# Heteroclinic Solutions for Nonautonomous EFK Equations

Authors: Y. L. Yeun
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We explore the nonautonomous fourth-order differential equation which has important applications in materials science. By variational approach, we find heteroclinic solutions of the equation. The conditions on the potential functionV(t,u) are mild enough to include a broad class of equations. We also consider a separate case where V(t,u) is periodic in t.


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## Body

## 1. Introduction

The goal of this paper is to study the nonautonomous extended Fisher-Kolmogorov (EFK) equation(1)q′′′′(t)-q′′(t)+Vq(t,q)=0,
where V(t,q) is a time-dependent potential function and Vq denotes the partial derivative with respect to the second variable q.Fourth-order differential equations, which often appear in nonlinear elasticity, fluid mechanics, and relating physical problems, have received growing attention from researchers. For example, Peletier and Troy [1–4] studied the EFK equation with odd nonlinearity
(2)μq′′′′-λq′′-q+q3=0,(μ,λ>0),
mainly using shooting argument and variational method. They found the existence of periodic, heteroclinic, and homoclinic solutions and proved the existence of chaotic solutions oscillating between 1 and −1 with all critical points being either minima or maxima. They also explored the boundedness, monotonicity, and other quantitative properties of solutions. Part of Peletier and Troy’s work was generalized by Kalies and VanderVorst [5]; they considered the EFK equation with general nonlinear term
(3)μq′′′′-λq′′+V′(q)=0,(μ,λ>0),
where V(q) is supposed to have symmetric wells with equal depth and grow superquadratically. Then, they investigated the multitransition structure of heteroclinic and homoclinic solutions. Primary examples are
(4)V1(q)=14(q2-1)2,V2(q)=2π2(1+cosπq).
The symmetric property of V(q) is rather restrictive, and many physical problems do not satisfy such conditions. Later, Kalies et al. [6, 7] obtained heteroclinic and homoclinic solutions connecting saddle-focus equilibria, and the potential function V(q) is assumed to grow superquadratically and has at least two nondegenerate global minima, not necessarily symmetric. In case of (3) with V(q)=V1(q), the presence of saddle-focus equilibria amounts to the requiring of 4μ>λ2/V1′′(±1). Periodic and chaotic solutions of (3) are also explored. These results were generalized in recent works [8, 9], where the author obtained heteroclinic solutions for (3) under very mild conditions of V; in particular, they do not assume that V is symmetric or that saddle-focus equilibria are present.We have just mentioned a few works that are closely related to our results. A good summary of typical researches in fourth-order differential equations can be found in [10].In this paper we step forward to study heteroclinic solutions of the nonautonomous EFK equation (1). We are inspired by Yeun [9], Rabinowitz [11, 12], and Izydorek and Janczewska [13]. However, we are emphasizing that the argument of [9, 11] relies on the fact that the equation is autonomous; therefore, the method there cannot be reproduced here to tackle the time-dependent version (1) which is no longer autonomous. The nonautonomous case necessitates careful analysis. Another point should be made is that we are working on the (q,q′) phase plane; this is essentially different from [11, 13]. In our argument, we also benefit from analysis of [3, 4] and comments of [5].In spite of its practical significance [14], there has been few researches on (1) with time-dependent potentials; the paper seems to be the first attempt in the direction and the methods here can be applied to similar equations.


## 2. Heteroclinic Solutions

By a heteroclinic solution connectinga- and a+, we mean a solution q(t) verifying
(5)(q(t),q′(t),q′′(t),q′′′(t))⟶(a+,0,0,0),t⟶+∞,(q(t),q′(t),q′′(t),q′′′(t))⟶(a-,0,0,0),t⟶-∞.Throughout the paper, letBε(C) denote the neighborhood of a set C⊂ℝn defined as below
(6)Bε(C)={w∈ℝn∣infv∈C|w-v|<ε}.We will make the following assumptions:(A1)
V(t,q)∈C2(ℝ×ℝ,ℝ), V⩾0;(A2)
let𝒱 be a set which has at least two points and we assume the points in 𝒱 do not collapse together,
(7)γ≡13inf{|x-y|∣x≠y∈𝒱}>0;(A3)
one of the following is satisfied:
(a)
there isε*∈(0,γ], such that σε≜infz∉Bε(𝒱),t∈ℝV(t,z)>0 for any ε∈(0,ε*];(b)
liminf|z|→∞|z|V(t,z)>0 uniformly for t∈ℝ.(A4)
lim|t|→+∞V(t,z)=+∞ uniformly for z in any compact subsets of ℝ∖𝒱.(A5)
For eachz∈𝒱,
(8)∫-∞∞‍V(t,z)dt<γσγ1/2,whereσγ is defined in (A3) with ε=γ.Theorem 1.
