TRANSITION LAYER SOLUTIONS OF A

HIGHER ORDER EQUATION IN AN

INFINITE TUBE

Peter W. Bates∗

Department of MathematicsBrigham Young University

Provo, Utah 84602

Xiaofeng Ren†

Department of MathematicsBrigham Young University

Provo, Utah 84602

1 Introduction

We are concerned with the higher order variational problem

I = infu∈A

∫Ω

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβu+W (u)]dx. (1.1)

The integral in the right side of (1.1) is denoted by E(u), the energy of u, inthis paper. This problem has its origin in phase transition theory. The functionu is the phase field of a binary material, and E(u) represents the total energy ofthe material, assuming the temperature is constant. This energy has two parts:the interaction energy and the bulk energy.∫

Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβudu (1.2)

represents the interaction energy of the phase field and∫Ω

W (u)dx (1.3)

represents the bulk energy of the phase field. Here we use multi-index notation:α = (α1, α2, ..., αd), Dα = ∂α1

x1∂α2x2...∂αdxd . We assume the matrix [Aαβ ] is sym-

metric and positive definite, which, as we show below, arises naturally in thederivation of (1.1).

∗Supported in part by NSF grant DMS-9305044.†Current address: IMA, University of Minnesota, Minneapolis, MN 55455.

1

The bulk energy density function W is assumed to be a balanced double wellpotential function, e.g. W (u) = (u2 − 1)2. W has two global minima scaledto be at −1 and 1. We assume W is continuous on R and C2 near −1 and 1,W ≥ 0, −1 and 1 are the only global minima of W , and W (−1) = W (1) = 0,W ′′(−1) > 0 and W ′′(1) > 0. Concerning the behavior of W near infinity, weassume that

lim infy→−∞

W (y) > 0 and lim infy→∞

W (y) > 0.

We also need a key assumption

N >d+ 2

4(1.4)

on the order of the problem.In this paper we study the variational problem in a d−dimensional tube of

infinite length, so we will take the physical domain Ω to be an infinite tube.Because we are interested in the transitions of the phase field, we assume thatA is a class of functions that take values −1 at one end of the tube and 1 at theother end. The precise definitions of Ω and A will be given later. Let us firstrecall how (1.1) is derived from a microscopic viewpoint.

In the theory of phase transitions for binary materials from a microscopicviewpoint, one considers a state to be a distribution u over a lattice so thatu(r) is the probability that a solute atom occurs at site r. The Helmholtz freeenergy of a state is given by

F = H − TSwhere H(u) is the internal interaction energy, T is the absolute temperatureand S(u) is the entropy. Following Khachaturyan (see [8] and [9] for instance),the internal energy takes the form

H(u) = −12

∑r,r′

J(r − r′)u(r)u(r′)

where the sum is over the lattice sites and J(r − r′) = J(r′ − r) > 0 gives thepairwise interaction energies.

The total entropy takes the form of a sum over the lattice of

−K[u(r) log u(r) + (1− u(r)) log(1− u(r))],

where K is a positive constant, which has minima at 0 and 1 and which naturallypenalizes values of u outside the interval [0, 1].

In the analysis to follow we will consider a rearrangement of (1.1) changingH to

14

∑r,r′

J(r − r′)(u(r)− u(r′))2

so that F is written14

∑r,r′

J(r − r′)(u(r)− u(r′))2

+∑r

(TK[u(r) log u(r) + (1− u(r)) log(1− u(r))]− cu2(r)),

2

where c = (1/4)∑r J(r), which we assume is finite. Note that for TK small

the second term is the sum over the lattice of a double well function of u. Wewill write this term as

14

∑r

W (u(r))

and assume that W is everywhere defined. For computational ease later, were-scale u so that the minima of the double well function occur at u = −1, 1.Taking the continuum limit gives

F (u) =14

[∫ ∫

J(x− y)(u(x)− u(y))2dxdy +∫W (u(x))dx].

We assume J is non-negative, symmetric about 0, and∫J(x)dx = 1.

As was done in [2], we change variables in the first integral using η = (x −y)/2, ξ = (x+ y)/2, then expand u(x) = u(ξ + η) and u(y) = u(ξ − η) about ξto get

2d∫ ∫

J(2η)(∞∑k=1

∑|α|=2k−1

1α!Dαu(ξ)ηα)2dξdη,

where both integrals are over Rd, α = (α1, α2, ..., αd), |α| = α1 + α2 + ...+ αd,Dα = ∂α1

x1∂α2x2...∂αdxd , α! = α1!α2!...αd!, and ηα = ηα1

1 ηα22 ...ηαdd .

We truncate the infinite sum at k = N and integrate with respect to η. This“truncated ” free energy is given by (1.1) where

Aαβ = 2d∫J(2η)ηα+β

α!β!dη.

Note that [Aαβ ] is positive definite since

N∑i,j=1

∑|α|=2i−1|β|=2j−1

Aαβξαξβ = 2d∫J(2η)(

N∑i=1

∑|α|=2i−1

ηαξαα!

