The linear and nonlinear Rayleigh–Taylor instability for the quasi-isobaric profile

Physica D 237 (2008) 1602–1639 www.elsevier.com/locate/physd

The linear and nonlinear Rayleigh–Taylor instability for the quasi-isobaric profile Olivier Lafitte ∗ LAGA, Institut Galil´ee, Universit´e de Paris XIII, 93 430 Villetaneuse, France CEA/DM2S, Centre d’Etudes de Saclay, 91191 Gif sur Yvette Cedex, France Available online 21 March 2008

Abstract We study the 2D system of incompressible gravity driven Euler equations in the neighborhood of a particular smooth density profile ρ0 (x) such that ρ0 (x) = ρa ξ( Lx ), where ξ is a nonconstant solution of ξ˙ = ξ ν+1 (1 − ξ ), L 0 > 0 is the width of the ablation region, ν > 1 is the 0 thermal conductivity exponent, and ρa > 0 is the maximum density of the fluid. The linearization of the equations around the stationary solution E p0 ), ∇ p0 = ρ0 gE leads to the study of the Rayleigh equation for the perturbation of the velocity at the wavenumber k: (ρ0 , 0,     d du g − ρ0 (x) + k 2 ρ0 (x) − 2 ρ00 (x) u = 0. dx dx γ We denote by the terms ‘eigenmode and growth rate’ an L 2 (R) solution of the Rayleigh equation associated with a value of γ . The purpose of this paper is twofold: h i • derive the following expansion in k L 0 , for small k L 0 , of the unique reduced linear growth rate √γ ∈ 14 , 1 gk

gk 2 =1+ 2 γ 0(1 + ν1 )



2k L 0 ν

1 ν

2

+ a2 (k L 0 ) ν + O(k L 0 )

where a2 is explicitly known, provided ν > 2,

ρ 0 (x)

• prove the nonlinear instability result for small times in the neighborhood of a general profile ρ0 (x) such that k0 (x) = ρ0 (x) is regular enough, 0 1 bounded, and k0 (x)(ρ0 (x))− 2 bounded (which is the case for ρa ξ( Lx )), thanks to the existence of Λ such that γ ≤ Λ for all possible growth 0

rates and at least one growth rate γ belongs to ( Λ 2 , Λ). This generalizes the result of Guo and Hwang [Y. Guo, H.J. Hwang, On the dynamical Rayleigh–Taylor instability, Arch. Ration. Mech. Anal. 167 (3) (2003) 235–253], which was obtained in the case ρ0 (x) ≥ ρl > 0. c 2008 Elsevier B.V. All rights reserved.

Keywords: Hydrodynamic instabilities; Evans function; Singular differential equations; Weakly nonlinear solutions; Semiclassical analysis; Fluid mechanics

0. Statement of the problem and main result In this paper, we study a theoretical system of equations deduced from the fluid dynamics analysis of an ablation front model. Such models have been studied from a physical point of view by many authors (see Kull and Anisimov [18], Goncharov, [12], Clavin and Masse [7], Bychkov et al. [3], Piriz et al. [24,25], Sanz et al. [29], Almarcha et al. [2] and the references therein). They ∗ Corresponding address: LAGA, Institut Galil´ee, Universit´e de Paris XIII, 93 430 Villetaneuse, France. Tel.: +33 149403570; fax: +33 149403568.

E-mail address: [email protected]. c 2008 Elsevier B.V. All rights reserved. 0167-2789/$ - see front matter doi:10.1016/j.physd.2008.03.017

O. Lafitte / Physica D 237 (2008) 1602–1639

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can be considered as a generalization in the ablation case of the Rayleigh–Taylor instability, studied in the pioneering works of Strutt (Lord Rayleigh) [27] and Taylor [31]. A full self-consistent model is described in [3], coupling the Euler equations with an equation for the temperature T . The equation ((17) of [3] for example, but many authors have obtained this equation) for the dimensionless temperature (that they call Θ) is given. This self-consistent model was widely studied from a linear point of view, in which one derives a formula for the instability growth rate for each Fourier mode [3–25,12,18], generalization of the so-called experimental Takabe formula [30]. Other models appear in the literature: the sharp boundary model (where the density profile is constant in each physically meaningful region [25]), and models of ablation fronts (with boundary conditions on the front as well as a transport equation for the front [7,2]). The sharp boundary model and the front model were also studied from a nonlinear point of view, using the so-called weakly nonlinear approximation for a statistical repartition of modes [11], or the potential model for two fluids leading to the Birkhoff–Rott equations [29,2]. Other weakly nonlinear expansions can be performed, see Hasegawa et al. [15], Sanz et al. [28] in which an expansion in harmonic modes is assumed, or Garnier et al. [10] in which three terms of the weakly nonlinear expansion are given. The linear analysis of the full Euler with thermal conduction system of equations in the neighborhood of the temperature profile solution of Eq. (17) of Bychkov is performed in [21] (under particular assumptions on the Froude number). We obtained the Evans function of the problem and we deduce the properties of a bounded ablation growth rate in the sharp boundary limit. It is not the purpose of the present paper to study this full system, but to reduce the study to a simpler model on which we can perform rigorously both the linear and the nonlinear analysis. From the temperature profile solution of the ablation front model, we deduce a density profile ρ0 (called the Kull–Anisimov profile as in [17]) through the Boyle and Marriot law in a quasi-isobaric set-up ρΘ = cste. We study the system of incompressible Euler equations in the neighborhood of the following solution of the Euler equations: (E u 0 = 0, ρ0 , p0 )

with ∇ p0 = ρ0 gE.

Note that the temperature equation rewrites as the following equation for the density ρ0 (where K is constant, ν is called the thermal conduction index, ρa denotes the density of the ablated fluid) dρ0 = Kρ0 (x)ν+1 (ρa − ρ0 (x)). dx

(1)

We introduce a density scalelength L 0 equal to ρaν+2 K −1 , hence denoting by ξ the function given by ρ0 (x) = ρa ξ( Lx0 ), one has ξ˙ = ξ ν+1 (1 − ξ ).

(2)

In this paper, we shall use the quantity k0 (x) =

ρ00 (x) ρ0 (x)

(3)

and we define L eff through max k0 (x) =

1 νν = L −1 . 0 L eff (ν + 1)ν+1

The system of incompressible Euler equations writes  ∂t ρ + ∂x (ρU ) + ∂z (ρV ) = 0    ∂t (ρU ) + ∂x (ρU 2 + P) + ∂z (ρU V ) = −ρg  ∂ (ρV ) + ∂x (ρU V ) + ∂z (ρV 2 + P) = 0   t ∂x U + ∂z V = 0. Consider ρ = ρ0 + σ , U = v1 , V = v2 , P = p0 + p, the linearized system is  ∂t σ + ρ00 (x)v1 = 0    ρ0 (x)∂t v1 + ∂x p = −σ g ρ0 (x)∂t v2 + ∂z p = 0    ∂x v1 + ∂z v2 = 0.

(4)

(5)

(6)

We then consider a Fourier mode for the first component of the velocity v1 = ue ˜ ikz , from which we deduce (through Fourier i ikz analysis) the second component of the velocity v2 = k ∂x ue ˜ , the perturbation of the pressure p=−

ρ0 (x) ∂t ∂x ue ˜ ikz , k2

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and finally the perturbation of the density   1 1 ρ0 (x)∂t u˜ − 2 ∂x (ρ0 (x)∂t ∂x u) σ =− ˜ eikz . g k Note that the analogous can be derived for real components of the velocity, pressure and density (using the cosine and sine functions, see (16)). One thus deduces the partial differential equation on u: ˜   ∂ ∂ ˜ (7) − ρ0 (x) ∂t22 u˜ + k 2 ρ0 (x)∂t22 u˜ = gk 2 ρ00 (x)u. ∂x ∂x Another formulation of (5) writes as follows. Introduce T (x, z, t) =

ρ0 (x) , ρ(x, z, t)

Q(x, z, t) =

p(x, z, t) − p0 (x) , ρ0 (x)

system (5) is equivalent to the following quadratic equation   ∂t T + UE .∇T − k0 (x)u ∂t UE + (UE .∇)UE + T ∇ Q + T Qk0 (x)Ee1 − (1 − T )E g  = 0. divUE

(8)

It is a consequence of the equality Tρ0−1 ∇ p = T ∇ Q + k0 T Q eE1 + T gE and of ∂t T + UE .∇T = −ρ −2 (∂t ρ + UE .∇ρ). We shall sometimes in what follows denote by (Emod) the quadratic operator appearing on the left-hand side of (8). The associated linearized system in the neighborhood of U0 (x) = (T, UE , Q) = (1, 0, 0, 0) is  ∂t T˜ = k0 (x)u˜ divuE˜ = 0  ˜ ˜ + T˜ gE = 0. ∂t uE + ρ0−1 ∇(ρ0 Q) It is of course equivalent to (6). In this paper we shall study the linear and the nonlinear solutions of (8). We study in a first part a perturbation such that u(x, ˜ t) = u(x, k, L 0 )eγ t , which is written as (the real part of) a normal mode eγ t eikz u(x, k, L 0 ), where γ is the growth rate in time of this perturbation. We obtain that u is the solution of the Rayleigh equation (9) (see many authors like Strutt [27], Taylor [31], Chandrasekhar [4], Kull and Anisimov [18] and used subsequently by Mikaelian [23], Cherfils and Lafitte [5], Lafitte [19], Cherfils et al. [6], then by Guo and Hwang [13], and Helffer and Lafitte [16] from a mathematical point of view):     gk 2 du d + k 2 ρ0 (x) − 2 ρ00 (x) u(x) = 0. ρ0 (x) (9) − dx dx γ Of course, when the wavenumber k is real, all values of γ such that this equation has an L 2 solution are either real or purely imaginary. The physical set-up can be understood from a mathematical point of view through a family of density profiles ρ0 (x) such that ρ0 (x) = ρa ξ( Lx0 ), where L 0 is a characteristic length of the base solution. In one of the physical applications, namely the case of the ICF, its magnitude is 10−6 m. We study (in a spherical model) perturbations of the form eimθ (where the mode m is less than 24), one deduces that the wavenumber of the perturbation is k = mR , where R is the radius of the ignited capsule. Hence we are in a regime (with R = 10−3 m) where k ≤ 2.4 × 104 , hence k L 0 ≤ 2.4 × 10−2 , allowing us to consider the mathematical limit L 0 → 0. Let u(y, ε) = u(L 0 y, k, L 0 ). The Rayleigh equation rewrites     du d − ξ(y) + ε 2 ξ(y) − λεξ 0 (y) u(y) = 0. (10) dy dy Hence the relevant parameter for the linear study is not L 0 but the dimensionless number ε = k L 0 . The good interpretation of the limit L 0 → 0 is ε → 0. Here we develop a constructive method for the study of the modes associated with the Kull–Anisimov density profile. Note that a nonconstant solution of (1) converges to ρa when x → +∞ and the convergence is exponential and the limit of ρ0 at −∞ is zero,

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 1 where the convergence is slow (−x) ν ρ0 (x) → C0 > 0 when x → −∞ . The associated Atwood number is thus 1. Remark also that all nonconstant solutions of (2) differ from a translation. This case may be related to the case of the water waves (the density of air being much smaller than the density of water). It is thus a limit case in all the theoretical set-ups used for the study of Euler equations for fluids of different densities. The following characteristic inequality for any possible growth rate γ holds: γ ≤

r

g = L eff

r

g L0



νν (ν + 1)ν+1

1 2

= Λ ⇐⇒

γ 2 L0 νν . ≤ g (ν + 1)ν+1

(11)

The profile ξ being given, we prove that, for each k, we obtain at least one value of γ depending on k and on the physical h i ∗ characteristics of the system (given by g and L 0 ). There is uniqueness of this value, denoted by γ∗ , when in addition √γgk ∈ 14 , 1 , and when ε = k L 0 , dimensionless number, is small enough. Thanks to (10), λ(ε) = when ε is small, that we calculate rigorously. We shall introduce two equivalent versions of (10), which are:

gk γ∗2

depends only on ε. It has an expansion

1. the system on (U+ , V+ ) such that U+ (y, ε) = u(y, ε)eεy and V+ (given by the first equation of the system below), v(y, ε) = V+ (yε)e−εy :  ε dU+    dy = ε(1 − λ)U+ + ξ(y) V+ (12) dV    + = ε(λ + 1)V+ + ε(1 − λ2 )ξ(y)U+ , dy v 2. using t = −εy, w = ξ(y) , the system on (u, w) is  du   = λu − w  dt   (13) 1 dw 1  2 0  ν  = (λ − 1)u − λw + + ε S (t, ε) w, dt νt where S is defined below by (23). 1

The first part of the main result of this paper was presented in [19], and a similar case (the case where ξ(y) = ξ(1)(y + 1)− ν for y ≥ 0) was solved explicitly in [6] using the confluent hypergeometric functions which appear in the present paper. The case of the global linearized ablation system appears in [21]. We finally recall that, if there exists a solution in L 2 (R) of (10), then λ satisfies the inequality   g 2 γ ≤ min gk, L eff that is λ=

  gk (ν + 1)ν+1 . ≥ max(1, k L ) = max 1, ε eff νν γ2

(14)

The main result of the first part of this paper is Theorem 1. 1. There exist ε0 > 0, and C 0 > 0 such that, for all ε ∈ (0, ε0 ) there exists λ(ε) ∈ [1, 32 ] such that the Rayleigh equation q (10) admits a bounded solution u for λ = λ(ε), corresponding to the eigenmode u. The associated growth rate is γ∗ =

gk λ(ε) ,

where λ(ε) satisfies 1

0 ≤ λ(ε) − 1 ≤ C 0 ε ν . 2. We have the estimate λ(ε) = 1 + = 1+

2 0(1 +

with α = min(1, ν2 ).

2ε ν

1



2ε ν

1

1 ν)

2 0(1 +



1 ν)

ν

ν

1

+ o(ε ν ) + O(εα )

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3. We have the expansion (up to a term in o(ε ν )), ψ denoting the Logarithmic derivative of the Gamma function  1 2ε ν B0 (0) 2   1 − λ(ε) = 1 1 0(1 + ν ) ν ln 2+ψ(1) ε ν1 εν lim R0 (t) − B0 (0) + ( ν ) + ν−1 ν t→0

where B0 (0) = −2 in formula (61).

