Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 138623, 9 pages
http://dx.doi.org/10.1155/2013/138623
Research Article
Heteroclinic Solutions for Nonautonomous EFK Equations
Y. L. Yeun1,2
1
2
School of Mathematics and Systems Science, Beijing University of Aeronautics and Astronautics, Beijing 100191, China
School of Mathematical Sciences, Peking University, Beijing 100875, China
Correspondence should be addressed to Y. L. Yeun; ruanyl@math.pku.edu.cn
Received 30 August 2012; Accepted 12 February 2013
Academic Editor: Nikolaos Papageorgiou
Copyright © 2013 Y. L. Yeun. This is an open access article distributed under the Creative Commons Attribution License, which
permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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 function 𝑉(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.
1. Introduction
The goal of this paper is to study the nonautonomous
extended Fisher-Kolmogorov (EFK) equation
𝑞󸀠󸀠󸀠󸀠 (𝑡) − 𝑞󸀠󸀠 (𝑡) + 𝑉𝑞 (𝑡, 𝑞) = 0,
(1)
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
𝜇𝑞󸀠󸀠󸀠󸀠 − 𝜆𝑞󸀠󸀠 − 𝑞 + 𝑞3 = 0,
(𝜇, 𝜆 > 0) ,
(2)
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
𝜇𝑞󸀠󸀠󸀠󸀠 − 𝜆𝑞󸀠󸀠 + 𝑉󸀠 (𝑞) = 0,
(𝜇, 𝜆 > 0) ,
(3)
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
2
1
2
𝑉2 (𝑞) = 2 (1 + cos 𝜋𝑞) . (4)
𝑉1 (𝑞) = (𝑞2 − 1) ,
4
𝜋
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
2
Abstract and Applied Analysis
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 connecting 𝑎− and 𝑎+ , we mean a
solution 𝑞(𝑡) verifying
󸀠
󸀠󸀠
󸀠󸀠󸀠
(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 󳨀→ −∞.
(5)
Throughout the paper, let 𝐵𝜀 (𝐶) denote the neighborhood
of a set 𝐶 ⊂ R𝑛 defined as below
𝐵𝜀 (𝐶) = {𝑤 ∈ R𝑛 | inf |𝑤 − V| < 𝜀} .
V∈𝐶
(𝐴 2 ) let V be a set which has at least two points and we
assume the points in V do not collapse together,
1
󵄨
󵄨
inf {󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 | 𝑥 ≠ 𝑦 ∈ V} > 0;
3
(7)
∗
(a) there is 𝜀 ∈ (0, 𝛾], such that 𝜎
≜𝜀 inf 𝑧∉𝐵𝜀 (V),𝑡∈R
𝑉(𝑡, 𝑧) > 0 for any 𝜀 ∈ (0, 𝜀∗ ];
(b) lim inf |𝑧| → ∞ |𝑧|𝑉(𝑡, 𝑧) > 0 uniformly for 𝑡 ∈ R.
(𝐴 4 ) lim|𝑡| → +∞ 𝑉(𝑡, 𝑧) = +∞ uniformly for 𝑧 in any compact subsets of R \ V.
(𝐴 5 ) For each 𝑧 ∈ V,
𝑉 (𝑡, 𝑧) 𝑑𝑡 <
𝑠
2
2
1
1
𝐼𝑟,𝑠 (𝑞) = ∫ { (𝑞󸀠󸀠 ) + (𝑞󸀠 ) + 𝑉 (𝑡, 𝑞)} 𝑑𝑡.
2
2
𝑟
(10)
We will work on the space
∞
󵄨 󵄨2 󵄨 󵄨2
(󵄨󵄨󵄨󵄨𝑞󸀠󸀠 󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨𝑞󸀠 󵄨󵄨󵄨󵄨 ) 𝑑𝑡 < ∞} ,
−∞
2,2
𝐸 = {𝑞 ∈ 𝑊loc
(R, R) | ∫
(11)
𝛾𝜎𝛾1/2 ,
(12)
It is readily seen that (𝐸, ‖ ⋅ ‖) is a Hilbert space and 𝑞 ∈ 𝐸
implies 𝑞 ∈ 𝐶1 (R).
̃ \ {0̃}, 𝜀 ∈ (0, 1). Denote by Γ𝜀 (𝜗) the set of 𝑞 ∈ 𝐸
Let 𝜗̃ ∈ V
verifying
(i) lim𝑡 → −∞ 𝑞(𝑡) = 0, lim𝑡 → +∞ 𝑞(𝑡) = 𝜗;
(𝐴 3 ) one of the following is satisfied:
−∞
(9)
Denote the energy on a bounded interval [𝑟, 𝑠] by
∞
󵄨 󸀠󸀠 󵄨2 󵄨 󸀠 󵄨2
󵄨2
󵄩󵄩 󵄩󵄩2
󵄨
󵄩󵄩𝑞󵄩󵄩 = ∫ (󵄨󵄨󵄨󵄨𝑞 󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨𝑞 󵄨󵄨󵄨󵄨 ) 𝑑𝑡 + 󵄨󵄨󵄨𝑞 (0)󵄨󵄨󵄨 .
−∞
(𝐴 1 ) 𝑉(𝑡, 𝑞) ∈ 𝐶 (R × R, R), 𝑉 ⩾ 0;
∫
∞
2
2
1
1
{ (𝑞󸀠󸀠 ) + (𝑞󸀠 ) + 𝑉 (𝑡, 𝑞)} 𝑑𝑡.
2
−∞ 2
𝐼 (𝑞) = ∫
equipped with the norm
2
∞
̃ = {(𝑧, 0) ∈ R2 | 𝑧 ∈ V}. For an
Notation. Define V
̃ we denote it by 𝑧̃ = (𝑧, 0) for brevity.
element (𝑧, 0) in V,
Whence, without further notice, we will use 𝑧 and 𝑧̃ 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
(6)
We will make the following assumptions:
𝛾≡
contain other points which are not global minima of 𝑉(𝑡, 𝑢)
but still verify (𝐴 5 ). In the special case V = {𝑧 ∈ R | 𝑉(𝑡, 𝑧) =
0 for all 𝑡}, (𝐴 5 ) is automatically satisfied.
(8)
where 𝜎𝛾 is defined in (𝐴 3 ) with 𝜀 = 𝛾.
Theorem 1. Let 𝑉(𝑡, 𝑞) satisfy (𝐴 1 )–(𝐴 4 ). 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 assume 𝑉 has wells of equal
depth. By the assumptions of the theorem, we necessarily
have {𝑧 ∈ R | 𝑉(𝑡, 𝑧) = 0 for all 𝑡} ⊂ V. However, V may
̃ for all 𝑡 ∈ R.