)2dη ≥ 0,

and the equality holds only if ξα = 0 for all α.The Euler-Lagrange equation of (1.1) is

N∑i,j=1

∑|α|=2i−1|β|=2j−1

2AαβDα+βu−W ′(u) = 0 in Ω,

Bj(u) = 0 on ∂Ω, j = 1, 2, ..., 2N − 1;

(1.5)

where the Bj ’s are the natural boundary conditions. The expressions of theBj ’s are complicated. In the case d = 2, N = 2 and the material is isotropic,i.e., J(x) = J(|x|), we have

B1(u) =b

16∂u

∂ν+c

3ν2

1

∂uyy∂ν

+c

3ν2

2

∂uxx∂ν

3

+ν21(

5c9ν1uxxx + cν2uxxy) + ν2

2(5c9ν2uyyy + cν1uxyy)

B2(u) =∂

∂τ[−10c

9ν2

1ν2uxxx − 2cν1ν22uxxy +

10c9ν2

1ν2uyyy

+2cν21ν2uxyy +

2c3ν1ν2

∂uxx∂ν− 2c

3ν1ν2

∂uyy∂ν

]

−ν21(

5c9uxxx + cuxxy)− ν2

2(5c9uyyy + cuxyy)− c

3ν2

2∆uxx −c

3ν2

1∆uyyν21 −

b

16∆u

B3(u) =∂2

∂τ2[ν2

2(5c9ν1uxxx + cν2uxxxyν2) + ν2

1(5c9ν2uyyy + cν1uxyy)

+ν21(c

3∂uxx∂ν

) + ν22(c

3∂uyy∂ν

) +b

16∂u

∂ν]

+∂

∂τ[ν1ν2(

5c9uxxx + cuxxy)− ν1ν2(

5c9uyyy + cuxyy) +

c

3ν1ν2(−∆uxx + ∆uyy)]

+ν1(5c9uxxxxx + cuxxxyy) + ν2(

5c9uyyyyy + cuxxyyy)

+c

3∆uxxyν2 +

c

3∆uxyyν1 +

b

8∂(∆u)ν

+a

2∂u

∂ν

where a, b, c are three constants, ν = (ν1, ν2) is the normal vector to ∂Ωpointing outward, and τ = (ν2,−ν1) is the unit tangential vector that makes(τ, ν) positively oriented.

This paper is devoted to the existence of the energy minimizers of (1.1), orthe solutions of (1.5). The simplest case Ω = R and N = 1 gives the equation

u′′ −W ′(u) = 0 in Rlim

x→−∞u(x) = −1, lim

x→∞u(x) = 1. (1.6)

It was shown by Fife and McLeod in [6] that this equation has a unique solutionup to translation and the manifold of all the translates is globally stable. Indeedthey considered the more general case where W is not necessarily balancedand they considered traveling wave solutions. Note that if W is balanced, thetraveling wave solutions become standing waves (the speed is 0).

Gardner and Jones studied the case where Ω = R and N = 2 in [7]. Theyalso considered traveling wave solutions when W is not necessarily balanced.Bates, Fife, Gardner and Jones [3] generalized this case to arbitrary N . Inboth papers the higher order equations were treated as the perturbations of thesecond order equation and the existence and stability of traveling wave solutionswas obtained if the coefficients of the higher order derivatives were “small”.

In [5], the authors studied the case where W is balanced, Ω = R, and N isarbitrary, but without the smallness condition on the higher order derivatives.We proved the existence of energy minimizing solutions having limits of −1,1 at −∞, ∞, by using the concentration-compactness principle to capture anenergy minimizer from any minimizing sequence.

4

In this paper we consider a domain Ω which is a tube of infinite length inRd. We will consider two kinds of tubes. In section 3, we study tubes whichare periodic in one direction. The discrete translational symmetry allows us toprove the existence of the minimizers by taking appropriate translates of termsin a minimizing sequence. Section 4 contains the proof of a lemma used insection 3. Then in section 5, we consider tubes whose tails are “half-periodic”,and we give a sufficient condition on the geometry of the tube for the existenceof energy minimizers. We also give various examples where such a condition issatisfied.

2 Concentration functions of W (un)Ω is assumed to be a sufficiently smooth open connected set in Rd. We willsometimes treat Ω as a metric space equipped with a distance function ρ :Ω× Ω→ R defined by

ρ(x, y) = infγ∈Γ

∫ 1

0

|γ′(t)|dt (2.1)

where Γ is the class of all smooth paths γ : [0, 1] → Ω with γ(0) = x andγ(1) = y. We define

BΩ(x, r) = y ∈ Ω : ρ(y, x) < r (2.2)

to be the ball centered at x with radius r in the metric space (Ω, ρ). If wedenote the usual ball in Rd centered at x with radius r by B(x, r), in generalBΩ(x, r) 6= B(x, r) ∩ Ω.

To say Ω is a tube, we assume that: (1) there is a large ball B in Rd suchthat Ω\B consists of two unbounded connected open sets Ω1 and Ω2; (2) thereexist Pi and Qi such that Pi ⊂ Ωi ⊂ Qi where Pi and Qi are diffeomorphic tothe half infinite cylinder [0,∞)×D ⊂ Rd, with D being the unit ball in Rd−1;(3) the diffeomorphisms have bounded gradients. This means that neither endof the tube will shrink to a point at infinity. We assume throughout this paperthat Ω is a tube.

To prescribe the condition of the phase fields at the two ends of the tubeΩ, we denote the two ends of Ω by E1 and E2. We use x → Ei to denote xapproaching Ei uniformly.

To define the function class A in (1.1), we first state the following lemma.

Lemma 2.1 If u is a function in L1loc(Ω) with W (u) ∈ L1(Ω), Dαu ∈ L2 for

all α, |α| = 2i − 1, i = 1, 2, ..., N , then limx→E1 u(x) and limx→E2 u(x) existand equal −1 or 1.

