# A High-Order Melnikov Method for Heteroclinic Orbits in Planar Vector Fields and Heteroclinic Persisting Perturbations

Authors: Yi Zhong
Journal: Journal of Mathematics
 (2021)

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## Abstract

This work extends the high-order Melnikov method established by FJ Chen and QD Wang to heteroclinic orbits, and it is used to prove, under a certain class of perturbations, the heteroclinic orbit in a planar vector field that remains unbroken. Perturbations which have this property together form the heteroclinic persisting space. The Van der Pol system is analysed as an application.


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## Body

## 1. Introduction

Since the pioneering work of Poincaré on celestial mechanics [1, 2], homoclinic tangles created by transversal intersection of the stable and unstable manifolds have been playing a prominent role in the progress of chaos theory. In 1967, Smale observed that horseshoe chaos occurs in all homoclinic tangles [3]. For the past few years, the studies in [4–7] presented that periodic sinks and H e´ nonlike attractors are also residing in homoclinic and heteroclinic tangles. The H e´ nonlike attractors provide observable chaos in the sense of Sinai–Ruelle–Bowen (SRB) measure [8, 9].In general, the standard Melnikov function is enough to verify whether the homoclinic tangle or heteroclinic tangle occurs [10, 11]. However, there exist perturbations which make the standard Melnikov function identically equal to zero. In such a case, no simple zero exists. As a result, we need to calculate high-order Melnikov functions for further investigation. There have been many previous works to refer high-order Melnikov method (see [12–19]). We focus on the method presented in [13] and extends the high-order Melnikov functions for homoclinic orbits to heteroclinic orbits.Naturally, another question arises: does there exist a perturbation such that all the high-order Melnikov functions identically equal to zero? This means the homoclinic or heteroclinic solution remains unbroken. Through the high-order Melnikov method, we prove that the answer of the above question is positive. The union of such perturbations will be defined as heteroclinic persisting space.This paper is built up as follows: In Section2, we outline the main results. Section 3 is devoted to focus on the derivation of high-order Melnikov functions for heteroclinic orbits. Then, in Section 4, we prove the heteroclinic orbit remains unbroken under a certain class of autonomous perturbations. Finally, Van der Pol system is considered in Section 5.


## 2. Statement of Results

We start with a planar system:(1)dxdt=fx,y,dydt=gx,y.Assume that (1) has two hyperbolic saddle fixed points O1x1,y1 and O2x2,y2 with a heteroclinic solution(2)ℓ=at,bt:t∈−∞,+∞,such that(3)limt⟶−∞at,bt=x1,y1,limt⟶+∞at,bt=x2,y2.The functionsfx,y and gx,y are real analytic near ℓ∪O1,O2.Adding time-periodic perturbations to (1) yields(4)dxdt=fx,y+εPx,y,t,dydt=gx,y+εQx,y,t,where ε is a small parameter. The periodic functions Px,y,tandQx,y,t are real analytic with respect to x,y, and they are terms of second order and higher at O1 and O2. Let WO1 be the unstable manifold of O1 and WεsO2 be the stable manifold of O2. In general, WεuO1≠WεsO2 for ε≠0. We let Dε,t0 be the splitting distance, which can be expressed in the form of(5)Dε,t0=M0t0ε+M1t0ε2+⋯+Mkt0εk+1+⋯,where M0t0 is equivalent to the standard Melnikov function in [10, 11] and Mkt0, k=1,2,…, are the high-order Melnikov functions. Moreover, we write a=at, b=bt, a˙=da/dt, b˙=db/dt, a¨=da˙/dt, and b¨=db˙/dt for easy of notations. We also introduce(6)At,0=−A0ta˙2+b˙2,A1,0t=dAt,0dt,A0,1t=−A1ta˙2+b˙2+A0tHta˙2+b˙22,Bt,0=B0ta˙2+b˙2,Ut,0,t+t0=−U0t,t0a˙2+b˙2,Vt,0,t+t0=V0t,t0a˙2+b˙2,U1,0t,t0=dUt,0,t+t0dt,U0,1t,t0=−U1t,t0a˙2+b˙2+U0t,t0Hta˙2+b˙22,Z0±t,t0=∫0±∞Uτ,0,τ+t0e∫0τAω,0dωdτ,S0±t,t0=∫0tBτ,0Z0±τ,t0+Vτ,0,τ+t0dτ,where(7)A0t=fxa,b−gya,bb˙2−a˙2−2fya,b+gxa,ba˙b˙,U0t,t0=b˙Pa,b,t+t0−a˙Qa,b,t+t0,B0t=fya,b+gxa,bb˙2−a˙2+2fxa,b−gya,ba˙b˙,V0t,t0=a˙Pa,b,t+t0+b˙Qa,b,t+t0,A1t,t0=b˙2fxxa,bb˙2+fyya,ba˙2−2fxya,ba˙b˙−a2gxxa,bb˙2+gyya,ba˙2−2gxya,ba˙b˙−fx2a,b−fy2a,b−gy2a,b+gx2a,ba˙b˙+fxa,bfya,b+gxa,bgya,ba˙2−b˙2,U1t,t0=a˙2Qya,b,t+t0+b˙2Pxa,b,t+t0−Pya,b,t+t0+Qxa,b,t+t0a˙b˙−a¨Pa,b,t+t0−b¨Qa,b,t+t0,Ht=a˙b¨−a¨b¨.In what follows, we give the formulas ofM0t0 and M1t0.Theorem 1.
In (5), we have the following:(i)
Integral formula forM0t0:(8)M0t0=∫−∞+∞Ut,0,t+t0e∫0tAω,0dωdt.(ii)
Integral formula forM1t0:(9)M1t0=∫0+∞A1,0tZ0+t,t0S0+t,t0+U1,0t,t0S0+t,t0+A0,1tZ0+t,t02+U0,1t,t0Z0+t,t0e∫0tAω,0dωdt−∫0−∞A1,0tZ0−t,t0S0−t,t0+U1,0t,t0S0−t,t0+A0,1tZ0−t,t02+U0,1t,t0Z0−t,t0e∫0tAω,0dωdt.On the other hand, we consider the question whether there exist any perturbations such that the heteroclinic solution remains unbroken, which implies the splitting distanceDε,t0 vanishes identically. By using the high-order Melnikov functions above, such class of perturbations are obtained. Before giving our main theorem, we first introduce some definitions.Definition 1.
LetX be a smooth manifold and F and G be vector fields on it. We assume there exists a heteroclinic orbit ℓ of F and a heteroclinic orbit ℓε of F+εG, where ε is a small parameter. We call G a heteroclinic persisting perturbation of ℓ if(10)limε⟶0ℓε=ℓ,in the sense of Hausdorff limit.Definition 2.
All heteroclinic persisting perturbations ofℓ are referred to as the heteroclinic persisting space of ℓ, which we denote by HPℓ.It is not difficult to find a heteroclinic persisting perturbation. LetPx,y,t,Qx,y,t in (4) be satisfying(11)Pat,bt,t=Qat,bt,t=0,t∈−∞,+∞.Then, we have(12)datdt=fat,bt+εPat,bt,t,dbtdt=gat,bt+εQat,bt,t.It follows thatat,bt is also a heteroclinic solution of (4). By Definition 1, we claim Px,y,t,Qx,y,t is a heteroclinic persisting perturbation.However, this heteroclinic persisting perturbation is trivial, since the heteroclinic orbitat,bt is just the previous one. We would like to quest for more general ones. For this purpose, we focus on an autonomous perturbed system:(13)dxdt=fx,y+εPx,y,dydt=gx,y+εQx,y.Then, splitting distance (5) can be rewritten as(14)Dε,P,Q=M0P,Qε+M1P,Qε2+⋯+MkP,Qεk+1…,where MkP,Q is the k+1-th order Melnikov integral, k=0,1,2,… The following fact is trivial.The fact is thatPx,y,Qx,y is a heteroclinic persisting perturbation if and only if Dε,P,Q=0 for all ε<ε0, where ε0>0 is sufficiently small.Note that the geometric shape of heteroclinic orbits in planar vector fields are often symmetric which contributes to the computations of high-order Melnikov integrals. We obtain the following theorem.Theorem 2.
For the given vector field defined by (1), we assume at is odd and bt is even in t and fx,y is even and gx,y is odd in x. Then, we have(15)P,Q:Px,yis even inx,Qx,yis odd inx⊂HPℓ.The example considered in Section5 will show that the parity assumption made in Theorem 2 is without loss of generality.


