Appolinaire Abourou
email: appolinaire.abourou.ella@math.univ-poitiers.fr

Arnaud Rougirel
email: arnaud.rougirel@math.univ-poitiers.fr

STEADY STATE BIFURCATIONS FOR PHASE FIELD CRYSTAL EQUATIONS WITH UNDERLYING TWO DIMENSIONAL KERNEL

Keywords: Mathematics Subject Classification. 35Q20, 35J40, 35B32. Key words and phrases. Phase Field Crystal Equation, Bifurcation Theory, Two Dimensional Kernel, Higher Order Elliptic Equations, Stability

Steady State Bifurcations for Phase Field Crystal Equations with underlying two Dimensional Kernel

Arnaud Rougirel, Appolinaire Abourou Ella

Introduction

During the last decades, Pattern Formation Equations have attracted much attention from researchers in applied sciences; see for instance [Hoy06, CH93, CGP00, TAR + 13].

In materials sciences, Pattern Formation Equations (as Allen-Cahn or Cahn-Hilliard equations) are obtained by phase field methods. In 2004, K. Elder and M. Grant have extended these methods by introducing the so-called Phase Field Crystal modelling in order to describe liquid/solid phase transitions in pure materials or alloys ( [START_REF] Elder | Modeling elastic and plastic deformations in nonequilibrium processing using phase field crystals[END_REF]). The solid phase, which can be a crystal, is represented by a periodic field whose wavelength accounts for the distance between neighbouring atoms. The liquid state is described by a (spatially) uniform field. We refer the reader to [PE10, EG04, EHP10, SWWV13, PACI + 13] for a more comprehensive exposition of the Phase Field Crystal method.

The simplest Phase Field Crystal model is the following sixth order evolution equation:

∂ t u -∂ xx ∂ xxxx u + 2∂ xx u + f (u) = 0, t > 0, x ∈ (0, L). (1.1)
Here L is the length of the domain and f is the derivative of a double-well potential. This equation can be viewed as a conservative Swift-Hohenberg equation exactly as the Cahn-Hilliard equation is a conservative version of the Allen-Cahn equation. Performing a linear change of variable mapping (0, L) onto (0, 1), Equation (1.1) can be rewritten as

∂ t u -ε∂ xx ε 2 ∂ xxxx u + 2ε∂ xx u + f (u) = 0, t > 0, x ∈ (0, 1), (1.2)
with ε = 1/L 2 . This paper focuses on the stationary solutions to (1.2) complemented with initial and boundary conditions (see (2.7)). In order to gain insight properties of stationary solutions, we use a bifurcation approach.

It is well known that bifurcations occur only if the kernel of the underlying linearized operator is non trivial. For the Phase Field Crystal Equation (1.2), the case of a one dimensional kernel has been investigated in [START_REF] Pierre | Stationary solutions to phase field crystal equations[END_REF]. In this paper, we focus on two dimensional kernels. We have choosen the multiparameter method (see [START_REF] Kielhöfer | Bifurcation theory[END_REF]Chapter I]) in order to obtain a wild set of solutions; that is to say, a one parameter family of branches of solutions. Notice that one or two branches can also be obtained (see [START_REF] Pierre | Stationary solutions to phase field crystal equations[END_REF]Subsection 5.3]) but this is indeed a smaller solution set.

Since two dimensional kernels are considered, we need two independent parameters. It turns out that ε which is the inverse of the domain size square, and the mass M of the initial condition (which is conserved by the dynamics) match the independence condition.

The first step is to characterize the parameters values that give rise to two dimensional kernels. The phase diagramm of Figure 1 features a simple geometric criterion for this (see also Proposition 3.1 for a analytic result).

Then we implement the multiparameter method in order to get bifurcation branches and expansions of solutions to (1.2). According to [START_REF] Kielhöfer | Bifurcation theory[END_REF], we have to choose a direction ( α β ) in the kernel which will be tangent to a branch of solutions. Let us denote by y → v(y) this branch, where y ∈ R, y ≃ 0. The parameters ε and M are also parametrized by y; this gives a parameter curve y → (ε(y), M (y)) in R 2 . Theorem 4.1 states an existence result for these bifurcation branches and gives second order expansions of ε(•), M (•) and v(•). In an explicit way, for y ≃ 0, the function x → v(y)(x) is solution to

ε(y) 2 ∂ xxxx v(y) + 2ε(y)∂ xx v(y) + f (M (y) + v(y)) = Ω f (M (y) + v(y)) dx
a.e. in (0, 1).

We are then led to study two curves: the parameter curve y → (ε(y), M (y)) and the function valued curve y → v(y). The former is studied by considering its oriented tangent at y = 0. This tangent will be denoted by T (α). We show how T (α) behaves w.r.t. α: see Propositions 4.5, 4.9 and Figure 3.

In Proposition 4.10, we state a monotonicity result for α → T (α). More precisely, in a well identified region of the parameter space, T (α) turns clockwise when α goes from 0 to 1. In a quiet surprising way, this monotonicity result is related to the stability of the bifurcating solutions what we will introduce now.

The main result of this paper is stated in Theorem 5.3 and concerns the stability of the bifurcating stationary solutions to the Phase Field Crystal Equation. If the wave numbers of the interactive modes (i.e k * and k * * in the sequel) are not consecutive integers then the bifurcating solutions are unstable. This is easily proved. In order to show stability, we use the principle of reduced stability from [Kie04, Section I-18] (see also [START_REF] Mielke | A new approach to sideband-instabilities using the principle of reduced instability[END_REF]). It allows us to reduce some infinite dimensional eigenvalue problem to a two dimensional one. As evoked above, it appears that the bifurcating solutions are stable exactly when the tangent T (α) turns clockwise. So we connect the issue of stability in the PDE (2.7) with the variation of a one dimensional object (the angle between T (α) and the horizontal axis).

Finally, we use a truncated bifurcation equation and symmetries to recover a bifurcation diagramm obtained originally in [START_REF] Pierre | Stationary solutions to phase field crystal equations[END_REF] by numerical integration: see Figure 4.

Equations and Functional Setting

Let Ω denote the interval (0, 1) ⊂ R and r be a real number. We define

f : R → R, u → (1 + r)u + u 3
(2.1)

V 2 = u ∈ H 2 (Ω) | u ′ (0) = u ′ (1) = 0 (2.2) V 4 = u ∈ H 4 (Ω) | u ′ = u ′′′ = 0 on ∂Ω (2.3) L2 (Ω) = v ∈ L 2 (Ω) | Ω v dx = 0 V4 = V 4 ∩ L2 (Ω), V2 = V 2 ∩ L2 (Ω).
(2.4)

The space V4 is equipped with the bilinear form

(u, v) V0 = Ω u (4) v (4) dx,
which becomes in turn a Hilbert space since every v ∈ V4 satisfies

v 2 ≤ 1 σ 2 1 v (4) 2 , (2.5)
where • 2 denotes the standard L 2 (Ω)-norm. Indeed,

v 2 ≤ 1 √ σ 1 v ′ 2 ≤ 1 σ 1 v ′′ 2 (2.6)
by Poincaré-Wirtinger and Poincaré's inequalities. Here σ 1 := π 2 denotes the first eigenvalue of the one-dimensional Laplace operator with homogeneous Dirichlet boundary conditions on Ω. Moreover v ′′ belongs to V2 thus the same estimates give v ′′ 2 ≤ 1 σ 1 v (4) 2 . Then (2.5) follows. In the same way, if (u, v) V2 := Ω u (2) v (2) dx then ( V2 , (•, •) V2 ) is a Hilbert space. Of course, u (2) stands for the second derivative of u.

Given initial data u 0 = u 0 (x) and a positive parameter ε, the Phase Field Crystal Equation with homogeneous Neumann boundary condition reads

     ∂ t u -ε∂ xx ε 2 ∂ xxxx u + 2ε∂ xx u + f (u) = 0 in Ω × (0, ∞) ∂ x u = ∂ xxx u = ∂ xxxxx u = 0 on ∂Ω × (0, ∞) u(0) = u 0 in Ω.
(2.7)

Since every solution u = u(t, x) to (2.7) satisfies

Ω u(x, t) dx = Ω u 0 (x) dx ∀t > 0,
the stationary solutions to the problem above solve

u ∈ M + V4 , ε 2 u (4) + 2εu (2) + f (u) = Ω f (u) dx in L 2 (Ω), (2.8)
where

M := Ω u 0 (x) is a real parameter. Introducing the new function v defined by u = M + v, (2.8) is equivalent to v ∈ V4 , ε 2 v (4) + 2εv (2) + f (M + v) = Ω f (M + v) dx in L 2 (Ω).
(2.9)

The bifurcation problem. We will formulate a bifurcation problem in order to get non trivial solutions of (2.9). To this end, we will introduce some notation. Let ε > 0, ε * > 0 and M , M * be real parameters. We put

δ := (ε, M ), δ * := (ε * , M * ), µ = (µ 1 , µ 2 ) := δ -δ * ∈ R 2 .
Let also

L(δ, •) : V4 → L2 (Ω), v → ε 2 v (4) + 2εv (2) + f ′ (M )v (2.10) L := L(δ * , •).
(2.11)

In the sequel, δ * = (ε * , M * ) is the bifurcation point and is fixed; the parameter δ will be closed to δ * . Then we define

F : (-ε * , ∞) × R × V4 → L2 (Ω) through F (µ 1 , µ 2 , v) = (ε * + µ 1 ) 2 v (4) + 2(ε * + µ 1 )v (2) + f (M * + µ 2 + v) - Ω f (M * + µ 2 + v) dx -Lv.
With these notation, we will consider the following bifurcation problem

µ ∈ (-ε * , ∞) × R, v ∈ V4 , Lv + F (µ, v) = 0 in L2 (Ω) (2.12) or equivalently δ = (ε, M ) ∈ (0, ∞) × R, v ∈ V4 , Lv + F (δ -δ * , v) = 0 in L2 (Ω). (2.13)
Remark that the equations in (2.9) and (2.13) are equivalent. We Taylor expend F (µ, v) w.r.t. µ and v at (µ, v) = (0, 0). For this, we write