Proof. Recall an interpolation inequality for intermediate derivatives (see theo-rem 4.14 [1]): Let Ω ⊂ Rd have the uniform cone property, and let ε0 be a finite

5

positive number. Then there exists a constant K = K(ε0,m, p,Ω) such that forany ε, 0 < ε ≤ ε0, any integer j, 0 ≤ j ≤ m− 1, and any v ∈Wm,p(Ω)

|v|j,p ≤ Kε|v|m,p +Kε−j(m− j)|v|0,p (2.3)

where|v|j,p =

∑|α|=j

∫Ω

|Dαv|pdx1/p.

With the help of this inequality, a standard smoothing argument implies thatfor every u in lemma 2.1, Dαu ∈ L2(Ω), |α| = 1, 2, ..., 2N − 1. We now applythe Sobolev embedding theorem (see theorem 5.4 [1]) to conclude that there isσ, 0 < σ < 1, such that

supx,y∈Ω

|u(x)− u(y)||x− y|σ

<∞. (2.4)

Here we used our key assumption (1.4). Therefore we have a uniform boundon the modulus of continuity of u. Since

∫ΩW (u) < ∞, we deduce that

limx→E1 u(x), limx→E2 u(x) exist and equal −1 or 1.We now define

A∗ = u ∈ L1loc(Ω) : Dαu ∈ L2(Ω), |α| = 2i− 1, i = 1, ..., N,

∫Ω

W (u) <∞.

In view of lemma 2.1, we can decompose A∗ in to four mutually disjoint classes,A, A′, A′′ and A′′′, where

A = u ∈ A : limx→E1

u(x) = −1 limx→E2

u(x) = 1,A′ = u ∈ A : lim

x→E1u(x) = 1 lim

x→E2u(x) = −1,

A′′ = u ∈ A : limx→E1

u(x) = 1 limx→E2

u(x) = 1,A′′′ = u ∈ A : lim

x→E1u(x) = −1 lim

x→E2u(x) = −1.

(2.5)

We are only interested in the first class, A.We now give a necessary condition for a phase transition to occur.

Lemma 2.2 For a given energy minimizing sequence un and every ε1 > 0,L > 0 there exists δ > 0 so that for every n and BΩ(y, r) ⊂ Ω, with L < r ≤ ∞,∫

BΩ(y,r)

W (un)dx < δ

implies that eitherun(x) ∈ (−1− ε1,−1 + ε1)

for all x ∈ BΩ(y, r) orun(x) ∈ (1− ε1, 1 + ε1)

for all x ∈ BΩ(y, r).

6

Proof. Suppose the contrary. There exist ε1, L, δk, δk →∞ as k →∞, unk fromthe minimizing sequence, BΩ(yk, rk), L ≤ rk ≤ ∞, and xk, xk ∈ BΩ(yk, rk),such that ∫

BΩ(yk,rk)

W (unk)dx ≤ δk → 0 as k →∞ (2.6)

and unk(xk) 6∈ (−1 − ε1,−1 + ε1) ∪ (1 − ε1, 1 + ε1). Because we have a boundindependent of k on the moduli of continuity of the unk ’s, (2.4), we can findη independent of k such that for all x ∈ BΩ(xk, η), |unk(x) − 1| ≥ ε1/2 and|unk(x) + 1| ≥ ε1/2, so we deduce∫

BΩ(yk,rk)

W (unk)dx ≥∫BΩ(xk,η)∩BΩ(yk,rk)

W (unk)dx

≥ Cηdminx∈BΩ(xk,η)∩BΩ(yk,rk)W (unk(x))

≥ Cηdminy∈R\(−1−ε1/2,−1+ε1/2)∪(1−ε1/2,1+ε1/2)W (y) ≥ C ′ > 0

where C depends on d only. This contradicts (2.6).Our next lemma asserts that

∫ΩW (un) does not vanish.

Lemma 2.3 For any energy minimizing sequence un ⊂ A there exists Λ ∈(0,∞) such that for all n ∫

Ω

W (un)dx ≥ Λ.

Proof. Since I < ∞, lim supn→∞∫

ΩW (un)dx < ∞. For every u ∈ A there

is xn ∈ Ω such that un(xn) = 0. (2.4) yields a bound independent of n onthe moduli of continuity of the un’s. Hence there exist ε1 > 0, δ > 0, bothindependent of n, such that un(x) ∈ (−ε1, ε1) for x ∈ BΩ(xn, δ). This impliesΛ > 0.

We now apply the concentration-compactness principle to W (un). Inlemma 2.3, we may take

Λ = lim infn→∞

∫Ω

W (un)dx (2.7)

and then take un to be a subsequence along which this is achieved. Wetreat W (un) as a sequence of non-negative functions in L1(Ω). Define theconcentration functions

Qn(t) = supy∈Ω

∫BΩ(y,t)

W (un(x))dx (2.8)

for t ≥ 0. Qn is a sequence of non-negative, non-decreasing, uniformlybounded functions on [0,∞) and

limn→∞

limt→∞

Qn(t) = Λ. (2.9)

7

By a classical result about monotone functions, there exist a subsequence of un,again denoted by un, and a non-negative, non-decreasing function Q(t) suchthat

limn→∞

Qn(t) = Q(t)

for all t ≥ 0. Letλ = lim

t→∞Q(t). (2.10)

Then λ ∈ [0,Λ]. Furthermore by (2.9) and (2.10) we may make the followingthree statements:

I: If λ = Λ, there exist a subsequence of un, again denoted by un, and yn ∈ Ωsuch that for every ε > 0 there exist r > 0 and a positive integer n1 so that∫

BΩ(yn,r)

W (un(x))dx ≥ Λ− ε

for n ≥ n1.

