Interacting Near-Solutions of aHamiltonian System

Gregory S. Spradlin

Department of Mathematics

Embry-Riddle Aeronautical University

Daytona Beach, Florida 32114-3900

USA

spradlig@erau.edu

Abstract: A Hamiltonian system with a superquadratic potential is examined.

The system is asymptotic to an autonomous system. The difference between the

Hamiltonian system and the “problem at infinity,” the autonomous system, may be

large, but decays exponientially. The existence of a nontrivial solution homoclinic

to zero is proven. Many results of this type rely on a monotonicity condition on the

nonlinearity, not assumed here, which makes the problem resemble in some sense

the special case of homogeneous (power) nonlinearity.

The proof employs variational, minimax arguments. In some similar results

requiring the monotonicity condition, solutions inhabit a manifold homeomorphic

to the unit sphere in a the appropriate Hilbert space of functions. An important

part of the proof here is the construction of a similar set, using only the mountain-

pass geometry of the energy functional. Another important element is the interaction

between functions resembling widely separated solutions of the autonomous problem.

Key Words: Mountain Pass Theorem, variational methods,

concentration-compactness, Nehari manifold, homoclinic solutions

AMS Subject Classification: 34C37, 47J30

1. Introduction

This paper is inspired by a result of Bahri and Li ([BL]). The authors studied

an elliptic partial differential equation of the form −∆u+u = b(x)up, x ∈ RN , with

1

b(x) → b∞ > 0 as |x| → ∞ and the negative part of b(x)−b∞ decaying exponentially

to zero as |x| → ∞. They showed that a positive solution v exists with v(x) → 0 as

|x| → ∞.

The proof employed a minimax argument. The “problem at infinity,”

−∆u + u = b∞up, has a unique (modulo translation) positive solution w with

w(x) → 0 as |x| → ∞. The minimax argument uses sums of translates of w,

and exploits how the “tails” of these translates interact. A similar concept is found

in [WX].

A natural problem is to generalize the result of [BL] to the case of a nohomo-

geneous nonlinearity. A step in this direction was taken by Adachi ([A]), in which

up is replaced by a more general superlinear function of u. The coefficient function

b(x) is assume to possess symmetry, however.

[S] examines an N = 1 version of the problem, −u′′ + u = b(t)f(u). Like in the

PDE problem, f(u) is a superlinear function of u, b(t) → b∞ > 0 as |t| → ∞, and

the negative part of b(t)−b∞ decays to zero suitably rapidly as |t| → ∞. It is proven

that this Hamiltonian system has a positive solution homoclinic to zero. The proof

involves translates of the unique positive, even solution of the problem at infinity,

−u′′ + u = b∞f(u), and exploits the interaction of their “tails.” The function f

satisfies the Ambrosetti-Rabinowitz condition (described later), which ensures that

f(u) behaves like a superlinear power of u. Another critical assumption is that

f(q)/q is a nondecreasing function of q for positive q. This popular assumption, also

found in [AM],[CMN], [STT] and elsewhere, has many helpful implications.

The present paper proves a result similar to [S], while dispensing with the

assumption that f(q)/q be nondecreasing. We examine a Hamiltonian system of the

form

−u′′ + u = W ′(t, u), (1.1)

and prove the following:

Theorem 1.2 Let N ∈ N+, and let V and W satisfy

(V1) V ∈ C1,1(R+,R+)

(V2) V (0) = 0

2

(V3) There exists µ > 2 such that V ′(q)q ≥ µV (q) > 0 for all q > 0

(W1) W ∈ C1,1(R×RN ,RN )

(W2) W (t, 0) = 0 for all t ∈ R

(W3) W ′(t, u)u ≥ µW (t, u) > 0 for all t ∈ R,u ∈ RN \ {0}, where

µ is from (V3) and W ′(t, u) = < ∂∂u1

W (t, u), ∂∂u2

W (t, u), . . . , ∂∂uN

W (t, u) >.

(W4) (W (t, u)− V (|u|))/V (|u|) → 0 as |t| → ∞, uniformly in u ∈ RN \ {0} .

(W5) W ′(t, u)u/|u|2 → 0 as |u| → 0, uniformly in t ∈ R .

(W6) There exist δ > 2µ/(µ− 2) and A > 0 with W (t, u)− V (u) ≥ −AV (|u|)e−δ|t|

for all t ∈ R, u ∈ RN , or W (t, u)− V (u) ≥ −A|u|µe−δ|t|

for all t ∈ R, u ∈ RN .

Then (1.1) has a nontrivial solution v homoclinic to zero, with I(v) ∈ (0, 2c0), where

c0 is the mountain pass value associated with the function J (see (1.7)-(1.9)).

The variational framework

Define the C2 functional I : W 1,2(R,RN ) → R by

I(u) =12‖u‖2 −

∫R

W (t, u(t)) dt, (1.3)

where ‖ ‖ is the standard norm, ‖u‖2 =∫∞−∞|u

′(t)|2 + u(t)2 dt. Critical points of

I correspond exactly to solutions of (1.1) homoclinic to zero. The conditions (V3)

and (W3) ensure that V and W are “superquadratic” functions, with, for example,

V (q)/q2 → 0 as q → 0 and V (q)/q2 → ∞ as q → ∞. Thus (W1) − (W3) imply I

has “mountain pass” geometry. That is, I(u) = 12‖u‖

2 + o(‖u‖2) as ‖u‖ → 0, and

I(u) < 0 for some u ∈ W 1,2(R,RN ). Therefore the set of mountain pass curves

Γ = {γ ∈ C([0, 1],W 1,2(R,RN )) | γ(0) = 0, I(γ(1)) < 0} (1.4)

is nonempty, and the “mountain pass” value c defined by

c = infγ∈Γ

maxθ∈[0,1]

I(γ(θ)) (1.5)

is positive.