F (µ, v) = F 1 (µ)v + F 2 (µ)v 2 + F 03 v 3 ,
where

F 1 (µ)v = L(δ * + µ, v) -Lv = (ε * + µ 1 ) 2 -ε 2 * v (4) + 2µ 1 v (2) + f ′ (M * + µ 2 ) -f ′ (M * ) v, so that v → F 1 (µ)v is a linear operator. Expending F 1 (µ)v w.r.t. µ, we get F 1 (µ)v = F 11 (µ)v + F 21 (µ)v with (see (2.1)) F 11 (µ)v = µ 1 (2ε * v (4) + 2v (2) ) + 6µ 2 M * v F 21 (µ)v = µ 2 1 v (4) + 3µ 2 2 v. Above, F 2 (µ) : V4 × V4 → L2 (Ω) is a continuous bilinear symmetric operator and F 2 (µ)v 2 stands for F 2 (µ)(v, v).
We proceed in the same way for F 2 (µ)v 2 , so that

F 2 (µ)v 2 = F 02 v 2 + F 12 (µ)v 2 with F 02 (µ)v 2 = 3M * (v 2 - Ω v 2 dx) F 12 (µ)v 2 = 3µ 2 (v 2 - Ω v 2 dx). 0 0 1 1 0 0 1 1 0 1 0 1 0 0.5 1 1.5 2 0 ε ε → -(σ 2 ε -1) 2 -r
The constant is stable

The constant is unstable 2D kernel The last term is

X := 3M 2 ε → -(σ 1 ε -1) 2 -r
F 03 v 3 = v 3 - Ω v 3 dx.
Solutions to (2.9) are critical point of E(M +•, ε) in V4 where the energy E is defined through

E : V 2 × (0, ∞) → R, (u, ε) → Ω 1 2 (εu ′′ + u) 2 + r 2 u 2 + 1 4 u 4 dx.
(2.14)

The Linearised Equation

For δ = (ε, M ) ∈ (0, ∞) × R, we study the eigenvalue problem (see the previous section and in particular (2.10), for notation)

L(δ, ϕ) = λϕ in L 2 (Ω) ϕ ∈ V4 \ {0}, λ ∈ R. (3.1)
The eigenvalues of (3.1) are

λ k := (εσ k -1) 2 + r + 3M 2 , where σ k := (kπ) 2 , k = 1, 2, . . . , (3.2)
with corresponding eigenfunctions

ϕ k : Ω → R, x → cos(kπx).
Then 0 is an eigenvalue of (3.1) iff there exists a positive integer k such that

3M 2 = -(εσ k -1) 2 -r.
That is to say, the point (ε, 3M 2 ) is on the parabola given by the function

ε → -(εσ k -1) 2 -r. (3.3)
Thus the operator L(•, δ) will have a two dimensional kernel iff the point (ε, 3M 2 ) lies at the intersection of two such parabolas: see Figure 1. If we express this geometric property in an analytical language, we obtain the following result whose proof is straightforward and will be omitted.

Proposition 3.1. Let δ = (ε, M ) ∈ (0, ∞) × R and k * , k * * be integers satisfying 1 ≤ k * * < k * . Then ker L(δ, •) = ϕ * , ϕ * * ⇐⇒ (εσ k * -1) 2 + r + 3M 2 = 0 (εσ k * * -1) 2 + r + 3M 2 = 0 . Moreover, if ker L(•, δ) = ϕ * , ϕ * * then ε = 2 σ k * + σ k * * . (3.4)
In the statement above, ϕ * := ϕ k * , ϕ * * := ϕ k * * and ϕ * , ϕ * * denotes the real vector space generated by ϕ * and ϕ * * .

Stability of the trivial solution. The trivial solution v = 0 of (2.9) is said to be linearly stable if (3.1) has only positive eigenvalues. In Figure 1, this corresponds to the case where the point (ε, 3M 2 ) is above all parabolas of the form (3.3). If (3.1) has at least one negative eigenvalue then v = 0 is linearly unstable.

Besides, The trivial solution is called neutrally stable if 0 is an eigenvalue of (3.1) and all the other eigenvalues of (3.1) are positive. In order to have stability of solutions to (2.12) bifurcating from v = 0, it is necessary that the trivial solution is neutrally stable. The next result gives a simple criterion for neutrally stability of v = 0 in the case of a 2D kernel.

Proposition 3.2. Let δ = (ε, M ) ∈ (0, ∞) × R and k * , k * * be integers such that 1 ≤ k * * < k * and ker L(δ, •) = ϕ * , ϕ * * . Then v = 0 is neutrally stable iff k * = k * * + 1.
Proof. For every k ≥ 1, we have with (3.2), (3.4)

λ k -λ k * = ε(σ k -σ k * ) ε(σ k + σ k * ) -2 = 2ε σ k * + σ k * * (σ k -σ k * )(σ k -σ k * * ) = 2π 4 ε σ k * + σ k * * (k -k * )(k -k * * ). (3.5) If v = 0 is neutrally stable and k = k * , k * * then λ k > 0 = λ k * . Hence (k -k * )(k -k * * ) > 0, ∀k = k * , k * * . (3.6)
The value of (k Figure 1 points out two values of the parameter δ for which the kernel of L(•, δ) has dimension two. One of these values corresponds to the case where k * = 2 and k * * = 1 and lies at the intersection of the green and red parabolas. By Proposition 3.2, the trivial solution is neutrally stable for this value of δ. The other value corresponds to the case where k * = 3 and k * * = 1. In this situation, v = 0 is not neutrally stable. Thus bifurcating solutions will be unstable.

-k * )(k -k * * ) at k = k * * + 1 is k * * + 1 -k * .

Bifurcation with 2D kernel

The case where the kernel of L(•, δ * ) has dimension one has been investigated in [START_REF] Pierre | Stationary solutions to phase field crystal equations[END_REF].

Here, we focus on the case where this kernel is two dimensional. More precisely, let

δ * := (ε * , M * ) ∈ (0, ∞) × R, p := -r -3M 2
* and k * , k * * be integers satisfying 1 ≤ k * * < k * . We assume

M * = 0 (4.1) (ε * σ k * -1) 2 + r + 3M 2 * = (ε * σ k * * -1) 2 + r + 3M 2 * = 0 (4.2) k * k * * = 2, k * k * * = 3. (4.3)
From the geometrical point of view of Figure 1, (ε * , 3M 2 * ) is a specified point at the intersection of two parabolas. According to Proposition 3.1, it follows that

ker L(•, δ * ) = ϕ * , ϕ * * . Consequently ε * σ k * = 1 + √ p, ε * σ k * * = 1 - √ p.
(4.4) We will implement the multiparameter method (see for instance [Kie04, Chapter I]). In view of Equation (2.9), the parameters will be ε and M . Moreover we will assume that δ = (ε, M ) is closed to δ * := (ε * , M * ) so that (see Section 2), µ := δ -δ * will be closed to zero. We recall that we have put

ϕ * := ϕ k * , ϕ * * := ϕ k * * .
(4.5)

Since the operator L (defined by (2.11)) is self-adjoint with compact resolvent, the set

ϕ k ϕ k 2 | k ∈ N \ {0}
is a spectral basis of L2 (Ω). Thus

dim ker L = codim R(L),
where R(L) ⊂ L2 (Ω) denotes the range of L, and

L2 (Ω) = R(L) ⊕ ker L V4 = R(L) ∩ V4 ⊕ ker L.
This decomposition of L2 (Ω), in turn, defines the projection

P : L2 (Ω) → ker L along R(L). (4.6)
Denoting by (•, •) 2 the L 2 -scalar product, there holds

P v = 2(v, ϕ * ) 2 ϕ * + 2(v, ϕ * * ) 2 ϕ * * , ∀v ∈ L2 (Ω). (4.7)
We decompose every v ∈ V4 in a unique way into

v = u 0 + u 1 , (4.8)
where

u 0 ∈ ker L, u 1 ∈ R(L) ∩ V4 . Moreover (u 0 , u 1 ) 2 = 0 since L is selfadjoint.
Projection on R(L). Applying I -P to (2.12) and using the notation of Section 2, we obtain

(I -P ) (L + F 1 (µ))u 1 + F 2 (µ)(u 0 + u 1 ) 2 + F 03 (u 0 + u 1 ) 3 = 0 in L2 (Ω). (4.9)
By the Implicit Function Theorem, for (u 0 , u 1 , µ) close to (0, 0, 0), this equation is equivalent to

u 1 = U (µ, u 0 ). (4.10) Moreover, U (µ, 0) = 0 and U (µ, u 0 ) = O(u 2 0 ), for (µ, u 0 ) ≃ (0, 0). (4.11)
This means that there exist C 0 , C such that

U (µ, u 0 ) V4 ≤ C u 0 2 V4 , ∀|µ| ≤ C 0 , u 0 2 V4 ≤ C 0 . Thus U (µ, u 0 ) = a 02 u 2 0 + O(µu 2 0 + u 3 0 ), ( 4
.12) where a 02 : ker L × ker L → R(L) ∩ V4 is a continuous bilinear symmetric operator independent of µ and a 02 u 2 0 := a 02 (u 0 , u 0 ). The equality (4.12) means that

U (µ, u 0 ) -a 02 u 2 0 V4 ≤ C |µ| u 0 2 V4 + u 0 3 V4 , for (µ, u 0 ) ≃ (0, 0).
Computation of a 02 u 2 0 . For α, β ∈ R, we put u 0 = αϕ * + βϕ * * and v 2 := a 02 u 2 0 . At order u 2 0 , (4.9) reads (I -P ) (Lv 2 + F 02 u 2 0 = 0 in L2 (Ω). (4.13) Since k * = 2k * * , we get

a 02 u 2 0 = v 2 = 1 2 x 2k * ϕ 2k * + x k * +k * * ϕ k * +k * * + x k * -k * * ϕ k * -k * * + x 2k * * ϕ 2k * * , (4.14) with x 2k * = - 3M * λ 2k * α 2 x k * +k * * = - 6M * λ k * +k * * αβ x k * -k * * = - 6M * λ k * -k * * αβ x 2k * * = - 3M * λ 2k * * β 2 . (4.15)
Computation of a 02 (u 0 , •). This quantity will be useful later on. Since

a 02 (u 0 , •) = 1 2 D u 0 v 2 = 1 2 D u 0 a 02 u 2 0 ,
we differentiate (4.13) w.r.t. u 0 to get