## 3. High-Order Melnikov Method

This section is intended to derive the high-order Melnikov functions for equation (4). The method here is based on the homoclinic case introduced in [13], but there still exist some nonnegligible differences between them.


### 3.1. Canonical Equation around the Heteroclinic Orbit

We introduce new phase variabless,z by letting(16)x,y=as,bs+zb˙s,−a˙s.Differentiating both sides of (16) with respect to t gives(17)dxdt=a˙s+zb¨sdsdt+b˙sdzdt,dydt=b˙s−za¨sdsdt−a˙sdzdt.By Cramer’s Rule and (4), we have(18)dzdt=−As,zz−εUs,z,t,dsdt=1+Bs,zz+εVs,z,t,where(19)As,z=−1zb˙−za¨fa+zb˙,b−za˙−a˙+zb¨ga+zb˙,b−za˙a˙2+b˙2+za˙b¨−a¨b˙Us,z,t=−b˙−za¨Pa+zb˙,b−za˙,t−a˙+zb¨Qa+zb˙,b−za˙,ta˙2+b˙2+za˙b¨−a¨b˙Bs,z=1za˙fa+zb˙,b−za˙+b˙ga+zb˙,b−za˙a˙2+b˙2+za˙b¨−a¨b˙−1Vs,z,t=a˙Pa+zb˙,b−za˙,t+b˙Qa+zb˙,b−za˙,ta˙2+b˙2+za˙b¨−a¨b˙.We supposes^t,z^t to be a stable solution or an unstable solution of system (18) such that s^t0=0 and z^t0=z0. Let st=s^t+t0 and zt=z^t+t0; then, st,zt is well-defined for all t∈0,+∞ or t∈−∞,0, and it is the solution of the system(20)dzdt=−As,zz−εUs,z,t+t0,dsdt=1+Bs,zz+εVs,z,t+t0,satisfying s0=0andz0=z0. To differ from the other solutions, we call them the primary stable and unstable solutions.