̃ \ {0̃, 𝜗}),
(ii) (𝑞(𝑡), 𝑞󸀠 (𝑡)) ∉ 𝐵𝜀 (V
That is, Γ𝜀 (𝜗) consists of functions which join 0 and 𝜗, and
̃ Clearly, Γ (𝜗)
̃ \ {0,
̃ 𝜗}.
at least keep a distance of 𝜀 to the set V
𝜀
is not empty, we may consider the minimization problem
𝑐𝜀 (𝜗) = inf 𝐼 (𝑞) .
𝑞∈Γ𝜀 (𝜗)
(13)
We will find for each 𝜀 > 0 a candidate in Γ𝜀 (𝜗) and then
send 𝜀 to zero to obtain the minimizer, which is actually a
heteroclinic solution.
For any 𝜀 ∈ (0, 𝛾), V > 0 define
𝜎𝜀,V =
inf
𝑧∉𝐵𝜀 (V),|𝑧|⩽V,
𝑡∈R
𝑉 (𝑡, 𝑧) ,
(14)
then 𝜎𝜀,V > 0 by our assumptions. For V = +∞, we take
𝜎𝜀,+∞ = 𝜎𝜀 which is defined in (𝐴 3 ).
From now on, we always assume that
0 < 𝜀 < min {𝜀∗ , 1} .
(15)
Abstract and Applied Analysis
3
Lemma 3. Let 𝑞 ∈ 𝐸. Then for any V > 0, 𝑟 < 𝑠 ∈ R such that
𝑞(𝑡) ∉ 𝐵𝜀 (V) and |𝑞(𝑡)| ⩽ V for any 𝑡 ∈ [𝑟, 𝑠],
󵄨
󵄨
𝐼 (𝑞) ⩾ √2𝜎𝜀,V 󵄨󵄨󵄨𝑞 (𝑟) − 𝑞 (𝑠)󵄨󵄨󵄨 .
(16)
In particular, if 𝜎𝜀 > 0, then for any 𝑟 < 𝑠 ∈ R such that
𝑞(𝑡) ∉ 𝐵𝜀 (V) for any 𝑡 ∈ [𝑟, 𝑠],
󵄨
󵄨
𝐼 (𝑞) ⩾ √2𝜎𝜀 󵄨󵄨󵄨𝑞 (𝑟) − 𝑞 (𝑠)󵄨󵄨󵄨 .
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⩾𝑛 with |𝑞(𝑠𝑗 )| = 𝑗,
𝑞
|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 inf 𝑗⩾𝑛𝑞 |𝑡𝑗 − 𝑠𝑗 | > 0.
Indeed, for 𝑗 ⩾ 𝑛𝑞 ,
󵄨 󵄨
󵄨󵄨
󵄨󵄨
1 = 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑞 (𝑡𝑗 )󵄨󵄨󵄨󵄨 − 󵄨󵄨󵄨󵄨𝑞 (𝑠𝑗 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨
󵄨
󵄨
⩽ 󵄨󵄨󵄨󵄨𝑞 (𝑡𝑗 ) − 𝑞 (𝑠𝑗 )󵄨󵄨󵄨󵄨
(17)
Proof. Denote 𝑙 = |𝑞(𝑟) − 𝑞(𝑠)| and 𝜏 = |𝑟 − 𝑠|. Then
𝑡𝑗
󵄨
󵄨
⩽ ∫ 󵄨󵄨󵄨󵄨𝑞󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑑𝑡
𝑠
1/2
󵄨󵄨 𝑠
󵄨󵄨
𝑠
𝑠
󵄨
󵄨2
󵄨
󵄨
󵄨
󵄨
𝑙 = 󵄨󵄨󵄨∫ 𝑞󸀠 (𝑡) 𝑑𝑡󵄨󵄨󵄨 ⩽ ∫ 󵄨󵄨󵄨󵄨𝑞󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑑𝑡 ⩽ 𝜏1/2 (∫ 󵄨󵄨󵄨󵄨𝑞󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑑𝑡) .
󵄨󵄨 𝑟
󵄨󵄨
𝑟
𝑟
(18)
𝑗
󵄨
󵄨2
󵄨1/2 𝑡𝑗 󵄨
⩽ 󵄨󵄨󵄨󵄨𝑡𝑗 − 𝑠𝑗 󵄨󵄨󵄨󵄨 (∫ 󵄨󵄨󵄨󵄨𝑞󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑑𝑡)
𝑠
Since 𝑉 ⩾ 0, by assumptions, we have
𝑙2
+ 𝜎𝜀,V 𝜏 ⩾ √2𝜎𝜀,V 𝑙.
2𝜏
,
so
󵄨
󵄨2
󵄨 𝑠󵄨
1 ⩽ 󵄨󵄨󵄨󵄨𝑡𝑗 − 𝑠𝑗 󵄨󵄨󵄨󵄨 ∫ 󵄨󵄨󵄨󵄨𝑞󸀠 (𝑡)󵄨󵄨󵄨󵄨 𝑑𝑡
(19)
𝑟
󵄨
󵄨
⩽ 2 󵄨󵄨󵄨󵄨𝑡𝑗 − 𝑠𝑗 󵄨󵄨󵄨󵄨 𝐼 (𝑞) ,
Proof for the second part is similar.
(23)
∀𝑗 ⩾ 𝑛𝑞 ,
which implies
Remark 4. The proof of Lemma 3 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. Let 𝑞 ∈ 𝐸. 𝐼(𝑞) < +∞ implies 𝑞 ∈ 𝐿∞ (R, R).
Proof. Suppose to the contrary that 𝑞 ∉ 𝐿 (R). Let 𝜀 > 0
satisfy (15).
Case 1 (𝜎𝜀 > 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,
𝑛
(24)
Hence,
𝑛
𝑡𝑗
𝑗=𝑛𝑞 𝑠𝑗
𝑛
𝑡𝑗
1
󵄨󵄨
󵄨 𝑛 1
󵄨󵄨𝑡𝑗 − 𝑠𝑗 󵄨󵄨󵄨) ∑
⩾ 𝑐1 ∑ ∫ 󵄨󵄨
,
󵄨󵄨 𝑑𝑡 ⩾ 𝑐1 (inf
󵄨
𝑗⩾𝑛𝑞 󵄨
𝑗=𝑛𝑞 𝑠𝑗 󵄨󵄨𝑞 (𝑡)󵄨󵄨
𝑗=𝑛𝑞 𝑗 + 1
(25)
for any 𝑛 ⩾ 𝑛𝑞 , a contradiction.
Therefore, in either case, we must have 𝑞 ∈ 𝐿∞ (R).
The preceding lemma has an important corollary which
will be used later; we prove it here.
𝐼 (𝑞) > ∑𝐼𝑠𝑖 ,𝑡𝑖 (𝑞)
𝑖=1
󵄨
󵄨
inf 󵄨󵄨󵄨󵄨𝑡𝑗 − 𝑠𝑗 󵄨󵄨󵄨󵄨 > 0.