(I -P ) La 02 (u 0 , •) + F 02 (u 0 , •) = 0. Since u 0 = αϕ * + βϕ * * , F 02 (u 0 , ϕ * ) = 3 2 M * αϕ 2k * + βϕ k * +k * * + βϕ k * -k * * . Hence a 02 (u 0 , ϕ * ) = - 3 2 M * α λ 2k * ϕ 2k * + β λ k * +k * * ϕ k * +k * * + β λ k * -k * * ϕ k * -k * * . (4.16)
In a same way,

a 02 (u 0 , ϕ * * ) = - 3 2 M * α λ k * +k * * ϕ k * +k * * + α λ k * -k * * ϕ k * -k * * + β λ 2k * * ϕ 2k * * . (4.17) Projection on ker L. Since k * = 2k * * , P F 02 u 2 0 = 0, ∀u 0 ∈ ker L. (4.18)
Then, with u 1 given by (4.10), the bifurcation equation reads

P F 1 (µ)u 0 + 2F 2 (µ)(u 0 , u 1 ) + F 2 (µ)u 2 1 + F 03 (u 0 + u 1 ) 3 = 0 or equivalently (see (4.12)), P F 1 (µ)u 0 + 2F 02 (u 0 , a 02 u 2 0 ) + F 03 u 3 0 + O(µu 3 0 + u 4 0 ) = 0. (4.19)
According to Lyapunov-Schmidt's method, every solution (µ, u 0 ) to the bifurcation equation (4.19) provides a solution to (2.12). In order to solve (4.19), we use the Newton polygon method. Namely, we fix α, β such that α 2 + β 2 = 1, set

ϕ 0 = αϕ * + βϕ * * , u 0 = yϕ 0 , for y ∈ R, y ≃ 0 (4.20)
and rescale the parameter µ by setting

µ = y 2 μ. If y = 0 then (4.19) is equivalent to P F 11 (μ)ϕ 0 + 2F 02 (ϕ 0 , a 02 ϕ 2 0 ) + F 03 ϕ 3 0 + O(y) = 0.
We recast this equation under the form G(μ, y) = 0.

(4.21)

There holds in view of (4.4)

D μG( μ, 0) = F 11 (•)ϕ 0 = 2 p+ √ p ε * α 6M * α 2 p- √ p ε * β 6M * β . (4.22)
The above matrix is the matrix of the linear mapping D μG( μ, 0) : R 2 → kerL, expressed in the canonical basis of R 2 and in the basis (ϕ * , ϕ * * ) of ker L. If α, β and M * are non zero then D μG( μ, 0) is an isomorphism. Remark that p = 0 due to (4.2) and k * = k * * . In order to apply the Implicit Function Theorem, it is enough to find μ0 = 0 such that

G(μ 0 , 0) = 0, with μ0 = (μ 1 , μ2 ) ∈ R 2 . (4.23)
For this, we notice that

G(μ 0 , 0) = P F 11 (μ)ϕ 0 + 2F 02 (ϕ 0 , a 02 ϕ 2 0 ) + F 03 ϕ 3 0 . Moreover, by (4.22), F 11 (μ)ϕ 0 = 2α μ1 p + √ p ε * + 3M * μ2 ϕ * + 2β μ1 p - √ p ε * + 3M * μ2 ϕ * * (4.24)
and, since k * = 3k * * ,

P F 02 (ϕ 0 , a 02 ϕ 2 0 ) = 3M * 4 x 2k * α + (x k * +k * * + x k * -k * * )β ϕ * + 3M * 4 (x k * +k * * + x k * -k * * )α + x 2k * * β ϕ * * P F 03 ϕ 3 0 = 3 2 1 2 α 3 + αβ 2 ϕ * + 3 2 1 2 β 3 + α 2 β ϕ * * .
If we assume α = 0 and β = 0 then (4.23) is equivalent to

       2 p + √ p ε * μ1 + 6M * μ2 = - 3M * 2 x 2k * + (x k * +k * * + x k * -k * * ) β α - 3 4 α 2 - 3 2 β 2 2 p - √ p ε * μ1 + 6M * μ2 = - 3M * 2 (x k * +k * * + x k * -k * * ) α β + x 2k * * - 3 4 β 2 - 3 2 α 2 . (4.25)
Since M * = 0 and p = 0, it is clear that (4.25) has a unique solution (μ 1 , μ2 ). Thus for every y ≃ 0, we have a bifurcating solution (δ(y), v(y)) of (2.13). Moreover

δ(y) = (ε(y), M (y)) = (ε * + μ1 y 2 , M * + μ2 y 2 ) + O(y 3 ).
Next we would like to compute (ε(0), M (0)) where ε(0) is the value at y = 0 of the second derivative of ε(•). We readily have

ε(0) = 2μ 1 , M (0) = 2μ 2 .
Hence we obtain from (4.25) the following equations.

       p + √ p ε * ε(0) + 3M * M (0) = - 3M * 2 x 2k * + (x k * +k * * + x k * -k * * ) β α - 3 4 α 2 - 3 2 β 2 p - √ p ε * ε(0) + 3M * M (0) = - 3M * 2 (x k * +k * * + x k * -k * * ) α β + x 2k * * - 3 4 β 2 - 3 2 α 2
. (4.26)

We will rewrite these equations in a more convenient form. For this, we put

f * := 9 2 M 2 * λ 2k * - 3 4 , f * * := 9 2 M 2 * λ 2k * * - 3 4 (4.27) C S := 9M 2 * 1 λ k * +k * * + 1 λ k * -k * * - 3 2 . (4.28)
In view of (4.15), Equations (4.26) reads Also we obtain

       p + √ p ε * ε(0) + 3M * M (0) = (f * -C S )α 2 + C S p - √ p ε * ε(0) + 3M * M (0) = (-f * * + C S )α 2 + f * * . ( 4
= Xϕ * + Y ϕ * * , that is X = yα, Y = yβ, this equation becomes -f * X 3 -C S XY 2 + (εσ k * -1) 2 + r + 3M 2 X = 0 -C S X 2 Y -f * * Y 3 + (εσ k * * -1) 2 + r + 3M 2 Y = 0 . ( 4
D = 8 3 C S -( √ p + 1)B (4.37) 3 4 A = f * + f * * -2C S (4.38) C = 8 3 (f * -C S ) -( √ p + 1)A. (4.39)
In order to express A, B, C, D more simply, we put

x := k * k * * 2 .
Then, in view of (4.4),

√ p = x -1 x + 1 and, since ε * σ k * = 1 + √ p, λ 2k * = (4ε * σ k * -1) 2 -p = (4(1 + √ p) -1) 2 -p = 12x 4x -1 (x + 1) 2 . (4.40)
Similarly,

λ 2k * * = -12 x -4 (x + 1) 2 (4.41) λ k * +k * * = 4 √ x (2 √ x + 1)( √ x + 2) (x + 1) 2 (4.42) λ k * -k * * = -4 √ x (2 √ x -1)( √ x -2) (x + 1) 2 . (4.43) Then A = 2 -M 2 * (x + 1) 2 (2x 2 -61x + 2) x(4x -1)(x -4) (4.44) B = -1 + M 2 * (x + 1) 2 (4x -61) 2(4x -1)(x -4) (4.45) C = -2 x -1 x + 1 + M 2 * (x 2 -1)(4x 2 -57x + 4) x(4x -1)(x -4) (4.46) D = -2 x + 2 x + 1 -M 2 * (x + 1)(4x 2 -x + 60) (4x -1)(x -4) . (4.47)
Then we can state a bifurcation result whose proof comes from the above analysis.

Theorem 4.1. Let δ * = (ε * , M * ) ∈ (0, ∞) × R, (α, β) ∈ (-1, 1) 2 and k * , k * * be integers satisfying 1 ≤ k * < k * * .
We assume that (4.1)-(4.3) hold and

α 2 + β 2 = 1, α = 0, β = 0. (4.48) Then (δ, v) = (δ * , 0) ∈ R 2 ×
V4 is a bifurcation point for Equation (2.13). More precisely, there exist r 1 > 0 (close to zero) and smooth functions

δ = (ε(•), M (•)) : (-r 1 , r 1 ) → R 2 , v : (-r 1 , r 1 ) → V4
depending on α, β, k * and k * * such that for every y ∈ (-r 1 , r 1 ), one has

ε(y) 2 v(y) (4) + 2ε(y)v(y) (2) + f (M (y) + v(y)) = Ω f (M (y) + v(y)) dx in L 2 (Ω). Moreover ε(y) = ε * + ε(0) 2 y 2 + O(y 3 ) (4.49) M (y) = M * + M (0) 2 y 2 + O(y 3 ) (4.50) v(y) = yϕ 0 + y 2 v 2 + O(y 3 ), (4.51)
where ε(0), M (0) are defined through (4.31) and (4.44)-(4.47), ϕ 0 = αϕ * + βϕ * * and v 2 is given by (4.14), (4.15).

4.1. Sign of ε(0) and M (0). For every α ∈ (0, 1), Theorem 4.1 gives us the parameter curve

y → δ(y) = (ε(y), M (y)).
The tangent to this curve at y = 0 is given by δ(0) = (ε(0), M (0))

provided that this vector do not vanish. In the sequel, we will compute the signs of ε(0) and M (0) in order to have informations on the profile of the above curve near y = 0. For instance, if ε(0) and M (0) are positive then δ(y) belongs to the first quadrant of the plane (ε, M ). We denote by A = A(x, M * ) the function defined on (1, ∞) \ {4} × R by the left hand side of (4.44). In the same way, we define the functions B = B(x, M * ), C = C(x, M * ) and D = D(x, M * ) with (4.45)-(4.47).