II: If λ = 0, then for all r > 0,

limn→∞

supy∈Ω

∫BΩ(y,r)

W (un(x))dx = 0.

III: If λ ∈ (0,Λ), then there exists a subsequence of un again denoted by unsuch that for every ε > 0 there exist arbitrarily large r > 0, a sequence rn →∞as n→∞, and a positive integer n1 such that for n ≥ n1

λ− ε < Qn(r), Qn(rn) < λ+ ε.

These three statements can be found in the proof of lemma I.1 [10].

3 An existence theorem for periodic Ω

Throughout section 3, we assume the tube Ω is periodic in one direction. Thistranslation invariance enables us to construct an energy minimizer.

We suppose there is T ∈ Rd such that Ω + T = Ω where Ω + T = x + T :x ∈ Ω is the translation of Ω by T . To say Ω is a tube, we assume the crosssections of Ω perpendicular to T are bounded (see figure 1). Our main result inthis section is the following theorem.

Theorem 3.1 If the tube Ω is periodic, every energy minimizing sequence of(1.1) gives rise to an energy minimizer in A. Indeed if un is an energyminimizing sequence, there exist a subsequence of un, again denoted by un,and a sequence of integers in such that un(·+ inT ) converges in a suitablesense to an energy minimizer.

8

Figure 1: An example of periodic Ω.

We will prove theorem 3.1 by showing that only the case λ = Λ in statementI in section 2 can occur when un is an energy minimizing sequence.

Lemma 3.2 The case λ = 0 can not occur.

Proof. As in the proof of lemma 2.3, we can find xn, and δ independent of n,such that

∫BΩ(xn,δ)

W (un)dx ≥ C for some C independent of n.

Lemma 3.3 The case λ ∈ (0,Λ) can not occur.

The proof of this lemma is postponed to section 4.

Lemma 3.4 The case λ = Λ implies that there is u ∈ A such that

E(u) = infv∈A

E(v).

Proof. Statement I in section 2 implies that for every ε > 0 there exist yn,r > 0 and a positive integer n1 such that for n ≥ n1∫

BΩ(yn,r)

W (un) ≥ Λ− ε.

Using the periodicity of Ω and replacing r by r+DT , we can translate the un’sby inT where the in’s are some integers, T is the period of the tube and DT isthe diameter of a piece of the tube of “length” T , and therefore take yn to bea point y independent of n. By further enlarging r, we may assume that Ω isdivided into mutually disjoint open sets Ω1, Ω2 and BΩ(y, r) with

Ω = Ω1 ∪ Ω2 ∪BΩ(y, r).

Assume E1 is associated to Ω1 and E2 is associated to Ω2. For large n∫Ω1

W (un(x+ inT ))dx < 2ε,∫

Ω2

W (u′n(x+ inT ))dx < 2ε.

Lemma 2.2 then implies that the un(·+ inT )’s stay close to −1 in Ω1 and closeto 1 in Ω2 uniformly in n. We pass to the limit along a subsequence and obtainu ∈ A such that

un(·+ inT )→ u in C(K ∩ Ω) for every compact K ∈ Rd, (3.1)

9

Dαun(·+ inT )→ u weakly in L2(Ω), |α| = 2i− 1, i = 1, 2, ..., N. (3.2)

The positivity of [Aαβ ] and Fatou’s lemma imply

E(u) ≤ limn→∞

E(un) = I,

so u is an energy minimizer.Once we have an energy minimizer u, we know that u solves the correspond-

ing Euler-Lagrange equation and the proof of theorem 3.1 is complete.

Corollary 3.5 The convergence in (3.1) and (3.2) is indeed

un(·+ inT )→ u in C(Ω),

Dαun(·+ inT )→ Dαu strongly in L2(Ω), |α| = 2i− 1, i = 1, 2, ..., N.

Proof. The uniform estimate of the un(· + inT )’s in Ω1 and Ω2 gives the con-vergence in C(Ω). To see the strong convergence in L2(Ω), one notes that

limn→∞

∫Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βundx

=∫

Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβudx. (3.3)

We then consider∫Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDα(un − u)Dβ(un − u)dx

=∫

Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βundx+∫

Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβudx

−∫

Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βudx−∫

Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβundx.

(3.2) implies

limn→∞

∫Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βudx =∫

Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβudx,

limn→∞

∫Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβundx =

∫Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβudx.

10

Therefore with the help of (3.3) we deduce

limn→∞

∫Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDα(un − u)Dβ(un − u)dx = 0. (3.4)

The positivity of [Aαβ ] implies that there exists c > 0 such that

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDα(un − u)Dβ(un − u) ≥ c

N∑i=1

∑|α|=2i−1

|Dα(un − u)|2

With the help of (3.4) we obtain

limn→∞

∫Ω

N∑i=1

∑|α|=2i−1

|D2i−1(un − u)|2 = 0

which gives the strong convergence.

4 Proof of lemma 3.3

Suppose λ ∈ (0,Λ). From statement III in section 2, we have for every smallε > 0, numbers r > 0, and rn, with rn →∞ as n→∞, satisfying

λ− ε < Qn(r), Qn(rn) < λ+ ε

for large n. We can find yn such that∫BΩ(yn,r)

W (un)dx ∈ (λ− ε, λ+ ε),

∫Ω\BΩ(yn,rn)

W (un)dx ∈ (Λ− λ− 2ε,Λ− λ+ 2ε)

(4.1)

for large n. If we choose r large enough we can write

BΩ(yn, rn)\BΩ(yn, r) = D1n ∪D2n (4.2)

where D1n and D2n are two connected open sets. We also denote

Ω\BΩ(yn, r) = M1n ∪M2n (4.3)

where M1n and M2n are again two connected open sets. We assume D1n (D2n,respectively) is contained in M1n (M2n, respectively) which has tail E1 (E2,respectively). Clearly

mes(D1n),mes(D2n)→∞

11

where mes(D) denotes the measure of D. For large n∫D1n

W (un)dx < 2ε,∫D2n

W (un)dx < 2ε. (4.4)

If we choose ε small enough, we see the un’s stay close to −1 or 1 in D1n andD2n in view of lemma 2.2. We will consider the three cases.