### 3.2. Integral of Primary Solutions

We let(21)Z=ε−1z,S=ε−1s−t,as one more change of variables for system (20). Then, we have(22)dZdt=−At+εS,εZZ−Ut+εS,εZ,t+t0,dSdt=Bt+εS,εZZ+Vt+εS,εZ,t+t0.ExpandingAt+εS,t+εZ, Ut+εS,t+εZ,t+t0, Bt+εS,t+εZ, and Vt+εS,t+εZ,t+t0 at t,0, we obtain(23)At+εS,t+εZ=At,0+∑n+m≥1An,mtεn+mSnZm,Ut+εS,t+εZ,t+t0=Ut,0,t+t0+∑n+m≥1Un,mt,t0εn+mSnZm,Bt+εS,t+εZ=Bt,0+∑n+m≥1Bn,mtεn+mSnZm,Vt+εS,t+εZ,t+t0=Vt,0,t+t0+∑n+m≥1Vn,mt,t0εn+mSnZm.It then follows that(24)dZdt=−At,0Z−Ut,0,t+t0−∑n+m≥1εn+mAn,mtSnZm+1+Un,mt,t0SnZm,dSdt=Bt,0Z+Vt,0,t+t0+∑n+m≥1εn+mBn,mtSnZm+1+Vn,mt,t0SnZm.We solve (24) to obtain the primary stable solution(25)Z+t,t0=e−∫0tAτ1,0dτ1Z0+−∫0tUτ,0,τ+t0e∫0τAτ1,0dτ1dτ−∑n+m=1+∞εn+m∫0tAn,mτZ+τ,t0+Un,mτ,t0S+τ,t0nZ+τ,t0me∫0τAτ1,0dτ1dτ,S+t,t0=∫0tVτ,0,τ+t0+Bτ,0Z+τ,t0dτ+∑n+m=1+∞εn+m∫0tBn,mτZ+τ,t0+Vn,mτ,t0S+τ,t0nZ+τ,t0mdτ,and the primary unstable solution(26)Z−t,t0=e−∫0tAω,0dωZ0−−∫0tUτ,0,τ+t0e∫0τAτ1,0dτ1dτ−∑n+m=1+∞εn+m∫0tAn,mτZ−τ,t0+Un,mτ,t0S−τ,t0nZ−τ,t0me∫0τAτ1,0dτ1dτ,S−t,t0=∫0tVτ,0,τ+t0+Bτ,0Z−τ,t0dτ+∑n+m=1+∞εn+m∫0tBn,mτZ−τ,t0+Vn,mτ,t0S−τ,t0nZ−τ,t0mdτ,in which Z0+ and Z0− are the initial values.Before proceeding further, we need to consider the asymptotic behaviors ofat and bt as t⟶±∞ which are crucial to get the integral equations for primary solutions. Observe that(27)limt⟶−∞at,bt=x1,y1,limt⟶+∞at,bt=x2,y2,and due to the local linearization theory, there exists a small number r>0 such that(28)fx,ygx,y=fxx1,y1fyx1,y1gxx1,y1gyx1,y1x−x1y−y1+f1x,yg1x,y,x−x1y−y1<2r;fx,ygx,y=fxx2,y2fyx2,y2gxx2,y2gyx2,y2x−x2y−y2+f2x,yg2x,y,x−x2y−y2<2r,where f1x,y and g1x,y are terms of second order and higher at O1x1,y1 and f2x,y and g2x,y are terms of second order and higher at O2x2,y2. We assume that λ1<0<μ1 are the eigenvalues of O1x1,y1 and λ2<0<μ2 are the eigenvalues of O2x2,y2. There exist two invertible matrices Γ1=α1β1γ1δ1 and Γ2=α2β2γ2δ2 such that(29)Γ1fxx1,y1fyx1,y1gxx1,y1gyx1,y1Γ1−1=λ100μ1,(30)Γ2fxx2,y2fyx2,y2gxx2,y2gyx2,y2Γ2−1=λ200μ2.In what follows, we introduce two coordinate transformations:(31)u1v1=Γ1x−x1y−y1,u2v2=Γ2x−x2y−y2.Unperturbed system (1) is transformed to(32)du1dt=λ1u1+f˜u1,v1,dv1dt=μ1v1+g˜u1,v1,du2dt=λ2u2+f˜u2,v2,dv2dt=μ2v2+g˜u2,v2.In the caset⟶−∞, our problem is about the unstable solution in a small neighborhood of O1. Let T0>0 be sufficiently large. For t<−T0, we can find a near-identity coordinate transformation:(33)u1=X+∑j=2+∞ϕj1X,Y,v1=Y+∑j=2+∞φj1X,Y,where ϕj1X,Y and φj1X,Y are homogeneous polynomials of degree j in X,Y, such that the unperturbed system near O1 is linearized to(34)dXdt=λ1X,dYdt=μ1Y.Assume that the power series in (33) are convergent on X,Y<2r. It is not difficult to verify that(35)α1β1γ1δ1at−x1bt−y1=reμ1t−t0+∑j=2+∞ϕj1reμ1t−t0,0∑j=2+∞φj1reμ1t−t0,0,where t0 be such that Xt0=r and Yt0=0. We solve (35) to obtain(36)at−x1=1detΓ1−β1reμ1t−t0+∑j=2+∞δ1ϕj1reμ1t−t0,0−β1φj1reμ1t−t0,0,bt−y1=1detΓ1α1reμ1t−t0−∑j=2+∞γ1ϕj1reμ1t−t0,0−α1φj1reμ1t−t0,0.On the other hand, lett>T0. The problem is then about the stable solution in a small neighborhood of O2. We similarly have(37)at−x2=1detΓ2δ2reλ2t−t0+∑j=2+∞δ2ϕj2reλ2t−t0,0−β2φj2reλ2t−t0,0,bt−y2=1detΓ2−γ2reλ2t−t0−∑j=2+∞γ2ϕj2reλ2t−t0,0−α2φj2reλ2t−t0,0,where ϕj2 and φj2 are homogeneous polynomials of degree j.Lemma 1.
For the above heteroclinic solutionℓ, we have(38)limt⟶−∞At,0>0,limt⟶+∞At,0<0.Proof.
We recall that(39)As,z=−1zb˙−za¨fa+zb˙,b−za˙−a˙+zb˙ga+zb˙,b−za˙a˙2+b˙2+za˙b¨−a¨b˙,and let(40)As,z=−za˙2+b˙2+za˙b¨−a˙b˙As,z.
Then,(41)As,z=b˙−za¨fa,b+zb˙fxa,b−za˙fya,b+Oz2−a˙+zb¨ga,b+zb˙gxa,b−za˙gya,b+Oz2=b˙fa,b−a′ga,b−za¨fa,b−zb¨ga,b+zb˙2fxa,b−za˙b˙fya,b−za˙b˙gxa,b+za˙2gya,b+Oz2=−za¨fa,b−zb¨ga,b+zb˙2fxa,b+za˙2gya,b−za˙b˙fya,b−za˙b˙gxa,b+Oz2.
This is to imply(42)As,0=d/dsfa,bfa,b+d/dsga,bga,b−g2a,bfxa,bf2a,b+g2a,b+−f2a,bgya,b+fa,bga,bfya,b+gxa,bf2a,b+g2a,b=fxf+fygf+gxf+gygg−g2fx−f2gy+fgfy+gxf2+g2=fx−gyf2−g2+2fy+gxfgf2+g2.
Ass<−T0, we infer from (36) that(43)fa,b=−1detΓ1β1μ1reμ1s−t0+Or2,ga,b=1detΓ1α1μ1reμ1s−t0+Or2.
Accordingly, we have(44)lims⟶−∞As,0=lims⟶−∞fx−gyf2−g2+2fy+gxfgf2+g2as,bs=lims⟶−∞fxas,bs−gyas,bsβ12−α12μ12r2e2μ1s−t0+Or3α12+β12μ12r2e2μ1s−t0+Oe3μ1s−t0−2fyas,bs+gxas,bsα1β1μ12r2e2μ1s−t0+Or3α12+β12μ12r2e2μ1s−t0+Oe3μ1s−t0.
Recall (29), and we have(45)lims⟶−∞fxas,bsfyas,bsgxas,bsgyas,bs=α1β1γ1δ1−1λ100μ1α1β1γ1δ1=1detΓ1λ1α1δ1−μ1β1γ1λ1−μ1β1δ1μ1−λ1α1γ1−λ1β1γ1+μ1α1δ1.
Substituting (45) into (44) gives(46)lims⟶−∞As,0=λ1−μ1α1δ1+β1γ1β12−α12−2λ1−μ1β1δ1−α1γ1α1β1detΓ1α12+β12=λ1−μ1β1γ1−α1δ1α12+β12detΓ1α12+β12=μ1−λ1>0.
Ass>T0, we infer from (37) that(47)fa,b=1detΓ2λ2δ2reλ2s−t0+Or2,ga,b=−1detΓ2λ2γ2reλ2s−t0+Or2.
After similar derivation, we have(48)lims⟶+∞As,0=λ2−μ2<0.
This completes the proof.By using Lemma2, one can easily verify that(49)limt⟶−∞e−∫0tAτ1,0dτ1=limt⟶+∞e−∫0tAτ1,0dτ1=+∞.Observe thatZ+t,t0 and Z−t,t0 should be bounded as t⟶±∞, and the initial values in (25) and (26) are claimed to be(50)Z0+=∫0+∞Uτ,0,τ+t0e∫0τAτ1,0dτ1dτ+∑n+m=1+∞εn+m∫0+∞An,mτZ+τ,t0+Un,mτ,t0S+τ,t0nZ+τ,t0me∫0τAτ1,0dτ1dτ(51)Z0−=∫0−∞Uτ,0,τ+t0e∫0τAτ1,0dτ1dτ+∑n+m=1+∞εn+m∫0−∞An,mτZ−τ,t0+Un,mτ,t0S−τ,t0nZ−τ,t0me∫0τAτ1,0dτ1dτ.Substituting (50) and (51) into (25) and (26), we have the integral equations of the primary stable solution(52)Z+t,t0=∫t+∞Uτ,0,τ+t0e∫tτAτ1,0dτ1dτ+∑n+m≥1εn+m∫t+∞An,mτZ+τ,t0+Un,mτ,t0S+τ,t0nZ+τ,t0mdτ,S+t,t0=∫0tVτ,0,τ+t0dτ+∫0tBτ,0Z+τ,t0dτ+∑n+m≥1εn+m∫0tBn,mτZ+τ,t0+Vn,mτ,t0S+τ,t0nZ+τ,t0mdτ,and the primary unstable solution(53)Z−t,t0=∫t−∞Uτ,0,τ+t0e∫tτAτ1,0dτ1dτ+∑n+m≥1εn+m∫t−∞An,mτZ−τ,t0+Un,mτ,t0S−τ,t0nZ−τ,t0mdτ,S−t,t0=∫0tVτ,0,τ+t0dτ+∫0tBτ,0Z−τ,t0dτ+∑n+m≥1εn+m∫0tBn,mτZ−τ,t0+Vn,mτ,t0S−τ,t0nZ−τ,t0mdτ.


### 3.3. Derivation of High-Order Melnikov Functions

The properties ofZ±t,t0,S±t,t0 here are coincident with the ones in [13]. We have the following power series:(54)Z±t,t0=Z0±t,t0+∑k=1∞εkZk±t,t0,S±t,t0=S0±t,t0+∑k=1∞εkSk±t,t0.In what follows, we substitute (54) into (52) and (53) to determine Zk±t,t0k=0,1,…, and Sk±t,t0k=0,1,…, inductively. They are(55)Z0±t,t0=∫0±∞Uτ,0,τ+t0E0,τdτ,S0±t,t0=∫0±tVτ,0+Bτ,0Z0±τdτ,(56)Zk±t,t0=∑n+m+i+ρ=k∫t±∞An,mτZi±τ,t0+Un,mτ,t0Cρ,±n,mτ,t0e∫tτAτ1,0dτ1dτ,Sk±t,t0=∑n+m+i+ρ=k∫0tBn,mτZi±τ,t0+Vn,mτ,t0Cρ,±n,mτ,t0dτ,for k=1,2,…,. Here, n,m,i,andρ are nonnegative integrals, and n+m≥1, i≥0, and ρ≥0. Cρ,+n,mt,t0 and Cρ,−n,mt,t0 are coefficient functions of ερ, which have the form of(57)Cρ,±n,mt,t0=∑ℓ0+⋯+ℓρ=n,l0+⋯+lρ=m,ℓ1+l1+⋯+ρ⋅ℓρ+lρ=ρK∏j=0ρSj±t,t0ℓj⋅Zj±t,t0lj.Here,K is a constant depending on the specific term. In the next section, these notations contribute to the proof of Theorem 2.Recall the definition ofDε,t0, and we have(58)Dε,t0=εZ+0,t0−Z−0,t0=∑k=0+∞εk+1Zk+0,t0−Zk−0,t0.Then,(59)M0t0=∫−∞+∞Ut,0,t+t0e∫0tAω,0dωdt,and for k=1,2,…, the k+1-th order Melnikov function is(60)Mkt0=Zk+0,t0−Zk−0,t0,where Zk+0,t0 and Zk−0,t0 are the values by letting t=0 in (56). Particularly, we have(61)Z1+0,t0=∑n+m+i+ρ=1∫0+∞An,mτZi+τ,t0+Un,mτ,t0Cρ,+n,mτ,t0e∫0τAτ1,0dτ1dτ,Z1−0,t0=∑n+m+i+ρ=1∫0−∞An,mτZi−τ,t0+Un,mτ,t0Cρ,−n,mτ,t0e∫0τAτ1,0dτ1dτ.Up to now, the formula ofM1t0 can be obtained.