3

We will show that the scalar differential equation

−u′′ + u = V ′(u) (1.6)

can be regarded as the “problem at infinity” for (1.1). This equation has a unique

(modulo translation) positive solution ω. ω is even, increasing on (−∞, 0), and

decreasing on (0,∞).

Extend V to the negative reals by making V even, that is, V (−q) = V (q) for

all q ∈ R. Then the functional J ∈ C2(W 1,2(R,R),R) corresponding to (1.6) is

J(u) =12‖u‖2 −

∫ ∞

−∞V (u(t)) dt, (1.7)

where ‖u‖2 =∫∞−∞u′(t)2 + u(t)2 dt. Like I, J has mountain-pass geometry, with

Γ0 = {γ ∈ C([0, 1],W 1,2(R,R)) | γ(0) = 0, J(γ(1)) < 0} (1.8)

nonempty, and the mountain pass value c0 defined by

c0 = infγ∈Γ0

maxθ∈[0,1]

I(γ(θ)) (1.9)

positive.

The Missing Monotonicity Assumption

An important feature of Theorem 1.2 is a condition that is not assumed. We

do not assume that

W (t, su)/s2 is a nondecreasing function of s (1.10)

for positive s and all t ∈ R and u ∈ RN \ {0},

or that

V (q)/q2 is a nondecreasing function of q for q > 0. (1.11)

The implications of such an assumption have been studied by Nehari. The first

assumption would imply that for any u ∈ W 1,2(R,RN ), the mapping s 7→ I(su)

starts at 0 when s = 0, increases to a positive maximum, then decreases to −∞ (see

[CR]). Then we could define the “Nehari manifold”

S = {u ∈ W 1,2(R,RN ) | I ′(u)u = 0, u 6= 0}. (1.12)

4

S would be a codimension-one manifold of W 1,2(R,RN ), homeomorphic to the unit

sphere in W 1,2(R,RN ) via radial projection. All nonzero critical points of I would

be in S. The minimax value c could be described more simply as

c = infu∈S

I(u). (1.13)

Also, for any u ∈ RN \ {0}, the function γ, defined by γ(θ) = Tθu for some suitable

scaling factor T , would belong to Γ. Therefore we could work with nonzero functions

in W 1,2(R,RN ) instead of curves in Γ. [S] takes advantage of these facts.

In this paper (1.10) is not assumed, so S may not have these properties. Instead,

the mountain pass geometry of I is used to constuct a set with similar properties

to S. The set is defined as follows: let ∇I denote the gradient of I. That is,

(∇I(u), w) = I ′(u)w for all u, w ∈ W 1,2(R,RN ). Here, (·, ·) is the usual inner

product defined by (u, w) =∫∞−∞u′(t)·v′(t)+u(t)v(t) dt. ∇I(u) exists because of the

Riesz Representation Theorem, and ∇I is a C1 vector field. Let ϕ : W 1,2(R,RN ) →

[0, 1] be a suitable locally Lipschitz cutoff function, and let η be the solution of the

initial value problemdη

ds= −ϕ(η)∇(η), η(0, u) = u. (1.14)

ϕ(η) is introduced and defined so that η is defined on R+ ×W 1,2(R,RN ). Let B

be the basin of attraction of 0 under the flow η, that is,

B = {u ∈ W 1,2(R,RN ) | η(s, u) → 0 as s →∞}. (1.15)

B is a connected, nonempty open neighborhood of 0. Let ∂B denote the topological

boundary of B, under the usual metric on W 1,2(R,RN ). As with S, any mountain

pass curve γ ∈ Γ intersects ∂B at least once. Also, ∂B is forward-η-invariant. That

is, for any u ∈ ∂B and s > 0, η(s, u) ∈ ∂B.

Sketch of the Proof

To prove Theorem 1.2, we assume that I has no critical point v with 0 < I(v) ≤

c0, then show that this implies I has a critical value in the interval (c0, 2c0). To

do this, we construct a minimax class with a minimax value m strictly between

c0 and 2c0. There then exists a Palais-Smale sequence (un) ⊂ W 1,2(R,RN ) with

5

I ′(un) → 0 and I(un) → m as n →∞. By a concentration-compactness argument,

(un) converges along a subsequence to v, a critical point of I with c0 < I(v) < 2c0,

proving Theorem 1.2.

The minimax class is constructed as follows (more details are in Sections 2-

4). Construct γ ∈ Γ0 satisfying maxθ∈[0,1] J(γ(θ)) = c0 and some other helpful

properties. Define the translation operator τ as follows: for a function u over the

reals and a ∈ R, let τau be u shifted a units to the right, that is, τau(t) = u(t− a)

for all t ∈ R. Let R1 > 0 be a suitably large constant, and define G to be a family

of functions from the unit square to W 1,2(R,RN ): let e1 = < 1, 0, 0, . . . , 0 > ∈ RN ,

and define

G = {G ∈ C([0, 1]2,W 1,2(R,RN )) | for all x, y ∈ [0, 1], (1.16)

G(x, 0) = 0, I(G(x, 1)) < 0,

G(0, y) = τ−R1γ(y)e1, G(1, y) = τR1γ(y)e1}.

Then define

m = infG∈G

max(x,y)∈[0,1]2

I(G(x, y)). (1.17)

We will prove that, assuming I has no critical values in (0, 2c0], then c0 < m < 2c0,

and there exists a Palais-Smale sequence (un) with I ′(un) → 0 and I(un) → m,

which converges along a subsequence. Therefore I has a critical value between c0

and 2c0.

To prove m < 2c0, we must construct a G0 ∈ G with I(G0(x, y)) < 2c0 for all

(x, y) ∈ [0, 1]2. Such a G0 is given by

G0(x, y) = max(τ−R1γ((1− x)y), τR1γ(xy))e1. (1.18)

Proving I(G0(x, y)) < 2c0 is most difficult to prove if J(γ((1−x)y)) and J(γ(xy)) are

both close to c0. By the construction of γ, γ((1− x)y) and γ(xy) both have “tails”

that are translates of the left or right half of ω. That is, for some large M , there exists

t ≥ 0 with γ(xy)(t) = ω(|t| −M + t) for all |t| > M (and similarly for γ((1− x)y)).

