Walls in infinite bent ferromagnetic nanowires

ANNALES

DE LA FACULTÉ DES SCIENCES

Mathématiques

ABDELKADERALSAYED ANDGILLESCARBOU

Walls in infinite bent ferromagnetic nanowires

Tome XXVII, n^o5 (2018), p. 897-924.

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Annales de la faculté des sciences de Toulouse Volume XXVII, n 5, 2018 pp. 897-924

Walls in infinite bent ferromagnetic nanowires

Abdel Kader Al Sayed⁽¹⁾ and Gilles Carbou⁽²⁾

ABSTRACT. — We study a one-dimensional model for a bent ferromagnetic nanowire. We prove the existence of static solutions describing either one domain or two domains separated by a wall. We address the stability of these solutions. In particular, we show that the asymptotically stable wall profiles are pinned at the bend even in presence of a small applied magnetic field.

RÉSUMÉ. — Dans cet article, on étudie un modèle monodimensionnel de fil fer- romagnétique présentant un coude. On explicite toutes les solutions stationnaires décrivant soit un domaine soit deux domaines séparés par un mur. On étudie ensuite la stabilité de ces solutions. On montre en particulier que certains profils de murs sont asymptotiquement stables, l’interprétation physique de ce résultat étant que les murs restent bloqués au niveau du coude, et ce même en présence d’un champ magnétique appliqué.

1. Introduction

Ferromagnetic nanowires are used in a wide range of applications such as microelectronics, paints for radar stealth, transformers, and computers. In particular, ferromagnetic nanowires can be used to record and store data on racetrack memories (see [13]). In such devices, the magnetization tends to be aligned in the direction of the wire, in one sense or the other. As a result, one observes in nanowires the formation of domains (large zone in which the

(*)Reçu le 24 mars 2016, accepté le 25 janvier 2017.

Keywords:ferromagnetism, Landau–Lifschitz equation, stability, domain walls.

2010Mathematics Subject Classification:35K55, 35Q60.

(1) CNRS / Université de Pau et des Pays de l’Adour, LMAP - UMR CNRS 5142, IPRA, E2S UPPA, Avenue de l’Université - BP 1155, 64013 PAU, France —

[email protected]

(2) CNRS / Université de Pau et des Pays de l’Adour, LMAP - UMR CNRS 5142, IPRA, E2S UPPA, Avenue de l’Université - BP 1155, 64013 PAU, France —

[email protected]

Article proposé par Radu Ignat.

magnetic moment is in the wire axis) and domain walls (thin zones in which the magnetic moment presents large variations).

In the framework of data storage, the stability of the walls position is crucial. Indeed, a non-desired movement of a wall may induce a degradation of the data.

Many papers address the stability of walls in ferromagnetic nanowires (cf. [4,5,6,7,10,11,15]). In [5], the stability of wall profiles is proved in the case of an infinite straight nanowire (i.e. without curvature). We remark that we do not have asymptotic stability because of the possible translations and rotations of the wall. In addition, even a small applied magnetic field can produce a displacement of the wall. In the case of finite-wires, walls profiles are linearly unstable (see [6]).

In this paper, we prove that a bend in a wire attracts the walls so that profiles for walls centered at the bend are asymptotically stable. This prop- erty is well-known in Physics literature (see [8, 9] for example) but to our knowledge, this is the first mathematical work about the effects of curvature on the walls stability in nanowires.

Let us recall the 3d model for ferromagnetic materials (see [3,12]). We consider a ferromagnetic body occupying the volume Ω ⊂ R³. We denote byM(t,x) the magnetization distribution at the time t and the point x∈ Ω. The material is supposed to be saturated so that the norm of M(t,x), denoted byMs, does not depend ontandx. The variations ofM satisfy the Landau–Lifschitz equation:

∂M

∂t =−γM×Heff − αγ

M_sM×(M ×Heff)

where × is the cross product in R³, γ is the gyromagnetic ratio, α is the damping coefficient and where the effective fieldHeff is given by:

Heff = A

µ0M_s²∆M +Hd(M) +Ha. (1.1) In (1.1),Ais the exchange constant, µ0 is the permeability of the vacuum, Ha is the applied magnetic field and Hd(M) is the demagnetizing field.

This last field is deduced fromM by the law of Faraday: divB = 0 (B is the magnetic induction), the constitutive relation:B=µ0(H+ ¯M) ( ¯M is the extension ofM by zero outside Ω), and by the static Maxwell equation:

curlH= 0. So, Hd(M) is obtained fromM by solving the system:

curlHd(M) = 0 and div(Hd(M) + ¯M) = 0. (1.2)

We rescale the time variable: t = γM_st, and the space variables x =

µ₀M_s²

A x. RewritingM as

M(t,x) =M_sm γM_st,

rµ₀M_s² A x

! ,

we obtain the following dimensionless model:

∂_tm=−m× H −αm×(m× H) (1.3) with

H= ∆m+Hd(m) +ha, where the rescaled applied field isha =_M^Ha

s.

