SPATIAL PATTERNS DESCRIBED BY THE EXTENDEDmath.cmu.edu/~tblass/CNA-PIRE/Peletier-Troy1997.pdfSPATIAL PATTERNS DESCRIBED BY THE EXTENDED FISHER{KOLMOGOROV EQUATION: PERIODIC SOLUTIONS

SPATIAL PATTERNS DESCRIBED BY THE EXTENDEDFISHER–KOLMOGOROV EQUATION: PERIODIC SOLUTIONS∗

L. A. PELETIER† AND W. C. TROY‡

SIAM J. MATH. ANAL. c© 1997 Society for Industrial and Applied MathematicsVol. 28, No. 6, pp. 1317–1353, November 1997 004

Abstract. Stationary antisymmetric single-bump periodic solutions of a fourth-order general-ization of the Fisher–Kolmogorov (FK) equation are analyzed. The coefficient γ > 0 of the additionalfourth-order spatial derivative is found to be a critical parameter. If γ ≤ 1

8, the family of periodic so-

lutions is still very similar to that of the FK equation. However, if γ > 18, it is possible to distinguish

different families of periodic solutions and the structure of such solutions is much richer.

Key words. differential equations, nonlinear, periodic solutions, phase transitions

AMS subject classifications. 34C15, 34C25, 35Q35

PII. S0036141095280955

1. Introduction. In this paper we shall study the formation of spatially periodicpatterns in bistable systems described by the extended Fisher–Kolmogorov (EFK)equation

(1.1)∂u

∂t= −γ ∂

4u

∂x4+∂2u

∂x2+ u− u3, γ > 0.

This equation was proposed in 1987 by Coullet, Elphick, and Repaux [8] and in 1988by Dee and van Saarloos [11] as a generalization of the classical Fisher–Kolmogorov(FK) equation

(1.2)∂u

∂t=

∂2u

∂x2+ u− u3,

which has played an important role in the studies of pattern formation in bistablesystems (cf. [3, 13, 14, 16, 20, 25, 26]). The term “bistable” refers here to the factthat the uniform states u = ±1 are stable as solutions of the equation

du

dt= u− u3.

The EFK equation arises in the study of singular points (so-called Lifshitz points[14]) in phase transitions and as the evolution equation in gradient systems describedby the energy functional

(1.3) I(u) =

∫ {γ2(u′′)2 +

β

2(u′)2 + F (u)

}dx, γ > 0, β ∈ R,

where F denotes the double-well potential

(1.4) F (u) =1

4(1− u2)2.

∗Received by the editors February 3, 1995; accepted for publication (in revised form) August 15,1996. This research was supported by NSF grant DMS9002028.

http://www.siam.org/journals/sima/28-6/28095.html†Mathematical Institute, Leiden University, Leiden, The Netherlands ([email protected]).‡Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260 ([email protected].

pitt.edu).

1317

1318 L. A. PELETIER AND W. C. TROY

We then obtain the EFK equation when we choose β = 1. Another important appli-cation of this equation is found in the theory of instabilities in nematic liquid crystals[5, 27].

In studies of second-order materials [10, 17, 18] one also finds the functional I(u).Here β < 0. The stationary points of I(u) are then equivalent to the equilibriumsolutions of the Swift–Hohenberg equation

(1.5)∂u

∂t= −

(1 +

∂2

∂x2

)2

u+ αu− u3, α > 0

when α > 1 through a simple scaling of x, t, and u (see, for instance, [7, 9] and thereferences therein).

In this paper we are interested in stationary spatial patterns which can be de-scribed by the EFK equation and in particular in periodic patterns. Thus, we areconcerned with bounded solutions u(x) of the equation

(1.6) γuiv = u′′ + u− u3 on R.

Our main objectives are to determine the effect of the added higher-order gradientterm and the value of the coefficient γ, on the class of possible stationary periodicpatterns, and the qualitative properties of these patterns.

For convenience we recall below the types of stationary patterns that can bedescribed by the FK equation. However, before doing so we introduce some notation.We observe that equation (1.6) has a constant of integration: if u is a solution ofequation (1.6), then

(1.7) E(u)def= 2γu′u′′′ − γ(u′′)2 − (u′)2 +

1

2(1− u2)2 = constant

def=

µ

2.

To eliminate arbitrary shifts, we always place the origin at a zero of u:

(1.8) u(0) = 0

whenever a zero exists.For the special value γ = 0, equation (1.6) reduces to the stationary FK equation

(1.9) u′′ + u− u3 = 0.

Proposition 1.1. The FK equation (γ = 0) has the following bounded stationarysolutions.

(a) There exists a unique solution u of equation (1.9) (a kink), such that

(u, u′) → (±1, 0) as x→ ±∞.

It is antisymmetric and monotone and is given explicitly by

u(x) = tanh( x√

2

).

This solution corresponds to the values γ = 0 and µ = 0 in (1.7).(b) For each µ ∈ (0, 1), there exists a unique periodic solution u of (1.9) such that

u′(0) > 0. It is (i) antisymmetric with respect to its zeros, (ii) symmetric with respect

EXTENDED FISHER–KOLMOGOROV EQUATION 1319

to the location of its maxima and its minima, (iii) concave where it is positive andconvex where it is negative, and

(iv) max{|u(x)| : x ∈ R} =√

1−√µ.

(c) There exist no bounded periodic solutions of (1.9) when µ /∈ (0, 1) and nobounded solutions of (1.9) when µ /∈ [0, 1].

In two earlier papers [21, 24] we investigated the existence and properties ofkinks of the EFK equation. They are solutions u(x) of equation (1.6) which have theproperties

(1.10) (u, u′, u′′, u′′′) → (±1, 0, 0, 0) as x→ ±∞.

We found that kinks exist for every γ > 0 but that their character and number changeabruptly at γ = 1

8 . When γ ≤ 18 , there exists a unique odd monotone kink with the

same characteristic properties as the kink of the FK equation. However, we haverecently shown that for γ > 1

8 , there exists a countably infinite number of odd kinks,and none of these is monotone in its approach to ±1 as x → ±∞ [22]. We collectthese results in the following proposition.

Proposition 1.2. (a) For each γ > 0, there exists an odd solution of equation(1.6) which satisfies (1.10).

(b) If γ ≤ 18 , there exists one and only one kink which is odd and monotone.

(c) If γ > 18 , there exists for every integer n ≥ 0 a kink with 2n+ 1 zeros.

The critical value γ = 18 is related to the linearization of (1.6) around u = 1 or

u = −1. For γ ≤ 18 the corresponding eigenvalues are all real, whilst for γ > 1

8 theyare all complex.

In the present paper we investigate stationary periodic solutions of the EFK equa-tion and we shall inquire how parts (b) and (c) of Proposition 1.1 generalize whenwe take γ > 0. We shall restrict this study to periodic solutions u which (i) areodd, (ii) are symmetric with respect to the location of their extrema, and (iii) have asingle relative maximum between consecutive zeros. Thus, let ζ be the first positivezero of u′. Then we are concerned with periodic solutions of (1.6) with the followingproperties:

(1.11)u(−x) = −u(x) for x ∈ R,

u(ζ − y) = u(ζ + y) for y ∈ R.

We refer to solutions which satisfy (1.11) as single-bump periodic solutions. In thispaper we shall only discuss such periodic solutions.

As we do with the FK equation we only consider solutions for which

E(u) ≥ 0 or µ ≥ 0.

We conjecture that if γ > 18 , there exist periodic solutions for some negative values of

µ as well. However, we leave their analyses to a further study.In our first result, we find that when 0 < µ < 1 the periodic solutions of the FK

equation continue to exist for all γ > 0.THEOREM A. Let 0 < µ < 1 and γ > 0. Then there exists a periodic solution

u(x) of (1.6) such that

max{|u(x)| : x ∈ R} < 1.


When µ ≥ 1 the FK equation has no periodic solutions and we see below thatthis also continues to be true for all γ > 0.

THEOREM B. Let µ ≥ 1 and γ > 0. Then there exists no periodic solution ofequation (1.6).

When µ = 0 and γ > 18 the situation becomes very different from that of the FK

equation. Whereas the FK equation has no periodic solutions for this value of µ, theEFK equation has two branches of periodic solutions bifurcating from the unique oddmonotone kink U(x) at γ = 1

8 . This is the content of the following two theorems.THEOREM C. Let µ = 0.(a) If 0 < γ ≤ 1

8 , then there exist no periodic solutions.(b) If γ > 1

8 , then there exist a periodic solution u1(x) such that

max{|u1(x)| : x ∈ R} < 1

and a periodic solution u2(x) such that

max{|u2(x)| : x ∈ R} ∈ (1,√

2).

THEOREM D. Let µ = 0 and let {γi} be a sequence such that

γi ↘ 1

8as i→∞.

For each i ≥ 1, let ui be a periodic solution corresponding to γi. Then

ui(x) → U(x) as i→∞,

where U is the unique odd monotone kink corresponding to γ = 18 . The convergence

is uniform on compact intervals.We conjecture that in addition to the single-bump periodic solutions of Theorem

C, infinitely many multibump periodic solutions bifurcate from U at γ = 18 .

A result which is analogous to Theorem D holds for periodic solutions when0 < µ < 1 and γ → 0 (cf. Lemma 6.3). They converge to the periodic solution of theFK equation for the given value of µ.

Periodic solutions with amplitude larger than 1 continue to exist when γ > 18 and

µ > 0 is sufficiently small. However, if either γ ≤ 18 or µ ≥ 4

9 , they cannot exist.THEOREM E. Let µ ≥ 0 and γ > 0. There exist no periodic solutions u(x) such

that

max{|u(x)| : x ∈ R} > 1

when one of the following conditions is satisfied:(a) 0 < γ ≤ 1

8 ,(b) µ ≥ 4

9 .For the global behavior of the branches of periodic solutions we obtain several

bounds. We begin with a universal upper bound.THEOREM F. Let 0 ≤ µ < 1 and γ > 0. Then any periodic solution satisfies

|u(x, γ)| <√

2 for x ∈ R.

We also prove the following lower bound.


THEOREM G. Let 0 ≤ µ < 1 and γ > (25 )4. Then any periodic solution satisfies

max{|u(x, γ)| : x ∈ R} > 1

50

√1− µ

log 2.