ω(t) decays exponentially to zero as |t| → ∞. γ is constructed so that γ(θ)(t) decays

to zero exponentially as |t| → ∞ for all θ ∈ [0, 1], uniformly in θ. As the title of this

paper suggests, the interaction of the exponentially decaying “tails” of γ((1 − x)y)

6

and γ(xy), along with the exponential decay of the negative part of W (t, u)−V (|u|),

imply that I(G0(x, y)) = I(max(t−R1γ((1− x)y), tR1γ(xy))e1) < 2c0.

The paper is organized as follows: in Section 2 a suitable mountain pass curve

γ is constructed. Section 3 proves some properties of ∂B. Section 4 contains the

minimax argument, proving that G, m, and G0 have the properties claimed in this

Introduction.

2. A Mountain-Pass Curve for the Scalar Equation

By [JT], there exists γ0 ∈ Γ0 with maxθ∈[0,1] J(γ0(θ)) = c0. This is proven

under weaker assumptions than here, and is not obvious from the definition of c0.

Instead of using this result, we will construct such a γ directly, using an argument

similar to that in [C]. We will prove

Proposition 2.1There exist M = M(V ) and γ ∈ Γ0 such that for all θ ∈ [0, 1] and

t ∈ R,

(i) J(γ(θ)) ≤ c0

(ii) θ ≥ 12⇒ J(γ(θ)) < −2c0

(iii) γ(θ)(t) > 0 if θ > 0

(iv) γ(θ)(t) = γ(θ)(−t)

(v) θ1 < θ2 ⇒ γ(θ1)(t) ≤ γ(θ)(t)

(vi) 0 ≤ t1 ≤ t2 ⇒ γ(θ)(t1) ≥ γ(θ)(t2)

(vii) θ > 0 ⇒ There exists tθ ≥ 0 with γ(θ)(t) = ω(|t| −M + tθ)

for all |t| ≥ M. tθ is a continuous function of θ.

Proof: The Hamiltonian 12ω′(t)2 − 1

2ω(t)2 + V (ω(t)) equals zero for all t, so

ω(0)2 = 2V (ω(0)). (2.2)

By (W1) − (W3), q2 < 2V (q) for all 0 < q < ω(0), so the integrand in I(ω) =∫∞−∞

12 (ω′)2 + 1

2ω2 − V (ω) dt is positive for all t 6= 0. Define

γ(θ)(t) ={

0, θ = 0;ω( 1

θ − 6 + |t|), 0 < θ ≤ 16 . (2.3)

7

γ is continuous on [0, 1/6], γ(1/6) = ω, and for all θ ∈ (0, 1/6),

I(γ(θ)) = 2∫ ∞

1θ−6

12ω′

2 +12ω2 − V (ω) < I(ω). (2.4)

By (V3) and (2.2),

ω(0)− V ′(ω(0)) ≤ ω(0)− µV (ω(0))ω(0)

= (2.5)

=ω(0)2 − µV (ω(0))

ω(0)= −1

2(µ− 2)ω(0) < 0.

Let α > 0 be small enough so that for all ω(0) ≤ q ≤ ω(0) + α,

q − V ′(q) ≤ −14(µ− 2)ω(0). (2.6)

ω(0)2 − V (ω(0)) = 0, so the Mean Value Theorem implies that for all

ω(0) ≤ q ≤ ω(0) + α,

12q2 − V (q) ≤ −1

4(µ− 2)ω(0)(q − ω(0)). (2.7)

Define

M = max(√ 8α

(µ− 2)ω(0),

24c0

(µ− 2)ω(0)α). (2.8)

For θ ∈ [1/6, 1/3], let γ(θ) be ω with a horizontal segment of height ω(0) inserted

in the center, with the segment growing from length 0 to M as θ increases from 1/6

to 1/3. That is,

γ(θ)(t) ={

ω(0), |t| ≤ (6θ − 1)M ;ω(|t| − (6θ − 1)M), |t| ≥ (6θ − 1)M.

(2.9)

Since ω(0)2 = 2V (ω(0)), for each θ ∈ [1/6, 1/3] we have

I(γ(θ)) =∫ −(6θ−1)M

−∞

12γ(θ)′(t)2 +

12γ(θ)(t)2 − V (γ(θ)(t)) dt + (2.10)

+∫ (6θ−1)M

−(6θ−1)M

12γ(0)2 − V (γ(0)) dt +

+∫ ∞

(6θ−1)M

12γ(θ)′(t)2 +

12γ(θ)(t)2 − V (γ(θ)(t)) dt =

=∫ 0

−∞

12ω′

2 + ω2 − V (ω) +∫ ∞

0

12ω′

2 + ω2 − V (ω) =

= I(ω) = c0.

8

Next, we will deform γ(1/3) so that it is only piecewise linear on [−M,M ]. For ease

in notation, for s ∈ [0, α] let us temporarily define us by

us(t) ={

ω(|t| −M), |t| ≥ M ;ω(0) + s− s|t|/M, |t| ≤ M . (2.11)

Now

I(us) =∫ −M

−∞

12u′s

2 +12u2

s − V (us) +∫ M

−M

12u′s

2 +12u2

s − V (us)+ (2.12)∫ ∞

M

12u′s

2 +12u2

s − V (us) =

=∫ 0

−∞

12ω′

2 +12ω2 − V (ω) +

∫ M

−M

12u′s

2 +12u2

s − V (us)+

+∫ ∞

0

12ω′

2 +12ω2 − V (ω) =

= I(ω) +∫ M

0

u′s2 + u2

s − 2V (us) =

= c0 +∫ M

0

(s

M)2

+ (ω(0) + s− s

Mt)2 − 2V (ω(0) + s− s

Mt) dt

= c0 +s2

M+

2M

s

∫ ω(0)+s

ω(0)

12x2 − V (x) dx ≤

= c0 +s2

M− (µ− 2)ω(0)M

2s

∫ ω(0)+s

ω(0)

x− ω(0) dx =

(by (2.7))

= c0 +s2

M− 1

4(µ− 2)ω(0)sM.