In this paper we are interested in an infinitely long wire with one bend.

We consider the following one-dimensional model justified by asymptotic process in [2,5, 6,16].

The wire is parametrized by:

x7−→

(x~u ifx60,

x ~e₁ ifx>0, (1.4) where (e~1, ~e2, ~e3) is the canonical basis ofR³ and~u= cosβ

−sinβ 0

is a unitary vector in the plane (e~₁, ~e₂). The angleβ = (~u, ~e₁) is supposed to be in ]0, π[.

Figure 1.1. Bent nanowire

Using the parametrization (1.4), the magnetic momentm is defined on R^∗+×Rwith values inR³and satisfies the saturation constraint|m(t, x)|= 1.

As it is proved in [5,6,14], the equivalent 1d demagnetizing field reduces to the following local operator:

hd(m)(x) =1

2(−m(x) + (m(x)|~τ(x))~τ(x)),

where (· | ·) is the usual scalar product in R³,and ~τ(x) is the direction of the wire at the pointx(with|~τ|= 1). In our case,~τ is given by

~ τ(x) =

(~u forx <0,

~

e1 forx >0. (1.5) We remark that sincem×hd(m) = ¹₂m×((m|~τ)~τ), we can replacehd(m) by ¹₂((m|~τ)~τ) in the Landau–Lifschitz equation. In addition, rescaling the space and time variables, we get rid of the coefficient ¹₂ in front of the de- magnetizing field so that we obtain the following model for our bent wire:

(_∂m

∂t =−m×He(m)−αm×(m×He(m)) fort>0 andx∈R, H_e(m) =∂_xxm+ (m|~τ(x))~τ(x) +h_a(x), (1.6) where~τ is defined by (1.5) andha:R→R³is the 1d resulting applied field.

Remark 1.1. — In our model, there is no jump for m and ∂xm at the bend:

[|m|] :=m(t,0⁺)−m(t,0⁻) = 0

and [|∂xm|] =∂xm(t,0⁺)−∂xm(t,0⁻) = 0. (1.7) For vanishing applied field, we deal with stationary solutions separating a

±~u-domain inR⁻~uand a±e~₁-domain inR⁺e~₁. So that we look for solutions satisfying:

m(x)−−−−→

x→−∞ ±~u and m(x)−−−−→

x→+∞ ±~e₁. (1.8) We first exhibit all the solutions for (1.6)–(1.8). We denote~v=sinβ

cosβ 0

. Theorem 1.2. — For β ∈ ]0, π[, there are eight stationary solutions for (1.6)withha= 0 satisfying the limit conditions (1.8).

The solutions satisfying the limit conditionm(x)→ −~u when x→ −∞

are given by:

m1(x) =

(tanh(x−c)~u+_cosh(x−c)¹ ~v if x60, tanh(x+c)e~₁+_cosh(x+c)¹ e~₂ if x>0,

with c= artanh(sin^β₂), (1.9)

m2(x) =

(tanh(x+c)~u+_cosh(x+c)¹ ~v ifx60,

−tanh(x−c)e~1−_cosh(x−c)¹ e~2 ifx>0,

with c= artanh(cos^β₂), (1.10)

m3(x) =

(tanh(x+c)~u−_cosh(x+c)¹ ~v ifx60, tanh(x−c)e~₁−_cosh(x−c)¹ e~₂ ifx>0,

withc= artanh(sin^β₂), (1.11)

m4(x) =

(tanh(x−c)~u−_cosh(x−c)¹ ~v if x60,

−tanh(x+c)e~₁+_cosh(x+c)¹ e~₂ if x>0.

with c= artanh(cos^β₂). (1.12) The solutions satisfying the limit condition m(x) → ~u when x→ −∞ are given by−m1,−m2,−m3 and−m4.

It is worth noting that the solutions m1 and m3 correspond to a wall profile in the case of a straight wire (case β = 0). The solutionm4 corre- sponds to a +~e1-domain in a straight wire. We remark also that the solution m2 is specific to the bent-wire case and has no equivalent in the case of a straight wire. Theorem1.2is proved in Section2.

We address now the stability of these solutions. We recall that in the case of straight wire, a +e~₁-domain and−e~₁-domain are asymptotically sta- ble while a wall profile is stable but not asymptotically stable because of the invariance of the system by translation in thex-variable and rotation around the wire axis. We obtain the following result concerning the asymptotic sta- bility ofm1 andm4.

Theorem 1.3. — Let β6= 0 modπ. Then m₁ given by Theorem1.2is asymptotically stable for Equation(1.6)with vanishingha, i.e. for allε >0, there existsη >0 such that for all initial datam0 such that

m0−m1∈H¹(R)and|m0(x)|= 1for all x∈R,

ifkm₀−m₁k_H1(R)6η, then the solution of (1.6)with initial datam(0, x) = m₀(x)and withha= 0 satisfies:

• ∀t >0,km(t)−m1k_H1(R)6ε,

• km(t)−m1k_H1(R)tends to zero whent tends to+∞.

