Jean-Francois Coulombel and Paolo Secchi- The Stability of Compressible Vortex Sheets in Two Space Dimensions

8/3/2019 Jean-Francois Coulombel and Paolo Secchi- The Stability of Compressible Vortex Sheets in Two Space Dimensions

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The Stability of Compressible Vortex Sheetsin Two Space Dimensions

JEAN-FRANCOIS COULOMBEL & PAOLO SECCHI

ABSTRACT. We study the linear stability of compressible vortexsheets in two space dimensions. Under a supersonic conditionthat precludes violent instabilities, we prove an energy estimatefor the linearized boundary value problem. Since the problemis characteristic, the estimate we prove exhibits a loss of controlon the trace of the solution. Furthermore, the failure of the uni-form Kreiss-Lopatinskii condition yields a loss of derivatives in

the energy estimate.

1. INTRODUCTION

A velocity discontinuity in an inviscid flow is called a vortex sheet. In three-spacedimensions, a vortex sheet has vorticity concentrated along a surface in the space.In two-space dimensions, the vorticity is concentrated along a curve in the plane.The present paper deals with compressible vortex sheets, i.e., vortex sheets in acompressible flow.

If the solution is piecewise constant on either side of the interface of discon-tinuity, one has planar vortex sheets in the three dimensional case and rectilinearvortex sheets in the two dimensional case, respectively. The linear stability of pla-nar and rectilinear compressible vortex sheets has been analyzed a long time ago,see [12, 27]. In three space dimensions, planar vortex sheets are known to be vio-lently unstable (see e.g. [30]). In the two dimensional case, subsonic vortex sheetsare also violently unstable, while supersonic vortex sheets are neutrally linearly sta-

ble, see e.g. [27, 30]. This result formally agrees with the theory of incompressiblevortex sheets. In fact, in the incompressible limit, the speed of sound tends to in-finity, with the result that two-dimensional vortex sheets are always unstable. Thiskind of instability is usually referred to as the Kelvin-Helmholtz instability. Forthe incompressible theory of two-dimensional vortex sheets, we refer the reader to


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2 JEAN -FRANCOI S COULOMBEL & PAOLO SECCHI

the books [7, 22]. Moreover, we refer to [14] for the study of the instability of

vortex sheets when heat conduction is taken into account.However, the normal modes analysis performed to derive the linear stabilityof supersonic vortex sheets is by far not sufficient to guarantee the existence ofnonconstant vortex sheets (that is, contact discontinuities) solutions to the com-pressible isentropic Euler equations. In this paper, we first show that supersonicconstant vortex sheets are linearly stable, in the sense that the linearized system(around these particular piecewise constant solutions) obeys an energy estimate.Then we consider the linearized equations around a perturbation of a constantvortex sheet, and we show that these linearized equations obey the same energyestimate. This is a first crucial step towards proving the existence of nonconstant

compressible vortex sheets.Several points need to be highlighted. First of all, the existence of compress-

ible vortex sheets is a free boundary nonlinear hyperbolic problem. Moreover, thefree boundary is characteristic with respect to both left and right states since wedeal with contact discontinuities. This is one of the reasons why one can not applyMajdas analysis on shock waves (see [20,21]), that are noncharacteristic interfaces.In some previous works devoted to weakly stable shock waves, see [10,11], the firstauthor has considered noncharacteristic hyperbolic Initial Boundary Value Prob-lems that did not meet the uniform Kreiss-Lopatinskii condition. In the case of

vortex sheets, the analysis is closely related, with the additional difficulty that theboundary is characteristic (the present analysis thus relies more on the work ofMajda and Osher [23] rather than on the work of Kreiss [6, 17]). The connection

with [10,11] is that in both cases, the analogue of the Kreiss-Lopatinskii conditionis fulfilled but not in a uniform way. Furthermore, in the case of vortex sheets asin the case of shock waves, the linearized Rankine-Hugoniot conditions form anelliptic system for the unknown front. This property is a key point in our worksince it allows to eliminatethe unknown front and to consider a standard Bound-ary Value Problem with a symbolic boundary condition (this ellipticity propertyis also crucial in Majdas analysis on shock waves [20, 21]).

Regarding the energy estimates for the linearized problems, the failure of theuniform Kreiss-Lopatinskii condition yields a loss of derivatives with respect tothe source terms. Furthermore, because the boundary is characteristic, we expectto lose some control on the trace of the solution at the boundary. As a matter offact, we shall see that the only loss of control is on the tangential velocity (whichcorresponds to the characteristic part of the solution). The good point is thatthe ellipticity of the boundary conditions for the unknown front enables us togain one derivative for it, as in Majdas work on shock waves [20]. Going slightlymore into the details, we shall prove that the only frequencies for which we lose

some control on the solution correspond to bicharacteristic curves. Those curvesoriginate from those points at the boundary of the space domain where the so-called Lopatinskii determinant vanishes. In the interior of the space domain, thesesingularities propagate along two bicharacteristics associated with the incomingmodes.


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Stability of Compressible Vortex Sheets 3

Let us now describe the content of the paper. In Section 2, we present the

nonlinear equations describing the evolution of compressible vortex sheets andintroduce some notations. Then, in Section 3, we shall consider the linearizedequations around a constant (stationary) vortex sheet. The main result for theconstant coefficient linearized problem is given in Theorems 3.1 and 3.2. Afterseveral reductions, we shall detail in Section 4 the normal modes analysis of thelinearized problem and construct a degenerateKreiss symmetrizers in order to de-rive our energy estimate. In Section 5, we first present the variable coefficientslinearized problem and introduce Alinhacs good unknown. Then we paralinearizethe equations, in order to use symbolic calculus and derive the energy estimate. Aprecise estimate of the paralinearization errors is given. Eventually, we show how to

control the different pieces of the solution, depending on their microlocalization.The main result for the variable coefficients linearized problem is given in Theo-rem 5.1. In Section 6, we make some remarks about possible future achievements.

Appendix A is devoted to the proof of several technical lemmas and AppendixB gathers the main results on paradifferential calculus that are used throughoutSection 5.

2. THE NONLINEAREQUATIONS

We consider Euler equations of isentropic gas dynamics in the whole plane R2.

Denoting byu the velocity of the fluid and the density, the equations read:

(2.1)

t + (u) = 0,t(u) + (uu) + p = 0,

where p = p() is the pressure law. In all this paper, p is assumed to be a strictlyincreasing function of , defined on ]0, +[. We also assume that p is a Cfunction of. The speed of sound c() in the fluid is then defined by the relation

c() := p().

Let (,u)(t,x1, x2) be a smooth function on either side of a smooth hyper-surface := {x2 = (t,x1)}. Then (,u) is a (weak) solution of (2.1) if andonly if (,u) is a classical solution of (2.1) on both sides ofand the Rankine-Hugoniot conditions hold at each point of:

t[] [u ] = 0,(2.2a)t[u] [(u )u] [p] = 0,(2.2b)

where :=

(

x1 , 1) is a (space) normal vector to . As usual, [q] = q+ qdenotes the jump of a quantity q across the interface (see [29]).Following Lax [18], we shall say that (,u) is a contact discontinuity if the

Rankine-Hugoniot conditions (2.2) are satisfied in the following way:

t = u+ = u , p+ = p.


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Because p is monotone, the previous equalities read

(2.3) t = u+ = u , + = .

Since the density and the normal velocity are continuous across the interface ,the only jump experimented by the solution is on the tangential velocity. (Here,normal and tangential mean normal and tangential with respect to ). For thisreason, a contact discontinuity is a vortex sheet and we shall make no distinctionin the terminology we use.

Note that the first two equalities above are nothing but eikonal equations: ifx2 = (t,x1) is the equation of the interface

, then satisfies

t + 2(+,u+, x1 ) = 0 and t + 2(,u, x1 ) = 0,

on {x2 = 0}, where

2(,u, ) := u

1

, R,

is the second characteristic field of the system (2.1). It is linearly degenerate sincethe corresponding eigenvector, in the quasilinear form of (2.1), is given by

r2(,u, ) :=01

.Recall that the space dimension equals 2.

The interface , or equivalently the function , is part of the unknowns ofthe problem. We thus deal with a free boundary problem. As it is common in thiskind of situation, we first straighten the unknown front in order to work in a fixed

domain. More precisely, the unknowns (,u), that are smooth on either side of{x2 = (t,x1)}, are replaced by the functions

(+ ,u+ )(t,x1, x2) := (,u)(t,x1,(t,x1, x2)),

( ,u )(t,x1, x2) := (,u)(t,x1,(t,x1, x2)),

where is a smooth function satisfyingx2

(t,x1, x2) > 0,

(t,x1, 0) = (t,x1).

With these requirements for , all functions ,u are smooth on the fixed do-main {x2 > 0}. For convenience, we drop the index and only keep the + and exponents. We also define the functions

(t,x1, x2) := (t,x1, x2),


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which are both smooth on the half-space {x2 > 0}.

Let us denote by v and u the two components of the velocity, that is, u =(v,u). Then the existence of compressible vortex sheets amounts to proving theexistence of smooth solutions to the following system:

(2.4a) t+ + v+x1 + + (u+ t+ v+x1+) x2 +x2++ +x1 v+ + +

x2 u+

x2

+ +

x1+x2

+ x2 v

+ = 0,

(2.4b) tv+ + v+x1 v+ + (u+ t+ v+x1+) x2 v+x2++ p

(+)+

x1 + p

(+)+

x1+x2+ x2 + = 0,

(2.4c) tu+

+v+x1 u

+

+(u+

t

+

v+x1

+)x2 u

+

x2++ p

(+)+

x2 +

x2+ = 0,

(2.4d) t + vx1 + (u t vx1) x2 x2

+x1 v

+

x2 u

x2 x1

x2 x2 v

=0,

(2.4e) tv + vx1 v + (u t vx1) x2 vx2+ p

()

x1 p

()

x1x2 x2 = 0,

(2.4f) tu + vx1 u + (u t vx1) x2 ux2+ p

()

x2

x2 = 0,


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in the fixed domain {x2 > 0}, together with the boundary conditions

+|x2=0 = |x2=0 = ,t = v+|x2=0 x1 + u

+|x2=0 = v

|x2=0 x1 + u

|x2=0 ,

+|x2=0 = |x2=0 .

