Study of the linear ablation growth rate for the quasi isobaric model of Euler equations with thermal conductivity
aa r X i v : . [ m a t h . A P ] J u l Study of the linear ablation growth rate for thequasi-isobaric model of Euler equations withthermal conductivity
Olivier Lafitte ∗† November 30, 2006
Abstract
In this paper, we study a linear system related to the 2d system of Eu-ler equations with thermal conduction in the quasi-isobaric approximationof Kull-Anisimov [14]. This model is used for the study of the ablationfront instability, which appears in the problem of inertial confinement fu-sion. The heat flux ~Q is given by the Fourier law T − ν ~Q proportional to ∇ T , where ν > ~g = − g~e x . This physical system contains a mixing region,in which the density of the gaz varies quickly, and one denotes by L anassociated characteristic length. The fluid velocity in the denser region isdenoted by V a .The system of equations is linearized around a stationary solution, andeach perturbed quantity ˜ u is written using the normal modes method˜ u ( x, z, t ) = ℜ (¯ u ( x, k, γ ) e ikz + γ √ gkt )in order to take into account an increasing solution in time.The resulting linear system is a non self-adjoint fifth order system. Itscoefficients depend on x and on physical parameters α, β , α and β beingtwo dimensionless physical constants, given by αβ = kL and αβ = gL V a (introduced in [5]). We study the existence of bounded solutions of thissystem in the limit α →
0, under the condition β ∈ [ β , β ], and theassumption ℜ γ ∈ [0 , β ] , | γ | ≤ β (regime that we studied for a simplermodel in [5]) calculating the Evans function Ev ( α, β, γ ) associated withthis linear system.Using rigorous constructions of decreasing at ±∞ solutions of systems ofODE, we prove that, for β ∈ [ β , β ], ℜ γ ∈ [0 , β ], | γ | ≤ β , there exists α > < α ≤ α . ∗ CEA/DM2S, Centre d’Etudes de Saclay, 91191 Gif sur Yvette Cedex † Universit´e de Paris XIII, LAGA, 93 430 Villetaneuse ecessarily, for any M > β > α > < α ≤ α and β ∈ [ β , β ], an admissible value γ ( α, β ) such that thereexists a bounded solution of the linearized system satisfying | γ | ≤ M issuch that ℜ γ / ∈ [0 , M ]. Contents α . . . . . . . . . . . . . . . . . . 263.3 Uniform estimates of the solution w (2)+ for ξ ∈ [ ζ α ν ,
1] . . . . . . . 283.4 Behavior of the terms of the expansion . . . . . . . . . . . . . . . 30 Introduction
This paper is devoted to the precise calculus of the Evans function Ev ( α, β, γ )of the normal mode formulation of the linearized system of equations associatedwith the quasi-isobaric (low Mach number) model. The calculus of this Evansfunction is not classical, because the matrix of the differential system has sin-gular coefficients,and because these coefficients do not behave exponentially inthe spatial variable. However, usual techniques of ordinary differential equa-tions and introduction of a Fuchsian problem allow us to calculate this Evansfunction under certain assumptions on the parameters α, β, γ introduced in theAbstract.In this Introduction, we first describe the physical model (5), define what iscalled a linear growth rate of the linearized system associated with a stationarysolution of this physical model, then finally describe the contents of each stepof the proof of the main Theorem (Theorem 4). We consider a compressible fluid characterized by its density ρ , its velocity ( u, v )and its temperature T in a gravity field ~g = −| g | ~e x . We assume that this fluidhas the following properties:a) when x goes to + ∞ , for all z we have ρ → ρ a , ( u, v ) → ( − V a ,
0) and T → T a .b) the functions ( ρ, u, v, p, T ), where p is linked to the pressure in the fluid,satisfy the system of the Euler equations in two dimensions ( x, z ) with thermalconduction in the quasi-isobaric approximation for a perfect gaz:div( C p ρT ~u + ~Q ) = 0 , (1)where C p is the calorific capacity of the fluid, the heat conduuction flux ~Q beinggiven by the Fourier law ~Q = − k ( T ) ∇ T, (2)the thermal conduction law is k ( T ) = K T ν , (3)where ν is the thermal conduction indice.We introduce a characteristic length L associated with the thermal propertiesof the fluid L = K T νa C p ρ a V a . (4)Physical values of L for the case of the ICF are of order 10 − meters.Under a quasi-isobaric assumption, the system modelizing the ablation model3as given by H.J. Kull [14] and appears for example in P. L. Lions [17]. It states ∂ t ρ + div( ρ~u ) = 0 ∂ t ( ρ~u ) + div( ρ~u ⊗ ~u + p ) = ρ~gρT = ρ a T a ~Q = − k ( T ) ∇ T div( C p ρT ~u + ~Q ) = 0 . (5)This model can be derived either from the low Mach approximation (Majda [19],Dellacherie [6]) or the quasi-isobaric approximation (Kull [14], Kull-Anisimov[15], Masse [20]). See a short analysis in Section 1. A stationary laminar solutionof the system (5) is ( ρ ( x ) , u ( x ) , , p ( x ) , T ( x )), where we introduce a function ξ such that ρ ( x ) = ρ a ξ ( xL ) , u ( x ) = − ρ a V a ρ ( x ) = − V a ξ ( xL ) , T ( x ) = ρ a T a ρ ( x ) = T a ξ ( xL )and p ( x ) satisfies p ( x ) + ρ a V a ρ ( x ) + g Z xx ρ ( s ) ds = p ( x ) + ρ a V a ρ ( x ) . The function ξ is the solution of the differential equation dξdy = ξ ν +1 (1 − ξ ) (6)such that ξ (0) = ν +1 ν +2 . Note that, in this case p ( L y ) = p (0) + ν + 2 ν + 1 ρ a V a − ρ a V a [ 1 ξ ( y ) + gL V a Z y ξ ( t ) dt ] . Introduce Z ( ρ ) = ρ ν +1 a ( ν + 1) ρ ν +1 . (7)The system of unknowns that we consider is˜ U = ρuρu + pρuvZ ( ρ ) u − L V a ∂ x ( Z ( ρ )) . From ˜ U , we recover ρ from Z ( ρ ), u = ρuρ , v = ρuvρu and p = p + ρu − ( ρu ) ρ .For this choice of unknowns, we introduce F ( ˜ U ) = ˜ U . There exists threeexplicit functions F , F and F such that the system (5) is equivalent to thesystem on ˜ U ( ∂ t F ( ˜ U ) + ∂ x F ( ˜ U ) + ∂ z ( F ( ˜ U , ∂ z ˜ U )) = F ( ˜ U ) T = ρ a T a Z − ( ˜ U ) (8)4 stationnary laminar solution U of this system depending only on x is U ( x )such that ddx ( F ( U ( x ))) = F ( U ( x )) . The identity F ( ˜ U ) = ˜ U is the natural choice when one studies a basic solutiondepending only on the variable x . We linearize (8) around U ( x ). Denote by U the unknowns ˜ U − U ( x ). Thelinearized system writes ( ∇ and ∇ denotes the gradient of F with respect tothe first and second set of variables ˜ U and ∂ z ˜ U ): ∂ x U + ∇ F ( U ( x )) ∂ t U + ∇ F ( U ( x ) , ∂ z U + ∇ F ( U ( x ) , ∂ z U = ∇ F ( U ( x )) U (9)which can be rewritten M ( x, ∂ x , ∂ z , ∂ z , ∂ t ) U = 0. Note that its coefficientsdepend on x through the stationnary solution.We are now ready to introduce the definition of a linear growth rate for anon linear system around a laminar solution: Definition 1
Let M ( x, ∂ x , ∂ z , ∂ z , ∂ t ) U = 0 be the linearized systemWe call a linear growth rate of this system for the wave number k a valueof σ (depending on k ) such that ℜ σ ≥ and there exists a non-trivial solution U ( x, k, σ ) of the system M ( x, ddx , ik, − k , σ ) U ( x, k, σ ) = 0 (10) such that U is bounded and going to 0 when x goes to ±∞ . The function U ( x, k, σ ) e ikz e σt is called a normal mode solution of the system. The normal mode system associated with (9) is dVdx + ∇ F ( U ( x )) σV + ik ∇ F ( U ( x ) , V − k ∇ F ( U , V = ∇ F ( U ( x )) V. (11)The scope of this paper is to find bounded non trivial solutions of (11), andassociated values of σ if any. If such a solution exist, it will lead to a normal modesolution of the linearized system. Note that, in the set-up we described, differentphysical parameters appear, namely k, L , V a , g . As the classical growth rate ofRayleigh is equal to ( ρ − ρ ρ + ρ gk ) for the discontinuity model [22], and as weproved ([5], [12]) that this value was the limit of the growth rate when kL goesto zero, we are led to introducing the following quantities ε = kL , F r = V a gL (12)5nd α = r εF r , β = √ εF r, γ = σ √ gk . (13)The aim of this paper is to study the existence of a growth rate γ , ℜ γ ≥
0, inthe limit L → F r is of order L , which means that α → , β > . Remark
Other regimes rely on different assumptions on α and β : we refer to[13] for the results that can be obtained for this model in the high frequencyregime k → + ∞ . In this other regime the scaling writes ε large, ε F r ≤ C ′ , where C ′ is a constant.In the first section, we study the physical origin of the model and derive theproperties of the stationary solution, where the associated density profile satis-fies: ρ ( x ) → x → −∞ ρ ( x ) − ρ a ≃ Ce − xL , x → + ∞ ρ ( x ) | x | ν → ρ a ( L ν ) ν , x → −∞ . We then derive the linearized system, which is a fifth order differential systemwhose coefficients depend on ρ ( x ) and are singular when ρ ( x ) go to zero. Notethat this is not a classical case for the study of such systems and that this leadsto rather tricky methods.In the second section, we recall the general set-up for the calculation of theEvans function of the linearized system, and we give the induced differentialsystems in Λ n ( K ) for n = 2 ,
3. Note that the field K is IR for real values of γ and K = C | for complex values of γ . This Evans function is (related to) thevectorial product of the normalized solution in Λ ( K ) which has the greatestdecay when x → + ∞ and of the normalized solution in Λ ( K ) which has thegreatest decay at x → −∞ .In the third section, we identify the solutions of the system in Λ ( K ) deducedfrom (10) for x → + ∞ . In this region, we use the exponential behavior of theprofile to obtain the classical analytic expansion of the normalized solution ofthe system in Λ ( K ). There exists ξ ∈ ]0 ,
1[ (corresponding to y ∈ R through ξ ( y ) = ξ ) such that this analytic expansion is valid for ξ ( xL ) ≥ ξ , that is x ≥ L y . Note that, however, the expansion of this solution cannot be obtainedby the techniques developed in Zumbrun et al [4], because the Gap lemma as-sumptions are not fulfilled.A general feature in the calculation of the Evans function is to obtain an over-lapping region of definition between the solution well behaved at + ∞ and thesolution well behaved at −∞ . A first step to achieve this overlap is then toprove that there exists α > R > α < ζ < R thesolution obtained for x ≥ L y can be extended in [ X ∗ ( α, ζ ) , L y ] where ξ ( X ∗ ( α, ζ ) L ) = α ν ζ − ν . α ν ζ − ν , ξ ] when α → α for y ∈ [ y , + ∞ [ and it is the aim of Sections 3.3 and 3.4.Once this extension is done, an easy calculus is the calculus of a growth rateassociated with the following stationnary solution, characterized by its densityprofile, for a ζ such that ζ < R : ρ ∗ ( x ) = ( ρ a ξ ( xL ) , x ≥ X ∗ ( α, ζ ) ρ a α ν ζ − ν . (14)This calculus is an improvement of the discontinuity model of Piriz, Sanz andIbanez [21] and it is the aim of Section 4.When the profile is not constant in the region ] − ∞ , X ∗ ( α, ζ )] (that is for thefull model), the system leads to a fuchsian problem in the region x → −∞ , andwe use the hypergeometric equation (see [16]). The solution of the system inΛ ( K ) deduced from (10) is identified in any region of the form xL ∈ ] −∞ , − t αβ ]for every t , which means that x ∈ ] − ∞ , − t k ]. The results of the analysis ofthese solutions is summarized in Theorem 1.The study of the roots of the Evans function is the aim of Section 6 and wesummarize the method here. From the relation ξ ( y ) | y | ν → ν − ν when y → −∞ ,we deduce that − αβX ∗ ( α ) → βζ ν > α →
0. Hence for all < α ≤ α there exists t > − ∞ , − t αβ ] and [ X ∗ ( α ) L , y ], where ξ ( y ) = ξ , overlap.We then express the Evans function Ev ( α, β, γ ) of the system at a point of[ − t αβ , X ∗ ( α ) L ]. The limit when α → t small exists and we write itsexpression in terms of r = γβ , β and t >
0. As it does not depend on t westudy the limit when t → + ∞ , hence proving that the only positive value of r which is admissible is r = 1. We deduce a contradiction, proving that there isno growth rate ( of positive real part) for the system. This can be stated as Theorem Let M be given. There exists α ∗ > such that, for <α < α ∗ , β ∈ [ M , M ] , the Evans function Ev ( α, β, γ ) of the system has noroot for | γ | ≤ M , ℜ γ ∈ [0 , M ] . The general equations are the thermal hydrodynamic equations, written in anon conservative form: ∂ t ρ + div( ρ~u ) = 0 ∂ t ( ρ~u ) + div( ρ~u ⊗ ~u + p ) = ρ~gρ ( ∂ t + ~u. ∇ ) h − ( ∂ t + ~u. ∇ ) p = − div( ~Q + ~J l ) . (15)7here C p and C v are the classical tehrmodynamic calorific capacities at constantpressure and at constant volume, h is the enthalpy h = C p T , the pressureand the density being given by the equation of state p = ( C p − C v ) ρT , ~Q = − k ( T ) ∇ T , ~J l = 0 (in our assumption the energy given to the system is 0). Thequasi-isobaric approximation writes C p − C v C v δpp << δTT . Following L. Masse [20], this relies on the two hypotheses M << , M F r << . (16)Hence the quasi-isobaric model relies on a low Mach hypothesis.The equation of the energy rewritesdiv ~u + 1 ρh div ~Q = − ( ∂ t + ~u. ∇ )( ρh − p ) ρh . This equation is approximated by (1) as we will see below. What follows is a formal derivation of the quasi-isobaric model under a low mach hypothesis. Itis closely related to the method given by Majda [19], Dellacherie [6].