R∞ 0

1 ν

s e−2s ds = −2

− ν1

0(1 + ν1 ), −ψ(1) being the Euler constant1 and R0 (t) is given in Proposition 4 and

Note that k is fixed and we deal with the limit L 0 → 0. We deduce, for k fixed and for L 0 < v u gk .  γ∗ = u t 1 2k L 0 ν1 2 ν ( ) 1+ + o (k L ) 0 1 ν

ε0 k ,

that (15)

0(1+ ν )

√ Note that, in this case, we do not have the estimate γ∗ − gk = O(k L 0 ) as usually guessed in the traditional analysis of the 1 Rayleigh–Taylor instability with a gradient scalelength. Item 1 of Theorem 1 appeared in [6] for the profile ρa ( Lx0 + 1) ν and

in [19]. Observe that the result of [16], based on ρ0 − ρa 1x>0 ∈ L ν+θ for all θ 0 > 0, is pertinent also in this set-up, except that the classical Rayleigh problem is not relevant when ρ = 0 on x < 0, that is when the Atwood number is 1. Closely following Helffer and OL [16], and generalizing the semiclassical result to the case where ρ0 (x) ≥ ρl > 0, we have 0

Proposition 1. Let us denote by λn (ε) the sequence, (possibly empty or finite) decreasing in n (λn+1 (ε) < λn (ε)), going to zero when n goes to +∞, of generalized eigenvalues of the Rayleigh equation. For any n, λn (ε) exists for ε large enough, and satisfies λn (ε) L eff = . ε→+∞ ε L0 lim

Remark that formula (15) and Proposition 1 are not in contradiction. They lead to two different stabilizing mechanisms induced by the transition region: one is a low frequency stabilization when L 0 → 0 and the other one is a high frequency stabilizing mechanism when k → +∞. This could also be rightfully written as the ε → 0 limit and the ε → +∞ limit. In a second part, we use a particular solution u(x, ˆ k) of the linear Rayleigh equation, associated with the mode k and a growth rate γ (k, ε) such that γ (k, ε) ∈ [ Λ2 , Λ]. It is constructed through Proposition 1. Note that it appears that γ (k, ε) is not, in general, equal to γ∗ . Multiple values of γ exist, see [5,16,23]. Contrary to the uniqueness of γ∗ , γ (k, ε) is not unique, hence we make a choice. Remark also that, for L 0 fixed and k → +∞, all eigenvalues of the Rayleigh equations lie in the region ( Λ2 , Λ). Remember that the modified Euler system (8) has the stationary solution T = 1, uE = 0, Q = 0. From this base solution and the particular solution of the linear Rayleigh equation, we construct an approximate solution of the nonlinear system (8) such that its initial value is   T1  u1   U0 (x) + δU1 (x, z) = U0 (x) + δ   v1  (x, z, 0) Q1    k (x)  0 0    γ (k, ε)  0 sin kz     = U0 (x) − δ uˆ 0 (x, k) ˆ k) cos kz  1  (16)  1  + δ u(x,  0  k  γ (k, ε)  0 k P We construct a function V N = (1, 0, 0, 0)(L 0 x) + Nj=1 δ j V j (x, y, t) satisfying  Emod(V N ) = δ N +1 R N +1 V N (x, z, 0) − (1, 0, 0, 0)(L 0 x) = δU (x, z, 0). It can be considered as a weakly nonlinear solution of the nonlinear system (8). We also construct the solution V (x, y, t) of the modified Euler system such that  Emod(V ) = 0 V (x, z, 0) − (1, 0, 0, 0)(L 0 x) = δV1 (x, y, 0). 1 That we cannot denote in our set-up through the letter γ which denotes the growth rate!

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Introduce finally V d (x, y, t) = V (x, y, t) − V N (x, y, t). The function V N is an approximate solution of the nonlinear system and V is the solution of the nonlinear system with the same initial value. We have the Theorem 2. 1. (Control on the weakly nonlinear solution) There exist two constants A and C0 , depending only  on the properties  1 of the Euler system, on the stationary solution and on the solution u(x), ˆ such that, for all θ < 1, for all t ∈ 0, γ (k,ε) ln δCθ0 A , one has the control of the approximate solution V N in H s , namely δ AC0 eγ (k,ε)t . 1 − δ AC0 eγ (k,ε)t The leading order term of the approximate solution is the solution of the linear system. It is proven through the inequality kT N − 1k H s + kE u N k H s + kQ N k H s ≤ C

kT N − 1k L 2 ≥ δkT1 (0)k L 2 eγ (k,ε)t − AC02 C3

eγ (k,ε)t . 1 − δ AC0 eγ (k,ε)t

2. (Control of the remainder term) There exists N0 such that for any N ≥ N0 , the function V d is well defined for t < and satisfies the inequality   1 1 kV d k ≤ δ N +1 e(N +1)γ (k,ε)t , ∀t ∈ 0, . ln γ (k, ε) δ

1 γ (k,ε)

ln 1δ

3. (The remainder term is large when t is large) We have the inequality, for ε0 < C0 A

 

1 ε0

uE

≥ ε0 kE ln u 1 (0)k L 2 .

γ (k, ε) C Aδ 2 2 0

L

This paper is organized as follows. Sections 1–3 study the linear system and identify the behavior of the growth rate γ (k, ε) when L 0 → 0 by constructing the Evans function, and in Section 4 we construct an approximate solution of the nonlinear system of Euler equations. We identify in a first section the family of solutions of (10) which are bounded when y → +∞ and we extend such solutions, for (ε, λ) in a compact B, on [ξ −1 (ε R), +∞), where R is a constant depending only on B (Proposition 2). In the second section, for all t0 > 0, we calculate a solution of (10) which is bounded on (−∞, − tε0 ] (Proposition 3). A solution u of (10) which is in L 2 (R) goes to zero when y → +∞ as well as when y → −∞. Moreover, as ρ0 (x) is a C ∞ ∞ function on R, any solution  u 1of (9) is also in C . h  i   1 Since limε→0 εξ −1 (ε R) ν = − ν1R , for each t0 ∈ 0, 12 limε→0 −εξ −1 (ε R) ν , the regions −∞, − tε0 and h    1 ξ −1 (ε R) ν , +∞ overlap for ε small enough and 

ξ

−1

     1 1 t0 3 ν (ε R) , − ,− ⊂ − . ε 4εν R 2εν R

Hence the solution u belongs to the family of solutions described in Proposition 2 (of the form C∗ u + (y, ε)) and belongs to the family of solutions described in Proposition 3 (of the form C∗∗ U (−εy, ε)), that is    1 C∗ u + (y, ε), y ≥ ξ −1 (ε R) ν u(y) = C∗∗ U (−εy, ε), y < − t0 . ε From the continuity of u and of u 0 , one deduces that, for all y⊥ ∈ [− 4εν3 R , − 2εν1 R ] (corresponding to t⊥ = −εy⊥ ∈ [ 2ν1R , 4ν3R ]), d we have C∗ u + (y⊥ , ε) = C∗∗ U (t⊥ , ε), C∗ dy u + (y⊥ , ε) = −C∗∗ εU 0 (t⊥ , ε). Introduce the Wronskian (where ε−1 has been added for normalization purposes)   d d −1 W(y, ε) = ε u + (y, ε) [U (−εy, ε)] − [u + (y, ε)]U (−εy, ε) . dy dy It is zero at y⊥ = −εt⊥ . Conversely, if λ and ε are chosen such that the Wronskian is zero (in particular at a point y⊥ = − tε⊥ ), the function  U (−εy, ε), y ≤ y⊥ u(y) ˜ = C∗∗ U (−εy⊥ , ε) (17) u + (y, ε), y ≥ y⊥  u + (y⊥ , ε)

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is, thanks to the Cauchy–Lipschitz theorem, a solution of (9). Moreover, it belongs to L 2 (R) owing to the properties of u + and of U . In Section 3, we compute the function W. As U and u + are solutions of the Rayleigh equation, which rewrites   ξ 0 (y) du + d2 ξ 0 (y) 2 [u + (y, ε)] = − u + (y, ε) + ε − ελ ξ(y) dy ξ(y) dy 2 the function W is a solution of

d dy W

ξ(y)W(y) = ξ(y0 )W(y0 )

(y) = − ξξ(y) W, which implies the equality 0

for all y, y0

(18)

This Wronskian can be computed for y⊥ ∈ [− 4εν3 R , − 2εν1 R ] using the expressions obtained for U and u + . We prove that it admits a unique root for 0 < ε < ε0 and λ in a fixed compact, and we identify the expansion of this root in ε, hence proving Theorem 1. Precise estimates of this solution are given in Section 3. Section 4 is devoted to the study of the nonlinear model. It generalizes the result of Guo and Hwang to a density profile not satisfying ρ0 (x) ≥ ρl > 0 for all x, and it uses the quadratic form of the nonlinearity in (8) to give more precise estimates. When the nonlinearity is not quadratic, one has to use caution, and if there is a true nonlinearity in the system (as it was in system (5)), the expansion in δ p of the approximate solution V N is not a converging expansion for N → +∞. Generalization of the result in a “quadratic type” set-up (which means that the Hessian of the nonlinearity is bounded, which is not the case, for example, for the term ρuv) is to be found in [26]. The last section is organized as follows. After proving a H s result on a general solution of the linear system (taking into account a mixing of modes), we use a linear eigenmode associated with a growth rate γ (k, ε) ∈ ( Λ2 , Λ). We calculate (using the method of Grenier [14], and Cordier, Grenier and Guo [8]) all the terms V j of the expansion of an approximate solution (as described before stating Theorem 2). We prove that kV j (x, y, t)k L 2 ≤ (AC0 ) j Q −1 e jγ (k,ε)t . We thus compare the solution of (8) with the same initial condition with the term V N and deduce the estimation of the difference. 1. Construction of the family of bounded solutions in the dense region d E System (12) writes dy U+ = εM0 (ξ(y), λ)UE+ . When y → +∞, the matrix converges exponentially towards M0 (1, λ), whose eigenvalues are 0 and 2, of associated eigenvectors (1, λ − 1) and (1, λ + 1). It is classical that

Lemma 1. There exists a unique solution (U+ , V+ ) of (12) whose limit at y → +∞ is (1, λ − 1). Moreover, there exists ξ0 > 0 such that this solution admits an analytic expansion in ε for ξ(y) ∈ [ξ0 , 1). The proof of this result is for example a consequence of Levinson [22]. Note that the same result holds for the other eigenvalue of the system at +∞: there exists a unique solution (U˜ , V˜ ) such that (U˜ , V˜ )e−2εy → (1, λ + 1) at +∞. The aim of this section is to express precisely the coefficients of this expansion when ξ(y) → 0 and to deduce that one can extend the expression obtained for ξ ∈ [ξ(ε R), ξ0 ]. We consider, in what follows, the change of variable ζ =

ε . ξ(y)ν

(19)

Begin with the definition of the following functions (a j , b j )(ξ ) given through the recurrence relations Z 1 b j (1 + θ (ξ − 1)) − (λ − 1)a j (1 + θ (ξ − 1)) a j+1 (ξ ) = ξ ν( j+1)+1 dθ (1 + θ (ξ − 1))ν( j+1)+2 0 Z 1 b j (1 + θ (ξ − 1)) − (λ − 1)a j (1 + θ (ξ − 1)) b j+1 (ξ ) = (λ + 1)ξ ν( j+1) dθ (1 + θ (ξ − 1))ν( j+1)+1 0 with the initial value a1 (ξ ) =

1−ξ ν+1 (ν+1)(1−ξ ) , b1 (ξ )

ν

= (1 + λ) 1−ξ 1−ξ . We have

Lemma 2. Let ξ0 > 0 be given. The functions a j and b j are bounded, analytic functions of ξ , for ξ ∈ [ξ0 , 1]. They satisfy, for all ξ ∈ [0, 1] |a j (ξ )| ≤ A R j ,

|b j (ξ )| ≤ A R j ,

(20)

where R depends only on λ. In addition, we have |a j (ξ ) − a j (0) − ξ a 0j (0)| ≤ Aξ 2 R j ,

|b j (ξ ) − b j (0) − ξ b0j (0)| ≤ Aξ 2 R j .

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Introduce A and B through  1 ! ∞ X ε ν ζ j, ζ A(ζ, ε) = aj ζ j=1

ζ B(ζ, ε) =

∞ X

bj

j=1

 1 ! ε ν ζ j, ζ

|ζ | <

1 . R

We prove in this section the Proposition 2. Let K be a compact set and λ ∈ K . For ξ0 given, ε0 =

ξ0ν R,

for all ε such that 0 < ε < ε0 , the family of solutions 1

of (12) which is bounded when y → +∞ is characterized, for y such that ξ(y) ≥ (ε R) ν , by  (1 − ξ(y))(1 − λ)  U+ (y, ε) = 1 + ζ A(ζ, ε) ξ(y) V (y, ε) = λ − 1 + (1 − λ)(1 − ξ )ζ B(ζ, ε). +

The associated solution of (10) is u + (y, ε) = U+ (y, ε)e−εy . Note that all the solutions of (12) bounded at +∞ are given by K + (U, V ) where K + is a constant. Proof of Proposition 2. We assume that U and V are given through the following formal expansion in ε: X X U =1+ εju j, V =λ−1+ εjvj. j≥1

j≥1

We deduce, in particular,  du λ−1 1  = (1 − ξ(y))  dy ξ(y) dv   1 = (λ2 − 1)(1 − ξ(y)) dy hence assuming that u 1 , v1 → 0 when ξ → 1 (which is equivalent to dividing the solution U by its limit when ξ → 1) we get  1 − λ 1 − ξ ν+1   u 1 = ν + 1 ξ ν+1 1 − λ2 1 − ξ ν    v1 = ν ξν and deduce the following recurrence system for j ≥ 1:  du 1 j+1  = (v j − (λ − 1)ξ u j )  dy ξ dv   j+1 = (λ + 1)(v j − (λ − 1)ξ u j ). dy

(21)

Usual methods for asymptotic expansions would lead to the estimates (which are not sufficient for the proof of Proposition 2) |u j (y)| + |v j (y)| ≤

MAj (ν+1) j

ξ0

.

However, using the relation 1 − ξ = u j (y) =

ξ˙ ξ ν+1

(and it is the key of our study), we obtain the following equalities

(1 − ξ(y))(1 − λ) a j (ξ(y)), ξ ν j+1

v j (y) =

(1 − ξ(y))(1 − λ) b j (ξ(y)) ξνj

hence better estimates for u j and v j , which allow us to obtain the equalities, for all y such that ξ(y) ≥ ξ0 :  j  ε (1 − ξ(y))(1 − λ) X  U+ (y, ε) = 1 + a j (ξ(y))   ξ(y) ξ(y)ν  j≥1      j   X (1 − ξ(y))(1 − λ) ε ε    = 1 + a (ξ(y)) j+1   ξ(y) ξ(y)ν j≥0 ξ(y)ν   j X  ε   V (y, ε) = λ − 1 + (1 − λ)(1 − ξ(y)) b (ξ(y))  + j  ξ(y)ν   j≥1      j  X  ε ε   = λ − 1 + (1 − λ)(1 − ξ(y)) b j+1 (ξ(y)) .  ξ(y)ν j≥0 ξ(y)ν

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O. Lafitte / Physica D 237 (2008) 1602–1639

Using the estimates (20) and the change of variable (19), for ζ < R −1 the series following functions are well defined  U˜ (y, ε) = 1 + (1 − λ)(1 − ξ(y)) ζ A(ζ, ε) ξ(y) ˜ V (y, ε) = λ − 1 + (1 − λ)(1 − ξ(y))ζ B(ζ, ε).

P

aj

 1 ε ζ

ν

ζ j is normally convergent and the

ν

ξ It is straightforward to check that U˜ and V˜ solve system (12) and that we have, for ξ(y) ≥ ξ0 , ζ (ξ ) ≤ ξεν , hence for2 ε < ε0 = 2R0 0 and ξ(y) ≥ ξ0 we have U˜ (y, ε) = U+ (y, ε) and V˜ (y, ε) = V+ (y, ε). We extended the solution constructed for ξ(y) ∈ [ξ0 , 1) to the region ζ < R1 . This proves Proposition 2.