LetV(t,q) satisfy (A1)–(A4). Then for every z∈𝒱, there is a pair of heteroclinic solutions of (1), one emanating from z and the other terminating at z.Remark 2.
To be noted, we do not assumeV has wells of equal depth. By the assumptions of the theorem, we necessarily have {z∈ℝ∣V(t,z)=0forallt}⊂𝒱. However, 𝒱 may contain other points which are not global minima of V(t,u) but still verify (A5). In the special case 𝒱={z∈ℝ∣V(t,z)=0forallt}, (A5) is automatically satisfied.Notation. Define 𝒱~={(z,0)∈ℝ2∣z∈𝒱}. For an element (z,0) in 𝒱~, we denote it by z~=(z,0) for brevity. Whence, without further notice, we will use z and z~ where it is appropriate to denote element in 𝒱 and its counterpart in 𝒱~.We will obtain heteroclinic solutions of (1) by minimizing the energy functional I(·) on an appropriate subset
(9)I(q)=∫-∞∞‍{12(q′′)2+12(q′)2+V(t,q)}dt.
Denote the energy on a bounded interval [r,s] by
(10)Ir,s(q)=∫rs‍{12(q′′)2+12(q′)2+V(t,q)}dt.We will work on the space(11)E={q∈Wloc2,2(ℝ,ℝ)∣∫-∞∞‍(|q′′|2+|q′|2)dt<∞},
equipped with the norm
(12)∥q∥2=∫-∞∞‍(|q′′|2+|q′|2)dt+|q(0)|2.
It is readily seen that (E,∥·∥) is a Hilbert space and q∈E implies q∈C1(ℝ).Letϑ~∈𝒱~∖{0~}, ε∈(0,1). Denote by Γε(ϑ) the set of q∈E verifying(i)
limt→-∞q(t)=0,limt→+∞q(t)=ϑ;(ii)
(q(t),q′(t))∉Bε(𝒱~∖{0~,ϑ~}), for all t∈ℝ.That is,Γε(ϑ) consists of functions which join 0 and ϑ, and at least keep a distance of ε to the set 𝒱~∖{0~,ϑ~}. Clearly, Γε(ϑ) is not empty, we may consider the minimization problem
(13)cε(ϑ)=infq∈Γε(ϑ)I(q).We will find for eachε>0 a candidate in Γε(ϑ) and then send ε to zero to obtain the minimizer, which is actually a heteroclinic solution.For anyε∈(0,γ), v>0 define
(14)σε,v=infz∉Bε(𝒱),|z|⩽v,t∈ℝV(t,z),
then σε,v>0 by our assumptions. For v=+∞, we take σε,+∞=σε which is defined in (A3).From now on, we always assume that(15)0<ε<min{ε*,1}.Lemma 3.
Letq∈E. Then for any v>0, r<s∈ℝ such that q(t)∉Bε(𝒱) and |q(t)|⩽v for any t∈[r,s],
(16)I(q)⩾2σε,v|q(r)-q(s)|.
In particular, if σε>0, then for any r<s∈ℝ such that q(t)∉Bε(𝒱) for any t∈[r,s],
(17)I(q)⩾2σε|q(r)-q(s)|.Proof.
Denotel=|q(r)-q(s)| and τ=|r-s|. Then
(18)l=|∫rs‍q′(t)dt|⩽∫rs‍|q′(t)|dt⩽τ1/2(∫rs‍|q′(t)|2dt)1/2.
SinceV⩾0, by assumptions, we have
(19)I(q)⩾∫rs‍{12(q′(t))2+V(t,q(t))}dt⩾l22τ+σε,vτ⩾2σε,vl.
Proof for the second part is similar.Remark 4.
The proof of Lemma3 indicates that any function q∈E with I(q)<+∞ cannot oscillate too often between different points in 𝒱 when the time approaches infinity; precisely, we will show q(±∞) exist and have limits in 𝒱.Lemma 5.
Letq∈E.I(q)<+∞ implies q∈L∞(ℝ,ℝ).Proof.
Suppose to the contrary thatq∉L∞(ℝ). Let ε>0 satisfy (15).
Case  1 (
σ
ε
>
0
). If 𝒱 is an infinite set, then we can find infinitely many nonoverlapping time intervals {(si,ti)}i⩾1 such that |q(si)-q(ti)|⩾γ for i⩾1 and q(t)∉Bε(𝒱) for t∈∪i⩾1(si,ti), by Lemma 3,
(20)I(q)>∑i=1n‍Isi,ti(q)⩾2σε∑i=1n‍|q(si)-q(ti)|⩾nγ2σε,
for any n⩾1, a contradiction.