For convenience, we rewrite the boundary conditions in the following way:

+

|x2

=0

=

|x2

=0

=,(2.5a)

(v+ v)|x2=0 x1 (u+ u)|x2=0 = 0,(2.5b)t + v+|x2=0 x1 u

+|x2=0 = 0,(2.5c)

(+ )|x2=0 = 0.(2.5d)

The functions + and should also satisfy(2.6) x2

+(t,x1, x2) , x2

(t,x1, x2) ,

for a suitable constant > 0.In [20, 21], Majda makes the particular choice

(t,x1, x2) := x2 + (t,x1).This choice is appropriate in the study of shock waves because these are nonchar-acteristic discontinuities. In the study of contact discontinuities, it seems rathernatural to choose the change of variables such that the eikonal equations

t+

+2(

+,u+, x1+)

=t

+

+v+ x1

+

u+

=0,

t + 2(,u, x1) = t + v x1 u = 0,are satisfied in the whole closed half-space {x2 0}. This choice, that is in-spired from [13], has several advantages. First, it simplifies much the expressionof the nonlinear equations (2.4). But it also implies that the so-called boundarymatrix has constant rank in the whole space domain {x2 0}, and not only onthe boundary {x2 = 0}. This will enable us to develop a Kreiss symmetrizerstechnique, in the spirit of [23]. We shall go back to this feature later on.

The problem is thus the construction of (local in time) smooth solutions to

(2.4)(2.5)(2.6), once initial data have been prescribed. Of course, such initialdata will have to fulfill a certain number of compatibility conditions. The first stepin proving such an existence result is the study of the linearized problem arounda particular constant solution, and this is our first main result, see Theorems 3.1and 3.2. The second step is the study of the linearized problem around a (variable


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coefficients) perturbation of the constant solution. The extension to the variable

coeffi

cients linearized problem is addressed in the second part of the paper. Oursecond main result states that the constant coefficients energy estimate still holdswhen one considers a variable coefficients linearized problem, see Theorem 5.1.

To avoid overloading the paper, we introduce some compact notations for thenonlinear equations (2.4). For all U := (,v,u)T, we define

A1(U) := v 0p()/ v 0

0 0 v

, A2(U) := u 0 0 u 0

p()/ 0 u

.Then the nonlinear equations (2.4) read

(2.7a) tU+ + A1(U+)x1 U+

+ 1x2+ (A2(U+) t+ x1+A1(U+)) x2 U+ = 0,

(2.7b) tU + A1(U)x1 U

+1

x2 (A2(U) t x1A1(U)) x2 U = 0.With an obvious definition for the differential operator L, the system (2.7) alsoreads

(2.8) L(U+, +)U+ = 0, L(U , )U = 0.When no confusion is possible, we also write this system under the form

L(U,)U = 0,

where U stands for the vector (U+, U) and for (+,). One should alwaysremember that the interior equations (2.7) are entirely decoupled. The couplingbetween the right and left states is made by the boundary conditions (2.5).

There exist many simple solutions of (2.8)(2.5)(2.6), that correspond, inthe original variables, to rectilinear vortex sheets:

(,u) =

(,ur) ifx2 > t + nx1,(,ul) ifx2 < t

+nx1.

Here above, ur, ul are fixed vectors in R2, and > 0, and n are fixed realnumbers. These quantities are linked by the Rankine-Hugoniot conditions:

= vrn + ur = vln + ul.


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Changing observer if necessary, we may assume without loss of generality

= n = ur = ul = 0 and vr + vl = 0 (vr 0).

In the new variables, this corresponds to the following regular solution of (2.8)(2.5)(2.6):

(2.9) Ur

vr0

, Ul

vl0

,

r ,l(t,x1, x2) x2,

with the relation vr + vl = 0. We only consider the case vr 0, and withoutloss of generality, we assume vr > 0. In the next section, we study the linearizedequations around the particular solution defined by (2.9). Under a certain super-sonic assumption, we shall show that the linearized equations satisfy an a priorienergy estimate.

3. THE CONSTANT COEFFICIENTS LINEARIZED SYSTEM3.1. The linearized equations. Let us denote by, u, some small per-

turbations of the exact solution given by (2.9). Up to second order, the perturba-tions U = (, v, u)T satisfy

tU+ + A1(Ur) x1 U+ + A2(Ur) x2 U+ = 0,(3.1a)tU + A1(Ul) x1 U A2(Ul) x2 U = 0,(3.1b)

in the domain {x2 > 0}, together with the linearized Rankine-Hugoniot relations

+ = = ,(3.2a)(vr vl) x1 (u+ u) = 0,(3.2b)

t + vr x1 u+ = 0,(3.2c)+ = 0,(3.2d)

on the boundary{x2 = 0}. In short, equations (3.1)(3.2) read

(3.3)

LU = 0, ifx2 > 0,B(U,) = 0, ifx2 = 0,


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with U := (U+, U), and obvious definitions for the operators L and B:

LU := t

U+U

+

A1(Ur) 0

0 A1(Ul)

x1

U+U

+

A2(Ur) 0

0 A2(Ul)

x2

U+U

,

B(U,) :=

(vr vl) x1 (u+ u)

t + vr x1 u++

.

It is important to note that the interior equations do not involve the perturbation, so L is an operator that only acts on U. This property also holds when onestudies the linearized shock wave equations around a planar shock, see [20, 25].

Proving an energy estimate for the linearized equations amounts to workingwith source terms, both in the interior domain and on the boundary. From nowon, we thus consider the linear equations

(3.4) LU = f , ifx2 > 0,

B(

U,) = g, ifx2 = 0,

and try to estimate U and in terms off and g (in appropriate functional spaces).In order to simplify the subsequent calculations, we introduce some new unknownfunctions, and define the following quantities:

(3.5)W1 := v+, W2 := 12 (+/ + u+/c) , W3 := 12 (+/ + u+/c) ,W4 :

=v

, W5 :

=1

2

(

/+

u

/c) , W6 :=

1

2

(

/+

u

/c) .

We also define the following vectors:

W := (W1, W2, W3, W4, W5, W6)T,Wc := (W1, W4)T,

Wnc := (W2, W3, W5, W6)T.

The notations Wc and Wnc are introduced in order to separate the characteristicpart of the vector W and the noncharacteristic part of W. We shall go backto this decomposition later on. It is obvious that estimating W is equivalent toestimating U.


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Let us define the following 6 6 symmetric matrices:

A0 :=

1 0 00 2c2 0 O0 0 2c2

1 0 0O 0 2c2 0

0 0 2c2

,(3.6)

A1 :=

vr c2 c2c2 2c2vr 0 Oc

2

0 2c2

vr vl c2 c2O c2 2c2vl 0

c2 0 2c2vl

,(3.7)

A2 :=

0 0 00 2c3 0 O0 0 2c3

0 0 0O 0 2c3 0

0 0 2c3

,(3.8)

where O stands for the 3 3 null matrix, as well as the following

(3.9) b :=0 vr vl1 vr

0 0

=0 2vr1 vr

0 0

, M :=c c c cc c 0 0

1 1 1 1

.Then, using (3.5)(3.6)(3.9), the linear equations (3.4) equivalently read

(3.10) LW = f , ifx2 > 0,B(Wnc, ) = g, ifx2 = 0,with newf and g, and where we have let

LW := A0 tW + A1 x1 W+ A2 x2 W ,

B(Wnc, ) := MWnc|x2=0 + b

tx1

.

Note that the kernel of

A2 is exactly the set of those W such that Wnc

=0 (and

Wc is arbitrary). The boundary{x2 = 0} is thus characteristic with multiplicity2. As already noted in earlier works, see e.g. [19, 23], we expect to lose control ofthe trace ofWc, that is, on the trace of the tangential velocities (v+, v). At theopposite, we expect to control the trace ofWnc on {x2 = 0}, that is, we expect tocontrol the trace of(+, , u+, u).


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3.2. The main result for the constant coefficients case. Before stating our

energy estimate for the system (3.10), we need to introduce some Sobolev weightednorms. First define the half-space

:= {(t,x1, x2) R3 s.t. x2 > 0} = R2 R+.The boundary is identified to R2.

For all real number s and all 1, define the space

Hs (R2) := {u D(R2) s.t. exp(t)u Hs (R2)}.

It is equipped with the norm

uHs (R2) := exp(t)uHs (R2).

Letting u := exp(t)u, one has

uHs (R2) us, , where v

2s, :=

1(2 )2 R2

(2 + ||2)s|v()|2 d,

where v is the Fourier transform of any function v defined on R2.For all integers k, one can define the space Hk () in an entirely similar way.The space L2(R+; Hs (R2)) is equipped with the norm

v2L2(Hs ) :=+

0

v(, x2)2Hs (R2) dx2.In the sequel, the variable in R

2

is (t,x1), while x2 is the variable in R+.Our first main result is stated as follows.

Theorem 3.1. Assume that the particular solution defined by(2.9) satisfies

(3.11) vr vl > 2

2 c.

Then there exists a positive constant C such that for all 1 and for all(W,) H2 (

) H2 (R2), the following estimate holds:

W2L2 () + Wnc|x2=02L2 (R2) + 2H1 (R2)(3.12)

C

13LW2L2(H1 ) + 12 B(Wnc, )2H1 (R2)

.


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Introducing

W := exp(t)W and

:= exp(t), we easily find that

(3.10) is equivalent to

(3.13)

LW := A0W + LW= exp(t)f, ifx2 > 0,

B (Wnc, ) := MWnc|x2=0 + b + t

x1

= exp(t)g, ifx2 = 0.

Consequently, Theorem 3.1 admits the following equivalent formulation.

Theorem 3.2. Assume that (3.11) holds. Then there exists a positive constantCsuch that for all 1 and for all(W , ) H2()H2(R2), the following estimateholds:

(3.14) W20 + Wnc|x2=020 + 21,

C

13

L

W

21, +

12

B (

Wnc,

)

21,

.

In (3.14), we have used the following notations for anyv L2(R+; Hs (R2)):v2s, := +

0

v(, x2)2s, dx2.For instance, ||| |||0, is the usual norm on L2() and does not involve , so

we shall denote it by ||| |||0. The norm ||| |||1, is the weighted norm onL2(R+, H1(R2)).

4. PROOF OF THEOREM 3.24.1. Some preliminary reductions. In this paragraph, we show that it is suf-

ficient to prove Theorem 3.2 in the particular case LW 0. This first reductionsimplifies many subsequent calculations. The argument we use was introduced in[23, page 636].

In order to simplify notations, we drop the tildas. Let W H2() and H2(R2). Then we define:

f := L W H1(

), g := B (Wnc, ) H1(R2).

Consider the following auxiliary problem:

(4.1)

L W1 = f , ifx2 > 0,MauxWnc1 |x2=0 = 0, ifx2 = 0,


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where Maux is defined by

Maux :=0 1 0 0

0 0 1 0

.