2. The solution in the low density region 2.1. Construction of the bounded solution In this section, we obtain the family of solutions of (10) bounded by |y| A eεy when y → −∞, that is in the low density region ξ → 0. Introduce the new variable t = −εy. Commonly, I call this solution the hypergeometric solution, because it has been 1 observed that, in the model case ρ0 (x) = (−x − 1)− ν studied in [6] as well as in [12], the Rayleigh equation rewrites as the hypergeometric equation. Define the functions r (t, ε) and τ (t, ε) through:  ν    d (ξ( −t )) 1 1 −t 1 −t 1 τ (t, ε) = − dt −tε = ξ + ε ν t −1− ν r (t, ε). (22) 1−ξ = ε ε ε νt ξ( ε ) 1

There exists t0 > 0 and ε0 > 0 such that r (t, ε) is bounded for t ≥ t0 , 0 ≤ ε ≤ ε0 , and has a C ∞ expansion in ε, ε ν . Define S through 1

ε ν S 0 (t, ε) = ε −1

ξ 0 (− εt ) ξ(− εt )

−

1 , νt

lim S(t, ε) = 0.

t→+∞

We have the identity 

t ξ − ε



νt ε

1 ν

  1 exp ε ν S(t, ε) = 1

(23)

which implies that there exists a function r bounded for t ≥ t0 and ε ≤ ε0 such that   1 1 1 exp −νε ν S(t, ε) = 1 + ε ν t − ν r (t, ε). We define the operators Rε , K ε and K˜ ελ through   y −λ  s λ Z ∞ Rε (g)(s) = ξ − τ (y, ε)e−2y ξ − g(y, ε)dy e2s , ε ε s Z 1 − λ2 +∞ τ (s, ε)Rε (g)(s, ε)ds. K ε (g)(t) = (1 − λ) K˜ ελ (g)(t) = 4 t

(24) (25)

These operators rewrite  Z +∞  1 1 λ λ 1 ν + ε ν S 0 (y, ε) e−2(y−s) s − ν y ν eε λ(S(y)−S(s)) g(y, ε)dy. Rε (g)(s, ε) = νy s  Z +∞  2 1 1−λ 1 K ε (g)(t, ε) = + ε ν S 0 (s, ε) Rε (g)(s, ε)ds. 4 νs t We have the inequalities, for g uniformly bounded, (and λ < ν, which implies ξ(− εs )ν−λ ≤ ξ(− εt )ν−λ for t ≥ s) Z +∞  1 ν−λ ξν −2y |Rε (g)(s)| ≤ kgk∞ ξ (1 − ξ )e dy e2s ξ λ ≤ kgk∞ ε ε s 2 Note that these inequalities depend on a given arbitrary ξ > 0. 0

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O. Lafitte / Physica D 237 (2008) 1602–1639

|K ε (g)(t)| ≤

|λ2 − 1| kgk∞ 4

∞

Z

τ (s, ε) t

ξν |λ2 − 1| ξν ds ≤ kgk∞ . ε 4ν ε

1611

(27)

Moreover, the following inequality is true: |g(s, ε)| ≤ C p



ξν ε

p

|λ2 − 1| H⇒ |K ε (g)(t, ε)| ≤ Cp 8ν( p + 1)



ξν ε

 p+1

.

(28)

In a similar way, we introduce Z 1 − λ2 +∞ 1 λ λ K 0 (g)(t) = R (g)(s)ds 4 νs 0 t Z +∞ 1 −2(y−s) − λ λ R0λ (g)(s) = e s ν y ν g(y)dy. νy s Let ε0 > 0 be fixed and 0 < ε < ε0 . Under suitable assumptions on g (we can for example consider g in C ∞ ([t0 , +∞[) such that |∂ p g| ≤ C p y α− p for all p), the operators K ε , Rε , K 0 , R0 are well defined. Moreover, one proves that X g(t, λ, ε) = K ε(n) (1)(t, ε) (29) n≥0

g0 (t, λ) =

X

(n)

K 0 (1)(t)

(30)

n≥0

are normally converging series on [t0 , +∞[, and that we have: g = 1 + K ε (g),

g0 = 1 + K 0 (g0 ).

(31)

P (|λ2 −1|A) p ξ ν p ν Moreover, we know that g is defined on R, because the series ( ε ) converges and is majorated by exp(|λ2 − 1|A ξε ), p! from inequality (28). We obtain the inequalities   2   2 |λ − 1| −1 |λ − 1| ζ . (32) |g0 (t, λ)| ≤ exp , |g(t, λ, ε)| ≤ exp 8ν 4ν 2 t Thus we cannot consider the limit ζ → 0 in the equalities containing g as (32). We shall assume that λ belongs to a compact set and that λ ≥ 12 . We prove Proposition 3. Let g be defined through (29). The family of solutions of system (13) on (u, w) which is bounded by |y| A eεy when y → −∞ is given by U (y, ε) = C(F(t, λ, ε) + G(t, λ, ε)), ξ(y)W (y, ε) = V (y, ε) = Cξ(y)[(λ − 1)F(t, λ, ε) + (λ + 1)G(t, λ, ε)] where C is a constant, t ∈ [t0 , +∞), t = −εy and F and G are given by equalities (34) and (35) below. We have the estimates, for t ∈ [t0 , ε[ 1

|g(t, λ, ε) − g0 (t, λ)| ≤ C0 ε ν |g0 (t, λ)|       u − t , ε − u 0 − t , ε ≤ C0 ε ν1 u 0 − t , ε ε ε ε       v − t , ε − v0 − t , ε ≤ C0 ε ν1 v0 − t , ε . ε ε ε Proof. System (13) rewrites on F and G given by Proposition 3:    1 1 1  0 0  ν + ε S (t, ε) [(λ − 1)F(t, λ, ε) + (λ + 1)G(t, λ, ε)]  F (t, λ, ε) = F(t, λ, ε) − 2 νt   1 1 1  0 0  ν + ε S (t, ε) [(λ − 1)F(t, λ, ε) + (λ + 1)G(t, λ, ε)]. G (t, λ, ε) = −G(t, λ, ε) + 2 νt

(33)

A nonexponentially growing solution of system (33) is obtained through the following procedure. We denote by g(t, λ, ξ ) the function

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O. Lafitte / Physica D 237 (2008) 1602–1639



 λ+1   λ+1 2 ε − 2ν ν   1 1+λ exp −ε ν S(t, ε) . 2

t g(t, λ, ε) = G(t, λ, ε)e ξ − ε t

= G(t, λ, ε)et t −

1+λ 2ν

(34)

We first get, from the fact that F is bounded when t → +∞, that   1−λ 1 λ−1 t 2  ε  λ−1 2ν ν λ−1 F(t, λ, ε)e−t t 2ν eε 2 S(t,ε) = F(t, λ, ε)e−t ξ − ε ν   Z 1 λ−1 1 λ + 1 +∞ 1 ν λ−1 = + ε ν S 0 (s, ε) s 2ν eε 2 S(s,ε) e−s G(s, λ, ε)ds 2 νs t Z +∞ ε λ λ+1 d ν =− ξ −1 (ξ )g(s, λ, ε)e−2s ξ −λ ds 2 ds ν t Z  ε  λ−1 1−λ λ + 1 +∞ −1 d 2ν =− ξ g(s, λ, ε)ds. (ξ )ξ 2 2 ds ν t

(35)

We deduce from system (33) the equality        1 1+λ 1 1+λ 1+λ 1 1+λ λ−1 1 d S(t, ε) = + ε ν S 0 et t − 2ν exp −ε ν S(t, ε) F(t, λ, ε). G(t, λ, ε)et t − 2ν exp −ε ν dt 2 2 νt 2 Under the assumptions g bounded and satisfies the condition lim g(t, λ, ε) = 1

(36)

t→∞

one gets the equality g(t, λ, ε) − 1 = K ε (g)(t, ε).

(37)

Using the usual Volterra method and inequalities (27), (28) and (32), we deduce that the only solution of (37) satisfying assumptions (36) is given through (29). One gets G through (34) then F thanks to ε λ λ + 1 Z ∞  ε  λ−1 1−λ ν 2ν = F(t, λ, ε)e−t ξ 2 τ (s, ε)e−2s ξ −λ g(s, λ, ε)ds. (38) ν ν 2 t The first part of Proposition 3 is proven. Denote by (u 0 , w0 ) the leading order term in ε of (u, w) when t and λ are fixed. Introduce F0 (t, λ) and G 0 (t, λ) through the equalities u 0 (t, λ) = F0 (t, λ) + G 0 (t, λ), w0 (t, λ) = (λ − 1)F0 (t, λ) + (λ + 1)G 0 (t, λ). The functions (F0 (t, λ), G 0 (t, λ)) are solutions of  dF λ−1 λ+1 0  (t, λ) = F0 (t, λ) − F0 (t, λ) − G 0 (t, λ)  dt 2νt 2νt   dG 0 (t, λ) = −G (t, λ) + λ − 1 F (t, λ) + λ + 1 G (t, λ). 0 0 0 dt 2νt 2νt The second part of Proposition 3 comes from the following estimates on the operators Rε and K ε , valid for ε ≤ ε0 and t ≥ t0 > 0: |Rε ( f ) − R0λ ( f )| ≤ C1 ε ν |R0λ ( f )|, 1

|K ε (g) − K 0λ (g)| ≤ C2 ε ν |K 0λ (g)|, 1

(39)

from which we deduce the uniform estimates for g given by (34) solution of (37) 1

|g(t, λ, ε) − g0 (t, λ)| ≤ C3 ε ν |g0 (t, λ)|,

t ≥ t0 , ε ≤ ε0

because the Volterra series associated with K 0 is normally convergent in [t0 , +∞). This ends the proof of Proposition 3. 2.2. Construction of the hypergeometric solution for ε = 0 We prove in this section

(40) 

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O. Lafitte / Physica D 237 (2008) 1602–1639

Lemma 3. The solution (F0 (t, λ), G 0 (t, λ)) constructed through (34), (37) and (38) for ε = 0 is given by    1 dU0  −t  U0 (t, λ) + (t, λ)  F0 (t, λ) = e 2 dt   1 dU0  −t  U0 (t, λ) − (t, λ) , G 0 (t, λ) = e 2 dt λ+1

1 where U0 (t) = 2− 2ν U (− 1+λ 2ν , − ν , 2t) the function U (a, b, T ) being the Logarithmic Kummer’s solution of the confluent hypergeometric equation (see [1]).

This leads to the following calculus of the limit of (F0 (t, λ), G 0 (t, λ)) when t → 0. The equation satisfied by U0 (t, λ) = u 0 (t)et is   1 λ+1 tU000 − 2t + U00 + U0 = 0. (41) ν ν Introducing T = 2t, we recognize (see [1]) the equation for hypergeometric confluent functions for b = − ν1 and a = − 1+λ 2ν : d2 U0 T − dT 2



1 +T ν



dU0 1+λ + U0 = 0. dT 2ν

The family of solutions of this Kummer’s equation is generated by two functions M(a, b, T ) and U (a, b, T ). Note that T 1−b M(1 + a − b, 2 − b, T ) is also a solution of (41), independent of M(a, b, T ), hence U (a, b, T ) can be expressed using M(a, b, T ) and T 1−b M(1 + a − b, 2 − b, T ). The family of solutions of (41) which goes to zero when T → +∞ is generated by U (a, b, T ), is called the logarithmic solution. It is the subdominant solution of the hypergeometric equation. The expression of the subdominant solution U (a, b, T ) is the following:   M(a, b, T ) M(1 + a − b, 2 − b, T ) π − T 1−b U (a, b, T ) = sin πb 0(1 + a − b)0(b) 0(a)0(2 − b) R ∞ s−1 −t
where 0 is the usual Gamma function 0(s) = 0 t e dt . The relation between U (a, b, 0) and U 0 (a, b, 0) characterizes the subdominant solution of the ordinary differential equation, and this particular solution has been chosen through the limit3 when z → +∞: U (a, b, 0) =

0(1 − b) , 0(1 + a − b)

lim z a U (a, b, z) = 1.

(42)

z→+∞

As we imposed that g(t, λ, ε) → 1 when t → +∞ for all λ, , we get that G 0 (t, λ)et t − λ+1 exists a constant C˜ such that F0 (t, λ)et t 1− 2ν → C˜ when t → +∞. Hence (F0 (t, λ) + G 0 (t, λ))et t −

λ+1 2ν

λ+1 2ν

→ 1 when t → +∞ and that there

→ 1.

As T a U (a, b, T ) → 1, we get that t −

1+λ 2ν

λ+1

1 2ν . We thus obtain the equality U (− 1+λ 2ν , − ν , 2t) → 2   1+λ λ+1 1+λ 1+λ 1 , − , 2t , t − 2ν et (F0 (t, λ) + G 0 (t, λ)) = 2− 2ν t − 2ν U − 2ν ν

hence one deduces 2U0 (t, λ) = 2−

λ+1 2ν

  1+λ 1 U − , − , 2t . 2ν ν

(43)

Introduce  1+λ 1 π B1 (λ) = U − ,− ,0 = − π 1 2ν ν sin ν 0(− ν )0(1 + 

π 3 0(1−b) = 0(1+a−b) sin πb0(b)0(1+a−b) .

λ−1 2ν )

=

0(1 + ν1 ) 0(1 +

λ−1 2ν )

.

(44)

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O. Lafitte / Physica D 237 (2008) 1602–1639 λ+1

1 −t We get that u 0 (t) = 2−1− 2ν U (− 1+λ 2ν , − ν , 2t)e . As w0 = λu 0 −   1 dU0 (t, λ) e−t , G 0 (t, λ) = U0 (t, λ) − 2 dt   1 dU0 F0 (t, λ) = U0 (t, λ) + (t, λ) e−t . 2 dt

du 0 dt

= ((λ + 1)U0 −

dU0 −t dt )e

one deduces

(45)

Using [1] and (41), we finally obtain G 0 (t, λ) → 2−1−

λ+1 2ν

B1 (λ),

F0 (t, λ) → 2−1−

λ+1 2ν

B1 (λ)

when t → 0.

(46)

Note that we can deduce the expressions of F0 + G 0 and of G 0 . We thus check that      λ+1 1 1+λ 1 1−λ 1 (F0 + G 0 )(t, λ)et = 2− 2ν B1 (λ) M − , − , 2t − C∗ (2t) ν +1 M 1 + , 2 + , 2t 2 ν 2ν ν      1 1+λ 1 1−λ 1 0 − λ+1 +1 0 t ν 2ν B1 (λ) (M − M ) − , − , 2t − C∗ (2t) (M − M ) 1 + , 2 + , 2t e G 0 (t, λ) = 2 ν 2ν ν     2 1 1 1 1 1−λ tνM 1+ − C0 C∗ 2 ν 1 + , 2 + , 2t . ν 2ν ν

(47)

(48)

In Section 3, we combine the results of Sections 1 and 2. 3. Precise calculus of the Evans function The Wronskian is related to a function independent of the variable t, called the Evans function, introduced below in (49) and denoted by Ev(λ, ε). In the present section, we shall identify the leading order term in ε of the Evans function, and all the terms of 2 1 the form ε ν (λ − 1) of the Evans function. We shall finish by the calculation of the term of the form ε ν . More precisely, we prove Lemma 4. The function Ev(λ, ε) = ξ(y0 )W(y0 )

(49)

is independent of y0 . It is analytic in λ and in ε , ε. Moreover, one has Ev(1, ε) = 2( νε ) and ∂λ Ev(1, 0) = 2 function is called the Evans function of Eq. (10). 1 ν

1 ν

Using the expressions of

d dy (U (−εy, ε))

and

d dy u + ,

0(1 + ν1 ). This

we have

 εW(y, ε) = u + (y, ε)[−ελU (−εy, ε) + εW (−εy, ε)] − U (−εy, ε) −ελu + (y, ε) + =

1− ν1

 ε  ξ(y)u + (y, ε)W (−εy, ε) − U (−εy, ε)v+ (y, ε) . ξ(y)

 ε v+ (y, ε) ξ(y)

Hence we have the following constant function to study, which depends only on λ, ε: Ev(λ, ε) = ξ(y)W(y) = [ξ(y)u + (y, ε)V (−εy, ε) − ξ(y)v+ (y, ε)U (−εy, ε)]. We shall use the equalities, valid for all y0 (and t0 ) such that both solutions are defined (which means y0 ∈ [− 4εν3 R , − 2εν1 R ])   t0 Ev(λ, ε) = ξ(y0 )W(y0 ) = (ξ W) − . ε We begin with the 1

Lemma 5. The Evans function has an analytic expansion in λ, whose coefficients depend analytically on ε and ε ν . For the precise study of the different terms of Ev(λ, ε), we introduce   ε t ξ =ξ − , ζ = ν, ζ0 = νt = ζ (t, 0), for t ≥ t0 > 0. ε ξ 1

We check that the function Ev(λ, ε) is analytic in λ and has an analytic expansion in ε ν and ε thanks to the equality [ν] X

1

p=0

ξ ν+1− p

+

1 1 − ξ ν−[ν] 1 + ν−[ν] = ν+1 1−ξ ξ (1 − ξ ) ξ (1 − ξ )

1615

O. Lafitte / Physica D 237 (2008) 1602–1639 1

which implies that the relation between t and ζ is analytic in ε and ε ν . Assume from now on that λ ≥ 1 ν

− ν1

1 2

and ν > 2 and replace

ξ(y) by ε ζ . Using Lemma 5, there exist two functions B0 (ε) and C0 (λ, ε) such that Ev(λ, ε) = Ev(1, ε) + B0 (ε)(λ − 1) + C0 (λ, ε)(λ − 1)2 .