Use the reference density ρ a , the reference velocity V a , and the reference pressure p s associated with the sound velocity c s such that c s C p C v ρ a = p s . The Machnumber is thus M = V a c s .Write ρ = ρ a ρ ′ , ~u = V a ~u ′ , p = p s p ′ . The system of equations (15) rewrites V − a ∂ t ρ ′ + div( ρ ′ ~u ′ ) = 0 V − a ∂ t ( ρ ′ ~u ′ ) + div( ρ ′ ~u ′ ⊗ ~u ′ + γ p ′ M ) = gV a ρ ′ div ~u ′ + V − a C p ρT div ~Q = − ( V − a ∂ t + ~u ′ . ∇ ) pp If we assume that all the quantities have an asymptotic expansion in M , inparticular p ′ = p ′ ( x, z, t ) + M p ( x, z, t, M ) we have the following relations fromthe momentum equations ∂ x p ′ = 0 , ∂ z p ′ = 0hence p ′ depends only on t . This is the same result as in the analysis of Del-lacherie [6]. In this model, we assume that the pressure p ′ is constant, becausewe assume that the ground state for the equations is stationnary.Replacing the relation p ′ ( x, z, t, M ) = p ′ + M p in the energy equation weobtain div ~u ′ + V − a C p − C v C p ( p ′ + M p ) div ~Q = − M ( V − a ∂ t + ~u ′ . ∇ ) pp ′ + M p ~Q = K ( p ′ + M p ( C p − C v ) ρ ′ ) ν ∇ p ′ + M p ( C p − C v ) ρ ′ , we deduce that ~Q = K ( p ′ C p − C v ) ν +1 ∇ Z ( ρ ) + O ( M )hence the formal analysis leads to the equationdiv( V − a ~u ′ + L ∇ Z ( ρ )) = O ( M )where we used p ′ C p − C v = ρ a T a p s deduced from the relation p = ( C p − C v ) ρT . Theresulting equation can be written div( C p ρT ~u + ~Q ) = 0, ρT = ρ a T a hence (1).Finally, in the momentum equations, rewriting p ′ M = p ′ M + p and using p ′ constant, we obtain the equations V − a ∂ t ( ρ ′ ~u ′ ) + div( ρ ′ ~u ′ ⊗ ~u ′ + pId ) = gL a ρ ′ . Note finally that the relation (1) and the relation ρT = ρ a T a lead to the equationon ρ : ( ∂ t + ~u. ∇ ) Z ( ρ ) − ( ν + 1) L V a Z ( ρ )∆ Z ( ρ ) = 0 . (17) The resulting system of equations that models our phenomenon is thus ∂ t ρ + div( ρ~u ) = 0 ∂ t ( ρ~u ) + div( ρ~u ⊗ ~u + pId ) = ρ~g div( ~u + L V a ∇ Z ( ρ )) = 0 . (18)A stationnary laminar solution satisfies ρ ( x ) u ( x ) = − ρ a V addx ( ρ ( x ) u ( x ) + p ( x )) = − ρ ( x ) g ddx ( u ( x ) − L V a Z ( ρ ( x ))) = 0 . where the first relation is a consequence of the mass conservation equation, theconstant ρ u being identified through its limit at x → + ∞ . Hence u ( x ) = − ρ a V a ρ ( x ) leading to the equation on ρ : − V a ρ a ρ ( x ) + L V a dρ ( x ) dx ρ ( x ) − ν − ρ ν +1 a = C . As y = xL this equation rewrites dξdy = C V a ξ ν +2 + ξ ν +1 . If C = 0, the equation becomes ddy ( ξ − ν ) = − ν , hence ξ − ν = D − νy hence ξ is not defined for y > D ν . We cannot consider this solution.9f C > ξ is increasing, hence if it is majorated, it has a limit l > y → + ∞ , this limit l satisfies l ν +1 + C V a l ν +2 = 0, which is impossible. We deducethat C <
0, hence the equation is dξdy = ξ ν +1 (1 − | C | V a ξ )hence from the resolution of the equation we deduce ξ → V a | C | , and as ξ → | C | = V a and the resulting equation is (6).This equation has a unique constant solution ξ = 1. The low Mach approxi-mation of S. Dellacherie [6] for a bubble model uses this stationnary solution asbase solution. When we consider a non constant solution, we have the followingLemma which gives the behavior of the solution when x goes to −∞ . Introducethe function h { ν } such that h { ν ( ξ ) } = R ξ η ν − [ ν ] (1 − η ) , and let n = [ ν ], ν +1 ν +2 = ξ ∗ and y ∗ = P np =0 1 ν − p ξ ν − p ∗ − h { ν } ( ξ ∗ ). Lemma 1
There exists t ∗ > such that for ≤ t ≤ t ∗ there exists a uniquecontinuous function g ( t ) , g (0) = 1 solution of ( g ( t )) ν = − t ν g ( t ) ν ( h ν ( tg ( t )) + y ) + n X p =0 t p ( g ( t )) p νν − p . The function g has a Taylor expansion at t = 0 .1. Behavior when y → −∞ For y ≤ − ν ( t ∗ ) ν , we have the identity ξ ( y ) = ( − νy ) ν g (( − νy ) ν ) . There exists a function r ( t, ε ) such that ε ( ξ ( − tε )) ν = 1 νt + ε ν t − − ν r ( t, ε ) . (19) There exists t > and ε > such that r ( t, ε ) is bounded for t ≥ t , ≤ ε ≤ ε
2. Auxiliary function at −∞ : We introduce S ( t, ε ) = − R + ∞ t s − − ν r ( s, ε ) ds .This function satisfies S ( t, ε ) t ν uniformy bounded for t ≥ t and ≤ ε ≤ ε .3. Behavior when y → + ∞ We have − ξ ( y ) = C ( y ) e − y , with C ( y ) → exp ( y − ν − .. − ν − n + R
10 (1 − η ν − n ) dηη ν − n (1 − η ) ) . − n X p =0 ν − p ξ − ν + p + h { ν } ( ξ ) = y − y ∗ The equality yields y = − νξ ν [ n X p =0 νν − p ξ p − νξ ν ( h { ν } ( ξ ) + y ∗ )] . Introduce ξ = tg and t = ( − νy ) ν , such that y = − νt ν . We obtain the equality g ν = n X p =0 νν − p t p g p − νt ν g ν [ h { ν } ( tg ) + y ∗ ] . We introduce Ψ( t, g ) = g ν − P np =0 νν − p t p g p − νt ν g ν [ h { ν } ( tg ) + y ]. We haveΨ(0 ,
1) = 0 and ∂ g Ψ(0 ,
1) = ν > t, g ) = 0 in the neighborhood of (1 , ∞ ,thanks to the equalityln(1 − ξ ) = y − n X p =0 ν − p ) ξ ν − p + Z ξ dη (1 − η ν − n ) η ν − n (1 − η ) − y hence 1 − ξ ( y ) = e C ( y ) e − y with C ( y ) → y − P np =0 1( ν − p ) + R dη (1 − η ν − n ) η ν − n (1 − η ) when y → + ∞ . The Lemma isproven. The particular case ν = 2 . y − C = ln 1 + √ ξ − √ ξ − ξ (1 + 53 ξ + 5 ξ ) . Hence, for y = − tε and ξ = ηε we get t = 25 η (1 + ε η + ε η ) − εC − ε ln 1 + ε √ η − ε √ η . We write η ( ε, t ) = ( t ) − g ( t, ε ). Considering the limit, for t > ε → g ( t, ε ) = 1and we construct step by step the expansion of g ( t, ε ) in ε .11 .4 Physical interpretation of the model We insist finally on the fact that this system of equations is only a theoreticalmodel: the equation satisfied by p is dp dy = − ρ a V a [ 1 F r ξ ( y ) + ξ ν − (1 − ξ )]hence the pressure is not bounded. The total pressure writes P ( x ) = ( C p − C v ) ρ a T a + M p ( L x )which leading order term is constant and low order term in Mach is not bounded.Hence a good way of calling this model could be to call it a relative isobaricmodel or a relative low mach model. See Majda and Sethian [19], Embid[7], or P.L. Lions [17] for other remarks on this modelling.Introduce the function M ( y ) such that M ( y ) = | ~u ( y ) | C s ( y ) , where C s ( y ) is thesound velocity at a point of the fluid given by C s ( y ) = C p C v p ( y ) ρ ( y ) , that we stillcall the Mach number. We have Lemma 2
The Mach number of the stationnary solution is bounded for ν ≥ and y ≤ − C . Proof As M ( y ) = ρ a V a ξ ( y ) C s ( y ) = ρ a V a γp ( y ) ξ ( y ) = ρ a V a ( C p − C v ) ρ a T a ξ ( y )+ M p ( y ) ξ ( y ) M which has a finite limit when y → −∞ under the condition ν >
2. More-over, as p ( y ) → −∞ when y → + ∞ , there exists a point where the pressurevanishes. There exists a constant C such that P ( y ) is bounded below ε > − ∞ , − C ] and − C is a positive constant of order ( ρ a T a M ) νν − .The low Mach number assumption is relevant (in particular when looking atthe temperature of ablation and the density in the ablated fluid). To simplify the notations of what follows, we denote by ˜ f a quantity appearingin the Euler system of equations, by f its stationnary leading order term, andby f the perturbation of order 1 normalized by the physical quantity ρ a for thedensity, ρ a V a for the impulsion, and ρ a V a for a pressure term. In the normalmode study in the vicinity of a profile depending on x , it is pertinent to linearizethe variables on which acts the derivative ddx , that is ˜ ρ ˜ u , ˜ ρ (˜ u ) + p and ˜ ρ ˜ u ˜ v ,which writes ˜ ρ ˜ u = − ρ a V a + ρ a V a x ˜ ρ (˜ u ) + p = p ( x ) + ρ ( x )( u ( x )) + ρ a V a x ˜ ρ ˜ u ˜ v = − iρ a V a x (20)The coupling with the equations of the energy is made through the pertur-bation of density ρ . Introduce ~τ = ~ ˜ u − L V a ∇ ( Z ( ρ )) − L V a ∇ ( Z (˜ ρ ) − Z ( ρ )) . x = Z ( ρ ) − Z (˜ ρ ) , x = 1 V a ( τ + V a ) . (21)With the choice of unknowns ( x , x , x , x , x ) (which corresponds to the un-knowns ρu, ρu + p, ρuv , − Z ( ρ ) , τ ), denoting by Z − the inverse function of Z , Z − ( f ) = ρ a (( ν + 1) f ) ν +1 , the non linear system (18) is equivalent to: V − a ∂ t ( Z − ( Z ( ρ ) − x )) + ρ a ∂ x x + i∂ z ( Z − ( Z ( ρ ) − x ) x − x ) = 0 V − a ∂ t x + ρ a ∂ x x − iρ a ∂ z x = − gV a ( Z − ( Z ( ρ ) − x )) + gV a ρ ( x ) V − a ∂ t ( Z − ( Z ( ρ ) − x ) x − x ) − ρ a ∂ x x + ρ a ∂ z ( x + x (1 − x ) Z − ( Z ( ρ ) − x ) ρ a + ρ a ρ ( x ) − ρ a ( − x ) Z − ( Z ( ρ ) − x ) ) = 0 L ∂ x x + ρ a (1 − x ) Z − ( Z ( ρ ) − x ) − ρ a ρ ( x ) − x = 0 ∂ x x + ∂ z ( L ∂ z x + i x − x ) = 0 . (22)In the system (18), one needs to obtain the linearization of ˜ ρ ˜ v and of ˜ p + ˜ ρ (˜ v ) in terms of x , x , x . It is a consequence of˜ v = iV a x − x , ˜ ρ = ρ a ( ξ + ρ ) , ˜ p = x − (1 − x ) ξ + ρ + 1 ξ , ˜ u = − V a ξ + ρ )hence the approximations (dropping the terms of order 2 at least)˜ v = iV a x , ˜ p = p + ρ a V a ( x + 2 ξ x + ρξ ) , ˜ u = u + V a ξ ρ. Moreover, the identity x = Z ( ρ a ξ ) − Z ( ρ a ( ξ + ρ )) leads to x = − Z ′ ( ρ ) ρ a ρ + O ( ρ ), hence x = ρξ ν +2 + O ( ρ ξ ν +4 ) . Moreover, as τ = ˜ u − u − V a − L V a ∂ x ( Z ( ρ a ( ξ + ρ )) − Z ( ρ a ξ ))one deduces x = V − a (˜ u − u ) − ρ a L ∂ x ( Z ′ ( ρ a ξ ) ρ ) . The linearized system is V − a ∂ t ρ + ∂ x x + ξ∂ z ( ix ) = 0 V − a ∂ t x + ∂ x x − ∂ z ( ix ) = gV a ρV − a ∂ t ( iξx ) − ∂ x ( ix ) + ξ − ∂ z (2 x − ξ − ρ ) = 0 ρ = − ( ρ a Z ′ ( ρ a ξ )) − x L ∂ x ( ρ a Z ′ ( ρ a ξ ) ρ ) + x + V a ξ ρ = 0 ∂ x x + ∂ z ( L ∂ z x + ix ) = 0 (23)13sing the new variable y such that x = L y , one deduces the relation x = ∂ y x + ρ + ξ ( y ) x ξ ( y )( ξ ( y ) + ρ ) = ∂ y x + ξ ν x + x ξ ( y ) + O ( x ρ + ρ ξ ) . Write the following normal mode expression: x x ix x x = ℜ ( x x ix x x e ikz + γ √ gkt ) . (24)Assume x x ix x x e ikz + γ √ gkt is a solution of the linearized system. This rewritesas a system of ordinary differential equations on x x ix x x . If x x ix x x is solu-tion of this linear system of ODE, then ( x , x , ix , x , x ) t is solution of (23).Note that, in this case ix is real. Remark on complex growth rates
Note that, if γ is complex, the solution x x ix x x is also complex and depend on y, k, γ . More precisely, introduce y j , z j such that x j = y j + iz j . We have thus ℜ ( x x ix x x e ikz + γ √ gkt ) | t =0 = ( y cos kz − z sin kzy cos kz − z sin kz − z cos kz − y sin kzy cos kz − z sin kzy cos kz − z sin kz and y j and z j are known through the decomposition of the initial perturbationin the even part and the odd part in z .Hence, from a complex solution ( x , ..., x ) t ( x, k, γ ) of the normal mode system,one deduces a solution of the perturbation system with a known initial condition.14 ormal modes system The equations for the normal modes associated withthe mass conservation and the momentum equation are αγρ + dx dy + αβξx = 0 dx dy − αβx + αγx + αβ ρ = 0 dx dy − αγξx + αβ ( x + ξ x + ρξ ) = 0 . (25)The normal mode formulation of the linearized energy equation is dx dy + iαβ ( − ix + iαβx ) = 0 . Let X be given by X = x x x x x . The linearized system on X is dXdy + M ( ξ, α, β, γ ) X = 0 (26)where the matrix M is given by M ( ξ, α, β, γ ) = αβξ αγξ ν +2 αγ − αβ αβ ξ ν +2 αβξ αβ − αγξ αβξ ν ξ ξ ν −
10 0 αβ − α β . (27) From now on, we will call this system the Kull-Anisimov system.
