Closed curves in R^3 with prescribed curvature and torsion in perturbative cases

Closed curves in R3 with prescribed curvature and torsion in perturbative cases∗ Paolo Caldiroli and Michela Guida Dipartimento di Matematica, Universit`a di Torino via Carlo Alberto, 10 – 10123 Torino, Italy Abstract Abstract. We study the existence of closed curves in the three-dimensional Euclidean space, with prescribed curvature and torsion, in a perturbative setting. In particular we prove existence of branches of closed curves emanating from circles which correspond to stable zeroes of a suitable vector field M : T2 × R3 → R5 . AMS Subject Classification: 53A04 (47A53).

1

Introduction

Recent years have seen a growing interest towards some geometrical problems concerning the existence and possibly location of k-dimensional manifolds embedded into RN with given topological type and prescribed curvature. In this paper we investigate a problem in low dimension. More precisely we study the existence of closed curves in the three-dimensional Euclidean space, with prescribed curvature and torsion. The problem can be stated as follows: given smooth functions κ: R3 → (0, +∞) and τ : R3 → R, find closed curves Γ in R3 such that at every point p ∈ Γ the curvature of Γ equals κ(p) and the torsion is τ (p). We shall call (κ, τ )-loops such curves. A special relevant case corresponds to the choice κ ≡ κ0 and τ ≡ 0, where κ0 is a positive constant. In this situation the only closed curves with such curvature and torsion are circles of radius 1/κ0 anywhere placed in R3 (see Lemma 3.1). We remark that the set of closed curves with constant curvature κ0 and torsion 0 defines a manifold Z of dimension 5, diffeomorphically parametrized by P2 × R3 , where P2 := R3 /R∗ denotes the two-dimensional projective space, namely the space of the directions in R3 (every pair (n, p) ∈ P2 × R3 corresponds to the circle of radius 1/κ0 centered at p and lying on the plane orthogonal to n). In this paper we will focus on some perturbative cases. In particular we will study the existence of branches of closed curves emanating from some circle when the curvature and the torsion are ∗ The

first author is supported by MIUR-PRIN project “Metodi variazionali ed equazioni differenziali nonlineari”

1

perturbations of the constants κ0 > 0 and 0 respectively, and depend on a smallness parameter ε in the following way:  κ(p) ≡ κε (p) := κ0 + K(ε, p) τ (p) ≡ τε (p) := T (ε, p) where K, T : R × R3 → R are smooth functions such that K(0, ·) ≡ 0

and

T (0, ·) ≡ 0 .

(1.1)

Let us observe that κε is admissible as a prescribed curvature, since there results κε > 0 on compact sets of R3 as |ε| is small enough. In general some conditions on K and T are needed for the existence of (κε , τε )-loops. Indeed, considering the case K ≡ 0 and T ≡ 1, one can see that for every ε 6= 0 the only curves with constant curvature κ0 and constant torsion ε are portions of helicoids. Hence in this case there is no closed curve. Also when T ≡ 0, namely one deals with planar curves, some restrictions on K are necessary (see [7]). Hereafter we shall assume for simplicity κ0 = 1, which is not restrictive, by obvious normalization. Henceforth, for all ε ∈ R and p ∈ R3 κε (p) = 1 + K(ε, p)

and τε (p) = T (ε, p) .

(1.2)

We shall see that the existence and nonexistence of (κε , τε )-loops is strongly related to the property of the zero set of the mapping M : T2 × R3 → R5 defined as follows:  R1  ∂ε K(0, Rφ z(t) + p) cos(2πt) dt 0  R 1 ∂ K(0, R z(t) + p) sin(2πt) dt  φ  R0 ε   1  M (φ, p) :=  0 ∂ε T (0, Rφ z(t) + p) cos(2πt) dt  for (φ, p) ∈ T2 × R3 (1.3)  R1    ∂ T (0, Rφ z(t) + p) sin(2πt) dt 0 ε R1 ∂ T (0, Rφ z(t) + p) dt 0 ε where T2 := (R/2πZ)2 is the two-dimensional torus, 

cos φ2 Rφ :=  sin φ2 0

− sin φ2 cos φ1 cos φ2 cos φ1 sin φ1

 sin φ1 sin φ2 − sin φ1 cos φ2  ∈ SO(3) cos φ1

for every φ = (φ1 , φ2 ) ∈ T2

(1.4)

and 

 cos(2πt) z(t) :=  sin(2πt)  0

for every t ∈ R.

(1.5)

By natural periodic extension, we shall also mean M : R2 × R3 → R5 . Denoting (e1 , e2 , e3 ) the canonical basis of R3 , z is a uniform1 parametrization of the unit circle centered at the origin and lying on the plane orthogonal to e3 . Moreover, Rφ z + p parametrizes the 1A

parametrization u of a curve Γ is called uniform if |u0 | is constant.

2

unit circle centered at p and lying on the plane orthogonal to Rφ e3 . Viceversa, any solution of the unperturbed problem, i.e., the problem corresponding to ε = 0, admits such a parametrization, so that Z = {Rφ z(R) + p | (φ, p) ∈ T2 × R3 }. Hence the mapping M establishes a link between the perturbation (K, T ) and the unperturbed manifold Z and, borrowing a notion from perturbation theory for dynamical systems [11], it can be interpretated as the Poincar´e-Melnikov vector associated to the problem. We point out that defining M in terms of the coordinates (φ, p) ∈ T2 × R3 we can ensure enough regularity for M , as we need, since the mapping φ 7→ Rφ from T2 into SO(3) is of class C ∞ . Parametrizing Z by means of global coordinates (n, p) ∈ P2 × R3 , even continuity is lost because of the Hairy Ball Theorem which prevents the existence of continuous mappings n 7→ R(n) from P2 into SO(3) such that R(n)e3 has direction n. Before stating our main results, let us introduce the following additional definitions: let {Γδ }δ∈I be a family of closed curves depending on a parameter δ varying in an open interval I of R; for δ0 ∈ I fixed, we say that Γδ → Γδ0 in C k as δ → δ0 if for every δ ∈ I the curve Γδ admits a 1-periodic, uniform parametrization uδ ∈ C 3 ∩ C k (R, R3 ) and kuδ − uδ0 kC k → 0 as δ → δ0 . Moreover we say that the mapping δ 7→ Γδ is of class C 1 (I, C k ) if the mapping δ 7→ uδ belongs to C 1 (I, C k ([0, 1], R3 )). Our first result states that the fact that M vanishes somewhere constitutes a necessary condition for the existence of a bounded sequence of (κε , τε )-loops with |ε| small. Theorem 1.1 Let K, T ∈ C 1 (R × R3 ) satisfy (1.1) and let κε and τε be as in (1.2). If there is a sequence εn → 0, εn 6= 0, and a corresponding sequence (Γn ) of (κεn , τεn )-loops such that for every n ∈ N one has: 0 < C0 ≤ length(Γn ) ≤ C and dist(0, Γn ) ≤ C for some constants C0 and C independent of n ∈ N, then, up to a subsequence, Γn → Rφ z(R) + p in C 1 as n → +∞, for some (φ, p) ∈ T2 × R3 and M (φ, p) = 0, being M defined by (1.3). On the other hand, the presence of “stable” zeroes for M is a sufficient condition for the existence of a branch of (κε , τε )-loops with |ε| small. The notion of “stable” zero is expressed by means of the topological degree, as follows. Theorem 1.2 Let K, T ∈ C 1 (R × R3 ) satisfy (1.1), let κε and τε be as in (1.2), and M as in (1.3). If there exists a nonempty bounded open set O in R5 such that deg(M, O, 0) 6= 0 then for |ε| small enough there exists a simple (κε , τε )-loop Γε . Moreover every sequence εn → 0 admits a subsequence, still denoted (εn ), such that Γεn → Γ0 in C 2 as n → +∞, where Γ0 = Rφ z(R) + p for some (φ, p) ∈ O such that M (φ, p) = 0. In presence of nondegenerate zeroes of M we gain regularity on the branch ε 7→ Γε of (κε , τε )loops, and the following result holds. Theorem 1.3 Let K, T ∈ C 2 (R × R3 ) satisfy (1.1), let κε and τε be as in (1.2), and M as in (1.3). If (φ, p) ∈ R2 × R3 is a nondegenerate zero of M (i.e. M (φ, p) = 0 and DM (φ, p) is invertible), then there is ε¯ > 0 and, for |ε| < ε¯, a simple (κε , τε )-loop Γε of class C 4 . Moreover the mapping ε 7→ Γε is of class C 1 ((−¯ ε, ε¯), C 2 ) and Γ0 = Rφ z(R) + p. 3