(50)

First check that, for λ = 1, U+ = 1 and V+ = 0. Hence g(t, 1, ε) = 1, which implies G(t, 1, ε) = e

−t

 ν − 1 ν

ε

ξ

−1

=e

−t

 1 ζ ν ν

(51)

from which one deduces Ev(1, ε) = 2et G(t, 1, ε)ξ = 2

ε 1 ν

. ν From (50), the unique root λ(ε) of Ev(λ, ε) in the neighborhood of λ = 1 satisfies λ(ε) − 1 = −

(52)

Ev(1, ε) . B0 (ε) + C0 (λ(ε), ε)(λ(ε) − 1)

(53)

This equality rewrites λ(ε) − 1 +

Ev(1, ε) C0 (λ(ε), ε)(Ev(1, ε))2 + =0 B0 (ε) B0 (ε)(B0 (ε) + C0 (λ(ε), ε)(λ(ε) − 1))2

hence the approximation C0 (1, 0) Ev(1, ε) + (Ev(1, ε))2 + o((Ev(1, ε))2 ). B0 (ε) (B0 (0))3

1 − λ(ε) =

The aim of what follows is to deduce C0 (1, 0) and B0 (ε). We rewrite the Evans function as   Ev(λ, ε) = (λ − 1)et (F + G)(t, λ, ε) + 2et G(t, λ, ε) [ξ(y) + (1 − λ)(1 − ξ )ζ A(ζ, ε)] − et (F + G)(t, λ, ε)(λ − 1 + (1 − λ)(1 − ξ(y))ζ B(ζ, ε)).

(54)

Considering the limit in (54) for ε = 0, we obtain Ev(λ, 0) = (λ − 1)et (F0 + G 0 )(t, λ)[−1 + (1 − λ)νt A(νt, 0) + νt B(νt, 0)]. As this quantity is independent of t, we consider the limit when t → 0, hence, using (46), we deduce that Ev(λ, 0) = −(λ − 1)2−

λ+1 2ν

B1 (λ).

(55)

Remark that this implies the identity (λ − 1)et (F0 + G 0 )(t, λ)[−1 + (1 − λ)νt A(νt, 0) + νt B(νt, 0)] = −2−

λ+1 2ν

B1 (λ)(λ − 1)

which rewrites et (F0 + G 0 )(t, λ)[−1 + (1 − λ)νt A(νt, 0) + νt B(νt, 0)] = −2−

λ+1 2ν

B1 (λ).

Moreover, from Ev(λ, 0) = B0 (0)(λ − 1) + C0 (λ, 0)(λ − 1)2 , we obtain the identity B0 (0) + C0 (λ, 0)(λ − 1) = −2−

λ+1 2ν

B1 (λ) = −2−

λ+1 2ν

0(1 + ν1 ) 0(1 +

λ−1 2ν )

"

2

hence the relations − ν1

B0 (0) = −2



 1 0 1+ , ν

C0 (λ, 0)(λ − 1) = B0 (0)

1−λ 2ν

0(1 +

λ−1 2ν )

# −1 .

Hence one deduces the value of C0 (1, 0) using only the usual Gamma function and its derivative ψ(a)0(a), namely 1−λ ln 2 + o(λ − 1), 2ν     λ−1 λ−1 0 1+ = 0(1) 1 + ψ(1) + o(λ − 1) 2ν 2ν 2

1−λ 2ν

=1+

(56)

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O. Lafitte / Physica D 237 (2008) 1602–1639

hence (λ − 1)

1−λ C0 (λ, 0) = (ln 2 + ψ(1)) + o(λ − 1) B0 (0) 2ν

which leads to B0 (0) (ln 2 + ψ(1)). 2ν For the calculus of B0 (ε), one identifies the terms of order 0 and 1. Using (54), we rewrite C0 (1, 0) = −

2et G(t, λ, ε) − 2et G(t, 1, ε) λ−1  t  − (1 − ξ(y)) e (F + G)(t, λ, ε)(1 − ζ B(ζ, ε)) + 2et G(t, λ, ε)ζ A(ζ, ε)

B0 (ε) + C0 (λ, ε)(λ − 1) = ξ(y)

+ (1 − λ)(1 − ξ(y))ζ A(ζ, ε)et (F + G)(t, λ, ε) hence with G(t, λ, ε)et = ( ζν )

λ+1 2ν

g(t, λ, ε), g(t, 1, ε) = 1 and g − 1 = (1 − λ) K˜ ελ (g), one checks that # "   λ−1  ε  1 ( ζ ) λ−1 2ν − 1 ζ 2ν ˜ λ 2et G(t, λ, ε) − 2et G(t, 1, ε) ν ν ξ(y) =2 − K ε (g) λ−1 ν λ−1 ν

1 1 ln( ζν ) − K˜ ε1 (1)]. Hence we get the identity whose limit when λ → 1 is 2( νε ) ν [ 2ν  ε 1  1 ζ  1 1 ν 1 ˜ B0 (ε) = 2 ln − K ε (1)(t) − (1 − ε ν ζ − ν )et (F + G)(t, 1, ε)(1 − ζ B1 (ζ, ε)) ν 2ν ν 1

1

− 2et G(t, 1, ε)ζ A1 (ζ, ε)(1 − ε ν ζ − ν )

(57)

and the right-hand side is independent of t. In this equality, the subscript 1 on A and B denotes the calculus for λ = 1. We write  ε 1  1 ζ  ν 1 ˜ ln − K ε (1)(t) = −(1 − ξ ) [R1 (t, ε) + R2 (t, ε)] (58) B0 (ε) − 2 ν 2ν ν where     R1 (t, ε) = (1 − ζ B1 (ζ, ε)) et (F + G)(t, 1, ε) − et (F0 + G 0 )(t, 1) + 2et G(t, 1, ε) − 2et G 0 (t, 1) ζ A1 (ζ, ε), R2 (t, ε) = et (F0 + G 0 )(t, 1)(1 − ζ B1 (ζ, ε)) + 2et G 0 (t, 1)ζ A1 (ζ, ε). 1

In these two terms, we keep only the terms of order ε ν and of order 0. For this purpose, introduce 1 1 ln t − 2 K˜ 01 (1)(t) − t − ν B0 (0). ν One checks the identity

R0 (t) =

B0 (ε) −

ε 1 ν

ν

    1 ε 1  1 ζ  1  ε  ν1 ζ ε ν ν 1 ˜ R0 (t) = B0 (ε) − 2 ln − K ε (1)(t) + ln + B0 (0). ν 2ν ν ν ν νt νt

Considering the limit ε → 0 in (58), using R1 (t, 0) = 0, we obtain B0 (0) = −R2 (t, 0). Using (58) and (59), we deduce  ε 1  ε 1  ν ν B0 (ε) − B0 (0) − R0 (t) = −(1 − ξ )R1 (t, ε) − (1 − ξ )R2 (t, ε) + 1 − R2 (t, 0) ν νt    ε 1   ε 1  ν ν = −(1 − ξ )R1 (t, ε) + ξ − R2 (t, ε) + 1 − [R2 (t, 0) − R2 (t, ε)] . νt νt We now use the Lemma 6. Denote by ζ0 (t) = νt. The following relations are true 1

et (F + G)(t, 1, ε) − et (F0 + G 0 )(t, 1) = −

2 εν + O(ε ν ) ν−1

(59)

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O. Lafitte / Physica D 237 (2008) 1602–1639 1

2et G(t, 1, ε) − 2et G 0 (t, 1) = −2 ζ (t, ε) − ζ0 (t) = −ξ

ν ζ. ν−1

2 εν + O(ε ν ) ν−1

 2 Using Lemma 6, we obtain that the term ξ R1 (t, ε) is of order ε ν and the term ξ −  
1 is left with the two terms 1 − νtε ν [R2 (t, 0) − R2 (t, ε)] and R1 (t, ε). We have

ε νt


1 ν



2

R2 (t, ε) is also of order ε ν , hence one

1

lim R1 (t, ε) = −

t→0

εν ν−1

because ζ A1 (ζ, ε) and ζ B1 (ζ, ε) go to zero when t → 0 according to A0 (ξ ) = A1 (ζ ) =

∞ X

a j (0)ζ j ,

B 0 (ζ ) =

∞ X

j=1

j=1

∞ X

∞ X

a 0j (0)ζ j ,

B 1 (ζ ) =

j=1

b j (0)ζ j b0j (0)ζ j ,

j=1

which go to zero when t goes to zero. This is uniform because of the estimates |ζ (a j (ξ ) − a j (0) − ξ a 0j (0))| ≤ ζ ξ 2 A R j under the hypothesis ν > 2. One may thus replace in the approximation of B0 (ε) the expansions of A and of B by the two first terms in a j and b j , hence the behavior when t → 0. Hence one gets   1 ε 1 1 2 εν 1 ν + ln t − 2 K˜ 01 (1)(t) − t − ν B0 (0) + O(ε ν ). lim t→0 ν ν−1 ν

B0 (ε) = B0 (0) +

(60)

These results are summarized in 1

Proposition 4. Introduce the function R0 (t) = ν1 ln t − 2 K˜ 01 (1)(t) − B0 (0)t − ν , where K˜ 01 (1) has been introduced in (25) and B0 (0) is given by   Z +∞ 1 1 1 B0 (0) = −2 . s ν e−2s ds = −2− ν 0 1 + ν 0 We have, B0 (ε)C0 (1, 0) being introduced in (50): 1

ε 1 εν ν +2 lim R0 (t), B0 (ε) = B0 (0) + t→0 ν−1 ν and C0 (1, 0) = −

B0 (0) [ln 2 + ψ(1)] 2ν

where ψ denotes the Logarithmic derivative of the Gamma function. Note that −ψ(1) is the Euler constant. Let us note that (53) leads to the following expression of λ(ε): 1 − λ(ε) = 2



2ε ν

1 ν

1 0(1 +

1 ν)

B0 (0)
1 B0 (ε) + 2 νε ν

C0 (1,0) B0 (0)

+ o(εα ).

Note also that the limit of R0 (t) exists thanks to the equality 1

R 0 (1)(s) = −1 − B0 (0)s − ν + M(s),   1 R +∞ 1 1 where M(s) = 2 0 e−2y (y + s) ν − y ν s − ν dy, hence R0 (t) = −

+∞

Z 1

ds R0 (1)(s) − νs

1

Z t

M(s) ds. νs

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1 R +∞ 1 −1+ ν1 −2(y−s) The equality is obtained through R 0 (1)(s) = s − ν s e dy, an integration by parts leads to R 0 (1)(s) = −1 + νy R 1 R +∞ 1 1 1 +∞ 2s − ν 0 (y + s) ν e−2y dy, and we use B0 (0) = − 0 y ν e−2y dy. Note that for s small we see that M(s) is equivalent to s 1− ν hence the integral of s −1 M(s) is defined in the neighborhood of s = 0.

4. Semiclassical limit of the growth rate Eq. (10) in which only λ and ε appear leads to a solution λ which depends only on ε. In the first part of the present paper, we dealt with a λ which goes to 1 as ε goes to zero. In this section, we will concentrate on solutions of Eq. (10) which verify λ ' ε as ε → +∞. This is called (following Helffer and many authors) the semiclassical limit. It corresponds to values of γ which stays bounded as k L 0 → +∞. Even though λ(ε) depends only on ε, it is easier to consider the growth rate γ and consider Eq. (9), and thus revert to the variable x = L 0 y. In this section, we generalize a result obtained by Helffer et al. [16] for the Rayleigh equation (10) under the assumption (1.2) of [16]: ∃α0 > 0,

∀x,

ρ0 (x) ≥ ρl .

This generalization takes into account density profiles similar to the profile studied in this paper. ρ 0 (x) The assumptions needed in this generalization are the following on k0 (x) = ρ00 (x) : k0 and k00 bounded,

k0 regular,

(62)

k0 admits an unique nondegenerate maximum. Recall that we introduced L eff such that sup k0 (x) =

1 L eff

ν

ν = L −1 0 (ν+1)ν+1 . 1

Lemma 7. Assume that (62) are fulfilled. Assume moreover that k0 (x)ρ0 (x)− 2 is bounded. There exists k∗ > 0 such that, for all k ≥ k∗ , there exists a real γ (k, ε) and a nonzero solution u(x)eikz+γ (k,ε)t of the Rayleigh equation (9) such that Λ < γ (k, ε) < Λ. 2 We have the following behavior of the eigenmode



12 21 0

ρ u + ρ u + kuk + ku 0 k + ku 00 k < +∞.

0 0 As the result of this Lemma is important for the nonlinear analysis, we rewrite an idea of the proof, based on Remark 8.1 of [16]. 1

We denote by L 2 1 the space of functions u such that ρ02 u ∈ L 2 (R). ρ02

Finding γ is equivalent to finding 0 as an eigenvalue (in L 2 (R) ) of   g 1 −1 d d −1 − 2 ρ0 2 ρ0 ρ0 2 + 1 − 2 k0 (x). dx dx k γ 2

g d −2 W (x) where W (x) = This operator rewrites − k12 dx 0 0 2 + 1 − γ 2 k 0 (x) + k 0 bounded (or equivalently when k0 is bounded).   − 21 d dx

We introduce the operator Q = − k12 ρ0

− 12

d ρ0 dx ρ0

1 0 2 k0 (x)

+ 14 (k0 (x))2 , which is bounded when

ρ000 ρ0

is

+ 1, which is coercive, thanks to the Poincare estimates, for k large

enough. The eigenvalue problem rewrites   1 1 γ2 ∈ σ p Q − 2 k0 Q − 2 . g Under the hypothesis that k0 has a nondegenerate maximum L eff , one deduces that for k large enough one has at least a value of g γ (k, ε) such that L 0 < γ (k,ε) 2 < 4L 0 using usual results on semiclassical Schrodinger operators whose potential has a well. 1

1

We thus constructed v ∈ L 2 (R) and γ (k, ε) such that v is the eigenvector of Q − 2 k0 Q − 2 associated with the eigenvalue −1

1

To v is associated a solution of (9) which is u = ρ0 2 Q − 2 v, such that u 0 ∈ L 2 1 , u ∈ L 2 1 . Remembering that u solves ρ02

−u 00 + k 2 u − k0 (x)u 0 −

gk 2 u = 0, γ (k, ε)2

ρ02

γ (k,ε)2 . g

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O. Lafitte / Physica D 237 (2008) 1602–1639

multiplying this equation by u and integrating, one gets (kuk1 defined in this equality is a weighted norm in H 1 (R))   Z Z h i 1 1 gk 2 − 12 2 2 0 2 2 2 2 0 k u + (u ) dx = k0 (x)ρ0 u ρ u + ρ0 u dx kuk1 = γ (k, ε)2 0 hence, using the hypothesis − 23

ρ00 ρ0

≤ M,

one obtains  kuk1 ≤ M



1

1 gk 2

ρ 2 u + ρ 2 u 0 , γ (k, ε)2 0 0

hence a control on the H 1 2norm of u. gk 0 2 00 2 r Moreover, as u 00 = γ (k,ε) 2 k 0 (x)u + k 0 (x)u − k u, one deduces that u ∈ L , and we have iteratively the control of u in H , r integer (r ≤ rmax , according to the number of derivatives of k0 that we consider). 5. The nonlinear analysis We show in this section that the result of Guo and Hwang [13] can be extended in our set-up, even if the density profile ρ0 (x) does ρ 0 (x)

not satisfy the coercivity assumption (3) of [13]. The quantity k0 (x) = ρ00 (x) plays a crucial role. It has a physical interpretation, being the inverse of a length: it is called the inverse of the density gradient scalelength. We need the assumptions (H)

− 21

k0 (x) bounded, k0 (x)ρ0

bounded.