If𝒱 is a finite set, let zv=maxz∈𝒱|z|, then zv∈(0,∞). Since q∉L∞(ℝ), there is a sequence of nonoverlapping time intervals, also denoted by {(si,ti)}i⩾1, such that |q(si)-q(ti)|⩾1 for i⩾1 and |q(t)|>max{|q(0)|,zv+1} for t∈∪i⩾1(si,ti). Again Lemma 3 shows
(21)I(q)⩾n2σε,
for any n⩾1, also a contradiction.
Case  2 (
liminf
|
z
|
→
∞
|
z
|
V
(
t
,
z
)
>
0
uniformly in
t
∈
ℝ
). There are z0>0 and c1>0 such that V(t,z)⩾c1|z|-1 for |z|>z0 and t∈ℝ. Let nq=max{[|q(0)|]+1,[z0]+1} ([x] denotes the integral part of x). Since q∉L∞(ℝ), there is a sequence of nonoverlapping time intervals {(sj,tj)}j⩾nq with |q(sj)|=j,  |q(tj)|=j+1 and |q(t)|∈[j,j+1] on (sj,tj) this is possible via the continuity of q(t). We claim that infj⩾nq|tj-sj|>0. Indeed, for j⩾nq,
(22)1=||q(tj)|-|q(sj)||⩽|q(tj)-q(sj)|⩽∫sjtj‍|q′(t)|dt⩽|tj-sj|1/2(∫sjtj‍|q′(t)|2dt)1/2,
so
(23)1⩽|tj-sj|∫rs‍|q′(t)|2dt⩽2|tj-sj|I(q),∀j⩾nq,
which implies
(24)infj⩾nq|tj-sj|>0.
Hence,
(25)I(q)⩾∑j=nqn‍∫sjtj‍V(t,q(t))dt⩾c1∑j=nqn‍∫sjtj‍1|q(t)|dt⩾c1(infj⩾nq|tj-sj|)∑j=nqn‍1j+1,
for any n⩾nq, a contradiction.
Therefore, in either case, we must haveq∈L∞(ℝ).The preceding lemma has an important corollary which will be used later; we prove it here.Lemma 6.
ForM>0 and u∈𝒱, define
(26)Qu,M={q∈E∣q(-∞)=u,I(q)⩽M}.
Then Qu,M is a bounded set in C(ℝ) with the usual uniform norm.Proof.
Without loss of generality, assumeu=0. Let ε>0 satisfy (15). Suppose to the contrary that, for any n∈ℕ, there is an orbit qn such that qn(-∞)=0,I(qn)⩽M and supt∈ℝ|qn(t)|>n. Since qn(t) is continuous, there is tn*∈ℝ for which |qn(tn*)|=n and |qn(t)|⩽n for t⩽tn*; this can be accomplished by taking as tn* the first hitting time of |qn(t)| on the level n.
Case  1 (
σ
ε
>
0
). If 𝒱 is an infinite set, then for each n⩾[3γ]+1, n∈ℕ, there are ln nonoverlapping time intervals {(sn,i,tn,i)}i⩽ln such that |qn(sn,i)-qn(tn,i)|⩾γ for 1⩽i⩽ln, qn(t)∉Bε(𝒱) for t∈∪i=1ln(sn,i,tn,i); moreover, the number of intervals ln→∞ as n→∞. Then by Lemma 3,
(27)I(qn)>∑i=1ln‍Isn,i,tn,i(qn)⩾2σε∑i=1ln‍|qn(sn,i)-qn(tn,i)|⩾lnγ2σε,
for any n⩾[3γ]+1, a contradiction.
If𝒱 is a finite set, let zv=maxz∈𝒱|z|, then zv∈(0,∞). For each n⩾[zv]+1, n∈ℕ, there is an sn<tn* with |qn(sn)|=zv+1 and |qn(t)|zv+1 for t∈(sn,tn*). Then Lemma 3 indicates
(28)I(qn)⩾(n-(zv+1))2σε,
for any n⩾[zv]+1, also a contradiction.
Case  2 (
liminf
|
z
|
→
∞
|
z
|
V
(
t
,
z
)
>
0
uniformly in
t
∈
ℝ
). There are z0>0 and c1>0 such that V(t,z)⩾c1|z|-1 for |z|>z0 and t∈ℝ. For each n⩾2, there are n-1 nonoverlapping time intervals {(sn,i,tn,i)}2⩽i⩽n such that |qn(sn,i)|=i-1,|qn(tn,i)|=i and |qn(t)|∈[i-1,i] on (sn,i,tn,i). The same argument that we used to obtain (24) shows
(29)infn⩾2infn⩾j⩾2|tn,j-sn,j|>0.