𝑗⩾𝑛𝑞
𝐼 (𝑞) ⩾ ∑ ∫ 𝑉 (𝑡, 𝑞 (𝑡)) 𝑑𝑡
∞
𝑛
1/2
𝑗
𝑠
2
1
𝐼 (𝑞) ⩾ ∫ { (𝑞󸀠 (𝑡)) + 𝑉 (𝑡, 𝑞 (𝑡))} 𝑑𝑡
𝑟 2
⩾
(22)
(20)
󵄨
󵄨
⩾ √2𝜎𝜀 ∑ 󵄨󵄨󵄨𝑞 (𝑠𝑖 ) − 𝑞 (𝑡𝑖 )󵄨󵄨󵄨 ⩾ 𝑛𝛾√2𝜎𝜀 ,
Lemma 6. For 𝑀 > 0 and 𝑢 ∈ V, define
𝑄𝑢,𝑀 = {𝑞 ∈ 𝐸 | 𝑞 (−∞) = 𝑢, 𝐼 (𝑞) ⩽ 𝑀} .
𝑖=1
for any 𝑛 ⩾ 1, a contradiction.
If V 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
(26)
Then 𝑄𝑢,𝑀 is a bounded set in 𝐶(R) with the usual uniform
norm.
for any 𝑛 ⩾ 1, also a contradiction.
Proof. Without loss of generality, assume 𝑢 = 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 2 (lim inf |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
Case 1 (𝜎𝜀 > 0). If V is an infinite set, then for each 𝑛 ⩾
[3𝛾] + 1, 𝑛 ∈ N, there are 𝑙𝑛 nonoverlapping time intervals
𝐼 (𝑞) ⩾ 𝑛√2𝜎𝜀 ,
(21)
4
Abstract and Applied Analysis
{(𝑠𝑛,𝑖 , 𝑡𝑛,𝑖 )}𝑖⩽𝑙 such that |𝑞𝑛 (𝑠𝑛,𝑖 ) − 𝑞𝑛 (𝑡𝑛,𝑖 )| ⩾ 𝛾 for 1 ⩽ 𝑖 ⩽ 𝑙𝑛 ,
𝑛
𝑙𝑛
∪𝑖=1
(𝑠𝑛,𝑖 , 𝑡𝑛,𝑖 );
𝑞𝑛 (𝑡) ∉ 𝐵𝜀 (V) for 𝑡 ∈
moreover, the number of
intervals 𝑙𝑛 → ∞ as 𝑛 → ∞. Then by Lemma 3,
𝑙𝑛
𝑙𝑛
𝑖=1
𝑖=1
󵄨
󵄨
𝐼 (𝑞𝑛 ) > ∑𝐼𝑠𝑛,𝑖 ,𝑡𝑛,𝑖 (𝑞𝑛 ) ⩾ √2𝜎𝜀 ∑ 󵄨󵄨󵄨𝑞𝑛 (𝑠𝑛,𝑖 ) − 𝑞𝑛 (𝑡𝑛,𝑖 )󵄨󵄨󵄨
(27)
⩾ 𝑙𝑛 𝛾√2𝜎𝜀 ,
for any 𝑛 ⩾ [3𝛾] + 1, a contradiction.
If V 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
𝐼 (𝑞𝑛 ) ⩾ (𝑛 − (𝑧V + 1)) √2𝜎𝜀 ,
(28)
for any 𝑛 ⩾ [𝑧V ] + 1, also a contradiction.
Case 2 (lim inf |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
󵄨
󵄨
inf inf 󵄨󵄨󵄨󵄨𝑡𝑛,𝑗 − 𝑠𝑛,𝑗 󵄨󵄨󵄨󵄨 > 0.
𝑛⩾2 𝑛⩾𝑗⩾2
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 any 𝜀 ∈ (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),
𝐼 (𝑞𝜀,𝜗,𝑛 ) 󳨀→ 𝑐𝜀 (𝜗) = inf 𝐼 (𝑢) .
𝑢∈Γ𝜀 (𝜗)
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
1
𝑞𝑛 → 𝑞 in 𝐶loc
(R). To show 𝑞 is the minimizer we are
looking for, we need firstly to prove
(29)
Hence
(33)
𝐼 (𝑞) < +∞.
(34)
Indeed, for any 𝑟 < 𝑠 ∈ R,
𝐼 (𝑞𝑛 ) ⩾
𝑛
∑ ∫
𝑡𝑛,𝑗
𝑗=[𝑧0 ]+2 𝑠𝑛,𝑗
𝑛
⩾ 𝑐1 ∑ ∫
𝐼𝑟,𝑠 (𝑞𝑛 ) ⩽ 𝐼 (𝑞𝑛 ) ⩽ 𝑀,
𝑉 (𝑡, 𝑞𝑗 (𝑡)) 𝑑𝑡
𝑡𝑛,𝑗
𝑗=[𝑧0 ]+2 𝑠𝑛,𝑗
by the weakly lower semicontinuity of 𝐼𝑟,𝑠 (⋅),
1
󵄨 𝑑𝑡
󵄨󵄨
󵄨󵄨𝑞𝑗 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
(30)
for any 𝑛 ⩾ [𝑧0 ] + 2, 𝑛 ∈ N, a contradiction.
Therefore, a constant 𝐶0 > 0 depending only on 𝑀 and 𝑢
exists, for which sup𝑡∈R |𝑞(𝑡)| ⩽ 𝐶0 for all 𝑞 ∈ 𝑄𝑢,𝑀.
Lemma 7. Let 𝑞 ∈ 𝐸. If 𝐼(𝑞) < +∞, then there are 𝛼, 𝛽 ∈ V
such that lim𝑡 → −∞ 𝑞(𝑡) = 𝛼 and lim𝑡 → +∞ 𝑞(𝑡) = 𝛽.
∞
Proof. By Lemma 5, 𝑞 ∈ 𝐿 (R, R). Since 𝑞 ∈ 𝐶(R), then
sup𝑡∈R |𝑞(𝑡)| ⩽ 𝑄 for some 𝑄 > 0. Therefore, the set
Ω (𝑞, −∞) = {𝑥 : 𝑞 (𝑠𝑘 ) 󳨀→ 𝑥 for some 𝑠𝑘 󳨀→ −∞} (31)
of accumulation points of 𝑞 at −∞ is not empty. Moreover,
Ω(𝑞, −∞) ∩ V ≠ 0. 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,
−∞
{𝑉 (𝑡, 𝑞)} 𝑑𝑡 > ∫
−𝑟0
−∞
𝜎𝛿,𝑄𝑑𝑡,
𝑛 → +∞
𝑛 → +∞
𝑢∈Γ𝜀 (𝜗)
Clearly, the constant 𝑀 is independent of 𝑟 and 𝑠. Since 𝑞 ∈ 𝐸,
the above inequality implies 𝑉(⋅, 𝑞(⋅)) ∈ 𝐿1 (R, R), and
0
−𝑟0
𝐼𝑟,𝑠 (𝑞) = lim inf 𝐼𝑟,𝑠 (𝑞𝑛 ) ⩽ lim inf 𝐼 (𝑞𝑛 ) = inf 𝐼 (𝑢) ⩽ 𝑀.