Sign of ε(0). Under the assumptions and notation of Theorem 4.1, it follows from (4.31) that ε(0) and Aα 2 + B have the same sign. So for every (x, M, α) ∈ (1, ∞) \ {4} × R × (-1, 1), we will compute the sign of A(x, M )α 2 + B(x, M ). Taking advantage of the monotonicity of A(x, M )α 2 + B(x, M ) w.r.t. α 2 , we will look at the sign of B(x, M ) and (A + B)(x, M ). For this, we introduce the so-called cancellation functions of B and A + B, namely

B 0 : (1, ∞) \ {61/4} → R, x → 2 (4x -1)(x -4) (x + 1) 2 (4x -61) (4.52) (A + B) 0 : (1, ∞) → R, x → -2 x(4x -1)(x -4) (x + 1) 2 (61x -4) . (4.53)
These functions satisfy

M 2 = B 0 (x) ⇐⇒ B(x, M ) = 0 M 2 = (A + B) 0 (x) ⇐⇒ (A + B)(x, M ) = 0.
The sign of B(x, M ) is given in the following result.

Lemma 4.2. If x ∈ (1, 4) ∪ ( 61 4 , ∞) then B 0 (x) > 0 and M 2 < B 0 (x) =⇒ B(x, M ) < 0 B 0 (x) < M 2 =⇒ B(x, M ) > 0.
If x ∈ (4, 61 4 ) then B(x, M ) < 0 for every M . The statement of Lemma 4.2 and those of the seven forcoming results are easily proved; thus their proof will be omitted. Regarding the sign of A + B, we have the lemma below.

M 2 < (A + B) 0 (x) =⇒ (A + B)(x, M ) > 0 (A + B) 0 (x) < M 2 =⇒ (A + B)(x, M ) < 0. If x ∈ (4, ∞) then (A + B)(x, M ) > 0 for every M .
Moreover the cancellation functions are ordered or are simultaneously negative.

Lemma 4.4. For every x ∈ (1, 4), 0 < (A + B) 0 (x) < B 0 (x).

For every x ∈ (4, 61 4 ), one has (A + B) 0 (x) < 0 and B 0 (x) < 0. For every x ∈ ( 61 4 , ∞), one has (A + B) 0 (x) < 0 < B 0 (x). Then we can give the sign of ε(0). Proposition 4.5. Under the assumptions and notation of Theorem 4.1, the sign of

ε(0) = 3ε * 8 √ p (A(x, M * )α 2 + B(x, M * )) is as follows. (i) If x ∈ (1, 4) then (a) if M 2 * < (A + B) 0 (x) then α → ε(0)
is increasing on (0, 1) and changes its sign on (0, 1);

(b) if (A + B) 0 (x) ≤ M 2 * ≤ B 0 (x) then ε(0) < 0 for every |α| ∈ (0, 1); (c) if B 0 (x) ≤ M 2
* then α → ε(0) is decreasing on (0, 1) and changes its sign on (0, 1).

(ii) If x ∈ (4, 61 4 ) then α → ε(0) is increasing on (0, 1) and changes its sign on (0, 1).

(iii) If x ∈ ( 61 4 , ∞) then (a) if M 2 * < B 0 (x) then α → ε(0)
is increasing on (0, 1) and changes its sign on (0, 1);

(b) if B 0 (x) ≤ M 2
* then ε(0) > 0 for every |α| ∈ (0, 1).

Sign of M (0). We proceed as above. We define the cancellation functions D 0 and (C + D) 0 of D and C + D, namely

D 0 : (1, ∞) → R, x → 2 (x + 2)(4x -1)(x -4) (x + 1) 2 (-4x 2 + x -60) (4.54) (C + D) 0 : (1, ∞) → R, x → 2 (2x + 1)x(4x -1)(x -4) (x + 1) 2 (-60x 2 + x -4) . (4.55)
The signs of D(x, M ) and (C +D)(x, M ) are easily determinated with the use of the cancellation functions. More precisely, the following lemmas hold.

Lemma 4.6. If x ∈ (1, 4) then D 0 (x) > 0 and

M 2 < D 0 (x) =⇒ D(x, M ) < 0 D 0 (x) < M 2 =⇒ D(x, M ) > 0. If x ∈ (4, ∞) then D 0 (x) < 0 and D(x, M ) < 0 for every M . Lemma 4.7. If x ∈ (1, 4) then (C + D) 0 (x) > 0 and M 2 < (C + D) 0 (x) =⇒ (C + D)(x, M ) < 0 (C + D) 0 (x) < M 2 =⇒ (C + D)(x, M ) > 0.
If x ∈ (4, ∞) then (C + D) 0 (x) < 0 and (C + D)(x, M ) < 0 for every M .

Lemma 4.8. For every x ∈ (1, 4),

0 < (A + B) 0 (x) < B 0 (x) < (C + D) 0 (x) < D 0 (x).
For every x ∈ (4, ∞), one has (C + D) 0 (x) < 0 and D 0 (x) < 0.

Proposition 4.9. Under the assumptions and notation of Theorem 4.1, the sign of

M (0) = 1 8M * (C(x, M * )α 2 + D(x, M * )) is as follows. (i) If x ∈ (1, 4) then (a) if M 2 * ≤ (C + D) 0 (x) then M (0) < 0 for every |α| ∈ (0, 1); (b) if (C + D) 0 (x) < M 2 * < D 0 (x) then α → M (0)
is increasing on (0, 1) and changes its sign on (0, 1);

(c) if D 0 (x) ≤ M 2 * then M (0) > 0 for every |α| ∈ (0, 1). (ii) If x ∈ (4, ∞) then M (0) < 0 for every |α| ∈ (0, 1).
The result of Propositions 4.5 and 4.9 are illustrated by Figure 2 for x ranging in (1, 4). ε(0) . We will prove a monotonicity result for the function α → M (0) ε(0) . In the parameter space (ε, M ), M (0) ε(0) is the tangent of the angle between the unit vector (1, 0) and (ε(0), M (0)). We recall that (ε(0), M (0)) is tangent to the curve (ε(•), M (•)) at y = 0.

ε(0) -ր + M (0) -ր + M 2 ε(0)<0 M (0)<0 M (0)>0 ε(0) + ց -
As we will see later on, the monotonicity of α → M (0) ε(0) is related to the stability of bifurcating solutions given by Theorem 4.1.

If A(x, M * )α 2 + B(x, M * ) = 0 then

M (0) ε(0) = √ p 3M * ε * Cα 2 + D Aα 2 + B (4.56)
and

∂ ∂X CX + D AX + B = BC -AD (AX + B) 2 .
Thus it is enough to compute the sign of BC -AD. For this, we write

BC -AD = B(C + D) -D(A + B).
The previous results give informations on the signs of B, C + D, D and A + B. Thus the following assertions hold.

(i) If x ∈ (1, 4) then (a) if M 2 * < (A + B) 0 (x) or D 0 (x) < M 2 * then BC -AD > 0; (b) B 0 (x) ≤ M 2 * < (C + D) 0 (x) then BC -AD < 0. (4.57) (ii) If x ∈ (4, 61 4 ) then BC -AD > 0. (iii) If x ∈ ( 61 4 , ∞) and M 2 * < B 0 (x) then BC -AD > 0.
In the other cases, we have to push the analysis a little bit further. (i) If x ∈ (1, 4) then there remains to consider two cases. The first one is when (A + B) 0 (x) < M 2 * < B 0 (x). Let us denote by M 1 the positive number satisfying M 2 1 = (A + B) 0 (x). We have (BC -AD)(x, M 1 ) = B(C + D)(x, M 1 ). Moreover B(x, M 1 ) < 0 (according to Lemmas 4.2 and 4.4) and (C + D)(x, M 1 ) < 0 (due to Lemmas 4.7 and 4.8). Thus

(BC -AD)| M 2 =(A+B) 0 (x) > 0.
(4.58)

Besides, by (4.57),

(BC -AD)| M 2 =B 0 (x) < 0. (4.59) Moreover, 1 2 (BC -AD)(x, M ) = 3 + (x + 1) 2 b 1 (x) x(4x -1)(x -4) M 2 + (x + 1) 4 b 2 (x) 4x 2 (4x -1) 2 (x -4) 2 M 4 , (4.60) with b 1 (x) = -2x 2 + 121x -2 b 2 (x) = x(4x 2 + 3583x + 4).
In particular, (BC -AD)(x, M ) is a quadratic polynomial w.r.t. the variable X := M 2 . With (4.58), (4.59), we deduce that, for every x ∈ (1, 4), there exists a unique

J 1 (x) ∈ ((A + B) 0 (x), B 0 (x)) such that (BC -AD)(x, J 1 (x)) = 0. (4.61)
The second case we will consider is when (C + D) 0 (x) < M 2 * < D 0 (x). Arguing as above, we prove that there exists a unique J 2 (x) ∈ ((C + D) 0 (x), D 0 (x)) such that (BC -AD)(x, J 2 (x)) = 0.

(4.62)

Then the sign of BC -AD follows easily by using (4.60), J 1 (x) and J 2 (x). (ii) If x ∈ ( 61 4 , ∞) and B 0 (x) < M 2 then in view of (4.60), the discriminant ∆(x) of the polynomial

X → 3 + (x + 1) 2 b 1 (x) x(4x -1)(x -4) X + (x + 1) 4 b 2 (x) 4x 2 (4x -1) 2 (x -4) 2 X 2 (4.63) satisfies ∆(x) = 4(x + 1) 4 x 2 (4x -1) 2 (x -4) 2 (x 4 -124x 3 + 975x 2 -124x + 1).
We check that there exists x 3 ∈ (115, 116) such that

∆(x) < 0 on ( 61 4 , x 3 ) (4.64) ∆(x) > 0 on (x 3 , ∞) ∆(x 3 ) = 0.
From (4.60), (4.64), we immediately obtain that

(BC -AD)(x, M ) > 0 ∀(x, M 2 ) ∈ ( 61 4 , x 3 ) × (B 0 (x), ∞).
There remains to consider the case where x > x 3 . With straightforward computations, we prove that the polynomial (4.63) has two positive roots J 1 (x) < J 2 (x) in the interval (B 0 (x), ∞). Hence for every x > x 3 , there holds

(BC -AD)(x, M ) < 0 if M 2 ∈ (J 1 (x), J 2 (x)) > 0 if M 2 ∈ (B 0 (x), J 1 (x)) ∪ (J 2 (x), ∞) .
We may summarize the above results in the proposition below.

Proposition 4.10. Under the assumptions and notation of Theorem 4.1, there exists x 3 ∈ (115, 116) such that for every x ∈ (1, 4) ∪ (x 3 , ∞), the polynomial (4.63) has two positive roots 0) is decreasing on the positive intervals of its domain of definition, i.e. on (0, 1) if AB(x, M * ) > 0 or on (0, |B/A|) and on 0) is increasing on the positive intervals of its domain of definition.