For s ∈ [0, α],

I(us) ≤ c0 +s

M(α− 1

4(µ− 2)ω(0)M2) ≤ c0 −

sα

M≤ c0. (2.13)

Also,

I(uα) ≤ c0 +α2

M− 1

4(µ− 2)αω(0)M (2.14)

≤ c0 +18(µ− 2)αω(0)M − 1

4(µ− 2)αω(0)M

= c0 −18(µ− 2)αω(0)M ≤ c0 − 3c0 = −2c0.

Now we can finish defining γ. For θ ∈ [1/3, 1/2], define

γ(θ) = u6α(θ− 13 ) (2.15)

9

and for θ ∈ [1/2, 1], define

γ(θ) = uα. (2.16)

The construction of γ is complete.

3. A Substitute for the Nehari Manifold

Let the gradient vector flow η be as described in the Introduction. That is, let

∇I denote the gradient of I; (∇I(u), w) = I ′(u)w for all u, w ∈ W 1,2(R,RN ). Here,

(·, ·) is the usual inner product defined by (u, w) =∫∞−∞u′(t) · v′(t) + u(t)v(t) dt.

∇I(u) exists because of the Riesz Representation Theorem, and ∇I is a C1 vector

field. Let ϕ : W 1,2(R,RN ) → [0, 1] be locally Lipschitz, with I(u) ≥ −1 ⇒ ϕ(u) = 1

and I(u) ≤ −2 ⇒ ϕ(u) = 0. Let η be the solution of the initial value problem

dη

ds= −ϕ(η)∇J(η), η(u, 0) = u. (3.1)

Lemma 3.2 η is well-defined on R+ ×W 1,2(R,RN )

Proof: Suppose that u ∈ W 1,2(R,RN ) and η(s, u) is not defined for all s > 0.

Since ϕ and ∇I are locally Lipschitz, there exist s > 0 and a sequence (sn)n≥1

with sn → s and ‖∇I(η(sn, u)‖ → ∞. Since I ′ is bounded on bounded subsets of

W 1,2(R,RN ), ‖η(sn, u)‖ → ∞. Clearly I(u) ≥ −2.

Let

P = ‖u‖+ 1 +√

4µ

µ− 2|I(u)|+ 8(I(u) + 2)

µ− 2(3.3)

Let η ≡ η(s) ≡ η(s, u). There exist 0 < s1 < s2 with ‖η(s1)‖ = P , ‖η(s2)‖ ≥ 2P ,

and P < ‖η(s)‖ < 2P for all s1 < s < s2. For all s1 < s < s2,

I ′(η)η = ‖η‖2 −∫ ∞

−∞V ′(η)η ≤ ‖η‖2 − µ

∫ ∞

−∞V (η) = (3.4)

= µI(η)− 12(µ− 2)‖η‖2 ≤ µ|I(η)| − 1

2(µ− 2)‖η‖2 <

<14(µ− 2)P 2 − 1

2(µ− 2)P 2 = −1

4(µ− 2)P 2,

so

‖I ′(η)‖ ≥ −I ′(η)η‖η‖

>18(µ− 2)P. (3.5)

10

Now

I(u) + 2 ≥ I(η(s1))− I(η(s2)) = −∫ s2

s1

d

dsI(η(s)) ds = (3.6)

=∫ s2

s1

ϕ(η)‖I ′(η(s))‖2 ds ≥ 18(µ− 2)P

∫ s2

s1

ϕ(η)‖I ′(η)‖ ds ≥

≥ 18(µ− 2)

∫ s2

s1

‖dη

ds‖ ds ≥ 1

8(µ− 2)

∥∥∫ s2

s1

dη

dsds

∥∥ =

=18(µ− 2)‖η(s2)− η(s1)‖ ≥

18(µ− 2)P.

Therefore

P ≤ 8(I(u) + 2)µ− 2

. (3.7)

This contradicts the definition of P . Lemma 3.2 is proven.

♠

Let B and ∂B be as defined in (1.14)-(1.15) in the Introduction. Here some

properties of ∂B are proven. First, it is well-known that any Palais-Smale sequence

for I is bounded in norm. The following lemma gives a formula that we will neeed

for the bound.

Lemma 3.8 For all u ∈ E,

‖u‖ ≤2‖I ′(u)‖+

√2µ(µ− 2) max(0, I(u))

µ− 2.

Proof:

−‖I ′(u)‖ ‖u‖ ≤ I ′(u)u = ‖u‖2 −∫R

h(t)W ′(t, u)u dt ≤ (3.9)

≤ ‖u‖2 − µ

∫R

W (t, u) dt = µI(u)− (µ− 2

2)‖u‖2,

so

(µ− 2

2)‖u‖2 − ‖I ′(u)‖ ‖u‖ − µI(u) ≤ 0. (3.10)

Applying the quadratic formula to (3.10), and the inequality√

A2 + B2 ≤ |A|+ |B|,

yields

‖u‖ ≤‖I ′(u)‖+

√‖I ′(u)‖2 + 2µ(µ− 2) max(0, I(u))

µ− 2≤ (3.11)

≤2‖I ′(u)‖+

√2µ(µ− 2) max(0, I(u))

µ− 2.

11

♠

Call a set A ⊂ W 1,2(R,RN ) forward-η-invariant if for all s > 0 and u ∈ A,

η(s, u) ∈ A. Now

Lemma 3.12

(i) B is an open neighborhood of 0 ∈ W 1,2(R,RN ).

(ii) inf∂B I > 0

(iii) ∂B is forward-η-invariant.