Hence, if the function M admits only nondegenerate zeroes, by Theorems 1.1 and 1.3, as soon as one switches on the perturbation (K, T ), for small |ε|, C 1 branches of (κε , τε )-loops come out from the manifold Z emanating exactly from those circles in Z corresponding to zeroes of M . Let us sketch the argument used to prove Theorems 1.2 and 1.3. As a first step we introduce the analytical statement of the problem, by identifying curves with corresponding uniform parametrizations. In this way looking for closed curves with prescribed curvature and torsion turns out to be equivalent to study the existence of nonconstant periodic solutions of the Frenet system (see Section 2). Then we introduce a functional setting in such a way that periodic solutions of the Frenet system can be found as zeroes of a pair of suitable nonlinear operators (Fε , J). More precisely the operator Fε acts between spaces of periodic functions and it is naturally defined through the Frenet equations; the operator J takes account of the orthonormality conditions for the Frenet trihedron. The operator Fε can be written as the sum of the “unperturbed” operator F0 , which corresponds to the problem with constant curvature 1 and constant torsion 0, with the perturbation operator G(ε, ·), which exhibits a dependence on K(ε, ·) and T (ε, ·). We point out that the manifold Z made by injective, uniform parametrizations of the circles in Z is (essentially) a set of zeroes of F0 . As a next step we show that the linearized problem of F0 at any point of Z is a Fredholm operator of index zero. This is proved in Section 3. With this information we tackle the perturbed problem making a finite dimensional reduction according to the Lyapunov-Schmidt method (see [14], [15], [18]), based on the Implicit Function Theorem. In this way we construct a function fε : T2 × R3 → R5 whose zeroes correspond, according to a suitable procedure, to zeroes of the pair (Fε , J) and thus to parametrizations of (κε , τε )-loops. This is developed in Section 4. The final step consists in showing that the mapping fε admits zeroes. This is obtained by studying the asymptotic behaviour of fε at ε = 0. The Poincar´e-Melnikov function M defined in (1.3) is essentially the first order term in the Taylor expansion of fε with respect to the smallness parameter ε. Thus the existence of zeroes for fε is related to the presence of stable zeroes of M . This part is discussed in Section 5 where we also prove Theorem 1.1 and we provide a couple of examples in which our general results apply. We conclude by observing that the techniques used for this problem, as well as the kind of results stated in the above theorems are common to a wide class of perturbative problems in different contexts, like Hamiltonian systems (see, e.g., [1], [13], [6]), nonlinear Schr¨odinger type equations (see [5] and the references therein), Yamabe’s problem ([3], [17]), H-bubbles [8], and other problems in conformal geometry (see, e.g., [2] and [9], for the scalar curvature problem for the standard sphere). See also the recent monograph [4]. However, as a technically relevant difference with respect to the above listed problems, here we deal with a nonvariational problem, even in the unperturbed case, and this fact makes more subtle the study of the linearized equation for the unperturbed problem. Moreover, as a consequence of the nonvariational character of our problem, in general, the Poincar´e-Melnikov function M cannot be expressed as a potential. In fact this holds true for the problem of closed curves in the plane,

4

with prescribed curvature; for this case we refer to [7].

2

Preliminaries

Let Γ be a closed, regular, parametric curve in R3 of class C 3 and let p: R → R3 be a parametrization of Γ by arc length, namely |p0 (s)| = 1 for all s ∈ R. The curvature of Γ at the point p(s) is given by the value κ(p(s)) := |p00 (s)|. If κ(p(s)) 6= 0 one defines the normal and the binormal vectors to the curve at the point p(s) as n(s) := p00 (s)/κ(p(s)) and b(s) := p0 (s) ∧ n(s) respectively. The triple {p0 (s), n(s), b(s)} constitutes a set of orthogonal versors at the point p(s) defining the so called Frenet trihedron and the value τ (p(s)) := b0 (s) · n(s) yields the torsion of Γ at the point p(s). We point out that the curvature κ and the torsion τ are geometrical entities associated to the curve which in fact depend on the point p(s) (and not on the parametrization). According to the classical theory of parametric curves in R3 (see [10]), the triple {p0 , n, b} satisfies the following equations, known as Frenet formulas:  00  p = κn (2.1) n0 = −κp0 − τ b  0 b = τn and the orthonormality conditions: |p0 | = |n| = |b| = 1, p0 · n = p0 · b = n · b = 0.

(2.2)

In fact, in (2.1) only two equations are independent because b = p0 ∧ n. In particular, since n = b ∧ p0 , (2.1) and (2.2) hold true if and only if  00 p = κb ∧ p0 (2.3) b0 = τ b ∧ p0 and |p0 | = |b| = 1, p0 · b = 0

(2.4)

are satisfied. Moreover, as p parametrizes a closed curve, p is a nonconstant periodic function. The system (2.3) together with the conditions (2.4) and the periodicity conditions provides the analytical formulation of the problem of finding closed curves with prescribed curvature κ and torsion τ , namely (κ, τ )-loops. Since in general the length of the curve (or, equivalently, the period of solutions of (2.3)) is also unknown, it is convenient to write the system (2.3) in an equivalent way as suggested by the next lemma. k Let us introduce some preliminary notation: Cper will denote the space of C k functions from R 3 into R which are periodic with period 1. Moreover, set 2 1 Ω := {(u1 , u2 ) ∈ Cper × Cper | u1 nonconstant, u2 6= 0}

5

0 and for every u ∈ Cper denote

s Z N (u) :=

1

|u|2 .

0

Lemma 2.1 Let κ ∈ C 1 (R3 ) and τ ∈ C 0 (R3 ), with κ > 0 in R3 . A pair (u1 , u2 ) ∈ Ω solves ( ` u001 = κ(u1 )u2 ∧ u01 ν u02 = τ (u1 )u2 ∧ u01

(2.5)

with ` = N (u01 ) and ν = N (u2 ) if and only if the mappings p(s) := u1 (s/`) and b(s) := (1/ν)u2 (s/`) are nonconstant periodic solutions of (2.3). In this case |p0 (s)| = |b(s)| = 1 for all s ∈ R and p0 (s) · b(s) is constant. If in addition u01 (t0 ) · u2 (t0 ) = 0 for some t0 ∈ R, then p parametrizes by arc-length a (κ, τ )-loop Γ and ` is a multiple of the length of Γ. Proof. By direct computations, one checks the equivalence between the systems (2.3) and (2.5). Moreover, by (2.3), one also obtains that (|p0 |2 )0 = (|b|2 )0 = (p0 · b)0 = 0 namely |p0 |, |b| and p0 · b are R` R1 constant. In particular the equality 0 |p0 |2 = `−1 0 |u01 |2 = ` yields |p0 (s)| ≡ 1. In a similar way one gets |b(s)| ≡ 1. If u01 (t0 ) · u2 (t0 ) = 0 for some t0 ∈ R then p0 (s0 ) · b(s0 ) = 0 for s0 = `t0 and consequently p0 (s) · b(s) = 0 for every s ∈ R. Hence the orthonormality conditions (2.4) are fulfilled and the conclusion follows. By virtue of Lemma 2.1, we look for pairs (u1 , u2 ) ∈ Ω solving the system  N (u01 )  00 (1 + K(ε, u1 )) u2 ∧ u01 u1 = (P )ε N (u2 )  0 u2 = T (ε, u1 )u2 ∧ u01 . Moreover, in order that a nonconstant periodic solution (u1 , u2 ) of (P )ε is geometrically meaningful, it has to satisfy the orthogonality condition u01 · u2 ≡ 0 which in fact is equivalent to the condition u01 (0) · u2 (0) = 0

(2.6)

because, by (P )ε , the mapping t 7→ u01 (t) · u2 (t) is constant. We remark that (P )ε as well as (2.6) is homogeneous with respect to u2 and are invariant with respect to translation in t. k+2 Finally notice that if K and T are of class C k and (u1 , u2 ) ∈ Ω solves (P )ε , then u1 ∈ Cper and k+1 u2 ∈ Cper .

3

The unperturbed problem

In this Section we study the unperturbed system, corresponding to ε = 0, and given by  N (u01 )  00 u2 ∧ u01 u1 = (P )0 N (u2 )  0 u2 = 0 . 6

The main goal is to prove that the linearized problem at any solution (u1 , u2 ) ∈ Ω of (P )0 defines 2 1 0 0 a Fredholm operator of index zero from Cper × Cper into Cper × Cper . As a first step, let us explicitely describe the set of nonconstant periodic solutions of (P )0 . Lemma 3.1 Any solution (u1 , u2 ) ∈ Ω of (P )0 can be written in the following form: u1 (t) = Rz(jt) + p u2 (t) = λRe3

(3.1)

with j ∈ N, R ∈ SO(3), p ∈ R3 , λ > 0 and z defined in (1.5). Notice that all the solutions (u1 , u2 ) ∈ Ω of (P )0 automatically satisfy the orthogonality condition (2.6). Proof. Firstly one has that u2 (t) = λa with λ > 0 and a ∈ S2 . Thus one is lead to look for 1-periodic solutions of the linear equation u001 = `a ∧ u01 (3.2) with ` = N (u01 ). Integrating (3.2) once, one obtains u01 (t) = sin(`t)a ∧ b + (1 − cos(`t))(a · b)a + cos(`t)b with b ∈ R3 arbitrary. Then the general solution of (3.2) is u1 (t) =

1 − cos(`t) sin(`t) a ∧ b + t(a · b)a + (b − (a · b)a) + c ` `

with c ∈ R3 arbitrary. From the equation (3.2) it follows that (|u01 (t)|2 )0 ≡ 0, namely |u01 | is constant. In particular |u01 (t)| = |u01 (0)| = |b| and then ` = N (u01 ) = |b|. Therefore b 6= 0 and one can write b = `ˆb with ˆb ∈ S2 . Now let us impose the periodicity condition u1 (0) = u1 (1). On one hand the equation u1 (0) · a = u1 (1) · a implies a · b = 0. On the other hand, from |u1 (1)| = |u1 (0)| it follows that cos ` = 1 that is ` = 2jπ for some j ∈ N. Hence u1 takes the form u1 (t) = − cos(2jπt)a ∧ ˆb + sin(2jπt)ˆb + a ∧ ˆb + c with a · ˆb = 0. Setting p1 = −a ∧ ˆb, p2 = ˆb and p = a ∧ ˆb + c one writes u(t) = cos(2jπt)p1 + sin(2jπt)p2 + p with |p1 | = |p2 | = 1, p1 · p2 = 0 and u2 (t) ≡ λa = λp1 ∧ p2 . Equivalently (3.1) holds for some R ∈ SO(3). Remark 3.2 Representing a matrix R ∈ SO(3) by means of Euler angles, every solution (u1 , u2 ) ∈ Ω of (P )0 can be equivalently written in the following form: u1 (t) = Rφ z(jt + φ0 ) + p u2 (t) = λRφ e3 7

with j ∈ N, p ∈ R3 , λ > 0, φ0 ∈ R/Z, φ ∈ T2 and Rφ and z defined as in (1.4) and (1.5), respectively. The parameters p, λ, φ0 and φ reflect corresponding simmetries for the problem (P )0 . Some simmetries are of analytical type and arise from the formulation of the problem in terms of a system of ode’s. This is the case for the invariance under dilation with respect to the second component u2 and the invariance under the change t 7→ jt + φ0 . These invariances are exhibited also by the perturbed problem (P )ε . The more meaningful simmetries are those ones of geometrical type, expressed by the parameters φ ∈ T2 and p ∈ R3 , and which are broken by the perturbation. In the following we are allowed to neglect the invariances corresponding to the parameters j, λ and φ0 , and we focus on the set Z := {(Rφ z + p, Rφ e3 ) | (φ, p) ∈ T2 × R3 } . As a next step we study the linearized problem of (P )0 at (u1 , u2 ) ∈ Z. To this purpose, let 2 1 X := Cper × Cper ,