Let u be a periodic solution which satisfies (1.11). Then for its slope u′(0) at theorigin we prove an upper and a lower bound:

(1.12)1

5

√1− µ < γ1/4u′(0, γ) < {8(1− µ) log 2}1/4

for γ > 18 if µ = 0 and for γ > (2

5 )4 if 0 < µ < 1. These bounds enable us toobtain information about the behavior of periodic solutions for large values of γ. Thedescription of their limiting behavior involves the reduced problem

(1.13a)

(1.13b)

(1.13c)

viv = v − v3,

v(0) = 0, v′′(0) = 0,

v′(0) = ω, v′′′(0) = −1− µ

4ω

in which ω is a positive number.THEOREM H. Let 0 ≤ µ < 1. Suppose that {γi} is a sequence which tends to

infinity and {ui} is a sequence of periodic solutions which satisfy (1.11). Then thereexist a subsequence, which we also denote by {γi}, and a periodic solution V of problem(1.13) such that

ui(γ1/4i s, γi) → V (s) as i→∞

uniformly on compact sets.Outline of the shooting method. We now give a brief description of the topological

shooting method used to prove the existence of the families of odd periodic solutionsdescribed in Theorem A and Theorem C(b).

Since we are looking for odd solutions of (1.6), it is sufficient to consider theequation on R+ only and with initial conditions of the form

(1.14)(u(0), u′(0), u′′(0), u′′′(0)

)= (0, α, 0, β),

where α and β are real. It is easily verified that a solution u of (1.6) satisfying (1.14)must be odd, and that −u is also a solution. We shall see that α cannot be zero, andso we may choose α to be positive. It follows from (1.7) that

(1.15) β =1

2αγ

{α2 − 1− µ

2

}.

In Theorem A we assume that γ and µ are fixed, with γ > 0 and 0 < µ < 1. Weare then free to vary α, our shooting parameter. We begin our analysis by showingthat for each α > 0 there exists a finite value ξ(α) > 0, which depends continuouslyon α, such that

(1.16) u′(x, α) > 0 for 0 < x < ξ(α) and u′(ξ(α), α) = 0.

We shall show that u′′′(ξ(α), α) < 0 for small values of α > 0, that there is an α > 0for which u′′′(ξ(α), α) > 0, and that

(1.17) 0 < u(ξ(α), α) <√

1−√µ for 0 < α < α.


From these observations and the continuity of ξ(α) with respect to α we conclude thatthere is an intermediate value α− ∈ (0, α) such that u′′′(ξ(α−), α−) = 0. At α = α−we have

0 < u(ξ(α−), α−) <√

1−√µ, u′(ξ(α−), α−) = 0, and u′′′(ξ(α−), α−) = 0,

and it follows from (1.7) and the definition of ξ(α−) that u′′(ξ(α−), α−) < 0. Byreflecting the graph {(x, u(x, α−)) : 0 ≤ x ≤ ξ(α−)} with respect to the endpointsx = 0 and x = ξ(α−), we obtain the desired periodic solution. The details of theanalysis described above are given in sections 2 and 3.

In section 5 we turn to the proof of Theorem C(b) in which we set µ = 0 andrestrict γ to satisfy γ > 1

8 . For γ ∈ (0, 18 ], part (a) of Theorem C states that no

periodic solutions exist, and this result is proved in section 4. However, as γ passesthrough 1

8 from below, a linearization of (1.6) around the constant solution u = 1shows that the eigenvalues change from real to complex, two of the eigenvalues havepositive real part and two have negative real part. Because of this change in characterof the eigenvalues, Coullet, Elphick, and Repaux [8] conjectured that the range γ > 1

8is where one would expect to observe complex pattern formation. Such patternscould include families of periodic and aperiodic solutions, multibump heteroclinic andhomoclinic orbits (kinks, respectively, solitons), and possibly chaos.

One of the main goals of our investigation of the EFK equation is to resolve thisconjecture of Coullet, Elphick, and Repaux and to determine the different types ofsolutions that exist for the parameter regime µ = 0 and γ > 1

8 . In Theorem C(b) westate our first existence result for this range of parameters. We prove that there areat least two families of periodic solutions; one of these is characterized by the factthat the relative maxima all lie below u = 1, while the relative maxima of the secondfamily all lie above u = 1. In addition we show that both families of periodic solutionsbifurcate from the odd, monotone kink (the heteroclinic orbit connecting u = −1 andu = +1) at γ = 1

8 and continue to exist for all γ > 18 .

The proof of Theorem C(b) is based on the same shooting technique we usedfor the proof of Theorem A. Again, for each α > 0 we find that there exists a firstξ(α) ∈ R+ for which u′(ξ(α), α) = 0. In the proof of Theorem A we found thatu(ξ(α), α) 6= 1 because µ 6= 0. However, now we have µ = 0 and indeed we find thatthere is a critical value α for which u(ξ(α), α) = 1. It follows from (1.6), (1.7), anduniqueness that

(1.18) u(ξ(α), α) = 1, u′(ξ(α), α) = 0, u′′(ξ(α), α) = 0, and u′′′(ξ(α), α) > 0.

To proceed with our shooting argument we need to know that ξ(α) is continuous.When u′′ 6= 0 this is an easy consequence of the implicit function theorem, but whenu′′ = 0, as it is at ξ(α), this is no longer obvious and the situation is much moredelicate. In Lemma 5.8 we further refine a method originally developed in [21], whichuses u as an independent variable, to prove this important property.

Having established continuity of ξ(α) we conclude that there exists an α− ∈ (0, α)for which u′′′(ξ(α−), α−) = 0. As before, it follows that u(·, α−) is a periodic solutionand by construction its relative maxima lie below u = 1.

For our second periodic solution, whose relative maxima are all greater than 1,we show that for α > α sufficiently large,

u(ξ(α), α) > 1 and u′′′(ξ(α), α) < 0.


Remembering that u′′′(ξ(α), α) > 0, we conclude from the continuity of ξ(α) thatthere exists an α+ > α for which

u(ξ(α+), α+) > 1, u′(ξ(α+), α+) = 0, and u′′′(ξ(α+), α+) = 0.

Again, we conclude that u(·, α+) is a periodic solution, but now the constructionensures that its relative maxima lie above u = 1.

We emphasize that the key result in the proof of the existence of both familiesof periodic orbits is Lemma 5.8, which implies that ξ(α) is continuous for all α > 0if µ = 0 and γ > 1

8 . We have recently extended this property [23] and provedthat all subsequent relative maxima ξi(α) and minima ηi(α), i = 1, 2, . . . , have thesame continuity property. In turn this has allowed us to further refine our shootingmethod to establish the existence of complicated types of solutions, such as multibumpheteroclinic orbits [22] and chaotic patterns [23]. The topological shooting argumentdeveloped here and extended in [22] and [23] enables one to prove chaos withouthaving to verify the typical transversality condition (see, for instance, [12]) requiredby the dynamical systems approach. This verification can be very difficult.

Preliminary results suggest that the method developed here presents a frameworkwhich can be used to investigate stationary spatial patterns described by a large classof model equations, such as the Swift–Hohenberg equation (1.5), and equations de-scribing soliton solutions in nonlinear optical fibers [1], traveling waves in a suspensionbridge [19], and the deflections of an asymmetrically supported strutt:

(1.19) uiv + Pu′′ + u− u2 = 0, P ∈ R.

In a series of papers [2, 6, 4] this equation has been studied by completely different(Hamiltonian) methods. Although the nonlinearity in this equation is not cubic,and the emphasis here lies on homoclinic orbits, it is interesting to compare the twodifferent methods and the different types of results they generate. Finally, we notethat in [15] a variational approach in combination with a partition of function spacesinto topological subclasses has proved to be successful in analyzing equations such as(1.6) and (1.9).

2. Preliminaries. Our basic method for proving existence of odd periodic so-lutions is a shooting technique, so we consider the initial value problem

(2.1a)

(2.1b)

{γuiv = u′′ + u− u3, x > 0,

u(0) = 0, u′(0) = α, u′′(0) = 0, u′′′(0) = β.

We may restrict our attention to α > 0; if α = 0, then (1.7) implies that µ = 1 and weshall show in Lemma 2.2 that in this case no periodic solution which satisfies (1.11)can exist. Thus, if we fix µ ≥ 0 and γ > 0, then (1.7) yields

(2.2) β = β(α) =1

2γα

{α2 − 1− µ

2

}.

We seek a positive value of α such that the solution u(x, α) has the properties(1.11). That is,

u′(x, α) > 0 for 0 ≤ x < ξ,(2.3a)

u′(ξ, α) = 0 and u′′′(ξ, α) = 0(2.3b)


for some finite ξ = ξ(α) > 0. It is easily verified that a solution defined on [0, ξ],which satisfies (2.1)–(2.3), can be extended to yield a periodic solution of period 4ξ.Thus, we define

ξ(α) = sup{x > 0 : u′(·, α) > 0 on [0, x)}.In this section we will show that for all values α > 0, except those for which thecorresponding solution u is a monotone kink, ξ(α) is finite and that u is bounded on[0, ξ] with horizontal slope at ξ. Therefore, to satisfy (2.3) we must still determineα > 0 so that

u′′′(ξ(α), α) = 0.

At times we shall find it convenient to adopt a different formulation for the initialvalue problem (2.1). Since we construct periodic solutions from strictly monotonesegments defined on [0, ξ], we may introduce u as an independent variable, as wasdone in [21] for the study of kinks. Denoting the inverse function of u(x) by x(u), weset

(2.4) t = u and z(t) = (u′)2(x(t)).

This yields

(2.5) z′(t) = 2u′′(x) and z′′(t) = 2u′′′(x)

u′(x).

Hence, upon substitution into (1.7), we obtain

(2.6a)

(2.6b)

zz′′ =

(z′)2

4+

1

γ{z − fµ(t)}, t > 0,

z(0) = α2 and z′(0) = 0,

where

(2.7) fµ(t) =1

2{(t2 − 1)2 − µ}.

We denote the solution by z(t, α) and write

τ(α) = sup{t > 0 : z(·, α) > 0 on [0, t)}.From (2.4) and the definition of ξ(α) it follows that

τ(α) = limx→ξ(α)−

u(x, α).

To be assured that z(·, α) corresponds to a periodic solution u(·, α), we need to provethe existence of a positive α for which

(2.8a) 0 < ξ(α) <∞, 0 < τ(α) <∞,

and

(2.8b) limt→τ(α)−

z(t, α) = 0, limt→τ(α)−

√z(t, α)z′′(t, α) = 0.