Theeigenvalues of − M are given by the classical result (see [14], [21]), and helpus to study the solution at ±∞ : Proposition 1 • The eigenvalues of − M are λ ( ξ ) = αγξ, λ a, + = αβ, λ a, − = − αβ, λ + ( ξ ) , λ − ( ξ ) (28) where (the square root is chosen of positive real part) λ ± ( ξ ) = − ξ ν ± r ξ ν αγξ ν +1 + α β (29) The eigenvalues λ , λ a, ± are called the hydrodynamic modes, the eigenval-ues λ ± ( ξ ) are called the thermal modes. A change of unknowns (that is a general ˜ X = R ( y ) X ) lead to a different set of eigenvalues,however if the matrix R ( y ) depend only on ξ ( y ) and is a C function of ξ ∈ [0 ,
1] then thelimit of the eigenvalues when y → ±∞ is the same as the limit of the eigenvalues describedabove. For ℜ γ ≥ , one has ℜ λ ( ξ ) ≥ , ±ℜ ( λ ± ( ξ )) > , and for ℜ γ > onehas ℜ λ (1) > . The matrix − M ( ξ ) has three eigenvalues of positive realpart, and two eigenvalues of negative real part. • The associated eigenvectors are given by E ( ξ ) = βi − βi − γξi E a, + ( ξ ) = − βi + ( β + γξ ) i + βi E a, − ( ξ ) = − βi + ( β − γξ ) i − βi (30) If we introduce ˜ R = γξ ξ β β , ˜ T = ξ ( γ ξ − β ) β − γ ξ + γξ β − ξ the eigenvectors F ± associated with λ ± ( ξ ) are given by F ± = i + α ˜ R + α λ ± − αγξ ˜ T + α ξ ( λ ± − αγξ ) E ( ξ ) . (31)The proof of this Proposition is straightforward, except for the sign of the realpart of λ ± ( ξ ). For this, we use( λ + ξ ν = ξ ν α β + αξ ( ℜ γ + i ℑ γ ) = ( A + iB ) where A >
A < A + iB ) = ( − A − iB ) , and A = 0 leading to( iB ) = − B is not possible when ℜ γ ≥ A − B = ξ ν α β + αξ ℜ γ, A + B = (( ξ ν α β + αξ ℜ γ ) + α ξ ( ℑ γ ) ) As λ ± ( ξ ) = − ξ ν ± ( A + iB ), ℜ ( λ + ( ξ ) λ − ( ξ )) = ( − ξ ν + A )( − ξ ν − A ) + B ,and ℜ λ + ( ξ ) ℜ λ − ( ξ ) = ℜ ( λ + ( ξ ) λ − ( ξ )) − B = − αξ ν +1 ℜ γ − α β − B <
0, oneobtains that the product ℜ λ + ( ξ ) ℜ λ − ( ξ ) is strictly negative, hence the real partsare of opposite sign, hence ℜ λ + ( ξ ) = − ξ ν + A .This calculus also defines uniquely in the case ℜ γ > λ + ( ξ )and λ − ( ξ ).Note also A ( α, β, γ ) and B ( α, β, γ ) the quantities such that A ( α, β, γ ) > A ( α, β, γ ) + iB ( α, β, γ )) = + α β + αγ .Using dξdy = ξ ν +1 (1 − ξ ), we also introduce a new system of unknowns : Y = T X = x − αγξx x − αβ ξx x − αβx ξ − ξ αβx x (32)16nd the system on Y , equivalent to (26) is dy dy − αγy + αβξy + α (1 − ξ )( β − γ β ) z + αγξx = 0 dy dy + α ( γ − β ) y − αβy + α (1 − ξ )( γ β − γβ − βξ ) z + α ξβ x = 0 dy dy + αβξ y + αβy − αγξy + α (1 − ξ )( β γ (1 − ξ ) ξ ) z + αβx = 0 dz dy + αγz αβ − ξ ( y − ξx ) = 0 dx dy + αβy = 0 , namely dYdy + αB ( ξ, β, γ ) Y = 0 (33)where Bi = − γi + ( γ − β ) i + βξ i + β − ξ i Bi = βi Bi = βξi − βi − γξi + βi Bi = (1 − ξ )[( β − γ β ) i + ( γβ ( γ − β ) − βξ ) i + ( β + γ − ξξ ) i ] + γi Bi = γξi + ξβ i + βi − βξ − ξ i . The next section is devoted to the statement of the methods used to find thesolutions at infinity for systems of ODE which coefficients depend on y , and ofthe general set-up to find solutions bounded at ±∞ . This section recalls results of the paper of Alexander, Gardner et Jones [2], aswell as the methods developed by K. Zumbrun [4], D. Serre [24], S. Benzoni-Gavage [3] and other authors. Its purpose is to study solutions with a prescribedbehavior at infinity of an ordinary linear system of differential equations. It isused in particular to identify solutions going to 0 as y → ±∞ .In the general case, we consider the ordinary differential system dydt = A ( t, α ) y, (34)when A is a regular matrix (for example analytic in α ).We notice that the vectorial product y ∧ y of two solutions y and y of (34)is solution of a new differential system on Λ ( K n ) which matrix is denoted by A (2) , because ddt ( y ∧ y ) = ( Ay ∧ y + y ∧ Ay ) . y ∧ y ∧ ... ∧ y k is solution of an ordinary differential system on Λ k (IR n )whose matrix is denoted by A ( k ) . The matrix A ( k ) is given by A ( k ) ( e i ∧ e i ... ∧ e i k ) = k X l =1 e i ∧ .. ∧ Ae i l ∧ ..e i k . (35)When the matrix A is diagonalizable, with eigenvalues λ ≤ λ < ... ≤ λ d then A ( k ) is diagonalizable and its eigenvalues are X i ∈ I,I ⊂{ ,...,d } , Card ( I )= k λ i . The largest eigenvalue of A ( k ) is P kp =1 λ d +1 − p , and its smallest eigenvalue is P kp =1 λ p .Under the hypothesis that λ d − k < λ d +1 − k , the largest eigenvalue of A ( k ) issimple. Its associated eigenvector is the vectorial product of the eigenvectorsassociated with ( e d +1 − k , .., e d ).Recall that the space Λ ( d ) (IR d ) is of dimension 1, hence the matrix associatedwith A ( d ) is a number, which is equal to Tr( A ( t, α )). The associated differentialequation is ddt ( y ∧ ... ∧ y d ) = Tr( A ( t, α ))( y ∧ ... ∧ y d ) . Hence the vectorial product of d solutions of (34) satisfies y ∧ y ∧ .. ∧ y d ( t, α ) = y ∧ y ∧ .. ∧ y d ( t , α ) exp ( Z tt Tr( A ( s, α )) ds ) . (36)whose solution is the Wronskian of d solutions of the system. Remark that, when γ is complex, one has to replace IR by K = C | butnothing will change as what is important is that we study objects ona field, which can be IR when γ is real, and which is C | when γ iscomplex. In the set-up of this paper, the matrix − M admits three eigenvalues ofpositive real part, which may be associated with the solutions going to 0 when y goes to + ∞ , and has two eigenvalues of negative real part, which help tounderstand the solutions going to 0 when y goes to + ∞ .It is in general hard to compute the solutions associated with an eigenvalueof the matrix − M . However, we may compute the solution for the matrix M (2)0 associated with the smallest eigenvalue λ a, − + λ − ( ξ ), and the solution for thematrix M (3)0 associated with the largest eigenvalue λ ( ξ ) + λ + ( ξ ) + λ a, + .We introduce from now on the base vectors in Λ ( K ) which take into accountthe role of − ξ in B : 18 = i ∧ i , f = i ∧ i , f = i ∧ i , f = i ∧ i g = i ∧ i , g = i ∧ i , g = i ∧ i g = i ∧ i , g = i ∧ i , g = i ∧ i (37)To these vectors are associated the following vectors in Λ ( K ) such that f i ∧ f ⊥ i = i ∧ i ∧ i ∧ i ∧ i = g j ∧ g ⊥ j , ∀ i, j . We have f ⊥ = i ∧ i ∧ i , f ⊥ = − i ∧ i ∧ i , f ⊥ = i ∧ i ∧ i , f ⊥ = i ∧ i ∧ i g ⊥ = i ∧ i ∧ i , g ⊥ = − i ∧ i ∧ i , g ⊥ = − i ∧ i ∧ i g ⊥ = i ∧ i ∧ i , g ⊥ = i ∧ i ∧ i , g ⊥ = − i ∧ i ∧ i . (38)Note that we shall use in the sequel the eigenvector of the matrix M (2)0 (1)associated with the eigenvalue − αβ + λ − (1): W + = F − (1) ∧ E a, − (1) = βf + ( γ − β ) f + βf − βf + G (39)where G belongs to the space generated by g j , j = 1 ..
6, and W + , = β , W + , = γ − β , W + , = β , W + , = − β . We also introduce µ ( α ) = − β + γ + αβ λ − (1) such that λ − (1) − αβ = − αµ ( α ) . (40)We notice that µ ( α ) = − β − γ + αβ + A + iB , hence ∀ α ∈ [0 , α ] , β ∈ [ β , β − ] , | γ | ≤ β − , ℜ γ ≥ , | µ ( α ) | ≤ α β − + 3 β − . (41)We construct a solution of the system dX (2) dy + M (2)0 X (2) = 0 (42)which belongs to the family of its most decreasing solutions when y goes to + ∞ .Similarily, we construct a solution of the system dX (3) dy + M (3)0 X (3) = 0 (43)belonging to the family of its most decreasing solution when y goes to −∞ .It is useful to introduce the following transformation for the study of the solutionwhen y → + ∞ : Y (2) = T (2) X (2) (44)where Lemma 3
Let P x j i j and P t j i j be two solutions of (26). Ve denote by X (2) = v f + v f + v f + v f + P j =1 w j g j a solution of (33). The associated solution Y (2) = T (2) X (2) writes Y (2) = Z f + Z f + Z f + Z f + X j =1 M j g j ith Z j = αβ ξ − ξ v j M = αβ ξv − αγξv + w M = αβv − αγξv + w M = αγξv + w M = αβv − αβ ξv + w M = αβ ξv + w M = αβv + w . For the construction of the solutions when y → −∞ , we use the followingtransformation of the unknowns ξz = x , z = αβx . (45)The system deduced from (26) is thus dz dy + ξ ν (1 − ξ ) z + αβx + γβ ξ ν +1 z = 0 dx dy + αγξz − αβx + β ξ ν +2 z = 0 dx dy + 2 αβz + αβx − αγξx + ξ ν z = 0 dz dy + αβz + ξ ν z − αβx = 0 dx dy + αβx − αβz = 0 (46)Introduce t = − αβy. We have
Lemma 4
Let Z ( t, α ) , Z ( t, α ) , Z ( t, α ) be three solutions of (46). Write Z (3) ( t, α ) = Z ( t, α ) ∧ Z ( t, α ) ∧ Z ( t, α ) = X j =1 f j ( t, α ) f ⊥ j + X p =1 g p ( t, α ) g ⊥ p . (47) The solution of (43) associated with Z (3) is w (3) ( y, α ) = f ( − αβy, α ) f ⊥ + ξ ( y ) P j =2 f j ( − αβy, α ) f ⊥ j + αβ [ P p =1 g p ( − αβy, α ) g ⊥ p + ξ P p =4 g p ( − αβy, α ) g ⊥ p ] . The proof of these two lemmas is straightforward.
Recall that the Evans function is characterized by the vectorial product of fivesolutions of the system. As it writes X (2) ∧ w (3) , we notice that, from Lemma3 and Lemma 4, we have αβξX (2) ∧ w (3) = ξ ( M g + M g + M g ) + ξ ( M g + M g + M g )+(1 − ξ )[ Z ( f − ξβ g − g ) + ξZ ( f + γβ g − g )+ ξZ ( f + γβ g − ξβ g ) + ξZ ( f − γβ g − ξβ g − g )] . (48)20et C be the limit of (1 − ξ ( y )) e y when y → + ∞ . To ensure uniqueness forthe systems (42) and (43) and to adapt the constant in the system on Z, M , weconsider the solution w (2)+ of the system (42) such that w (2)+ ( y ) e ( αβ − λ − (1)) y (49)converges to W + C αβ when y → + ∞ . Similarily, we consider the solution w (3) − ofthe system (43) such that w (3) − ( − tαβ , α ) e t + γβ R tt ∗ ξ ( − sαβ ) ds → f ⊥ + f ⊥ + f ⊥ − g ⊥ − g ⊥ − g ⊥ − g ⊥ − g ⊥ = S. (50)Let B (2) be given by B (2) f = ( γ − β ) f + βξ f +(1 − ξ )[( γ β − βξ − γβ ) g +( β + γ − ξξ ) g ] B (2) f = γf + βf +(1 − ξ )[( γ β − β ) g +( β + γ − ξξ ) g ] B (2) f = βξf − βf + γ (1 − ξ ) f − βf +(1 − ξ )[( − β + γ β ) g +( βξ + γβ − γ β ) g ] B (2) f = − γξf − ξβ f − βf + γf +(1 − ξ )[( β − γ β ) g +( − βξ − γβ + γ β ) g +( β + γ − ξξ ) g ] B (2) g = − β − ξ f − γg + βg − βξ g B (2) g = − β − ξ f − βg − γ (1 + ξ ) g + βg +( γ − β ) g B (2) g = β − ξ ( − ξf + f ) + ξβ g + βg − γg +( γ − β ) g + βξ g B (2) g = − βξg − γξg + βg B (2) g = − βξ − ξ f − γξg + βg + βg B (2) g = − βξ − ξ f − γξg + βξg − ξβ g − βg − γξg . (51)The system on Y (2) is dY (2) dy + αB (2) Y (2) = 0 (52)The Evans function that we shall use is given by: Definition 2
Introduce Ev ( α, β, γ ) = αβw (2)+ ( y ) ∧ w (3) − ( y ) e − R y ( αγξ ( y ′ ) − ( ξ ( y ′ )) ν ) dy ′ . This function, independant of y , is called the Evans function of the problem.
21t is easy to derive the
Proposition 2
The complex number γ is an instability growth rate accordingto Definition 1 if and only if Ev ( α, β, γ ) = 0 . The proof of Proposition 2 is to be found in [2].