0 0 Y := Cper × Cper

be the Banach spaces endowed with their own standard norms, and let us introduce the operator F0 : Ω ⊂ X → Y defined by   N (u01 ) F0 (u1 , u2 ) := −u001 + u2 ∧ u01 , −u02 for every (u1 , u2 ) ∈ Ω . (3.3) N (u2 ) Notice that F0 is of class C ∞ on its domain. In particular, for fixed (u1 , u2 ) ∈ Ω, the differential F00 (u1 , u2 ) is a bounded linear operator from X into Y acting in the following way:  hu01 , x01 i u2 ∧ u01 F00 (u1 , u2 )[x1 , x2 ] = −x001 + N (u01 )N (u2 )  N (u01 ) N (u01 )hu2 , x2 i 0 0 + (u2 ∧ x01 + x2 ∧ u01 ) − u ∧ u , −x (3.4) 2 1 2 N (u2 ) N (u2 )3 for every (x1 , x2 ) ∈ X, where in general Z hu, vi =

1

u·v . 0

In the following, X and Y will be equipped with the L2 inner product: Z 1 h(u1 , u2 ), (v1 , v2 )i = hu1 , v1 i + hu2 , v2 i = (u1 · v1 + u2 · v2 ) .

(3.5)

0

The notion of orthogonality we will consider will be always refered to the above inner product. Lemma 3.3 For every (u1 , u2 ) ∈ Z one has ker F00 (u1 , u2 ) = {(a ∧ u1 + b, a ∧ u2 + λu2 ) | a, b ∈ R3 , λ ∈ R}.

8

Proof. Firstly let us prove the lemma taking (φ, p) = (0, 0), namely (u1 , u2 ) = (z, e3 ). Notice that (x1 , x2 ) ∈ ker F00 (z, e3 ) if and only if (x1 , x2 ) is a 1-periodic solution of
 00 x1 = 2πe3 ∧ x01 + 2πx2 ∧ z 0 − hz 0 , x01 i − (2π)2 he3 , x2 i z (3.6) x02 = 0. First, observe that x2 has to be constant. Hence x2 (t) ≡ a2 ∈ R3 and one is lead to look for 1-periodic solutions of x001 = 2πe3 ∧ x01 + 2πa2 ∧ z 0 − αz (3.7) with α = hz 0 , x01 i − (2π)2 e3 · a2 .

(3.8)

Integrating (3.7) once, one gets x01 (t) = L(t)b1 + L(t)

Z

t

L(−s)q(s) ds 0

where L(t)p q(t)

= =

sin(2πt)e3 ∧ p + (1 − cos(2πt))(e3 · p)e3 + cos(2πt)p

(3.9)

0

2πa2 ∧ z (t) − αz(t)

and b1 ∈ R3 is arbitrary. Making computations one finds L(−s)q(s) = −(α + (2π)2 a23 )e1 + (2π)2 (sin(2πs)a22 + cos(2πs)a21 )e3 where we denote a2i = a2 · ei for i = 1, 2, 3. Therefore one has Z t L(−s)q(s) ds = −(α + (2π)2 a23 )te1 + 2π((1 − cos(2πt))a22 + sin(2πt)a21 )e3 0

and then x01 (t) = L(t)b1 − (α + (2π)2 a23 )tz(t) + 2π((1 − cos(2πt))a22 + sin(2πt)a21 )e3 . Observing that x01 (0) = b1 and x01 (1) = b1 −(α +(2π)2 a23 )e1 , and imposing the periodicity condition x01 (0) = x01 (1) one obtains α + (2π)2 a23 = 0 . (3.10) Moreover, after computations, (3.8) and (3.10) imply Z 0=

1

z 0 · x01 = 2πb1 · e2 .

0

Hence, it remains x01 (t) = (b13 + 2πa22 )e3 + sin(2πt)(b11 e2 + 2πa21 e3 ) + cos(2πt)(b11 e1 − 2πa22 e3 ) 9

where, as before, b11 = b1 · e1 and b13 = b1 · e3 . Thus     b11 b11 x1 (t) = a1 + (b13 + 2πa22 )te3 + (1 − cos(2πt)) e2 + a21 e3 + sin(2πt) e1 − a22 e3 2π 2π where a1 ∈ R3 is arbitrary. Since x1 (0) = a1 and x1 (1) = a1 + (b13 + 2πa22 )e3 , in order that x(t) is 1-periodic, it must be b13 + 2πa22 = 0. Hence, 1-periodic solutions of (3.6) are given by     b11 b11 x1 (t) = a1 + (1 − cos(2πt)) e2 + a21 e3 + sin(2πt) e1 − a22 e3 (3.11) 2π 2π x2 (t) = a2 where a1 , a2 ∈ R3 and b11 ∈ R are arbitrary. Setting a = e3 ∧ a2 − b11 /(2π)e3 , b = a1 − a ∧ e1 and λ = a23 , the solution (3.11) takes the form x1 (t) = a ∧ z(t) + b x2 (t) = a ∧ e3 + λe3 with arbitrary a, b ∈ R3 and λ ∈ R. Finally, we recover the result for any (u1 , u2 ) ∈ Z. For every R ∈ SO(3) and (p1 , p2 ) ∈ R3 × R3 set R(p1 , p2 ) := (Rp1 , Rp2 ). Using this notation and (3.4) one can check that F00 (Rz + p, Re3 )[Rx1 , Rx2 ] = R (F00 (z, e3 )[x1 , x2 ]) . (3.12) Hence, taking (u1 , u2 ) = (Rφ z + p, Rφ e3 ) ∈ Z, thanks to the result proved in case (φ, p) = (0, 0), we have that ker F00 (Rφ z + p, Rφ e3 ) = {(Rφ (a ∧ z + b), Rφ (a ∧ e3 + λe3 )) | a, b ∈ R3 , λ ∈ R} which, up to an obvious equivalence, yields the statement of the lemma. Given any (u1 , u2 ) ∈ Z, let us introduce the linear subspace of Y given by: Y0 (u1 , u2 ) := {(y1 , y2 ) ∈ Y | hF00 (u1 , u2 )[x1 , x2 ], (y1 , y2 )i = 0 for all (x1 , x2 ) ∈ X} . For next purposes, the following more explicit characterization of Y0 (u1 , u2 ) is useful. Lemma 3.4 For every (u1 , u2 ) ∈ Z one has Y0 (u1 , u2 ) = {(λu01 +a, 2πa∧u1 +b) | λ ∈ R, a, b ∈ R3 }. Proof. Let (u1 , u2 ) = (Rφ z + p, Rφ e3 ) ∈ Z. Thanks to (3.12) one has that Y0 (Rφ z + p, Rφ e3 ) = {(Rφ y1 , Rφ y2 ) | (y1 , y2 ) ∈ Y0 (z, e3 )}

(3.13)

and then we can limit ourselves to prove the lemma for (u1 , u2 ) = (z, e3 ). For every (x1 , x2 ) ∈ X set α(x01 , x2 ) = hz 0 , x01 i − (2π)2 he3 , x2 i . (3.14) Hence (y1 , y2 ) ∈ Y0 (z, e3 ) if and only if (y1 , y2 ) is a 1-periodic solution of h(−x001 − α(x01 , x2 )z + 2π(e3 ∧ x01 + x2 ∧ z 0 ), −x02 ) , (y1 , y2 )i = 0 10

for all (x1 , x2 ) ∈ X .

(3.15)

In particular, taking x2 = 0, it must be 2 h−x001 − α(x01 , 0)z + 2πe3 ∧ x01 , y1 i = 0 for all x1 ∈ Cper .

Observing that α(x01 , 0) = (2π)2 hz, x1 i, (3.16) is equivalent to   Z 1 Z 1 Z 1 Z 1 z · y1 = 0 z · x1 x01 · y1 ∧ e3 − (2π)2 x001 · y1 + 2π − It is standard to recognize that y1 ∈ of

0 Cper

2 for all x1 ∈ Cper . (3.17)

0

0

0

0

(3.16)

solves (3.17) if and only if y1 is a (weak) 1-periodic solution

y100 = 2πe3 ∧ y10 − βz β = (2π)2 hz, y1 i. Arguing as in the proof of Lemma 3.3 one finds 

y10 (t) = L(t)b1 − βtz(t) where L(t) is given by (3.9) and b1 ∈ R3 is an arbitrary vector. Imposing the periodicity condition y10 (0) = y10 (1) one infers that it must be β = 0, namely Z 1 z · y1 = 0. (3.18) 0

Integrating once more, one obtains y1 (t) = a1 +

1 − cos(2πt) sin(2πt) e3 ∧ b1 + (e3 · b1 )te3 − ((e3 · b1 )e3 − b1 ) 2π 2π

with a1 ∈ R3 arbitrary. The periodicity condition y1 (0) = y1 (1) yields e3 · b1 = 0 and thus sin(2πt) 1 − cos(2πt) e3 ∧ b1 + b1 . 2π 2π Now we impose (3.18) obtaining the further restriction e2 · b1 = 0. Therefore b1 = b11 e1 where b11 ∈ R is arbitrary, and then y1 (t) = a1 +

y1 (t) = −

b11 0 b11 z (t) + e2 + a1 . (2π)2 2π

Hence, up to redefine the constants one concludes that the general solution of (3.17) is given by y1 (t) = λz 0 (t) + a

(3.19)

with arbitrary λ ∈ R and a ∈ R3 . Now one plugs (3.19) into (3.15) finding the following equation for y2 : Z 1 Z 1 1 2πa · x2 ∧ z 0 = x02 · y2 for all x2 ∈ Cper , 0