Then (2.8), together with (2.4) and (2.5), implies that u(·, α) satisfies (2.3) so that uis periodic.

Lemma 2.1. Suppose that µ ≥ 0 and γ > 0.(a) For any α ∈ R+, we have

(2.9) u(ξ(α), α) <∞ and u′(ξ(α), α) = 0.

(b) If µ > 0, then

ξ(α) <∞ for any α ∈ R+.

(c) If µ = 0, then

ξ(α) <∞

for any α ∈ R+ if γ >1

8,

for any α ∈ R+ \ {α0} if 0 < γ ≤ 1

8.

Here α0 = U ′(0), and U is the unique odd monotone kink found in [21].Proof. (a) We write (2.6a) as

(2.10) (z3/4)′′ =3

4γ

z − f

z5/4,

where we have suppressed the subscript µ from f , and we define

τ0 = sup{t ∈ (0, τ) : z′ < 0 on (0, t)},if z′ < 0 in a right-neighborhood of the origin, and τ0 = 0 otherwise.

We distinguish two cases:

(i) τ0 = τ and (ii) τ0 < τ.

(i) In this case, z(t) < α2 for 0 < t < τ . Suppose that τ = ∞ (i.e., u(ξ(α), α) =∞). Then, since f(t) ∼ 1

2 t4 as t → ∞, there exists a T > 0 such that (2.10) reduces

to

(z3/4)′′ < −1 for t > T,

which implies that τ <∞, a contradiction. Thus, it must be the case that u(ξ(α), α) <∞. If limt→τ− z(t) > 0, then standard theory applied to (2.6) shows that z contin-ues to exist, with z > 0 on an interval [τ, τ + ε), contradicting the definition of τ .Therefore,

limt→τ−

z(t) = 0.

From this and (2.4) we conclude that u′(ξ(α), α) = 0.(ii) We now consider the case τ0 < τ , and again we suppose that τ = ∞. At

t = τ0 we have

z(τ0) > 0, z′(τ0) = 0, and z′′(τ0) ≥ 0.

We again distinguish two cases:

(ii∗) 0 ≤ τ0 < 1 and (ii∗∗) τ0 ≥ 1.


(ii∗) We claim that

(2.11) z(t) > f(t) and z′(t) > 0 for τ0 < t < τ0 + ε,

where ε is some small positive constant. If z′′(τ0) > 0, this follows immediatelyfrom (2.6a). If z′′(τ0) = 0, differentiation of (2.6a) yields z′′′(τ0) > 0 if τ0 > 0 andz′′′(τ0) = 0 if τ0 = 0. In the latter case one further differentiation of (2.6a) shows thatziv(τ0) > 0. Thus, in all cases (2.11) holds.

This enables us to define

τ1 = sup{t > τ0 : z′ > 0 on (τ0, t)}.

We shall show that

(2.12) τ1 <∞, z′(τ1) = 0, and z′′(τ1) < 0.

Suppose, to the contrary, that τ1 = ∞. Then, since f(t) > 0 for t > τ+(µ) =√1 +

√µ, it follows from (2.10) that

(z3/4)′′ <3

4γz−1/4 for t > τ+

or

(2.13) y′′ <3

4γy−1/3 for t > τ+,

where we have set y = z3/4. If we now multiply (2.13) by 2y′ and integrate over(τ+, t), we find that

y′ <3

2√γ

√y2/3 + C for t > τ+,

where C is a positive constant. Writing w = y2/3 = z1/2, this inequality translatesinto

w′ <1√γ

√w + C

wfor t > τ+.

Thus, since w is increasing, w′ is uniformly bounded on (τ+,∞), so that

(2.14) z(t) < A(1 + t)2 for t > 0

for some positive constant A.Remembering that f(t) ∼ 1

2 t4 as t→∞, it follows from (2.14) that z(t)− f(t) ∼

− 12 t

4 as t→∞. Hence, since z is increasing, there exists a constant K > 0 such that

(z3/4)′′ < −K(1 + t)3/2 for t > t1,

where t1 is some sufficiently large number. Two integrations show that z cannot keepincreasing indefinitely, contradicting our assumption that τ1 = ∞. Thus,

τ1 <∞ and z′(τ1) = 0.


To complete the proof of (2.12), in this case, we shall show that

(2.15) τ1 > 1 and z′′(τ1) < 0.

By (2.11), we can define

τ∗ = sup{t > τ0 : z > f on (τ0, t)}.It then follows from (2.6a) that z′′ > 0 on (τ0, τ

∗]. This implies that τ1 > τ∗. Also,since f ′ < 0 on (0,1), it must be that τ1 > 1.

To prove the second assertion in (2.15), we need to show that z′′(τ1) < 0. Suppose,to the contrary, that z′′(τ1) = 0. Then, since f ′ > 0 on (1,∞), it follows thatz′′′(τ1) < 0, which implies that z′′ > 0 and z′ < 0 on a left-neighborhood of τ1,contradicting the definition of τ1. This completes the proof of (2.15).

It now follows from (2.6a) that z′′ < 0 and z′ < 0 for t > τ1 until z = 0 at somefinite value τ . This contradicts our assumption that τ = ∞. As in case (i) above itfollows that u′(ξ(α), α) = 0.

(ii∗∗) We now assume that τ0 ≥ 1. The definition of τ0 implies that z′′(τ0) ≥ 0.If z′′(τ0) > 0, then z′ > 0 on a right-neighborhood of τ0 and we proceed as in theprevious case. If z′′(τ0) = 0, then z′′′(τ0) < 0 if τ0 > 1, and it follows from (2.10) thatz′′ < 0 and z′ < 0 to the right of τ0 until z = 0 at a finite τ , a contradiction. If τ0 = 1in this case, then z′′′(τ0) = 0 as well, but ziv(τ0) < 0. Again we find that z′′ < 0 andz′ < 0 to the right of τ0 until z = 0 at a finite τ , a contradiction.

This completes the proof of part (a).(b) Suppose, to the contrary, that there exist constants µ > 0, α > 0, and γ > 0

such that ξ(α) = ∞. Then u′(x) > 0 for all x > 0 and by part (a), u is uniformlybounded on R+. Hence,

limx→∞u(x, α) exists

def= `.

We distinguish three cases:

(i) ` > 1, (ii) 0 < ` < 1, and (iii) ` = 1.

(i) Since `(1− `2) < 0 if ` > 1, there exist a point x1 > 0 and a constant M > 0such that

γuiv − u′′ < −M for x > x1.

When we integrate this inequality twice over (x1, x), we obtain in turn

γu′′′(x)− u′(x) < A−Mx,

γu′′(x)− u(x) < Ax+B − 1

2Mx2,

where A and B are appropriate constants. Plainly, it is possible to choose x2 > x1 solarge that

Ax+B <1

4Mx2 for x > x2

and hence,

γu′′(x) < u(x)− 1

4Mx2 for x > x2.


Because u is uniformly bounded, this means that there exists a point x3 > x2 suchthat

γu′′(x) < −1

8Mx2 for x > x3.

One more integration shows that u′(x) → −∞ as x → ∞, which contradicts theassumption that ξ(α) = ∞.

(ii) Since `(1 − `2) > 0 if 0 < ` < 1, there exist a point y1 > 0 and a constantN > 0 such that

γuiv − u′′ > N for x > y1.

Proceeding as in the previous case, we can find a point y2 > y1 such that

γu′′(x) > +1

8Nx2 for x > y2,

and we conclude that u′(x) → ∞ as x → ∞, which contradicts the fact that u isuniformly bounded.

(iii) If ` = 1, we conclude from (1.7) that there exists a point x∗ > 0 such that

(2.16) u′(x)u′′′(x) ≥ µ

8γ> 0 for x > x∗.

Hence, u′′ has one sign on (x∗,∞), so limx→∞ u′(x) = m exists and m ≥ 0. If m > 0,then u(x) →∞ as x→∞, violating the boundedness of u. Hence m = 0, so that by(2.16), u′′′(x) → ∞ as x → ∞. Thus, u′′(x) → ∞, u′(x) → ∞, and u(x) → ∞ asx→∞, contradicting again the boundedness of u.

(c) Suppose, to the contrary, that there exist constants γ > 0 and α > 0 suchthat ξ(α) = ∞. Then u′(x) > 0 for all x > 0 and it follows as in part (b) that

(2.17) limx→∞u(x, α) = 1.

Hence, 0 < u < 1 for all x > 0, and we deduce from the differential equation that

(2.18) γuiv − u′′ > 0 for x > 0.

By the maximum principle, u′′ can have only one sign on R+, and therefore u′ tendsto a limit: limx→∞ u′(x) = `1 ≥ 0. If `1 > 0, the solution will be unbounded on R+,so

(2.19) limx→∞u′(x) = 0.

In view of (2.18), limx→∞(γu′′′ − u′) = `2 exists, and therefore,

limx→∞u′′′(x) =

`2γ.

Reasoning as before we find that

(2.20) limx→∞u′′′(x) = 0.

Thus, since µ = 0, it follows from (2.17)–(2.20) and the energy identity (1.7) that

(2.21) limx→∞(u, u′, u′′, u′′′) = (1, 0, 0, 0) and u′ > 0 on [0,∞).


If γ ∈ (0, 18 ], then (2.21) implies that α = α0, where α0 is the unique positive

value of α for which a monotone, antisymmetric kink U exists. But this value hasbeen excluded in the hypotheses.

If γ > 18 , then a linearization of (2.6a) around u = 1 shows that all four eigenvalues

are complex with nonzero real and imaginary parts. Therefore, u cannot approach u =1 monotonically, as asserted in (2.21). Thus, we have arrived at the final contradictionand Lemma 2.1 is proved.

We conclude this section with a lemma which restricts the admissible values of µand α.

Lemma 2.2. Let γ > 0 and suppose that either

(a) 0 ≤ µ < 1 and α ≥√

1− µ

2or

(b) µ ≥ 1 and α ≥ 0.

Then the solution u(·, α) of problem (2.1), (2.2) cannot satisfy (2.3).Proof. Suppose, to the contrary, that there exist values of µ and α for which

either (a) or (b) holds and which are such that the corresponding solution u(x, α) ofproblem (2.1), (2.2) does satisfy (2.3). We consider the two cases in succession.

(a) Observe that

z′′(0, α) =1

γα2

(α2 − 1− µ

2

).