Case 1: Suppose, taking a subsequence if necessary, the un’s stay close to −1in D2n.

We will show that in this case the un’s already have large energies in M2n

so they can not constitute an energy minimizing sequence. Without loss ofgenerality, we assume the period T of Ω is T = (T 1, 0, ..., 0). Let ξ be a non-negative smooth cut-off function such that

ξ(t) = 0 if t < 0,ξ(t) = 1 if t > 1, (4.5)

0 ≤ ξ(t) ≤ 1, and |Dkξ(x)| ≤ C1 for k = 1, 2, ...2N − 1. Set

vn(x) = (un(x) + 1)ξ(x1 − (y1

n + (r + rn)/2)h

)− 1 (4.6)

where x = (x1, ..., xd) and yn = (y1n, ..., y

dn) and h is a large number to be

specified later. Note vn ∈ A. We fix h and take n large enough so that h <(rn − r)/2.

We will compare∫M2n

W (un)dx with∫

ΩW (vn)dx. Since for x ∈ M2n, un

and vn only differ in D2n, and on that set vn is closer to −1 than un is, so wehave ∫

Ω

W (vn)dx ≤∫M2n

W (un)dx. (4.7)

We also compare∫M2n

AαβDαunD

βundx with∫

Ω

AαβDαvnD

βvndx.

Since the two quantities agree in M2n\D2n and Dαvn vanishes in Ω\M2n, weneed only to compare∫

D2n

AαβDαunD

βundx with∫D2n

AαβDαvnD

βvndx.

ConsiderAαβD

αvn(x)Dβvn(x) =

Aαβ [∑ι≤α

α!ι!Dα−ι(un(x) + 1)Dιξ(

x1 − (y1n + (r + rn)/2)

h)]

12

×[∑κ≤β

β!κ!Dβ−κ(un(x) + 1)Dκξ(

x1 − (y1n + (r + rn)/2)

h)]

= Aαβ [∑ι1≤α1

α!ι1!Dα−(ι1,0,...,0)(un(x) + 1)Dι1ξ(

x1 − (y1n + (r + rn)/2)

h)]

×[∑κ1≤β1

β!κ1!

Dβ−(κ1,0...,0)(un(x) + 1)Dκ1ξ(x1 − (y1

n + (r + rn)/2)h

)].

The first term from the product is

Aαβξ2(x1 − (y1

n + (r + rn)/2)h

)DαunDβun (4.8)

and the other terms assume the form

Aαβα!β!hι1+κ1ι1!κ1!

Dα−(ι1,0,...,0)(un(x) + 1)Dβ−(κ1,0...,0)(un(x) + 1)

ξ(ι1)(x1 − (y1

n + (r + rn)/2)h

)ξ(κ1)(x1 − (y1

n + (r + rn)/2)h

)

where at least one of ι1 and κ1 is not equal to 0. They can be bounded by

C

hι1+κ1|Dα−(ι1,0,...,0)(un(x) + 1)Dβ−(κ1,0...,0)(un(x) + 1)|.

Lemma 4.1 With un as above, there is C > 0 and a positive integer n1 so thatfor n ≥ n1∫

D2n

1hι1+κ1

|Dα−(ι1,0,...,0)(un(x) + 1)Dβ−(κ1,0...,0)(un(x) + 1)|dx ≤ C

h

for all α, β, |α| = 2i− 1, |β| = 2j − 1, i, j ∈ 1, 2, ..., N, ι1 ∈ 1, 2, ..., 2i− 1,κ1 ∈ 1, 2, ..., 2j − 1, ι1 + κ1 ≥ 1.

Proof. In view of lemma 2.2, we know un+1 is close to 0 uniformly for x ∈ D2n.According to our assumption W ′′(−1) > 0, we find

(un(x) + 1)2 ≤ CW (un(x))

for x ∈ D2n where C is independent of large n. We then obtain∫D2n

(un(x) + 1)2dx ≤ C

where C again is independent of large n. We also know that from the positivityof [Aαβ ] ∫

Ω

[Dαun(x)]2dx ≤ C, |α| = 2i− 1, i = 1, 2, ..., N.

13

Appealing to (2.3) again, we deduce∫Ω

[Dαun(x)]2dx ≤ C, |α| = 2i, i = 1, 2, ..., N − 1.

Lemma 4.1 follows from these two bounds and Holder’s inequality.We sum over α, β and find∫D2n

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαvnD

βvndx ≤∫D2n

ξ2N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βundx

+O(1h

) ≤∫D2n

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βundx+O(1h

).

Together with (4.7) and the fact vn ∈ A, we deduce

I ≤ E(vn) =∫

Ω

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαvnD

βvn +W (vn)]dx

≤∫M2n

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βun +W (un)]dx+O(1h

).

If we choose h large enough, we obtain

E(un) ≥∫

Ω\M2n

W (un)dx+∫M2n

[N∑

i,j=1

∑‖α |=2i−1|β|=2j−1

AαβDαunD

βun +W (un)]dx

≥ λ+ I − 3ε

which implies that un can not be an energy minimizing sequence.

