat+1-λ≫0, which satisfies the condition of Lemma 7. Since zt is a T-periodic solution of (40)z′′+wtz′+cz=0,it follows from Lemma 7 that zt≡0, which contradicts the fact that zt=1. This completes the boundedness of x∞ under condition (4).
Next, we show thatx∞ is bounded which is independent of λ∈0,1 under condition (5).
Ifλ<1, then z satisfies(41)z′′+1-λz′+cz=0.It follows from Lemma 7 that zt≡0, which contradicts zt=1.
Ifλ=1, then z satisfies (42)z′′+cz=0.By assumption, c is not the eigenvalue of Lx=-x′′. Obviously, zt≡0; we reach a contradiction. This shows that the solution of (1) is bounded.
Evidently, the periodic solution of (37) is equivalent to the planar system (43)x′=y-λft,x-1-λx,y′=hλt-gx.
A natural choice for the homotopy in applying Lemma6 is to take (44)ht,x,y,λ=y-λft,x-1-λx,hλt-λgx-1-λcx.Let xt,yt be the T-periodic solution of (43); in order to apply Continuation Theorem to (43), we have to show that yt is bounded. Directly from the first equation of (43) and the periodic condition, we see that there is τ∈0,T such that x′(τ)=0 which implies that yτ is bounded which is independent of λ. Integrating the second equation of (43) yields (45)yt=yτ+∫τthλs-gxds,which is bounded.
Letr1 and r2 be sufficiently large, and set (46)Ω=x,y∣x≤r1,y≤r2.It follows from the estimates obtained above that the equivalent planar system defined in (43) (47)x′=ht,x,y,λhas no solutions on ∂Ω for λ∈0,1. Since (48)ht,x,y,0=y-x,h¯-cx=Gx,y,then condition (2) in Lemma 6 reduces to (49)deg⁡G,Ω∩R2,0=sgn⁡c≠0.
By applying Lemma6, we see that (1) has at least one T-periodic solution.


### 3.2. Proof of Theorem2

Proof.
The idea of the proof of Theorem2 is essentially the same as above, so here we just outline the proof and explain how to use the resonance conditions and (8)-(9) to get desired a priori estimates.
Consider the parametrized equation(50)x′′t+λddtft,xt+1-λx′t+gt,xt=λht.
First, we will show that there isR>0 which is independent of λ∈0,1 such that x∞<R for any solutions x of (50).
If not, letxn be a sequence of T-periodic solutions such that xn→∞ and λn∈0,1 be the corresponding sequence. Let zn=xn/xn. Noting the resonance condition (6), the same procedure of the proof of Theorem 1 gives that there exists some subsequences such that znt→zt weakly and λn→λ. Passing to the limit, one obtains that zt satisfies (51)∫0Tzφ′′-1-λzφ′+2nπT2φzdt=0.Namely, (52)z′′+1-λz′+2nπT2z=0.
Ifλ<1, it follows from Lemma 7 that zt≡0, which contradicts zt=1. Hence it remains to focus on the case that λn→1 and z∈sin⁡2nπ/Tt,cos⁡2nπ/Tt.
InWT1,2, we introduce the following symbol:(53)xt=a0+∑k=1∞aksin⁡2kπTt+bkcos⁡2kπTt,x0t=aksin⁡2kπTt+bkcos⁡2kπTt,x⊥t=xt-x0t.Since zn→zt∈sin⁡2nπ/Tt,cos⁡2nπ/Tt, we have that xn0+xn⊥/xn→zt.
That is to say,xn0/xn→1 and xn⊥/xn→0. Then xn⊥/xn0→0. Thus, the sign of xn is the same as that of z for n large enough. Taking inner product of (50) with z and noting that gt,x=2nπ/T2+qt,x, where λ is replaced by λn and x by xn, we have (54)λn∫0Thzdt=-∫0Tλnft,xn0+xn⊥z′dt+∫0Tqt,xn0+xn⊥zdt.
Taking the limits in the above equation, we obtain(55)∫0Thzdt≤liminfn→∞⁡-∫0Tft,xn0+xn⊥z′dt+∫0Tqt,xn0+xn⊥zdt≤-∫z>0a′zdt+∫z<0b′zdt+liminfn→∞⁡∫z>0qt,xnzdt+liminfn→∞⁡∫z<0qt,xnzdt≤-∫z>0a′zdt+∫z<0b′zdt+∫z>0α+zdt+∫z<0β-zdt,which contradicts (9). This shows the boundedness of the periodic solutions of (50).
Similarly, by takinglimsup one can prove that periodic solutions of (50) are bounded under condition (10).
The rest is along the same line as the proof of Theorem1; we omit the detail.