0 0

namely y2 is a (weak) 1-periodic solution of z ∧ a +

1 0 2π y2

= 0. Hence

y2 (t) = 2πa ∧ z(t) + b with b ∈ R arbitrary. Finally one can check that any pair of the form (y1 , y2 ) = (λz 0 +a, 2πa∧z +b) solves (3.15). This concludes the proof. 3

11

⊥

Notice that, by definition, Y0 (u1 , u2 ) = (im F00 (u1 , u2 )) , where the orthogonality is meant with respect to the inner product (3.5). In fact we also have: ⊥

Lemma 3.5 For every (u1 , u2 ) ∈ Z one has im F00 (u1 , u2 ) = (Y0 (u1 , u2 )) . ⊥

⊥

Proof. Since by definition Y0 (u1 , u2 ) = (im F00 (u1 , u2 )) , the inclusion im F00 (u1 , u2 ) ⊆ (Y0 (u1 , u2 )) is trivial and we have just to prove the opposite one. Let us begin with (u1 , u2 ) = (z, e3 ). For any fixed (w1 , w2 ) ∈ (Y0 (z, e3 ))⊥ we look for (x1 , x2 ) ∈ X satisfying F00 (z, e3 )[x1 , x2 ] = (w1 , w2 ), that is  −x001 − α(x01 , x2 )z + 2π(e3 ∧ x01 + x2 ∧ z 0 ) = w1 (3.20) −x02 = w2 where α(x01 , x2 ) is given by (3.14). Since h(w1 , w2 ), (y1 , y2 )i = 0 for every (y1 , y2 ) ∈ Y0 (z, e3 ), the representation stated by Lemma 3.4 yields h(w1 , w2 ), (z 0 , 0)i = 0, i.e., hw1 , z 0 i = 0

(3.21) Z

h(w1 , w2 ), (0, ei )i = 0 for i = 1, 2, 3, i.e.,

1

w2 = 0

(3.22)

0

h(w1 , w2 ), (e1 , 2πe1 ∧ z)i = 0, i.e., hw1 , e1 i = −2πhw2 , (e2 · z)e3 i

(3.23)

h(w1 , w2 ), (e2 , 2πe2 ∧ z)i = 0, i.e., hw1 , e2 i = 2πhw2 , (e1 · z)e3 i

(3.24)

0

h(w1 , w2 ), (e3 , 2πe3 ∧ z)i = 0, i.e., hw1 , e3 i = −hw2 , z i .

(3.25)

Now, the second equation in (3.20) is solved by Z x2 (t) = d0 −

t

w2

(3.26)

0 1 thanks to (3.22). Integrating the first where d0 ∈ R3 is arbitrary. Notice that x2 belongs to Cper equation in (3.20) we obtain Z t x01 (t) = L(t)c1 + L(t) L(−s)f (s) ds (3.27) 0

where L(t) is given in (3.9), c1 ∈ R3 is an arbitrary constant vector which shall satisfy some restrictions and f (s) = 2πx2 (s) ∧ z 0 (s) − α(x01 , x2 )z(s) − w1 (s) . One can explicitely compute
−(2π)2 x2 (s) · e3 − α(x01 , x2 ) − w1 · z(s) e1
1 − w1 (s) · z 0 (s)e2 + (2π)2 x2 (s) · z(s) − w1 (s) · e3 e3 . 2π R1 The periodicity condition x01 (0) = x01 (1) is equivalent to 0 L(−s)f (s) ds = 0, namely: L(−s)f (s)

=

Z

1


−(2π)2 x2 · e3 − α(x01 , x2 ) − w1 · z = 0

0

12

(3.28)

Z

1

w1 · z 0 = 0

(3.29)

0

Z

1


(2π)2 x2 · z − w1 · e3 = 0 .

(3.30)

0

One has that (3.29) is (3.21) and thus it holds true. Also (3.30) is satisfied because, by (3.25) and by the second equation in (3.20), one has hw1 , e3 i = −hw2 , z 0 i = hx02 , z 0 i = −hx2 , z 00 i = (2π)2 hx2 , zi. Hence it suffices to check (3.28) which in fact, using (3.14), is equivalent to hz 0 , x01 i = −hw1 , zi .

(3.31)

By explicit computations, (3.27) gives x01 (t) = A1 (t)z(t) −

1 A2 (t)z 0 (t) + A3 (t)e3 2π

(3.32)

where t

Z

(2π)2 x2 · e3 + α(x01 , x2 ) + z · w1 0 Z t 1 A2 (t) = −c12 + z 0 · w1 2π 0 Z t
A3 (t) = c13 + (2π)2 x2 · z − e3 · w1 A1 (t)

= c11 −




(3.33)

0

and c1i = c1 · ei for i = 1, 2, 3. Thus (3.31) turns out to be equivalent to Z 1 1 ((1 − t)z 0 − z) · w1 . c12 = 2π 0

(3.34)

Integrating (3.32) one obtains Z t x1 (t) = c0 + 0

1 A2 z 0 + A3 e3 A1 z − 2π

 (3.35)

where c0 ∈ R3 is arbitrary. Using (3.23), (3.24) and the second equation in (3.20), one can check
R1 R1 1 A2 z 0 = 0, so that x1 is periodic if and only if 0 A3 = 0, i.e., upon explicit that 0 A1 z − 2π computations, Z Z 1

c13 = −d02 +

1

((1 − t)z 0 + z − e1 ) · w2

(1 − t)e3 · w1 + 0

(3.36)

0 3

where d02 = d0 · e2 . Hence for arbitrary c0 , c1 , d0 ∈ R with c12 and c13 satisfying (3.34) and (3.36) the pair (x1 , x2 ) given by (3.35) and (3.26) yields a periodic solution of (3.20). This concludes the proof in the case (u1 , u2 )= (z, e3 ). For  an arbitrary (u1 , u2 ) = (Rφ z + p, Rφ e3 ) ∈ Z one observes
⊥ ⊥ ⊥ that Y0 (u1 , u2 ) = Rφ (Y0 (z, e3 )) , by (3.13). Therefore, fixing (v1 , v2 ) ∈ (Y0 (u1 , u2 )) , by the ⊥

first part of the proof, there exists (x1 , x2 ) ∈ X such that F00 (z, e3 ) = (Rφ−1 v1 , Rφ−1 v2 ) ∈ (Y0 (z, e3 )) . Then, from (3.12) it follows that F00 (u1 , u2 )[Rφ x1 , Rφ x2 ] = Rφ (F00 (z, e3 )[x1 , x2 ]) = (v1 , v2 ). This completes the proof. 13

For future convenience, let us remark that, thanks to (3.32), (3.33), (3.36), (3.26) and (3.12), fixing (u1 , u2 ) = (Rφ z + p, Rφ e3 ) ∈ Z, if F00 (u1 , u2 )[x1 , x2 ] = (y1 , y2 ) then x01 (0)

Z

1

· Rφ e3 + 2πx2 (0) · Rφ e2 =

(1 − t)e3 · 0

Rφ−1 y1

Z + 0

1

((1 − t)z 0 + z − e1 ) · Rφ−1 y2 .

(3.37) ⊥

Remark 3.6 For every (u1 , u2 ) ∈ Z the operator F00 (u1 , u2 ) is bijective from (ker F00 (u1 , u2 )) onto im F00 (u1 , u2 ). In particular, if (u1 , u2 ) = (Rφ z +p, Rφ e3 ), denoting F00 (Rφ z +p, Rφ e3 )−1 the inverse ⊥ of the restriction of F00 (Rφ z + p, Rφ e3 ) to the subspace (ker F00 (Rφ z + p, Rφ e3 )) , thanks to (3.12), 0 0 one has that im F0 (Rφ z + p, Rφ e3 ) = Rφ (im F0 (z, e3 )) and F00 (Rφ z + p, Rφ e3 )−1 = Rφ F00 (z, e3 )−1 Rφ−1 .

4

(3.38)

Lyapunov-Schmidt reduction for the perturbed problem

In this section we study the perturbed problem Fε (u1 , u2 ) := F0 (u1 , u2 ) + G(ε; u1 , u2 ) = 0

(4.1)

where F0 : Ω ⊂ X → Y is the operator (3.3) corresponding to the unperturbed problem (P )0 , whereas G: R × Ω → Y is given by   N (u01 ) K(ε, u1 )u2 ∧ u01 , T (ε, u1 )u2 ∧ u01 for ε ∈ R and (u1 , u2 ) ∈ Ω (4.2) G(ε; u1 , u2 ) = N (u2 ) with K, T ∈ C 1 (R × R3 ) satisfying (1.1). 0 and Q ∈ C 0 (R × R3 ), writing Q(ε, u) Here and in the rest of the paper, given mappings u ∈ Cper 0 we mean both the function t 7→ Q(ε, u(t)) and the Nemitski operator u 7→ Q(ε, u) ∈ Cper . We point out that (u1 , u2 ) ∈ Ω solves (P )ε if and only if Fε (u1 , u2 ) = 0. Let us introduce some new short notation: θ := (φ, p) ∈ T2 × R3

and ωθ := (Rφ z + p, Rφ e3 ) ∈ Z .