(36)
𝑛
1
󵄨
󵄨
⩾ 𝑐1 (inf inf 󵄨󵄨󵄨󵄨𝑡𝑛,𝑗 − 𝑠𝑛,𝑗 󵄨󵄨󵄨󵄨) ∑
,
𝑛⩾2 𝑛⩾𝑗⩾2
𝑗
𝑗=[𝑧 ]+2
𝐼 (𝑞) > ∫
(35)
(32)
𝐼 (𝑞) ⩽ inf 𝐼 (𝑢) ⩽ 𝑀.
𝑢∈Γ𝜀 (𝜗)
(37)
Hence, (34) holds.
Next we show
lim 𝑞 (𝑡) = 0,
𝑡 → −∞
lim 𝑞 (𝑡) = 𝜗.
𝑡 → +∞
(38)
Inequality (34) and Lemma 7 imply that there are 𝛼, 𝛽 ∈ V
verifying
lim 𝑞 (𝑡) = 𝛼,
𝑡 → −∞
lim 𝑞 (𝑡) = 𝛽.
𝑡 → +∞
(39)
̃ for all 𝑡 ∈ R and
̃ \ {0,
̃ 𝜗})
But for each 𝑛, (𝑞𝑛 (𝑡), 𝑞𝑛󸀠 (𝑡)) ∉ 𝐵𝜀 (V
1
󸀠
̃ for all 𝑡 ∈
̃
̃ 𝜗})
𝑞𝑛 → 𝑞 in 𝐶loc (R), so (𝑞(𝑡), 𝑞 (𝑡)) ∉ 𝐵𝜀 (V \ {0,
R. Therefore, 𝛼, 𝛽 ∈ {0, 𝜗}. For each 𝑡 ∈ R, there is 𝑛(𝑡) ∈ R
such that
󵄨
󵄨󵄨
󵄨󵄨𝑞𝑛 (𝑡) − 𝑞 (𝑡)󵄨󵄨󵄨 < 𝜀,
∀𝑛 ⩾ 𝑛 (𝑡) .
(40)
Abstract and Applied Analysis
5
the system has a unique 𝐶4 solution, denoted by 𝜐. Then
We claim 𝛽 = 𝜗. Suppose to the contrary that
lim 𝑞 (𝑡) = 0.
Then, there is 𝑡0 > 0 such that
󵄨󵄨󵄨𝑞 (𝑡)󵄨󵄨󵄨 < 𝜀, ∀𝑡 ⩾ 𝑡0 .
󵄨
󵄨
Hence,
󵄨
󵄨 󵄨
󵄨 󵄨
󵄨󵄨
󵄨󵄨𝑞𝑛 (𝑡)󵄨󵄨󵄨 ⩽ 󵄨󵄨󵄨𝑞𝑛 (𝑡) − 𝑞 (𝑡)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑞 (𝑡)󵄨󵄨󵄨 < 2𝜀,
whenever 𝑡 ⩾ 𝑡0 and 𝑛 ⩾ 𝑛(𝑡). In particular
󵄨
󵄨󵄨
󵄨󵄨𝑞𝑛(𝑡) (𝑡)󵄨󵄨󵄨 ⩽ 2𝜀, ∀𝑡 ⩾ 𝑡0 .
Let
𝑠𝑛 = 𝑠𝑛 (𝜀) = sup {𝑡 ∈ R | 𝑞𝑛 (𝑡) ∈ 𝜕𝐵2𝜀 (0)} ,
𝑠𝑛 = 𝑠𝑛 (𝜀) = inf {𝑡 > 𝑠𝑛 | 𝑞𝑛 (𝑡) ∈ 𝜕𝐵4𝜀 (0)} .
(42)
𝑠𝑛(𝑡)
Combining (49) and (51) yields
𝑠
(43)
(44)
(45)
𝑉 (𝑠, 𝑞𝑛(𝑡) ) 𝑑𝑠 = 𝑐0 𝑉 (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 (𝐴 4 ), 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
G = G (𝜀, 𝜗)
(48)
Theorem 9. The minimizer 𝑞 = 𝑞𝜀,𝜗 obtained in Theorem 8 is
a classical solution on R \ G.
Proof. For any 𝑡 ∈ 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),
(𝐼󸀠 (𝑞) , 𝜑) = ∫
∞
−∞
{𝑞󸀠󸀠 𝜑󸀠󸀠 + 𝑞󸀠 𝜑󸀠 + 𝑉𝑞 (𝑡, 𝑞) 𝜑} 𝑑𝑡 = 0.
(49)
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:
𝑢󸀠󸀠󸀠󸀠 (𝑡) − 𝑢󸀠󸀠 (𝑡) + 𝑉𝑞 (𝑡, 𝑞) = 0,
𝑢 (𝑟) = 𝑞 (𝑟) ,
𝑢󸀠 (𝑟) = 𝑞󸀠 (𝑟) ,
𝑟 < 𝑡 < 𝑠;
𝑢 (𝑠) = 𝑞 (𝑠) ;
𝑢󸀠 (𝑠) = 𝑞󸀠 (𝑠) ;
(50)
(52)
𝑟
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
𝑐𝜀 = inf 𝑐𝜀 (𝜗) .
(53)
𝜗∈V\{0}
Since 𝑐𝜀 is nonincreasing as a function of 𝜀 > 0, inf 𝜀>0 𝑐𝜀 exists
and we may choose a decreasing sequence 𝜀𝑖 → 0 such that
𝑐𝜀𝑖 󳨀→ inf 𝑐𝜀
as 𝑖 󳨀→ ∞.
𝜀>0
(54)
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,
𝑐𝜀𝑖 = 𝑐𝜀𝑖 (𝜅𝑖 ) = inf 𝐼 (𝑞) .
(47)
󸀠
̃ .
̃ \ {0,
̃ 𝜗})}
= {𝑡 ∈ R | (𝑞𝜀,𝜗 (𝑡) , 𝑞𝜀,𝜗
(𝑡)) ∈ 𝜕𝐵𝜀 (V
(51)
∫ {(𝑞󸀠󸀠 − 𝜐󸀠󸀠 ) 𝜑󸀠󸀠 + (𝑞󸀠 − 𝜐󸀠 ) 𝜑󸀠 } 𝑑𝑡 = 0,
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,
𝐼 (𝑞𝑛(𝑡) ) ⩾ ∫
∫ {𝜐󸀠󸀠 𝜑󸀠󸀠 + 𝜐󸀠 𝜑󸀠 + 𝑉𝑞 (𝑡, 𝑞) 𝜑} 𝑑𝑡 = 0.