### 3.3. Proof of Theorem5

Proof.
Firstly, we will show that there exists a uniqueT-periodic solution.
The existence ofT-periodic solution of (1) has been obtained in the proof of Theorem 1. Hence it suffices to investigate the uniqueness of (1).
Letxt and yt be two distinct T-periodic solutions of (1), and vt=xt-yt. Then vt is a nontrivial T-periodic solution of the following equation: (56)x′′t+ddtqtxt+cxt=0, where qt=∫01fx′xt+syt-xt,tds.
We claim that, under the assumption of the theorem, there exists no nontrivialT-periodic solution of the linearized equation (28) associated with (1).
Supposing the opposite of the previous claim, multiplying (28) by y′t and integrating from 0 to T, we have(57)0=∫0Ty′ty′′t+ddtfx′xt,tyt+cytdt=∫0Ty′tdfx′xt,tyt=fx′xt,tyty′t0T-∫0Tfx′xt,tyty′′tdt=-∫0Tfx′xt,tyt-cyt-ddtfx′xt,tytdt=∫0Tcfx′xt,ty2tdt+∫0Tfx′xt,tyt·ddtfx′xt,tytdt=c∫0Tfx′xt,ty2tdt,which contradicts the fact that both c and fx′xt,t are positive. In the same manner, we can demonstrate that there exists no nontrivial 2T-periodic solutions.
By the similar argument as above, multiplying (56) by v′t and integrating by parts, a contradiction will be reached directly. Therefore, v≡0.
Secondly, we will show that the uniqueT-periodic solution is locally asymptotically stable.
For eachs with 0≤s≤1, we have (58)sc+fxx′′t,x0tx0′t+fxt′′t,x0t+1-s>0.Therefore, from the above conclusion, for each s∈0,1, there exists a unique periodic solution of period 2T of the linear homogeneous differential equation (59)u′′+fx′t,x0tu′+sc+fxx′′t,x0tx0′t+fxt′′t,x0t+1-su=0.
Under the hypothesis of theorem, we can conclude that there exists no nontrivial2T-periodic solution of (59) for all 0≤s≤1.
If(60)Pt,s=01-sc+fxx′′t,x0tx0′t+fxt′′t,x0t-1-s-fx′t,x0t,then y=col⁡y1,y2 is a 2T-periodic solution of (61)y′t=Pt,sytif and only if y1=u and y2=u′, where u is a nontrivial 2T-periodic solution of (59). Therefore, system (61) has no nontrivial 2T-periodic solution for 0≤s≤1.
LetYt,s denote the fundamental matrix associated with system (61) for 0≤s≤1; we have (62)Y′t,s=Pt,sYt,s,Y0,s=I,where I is the 2×2 identity.
If for somev≠0(63)YT,sv=-v,then yt≡Yt,sv is a solution of system (61) which satisfies yt=-y0. So by the fact that -yt and yt+T are both solutions of (61) which are equal at t=0 and the uniqueness theorem, it follows that yt+T≡-yt. Hence, yt is a nontrivial 2T-periodic solution. So yt≡0, and therefore v=0.
It follows that-1 is not an eigenvalue of YT,s for all 0≤s≤1.
A similar argument as above shows that ifYT,sv=v, then yt≡Yt,sv is a T-periodic solution, and hence a 2T-periodic solution of (61). Therefore, for 0≤s≤1, 1 is not an eigenvalue of YT,s.
For eachs with 0≤s≤1, let ρ1s and ρ2s denote the eigenvalues of YT,s. By standard results concerning continuous dependence of solutions of differential equations on parameters and Rouche’s theorem of complex analysis, the moduli ρ1s and ρ2s depend continuously on s. We will show that (64)ρis<1,i=1,2,0≤s≤1,and since XT=YT,1, where Xt is the fundamental matrix for the linear system (65)x′t=Atxt,and At is the matrix function (66)At=01-c+fxx′′t,x0tx0′t+fxt′′t,x0t-fx′t,x0t.