Reduction of the Evans function
We deduce from the systems (42) and(43) that w = w (2)+ ∧ w (3) − is solution of (36). From Tr M = − αγξ + ξ ν = − αγξ + ˙ ξξ + ˙ ξ − ξ we deduce that the derivative of w ( y ) e − R y ( αγξ ( y ′ ) − ( ξ ( y ′ )) ν ) dy ′ is zero, hence ξ − ξ w ( y ) e − R y αγξ ( y ′ ) dy ′ constant.Using (48), we obtain Ev ( α, β, γ ) = [ Z ( f − ξβ g − g ) + ξZ ( f + γβ g − g )+ ξZ ( f + γβ g − ξβ g ) + ξZ ( f − γβ g − ξβ g − g )+ ξ M g + M g + M g − ξ + ξ M g + M g + M g − ξ ] e − R y αγξ ( y ′ ) dy ′ − ξ (0) ξ (0) . (53) The aim of this paper is to compute the Evans function through the calculation of w (2)+ and w (3) − . Theorem 1 • For all ξ > , there exists α > , β > such that, for α ≤ α , β ∈ [ β , β ] , γ ∈ [0 , β ] and y such that ξ ( y ) ∈ [ ξ , there exists a uniquesolution w (2)+ of (42) satisfying (49). • Let ˜ w = T (2) w (2)+ ( y ) exp( − αµ ( α ) y ) − ( W + , f + W + , f + W + , f + W + , f ) . (54) The function ˜ w is analytic in ( y, α ) for ξ ( y ) ∈ [ ξ , and for α ≤ α . For all α > and for all ξ ∈ ]0 , there exists a constant C ( ξ , α ) such that ∀ y, ξ ( y ) ∈ ] ξ , , | ˜ w ( y ) | ≤ C ( ξ , α )(1 − ξ ( y )) . Note that the techniques of differential equations allow us only to compute the solutionof a differential system with non constant coefficients only when this solution is associatedwith the largest eigenvalue of the matrix in the neighborhood of −∞ and with the smallesteigenvalue of the matrix in the neighborhood of + ∞ . There exists α ≤ α and R > , depending only on ξ , such that for ζ < R , w admits an analytic extension for y such that y ∈ [( αζ ) ν , ξ ] . • For α > and β > given, and for all α ≤ α , β ∈ [ β , β ] , γ ∈ [0 , β ] ,there exists t > such that there exists a unique solution w (3) − ( y, α ) of (43) for − αβy ∈ [ t , + ∞ [ satisfying (50). Moreover we have w (3) − ( − tαβ , α ) exp( − Z − tαβ ( λ + + α ( β + γξ ( y ′ ))) dy ′ ) − W ( t ) e t t − ν = O ( α ν )(55) uniformly for t ∈ [ t , + ∞ [ . We see in this theorem the division in three regions for the computation of thesolutions of (42) and of (43). The first region ξ ( y ) ∈ [ ξ ,
1[ is the aim of Section3, the second region (which extends the result of [ ξ , −∞ is characterized in Section 5. Inthe next paragraph, we describe the systems that we shall use in what follows.One of the main problems is that the problem to solve is not a Cauchy problem,but we have informations at ±∞ for the solution of (26) that we want to study.Let us say a word on this system. As ξ goes to 0 when y goes to −∞ , we noticethat the matrix M is singular when y goes to −∞ .Moreover, the term αβ prevents us to have a result which is uniform when β goesto 0. A possible choice to overcome this difficulty is the choice β ∈ [ β , β ].Finally, even if we remove the singularity of M at y → −∞ by whatevermethod, another problem is induced by the behavior of ξ when y goes to −∞ ,because ξ goes to 0 as | y | − ν . This will lead to a fuchsian problem (Section 5).Note that in the theorem 1 we defined unique solutions of the problems (42)and (43) with the prescribed behavior at infinity.The behavior of the solutions induced by the theorem 1 lead to another expres-sion of the Evans function of relation (53).Assume that y < t = − αβy . We introduce the functions L p ( t, α ) , R j ( t, α ) given by the equalities g p ( t, α ) e t + γβ R tt ξ ( − t ′ αβ ) dt ′ t − ν = L p ( t, α ) , ≤ p ≤ f j ( t, α ) e t + γβ R tt ξ ( − t ′ αβ ) dt ′ t − ν = R j ( t, α ) , ≤ j ≤ . (56)The corrective factor e t + γβ R tt ξ ( − t ′ αβ ) dt ′ t − ν is induced by the relation (55).Similarily, we introduce z j ( y, α ) = Z j ( y, α ) e − αµ ( α ) y , m p ( y, α ) = M p ( y, α ) e − αµ ( α ) y . (57)23e obtain the relation Ev ( α, β, γ ) = e − (2+ µ ( α ) β ) t t ν exp( − R − t αβ αγξ ( y ′ ) dy ′ )[ z ( R − ξβ L − L ) + ξz ( R + γβ L − L )+ ξz ( R + γβ L − ξβ L ) + ξz ( R − γβ L − ξβ L − L )+ ξ m L + m L + m L − ξ + ξ m L + m L + m L − ξ ] − ξ (0) ξ (0) . (58)In what follows we shall describe the functions R , L , and z , m . The aim of this section is to obtain w (2)+ , which is the unique solution of thesystem (42) under the condition (49) in the region [ y , + ∞ [ for all y . Thefirst idea would be to try to apply the Gap Lemma. However, the eigenvalue ofsmallest real partof − M (2)0 (1) is λ min = λ − (1) − αβ , and the next eigenvalueis λ ∗ = λ − (1) + αγ . As λ ∗ − λ min = α ( γ + β ), the difference between twoeigenvalues of M (2)0 (1) is not uniformly bounded below for α ∈ ]0 , α [. Hencethe hypotheseses of the Gap lemma theorem are not fulfilled. Remark
As the coefficients of the differential system behave as 1 − e − y when y goes to + ∞ , one may obtain Y (2) through a Volterra expansion as Y (2) ( y ) = Ae ( λ − (1) − αβ ) y + e − y r ( y, α, β, γ ) r being a remainder term. We prove that the result that we obtain is analyticin α for β , γ in a certain compact set. Once we know that, we identify w (2)+ (deduced from Y (2) and obtained also through a Volterra expansion) and weexpress it with an expansion in powers of α , which is bounded by a geometricseries for ξ ( y ) ∈ [ ξ , α j in the expansion in α of the solution isof the form ξ νj +2 X j ( ξ ), the radius of convergence of the series depends on ξ ( y ).The aim of the explicit calculus is to deduce a new region of convergence of theseries using the behavior in α j ξ νj +2 X j ( ξ ). In this new region of convergence, wehave a converging series, which coincides with the original one for ξ ∈ [ ξ ( y , −∞ and the most decreasing solution at + ∞ .24e introduce z j and m p through T (2) w (2)+ ( y ) = αβ ( X j =1 z j f j + X p =1 m p g p ) e αµ ( α ) y where µ ( α ) has been given in (40). We recall that z j → W + ,j and m j → y → + ∞ .The following Theorem summarizes the results of this section. Let us introduce ξ > Theorem 2
1. The functions z j and m j have a normally convergent expan-sion in powers of α of the form: z ( y, α ) = P j ≥ α j ξ νj a ,j ( ξ ) + W + , z k ( y, α ) = P j ≥ α j ξ νj +1 a k,j ( ξ ) + W + ,k , k = 2 , , m l ( y, α ) = P j ≥ α j ξ νj +1 b l,j ( ξ ) , l = 1 , , m p ( y, α ) = P j ≥ α j ξ νj +2 b p,j ( ξ ) , p = 4 , , . We introduce δ = 0 , δ = δ = δ = d = d = d = 1 , d = d = d = 2 .Let K be a compact subset of IR ∗ + × IR . There exists R > and α ( ξ ) such that for β, γ in K and α < α ( ξ ) we have for ξ ∈ [ ξ , | ddξ ( a k,j ( ξ ) ξ νj + dk ) | ≤ R j ξ ν j + d k +1 | a k,j ( ξ ) | ≤ R j (1 − ξ ) , | ddξ ( b l,j ( ξ ) ξ νj + δl ) | ≤ R j ξ νj + δl +1 , | b l,j ( ξ ) | ≤ R j (1 − ξ ) . Assume this theorem is proven. We show that, for all ξ , there exists α ( ξ )such that, for α < α ( ξ ) the power series is convergent for ξ ∈ [ ξ , αξ ν R < αξ ν M <
1. The power series P j ≥ a p,j ( ξ ) α j ξ νj + δp defines ananalytic function which is the analytic extension of the sum of the normallyconvergent series P j ≥ α j ξ νj + dc a p,j ( ξ ) for α < α ( ξ ) and ξ ∈ [ ξ , P j ≥ b l,j ( ξ ) α j ξ νj + δl .The proof of item 1) of the theorem is in Annex 7. We calculate in the firstsubsection of the present section the first order terms of z and m .In the second subsection, we prove by recurrence the structure of the j − th termof the expansion of z and m , which is a consequence of the structure of thesystem. We deduce the precise estimates wich help us to extend the expansion. We introduce Z j , ≤ j ≤ m p , 1 ≤ p ≤ αβ T (2) w (2)+ ( y, α ) = y (2)+ ( y, α ) = X j =1 Z j f j + X p =1 M p g p = e αµ ( α ) y ( X j =1 z j f j + X p =1 m p g p ) . (59)The relation w (2)+ e y e − αµ ( α ) y → F − (1) ∧ E a, − (1) = W + imply that z j → W + ,j because ξ (1 − ξ ) e y → y → + ∞ . The system on Z, M writes ( dZdy + αξ J Z + αB ( ξ ) Z + αβ − ξ ( L + ξL ) M = 0 dMdy + αξ K M + αD ( ξ ) M + α − ξξ C ( ξ ) Z = 0 . (60)that is α dZ dy + βξZ − γξZ − βξM − ξ = 0 α dZ dy + ( γ − β ) Z + γZ − βZ − ξβ Z − β ( M + ξM )1 − ξ = 0 α dZ dy + βξ Z + βZ + γ (1 − ξ ) Z − βZ − β ( M + ξM )1 − ξ = 0 α dZ dy − βZ + γZ + βM − ξ = 0 α dM dy − γM − βM + ξβ M + ξ ( − βM − γM )+(1 − ξ )[( γ β − γβ − βξ ) Z + γ − β β Z ] = 0 α dM dy + βM − γ (1 + ξ ) M + βM − γξM +(1 − ξ )[( β + γ − ξξ ) Z + γ − β β Z ] = 0 α dM dy + βM − γM + βξM + (1 − ξ ) β − γ β Z = 0 α dM dy − βξ M + ( γ − β ) M − γξM + βM − ξβ M +(1 − ξ )[ Z ( β + γ − ξξ ) + Z ( βξ + γβ − γ β )] = 0 α dM dy +( γ − β ) M + βM − βM + (1 − ξ ) Z ( − βξ + γβ ( γ − β )) = 0 α dM dy + βξ M + βM − γξM + (1 − ξ ) Z ( β + γ − ξξ ) = 0 . (61)This system writes 1 α dUdy + βK ( r, ξ ( y ) , β ) U = 0 (62)where U t = ( Z , Z , Z , Z , m , m , m , m , m , m ). The system on ( z, m ) isdeduced from (62) by replacing the matrix βK ( r, ξ ( y )) by βK ( r, ξ ( y )) + µ ( α ) I .We verify that m p → y → + ∞ . In what follows, we describe the analyticexpansion of the solution in α when α is in a neighborhood of 0. α From z j → W + ,j , m j → z j = W + ,j , m p = 0, that is z = β, z = γ − β, z = β, z = − β . Wereplace in the system these relations to obtain the system on the term z j and m j . We get 26 dz dy + µ (0) z + βξz − γξz = 0 dz dy + µ (0) z + ( γ − β ) z + γz − βz − ξβ z = 0 dz dy + µ (0) z + βξ z + βz + γ (1 − ξ ) z − βz = 0 dz dy + µ (0) z − βz + γz = 0 dm dy + (1 − ξ )[( γ β − γβ − βξ ) z + γ − β β z ] = 0 dm dy + (1 − ξ )[( β + γ − ξξ ) z + γ − β β z ] = 0 dm dy + (1 − ξ ) β − γ β z = 0 dm dy + (1 − ξ )[ z ( β + γ − ξξ ) + z ( βξ + γβ − γ β )] = 0 dm dy + (1 − ξ ) z ( − βξ + γβ ( γ − β )) = 0 dm dy + (1 − ξ ) z ( β + γ − ξξ ) = 0As µ (0) = − β − γ we deduce dz dy − β ( β + γ )(1 − ξ ) = 0 dz dy + ξ − dz dy + β (1 − ξ ) ξ + γβ (1 − ξ ) = 0 dz dy = 0 dm dy + (1 − ξ )[( γ β − γβ − βξ ) β + γ − β β ( γ − β )] = 0 dm dy + (1 − ξ )[( β + γ − ξξ ) β + γ − β β β ] = 0 dm dy − (1 − ξ ) β − γ β β = 0 dm dy + (1 − ξ )[( γ − β )( β + γ − ξξ ) + β ( βξ + γβ − γ β )] = 0 dm dy − β (1 − ξ )( − βξ + γβ ( γ − β )) = 0 dm dy − β (1 − ξ )( β + γ − ξξ ) = 0We obtain z = β ( β + γ ) ξ ν − νξ ν z = ξ ν − νξ ν z = γβ − ξ ν νξ ν + β ν +1 1 − ξ ν +1 ξ ν +1 z = 0and the following system on m j : dm dy + ˙ ξξ ν +2 [( γ ξ − γβ ξ − β ) + γ − β β ( γ − β ) ξ ] = 0 dm dy + ˙ ξξ ν +2 [ ξ + γβ (1 − ξ ) + ( γ − β ) ξ ] = 0 dm dy − ˙ ξξ ν +1 ( β − γ ) = 0 dm dy + ˙ ξξ ν +2 [( γ − β )( ξβ + γ (1 − ξ )) + ( β + γβ ξ − γ ξ )] = 0 dm dy − ˙ ξξ ν +2 ( − β + γξ ( γ − β )) = 0 dm dy − ˙ ξξ ν +2 ( ξ + γβ (1 − ξ )) = 027ence we obtain the expansion of the solution for ξ ≥ ξ z = β − ανξ ν (1 − ξ ν ) β ( β + γ ) + O ( α ) z = γ − β − ανξ ν (1 − ξ ν ) + O ( α ) z = β + α ( ν +1) ξ ν +1 β (1 − ξ ν +1 ) + ανξ ν γβ (1 − ξ ν ) + O ( α ) z = − β + O ( α )The next subsection is dedicated to the precise study of the behavior of w (2)+ ,which depends on inverse powers of ξ and cannot be extended directly to ξ → w (2)+ for ξ ∈ [ ζ α ν , We write z = W + , f + W + , f + W + , f + W + , f + u. (64)We introduce the constant matrices J , L and K such that J = J + ξB ( ξ ) , L = L + ξL , K = K + ξD ( ξ ). The system (9) yields ( dzdy + αµ ( α ) z + αξ Jz + αβ − ξ Lm = 0 dmdy + αµ ( α ) m + αξ Km + α − ξξ Cz = 0 (65)The aim of this paragraph is to find a simpler formulation for the uniquesolution of this system.As the solution going to 0 at infinity of dfdy = − ξξ is f ( y ) = ν +1 (1 − ξ ν +1 ),direct estimates of the behavior in ξ of a coefficient u j or m j of (65) lead to amultiplying factor of the form αξ ν +1 . The behavior of α j +1 u j +1 or of α j +1 m j +1 in ξ is then given by α j +1 ξ ( ν +1) j for thenext coefficient. However, the structure of the system allows us to obtain alower inverse power of ξ ν in the expansion in α . For this purpose, we introducethe new unknowns a p,j , b q,j such that z = W + , + P Nj =1 α j ξ νj a ,j ( ξ ) + z N +11 ( α, ξ ) z p = W + ,p + P Nj =1 α j ξ νj +1 a p,j ( ξ ) + z N +1 p ( α, ξ ) , p = 2 , , m k = P Nj =1 α j ξ νj +1 b k,j ( ξ ) + m N +1 k ( α, ξ ) , k = 1 , , m l = P Nj =1 α j ξ νj +2 b l,j ( ξ ) + m N +1 l ( α, ξ ) , l = 4 , , a p,j +1 is of order ξ when every a k,j , b l,j is bounded when ξ goes to 0. This is not the case, and the crucialequality states as follows. We introduce the diagonal matrix T such that28 a a a a b b b b b b = a ξ − a ξ − a ξ − a ξ − b ξ − b ξ − b ξ − b ξ − b ξ − b . There exists a matrix ˆ C ( ξ, β, γ ), polynomial in ξ , such that ξ J β − ξ L − ξξ C ξ K ! T = T ˆ C. (67)We shall make use of the following fundamental Lemma, noting that at eachstep we solve an equation of the form dfdy = A ( ξ )(1 − ξ ) ξ α where α = νj + d , d = 0 , , Lemma 5
The unique solution going to 0 when ξ goes to 1 of dfdy = A ( ξ )(1 − ξ ) ξ α is f ( y ) = R ξ A ( η ) η α + ν +1 dη . We have the estimate | f ( y ) | ≤ − ξξ ν + α || A || ∞ . Proof
From ξ ν +1+ α = ddξ ( ν + α (1 − ξ ν + α )), we deduce | f ( y ) | ≤ || A || ∞ Z ξ ddξ ( 1 ν + α (1 − η ν + α )) dη = || A || ∞ − ξ ν + α ( ν + α ) ξ ν + α . The equality ξ ν + α − ξ − = R ( ν + α )(1 + t ( ξ − ν + α − dt implies | ξ ν + α − | ≤ ( ν + α )(1 − ξ ) . (68)hence the lemma. The indices defined in Theorem 2 will express the weight ofeach coordinate of the vector U defined in (62) T (cid:20) ξ νj (cid:18) a j b j (cid:19)(cid:21) = a p,j ξ νj + δj b q,j ξ νj + dq ! .