In the auxiliary problem (4.1), the boundary conditions are maximally dissipative(see the formula (3.6) defining the matrix A2). Since f L2(R+; H1(R2)), itfollows from [19] that there exists a function W1 L2(R+; H1(R2)), such that thetrace ofWnc1 on {x2 = 0} belongs to H1(R2), and that is a solution to (4.1). Inparticular, the function W1 satisfies the following estimates:

W120 C f20,(4.2a) Wnc1 |x2=021, C f21, = C

+0

f (, x2)21, dx2.(4.2b)Let us define W2 := W W1. It satisfies

L W2 = 0, ifx2 > 0,

B

(Wnc

2 , ) = g MWnc

1 |x2=0 , ifx2 = 0.Consequently, if we manage to prove that Theorem 3.2 holds true as long as theinterior source term is zero, we shall obtain

W220 + Wnc2 |x2=020 + 21, C2g MWnc1 |x2=021,

C2

g

21, +

Wnc1 |x2=0

21,

.

Then using (4.2) to estimate the H1 norm of the trace of Wnc1 as well as the L2

norm of W1, we shall derive our main energy estimate (3.14). Without loss ofgenerality, we thus assume from now on that W and satisfy

(4.3) A0W+ A0 tW + A1 x1 W + A2 x2 W = 0

in the interior domain , as well as the following boundary conditions(4.4) MWnc|x2=0 + b + tx1 = g, x2 = 0.

With slight abuse of notations, we still denote the source term in the boundaryconditions byg, instead ofg MWnc1 . This is of pure convenience.


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Recall that all matrices Aj are symmetric, and that A0 is positive definite, see

(3.6). Taking the scalar product of (4.3) with W and integrating over yields thefollowing inequality:W20 CWnc|x2=020.

Consequently, it is sufficient to derive an estimate of the form

(4.5)Wnc|x2=020 + 21, C2 g21,

in order to obtain (3.14).

We shall derive (4.5) by means of a Kreiss symmetrizer, whose constructionis detailed in the next paragraphs. Once performed a Fourier transform in (t,x1),the first step consists in eliminating the front in the boundary conditions(4.4). We emphasize that this operation is possible, even though the vortex sheetis a characteristic boundary. Then we shall detail the normal modes analysis andconstruct a symbolic symmetrizer.

4.2. Eliminating the front. As mentionned above, we focus on (4.3)(4.4),and perform a Fourier transform in (t,x1). The dual variables are denoted by(,). We also define :

=

+i. This is the Laplace dual variable (indeed, the

previous manipulations amount to performing a Laplace transform with respectto t). We obtain the following system of differential equations:

(A0 + iA1)W + A2 dWdx2

= 0, x2 > 0,(4.6a)

b(,) + MWnc(0) = g,(4.6b)where b(,) is simply defined by

(4.7) b(,) := b

i

=

2ivr

+ ivr0

.Recall that b and M are defined by (3.9). Observe that b(,) is homogeneous(of degree 1) with respect to (,). In order to take such homogeneity propertiesinto account, we define the hemisphere

:= {(,) CR s.t. ||2 + v2r2 = 1 and 0}.Recall that is the Laplace dual variable of the time variable t, while is theFourier dual variable of x1, so / is a velocity. Our definition of takes thisproperty into account.


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We denote by

the set

:= {(,,) [0, +[ R2 s.t. (,,) (0, 0, 0)} = ]0, +[ .It is the set of frequencies we shall consider in the sequel. We always identify(,) R2 and = + i C.

One crucial remark is that the symbol b(,) is elliptic, that is, it does notvanish on the closed hemisphere . More precisely, we have the following lemma.

Lemma 4.1. There exists a C mappingQ defined on

such thatQ has values

in GL3(C), is homogeneous of degree0, and satisfies

(,) , Q(, )b(, ) = 00

(,)

,where isC, homogeneous of degree1, and has the additional property:

min(,) |(,)| > 0.

Proof. We shall define the mapping Q on and then extend Q by homo-geneity. If we define

(,) , Q( , ) :=

0 0 1

+ ivr 2ivr 02ivr ivr 0

,then we check that Q has all the required properties. The corresponding is givenby

(,) , (, ) := | + ivr|2 + 4v2r2.Note that the last row ofQ(,) is nothing but b(,), when (,) .

Let us multiply the boundary conditions in (4.6) by the matrix Q(,). Weobtain:

(4.8)

00

(,)

+

(,)

(,)

Wnc(0) = Q(,)g,

where has two rows while has one row and(,)

(,)

:= Q(,)M.


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The exact expression of is useless, but we shall use the expression of :

(4.9)

(,) = 1 1 1 1

c( + ivl) c( + ivl) c( + ivr) c( + ivr)

,

(,) .Both and are homogeneous of degree 0 and C on .

The last equation in (4.8) is

(,) , ( , ) + (,)Wnc(0) = b(,)g,since b is the last row ofQ. Using the ellipticity property of (see Lemma 4.1)together with a uniform bound for and b on , we obtain

(,) , (||2 + v2r2)| |2 C(|Wnc(0)|2 + |g|2).Let us now integrate this last inequality with respect to (,) R2. (Recall that is the imaginary part of). Using Plancherels Theorem, we obtain

21, CWnc|x2=020 + g20(4.10) C

Wnc|x2=020 + 12 g21,

.

In order to derive (4.5), it is therefore sufficient to derive an estimate of thetrace ofWnc. Consequently, we focus on the reduced problem

(

A0

+i

A1)

W

+ A2

d

W

dx2 =

0, x2 > 0,

(,)Wnc(0) = h,and try to derive an estimate for Wnc(0). One has to remember that the sourceterm h C2 is easily estimated byg, see (4.8).

In the next paragraph, we recall that under the assumption made in Theorems3.1 and 3.2, the above boundary problem satisfies the Kreiss-Lopatinskii conditionbut violates the uniform Kreiss-Lopatinskii condition.

4.3. The normal modes analysis. Writing W=

(W1, W2, W3, W4, W5, W6)T,the two first equations in (4.6) are:

( + ivr)W1 ic2W2 + ic2W3 = 0,( + ivl)W4 ic2W5 + ic2W6 = 0.


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They do not involve derivation with respect to the normal variable x2. For >

0, we obtain an expression for W1 and W4 that we plug in the last four equations.This operation yields a system of ordinary differential equations of the followingform:

(4.11)

dWncdx2

= A(,)Wnc, ifx2 > 0,(,)Wnc(0) = h, ifx2 = 0.

The matrixA

(,) in (4.11) is given by

A(,) :=

r mr 0 0

mr r 0 00 0 l ml0 0 ml l

,(4.12)

r ,l := (1/c)( + ivr ,l)2 + (c/2)2

+ ivr ,l ,

mr ,l := (c/2)2 + ivr ,l .

A well-known result, that is due to Hersh [15] in the noncharacteristic case(see [23] for the extension to the characteristic case), asserts that the matrix A(,)has no purely imaginary eigenvalue as long as > 0. As a consequence, the sta-ble subspace ofA(,) has constant dimension when > 0. This dimensionequals the number of characteristics going out of the discontinuity. In our case,those theoretical results can be checked directly by computing the eigenvalues and

the stable subspace ofA(,). The following lemma gives an expression of thestable subspace.

Lemma 4.2. Let C and R, with > 0 and (,) . Theeigenvalues ofA(,) are the roots of the dispersion relations

2 = 2r m2r =1

c2( + ivr)2 + 2,(4.13a)

2

=2l

m2l

=1

c2(

+ivl)

2

+2.(4.13b)

In particular, (4.13a) (resp. (4.13b)) admits a unique rootr (resp. l) of negativereal part. The other root of (4.13a) (resp. (4.13b)) isr (resp. l), and haspositive real part. The stable subspaceE(,) ofA(,) has dimension 2, and is


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spanned by the two following vectors:

Er(,) :=c

22,

1c

( + ivr)2 + c2

2 ( + ivr)r, 0, 0T

,(4.14a)

El(,) :=

0, 0,1c

( + ivl)2 + c2

2 ( + ivl)l, c2

2T

.(4.14b)

Both r and l admit a continuous extension to any point (,) such that = 0 and(,) . This allows to extend both vectors Er andEl in (4.14) tothe whole hemisphere

. Those two vectors are linearly independent for any value of

(,)

.The symbolA(,) is diagonalizable as long as both r andl do not vanish,that is, when (vr c)i. Away from such points, A admits a C basis ofeigenvectors.

The proof follows from straightforward computations, and we shall omit it.We point out that the stable subspace E(,) is defined for all (,) ,

while the matrix A(,) has some poles on the boundary of, see (4.12). Thepoles are exactly those points (,) verifying = ivr ,l = ivr (recallthat we have vr = vl 0).

Following Majda and Osher [23], we define the Lopatinskii determinant as-sociated with the boundary conditions in the following way:

(4.15) (,) := det(,)Er( , ) E l(,) ,with defined by (4.9) and (Er, El) defined by (4.14). We emphasize that theLopatinskii determinant is defined on the whole hemisphere and is continu-ous with respect to (,). The first step in proving an energy estimate for (4.11)consists in determining whether

vanishes on

. The answer is given in the

following result.

Proposition 4.3. Assume that (3.11) holds. Then there exists a positive numberV1 such that for any(,) , one has(,) = 0 if and only if

= 0 or = iV1.

Each of these roots is simple. For instance, there exists a neighborhoodVof(0, 1/vr)in and a C function h defined on Vsuch that

(,) V,

(,) = h(,) and h(0, 1/vr) 0.

A similar result holds near(0, 1/vr) or near the points(iV1,) .The numberV1 is given byV21 := c2 + v2r c

c2 + 4v2r. In particular, one has:

0 < V1 < vr c. When = 0 or = iV1, both eigenmodes r andl arepurely imaginary.


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We postpone the proof of Proposition 4.3 to Appendix A. We simply note

here that the three critical speeds V1, 0, V1 are exactly the speeds of the kinkmodes exhibited in the work by Artola and Majda [2]. As a matter of fact, Artolaand Majda used a geometric optics approach, while we have followed here anormal modes analysis approach. However, the calculations are similar in bothcases (in [2], the number equals 1).

4.4. Constructing a symmetrizer: the interior points. We now turn to theconstruction of our degenerate Kreiss symmetrizer. The construction is microlo-cal and is achieved near any point (0, 0)

. The analysis is rather long since

one has to distinguish between five different cases. In the end, we shall consider a

partition of unity to patch things together and derive our energy estimate.In all the rest of the article, the letter denotes a generic positive constant(typically, though not necessarily, a rather small one).