Hence
(30)I(qn)⩾∑j=[z0]+2n‍∫sn,jtn,j‍V(t,qj(t))dt⩾c1∑j=[z0]+2n‍∫sn,jtn,j‍1|qj(t)|dt⩾c1(infn⩾2infn⩾j⩾2|tn,j-sn,j|)∑j=[z0]+2n‍1j,
for any n⩾[z0]+2, n∈ℕ, a contradiction.
Therefore, a constantC0>0 depending only on M and u exists, for which supt∈ℝ|q(t)|⩽C0 for all q∈Qu,M.Lemma 7.
Letq∈E. If I(q)<+∞, then there are α,β∈𝒱 such that limt→-∞q(t)=α and limt→+∞q(t)=β.Proof.
By Lemma5, q∈L∞(ℝ,ℝ). Since q∈C(ℝ), then supt∈ℝ|q(t)|⩽Q for some Q>0. Therefore, the set
(31)Ω(q,-∞)={x:q(sk)⟶xforsomesk⟶-∞}
of accumulation points of q at -∞ is not empty. Moreover, Ω(q,-∞)∩𝒱≠∅. Indeed, suppose there are δ∈(0,γ) and r0>0 such that q(t)∉Bδ(𝒱) for all t⩽-r0. Then, by definition of σδ,Q in (14) with ε=δ and v=Q,
(32)I(q)>∫-∞-r0‍{V(t,q)}dt>∫-∞-r0‍σδ,Qdt,
which shows I(q)=+∞, a contradiction. Hence, Ω(q,-∞)∩𝒱 is not empty and there exists an α∈Ω(q,-∞)∩𝒱. Finally we claim q(-∞)=α. Suppose otherwise, there must be some δ1∈(0,ε*) such that q(t) intersects ∂Bδ1(α) and ∂B2δ1(α) infinitely many times, then Lemma 3 implies that the energy along the orbit q(t) would tend to infinity, contradicting the hypothesis. Thus, we have shown q(t) tends to α as t→-∞. The other limit can be proved similarly.Theorem 8.
For anyε∈(0,ε*], ϑ∈𝒱∖{0}. There exists qε,ϑ∈Γε(ϑ) such that qε,ϑ minimizes the functional I(q) on the set Γε(ϑ), that is, I(qε,ϑ)=cε(ϑ).Proof.
Let{qε,ϑ,n}∈Γε(ϑ) be a minimizing sequence for (13),
(33)I(qε,ϑ,n)⟶cε(ϑ)=infu∈Γε(ϑ)I(u).
For notational convenience, we will suppress subscripts ε,ϑ in what follows and simply write qε,ϑ,n=qn. Now since {qn} is a minimizing sequence, we have I(qn)⩽M for some M>0. Moreover, Lemma 6 shows that {qn(0)} is a bounded sequence; therefore, {qn} is bounded in E. Going if necessary to a subsequence, we may suppose that {qn} converges weakly in E to an element q∈E. Sobolev imbedding theorem shows qn→q in Cloc1(ℝ). To show q is the minimizer we are looking for, we need firstly to prove
(34)I(q)<+∞.
Indeed, for any r<s∈ℝ,
(35)Ir,s(qn)⩽I(qn)⩽M,
by the weakly lower semicontinuity of Ir,s(·),
(36)Ir,s(q)=liminfn→+∞Ir,s(qn)⩽liminfn→+∞I(qn)=infu∈Γε(ϑ)I(u)⩽M.
Clearly, the constant M is independent of r and s. Since q∈E, the above inequality implies V(·,q(·))∈L1(ℝ,ℝ), and
(37)I(q)⩽infu∈Γε(ϑ)I(u)⩽M.
Hence, (34) holds.
Next we show(38)limt→-∞q(t)=0,limt→+∞q(t)=ϑ.
Inequality (34) and Lemma 7 imply that there are α,β∈𝒱 verifying
(39)limt→-∞q(t)=α,limt→+∞q(t)=β.
But for each n, (qn(t),qn′(t))∉Bε(𝒱~∖{0~,ϑ~}) for all t∈ℝ and qn→q in Cloc1(ℝ), so (q(t),q′(t))∉Bε(𝒱~∖{0~,ϑ~}) for all t∈ℝ. Therefore, α,β∈{0,ϑ}. For each t∈ℝ, there is n(t)∈ℝ such that
(40)|qn(t)-q(t)|<ε,∀n⩾n(t).
We claimβ=ϑ. Suppose to the contrary that
(41)limt→+∞q(t)=0.