(iv) For any K > 0, the set ∂B ∩ {u ∈ W 1,2(R,RN | I(u) < K} is bounded.

Proof: (i): let r0 > 0 be small enough so that for all t ∈ R and v ∈ RN with |v| ≤ r0,

W (t, v) ≤ 16|v|2 and W ′(t, v)v ≤ 1

2|v|2. (3.13)

Let ‖u‖ ≤ r0, and η ≡ η(s) ≡ η(s, u). Then ‖u‖L∞ ≤ r0, and

d

ds‖η‖2 = −2I ′(η)(η) = −2‖η‖2 + 2

∫ ∞

−∞W ′(t, η)η ≤ −‖η‖2, (3.14)

so ‖η(s)‖2 → 0 as s → ∞, and u ∈ B. Thus B contains the ball

Br0(0) ≡ {w ∈ W 1,2(R,RN ) | ‖w‖ < r0}, an open neighborhood of 0. Now let

u ∈ B. For some s∗ > 0, η(s∗, u) ∈ Br0(0). For small enough r > 0, ‖w − u‖ < r

implies η(s∗, w) ∈ Br0(0), and η(s(η(s∗, w)) = η(s+s∗, w) → 0 as s →∞, so w ∈ B.

So Br(u) ≡ {w | ‖w − u‖ < r} is an open neighborhood of u that is contained in B.

(ii): ∂B is nonempty, for if I(u) < 0, then u 6∈ B. Let u ∈ ∂B. There exists a

sequence (un) ⊂ B with un → u. Let r0 be as in the proof of (i). ‖un‖ ≥ r0 for large

n. Since η(s, un) → 0, there exists sn with ‖η(sn, un)‖ = r0. Then |η(sn, un)(t)| ≤ r0

for all t ∈ R. By the definition of r0,

I(η(sn, un)) =∫ ∞

−∞

12η(sn, un)′(t)2 +

12η(sn, un)(t)2 (3.15)

−W (t, η(sn, un)(t)) dt ≥ 13‖η(sn, un)‖2 = r2

0/3,

so I(un) ≥ r20/3, I(u) ≥ r2

0/3, and (ii) is proven.

(iii): Let u ∈ B and s1 > 0. Since η(s, u) → 0 as m → ∞, η(s + s1, u) =

η(s, η(s1, u)) → 0 as s →∞, and η(s1, u) ∈ B. Thus B is forward-η-invariant. Next,

12

let u ∈ ∂B and s > 0. Since B is open, u 6∈ B. η(s, u) is not in B, for if it were, the

definition of B would imply u ∈ B. u is in the closure of B, so let (um) ⊂ B with

um → u. η(s, um) → η(s, u) and η(s, um) ∈ B, so η(s, u) belongs to the closure of B.

(iv) If suffices to show that for any K > 0, the set B ∩ {u ∈ W 1,2(R,RN |

I(u) < K} is bounded. We use an “annulus” argument. Let K > 0, and let

P = 1 +2µK

µ− 2+

16K2

(µ− 2)2. (3.16)

Let u ∈ ∂B with I(u) ≤ K. Assume ‖u‖ > 2P . This will lead to a contradiction.

By the definition of B and the fact that B is open, it is clear that I(u) ≥ 0. For

any w ∈ E with I(w) ≤ 0 and ‖w‖ ≥ P , Lemma 3.8 gives

‖I ′(w)‖ ≥ 12((µ− 2)‖w‖ −

√2µ(µ− 2)I(w)

)≥ (3.17)

≥ 12((µ− 2)P − 2µ

√K

)≥ µ− 2

4P.

By Lemma 3.2, η(s, u) is well-defined for all s > 0. Since I(η(s, u)) > 0 for all

s > 0, and ddsI(η(s, u)) = −‖I ′(η(s, u))‖2, (3.17) implies that ‖η(s∗, u)‖ ≤ P for

some s∗ > 0. Let η ≡ η(s) ≡ η(s, u). There exist 0 < s1 < s2 with ‖η(s1)‖ = 2P ,

‖η(s2)‖ = P , and ‖η(s1)‖ ∈ (P, 2P ) for all s ∈ (P, 2P ). Then by (3.17),

K ≥ I(η(s1))− I(η(s2)) = −∫ s2

s1

d

dsI(η(s)) ds = (3.18)

=∫ s2

s1

‖I ′(η(s))‖2 ds ≥ (s2 − s1)(µ− 2)2

16P 2.

But

P ≤ ‖η(s1)− η(s2)‖ = ‖∫ s2

s1

dη

dsds‖ ≤

∫ s2

s1

‖dη

ds‖ ds = (3.19)

=∫ s2

s1

‖I ′(η)‖ ds ≤√

s2 − s1 ·

√∫ s2

s1

‖I ′(η)‖2 ds =

(by the Cauchy-Schwarz Inequality)

=√

s2 − s1 ·

√∫ s2

s1

d

dsI(η(s)) ds =

=√

s2 − s1 ·√

I(η(s1))− I(η(s2)) ≤√

(s2 − s1)K.

(3.18)-(3.19) give

P 2

K≤ s2 − s1 ≤

16K

(µ− 2)2P 2, P 4 ≤ 16K2

(µ− 2)2. (3.20)

13

This contradicts the definition of P . Lemma 3.12 is proven.

♠

Note: it is unclear whether ∂B must be homeomorphic to the unit ball of

W 1,2(R,RN ).

Define I0 ∈ C2(W 1,2(R,RN ),R) by

I0(u) =12‖u‖2 −

∫ ∞

−∞V (|u|) dt. (3.21)

Roughly, this “autonomous” functional satisfies I0(u) ≈ I(u) if the bulk of u is

supported far from 0. The reason that we can consider the scalar equation (1.6)

to be the problem at infinity for (1.1) is that any nonzero critical point of I0 has

the form τaω u for some a ∈ R and unit vector u ∈ RN . To prove this, it suffices

to show that all critical points u of I0 are radial, that is, u(t) = g(t)u for some

scalar function g and unit vector u ∈ RN . Let u be a nontrivial critical point of I0,

satisfying

−u′′ + u = ∇(V (|u|)) =V ′(|u|)u|u|

. (3.22)

Consider the quantity (u ·u′)2−|u|2|u′|2. This expression tends to zero as t → ±∞.