Case 2: Suppose, taking a subsequence if necessary, the un’s stay close to 1 inD1n.

This possibility can be ruled out by the same argument as in the previouscase. Instead of considering D2n and M2n, one considers D1n and M1n. Onetakes

vn(x) = (un(x)− 1)ξ(−x1 + (y1

n − (r + rn)/2)h

) + 1. (4.9)

Case 3: Suppose, taking a subsequence if necessary, the un’s stay close to −1in D1n and 1 in D2n.

14

We truncate un outside BΩ(yn, rn) and replace it by vn. These have approx-imately equal energies being at least close to I there, but the two tails of un haveextra energies bounded away from 0. Therefore such un’s can not constitute anenergy minimizing sequence. Set

wn = [(un(x) + 1)ξ(x1 − (y1

n − (r + rn)/2)h

)− 1]− 1

×ξ(−x1 + (y1

n + (r + rn)/2)h

) + 1 (4.10)

where h < (rn − r)/2 is a large number to be specified later.We first compare

∫BΩ(yn,rn)

W (un)dx with∫

ΩW (wn)dx. Observe that un

and wn only differ in M1n∪M2n for large n, and outside BΩ(yn, rn), W (wn) = 0.Also in D1n, where un is close to -1, wn is closer to -1, and in D2n, where un isclose to 1, wn is closer to 1. Therefore we deduce∫

Ω

W (wn)dx ≤∫BΩ(yn,rn)

W (un)dx. (4.11)

We then compare∫BΩ(yn,rn)

AαβDαunD

βundx with∫

Ω

AαβDαwnD

βwndx.

We consider the integrals on three different sets, D1n, D2n and BΩ(yn, r). Sinceun and wn agree in BΩ(yn, r) for large n. We only consider D1n and D2n.

In D1n,

ξ(x1 + (y1

n + (r + rn)/2)h

) = 1

for large n. Therefore

wn(x) = [un(x) + 1]ξ(x1 − (y1

n − (r + rn)/2)d

)− 1.

In D2n,

ξ(x1 − (y1

n − (r + rn)/2)h

) = 1

for large n. Therefore

wn(x) = [un(x)− 1]ξ(−x1 + (y1

n + (r + rn)/2)h

) + 1.

Using the same argument as the one in case 1, we find∫Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαwnD

βwndx

15

≤∫BΩ(yn,rn)

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βundx+O(1h

).

Together with (4.11) and the fact wn ∈ A we deduce

I ≤∫

Ω

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβD2i−1wnD

αwn +W (wn)]dx

≤∫BΩ(yn,rn)

[∑

|α|=2i−1|β|=2j−1

AαβDαunD

βun +W (un)]dx+O(1h

).

If we choose h large enough, we obtain

E(un) ≥∫

Ω\BΩ(yn,rn)

W (un)dx

+∫BΩ(yn,rn)

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βun +W (un)]dx ≥ Λ− λ+ I − 3ε

which implies that un can not be an energy minimizing sequence if we chooseε small. This rules out the third possibility and proves lemma 3.3.

5 A sufficient condition for non-periodic Ω

In this section we will study problem (1.1) in a non-periodic tube. The lack ofperiodicity causes the loss of translation. This brings a subtle difference betweenproblem (1.1) in a periodic tube and that in a non-periodic tube. We will onlyconsider non-periodic tubes that are “half-periodic” near the two ends. In thiscase, the concentration-compactness principle can give a sufficient conditionwhich ensures the existence of a transition layer solution.

Let Ω be an open connected set in Rd. Suppose Ω can be divided into threemutually disjoint connected open sets Ω1, Ω2 and Ω0 where Ω = Ω1 ∪ Ω2 ∪ Ω0.We further assume that Ω0 is bounded and Ω1 and Ω2 are “half periodic”. ThatΩ1 is half periodic means there exists T1 ∈ Rd such that for every x1 ∈ Ω1,x1 + T1 ∈ Ω1, and for every x1 ∈ Ω1 there exist y1 ∈ Ω0 and a non-negativeinteger n1 satisfying x1 = y1 +n1T1. The same condition applies to Ω2. Ω1 andΩ2 are indeed the two half periodic tails of Ω. We associate the end E1 to Ω1

and the other end E2 to Ω2 (see figure 2).The function class A is defined in (2.5). For u ∈ A we require

limx→E1

u(x) = −1, limx→E2

u(x) = 1.

16

Figure 2: An example of non-periodic tubes.

17

We define G1 (G2, respectively) to be the “periodic extension” of Ω1 (Ω2,respectively).

G1 = x ∈ Rd : x+ n1T1 ∈ Ω1 for some integer n1. (5.1)

We require that G1 (G2, respectively) has bounded cross sections perpendicularto T1 (T2, respectively, where T2 is the period of G2.) We denote the other endof G1 (G2, respectively) by E′1 (E′2, respectively) (see figure 2).

We now define two auxiliary variational problems:

I1 = infA1

∫G1

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβu+W (u)]dx,

I2 = infA2

∫G2

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβu+W (u)]dx.

(5.2)

The function classes A1 and A2 are defined in a way similar to (2.5). Werequire for u ∈ A1, limx→E1 u(x) = −1, limx→E′1 u(x) = 1, and for u ∈ A2,limx→E′2 u(x) = −1, limx→E2 u(x) = 1.

Theorem 5.1 Let Ω be a tube half-periodic near the two ends. We always have

I ≤ minI1, I2

where I is defined in (1.1) and I1, I2 are defined in (5.2). If the strict inequality

I < minI1, I2

holds, then problem (1.1) has an energy minimizer in A.