Thus, if α >√

12 (1− µ), then z′′(0, α) > 0, and if α =

√12 (1− µ), then z′′(0, α) = 0.

However, in the second case z′′′(0, α) > 0. Therefore, in both cases z′′ > 0 on aninterval (0, δ).

It follows by the arguments used in the proof of part (a) of Lemma 2.1 thatz′′(t) > 0 and z′(t) > 0 for all t ∈ (0, τ+(µ)]. In terms of the function u(·, α) thismeans that there exists a point x1 such that

u′ > 0, u′′ > 0, and u′′′ > 0 on (0, x1) and u(x1) > 1.

For condition (2.3b) to hold, there must be a first x2 > x1 such that

u′′(x2) = 0 and u′′′(x2) ≤ 0.

Thus, u′′ > 0 and, hence, also u′ > 0 on (0, x2), so that ξ(α) > x2. By equation(2.1a), we have uiv(x2) < 0, so

u′′′ < 0 and u′′ < 0 on (x2, ξ(α)].

In particular, u′′′(ξ(α), α) < 0, which contradicts (2.3b).(b) In this case, if α > 0 then z′′(0, α) > 0 and we can proceed using the same

arguments we used in part (a) to complete the proof. We omit the details for the sakeof brevity.

If α = 0, then by (1.7) we must have µ = 1, so that u(0) = 0, u′(0) = 0, u′′(0) = 0.If u′′′(0) = 0 as well, then u is the trivial solution, so we must assume that u′′′(0) 6= 0.Without loss of generality we may assume that u′′′(0) > 0. Then u′ > 0 and u′′ > 0in a right-neighborhood of the origin; hence, z(t) > 0 and z′(t) > 0 for small valuesof t. We may now proceed as in part (a) to show that u cannot be a periodic solutionwhich satisfies (2.3). Theorem B is an immediate consequence of part (b) of Lemma2.2.


3. Existence and uniqueness of periodic solutions: 0 < µ < 1, γ > 0.In this section we focus our attention on the parameter range 0 < µ < 1, γ > 0. Weprove Theorem A and a uniqueness theorem for a more restricted range of values ofµ, i.e., µ ∈ (0, 4

9 ].Theorem 3.1. Let µ ∈ (0, 1) and γ > 0. Then there exists a periodic solution

u(x) such that

max{|u(x)| : x ∈ R} < 1.

The proof proceeds via a sequence of lemmas.We define the shooting set

S = {α > 0 : u(ξ(α), α) < 1, u′′(ξ(α), α) < 0, and u′′′(ξ(α), α) < 0 for 0 < α < α}.Lemma 3.2. We have(a) ξ ∈ C1(S).(b) S is an open interval.

(c) u(ξ(α), α) <√

1−√µ if α ∈ S.

Proof. (a) Let α ∈ S. Then, at ξ = ξ(α) we have

u′(ξ(α), α) = 0 and u′′(ξ(α), α) < 0.

Hence, by the implicit function theorem, ξ ∈ C1(S).(b) Since the inequalities in the definition of S are strict, the assertion follows

immediately from part (a) and the continuous dependence of solutions on initial data.Part (c) follows at once from the energy identity (1.7).In the following lemma we show that S is nonempty. Define

α = min

{√1− µ

2,

√3(1− µ)

24γ + 7

}.

Lemma 3.3. (0, α) ⊂ S.Proof. Let α ∈ (0, α). Observe that (1.7) implies that u′′′(0) < 0, since 0 < µ < 1

and α < α. As we increase x and as long as u′′′ < 0, it follows that u′′ < 0, u′ < α,and u(x) < αx. Thus, as long as u ≥ 0 and u′′′ < 0, it follows from (2.1a) that

(3.1a) uiv(x) <α

γx.

We integrate this inequality three times to obtain

u′′′(x) < β +α

2γx2,(3.1b)

u′′(x) < βx+α

6γx3,(3.1c)

u′(x) < α+β

2x2 +

α

24γx4.(3.1d)

Since we assume that α < 12

√1− µ, the energy identity (1.7) implies that

β < −1− µ

8γα.


Hence, the right-hand sides of (3.1b) and (3.1c) are negative for 0 < x ≤ 1 and, sinceα < 1, it follows that u < 1 on [0, 1] as long as u′ ≥ 0.

On the other hand, because we have chosen α <√

3(1−µ)24γ+7 , the right-hand side of

(3.1d) is negative at x = 1. Thus, there must exist a first zero ξ of u′ on (0, 1), whereu < 1, u′′ < 0, and u′′′ < 0, so that α ∈ S.

Define

α∗ = supS.

Lemma 3.4. We have

α∗ ≤√

1− µ

2.

Proof. It follows from (1.7) that

2γu′u′′′ ≥ (u′)2 − 1

2{(1− u2)2 − µ}

> (u′)2 − 1− µ

2for 0 < u < 1.

Thus, if α2 > (1− µ)/2, then u′′′ > 0, u′′ > 0, and u′ > α > 0 as long as 0 < u ≤ 1,so that u′ cannot have a first zero ξ such that u(ξ) < 1.

Lemma 3.5. We have

0 < u(ξ(α∗), α∗) < 1, u′′(ξ(α∗), α∗) < 0, and u′′′(ξ(α∗), α∗) = 0.

Proof. Suppose that u(ξ(α∗), α∗) ≥ 1. Then by the continuous dependence ofu(·, α) on α on compact intervals, it follows that u(ξ(α), α) >

√1−√µ for all α in a

small enough neighborhood of α∗. Since (0, α∗) ⊂ S, this contradicts Lemma 3.2 andwe conclude that

(3.2) u(ξ(α∗), α∗) < 1.

Thus, in the limit as α increases toward α∗, the first inequality in the definitionof S continues to hold, and we wish to prove that the second one continues to hold aswell. Suppose that the second inequality fails. It follows from the definition of ξ thatu′′(ξ(α∗), α∗) ≤ 0. Hence, we suppose that

(3.3) u′′(ξ(α∗), α∗) = 0.

In what follows we shall write ξ∗ = ξ(α∗) and u∗ = u(ξ∗, α∗).To show that (3.3) leads to a contradiction we proceed via a series of steps.Step 1. We show that (3.3) implies that

(3.4) u′′′(ξ∗, α∗) > 0.

Suppose that u′′′(ξ∗, α∗) < 0. Then u′′ > 0 and u′ < 0 in a left-neighborhood of ξ∗,contradicting the definition of ξ∗.

Next, suppose that u′′′(ξ∗, α∗) = 0. Then, since u∗ ∈ (0, 1) by (3.2), it followsfrom the differential equation that uiv(ξ∗, α∗) > 0, so that u′′′ < 0, u′′ > 0, and u′ < 0


in a left-neighborhood of ξ∗. This means that u′ has a zero on (0, ξ∗) contradictingagain the definition of ξ∗.

Thus, if (3.3) holds, then so does (3.4).Step 2. We show that

(3.5) ξ(α) → ξ(α∗) as α→ α∗, α ∈ S.First, it follows from (3.3), (3.4), and (2.1a) that

u′′′(x, α∗) > 0, u′′(x, α∗) > 0, and u′(x, α∗) > 0

for x > ξ∗ until u(x0, α∗) = 1 at a finite x0 > ξ(α∗).

Next, let ε > 0 be small and arbitrarily chosen to satisfy

0 < ξ(α∗)− ε < ξ(α∗) < ξ(α∗) + ε < x0.

Since [0, x0] is compact, it follows from the definition of α∗, part (b) of Lemma 3.2,and continuity that there exists a δ = δ(ε) > 0 such that if 0 < α∗ − α < δ, thenα ∈ S,

(3.6a) u(x0, α) >√

1−√µ,

and

(3.6b) u′(x0, α) > 0 for all x ∈ [0, ξ(α∗)− ε] ∪ [ξ(α∗) + ε, x0].

From (3.6a), part (c) of Lemma 3.2, (3.6b), and the definition of ξ(α) we concludethat ξ(α) ∈ (ξ(α∗)− ε, ξ(α∗) + ε) if 0 < α∗ − α < δ. This implies (3.5).

Step 3. The contradiction. It follows from (3.5) that

u′′′(ξ(α), α) → u′′′(ξ(α∗), α∗) as α→ α∗, α ∈ S.Because u′′′(ξ(α), α) < 0 for all α ∈ S, this implies that

u′′′(ξ(α∗), α∗) ≤ 0,

which contradicts (3.4) and we conclude that (3.3) cannot hold. Thus,

(3.7) u′′(ξ(α∗), α∗) < 0.

To complete the proof we suppose that

u′′′(ξ(α∗), α∗) < 0.

Then α∗ ∈ S and since S is open, α∗ cannot be the supremum of S. Therefore,

u′′′(ξ(α∗), α∗) = 0

and the lemma is proved.Corollary 3.6. We have

u(ξ(α∗), α∗) <√

1−√µ and α∗ <

√1− µ

2.


Proof. The first inequality follows as in Lemma 3.2(c) from the energy identity(1.7). However, because we know from Lemma 3.5 that u′′(ξ∗, α∗) < 0, we now obtainstrict inequality.

The second inequality is proved if we can rule out equality from Lemma 3.4. Thus,suppose that (α∗)2 = (1 − µ)/2. Then u(i)(0) = 0 for i = 2, 3, 4 and u(5)(0) > 0.Hence, since u(i) > 0, i = 2, 3, 4 in a right-neighborhood of the origin, we concludethat u′ > 0 as long as u ≤ 1, so that u(·, α∗) cannot yield a periodic solution suchthat u(ξ∗, α∗) < 1.

Proof of Theorem 3.1. It follows from Lemma 3.5 that the solution u(x, α∗) ofproblem (2.1) satisfies the conditions (2.3) at ξ = ξ(α∗) and so can be continued toyield a periodic solution with period 4ξ(α∗).

Concerning uniqueness, we give the following partial result.

Lemma 3.7. Let γ > 0 and 49 ≤ µ < 1. Then there exists a unique periodic

solution u which satisfies (2.1), (2.2) and is such that max |u| < 1.

Proof. Suppose that there are values γ > 0 and µ ∈ [49 , 1) such that there existtwo distinct periodic solutions u1 and u2 with max |ui| < 1 (i = 1, 2). Let α1 and α2

be their respective slopes at x = 0. Since

dβ

dα=

1

2γ+

1− µ

4γα2> 0,

it follows that

α1 < α2 ⇒ u′′′1 (0) < u′′′1 (0) < 0.