If it equals zero for some t, then u′(t) and u(t) are parallel (this is the equality case

of the Cauchy-Schwarz Inequality). So it suffices to show that ddt [(u ·u

′)2−|u|2|u′|2]

is always zero.

d

dt

[(u · u′)2 − |u|2|u′|2

]= (3.23)

= 2(u · u′)(|u′|2 + u · u′′)− 2(u · u′)|u′|2 − 2|u|2(u′ · u′′) =

= 2[(u · u′)(u · u′′)− |u|2(u′ · u′′)

]=

= 2[(u · u′)(u · (u− V ′(|u|)u/|u|))− |u|2(u′ · (u− V ′(|u|)u/|u|))

]=

= 2[(u · u′)(|u|2 − |u|V ′(|u|))− |u|2(u′ · u− V ′(|u|)u · u′

|u|)]

= 0.

It is well-known that the functional I does not satisfy the Palais-Smale condi-

tion, that is, a Palais-Smale sequence need not have a convergent subsequence. A

Palais-Smale sequence is a sequence (un) ⊂ W 1,2(RN ) with I ′(un) → 0 and (I(un))

convergent. The proposition below states that a Palais-Smale sequence “splits” into

the sum of a critical point of I and translates of critical points of I0:

14

Proposition 3.24 If (un) ⊂ W 1,2(R,RN ) with I ′(un) → 0 and I(un) → a > 0,

then there exist k ≥ 0, v0, v1, . . . , vk ∈ W 1,2(R,RN ), and sequences (tim)1≤i≤km≥1 ⊂ R,

such that

I ′(v0) = 0(i)

I ′0(vi) = 0 for all i = 1, . . . , k(ii)

and along a subsequence (also denoted (un))

‖un − (v0 +k∑

i=1

τtinvi)‖ → 0 as n →∞(iii)

|tin| → ∞ as n →∞ for all i = 1, . . . , k(iv)

ti+1n − tin →∞ as n →∞ for all i = 1, . . . , k − 1(v)

I(v0) + kc0 = a(vi)

A proof for the case of t-periodic W is found in [CR], and essentially the same

proof works here. Similar propositions for nonperiodic coefficient functions, for both

ODE and PDE, are found in [CMN], [AM], and [S2], for example. All are inspired

by the “concentration-compactness” theorems of P. -L. Lions ([L]).

From now on, assume

I has no critical values in the interval (0, c0]. (3.25)

Define the (continuous) “location” functional L : L2(R,RN ) \ {0} → R by∫ ∞

−∞|u|2 tan−1(t− L(u)) dt = 0 (3.26)

Roughly, L tells where along the real line a nonzero function is concentrated. If u

is even, then L(u) = 0. Define

(3.28) c = inf{I(u) | u ∈ ∂B, L(u) = 0}.

Under the assumption (3.25), we claim:

c > c0. (3.28)

Proof: If c < c0, then there exists u ∈ ∂B with I(u) < c0. By arguments of

[CMN], the sequence (η(n, u)) is a Palais-Smale sequence. By Proposition 3.24,

15

(η(n, u)) converges along a subsequence to a critical point v of I with 0 < I(u) < c0,

contradicting (3.25). Next, suppose c = c0. Then there exists (un) ⊂ ∂B with

L(un) = 0 and I(un) → c. (un) is bounded, by Lemma 3.12(iv). If, along a

subsequence, ‖I ′(un)‖ > p > 0, then by arguments of [CMN], since I ′ is Lipschitz

on bounded subsets of W 1,2(R,RN ), I(η(1, un)) < c = c0 for large enough n. Then,

like above, for large n, (η(m,un))m≥1 is a Palais-Smale sequence converging to a

critical point v of I with 0 < I(v) < c0, contrary to (3.25). Thus ‖I ′(un)‖ → 0.

Since L(un) = 0 for all n, Proposition 3.24 implies that (un) converges strongly to a

critical point v of I with I(v) = c0. This contradicts assumption (3.25). Claim (3.28)

is proven.

4. The Minimax Argument: Interacting Tails

We are almost ready to define G, from (1.16). First we need to define R1

precisely. Let r1 > 0 be small enough so

0 ≤ q ≤ r1 ⇒ V (q) ≤ 118

q2. (4.1)

Let R0 > M where M is from Proposition 2.1, and big enough so that

γ(θ)(t) < r1 (4.2)

for all θ ∈ [0, 1] and |t| ≥ R0. Let R1 > R0 and let R1 be large enough so that for

all u ∈ RN ,

|t| ≥ R1 ⇒ W (t, u) ≤ 3V (u). (4.3)

This is possible by (W4). By Proposition 2.1(iii),

inf{γ(θ)(t) | J(γ(θ)) ≥ c0/2, |t| ≤ R0} > 0. (4.4)

Let R1 be big enough so that

sup{γ(θ)(t) | θ ∈ [0, 1], |t| ≥ R1} < inf{γ(θ)(t) | J(γ(θ)) ≥ c0/2, |t| ≤ R0}. (4.5)

By (W6), 2/δ < 1− 2/µ, so we may choose ε > 0 and d with

ε <12,

2δ

< d < 1− 2µ(1− ε)

. (4.6)

16

Let C be large enough so that for all θ ∈ [0, 1] and t ∈ R,

γ(θ)(t) ≤ Ce−(1−ε)|t|. (4.7)

This is possible by Proposition 2.1(v) and (vi), and since ω satisfies −ω′′+ω = V ′(ω)

with V ′(q) = o(q) as q → 0.