Since the trace ofPt,s is equal to -fx′t,x0t, applying Liouville’s theorem, we have (67)ρ1sρ2s=e∫0TtracePt,sdt=e-∫0Tfx′x0t,tdt<1.
Fors=0,P(t,s) is equal to the matrix (68)L=01-1-fx′x0t,t,so Yt,0=eLt and ρi0=eλiT,i=1,2, where λ1 and λ2 are the eigenvalues of L. Since fx′t,x0t>0, it is easy to verify that both λ1 and λ2 have negative real parts. Therefore, ρi0<1 for i=1,2.
If (64) were not true for 0≤s≤1, then, by continuity, there exists a s- with 0<s-≤1 such that ρis-=1, for either i=1 or i=2.
ρ
1
(
s
-
) and ρ2(s-) can not be complex conjugates, for this means ρ1s-=ρ2s-=1, which contradicts with (67). Therefore, both ρ1s- and ρ2s- would be real, so ρis-=±1 for either i=1 or i=2. This contradicts the fact that YT,s can not have ±1 as an eigenvalue for all s∈0,1.
Hence, the eigenvalues ofYT,s have moduli less than 1 for 0≤s≤1. In other words, the unique T-periodic solution is locally asymptotically stable.
Thirdly, we will show that the uniqueT-periodic solution is globally asymptotically stable.
We choose as a Lyapunov function(69)V=x′t+fxt,t-ϕ′t+fϕt,t2+cxt-ϕt2,where ϕt is the unique T-periodic solution satisfying the following equation: (70)x′′t+ddtfxt,t+cxt=ht.
A direct computation, byH1, yields (71)dVdt=2x′t-ϕ′t+fxt,t-fϕt,tx′′t-ϕ′′t+ddtfxt,t-fϕt,t+2cxt-ϕtx′t-ϕ′t=-2cxt-ϕtx′t-ϕ′t+fxt,t-fϕt,t+2cxt-ϕtx′t-ϕ′t=-2cxt-ϕt2∫01fx′xt+sϕt-xt,tds<0.Thus we display that the periodic solution is globally asymptotically stable.
Finally, we are in a position to prove the rate of decay of the solution to the uniqueT-periodic solution is 1/2∫0Tfx′t,x0tdt.
Consider the planar system associated with (1),(72)x′=y-fx,t,y′=ht-cx.
LetX0t=x0t,y0t be the unique T-periodic solution determined by the initial condition X0(0)=x0,y0. Then X0 corresponds to the unique fixed point of thepoincaré mapping PX=UT,X, where Ut,X is the initial value solution of (72) with U(0,X)=X.
LetMt be the fundamental matrix solution of the linearization (61) of (72), and by the differentiability of Xt with respect to the initial value, thepoincaré mapping can be expressed in terms of the initial value X by (73)PX-X0=MTX-X0+oX-X0.
By virtue of Lemmas11 and 13, we see that MT has a pair of conjugate eigenvalue ρ1 and ρ2. Thus PX is a contracting mapping. According to Lemma 9, there is a C1 diffeomorphism ϕ which is near enough to the identity that PX-X0 is conjugate equivalent to MT. There is an invertible constant matrix B such that (74)B-1MTB=ρ1o0ρ2=D,and we may suppose that (75)12X-X0<ϕX-ϕX0<2X-X0,for X-X0 small, since ϕ is near the identity.
So, the Lyapunov exponent is given by(76)μ=limn→∞⁡1nTln⁡PnX-X0=limn→∞⁡1nTln⁡ϕ∘MTn∘ϕ-1X-ϕ∘MTn∘ϕ-1X0=limn→∞⁡1nTln⁡Dn∘B-1ϕ-1X-ϕ-1X0=-12∫0Tfx′t,x0tdt.
Hence, the rate of decay of the solution to the uniqueT-periodic solution is 1/2∫0Tfx′t,x0tdt, independent of the initial value X. This completes the proof of Theorem 5.