The same result holds for−m1,m4 and−m4 given by Theorem 1.2.

The other solutions are unstable:

Theorem 1.4. — Forβ ∈]0, π[, the solutions m2,−m2,m3 and−m3

given by Theorem 1.2 are linearly unstable for Equation (1.6) with limit conditions (1.8).

Contrary to the straight-wire case, the wall is pinned at the bend, so that the profile m₁ is asymptotically stable. In addition, we lose the invariance by rotation around the wire axis so that only one chirality of the wall profile is relevant. This is the reason whym₃is unstable.

In order to consider only perturbations satisfying the saturation con- straint, we use in Part 3 the mobile frame method developed in [5, 6, 7].

This step is followed by a careful study of the linearized equation which en- sures the asymptotic stability form1andm4(see Section4) and the linear instability form2andm3(see Section 5).

After that, we study the behavior of asymptotically stable configurations when the wire is submitted to an applied magnetic field. In the case of a straight nanowire, a non-vanishing applied field in the direction of the wire induces a displacement of the wall (see [5] and [10]). In our bent-wire case, we only assume that the applied field is along the wire far from the origin:

letξ∈C⁰(R;R³) such that

∃A >0,∀x∈R, x > A⇒ξ(x) =e~₁ andξ(−x) =~u. (1.13) We assume that the applied fieldha is given by

h_a(x) =λξ(x), λ∈R. (1.14)

We prove that a small non-vanishing applied field defined by (1.14) does not induce wall motion: the wall remains pinned at the bend.

Theorem 1.5. — Let β ∈ ]0, π[, let m1 given by Theorem 1.2. There existhmax>0 and a one parameter familyλ7→m(λ)satisfying:

• m(0) =m₁,

• m(λ)is defined for|λ|6hmaxand is a stationary solution for (1.6) withha given by (1.14)

• λ7→m(λ)−m1 is inC¹([−hmax, hmax];H²(R)).

In addition, for allλ∈[−h_max, h_max],m(λ)is asymptotically stable for(1.6).

The same result holds for Solutions−m1,m4 and−m4 given by Theo- rem1.2.

Section 6 of the present paper is devoted to the proof of this Theorem using the implicit function theorem.

2. Stationary solution

We considerM₀a stationary solution for (1.6) withh_a= 0 satisfying the limit condition (1.8). WritingM₀ as:

M0(x) =











M₀⁻(x) = sinθ⁻(x)~u+ cosθ⁻(x) cosϕ⁻(x)~v

+ cosθ⁻(x) sinϕ⁻(x)~e₃ forxinR⁻, M₀⁺(x) = sinθ⁺(x)e~1+ cosθ⁺(x) cosϕ⁺(x)e~2

+ cosθ⁺(x) sinϕ⁺(x)e~₃ forxinR⁺, (2.1) where

~ u=



 cosβ

−sinβ 0



 and~v=



 sinβ cosβ 0



,

we obtain that M0 is a stationary solution for (1.6)–(1.8) if and only if the following four assertions are satisfied:

(i) M₀⁻× ∂_xxM₀⁻+ (M₀⁻|~u)~u , (ii) M₀⁺× ∂xxM₀⁺+ (M₀⁺|e~1)e~1

, (iii) M₀⁻(0) =M₀⁺(0) and ^dM

− 0(0)

dx = ^dM

+ 0(0)

dx (jump conditions (1.7)), (iv) M₀⁻−−−−→

x→−∞ ±~u, M₀⁺−−−−→

x→+∞ ±e~₁(limit condition (1.8)).

Plugging (2.1) in the first equation (i), we obtain that:







d²θ⁻ dx² +

dϕ⁻ dx

2

sinθ⁻cosθ⁻+ sinθ⁻cosθ⁻= 0 forx∈R⁻,

−^d_dx^2ϕ2⁻ cosθ⁻+ 2^dθ_dx^−dϕ_dx⁻sinθ⁻= 0 forx∈R⁻.

(2.2)

The second equation yields _dx^d(^dϕ_dx⁻cos²θ⁻) = 0, so that dϕ⁻

dx cos²θ⁻=cst. (2.3)

From (iv), we obtain thatθ⁻(x)−−−−→

x→−∞

π

2 modπ. This implies that the constant in (2.3) is zero, so that:

dϕ⁻

dx cos²θ⁻= 0. (2.4)

Therefore

∀x∈R⁻, dϕ⁻

dx = 0 orθ⁻(x) =π

2 modπ. (2.5)

Let us prove that we have either (∀ x ∈ R⁻, θ⁻(x) = ^π₂ modπ) or (∀x∈R⁻, ^dϕ_dx⁻ = 0).