Proof. Without the loss of generality, we assume I1 ≤ I2. We know fromtheorem 3.1 that the infimum I1 can be achieved by a function, say v, in A1. Weassume without losing the generality T1 = (T 1

1 , 0..., 0), T 11 < 0 and (0, 0, ..., 0) ∈

Ω1. Set

vn(x) =

[v(x+ nT1)− 1]ξ(−x1

h) + 1, x ∈ Ω1,

1, x ∈ Ω\Ω1

(5.3)

where ξ is defined in (4.5) and n is a positive integer. Note vn ∈ A. We comparethe energy of vn in A with that of v in A1. When n is large, vn is close to 1 forx1 ∈ [−h, 0], so ∫

Ω

W (vn)dx =∫x∈Ωx1<0

W (vn) ≤∫G1

W (v)dx. (5.4)

ConsiderAαβD

αvnDβvn =

18

Aαβ [∑ι1≤α1

α!ι1!Dα−(ι1,0,...,0)(v(x+ nT1)− 1)Dι1ξ(

−x1

h)]

×[∑κ1≤β1

β!κ1!

Dβ−(κ1,0,...,0)(v(x+ nT1)− 1)Dκ1ξ(−x1

h)].

The first term from the product is

AαβDαv(x+ nT1)Dβv(x+ nT1)ξ2(

−x1

h)

and the other terms assume the form

Aαβα!β!hι1+κ1ι1!κ1!

Dα−(ι1,...,0)v(x+ nT1)Dβ−(κ1,0,...,0)v(x+ nT1)

×ξ(ι1)(−x1

h)ξ(κ1)(

−x1

h)

where ι1 + κ1 ≥ 1. We sum over α, β and integrate over Ω. We find, as in theproof of lemma 3.3, ∫

Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβAαβDαvnD

βvndx

=∫G1∩x1>−1−nT 1

1

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαvDβvdx+O(

1h

).

Together with (5.4), we deduce

I ≤∫

Ω

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβAαβDαvnD

βvn +W (vn)]dx

≤∫G1∩x1>−1−nT 1

1

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαvDβvdx+

∫G1

W (v)dx+O(1h

).

If we choose h comparable with n and send n→∞, we find

I ≤∫G1

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαvDβv +W (v)]dx = I1.

This proves the first part of theorem 5.1.We now assume I < minI1, I2. We again consider the concentration func-

tions, the Qn’s and Q, of the W (un)’s where un is an energy minimizingsequence in A. We consider the three cases λ = 0, λ ∈ (0,Λ) and λ = Λ. As in

19

the proof of theorem 3.1, we can rule out the cases λ = 0 and λ ∈ (0,Λ). Weare left with the case λ = Λ which implies from statement I in section 2 thatthere exist a subsequence of un, again denoted by un, and yn ∈ Ω such that forevery ε > 0 there exists r > 0 so that∫

BΩ(yn,r)

W (un(x))dx ≥ Λ− ε

for large n.We will prove that the yn’s indeed stay in a bounded subset of Ω. Suppose

the contrary, so we can assume a subsequence of yn, still denoted by ynapproaches E1. Without the loss of generality, we also assume (0, 0, ..., 0) ∈ Ω1,T1 = (T 1

1 , 0, ..., 0), T 11 < 0. Define in G1

vn(x) =

(un(x)− 1)ξ(−x1

h) + 1, x ∈ Ω1

1, x ∈ G1\Ω1.(5.5)

Here we first choose h < −y1n − r.

We observe since the un’s are close to 1 in Ω0 ∪ Ω2,∫G1

W (vn)dx =∫

Ω1

W (vn)dx ≤∫

Ω

W (un)dx. (5.6)

As in the proof of lemma 3.3, we write

AαβDαvn(x)Dβvn(x) = AαβD

α[(un(x)− 1)ξ(−x1

h)]Dβ [(un(x)− 1)ξ(

−x1

h)]

= Aαβ [∑ι1≤α1

α!ι1!Dα−(ι1,0...,0)(un(x)− 1)Dι1ξ(

−x1

h)]

×[∑κ1≤β1

β!κ1Dβ−(κ1,0,...,0)(un(x)− 1)Dκ1ξ(

−x1

h)]

where the first term from the product is

Aαβξ2(−x1

h)DαunD

βun

and the other terms assume the form

Aαβα!β!hι1+κ1ι1!κ1!

Dα−(ι1,0...,0)(un − 1)Dβ−(κ1,0...,0)(un − 1)ξ(ι1)(−x1

h)ξ(κ1)(

−x1

h),

ι1 + κ1 ≥ 1. Summing over α, β and integrating over G1, we find, as in theproof of lemma 3.3,∫

G1

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαvnD

βvndx =∫

Ω1

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαvnD

βvndx

20

≤∫

Ω1

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βundx+O(1h

)

≤∫

Ω

N∑i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βundx+O(1h

).

Together with (5.5), we deduce

I1 ≤∫G1

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαvnD

βvn +W (vn)]dx

≤∫

Ω

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαunD

βun +W (un)]dx+O(1h

) = I +O(1h

).