We first consider the case (0, 0) with 0 > 0. Then the matrixA(,) is diagonalizable for all (,) close to (0, 0). A smooth (that is, C)basis of eigenvectors is given by the following family:

Er(, ), E l(,),

c

2

2,1

c

(

+ivr)

2

+

c

2

2

+(

+ivr)r, 0, 0

T

,0, 0,

1c

( + ivl)2 + c2 2 + ( + ivl)l, c2

2T

.

Both vectors Er and El are defined by (4.14). Therefore, there exists a C mappingT(,), defined on a neighborhood Vof (0, 0) in , with values in GL4(C),and such that

(,)

V, T (, )

A(,)T(,)1

=

r 0 0 00 l 0 0

0 0 r 00 0 0 l

.

The first two columns of the matrix T(,)1 are the vectors Er and El. In theneighborhood V, we define the symmetrizer r in the usual way:

(,) V, r (, ) :=

1 0 0 00 1 0 00 0 K 00 0 0 K

,

where K 1 is a positive real number, to be fixed large enough. In what follows,we use the standard notation

M := M+ M

2


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for all square matrix M with complex entries (M is the classical adjoint matrix).

The matrix r defined just above is hermitian, and we have

(4.16) (,) V, (r(,)T(,)A(,)T(,)1) I I,

for some positive constant . This is because r and l have negative real partwhen (,) V, and 1 when (,) .

We now simply need to fix K 1 in order to recover an estimate for the traceofWnc. We show that for K sufficiently large, the following inequality holds:

(4.17)

(,)

V, r (, )

+C((,))(,)

I,

where C is a positive constant and (,) := (,)T(,)1.Let Z = (Z, Z+)T C4, with Z, Z+ C2. We write

(,)Z= (,)

Z0

+ (,)

0

Z+

,

and recall that the first two columns of T(,)1 are Er and El. Because theLopatinskii determinant does not vanish at (0, 0), see (4.15) and Proposition

4.3, we obtain an estimate of the form

|Z|2 C0(|Z+|2 + |(,)Z|2),

for a suitable constant C0 that is independent of (,) V. With C0 satisfyingthis inequality, the definition ofr yields

r(,)Z,ZC4 + 2C0|(,)Z|2 = |Z|2 + K|Z+|2 + 2C0|(,)Z|2

|Z

|2

+(K

2C0)

|Z+

|2.

This gives (4.17) for K large enough (e.g., K= 2C0 + 1).4.5. Constructing a symmetrizer: the boundary points (case 1). We now

turn to the construction of the symmetrizer near those points (,) such that = 0. We first prove a general result on the behavior of the eigenmodes r ,l inthe neighborhood of such points.

Lemma 4.4. Let(0, 0)

so that0 = 0 and0 (vr c)i0. In

particular, r depends analytically on (,) near(0, 0). Then the two following

cases may occur:(1) The eigenmoder has negative real part at(0, 0), and, in a suitable neigh-

borhoodVof(0, 0), one has

r < 0.


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(2) The eigenmode r is purely imaginary at (0, 0). In this case, the derivative

r calculated at(0, 0) is a nonzero real number. In a suitable neighborhoodVof(0, 0) in , we haver .

A completely similar result holds forl near all points(0, 0) satisfying0 =0 and0 (vl c)i0.

Proof. Because 0 (vr c)i0, the eigenmode r is not zero at (0, 0)and it depends analytically on (,) by the implicit functions theorem.

The first case in Lemma 4.4 simply follows from the continuity of r withrespect to (,). In the second case, we use (4.13a) to derive

rr

= 1c2

( + ivr).

When = 0 and = 0, one has r iR \ {0} and (0 + ivr0) iR. Thisproves that the derivative r is real. We now remark that 0 ivr0 for, insuch a case, r has negative real part. Consequently, the derivative r is not

zero. The estimate on r is obtained by performing a Taylor expansion of rat (0, 0). According to Proposition 4.3, there are exactly four types of points on the

boundary of:(1) Those points (0, 0) where A(0, 0) is diagonalizable and the Lopatinskii

condition is satisfied at (0, 0).(2) Those points (0, 0) where A(0, 0) is diagonalizable and the Lopatinskii

condition breaks down at (0, 0).(3) Those points (0, 0) where

A(0, 0) is not diagonalizable, that is, 0

=(vr c)i0. In this case, Proposition 4.3 asserts that the Lopatinskii condi-tion is satisfied at (0, 0).

(4) Those points (0, 0) that are the poles ofA, that is, 0 = ivr0. At thosepoints, the Lopatinskii condition is satisfied.

As a matter of fact, an immediate consequence of Proposition 4.3 is that when-ever the Lopatinskii condition fails at (0, 0), then (0, 0) is not a pole and thesymbol A(,) is (smoothly) diagonalizable in a neighborhood of (0, 0). Thethree first categories of boundary points can thus be treated as in [6, 10, 17, 28],provided the technical assumptions of [10] near instability points hold. We are

going to show that such technical assumptions hold true. The last category ofboundary points (the poles of the symbol A) requires special attention.

We now deal with the construction of our symmetrizer in case 1: (0, 0) is such that A(0, 0) is diagonalizable and the Lopatinskii condition is satisfiedat (0, 0), that is, (0, 0) 0.


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Because A(0, 0) is diagonalizable, we have 0 (vr c)i0. Hence

there exists a neighborhood Vof(0, 0) in and a smooth basis of eigenvectorsofA defined on V. The smooth basis is the same as in the case of interior points(see the preceding paragraph). We thus have

(,) V, T (, )A(,)T(,)1 =

r 0 0 0

0 l 0 00 0 r 00 0 0 l

,

where, once again, the two first columns ofT(,)1

are exactlyEr and El. Wechoose r as in the case of interior points:

(,) V, r (, ) :=

1 0 0 00 1 0 00 0 K 00 0 0 K

,

with K 1 to be fixed large enough. Clearly, r is a hermitian matrix. Using

Lemma 4.4, we can already conclude that

(4.18) (,) V, (r(,)T(,)A(,)T(,)1) I.

Because the Lopatinskii condition is satisfied at (0, 0), it is possible to choose Klarge enough so that the following estimate holds:

(4.19) (,) V, r (, ) + C((,))(,) I.

In (4.19), we have let, as usual, (,) := (,)T(,)1. The analysis is thesame as what we have done for interior points. The estimates (4.18)(4.19) aresimilar (but not exactly identical) to (4.16)(4.17).

4.6. Constructing a symmetrizer: the boundary points (case 2). In thisparagraph, we consider a point (0, 0) such that the Lopatinskii determinant vanishes at (0, 0). From Proposition 4.3, we know that the symbol A issmoothly diagonalizable on a neighborhood Vof(0, 0) in

. In other words,

there exists a smooth mapping T(,) with values in GL4(C), and such that

(,) V, T (, )A(,)T(,)1 =

r 0 0 0

0 l 0 00 0 r 00 0 0 l

.


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The first two columns of T(,)1 are Er and El. In this case, we define our

symmetrizer r in the following (degenerate) way:

(,) V, r (, ) :=

2 0 0 0

0 2 0 00 0 K 00 0 0 K

,

with K 1 to be fixed large enough. The matrix r(,) is hermitian and we have

(4.20) (r(,)T(,)A(,)T(,)1)

2 0 0 00 2 0 00 0 1 00 0 0 1

, (,) V.

It only remains to fix K appropriately in order to recover an estimate on theboundary. The choice ofK relies on the following lemma.

Lemma 4.5. Let (0, 0) be a point where the Lopatinskii determinantvanishes. Then there exists a neighborhoodVof(0, 0) in and a constant0 > 0such that the following estimate holds for all(,) V:

Z C2,(,)Er( , ) E l(,)Z2 02|Z|2.

Before proving Lemma 4.5, let us first show that it enables us to obtain anestimate between r and the boundary matrix . More precisely, we are going toshow the following estimate:

(4.21) (,) V, r (, ) + C((,))(,) 2I,

for an appropriate positive constant C. The definition of(,) is the same as inthe preceding cases.

Let Z = (Z, Z+)T C4, where Z and Z+ belong to C2. Once again, wewrite

(,)Z

=(,)

Z

0 + (,)0

Z+ ,and we recall that the first two columns ofT(,)1 are Er and El. Using Lemma4.5, we obtain

02|Z|2 C0(|Z+|2 + |(,)Z|2).


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We thus derive

r(,)Z,ZC4 + 2C00 |(,)Z|2 = 2|Z|2 + K|Z+|2 + 2C0

0|(,)Z|2

2|Z|2 +

K 2C00

|Z+|2.

Choosing K= 2C0/0 + 1 yields

r(,)Z,ZC4 + 2C0

0|(,)Z|2 2|Z|2 + |Z+|2 2|Z|2.

Here we have much weakened our estimate for the last two components Z+. How-ever, an inequality like (4.21) is simpler to use since it does not distinguish betweenthe different coordinates of the vector Z. One should remember that the real lossof control is on the modes r and l, and not on the modes r and l.

The proof of Lemma 4.5 is postponed to Appendix A. It relies on the fact thatthe roots of the Lopatinskii determinant are simple (see Proposition 4.3).

4.7. Constructing a symmetrizer: the boundary points (case 3). In thisparagraph, we consider a point (0, 0)

such that 0 = ivr0 ic0. (The

case

0 = iv

l

0ic

0 is entirely similar, and we shall not detail it). Becausev

l =vr and vr > c2, we have 0 ivl0 ic0, and therefore, the eigenmodel depends smoothly on (,) in a neighborhood V of (0, 0). Oppositely,r is only continuous with respect to (,) near (0, 0), but 2r is C near(0, 0). This is because (4.13a) has a double root when (,) = (0, 0).

When (,) is close to (0, 0), the following family is a C basis ofC4:

(mr, mr, 0, 0)T, (i, 0, 0, 0)T, El(,),

0, 0,1

c(

+ivl)

2

+c

22

+(

+ivl)l,

c

22

T

.

Recall that mr is defined by (4.12). Let T(,)1 be the (regular) matrix whosecolumns are those four vectors. We compute

T(,)A(,)T(,)1 =ar O OO l 0

O 0 l

,where ar is the 2

2 matrix defined as follows:

ar(,) :=

c2r

+ ivr i2imrc2r + ivr

c2r + ivr

.


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In particular, we have

ar(0, 0) =0 i

0 0

=: iN.Moreover, we make the following observations:

For all (,) close to (0, 0) so that iR, ar has purely imaginary coeffi-cients.