Assume that we are not in the second case, that is that there exists x0 such that ^dϕ_dx⁻(x0) 6= 0. By continuity argument, this is also satisfied in a neighborhood of x₀. Therefore, by (2.5), θ⁻(x) = ^π₂ modπ in this neighborhood of zero, and by continuity argument, there existsk∈Zsuch that θ⁻(x) = ^π₂ +kπ in this neighborhood of zero (k is the same for allx in this neighborhood). So ^dθ_dx⁻(x₀) = 0. Therefore, θ⁻ is a solution for the Cauchy problem:







d²θ⁻ dx² +

dϕ⁻ dx

2

sinθ⁻cosθ⁻+ sinθ⁻cosθ⁻= 0 forx∈R⁻, θ⁻(x₀) = ^π₂+k₀π, ^dθ_dx⁻(x₀) = 0.

By uniqueness argument, we obtain that

∀x∈R⁻, θ⁻(x) =π 2 +k0π.

In the second case, we remark that ϕ⁻ is constant, and that θ⁻ is a solution of the pendulum equation

d²θ⁻

dx² + sinθ⁻cosθ⁻ = 0 for x∈R⁻. (2.6) Sinceθ⁻satisfies the limit condition (iv), eitherθ⁻is constant equal to ^π₂ modπorθ⁻ is a solution represented by a separatrix on the phase portrait.

Remark 2.1. — By solving (2.6), we obtain that the solutions θ repre- sented by a separatrix are on the formx7→kπ+arcsin(tanh(x+c)) where c is an arbitrary constant,=±1 and k∈Z. Thus for these solutions, we have sin(θ(x)) =atanh(x+c) and cos(θ(x)) =b_cosh(x+c)¹ wherea=±1 and b=±1.

From (ii) and (iv), the same analysis onR⁺ yields that we have: either θ⁺ = ^π₂ modπor θ⁺ is a solution on the separatrix of the phase portrait and in the last case,ϕ⁺ is constant onR⁺.

We will now discriminate the different cases by using the transmission condition (iii).

Case 1. — If θ⁻(x) ≡ ^π₂ modπ for x < 0 and θ⁺(x) ≡ ^π₂ modπ for x >0, then M₀⁻(x) = ±~u and M₀⁺ =±e~₁. So by jump conditions (iii) we have~u=±~e₁, which is impossible sinceβ 6= 0 modπ.

Case 2. —Ifθ⁻(x)≡^π₂ modπforx <0 and if onR⁺,θ⁺is a solution of (2.6) represented by a separatrix and ϕ⁺ is constant, thenM₀⁻ ≡ ±~uso

by (iii), ^dM

+ 0

dx (0) = 0. This last equation implies that ^dθ_dx⁺(0) = 0, which is impossible on the separatrix. In the same way, the case θ⁺ ≡ ^π₂ modπ is also impossible.

The analysis of the first two cases yields thatθ⁻ andθ⁺ are trajectories on the separatrix of the phase portrait andϕ⁻ andϕ⁺ are constant.

Case 3. — Let us assume thatϕ⁻ 6= 0 modπorϕ⁺6= 0 modπ. Then, forx >0,M₀⁺ is in the planeP⁺ given by

P⁺= span

~

e₁,cosϕ⁺e~₂+ sinϕ⁺e~₃ , thus, ^dM

+ 0

dx ∈P⁺ forx >0.

In the same way, forx <0,M₀ and ^dM

− 0

dx belong toP⁻ given by P⁻= span

~

u ,cosϕ⁻~v+ sinϕ⁻w~ . By the jump conditions (iii) we have

M₀⁺(0⁺) =M₀⁻(0⁻)∈P⁺∩P⁻ and dM₀⁻

dx (0⁻) =dM₀⁺

dx (0⁺)∈P⁺∩P⁻. Since one of the anglesϕ⁻or ϕ⁺ is different from 0 modπ,P⁺∩P⁻ is a straight line, so thatM₀⁺(0) and ^dM

+ 0

dx (0) are collinear. In addition, from the saturation constraint |M₀⁺| = 1, we obtain that M₀⁺(0) ⊥ ^dM_dx⁰⁺(0), so that ^dM

+ 0

dx (0) = 0. This implies that ^dθ_dx⁺(0) = 0, which is impossible since θ⁺ parametrizes a solution on the separatrix.

Therefore the only possible case is the following:

Case 4. —ϕ^± = 0 modπ (so that M0 takes its values in the plane Re~1+Re~2) andθ⁻ and θ⁺ are solutions of the pendulum equation on the separatrix. Therefore:

M₀⁻ =a⁻tanh(x+c⁻)~u+b⁻ 1

cosh(x+c⁻)~v,

where b⁻ ∈ {−1,1} anda⁻ =−1 or a⁻ = +1 if M₀⁻(x) tends to ~u or −~u respectively whenxtends to−∞, and

M₀⁺=a⁺tanh(x+c⁺)~e1+b⁺ 1

cosh(x+c⁺)e~2,

whereb⁺∈ {−1,1}and a⁺= 1 or a⁺ ifM₀⁻(x) tends toe~1 or −e~1 respec- tively whenxtends to +∞.