J 1 (x) < J 2 (x). Moreover, (i) If M 2 * ∈ (J 1 (x), J 2 (x)) then the function α → M(0) ε(
( |B/A|, 1) if AB(x, M * ) < 0. (ii) If M 2 * ∈ (0, J 1 (x)) ∪ (J 2 (x), ∞) then (BC -AD)(x, M * ) > 0 and α → M (0) ε(
If x ∈ (4, x 3 ) then α → M (0)
ε(0) is increasing on the positive interval of its domain of definition. Remark 4.1. • J 1 (x 3 ) = J 2 (x 3 ).

• We may compute explicitly J 1 (x) and J 2 (x) thanks (4.60) (see [START_REF] Ella | Etude des solutions stationnaires d'un modèle de champs de phase cristallin[END_REF]).

• ε(0) = 3ε * 8 √ p (Aα 2 + B) may vanish. This occurs for instance for some α 0 ∈ (0, 1) if x ∈ (1, 4) and M 2 * < (A + B) 0 (x). See Proposition 4.5. In this case, the function α → M (0) ε(0) is defined on (-1, 1) \ {-|B/A|, |B/A|}. • Fig 3 is a phase diagram in the parameter space (ε, 3M 2 ). We have choosen some x ∈ (1, 4) and M 2 * ∈ (J 1 (x), J 2 (x)). The vector u is positively colinear to the tangent to the curve (ε(•), 3M 2 (•)) at y = 0 for α = 0.6. If α = 0 or α = 1 then u = 0 since there is no bifurcation branch according to Theorem 4.1. When α goes from 0 to 1, the vector u turns clockwise; in accordance with the assertion (i) of Proposition 4.10. Its range is depicted by the blue curve.

Finally we can combine the above results on the signs of ε(0), M (0) and the variation of M (0) ε(0) to obtain a better insight of the behaviour of the curve y → δ(y) w.r.t. α. We will only investigate the cases useful in the sequel.

We assume that x ∈ (1, 4) and recall that

(A + B) 0 (x) < J 1 (x) < B 0 (x) < (C + D) 0 (x) < J 2 (x) < D 0 (x). (4.65) (a) If (C + D) 0 (x) < M 2 < J 2 (x)
then, due to Proposition 4.5 and (4.65), α → ε(0) has a unique zero α 0 in (0, 1) and α 0 = |B/A|. Also (see Proposition 4.9), α → M (0) has a unique zero α 1 ∈ (0, 1) and α 1 = |D/C|.

Let us show that α 0 < α 1 . By Proposition 4.10 and (4.65), α → M (0) ε(0) is decreasing on (0, α 0 ) and on (α 0 , 1). Moreover, Lemmas 4.2 and 4.6 imply that

M (0) ε(0) α=0 = D B < 0. (4.66) Thus M (0) ε(0) < 0 ∀α ∈ (0, α 0 ). (ε * , 3M 2 * ) u α = 0.6 α = 0.1 ε → -(εσ k * * -1) 2 -r ε → -(εσ k * -1) 2 -r Figure 3. Phase diagramm for k * = 3, k * * = 2 (thus x = 9/4) and M 2 * = 0.4(C + D) 0 (x) + 0.6J 2 (x). See Remark 4.1.
Hence M (0) = 0 for every α ∈ (0, α 0 ), and consequently, α 1 ≥ α 0 . That is to say -D/C ≥ -B/A. However, these two numbers are not equal since M 2 = J 1 (x), J 2 (x). Then α 0 < α 1 . Summing up, we get ε(0) ≥ 0, M (0) < 0 ∀α ∈ (0, α 0 ] (4.67)

ε(0) < 0, M (0) < 0 ∀α ∈ (α 0 , α 1 ) ε(0) < 0, M (0) > 0 ∀α ∈ (α 1 , 1].
This behaviour may be observed in Figure 3. (b) If J 2 (x) < M 2 < D 0 (x) then, as above, α 0 , α 1 belong to (0, 1) and α 0 = α 1 . Moreover,

α → M(0) ε(0) is increasing on (0, α 0 ), hence lim α→α 0 ,α<α 0 M (0) ε(0) = ∞.
Since (4.66) still holds, we deduce that M (0) vanishes in (0, α 0 ). Thus α 1 < α 0 . Summing up, we get

ε(0) > 0, M (0) < 0 ∀α ∈ (0, α 1 ) ε(0) ≥ 0, M (0) > 0 ∀α ∈ (α 1 , α 0 ] (4.68) ε(0) < 0, M (0) > 0 ∀α ∈ (α 0 , 1].

Properties of bifurcating solutions

5.1. Energy of the bifurcating solutions. In this section, we will compare the energy of the bifurcating solutions u = M (y) + v(y) given by Theorem 4.1 and the energy of the trivial solution u = M (y). Let us recall that, for (u, ε) ∈ V 2 × (0, ∞), the energy of u is given by (2.14). Moreover, δ(y) := (ε(y), M (y)) for y ≃ 0.

Theorem 5.1. Let y → (δ(y), v(y)) be a bifurcation branch given by Theorem 4.1. Then

E M (y) + v(y), ε(y) -E M (y), ε(y) = 1 8 G(α)y 4 + O(y 5 ),
where

G(α) := (2 √ pα 2 + p - √ p) ε(0) ε * + 3M * M (0). (5.1)
Proof. We put u(y) := M (y)+ v(y). For y ≃ 0, the derivative of the function S defined through

S(y) = E u(y), ε(y) -E M (y), ε(y) satisfies d dy S(y) = D u E u(y), ε(y) Ṁ (y) + v(y) + D ε E u(y), ε(y) ε(y) -D u E M (y), ε(y) Ṁ (y) -D ε E M (y), ε(y) ε(y). (5.2)
Since u(y) and M (y) solve (2.8) with M = M (y) and 1 0 v(y) dx = 0, we have

D u E M (y), ε(y) Ṁ (y) = 1 0 f M (y) dx Ṁ (y) D u E u(y), ε(y) Ṁ (y) + v(y) = 1 0 f u(y) dx Ṁ (y).
Furthermore, by Taylor expansions,

1 0 f u(y) -f M (y) dx = 3M (y) 1 0 v(y) 2 dx + 1 0 v(y) 3 dx = 3M * y 2 1 0 ϕ 2 0 dx + O(y 3 ),
in view of (4.50), (4.51). Thus

1 0 f u(y) -f M (y) dx Ṁ (y) = 3 2 M * M (0)y 3 + O(y 4 ).
Considering, in (5.2), the derivatives w.r.t. ε, we have

D ε E M (y), ε(y) = 0 D ε E u(y), ε(y) = 1 0 ε(y) v(y) xx 2 -v(y) x 2 dx.
Thanks to (4.51), we get the following expansions.

1 0 v(y) x 2 dx = y 2 1 0 (ϕ 0x ) 2 dx + O(y 3 ) 1 0 v(y) xx 2 dx = y 2 1 0 (ϕ 0xx ) 2 dx + O(y 3 ).
By (4.49), we have ε(y) = ε(0)y + O(y 2 ); and since ϕ 0 = αϕ * + βϕ * * , we infer

D ε E u(y), ε(y) ε(y) = y 3 ε(0) 2ε * (2 √ pα 2 + p - √ p).
By combining the above results, we prove the assertion of the theorem.

According to Theorem 5.1, it is enough to compute the sign of G(α). Before to state our result, we will introduce some notation. In view of (4.27), (4.40) and (4.41), we have

f * = f * (x, M ) = 3 8 M 2 * x(4x -1) (x + 1) 2 - 3 4 , f * * = f * * (x, M ) = - 3 8 M 2 * x -4 (x + 1) 2 - 3 4 .
Then we denote by (f * ) 0 the cancellation function of f * , i.e.

(f * ) 0 (x) := 2 x(4x -1) (x + 1) 2 .

(5.3)

Similarly (f * * ) 0 (x) := -2 x -4 (x + 1) 2 .
(5.4) Theorem 5.2. Let M * > 0, α ∈ (0, 1) and G(α) be given by (5.1).

(i) If x ∈ (1, 4) then (a) if M 2 * < J 2 (x) then G < 0 on (0, 1) and consequently, E u(y), ε(y) < E M (y), ε(y) for y ≃ 0; (b) if J 2 (x) < M 2 * < (f * * ) 0 (x) then G has two zeros 0 < α 1 < α 2 < 1 and G < 0 on (0, α 1 ) ∪ (α 2 , 1), G > 0 on (α 1 , α 2 ); (c) if (f * * ) 0 (x) < M 2 * < (f * ) 0 (x)
then G has one zeros 0 < α 1 < 1 and G > 0 on (0, α 1 ), G < 0 on (α 1 , 1);

(d) if (f * ) 0 (x) < M 2 * then G > 0 on (0, 1). (ii) If x ∈ (4, ∞) then (a) if M 2 * < (f * ) 0 (x) then G < 0 on (0, 1); (b) if (f * ) 0 (x) < M 2
* then G has one zeros 0 < α 1 < 1 and G < 0 on (0, α 1 ), G > 0 on (α 1 , 1).

Proof. By (4.29), we can write

G(α) = 2 √ p ε(0) ε * α 2 + (-f * * + C S )α 2 + f * * .
With (4.31) and (4.36), we get

G(α) = 3 4 Aα 4 + 3 2 Bα 2 + f * * .
Thus we will study the quadratic polynomial H(•) defined through

H(X) = 3 4 AX 2 + 3 2 BX + f * * .
We readily have

H ′ (α 2 ) = 4 √ p ε(0) ε * (5.5) H(0) = f * * .
Then H(0) > 0 ⇐⇒ x ∈ (1, 4) and M 2 * > (f * * ) 0 (x).

(5.6) Regarding H(1), it turns out that, with (5.1) and (4.29), there holds

H(1) = G(1) = (p + √ p) ε(0) ε * + 3M * M (0) α=1 = f * . Then H(1) > 0 ⇐⇒ M 2 * > (f * ) 0 (x).
(5.7) Moreover we prove easily that

D 0 (x) < (f * * ) 0 (x) < (f * ) 0 (x) ∀x ∈ (1, 4).
(5.8)

Hence we are in position to study the different cases appearing in the statement of the theorem and labelled from (i)(a) to (ii)(b).