The lower left coefficient r ofar satisfies

r

(0, 0) R

\ {0}.Here we have exhibited a basis in which Ralstons result [28] applies. Readers whoare familiar with the theory will have recognized the block structure condition,that is a consequence here of the strict hyperbolicity of (2.1), see [24].

We are looking for a symmetrizer r under the form

r(,) =

s(,) O O

O 1 0O

0K

,

where K 1 is a real number, to be fixed large enough, and s is some 2 2hermitian matrix, depending smoothly on (,). More precisely, we are lookingfor the matrix s under the following form

s(,) =

0 e1e1 e2

E

+

f(,) 00 0

F(,)

i

0 gg 0

G

,

where e1, e2 and g are real numbers, and f is a real valued C mapping thatvanishes at (0, 0), see [6, 17, 28]. We choose e1 in the following way:

e1 :=

r

(0, 0)

1 R \ {0},

where r is defined just above. This choice may look surprising but it will bejustified later on.

Now we observe that our choice ofs yields (0=

0):

r (0, 0) =

0 e1 0 0e1 e2 0 00 0 1 00 0 0 K

.


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Moreover, the first and third columns ofT (0, 0)1 are nothing but Er(0, 0)

and El(0, 0). This is due to the equalityr = mr when (,) = (0, 0).Because Lopatinskiis condition is satified at (0, 0), we can choose e2 and Klarge enough such that the following estimate holds:

r (0, 0) + C((0, 0))(0, 0) I.

As was done before, we have let (,) := (,)T(,)1. Up to shrinkingV, we have thus derived the following estimate

(4.22) (,) V, r (, ) + C(

(,))

(,) 1

2 I,

for a suitable constant C.Now, we show how to choose the real valued function f and the real number

g. We first write

ar(,) = ar(0, 0) + (ar(i,) ar(0, 0)) + (ar(,) ar(i, )),

and then use Taylors formula. We obtain

ar(,) = iN+ (ar(i,) ar(0, 0)) + ar

(i, ) + 2M(,),

for a suitable continuous function M. Because ar has purely imaginary coefficientswhen is purely imaginary, we have

ar(i,) ar(0, 0) =

ib1(i, ) 0ib2(i, ) ib3(i,)

=: iBr(i,),

for some C, real valued mappings b1, b2, b3. Those three mappings obviouslyvanish at (0, 0). We choose

f(,) := e1(b1(i,) b3(i,)) + e2b2(i,),

so that f is a C real valued mapping that vanishes at (0, 0). Moreover, thischoice off implies that the matrix

(E+ F(,))(N+ Br(i,))

is real and symmetric for all (,). Consequently, we obtain

(s(,)ar(,)) =

GN+ Ear

(i,)

+ L(,),


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where L is continuous (it is even C) and L(0, 0) = 0. Our choice for e1 yields

GN+ Ear

(0, 0)

=

0 00 g

+

1

,

where the are coefficients that only depend on e1, e2 (that have already beenfixed) and (0, 0). Choosing g large enough, and shrinking V if necessary, weend up with

(s(,)ar(,)) 1

4 I ,

and we thus obtain

(4.23) (,) V, (r(,)T(,)A(,)T(,)1) I.

4.8. Constructing a symmetrizer: the boundary points (case 4). We nowconsider the last case, which is (0, 0)

with 0 = ivr0. (We shall notdetail the case 0 = ivl0 that is entirely similar). The symbol A is not definedat (0, 0), while the stable subspace E ofA admits a continuous extension atthis point. The family(Er, El) is a C basis ofE near (0, 0), see Lemma 4.2,and Lopatinskiis condition is satisfied near (0, 0), see Proposition 4.3.

The eigenmode r has negative real part when (,) is close to (0, 0), see(4.13a). Oppositely, the eigenmode l is purely imaginary when (,) is closeto (0, 0) and = 0, see (4.13b). Furthermore, both r and l dependanalytically on (,) in a neighborhood Vof(0, 0).

The matrix A is not smoothly diagonalizable on V. This is because the eigen-vector associated with the eigenvalue r tends to be parallel to the eigenvectorassociated with the eigenvalue r. Consequently, Majda and Oshers construc-tion of a symmetrizer in this case involves a singularity in the symmetrizer, see[23]. We prefer to avoid this singularity and construct a smooth (that is, C)symmetrizer in the whole neighborhood V. This is possible if we go back to theoriginal system:

(A0 + iA1)

W + A2 d

W

dx2= 0, x2 > 0,(4.24a)

(,)Wnc(0) = h,(4.24b)The following analysis is inspired from [4]. For (,) in a neighborhood V of(0, 0), the following matrix is regular (that is, invertible):


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(4.25) T(,) :=

1 ic( + ivr cr) 0

0c

22 0 O

0 ( + ivr)(r r) 11 ic( + ivl cl) ic( + ivl + cl)

O 0 ( + ivl)(l l) ( + ivl)(l + l)

0 c2 2 c2

2

.

To avoid overloading the equations, we shall simply denote by O the 3 3 zeromatrix. The determinant ofT(,) is given by

det T(,) = c2

24( + ivl)l,

and it is easy to check that this quantity does not vanish near (0, 0). The matrix

T depends smoothly on(,)

in the whole neighborhood V. It has no pole.Let us define

(4.26a) r := ( + ivr)(r r) = 1c

( + ivr)2 + c2 2 ( + ivr)r,

(4.26b) l := ( + ivl)(l l) =1c

( + ivl)2 + c2

2 ( + ivl)l.

The main idea now is that (4.24a) has a simple expression if we decomposethe vector W in the basis defined by T. In other words, the matrices (A0 +iA1)T(,) and A2T(,) have a rather similar structure. Performing somemanipulations on the rows of these two matrices, we shall transform (4.24a) intoan almost diagonal system of differential equations.

After some simplifications, we obtain the following expressions:

(A0 + iA1)T

=

+ ivr 0 ic2

ic2

c4

2

r 0 Oic2 2c3rr 2c2( + ivr)3pt] + ivl 0 03pt] O ic2 2c3ll 2c3l+l

ic2 c42l c42l

,


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A2T =

0 0 0

0 c42 0 O0 2c3r 2c30 0 0

O 0 2c3l 2c3+l

0 c42 c42

,

where r and l are defined by (4.26a)(4.26b). These expressions may looksomehow complicated, but it is not so hard to simplify them multiplying on theleft by a suitable symmetrizer. Indeed, for all (,) in a neighborhood

Vof

(0, 0), let us define the following matrix:

(4.27) R1(,) :=

1 0 0

0 1 0 O

2ic4( + ivr cr) 2c3r c421 0 0

O

2ic4(

+ivl

+cl) c42 2c3+l

2ic4( + ivl cl) c42 2c3l

.

First we obtain

R1A2T = diag(0, c42, 2c72, 0, 4c72( + ivl)l, 4c72( + ivl)l),

where diag(a1, . . . , ap) stands for the diagonal matrix whose diagonal elementsare a1, . . . , ap. We also obtain the following expression:

(4.28) R1(A0 + iA1)T =

+ ivr 0 ic2ic2 c42r 0 O

0 0 2c72r + ivl 0 0

O 0 4c722l ( + ivl) 0

0 0 4c7

2

2l ( + ivl)

Recall that l, and (+ivl) do not vanish in the neighborhood Vof(0, 0),up to shrinking V. The following matrix is therefore a C mapping of(,) onV:


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(4.29) R2(,)

:= diag1, 1, 12c72

, 1, 14c72l( + ivl) ,

14c72l( + ivl)

.We define

S(,) := R2(,)R1(,),where R1 and R2 are defined by (4.27)(4.29). It is now easy to derive the follow-ing equalities for all (,) in V:

S(,)A2T(,) = diag(0, c42, 1, 0, 1, 1),(4.30a)S(,)(A0 + iA1)T(,)(4.30b)

=

+ ivr 0 ic2ic2 c42r 0 O

0 0 r + ivl 0 0

O 0 l 00 0 l

.

In particular, it is important to observe that both matrices S and

Tare

C on the

whole neighborhood V. Up to shrinking V, we may assume that r has negativereal part for all (,) V. This is possible because r = || at (0, 0).

Though a little complicated, the preceding calculations are based on the sim-ple idea that the differential equations (4.24a) have an easy expression if we replacethe standard coordinates by the coordinates on the stable subspace. The difficultycomes from the fact that we deal with the differential system satisfied byW and not

with the system satisfied byWnc. Because the boundary is characteristic, (4.24a) isnot an ordinarydifferential equation, and we thus need to diagonalize simultane-ously(A0 + iA1) and A2. Even though the matrix S(A0 + iA1)T is notdiagonal, we shall see that the reduced expressions (4.30a)(4.30b) are sufficientto derive energy estimates in such a neighborhood Vof the pole (0, 0).

4.9. Derivation of the energy estimate. We now turn to the derivation ofthe estimate (4.5). Recall that we are considering a function W H1() suchthat

(A0 + iA1)W + A2 dWdx2

= 0, x2 > 0,

(,)Wnc(0)

=h,

where h is obtained from the source term g in (4.6) after eliminating the unknownfront:

(,) , h = 0 0 1 + ivr 2ivr 0

g.


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32/72


We take the scalar product of this later ordinary differential equation with riUi

and integrate with respect to x2 onR

+. Then we take the real part of the obtainedequality and use (4.31). These operations yield the classical Kreiss estimate:

i

+0

|Ui(,,x2)|2 dx2 + 12 |Ui(,, 0)|2 Cii(,)2|h(,)|2.

Now we use the definition of Ui and a uniform bound for Ti(,)1 on thesupport ofi, to derive

i(,)2

+

0|Wnc(,,x2)|2 dx2 + i(,)2|

Wnc(,, 0)|2(4.32)

Cii(,)2|h|2.In the second case, Vi is a neighborhood of a zero of the Lopatinskii deter-

minant. On such a neighborhood, there exist two C mappings ri and Ti suchthat

ri is hermitian, Ti has values in GL4(C), the following estimates hold for all (,) Vi:

(ri(,)Ti(,)A(,)Ti(,)1

) i3

I,(4.33a)ri(,) + Ci((,)Ti(,)1)(,)Ti(,)1 2I.(4.33b)

These inequalities correspond to (4.20)(4.21).As was done before, we first extend ri and Ti as C mappings on the whole

hemisphere . Then we extend Ti and i as homogeneous mappings of degree 0with respect to (,), and we extend ri as a homogeneous mapping of degree 2with respect to (,). Thus (4.33) reads

(4.34a) (ri(,)Ti(,)A(,)Ti(,)1

) i3

I,

(4.34b) ri(,) + Ci(||2 + v2r2)((,)Ti(,)1)(,)Ti(,)1 2I,

for all (,) R+ Vi. Once again, we define

Ui(,,x2) := i(,)Ti(,)

Wnc(,,x2).