Let w = u1 − u2. Then, by the mean value theorem,

(3.8a)

(3.8b)

{γwiv = w′′ + (1− 3u2)w,

w(0) = 0, w′(0) < 0, w′′(0) = 0, and w′′′(0) < 0,

where u is a function whose values lie between those of u1 and u2. Since 49 ≤ µ < 1

and max |ui| <√

1−√µ, it follows that

|u(x)| ≤ 1√3

for x ∈ R

and, therefore,

1− 3u2(x) ≥ 0 for x ∈ R.

Thus, we conclude from (3.8a) and (3.8b) that

wiv < 0, w′′′ < 0, w′′ < 0, and w′ < 0 for x ∈ R.

This implies that w(x) → −∞ as x → ∞. Because |w(x)| ≤ |u1(x)| + |u2(x)| ≤ 2√3

for all x ∈ R, this is not possible and we have a contradiction.

We conjecture that for every γ > 0 and 0 < µ < 1, there exists a unique periodicsolution with maximum less than 1.


4. Nonexistence of periodic solutions. In this section we restrict our atten-tion to the parameter regime

(4.1) 0 ≤ µ < 1 and γ > 0.

In Lemma 2.2 we already proved that no periodic solution can exist if µ ≥ 1 andγ > 0. We wish to determine the largest possible range of values of µ and γ in thisregime in which no periodic solution exists. We emphasize again that by a periodicsolution we mean a solution of problem (2.1)–(2.2) which satisfies (2.3).

In the analysis below we extend Lemma 2.2 and prove two nonexistence theoremsfor 0 < γ ≤ 1

8 . In the previous section we found that there exist periodic solutionsfor every γ > 0 when 0 < µ < 1 and that their maxima lie below u = 1. In our firstnonexistence theorem we set µ = 0 and show that such solutions cease to exist when0 < γ ≤ 1

8 . This range of γ-values is optimal; in the next section we shall show thatwhen µ = 0 and γ > 1

8 such periodic solutions do exist.In section 5 it is shown that when µ = 0 and γ > 1

8 , there exist periodic solutionswith maxima above u = 1. In our second nonexistence theorem we shall show thatwhen µ = 0, this range of γ-values is also optimal and that no periodic solutions withmaxima above u = 1 exist when γ ≤ 1

8 .Before proving these two theorems, we establish three technical lemmas. For this

we recall that if u(·, α) is a periodic solution, then the corresponding solution z(·, α)of problem (2.6) has the properties

(2.8b) limt→τ(α)−

z(t, α) = 0 and limt→τ(α)−

√z(t, α)z′′(t, α) = 0,

where τ(α) is finite and defined by

τ(α) = sup{t > 0 : z(·, α) > 0 on [0, t)}.Lemma 4.1. Let 0 ≤ µ < 1 and γ > 0. If z corresponds to a periodic solution,

then

(a) z(0) <1− µ

2,

(b) z′(t) < 0 for 0 < t < τ.

Proof. (a) Since z(0) = α2, the assertion follows at once from Lemma 2.2.(b) It follows from part (a) and (2.6a) that z′′(0) < 0. Hence, z′ < 0 in an interval

(0, ε) for some small ε > 0. If z′ vanishes at a first τ0 ∈ (0, τ), then

(4.2) z(τ0) > 0, z′(τ0) = 0, and z′′(τ0) ≥ 0.

If z′′(τ0) = 0, then z(τ0) = fµ(τ0) by (2.6a) and a differentiation of (2.6a) yields

(4.3) z′′′(τ0) = − f ′µ(τ0)

γz(τ0)= −2τ0(τ

20 − 1)

γ z(τ0).

We shall discuss the cases (i) τ0 = 1, (ii) τ0 < 1, and (iii) τ0 > 1 in succession.(i) τ0 = 1. It follows from (2.6a) that

z(τ0) = fµ(τ0) = −1

2µ ≤ 0,


which contradicts (4.2).(ii) τ0 < 1. It follows from (4.3) that z′′′(τ0) > 0, so that z′′ < 0 and z′ > 0 in a

left-neighborhood (τ0 − ε, τ0) of τ0. But this contradicts the definition of τ0.(iii) τ0 > 1. It follows from (4.3) that z′′′(τ0) < 0. Therefore, z′′ < 0 and z′ < 0

in a small interval (τ0, τ0 + ε). When we differentiate (2.6a) and divide by√z, we

obtain

(4.4) (√z z′′)′ =

z′ − f ′µγ√z

.

Since τ0 > 1 and f ′µ > 0 on (1,∞), it follows from (4.4) that z′′ < 0 and z′ < 0 onthe entire interval (τ0, τ). Integration of (4.4) over (τ0, τ) shows that

limt→τ−

√z(t) z′′(t) =

∫ τ

τ0

z′ − f ′µγ√z

ds < 0,

contradicting (2.8b).Thus, because (i), (ii), and (iii) lead to contradictions, we must conclude that

z′′(τ0) > 0. This implies by (2.6a) that z(τ0) > fµ(τ0). Hence, z′ > 0 in a right-neighborhood of τ0. Because z(τ) = 0, there must exist a first τ1 > τ0 where z′(τ1) =0. We assert that τ1 > 1. To see this, note that (2.6a) implies that z′′ > 0; hence,z′ > 0 as long as z > fµ. Because f ′µ < 0 on (0, 1), it follows that

z(t) > z(τ0) > fµ(τ0) > fµ(t) on τ0 < t ≤ 1.

This means that z′ > 0 on (τ0, 1] and, hence, τ1 > 1.At t = τ1 we have z′′(τ1) ≤ 0. If z′′(τ1) < 0, then z′′ < 0 and z′ < 0 on an

interval (τ1, τ1 + ε). Reasoning as before, using (4.4) again, we find that z′ < 0 on theentire interval (τ1, τ) and that limt→τ−

√z(t) z′′(t) < 0, which contradicts (2.8b). If

z′′(τ1) = 0, then, as in (4.3), we find that

z′′′(τ1) = −2τ1(τ21 − 1)

γ z(τ1)< 0.

Therefore, z′′ < 0 and z′ < 0 on a right-neighborhood of τ1 and we can repeat theprevious argument to obtain a contradiction of (2.8b).

This leads us to the conclusion that z′ < 0 on (0, τ) and the lemma is proved.Lemma 4.2. Suppose that z corresponds to a periodic solution and that fµ(τ) 6= 0.

Then

(4.5) z′′(τ) = limt→τ−

z′′(t) =2

γ

{1 +

√γ

2

f ′µ(τ)√fµ(τ)

}.

Proof. Because u′ and u′′′ both vanish as x → ξ−, or t → τ−, it follows froml’Hopital’s rule that

(4.6) z′′(τ) = 2 limx→ξ−

u′′′(x)

u′(x)= 2 lim

x→ξ−

uiv(x)

u′′(x)=

2

γ

(1 +

u− u3

u′′),

where the last term is evaluated at x = ξ. By the energy identity (1.7) we have

(u′′)2 =1

γfµ(u) at x = ξ,


so that

u′′ = − 1√γ

√fµ(u) at x = ξ.

If we substitute this expression for u′′ into (4.6) and remember that u(ξ) = τ , theassertion follows.

Define

(4.7) H = z(z′′ − 1

γ

)− µ

2γ

and

τ0 = sup{t ∈ (0, τ) : z′ < 0 on (0, t)}.Lemma 4.3. Let 0 ≤ µ < 1 and 0 < γ ≤ 1

8 , and let z be the solution of problem(2.6). Then

H(t) < 0 for 0 ≤ t < τ∗ = min{τ0, 1}.

Proof. Observe that we can write H as

(4.8) H =(z′)2

4− (t2 − 1)2

2γ.

Hence

H(0) = − 1

2γ< 0

and it follows that H < 0 in a right-neighborhood of the origin. Suppose that H firstvanishes at a point t0 ∈ (0, τ∗). Then

(4.9) H(t0) = 0 and H ′(t0) ≥ 0.

We deduce from (4.8), (4.9), and part (b) of Lemma 4.1 that, since t0 < 1,

(4.10) z′(t0) = −√

2

γ(1− t20).

For H ′ we differentiate (4.8) and use (4.10) to obtain

(4.11) H ′(t0) = −(1− t20)

{z′′√2γ

− 2t0γ

}.

Since µ ≥ 0 and H vanishes at t0, it follows from (4.7) and (4.9) that

z(z′′ − 1

γ

)=

µ

2γ≥ 0 at t = t0.

Hence z′′(t0) ≥ 1γ , and we conclude from (4.11) that

H ′(t0) ≤ − 1

γ(1− t20)

{1√2γ

− 2t0

}< 0


because 0 < t0 < 1 and γ ≤ 18 . This contradicts (4.9) and the lemma is proved.

We are now ready to prove the two nonexistence theorems for 0 < γ ≤ 18 . As

was explained earlier, we know that there exist periodic solutions for these values ofγ when 0 < µ < 1 and that they do not exceed u = 1. In the first theorem we showthat such periodic solutions no longer exist when µ = 0. In the second theorem weshow that if µ ∈ [0, 1), then there exist no periodic solutions which exceed u = 1.

Theorem 4.4. Let µ = 0 and 0 < γ ≤ 18 . Then there exists no periodic solution

u(x) such that

max{|u(x)| : x ∈ R} < 1.

Proof. Suppose, to the contrary, that there exists a periodic solution u whosemaximum is less than 1. Let z correspond to u. Then τ < 1, Lemma 4.1 implies thatz′ < 0 on (0, τ), and we deduce from Lemma 4.3 that H < 0 on (0, τ). Thus, by (4.7),

(4.12) z′′(t) <1

γfor 0 < t < τ

and, in particular,

z′′(τ) ≤ 1

γ.

From this last inequality and Lemma 4.2, we conclude that

f ′(τ) ≤ − 1√γ

√f(τ).

Therefore, because of (2.7),

τ ≥ 1√8γ

.

Since γ ≤ 18 , this means that we must have τ ≥ 1, a contradiction.

Theorem 4.5. Let 0 ≤ µ < 1 and 0 < γ ≤ 18 . Then there exists no periodic

solution u(x) such that

max{|u(x)| : x ∈ R} > 1.

Proof. Suppose that there exists a periodic solution u whose maximum is greaterthan 1. Let z correspond to u. Then τ > 1. Since Lemma 4.1 implies that z′ < 0 on(0, τ), it follows that z′(1) < 0.