## 3.1. Canonical Equation around the Heteroclinic Orbit

We introduce new phase variabless,z by letting(16)x,y=as,bs+zb˙s,−a˙s.Differentiating both sides of (16) with respect to t gives(17)dxdt=a˙s+zb¨sdsdt+b˙sdzdt,dydt=b˙s−za¨sdsdt−a˙sdzdt.By Cramer’s Rule and (4), we have(18)dzdt=−As,zz−εUs,z,t,dsdt=1+Bs,zz+εVs,z,t,where(19)As,z=−1zb˙−za¨fa+zb˙,b−za˙−a˙+zb¨ga+zb˙,b−za˙a˙2+b˙2+za˙b¨−a¨b˙Us,z,t=−b˙−za¨Pa+zb˙,b−za˙,t−a˙+zb¨Qa+zb˙,b−za˙,ta˙2+b˙2+za˙b¨−a¨b˙Bs,z=1za˙fa+zb˙,b−za˙+b˙ga+zb˙,b−za˙a˙2+b˙2+za˙b¨−a¨b˙−1Vs,z,t=a˙Pa+zb˙,b−za˙,t+b˙Qa+zb˙,b−za˙,ta˙2+b˙2+za˙b¨−a¨b˙.We supposes^t,z^t to be a stable solution or an unstable solution of system (18) such that s^t0=0 and z^t0=z0. Let st=s^t+t0 and zt=z^t+t0; then, st,zt is well-defined for all t∈0,+∞ or t∈−∞,0, and it is the solution of the system(20)dzdt=−As,zz−εUs,z,t+t0,dsdt=1+Bs,zz+εVs,z,t+t0,satisfying s0=0andz0=z0. To differ from the other solutions, we call them the primary stable and unstable solutions.