Because Vi does not contain any pole ofA, we still havedUidx2

= Ti(,)A(,)Ti(,)1Ui, x2 > 0,

(,)Ti(,)1Ui(0) = ih.


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We perform the same calculations as above (in the first case), but now we use

(4.34) instead of (4.31). We obtain

i(,)2+

0|Wnc(,,x2)|2 dx2 + i(,)2|Wnc(,, 0)|2(4.35)

Ci2

i(,)2|h|2(||2 + v2r2).

In the third and last case, Vi is a neighborhood of a pole of the symbol A. Forinstance, Vi is a neighborhood of a point (ivr0, 0)

. In this case, we have

shown that there exists C mappings Ti andS

i defined on Vi such that Ti has values in GL6(C), both relations (4.30a)(4.30b) hold on Vi.Recall that r has negative real part on Vi.

Here, we extend i, Ti and Si as homogeneous mappings of degree 0 withrespect to (,). Moreover, we shall make as if Ti and Si were defined on the

whole hemisphere (this is of pure convenience since only the values on thesupport ofi are involved in what follows). We define

Ui(,,x2) := i(,)Ti(,)1W(,,x2) C6.The components of the vector Ui are denoted as follows

Ui = (Ui,1, Ui,2, . . . , U i,6)T,

and we also define

Unci :

=(Ui,2, Ui,3, Ui,5, Ui,6)

T

C

4.

Using the definition (4.25) ofTi(,), it is clear that the vector Unci is given by arelation of the form

Unci = Ti(,)1Wnc, with Ti(,) GL4(C).We also note that the first and third column vectors of the matrix Ti(,) areexactly the vectors Er(,) and El(,). With these notations and definitions,

we get

(A0 + iA1)Ti(,)Ui + A2Ti(,) dUidx2

= 0, x2 > 0,

(,)Ti(,)Unc

i (0) = i(,)h.


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Multiplying the equation in {x2 > 0} bySi(,) and using (4.30a)(4.30b), we

obtain the following system:( + ivr)Ui,1 + ic2Ui,3 = 0,(4.36a)

ic2Ui,1 + c42

||2 + v2r2rUi,2 c

42

||2 + v2r2dUi,2dx2

= 0,(4.36b)

rUi,3 + dUi,3dx2

= 0,(4.36c)( + ivl)Ui,4 = 0,(4.36d)

lUi,5 dUi,5dx2 = 0,(4.36e)

lUi,6 + dUi,6dx2

= 0.(4.36f)

Recall that when (,) belongs to the conical set R+ Vi, one has

r (||2 + v2r2)1/2 and l ,

for a suitable constant > 0. Because Ui,3(x2) and Ui,6(x2) belong to L2(R+),

for > 0, (4.36c) and (4.36f) implyUi,3 0 and Ui,6 0. Using (4.36a) and(4.36d), we also obtain Ui,1 0 and Ui,4 0. Eventually, (4.36b) and (4.36e)reduce to

rUi,2 dUi,2dx2

= 0 and lUi,5 dUi,5dx2

= 0.

Because the first and third columns ofTi(,) are nothing but Er and El definedin Lemma 4.2, the boundary conditions for Ui,2 and Ui,5 read:

(4.37) (,)Er( , ) E l(,)Ui,2(0)

Ui,5(0) = i(,) h.Using the properties ofr and l on the conical set R+ Vi, we derive

(||2 + v2r2)1/2+

0|Ui,2(,,x2)|2 dx2 C|Ui,2(,, 0)|2,

+0

|Ui,5(,,x2)|2 dx2 C|Ui,5(,, 0)|2.

Because the uniform Lopatinskii condition is satisfied on Vi, (4.37) yields thefollowing estimate:

Ui,2(,, 0)Ui,5(,, 0)

2

Ci(,)2|h|2.


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Eventually, we obtain

+

0|Unci (,,x2)|2 dx2 + |Unci (,, 0)|2 Ci(,)2|h|2.

We now use the definition of the vector Unci to derive

(4.38) 2i

+0

|Wnc(,,x2)|2 dx2 + 2i |Wnc(,, 0)|2 C2i |h|2.We now add up (4.32)(4.35)(4.38), and use that the is form a partition

of unity. We obtain

+0

|Wnc(,,x2)|2 dx2 + |Wnc(,, 0)|2 C2

|h|2(||2 + v2r2).

We have already recalled that h is simply obtained from the source term g in (4.6)by multiplying by a uniformly bounded matrix. Thus integrating the previousinequality with respect to (,) R2 and using Plancherels theorem yields thedesired estimate:

Wnc20 + Wnc|x2=020 C2 g21, .Combining with (4.10), we have finished to prove (4.5).

5. THE VARIABLE COEFFICIENTS LINEARIZED PROBLEM

5.1. The linearized equations and the main result. We introduce the lin-earized equations around a state given by a perturbation of the constant solutionin (2.9). More precisely, let us consider the functions

(5.1) Ur = vr

0

+ Ur, Ul = vr

0

+ Ul, r, l,where , vr are fixed positive constants (in this section we introduce the smallchange of notation vr vr for the piecewise constant solution) and where

Ur(t,x1, x2)=

r(t,x1, x2)vr(t,x1, x2)

ur(t,x1, x2) , Ur(t,x1, x2) = r(t,x1, x2)vr(t,x1, x2)

ur(t,x1, x2) ,Ul(t,x1, x2) =

l(t,x1, x2)vl(t,x1, x2)ul(t,x1, x2)

, Ul(t,x1, x2) =l(t,x1, x2)vl(t,x1, x2)

ul(t,x1, x2)

.


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The functions

r,

l are perturbations of the change of variables. The index r

(resp. l) denotes the state on the right (resp. on the left) of the interface (after thechange of variables). We assume that

Ur, Ul, r, l W2,(),(5.2a)(Ur, Ul)W2,() + (r, l)W2,() K0,(5.2b)

where K0 is a suitable positive constant, and that the perturbations Ur, Ul havecompact support. These quantities are linked by the Rankine-Hugoniot condi-tions and the continuity condition for the functions

r ,l that, written in the form

of (2.5), become r(t,x1, 0) = l(t,x1, 0) = (t,x1),(5.3a)(vr vl)|x2=0 x1 (ur ul)|x2=0 = 0,(5.3b)

t + vr|x2=0 x1 ur|x2=0 = 0,(5.3c)(r l)|x2=0 = 0.(5.3d)

The functions

r and

l should also satisfy

tr + vr x1r ur = 0,(5.4a)tl + vl x1l ul = 0,(5.4b)

together with

(5.5) x2r 0, x2l 0,for a suitable constant 0 > 0, in the whole closed half-space {x2 0}.

Let us consider the families Us=

Ur ,l

+sU

,

s

= r ,l + s, where s is asmall parameter. We compute the linearized equationsL(Ur ,l, r ,l)(U,) := d

dsL(Us , s )Us|s=0 = f.

We obtain

(5.6) t U+ + A1(Ur) x1 U+ +1

x2r (A2(Ur) tr x1rA1(Ur)) x2 U++

[dA1(Ur)U+

]x1 Ur

x2

+

(x2r)2 (A2(Ur) tr x1rA1(Ur)) x2 Ur+ 1

x2r

dA2(Ur)U+ t+ x1+A1(Ur) x1rdA1(Ur)U+x2 Ur = f+,in the domain {x2 > 0}, and a similar equation with U, , Ul, l, f insteadofU+, +, Ur, r, f+. Recall that, according to the definition in (2.7), (2.8), the


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first row in (5.6) may be simply denoted byL(Ur,

r)U+, namely we set:

L(Ur, r)U+ := t U+ + A1(Ur) x1 U++ 1

x2r (A2(Ur) tr x1rA1(Ur)) x2 U+.The equation (5.6) and the corresponding one for (U, ) may be simplified bythe introduction of la bonne inconnue (the good unknown) as in [1]:

(5.7) U+ :

=U+

+

x2r x2 Ur, U := U

x2l x2 Ul.A direct calculation shows that U+, U satisfy

L(Ur, r)U+ + C(Ur, Ur, r)U+ + +x2r x2 [L(Ur, r)Ur] = f+,

L(Ul, l)U + C(Ul, Ul, l)U + x2l x2 [L(Ul, l)Ul] = f,

where

C(Ur, Ur, r)U+ := [dA1(Ur)U+]x1 Ur+ 1

x2r [dA2(Ur)U+ x1rdA1(Ur)U+] x2 Ur,and with a similar expression for C(Ul, Ul, l)U. In view of the results in[1, 13], we neglect the zeroth order term in

+,

in the linearized equations and

thus consider the linear equations

LrU+ := L(Ur, r)U+ + C(Ur, Ur, r)U+ = f+,(5.8a)LlU := L(Ul, l)U + C(Ul, Ul, l)U = f.(5.8b)

We easily verify, using (5.2), that the coefficients of the operators L(Ur, r) andL(Ul, l) are in W2,(), that is

A1(Ur) W2,(

),

1x2

r

(A2(Ur) t

r x1

rA1(Ur)) W2,(

),

A1(Ul) W2,(), 1x2l (A2(Ul) tl x1lA1(Ul)) W2,().

Moreover, we have C(Ur ,l, Ur ,l, r ,l) W1,(). It is clear that the linearizedequations (5.8) form a symmetrizable hyperbolic system. For instance, a Friedrichs


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symmetrizer for the operator Lr is given by

Sr :=

(p(r)/r) 0 0

0 r 00 0 r

.Using (5.4), we compute

Srx2

r

(A2(Ur) t

r x1

rA1(Ur))

= 1x2r

0

p(r) x1r p(r)p(r) x1r 0 0

p(r) 0 0

,and we thus expect to control the traces of U+,1 and (U+,3 x1rU+,2) on theboundary{x2 = 0}. These considerations motivate the introduction of the fol-lowing operator:

(5.9) PU|x2=0 :=

U,1U

,3

x1 U

,2

|x2=0

.

We now turn to the linearized boundary conditions. The linearization of (2.5)gives

+|x2=0 = |x2=0 = ,(vr vl) x1 + (v+ v) x1 (u+ u) = g1,

t + vr x1 + v+x1 u+ = g2,+ = g3,

on the boundary{x2 = 0}. Let us introduce the vector b0 = (0, 1, 0)T and thematrices

b(t,x1) :=0 (vr vl)|x2=01 vr|x2=0

0 0

,

M(t,x1) :=

0 x1 1 0 x1 10 x1 1 0 0 0

1 0 0 1 0 0

.