## 3.1. Proof of Theorem1

Proof.
Without loss of generality, we may assume thatg0=0; otherwise, we can subtract g(0) from both sides of (1).
Consider the parametrized equation,(37)x′′t+λddtft,xt+1-λx′t+gxt=hλt,where hλt=λht+1-λ1/T∫0Thtdt.
First, we claim that there is anR>0 which is independent of λ∈0,1 such that x∞<R for any solution x of (37).
If there is not such anR, let xn be a sequence such that xn→∞ and λn∈0,1, and denote by zn the ratio xn/xn. Dividing (37) by xn, multiplying by φt∈CT2, and integrating from 0 to T gives (38)∫0Tznφ′′-φ′λnft,xnxn+1-λnzn+φgxnxndt=∫0Tφhλnxndt.
The condition of Theorem1 implies that λnft,xn+1-λnxn/xn is bounded. It is precompact in the weak∗ topology in L10,T. Thus there are subsequences such that ft,xn/xn→at and λn→λ. Passing to the limit in (38), we get (39)∫0Tzφ′′-wtzφ′+cφzdt=0,where wt=λat+1-λ.
If (4) holds, then wt=λat+1-λ≫0, which satisfies the condition of Lemma 7. Since zt is a T-periodic solution of (40)z′′+wtz′+cz=0,it follows from Lemma 7 that zt≡0, which contradicts the fact that zt=1. This completes the boundedness of x∞ under condition (4).
Next, we show thatx∞ is bounded which is independent of λ∈0,1 under condition (5).
Ifλ<1, then z satisfies(41)z′′+1-λz′+cz=0.It follows from Lemma 7 that zt≡0, which contradicts zt=1.
Ifλ=1, then z satisfies (42)z′′+cz=0.By assumption, c is not the eigenvalue of Lx=-x′′. Obviously, zt≡0; we reach a contradiction. This shows that the solution of (1) is bounded.
Evidently, the periodic solution of (37) is equivalent to the planar system (43)x′=y-λft,x-1-λx,y′=hλt-gx.
A natural choice for the homotopy in applying Lemma6 is to take (44)ht,x,y,λ=y-λft,x-1-λx,hλt-λgx-1-λcx.Let xt,yt be the T-periodic solution of (43); in order to apply Continuation Theorem to (43), we have to show that yt is bounded. Directly from the first equation of (43) and the periodic condition, we see that there is τ∈0,T such that x′(τ)=0 which implies that yτ is bounded which is independent of λ. Integrating the second equation of (43) yields (45)yt=yτ+∫τthλs-gxds,which is bounded.
Letr1 and r2 be sufficiently large, and set (46)Ω=x,y∣x≤r1,y≤r2.It follows from the estimates obtained above that the equivalent planar system defined in (43) (47)x′=ht,x,y,λhas no solutions on ∂Ω for λ∈0,1. Since (48)ht,x,y,0=y-x,h¯-cx=Gx,y,then condition (2) in Lemma 6 reduces to (49)deg⁡G,Ω∩R2,0=sgn⁡c≠0.
By applying Lemma6, we see that (1) has at least one T-periodic solution.


## 3.2. Proof of Theorem2

Proof.
The idea of the proof of Theorem2 is essentially the same as above, so here we just outline the proof and explain how to use the resonance conditions and (8)-(9) to get desired a priori estimates.
Consider the parametrized equation(50)x′′t+λddtft,xt+1-λx′t+gt,xt=λht.
First, we will show that there isR>0 which is independent of λ∈0,1 such that x∞<R for any solutions x of (50).
If not, letxn be a sequence of T-periodic solutions such that xn→∞ and λn∈0,1 be the corresponding sequence. Let zn=xn/xn. Noting the resonance condition (6), the same procedure of the proof of Theorem 1 gives that there exists some subsequences such that znt→zt weakly and λn→λ. Passing to the limit, one obtains that zt satisfies (51)∫0Tzφ′′-1-λzφ′+2nπT2φzdt=0.Namely, (52)z′′+1-λz′+2nπT2z=0.
Ifλ<1, it follows from Lemma 7 that zt≡0, which contradicts zt=1. Hence it remains to focus on the case that λn→1 and z∈sin⁡2nπ/Tt,cos⁡2nπ/Tt.
InWT1,2, we introduce the following symbol:(53)xt=a0+∑k=1∞aksin⁡2kπTt+bkcos⁡2kπTt,x0t=aksin⁡2kπTt+bkcos⁡2kπTt,x⊥t=xt-x0t.Since zn→zt∈sin⁡2nπ/Tt,cos⁡2nπ/Tt, we have that xn0+xn⊥/xn→zt.
That is to say,xn0/xn→1 and xn⊥/xn→0. Then xn⊥/xn0→0. Thus, the sign of xn is the same as that of z for n large enough. Taking inner product of (50) with z and noting that gt,x=2nπ/T2+qt,x, where λ is replaced by λn and x by xn, we have (54)λn∫0Thzdt=-∫0Tλnft,xn0+xn⊥z′dt+∫0Tqt,xn0+xn⊥zdt.
Taking the limits in the above equation, we obtain(55)∫0Thzdt≤liminfn→∞⁡-∫0Tft,xn0+xn⊥z′dt+∫0Tqt,xn0+xn⊥zdt≤-∫z>0a′zdt+∫z<0b′zdt+liminfn→∞⁡∫z>0qt,xnzdt+liminfn→∞⁡∫z<0qt,xnzdt≤-∫z>0a′zdt+∫z<0b′zdt+∫z>0α+zdt+∫z<0β-zdt,which contradicts (9). This shows the boundedness of the periodic solutions of (50).
Similarly, by takinglimsup one can prove that periodic solutions of (50) are bounded under condition (10).
The rest is along the same line as the proof of Theorem1; we omit the detail.