𝑟
Since 𝑞𝑛 ∈ Γ𝜀 (𝜗), 𝑠𝑛 < 𝑠𝑛 are well defined for each 𝑛. The same
argument that we used to obtain (24) shows
󵄨󵄨
󵄨
inf
󵄨𝑠 − 𝑠𝑛 󵄨󵄨󵄨 ⩾ 𝑐0 ,
(46)
𝑛 󵄨 𝑛
𝑠𝑛(𝑡) +𝑐0
𝑠
(41)
𝑡 → +∞
(55)
𝑞∈Γ𝜀𝑖 (𝜅𝑖 )
Theorem 8 shows, for each 𝑖, there is a 𝑞𝑖 ∈ Γ𝜀𝑖 (𝜅𝑖 ) such that
𝐼 (𝑞𝑖 ) = inf 𝐼 (𝑞) .
(56)
𝑞∈Γ𝜀𝑖 (𝜅𝑖 )
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
𝑐𝜀𝑖 = inf 𝐼 (𝑞) = 𝐼 (𝑞𝑖 ) ,
(57)
𝑞∈Γ𝜀𝑖 (𝜅)
where 𝑞𝑖 ∈ Γ𝜀𝑖 (𝜅).
Theorem 10. For 𝑖 sufficiently large, 𝑞𝑖 is a heteroclinic solution
connecting 0 and 𝜅.
Proof. To finish the proof, we show that 𝑞𝑖 never touches
̃ \ {0̃, 𝜅̃}) for 𝑖 large enough. Assume not, then there
𝜕𝐵𝜀𝑖 (V
are {𝜒𝑗 } ⊂ V \ {0, 𝜅} and {𝜏𝑗 } ∈ R such that (𝑞𝑗 (𝜏𝑗 ), 𝑞𝑗󸀠 (𝜏𝑗 )) ∈
𝜕𝐵𝜀𝑗 (𝜒̃𝑗 ) and (𝑞𝑗 (𝑡), 𝑞𝑗󸀠 (𝑡)) ∉ 𝜕𝐵𝜀𝑗 (𝜒̃𝑗 ) for 𝑡 < 𝜏𝑗 ; precisely, 𝜏𝑗
is the first hitting time of 𝑞𝑗 on 𝜕𝐵𝜀𝑗 (𝜒̃𝑗 ). 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,
̃ \ {𝜒})
̃ before
for infinitely many 𝑗, (i) (𝑞𝑗 , 𝑞𝑗󸀠 ) reaches 𝜕𝐵𝜀𝑗 (V
̃ 𝜅}) or (ii) touches 𝜕𝐵𝜀 (V\{̃
̃ 𝜅}) before 𝜕𝐵𝜀 (V\{
̃ 𝜒}).
̃
𝜕𝐵𝜀 (V\{̃
𝑗
𝑗
𝑗
6
Abstract and Applied Analysis
Since 𝑞𝑗 ∈ Γ𝜀𝑗 (𝜅), 𝜀𝑗 is a decreasing sequence of 𝑗 and 𝑐𝜀𝑗
is nonincreasing in 𝑗; hence,
𝐼 (𝑞𝑗 ) = 𝑐𝜀𝑗 ⩽ 𝑐𝜀1 < +∞,
𝜅) for all 𝑡 < 𝜏𝑗 ). Before stepping
Case 1 ((𝑞𝑗 (𝑡), 𝑞𝑗󸀠 (𝑡)) ∉ 𝐵𝜀𝑗 (̃
forward, we need to show {𝜏𝑗 } is bounded both from above
and below. Suppose {𝜏𝑗 } is unbounded from above. Let
𝑠𝑗 = inf {𝑡 > 𝑠𝑗 |
(𝑞𝑗 (𝑡) , 𝑞𝑗󸀠
𝜅)} .
(𝑡)) ∈ 𝜕𝐵𝛾 (̃
(64)
(𝑃𝑗 (𝜀𝑗1/3 ) , 𝑃𝑗󸀠 (𝜀𝑗1/3 )) = (𝜒, 0) .
(65)
In addition, lim𝑗 → ∞ 𝑎𝑗 = 0, lim𝑗 → ∞ 𝑏𝑗 = 0, lim𝑗 → ∞ 𝑐𝑗 = 0,
and {𝑑𝑗 } is bounded.
Comparing the energy of 𝑞𝑗 and 𝜍𝑗 , we have
𝐼 (𝑞𝑗 ) − 𝐼 (𝜍𝑗 ) = 𝐼𝜏𝑗 ,+∞ (𝑞𝑗 ) − 𝐼𝜏𝑗 ,+∞ (𝜍𝑗 )
= 𝐼𝜏𝑗 ,+∞ (𝑞𝑗 ) − 𝐼𝜏𝑗 ,𝜏𝑗 +𝜀1/3 (𝜍𝑗 ) − 𝐼𝜏𝑗 +𝜀1/3 ,+∞ (𝜍𝑗 )
𝑗
(59)
𝑗
for some positive constant 𝑐1 > 0. Hence,
By assumption (𝐴 5 ),
𝐼𝜏𝑗 +𝜀1/3 ,+∞ (𝜍𝑗 ) < ∫
𝑗
𝜏𝑗
𝑠𝑗
𝜏𝑗 +𝜀𝑗1/3
∫
𝜏𝑗
> 𝑐1 𝑉 (̂𝑠𝑗 , 𝑞𝑗 (̂𝑠𝑗 )) ,
𝜀𝑗1/3 ;
𝜒,
for 𝑡 ⩾ 𝜏𝑗 +
{
{
{
{
𝜍𝑗 (𝑡) = {𝑃𝑗 (𝑡 − 𝜏𝑗 ) , for 𝑡 ∈ [𝜏𝑗 , 𝜏𝑗 + 𝜀𝑗1/3 ) ;
{
{
{
for 𝑡 < 𝜏𝑗 ,
{𝑞𝑗 (𝑡) ,
(62)
𝑐𝑗 = −𝜀𝑗−2/3 [3 (𝑞𝑗 (𝜏𝑗 ) − 𝜒) + 2𝜀𝑗1/3 𝑞𝑗󸀠 (𝜏𝑗 )] ,
𝑑𝑗 = 𝜀𝑗−1 [2 (𝑞𝑗 (𝜏𝑗 ) − 𝜒) + 𝜀𝑗1/3 𝑞𝑗󸀠 (𝜏𝑗 )] .