## 3.2. Integral of Primary Solutions

We let(21)Z=ε−1z,S=ε−1s−t,as one more change of variables for system (20). Then, we have(22)dZdt=−At+εS,εZZ−Ut+εS,εZ,t+t0,dSdt=Bt+εS,εZZ+Vt+εS,εZ,t+t0.ExpandingAt+εS,t+εZ, Ut+εS,t+εZ,t+t0, Bt+εS,t+εZ, and Vt+εS,t+εZ,t+t0 at t,0, we obtain(23)At+εS,t+εZ=At,0+∑n+m≥1An,mtεn+mSnZm,Ut+εS,t+εZ,t+t0=Ut,0,t+t0+∑n+m≥1Un,mt,t0εn+mSnZm,Bt+εS,t+εZ=Bt,0+∑n+m≥1Bn,mtεn+mSnZm,Vt+εS,t+εZ,t+t0=Vt,0,t+t0+∑n+m≥1Vn,mt,t0εn+mSnZm.It then follows that(24)dZdt=−At,0Z−Ut,0,t+t0−∑n+m≥1εn+mAn,mtSnZm+1+Un,mt,t0SnZm,dSdt=Bt,0Z+Vt,0,t+t0+∑n+m≥1εn+mBn,mtSnZm+1+Vn,mt,t0SnZm.We solve (24) to obtain the primary stable solution(25)Z+t,t0=e−∫0tAτ1,0dτ1Z0+−∫0tUτ,0,τ+t0e∫0τAτ1,0dτ1dτ−∑n+m=1+∞εn+m∫0tAn,mτZ+τ,t0+Un,mτ,t0S+τ,t0nZ+τ,t0me∫0τAτ1,0dτ1dτ,S+t,t0=∫0tVτ,0,τ+t0+Bτ,0Z+τ,t0dτ+∑n+m=1+∞εn+m∫0tBn,mτZ+τ,t0+Vn,mτ,t0S+τ,t0nZ+τ,t0mdτ,and the primary unstable solution(26)Z−t,t0=e−∫0tAω,0dωZ0−−∫0tUτ,0,τ+t0e∫0τAτ1,0dτ1dτ−∑n+m=1+∞εn+m∫0tAn,mτZ−τ,t0+Un,mτ,t0S−τ,t0nZ−τ,t0me∫0τAτ1,0dτ1dτ,S−t,t0=∫0tVτ,0,τ+t0+Bτ,0Z−τ,t0dτ+∑n+m=1+∞εn+m∫0tBn,mτZ−τ,t0+Vn,mτ,t0S−τ,t0nZ−τ,t0mdτ,in which Z0+ and Z0− are the initial values.Before proceeding further, we need to consider the asymptotic behaviors ofat and bt as t⟶±∞ which are crucial to get the integral equations for primary solutions. Observe that(27)limt⟶−∞at,bt=x1,y1,limt⟶+∞at,bt=x2,y2,and due to the local linearization theory, there exists a small number r>0 such that(28)fx,ygx,y=fxx1,y1fyx1,y1gxx1,y1gyx1,y1x−x1y−y1+f1x,yg1x,y,x−x1y−y1<2r;fx,ygx,y=fxx2,y2fyx2,y2gxx2,y2gyx2,y2x−x2y−y2+f2x,yg2x,y,x−x2y−y2<2r,where f1x,y and g1x,y are terms of second order and higher at O1x1,y1 and f2x,y and g2x,y are terms of second order and higher at O2x2,y2. We assume that λ1<0<μ1 are the eigenvalues of O1x1,y1 and λ2<0<μ2 are the eigenvalues of O2x2,y2. There exist two invertible matrices Γ1=α1β1γ1δ1 and Γ2=α2β2γ2δ2 such that(29)Γ1fxx1,y1fyx1,y1gxx1,y1gyx1,y1Γ1−1=λ100μ1,(30)Γ2fxx2,y2fyx2,y2gxx2,y2gyx2,y2Γ2−1=λ200μ2.In what follows, we introduce two coordinate transformations:(31)u1v1=Γ1x−x1y−y1,u2v2=Γ2x−x2y−y2.Unperturbed system (1) is transformed to(32)du1dt=λ1u1+f˜u1,v1,dv1dt=μ1v1+g˜u1,v1,du2dt=λ2u2+f˜u2,v2,dv2dt=μ2v2+g˜u2,v2.In the caset⟶−∞, our problem is about the unstable solution in a small neighborhood of O1. Let T0>0 be sufficiently large. For t<−T0, we can find a near-identity coordinate transformation:(33)u1=X+∑j=2+∞ϕj1X,Y,v1=Y+∑j=2+∞φj1X,Y,where ϕj1X,Y and φj1X,Y are homogeneous polynomials of degree j in X,Y, such that the unperturbed system near O1 is linearized to(34)dXdt=λ1X,dYdt=μ1Y.Assume that the power series in (33) are convergent on X,Y<2r. It is not difficult to verify that(35)α1β1γ1δ1at−x1bt−y1=reμ1t−t0+∑j=2+∞ϕj1reμ1t−t0,0∑j=2+∞φj1reμ1t−t0,0,where t0 be such that Xt0=r and Yt0=0. We solve (35) to obtain(36)at−x1=1detΓ1−β1reμ1t−t0+∑j=2+∞δ1ϕj1reμ1t−t0,0−β1φj1reμ1t−t0,0,bt−y1=1detΓ1α1reμ1t−t0−∑j=2+∞γ1ϕj1reμ1t−t0,0−α1φj1reμ1t−t0,0.On the other hand, lett>T0. The problem is then about the stable solution in a small neighborhood of O2. We similarly have(37)at−x2=1detΓ2δ2reλ2t−t0+∑j=2+∞δ2ϕj2reλ2t−t0,0−β2φj2reλ2t−t0,0,bt−y2=1detΓ2−γ2reλ2t−t0−∑j=2+∞γ2ϕj2reλ2t−t0,0−α2φj2reλ2t−t0,0,where ϕj2 and φj2 are homogeneous polynomials of degree j.Lemma 1.
For the above heteroclinic solutionℓ, we have(38)limt⟶−∞At,0>0,limt⟶+∞At,0<0.Proof.
We recall that(39)As,z=−1zb˙−za¨fa+zb˙,b−za˙−a˙+zb˙ga+zb˙,b−za˙a˙2+b˙2+za˙b¨−a¨b˙,and let(40)As,z=−za˙2+b˙2+za˙b¨−a˙b˙As,z.
Then,(41)As,z=b˙−za¨fa,b+zb˙fxa,b−za˙fya,b+Oz2−a˙+zb¨ga,b+zb˙gxa,b−za˙gya,b+Oz2=b˙fa,b−a′ga,b−za¨fa,b−zb¨ga,b+zb˙2fxa,b−za˙b˙fya,b−za˙b˙gxa,b+za˙2gya,b+Oz2=−za¨fa,b−zb¨ga,b+zb˙2fxa,b+za˙2gya,b−za˙b˙fya,b−za˙b˙gxa,b+Oz2.
This is to imply(42)As,0=d/dsfa,bfa,b+d/dsga,bga,b−g2a,bfxa,bf2a,b+g2a,b+−f2a,bgya,b+fa,bga,bfya,b+gxa,bf2a,b+g2a,b=fxf+fygf+gxf+gygg−g2fx−f2gy+fgfy+gxf2+g2=fx−gyf2−g2+2fy+gxfgf2+g2.
Ass<−T0, we infer from (36) that(43)fa,b=−1detΓ1β1μ1reμ1s−t0+Or2,ga,b=1detΓ1α1μ1reμ1s−t0+Or2.
Accordingly, we have(44)lims⟶−∞As,0=lims⟶−∞fx−gyf2−g2+2fy+gxfgf2+g2as,bs=lims⟶−∞fxas,bs−gyas,bsβ12−α12μ12r2e2μ1s−t0+Or3α12+β12μ12r2e2μ1s−t0+Oe3μ1s−t0−2fyas,bs+gxas,bsα1β1μ12r2e2μ1s−t0+Or3α12+β12μ12r2e2μ1s−t0+Oe3μ1s−t0.
Recall (29), and we have(45)lims⟶−∞fxas,bsfyas,bsgxas,bsgyas,bs=α1β1γ1δ1−1λ100μ1α1β1γ1δ1=1detΓ1λ1α1δ1−μ1β1γ1λ1−μ1β1δ1μ1−λ1α1γ1−λ1β1γ1+μ1α1δ1.
Substituting (45) into (44) gives(46)lims⟶−∞As,0=λ1−μ1α1δ1+β1γ1β12−α12−2λ1−μ1β1δ1−α1γ1α1β1detΓ1α12+β12=λ1−μ1β1γ1−α1δ1α12+β12detΓ1α12+β12=μ1−λ1>0.
Ass>T0, we infer from (37) that(47)fa,b=1detΓ2λ2δ2reλ2s−t0+Or2,ga,b=−1detΓ2λ2γ2reλ2s−t0+Or2.
After similar derivation, we have(48)lims⟶+∞As,0=λ2−μ2<0.
This completes the proof.By using Lemma2, one can easily verify that(49)limt⟶−∞e−∫0tAτ1,0dτ1=limt⟶+∞e−∫0tAτ1,0dτ1=+∞.Observe thatZ+t,t0 and Z−t,t0 should be bounded as t⟶±∞, and the initial values in (25) and (26) are claimed to be(50)Z0+=∫0+∞Uτ,0,τ+t0e∫0τAτ1,0dτ1dτ+∑n+m=1+∞εn+m∫0+∞An,mτZ+τ,t0+Un,mτ,t0S+τ,t0nZ+τ,t0me∫0τAτ1,0dτ1dτ(51)Z0−=∫0−∞Uτ,0,τ+t0e∫0τAτ1,0dτ1dτ+∑n+m=1+∞εn+m∫0−∞An,mτZ−τ,t0+Un,mτ,t0S−τ,t0nZ−τ,t0me∫0τAτ1,0dτ1dτ.Substituting (50) and (51) into (25) and (26), we have the integral equations of the primary stable solution(52)Z+t,t0=∫t+∞Uτ,0,τ+t0e∫tτAτ1,0dτ1dτ+∑n+m≥1εn+m∫t+∞An,mτZ+τ,t0+Un,mτ,t0S+τ,t0nZ+τ,t0mdτ,S+t,t0=∫0tVτ,0,τ+t0dτ+∫0tBτ,0Z+τ,t0dτ+∑n+m≥1εn+m∫0tBn,mτZ+τ,t0+Vn,mτ,t0S+τ,t0nZ+τ,t0mdτ,and the primary unstable solution(53)Z−t,t0=∫t−∞Uτ,0,τ+t0e∫tτAτ1,0dτ1dτ+∑n+m≥1εn+m∫t−∞An,mτZ−τ,t0+Un,mτ,t0S−τ,t0nZ−τ,t0mdτ,S−t,t0=∫0tVτ,0,τ+t0dτ+∫0tBτ,0Z−τ,t0dτ+∑n+m≥1εn+m∫0tBn,mτZ−τ,t0+Vn,mτ,t0S−τ,t0nZ−τ,t0mdτ.


## 3.3. Derivation of High-Order Melnikov Functions

The properties ofZ±t,t0,S±t,t0 here are coincident with the ones in [13]. We have the following power series:(54)Z±t,t0=Z0±t,t0+∑k=1∞εkZk±t,t0,S±t,t0=S0±t,t0+∑k=1∞εkSk±t,t0.In what follows, we substitute (54) into (52) and (53) to determine Zk±t,t0k=0,1,…, and Sk±t,t0k=0,1,…, inductively. They are(55)Z0±t,t0=∫0±∞Uτ,0,τ+t0E0,τdτ,S0±t,t0=∫0±tVτ,0+Bτ,0Z0±τdτ,(56)Zk±t,t0=∑n+m+i+ρ=k∫t±∞An,mτZi±τ,t0+Un,mτ,t0Cρ,±n,mτ,t0e∫tτAτ1,0dτ1dτ,Sk±t,t0=∑n+m+i+ρ=k∫0tBn,mτZi±τ,t0+Vn,mτ,t0Cρ,±n,mτ,t0dτ,for k=1,2,…,. Here, n,m,i,andρ are nonnegative integrals, and n+m≥1, i≥0, and ρ≥0. Cρ,+n,mt,t0 and Cρ,−n,mt,t0 are coefficient functions of ερ, which have the form of(57)Cρ,±n,mt,t0=∑ℓ0+⋯+ℓρ=n,l0+⋯+lρ=m,ℓ1+l1+⋯+ρ⋅ℓρ+lρ=ρK∏j=0ρSj±t,t0ℓj⋅Zj±t,t0lj.Here,K is a constant depending on the specific term. In the next section, these notations contribute to the proof of Theorem 2.Recall the definition ofDε,t0, and we have(58)Dε,t0=εZ+0,t0−Z−0,t0=∑k=0+∞εk+1Zk+0,t0−Zk−0,t0.Then,(59)M0t0=∫−∞+∞Ut,0,t+t0e∫0tAω,0dωdt,and for k=1,2,…, the k+1-th order Melnikov function is(60)Mkt0=Zk+0,t0−Zk−0,t0,where Zk+0,t0 and Zk−0,t0 are the values by letting t=0 in (56). Particularly, we have(61)Z1+0,t0=∑n+m+i+ρ=1∫0+∞An,mτZi+τ,t0+Un,mτ,t0Cρ,+n,mτ,t0e∫0τAτ1,0dτ1dτ,Z1−0,t0=∑n+m+i+ρ=1∫0−∞An,mτZi−τ,t0+Un,mτ,t0Cρ,−n,mτ,t0e∫0τAτ1,0dτ1dτ.Up to now, the formula ofM1t0 can be obtained.