Then, there is t0>0 such that
(42)|q(t)|<ε,∀t⩾t0.
Hence,
(43)|qn(t)|⩽|qn(t)-q(t)|+|q(t)|<2ε,
whenever t⩾t0 and n⩾n(t). In particular
(44)|qn(t)(t)|⩽2ε,∀t⩾t0.
Let(45)sn=sn(ε)=sup{t∈ℝ∣qn(t)∈∂B2ε(0)},s-n=s-n(ε)=inf{t>sn∣qn(t)∈∂B4ε(0)}.
Since qn∈Γε(ϑ), sn<s-n are well defined for each n. The same argument that we used to obtain (24) shows
(46)infn|s-n-sn|⩾c0,
for some constant c0>0. Whence, for t⩾t0, by the definition of sn and c0, |qn(t)(τn(t))|∈(2ε,4ε), and the mean value theorem,
(47)I(qn(t))⩾∫sn(t)sn(t)+c0‍V(s,qn(t))ds=c0V(τn(t),qn(t)(τn(t))),
where τn(t)∈(sn(t),sn(t)+c0). Inequality (44) also implies t⩽sn(t)<τn(t). Hence, by (A4), V(τn(t),qn(t)(τn(t)))→+∞ as t→+∞, which leads to a contradiction that I(qn(t))→+∞ as t→+∞. Similar argument shows that limt→-∞q(t)=0 and the proof is completed.Denote(48)𝒢=𝒢(ε,ϑ)={t∈ℝ∣(qε,ϑ(t),qε,ϑ′(t))∈∂Bε(𝒱~∖{0~,ϑ~})}.Theorem 9.
The minimizerq=qε,ϑ obtained in Theorem 8 is a classical solution on ℝ∖𝒢.Proof.
For anyt∈ℝ∖𝒢, since ℝ∖𝒢 is an open set, there is a maximal open interval ℑ⊂ℝ∖𝒢 containing t. Let φ∈C0∞(ℑ), then for δ>0 sufficiently small q+δφ∈Γε(ϑ). Since q minimize I(·) on Γε(ϑ), then, for all φ∈C0∞(ℑ),
(49)(I′(q),φ)=∫-∞∞‍{q′′φ′′+q′φ′+Vq(t,q)φ}dt=0.
Fix any r<s∈ℑ, this equality also holds for any φ∈W02,2([r,s]). Hence, q is a weak solution of the following system:
(50)u′′′′(t)-u′′(t)+Vq(t,q)=0,r<t<s;u(r)=q(r),u(s)=q(s);u′(r)=q′(r),u′(s)=q′(s);
the system has a unique C4 solution, denoted by υ. Then
(51)∫rs‍{υ′′φ′′+υ′φ′+Vq(t,q)φ}dt=0.
Combining (49) and (51) yields
(52)∫rs‍{(q′′-υ′′)φ′′+(q′-υ′)φ′}dt=0,
for all φ∈W02,2([r,s]). Since q-υ belongs to E, it follows that q-υ is constant. Noting the boundary conditions that q coincides with υ at r and s, we have q=υ∈C4([r,s]). Since r,s∈ℑ are arbitrary, it follows that q∈C4(ℑ), which completes the proof.Define(53)cε=infϑ∈𝒱∖{0}cε(ϑ).
Since cε is nonincreasing as a function of ε>0, infε>0cε exists and we may choose a decreasing sequence εi→0 such that
(54)cεi⟶infε>0cεasi⟶∞.
By Lemma 6, for each i, the set of candidates for the infimum cεi is finite; hence, it is achieved by an element κi∈𝒱∖{0}, namely,
(55)cεi=cεi(κi)=infq∈Γεi(κi)I(q).
Theorem 8 shows, for each i, there is a qi∈Γεi(κi) such that
(56)I(qi)=infq∈Γεi(κi)I(q).
Again by Lemma 6, {κi,i⩾1}⊂𝒱∖{0} must be a finite set and thus contains a constant subsequence {κik,k⩾1}; in other words, there is a κ∈{κi,i⩾1} for which κik=κ, for all k⩾1. Without loss of generality, we also denote this constant sequence by {κi,i⩾1} (correspondingly {εi,i⩾1}), then
(57)cεi=infq∈Γεi(κ)I(q)=I(qi),
where qi∈Γεi(κ).Theorem 10.
Fori sufficiently large, qi is a heteroclinic solution connecting 0 and κ.Proof.