(68)
∫
𝜏𝑗 +𝜀𝑗1/3
𝜏𝑗
2
(𝜍𝑗󸀠󸀠 )
2
(𝜍𝑗󸀠 ) = ∫
𝜏𝑗 +𝜀𝑗1/3
=∫
𝜏𝑗
𝜏𝑗 +𝜀𝑗1/3
𝜏𝑗
2
(2𝑐𝑗 + 6𝑑𝑗 𝑡) 𝑑𝑡 󳨀→ 0,
2
(𝑏𝑗 + 2𝑐𝑗 𝑡 + 3𝑑𝑗 𝑡2 ) 𝑑𝑡 󳨀→ 0.
(69)
So
lim 𝐼
1/3
𝑗 → +∞ 𝜏𝑗 ,𝜏𝑗 +𝜀𝑗
(𝜍𝑗 ) = 0.
(70)
Therefore, for 𝑗 large enough,
𝐼 (𝑞𝑗 ) − 𝐼 (𝜍𝑗 ) ⩾ 𝛾√2𝜎𝛾 −
√2 − 1
𝛾√𝜎𝛾 − 𝛾√𝜎𝛾
2
(71)
which is impossible.
𝑃𝑗 (𝑡) = 𝑎𝑗 + 𝑏𝑗 𝑡 + 𝑐𝑗 𝑡2 + 𝑑𝑗 𝑡3 ,
𝑏𝑗 = 𝑞𝑗󸀠 (𝜏𝑗 ) ,
𝑉 (𝑡, 𝜍𝑗 ) 𝑑𝑡 󳨀→ 0 as 𝑗 󳨀→ ∞.
√2 − 1
>
𝛾√𝜎𝛾 ,
2
where 𝑃𝑗 is a polynomial defined as
(67)
Meanwhile, by construction,
= (𝑠𝑗 − 𝑠𝑗 ) 𝑉 (̂𝑠𝑗 , 𝑞𝑗 (̂𝑠𝑗 ))
where 𝑠𝑗 < 𝑠̂𝑗 < 𝑠𝑗 . Now assumption (𝐴 4 ) and the uniform
boundedness of {𝑞𝑗 } indicate that {̂𝑠𝑗 } 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 (0̃)} and
𝑟𝑗 = inf{𝑡 > 𝑟𝑗 | (𝑞𝑗 (𝑡), 𝑞𝑗󸀠 (𝑡)) ∈ 𝜕𝐵𝛾 (̃
𝜅)}. Hence, {𝜏𝑗 } is
bounded.
Define
𝑉 (𝑡, 𝜒) 𝑑𝑡 < 𝛾√𝜎𝛾 .
Moreover, since {𝜍𝑗 } is uniformly bounded on R and {𝜏𝑗 } is
bounded, it readily follows that
𝜏𝑗 +𝜀𝑗1/3
(61)
∞
−∞
∫
𝑠𝑗
𝑗
(66)
+∞ > 𝑐𝜀𝑗 = 𝐼 (𝑞𝑗 )
⩾ ∫ 𝑉 (𝑠, 𝑞𝑗 (𝑠)) 𝑑𝑠
𝑗
⩾ 𝛾√2𝜎𝛾 − 𝐼𝜏𝑗 ,𝜏𝑗 +𝜀1/3 (𝜍𝑗 ) − 𝐼𝜏𝑗 +𝜀1/3 ,+∞ (𝜍𝑗 ) .
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
󵄨
󵄨
inf 󵄨󵄨󵄨󵄨𝑠𝑗 − 𝑠𝑗 󵄨󵄨󵄨󵄨 > 𝑐1 ,
(60)
𝑗
𝑎𝑗 = 𝑞𝑗 (𝜏𝑗 ) ,
(𝑃𝑗 (0) , 𝑃𝑗󸀠 (0)) = (𝑞𝑗 (𝜏𝑗 ) , 𝑞𝑗󸀠 (𝜏𝑗 )) ,
(58)
which by Lemma 6 implies that {𝑞𝑗 } must be uniformly
bounded on R, say sup𝑠∈R |𝑞𝑗 (𝑠)| < 𝑀 for all 𝑗.
̃ ,
𝑠𝑗 = sup {𝑡 ∈ R | (𝑞𝑗 (𝑡) , 𝑞𝑗󸀠 (𝑡)) ∈ 𝜕𝐵𝛾 (𝜒)}
It is easy to verify the boundary conditions
(63)
𝜅}) for some 𝑠𝑗 < 𝜏𝑗 ). A contraCase 2 ((𝑞𝑗 (𝑡), 𝑞𝑗󸀠 (𝑡)) ∈ 𝜕𝐵𝜀𝑗 ({̃
diction can also be reached by similar arguments.
Either case leads to a contradiction; therefore, for 𝑖 sufficiently large, the orbit of 𝑞𝑖 = 𝑞𝜀𝑖 ,𝜅 must keep some distance
̃ \ {0̃, 𝜅̃}, and hence it is a classical solution, and
away from V
the proof is complete.
Abstract and Applied Analysis
7
Proof of Theorem 1. Theorem 10 shows for every 𝑧 ∈ V,
there is heteroclinic solution of (1) emanating from 𝑧. If we
consider the set Γ𝜀 (𝜗) ⊂ 𝐸 of function 𝑞(𝑡) for which 𝑞(+∞) =
̃ for
̃ \ {0,
̃ 𝜗}),
0, 𝑞(−∞) = 𝜗 ∈ V \ {0} and (𝑞(𝑡), 𝑞󸀠 (𝑡)) ∉ 𝐵𝜀 (V
all 𝑡 ∈ R; then the proof for Γ𝜀 (𝜗) works equally well for Γ𝜀 (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 where 𝑉(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:
Γ𝜀 (𝜁) is not empty, we may consider the minimization
problem
𝑐𝜀 (𝜁) = inf 𝐼 (𝑞) .
𝑞∈Γ𝜀 (𝜁)
We shall follow the lines of argument in previous section,
and similar proofs will be omitted. For example, Lemmas 3,
5, 6, and 7 remain valid in the setting of this section.
̃ \ {0̃}. There exists
Theorem 11. For any 𝜀 ∈ (0, 𝛾], 𝜁̃ ∈ V
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
1
some 𝑞 ∈ 𝐸 and 𝑞𝑛 → 𝑞 in 𝐶loc
(R). Moreover,
𝐼 (𝑞) < +∞.
(𝐻1 ) 𝑉(𝑡, 𝑞) ∈ 𝐶2 (R × R), 𝑉 ⩾ 0;
(𝐻2 ) the set V = {𝑧 ∈ R | for all 𝑡, 𝑉(𝑡, 𝑧) = 0, 𝑉𝑞 (𝑡, 𝑧) =
0 and 𝑉𝑞𝑞 (𝑡, 𝑧) ⩾ 0} is discrete and has at least two
points, and
1
󵄨
󵄨
𝛾 ≡ inf {󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 | 𝑥 ≠ 𝑦 ∈ V} > 0.