## 4. Heteroclinic Persisting Perturbations

This section serves as the proof of Theorem2. For autonomous perturbations, all the functions above having the variable t0 are now dropped off. Before starting our proof, we need some lemmas.Lemma 2.
Under the conditions of Theorem2, we have the following:(1)
Ifn is odd, then An,mt and Un,mt are even in t, while Bn,mt and Vn,mt are odd in t(2)
Ifn is even, then An,mt and Un,mt are odd in t, while Bn,mt and Vn,mt are even in t.Proof.
We will show the parity ofVn,mt, and the other three can be verified similarly. Recall that(62)Vs,z=a˙Pa+zb˙,b−za˙+b˙Qa+zb˙,b−za˙a˙2+b˙2+za˙b¨−a¨b˙.
Sinceas is an odd function while bs is even, then a˙s and b¨s are even and b˙s and a¨s are odd. It follows that(63)V−s,z=Vs,z.
DenoteVn,ms,z=∂n+mVs,z/∂sn∂zm. We have(64)V1,0−s,z=−∂V−s,z∂s=−∂Vs,z∂s=−V1,0s,z.
Moreover, the parity ins will not change by taking a derivative with respect to z. Inductively, we obtain(65)Vn,ms,z=−Vn,m−s,z,nis odd,Vn,m−s,z,nis even.
SinceVn,mt differs from Vn,mt,0 by a constant factor, the proof is now completed by letting s,z=t,0.At the end of this section, let us prove Theorem2 with the help of the preceding preparations.Proof of Theorem 2.
To start with, we prove, for allt≥0,(66)Zk+t=Zk−−t,Sk+t=−Sk−−t,k∈ℕ.
By using mathematical induction, we first consider the casek=0. Recall that(67)Z0−−t=∫−t−∞Uτ,0e∫tτAτ1,0dτ1dτ,S0−−t=∫0−tVτ,0+Bτ,0Z0−τdτ,where Uτ,0 is an odd function in τ and Bτ,0 and Vτ,0 are even functions in τ. Moreover, we have(68)e∫−t−τAτ1,0dτ1=e∫tτA−τ1,0d−τ1=e∫tτAτ1,0dτ1.
By means of variable substitution, we have(69)Z0−−t=∫t+∞U−τ,0e∫−t−τAτ1,0dτ1d−τ=∫t+∞Uτ,0e∫tτAτ1,0dτ1dτ=Z0+t,S0−−t=∫0tV−τ,0+B−τ,0Z0−−τd−τ=−∫0tVτ,0+Bτ,0Z0+τdτ=−S0+t.
Now, we letL be a positive integer, and suppose that (66) holds for all nonnegative integer k<L. According to (56), we have(70)ZL−−t=∑n+m+i+ρ=L∫−t−∞An,mτZi−τ+Un,mτCρ,−n,mτe∫−tτAτ1,0dτ1dτ,SL−−t=∑n+m+i+ρ=L∫0−tBn,mτZi−τ+Vn,mτCρ,−n,mτdτ.
Notice thatCρ,−n,mτ is the counterpart of Cρ,+n,mτ, changing “Zj+τ” and “Sj+τ” to “Zj−τ” and “Sj−τ,” respectively. Since ρ<L, we obtain(71)Cρ,−n,m−τ=∑ℓ0+⋯+ℓρ=n,l0+⋯+lρ=m,ℓ1+l1+⋯+ρ⋅ℓρ+lρ=ρK∏j=0ρSj−−τℓj⋅Zj−−τlj=∑ℓ0+⋯+ℓρ=n,l0+⋯+lρ=m,ℓ1+l1+⋯+ρ⋅ℓρ+lρ=ρK∏j=0ρ−Sj+τℓj⋅Zj+τlj=∑ℓ0+⋯+ℓρ=n,l0+⋯+lρ=m,ℓ1+l1+⋯+ρ⋅ℓρ+lρ=ρ−1nK∏j=0ρSj+τℓj⋅Zj+τlj=−1nCρ,+n,mτ.
That is,(72)Cρ,−n,m−τ=−Cρ,+n,mτ,nis odd,Cρ,+n,mτ,nis even.
From Lemma2, we have(73)An,m−τZi−−τ+Un,m−τCρ,−n,m−τe∫−t−τAτ1,0dτ1=−An,mτZi+τ+Un,mτCρ,+n,mτe∫tτAτ1,0dτ1,Bn,m−τZi−−τ+Vn,m−τCρ,−n,m−τ=Bn,mτZi+τ+Vn,mτCρ,+n,mτ.
Finally, we obtain(74)ZL−−t=ZL+t,SL−−t=−SL+t.
The proof of (66) is completed.
Recall (60), and for autonomous perturbation, we have(75)MkP,Q=Zk+0−Zk−0=0,k=0,1,2,….
Due to (14), we thus arrive the conclusion that Dε,P,Q=0 for all small ε. This means the stable manifold and unstable manifold coincide. As a result, there exists a heteroclinic orbit ℓε which satisfies(76)limε⟶0ℓε=ℓ.
Therefore,P,Q is the heteroclinic persisting perturbation of ℓ.


## 5. An Concrete Example: Van der Pol System

Let us consider the Van der Pol system on a torus:(77)dxdt=siny,dydt=−sinx,where x,y∈ℝ2. Without loss of generality, we fix a periodic region of the vector field in which we have four hyperbolic saddle fixed points:(78)O1−π,0,O20,π,O3π,0,O40,−π.We writefx,y=siny and gx,y=−sinx, and the heteroclinic solutions are(79)ℓ12:−2arctane−t,π−2arctane−t,t∈ℝ,ℓ23:2arctanet,π−2arctanet,t∈ℝ,ℓ34:2arctane−t,2arctane−t−π,t∈ℝ,ℓ41:−2arctanet,2arctanet−π,t∈ℝ.The vector field of (77) is displayed in Figure 1.Figure 1
Vector field of (77).In order to obtain the heteroclinic persisting perturbations of these heteroclinic orbits, we need some rotation transformations of coordinates to satisfy the preconditions in Theorem2.