Let B be large enough so that for all θ ∈ [0, 1] and t ∈ R,

V (γ(θ)(t)) ≤ BV (γ(θ)(t))µ. (4.8)

This is possible by (V3).

Let l > 0 be small enough so that for all θ ∈ [0, 1],

J(γ(θ)) ≥ c0/2 ⇒ γ(θ)(t) > le−t for all |t| > R0. (4.9)

This is possible by Proposition 2.1(vii), since ω satisfies ω′′ = ω − V (ω).

Now δd > 2 and µ(1− d)(1− ε) > 2, so we may choose R1 to be large enough

so13l2e−2R1 > 2ABCµ(e−δdR1 + e−µ(1−d)(1−ε)R1). (4.10)

Finally, assume that R1 is large enough that for all θ1, θ2 ∈ [0, 1],

I(max(τ−R1γ(θ1), τR1γ(θ2))e1) < J(γ(θ1)) + J(γ(θ2)) + min(c0

4,c− c0

2). (4.11)

Now let G, m, and G0 be defined as in (1.6)-(1.8). We will prove:

Proposition 4.12

(i) m > c0

(ii) m > supG∈G max(x,y)∈∂[0,1]2 I(G(x, y))

(iii) m < 2c0

Then, by standard deformation arguments as in [R], there exists a Palais-Smale

sequence (un) ⊂ W 1,2(R,RN ) with I ′(un) → 0 and I(un) → m. By Proposi-

tion 3.24, (un) has a subsequence converging to v, a critical point of I with I(v) = m,

and Theorem 1.2 follows.

To prove Proposition 4.12(i), let G ∈ G and suppose that g is an arbitrary path

from the bottom to the top of [0, 1]2. That is, g ∈ C([0, 1], [0, 1]2) with π2(g(0)) = 0

17

and π2(g(1)) = 1, where π2 denotes projection onto the second coordinate. Define

γg ∈ Γ by γg(θ) = G(g(θ)). Since γg ∈ Γ, γg(θ) ∈ ∂B for some θ ∈ [0, 1].

Since g is an arbitrary path from the bottom to the top of [0, 1]2, there exists

a connected set D ⊂ [0, 1]2 with (x, y) ∈ D ⇒ G(x, y) ∈ ∂B, connecting the left and

right sides of [0, 1]2. That is, D is a connected subset of [0, 1]2 with (0, y0), (1, y1) ∈ D

for some y0, y1 ∈ [0, 1], and (x, y) ∈ D ⇒ G(x, y) ∈ ∂B. G(0, y0) = τ−R1γ(y0) e1

with y0 > 0, and γ(y0) is a nonzero even function of t, so L(G(0, y0)) = −R1. Like-

wise, L(G(1, y1)) = R1. Since L is continuous and D is connected, L(G(x∗, y∗)) = 0

for some (x∗, y∗) ∈ D. I(G(x∗, y∗)) ≥ c by the definition of c. Since G is an arbitrary

element of G, m ≥ c > c0.

To prove Proposition 4.12(ii), note that for any G ∈ G and y ∈ [0, 1],

I(G(0, y)) = I(τ−R1γ(y)e1) = I(max(τ−R1γ(y), 0)e1) ≤ (4.13)

≤ J(γ(y)) + 0 + (c− c0)/2

≤ c0 + 0 + (c− c0)/2 =c0 + c

2< c.

Similarly, I(G(1, y)) ≤ (c0 + c)/2. For all x ∈ [0, 1], I(G(x, 0)) = I(0) = 0, and since

either x or 1− x is ≥ 1/2, (4.11) gives

I(G(x, 1)) = I(max(τ−R1γ(1− x), τR1γ(x))e1) ≤ (4.14)

≤ J(γ(1− x)) + J(γ(x)) +c0

4≤ c0 − 2c0 +

c0

4< 0.

Finally, we must prove Proposition 4.12(iii). Let G0 ∈ G be defined as in (1.18).

First, suppose that

J(γ((1− x)y)) ≤ c0/2 or J(γ(xy)) ≤ c0/2. (4.15)

Then by (4.11),

I(G0(x, y)) ≤ J(γ((1− x)y)) + J(γ(xy)) + c0/4 ≤ c0 + c0/2 + c0/4 < 2c0. (4.16)

So suppose from now on that

J(γ((1− x)y)) > c0/2 and J(γ(xy)) > c0/2. (4.17)

18

For ease of notation, let u1 = γ((1 − x)y) and u2 = γ(xy). We must show

2c0 − I(G0(x, y)) is positive:

2c0 − I(G0(x, y)) = 2c0 − I(max(τ−R1u1, τR1u2)e1) ≥ (4.18)

≥ J(τ−R1u1) + J(τR1u2)− I(max(τ−R1u1, τR1u2)e1) =(J(τ−R1u1)− I(τ−R1u1e1)

)+(

J(τR1u2)− I(τR1u2e1))

+(I(τ−R1u1e1) + I(τR1u2e1)− I(max(τ−R1u1, τR1u2)e1)

)≡

X1 + X2 + Y.

We will show X1+X2+Y > 0. By (W6) with W (t, u)−V (u) ≥ −AV (|u|)e−δ|t|,(4.8),

and (4.7),

X2 =∫ ∞

−∞W (t, τR1u2e1)− V (τR1u2) dt ≥ (4.19)

≥ −A

∫ ∞

−∞V (τR1u2)e−δ|t| dt ≥ −AB

∫ ∞

−∞(τR1u2)µe−δ|t| dt =

= −AB

∫ ∞

−∞u2(t−R1)µe−δ|t| dt ≥ −ABCµ

∫ ∞

−∞e−µ(1−ε)|t−R1| dte−δ|t| dt.

A similar integral is obtained if instead W (t, u) − V (u) ≥ −A|u|µe−δ|t| in (W6).