(i)(a) First we claim that H(0) and H(1) are negative due (5.6), (5.7), (5.8) and (4.65). Thus if ε(0) has a sign on (0, 1) then (5.5) implies that H < 0 on (0, 1). Otherwise ε(0) has a unique zero α 0 in (0, 1) and, by (5.5), X := α 2 0 is the unique critical point of H in (0, 1). Moreover by (5.1),

H(α 2 0 ) = G(α 0 ) = 3M * M (0) α=α 0 .
(5.9)

If M 2 * ≤ (C + D) 0 (x) then at α = α 0 , we have M (0) < 0 by Proposition 4.9. If (C + D) 0 (x) < M 2 * ≤ J 2 (x) then at α = α 0 , M (0)
is also negative by (4.67). Consequently, G(α) < 0 ∀α ∈ (0, 1). (i)(b) We still have H(0) and H(1) negative. If J 2 (x) < M 2 * < D 0 (x) then M (0) α=α 0 > 0 by (4.68). Thus H(α 2 0 ) > 0 in view of (5.9) and we are able to conclude in this case since X := α 2 0 is the unique critical point of H in (0, 1). If D 0 (x) ≤ M 2 * < (f * * ) 0 (x) then M (0) > 0 for every α ∈ (0, 1), according to Proposition 4.9. So the assertion follows in this case also. The other case can be proved easily by using the above methods together with Proposition 4.9. 5.2. Stability of bifurcation solutions. We refer to the appendix hereafter for the background concerning the one-dimensional Phase Field Crystal Equation (2.7). The main result of this paper is the following.

Theorem 5.3. Under the assumptions and notation of Theorem 4.1, let us suppose that

M 2 * = - (4x -1)(x -4) 15(x + 1) 2 .
(5.10)

Then the stability of the stationary solution v(y) to the Phase Field Crystal Problem (A.1) is as follows.

If

k * = k * * + 1 (i.e. x = (1 + 1 k * * ) 2 ) and M 2 * ∈ (J 1 (x), J 2 (x)) then v(y) is asymptotically stable in the sense of Lyapunov. If k * = k * * + 1 and M 2 * ∈ [J 1 (x), J 2 (x)] then v(y) is not stable in the sense of Lyapunov. If k * = k * * + 1 then v(y) is not stable in the sense of Lyapunov.
Remark 5.1. • For x ∈ (1, 4), J 1 (x) and J 2 (x) are defined by (4.61) and (4.62). Notice that they are the cancellation function of BC -AD.

• If x ∈ (1, 4) and M 2 * ∈ (J 1 (x), J 2 (x)) then the tangent at y = 0 (denoted by T (α)) to the parameter curve y → δ(y) turns clockwise when α goes from 0 to 1. See Proposition 4.10 and Figure 3.

• The above result was unexpected since it connects the stability of bifurcating solutions with the variation of the angle of T (α) with the horizontal axis.

• In view of (4.42), (4.43), C S defined by (4.28) satisfies

C S = - 45 2 (x + 1) 2 (4x -1)(x -4) M 2 * - 3 2 .
Thus (5.10) is equivalent to C S = 0.

• Let x ∈ (1, 4). Recalling that (f * * ) 0 (x) is defined by (5.4), let us suppose that M 2 ∈ (J 2 (x), (f * * ) 0 (x)). Then v(y) is unstable according to the above result. However, the energy of v(y) may be less than the energy of the trivial solution. More precisely, by Theorem 5.2, there exists α 1 ∈ (0, 1) such that for every α ∈ (0, α 1 ) and y ≃ 0,

E M (y) + v(y), ε(y) < E M (y), ε(y) .
This is a quiet unusual result. Let us recall that v(y) depends on α in the following way:

v(0) = αϕ * + βϕ * * .
Proof of Theorem 5.3. According to Proposition A.1, it is enough to consider the constrained Swift-Hohenberg Equation (A.2). As explain in Section 3, if the trivial solution v = 0 is not neutrally stable then v(y) is unstable. By Proposition 3.2, it follows that v(y) is unstable if

k * = k * * + 1.
Let us now assume that k * = k * * + 1. We use the principle of reduced stability from [Kie04, Section I-18] (see also [START_REF] Mielke | A new approach to sideband-instabilities using the principle of reduced instability[END_REF]). According to this result, it is enough to consider the twodimensional eigenvalue problem obtained from the linearization of the bifurcation equation (4.19) at u 0 = yϕ 0 . That is to say (by differentiating (4.19) w.r.t. u 0 ), we have to find λ ∈ R such that the following linear equation set on ker L, namely

P F 1 (µ) • +2F 02 (•, a 02 u 2 0 ) + 4F 02 (u 0 , a 02 (u 0 , •)) + 3F 03 (•, u 2 0 ) + O(µu 2 0 + u 3 0 ) = λ Id ker L (5.
11) has non trivial solutions. For the bifurcating solution (µ(y), v(y)) of (2.12), we have u 0 = yϕ 0 with ϕ 0 = αϕ * + βϕ * * and µ(y) = y 2 µ 2 + O(y 3 ). Here µ 2 stands for the vector 1 2 (ε(0), M (0)). Thus we rescale the eigenvalue λ into λ = y 2 λ and we denote by A : ker L → ker L, the linear operator defined by The principle of reduced stability states that the eigenvalue problem ε(y) 2 w (4) + 2ε(y)w (2) + f ′ (M (y) + v(y))w -

Aw := P F 11 (µ 2 )w + 2F 02 (w, a 02 ϕ 2 0 ) + 4F 02 (ϕ 0 , a 02 (ϕ 0 , w)) + 3F 03 (w, ϕ 2 0 ) . ( 5
Ω f ′ (M (y) + v(y))w dx = λw, w ∈ V4 , w = 0, λ ∈ R,
has two critical eigenvalues λ 1 (y), λ 2 (y) (i.e. eigenvalues close to zero for y ≃ 0) with the following expansions

λ 1 (y) = y 2 λ1 + O(y 3 ), λ 2 (y) = y 2 λ2 + O(y 3 ).
Hence it remains to compute the sign of the eigenvalues λ1 and λ2 of A.

If M 2 * ∈ [J 1 (x), J 2 (x)] then according to (4.56) and Proposition 4.10, we have (BC -AD)(x, M * ) > 0. So det A = λ1 λ2 < 0 by Lemma 5.5 below. Hence v(y) is unstable.

In the same way, if M 2 * ∈ (J 1 (x), J 2 (x)) then λ1 λ2 = det A > 0. Since (see (4.65), (5.8))

J 2 (x) < (f * * ) 0 (x) < * ) 0 (x), one has f * < 0, f * * < 0.
Thus, with Lemma 5.4,

λ1 + λ2 = trace(A) = -2(f * α 2 + f * * β 2 ) > 0.
(5.14) Therefore λ1 and λ2 are positive and v(y) is stable for y ≃ 0. This completes the proof of the theorem.

We now state and prove the lemmas used in the proof of Theorem 5.3.

Lemma 5.4. Let A : ker L → ker L be the linear operator defined by (5.12). Then the matrix M (A) of A in the basis (ϕ * , ϕ * * ) is

M (A) = 2 -f * α 2 -C S αβ -C S αβ -f * * β 2 .
(5.15)

Remark 5.2. The simple formula (5.15) can be obtained, at least at a formal level, by differentiation starting from (4.30). Since (4.30) has a Z 2 ⊕ Z 2 symmetry, (5.15) is in accordance with the results of [GS85, Chap X]. These results are obtained by using symmetries and universal unfolding theory. Moreover the analog of v(y) is obtained in [START_REF] Golubitsky | Singularities and groups in bifurcation theory[END_REF] as a secondary bifurcation.

Here, there is no Z 2 ⊕ Z 2 symmetry in (2.12) and v(y) is a primary bifurcating solution in the sense that it bifurcates from the trivial solution.

Proof. We compute the first column of the matrix.

Aϕ * = P F 11 (µ 2 )ϕ * + 2F 02 (ϕ * , a 02 ϕ 2 0 ) + 4F 02 ϕ 0 , a 02 (ϕ 0 , ϕ * ) + 3F 03 (ϕ * , ϕ 2 0 ) .

We have µ 2 = 1 2 (ε(0), M (0)), thus with (4.24),

P F 11 (µ 2 )ϕ * = p + √ p ε * ε(0) + 3M * M (0) ϕ * .
Due to (4.14), (4.15), we infer

2P F 02 (ϕ * , a 02 ϕ 2 0 ) = - 9 2 M 2 * α 2 λ 2k * ϕ * -9M 2 * αβ 1 λ k * +k * * + 1 λ k * -k * * ϕ * * .
In view of (4.16), we obtain

4P F 02 ϕ 0 , a 02 (ϕ 0 , ϕ * ) = -9M 2 * α 2 λ 2k * + β 2 ( 1 λ k * +k * * + 1 λ k * -k * * ) ϕ * + αβ 1 λ k * +k * * + 1 λ k * -k * * ϕ * * . Besides 3P F 03 (ϕ * , ϕ 2 0 ) = 9 4 α 2 + 3 2 β 2 ϕ * + 3αβϕ * * .
Let us denotes the entries of M (A) by

M (A) = a 11 a 12 a 21 a 22 .
Then, since α 2 + β 2 = 1,

a 11 = (p + √ p) ε(0) ε * + 3M * M (0) + 3 4 α 2 + 3 2 -9M 2 * 3 2λ 2k * -( 1 λ k * +k * * + 1 λ k * -k * * ) α 2 + 1 λ k * +k * * + 1 λ k * -k * * .
Recalling the notation (4.27), (4.28) and using (4.29), we obtain

a 11 = (f * -C S )α 2 + C s + (-3f * + C S )α 2 -C S = -2f * α 2 .
Regarding a 21 , we have

a 21 = -18M 2 * 1 λ k * +k * * + 1 λ k * -k * * αβ + 3αβ = (-2C S -3)αβ + 3αβ = -2C S αβ.
This gives the first column of M (A). By using symmetries resulting from the non raisonnant condition x = 4, 9, we may obtain the second column from the first one. More precisely, the second column is obtained by exchanging k * and k * * on one hand, and by exchanging α and β on the other hand. Thus a 12 = a 21 and

a 22 = -2f * * β 2 .
This completes the proof of the lemma.