Note that k0 bounded is fulfilled in the case studied by Guo and Hwang (where ρ0 is bounded below), and in the case of the striation model (studied by Poncet [26]) but is not automatically fulfilled by a profile h that ρ0 (x) i → 0 when x → −∞. However, for  such ν

x 1 − ξ Lx0 , hence it is bounded and belongs to the particular case of the ablation front profile, we have k0 (x) = L −1 0 ξ L0 i  1 νν 0, L −1 0 (ν+1)ν+1 . The inequality ν > 2 imply that (H) is fulfilled for the ablation model profile studied in Sections 1–3. Before starting the proof of Theorem 2, which is rather technical, let us describe our procedure. Firstly, we prove that the linear system reduces to an elliptic equation on the pressure, from which we obtain a general solution. We identify a normal mode solution of this system using the first part of the paper (Theorem 1 and Proposition 1). Once this normal mode solution U is constructed, with suitable assumptions on the growth rate, one introduces a perturbation solution of the nonlinear system, whose initial condition is δU |t=0 and an approximate solution V N of the nonlinear system which admits an expansion in δ N up to the order N with the same initial condition. Using the Duhamel principle for the construction of the jth term of the expansion in δ of V N , one obtains a control of all the terms of V N . 1 1 −1 −1 The natural energy inequalities are on the quantities ρ02 u j , ρ02 v j , ρ0 2 p j , ρ0 2 ρ j . We verify that the properties of ρ0 (x) imply that we can deduce inequalities on u j , v j , ρ0−1 p j and T j . Note that we have, as a consequence of the method that we chose, a control in t s eΛt of the H s norm of all solutions of the homogeneous linear system (with any initial condition U (x, z, 0)), and a control by e jγ (k,ε)t (with no additional power in t) of the H s norm of the jth term of the expansion.

Remark 1. When an initial value mixes eigenmodes, the H s norm of the solution behaves as t s eΛt . If one starts from a pure eigenmode with Λ2 < γ (k, ε) < Λ the exponential behavior comes at most from the growth of the pure eigenmode. 5.1. Obtention of a solution of the linear system Consider the system  0   ∂ t σ + ρ 0 v1 = f 0  ρ0 ∂t v1 + ∂x p = σ g + f 1 ρ0 ∂t v2 + ∂z p = f 2    ∂x v1 + ∂z v2 = 0. 1

−1

We know that the relevant quantities are ρ02 v1,2 , ρ0 2 σ , and we denote these three quantities by X, Y, τ . To have the same behavior −1

−1

when ρ0 → 0, consider ψ such that, once ψ is obtained, we revert to v1 and v2 using v1 = −∂z (ρ0 2 ψ), v2 = ∂x (ρ0 2 ψ). Introduce −1

b = ρ0 2 [∂z (ρ0 v1 ) − ∂x (ρ0 v2 )].

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O. Lafitte / Physica D 237 (2008) 1602–1639

The system on v1 , v2 , σ, p implies the two equations  −1  ∂t b = g∂z τ + ρ0 2 (∂z f 1 − ∂x f 2 ) ∂ τ + k (x)X = ρ − 21 f . t

0

0

(64)

0

We obtain ψ from b through the elliptic equation   1 0 1 1ψ − k0 + k02 ψ = −b. 2 4

(65)

We then revert to X through the equality X = −∂z ψ. Finally, the pressure p is obtained through the elliptic equation   1 1 1 −1 −1 ρ0 ∂x (ρ0−1 ∂x p) + ∂z22 p = ρ02 ρ02 ∂x (ρ0 2 τ )g + ρ02 ∂x (ρ0−1 f 1 ) + ρ0 2 f 2 which rewrites 1

1p − k0 ∂x p = ρ02



  1 −1 ∂x τ g − k0 τ g + ρ0 2 (div fE − k0 f 1 ) . 2

(66)

Hence we solve the system  − 21  f0 ∂ τ = k ∂ ψ(b) + ρ t z 0  0    − 21  ∂t b = g∂z τ + ρ0 (∂z f 1 − ∂x f 2 )   τ (0) = τ (x, z), b(0) = b (x, z) 0  0 1 2 1 0   k + k ψ = −b 1ψ −   2 0 4 0     1 1   1p − k0 ∂x p = ρ 2 ∂x τ g − 1 k0 τ g + ρ − 2 (div fE − k0 f 1 ) 0 0 2

(67)

which has the same properties as system (13) of [13], the Poincare estimate being still valid. From b and τ , one reverts to X and Y , hence a solution of the system. Moreover, one checks that (X, Y ) ∈ L 2 (R) (according to the energy equality), hence X ∈ H 1 (R) under the assumption that k0 is bounded. Proposition 5. Under the hypotheses (H), and under the hypothesis h j ∈ L 2 , j = 0, 1, 2, the functions u 1 , v1 , T1 , p1 solution of  ∂t T1 − k0 u 1 = h 0    ρ0 ∂t u 1 + ∂x p1 + ρ0 gT1 = h 1 ρ0 ∂t v1 + ∂z p1 = h 2    ∂ x u 1 + ∂ z v1 = 0 satisfy u 1 (t), v1 (t), T1 (t) ∈ L 2 when it is true for t = 0. Moreover, one has ρ0−1 p1 (t) ∈ L 2 (R2 ). 1

1

−1

1

Proof. The proof of this result follows two steps: first of all the assumption k0 bounded implies that ρ02 u 1 , ρ02 v1 , ρ0 2 ∇ p1 , ρ02 T1 belong to L 2 . We thus multiply the equality ∂t uE1 + ρ0−1 ∇ p1 + T1 gE = hE by ∇(ρ0−1 p). We get, integrating in x, y: Z Z 2 E 1 (∇q1 ) + k0 (x)q1 ∇q1 .Ee1 + T1 gE∇q1 = h∇q from which one deduces

− 1 21

E ρ k∇q1 k ≤ max(k0 ρ0 2 ) q

0 1 + gkT1 k∞ + khk. 1

−1

It is then enough to use the Poincare estimate between ρ02 q1 and ρ0 2 ∇ p1 to obtain the estimate on ∇q1 , from which one deduces the estimate on q1 .  Finally, from the estimate on q1 and on ∇q1 , multiplying the equation on the velocity by uE1 and integrating, we get the Gronwall type inequality d E + gkT1 k∞ kE u 1 k ≤ Ckq1 k H 1 + khk dt hence a control on kE u 1 k on [0, T ] for all t as soon as it is true for t = 0.

O. Lafitte / Physica D 237 (2008) 1602–1639

1621

5.2. Weakly nonlinear approximation of order N P PN In system (8) of the Introduction only quadratic terms appear. We consider T ∗,N = 1 + Np=1 δ p T p , uE∗,N = E p, p=1 u PN ∗,N p Q = p=1 δ Q p . This is generally called a weakly nonlinear approximation of order N of the solution. It is generally a formal expansion of a solution, and in the present case, we prove that the limit for N → +∞ exists for small times. We check that there exists G j , HE j such that  ∗,N      Gj Gj T N 2N X X (Emod) uE∗,N  = δ j  HE j  + δ j  HE j  . j=1 j=N +1 divE uj 0 Q ∗,N An approximate incompressible solution of order N is thus characterized by the 4N equations divE u j = 0, 1 ≤ j ≤ N .

HE j = 0,

G j = 0,

In particular, the system HE N = 0,

G N = 0,

divE uN = 0

rewrites  ∂t TN − u N k0 (x) = S N    ∂t u N + ∂x Q N + Q N k0 (x) + gTN = R1,N ∂t v N + ∂z Q N = R2,N    ∂ x u N + ∂z v N = 0

(68)

with SN =

N −1 X

u j TN − j k0 (x) − u j ∂x TN − j − v j ∂z TN − j

j=2

R1,N = −

N −1 X

u j ∂x u N − j + v j ∂z u N − j + T j ∂x Q N − j + T j Q N − j k0 (x)

j=2

R2,N = −

N −1 X

u j ∂ x v N − j + v j ∂z v N − j + T j ∂z Q N − j .

j=2

We shall consider (68) from now on. 5.3. The energy equalities One of the main tools that we have to use is the divergence free condition, in order to get rid of the pressure p or the reduced pressure Q when obtaining the energy inequality. Recall that system (68) rewrites  ∂t TN − u N k0 (x) = S N ρ ∂ uE + ∇(ρ0 Q N ) + gρ0 TN eE1 = ρ0 RE N  0 t N divE uN = 0 where RE N = (R1,N , R2,N ). E N = ρ0 ∂t RE N − gρ0 S N eE1 . Applying the operator ∂t ∂ nn to equation on the velocity and using the equation on the specific Denote G x volume, one obtains E N ). ∂xnn (ρ0 ∂t22 uE N ) + ∇∂t ∂xnn (ρ0 Q N ) + g∂xnn (ρ0 k0 u N ) = ∂xnn (G

(69)

One deduces the Lemma 8. For all n, one has the estimate

!

X

12 2 n

12 p

1

ρ ∂ 2 ∂ n uE N ≤ Cn

ρ ∂ p u N + ρ 2 ∂ pp G E

0 t x

0 x

0 x N . p≤n

Moreover, as the coefficients of the system depend only on x, this inequality is also true with the same constants when ∂xnn is replaced q by ∂xnn ∂z q for all q ≥ 0.

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O. Lafitte / Physica D 237 (2008) 1602–1639

Proof. One notices that (69) writes n−1 n−1 X X   p (n− p+1) p p (n− p) p 2 EN − C n ρ0 ∂x p u N . Cn ρ0 ∂x p ∂t 2 uE N − gE ρ0 ∂t22 ∂xnn uE N + ∇ ∂t ∂xnn (ρ0 Q) + gEk0 (x)ρ0 ∂xnn u N = G p=0

p=0

Multiplying by ∂t22 ∂xnn uE N and integrating, using the recurrence hypothesis that



X X 1



12 2 p

21 m N

ρ ∂ 2 ∂ p uE N ≤ C p

ρ 2 ∂ 22 ∂ mm uE N + 2g 2 Λ2

ρ ∂ m u N +

G n−1

0 t x



0 t x

0 x m≤ p−1

m≤ p−1

as well as the inequalities ( p)

∀x ∈ R, |k0 (x)g| ≤ Λ2 ,

|ρ0−1 ρ0 | ≤ Λ p

(which are true as soon as k0 is a C ∞ function whose derivatives are bounded, because ρ00 = k0 ρ0 ) one obtains the inequality



X 1

12 2 n

ρ 2 ∂ 22 ∂ mm uE N .

ρ ∂ 2 ∂ n uE N ≤ Cn

0 t x

0 t x

m≤n

Lemma 8 is proven.



Note that the system for the leading term of the perturbation is system (68) with a null source term. Owing to this remark, we shall treat the general case and apply the equality to the particular cases. Multiplying (69) by ∂t ∂xnn uE N and integrating, using the divergence free relation, one obtains Z Z Z ∂xnn (ρ0 ∂t22 uE N ) · ∂xnn ∂t uE N dxdz + g∂xnn (ρ0 k0 u N + ρ0 S1N )Ee1 · ∂t ∂xnn uE N dxdz = ∂xnn (ρ0 ∂t SE N ) · ∂t ∂xnn uE N dxdz. In this equality, we can consider (for Sobolev inequalities) the term containing the largest number of derivatives of uE N . We obtain, denoting by REnN = ∂xnn (ρ0 ∂t22 uE N ) − ρ0 ∂xnn ∂t22 uE N BnN = ∂xnn (ρ0 k0 u N ) − ρ0 k0 ∂xnn u N the equality Z Z Z Z ρ0 ∂xnn ∂t22 uE N · ∂xnn uE N dxdz + gρ0 k0 ∂xnn u N · ∂t ∂xnn u N dxdz + REnN · ∂xnn ∂t uE N dxdz + g BnN · ∂t ∂xnn u N dxdz Z Z n n E = ∂x n (ρ0 ∂t R N ) · ∂t ∂x n uE N dxdz − g∂xnn (ρ0 S N )Ee1 · ∂t ∂xnn uE N dxdz. The terms REnN and BnN contain only derivatives of order less than n − 1, hence it will appear as a source term in the application of the Duhamel principle later on. The two first terms of the previous equality are the exact derivatives in time of Z  Z 1 N n 2 n 2 E n (t) = ρ0 (∂x n ∂t uE N ) dxdz + gρ0 k0 (∂x n u N ) dxdz . 2 The energy equality is thus Z t N N E n (t) = E n (0) + gnN (s)ds 0

where Z ∂xnn (ρ0 ∂t RE N ) · ∂t ∂xnn uE N dxdz − g∂xnn (ρ0 S N )Ee1 · ∂t ∂xnn uE N dxdz Z  Z =− REnN · ∂xnn ∂t uE N dxdz + g BnN · ∂t ∂xnn u N dxdz .

gnN (t) =

Z

Note that this source term satisfies

21 n

N N

|gn (t)| ≤ ρ0 ∂x n ∂t uE N

K n (t) L2

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O. Lafitte / Physica D 237 (2008) 1602–1639

where one has K nN (t)



 1



− 21 N − 21 n

−2 N

n N N



E E ≤ ρ0 Rn + ρ0 ∂x n (ρ0 ∂t S ) + |g| ρ0 Bn + ∂x n (ρ0 S1 ) .

(71)

We are ready to prove the Duhamel inequality associated with this problem, using gk0 (x) ≤ Λ2 . 5.4. The Duhamel principle Two versions of the behavior of the evolution semigroup will be studied. The first one corresponds to the general case using an exponential bound for the source term of the Rayleigh equation (7), see the Appendix for the statement and the proof of this result. A general formulation of the Duhamel principle (taking into account nonzero initial values), comes from Proposition 7, which allows a mixing of modes and a weak nonlinear result. See Cherfils, Garnier, Holstein [11] for more details on a weak nonlinear analysis mixing modes. We are now ready to state and prove the second version of the behavior of the evolution semigroup: Proposition 6. The solution of Z    ρ00 1 d 2 2 ρ0 (∂t uE N ) − g ρ0 (u N ) dxdz = g(t, x, ∂t uE N ) 2 dt ρ0 with initial conditions ∂t uE N (0), uE N (0), with the assumption that



12

|g(t, x, ∂t uE N )| ≤ K (t) ρ0 ∂t uEn

L2

where K is a positive increasing function for t ≥ 0 satisfies the inequalities

1   Z tq

12 2 Λ −Λs ds e 2 t

ρ uE N ≤ C1 + K (s)e

0

0

 2 Z tq

12

− Λ s

ρ ∂t uE N ≤ C1 + K (s)e ds eΛt

0

0

where C1 depends on the initial data. Proof. We deduce from the energy equality the following inequality: Z Z Z 2 2 ρ0 (x)(∂t uE N ) dxdz − g k0 (x)ρ0 (x)u N dxdz ≤ C0,+ + 2

t 0



12

K (s) ρ0 ∂t uE N

(s)ds L2

R ρ0 (x)(∂t uE N )2 (0)dxdz − g k0 (x)ρ0 (x)u 2N (0)dxdz and C0,+ = max(C0 , 0). Now consider the function

Z t

21

12



u(t) = ρ0 uE N (0) +

ρ0 ∂t uE N (s)ds.

where C0 =

R

0



21

0 2

We notice that u (t) = ρ0 ∂t uE N

(t) ≥ 0. Recall that gk0 (x) ≤ Λ . The inequality implies Z t 2 0 2 2 u (t) ≤ Λ u(t) + C0,+ + 2 K (s)u 0 (s)ds 0

≤ Λ2 u(t)2 + C0,+ + 2K (t)u(t)   K (t) 2 K (t)2 ≤ Λu + + C0,+ − . Λ Λ2 1

Use now the inequality (a 2 + b2 + c2 ) 2 ≤ a + b + c for positive numbers a, b, c to obtain p p u 0 (t) ≤ Λu(t) + C0,+ + 2K (t)u(t). Introducing v(t) = u(t)e−Λt which satisfies v(t) ≥ u(0)e−Λt , we deduce q p −Λt 0 v (t) ≤ C0,+ e + 2K (t)e−Λt v(t).