Estimating the last integral,∫ ∞

−∞e−µ(1−ε)|t−R1|e−δ|t| dt ≤

∫ dR1

−∞e−µ(1−ε)|t−R1| dt +

∫ ∞

dR1

e−δ|t| dt = (4.20)

=1

µ(1− ε)e−µ(1−ε)(1−d)R1 +

1δe−δdR1 <

< e−µ(1−ε)(1−d)R1 + e−δdR1 ,

so

X2 ≥ −ABCµ(e−µ(1−ε)(1−d)R1 + e−δdR1). (4.21)

Similarly,

X1 ≥ −ABCµ(e−µ(1−ε)(1−d)R1 + e−δdR1). (4.22)

To estimate Y , we must work with the maximum of the functions τ−R1u1 and

τR1u2. First we establish the following claim.

There exists t∗ ∈ (−(R1 −R0), R1 −R0) such that (4.23)

τ−R1u1 ≥ τR1u2 on (−∞, t∗),

τ−R1u1(t∗) = τR1u2(t∗), and

τR1u2 ≥ τ−R1u1 on (t∗,∞).

19

Proof of claim: Recall that u1 = γ(xy), u2 = γ((1 − x)y) where

I(γ(xy)) ≥ c0/2, I(γ((1− x)y)) ≥ c0/2. To prove Claim (4.23), it suffices to prove

τ−R1u1 > τR1u2 on (∞,−(R1 −R0]), (4.24)

τR1u2 > τ−R1u1 on [R1 −R0,∞),

τ−R1u1 is nonincreasing on [−(R1 −R0), R1 −R0], and

τR1u2 is nondecreasing on [−(R1 −R0), R1 −R0].

u1 is nonincreasing on [0,∞) by Proposition 2.1(vi), so τ−R1u1 is nonincreasing

on [−R1,∞), an interval that includes [−(R1 − R0), R1 − R0]. Likewise, τR1u2 is

nondecreasing on [−(R1 −R0), R1 −R0].

Let t ∈ [R1−R0, R1+R0]. Then t−R1 ∈ [−R0, R0] and t+R1 > 2R1−R0 > R1,

so by (4.5),

τ−R1u1(t) = u1(t + R1) < u2(t−R1) = τR1u2(t). (4.25)

So τR1u2 > τ−R1u1 on [R1 −R0, R1 + R0].

On [R1 + R0,∞), τ−R1u1 and τR1u2 both equal right-hand “tails” of ω, with

τ−R1u1(R1 + R0) < τR1u2(R1 + R0). Since ω is decreasing on the positive reals,

τR1u2 > τ−R1u1 on [R1 +R0,∞). More precisely, there exist t1, t2 ≥ 0 such that for

all t ≥ R1+R0, τR1u2(t) = ω(t−(R1+R0)+t2) and τ−R1u1(t) = ω(t−(R1+R0)+t1).

Since τR1u2(R1 + R0) > τ−R1u1(R1 + R0), t2 > t1, and for all t ≥ R1 + R0,

τR1u2(t) = ω(t− (R1 + R0) + t2) > ω(t− (R1 + R0) + t1) = τ−R1u1(t). (4.26)

So τR1u2 > τ−R1u1 on [R1 + R0,∞). Similarly, τR1u2 > τ−R1u1 on

(−∞,−(R1 −R0)]. (4.23) and Claim (4.22) are proven.

Now I is defined by I(u) =∫∞−∞

12 |u

′(t)|2 + 12 |u(t)|2 −W (t, u(t)) dt.

max(τ−R1u1, τR1u2) agrees with τ−R1u1 on (−∞, t∗) and with τR1u2 on (t∗,∞).

Therefore,

Y = I(τ−R1u1e1) + I(τR1u2e1)− I(max(τ−R1u1, τR1u2)e1) = (4.27)

=∫ t∗

−∞

12τR1u

′2(t)

2 +12τR1u2(t)2 −W (t, τR1u2(t)) dt+

+∫ ∞

t∗

12τ−R1u

′1(t)

2 +12τ−R1u1(t)2 −W (t, τ−R1u1(t)) dt.

20

Both integrals are nonnegative: |t∗| < R1 − R0, so for all t ≥ t∗, t + R1 > R0, and

by (4.1)-(4.3), u1(t + R1) < r1, W (t, u1(t + R1)) ≤ 3V (u1(t + R1)) ≤ u1(t + R1)2/6,

and the integrand of the second integral is nonnegative. Similarly, the first integral

is nonnegative.

If t∗ < 0, then R0 ≤ t∗ + R1 ≤ R1, and we can estimate the second integral in

(4.27):∫ ∞

t∗

12τ−R1u

′1(t)

2 +12τ−R1u1(t)2 −W (t, τ−R1u1(t)) dt ≥ (4.28)

≥ 13

∫ ∞

t∗u′1(t + R1)2 + u1(t + R1)2 dt =

13

∫ ∞

t∗+R1

u′1(t)2 + u1(t)2 dt ≥

≥ 13‖u1‖2L∞(t∗+R1,∞) ≥

13u1(t∗ + R1)2 ≥

13l2e−2|t∗+R1| ≥ 1

3l2e−2R1

by (4.9). If t∗ > 0, then we can estimate the first integral in (4.27), similarly:∫ t∗

−∞

12τR1u

′2(t)

2 +12τR1u2(t)2 −W (t, τR1u2(t)) dt ≥ 1

3l2e−2R1 . (4.29)

Thus

Y ≥ 13l2e−2R1 . (4.30)

Putting together (4.18), (4.21)-(4.22), and (4.30), it follows that 2c0 > I(G0(x, y)),

Proposition 4.12(iii) is proven, and Theorem 1.2 follows.

Open Questions

Many open questions remain. For example, is the constant 2µ/(µ − 2) in

(W6) optimal? Proving so would require a counterexample. Can the Ambrosetti-

Rabinowitz condition (W3) be weakened? Does a PDE version of Theorem 1.2 hold,

with or without the monotonicity condition (1.10)?

21

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23