Lemma 5.5. Let M * ∈ R, k * , k * * be positive integers such that k * * < k * and f * , f * * , C S be defined by (4.27), (4.28). Let A, B, C, D be given by (4.36)-(4.39) and A be the operator defined through (5.12) (whose matrix is given by (5.15)). Then

det A = - 9 8 α 2 β 2 (BC -AD).
(5.16)

Remark 5.3. To our knowledge, the relation (5.16) is new. It explains why the statement of the stability of v(y) is quiet simple in the sense that the stability is linked to quantities relying on the parameter curve y → δ(y). In particular, if

α → M (0) ε(0)
is increasing then (5.16) implies that v(y) is unstable.

Proof. By using (4.36)-(4.39), we get

BC -AD = B 8 3 (f * -C S ) -( √ p + 1)A -A 8 3 C S -( √ p + 1)B = 8 3 B(f * -C S ) -AC S = 8 3 4 3 (-f * * + C S )(f * -C S ) - 4 3 (f * + f * * -2C S )C S = 32 9 (-f * f * * + C 2 S ).
However by (5.15), det A = 4α 2 β 2 (f * f * * -C 2 S ). Thus (5.16) follows. 5.3. Symmetries. We start to state precisely a local uniqueness result for bifurcating branches. In particular, we will emphasize the dependence of these solutions w.r.t. α and β.

Theorem 5.6. Under the assumptions and notation of Theorem 4.1, there exist

R 1 = R 1 (α, β) > 0, δ 1 = δ 1 (α, β) > 0 and a smooth function µ(•, α, β) : (-δ 1 , δ 1 ) → (-R 1 , R 1 )
such that for every y ∈ (-δ 1 , δ 1 ) and µ ∈ (-R 1 , R 1 ), one has P F µ, yϕ 0 + U (µ, yϕ 0 ) = 0 ⇐⇒ µ = µ(y, α, β).

Remark 5.4. • We recall that ϕ 0 := αϕ * + βϕ * * .

• δ 1 and R 1 may be choosen independently of α and β provided that α and β remain bounded away from 0.

• In this setting, the bifurcating solution v(y) of Theorem 4.1 will be denoted by v(y, α, β).

Let (α, β) ∈ (-1, 1) 2 satisfy (4.48). For y close to zero, we have four (distinct) solutions µ(y, ±α, ±β), v(y, ±α, ±β)

to Equation (2.12). In view of the remark above, we may suppose that the numbers δ 1 (y, ±α, ±β) are equal. So we will denote their common value by δ m . The goal of this subsection is to establish relations between these solutions. This is achieved by using a suitable translation of the space variable. Let us write k * and k * * under the form

k * = 2 r 1 ℓ * , k * * = 2 r 2 ℓ * * ,
(5.17)

where r 1 , r 2 are non negative integers and ℓ * , ℓ * * are positive odd integers. Let us denote by r the minimum of r 1 and r 2 .

This symmetry allows to relate solutions in some cases. However, if k * , k * * are even then it turns out that each of the four solutions µ(y, ±α, ±β), v(y, ±α, ±β)

is invariant under S. So S is useless in what case unlike T (see (5.18)).

Rough approximation of the 8-loop

The aim of this section is to recover by means of two analytical approximations the so-called 8-loops appearing in [PR11, Figure 15]. More precisely, we will use the truncated bifurcation equation (4.30) to approximate the bifurcating solutions (µ, v) to (2.12) given by Theorem 4.1. By suitable choices of parameters values, we will reconstruct analytically the first 8-loop on the left of [PR11, Figure 15], which was obtained by numerical integration.

Under the assumptions and notation of Theorem 4.1, we set for δ

= (ε, M ) ∈ (0, ∞) × R P * (δ) := (εσ k * -1) 2 + r + 3M 2 P * * (δ) := (εσ k * * -1) 2 + r + 3M 2 .
If we neglect the higher order terms in (4.19), we may assume that (X, Y ) solves (4.30). This is the first approximation. Thus, if y, α and β are not zero, we have

f * y 2 α 2 + C S y 2 β 2 = P * (δ) C S y 2 α 2 + f * * y 2 β 2 = P * * (δ) . (6.1)
In order to clarify things, we would like to highlight that, in (6.1), f * depends on δ * and not on δ, since in (4.27), we have

λ 2k * = (ε * σ 2k * -1) 2 + r + 3M 2 * .
Our second approximation is a second order approximation of the function v(y, δ, α), namely v(y, α, β) ≃ yϕ 0 + y 2 a 02 ϕ 2 0 , where ϕ 0 = αϕ * +βϕ * * and a 02 ϕ 2 0 is given by (4.14) and (4.15). Unlike to our previous analysis, we will no more assume that y is close to zero. This is why the above approximation is said to be rough.

In order to solve (6.1), we choose a suitable value M 0 of M , close to M * . The solutions are parametrized by ε ≃ ε * as in [START_REF] Pierre | Stationary solutions to phase field crystal equations[END_REF]. We obtain

y 2 α 2 = -f * * P * (ε, M 0 ) + C S P * * (ε, M 0 ) -f * f * * + C 2 S (6.2) y 2 = P * (ε, M 0 ) -(f * -C S )y 2 α 2 C S . (6.3)
If, in the above equations, y 2 ∈ (0, 1) and α 2 ∈ (0, 1) then we choose w.l.o.g. y, α to be positive and β := √ 1 -α 2 . So what for δ = (ε, M 0 ), we obtain four approximated solutions to (2.8) of the form M 0 + ṽ(y, ±α, ±β) with ṽ(y, α, β) = y(αϕ * + βϕ * * )

- 3M * 2 y 2 α 2 λ 2k * ϕ 2k * + 2 αβ λ k * +k * * ϕ k * +k * * + 2 αβ λ k * -k * * ϕ k * -k * * + β 2 λ 2k * * ϕ 2k * * . (ε k * , M 0 ) M 0 + v| x=0 α > 0, β > 0 α < 0, β < 0 α > 0, β < 0 α < 0, β > 0 α < 0, β > 0
ε → M 0 + ṽ(y, α, β)| x=0 ,
where the dependence of (y, α, β) w.r.t. ε is given by (6.2), (6.3).

• By implementing the Galerkin scheme, we prove that (A.1) admits a unique solution v (see [START_REF] Temam | Infinite-dimensional dynamical systems in mechanics and physics[END_REF], [START_REF] Grasselli | Well-posedness and longtime behavior for the modified phase-field crystal equation[END_REF]). Moreover, the map S P F C : [0, ∞) × V2 → V2 , (t, v 0 ) → v(t) is a semigroup in the sense of Temam. Thus S P F C (t)v 0 stands for v(t).

In the same way but referring to the above constrained Swift-Hohenberg Equation, the problem Notice that the linearized operator for (A.2) at any stationary solution v ∞ is self adjoint with compact resolvant. Thus its spectrum consists on an increasing sequence of eigenvalues.

The main result of this appendix is the following.

Proposition A.1. Let v ∞ be a stationary solution to (A.1). If v ∞ is linearly stable for the semigroup S cSH (i.e. the corresponding eigenvalues are positive) then v ∞ is asymptotically stable in the sense of Lyapunov, for the semigroup S P F C . If one of these eigenvalues is negative and 0 is not an eigenvalue then v ∞ is not stable in the sense of Lyapunov, for the semigroup S P F C .

Roughly speaking, the above results states that if 0 is not an eigenvalue then the stationary solution v ∞ has the same stability for the semigroup S P F C and for the semigroup S cSH .

Proof. Let us assume that v ∞ is linearly stable for S cSH . Expending the energy E(v) for v ∈ V2 , v ≃ v ∞ , we get

E(v) = E(v ∞ ) + 1 2 D 2 E(v ∞ )(v -v ∞ ) 2 + O( v -v ∞ 3 V2
).

With Lemma A.2 below, we deduce that there exist ε > 0 and r 2 > 0 such that

E(v) ≥ E(v ∞ ) + 2ε, ∀v ∈ V2 , v -v ∞ V2 = r 2 . (A.3)
Since v ∞ is linearly stable, we may assume without los of generality, that v ∞ is the only one stationary solution in the ball B(v ∞ , r 2 ) of V2 with radius r 2 and center v ∞ . Since (A.1) and (A.2) have the same steady states, v ∞ is also the unique stationary solution to (A.1) in B(v ∞ , r 2 ). Besides, by continuity of E(•), there exists r 1 ∈ (0, r 2 ) such that

E(v 0 ) ≤ E(v ∞ ) + ε, ∀v 0 ∈ B(v ∞ , r 1 ).
Since the energy is a Lyapunov function for the semigroup S P F C , there holds

E(S P F C (t)v 0 ) ≤ E(v ∞ ) + ε, ∀v 0 ∈ B(v ∞ , r 1 ), ∀t ≥ 0.
Hence by (A.3), S P F C (t)v 0 ∈ B(v ∞ , r 2 ), ∀t ≥ 0.

Moreover, by standard methods (see [START_REF] Temam | Infinite-dimensional dynamical systems in mechanics and physics[END_REF]), we can show that the trajectory {S P F C (t)v 0 | t ≥ 0} is relatively compact in V2 . Since v ∞ is a isolated stationary solution, we deduce from Lassalle's Invariance Principle (see [START_REF] Haraux | Systèmes dynamiques dissipatifs et applications[END_REF]) that

S P F C (t)v 0 ---→ t→∞ v ∞ in V2 .
Thus v ∞ is asymptotically stable equilibrium of S P F C .

Conversely, since 0 is not in the spectrum of the linearized operator for (A.2) at v ∞ , there exists r 3 > 0 such that v ∞ is the only one stationary solution to (A.1) in B(v ∞ , r 3 ).

Since v ∞ is linearly unstable, there exists v 0 ∈ V2 arbitrary close to v ∞ such that

E(v 0 ) < E(v ∞ ).
By continuity of E, there exists a positive number r 1 depending on v 0 such that E(v) > E(v 0 ), ∀v ∈ B(v ∞ , r 1 ).