Observe that the mapping θ 7→ ωθ is of class C ∞ from T2 × R3 into X. From Lemmata 3.3, 3.4 and 3.5 for every θ ∈ T2 × R3 one has that F00 (ωθ ) is a Fredholm operator of index zero from X into Y and the following decompositions hold: X Y

= =

⊥

ker F00 (ωθ ) ⊕ (ker F00 (ωθ ))

(4.3)

⊥ F00 (ωθ ))

(4.4)

im

F00 (ωθ )

⊕ (im

⊥

where all the subspaces are closed and dim ker F00 (ωθ ) = dim (im F00 (ωθ )) = 7. The next goal is to find a regular mapping (ε, θ) 7→ ηε (θ) ∈ X, defined for |ε| small and θ in a ⊥ compact set, such that Fε (ωθ + ηε (θ)) ∈ (im F00 (ωθ )) . As a consequence, the problem of searching solutions of (P )ε is essentially reduced to the finite-dimensional problem of looking for zeroes of the 14

mapping (ε, θ) 7→ Fε (ωθ + ηε (θ)), namely to study a system of seven equations in five unknowns, ⊥ after fixing a basis in (im F00 (ωθ )) . This goal will be achieved through a reduction procedure in the spirit of the Lyapunov-Schmidt method (see [14], [15], [18]). Let us begin by fixing an orthonormal basis {ζ1 (θ), . . . , ζ7 (θ)} ⊂ X for ker F00 (ωθ ) and an orthonormal basis {ξ1 (θ), . . . , ξ7 (θ)} ⊂ Y for (im F00 (ωθ ))⊥ . In particular we can take ζi (θ) = Rφ ζ¯i and ξi (θ) = Rφ ξ¯i where

for i = 1, . . . , 7 and θ = (φ, p) ,

ζ¯1 = (0, e3 ) ξ¯1 = (e3 ∧ z, 0) ¯ ζ1+i = q (ei , 0) ξ¯1+i = (0, ei ) for i = 1, 2, 3 1 (ei , 2πei ∧ z) for i = 1, 2 ζ¯4+i = 23 (ei ∧ z, ei ∧ e3 ) ξ¯4+i = √1+2π 2 1 ¯ ¯ √ (e3 , 2πe3 ∧ z) . ζ7 = (e3 ∧ z, 0) ξ7 = 1+4π 2

The following lemma, which gives the finite dimensional reduction of the problem, constitutes the main result of this Section and the key step for the proof of Theorems 1.2 and 1.3. Lemma 4.1 Let K, T ∈ C 1 (R × R3 ) satisfy (1.1). For every r > 0 there exist a value εr > 0 and unique mappings ε 7→ η ε ∈ C 0 (T2 ×B r , X) and ε 7→ µε ∈ C 0 (T2 ×B r , R7 ) of class C 1 from (−εr , εr ) into their target spaces, such that for every ε ∈ (−εr , εr ) and for every θ ∈ T2 × B r one has: kη ε (θ)kX < 2π 0

η (θ) = 0

and

Fε (ωθ + η ε (θ)) =

(4.5) 0

µ (θ) = 0 7 X

µεi (θ)ξi (θ)

(4.6) (4.7)

i=1

hη ε (θ), ζi (θ)i = 0

for i = 1, . . . , 7 .

Moreover for every θ ∈ T2 × B r there results ! 7 X dη ε (θ) = F00 (ωθ )−1 h∂ε G(0; ωθ ), ξi (θ)i ξi (θ) − ∂ε G(0; ωθ ) dε ε=0 i=1 dµεi (θ) = h∂ε G(0; ωθ ), ξi (θ)i for i = 1, . . . , 7 . dε ε=0

(4.8)

(4.9) (4.10)

In addition, if µε (θ) = 0 for some ε ∈ (−εr , εr ) and θ ∈ T2 × B r then (uε1 , uε2 ) := ωθ + η ε (θ) belongs to Ω and it solves problem (P )ε . If, furthermore, K, T ∈ C 2 (R × R3 ), then the mappings ε 7→ η ε and ε 7→ µε belong to C 2 ((−εr , εr ), C 0 (T2 × B r , X)) ∩ C 1 ((−εr , εr ), C 1 (T2 × B r , X)) and to C 2 ((−εr , εr ), C 0 (T2 × B r , R7 )) ∩ C 1 ((−εr , εr ), C 1 (T2 × B r , R7 )), respectively. Remark 4.2 (i) The right hand side in (4.9) is well defined because, according to (4.4), for any P7 0 (y1 , y2 ) ∈ Y one has that i=1 h(y1 , y2 ), ξi (θ)i ξi (θ) − (y1 , y2 ) belongs to im F0 (ωθ ) which is the 0 −1 domain of F0 (ωθ ) (see Remark 3.6). (ii) The mapping µε can be expressed in terms of η ε since by (4.7) one has that µεi (θ) = hFε (ωθ + η ε (θ), ξi (θ)i for all i = 1, . . . , 7 and θ ∈ T2 × B r . 15

Proof. For fixed r > 0, let us set Ur := T2 × Br and let us introduce the Banach spaces Xr0 := C 0 (U r , X),

Yr0 := C 0 (U r , Y ),

Rr0 := C 0 (U r , R7 )

endowed with their standard norms. Clearly the map θ 7→ ωθ , as well as ζ1 , . . . , ζ7 , belong to Xr0 whereas ξ1 , . . . , ξ7 ∈ Yr0 . Moreover, introduce the open subset of Xr0 given by: B := {η ∈ Xr0 | kηkXr0 < 2π} and the function F : R × B × Rr0 → Yr0 × Rr0 defined as follows: F (ε, η, µ)(θ) :=

Fε (ωθ + η(θ)) −

7 X

! µi (θ)ξi (θ), hη(θ), ζ1 (θ)i, . . . , hη(θ), ζ7 (θ)i

i=1

for all θ ∈ U r and for every (ε, η, µ) ∈ R × B × Rr0 . Observe that ωθ + η(θ) ∈ Ω for every θ ∈ U r since η ∈ B. Hence F is well defined on R × B × Rr0 . Our goal is to apply the Implicit Function Theorem to F at the point (0, 0, 0) in order to find mappings ε 7→ η ε and ε 7→ µε such that F (ε, η ε , µε ) = (0, 0) namely (4.7) and (4.8). Regularity of F . One has that F (ε, η, µ) = F (0, η, µ) + G (ε, η) where G (ε, η)(θ) := (G(ε; ωθ + η(θ)), 0, . . . , 0).

(4.11)

One can easily check that the mapping (η, µ) 7→ F (0, η, µ) is of class C ∞ because F0 is so. As far as concerns the regularity of G , by the definition (4.2) of G, it follows from the regularity of the Nemitski operators associated to K and T . More precisely, as K, T are of class C 1 , the mappings 2 0 K , T : R × C 0 (U r , Cper ) → C 0 (U r , Cper ) defined by: K (ε, χ)(θ) := K(ε, χ(θ))

and T (ε, χ)(θ) := T (ε, χ(θ))

(4.12)

are of class C 1 (see [12] for the details). As a consequence, also G turns out to be of class C 1 from R × B ⊂ R × Xr0 into Yr0 . Now let us study the linearized problem for F at (0, 0, 0). Clearly F (0, 0, 0) = (F0 (ωθ ), 0) = (0, 0) because G(0; ·) = 0. Moreover, considering the bounded linear operator L :=

∂F (0, 0, 0) : Xr0 × Rr0 → Yr0 × Rr0 , ∂(η, µ)

for every (ϕ, ν) ∈ Xr0 × Rr0 we have that L (ϕ, ν)(θ) =

F00 (ωθ )[ϕ(θ)] −

7 X

! νi (θ)ξi (θ), hϕ(θ), ζ1 (θ)i, . . . , hϕ(θ), ζ7 (θ)i

i=1

We will show that L is bijective from Xr0 × Rr0 onto Yr0 × Rr0 . Injectivity. Let (ϕ, ν) ∈ Xr0 × Rr0 be such that L (ϕ, ν) = 0, namely  0 P7 F0 (ωθ )[ϕ(θ)] = i=1 νi (θ)ξi (θ) hϕ(θ), ζi (θ)i = 0 i = 1, . . . , 7 16

.

Since ξi (θ) ∈ (im F00 (ωθ ))⊥ , the first equation implies νi (θ) = 0 for every i = 1, . . . , 7 and thus ⊥ ϕ(θ) ∈ ker F00 (ωθ ). On the other hand the second equation means that ϕ(θ) ∈ (ker F00 (ωθ )) . Hence ϕ(θ) = 0. As θ is arbitrary, we have injectivity. Surjectivity. Given (ψ, ρ) ∈ Yr0 × Rr0 we have to find (ϕ, ν) ∈ Xr0 × Rr0 such that L (ϕ, ν) = (ψ, ρ), namely, for every θ ∈ U r F00 (ωθ )[ϕ(θ)] −

7 X

νi (θ)ξi (θ) = ψ(θ)

(4.13)

i=1

hϕ(θ), ζi (θ)i = ρi (θ) i = 1, . . . , 7 .

(4.14)

Fix θ ∈ U r . According to the decomposition (4.3)–(4.4) and thanks to the orthonormality of the sets {ζi (θ)}7i=1 and {ξi (θ)}7i=1 we can write ϕ(θ)

=

7 X

hϕ(θ), ζi (θ)iζi (θ) + ϕ(θ) ¯

i=1

ψ(θ)

=

7 X

¯ hψ(θ), ξi (θ)iξi (θ) + ψ(θ)

i=1 ⊥

with ϕ(θ) ¯ ∈ (ker F00 (ωθ ))

¯ and ψ(θ) ∈ im F00 (ωθ ). Then (4.14) gives ϕ(θ) =

7 X

ρi (θ)ζi (θ) + ϕ(θ) ¯ .

(4.15)

i=1 ⊥

Moreover, by (4.13) and since ξi (θ) ∈ (im F00 (ωθ ))

we have that

νi (θ) = −hψ(θ), ξi (θ)i i = 1, . . . , 7 .

(4.16)

¯ In addition F00 (ωθ )[ϕ(θ)] ¯ = ψ(θ) because ζi (θ) ∈ ker F00 (ωθ ). According to Remark 3.6, we can take ¯ ϕ(θ) ¯ = F00 (ωθ )−1 [ψ(θ)] .

(4.17)

As θ varies, the expression (4.16) defines a continuous mapping from U r into R7 , that is ν ∈ Rr0 . As far as concerns the regularity of the function θ 7→ ϕ(θ) defined by (4.15) and (4.17), we observe that by (3.38) ¯ ϕ(θ) ¯ = Rφ F00 (z, e3 )−1 Rφ−1 [ψ(θ)] . This shows that ϕ¯ ∈ Xr0 and then, by (4.15), also ϕ ∈ Xr0 . Thus the surjectivity is proved. Hence we can apply the Implicit Function Theorem and, since F (0, 0, 0) = (0, 0), (4.6)–(4.8) follows. As regards (4.9) and (4.10), we recall that   ε dη dµε ∂F (0, 0, 0) , = −L −1 (4.18) dε ε=0 dε ε=0 ∂ε

17

Moreover, according to what found for the surjectivity, we have L

−1

(ψ, ρ)(θ) =

7 X

ρi (θ)ζi (θ) +

F00 (ωθ )−1

ψ(θ) −

i=1

7 X

!