We use now the transmission conditions (iii) in order to fix the constants a^±,b^± andc^±. At the bend,M₀⁻(0) =M₀⁺(0) and ^dM

− 0

dx (0) = ^M

+ 0

dx (0), so we have:











a⁻tanh(c⁻)~u+b⁻_cosh(c¹ −)~v=a⁺tanh(c⁺)e~₁+b⁺_cosh(c¹ ₊₎e~₂, and _cosh(c¹ −)

a⁻_cosh(c¹ −)~u−b⁻tanh(c⁻)~v

=_cosh(c¹ +)

a⁺_cosh(c¹ +)e~1−b⁺tanh(c⁺)e~2

.

(2.7)

The last equation induces that _cosh(c¹ −) = _cosh(c¹ ₊₎, thus c⁻ = c⁺ with =±1. System (2.7) is equivalent to the system:

Q₁X=X and Q₂X =X, (2.8)

where

X =

tanh(c⁻)

1 cosh(c⁻)

and where Q₁=

a⁻a⁺cosβ εa⁺b⁻sinβ

−a⁻b⁺sinβ b⁻b⁺cosβ

and Q2=

b⁻b⁺cosβ εa⁻b⁺sinβ

−a⁺b⁻sinβ a⁺a⁻cosβ

.

The matricesQ₁ andQ₂ are orthogonal (rotation of orthogonal symme- try). In addition, (2.8) induces that one is an eigenvalue ofQ₁ andQ₂. So, Q₁ and Q₂ are matrices of orthogonal symmetries (Q₁ = Id is impossible since sinβ 6= 0). So the determinant ofQ₁ equals−1, i.e.

a⁻a⁺b⁻b⁺=−1. (2.9)

Since both matrices have the same eigenvectorX associated to +1, since they are orthogonal symmetries,Q1=Q2. Therefore

a⁻b⁺=a⁺b⁻ and a⁺a⁻=b⁺b⁻. (2.10) Coupling (2.9) and (2.10), we obtain=−1, i.e.c⁺=−c⁻.

Now, we have four cases witha⁻= +1:

Case 1 (a⁻= 1,a⁺= 1,b⁻= 1 andb⁺ = 1). —We have in this case:

Q1=Q2=

−cosβ −sinβ

−sinβ cosβ

.

The eigenspace associated to the eigenvalue 1 is generated by_−sinβ 2

cos^β₂

and containsX =

tanh(c⁻)

1 cosh(c−)

. So there existsσ∈Rsuch thatX =σ_−sin^β

2

cos^β₂

. Since both vectors are unitary,σ=±1, and since the second coordinate is positive (we recall thatβ∈]0, π[,σ= 1. In particular, tanh(c⁻) =−sin^β₂, so c⁻=−artanh(sin^β₂). Thus, in this case, we obtain the following stationary solution:

m1(x) =

(tanh(x−c)~u+_cosh(x−c)¹ ~v ifx60,

tanh(x+c)~e1+_cosh(x+c)¹ e~2 ifx>0, withc= artanh(sin^β₂).

Figure 2.1. Graph ofm₁

Case 2 (a⁻ = 1, a⁺ = 1,b⁻ =−1 and b⁺ =−1). — We have in this case:

Q1=Q2=

cosβ sinβ sinβ −cosβ

.

Considering the eigenspace associated to the eigenvalue 1, we obtain cos^β₂

sin^β₂

=

tanh(c⁻)

1 cosh(c⁻)

,

soc⁻ = artanh(cos^β₂) and we obtain the following solution:

m2(x) =

(tanh(x+c)~u+_cosh(x+c)¹ ~v ifx60,

−tanh(x−c)e~₁−_cosh(x−c)¹ e~₂ ifx>0, withc= artanh(cos^β₂).

Case 3 (a⁻= 1,a⁺= 1,b⁻=−1 andb⁺=−1). — In this case, Q1=Q2=

−cosβ sinβ sinβ cosβ

,

Figure 2.2. Graph ofm2

so, by considering the eigenspace associated to 1 we obtain sin^β₂

cos^β₂

=

tanh(c⁻)

1 cosh(c⁻)

.

Thusc⁻= artanh(sin^β₂),and the associated stationary solution is:

m3(x) =

(tanh(x+c)~u−_cosh(x+c)¹ ~v ifx60,

tanh(x−c)e~1−_cosh(x−c)¹ e~2 ifx>0, withc= artanh(sin^β₂).