Let us also denote U = (U+, U)T, = (t, x1 )T and g = (g1, g2, g3)T.Then the boundary conditions equivalently read

+|x2=0 = |x2=0 = , b + MU|x2=0 = g.


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In terms of the good unknown U defined by (5.7), the linearized boundary condi-

tions read +|x2=0 = |x2=0 = ,(5.10a)B(U,) := b + M

x2 Ur/x2rx2 Ul/x2l

b

+ MU|x2=0 = g.(5.10b)

We observe that the linearized boundary conditions only involve the traces ofPU+and PU, with P defined by (5.9). With this notation, we can state our main result(the norms are the weighted norms defined in Section 3).

Theorem 5.1. Assume that the particular solution defined by(5.1) satisfies

(5.11) vr >

2c(),

and that the perturbations Ur ,l, r ,l have compact support and are small enoughin W2,(

). Then there exist some constantsC1 and 1 1, that only depend on

K0 and0 (defined in (5.2), (5.5)), such that for all

1 and for all(U,)H2 () H2 (R2) the following estimate holds:

(5.12) U2L2 () + PU|x2=02L2 (R2) + 2H1 (R2)

C1

13LU2L2(H1 ) + 12 B(U,)2H1 (R2)

.

The linearized operatorsL andB are defined in (5.8) and (5.10).

The remaining part of this section is devoted to the proof of Theorem 5.1.

5.2. Some preliminary transformation. Let us consider again the linearizedequations (5.8). After multiplication by the Friedrichs symmetrizer defined aboveand a straightforward integration by parts, we easily prove the following lemma.

Lemma 5.2. There exist two constants C > 0 and 0 1 such that for all 0, the following estimate holds:

U+

2L2 (

)

C

LrU+

2L2 (

) + C

PU+|x2=0

2L2 (R2)

,

where the operatorP is defined in (5.9). The estimate forU is the same, namely:

U2L2 () C LlU2L2 () + CPU|x2=02L2 (R2).


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As was done in the constant coefficients case, it remains to show an estimate

of the tracesP

U|x2=0 and the front function in terms of the source terms inthe interior domain and on the boundary. In order to prove such an estimate,it is convenient to transform further the interior equations (5.8) in order to deal

with a problem with a constant and diagonal boundary matrix (i.e., the matrixcoefficient of x2 in the differential operators L

r ,l). This is possible because the

boundary matrix has constant rank in the whole closed half-space. Namely, let usconsider the coefficients ofx2 U in (5.8). The coefficients are equal to

1x2

(A2(U) t

x1A1(U)),

where we forget for the moment the indices r, l. Under the assumption (5.4), thiscoefficient reduces to the matrix

A2(U, ) = 1x2

0 x1 p()/x1 0 0

p()/

0 0

which has eigenvalues

1 = 0, 2,3 = c()x1x2 .

Here we have introduced the notation x1 = 1 + (x1)2. In order to di-agonalize the above matrix we compute the eigenvectors associated to the aboveeigenvalues. We obtain respectively the vectors

0 1 x1T , x1 c() x1 c() T ,x1 c() x1 c()

T.

Observe that these eigenvectors are not orthonormal (because A2 is not symmet-ric). Thus, we may define the following (non orthogonal) matrix

T(U, ) :=

0

x1 x11 c()

x1 c() x1

x1 c() c()

,


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which permits to diagonalize the above matrix A2(U,

):

T1(U, )A2(U, )T(U, ) =0 0 00 2 0

0 0 3

.In order to obtain a constant boundary matrix in the differential operators, we alsointroduce the matrix

A0(U, ) := 1 0 00 12 00 0 13

=

1 0 0

0 x2c()x1 00 0 x2

c()x1

.

It follows that A0T1A2T = I2 := diag(0, 1, 1). Let us define the new unknownfunctions W+ := T1(Ur, r)U+, W := T1(Ul, l)U, and set

Tr ,l := T (Ur ,l,

r ,l), Ar ,l0 := A0(Ur ,l,

r ,l).

After multiplication on the left side of the equations in (5.8) byAr ,l0 T1r ,l , we see

that W solve the equations

Ar0 t W+ +Ar1 x1 W+ + I2 x2 W+ +Ar0CrW+ = F+,(5.13a)

Al0 tW +Al1 x1 W + I2 x2 W +Al0ClW = F,(5.13b)

where we have set (with slight abuse of notation)

Ar ,l1 := Ar ,l0 T1A1T (Ur ,l, r ,l),Cr ,l := [T1tT + T1A1x1 T + T1A2x2 T + T1CT](Ur ,l, Ur ,l, r ,l),

F = Ar ,l0 T1r ,l f.

The above equations (5.13) are equivalent to the linearized equations (5.8). Intro-ducing W := et W, the equations (5.13) become equivalent to

LrW

+ :=

Ar0W+

+Ar0 tW

+(5.14a)

+ Ar1 x1W+ + I2 x2W+ +Ar0CrW+ = et F+,Ll W := Al0W +Al0 tW(5.14b)

+ Al1 x1W + I2 x2W +Al0ClW = et F.


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Recall that we have Ar ,lj W2,(

), and Cr ,l W1,(

). Using the vector W =(W+, W)

T

as defined above, the boundary conditions (5.10) become equivalentto

+|x2=0 = |x2=0 = ,(5.15a)b + b + M

Tr 00 Tl

W|x2=0 = g.(5.15b)

Introducing W and := et, := et , the equations (5.15) are alsoequivalent to

+ = = ,(5.16a)B (W , ) := b0 + b + b + M

Tr 00 Tl

W|x2=0 = et g.(5.16b)From (5.2) we have

b W2,(R2), b W1,(R2),M W2,(R2), Tr ,l|x2=0 W

2,(R2).

The next step is to look for an a priori estimate of the solution to the (weighted)linearized problem (5.14), (5.16). In view of Lemma 5.2, we are looking for anestimate ofPU+ and PU. Of course, we shall derive this estimate using the newfunction W. The reader should keep in mind the relations

PU+|x2=0 = x1 (W

+2 + W+3 )|x2=0

crr

x1 2(W+2 W+3 )|x2=0

,

PU|x2=0 = x1 (W2 + W3 )|x2=0cl

lx1 2(W2 W3 )|x2=0

,from which we easily deduce the estimate

(5.17) PU+|x2=0L2 (R2) + PU|x2=0L2 (R2) C((W+2 , W+3 )|x2=0L2 (R2) + (W2 , W3 )|x2=0L2 (R2)).

We are thus led to estimating the trace of the vector (W+2 ,W+3 , W2 , W3 ), whenW is a solution to the (weighted) linearized equations (5.14), (5.16). From nowon, for the sake of simplicity, we drop the tildas and write W, , instead ofW, , . Observe that , are coupled to W only through the boundaryconditions.


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5.3. Paralinearization. We refer to Appendix B for the definition of parad-

iff

erential symbols and operators, where the reader will also find the main resultson paralinearization and symbolic calculus. We recall that the Fourier dual vari-ables of(t,x1) are (,), and that we always denote = + i the Laplace dualvariable of t. Recall we have introduced the positive constants K0, 0 in (5.2),(5.5). We now turn to the paralinearization of the linearized equations.

(1) The boundary conditions. Define the following symbols:

b0 :

=

01

0

,

b1(t,x1) :=vr vlvr

0

(t,x1, 0),b(t,x1,,,) := b0 + ib1(t,x1).

Because b0 is constant, we have

b0 + b0 t = Tb0 .

The main paralinearization estimate (Theorem B.9) yields

b1 x1 Tib1 1, Cb1W2,(R2) 0 C(K0)

1, .

We now easily obtain

(5.18) b0 + b0 t + b1 x1 Tb 1, C(K0) 1, .

We also have the following inequalities:

b Tb

1, CbW1,(R2) 0 C(K0, 0)

1, ,(5.19a)

Tb

1, CbL(R2) 1, C(K0, 0)1, ,(5.19b)

where b is defined by (5.10). Eventually, we define the symbol

M(t,x1) := M(t,x1, 0)

Tr 00 Tl

(t,x1, 0),


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with the matrices M, Tr, Tl defined above. Recall that the state around which the

equations are linearized satisfiesr(t,x1, 0) = l(t,x1, 0) = (t,x1), r(t,x1, 0) = l(t,x1, 0).A direct calculation yields

M =

0 crr

x1 2crr

x1 2 0cll

x1 2 cll

x1 2

0 crr

x1 2crr

x1 2 0 0 00

x1

x1

0

x1

x1

.

Thus the matrix M only acts on the noncharacteristic part of the vector W =(W+, W), that is, Wnc := (W+2 , W+3 , W2 , W3 ). Since M W2,(R2), we have

MW|x2=0 TMW|x2=01,

C

MW2,(R2) Wnc|x2=00(5.20)

C(K0)

Wnc|x2=00.

Adding (5.18)(5.19)(5.20), we obtain the paralinearization estimate for the

boundary operator:

(5.21) B (W,)Tb TMW|x2=01, C(K0, 0)

1, + 1

Wnc|x2=00

.

We recall that the boundary operator B is defined by (5.16). Observe that in theparalinearized version ofB , there is no more zeroth order term in .

(2) The interior equations. We first estimate the paralinearization error forfixed x2, and then integrate with respect to x2. For instance, we have

Ar0 W+ TAr0 W+21,=+

02Ar0 W+(, x2) TAr0 W+(, x2)21, dx2

C+

0

Ar0 (, x2)2W2,(R2)W+(, x2)20 dx2 CAr02W2,()W+20 C(K0)W+20.

In a completely similar way, we obtain the following estimates

|||Ar0 t W+ TiAr0 W+|||1, C(K0)|||W+|||0,

|||Ar1 x1 W+ TiAr1 W+|||1, C(K0)|||W+|||0,

|||Ar0CrW+ TAr0CrW+|||1, C(K0, 0)|||W+|||0.


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Adding these inequalities, we end up with the paralinearization estimate for the

interior equations:|||LrW+ TAr0 +iAr1 +Ar0CrW

+ I2 x2 W+|||1,(5.22) C(K0, 0)|||W+|||0,

where the linearized operator Lr is defined by (5.14). The estimate for the equa-tion on W is identical:

|||Ll W TAl0+iAl1+Al0Cl W I2 x2 W|||1,(5.23)

C(K0, 0)|||W|||0.(3) Eliminating the front. We proceed as in the constant coefficients case, and

show how to eliminate the front in the (paralinearized) boundary conditions. Ifthe perturbation is small enough (in the L norm), there exists a constant c > 0(depending only on K0) such that

|b(t,x1,,,)|2 c( 2 + 2 + 2).