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Three cases are to be studied: • the case where u(0) > 0. We obtain, denoting by h(t) = q p 2hh 0 ≤ C0,+ e−Λt + 2K (t)e−Λt h(t)

√

v(t)

hence 2h 0 ≤



C0,+ u(0)

1 2

e−

Λt 2

q +

2K (t)e−Λt .

We deduce the inequality 1

Z tq Λt 1 K (s)e−Λs ds. (1 − e− 2 ) + √ 2 0 which imply that there exist A and B such that " Z t q 2 # 2 Λt 2 Λt u(t) ≤ A e + B e K (s)e−Λs ds . h(t) ≤ h(0) + Λ−1



C0,+ u(0)

2

0

• the case where u(0) = u 0 (0) = 0. As C0,+ = 0, we have the inequality p u 0 (t) ≤ Λu(t) + 2K (t)u(t) from which one deduces, with the same notations as above, that r 1 0 h (t) ≤ K (t)e−Λt 2 hence with h(0) = 0 one obtains Z tr 1 K (s)e−Λs ds. h(t) ≤ 2 0 • the case where u(0) = 0 and u 0 (0) > 0. We obtain q p v 0 (t) ≤ C0,+ e−Λt + 2K (t)e−Λt v(t). p −Λt Introduce v(t) ˜ = v(t) − C0,+ 1−eΛ . We have s   p 1 − e−Λt 0 − Λ t v˜ (t) ≤ 2K (t)e v(t) ˜ + C0,+ Λ v ! p u u C0,+ ≤ t2K (t)e−Λt v(t) ˜ + Λ from which one deduces the inequality s sp p Z tq C0,+ C0,+ 2 v˜ + ≤2 + 2K (s)e−Λs ds. Λ Λ 0 The inequality  2 Z tq u(t) ≤ C1 + K (s)e−Λs ds eΛt 0

always holds. Finally using the relation



1

1

d

ρ 2 uE N ≤ ρ 2 ∂t uE N = u 0 (t)

0

dt 0 we get

12

ρ uE N ≤ u(t) − u(0).

0

These are the two estimates of Proposition 6.

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Of course, the proof is much simpler in the case we are interested in, that is ∂t uE N = 0, uE N = 0, where (using the notations of this paragraph, C0 = C0,+ = u(0) = u 0 (0) = 0), where one easily deduces Z t q q 1 2− 2 K (s)e−Λs ds.  u(t)e−Λt ≤ 0

5.5. H s estimates for a general solution of the linearized system 5.5.1. The H s inequalities for the solution of the homogeneous system We consider the system satisfied by the leading order term of the perturbation of the Euler system (which is system (8)), particular case of (68) for N = 1. We prove in this section the analogous of the Proposition 1 of [13], with a slightly better estimate which essentially shows that the relevant growth rate is, up to polynomial terms, Λ: Proposition 7. Let T1 (t), uE1 (t) be the solution of the modified linearized Euler system (8). There exists a constant Cs depending only on the characteristics of the system, that is on k0 and g, such that





 1

1



1

12

ρ T1 (t) + ρ 2 uE1 (t) ≤ Cs (1 + t)s exp(Λt) ρ 2 T1 (0) + ρ 2 uE1 (0)

s 0

s

0

s 0

s .

0 H

H

H

H

Note that in these inequalities (which are general) a power of t appears in the bound for the norm H s . This is the general case. Note that similar estimates were obtained independently by Poncet [26]. We introduce

Z t 1

2 n

12 n N

ρ0 ∂t ∂x n uE N (s) u n (t) = ρ0 ∂x n uE N (0) +

2 ds

2 L

0

L

and vnN (t) = u nN (t)e−Λt . An important feature of this result takes into consideration an initial condition which is not an eigenmode of the Rayleigh equation, and which is a combination of different eigenmodes. As we shall see in what follows, the interaction of these different eigenmodes leads to a linear growth of the form (1 + t)s eΛt for the H s norm of the solution. Proof. We prove in a first stage the H s inequality result for the system satisfied by (T1 , u 1 , v1 , Q 1 ). We use the pressure p1 in the analysis. The system implies the equation ρ0 (x)∂t22 uE1 + ∇∂t p1 = ρ0 gEk00 u 1 . We apply the operator Dm, p to this equation. The energy inequality deduced from (70) and from inequality (71) is ((u 1n )0 )2 ≤ Λ2 (u 1n )2 + C0 + K n1 (t)u 1n (t) where we have the estimate



1

− 21 1 En + |g| · ρ − 2 Bn1 . ρ K n1 (t) ≤ R

0

0

1. Principal term

12

The inequation on ρ0 uE1

writes

 1  2

1 2 d

ρ 2 uE1 ≤ Λ2 ρ 2 uE1 + C0

0 dt 0 hence one obtains the inequality s



2

12 21

1

ρ uE1 ≤ ρ uE1 (0) cosh Λt + C0 + ρ 2 uE1 (0) sinh Λt ≤ D0 eΛt . 0

0 0

2 Λ 2. Derivative of the principal term In the inequality obtained for D1, p uE1 , the source term g1 is bounded by M D0 eΛt kD1, p ∂t uE1 k because it contains only derivatives of order n − 1 = 0. We thus have the inequality ((u 11 )0 )2 ≤ Λ2 (u 11 )2 + C0 + 2M D0 eΛt u 11 (t) from which one deduces     M D0 Λt 2 M D0 2 2Λt 1 0 2 1 ((u 1 ) (t)) ≤ Λu 1 + e + C0 − e Λ Λ

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hence (u 11 )0 (t) ≤ Λu 11 (t) +

M D0 Λt p e + C0 Λ

that is d 1 −Λt M D0 p (u 1 e + C0 e−Λt )≤ dt Λ from which one deduces p  u 11 (t) ≤ Λ−1 C0 + M D0 t + Λu 11 (0) eΛt . 3. Greater order term We prove thus by recurrence that there exists An and Bn such that u 1n (t) ≤ (An + Bn t)n eΛt , according to the inequality  p d  1 (An−1 + t Bn−1 )n−1 u n (t)e−Λt ≤ Cn e−Λt + . dt 2Λ One deduces the same inequality for dtd u 1n (t). 4. In the derivative Dn, p , the only term which matters for the order of the power of t is n, hence one deduces that



 X 

12

1

ρ Dn, p uE N + ρ 2 Dn, p ∂t uE N ≤ (Cs + t Ds )s eΛt ,

0

0

n+ p=s

Proposition 7 is proven. Note that this improvement does not change the behavior of the approximate solution we intend to construct, because for a normal mode solution ikz+γ (k,ε)t u(x, y, t) = u(x)e ˆ

where γ (k, ε) has been calculated and where u(x) ˆ is solution of the Rayleigh equation, one has the following equalities:



1

12

ρ Dm, p uE1 (t) = ρ 2 Dm, p uE1 (0) eγ (k,ε)t

0

0

Dm, p T1 (t) = kT1 (0)k eγ (k,ε)t



12

1

ρ Dm, p Q 1 (t) = ρ 2 Dm, p Q 1 (0) eγ (k,ε)t .

0

0

Remark that we also have the relations



Dm, p uE1 (t) = Dm, p uE1 (0) eγ (k,ε)t



Dm, p Q 1 (t) = Dm, p Q 1 (0) eγ (k,ε)t , according to the equality ik Q 1 (x, y, t) =

γ (k,ε) ik ∂x u 1 (x, z, t)

and Lemma 7.



5.6. Inequalities for the following terms of the expansion Λ 2

Recall that from Lemma 7, there exists a normal mode solution of the linearized system of the form u(x, ˆ k)eikz+γ (k,ε)t where < γ (k, ε) < Λ. With this normal mode solution one constructs an approximate solution of the nonlinear system, of the form T N (x, z, t) = 1 +

N X

δ j T j (x, z, t)

j=1

u N (x, z, t) =

N X

δ j u j (x, z, t)

j=1

v N (x, z, t) =

N X

δ j v j (x, z, t)

j=1

Q N (x, z, t) =

N X j=1

δ j Q j (x, z, t).

O. Lafitte / Physica D 237 (2008) 1602–1639

1627

There is an important lemma, which depends on Hypothesis (H): Lemma 9. The functions u j , v j , Q j , T j belong to L 2 . The proof of this Lemma is a consequence of Proposition 5, which will lead to the control of the source term of the linear system on TN , u N , v N , Q N . We shall use the estimates of Cordier, Grenier and Guo [8], and the method of Guo and Hwang [13] to give an H s estimate of N T , u N , v N , Q N and a L 2 estimate of T N − T0 − δT1 , u N − δu 1 ,v N − δv1 to obtain a lower bound on T N , u N , v N .

12 − 21

We prove in this section the H s estimate uE N in the weighted norm

ρ0 f . Using the assumption that k0 ρ0 is bounded, we deduce estimates in H s for uE N . The first result reads as p

Proposition 8. There exist constants C0 and A p , depending only on the characteristics of the system (namely g, k0 (x) and its derivatives) and on the H p norm of the initial data such that u Np (t) ≤ (C0 ) N (A p ) N −1 e N γ (k,ε)t . p

Remark 2. This estimate relies heavily, as in [9], on the quadratic structure of the nonlinearity, and that we give the precise estimate on the constant C j which appears in (13) of [9]. This estimate could not be obtained in the set-up of Guo and Hwang [13] because the nonlinearity was written using ρ uE.∇ uE, hence a cubic nonlinearity. A second comment is the following: the inequality 2γ (k, ε) > Λ allows us to forget the coefficient (1 + t)s in the H s estimate for a general solution of the linear system (obtained in Proposition 7). This is a consequence, as we shall see below, of the relation Z t 1 eΛt e(N γ (k,ε)−Λ)s ds ≤ e N γ (k,ε)t N γ (k, ε) − Λ 0   Rt to be compared with the relation eΛt 0 e(Λ−Λ)s ds ≤ teΛt . Case N = 2 Recall that we have the following system  ∂t T2 − k0 (x)u 2 = −u 1 ∂x T1 − v1 ∂ y T1 − T1 u 1 ρ0 (x)∂t uE2 + ∇(ρ0 Q 2 ) + gT2 = −ρ0 (x)[E u 1 .∇ uE1 ] − ρ0 T1 ∂x Q 1 − ρ00 Q 1 T1  divE u 2 = 0. We thus have the estimates



12

1

ρ ∂t S 2 + ρ 2 S 2 ≤ C 2 e2γ (k,ε)t . j j

0 j

0 This means that K 02 (t) ≤ D2 e2γ (k,ε)t , hence

 2 Z tp

12

)s (γ (k,ε)− Λ

ρ ∂t uE2 ≤ C0 + 2 2D e ds eΛt 2

0

0

hence the inequality



12

1

ρ ∂t uE2 + ρ 2 uE2 ≤ M0 e2γ (k,ε)t .

0

0 We need to derive the estimates for the terms T2 and Q 2 . For the term T2 , one has Z Z Z d 1 ρ0 T22 dxdz = k0 (x)ρ0 u 2 T2 + S12 ρ0 T2 dxdz dt 2 from which one deduces the inequality





1

1 1 d

ρ 2 T2 ≤ M ρ 2 u 2 + ρ 2 S 2 ≤ (M M0 + C 2 )e2γ (k,ε)t 1

0 0 1 dt 0 hence the estimate



2

12

1

ρ T2 (t) ≤ ρ 2 T2 (0) + C1 + M M0 (e2γ (k,ε)t − 1).

0

0

2γ (k, ε)

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As for the estimate on Q 2 , one deduces ∂x (ρ0−1 ∂x (ρ0 Q 2 )) + ∂z22 Q 2 + g∂x T2 = div SE2 which imply estimates on Q 2 . From now on, we introduce Z t 1 kρ02 ∂t uE N (., s)kds u N (t) = 0

1

h N (t) = (u N (t)e−Λt ) 2 . Case N ≥ 3 We start with the induction hypothesis that, for j ≤ N − 1, there exists C0 and A such that











21 12

1

1

1

1

ρ uE j + ρ ∂x uE j + ρ 2 ∂z uE j + ρ 2 ∂x T j + ρ 2 ∂z T j + ρ 2 T j ≤ A j−1 C j e jγ (k,ε)t 0

0

0

0

0

0 0

(72)

and that the derivative in time of all quantities is bounded by jγ (k, ε)A j−1 C0 e jγ (k,ε)t . j

Thus there exists M (independent of the number of terms which appear in the source term and which depends only on the coefficients of the system) such that the source term of (71) for n = 0 is bounded by: K 0N (t) ≤ M A N −2 C0N N 2 γ (k, ε)e N γ (k,ε)t .

(73)

Note that in this estimate the N 2 term comes, one from the number of the terms in the expansion one4 from the derivative in time which appears in the source term ∂t SE N . We thus obtain, using Z tq h N (t) ≤ 2K 0N (s)e−Λs ds

P N −1 j=0

A j B N − j and a second

0

the inequality h N (t) ≤

q

2M A N −2 C0N N 2 γ (k, ε)

Z tp

e(N γ (k,ε)−Λ)s ds

0

which yields h N (t)2 ≤ A N −1 C0N e(N γ (k,ε)−Λ)t The choice of A is thus induced by

8M N 2 γ (k, ε) . (N γ (k, ε) − Λ)2 A

8M N 2 γ (k,ε) (N γ (k,ε)−Λ)2 A

≤ 1 for all N (forgetting that we have to be more precise to obtain estimates not

only on uE N but also on TN ) hence the simplest choice is A = term (T1 , u 1 , v1 , Q 1 ). The final estimate is

12

ρ uE N ≤ C N A N −1 e N γ (k,ε)t . 0

0

8Mγ (k,ε) . γ (k,ε)− Λ 2

The value of C0 is thus given by the norm of the leading

We proved the assumption (72) at the rank N , hence the estimate (73) on the source term of Proposition 6. We use this result and the estimates for a normal mode solution (on which no powers of t appear for the norms of the derivatives). We obtain 1  Z t 1 2 N γ (k,ε)−Λ 1 −1 C 0,+ s 2 2 2 h N (t) ≤ h N (0) + Λ + N C N γ (k, ε) e ds u(0) 0 hence as N γ (k, ε) > Λ one gets  1 1 2 1 N γ (k,ε)−Λ N −1 C 0,+ t 2 h N (t) ≤ h N (0) + Λ + C N2 γ (k, ε) 2 e . u(0) N γ (k, ε) − Λ 4 Note also that if we consider a cubic model, the number of terms in the source term is N (N − 1), hence adding a derivative in time we get N 3 in the estimate. As we can see in the following lines, this gives a less efficient estimate.