3
(72)
(𝐻3 ) There is a constant 𝑉0 > 0 such that lim inf |𝑞| → ∞
𝑉(𝑡, 𝑞) ⩾ 𝑉0 uniformly for 𝑡 ∈ R.
(𝐻4 ) For each 𝑧 ∈ V,
∫
∞
−∞
𝑉 (𝑡, 𝑧) 𝑑𝑡 < 𝛾𝜎𝛾1/2 .
(73)
Note that (𝐴 4 ) is no longer needed in the periodic case.
(𝐻3 ) is indeed a special case of (𝐴 3 ), we are using (𝐻3 ) instead
of (𝐴 3 ) only for illustrative purpose. As before, we work on
the space
2,2
𝐸 = {𝑞 ∈ 𝑊loc
(R, R) | ∫
∞
−∞
󵄨 󵄨2 󵄨 󵄨2
(󵄨󵄨󵄨󵄨𝑞󸀠󸀠 󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨𝑞󸀠 󵄨󵄨󵄨󵄨 ) 𝑑𝑡 < ∞} ,
(74)
with the norm
∞
󵄨 󸀠󸀠 󵄨2 󵄨 󸀠 󵄨2
󵄨2
󵄩󵄩 󵄩󵄩2
󵄨
󵄩󵄩𝑞󵄩󵄩 = ∫ (󵄨󵄨󵄨󵄨𝑞 󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨𝑞 󵄨󵄨󵄨󵄨 ) 𝑑𝑡 + 󵄨󵄨󵄨𝑞 (0)󵄨󵄨󵄨 .
−∞
(75)
We will use the shift operator 𝜎𝑗 defined for each 𝑗 as
𝜎𝑗 𝑞 = 𝑞 (𝑡 − 𝑗𝑇) .
It remains to show
lim 𝑞 (𝑡) = 0,
𝑡 → −∞
(76)
Clearly, if 𝑞 is a heteroclinic solution, so is 𝜏𝑗 𝑞 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;
̃ for all 𝑡 ∈ R; where V
̃ 0,
̃=
̃ 𝜗}),
(ii) (𝑞(𝑡), 𝑞󸀠 (𝑡)) ∉ 𝐵𝜀 (V/{
2
{(𝑧, 0) ∈ R | 𝑧 ∈ V}.
lim 𝑞 (𝑡) = 𝜁.
𝑡 → +∞
(78)
(79)
̃
Inequality (78) and Lemma 7 imply that there are 𝛼̃, 𝛽̃ ∈ V
verifying
lim 𝑞 (𝑡) = 𝛼,
We also assume without no loss of generality 0 ∈ V.
(77)
𝑡 → −∞
lim 𝑞 (𝑡) = 𝛽.
𝑡 → +∞
(80)
Since 𝑉(𝑡, 𝑢) is periodic in 𝑡, if {𝑞𝑛 } is the minimizing
sequence, so is 𝜎𝑗𝑛 𝑞𝑛 for any sequence {𝑗𝑛 } ⊂ N and 𝐼(𝑞𝑛 ) =
𝐼(𝜎𝑗𝑛 𝑞𝑛 ). Whence, with {𝑗𝑛 } being appropriately chosen and
{𝑞𝑛 } replaced with {𝜎𝑗𝑛 𝑞𝑛 }, 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 all 𝑡 ∈ R and
̃ \ {0,
̃ 𝜁})
for each 𝑛, (𝑞𝑛 (𝑡), 𝑞𝑛󸀠 (𝑡)) ∉ 𝐵𝜀 (V
1
󸀠
̃ for
̃ \ {0,
̃ 𝜁})
𝑞𝑛 → 𝑞 in 𝐶loc (R), we have (𝑞(𝑡), 𝑞 (𝑡)) ∉ 𝐵𝜀 (V
̃
̃
̃
all 𝑡 ∈ R and so 𝛼̃, 𝛽 ∈ {0, 𝜁}. Since 𝑞𝑛 (𝑡) ∈ 𝐵𝜀 (0) for 𝑡 ⩽ 0,
𝛼 ∈ {0, 𝜁} ∩ 𝐵𝜀 (0) = {0}, so 𝛼 = 0.
It remains to show 𝛽 = 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 →
1
𝑞 in 𝐶loc
(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
𝜀
𝐼 (𝑞𝑛 ) ⩾ √2𝜎𝜀/2 + 𝐼𝑡𝛿 ,+∞ (𝑞𝑛 ) .
(81)
2
Now construct a sequence as follows:
0,
for 𝑡 ⩽ 𝑡𝛿 − 𝑡𝑛 ;
{
{
{
{
𝜍𝑛 (𝑡) = {𝑃𝑛 (𝑡 − 𝑡𝛿 + 𝑡𝑛 ) , for 𝑡 ∈ (𝑡𝛿 − 𝑡𝑛 , 𝑡𝛿 ] ;
{
{
{
for 𝑡 > 𝑡𝛿 ,
{𝑞𝑛 (𝑡) ,
(82)
where
󵄨 󵄨
󵄨
󵄨 󵄨
󵄨
3 󵄨
3 󵄨
{
{√󵄨󵄨󵄨𝑞𝑛 (𝑡𝛿 )󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑞𝑛󸀠 (𝑡𝛿 )󵄨󵄨󵄨, if √󵄨󵄨󵄨𝑞𝑛 (𝑡𝛿 )󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑞𝑛󸀠 (𝑡𝛿 )󵄨󵄨󵄨 ≠ 0,
𝑡𝑛 = {
󵄨
󵄨 󵄨
󵄨
{
1,
if √3 󵄨󵄨󵄨𝑞𝑛 (𝑡𝛿 )󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑞𝑛󸀠 (𝑡𝛿 )󵄨󵄨󵄨 = 0,
{
(83)
8
Abstract and Applied Analysis
and 𝑃𝑛 (𝑡) = 𝑔𝑛 𝑡2 + ℎ𝑛 𝑡3 is a polynomial with coefficients
𝑔𝑛 = 𝑡𝑛−2 [3𝑞𝑛 (𝑡𝛿 ) − 𝑡𝑛 𝑞𝑛󸀠 (𝑡𝛿 )] ,
ℎ𝑛 = 𝑡𝑛−3 [𝑡𝑛 𝑞𝑛󸀠 (𝑡𝛿 ) − 2𝑞𝑛 (𝑡𝛿 )] .