29n the system (66), write (cid:18) zm (cid:19) = (cid:18) W + (cid:19) + X j ≥ α j ξ νj T (cid:18) a j b j (cid:19) we get, for the term in α j +1 : ddy ( 1 ξ ν ( j +1) T (cid:18) a j +1 b j +1 (cid:19) )+ µ j (cid:18) W + (cid:19) + j X l =1 µ j − l ξ νl T (cid:18) a l b l (cid:19) + 1 ξ νj T C (cid:18) a j b j (cid:19) = 0 . This system of equations becomes, for j ≥ ddy ( a p,j +1 ξ δp + ν ( j +1) ) + ( P jl =0 µ l a p,j − l ξ νl )+( C a j + C bj − ξ )) p ξ δp + νj + µ j − W + ,p = 0 ddy ( b p,j +1 ξ dp + ν ( j +1) ) + ( P jl =0 µ l b p,j − l )+ C (1 − ξ ) a j + C b j ) ξ δp + ν ( j +1) = 0 (69)and the equality for j = 1 ddy ( a p,j +1 ξ δp + ν ( j +1) ) + ( P jl =0 µ l a p,j − l ξ νl )+( C a j + C bj − ξ )) p ξ δp + νj + (1 − ξ ) h p = 0 ddy ( b p,j +1 ξ dp + ν ( j +1) ) + ( P jl =0 µ l b p,j − l )+ C (1 − ξ ) a j + C b j ) ξ dp + ν ( j +1) + (1 − ξ ) h p = 0 . (70)We have the identity C (1) b ′ j (1) = h j (1) + µ j − W + ,j , which is necessary toobtain that the source term in the equation on a p,j +1 vanishes at ξ = 1. The regularity of the quantities a p,j , b q,j is given by the following proposition,which gives precise estimates on the functions provided that β and γ stay in acompact set: Proposition 3
Assume that β, γ are in a compact set K of IR ∗ + × C | , namely β ≤ β ≤ β , | γ | ≤ β , ℜ γ ≥ . Assume that ξ ∈ ]0 , is given and that there exists α ( ξ ) such that the analyticexpansion of the solution of (65) is valid for α < α ( ξ ) and ξ ( y ) ∈ [ ξ , .There exists R and M depending only of α and β such that, forall p =1 , , , , forall q = 1 , ..., , for all j ≥ we have the estimates | a p,j ( ξ ) | ≤ R j (1 − ξ ) | b q,j ( ξ ) | ≤ R j (1 − ξ ) If we introduce α j = a j (0) and β j = b j (0) we have | a j ( ξ ) − α j (1 − ξ ) | ≤ M j ξ (1 − ξ ) | b j ( ξ ) − β j (1 − ξ ) | ≤ M j ξ (1 − ξ ) . Proposition 4
Under the same hypothesis as the proposition 3,i) there exists α ( ξ ) such that, for α < α ( ξ ) , the functions A p ( ζ, α ) = P ∞ j =1 a p,j (( αζ ) ν ) ζ j B q ( ζ, α ) = P ∞ j =1 b q,j (( αζ ) ν ) ζ j are analytic for ζ < R .2) For y such that α ν R ν < ξ ( y ) ≤ the functions u p ( y, α ) = W + ,p + ξ − δ p A p ( α ( ξ ( y )) ν , α ) , v q ( y, α ) = ξ − d q B q ( α ( ξ ( y )) ν , α ) are solution of the system (65) and extend the gap lemma solution.3) Introduce ˜ A p ( ζ, α ) = ∞ X j =1 ζ j a p,j (( αζ ) ν ) − α p,j (1 − ( αζ ) ν ) ζ j . ˜ B q ( ζ, α ) = ∞ X j =1 ζ j b q,j (( αζ ) ν ) − β q,j (1 − ( αζ ) ν ) ζ j . The functions ˜ A p and ˜ B q are analytic for ζ < M , and we have the inequalitiesfor ζ < M | ˜ A p ( ζ, α ) | ≤ ( αζ ) ν ζ − M ζ . | ˜ B q ( ζ, α ) | ≤ ( αζ ) ν ζ − M ζ = α ν ζ − ν − M ζ .
We deduce, for α ≤ α ( ξ ) and ζ ≤ M , the equalities U ( ζ, α ) = W + , + P ∞ j =1 α ,j ζ j + α ν R ( ζ, α ) α ν U p ( ζ, α ) = ζ ν P ∞ j =1 α p,j ζ j + α ν R p ( ζ, α ) , p = 2 , , α ν V q ( ζ, α ) = ζ ν P ∞ j =1 β q,j ζ j + α ν S q ( ζ, α ) , q = 1 , , α ν V q ( ζ, α ) = ζ ν P ∞ j =1 β q,j ζ j + α ν S q ( ζ, α ) , q = 4 , , roof of Proposition 4 Let ξ ∈ ]0 ,
1[ be given. There exists α ( ξ ) suchthat, for α ≤ α ( ξ ), the solution satisfies the conclusions of the gap lemma(which means that the solution of the system (65) is analytic in the region α ≤ α ( ξ ) for ξ ∈ [ ξ , ξ there exists R depending only on β such that for ζ < R the functions A p ( ζ, α ) and the functions B q ( ζ, α ) given inProposition 4 are analytic through their expansion in α for ζ < R .Introduce α ( ξ ) = min( α ( ξ ) , ξ ν R ). For α < α ( ξ ) and ζ < R , ξ = ( αζ ) ν ≥ ξ .This means that( w + ,p + ( ζα ) δpν A p (( αζ ) ν , α ) , ( ζα ) dqν B q (( αζ ) ν , α ))is solution of the system (65) for α ≤ α ( ξ ) when y is given by ξ ( y ) = ( αζ ) ν byconstruction of the analytic solution given by the gap lemma.By uniqueness of the solution which is analytic in α , we check that this func-tion is also solution of the system (65) for α < α ( ξ ) and ξ ( y ) ∈ [( αR ) ν , ξ ],because the analytic expansion defining the solution in the set-up of the gaplemma can be rearranged and the remainder term is regular enough (and uni-formly bounded), and because the two solutions are equal at the point ˜ y suchthat ξ (˜ y ) = ξ .Hence we extended the solution for ξ in the interval [( αR ) ν , Proof of Proposition 3
We prove by recurrence the inequalities (71), (72),(73) below: | ddξ ( a p,j ξ δ p + νj ) | ≤ R j ξ νj +1+ δ p . (71) | ddξ ( b q,j ξ d q + νj ) | ≤ R j ξ νj +1+ d q . (72) | ddξ ( b q,j ξ d q + νj )( ξ ) − ddξ ( b q,j ξ d q + νj )(1) | ≤ R j (1 − ξ ) ξ νj +1+ d q . (73)We assume that ( β, γ ) ∈ K ⊂ [ β , β ] × { γ, | γ | ≤ β − , ℜ γ ≥ } . First step:
From the identity a p,j ( ξ ) ξ − ξ δ p + νj Z ξ ddξ ( a p,j ξ δ p + νj )(1 + s ( ξ − ds and from the same identity on b j , relying on the fact that a j and b j are 0 at ξ = 1 we deduce the inequality | a p,j ( ξ ) | ≤ R j ξ δ p + νj Z ξ (1 − ξ ) ds (1 + s ( ξ − νj +1+ δ p .
32e use (68) to obtain | a p,j ( ξ ) | ≤ R j (1 − ξ ) , or | a p,j ( ξ )1 − ξ | ≤ R j (74) | b q,j ( ξ )1 − ξ | ≤ R j (75)and of course | db p,j dξ (1) | = | ddξ ( b p,j ξ d p + νj )(1) | ≤ R j . (76)We assume that (71), (72), (73) are true for l ≤ j . We deduce the inequalities(74), (75), (76).The equation on a p,j +1 rewrites ξ ν +1 (1 − ξ ) ddξ ( a p,j +1 ξ δp + ν ( j +1) ) + ξ νj + δp [ C ( ξ ) a j ( ξ ) + P jl =0 µ l ξ νl a p,j − l +( C b j − ξ + C (1) b ′ j (1))] . Recall that we have the equality − µ ( α ) = β + 2 γ + αβ p α ( γ + αβ )hence for ( β, γ ) in the compact K , the function µ ( α ) admits a DSE at α = 0,of radius of convergence greater than θ = min(1 , β ). Moreover, denoting by C = P ∞ l =0 | µ l | θ l , we have ∞ X l =0 | µ l | θ l ≤ C . Using the norm of the matrices C + µ I , C , C and C + µ I (with 0 ≤ ξ ≤
1) and the inequalities (74), (75), (76) we obtain | ddξ ( a p,j +1 ξ δp + ν ( j +1) ) | ≤ ξ ν ( j +1)+1+ δp [ P jl =1 µ l R j − l ξ νl + (3 | γ | + 6 β + | γ | +3 β + | γ | β ) R j ] ≤ R j ξ ν ( j +1)+1+ δp [ C + 3 | γ | + 6 β + | γ | +3 β + | γ | β ]as soon as R − ≤ min(1 , β . (77)We have the same type of estimates for b q,j +1 : | ddξ ( b q,j +1 ξ d q + ν ( j +1) ) | ≤ R j ξ ν ( j +1)+1+ d q [ C + 3 | γ | + 6 β + 2 | γ | + 3 β + | γ | β ] . D ( β ) depending only on the compact set K , suchthat for R satisfying (77) and assuming (71), (72), (73) at the order j , we obtainthe estimates | ddξ ( a p,j +1 ξ δ p + ν ( j +1) ) | ≤ R j D ( β ) ξ ν ( j +1)+1+ δ p | ddξ ( b q,j +1 ξ d q + ν ( j +1) ) | ≤ R j D ( β ) ξ ν ( j +1)+1+ d q . The last estimate that we need is based on the difference of derivatives and weuse the identity f ( ξ ) ξ α (1 − ξ ) + f ′ (1) = 1 ξ α Z ( f ′ (1) − f ′ (1 + s ( ξ − ds − f ′ (1) ξ α (1 − ξ α )for f (1) = 0 and f ∈ C . Assume the inequalities | f ′ ( ξ ) − f ′ (1) | ≤ C (1 − ξ ) ξ β +1 and | f ′ (1) | ≤ C . We obtain | f ( ξ ) ξ α (1 − ξ ) + f ′ (1) | ≤ (1 − ξ ) C (1 + α ) ξ α + β . We apply this estimate for f ( ξ ) = b p,j − l ξ dp + νl and α = ν + 1 ( β = d p + ν ( j − l )) toobtain the inequality | b p,j − l (1 − ξ ) ξ ν ( j − l )+ d p + ν +1 + b ′ p,j − l (1) | ≤ ξ ν +1 R j − l (1 − ξ ν ( j − l )+ d p )( ν ( j − l ) + d p ) ξ ν ( j − l )+ d p + R j − l ( ν +1) 1 − ξξ ν +1 . We thus deduce the estimate | ddξ ( b q,j +1 ξ d q + ν ( j +1) )( ξ ) − ddξ ( b q,j +1 ξ d q + ν ( j +1) )(1) | ≤ R j D ( β )(1 − ξ ) ξ ν ( j +1)+1+ d q . We use the estimate for j = 1, for which there exists a constant D ( β ) suchthat | ddξ ( a p, ξ δ p + ν ) | ≤ D ( β ) ξ ν +1+ δ p , | ddξ ( b q, ξ d q + ν ) | ≤ D ( β ) ξ ν +1+ d q | ddξ ( b q, ξ d q + ν )( ξ ) − ddξ ( b q, ξ d q + ν )(1) | ≤ D ( β )(1 − ξ ) ξ ν +1+ d q . It is then enough to consider R = max(1 , β , D ( β ) , D ( β ))to obtain the inequalities (71), (72), (73) for all j . We write each term a p,j = α p,j + ξc p,j , b q,j = β q,j + ξc q,j , and we have similar inequalities for the terms c p,j and d q,j , with a coefficient M depending only on the compact set K . Toobtain the estimate of the rest (versus the leading order term), we denote by34 j and d j the functions such that a j,p ( ξ ) = (1 − ξ )( α j,p + ξc j,p ( ξ )), b q,j ( ξ ) =(1 − ξ )( β j,q + ξd q,j ( ξ )). We prove in a similar fashion the inequalities | ddξ ( c p,j ( ξ ) ξ δ p + νj ) | ≤ M j ξ νj + δ p (78) | ddξ ( d j,q ( ξ ) ξ d q ) | ≤ M j ξ νj + d q (79) | ddξ ( d j,q ( ξ ) ξ d q )( ξ ) + d ′ j,q (1) | ≤ M j (1 − ξ ) ξ νj + d q . (80)This ends the proof of Proposition 3. Before studying the coupling between the hypergeometric region and the over-dense region, we shall in this section study a simple model where the profile ofdensity is ξ ( y ) = (cid:26) ξ ( y ) , y ∈ ] − ∞ , y ] ξ ( y ) , y ∈ ] y , + ∞ [ . It is a slightly better model than the discontinuity model (see [21]) for tworeasons:i) we assume that the density profile is continuous,ii) it corresponds to a simple form of the energy equation.The stationary associated quantities ρ ∗ and u ∗ are solution of: ρ ∗ ( x ) u ∗ ( x ) = − ρ a V addx ( ρ ∗ ( x ) u ∗ ( x ) + p ∗ ( x )) = − ρ ∗ ( x ) g ddx ( u ( x ) − L V a ddx ( Z ( ρ ∗ ( x )))) = − V a Z ′ ( ρ a ζ ) ρ a ξ ′ ( y ) δ x − y L coming from ∂ t ρ + div( ρ~u ) = 0 ∂ t ( ρ~u ) + div( ρ~u ⊗ ~u + pId ) = ρ~g div( ~u + L V a ∇ Z ( ρ )) = V a − ξ ξ δ x − y L . (81)It is easy to see that the stationary solution is thus given by u ( y ) = ( − V a ξ , y ∈ ] − ∞ , y ] − V a ξ ( y ) , y ∈ ] y , + ∞ [ p ( y ) = ( − ρ a V a ξ − ρ a gξ L ( y − y ) , y ∈ ] − ∞ , y ] − ρ a V a ξ ( y ) − ρ a g R yy ξ ( s ) ds, y ∈ [ y , + ∞ [We have 35 roposition 5 The linear growth rate associated with the system (81) in thecase y fixed independant of α in the regime α → and β, γ fixed is γ = s − ξ ξ − βξ which corresponds to ¯ γ = r gk ρ a − ρ ρ a + ρ − kV blowoff . We recall that for all ξ c : E ( ξ c ) ∧ E a, + ( ξ c ) ∧ F + ( ξ c ) γξ c − β = E ∧ E a, + γξ c − β ∧ ( i + α ˜ R + α λ + − αγξ c )= − β ( ξ c g ⊥ + g ⊥ + ξ c g ⊥ + g ⊥ ) − ( γξ c + 2 β ) g ⊥ + α [( ξ c + β ) f ⊥ + ξ c ( ξ c + ( β + γξ c ) ) f ⊥ − βγξ c ( f ⊥ + f ⊥ )]+ α λ + ( ξ c ) − αγξ c [ ξ c ( γξ c + β )( f ⊥ + ξ c f ⊥ ) + ( β − γ ξ c )( − ξ c ( γξ c + β ) f ⊥ + β ( f ⊥ + ξ ( f ⊥ + f ⊥ )))]Note that the leading order term in α comes from the coefficient of i and of˜ T . As the leading order term in α of the solution in the region [ y , + ∞ [ inthe case where y is independant of α is given by the leading order term of f − (1) ∧ E a, − (1), we find i ∧ E a, − (1) = W + as leading order term. The Evansfunction is Ev (0 , β, γ, ξ ) = β ( β − γξ ) ξ ν β − γ ξ [ β ( ξ − T + T ) + (2 β + γ ( ξ + 1)) T ]hence Ev (0 , β, γ, ξ ) = β ( β − γξ ) ξ ν β − γ ξ [ ξ (1 − ξ ) − (1 + ξ )( β + γξ ) ] . The theorem 5 is proven.