!

hψ(θ), ξi (θ)iξi (θ) , −hψ(θ), ξi (θ)i

i=1

(4.19) for every (ψ, ρ) ∈ Yr0 × Rr0 . In addition ∂F (0, 0, 0) (θ) = (∂ε G(0; ωθ ), 0) . ∂ε

(4.20)

In conclusion (4.9) and (4.10) follow from (4.18)–(4.20). The fact that if µε (θ) = 0 then ωθ + η ε (θ) provides a solution of (P )ε immediately follows from (4.7) and from the definition of Fε . Finally let us discuss the last part of the lemma, concerning the regularity of the mappings ε 7→ η ε and ε 7→ µε when K, T ∈ C 2 (R × R3 ). We just give a sketch and we refer again to [12] for a more detailed proof. As K, T are of class C 2 , the operator G : R × B → Yr0 defined in (4.11) turns out to be of class C 2 and then one readily finds that the mappings ε 7→ η ε and ε 7→ µε are of class C 2 from (−εr , εr ) into C 0 (U r , X) and C 0 (U r , R7 ), respectively. The further regularity is accomplished by repeating the same argument of the C 1 regularity, by making a different choice of the spaces. More precisely, instead of Xr0 , Yr0 and Rr0 we take Xr1 := C 1 (U r , X),

Yr1 := C 1 (U r , Y ),

Rr1 := C 1 (U r , R7 )

endowed with their standard norms. Clearly, in this case, the set B is given by {η ∈ Xr1 | kηkXr1 < 2π}. The proof goes on exactly as before without substantial differences. The only remark concerns the proof of the regularity of the mapping F and, in particular, the regularity of G . Although 2 the Nemitski operators K and T defined by (4.12) are of class C 2 from R × C 0 (U r , Cper ) into 0 0 C (U r , Cper ), the regularity of the operator G defined in (4.11)(ε, η) 7→ G (ε, η) is just C 1 (as needed in order to apply the Implicit Function Theorem) because the existence of the differential dG (ε, η) as a bounded linear operator from R × Xr1 into Yr1 , as well as its continuous dependence on (ε, η) involve the second order partial derivatives of K and T . Hence we can find C 1 functions ε 7→ η ε (θ) ∈ C 1 (U r , X) and ε 7→ µε (θ) ∈ C 1 (U r , R7 ) satisfying F (ε, η ε , µε ) = 0 and the conclusion follows. In fact we are interested in solutions (u1 , u2 ) ∈ Ω of (4.1) satisfying the additional condition · u2 ≡ 0 or equivalently (2.6) which, according to Lemma 2.1, guarantees that (u1 , u2 ) identifies a (κε , τε )-loop. To this purpose let us introduce the functional J: X → R defined as follows: u01

J(u1 , u2 ) := u01 (0) · u2 (0)

for every (u1 , u2 ) ∈ X .

(4.21)

Observe that J ∈ C ∞ (X) and in particular J 0 (u1 , u2 )[x1 , x2 ] = u01 (0) · x2 (0) + u2 (0) · x01 (0)

18

for every (u1 , u2 ), (x1 , x2 ) ∈ X .

(4.22)

Now, fixing r > 0 let εr > 0, η ε and µε be given according to lemma 4.1. For every ε ∈ (−εr , εr ) let us introduce the mapping f¯ε : T2 × B r → R7 × R given by f¯ε (θ) := (µε (θ), J(ωθ + η ε (θ))) .

(4.23)

Our goal is to look for zeroes of f¯ε since, by the last part of Lemma 4.1 and by the above discussion, if f¯ε (θ) = 0 (for some ε and θ) then ωθ +η ε (θ) is a nonconstant periodic solution of (P )ε and satisfies (2.6), namely it corresponds to a (κε , τε )-loop. Notice that, according to Lemma 4.1, f¯ε ∈ C 0 (T2 × B r , R8 ) for every ε ∈ (−εr , εr ), f0 ≡ 0 and the mapping ε 7→ f¯ε is of class C 1 from (−εr , εr ) into C 0 (T2 × B r , R8 ). If K, T ∈ C 2 (R × R3 ) (and not just of class C 1 ) then the mapping ε 7→ f¯ε belongs both to C 2 ((−εr , εr ), C 0 (T2 × B r , R8 )) and to C 1 ((−εr , εr ), C 1 (T2 × B r , R8 )). The next result makes clear the relationship between the function M defined in (1.3) and the first order term in the expansion of f¯ε with respect to ε. Lemma 4.3 There exist Φ ∈ GL(8, R) and Ψ ∈ GL(5, R) such that, setting Φf¯ε =: (fε , f˜ε ): T2 × B r → R5 × R3 , as ε → 0 one has: fε = εΨM + o(ε) in C 0 (T2 × B r , R5 ) f˜ε = o(ε) in C 0 (T2 × B r , R3 ) where M is defined in (1.3). The convergences hold in C 1 (T2 × B r , R5 ) and in C 1 (T2 × B r , R3 ) respectively, if K, T ∈ C 2 (R × R3 ). Moreover if fε (θ) = 0 for some ε ∈ (−εr , εr ) and θ ∈ T2 × B r , then f¯ε (θ) = 0. Proof. By (4.6), (4.9) and (4.10), the first order Taylor expansion of ε 7→ f¯ε ∈ C 0 (T2 × B r , R8 ) at ε = 0 is !! 7 X 0 0 −1 fε (θ) = ε h∂ε G(0; ωθ ), ξi (θ)i , J (ωθ )F0 (ωθ ) h∂ε G(0; ωθ ), ξi (θ)i ξi (θ) − ∂ε G(0; ωθ ) + o(ε) i=1

where o(ε)/ε → 0 in C 0 (T2 ×B r , R8 ). According to the last part of Lemma 4.1, if K, T ∈ C 2 (R×R3 ) one can take C 1 (T2 × B r , R8 ) instead of C 0 (T2 × B r , R8 ). Since ∂ε G(0; ωθ ) = −2π(2π∂ε K(Rφ z + p)Rφ z, ∂ε T (Rφ z + p)Rφ z) for any θ = (φ, p) ∈ T2 × B r , we readily get h∂ε G(0; ωθ ), ξi (θ)i =

0

for i = 1, 4, 7 Z 1 h∂ε G(0; ωθ ), ξ2 (θ)i = −2π ∂ε T (0, Rφ z + p)z · e1 = −2πM3 (θ)

(4.24) (4.25)

0

Z h∂ε G(0; ωθ ), ξ3 (θ)i = −2π

1

∂ε T (0, Rφ z + p)z · e2 = −2πM4 (θ)

(4.26)

0

Z 1 4π 2 4π 2 ∂ε K(0, Rφ z + p)z · e1 = − √ M1 (θ) (4.27) 1 + 2π 2 0 1 + 2π 2 Z 1 4π 2 4π 2 h∂ε G(0; ωθ ), ξ6 (θ)i = − √ ∂ε K(0, Rφ z + p)z · e2 = − √ M2 (θ) . (4.28) 1 + 2π 2 0 1 + 2π 2 h∂ε G(0; ωθ ), ξ5 (θ)i = − √

19

Set (x1 (θ), x2 (θ))

:=

(y1 (θ), y2 (θ))

:=

7 X

h∂ε G(0; ωθ ), ξi (θ)iξi (θ) i=1 F00 (ωθ )[x1 (θ), x2 (θ)]

− ∂ε G(0; ωθ )

⊥

and note that, by (4.8), (x1 (θ), x2 (θ)) ∈ (ker F00 (ωθ )) . Moreover, using (4.9) and (4.24)–(4.28), one can compute y1 (θ) y2 (θ)

4π 2 4π 2 M (θ)R e − M2 (θ)Rφ e2 + 4π 2 ∂ε K(0, Rφ z + p)Rφ z 1 φ 1 1 + 2π 2 1 + 2π 2 8π 3 M1 (θ)Rφ (e1 ∧ z) = −2πM3 (θ)Rφ e1 − 2πM4 (θ)Rφ e2 − 1 + 2π 2 8π 3 − M2 (θ)Rφ (e2 ∧ z) + 2π∂ε T (0, Rφ z + p)Rφ z 1 + 2π 2 = −

so that, by (3.37) and (4.22) J 0 (ωθ )[x1 (θ), x2 (θ)]

= (Rφ z + p)0 (0) · x2 (θ)(0) + Rφ e3 · x1 (θ)0 (0) 2πRφ e2 · x2 (θ)(0) + Rφ e3 · x1 (θ)0 (0) Z 1 Z 1 −1 = (1 − t)e3 Rφ y1 (θ)(t) dt + ((1 − t)z 0 (t) + z(t) − e1 )Rφ−1 y2 (θ)(t) dt

=

0

0

Z =

1

(1 − z(t) · e1 )∂ε T (0, Rφ z(t) + p) dt

4πM3 (θ) + 2π 0

=

2π (M5 (θ) + M3 (θ)) .