### 5.1. Persistence ofℓ12 and ℓ34

Introducing the change of variables(80)u1v1=22−222222xy,we obtain(81)du1dt=22sinu1+v12−sinu1−v12,dv1dt=−22sinu1+v12+sinu1−v12.In such a case, the saddle points are transformed to(82)O1′−22π,−22π,O2′−22π,22π,O3′22π,22π,O4′22π,−22π.Denote(83)f1u1,v1=22sinu1+v12−sinu1−v12,g1u1,v1=−22sinu1+v12+sinu1−v12.The new vector field and heteroclinic orbits are plotted in Figure2.Figure 2
Vector field of (81).We note that the image ofℓ12 is(84)ℓ12′:u112t,v112t=224arctanet−π,22π,t∈ℝ.One can easily verify thatu112t is odd in t while v112t is even and f1u1,v1 is even in u1 while g1u1,v1 is odd. For the convenience of derivation, we denote(85)Λℓ12′=P1,Q1:P1u1,v1is even inu1,Q1u1,v1is odd inu1.Applying Theorem2, we have Λℓ12′⊂HPℓ12′. Now, we consider the following equations:(86)du1dt=22sinu1+v12−sinu1−v12+εP1u1,v1,dv1dt=−22sinu1+v12+sinu1−v12+εQ1u1,v1.where P1u1,v1,Q1u1,v1∈Λℓ12′. Let us go back to the x,y-plane and replace ε/2 by ε for easy computation. It follows that(87)dxdt=siny+εP1x−y2,x+y2+Q1x−y2,x+y2,dydt=−sinx+ε−P1x−y2,x+y2+Q1x−y2,x+y2.We introduceΛℓ12 to be the set obtained by Λℓ12′ through the transformation of coordinates (80), which means given Px,y,Qx,y∈Λℓ12, there exists a P1u1,v1,Q1u1,v1∈Λℓ12′ such that(88)Px,y=P1x−y2,x+y2+Q1x−y2,x+y2,Qx,y=−P1x−y2,x+y2+Q1x−y2,x+y2.Thus, we have(89)Λℓ12=P,Q:∃P1,Q1∈Λℓ12′s.t.88holds..Obviously,HPℓ12=HPℓ34. We also have Λℓ12⊂HPℓ34. In particular, we set(90)P1u1,v1=sinu1+v12sinu1−v12,Q1u1,v1=0.This implies(91)Px,y=−sinxsiny,Qx,y=sinxsiny.We consider the following system:(92)dxdt=siny−εsinxsiny,dydt=−sinx+εsinxsiny.It follows that there exist two heteroclinic solutions which satisfy equality (10) in Definition 1. The perturbed vector field and the two heteroclinic orbits ℓ12ε and ℓ34ε are plotted in Figure 3. Figure 4 shows the other two are broken.Figure 3
Vector field of92ε=0.5 and ℓ12ε and ℓ34ε.Figure 4
Vector field of92ε=0.5 and ℓ23ε and ℓ41ε.


### 5.2. Persistence ofℓ23 and ℓ41

For the heteroclinic orbitsℓ23 and ℓ41, we introduce another transformation of coordinates:(93)u2v2=2222−2222xy.It follows that(94)du2dt=22sinu2+v22−sinu2−v22,dv2dt=−22sinu2+v22+sinu2−v22.Denote(95)f2u2,v2=22sinu2+v22−sinu2−v22,g2u2,v2=−22sinu2+v22+sinu2−v22.Similarly, the image ofℓ23 is transformed to(96)ℓ23″:u223t,v223t=22π−4arctane−t,22π,t∈ℝ.The new saddle points of (94) are(97)O1″−22π,22π,O2″22π,22π,O3″22π,−22π,O4″−22π,−22π.One can easily verify thatu123t is odd in t while v123t is even and f2u2,v2 is even in u2 while g2u2,v2 is odd. Denote(98)Λℓ23″=P2,Q2:P2u2,v2is even inu2,Q2u2,v2is odd inu2It then follows from Theorem2 that Λℓ23″⊂HPℓ23″. We take P2u,v,Q2u,v∈Λℓ23″, and consider the following perturbed system:(99)du2dt=22sinu2+v22−sinu2−v22+εP2u2,v2,dv2dt=−22sinu2+v22+sinu2−v22+εQ2u2,v2.Back tox,y−plane, we have(100)dxdt=siny+εP2x+y2,−x−y2−Q2x+y2,−x−y2,dydt=−sinx+εP2x+y2,−x−y2+Q2x+y2,−x−y2.Similarly, we obtain the setΛℓ23 through transformation (93). Then, for any Px,y,Qx,y∈Λℓ23, we obtain(101)Px,y=P2x+y2,−x−y2−Q2x+y2,−x−y2,Qx,y=P2x+y2,−x−y2+Q2x+y2,−x−y2.Consequently, we have(102)Λℓ23=P,Q:∃P2,Q2∈CPℓ23″s.t.101holds⋅.SinceHPℓ23=HPℓ41, we also have Λℓ23⊂HPℓ41.We take(103)P2u2,v2=sinu2+v22sinu2−v22,Q2u2,v2=0.It follows that(104)Px,y=sinxsiny,Qx,y=sinxsiny.Hence, we get the following perturbed system:(105)dxdt=siny+εsinxsiny,dydt=−sinx+εsinxsiny.Figures5 and 6 are the plots of the perturbed vector field of (105) and the two heteroclinic orbits ℓ23ε and ℓ41ε. The other two are broken.Figure 5
Vector field of105ε=0.5 and ℓ23ε and ℓ41ε.Figure 6
Vector field of105ε=0.5 and ℓ12ε and ℓ34ε.


### 5.3. Persistence of the Heteroclinic 4-Cycle

We note that in unperturbed system (77), ℓ12∪ℓ23∪ℓ34∪ℓ41∪O1,O2,O3,O4 forms a heteroclinic 4-cycle. It is natural to ask when the heteroclinic 4-cycle remains unbroken. We rewrite Λℓ12 and Λℓ23 as(106)Λℓ12=P,Q:Px,y=P1u1,v1+Q1u1,v1,Qx,y=−P1u1,v1+Q1u1,v1,P1,Q1∈Λℓ12′,Λℓ23=P,Q:Px,y=P2u2,v2−Q2u2,v2,Qx,y=P2u2,v2+Q2u2,v2,P2,Q2∈Λℓ23′.From (80) and (93), we have the relation between phase variables u1,v1 and u2,v2 as(107)u1v1=0−110u2v2=−v2u2.Thus, we have(108)Λℓ12=P,Q:Px,y=P1−v2,u2+Q1−v2,u2,Qx,y=−P1−v2,u2+Q1−v2,u2,P1,Q1∈Λℓ12′=P,Q:Px,y=P1v2,u2−Q1v2,u2,Qx,y=−P1v2,u2−Q1v2,u2,P1,Q1∈Λℓ12′.Our goal is to figure out the elements inCPℓ12∩CPℓ23 and so we let(109)P1v2,u2−Q1v2,u2=P2u2,v2−Q2u2,v2,−P1v2,u2−Q1v2,u2=P2u2,v2+Q2u2,v2.It then follows that(110)P1v2,u2=−Q2u2,v2,Q1v2,u2=−P2u2,v2.Depending on the parities ofP1, P2, Q1, and Q2 with respect to the first variable, we can use (110) to obtain their parities with respect to the second variable. Let(111)Ω=P2,Q2:P2u2,v2is even inu2and odd inv2,Q2u2,v2is odd inu2and even inv2.We finally get(112)Λℓ12∩Λℓ23=P,Q:Px,y=P2u2,v2−Q2u2,v2,Qx,y=P2u2,v2+Q2u2,v2,P2,Q2∈Ω,in which u2=2/2x+y and v2=−2/2x−y.In order to visualize this result, we consider the following perturbations:(113)Px,y=sinxsinysinx−y2+sinx+y2,Qx,y=sinxsinysinx−y2−sinx+y2.Adding them to system (77) gives(114)dxdt=siny+εsinxsinysinx−y2+sinx+y2,dydt=−sinx+εsinxsinysinx−y2−sinx+y2.SincePx,y,Qx,y∈Λℓ12∩Λℓ23, all the four heteroclinic orbits of (114) remain unbroken (see Figure 7).Figure 7
Vector field of105ε=0.5.