Applying Gardings inequality (Theorem B.7), we obtain

Tbb, L2(R2) c

2

21, ,for all 0 (where 0 only depends on K0). Using the rules of symbolic cal-culus (Theorem B.6), we have T

bb = (Tb )Tb + R , where R is of order 1.

Consequently, we have an estimate of the form

1, C(K0)Tb 0.

For all 0, we thus obtain

1, C(K0)(Tb + TMW|x2=0 0 + Wnc|x2=0 0)(5.24)

C(K0)

1

Tb + TMW|x2=0 1, + Wnc|x2=0 0

.

From (5.21) and (5.24) we deduce for 0 large enough (depending on K0)the estimate

1, C(K0) 1

B (W,)1, + Wnc|x2=0 0 ,which shows that it only remains to prove an estimate of Wnc|x2=0 in terms of thesource terms.


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For all (,) in the hemisphere

, we define the matrix

(t,x1,,,) := 0 0 1 + ivr(t,x1, 0) i(vr vl)(t,x1, 0) 0 ,and we extend as a homogeneous mapping of degree 0 with respect to (,).

We have b 0, and 02 . Applying Theorem B.6, we thus obtain:TTb 1, = TTb Tb1, C(K0)1, ,T

MW|x2=0 T

T

MW|x2=01, C(K0)Wnc|x2=00.

Using the decomposition

TMW|x2=0 = (TM TTM)W|x2=0 + T (TMW|x2=0 + Tb ) TTb ,

we get the following estimate

TMW|x2=01,(5.25) C(K0)(Wnc|x2=00 + T

b + TMW|x2=01, + 1, ).

As was done in the constant coefficients case, we define the symbol of thereduced boundary conditions:

(t,x1,,,) R4 R+, (t,x1,,,) := (t,x1,,,)M(t,x1).We now focus on the paralinearized system with reduced boundary conditions:

(5.26) T

Ar0 +iAr1 +Ar0CrW

+ + I2 x2 W+ =

F+, x2 > 0,

T

Al0+iAl1+Al0ClW

+I

2

x2W

= F, x2 > 0,T

W|x2=0 = G, x2 = 0.

Our aim is to prove an energy estimate for the paralinearized equations (5.26).Once this is done, we shall obtain an energy estimate for the linearized equations.More precisely, we have the following proposition.

Proposition 5.3. Assume that there exists a constantC0, depending only on K0and0, such that the solution W to (5.26) satisfies

(5.27) Wnc|x2=020 C0 13 F21, + 12 G21, ,for all 0 (where0 only depends on K0 and0). Then the thesis of Theorem 5.1holds.


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Proof. The proof is straightforward. We first write

TAr0 +iAr1 +Ar0CrW

+ + I2 x2 W+ = LrW+ + error,T

Al0+iAl1+Al0ClW + I2 x2 W = Ll W + error,

and estimate the error terms with the help of (5.22)(5.23). We use (5.27) toderive

Wnc|x2=0

20 C0

1

3L W

21, +

13

W

20 +

12

T W|x2=0

21,

,

where, as usual, L W = (LrW+, Ll W). Using (5.24) and (5.25), and choosing large enough, we obtain the following inequality:

Wnc|x2=020 + 21, C0

1

3L W21, + 13 W20 + 12 Tb + TMW|x2=021,

.

Eventually, we use (5.21) to derive (up to choosing large enough):Wnc|x2=020 + 21, C0

1

3L W21, + 13 W20 + 12 B (W,)21,

.

Then one uses the definitions

et

U+ = TrW+, et

U = TlW,etAr0 T

1r L

rU+ = LrW+, etAl0T1l LlU = Ll W,

as well as (5.17) and Lemma 5.2 to derive (5.12). The reader will easily check thatthe constants C0, C

0 etc. involved in the energy estimates only depend on K0 and

0.

Thanks to Proposition 5.3, we only need to prove the estimate (5.27) for theparalinearized system (5.26). This will be done in the next paragraphs.

Recall that the boundary matrix in (5.26) only acts on Wnc

= (W+2 , W+3 , W2 ,W3 ) and not on the full vector W. Namely, the first and fourth columns of van-ish. Consequently, we feel free to write the boundary conditions under the formT

W

nc

|x2=0 = G, that is, we consider as a matrix with only four columns and tworows.


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5.4. Microlocalization. To derive the desired energy estimate for (5.26), we

follow the general strategy of the constant coeffi

cients case. Namely, we first con-sider the two equations that do not involve any x2 derivative:

T+ivrW

+1 + Tic2r/rx1rW+2+ T

ic2r/rx1rW+3 + order 0 terms = F+1 ,T

+ivlW

1 + Tic2l /lx1lW2+ T

ic2l /lx1

lW

3 + order 0 terms = F1 .

Formally, the idea is to invert the operators T+ivr ,l and to substitute the cor-responding value of W1 into the four remaining equations. We shall thus get asystem of the formx2 W

nc = TAWnc + TE Wnc + source term, x2 > 0,T

W

nc

|x2=0 = source term, x2 = 0,

where A is of degree 1 and E is of degree 0. (Both matrices A and E are block

diagonal since the equations for W+ and W are decoupled). An important issueis to show that this operation can be achieved. Namely, the zeroth order terms inthe two scalar equations above involve W+1 and W

1 . When inverting the opera-

tors T+ivr ,l, one needs to take the zeroth order terms into account, in order to

avoid introducing W1 in the final equation for Wnc. We shall show that such aninversion is possible. But in this paragraph, we focus on the first order term andexplicit the symbol A. Consider the following 2 2 matrix:

(5.28) Ar :=

Ar1 Ar3

Ar3 Ar2 ,with

(5.29a) Ar1 := cr2 x2r

2( + ivr)x1r3 ( + ivr) x2rcrx1r + x2r x1rix1r2 ,

(5.29b) Ar2

:=

cr2 x2r

2( + ivr)x1r3 +( + ivr) x2

r

crx1r +x2

r x1ri

x1r2 ,

(5.29c) Ar3 :=cr2 x2r

2( + ivr)x1r3 .


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The definition ofAl is completely similar, changing the r index byl. The symbolA

mentionned above is nothing but the block diagonal matrix

(5.30) A :=Ar 00 Al

.

The set of poles ofA is denoted byp, that is,p := {(t,x1, x2,,) such that = i vr ,l(t,x1, x2)}.

As was done in the constant coefficients case, we denote byE(t,x1, x2,,) the

stable subspace ofA(t,x

1, x

2,,)

. This stable subspace is well defined when > 0, and admits a continuous extension up to any (,) such that iRand (,) (0, 0).

At each point (t,x1, 0) of the boundary, the subspace{Z E(t,x1, 0,,) s.t. (t,x1,,)Z= 0}

is nontrivial (that is, not reduced to {0}) if and only if

= i vr(t,x1, 0) + vl(t,x1, 0)

2

or = iV1(t,x1) or = iV2(t,x1),

for suitable functions V1,2 W2,(R2). This is just because when one freezesthe coefficients on the boundary and computes the associated Lopatinskii deter-minant, the calculations yield the same result as in Section 4. Recall that thestate (Ur ,l, r ,l) around which the equations are linearized satisfy the Rankine-Hugoniot conditions, see (5.3). As in [11], we define the critical set of space-frequency variables in the following way:

0c :

= (t,x1,,) s.t.

i (vr + vl)(t,x1, 0)

2,iV1(t,x1),iV2(t,x1)

.

This is exactly the set of space variables on the boundary and frequencies sothat the Lopatinskii determinant vanishes. Moreover, if the perturbation (Ur ,l, r ,l)is sufficiently small (in the L norm), we have

vl(t,x1, 0) > V1(t,x1) > (vr + vl)(t,x1, 0)2 > V2(t,x1) > vr(t,x1, 0),

(t,x1) .Therefore, the critical set 0c does not intersect the set of poles on the boundary ofthe space domain: 0c (p {x2 = 0}) = .


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Another important feature of the critical set

0c is that it admits a neighborhood in

which the symbolA

is diagonalizable. To be more precise, there exists a neighbor-hood V0c of0c in R2 and a mapping Q0 on V0c (with values in the set of 4 4invertible matrices and homogeneous of degree 0 with respect to (,)) such that

(5.31) Q0(z)A(z)Q0(z)1 = diag(r(z),+r(z),l (z),+l (z)), z = (t,x1,,) V0c ,

where r (resp. +r) is the eigenvalue with negative (resp. positive) real part ofAr when > 0. (The definition ofl is similar). Note that the matrix Q0 has

the same block diagonal structure asA

, see (5.30). The symbolQ

0 belongs to theclass 02 (see Appendix B for the precise definition).Since 0c does not intersect p {x2 = 0}, we may assume that the neighbor-

hood V0c does not intersect p {x2 = 0} either.The key point in the derivation of an energy estimate is to understand how

the singularities at the boundary (that is, the set 0c ) propagate in the interior do-main. Following [11], we shall show that the singularities propagate along thetwo bicharacteristic curves associated with the (real) symbols r ,l, provided thatthese curves do not reach the poles ofA or the points where A stops being diago-nalizable. These ideas motivate the following result.

Proposition 5.4. Assume that the perturbation (Ur ,l, r ,l) is small in W2,()and has compact support. Then one can choose the neighborhoodV0c such that thereexists an open setVc satisfying the following properties: Vc {x2 = 0} = V0c andVc p = . The symbolA defined by(5.28)(5.29)(5.30) is diagonalizable on the setVc. In

other words, (5.31) holds on allVc, and not only on the traceV0c . For allz = (t,x1, x2,,) Vc, one has

r(z)

+r(z) and l (z)

+l (z). The solutions of the hamiltonian system of ODEs

dt

dx2= h

(t,x1, x2,,),(5.32a)

dx1dx2

= h

(t,x1, x2,,),(5.32b)

d

dx2= h

t(t,x1, x2,,),(5.32c)

ddx2

= hx1

(t,x1, x2, , ), (t, x1,,,)|x2=0 V0c ,(5.32d)

are defined for allx2 0 and remain in Vc, both forh = r andh = l .These solutions are referred to as bicharacteristic curves.


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Following [11], we now construct a (real) weight that vanishes on the bichar-

acteristic curves, and that satisfies a linear transport equation. For all z =(t,x1,,) R2 , define(5.33) (z) :=

+ (vr + vl)(t,x1, 0)

2

( V1(t,x1))( V2(t,x1)),

and extend to the whole set R2 as a homogeneous mapping of degree 1 withrespect to (,). The velocities V1,

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