(84)
It is easy to check that 𝑃𝑛 (𝑡) satisfies
(𝑃𝑛 (0) , 𝑃𝑛󸀠
𝑞󸀠󸀠󸀠󸀠 − 𝑞󸀠󸀠 + 𝑉𝑧 (𝑡, 𝑞) = 0,
(85)
∫
+∞
𝑡∗
󵄨󵄨 󸀠󸀠󸀠󸀠 󵄨󵄨2
󵄨󵄨𝑞 (𝑡)󵄨󵄨 𝑑𝑡
󵄨
󵄨
+∞
⩽ 2∫
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 ,
𝑡∗
+∞
⩽ 2∫
𝑡∗
𝐼𝑡𝛿 −𝑡𝑛 ,𝑡𝛿 (𝜍𝑛 )
+∞
𝑡𝛿
2
1
1
2
{ (2𝑔𝑛 +6ℎ𝑛 𝑡) + (2𝑔𝑛 𝑡+3ℎ𝑛 𝑡2 ) +𝑉 (𝑡, 𝜍𝑛 )} 𝑑𝑡,
2
𝑡𝛿 −𝑡𝑛 2
=∫
𝜀
𝐼𝑡𝛿 −𝑡𝑛 ,𝑡𝛿 (𝜍𝑛 ) < √2𝜎𝜀/2 .
4
= 2∫
𝑡∗
+∞
⩽ 2∫
𝑡∗
(86)
= 𝐼−∞,𝑡𝛿 (𝑞𝑛 ) − 𝐼𝑡𝛿 −𝑡𝑛 ,𝑡𝛿 (𝜍𝑛 )
∫
(87)
+∞
𝑐22
+∞
󵄨󵄨 󸀠󸀠 󵄨󵄨2
󵄨󵄨𝑞 (𝑡)󵄨󵄨 𝑑𝑡 + 2 ∫
󵄨
󵄨
𝑐1 𝑡∗
(91)
󵄨2
󵄨
𝑐1 󵄨󵄨󵄨𝑞 (𝑡) − 𝜁󵄨󵄨󵄨 𝑑𝑡
𝑐2 +∞
󵄨󵄨 󸀠󸀠 󵄨󵄨2
󵄨󵄨𝑞 (𝑡)󵄨󵄨 𝑑𝑡 + 2 2 ∫ 𝑉 (𝑡, 𝑞 (𝑡)) 𝑑𝑡
󵄨
󵄨
𝑐1 𝑡∗
𝑡∗
󵄨󵄨 󸀠󸀠󸀠 󵄨󵄨2
󵄨󵄨𝑞 (𝑡)󵄨󵄨 𝑑𝑡
󵄨
󵄨
+∞
⩽ 𝐾∫
𝑡∗
+∞
󵄨2
󵄨󵄨 󸀠󸀠 󵄨󵄨2
󵄨
󵄨󵄨𝑞 (𝑡)󵄨󵄨 𝑑𝑡 + 𝐾 ∫ 󵄨󵄨󵄨𝑞󸀠󸀠󸀠󸀠 (𝑡)󵄨󵄨󵄨 𝑑𝑡 < ∞,
󵄨
󵄨
󵄨
󵄨
∗
𝑡
(92)
where 𝐾 is a positive constant.
Therefore, we have proved that
lim 𝑞󸀠󸀠 (𝑡) = lim 𝑞󸀠󸀠󸀠 (𝑡) = 0.
𝑛 → +∞
⩾ lim inf 𝐼 (𝜍𝑛 )
⩾ inf 𝐼 (𝑞)
+∞
󵄨󵄨 󸀠󸀠 󵄨󵄨2
󵄨󵄨𝑞 (𝑡)󵄨󵄨 𝑑𝑡 + 2𝑐22 ∫ 󵄨󵄨󵄨󵄨𝑞 (𝑡) − 𝜁󵄨󵄨󵄨󵄨2 𝑑𝑡
󵄨
󵄨
𝑡∗
An application of the interpolation inequality yields
𝐼 (𝑞𝑛 ) − 𝐼 (𝜍𝑛 )
𝑛 → +∞
+∞
󵄨󵄨 󸀠󸀠 󵄨󵄨2
󵄨󵄨𝑞 (𝑡)󵄨󵄨 𝑑𝑡 + 2 ∫ 󵄨󵄨󵄨󵄨𝑉𝑧 (𝑡, 𝑞)󵄨󵄨󵄨󵄨2 𝑑𝑡
󵄨
󵄨
𝑡∗
< ∞.
Then
𝜀
⩾ √2𝜎𝜀/2 − 𝐼𝑡𝛿 −𝑡𝑛 ,𝑡𝛿 (𝜍𝑛 )
2
𝜀
> √2𝜎𝜀/2 ,
4
for 𝑛 > 𝑛𝛿 . This implies
𝜀
lim inf 𝐼 (𝑞𝑛 ) − √2𝜎𝜀/2
𝑛 → +∞
4
(90)
we have
(0)) = (0, 0) ,
(𝑃𝑛 (𝑡𝑛 ) , 𝑃𝑛󸀠 (𝑡𝑛 )) = (𝑞𝑛 (𝑡𝛿 ) , 𝑞𝑛󸀠 (𝑡𝛿 )) .
whenever |𝑢 − 𝜁| ⩽ 𝑟∗ . Since 𝑞(+∞) = 𝜁, |𝑞(𝑡) − 𝜁| ⩽ 𝑟∗ for 𝑡
larger than some 𝑡∗ > 0.
From (89) and the Euler equation,
(88)
𝑞∈Γ𝜀 (𝜁)
= lim inf 𝐼 (𝑞𝑛 ) ,
𝑛 → +∞
which is impossible. The proof is complete.
Theorem 12. There is a pair of orbits 𝑞0,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 solution 𝑞 = 𝑞0,𝜁 or 𝑞𝜁,0 obtained above,
lim𝑛 → ±∞ 𝑞󸀠󸀠 (𝑡) = 0, lim𝑛 → ±∞ 𝑞󸀠󸀠󸀠 (𝑡) = 0.
Proof. Consider 𝑞 = 𝑞0,𝜁 . By (𝐻2 ), there are positive constants
𝑟∗ , 𝑐1 , 𝑐2 such that
󵄨2
󵄨
𝑉 (𝑡, 𝑢) 𝑐1 ⩾ 󵄨󵄨󵄨𝑢 − 𝜁󵄨󵄨󵄨 ,
(89)
󵄨
󵄨󵄨
󵄨
󵄨
󵄨󵄨𝑉𝑧 (𝑡, 𝑢)󵄨󵄨󵄨 ⩽ 𝑐2 󵄨󵄨󵄨𝑢 − 𝜁󵄨󵄨󵄨 ,
𝑛 → +∞
(93)
The limit for negative infinity can be derived similarly. The
proof is complete.
Acknowledgment
This work was supported by NSFC under Grant no. 11126035
and partly NSFC Grant no. 11201016.
References
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extended Fisher-Kolmogorov (EFK) equation: kinks,” Differential and Integral Equations, vol. 8, no. 6, pp. 1279–1304, 1995.
[2] L. A. Peletier and W. C. Troy, “A topological shooting method
and the existence of kinks of the extended Fisher-Kolmogorov
equation,” Topological Methods in Nonlinear Analysis, vol. 6, no.
2, pp. 331–355, 1995.
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