Hence, setting Φq := (q5 , q6 , q2 , q3 , q8 , q1 , q4 , q7 ) for all q = (q1 , . . . , q8 ) ∈ R8 , we have   2π 2π Φf¯ε (θ) = −2πε √ M1 (θ), √ M2 (θ), M3 (θ), M4 (θ), −M5 (θ) − M3 (θ), 0, 0, 0 + o(ε) 1 + 2π 2 1 + 2π 2 as ε → 0 and we conclude by obvious definition of Ψ, fε and f˜ε . Now assume that fε (θ) = 0 for some fixed ε ∈ (−εr , εr ) and θ = (φ, p) ∈ T2 × B r . In particular we have µεi (θ) = 0 for i = 2, 3, 5, 6. Setting (u1 , u2 ) = ωθ + η ε (θ), from (4.7) we deduce that Fε (u1 , u2 ) =

1 ε 1 µ (θ)(Rφ z 0 , 0) + µε4 (θ)(0, n) + √ µε7 (θ)(Rφ e3 , Rφ z 0 ) 2π 1 1 + 4π 2

namely N (u01 ) (1 + K(ε, u1 ))u2 ∧ u01 = µ ¯ 1 Rφ z 0 + µ ¯7 Rφ e3 N (u2 ) −u02 + T (ε, u1 )u2 ∧ u01 = µ ¯4 Rφ e3 + µ ¯ 7 Rφ z 0 −u001 +

20

(4.29) (4.30)

where µ ¯1 µ ¯4 µ ¯7

1 ε µ (θ) 2π 1 := µε4 (θ) 1 µε7 (θ) . := √ 1 + 4π 2 :=

Letting now η ε (θ) =: (η1 , η2 ), we multiply both (4.29) and (4.30) by u01 = Rφ z 0 + η10 and u2 = Rφ e3 + η2 , and thus we get −u001 · u01 = 4π 2 µ ¯1 + µ ¯1 Rφ z 0 · η10 + µ ¯7 Rφ e3 · η10

(4.31)

−u001 −u02 −u02

(4.32)

0

· u2 = µ ¯1 Rφ z · η2 + µ ¯7 + µ ¯7 Rφ e3 · η2 u01

·

=µ ¯4 Rφ e3 ·

η10

0

2

+ 4π µ ¯7 + µ ¯ 7 Rφ z ·

η10

0

· u2 = µ ¯4 + µ ¯4 Rφ e3 · η2 + µ ¯7 Rφ z · η2 .

(4.33) (4.34)

By the periodicity of η1 , u01 and u2 and since hη2 , Rφ e3 i = hη ε (θ), ξ4 (θ)i = 0, by (4.8), upon integrating (4.31), (4.34) and the sum of (4.32) and (4.33), we respectively obtain   Z 1 µ ¯1 4π 2 + Rφ z 0 · η10 = 0 (4.35) 0

Z

1

Rφ z 0 · η2 = 0  Z 1 Z 0 2 µ ¯1 Rφ z · η2 + µ ¯7 1 + 4π +

µ ¯4 + µ ¯7

(4.36)

0

0

1

Rφ z 0 · η10

 =0.

(4.37)

0

Since (4.5) yields Z

1

0

Rφ z 0 · x01 ≤ 2π max |η10 (t)| ≤ 2πkη ε (θ)kX < 4π 2 , t∈[0,1]

(4.35) implies µ ¯1 = 0. Then, in turn, (4.37) implies µ ¯7 = 0 and finally (4.36) gives µ ¯4 = 0. Therefore µεi (θ) = 0 also for i = 1, 4, 7 and hence f¯ε (θ) = 0. Remark 4.4 Notice that the implication fε (θ) = 0 ⇒ f¯ε (θ) = 0 in the previous lemma has been proved without using the vanishing of J(ωθ + η ε (θ)), but only the fact that µεi (θ) = 0 for i = 2, 3, 5, 6 was needed.

5

Main results

In this section we prove Theorems 1.1, 1.2 and 1.3. 3 Proof of Theorem 1.1. Let un ∈ Cper be a uniform parametrization of Γn , with |u0n | = cn . Notice that |u00 | κεn (un ) = 2n . (5.1) cn

21

Define u1,n

= un u0n ∧ u00n = . c3n κεn (un )

u2,n Then (u1,n , u2,n ) ∈ Ω solves 

u001,n = cn κεn (u1,n )u2,n ∧ u01,n . u02,n = τεn (u1,n )u2,n ∧ u01,n

(5.2)

Moreover |u01,n | = cn and thus N (u01,n ) = cn . In addition, by the definition of u2,n , using (5.1) and the fact that u0n · u00n = 0 (because |u0n | is constant), one also deduce that |u2,n | = 1 and thus 1 N (u2,n ) = 1. By hypothesis, the sequence (u1,n ) is bounded in Cper . Moreover the sequence (u2,n ) 0 2 is bounded in Cper . Thanks to (5.2), the sequences (u1,n ) and (u2,n ) are bounded in Cper and in 1 Cper , respectively. By the Ascoli-Arzel`a theorem, passing to subsequences, we may assume that 1 in Cper

u1,n → u1

and u2,n → u2

0 in Cper

0 1 . In particular cn = N (u01,n ) → N (u01 ) =: c and N (u2 ) = 1. By × Cper for some (u1 , u2 ) ∈ Cper hypothesis c 6= 0, namely u1 is nonconstant. In addition, by the uniform continuity, κεn (u1,n ) → 1 and τεn (u1,n ) → 0 uniformly on [0, 1]. By standard arguments we can pass to the limit in (5.2), finding that (u1 , u2 ) is a nonconstant solution of  00 u1 = cu2 ∧ u01 u02 = 0

with c = N (u01 ). Then, by Lemma 3.1 and Remark 3.2, u1 (t) = Rφ z(jz + φ0 ) + p and u2 (t) = Rφ e3 for some φ ∈ T2 , p ∈ R3 , j ∈ N, and φ0 ∈ R/Z. Now we show that M (n, p) = 0. Set ( ( T (ε, p) K(ε, p) ˆ p) = ∂ε K(0, p) − as ε 6= 0 and Tˆ(ε, p) = ∂ε T (0, p) − as ε 6= 0 K(ε, ε ε 0 as ε = 0 0 as ε = 0 . ˆ p) → 0 and Tˆ(ε, p) → 0 as ε → 0 uniformly Since ∂ε K, ∂ε T ∈ C 0 (R × R3 ), one has that K(ε, 3 on compact sets of R . As a consequence, since u1,n → u1 uniformly on [0, 1], one obtains that ˆ n , u1,n ) → 0 and Tˆ(εn , u1,n ) → 0 uniformly on [0, 1]. Then, since the sequence (u2,n ∧ u0 ) is K(ε 1,n uniformly bounded on [0, 1], ˆ n , u1,n )u2,n ∧ u01,n → 0 K(ε

and Tˆ(εn , u1,n )u2,n ∧ u01,n → 0

Then, by (5.3), one has Z 1 Z ∂ε T (0, u1,n )u2,n ∧ u01,n = 0

and Z

1

∂ε K(0, u1,n )u2,n ∧ 0

0

u01,n

Z =

1

uniformly on [0, 1] .

1 Tˆ(εn , u1,n )u2,n ∧ u01,n + εn

1

ˆ n , u1,n )u2,n ∧ u01,n + K(ε

0

22

1 cn εn

Z 0

Z

(5.3)

1

u02,n → 0

0

1

u001,n

1 − εn

Z 0

1

u2,n ∧ u01,n

Z 1 1 u0 ∧ u1,n = o(1) + εn 0 2,n Z 1  ∂ε T (0, u1,n ) − Tˆ(εn , u1,n ) (u2,n ∧ u01,n ) ∧ u1,n = o(1) + 0

Z

1

= o(1) +

∂ε T (0, u1 )(u2 ∧ u01 ) ∧ u1 .

0

Knowing explicitely u1 and u2 one can compute (u2 ∧ u01 ) ∧ u1 = 2πjp ∧ u1 and thus one obtains that Z 1 Z 1 ∂ε T (0, u1 )p ∧ u1 . (5.4) ∂ε K(0, u1,n )u2,n ∧ u01,n → 2πj 0

0

On the other hand, since u2,n ∧

u01,n

→ u2 ∧

u01

= 2πj(p − u1 ) uniformly on [0, 1], one has that

1

Z

Z

∂ε T (0, u1,n )u2,n ∧ u01,n → 2πj

1

∂ε T (0, u1 )(p − u1 )

0

0

and then Z

1

∂ε T (0, u1 )(p − u1 ) = 0 ,

(5.5)

0

namely Z

1

0=

Z ∂ε T (0, Rφ z(jt + φ0 ) + p)z(jt + φ0 ) dt =

0

1

∂ε T (0, Rφ z + p)z , 0

that is M3 (φ, p) = M4 (φ, p) = 0. In a similar way one has that Z 1 Z 1 ∂ε K(0, u1,n )u2,n ∧ u01,n → 2πj ∂ε K(0, u1 )(p − u1 ) 0

0

and then, from (5.4), one deduces that Z 1 Z ∂ε K(0, u1 )(p − u1 ) = 0

1

∂ε T (0, u1 )p ∧ u1 = 0

0

where the last equality follows by the equality p ∧ u1 (t) = (p ∧ Rφ e1 )z(jt + φ0 ) · e1 + (p ∧ Rφ e2 )z(jt + φ0 ) · e2 and by the fact that M3 (φ, p) = M4 (φ, p) = 0. Hence, arguing as before, one infers that also M1 (φ, p) = M2 (φ, p) = 0. Finally, using the second equation in (5.2) and the fact that u01,n ·u2,n = 0, we obtain that Z 1