Moreover, E(S P F C (t)v 0 ) ≤ E(v 0 ), ∀t ≥ 0, since E is a Lyapunov function for the semigroup S P F C . Thus where E denotes the set of all stationary solutions to (A.1). Thus for some positive time t 1 and w ∞ ∈ E, there holds

S P F C (t 1 )v 0 -w ∞ V2 ≤ r 3 2 .
However, since v ∞ is the only steady states of (A.1) in B(v ∞ , r 3 ), we have v ∞ -w ∞ V2 ≥ r 3 .

Thus S P F C (t 1 )v 0 -v ∞ V2 ≥ v ∞ -w ∞ V2 -S P F C (t 1 )v 0 -w ∞ V2 ≥ r 3 2 ,
which means that v ∞ is not stable in the sense of Liapunov. This completes the proof of the proposition.

Figure 1 .

 1 Figure 1. Phase diagramm for r = 2 with four parabolas corresponding to k = 1, . . . , 4.

  This number is non positive since by assumption k * * < k * . Thus with (3.6) we get k * = k * * + 1. Conversely, if k * = k * * + 1 then (3.5) imply that λ k -λ k * > 0 for k = k * , k * * . Thus v = 0 is neutrally stable.

Lemma 4. 3 .

 3 If x ∈ (1, 4) then (A + B) 0 (x) > 0 and

Figure 2 .

 2 Figure 2. The sign of ε(0) and M (0) in the parameter space (x, M 2 ) with x :=(k * /k * * ) 2 . 4.2. Variations of α → M (0)ε(0) . We will prove a monotonicity result for the function α → M (0) ε(0) . In the parameter space (ε, M ), M (0) ε(0) is the tangent of the angle between the unit vector (1, 0) and (ε(0), M (0)). We recall that (ε(0), M (0)) is tangent to the curve (ε(•), M (•)) at y = 0.

  .12) So that (5.11) reads A + O(y) = λ Id ker L . (5.13) Then eigenvalues λ satisfy g(y, λ) := det(A + O(y) -λ Id ker L ) = 0.In view of Lemma 5.4 below, A is a symmetric and non diagonal operator since (5.10) is equivalent to C S = 0. Hence A possesses two distinct real eigenvalues λ1 < λ2 . Then, for i = 1, 2,g(0, λi ) = 0, ∂ ∂ λ g(0, λi ) = d d λ det(A -λ) λ= λi = 0,since λi is a simple eigenvalue. Therefore, by the Implicit function theorem, we get for y ≃ 0, two eigenvalues of (5.13), namely λ1 (y) = λ1 + O(y), λ2 (y) = λ2 + O(y).

Figure 4 .

 4 Figure 4. The blue curves represent four branches bifurcating from the point (ε * , M * ) with 2D kernel. The two red curves show bifurcation branches with 1D kernel, fixed mass M 0 and bifurcation points ε k * (pointed on the figure) and ε k * * . We have choosen k * = 4, k * * = 3, r = -0.5 and M 0 = 2/15. See also Remark 6.1.

Following [ PR11 ,

 PR11 Figure 15], we have choosen k * = 4, k * * = 3, r = -0.5 and M 0 = 2/15. Then by (4.4),Let v 0 ∈ V2 and T ∈ (0, ∞). We say that v is a weak solution to(2.7) in Ω × (0, T ) if v ∈ L 2 (0, T, V5 ) ∩ C([0, T ], V2 ), d dt v ∈ L 2 (0, T, Ḣ-1 (Ω)) d dt v -ε∆ ε 2 ∆ 2 v + 2ε∆v + f (M + v) -Ω f (M + v) dx = 0 in L 2 (0, T, Ḣ-1 (Ω))v| t=0 = v 0 in V2 .(A.1)

v

  ∈ L 2 (0, T, V4 ) ∩ C([0, T ], V2 ), d dt v ∈ L 2 (0, T, L2 (Ω)) d dt v + ε 2 ∆ 2 v + 2ε∆v + f (M + v) -Ω f (M + v) dx = 0 in L 2 (0, T, L2 (Ω)) v| t=0 = v 0 in V2 (A.2)has a unique (strong) solution and infer also a semigroup denoted by S cSH . Without los of generality, we may assume ε = 1 and M = Ω u 0 dx = 0. It is clear that (A.1) and (A.2) have the same steady states. Moreover, the energy E -see (2.14) -defined through Lyapunov functional for S P F C and S cSH . It is clear for the later. For the former, we test the equation of (A.1) with (-∆ -1 ) d dt v where -∆ : H 1 (Ω) ∩ L2 (Ω) → Ḣ-1 (Ω). We find d dt E v(t) = -

S

  P F C (t)v 0 ∈ B(v ∞ , r 1 ), ∀t ≥ 0.Moreover, the Lassalle Invariance Principle implies thatd S P F C (t)v 0 , E := inf w∈E S P F C (t)v 0 -w V2 ---→ t→∞ 0,

  , k * * are fixed and α is a parameter ranging in (0, 1).Let us notice that f * , f * * and C S appear naturally if we consider the truncated bifurcation equation of (4.19). Indeed if u 0

	.29)
	In (4.29), the unknown is (ε(0), M (0)). ε

* , M * , k *

  .30) It turns out that this truncated equation has a Z 2 ⊕ Z 2 symmetry; unlike the bifurcation equation (since the fact that u 0 is a solution to (4.19) does not implies that -u 0 is a solution too).

	Thus we derive from (4.29)
	ε(0) =	3ε * 8 √ p	(Aα 2 + B),		M (0) =	1 8M *	(Cα 2 + D),	(4.31)
	where A, B, C, D satisfy
		3 8	( √ p + 1)A +	3 8	C = f * -C S	(4.32)
		3 8	( √ p + 1)B +	3 8	D = C S	(4.33)
		3 8	( √ p -1)A +	3 8	C = -f * * + C S	(4.34)
		3 8	( √ p -1)B +	3 8	D = f * * .	(4.35)
	Subtracting (4.35) from (4.33), we obtain
		3 4	B = -f

* * + C S . (4.36)

Acknowledgements. The authors would like to thank Prof. Edgar Knobloch and Prof. Mariana Haragus for interesting discussions and comments regarding Phase Field Crystal models and bifurcation theory.

The above mentioned translation consists, roughly speaking, in the translation x → x + 2 -r . To be more specific, for every v ∈ L 2 (Ω), let us denote by Jv : R → R the 2-periodic and even function satisfying Jv = v a.e. in [0, 1].

We put T v := (Jv)(• + 2 -r ) Ω , which means that T v is the restriction to Ω of the function (Jv

. Then we infer that Theorem 4.1 still holds if we restrict our analysis to 2 1-r -periodic functions. Thus, each bifurcating solution given by Theorem 4.1 satisfies

Thus T commutes with L + F (µ, •).

Since ℓ * and ℓ * * are odd, we have

We then deduce in a standard way that, for every y ∈ (-δ m , δ m ), v y, (-1) 2 r 1 -r α, (-1) 2 r 2 -r β = T v(y, α, β), µ(y, (-1) 2 r 1 -r α, (-1) 2 r 2 -r β) = µ(y, α, β).

(5.18) Thus, among the four (distinct) solutions (µ, v)(y, ±α, ±β) to (2.12), only two are essentially different. The other ones are obtained through T (since at least one of the numbers r

) is even. Indeed, by Theorem 5.6 and (5.18), we have

However, if k * = 4k * * then we can prove that µ(•, α, β) is not even. Let us notice that (2.12) has the trivial symmetry

So M 0 is close to M * as required by the theory. Moreover, M 0 must be choosen smaller than M * . Indeed, one has (x, M 2 * ) ≃ (1.8, 0.14). Thus, in view of Figure 2, we have M (0) < 0 for all |α| ∈ (0, 1) and, by (4.50), each bifurcating branch of mass y → M (y) satisfies

This estimate can also be obtained by analytical arguments thanks Proposition 4.9.

• The translation T of subsection 5.3 acts on the approximate solutions ṽ(y, ±α, ±β) as well. Indeed, we have 0 = r = r 2 < r 1 = 2. Thus ṽ(y, α, -β) = T ṽ(y, α, β), ṽ(y, -α, -β) = T ṽ(y, -α, β).

Thus, in Figure 4, the curve corresponding to α > 0, β > 0 can be related to the curve corresponding to α > 0, β < 0 by means of T . Also the curve corresponding to α < 0, β < 0 can be related to the curve corresponding to α < 0, β > 0 by means of T .

• We can see secondary secondary bifurcations between interactive modes solutions and single modes solutions. At the bifurcation point, we have α = 0 or β = 0.

• Let k * = 4, k * * = 3 and r = -0.5. If y is small enough then v(y, α, β) is an asymptotically stable solution to

Indeed, (4.65) implies that the graphs of x → J 1 (x) and x → J 2 (x) lie respectively between the red curves and the blue curves of Figure 2. Thus in view of the remark above M 2 * ∈ (J 1 (x), J 2 (x)). The claim follows then from Theorem 5.3. This stability result is in accordance with the numerical simulations featured in [PR11, Figure 15].

Appendix A. Phase Field Crystal Equation and stability

The aim of this appendix is to show that stability for the Phase Field Crystal Equation (2.7) and for the following constrained Swift-Hohenberg Equation,

are essentially the same. Since the linearized operator corresponding to (2.7) is not symmetric, we will consider for this equation, asymptotic stability in the sense of Lyapunov ; see for instance [START_REF] Haraux | Systèmes dynamiques dissipatifs et applications[END_REF]Chap 3]. We will first define the semigroup associated to (2.7). For simplicity, the derivatives ∂ x and ∂ xx will be denoted by ∇ and ∆. Moreover, notice that all of the results below still hold if the interval Ω is replaced by a smooth bounded domain of R 2 or R 3 . Recalling the notation (2.3), we put

Let us assume that, for some positive constant c, we have

where • 2 denote the standard norm in L 2 (Ω). Then the bilinear mapping

Proof. Let ε > 0 to be choosen later. Since v ∞ ∈ L ∞ (Ω), there exists C > 0 such that for every v ∈ V2 ,

With (A.4),

We use the interpolation inequality

We obtain (A.5) by choosing ε small enough.