Remark
This result is somewhat surprising: in fact when ξ → ξ must verify ξ + ( β − ξ + β > β > max ξ ∈ ]0 , ( ξ
20 1 − ξ ξ ) and is satisfied only in a region ξ − ( β ) , ξ + ( β ) for β < max ξ ∈ ]0 , ( ξ
20 1 − ξ ξ ).We may thus conclude that this model is not relevant according to the phys-ical results. 36 Solution in the hypergeometric region
In this section, we identify the solution when y goes to −∞ . We show a result for t ≤ t finite. As the coefficients behave as | y | − ν , we call this problem a Fuchsianproblem, and, as the leading term of the equation leads to a hypergeometricalequation, we call this the hypergeometrical set-up. We introduce η ( t ) = ξ ( − tαβ ) α ν , p ( t ) = α ν γβ η ( t ) . (82)Let h ( p ) = p X k =1 νν − k ( βγ )) k p k − − ν ( βγ ) ν p ν − [ h { ν } ( βpγ ) + y ] . (83)Note that h is a complex valued function because p is complex, but as pγ − and γh ( p ) are real, ph ( p ) is also real. From lemma 1, we deduce( η ( t )) ν β = 1 νt (1 + ph ( p )) . (84)The system of equations on Z = z x x z x obtained from (46) is − dz dt + η ν β (1 − α ν η ) z + x + α ν γβ η ν +1 z = 0 − dx dt + pz − x + α ν β η ν +2 z = 0 − dx dt + 2 z + x − px + η ν β z = 0 − dz dt + z + η ν β z − x = 0 − dx dt + x − z = 0 . (85)Introduce the matrices M ( p ) = p − − p −
10 0 1 − , N Z = z z z . Using the equalities η ν β = νt + α ν ( νt )
1+ 1 ν γβ β ν h ( p )(1 + ph ( p )) ν α ν η ν +1 β = α ν ( νt )
1+ 1 ν (1 + ph ( p )) ν β ν − α ν β η ν +2 = pγ α ν η ν +1 β M ( p, β, γ ) such that (85) rewrites − dZdt + M ( p ) Z + 1 νt N Z + α ν ( νt ) ν M ( p, β, γ ) Z = 0 . It is then natural to introduceWe associate with the system (85) the model system d ˜ Zdt = M (0) ˜ Z + 1 νt N ˜ Z (86)and the model system of the vectorial product of three solutions of (86) d ˆ Zdt = M (0) (3) ˆ Z + 1 νt N (3) ˆ Z. (87)We need the following Lemma 6
The matrix M ( p ) has three eigenvalues − p, , − . The eigenvectorasociated with λ ( p ) = − p is e ( p ) = (1 , − , − p, , . The eigenvectors as-sociated with the eigenvalue of multiplicity 2 λ − = − are e − ( p ) = (1 , − − p, − , , and f − = (0 , , , , . The eigenvectors associated with the eigen-value of multiplicity 2 λ + = 1 are e + ( p ) = (1 , p − , , , and f + = (0 , , , , − . We notice that
N e ( p ) = N e + ( p ) = N e − ( p ) = i = e + ( p ) + e − ( p ) − e ( p ) + f + − f − and N f + = N f − = 12 ( i + i ) = 14 ( e − ( p ) − e + ( p )) + 12 ( f + + f − ) . We also introduce the function ψ ( t ) = γβ α ν Z tt η ( s ) ds. (88)which is the integral of p and corresponds to the eigenvalue of largest real partof − M (3)0 . This eigenvalue is associated with the eigenvector P ( p ) such that e ( p ) ∧ e − ( p ) ∧ f − = − p P ( p ) and we find P ( p ) = (2 + p ) f ⊥ + f ⊥ + f ⊥ − g ⊥ − g ⊥ − (2 + p ) g ⊥ − g ⊥ − g ⊥ . (89)The aim of this section is to prove the Theorem 3
1) There exists a unique solution U ( t, α ) of the system (85) and t > such that there exists C > such that, for all t ≥ t we have | U ( t, α ) e t + ψ ( t,α ) | ≤ Ct ν . lim t → + ∞ U ( t, α ) e t + ψ ( t,α ) t − ν = P (0) , (90)38 he vector P (0) being given by (89). Note that this condition is equivalent to(50).2) There exists a unique solution of the system (87) such that ˜ Z ( t ) e t t − ν → P (0) when t → + ∞ .3) For all t > , there exists C ( t ) > such that for t ≥ t the estimate | ˜ Z ( t ) e t t − ν − U ( t, α ) e t + ψ ( t,α ) t − ν | ≤ C ( t ) α ν . (91)In a first paragraph, me make a reduction of the system to simplify itsresolution. The system (85) rewrites dZdt = M ( p ) Z + 1 νt N Z + α ν t − − ν R ( p ) Z (92)where M ( p ) are analytic functions of p for p ≤ p and N is a constant matrix.The system satisfied by the vectorial product of three solutions of (85) is dZ (3) dt = M ( p ) (3) Z (3) + 1 νt N (3) Z (3) + α ν t − − ν R ( p ) (3) Z (3) . (93)to which we associate its model system (87).Introduce the eigenvalue of smallest real part of M ( p ) (3) , (which is λ ( p ) = − − p ). Consider the matrixΛ( p ) = M ( p ) (3) − λ ( p ) I as well as the unknown Y ∗ ( t ) = Y (3) e − R tt λ ( p ( s )) ds . We obtain the system dY ∗ dt = Λ( p ) Y ∗ + 1 νt N (3) Y ∗ + α ν t − − ν R ( p ) (3) Y ∗ ( GS ) (94)associated with the model system ddt ( ˜ Y e t ) = Λ(0)( ˜ Y e t ) + 1 νt N (3) ( ˜ Y e t ) ( M S ) (95)The aim of what follows is to identify the family of solutions of (GS) such that Y ∗ ( t ) ≃ ct M E when t → + ∞ , where M = N ν and E is the eigenvector asso-ciated with λ ( p ). 39 emark The result that we obtain here depends heavily on the fact that thenon zero eigenvalues of Λ( p ) denoted by λ i ( p ) satisfy(ln t ) − Z tt λ i ( p ) ds → + ∞ , t → + ∞ . The first transformation uses Λ( p ) = ( P ( p )) − D ( p ) P ( p ), where P ( p ) isa transfer matrix and D ( p ) is the matrix of eigenvalues of λ ( p ), such that D ii ( p ) = λ i ( p ), λ i ( p ) ≤ λ i +1 ( p ) and all the eigenvalues are positive. We in-troduce U ( t, α ) = P ( p ) Y ∗ . The system is dUdt = D ( p ) U + 1 νt P ( p )[ N (3) + ν ( αt ) ν R ( p ) (3) − P ( p ) − dPdp ( νt dpdt )] P ( p ) − U, and using the relation dpdt = − α ν ( νt ) − − ν γβ − ν (1 − βγ p )(1 + ph ( p )) ν we end-up with the system dUdt = D ( p ) U + 1 νt M ( t, α ) U. (96)where the matrix M writes M ( t, α ) = N (3) + R ( t, α ) α ν t − ν , with the following estimate on R : ∃ α , T > , ∀ β ∈ [ β , β ] , ∀ γ, ℜ γ ≥ , | γ | ≤ β − , | R ( t, α ) | ≤ C ( β ) . On each eigenspace of D ( p ) we assume that N is diagonal.We denote by E i the eigenspaces of D ( p ), 1 ≤ i ≤ m , E and E m being ofdimension 1. The unknowns U are written U i ∈ K dim E i .The aim of the next paragraph is to construct iteratively the solution of thesystem (96). We use the methods of Levinson [18] and Hartmann [11]. It is necessary to begin with the computation of U m , solution of dU m dt = λ m ( p ) U m + νt ( N mm + α ν t − ν R mm ) U m + νt P m − j =1 M mj U j . We consider the differential equation dU m dt = λ m ( p ) U m + 1 νt ( N mm + α ν t − ν R mm ) U m − f
40s we want to obtain a bounded solution of (96), if we introduce φ m ( t ) = R tt λ m ( p ( s )) ds , this differential equation becomes ddt ( U m ( t ) e − φ m ( t ) t − Nmmν ) = 1 νt α ν t − ν R mm ( U m ( t ) e − φ m ( t ) t − Nmmν ) − f ( t ) e − φ m ( t ) t − Nmmν . If the function U m ( t ) e − φ m ( t ) t − Nmmν goes to a non zero finite limit when t goes to+ ∞ , then U m ( t ) goes to infinity when t goes to infinity under the sufficientcondition ℜ φ m ( t )ln t → + ∞ for t → + ∞ , which is contradictory with the factthat we seek a bounded solution. Hence it is necessary (but not sufficient) that U m ( t ) e − φ m ( t ) t − Nmmν goes to 0 when t goes to infinity.The system rewrites ddt ( U m ( t ) e − φ m ( t ) t − Nmmν e R + ∞ t α ν s − − ν R mm ( s ) ds ) = − f ( t ) e − φ m ( t ) t − Nmmν e R + ∞ t α ν s − − ν R mm ( s ) ds . We introduce the operator T ( m ) ( f )( t ) = e φ m ( t ) t Nmmν e − R + ∞ t α ν s − − ν R mm ( s ) ds Z + ∞ t f ( s ) e − φ m ( s ) s − Nmmν e R + ∞ s α ν l − − ν R mm ( l ) dl . (97)We verify ddt ( T ( m ) ( f )) = λ m ( p ) T ( m ) ( f )) + 1 νt ( N mm + α ν t − ν R mm ) T ( m ) ( f )) − f ( t ) . (98)Hence the equation on U m leads to the necessary relation U m ( t ) = − T ( m ) [ 1 νt m − X j =1 M mj U j ]which rewrites U m ( t ) = 1 νt m − X j =1 T ( m ) j ( U j ) (99)with T ( m ) j ( U j ) = − νtT ( m ) ( 1 νt M mj U j ) . We replace this equality in the system satisfied by ( U j ) ≤ j ≤ m . We obtain1 ≤ j ≤ m − dU j dt = λ j ( p ) + 1 νt m − X k =1 M (1) jk ( U k ) (100)where M (1) jk ( U k ) = M jk ( U k ) + 1 νt M jm T ( m ) k ( U k ) . (101)41f course, we notice that M (1) jk ( U k ) − n jk U k = ( M jk − N jk ) U k + 1 νt M jm T ( m ) k ( U k ) = O ( α ν t − ν ) (102)hence its contribution is a regularizing operator.The scheme of the proof is the same for all the terms of the vector U . Ateach stage, we obtain the system1 ≤ j ≤ m − e dU j dt = λ j ( p ) U j + 1 νt m − e X k =1 M ( e ) jk ( U k ) (103)where M ( e ) jk ( U k ) = M ( e − jk ( U k ) + 1 νt M ( e − jm − e +1 T ( m − e +1) k ( U k ) . (104)The operator T ( m − e ) of the following step is given by the solution T ( m − e ) ( f )going to 0 at + ∞ of the equation dU m e dt = λ m − e ( p ) U m − e ( t ) + 1 νt M ( e − m − em − e ( U m − e )( t ) − f ( t )and the operators T ( m − e ) k , 1 ≤ k ≤ m − e − T ( m − e ) k ( U k ) = − νtT ( m − e ) [ 1 νt M ( e ) m − ek ( U k )] . (105)The construction of the formal solution is done. We end up with the remain-ing equation on U , to which we cannot apply the previous method because theassociated eigenvalue is 0. This system writes ddt ( t − N ν U ( t )) = 1 νt t − N ν ( M ( m − − N ) U . (106)We deduce that there exists a constant A such that U ( t ) − At N ν = − t N ν Z + ∞ t νs s − N ν ( M ( m − − N )( U )( s ) ds. (107) The resolution ends up with the construction of the solution of (107). For thisconstruction, it is necessary to study the regularity of all the operators T ( m − e ) (and of all the induced operators T ( m − e ) k and M ( e ) jk ).For a given function ψ , such that ψ ( t ) is increasing, going to + ∞ at + ∞ ,we introduceΛ Kψ ( t ) = { f ∈ C ∞ ([ t , + ∞ [) , ∃ C, | f ( t ) | ≤ Ct K e ψ ( t ) } .
42e say that ψ ∈ L ε m − e ( t ) if we have ψ ∈ C ∞ ([ t , + ∞ [) , ψ ′ ( t ) −ℜ φ ′ m − e ( t ) ≤ − ε < , t ≥ t , t | ψ ′′ ( t ) | bounded on [ t , + ∞ [ . These notations being introduced, we consider the operator K (1) ( f ) = − t N ν Z + ∞ t νs s − N ν ( M ( m − − N )( f )( s ) ds. The equation that we intend to solve is U − At N ν = t N ν K (1) ( U ) . We notice that νs s − N ν ( M ( m − − N )( f ) ∈ L ([ t , + ∞ [) as soon as f ∈ Λ N ν ( t ). Moreover, for t given, there exists a constant C such that | νs s − N ν ( M ( m − − N )( f ) ≤ C α ν s − − ν max l ∈ [ s, + ∞ [ ( | f ( l ) l − N ν ) . Hence we get the inequality | A − K (1) ( At N ν ) | = | K (1) ( t N ν ) | ≤ α ν C Z + ∞ t s − − ν ds = C να ν t − ν . Assume α ≤ α given. There exists a value of t , given by t = max( t , (2 C ν ) ν α )such that for t ≥ t we have | K (1) ( t N ν ) | ≤ t N ν . Hence, by induction, we get that | ( K (1) ) ( l ) ( t N ν ) | ≤ l t N ν , t ≥ t (108)from which we deduce the convergence of the series P ( K (1) ) ( l ) ( t N ν ) and itsbound by 2 t N ν . More precisely, we have, for all α ≤ α | ( K (1) ) ( l ) ( t N ν ) | ≤ l ( αα ) l t N ν , t ≥ t (109)hence the behavior when α →
0. 43
The instability growth rate
This section relies on the relation (58) that we obtained in the third section.The scope of the present section is to derive a limit, when α → Ev ( α, β, γ ). As the right hand side of (58) depends on t , and the lefthand side of (58) is independant of t , we will study, for a given (suitable) t > α → α → z , ξ ( y ) z p ( p = 2 , , ξ ( y ) m l ,( l = 1 , , ξ ( y ) m l , l = 4 , , R j ( t, α ), L k ( t, α ) for y = − tαβ . Recall thatwe restrict ourselves to the regime α → β , γ in a fixed compact set.In all what follows, we introduce r = γβ . (110) Introduce t ∗ > νt ∗ β < R .