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O. Lafitte / Physica D 237 (2008) 1602–1639

We deduce the inequality (using (a + b)2 ≤ 2(a 2 + b2 )) 2  C0,+ Λt N C N γ (k, ε)e N γ (k,ε)t . u(t) ≤ 2[h(0) + u(0)] + Λ−1 e +2 u(0) N γ (k, ε) − Λ 1

Remark. If the system has a cubic source term, at each stage of the construction one gets N 2 M N −1 C N as estimate, hence the convergence of the infinite series is not ensured by these estimates. 5.7. Estimates for the weakly nonlinear approximation In this paragraph, we derive estimates on the approximation (T N , uE N , Q N ). Throughout what follows we shall use the Moser estimates kD α ( f g)k L 2 ≤ C (k f k∞ kgks + kgk∞ k f ks )

(74)

kD α ( f g) − f D α gk L 2 ≤ C(kD f k∞ kgks−1 + kgk∞ k f ks )

(75)

and

and the Sobolev embedding k f k∞ ≤ Ck f ks for s > Proposition 9. Introduce Tδθ =

1 γ (k,ε)

d 2

and k∇ f k∞ ≤ Ck f ks for s >

d 2

+ 1. More precisely, we prove that

ln δCθ0 A . For all θ < 1 and for all t < Tδθ , we have

δ AC0 eγ (k,ε)t 1 − δ AC0 eγ (k,ε)t eγ (k,ε)t kT N − 1k L 2 ≥ kT1 (0)k L 2 δeγ (k,ε)t − AC02 C3 δ 2 1 − δ AC0 eγ (k,ε)t eγ (k,ε)t ku N k L 2 ≥ ku 1 (0)k L 2 δeγ (k,ε)t − AC02 C3 δ 2 1 − δ AC0 eγ (k,ε)t eγ (k,ε)t . kv N k L 2 ≥ kv1 (0)k L 2 δeγ (k,ε)t − AC02 C3 δ 2 1 − δ AC0 eγ (k,ε)t kT N − 1k H s + kE u N k H s + kQ N − q0 k H s ≤ C

We have also the following estimates for the remainder terms k RE N k H s + kS N k H s ≤ Mδ N +1 (N + 1)2 A N −1 C0N +2 δ N +1 e(N +1)γ (k,ε)t . Proof. We have proven the H s estimates for all the terms of the expansion u j , v j , T j , Q j . It is thus easy to deduce, using (73), the estimate for the remainder terms. This comes from the inequality (1 ≤ j ≤ N − 1) kD α (u j ∂1 u N − j )k ≤ C(ku j k∞ ku N − j k H |α|+1 + ku j k|α| k∂1 u N − j k∞ ) (and subsequent inequalities), the Sobolev embedding k f k∞ ≤ k f k2 and the H s estimate for s = 2, 3 for all the terms of the j expansion, using also that the norm H s of the terms of the expansion in δ j of order less than N is bounded by C0 A j−1 e jγ (k,ε)t . We thus deduce that

N N

X

X

j Tj ≤ C A j−1 C0 δ j e jγ (k,ε)t

j=1 j=1 Hs

≤ CC0 δeγ (k,ε)t

1 − (C0 Aδ) N −1 e(N −1)γ (k,ε)t . 1 − C0 Aδeγ (k,ε)t

When t < Tδθ , we obtain 1 − C0 Aδeγ (k,ε)t ≥ 1 − θ, hence we deduce the estimate

N

X

CC0 γ (k,ε)t

kT N − 1k H s = Tj ≤ δe .

j=1 1−θ Hs

Moreover, one has kT N − 1k L 2 ≥ δkT1 k L 2 −

N X j=2

δ j kT j k

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hence using N X

δ j kT j k L 2 ≤

j=2

N X

δ j kT j k L 2 ≤

j=2

C02 AC 2 2γ (k,ε)t δ e 1−θ

one obtains kT N − 1k L 2 ≥ δkT1 (0)k L 2 eγ (k,ε)t − One may thus consider C3 = kT N − 1k L 2 ≥ for t <

1 γ (k,ε)

ln

C 1−θ .

C02 AC 2 2γ (k,ε)t δ e . 1−θ

We thus deduce that we obtain

1 δkT1 (0)k L 2 eγ (k,ε)t , 2

kT1 (0)k L 2 (1−θ) . C AC02

Similar estimates hold for kE u N kL 2 .



Note that this proves that the first term of the expansion is the leading term of the approximate total solution. For all that follows, we introduce I (t) =

AC0 eγ (k,ε)t 1 − δ AC0 eγ (k,ε)t

(76)

I N +1 (t) = N 2 A N −1 C0N +1 e(N +1)γ (k,ε)t .

(77)

5.8. Estimates of the (nonlinear) solution We constructed in the previous section a solution T N , uE N , Q N such that  N N N N N N ∂t T + uE ∇T − k0 (x)u T = S ∂ uE N + uE N .∇ uE N + T N ρ0−1 ∇(ρ0 Q N ) = gE + RE N  t N divE u =0 with the following properties for the remainder terms:



21 n N 21 n N

ρ ∂ n R + ρ ∂ n S ≤ Cn δ N +1 I N +1 (t)

0 x j 0 x

(78)

(79)





n N n N

∂x n R j + ∂x n S ≤ Cn δ N +1 I N +1 (t)

(80) 1

the constant Cn depending on the Sobolev norm with weight ρ02 of the initial value of the normal mode solution and of the characteristic constants of the problem. −1 We deduced from this equality and the additional assumption k0 ρ 2 bounded that we have identical estimates on RE N and S N : 0

k∂xnn R Nj k + k∂xnn S N k ≤ Cn δ N +1 I N +1 (t).

(81)

We study in this section the global solution of the Euler system (8) to obtain Sobolev estimates on the difference between the approximate solution and the full solution. Let T d = T − T N , uEd = uE − uE N , Q d = Q − Q N . We have the following system of equations:  d N d d N d d N N ∂t T + uE ∇T + uE ∇T = k0 (uT + u T ) − S d d N d N (82) ρ (∂ uE + uE ∇ uE + uE ∇ uE ) + T ∇(ρ0 Q) − T ∇(ρ0 Q N ) = −ρ0 RE N  0 dt divE u = 0. Before stating the results on the difference quantities according to the system, we use the properties of T N − 1, uE N , Q N : Lemma 10. Let t ∈ [0, Tδθ ]. For all α, there exists a constant C(|α|) such that kD α (E u d .∇ uE N )k ≤ C(|α|)kE u d k|α|

CC0 δ γ (k,ε)t e 1−θ

O. Lafitte / Physica D 237 (2008) 1602–1639

CC0 δ γ (k,ε)t e 1−θ CC0 δ γ (k,ε)t e kD α (k0 u d (T N − 1))k ≤ C(|α|)kE u d k|α| 1−θ CC0 δ γ (k,ε)t e kD α (T N − 1)(∇ Q d + k0 Q d eE1 )k ≤ C(|α|)kQ d k|α|+1 . 1−θ kD α (T d (∇ Q N + k0 Q N ))k ≤ C(|α|)kT d k|α|

The proof is written in the Appendix. We shall also use the following estimates kD α (E u d .∇ uEd )k ≤ C(|α|)kE u d k4 kE u d k|α|+1 kD α (E u d .∇ uEd ) − uEd .∇ D α uEd k ≤ C(|α|)kE u d k4 kE u d k|α| . These equalities come respectively from (74) and (75). E = VE + T d ρ −1 ∇(ρ0 Q N ). We have the estimates for all α Introduce in what follows VE = uEd .∇ uE N + uE N .∇ uEd , W 0 u d k|α|+1 kD α VE k ≤ M|α| I (t)δkE E k ≤ M|α| I (t)δ(kE u d k|α|+1 + kT d k|α| ), kD α W and for |α| = 2 kD α (T d ∇ Q d ) − T d ∇ D α Q d k ≤ C(kT d k3 kQ d k2 + kT d k4 kQ d k1 ) −1

kT d ρ0 2 ∇(ρ0 Q N )k ≤ δ I (t)kT d k. We have the following result: Lemma 11. 1. Estimate on the pressure: kQ d k|α|+1 ≤ M|α|+1 (1 + kE u d k3 )kE u d k|α|+1 + δ N +1 I N +1 (t) + kT d k|α| (1 + kQ d k3 ) + (1 + kT d k3 )kQ d k|α| . 2. Estimate on the density: d kD α T d k ≤ C(k∇ uEd k∞ kT d k|α| + kE u d k|α| kT d k∞ ) + δ N +1 I N +1 M + Cδ I (t)(kE u d k|α| + kT d k|α| ). dt 3. Estimate on the velocity h i d kD α uEd k ≤ C (1 + kT d k3 )kQ d kα + (1 + kE u d k3 )kE u d k|α| + δ N +1 I N +1 (t) . dt The proof is given in the Appendix. We are now ready to obtain a global control on the remainder terms, as follows. 5.9. End of the proof We thus know that, for t ≤ T δ , we have δ N +1 I N +1 (t) ≤ 1 hence an inequality of the form h i d H (t) ≤ C (1 + H (t)4 )H (t) + 1 dt
1 where H (t) = kT d k24 + kE u d k24 2 . As we have H (0) = 0, one deduces that Z H (t) ds ≤ Ct. 4 )s + 1 (1 + s 0 h R  RH +∞ ds Since the function H 7−→ 0 (1+sds4 )s+1 is a bijection from [0, +∞) onto 0, 0 , one deduces for H (t) ≥ 1 (1+s 4 )s+1 Z

1

ds , 4 )s + 1 (1 + s 0 R1 hence for t < C1 0 (1+sds4 )s+1 = T1 , one obtains H (t) ≤ 1. The set of points t such that t > 0 and H (t) ≤ 1 is not empty. Ct ≥

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O. Lafitte / Physica D 237 (2008) 1602–1639

Once this set is not empty (and once we proved that the solution exists for a time T1 ), we obtain Lemma 12. Let h be a function such that dh ≤ C(1 + h(t))4 h(t) + Cδ N +1 e(N +1)γ (k,ε)t , h(0) = 0. dt For δ < 1 and (N + 1)γ (k, ε) > 17C, denoting by T0δ =

1 γ (k,ε)

ln 1δ , one has

∀t ∈ [0, T0δ ], h(t) ≤ δ N +1 e(N +1)γ (k,ε)t . Proof. The inequality we start with is d h(t) ≤ C(1 + h(t))4 h(t) + Cδ N +1 e(N +1)γ (k,ε)t . dt We consider N such that (N + 1)γ (k, ε) > 17C. We study the interval where h(t) ∈ [0, 1], knowing that h(0) = 0. Consider t0 the first time (if it exists) where h(t0 ) = 1. If it does not exist, then h(t) ≤ 1 for t ∈ [0, T0δ ] and we have, for all t ∈ [0, T0δ ] the inequality h 0 (t) ≤ 16Ch(t) + Cδ N +1 e(N +1)γ (k,ε)t from which one deduces h(t) ≤

Cδ N +1 e(N +1)γ (k,ε)t < δ N +1 e(N +1)γ (k,ε)t (N + 1)λ − 16C

hence h(T0δ ) < 1. If t0 exists, we have, for all t ∈ [0, t0 ], the inequality d (h(t)e−16Ct ) ≤ −C(1 − h(t))h(t)R(h(t))e−16Ct + Cδ N +1 e(N +1)γ (k,ε)t−16Ct dt where R(x) = (1 + x)3 + 2(1 + x)2 + 4(1 + x) + 8, from which one deduces that h(t0 )e−16Ct0 ≤

C δ N +1 e(N +1)γ (k,ε)t0 −16Ct0 (N + 1)γ (k, ε) − 16C

< δ N +1 e(N +1)γ (k,ε)t0 −16Ct0 hence h(t0 ) < 1, contradiction. We thus deduce that h(t) ≤ 1 for t ∈ [0, T0δ ], hence h(t) ≤ δ N +1 e(N +1)γ (k,ε)t , t ∈ [0, T0δ ]. 

Lemma 12 is proven.

We thus have the inequalities kE u k ≥ kE u N k − kE u d k ≥ δkE u 1 (0)k − C02 Aδ 2 Choose t = T1δ =

1 γ (k,ε)

γ (k,ε)t

kE u k ≥ δe

e2γ (k,ε)t − δ N +1 e(N +1)γ (k,ε)t . 1 − C0 δ Aeγ (k,ε)t

ln C0θAδ . We have

 kE u 1 (0)k − C0

 θ N −θ . 1−θ

We thus check that there exists ε0 ≤ 56 such that θ < ε0 implies   θ 1 kE u 1 (0)k − C0 − θ N ≥ kE u 1 (0)k. 1−θ 2 Hence for t ≤

1 γ (k,ε)

kE u (t)k ≥

ln C0ε0Aδ , one has

1 kE u 1 (0)kδeγ (k,ε)t . 2

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In particular

 

ε0

1 ε0

uE

≥ kE ln u 1 (0)k.

γ (k, ε) C0 Aδ 2 We proved the second and the third items of Theorem 2. It is then clear that, for 0 ≤ t ≤ T1δ , this term is smaller than θ, as small as one wants, hence the inequality on T d , uEd . As T = T N + T d , uE = uE N + uEd , one obtains kT − 1k∞ ≥ kT N k − kT d k which imply the first result of Theorem 2. Appendix. Technical proofs A.1. Proof of Lemma 2 We prove Lemma 2 by recurrence. Assume that this relation is true for j. We have the relations  du j+1 ξ˙   = (1 − λ)(b j − (λ − 1)a j ) ν( j+1)+2  dy ξ dv j+1 ξ˙    = (1 − λ)(λ + 1)(b j − (λ − 1)a j ) ν( j+1)+1 dy ξ from which we deduce, using the limit 0 at ξ → 1 Z ξ(y) b j (η) − (λ − 1)a j (η) dη u j+1 (y) = (1 − λ) ην( j+1)+2 1 and v j+1 (y) = (1 − λ)(λ + 1)

ξ(y)

Z 1

b j (η) − (λ − 1)a j (η) dη. ην( j+1)+1

ξ ν( j+1) v

ν( j+1)+1 u We thus deduce that j+1 (y) and ξ j+1 (y) are bounded functions when ξ ∈ ]0, 1]. Moreover, if we assume that j j |b j | ≤ A R and |a j | ≤ A R , then Z 1 dη j |u j+1 | ≤ A R |1 − λ|(|λ − 1| + 1) ν( j+1)+2 ξ η Z 1 dη |v j+1 | ≤ A R j |1 − λ||λ + 1|(|λ − 1| + 1) . ν( j+1)+1 η ξ

We end up with (|λ − 1| + 1) 1 − ξ ν( j+1)+1 , ξ ν( j+1)+1 ν( j + 1) + 1 (|λ − 1| + 1) 1 − ξ ν( j+1) |v j+1 | ≤ |λ − 1|A R j |λ + 1| . ν( j + 1) ξ ν( j+1) |u j+1 | ≤ |λ − 1|A R j

As

1−ξ a a

≤ 1 − ξ, ξ ∈ [0, 1], we get

(|λ − 1| + 1)(1 − ξ(y)) , ξ ν( j+1)+1 (|λ − 1| + 1)(1 − ξ(y)) |v j+1 | ≤ A R j |λ − 1||λ + 1| . ξ ν( j+1) |u j+1 | ≤ |λ − 1|A R j

Consider Rλ = (|λ − 1| + 1) max(1, |λ + 1|).