T (εn , u1,n )u2,n ∧ u01,n · u1,n = 0

0

which, using also (5.3), implies that Z 1 ∂ε T (0, u1,n )u2,n ∧ u01,n · u1,n → 0 0

and then Z

1

∂ε T (0, u1 )u2 ∧

0=

u01

Z · u1 = 2πj

0

∂ε T (0, u1 )u1 · (p − u1 ) . 0

23

1

Therefore, using (5.5), we obtain Z 0=−

1

Z ∂ε T (0, u1 )u1 · (p − u1 ) + p ·

0

1

Z ∂ε T (0, u1 )(p − u1 ) =

0

1

∂ε T (0, u1 ) 0

namely M5 (φ, p) = 0. Now let us discuss Theorems 1.2 and 1.3. In order to complete the proof of Theorem 1.2, the following lemma will be useful. Lemma 5.1 If there exists a nonempty, bounded open set O in R5 such that deg(M, O, 0) 6= 0 then for |ε| small enough there is θε ∈ O such that f¯ε (θε ) = 0. Proof. Let Ψ ∈ GL(5, R) be given by Lemma 4.3. Since det Ψ 6= 0, one has that 0 6∈ ΨM (∂O) and |deg(ΨM, O, 0)| = |deg(M, O, 0)|. As O is bounded, there exists r > 0 such that O ⊂ R2 × B r and d := inf θ∈∂O |ΨM (θ)| > 0. Let εr be given by Lemma 4.1. For ε ∈ (−εr , εr ) let us define the homotopy Hε : O × [0, 1] → R5 by setting Hε (θ, s) := sfε (θ) + (1 − s)εΨM (θ)

for (θ, s) ∈ O × [0, 1]

where fε is defined in Lemma 4.3. We claim that the homotopy Hε is admissible for |ε| small enough. Indeed, since ∂O ⊂ R2 × B r , using Lemma 4.3, we have that fε (θ) − εΨM (θ) = o(ε) as ε → 0, uniformly in θ ∈ ∂O. Hence, for every (θ, s) ∈ ∂O × [0, 1] and ε 6= 0 there results o(ε) 1 1 |Hε (θ, s)| >0 = |εΨM (θ) + s (fε (θ) − εΨM (θ))| ≥ |ε| |ΨM (θ)| − s|o(ε)| ≥ d − |ε| |ε| |ε| ε provided that |ε| is small enough. Thus the claim is proved. Finally the homotopy invariance property of Brouwer’s degree gives deg(fε , O, 0) = deg(εΨM, O, 0) = deg(ΨM, O, 0) 6= 0 and the conclusion follows, using again Lemma 4.3. Proof of Theorem 1.2. By Lemma 2.1 and by the definitions (4.1) and (4.21), a pair (u1 , u2 ) ∈ Ω identifies a (κε , τε )-loop if and only if  Fε (u1 , u2 ) = 0 (5.6) J(u1 , u2 ) = 0. By Lemma 4.1 and by the definition (4.23), (uε1 , uε2 ) = ωθ + η ε (θ) solves (5.6) if f¯ε (θ) = 0. Lemma 5.1 ensures that there exists a mapping ε 7→ θε ∈ O defined for |ε| small, such that f¯ε (θε ) = 0. Hence, for |ε| small, the function uε := Rφε z + pε + η1ε (θε ) is a parametrization of a (κε , τε )-loop, 2 being (φε , pε ) = θε and η1ε (θ) ∈ Cper the first component of η ε (θ). Moreover, since θε ∈ O and O is compact, fixing an arbitrary sequence εn → 0, there exists a subsequence, denoted again (εn ), such ¯ p¯) ∈ O. By Lemma 4.1, kη εn (θε )kX → 0 and then uε → R ¯z + p¯ that θεn → θ¯ for some θ¯ = (φ, φ n n 2 in Cper . In particular, since the parametrization Rφ¯z + p¯ is injective on R/Z, for |ε| small, Γε is a ¯ p¯) = 0. simple curve. Moreover, by Theorem 1.1, M (φ,

24

As far as concerns the proof of Theorem 1.3, first let us state the following lemma. ¯ p¯) ∈ Lemma 5.2 Assuming K and T of class C 2 , if M admits a nondegenerate zero at some θ¯ = (φ, 2 3 1 2 3 T × R , then there exists a C mapping ε 7→ θε ∈ T × R , defined on some interval (−¯ ε, ε¯), such ¯ ¯ that fε (θε ) = 0 for |ε| < ε¯ and θ0 = θ. Proof. Fix r > 0 such that θ¯ ∈ Ur := R2 × Br and let εr be given by Lemma 4.1 and, for |ε| < εr let fε be given by Lemma 4.3. Let us introduce the mapping fˆ: (−εr , εr ) × Ur → R5 defined by setting:  1   fε (θ) if ε 6= 0 ˆ f (ε, θ) := ε   ΨM (θ) if ε = 0 , where Ψ ∈ GL(5, R) is given by Lemma 4.3. Our goal is to find the mapping ε 7→ θε satisfying the statement of the lemma, by applying the Implicit Function Theorem with respect to the equation ¯ By the hypothesis, fˆ(0, θ) ¯ = 0 and ∂θ fˆ(0, θ) ¯ = ΨDM (θ) ¯ fˆ(ε, θ) = 0 in a neighbourhood of (0, θ). 1 ˆ ˆ ˆ is invertible. We claim that f is of class C on its domain. Clearly ∂ε f and ∂θ f are well defined and continuous in ((−εr , εr ) \ {0}) × Ur , since they are continuous with respect to the variable θ and with respect to ε uniformly in θ ∈ Ur . Being K and T of class C 2 , according to Lemma 4.1 the mapping ε 7→ fε is of class C 2 from (−εr , εr ) into C 0 (U r , R5 ) and thus, using also Lemma 4.3, we can write ε2 fε (θ) = εΨM (θ) + g(θ) + h(ε, θ) for every (ε, θ) ∈ (−εr , εr ) × U r (5.7) 2 where d2 fε g := ∈ C 0 (U r , R5 ) , dε2 ε=0 and h(ε, ·) ∈ C 0 (U r , R5 ) is such that h(ε, θ) = o(ε2 )

as ε → 0 uniformly in θ ∈ U r .

(5.8)

On the other hand, by Lemma 4.3, we also have ˜ θ) fε (θ) = εΨM (θ) + h(ε,

for every (ε, θ) ∈ (−εr , εr ) × U r

(5.9)

˜ ·) ∈ C 1 (U r , R5 ) is such that where h(ε, ˜ θ) = o(ε) ∂θ h(ε,

as ε → 0 uniformly in θ ∈ U r .

Then (5.9) and the definition of fˆ yield ( 1 ˜ ˆ ∂θ f (ε, θ) = ΨDM (θ) + ε ∂θ h(ε, θ) if ε 6= 0 ΨDM (θ) if ε = 0.

25

(5.10)

Hence, thanks to (5.10), ∂θ fˆ is continuous also at every point (0, θ) with θ ∈ Ur . Moreover, using again the definition of fˆ and (5.7), we deduce that ∂ε fˆ(ε, θ) ∂ε fˆ(0, θ)

1 1 1 g(θ) + ∂ε h(ε, θ) − 2 h(ε, θ) for ε 6= 0 2 ε ε   fˆ(ε, θ) − fˆ(0, θ) 1 1 h(ε, θ) = lim = g(θ) = lim g(θ) + ε→0 ε→0 2 ε ε2 2

=

where the last equality follows from (5.8). In addition, by the definition of g, we have     ∂ε h(ε, θ) 1 dfε 1 dfε dfε (θ) − g(θ) → 0 = (θ) − ΨM (θ) − εg(θ) = (θ) − ε ε dε ε dε dε ε=0 as ε → 0, uniformly with respect to θ ∈ Ur . Therefore, using (5.8), also ∂ε fˆ is continuous at every point (0, θ) with θ ∈ Ur . In conclusion fˆ is of class C 1 and the thesis of the lemma can be obtained as an application of the Implicit Function Theorem. Proof of Theorem 1.3. One argues as in the proof of Theorem 1.2, by exploiting Lemma 5.2 instead of Lemma 5.1. We conclude this Section with a couple of examples. Let κε (p) = 1 + εK(p) and τε (p) = εT (p). Hence the mapping M is  R1  K(Rφ z(t) + p) cos(2πt) dt 0  R 1 K(R z(t) + p) sin(2πt) dt  φ  R0   1  M (φ, p) =  0 T (Rφ z(t) + p) cos(2πt) dt  .  R1   T (R z(t) + p) sin(2πt) dt  0 R φ 1 T (Rφ z(t) + p) dt 0 Example 1. Let K, T ∈ C 2 (R3 ) be such that K(p) = p2 p3

and T (p) = p1

for p = (p1 , p2 , p3 ) ∈ R3 with |p| < r,

for some r > 1. One can check that for |p| < r − 1   1  M (φ, p) =  2 

p3 sin φ2 p3 cos φ2 cos φ1 + p2 sin φ1 cos φ2 − sin φ2 cos φ1 2p1

    .  

(5.11)

¯ p¯) turns out to be a nondegenerate zero of M . Setting φ¯ = ( π2 , π2 ) and p¯ = (0, 0, 0), the point (φ, Then Theorem 1.3 applies. Example 2. Let K, T ∈ C 2 (R3 ) be such that K(p) = p2 p3

and T (p) = p1

for p = (p1 , p2 , p3 ) ∈ R3 with |p| ≥ r, 26

for some r > 0. Let O be the set of pairs (φ, p) ∈ R2 × R3 such that: π 3π < φi < (i = 1, 2) 4 4 −ρ < pi < ρ (i = 1, 3) −ρ2 < p2 < ρ2 with ρ ≥ r + 1 large enough. Then, for (φ, p) ∈ ∂O the vector M (φ, p) is given by (5.11) and, denoting Mi the i-th component of M , one can check that, for ρ sufficiently large: 1 ρ sin φ2 > 0 for (φ, p) ∈ ∂O with p3 = ρ 2 1 M1 (φ, p) = − ρ sin φ2 < 0 for (φ, p) ∈ ∂O with p3 = −ρ 2 1 M2 (φ, p) = (p3 cos φ2 cos φ1 + ρ2 sin φ1 ) > 0 for (φ, p) ∈ ∂O with p2 = ρ2 2 1 M2 (φ, p) = (p3 cos φ2 cos φ1 − ρ2 sin φ1 ) < 0 for (φ, p) ∈ ∂O with p2 = −ρ2 2√ π 2 M3 (φ, p) = for (φ, p) ∈ ∂O with φ2 = 4√ 4 3π 2 M3 (φ, p) = − for (φ, p) ∈ ∂O with φ2 = 4 4 √ π 2 sin φ2 < 0 for (φ, p) ∈ ∂O with φ1 = M4 (φ, p) = − 4 √4 2 3π M4 (φ, p) = sin φ2 > 0 for (φ, p) ∈ ∂O with φ1 = 4 4 M5 (φ, p) = ρ for (φ, p) ∈ ∂O with p1 = ρ M1 (φ, p) =

M5 (φ, p) = −ρ for (φ, p) ∈ ∂O with p1 = −ρ . Hence, by the Miranda theorem [16], deg(M, O, 0) 6= 0 and Theorem 1.2 applies.

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