We introduce ζ ( t, α ) = ζ ( ξ ( − tαβ ) , α ) = α ( ζ ( ξ ( − tαβ )) ν . We have ξ ( − tαβ ) = α ν ( ζ ( t, α )) − ν . The equation on ξ gives ζ + O ( α ν ) = νtβ , hence ζ ( t,
0) = νtβ . (111)Hence we get (and it is the same for all other quantities)lim α → z ( − tαβ , α ) = ζ ( t,
0) ¯ A ( ζ ( t, , . We have, moreover exp ( − Z − t αβ αγξ ( y ′ ) dy ′ ) = exp ( α ν Z t γβ ( ζ ( s, α )) ν ds )hence the limit when α → β given and ξ = , we identify α such that, for α ≤ α and ζ < R ,the asymptotic series defining z j , m p converges to an analytic function. As ζ ( t ∗ ,
0) = νt ∗ β , if one chooses 0 < νt ∗ β < R , that is 0 < t ∗ < β νR , there exists α > α ≤ α , ζ ( t ∗ , α ) ≤ R < R .For t ∈ [ t ∗ , t ∗ ], and α ≤ α , ζ ( t, α ) ≤ R , hence the functions ( ξ ( − tαβ )) δpν z p ( − tαβ )and ( ξ ( − tαβ )) dqν m q ( − tαβ ) are well defined through the expansion of section44. Moreover, for t ≥ t ∗ there exists C ( t ∗ ) such that (91) holds. Hence for t ∈ [ t ∗ , t ∗ ] we have the limit of the functions R p and L k when α →
0. We arenow ready to prove our main Theorem:
Theorem 4
Let M be given. There exists α ∗ > such that, for < α < α ∗ , β ∈ [ M , M ] and | γ | ≤ M, ℜ γ ≥ , the Evans function of the system Ev ( α, β, γ ) does not vanish. Proof
Recall that we proved that Ev ( α, β, γ ) = Ev ( β, γ ) + α ν Ev ( β, γ, α ) . The value of Ev ( β, γ ) (which is expressed through (112), (114)) does not de-pend on t ∗ . Letting the leading order term of Ev ( β, γ ) go to 0 when t ∗ → + ∞ (which gives, of course, the value of Ev ( β, γ ) because it does not depend on t ∗ ) yields r = 1 as only possible positive solution for Ev (0 , β, γ ) = 0 (see (126)).However, for r = 1 the remaining leading order term in t ∗ of Ev (0 , β, γ ) is notzero (see (127)), hence a contradiction. The system of Kull-Anisimov has no bounded complex growth rate γ when F r = O ( ε ) in the limit ε → . To be more precise, recall that Ev ( α, β, γ ) = e − (2+ µ ( α ) β ) t t ν exp( − R − t αβ αγξ ( y ′ ) dy ′ ) × [ z ( R − ξβ L − L ) + ξz ( R + γβ L − L )+ ξz ( R + γβ L − ξβ L )+ ξz ( R − γβ L − ξβ L − L )+ ξ − ξ ( m L + m L + m L ) + ξ − ξ ( m L + m L + m L )] . (112)We introduce ¯ A p ( ζ, α ) , ¯ B q ( ζ, α ), 1 ≤ p ≤
4, 1 ≤ q ≤ βζ ¯ A p ( ζ, α ) = A p ( ζ, α ) , βζ ¯ B q ( ζ, α ) = B q ( ζ, α ) . (113)We proved in Section 3 that the solution w (2)+ of (42) which behaves as e ( λ − (1) − αβ ) y when y → + ∞ , satisfying the condition (49) is given through the relation (59) P j =1 z j f j + P p =1 m p g p = e − αµ ( α ) y T (2) w (2)+ , where ( z j , m p ) are given by Propo-sition 4 through z = β + βζ ¯ A ( ζ, α ) z = γ − β + β ζξ ¯ A ( ζ, α ) z = β + β ζξ ¯ A ( ζ, α ) z = − β + β ζξ ¯ A ( ζ, α ) m = β ζξ ¯ B ( ζ, α ) m = β ζξ ¯ B ( ζ, α ) m = β ζξ ¯ B ( ζ, α ) m = β ζξ ¯ B ( ζ, α ) m = β ζξ ¯ B ( ζ, α ) m = β ζξ ¯ B ( ζ, α ) . (114)45quality (58) for a t such that ζ ( t, α ) < R , along with ξ = α ν ζ ( t, α ) − ν yields Ev ( α, β, γ ) = e − (2+ µ ( αβ ) t t ν exp ( − R − t αβ αγξ ( y ′ ) dy ′ ) − ξ (0) ξ (0) β [(1 + ζ ¯ A )( R − ξβ L − L ) + ( ξ ( r −
1) + ζ ¯ A )( R + γβ L − L )+( ξ + ζ ¯ A )( R + γβ L − ξβ L ) + ( − ξ + ζ ¯ A )( R − γβ L − ξβ L − L )+ ζ − ξ [ ¯ B L + ¯ B L + ¯ B L + ¯ B L + ¯ B L + ¯ B L ] . We proved in Section 5 that the unique solution U ( t, α ) of the system (85)satisfying the uniqueness condition (90)is U ( t, α ) = X p =1 f p ( t, α ) f ⊥ p + X q =1 g q ( t, α ) g ⊥ q and the functions R j and L p given by (56) satisfy the estimates X j =1 | R j ( t, α ) − R j ( t ) | + X p =1 | L p ( t, α ) − L p ( t ) | ≤ C ( t ) α ν , t ≥ t . Note that the limit of the quantities R j (as well as R j when t → + ∞ ) is known.We now consider the equality on Ev ( α, β, γ ) when α →
0. The right hand sideis independant of t because the left hand side is independant of t . Hence itsvalue can be considered at t ∗ . Once t ∗ is fixed, we get the limit by taking α = 0,hence, using µ (0) β = − r − Ev (0 , β, γ ) = e ( r − t ∗ t ν ∗ − ξ (0) ξ (0) β [(1 + ζ ¯ A ( ζ ))( R ( t ∗ ) − L ( t ∗ )) + ζ ¯ A ( R + rL − L )+ ζ ¯ A ( R + rL ) + ζ ¯ A ( R − rL − L )+ ζ [ ¯ B L + ¯ B L + ¯ B L + ¯ B L + ¯ B L + ¯ B L ]] (115)where the relations are written at t = t ∗ and at ζ = ζ ( t ∗ ,
0) = νt ∗ β .Let A p,j and B q,j being given through βζ ¯ A p ( ζ ) = β ∞ X j =1 ( βζ ) j A p,j , βζ ¯ B q ( ζ ) = β ∞ X j =1 ( βζ ) j B q,j . (116)By keeping only the leading order term in ξ ( y ) for each equation in the system(61) (which means that we consider the order of each quantity Z p and M q ), weobtain the recurrence system (117): 46 ν ( j + 1) A ,j +1 = A ,j − rA ,j − B ,j ( ν ( j + 1) + 1) A ,j +1 = rA ,j − A ,j − B ,j − B ,j ( ν ( j + 1) + 1) A ,j +1 = A ,j + A ,j + rA ,j − A ,j − B ,j − B ,j ( ν ( j + 1) + 1) A ,j +1 = − A ,j + rA ,j + B ,j ( ν ( j + 1) + 1) B ,j +1 = − rB ,j − B ,j − B ,j − rB ,j − A ,j + ( r − A ,j ( ν ( j + 1) + 1) B ,j +1 = B ,j − rB ,j + B ,j − rB ,j + rA , j + ( r − A ,j ( ν ( j + 1) + 1) B ,j +1 = B ,j − rB ,j + B ,j + (1 − r ) A ,j ( ν ( j + 1) + 2) B ,j +1 = − B ,j + B ,j + rA ,j + A ,j ( ν ( j + 1) + 2) B ,j +1 = B ,j − B ,j − A ,j ( ν ( j + 1) + 2) B ,j +1 = B ,j + B ,j + rA ,j . (117)It is easy from this system to deduce that, under the hypothesis | r | ≤ M − ,there exists a constant C > P p =1 | A p,j | + P q =1 | B q,j | ≤ ( Cν ) j j ! ,hence ensuring that the analytic expansions defining ¯ A p and ¯ B q are extendiblefor all ζ . The study of this recurrence system is the aim of the next paragraph. In what follows, we study the recurrence system.Consider B ( ξ, r ) given by B ( ξ, r ) i = − ri + ξ i + i B ( ξ, r ) i = i B ( ξ, r ) i = ξi − i + i B ( ξ, r ) i = ri + (1 − r ) i − ξ i + rξ i B ( ξ, r ) i = i − ξi + rξi . associated with the differential equation dY dt = B ( ξ, r ) Y . It is easy to checkthat B (2)0 is associated with the system (118), obtained from (61) by taking intoaccount the behavior of Z p and M q that we obtained in Section 3: dZ dt = ξZ − rξZ − ξM dZ dt = rZ − Z − ( M + ξM ) dZ dt = ξ Z + Z + γZ − Z − ( M + ξM ) dZ dt = − Z + rZ + M dM dt = − rM − M − ξ ( M + rM ) − ξ ) Z + ( r − Z dM dt = M − rM + M − rξM + rξ Z + ( r − Z dM dt = M − rM + ξM + (1 − r ) Z dM dt = − ξ M + M + rξ Z + ξ Z dM dt = M − M − ξ Z dM dt = ξ M + M + rξ Z . (118)We introduce F ( y ) = − r + 1 ν ξ − ν f + 1 ν + 1 ξ − ν − ( f − g + rg ) . (119)47e have Lemma 7
Let H be the unique solution going to zero when y → of dHdy = εB (2)0 ( ξ ( y ) , r ) H ( y ) + ε ddy ( F ( y )) The coefficients A p,j , B q,j are the coefficients in the expansion in ε ( ξ ( y )) − ν of H , where ξ ( y ) − ν = νy . The vector ( ζ ¯ A p , ζ ¯ B q ) is equal to H ( βζνε ) for ζ = ξ ( y ) − ν . We need to obtain the relation with the asymptotic expansion in αβ ofSection 3. For this purpose, we introduce t = εy . We consider the followingfunction, which will be tentatively a solution of (118) Z ( y ) = P ∞ j =1 A ′ ,j ε j ( ξ ( y )) − νj Z p ( y ) = P j ≥ A ′ p,j ε j ( ξ ( y )) − νj − , p = 2 , , M q ( y ) = P j ≥ B ′ q,j ε j ( ξ ( y )) − νj − , q = 1 , , M l ( y ) = P j ≥ B ′ l,j ε j ( ξ ( y )) − νj − , l = 4 , , . (120)Consider for example the first equation of (120). It rewrites ∞ X j =1 ε j A ′ ,j ddy ( ξ − νj ) = ε ( X j ≥ ( A ′ ,j − rA ′ ,j − B ′ ,j ) ξ − νj ε j )hence − dξdy X j =0 ν ( j +1) A ′ ,j +1 ε j +1 ( ξ ( y )) − ν ( j +1) − = X j ≥ ε j +1 ( ξ ( y )) − νj ( A ′ ,j − rA ′ ,j − B ′ ,j ) . If we want to obtain ( A ′ p,j , B ′ q,j ) independant on ξ ( y ), it is a natural choice towrite − dξdy = ξ ν +1 (121)and, up to a change of origin in y , we obtain ξ ( y ) − ν = νy. (122)Even with this equation, there will still be an additional term in the relation,related with j = 0. The resulting equation on Z ( y ) given in (120) is ddy [ Z ( y ) − εA ′ , ( ξ ( y )) − ν ] = ξZ − rξZ − ξM . With these two relations, we obtain the recurrence relation ν ( j + 1) A ′ ,j +1 = ( A ′ ,j − rA ′ ,j − B ′ ,j ) , j ≥ . A ′ , = − r +1 ν , A ′ , = A ′ , = 0 , A ′ , = ν +1 ,B ′ , = − ν +1 , B ′ , = rν +1 , B ′ , = B ′ , = B ′ , = B ′ , = 0 (123)which corresponds to the source term F ( y ) given by (119), we have the identity A p,j = A ′ p,j , B q,j = B ′ q,j , ∀ j ≥ , ∀ p = 1 , , , , ∀ q = 1 , , , , , A ′ p,j , B ′ q,j ) are the coefficients of the expansion in ε of the solution of(118) with the source term (119) whereas ( A p,j , B q,j ) are the coefficients of theexpansion in ζ of ( ¯ A p , ¯ B q ). Lemma 7 is proven. Reduction of the problem
Let us study the matrix B . Its eigenvalues are0 ,
1, and − e = ( ξ, − , , , , e = (1 − r, − ξ , , , , F = ( ξ, − , , , ,e − = ( r + 1 , − ξ , , − , , F − = ( ξ, − , − , , . The inverse matrix is given by i = e + e − + ξ ( F + F − − e )2 i = F + F − − e i = F − F − i = (1 + r ) e + ( r − e − + rξ ( F + F − − e )2 i = F + F − − ξ ( e + e − ) . We rewrite a solution V of dVdy = εB V as V = V e + V e + W F + V − e − + W − F − . We obtain dVdy = dV dy e + dV dy e + dW dy f + dV − dy e − + dW − dy f − +( V + W + W − ) dξdy i + dξdy ξ − ( V + V − ) i , hence the associated system is dV dy = ε dξdy [ ξ − ( V + W + W − ) + ξ − ( V + V − )] dV dy = εV − ε dξdy ( V + W + W − ) dW dy = εW − ε dξdy [ ξ − ( V + W + W − ) + ξ − ( V + V − )] dV − dy = − εV − − ε dξdy ( V + W + W − ) dW − dy = − εW − − ε dξdy [ ξ − ( V + W + W − ) + ξ − ( V + V − )] . (124)49rom the relation4( A f + A f + A f + A f + B g + B g + B g + B g + B g + B g )= A ( e + e − + ξ R ) ∧ ( e − e − ) + ( rA − B ) R ∧ ( e + e − ) + A R ∧ ( e − e − )+( rA − B )( F − F − ) ∧ ( e + e − + ξ R ) + A ( F − F − ) ∧ ( e − e − )+( rA + B )( e + e − + ξ R ) ∧ ( F − F − − ξ ( e + e − ))+ A ( e + e − ) ∧ ( F + F − − ξ ( e + e − ))+ B R ∧ ( F − F − ) + B R ∧ ( F + F − − ξ ( e + e − ))+ B ( F − F − ) ∧ ( F + F − − ξ ( e + e − )) . we deduce a new basis of Λ (IR ) in which the coefficients are A , rA − B , A , rA − B , A , rA + B , A , B , B , B . The source term (119) in the basis ofΛ ( R ) associated with e , e ± , F ± we find F ( y ) = − r +1 ν ξ − ν [( e + e − ) ∧ ( e − e − ) − ξ ( e − e − ) ∧ ( F + F − − e )]+ ν +1) ξ − ν − [( F − F − ) ∧ ( e − e − ) − ξ ( e + e − ) ∧ ( F + F − − e )] . (125) End of the proof