## 3.3. Proof of Theorem5

Proof.
Firstly, we will show that there exists a uniqueT-periodic solution.
The existence ofT-periodic solution of (1) has been obtained in the proof of Theorem 1. Hence it suffices to investigate the uniqueness of (1).
Letxt and yt be two distinct T-periodic solutions of (1), and vt=xt-yt. Then vt is a nontrivial T-periodic solution of the following equation: (56)x′′t+ddtqtxt+cxt=0, where qt=∫01fx′xt+syt-xt,tds.
We claim that, under the assumption of the theorem, there exists no nontrivialT-periodic solution of the linearized equation (28) associated with (1).
Supposing the opposite of the previous claim, multiplying (28) by y′t and integrating from 0 to T, we have(57)0=∫0Ty′ty′′t+ddtfx′xt,tyt+cytdt=∫0Ty′tdfx′xt,tyt=fx′xt,tyty′t0T-∫0Tfx′xt,tyty′′tdt=-∫0Tfx′xt,tyt-cyt-ddtfx′xt,tytdt=∫0Tcfx′xt,ty2tdt+∫0Tfx′xt,tyt·ddtfx′xt,tytdt=c∫0Tfx′xt,ty2tdt,which contradicts the fact that both c and fx′xt,t are positive. In the same manner, we can demonstrate that there exists no nontrivial 2T-periodic solutions.
By the similar argument as above, multiplying (56) by v′t and integrating by parts, a contradiction will be reached directly. Therefore, v≡0.
Secondly, we will show that the uniqueT-periodic solution is locally asymptotically stable.
For eachs with 0≤s≤1, we have (58)sc+fxx′′t,x0tx0′t+fxt′′t,x0t+1-s>0.Therefore, from the above conclusion, for each s∈0,1, there exists a unique periodic solution of period 2T of the linear homogeneous differential equation (59)u′′+fx′t,x0tu′+sc+fxx′′t,x0tx0′t+fxt′′t,x0t+1-su=0.
Under the hypothesis of theorem, we can conclude that there exists no nontrivial2T-periodic solution of (59) for all 0≤s≤1.
If(60)Pt,s=01-sc+fxx′′t,x0tx0′t+fxt′′t,x0t-1-s-fx′t,x0t,then y=col⁡y1,y2 is a 2T-periodic solution of (61)y′t=Pt,sytif and only if y1=u and y2=u′, where u is a nontrivial 2T-periodic solution of (59). Therefore, system (61) has no nontrivial 2T-periodic solution for 0≤s≤1.
LetYt,s denote the fundamental matrix associated with system (61) for 0≤s≤1; we have (62)Y′t,s=Pt,sYt,s,Y0,s=I,where I is the 2×2 identity.
If for somev≠0(63)YT,sv=-v,then yt≡Yt,sv is a solution of system (61) which satisfies yt=-y0. So by the fact that -yt and yt+T are both solutions of (61) which are equal at t=0 and the uniqueness theorem, it follows that yt+T≡-yt. Hence, yt is a nontrivial 2T-periodic solution. So yt≡0, and therefore v=0.
It follows that-1 is not an eigenvalue of YT,s for all 0≤s≤1.
A similar argument as above shows that ifYT,sv=v, then yt≡Yt,sv is a T-periodic solution, and hence a 2T-periodic solution of (61). Therefore, for 0≤s≤1, 1 is not an eigenvalue of YT,s.
For eachs with 0≤s≤1, let ρ1s and ρ2s denote the eigenvalues of YT,s. By standard results concerning continuous dependence of solutions of differential equations on parameters and Rouche’s theorem of complex analysis, the moduli ρ1s and ρ2s depend continuously on s. We will show that (64)ρis<1,i=1,2,0≤s≤1,and since XT=YT,1, where Xt is the fundamental matrix for the linear system (65)x′t=Atxt,and At is the matrix function (66)At=01-c+fxx′′t,x0tx0′t+fxt′′t,x0t-fx′t,x0t.