To finish the proof, we show thatqi never touches ∂Bεi(𝒱~∖{0~,κ~}) for i large enough. Assume not, then there are {χj}⊂𝒱∖{0,κ} and {τj}∈ℝ such that (qj(τj),qj′(τj))∈∂Bεj(χ~j) and (qj(t),qj′(t))∉∂Bεj(χ~j) for t<τj; precisely, τj is the first hitting time of qj on ∂Bεj(χ~j). An application of Lemma 6 shows that {χj} is bounded, so it must contain a constant subsequence which we still denote by {χj}, that is, χj≡χ∈𝒱∖{0,κ} for j⩾1. Now, two situations may occur, for infinitely many j, (i) (qj,qj′) reaches ∂Bεj(𝒱~∖{χ~}) before ∂Bεj(𝒱~∖{κ~}) or (ii) touches ∂Bεj(𝒱~∖{κ~}) before ∂Bεj(𝒱~∖{χ~}).
Sinceqj∈Γεj(κ), εj is a decreasing sequence of j and cεj is nonincreasing in j; hence,
(58)I(qj)=cεj⩽cε1<+∞,
which by Lemma 6 implies that {qj} must be uniformly bounded on ℝ, say sups∈ℝ|qj(s)|<M for all j.
Case  1 (
(
q
j
(
t
)
,
q
j
′
(
t
)
)
∉
B
-
ε
j
(
κ
~
)
for all
t
<
τ
j
). Before stepping forward, we need to show {τj} is bounded both from above and below. Suppose {τj} is unbounded from above. Let
(59)sj=sup{t∈ℝ∣(qj(t),qj′(t))∈∂Bγ(χ~)},s-j=inf{t>sj∣(qj(t),qj′(t))∈∂Bγ(κ~)}.
For j large enough, γ>εj, so sj>τj by definition of τj. Hence, {sj} is unbounded from above. Without loss of generality; we may suppose limj→∞sj=+∞. Similar to (46) it can be shown that
(60)infj|sj-s-j|>c1,
for some positive constant c1>0. Hence,
(61)+∞>cεj=I(qj)⩾∫sjs-j‍V(s,qj(s))ds=(s-j-sj)V(s^j,qj(s^j))>c1V(s^j,qj(s^j)),
where sj<s^j<s-j. Now assumption (A4) and the uniform boundedness of {qj} indicate that {s^j} is bounded, which is a contradiction. Thus, {τj} must have an upper bound. Similarly it can be proved that {τj} is also bounded from below by considering rj=sup{t∈ℝ∣(qj(t),qj′(t))∈∂Bγ(0~)} and r-j=inf{t>rj∣(qj(t),qj′(t))∈∂Bγ(κ~)}. Hence, {τj} is bounded.
Define(62)ςj(t)={χ,fort⩾τj+εj1/3;Pj(t-τj),fort∈[τj,τj+εj1/3);qj(t),fort<τj,
where Pj is a polynomial defined as
(63)Pj(t)=aj+bjt+cjt2+djt3,aj=qj(τj),bj=qj′(τj),cj=-εj-2/3[3(qj(τj)-χ)+2εj1/3qj′(τj)],dj=εj-1[2(qj(τj)-χ)+εj1/3qj′(τj)].
It is easy to verify the boundary conditions
(64)(Pj(0),Pj′(0))=(qj(τj),qj′(τj)),(65)(Pj(εj1/3),Pj′(εj1/3))=(χ,0).
In addition, limj→∞aj=0,limj→∞bj=0,limj→∞cj=0, and {dj} is bounded.
Comparing the energy ofqj and ςj, we have
(66)I(qj)-I(ςj)=Iτj,+∞(qj)-Iτj,+∞(ςj)=Iτj,+∞(qj)-Iτj,τj+εj1/3(ςj)-Iτj+εj1/3,+∞(ςj)⩾γ2σγ-Iτj,τj+εj1/3(ςj)-Iτj+εj1/3,+∞(ςj).
By assumption (A5),
(67)Iτj+εj1/3,+∞(ςj)<∫-∞∞‍V(t,χ)dt<γσγ.
Moreover, since {ςj} is uniformly bounded on ℝ and {τj} is bounded, it readily follows that
(68)∫τjτj+εj1/3‍V(t,ςj)dt⟶0asj⟶∞.
Meanwhile, by construction,
(69)∫τjτj+εj1/3‍(ςj′′)2=∫τjτj+εj1/3‍(2cj+6djt)2dt⟶0,∫τjτj+εj1/3‍(ςj′)2=∫τjτj+εj1/3‍(bj+2cjt+3djt2)2dt⟶0.
So
(70)limj→+∞Iτj,τj+εj1/3(ςj)=0.
Therefore, for j large enough,
(71)I(qj)-I(ςj)⩾γ2σγ-2-12γσγ-γσγ>2-12γσγ,
which is impossible.