The coefficient of e ∧ F in the source term is thus − ξ − ν − ( r + 1 ν − ν + 1 − ξ ) . The theory of Fuchsian systems (see Hartmann [11]) shows that there ex-ists a constant α ∗ such that the projection of H ( y ) on e ∧ F behaves as − t α ∗ e t ( r +1 ν − ν +1 ) e ∧ F . The leading order term in t ∗ of the Evans func-tions Ev (0 , β, γ ) writes Ev (0 , β, γ ) = e ( r +1) t ∗ t ν ∗ t α ∗ ∗ − ξ (0) ξ (0) β [( R − L ) + ζ ( t ∗ , A ( R − L ) + ˜ A ( R + rL − L )+ ˜ A ( R + rL ) + ˜ A ( R − rL − L )+ ˜ B L + ˜ B L + ˜ B L + ˜ B L + ˜ B L + ˜ B L ]]where ˜ A p ( t ∗ ) = ¯ A p ( ζ ( t ∗ , e − t ∗ t − α ∗ ∗ , ˜ B q ( t ∗ ) = ¯ B q ( ζ ( t ∗ , e − t ∗ t − α ∗ ∗ . Hence,as − e ∧ F = − ξf + f − f + f + rg + (1 − r )( g + g ) − ξ ( g + g )and the limit of ( R p , L q ) when t ∗ → + ∞ is given by (50), the limit of Ev (0 , β, γ )( e ( r +1) t ∗ t ν ∗ t α ∗ ∗ − ξ (0) ξ (0) β ) − is ( r − − r + 1 ν + 1 ν + 1 ) . (126)50s we seek positive values of r , the factor e ( r +1) t ∗ goes to infinity when t ∗ → + ∞ , hence this limit is necessarily 0. The two possible values of r are thus r = 1 and r = − ν +1 , hence r = 1.We also check that, for r = 1, the projection of F ( y ) on the space associatedwith the eigenvalue +1 of B (2)0 , space generated by e ∧ e and F ∧ e , is( ν ξ − ν − + ν +1) ξ − ν − ) e ∧ e . We have the relations e ∧ e ∧ S = 0 , F ∧ e ∧ S = 0because e ∧ e = − ξf +2 f + f + g − ξ g and F ∧ e = − ξg − ξg +2 g + g + g .The leading order term of the projection on this space is 0. We get e ∧ e − ∧ S = − i ∧ i ∧ i ∧ i ∧ i because e ∧ e − = − f + ξ f + ξ g , and similarily e ∧ F − ∧ S = 0, e − ∧ F ∧ S = − i ∧ i ∧ i ∧ i ∧ i , F ∧ F − ∧ S = 0. Asthe vector F ( y ) has the following projection on the eigenspace associated withthe eigenvalue 0 P r ( F ( y )) = − ν ξ − ν (2 e ∧ e − + ξ e − ∧ F − ξ e ∧ F − )+ ν +1 ( e − ∧ F + ξ e − ∧ F + (1 − ξ ) e ∧ F − ) (127)the associated leading order term gives a non-zero contribution, hence a contra-diction. The main theorem is proven. We consider a solution of (33) associated with the growth rate λ − (1) − αβ ,which is the eigenvalue of smallest real part − M (2)0 (+ ∞ ). We recall that thereexists a regular function µ ( α ) such that λ − (1) − αβ = − αµ ( α ) . (128)We will prove that there exists a unique solution w (2)+ of (33) satisfying (49). Inorder to prove the existence and uniqueness of w (2)+ we prove the following Proposition 6
Let U be the solution going to (1 , , ..., at + ∞ of the modelsystem: dU j dy = λ j U j + (1 − ξ ( y )) d X k =1 N jk ( ξ ( y )) U k ( y ) , ≤ j ≤ d where the properties of the complex numbers λ j and of the functions N jk are thefollowing | N jk ( ξ ) | ≤ M , ξ ≥ ξ
51 = λ < ℜ λ ... ≤ ℜ λ d − , N dd = 0 .The function U is given by U ( y ) = (1 , , ....,
0) + (1 − ξ ( y )) w ( y ) . Note that, in the hypothesis, N dd = 0 is only there for simplicity purposesand is obtained by considering in the last equation (in which we should have(1 − ξ ) N dd = ξ ′ N dd ξ ν +1 ) the conjugation by the exponential of the primitive of N dd ξ ν +1 . Construction by recurrence of the Volterra operators
We prove Propo-sition 6 by recurrence. Consider the last equation (line d of the previous system).We have dU d dy = λ d U d + d − X k =1 (1 − ξ ( y )) N dk U j ( y ) . This equation is equivalent to ddy ( U d e − λ d y ) = d − X k =1 e − λ d y (1 − ξ ( y )) N dk U j ( y ) . As U d is bounded, the limit of U d e − λ d y is zero, otherwise U d would not bebounded. Integrating from y to y and letting y go to infinity, we have U d ( y ) = d − X k =1 e λ d y Z + ∞ y ( ξ ( s ) − N dk ( ξ ( s )) e − λ d s U j ( s ) ds, and defining K (1) dj through K (1) dj ( U ) = (1 − ξ ( y )) − e λ d y Z + ∞ y ( ξ ( s ) − N dj ( ξ ( s )) e − λ d s U ( s ) ds (129)we obtain U d ( y ) = (1 − ξ ( y )) d − X k =1 K (1) dj ( U j )( y ) . (130)Let us study the properties of K (1) dj .Consider U such that, for ξ ≥ ξ (and y such that ξ = ξ ( y )) we have theestimate ∃ N, C, ∀ y ≥ y , | U ( y ) | ≤ C (1 − ξ ( y )) N . (131)We obtain easily the estimate (thanks to λ d − p ≥ ∀ y ≥ y , | K (1) dj ( U )( y ) | ≤ M C ( N + 1) ξ ν +10 (1 − ξ ( y )) N . d − ≤ j ≤ d − , dU j dy = λ j U j + (1 − ξ ( y )) d − X k =1 N (1) jk ( U k )where N (1) jk ( U k ) = N jk .U k + N jd K (1) dj . (132)We get the estimate, under the assumption (131) on Uy ≥ y , | N (1) jk ( U )( y ) | ≤ M C (1 + M ( N + 1) ξ ν +10 )(1 − ξ ( y )) N . This rewrites y ≥ y , | N (1) jk ( U )( y ) | ≤ M C (1 − ξ ( y )) N where M = M (1 + M ( N +1) ξ ν +10 ). The procedure proceeds as follows:Let p be an element of 0 ..d −
2. We write the sequence of operators K ( p +1) d − p,k ,1 ≤ k ≤ d − p −
1, and N ( p +1) jk , j ≤ d − p − k ≤ d − p − U d − p ( y ) = (1 − ξ ( y )) d − p − X k =1 K ( p +1) d − p,k ( U k )( y ) (133) dU j dy = λ j U j + (1 − ξ ( y )) d − p − X k =1 N ( p +1) jk ( U k )( y ) (134)where N ( p +1) jk ( U k ) = N ( p ) jk ( U k ) + (1 − ξ ( y )) − N ( p ) j,d − p ((1 − ξ ) K ( p +1) d − p,k ( U k )) (135)and the operators K ( p +1) d − p,k are constructed as follows using the auxiliary problem dVdy = λ d − p V + (1 − ξ ( y )) N ( p ) d − p,d − p ( V ) + f. (136)Bounded solutions when y → + ∞ of (136) are obtained, for d − p ≥
2, through V ( y ) e − λ d − p y = Z + ∞ y ( ξ ( s ) − e − λ d − p s N ( p ) d − p,d − p ( V )( s ) ds − Z + ∞ y f ( s ) e − λ d − p s ds. This can be written, using g ( y ) = V ( y ) e − λ d − p y , under the form g = K d − p ( g ) − Z + ∞ y f ( s ) e − λ d − p s ds. (137)53ence we may write, for f satisfying (131) g = ∞ X l =0 ( K d − p ) l ( Z y + ∞ f ( s ) e − λ d − p s ds )which defines T d − p through V = T d − p ( f ) . (138)Replacing f by (1 − ξ ( y )) P d − pk =1 N ( p ) d − p,k ( U k )( y ), we obtain the expression of U d − p in function of U k , ≤ k ≤ d − p − U d − p ( y ) = T d − p ((1 − ξ ) X k The last step of this recurrence is to constructthe solution U . The equation on U writes then dU dy = (1 − ξ ( y )) N ( d − ( U ) , (140)where N ( d − satisfy the relations of Proposition 7. This is equivalent to U ( y ) − U ( y ) = Z yy (1 − ξ ( s )) N ( d − ( U )( s ) ds. Using the limit U ( y ) → y → + ∞ , we obtain U ( y ) = 1 + Z y + ∞ (1 − ξ ( s )) N ( d − ( U )( s ) ds. We introduce the operator K such that U = 1 + K ( U ). For g satisfying theestimate (131), a consequence of proposition 7 is that | K ( g )( y ) | ≤ M Cξ ν +10 ( N + 1) (1 − ξ ( y )) N +1 . Introduce U ( y ) = 1, and the sequence U N = K ( U N − ). It is straightforwardto show the inequality for y ≥ y | U N ( y ) | ≤ ( Mξ ν +10 ) N N ! (1 − ξ ( y )) N hence as the series P ∞ N =0 ( Mξ ν +10 ) N N ! (1 − ξ ( y )) N is normally convergent for y ≥ y , the series P ∞ n =0 U N ( y ) is normally convergent and is the only solution goingto 1 as y goes to + ∞ of (140). We have the estimate | U ( y ) | ≤ exp( Mξ ν +10 ) . which is of the form (131) for N = 0. 54 nd of the construction As U satisfies an estimate of the form (131) for N = 0, from the definition of the operators K ( p +1) d − p,k we deduce that U ( y ) = (1 − ξ ( y )) w ( y )where | w ( y ) | ≤ M exp( Mξ ν +10 ). Replacing in the equality (133) for p = d − U and on U we deduce that U ( y ) ≤ M (1 − ξ ( y )) exp( Mξ ν +10 ) . By recurrence on p we obtain the inequalities on U j for j = 1, which provesProposition 6. Estimates We prove in this Section the following Proposition 7 The operators N ( p ) jk and K ( p +1) d − p,k satisfy the following estimatesfor g satisfying (131): ∀ y ≥ y , | T ( g )( y ) | ≤ CMξ ν +10 ( N + 1) (1 − ξ ( y )) N . This is a consequence of the more precise proposition Proposition 8 Under the assumption g satisfy the estimate (131), the opera-tors N ( p ) jk satisfy the estimate | N ( p ) jk ( g )( y ) | ≤ CM p (1 − ξ ( y )) N where M p +1 = M p (1 + M p ξ ν +10 exp( M p ξ ν +10 )) and | T d − p ( f )( y ) | ≤ Cξ ν +10 exp( M p ξ ν +10 )(1 − ξ ( y )) N . We prove the second proposition by recurrence. The first estimate that we haveto deduce from the recurrence assumption on N ( p ) jk is the estimate on K ( p +1) d − p,k .To obtain this result, we have to study the behavior of K d − p through the esti-mate on N ( p ) jk . We have, for f satisfying (131) for all N , the inequality | K d − p ( f )( y ) | ≤ Z ∞ y (1 − ξ ( s )) N +1 M e λ d − p ( y − s ) ds. As N + 1 ≥ 1, we use y − s ≤ λ d − p ≥ ∀ y ≥ y , | K d − p ( f )( y ) | ≤ M Z ∞ y (1 − ξ ( s )) N +1 ds ≤ Mξ ν +10 ( N + 1) (1 − ξ ( y )) N +1 . − ξ ( y )) in the result.The second estimate is based on the expression of T d − p obtained through (138).For N ≥ f satisfying (131), we have | Z ∞ y f ( s ) e − λ d − p s ds | ≤ Ce − λ d − p y (1 − ξ ( y )) N N ξ ν +10 . The estimate on K d − p , deduced from the recurrence hypothesis is | K d − p ( Z ∞ y f ( s ) e − λ d − p s ds ) | ≤ C Z ∞ y (1 − ξ ( s )) N +1 M p ds ≤ M p Cξ ν +10 (1 − ξ ( y )) N +1 . Hence we obtain | ( K d − p ) l ( Z ∞ y f ( s ) e − λ d − p s ds ) | ≤ Ce − λ d − p y ( M p ξ ν +10 ) l (1 − ξ ( y )) N N + l l !hence the series P l ≥ e λ d − p y ( K d − p ) l ( R ∞ y f ( s ) e − λ d − p s ds ) is normally convergent.It defines the function T d − p ( f ) and we have | T d − p ( f )( y ) | ≤ C (1 − ξ ( y )) N exp( M p ξ ν +10 ) . Finally, using this estimate on T d − p as well as the relation (139), for f satisfying(131) we have | (1 − ξ ( y )) K ( p +1) d − p,k ( f ) | ≤ C (1 − ξ ( y )) N +1 exp( 1 ξ ν +10 )which gives | K ( p +1) d − p,k ( f ) | ≤ C (1 − ξ ( y )) N exp( M p ξ ν +10 ) . Using the estimate on N ( p ) j,d − p we obtain | (1 − ξ ) − N ( p ) j,d − p ((1 − ξ ) K ( p +1) d − p,k ( f )) | ≤ M p C (1 − ξ ( y )) N exp( M p ξ ν +10 )hence | N ( p +1) jk ( f )( y ) | ≤ C ( M p + M p exp( M p ξ ν +10 ))(1 − ξ ( y )) N . If we introduce M p +1 = M p (1 + M p ξ ν +10 exp( M p ξ ν +10 )), we thus deduce | N ( p +1) jk ( f )( y ) | ≤ CM p +1 (1 − ξ ( y )) N . The proposition 8 is proven. As we used the assumption that ℜ λ d − p > dy ( V e − λ d − p y ) = e − λ d − p y (1 − ξ ( y )) N ( p ) d − p,d − p ( V ) + f e − λ d − p y hence V ( y ) e − λ d − p y − V ( y ∗ ) e − λ d − p y ∗ = Z yy ∗ e − λ d − p s (1 − ξ ( s )) N ( p ) d − p,d − p ( V )( s )+ f ( s ) e − λ d − p s ds. If V e − λ d − p y has a limit, then this is in contradiction with the fact that V isbounded for ℜ λ d − p > 0, and this argument is no longer valid if λ d − p = 0.Hence this proves that the recurrence stops at p such that λ d − p = 0 hence p = d − 1. The estimates of Proposition 7 are valid for N ( p +1) jk , hence therecurrence proceeds till ℜ λ d − p > 0. This recurrence processus stops for λ d − p =0 because we cannot assert that the equation (136) has bounded solutions goingto a constant for y → + ∞ . References [1] M. Abramovitz and I.A. Stegun : Handbook of mathematical func-tions Dover Publications, NY, 9th printing. 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