Since the trace ofPt,s is equal to -fx′t,x0t, applying Liouville’s theorem, we have (67)ρ1sρ2s=e∫0TtracePt,sdt=e-∫0Tfx′x0t,tdt<1.
Fors=0,P(t,s) is equal to the matrix (68)L=01-1-fx′x0t,t,so Yt,0=eLt and ρi0=eλiT,i=1,2, where λ1 and λ2 are the eigenvalues of L. Since fx′t,x0t>0, it is easy to verify that both λ1 and λ2 have negative real parts. Therefore, ρi0<1 for i=1,2.
If (64) were not true for 0≤s≤1, then, by continuity, there exists a s- with 0<s-≤1 such that ρis-=1, for either i=1 or i=2.
ρ
1
(
s
-
) and ρ2(s-) can not be complex conjugates, for this means ρ1s-=ρ2s-=1, which contradicts with (67). Therefore, both ρ1s- and ρ2s- would be real, so ρis-=±1 for either i=1 or i=2. This contradicts the fact that YT,s can not have ±1 as an eigenvalue for all s∈0,1.
Hence, the eigenvalues ofYT,s have moduli less than 1 for 0≤s≤1. In other words, the unique T-periodic solution is locally asymptotically stable.
Thirdly, we will show that the uniqueT-periodic solution is globally asymptotically stable.
We choose as a Lyapunov function(69)V=x′t+fxt,t-ϕ′t+fϕt,t2+cxt-ϕt2,where ϕt is the unique T-periodic solution satisfying the following equation: (70)x′′t+ddtfxt,t+cxt=ht.
A direct computation, byH1, yields (71)dVdt=2x′t-ϕ′t+fxt,t-fϕt,tx′′t-ϕ′′t+ddtfxt,t-fϕt,t+2cxt-ϕtx′t-ϕ′t=-2cxt-ϕtx′t-ϕ′t+fxt,t-fϕt,t+2cxt-ϕtx′t-ϕ′t=-2cxt-ϕt2∫01fx′xt+sϕt-xt,tds<0.Thus we display that the periodic solution is globally asymptotically stable.
Finally, we are in a position to prove the rate of decay of the solution to the uniqueT-periodic solution is 1/2∫0Tfx′t,x0tdt.
Consider the planar system associated with (1),(72)x′=y-fx,t,y′=ht-cx.
LetX0t=x0t,y0t be the unique T-periodic solution determined by the initial condition X0(0)=x0,y0. Then X0 corresponds to the unique fixed point of thepoincaré mapping PX=UT,X, where Ut,X is the initial value solution of (72) with U(0,X)=X.
LetMt be the fundamental matrix solution of the linearization (61) of (72), and by the differentiability of Xt with respect to the initial value, thepoincaré mapping can be expressed in terms of the initial value X by (73)PX-X0=MTX-X0+oX-X0.
By virtue of Lemmas11 and 13, we see that MT has a pair of conjugate eigenvalue ρ1 and ρ2. Thus PX is a contracting mapping. According to Lemma 9, there is a C1 diffeomorphism ϕ which is near enough to the identity that PX-X0 is conjugate equivalent to MT. There is an invertible constant matrix B such that (74)B-1MTB=ρ1o0ρ2=D,and we may suppose that (75)12X-X0<ϕX-ϕX0<2X-X0,for X-X0 small, since ϕ is near the identity.
So, the Lyapunov exponent is given by(76)μ=limn→∞⁡1nTln⁡PnX-X0=limn→∞⁡1nTln⁡ϕ∘MTn∘ϕ-1X-ϕ∘MTn∘ϕ-1X0=limn→∞⁡1nTln⁡Dn∘B-1ϕ-1X-ϕ-1X0=-12∫0Tfx′t,x0tdt.
Hence, the rate of decay of the solution to the uniqueT-periodic solution is 1/2∫0Tfx′t,x0tdt, independent of the initial value X. This completes the proof of Theorem 5.



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