If we choose h comparable with −y1n and send n→∞, we obtain I1 ≤ I which

contradicts the assumption I < minI1, I2.Once we know the yn’s stay in a bounded subset of Ω, we conclude that

there exists u ∈ A such that passing to a subsequence if necessary

un → u in C(K) for every compact K ⊂ Ω, (5.7)

Dαun → Dαu weakly in L2(Ω), |α| = 2i− 1, i = 1, 2, ..., N. (5.8)

As in the proof of lemma 3.4, this u is an energy minimizer of (1.1).Similar to corollary 3.5, we have

Corollary 5.2 If I < minI1, I2, the convergence in (5.7) and (5.8) is indeed

un → u in C(Ω),

Dαun → Dαu strongly in L2(Ω), |α| = 2i− 1, i = 1, 2, ..., N.

6 Some applications of theorem 5.1

We will construct a few examples of Ω to illustrate theorem 5.1 in this section.We assume that equation (1.5) has the property of unique continuation, namelyif a solution u of (1.5) is equal to a constant in an open subset of Ω, then u isnecessarily a constant in Ω. If W is real analytic, then the solutions of (1.5) arereal analytic and (1.5) in turn has the property of unique continuation.

Our first example is a near periodic tube ΩA in R2. The correspondingperiodic extensions G1, G2 of the tails are congruent to GA (see figure 3). Notethat the cross sections of ΩA have the same lengths as those of the correspondingcross sections of GA except in a middle region where the cross sections of ΩAhave shorter lengths than that of GA.

21

Figure 3: The picture of ΩA.

Let uA be an energy minimizer of problem (1.1) in GA. The existence ofsuch an energy minimizer is guaranteed by theorem 3.1. Denote this problemby

IG = infu∈AG

∫GA

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuDβu+W (u)]dx. (6.1)

Clearly IG = I1 = I2 where I1 and I2 are the infima of the correspondingauxiliary problems. Take uA to be a test function and put it in ΩA. Becausethe middle region of ΩA is thinner than that of GA, we have

I ≤∫

ΩA

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuAD

βuA +W (uA)]dx

<

∫GA

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuAD

βuA +W (uA)]dx = IG. (6.2)

The strict inequality in (6.2) follows from the unique continuation which ensuresthat uA is not a constant in any open subset of GA. Therefore in this tube thereis an energy minimizer.

The second example is a near perfect tube except that somewhere in themiddle there is a hole, but the lengths of the cross sections are identical. Wedenote the tube, its tails and their congruent periodic extension by ΩB , ΩB,1,

22

Figure 4: The picture of ΩB .

ΩB,2 and GB (see figure 4). We take uB to be an energy minimizer of (1.1)in GB . It is not clear to us whether uB is necessarily one-dimensional, i.e., uBis a function of x1 only. If we are willing to make the assumption that uB isone-dimensional, then we extend uB to all R2 in the obvious way and put it inΩB . Because the cross sections of Ω have the same length, we have∫

ΩB

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuBD

βuB +W (uB)]dx

=∫GB

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuBD

βuB +W (uB)]dx = IG.

However uB is not an energy minimizer in ΩB since it does not satisfy theboundary conditions of the Euler-Lagrange equation (1.5) in ΩB . Therefore

I <

∫ΩB

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

AαβDαuBD

βuB +W (uB)]dx = IG (6.3)

which again yields an energy minimizer in ΩB .

23

Figure 5: The picture of ΩC .

The third example will show that the sufficient condition is not always valid.Indeed in this example, problem (1.5) does not have an energy minimizer. LetΩC be a near perfect tube. Unlike the first example, we assume that somewherein the tube the cross sections have longer lengths than that of the correspondingGC (see figure 5). If u were an energy minimizer of (1.1) in ΩC , then we wouldhave

I =∫

ΩC

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

DαuDβu+W (u)]dx

>

∫GC

[N∑

i,j=1

∑|α|=2i−1|β|=2j−1

DαuDβu+W (u)]dx ≥ IG = I1 = I2.

This contradicts the first part of theorem 5.1.

Note added in proof. In a forthcoming paper, [11], the second named author isable to prove theorems 3.1 and 5.1 without condition (1.4).

References

[1] R.A. Adams, Sobolev Spaces, Academic Press, New York, 1975.

[2] P.W. Bates, P.C. Fife, R. Gardner and C.K.R.T. Jones, Phase field modelsfor hypercooled solidification, preprint.

[3] P.W. Bates, P.C. Fife, R. Gardner and C.K.R.T. Jones, The existence oftraveling wave solutions of a generalized phase-field model, preprint.

[4] P.W. Bates, P.C. Fife, X. Ren and X. Wang, Traveling waves in a convolu-tion model for phase transitions, preprint.

[5] P.W. Bates and X. Ren, Heteroclinic orbits for a higher order phase tran-sition problem, preprint.

[6] P. Fife and J.B. McLeod, The approach of solutions of nonlinear diffusionequations to traveling front solutions, Arch. Rat. Mech. Anal. 65 (1977),335-361.

24

[7] R. Gardner and C.K.R.T. Jones, Traveling waves of a perturbed diffusionequation arising in a phase field model, Indiana Math. J. Vol 38, no. 4(1989), 1197-1222.

[8] A.G. Khachaturyan, Microscopic theory of diffusion in crystalline solid so-lutions and the time evolution of the diffuse scattering of X rays and thermalneutrons, Soviet Phys. Solid State, Vol. 9 (1968), 2040-2046.

[9] A.G. Khachaturyan, Theory of Structural Transformations in Solids,Wiley-Interscience Publications, New York, 1983.

[10] P.L. Lions, The concentration-compactness principle in the calculus ofvariations: The locally compact case. Part I and Part II, Ann. Inset. H.Poincare Sect. A (N.S.) 1. (1984), 109-145, 223-283.

[11] X. Ren, Variational approach to multiple layers of the bistable equation inlong tubes, preprint.

25