## 5.1. Persistence ofℓ12 and ℓ34

Introducing the change of variables(80)u1v1=22−222222xy,we obtain(81)du1dt=22sinu1+v12−sinu1−v12,dv1dt=−22sinu1+v12+sinu1−v12.In such a case, the saddle points are transformed to(82)O1′−22π,−22π,O2′−22π,22π,O3′22π,22π,O4′22π,−22π.Denote(83)f1u1,v1=22sinu1+v12−sinu1−v12,g1u1,v1=−22sinu1+v12+sinu1−v12.The new vector field and heteroclinic orbits are plotted in Figure2.Figure 2
Vector field of (81).We note that the image ofℓ12 is(84)ℓ12′:u112t,v112t=224arctanet−π,22π,t∈ℝ.One can easily verify thatu112t is odd in t while v112t is even and f1u1,v1 is even in u1 while g1u1,v1 is odd. For the convenience of derivation, we denote(85)Λℓ12′=P1,Q1:P1u1,v1is even inu1,Q1u1,v1is odd inu1.Applying Theorem2, we have Λℓ12′⊂HPℓ12′. Now, we consider the following equations:(86)du1dt=22sinu1+v12−sinu1−v12+εP1u1,v1,dv1dt=−22sinu1+v12+sinu1−v12+εQ1u1,v1.where P1u1,v1,Q1u1,v1∈Λℓ12′. Let us go back to the x,y-plane and replace ε/2 by ε for easy computation. It follows that(87)dxdt=siny+εP1x−y2,x+y2+Q1x−y2,x+y2,dydt=−sinx+ε−P1x−y2,x+y2+Q1x−y2,x+y2.We introduceΛℓ12 to be the set obtained by Λℓ12′ through the transformation of coordinates (80), which means given Px,y,Qx,y∈Λℓ12, there exists a P1u1,v1,Q1u1,v1∈Λℓ12′ such that(88)Px,y=P1x−y2,x+y2+Q1x−y2,x+y2,Qx,y=−P1x−y2,x+y2+Q1x−y2,x+y2.Thus, we have(89)Λℓ12=P,Q:∃P1,Q1∈Λℓ12′s.t.88holds..Obviously,HPℓ12=HPℓ34. We also have Λℓ12⊂HPℓ34. In particular, we set(90)P1u1,v1=sinu1+v12sinu1−v12,Q1u1,v1=0.This implies(91)Px,y=−sinxsiny,Qx,y=sinxsiny.We consider the following system:(92)dxdt=siny−εsinxsiny,dydt=−sinx+εsinxsiny.It follows that there exist two heteroclinic solutions which satisfy equality (10) in Definition 1. The perturbed vector field and the two heteroclinic orbits ℓ12ε and ℓ34ε are plotted in Figure 3. Figure 4 shows the other two are broken.Figure 3
Vector field of92ε=0.5 and ℓ12ε and ℓ34ε.Figure 4
Vector field of92ε=0.5 and ℓ23ε and ℓ41ε.


## 5.2. Persistence ofℓ23 and ℓ41

For the heteroclinic orbitsℓ23 and ℓ41, we introduce another transformation of coordinates:(93)u2v2=2222−2222xy.It follows that(94)du2dt=22sinu2+v22−sinu2−v22,dv2dt=−22sinu2+v22+sinu2−v22.Denote(95)f2u2,v2=22sinu2+v22−sinu2−v22,g2u2,v2=−22sinu2+v22+sinu2−v22.Similarly, the image ofℓ23 is transformed to(96)ℓ23″:u223t,v223t=22π−4arctane−t,22π,t∈ℝ.The new saddle points of (94) are(97)O1″−22π,22π,O2″22π,22π,O3″22π,−22π,O4″−22π,−22π.One can easily verify thatu123t is odd in t while v123t is even and f2u2,v2 is even in u2 while g2u2,v2 is odd. Denote(98)Λℓ23″=P2,Q2:P2u2,v2is even inu2,Q2u2,v2is odd inu2It then follows from Theorem2 that Λℓ23″⊂HPℓ23″. We take P2u,v,Q2u,v∈Λℓ23″, and consider the following perturbed system:(99)du2dt=22sinu2+v22−sinu2−v22+εP2u2,v2,dv2dt=−22sinu2+v22+sinu2−v22+εQ2u2,v2.Back tox,y−plane, we have(100)dxdt=siny+εP2x+y2,−x−y2−Q2x+y2,−x−y2,dydt=−sinx+εP2x+y2,−x−y2+Q2x+y2,−x−y2.Similarly, we obtain the setΛℓ23 through transformation (93). Then, for any Px,y,Qx,y∈Λℓ23, we obtain(101)Px,y=P2x+y2,−x−y2−Q2x+y2,−x−y2,Qx,y=P2x+y2,−x−y2+Q2x+y2,−x−y2.Consequently, we have(102)Λℓ23=P,Q:∃P2,Q2∈CPℓ23″s.t.101holds⋅.SinceHPℓ23=HPℓ41, we also have Λℓ23⊂HPℓ41.We take(103)P2u2,v2=sinu2+v22sinu2−v22,Q2u2,v2=0.It follows that(104)Px,y=sinxsiny,Qx,y=sinxsiny.Hence, we get the following perturbed system:(105)dxdt=siny+εsinxsiny,dydt=−sinx+εsinxsiny.Figures5 and 6 are the plots of the perturbed vector field of (105) and the two heteroclinic orbits ℓ23ε and ℓ41ε. The other two are broken.Figure 5
Vector field of105ε=0.5 and ℓ23ε and ℓ41ε.Figure 6
Vector field of105ε=0.5 and ℓ12ε and ℓ34ε.


## 5.3. Persistence of the Heteroclinic 4-Cycle

We note that in unperturbed system (77), ℓ12∪ℓ23∪ℓ34∪ℓ41∪O1,O2,O3,O4 forms a heteroclinic 4-cycle. It is natural to ask when the heteroclinic 4-cycle remains unbroken. We rewrite Λℓ12 and Λℓ23 as(106)Λℓ12=P,Q:Px,y=P1u1,v1+Q1u1,v1,Qx,y=−P1u1,v1+Q1u1,v1,P1,Q1∈Λℓ12′,Λℓ23=P,Q:Px,y=P2u2,v2−Q2u2,v2,Qx,y=P2u2,v2+Q2u2,v2,P2,Q2∈Λℓ23′.From (80) and (93), we have the relation between phase variables u1,v1 and u2,v2 as(107)u1v1=0−110u2v2=−v2u2.Thus, we have(108)Λℓ12=P,Q:Px,y=P1−v2,u2+Q1−v2,u2,Qx,y=−P1−v2,u2+Q1−v2,u2,P1,Q1∈Λℓ12′=P,Q:Px,y=P1v2,u2−Q1v2,u2,Qx,y=−P1v2,u2−Q1v2,u2,P1,Q1∈Λℓ12′.Our goal is to figure out the elements inCPℓ12∩CPℓ23 and so we let(109)P1v2,u2−Q1v2,u2=P2u2,v2−Q2u2,v2,−P1v2,u2−Q1v2,u2=P2u2,v2+Q2u2,v2.It then follows that(110)P1v2,u2=−Q2u2,v2,Q1v2,u2=−P2u2,v2.Depending on the parities ofP1, P2, Q1, and Q2 with respect to the first variable, we can use (110) to obtain their parities with respect to the second variable. Let(111)Ω=P2,Q2:P2u2,v2is even inu2and odd inv2,Q2u2,v2is odd inu2and even inv2.We finally get(112)Λℓ12∩Λℓ23=P,Q:Px,y=P2u2,v2−Q2u2,v2,Qx,y=P2u2,v2+Q2u2,v2,P2,Q2∈Ω,in which u2=2/2x+y and v2=−2/2x−y.In order to visualize this result, we consider the following perturbations:(113)Px,y=sinxsinysinx−y2+sinx+y2,Qx,y=sinxsinysinx−y2−sinx+y2.Adding them to system (77) gives(114)dxdt=siny+εsinxsinysinx−y2+sinx+y2,dydt=−sinx+εsinxsinysinx−y2−sinx+y2.SincePx,y,Qx,y∈Λℓ12∩Λℓ23, all the four heteroclinic orbits of (114) remain unbroken (see Figure 7).Figure 7
Vector field of105ε=0.5.



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