(A.1)

The previous inequalities become j+1 (1 − ξ(y))|λ − 1| , ξ ν( j+1)+1

|u j+1 | ≤ A Rλ

hence we proved the inequality for j + 1.

j+1 (1 − ξ(y))|λ − 1| , ξ ν( j+1)

|v j+1 | ≤ A Rλ

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The inequality is true for j = 1, hence the end of the proof of Lemma 2, where we may choose the value of R for λ ∈ [ 21 , 23 ] as R = 15 4 . Note that, from the above expansions and the analyticity of all expressions, it is also easy to have the same estimate for |

a j (ξ )−a j (0)−ξ a 0j (0) ξ2

| and |

b j (ξ )−b j (0)−ξ b0j (0) ξ2

|.

A.2. Proof of Lemma 6 It is enough to prove that the relation giving ζ is t 1 1 − =C− ν − − ξ 2−ν R(ξ ) ε νξ (ν − 1)ξ ν−1 hence we deduce t = −Cε +

ζ ζ +ξ − R(ξ )ξ 2 ζ. ν ν−1

We thus obtain t =

ζ0 ν,

hence

ζ ζ0 − ζ =ξ + O(ξ 2 )ζ. ν ν−1 We deduce that w(ζ ) − w(ζ0 ) = (ζ − ζ0 )w0 (ζ0 ) + O((ζ − ζ0 )2 ), hence 2

w(ζ ) − w(ζ0 ) − (ζ − ζ0 )w0 (ζ0 ) = O(ε ν ) and 2

w(ζ ) − w(ζ0 ) − (ζ − ζ0 )w0 (ζ ) = O(ε ν ). We use et G(t, λ, ε) = ( νε )

λ+1 2ν

ξ−

λ+1 2

g(t, λ, ε), hence for λ = 1 we obtain

et G(t, 1, ε) = η−1 . The equality giving F(t, 1, ε) being Z Z +∞ e−t F(t, 1, ε) = τ (s, ε)e−2s η(s, ε)−1 ds = −e−2t η(t, ε)−1 + 2 t

+∞

e−2s η(s, ε)−1 ds, t

one obtains e (F + G)(t, 1, ε) − e (F0 + G 0 )(t, 1) = 2e t

t

2t

Z

∞

e

−2s

t

Similarly et [2G(t, 1, ε) − 2G 0 (t, 1)] =

  2 η(s, ε) 1− . η(s, ε) η(s, 0)

Using the relation   1 1 ν 1 ν η(t, ε) + O(ε α ) 1 + ε = ν η(t, ε) ν−1 η(t, 0)ν one obtains 1 1 η(t, ε) −1= ε ν η(t, 0) + O(ε α ). η(t, 0) ν−1

This directly gives the two equalities of Lemma 6.



 1 1 − ds. η(s, ε) η(s, 0)

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A.3. General Duhamel inequality We consider the (general) system E E + ∇(ρ0 ∂t Q) + gk0 ρ0 wEe1 = M, ρ0 (x)∂t22 w

divw E =0

(A.2)

with the initial conditions w| E t=0 = 0,

∂t w| E t=0 = 0.

(A.3)

Note that this system is easily deduced from the system obtained for the N th term of the expansion in δ of the solution. Proposition 10. Assume that there exist two constants K and L, with L > Λ, such that

− 21

ρ M E ≤ K e Lt .

0

(A.4)

The unique solution of the linear system (A.2) with initial Cauchy conditions (A.3) satisfies the estimate 

  21 2

21  2K Λ 

ρ w  e Lt 

0 E ≤ L(L − Λ) 1 + (L − Λ)2      2K   ≤ e Lt  (L − Λ)2

  21 2 

12

Λ 2K  

ρ ∂t w

 e Lt 

0 E ≤ L − Λ 1 + (L − Λ)2  

     21 2 2Λ2  Lt

 ρ ∂ 2 w

0 t E ≤ K 1 + (L − Λ)2 e . Proof. We begin by multiplying Eq. (A.2) by ∂t22 w E and integrate in space. One deduces that



1

21 2

ρ ∂ 2 w

≤ Λ2 ρ 2 w + K e Lt . E

0

0 t We will make use of this inequality later.

(A.5)



Let us multiply Eq. (A.2) by ∂t w. E We obtain the identity   Z Z Z d 1 1 ρ0 (∂t w) E 2 dxdz + k0 ρ0 k0 w 2 dxdz = M(x, y, t)∂t wdxdz. E dt 2 2 Integrating in time and using the initial condition (A.3) as well as the estimate (A.4), we obtain the inequality Z t Z Z 1 1 ρ0 (∂t w) E 2 dxdz ≤ Λ2 k0 ρ0 w 2 dxdz + 2K e Ls kρ02 ∂t wk(s)ds. E 2 0

R R t 12 d

Let us now introduce the function u(t) = 0 ρ0 ∂t w E E 2 dxdz, that

(s)ds. We obtain, considering dt ρ0 w

Z

12

ρ w

(t) ≤ E

0

0

t



21

ρ ∂t w

(s)ds = u(t). E

0

Inequality (A.6) transforms into Z t 2 0 2 2 u (t) ≤ Λ u(t) + 2K e Ls u 0 (s)ds 0

≤ Λ2 u(t)2 + 2K e Lt u(t) p ≤ (Λu(t) + 2K e Lt u(t))2 . Introduce h > 0 such that u(t) = h(t)2 eΛt . We obtain the inequality √ L+Λ 2hh 0 eΛt ≤ 2K he 2 t

(A.6)

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hence one deduces √ 2K L−Λ t h(t) ≤ e 2 L −Λ which leads to u(t) ≤

2K e Lt . (L − Λ)2

The estimate on u 0 (t) follows, using (u 0 )2 ≤ Λ2 u 2 + 2K e Lt u. Thus, by integration, we deduce a better estimate on u. The estimate E is the consequence of (A.5). on ρ0 ∂t22 w A.4. Proof of Lemma 10 The proof of this Lemma comes from the fact that X Dα ( f g N ) = Cαβ D β f D α−β g N PN j and we use the estimate kD α−β g N k∞ ≤ Ckg N k2+|α|−|β| , as well as the H s result on any term of the form g N = j=1 δ g j , where g j = u j , v j , T j , Q j to conclude for any term studied in the Lemma. Moreover, we use the Moser estimates to obtain kD α (E u · ∇ f ) − uE.∇ D α f k ≤ C(k∇ uEk∞ k∇ f k|α|−1 + k∇ f k∞ kE u k|α| ) hence, using uE = uE N + uEd , one deduces   kD α (E u .∇ f ) − uE.∇ D α f k ≤ C k∇ uEd k∞ k f k|α| + δ I (t)k f k|α| + k∇ f k∞ kE u k|α| + k∇ f k∞ δ I (t) and, similarly kD α (E u .∇ f )k ≤ Cδ I (t)(k∇ f k∞ + k f k|α|+1 ) + kE u d k∞ k f k|α|+1 + kE u d kα k∇ f k∞ , kD α (T ∇ Q d ) − T ∇ D α Q d k ≤ Cδ I (t)kQ d k|α| + k∇T d k∞ k∇ Q d k|α|−1 + k∇ Q d k∞ kT d k|α| according to the equality D α (T ∇ Q d ) − T ∇ D α Q d = D α ((T − 1)∇ Q d ) − (T − 1)∇ D α Q d . A.5. Proof of the estimates on density and pressure A.5.1. Estimates on the density The equation on the density yields ∂t T d + uE.∇T d − k0 uT d = k0 u d T N − uEd .∇T N − S N . Apply the operator D α and denote Wα1 = D α (E u .∇T d ) − uE.∇ D α T d . This equation rewrites ∂t D α T d + uE.∇ D α T d + Wα1 − D α (k0 uT d ) + D α (E u d .∇T N ) − D α (k0 u d T N ) = 0. We can decompose Wα1 − D α (k0 uT d ) into two parts, one with uE N , the other one with uEd , denoted respectively by Wα and WαN . It is clear that k(WαN − D α (k0 uT d ) + D α (E u d .∇T N ) − D α k0 u d T N )k ≤ Cδ(kE u d k|α| + kT d k|α| ) it is also clear that, using Moser estimates, kWα k ≤ C(k∇ uEd k∞ kT d k|α| + kE u d k|α| kT d k∞ ). One is thus left with the inequality d kD α T d k ≤ kWα k + kD α S N k + Cδ I (t)(kE u d k|α| + kT d k|α| ). dt We thus have the estimate d kD α T d k ≤ C(k∇ uEd k∞ kT d k|α| + kE u d k|α| kT d k∞ ) + δ N +1 I N +1 M + Cδ I (t)(kE u d k|α| + kT d k|α| ). dt

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A.5.2. Estimates on the pressure We obtained the relations k∇ Q d k ≤ M1 (kE u d .∇ uEd k + δ I (t)(kE u d k1 + kT d k) + δ N +1 I N +1 (t)). i h X X u d .∇ uEd k k∇ D α Q d k ≤ M2 kD α (E u d .∇ uEd )k + δ I (t) kE u d k2 + kT d k1 + δ N +1 I N +1 (t) + (1 + δ I (t) + kT d k3 )(kE |α|=1

|α|=1

+ δ I (t)(kT d k + kE u d k1 ) + δ N +1 I N +1 (t)). Using the fact that t ≤ T δ , one obtains k∇ Q d k ≤ M1 (kE u d .∇ uEd k + kE u d k1 + kT d k + δ N +1 I N +1 (t)), X X k∇ D α Q d k ≤ M2 kD α (E u d .∇ uEd )k + kE u d k2 + kT d k1 + (1 + kT d k3 )(kE u d .∇ uEd k + kT d k + kE u d k1 ) |α|=1

|α|=1

! + δ N +1 I N +1 (t) . In what follows, we introduce E αN = D α (T N ∇ Q d ) − T N ∇ D α Q d + D α (T N Q d k0 eE1 ) + D α (T d ∇ Q N + k0 T d Q N eE1 ) + D α RE N , G E α = D α (T d ∇ Q d ) − T d ∇ D α Q d + D α (T d Q d k0 eE1 ). G The equation on D α uEd is Eα + G E αN + T ∇ D α Q d = 0. ∂t D α uEd + D α (E u d .∇ uEd ) + D α RE N + G When one multiplies by ∇ D α Q d , one uses the divergence free condition on D α uEd to get the estimate 2 E α k + kG E αN k. k∇ D α Q d k ≤ kD α (E u d .∇ uEd )k + kD α RE N k + kG 3 We use E αN k ≤ C(1 + t)|α|+3 (kQ d k|α| + kT d k|α| + kE kG u d k|α|+1 )δ I (t) and E α k ≤ C(k∇T d k∞ kQ d k|α| + kT d k|α| k∇ Q d k∞ + kT d k∞ kQ d k|α| ). kG Hence we obtain (and it is pertinent for |α| > 2) k∇ D α Q d k ≤ C 0 (kD α (E u d .∇ uEd )k + kD α RE N k) + Cδ I (t)(kQ d k|α| + kT d k|α| + kE u d k|α|+1 ) + C(kT d k3 kQ d k|α| + kT d k|α| kQ d k3 ). For |α| = 2, we will obtain kQ d k3 , which is important. We use the equality, for |α| = 2



X

α d α d β d α−β d β E D T ∇D Q Cα kG k = D T ∇ Q +

0<β<α which leads to the inequality E α k ≤ D0 (kT d k4 kQ d k1 + kT d k3 kQ d k2 ). kG Replacing this estimate in the inequality for α such that |α| = 2, one gets kD α ∇ Q d k ≤ C1 (kD α (E u d .∇ uEd )k) + kT d k2 + kE u d k3 + (1 + kT d k3 )kQ d k2 + kT d k4 kQ d k1 + δ N +1 I N +1 (t). Using the inequalities on kQ d k1 and kQ d k2 , one gets kQ d k1 ≤ M1 (kE u d .∇ uEd k + kE u d k1 + kT d k + δ N +1 I N +1 (t)) d d d kQ k2 ≤ M2 (kE u .∇ uE k1 + kE u d k2 + kT d k1 + δ N +1 I N +1 (t)(1 + kT d k3 ) + (1 + kT d k3 )(1 + kT d k + kE u d .∇ uEd k)) kQ d k3 ≤ M3 (kE u d .∇ uEd k2 + kE u d k3 + kT d k2 + (1 + kT d k23 + kT d k4 )kE u d .∇ uEd k + (1 + kT d k3 )kE u d k2 + kT d k4 kE u d k + δ N +1 I N +1 (t)(1 + kT d k4 + (1 + kT d k3 )2 )).

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We then use the inequalities u d k|α|+1 + (1 + kT d k3 )kQ d k|α| ) k∇ D α Q d k ≤ C(kD α (E u d .∇ uEd )k + kD α RE N k + kQ d k|α| + kT d k|α| (1 + kQ d k3 ) + kE from which one obtains kQ d k|α|+1 ≤ M|α|+1 (kE u d .∇ uEd k|α| + kE u d k|α|+1 + δ N +1 I N +1 (t) + kT d k|α| (1 + kQ d k3 ) + kQ d k|α| (1 + kT d k3 )). Note that we have the estimate kE u d .∇ uEd k|α| ≤ CkE u d k3 kE u d k|α|+1

(A.7)

kQ d k|α|+1 ≤ M|α|+1 ((1 + kE u d k3 )kE u d k|α|+1 + δ N +1 I N +1 (t) + kT d k|α| (1 + kQ d k3 ) + (1 + kT d k3 )kQ d k|α| ).

(A.8)

hence

It is then enough to use a recurrence argument to control the norm of Q d in H s+1 using the control of the norm of Q d in H s . For the control on uEd , let us rewrite the equation on D α uEd . We introduce VEα = D α (E u .∇ uEd ) − uE.∇ D α uEd ,

E α = D α (T.∇ Q d ) − T.∇ D α Q d . W

We have the estimates kVEα k ≤ C(1 + kE u d k3 )kE u d k|α| ,

E α k ≤ C(1 + kT d k3 )kQ d k|α| . kW

Using the relation Z Z T ∇ D α Q d D α uEd dxdz = − D α Q d (∇(T N − 1) + ∇T d )D α uEd dxdz thanks to the divergence free condition, as well as Z uE.∇ D α uEd .D α uEd dxdz = 0 one obtains the estimate d E α k + kD α RE N k + kD α (k0 T Q d )k + kD α Q d k(1 + kT d k3 ), kD α uEd k ≤ kVEα k + kW dt hence the inequality h i d kD α uEd k ≤ C (1 + kT d k3 )kQ d kα + (1 + kE u d k3 )kE u d k|α| + δ N +1 I N +1 (t) . dt

(A.9)

For |α| ≥ 3, this inequality is an a priori inequality. We have to state the identical inequalities for |α| = 0, 1, 2. We have the following inequalities: d d kE u k ≤ C0 ((1 + kT d k3 )kQ d k + δ N +1 I N +1 (t)) dt

(A.10)

E α = 0, because VEα = W d k∇ uEd k ≤ C((1 + kT d k3 )kQ d k1 + (1 + kE u d k3 )kE u d k1 + (1 + kT d k1 )kQ d k3 + δ N +1 I N +1 (t)) dt

(A.11)

d d kE u k2 ≤ C((1 + kT d k3 )kQ d k2 + (1 + kT d k4 )kQ d k1 + (1 + kE u d k3 )kE u d k2 + δ N +1 I N +1 (t)). dt

(A.12)

and

We thus deduce an estimate of the form 1 d (kT d k24 + kE u d k24 ) ≤ C((1 + kT d k4 )4 + kE u d k3 )(kT d k24 + kE u d k24 ) + δ N +1 I N +1 (t)(kT d k24 + kE u d k24 ) 2 dt

from which one deduces an estimate of the form 1 1 d (kT d k24 + kE u d k24 ) 2 ≤ C((1 + kT d k4 )4 + kE u d k3 )(kT d k24 + kE u d k24 ) 2 + Cδ N +1 I N +1 (t). dt

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