Figure 2.3. Graph ofm3

Case 4 (a⁻= 1,a⁺=−1,b⁻=−1 andb⁺= 1). — In this case, Q1=Q2=

cosβ −sinβ

−sinβ −cosβ

,

so, the eigenspace associated to 1 is generated by:

−cos(^β₂) sin(^β₂)

=

tanh(c⁻)

1 cosh(c⁻)

,

so thatc⁻=−artanh(cos^β₂).The associated stationary solution is:

m4(x) =

(tanh(x−c)~u−_cosh(x−c)¹ ~v ifx60,

−tanh(x+c)e~1+_cosh(x+c)¹ e~2 ifx>0, withc= artanh(sin^β₂).

Figure 2.4. Graph ofm4

By symmetry, the solutions to (1.6)–(1.8) satisfying the limit conditions M₀⁻(x)→~uwhenx→ −∞ (that is witha⁻=−1) are −m₁,−m₂,−m₃, and−m₄.

3. Equation for the perturbations

LetM0be one of the static solutions for (1.6) with vanishing applied field given by Theorem1.2:

M₀(x) =

(M₀⁻(x) = sinθ⁻~u+ cosθ⁻(x)~v ifx60, M₀⁺(x) = sinθ⁺e~1+ cosθ⁺(x)e~2 ifx>0.

We aim to address the stability ofM0for (1.6). we denote byJ the matrix J =





0 −1 0

1 0 0

0 0 1



.

As in [4, 5, 6, 10], in order to consider only perturbations m satis- fying the saturation constraint |m| = 1, we introduce the mobile frame (M₀(x), M₁(x), M₂) with

M1(x) =J M0(x) and M2=e~3

and we describemin this mobile frame writing:

m(t, x) =M₀(x) +r₁(t, x)M₁(x) +r₂(t, x)M₂+µ(r)M₀(x), (3.1) wherer= (r₁, r₂)∈R²andµ(r) =p

1−r²₁−r²₂−1, so that the saturation constraint is satisfied.

We remark that [|M0|] = [|M1|] = [|^dM_dx⁰|] = [|^dM_dx¹|] = 0, then the jump conditions [|m|] = [|∂xm|] = 0 at x = 0 are equivalent to the null-jump condition onr:

[|r|] = [|∂xr|] = 0 at x= 0.

Now, we plug (3.1) in (1.6). By projection ontoRM₁andRM₂, we obtain thatmsatisfies (1.6) if and only ifrsatisfies

∂_tr= Λr+F onR. (3.2)

The linear part of (3.2) is given by:

Λ =

−αL −L

L −αL

withL=−∂xx+fβ, (3.3) with

fβ(x) = sin²θ(x)−cos²θ(x), whereθ(x) =θ⁻(x) forx <0 andθ(x) =θ⁺(x) forx >0.

The nonlinear partF :R⁺×B(0,1)×R²×R²→R² of (3.2) is defined by

F(x, r, ∂xr, ∂xx) =A(r)∂xxr+B(r)(∂xr, ∂xr) +C(x, r)∂xr+D(x, r), (3.4) where, forx∈R,r= (r₁, r₂)∈R² andζ= (ζ₁, ζ₂)∈R²,

• A∈C^∞(B(0,1),M2(R)) (M2(R) is the set of the real 2×2 matri- ces) with

A(r)ζ=

−α(r₁)² −αr₁r₂+µ(r)

−αr1r2−µ(r) −α(r2)²

ζ

−

αr1(µ(r) + 1) +r2

αr2(µ(r) + 1)−r1

µ⁰(r)(ζ), (3.5)

• B ∈ C^∞(B(0,1),L2(R²)) (L2(R²) denotes the set of the bilinear functions defined onR²×R² with values inR²) given by

B(r)(ζ, ζ) =−

αr1(µ(r) + 1) +r2

αr₂(µ(r) + 1)−r₁

µ⁰⁰(r)(ζ, ζ). (3.6)

• C∈C^∞(R×B(0,1),M2(R)) with C(x, r)ζ=−2θ⁰(x)

αr1(µ(r) + 1) +r2

αr2(µ(r) + 1)−r1

ζ1

+ 2θ⁰(x)

α(r₁²−1) 1 +µ(r) +αr1r2

µ⁰(r)(ζ) (3.7)

• D∈C^∞(R×B(0,1),R²) such that D=

2 sinθ(x) cosθ(x)r1− µ(r)(sin²θ(x)−cos²θ(x))

×

αr1(µ(r) + 1) +r2

αr2(µ(r) + 1)−r1

. (3.8) We remark that this equation is valid whilertakes its values in a neigh- borhood of zero since µ is singular for |r| >1. In order to obtain uniform estimates, we will consider below perturbations such that krkL^∞ 6 ¹2. We remark also that the asymptotic stability ofM0for (1.6) is equivalent to the asymptotic stability of zero for (3.2).

4. Proof of Theorem 1.2

4.1. Stability for m₁

For the first solutionm1 given by (1.9), the linear part in (3.2) is given by

Λ1=

−αL1 −L1

L1 −αL1

and L1=−∂xx+f1,β(x), (4.1) where

f_1,β(x) =

(f(x+c) ifx>0, f(x−c) ifx60,

withf(x) = 2 tanh²(x)−1 andc= artanh(sin(^β₂)).