To force a contradiction we shall show that z′(1) > 0. The proof of Lemma 3.4shows that this is the case when α > αµ =

√(1− µ)/2, and by continuity this will

remain so until z′(1) = 0 for some α < αµ. When z′(1) = 0, it follows from (4.8) that

H(1) = 0 and H ′(1) = 0.

In addition,

H ′′(1) ≥ 1− 8γ

2γ2+

µ

2γ2z,


where strict inequality holds if µ > 0 and equality holds if µ = 0. Thus,

H ′′(1) > 0 if (µ, γ) 6=(

0,1

8

),

in which case H ′ < 0 and H > 0 in a left-neighborhood of t = 1. Since H < 0 on(0, 1) by Lemma 4.3, we have a contradiction.

On the other hand, if µ = 0 and γ = 18 , then H ′′(1) = 0 and we have to consider

higher derivatives. We find that

H ′′′(1) = −12

γ< 0.

Therefore, in this case, H ′′ > 0, H ′ < 0, and H > 0 in a left-neighborhood of t = 1and by Lemma 4.3, we have once again arrived at a contradiction.

5. Existence of periodic solutions: µ = 0, γ > 18. In the previous section

we saw that if µ = 0, then there are no periodic solutions for 0 < γ ≤ 18 . In this

section we shall show that there do exist periodic solutions when γ > 18 , both with a

maximum less than 1 and with a maximum greater than 1. The method of proof issimilar to the one used in section 3.

Theorem 5.1. Let µ = 0 and γ > 18 . Then there exists a periodic solution u

such that

max{|u(x)| : x ∈ R} < 1.

We recall from Lemma 2.1 that if µ = 0 and γ > 18 , then

ξ(α) <∞ and u′(ξ(α), α) = 0 for every α > 0.

Continuing as in section 3, we set

S = {α > 0 : u(ξ(α), α) < 1, u′′(ξ(α), α) < 0, and u′′′(ξ(α), α) < 0 for 0 < α < α}.Reproducing the proofs of Lemmas 3.2–3.4, we establish the following properties of ξand S.

Lemma 5.2. Let µ ∈ [0, 1) and γ > 0.(a) ξ ∈ C1(S).(b) The set S is a nonempty, open interval of the form (0, α∗).

(c) min

{√1− µ

2,

√3(1− µ)

24γ + 7

}≤ α∗ ≤

√1− µ

2.

We must still determine the properties of u(·, α∗). This will be done in the nextlemma.

Lemma 5.3. Let µ = 0 and γ > 0. Then

u(ξ(α∗), α∗) < 1, u′′(ξ(α∗), α∗) < 0, and u′′′(ξ(α∗), α∗) = 0.

Proof. We first show that u′′(ξ∗, α∗) < 0, where we have written ξ∗ = ξ(α∗).From the definition of ξ we conclude that u′′(ξ∗, α∗) ≤ 0. We claim that

u′′(ξ∗, α∗) < 0. Thus, suppose to the contrary that

(5.1) u′′(ξ∗, α∗) = 0.


Then, by the energy identity (1.7),

u(ξ∗, α∗) = 1.

We assert that (5.1) implies that

(5.2) u′′′(ξ∗, α∗) > 0.

Suppose that u′′′(ξ∗, α∗) < 0. Then u′′ > 0 and u′ < 0 in a left-neighborhood ofξ∗, contradicting the definition of ξ∗. On the other hand, if u′′′(ξ∗, α∗) = 0, then byuniqueness, u(x) = 1 for all x ∈ R, a contradiction. Thus, indeed, (5.2) holds.

As in the proof of Lemma 3.5 it follows that

ξ(α) → ξ(α∗) as α→ α∗, α ∈ S,which means that

u′′′(ξ(α), α) → u′′′(ξ(α∗), α∗) as α→ α∗, α ∈ S.Because u′′′(ξ(α), α) < 0 for all α ∈ S, we arrive in the limit as α→ α∗ at

(5.3) u′′′(ξ(α∗), α∗) ≤ 0,

which contradicts (5.2). Thus, (5.1) must be false; hence,

(5.4) u′′(ξ(α∗), α∗) < 0.

It follows from (5.4) and the energy identity (1.7) that

u∗ = u(ξ(α∗), α∗) 6= 1

and that ξ(α) is continuous at α = α∗. Therefore, by the continuous dependence oninitial data, if u∗ > 1, then u(ξ(α), α) > 1 for α in a left-neighborhood of α∗. Since(0, α∗) ⊂ S, the definition of S shows that this is impossible. Thus,

(5.5) u(ξ(α∗), α∗) < 1.

Finally, regarding u′′′, we must have equality in (5.3). For if

u′′′(ξ(α∗), α∗) < 0,

then continuity implies that α∗ < supS, a contradiction.Thus, we have shown that the solution u(x, α∗) of problem (2.1) satisfies the

properties (2.3) at x = ξ(α∗) and this yields a periodic solution of which, by (5.5),the maximum is less than 1. This completes the proof of Theorem 5.1.

In the next theorem we find periodic solutions whose maxima exceed unity.Theorem 5.4. Let µ = 0 and γ > 1

8 . Then there exists a periodic solution usuch that

max{|u(x)| : x ∈ R} > 1.

We now take the shooting set from those values of α for which the maximum ofu exceeds 1. Specifically, we define

T = {α > 0 : u(ξ(α), α) > 1, u′′(ξ(α), α) < 0, and u′′′(ξ(α), α) < 0 for α > α}.


Lemma 5.5. We have (1√2,∞)⊂ T .

Proof. It follows from (1.7) that

2γu′u′′′ ≥ (u′)2 − 1

2(1− u2)2.

Thus, if α2 > 12 , then u′′′ > 0, u′′ > 0, and u′ > 0 as long as 0 < u ≤ 1. Hence, u

first reaches 1 at a finite value x1, where

u′(x1) > 0, u′′(x1) > 0, u′′′(x1) > 0.

Therefore, at x = ξ we have

(5.6) u(ξ) > 1, u′(ξ) = 0, u′′(ξ) < 0,

where the last inequality is strict because of the energy identity (1.7). Hence, u′′ hasa first zero at a point x2 ∈ (x1, ξ). At this point we have

u(x2) > 1, u′′(x2) = 0, u′′′(x2) ≤ 0.

Since, by equation (2.1a), uiv < 0 when both u > 1 and u′′ ≤ 0, it follows that

(5.7) u′′′(ξ) < 0.

From (5.6) and (5.7) we deduce that for any α > 1√2,

u(ξ(α), α) > 1, u′′(ξ(α), α) < 1, and u′′′(ξ(α), α) < 0,

so that ( 1√2,∞) ⊂ T .

As with Lemma 5.2, we can prove the following properties of ξ.Lemma 5.6. We have(a) ξ ∈ C1(T ),(b) the set T is an open interval of the form (α∗,∞),

(c) α∗ ≥ α∗,

where α∗ = supS as defined in Lemma 5.2, part (b).In the next lemma we list again the important properties of u(ξ(α), α) at α = α∗.Lemma 5.7. We have

u(ξ∗, α∗) > 1, u′′(ξ∗, α∗) < 0, and u′′′(ξ∗, α∗) = 0,

where we have written ξ∗ = ξ(α∗).Proof. From the definition of ξ we conclude that u′′(ξ∗, α∗) ≤ 0. Let us first

suppose that

(5.8) u′′(ξ∗, α∗) = 0.

Then, by the energy identity (1.7),

u(ξ∗, α∗) = 1.


We assert that (5.8) implies that

(5.9) u′′′(ξ∗, α∗) > 0.

For if u′′′(ξ∗, α∗) < 0, then u′′ > 0 and u′ < 0 in a left-neighborhood of ξ∗, whichcontradicts the definition of ξ∗. If u′′′(ξ∗, α∗) = 0, it follows from uniqueness thatu(x) = 1 for all x ∈ R, which contradicts the condition at x = 0. Therefore, (5.9)holds (see also [21, Lemma 3.10]).

To complete the proof of Lemma 5.7, we need the following lemma in which weestablish continuity of ξ at α∗ under the above conditions.

Lemma 5.8. Suppose that for some α0 > 0 we have

u(ξ(α0), α0) = 1, u′′(ξ(α0), α0) = 0, and u′′′(ξ(α0), α0) > 0.

Then

ξ(α) → ξ(α0) as α→ α0.

Accepting Lemma 5.8 for the moment, we conclude that

u′′′(ξ(α), α) → u′′′(ξ(α∗), α∗) as α→ α∗, α ∈ T .Because u′′′(ξ(α), α) < 0 for all α ∈ T , it follows that

(5.10) u′′′(ξ(α∗), α∗) ≤ 0,

which contradicts (5.9). Thus, (5.8) cannot be true and we conclude that

(5.11) u′′(ξ(α∗), α∗) < 0.

It follows from (5.11) and the energy identity (1.7) that

either u(ξ∗, α∗) > 1 or u(ξ∗, α∗) < 1

and that ξ(α) is continuous at α = α∗. Hence, by continuous dependence on initialdata, if u(ξ∗, α∗) < 1, then u(ξ(α), α) < 1 for α in a right-neighborhood of α∗. Since(α∗,∞) ⊂ T , the definition of T implies that this is impossible. Thus,

(5.12) u(ξ∗, α∗) > 1.

Finally, regarding u′′′, we must have equality in (5.10). For if

u′′′(ξ(α∗), α∗) < 0,

then continuity implies that α∗ > inf T , a contradiction. Therefore,

(5.13) u′′′(ξ(α∗), α∗) = 0.

We conclude from (5.13) that the solution u(x, α∗) of problem (2.1) satisfies theproperties (2.3) at x = ξ(α∗) and, thus, yields a periodic solution of which, by (5.12),the maximum is greater than 1. This completes the proof of Lemma 5.7.

The proof of Theorem 5.4 is complete once we have proved Lemma 5.8.Proof of Lemma 5.8. Fix ε > 0 and small. Then by the assumptions on u(·, α0)

there exists a δ > 0 such that

(5.14a) u(ξ0 − ε, α0) < 1− 2δ, u(ξ0 + ε, α0) > 1 + 2δ,


and

(5.14b) u′(x, α0) > δ for all x ∈ [0, ξ0 − ε].

We wish to prove that there exists a ν > 0 such that if |α−α0| < ν, then u′(·, α) hasa zero on (ξ0 − ε, ξ0 + ε).