Case  2 (
(
q
j
(
t
)
,
q
j
′
(
t
)
)
∈
∂
B
ε
j
(
{
κ
~
}
)
for some
s
j
<
τ
j
). A contradiction can also be reached by similar arguments.
Either case leads to a contradiction; therefore, fori sufficiently large, the orbit of qi=qεi,κ must keep some distance away from 𝒱~∖{0~,κ~}, and hence it is a classical solution, and the proof is complete.Proof of Theorem1.
Theorem10 shows for every z∈𝒱, there is heteroclinic solution of (1) emanating from z. If we consider the set Γ-ε(ϑ)⊂E of function q(t) for which q(+∞)=0, q(-∞)=ϑ∈𝒱∖{0} and (q(t),q′(t))∉Bε(𝒱~∖{0~,ϑ~}), for all t∈ℝ; then the proof for Γε(ϑ) works equally well for Γ-ε(ϑ), and so, for every z∈𝒱, there is heteroclinic solution of (1) terminating at z.


## 3. Heteroclinic Solution in Periodic Case

In the last section, we consider the case whereV(t,z) is periodic in t with period T>0. In this case, the assumptions and proof can be simplified. For completeness, we devote this section to the periodic case. Since most results in the last section can be carried out verbatim, we just present those that are different. We make the following assumptions:(H1)
V(t,q)∈C2(ℝ×ℝ), V⩾0;(H2)
the set𝒱={z∈ℝ∣forallt,V(t,z)=0,Vq(t,z)=0andVqq(t,z)⩾0} is discrete and has at least two points, and
(72)γ≡13inf{|x-y|∣x≠y∈𝒱}>0.
We also assume without no loss of generality 0∈𝒱.(H3)
There is a constantV0>0 such that liminf|q|→∞V(t,q)⩾V0 uniformly for t∈ℝ.(H4)
For eachz∈𝒱,
(73)∫-∞∞‍V(t,z)dt<γσγ1/2.Note that(A4) is no longer needed in the periodic case. (H3) is indeed a special case of (A3), we are using (H3) instead of (A3) only for illustrative purpose. As before, we work on the space
(74)E={q∈Wloc2,2(ℝ,ℝ)∣∫-∞∞‍(|q′′|2+|q′|2)dt<∞},
with the norm
(75)∥q∥2=∫-∞∞‍(|q′′|2+|q′|2)dt+|q(0)|2.We will use the shift operatorσj defined for each j as
(76)σjq=q(t-jT).Clearly, ifq is a heteroclinic solution, so is τjq for all j. Recall Γε(ζ),ζ∈𝒱∖{0} is the set of q∈E verifying(i)
limt→-∞q(t)=0,limt→+∞q(t)=ζ;(ii)
(q(t),q′(t))∉Bε(𝒱~/{0~,ϑ~}), for all t∈ℝ; where 𝒱~={(z,0)∈ℝ2∣z∈𝒱}.Γ
ε
(
ζ
) is not empty, we may consider the minimization problem
(77)cε(ζ)=infq∈Γε(ζ)I(q).We shall follow the lines of argument in previous section, and similar proofs will be omitted. For example, Lemmas3, 5, 6, and 7 remain valid in the setting of this section.Theorem 11.
For anyε∈(0,γ], ζ~∈𝒱~∖{0~}. There exists qε,ζ∈Γε(ζ) such that qε,ζ minimizes the functional I(q) on the set Γε(ζ), that is, I(qε,ζ)=cε(ζ).Proof.
Let{qn} be a minimizing sequence for (77). As in the preceding section, it is bounded in E and we may choose a subsequence, also denoted by {qn}, converging weakly in E to some q∈E and qn→q in Cloc1(ℝ). Moreover,
(78)I(q)<+∞.
It remains to show(79)limt→-∞q(t)=0,limt→+∞q(t)=ζ.
Inequality (78) and Lemma 7 imply that there are α~,β~∈𝒱~ verifying
(80)limt→-∞q(t)=α,limt→+∞q(t)=β.
Since V(t,u) is periodic in t, if {qn} is the minimizing sequence, so is σjnqn for any sequence {jn}⊂ℕ and I(qn)=I(σjnqn). Whence, with {jn} being appropriately chosen and {qn} replaced with {σjnqn}, we are safe to suppose that qn(t)∈Bε(0) for t⩽0 and qn(s)∈∂Bε(0) for some s∈[0,T]. Noting for each n, (qn(t),qn′(t))∉Bε(𝒱~∖{0~,ζ~}) for all t∈ℝ and qn→q in Cloc1(ℝ), we have (q(t),q′(t))∉Bε(𝒱~∖{0~,ζ~}) for all t∈ℝ and so α~,β~∈{0~,ζ~}. Since qn(t)∈Bε(0) for t⩽0, α∈{0,ζ}∩B-ε(0)={0}, so α=0.