Sincef is strictly decreasing onR⁻and increasing onR⁺, asc >0 since β∈]0, π[, we remark that:

∀x∈R, f1,β(x)> f(x). (4.2)

4.1.1. First step: coercivity ofL₁

We denote byh · | · ithe usualL²-inner product inL²(R).

We recall that the operatorL =−∂xx+f(x) is a self-adjoint operator acting on H²(R), its essential spectrum is [1,+∞[ and zero is its unique eigenvalue which eigenspace is generated by _cosh¹ _x (see [4,5]). Therefore, for allw∈H²(R) satisfyinghw|_cosh¹ _xi= 0, we have

kwk²_L2 6hLw|wi6kLwk²_L2. (4.3) In addition there exists constants c₁ > 0 and c₂ > 0 such that for all w∈ _coshx¹ ⊥

,

c1kwk²_H1 6hLw|wi6c2kwk²_H1. (4.4) We claim the coercivity ofL₁ in the following proposition:

Proposition 4.1. — There exists c > 0, such that for all u∈ H²(R), we havehL1u|ui>ckuk²_L2.

Proof. — Suppose that there exists (un)n, such that kunkL² = 1 and hL1u_n|u_ni<_n¹.

We writeun=wn+_cosh^σn_x,where wn ∈(_cosh¹ _x)^⊥ andσn∈R.Thus hL1un|uni=h−∂xxun+f1,βun|uni=h−∂xxun+f un+ (f1,β−f)un|uni

=hLu_n|u_ni+ Z

R

(f_1,β−f)|u_n|²< 1 n.

MoreoverLun=Lwn sinceL(_coshx¹ ) = 0. So, sinceLis self-adjoint, hLu_n|u_ni=

Lw_n

w_n+σ_n 1

coshx

=hL(wn)|wni+D L(wn)

σn

coshx E

=hLwn|wni.

Therefore we have

hLwn|w_ni+ Z

R

(f_1,β−f)|un|²< 1 n.

So, with (4.4) we conclude that kwnk_H1 tends to zero when n tends to infinity.

On the other hand,

1 =kunk²_L2 =kwnk²_L2+ 2D wn

σ_n coshx

E +

σ_n coshx

2 L²

=kw_nk²_L2+|σ_n|²

1 coshx

2

L²

.

Thus, we have lim_n→∞|σn|²= ¹₂.Consider now the second term Z

R

(f1,β−f)|un|²dx=I1+I2+I3, (4.5) with

• I1=R

R(f1,β−f)|wn|²dx6kf1,β−fkL^∞kwnk²_L2 →0 asn→ ∞,

• I₂= 2R

R(f_1,β−f)w_ncoshx^σn dx62kf1,β−fkL^∞kwnkL²k_cosh^σn_xkL² → 0 asn→ ∞,

• I₃=|σ_n|²R

R(f_1,β−f)_cosh¹2xdx→ ¹₂R

R(f_1,β−f)_cosh¹2xdx.

In view of the above analysis, we obtain Z

R

(f_1,β−f)|u_n|²dx→ 1 2

Z

R

(f_1,β−f) 1

cosh²xdx >0 by (4.2).

On the other hand, sinceR

R(f_1,β−f)|un|²dx < _n¹,we obtain that Z

R

(f1,β−f)|un|²dx→0 asn→ ∞,

which leads to a contradiction. So the assumption is false and therefore there existsc >0 such that for allu∈H²(R), we have

hL1u|ui>ckuk²_L2. Corollary 4.2. — There exist two constantsK1 andK2 such that for everyu∈H²(R)we have

K₁kuk²_H1 6hL1u|ui6K₂kuk²_H1, K₁kuk_H2 6kL₁uk_L2 6K₂kuk_H2. Proof. — Thanks to Proposition4.1, we have

k∂_xuk²_L2=h∂_xu|∂_xui=h−∂_xxu|ui=hL₁u−f_1,βu|ui

6hL₁u|ui+kf_1,βk_∞kuk²_L2. Sincekf1,βkL^∞ = 1, we obtain by Proposition4.1that

k∂xuk²_L26

1 + 1 c

hL1u|ui.

Therefore,

kuk²_H1 6

2 + 1 c

hL1u|ui.

In addition, we have:

hL1u|ui= Z

R

|∂xu|²+ Z

R

f1,β|u|²6kuk²_H1 sincekf1,βk∞= 1.

This proves the equivalence of norms inH¹(R).

From Proposition4.1we have also kuk_L26 1

ckL₁uk_L2. Furthermore,

k∂_xxuk_L2=k −∂_xxu+f_1,βu−f_1,βuk6kL₁uk_L2+kf_1,βk_∞kuk_L2

6

1 + 1 c

kL1uk_L2.