By the continuous dependence of solutions on initial data it follows from (5.14)that there exists a ν1 > 0 such that

(5.15) u(ξ0−ε, α) < 1−δ, u(ξ0+ε, α) > 1+δ, and u′(x, α) > 0 for all x ∈ [0, ξ0−ε]if |α− α0| < ν1 so that

(5.16) ξ(α) > ξ0 − ε if |α− α0| < ν1.

To show that ξ(α) < ξ0 + ε for α sufficiently close to α0, it is sufficient to provethat

(5.17) τ(α) → τ(α0) = 1 as α→ α0,

where we recall that τ(α) = u(ξ(α), α). For (5.17) implies that there exists a ν2 > 0such that

τ(α) < 1 + δ if |α− α0| < ν2,

and, because u′ > 0 on (0, ξ), we conclude from (5.15) that

(5.18) ξ(α) < ξ0 + ε if |α− α0| < ν = min{ν1, ν2}.Thus, (5.15) and (5.18) yield the continuity of ξ(α) at α0.

Let us now prove (5.17). Let z0(t) = z(t, α0) be the solution of problem (2.6)which corresponds to u(x, α0). Then

z0(t) → 0 and√z0(t)z

′′0 (t) → A as t→ 1−,

where by assumption A = 2u′′′(ξ(α0), α0) is a positive constant. It is readily shown

that this implies that the function y0(t) = z3/40 (t) has the properties

(5.19)y0(t)

1− t→ B and y′0(t) → −B as t→ 1−,

where B = 32

√A.

We can write the equation (2.6a) for z as

(z−1/4z′)′ =z − f

γz5/4,

so that, since f(t) ≥ 0 for all t ≥ 0, the function y(t) = z3/4(t) satisfies

(5.20) y′′ ≤ 3

4γy−1/3.

Fix ρ ∈ (0, 1). Then y0(1 − ρ) > 0 and it follows from the continuous dependenceon initial data on [0, ρ] that there exists a ϑ1 > 0 such that τ(α) > 1 − ρ when|α− α0| < ϑ1. Since ρ may be chosen as small as we wish, we conclude that

(5.21a) lim infα→a0

τ(α) ≥ 1.


It remains to prove that

(5.21b) lim supα→a0

τ(α) ≤ 1.

Fix ε > 0 and t0 ∈ (0, 1). By (5.19), it is possible to choose t0 so close to 1 that

y′0(t0) ≤ −√

3

2B and y0(t0) ≤ B

8ε.

By continuity we can find a constant ϑ2 > 0 so small that if |α− α0| < ϑ2, then

(5.22) y′(t0) ≤ −B

2and 0 < y(t0) ≤ B

4ε.

Thus, in a neighborhood of t0 we have y′ < 0, so that when we multiply (5.20) by y′

we obtain

(y′2)′ ≥ 9

4γ(y2/3)′

for t > t0 as long as y > 0 and y′ < 0. This yields, upon integration over (t0, t),

y′2(t) ≥ y′2(t0) +9

4γ{y2/3(t)− y2/3(t0)}

≥ y′2(t0)− 9

4γy2/3(t0)

≥ B2

4− 9

4γ

(Bε4

)2/3

>B2

16,

if we choose ε < ε0 = 12 (γ3 )3/2B2. Therefore,

y′(t) < −B

4for t0 ≤ t ≤ τ.

Thus, when 0 < ε < ε0, integration over (t0, τ) yields

τ ≤ t0 +4

By(t0) < t0 + ε < 1 + ε,

where we have used (5.22). Since ε can be chosen arbitrarily small, this proves (5.21b)and the proof of Lemma 5.8 is complete.

We conclude this section with a few observations about the existence of periodicsolutions with amplitude greater than 1 when 0 < µ < 1.

Because the initial data, and hence the solution u of problem 2.1, depend con-tinuously on µ, it is evident from the proof of Theorem 5.1 that there exist periodicsolutions with amplitude greater than 1 when µ is sufficiently small. In the followingtheorem we shall show that this is no longer true when µ ≥ 4

9 .Theorem 5.9. If µ ≥ 4

9 and γ > 0 is arbitrary, then there exists no periodicsolution u for which

max{|u(x)| : x ∈ R} > 1.


Proof. Let µ ∈ (0, 1) and suppose that z corresponds to a periodic solution suchthat

τ = sup{u(x) : x ∈ R} > 1.

Then (1.7) implies that

τ >√

1 +√µ.

From Lemma 4.1(b) we know that

z′(t) < 0 and 0 < t < τ,

and by multiplying equation (2.6a) by 34z−5/4 we obtain

(2.10) (z3/4)′′ =3

4γ

z − fµz5/4

for 0 < t < τ,

where

fµ =1

2{(1− t2)2 − µ}.

Let us denote the zeros of fµ by a and b:

a =√

1−√µ and b =√

1 +√µ.

Because τ > b, we can integrate (2.10) over (0, b) to obtain

3

4γ

(∫ a

0

+

∫ b

a

)z − fµz5/4

dt = (z3/4)′∣∣∣b0< 0,

where the inequality is clear when we remember that z′(0) = 0 and z′(b) < 0. Thus,writing

I1 =

∫ a

0

z − fµz5/4

dt and I2 =

∫ b

a

z − fµz5/4

dt,

we have

(5.23) I1 + I2 < 0.

Recall that z > 0 and z′ < 0 on (0, τ). Hence

(5.24) I1 > −∫ a

0

fµz5/4

dt > − 1

z5/4(a)

∫ a

0

fµ(t) dt

and

(5.25) I2 =

∫ b

a

z + |fµ|z5/4

dt >1

z5/4(a)

∫ b

a

|fµ(t)| dt > 0,

because fµ < 0 on (a, b). Putting (5.24) and (5.25) into (5.23) we find that

−∫ a

0

fµ(t) dt+

∫ b

a

|fµ(t)| dt < 0,


or, equivalently,

(5.26)

∫ b

0

fµ(t) dt > 0.

An elementary computation shows that (5.26) holds if and only if

µ <4

9.

Thus, if µ ≥ 49 , there can be no periodic solution with maxima above 1.

6. Qualitative properties. In this section we prove several qualitative prop-erties of periodic solutions. We begin with a convexity lemma and then we establishuniversal global bounds for periodic solutions. This is followed by an analysis of thebehavior of periodic solutions as γ → 0 (when 0 < µ < 1), as γ → 1

8 (when µ = 0),and as γ →∞.

We begin with a convexity property.Lemma 6.1. Let u(x) be a periodic solution which has a single critical point

between zeros and has the symmetry properties (1.11). Then

u′′(x) < 0 when u(x) > 0.

Proof. By Lemma 4.1(b), if z(t) is the solution of problem (2.6) which correspondsto u(x), then z′(t) < 0 for 0 < t < τ , and hence,

u′′(x) =1

2z′(t(x)) < 0 for 0 < x < ξ.

Since u′′(ξ) < 0 by the energy identity, the assertion follows.A remarkable feature of all single-bump periodic solutions is that they are bounded

by a constant which does not depend on either γ or µ. This is shown in the nextlemma.

Lemma 6.2. Let 0 ≤ µ < 1 and γ > 0, and let u(x) be a periodic solution thathas the symmetry properties (1.11). Then

|u(x)| <√

2 for x ∈ (−∞,∞).

Proof. Suppose that for some a ∈ R we have |u(a)| ≥ √2. Without loss of

generality we may assume that u has a maximum at x = a. Thus, we have

(6.1) u(a) ≥√

2, u′(a) = 0, u′′(a) ≤ −√

1− µ

2γ, u′′′(a) = 0,

where the upper bound for u′′ follows from the energy identity (1.7). From (1.6) wesee that uiv(a) < 0.

Thus, as x increases above a, u′′′ and u′′ decrease. Thus, u′′′ < 0 and u′′ < 0 aslong as uiv < 0. Furthermore,

uiv = u′′ + u− u3 < 0 as long as u > 1.

Thus, if

b = sup{x > a : u > 1 on [a, x)},


then

(6.2) u(b) = 1, u′(b) < 0, u′′(b) < −√

1− µ

2γ, u′′′(b) < 0.

We claim that

(6.3) u′′(x) < −√

1− µ

2γas long as |u| < 1.

Suppose that (6.3) does not hold. Then u′′′ must have a zero at a point where |u| < 1.Let y > b be the first zero of u′′′. Then by (1.7),

−γ(u′′)2 − (u′)2 +1

2(1− u2)2 =

µ

2at y.

Because of our assumption that |u(y)| < 1,

{u′′(y)}2 < 1− µ

2γ.

This means that

u′′(y) > −√

1− µ

2γ.

However, since u′′′ < 0 on (b, y), it follows from (6.2) that

u′′(y) < −√

1− µ

2γ,

so we have a contradiction.Thus, (6.3) holds. In particular, we have that u′′ < 0 at the first zero of u and,

by (1.11), at all zeros of u. This contradicts the fact that because u is odd, u′′ = 0whenever u = 0. Thus, we must conclude that the assertion holds.

We now turn to a discussion of the behavior of periodic solutions for values of γclose to γ = 0, or γ = 1

8 when µ = 0, and for large values of γ.Lemma 6.3. Let {γi} be a sequence such that

γi ↘ θ =

1

8when µ = 0

0 when 0 < µ < 1as i→∞.

For each i ≥ 1, let ui be a periodic solution corresponding to γi. Then

ui(x) → U(x) as i→∞, uniformly on compact intervals,

where(i) if µ = 0, then U is the unique monotone symmetric kink corresponding to

γ = 18 ;(ii) if 0 < µ < 1, then U is the unique periodic solution of the FK equation with

energy µ.Proof. Let αi = u′i(0). If 0 < µ < 1, then Lemmas 3.3 and 3.4 imply that

(6.4)

√1− µ

4≤ αi ≤

√1− µ

2


for i sufficiently large and γi − θ > 0 sufficiently small. If µ = 0, then (6.4) followsfrom Lemmas 5.2, 5.5, and 5.6. Hence, there exists a convergent subsequence, whichwe also denote by {γi}, such that

αi → α as i→∞,

where α satisfies (6.4).We consider the cases µ = 0 and 0 < µ < 1 in succession.Case I. Let µ = 0 and let α0 = U ′(0), where U is the kink for γ = θ = 1

8 . Supposethat α > α0. Then, by [21, Lemma 3.6 and Theorem 3.7],

(6.5) u

(ξ

(α,

1

8

), α,

1

8

)> 1 and u′′

(ξ

(α,

1

8

), α,

1

8

)< 0.