It remains to showβ=ζ. Note β=q(+∞)∈{0,ζ}. Suppose that q(+∞)=0. Choose δ>0, for the time being we only assume that 4δ<ε. Since q′(+∞)=0, there is a tδ>T such that q(t),q′(t)∈Bδ(0) for all t>tδ. Since qn→q in Cloc1(ℝ), so there is nδ, which depends on δ, such that qn(tδ),qn′(tδ)∈B2δ(0) for n larger than nδ. But qn(s)∈∂Bε(0) for some s∈[0,T], so qn intersects ∂B2δ(0) and ∂Bε(0). By Lemma 3, we have
(81)I(qn)⩾ε22σε/2+Itδ,+∞(qn).
Now construct a sequence as follows:
(82)ςn(t)={0,fort⩽tδ-tn;Pn(t-tδ+tn),fort∈(tδ-tn,tδ];qn(t),fort>tδ,
where
(83)tn={|qn(tδ)|+|qn′(tδ)|3,if|qn(tδ)|+|qn′(tδ)|3≠0,1,if|qn(tδ)|+|qn′(tδ)|3=0,
and Pn(t)=gnt2+hnt3 is a polynomial with coefficients
(84)gn=tn-2[3qn(tδ)-tnqn′(tδ)],hn=tn-3[tnqn′(tδ)-2qn(tδ)].
It is easy to check that Pn(t) satisfies
(85)(Pn(0),Pn′(0))=(0,0),(Pn(tn),Pn′(tn))=(qn(tδ),qn′(tδ)).
Moreover, qn(tδ),qn′(tδ) tends to zero if δ→0 and n>nδ. By construction ςn∈Γε(ζ). So far, we only used the fact 0<4δ<ε, we can fix δ smaller still such that, for all n>nδ,
(86)Itδ-tn,tδ(ςn)=∫tδ-tntδ‍{12(2gn+6hnt)2+12(2gnt+3hnt2)2+V(t,ςn)}dt,Itδ-tn,tδ(ςn)<ε42σε/2.
Then
(87)I(qn)-I(ςn)=I-∞,tδ(qn)-Itδ-tn,tδ(ςn)⩾ε22σε/2-Itδ-tn,tδ(ςn)>ε42σε/2,
for n>nδ. This implies
(88)liminfn→+∞I(qn)-ε42σε/2⩾liminfn→+∞I(ςn)⩾infq∈Γε(ζ)I(q)=liminfn→+∞I(qn),
which is impossible. The proof is complete.Theorem 12.
There is a pair of orbitsq0,ζ,qζ,0 which connect 0 and ζ, one emanating from 0 and the other terminating at 0.Proof.
The proof is similar to Theorems 8 and 10 in [9].Lastly, we observe the following.Theorem 13.
For the solutionq=q0,ζ or qζ,0 obtained above, limn→±∞q′′(t)=0,limn→±∞q′′′(t)=0.Proof.
Considerq=q0,ζ. By (H2), there are positive constants r*,c1,c2 such that
(89)V(t,u)c1⩾|u-ζ|2,|Vz(t,u)|⩽c2|u-ζ|,
whenever |u-ζ|⩽r*. Since q(+∞)=ζ, |q(t)-ζ|⩽r* for t larger than some t*>0.
From (89) and the Euler equation,
(90)q′′′′-q′′+Vz(t,q)=0,
we have
(91)∫t*+∞‍|q′′′′(t)|2dt⩽2∫t*+∞‍|q′′(t)|2dt+2∫t*+∞‍|Vz(t,q)|2dt⩽2∫t*+∞‍|q′′(t)|2dt+2c22∫t*+∞‍|q(t)-ζ|2dt=2∫t*+∞‍|q′′(t)|2dt+2c22c1∫t*+∞‍c1|q(t)-ζ|2dt⩽2∫t*+∞‍|q′′(t)|2dt+2c22c1∫t*+∞‍V(t,q(t))dt<∞.
An application of the interpolation inequality yields
(92)∫t*+∞‍|q′′′(t)|2dt⩽K∫t*+∞‍|q′′(t)|2dt+K∫t*+∞‍|q′′′′(t)|2dt<∞,
where K is a positive constant.
Therefore, we have proved that(93)limn→+∞q′′(t)=limn→+∞q′′′(t)=0.
The limit for negative infinity can be derived similarly. The proof is complete.



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*Source: 138623-2013-04-04.xml*