Thus, we conclude that there exists a constant K such that kukH²6KkL1ukL².

In addition,

kL1ukL² 6k∂xxukL²+kf1,βkL^∞kukL²62kukH².

This concludes the proof of Corollary4.2.

4.1.2. Second step: estimate for the nonlinear term

In this section we estimate theL²(R)-norm of the nonlinear termF given by:

F(x, r, ∂xr, ∂xx) =A(r)∂xxr+B(r)(∂xr, ∂xr) +C(x, r)∂xr+D(x, r), where the right-hand-side terms are defined by (3.5)–(3.8).

Using the Sobolev injection ofH¹(R) in L^∞(R) and the equivalence of norms claimed in Corollary 4.2, we introduce η0 >0 such that for all r ∈ H²(R), hL₁r|ri¹² 6η₀ ⇒ krk_L∞(R) 6 ¹2. We prove the following estimate for the nonlinear partF:

Proposition 4.3. — There exists a constant k0 such that for all r ∈ H²(R)withhL1r|ri¹² 6η₀ then:

kFk_L26k0hL1r|ri¹²kL1rk_L2. (4.6)

Proof. — We estimate each term ofF separately. The same notationK is used for different constants independent of β and r ∈ H²(R) satisfying thatkrk_L^∞ 6 ¹₂.

We first remark that forr∈R²in a neighborhood of zero,µ(r) =O(|r|²), µ⁰(r) =O(|r|) andµ⁰⁰(r) =O(1).

By (3.5) we remark thatA(r) =O(|r|²) so that:

|A(r)∂xxr|6C|r|²|∂xxr|.

So using classical Sobolev embedding, we obtain that

kA(r)∂xxrk_L26KkrkL^∞k∂xxrk_L2 6Kkrk_H1krk_H2.

Concerning the second term defined by (3.6), sinceB(r) is bounded for r∈B(0,¹₂), we have

|B(r)(∂_xr, ∂_xr)|6K|r||∂_xr|². So,

kB(r)(∂xr, ∂xr)k_L26Kk∂xrk_L2k∂xrkL^∞, thus, by Sobolev embeddings,

kB(r)(∂xr, ∂xr)k_L2 6Kkrk_H1krk_H2.

By (3.7), sinceC(x, r) =O(|r|), we have

|C(x, r)∂xr|6K|r| |∂xr|, so we obtain:

kC(x, r)∂_xrk_L26Kkrk_L^∞k∂_xrk_L2, thus

kC(x, r)∂xrk_L2 6Kkrk²_H1.

Concerning the last term, we haveD(x, r) =O(|r|²),thus kD(x, r)kL² 6krkL²krkL^∞ 6Kkrk²_H1.

Finally, there exists a constantKsuch that ifr∈H²(R) satisfieskrkL^∞ 6

1

2, we have

kFkL² 6KkrkH¹krkH². (4.7) Using Corollary4.2 and the definition ofη0, we obtain that there exists k0 such that for allr∈H²(R) withhL1r|ri¹² 6η0then:

kFkL²6k0hL1r|ri¹²kL1rkL².

4.1.3. Stability Proof

We recall that the asymptotic stability of m1 for Equation (1.6) as claimed in Theorem 1.2 is equivalent to the asymptotic stability of 0 for Equation (3.2). We consider an initial data r₀ ∈ H²(R) such that hL1r0|r0i¹² 6η0and we denote byrthe solution of (3.2) with initial datar0.

We take theL²-inner product of (3.2) withL1r, and we obtain that:

1 2

d

dthL1r|ri=−αhL1r|L1ri+hF|L1ri.

Therefore using (4.6), whilehL1r(t)|r(t)i¹² 6η0, we have 1

2 d

dthL1r|ri+αkL1rk²_L2 6k0hL1r|ri¹²kL1rk²_L2, then,

1 2

d

dthL1r|ri+ (α−k₀hL1r|ri¹²)kL1rk²_L260. (4.8) Thus, whilehL₁r|ri¹²(t)6min{η₀,_2k^α

0}, we obtain thatα−k₀hL₁r|ri¹²>

α 2.Then,

1 2

d

dthL1r|ri+α

2kL1rk²_L260.

By Corollary4.2, there exists a constant c₀>0 such that kL1rk²_L2 >c0hL1r|ri.

Therefore we obtain that whilehL1r|ri¹²(t)6min{η0,_2k^α

0}, d

dthL1r|ri+c₀αhL1r|ri60 which implies by comparison lemma that

hL₁r(t)|r(t)i6hL₁r₀|r₀iexp(−αc₀t). (4.9) We setη₁= ¹₂inf{η0,_2k^α

0}. We assume that the initial datar₀ satisfies hL1r0|r0i¹² 6η1. (4.10) Let us show that for allt>0,

hL1r(t)|r(t)i¹² <min

η0, α 2k0

. (4.11)