Since by Theorem 4.5, u(·, α, 18 ) cannot be a periodic solution, it also follows that

(6.6) u′′′(ξ

(α,

1

8

), α,

1

8

)6= 0.

The inequality in (6.5) implies that the function ξ(α, γ) is continuous at (α, 18 ) so, by

the continuous dependence of u(·, α, γ) on α and γ, it follows that for i large enough,

(6.7) u′′′(ξ(αi, γi), αi, γi) 6= 0

as well. However, since u(·, αi, γi) is a periodic solution for every i, we must have

u′′′(ξ(αi, γi), αi, γi) = 0

for every i, which contradicts (6.7).If α < α0, then by [21, Lemma 3.6 and Theorem 3.7],

u

(ξ

(α,

1

8

), α,

1

8

)< 1 and u′′

(ξ

(α,

1

8

), α,

1

8

)< 0.

It follows from Theorem 4.4 that (6.6) holds again and, as before, we arrive at acontradiction.

Thus, for the subsequence we have αi → α0 as i → ∞. Because the limit isuniquely determined, it follows that the entire sequence {αi} converges to α0. There-fore, ui → U uniformly on compact sets of the form [0, L], L > 0.

Case II. Let 0 < µ < 1 and let αµ = U ′(0), where U is the periodic solution ofthe FK equation with energy equal to µ. From the energy identity (1.7) in which weset γ = 0, we conclude that

αµ =

√1− µ

2.

It follows from Corollary 3.6 that

α ≤ αµ.

In the remainder of the proof we shall show that

α ≥ αµ


as well. This proves that αi → αµ as i → ∞, and therefore ui → U uniformly oncompact sets, as asserted.

We shall show that for each ε > 0 there exists a γε > 0 such that if 0 < γ < γεand u(·, α, γ) is a periodic solution, then α > αµ − ε so

lim infi→∞

αi ≥ αµ.

Remember that the initial conditions for u are

u(0) = 0, u′(0) = α, u′′(0) = 0, u′′′(0) = β(α) =α2 − α2

µ

2αγ.

Because β′(α) > 0 it follows that

β(α) ≤ β(αµ − ε) = −δ(ε)

γfor 0 < α ≤ αµ − ε,

where δ(ε) ∼ ε as ε→ 0.We now proceed as in the proof of Lemma 3.3. Because u(0) = 0 and u′′′(0) < 0,

it follows that u < 1 and u′′′ < 0 in a neighborhood of the origin. As long as theseinequalities do not change it follows from the equation for u that

uiv(x) <α

γx,(6.8a)

u′′′(x) < − δ

γ+

α

2γx2,(6.8b)

u′′(x) < − δ

γx+

α

6γx3,(6.8c)

u′(x) < α− δ

2γx2 +

α

24γx4.(6.8d)

Set

xγ = γ1/4.

Then, when γ < γ1 = (2δ/αµ)2, it follows from (6.6b) that u′′′ < 0 on (0, xγ ]. More-over, the right-hand side of (6.6d) will be negative at xγ if γ < γ2 = {12δ/(25αµ)}2.

Thus, if we set γε = min{γ1, γ2} and denote as usual the first zero of u′ by ξ, then

ξ ∈ (0, xγε) and u′′′(ξ) < 0 if 0 < γ < γε.

This means that if α ≤ αµ − ε and γ ∈ (0, γε), then u(·, α, γ) cannot be a periodicsolution. Therefore, if it is given that u(·, α, γ) is a periodic solution, then we mustconclude that α > αµ − ε, and the proof is complete.

To determine the behavior of periodic solutions as γ →∞, we first need an upperand a lower bound for the slope at the origin.

Lemma 6.4. Let 0 ≤ µ < 1 and γ > 0, and let u(x) be a periodic solution whichsatisfies (1.11). Then

u′(0) ≤ {8(1− µ) log 2}1/4γ−1/4 for γ > θ,

where θ = 0 if 0 < µ < 1 and θ = 18 if µ = 0.


Proof. It is convenient to prove this lemma using the function z(t) introduced insection 3. We then need to show that

(6.9) z(0) ≤ {8(1− µ) log 2}1/2γ−1/2.

It follows from (2.6a) that

zz′′ =(z′)2

4+

1

γ{z − fµ(t)} > −1− µ

2γfor 0 ≤ t <

√2.

Because τ <√

2 by Lemma 6.2, it follows that

z′′ > −1− µ

2γzfor 0 ≤ t < τ.

We multiply by z′ < 0, integrate over (0, t), and obtain

(6.10) z′(t) > −{

1− µ

γlog

z(0)

z(t)

}1/2

for 0 ≤ t < τ.

Let

t0 = sup

{t > 0 : z >

1

2z(0) on [0, t)

}.

Then

0 < t0 <√

2 andz(0)

z(t)≤ 2 for 0 ≤ t ≤ t0.

Hence, by (6.10),

z′(t) > −(

(1− µ) log 2

γ

)1/2

for 0 < t < t0

and we obtain, after an integration over (0, t0),

(6.11) z(t0)− z(0) > −(

(1− µ) log 2

γ

)1/2

t0.

Since t0 <√

2, this implies that

1

2z(0) <

((1− µ) log 2

γ

)1/2

,

from which (6.9) follows.Next we establish a lower bound for u′(0).Lemma 6.5. We have

u′(0) >1

5

√1− µγ−1/4 for γ >

(2

5

)4

.


Proof. In light of the upper bound obtained in Lemma 6.3, we scale the variablesand set

(6.12) s =x

γ1/4and v(s) = u(x).

We then obtain the problem

(6.13a)

(6.13b)

(6.13c)

viv =v′′√γ

+ v − v3,

v(0) = 0, v′′(0) = 0,

v′(0) = ω, v′′′(0) =1

2ω

( ω2

√γ− 1− µ

2

)

in which

ω = αγ1/4

and we need to prove that

ω >1

5

√1− µ.

Suppose, to the contrary, that

ω ≤ 1

5

√1− µ.

Then for γ > (2/5)4, we have

v′′′(0) < −1− µ

8ω.

As long as v > 0 and v′′′ < 0, we have

v(s) < ωs

and

viv(s) < ωs,

which yields, upon integration over (0, s),

(6.14) v′′′(s) < −1− µ

8ω+

1

2ωs2.

One verifies that the right-hand side of (6.14) is negative for all s ∈ [0, 1]. Two moreintegrations yield

v′(s) < ω − 1− µ

16ωs2 +

1

24ωs4.

It follows that the first zero σ = σ(ω, γ) of v′ has the properties

0 < σ < 1 and v′′′(σ) < 0,


so v, and hence u, cannot be periodic solutions, a contradiction.With the lower bound we now have in hand we can return to the argument used

in the proof of Lemma 6.1 to obtain a lower bound for the maximum of |u(x)| on R.Lemma 6.6. Let u(x, γ) be a periodic solution. Then

max{|u(x, γ)| : x ∈ R} > 1

50

√1− µ

log 2if γ >

(2

5

)4

.

Proof. If u is a periodic solution, then by Lemma 6.5,

z(0) >1− µ

25√γ

if γ >(2

5

)4

.

Therefore, by (6.11),

((1− µ) log 2

γ

)1/2

t0 >1− µ

50√γ.

This means that

t0 >1

50

√1− µ

log 2.

Now, because

max{|u(x, γ)| : x ∈ R} = τ(γ) > t0,

the assertion follows.From Lemmas 6.4 and 6.5 we conclude that if u is a periodic solution which

satisfies (1.11), then for γ large enough,

1

8

√1− µ < ω < {8(1− µ) log 2}1/4.

Let {γi} be a sequence tending to infinity and let ui be a corresponding sequence ofperiodic solutions, with initial slopes αi. Let vi and ωi be the solutions of problem(6.13) corresponding to these periodic solutions. Then by compactness there existsa subsequence, which we also denote by {ωi}, which converges to a number ω∗ < ∞as i → ∞. Plainly, it must be the case that vi → V , uniformly on compact sets, asi→∞, where V satisfies

(6.15a)

(6.15b)

(6.15c)

V iv = V − V 3,

V (0) = 0, V ′′(0) = 0,

V ′(0) = ω∗, V ′′′(0) = −1− µ

4ω∗.

We assert that the sequence of maxima {σi}, where σi = σ(ωi, γi), remains bounded:

lim supi→∞

σ(ωi, γi) <∞.

For if not, then there exists a subsequence along which σi tends to infinity. Hence,

V ′(s) > 0 for 0 ≤ s <∞.


An argument like the one used in the proof of Lemma 2.1 shows that this is impos-sible. Therefore, the sequence {σi} is bounded and there exists a subsequence whichconverges to some σ∗ < ∞ as i → ∞. Since v′′′i (σi) = 0 for every i, it easily followsthat V ′′′(σ∗) = 0, so that V is a periodic solution of problem (6.15).

Thus we have the following lemma.Lemma 6.7. Let 0 ≤ µ < 1. Suppose that {γi} is a sequence which tends to

infinity and {ui} is a sequence of periodic solutions which correspond to γi and whicheach satisfy (1.11). Then there exists a subsequence and a periodic solution V ofproblem (6.15) which also satisfies (1.11) such that

(6.16) ui(γ1/4i s, γi) → V (s) as i→∞,

uniformly on compact sets.As a by-product, the above argument yields the existence of periodic solutions

V of problem (6.15) which satisfy (1.11) for every µ ∈ [0, 1). This result can alsobe proved by means of the method used in sections 3 and 5. Since the proofs arevery close to those already presented, we omit them here. Summarizing, we have thefollowing result.

Theorem 6.8. If 0 ≤ µ < 1, then problem (6.15) has a periodic solution V1 suchthat

max{|V1(x)| : x ∈ R} < 1.

If µ = 0, then problem (6.15) has a periodic solution V2 which satisfies (1.11) suchthat

max{|V2(x)| : x ∈ R} ∈ (1,√

2).

Acknowledgment. It is a pleasure to thank J. Serrin for numerous stimulatingdiscussions. We thank the School of Mathematics, as well as the Institute of Math-ematics and its Application, at the University of Minnesota and the University ofPittsburgh for providing hospitality during the preparation of this paper. Finally theauthors are much indebted to the referee, whose questions and comments led to asubstantially improved manuscript.

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