Well-posedness of hyperbolic Initial Boundary Value Problems

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1 Well-poseness of hyperbolic Initial Bounary Value Problems Jean-François Coulombel CNRS & Université Lille 1 Laboratoire e mathématiques Paul Painlevé Cité scientifique VILLENEUVE D ASCQ CEDEX, France jfcoulom@math.univ-lille1.fr March 16, 2004 Abstract Assuming that a hyperbolic initial bounary value problem satsifies an a priori energy estimate with a loss of one tangential erivative, we show a well-poseness result in the sense of Haamar. The coefficients are assume to have only finite smoothness in view of applications to nonlinear problems. This shows that the weak Lopatinskii conition is roughly sufficient to ensure well-poseness in appropriate functional spaces. AMS subject classification: 35L50, 35L40 1 Introuction In this paper, we consier hyperbolic Initial Bounary Value Problems (IBVPs) in several space imensions. Such problems typically rea: t U + A j(t, x) xj U + D(t, x)u = f(t, x), t ]0, T [, x R +, B(t, y) U x =0 = g(t, y), t ]0, T [, y R 1, (1) U t=0 = U 0 (x), x R +. The space variable x lies in the half-space R + := {x = (x 1,..., x ) R /x > 0}, y = (x 1,..., x 1 ) enotes a generic point of R 1, an t = x 0 is the time variable. The A j s an D are square n n matrices, while B is a p n matrix of maximal rank (the integer p is given below). For simplicity, we shall only eal with noncharacteristic problems, but we point out that the analysis can be reprouce with only minor changes for uniformly characteristic problems (we shall go back to this in our final remarks). To prove the well-poseness of (1), there are basically four steps (see e.g. [3] for a complete escription). One first proves a priori energy estimates for smooth solutions. Then one efines a ual problem an shows the existence of weak solutions (this works because the original an the ual problems usually share the same stability properties). The thir step is to show that weak solutions are strong solutions an thus satisfy the energy estimate. Eventually, one constructs solutions of the IBVP. The first step of this analysis is linke to the so-calle (uniform) Lopatinskii conition (or uniform Kreiss-Lopatinskii conition), see [10]. Namely, the uniform Lopatinskii conition yiels an energy estimate in L 2, with no loss of erivative from the source terms (f, g) to the solution U. The secon step relies on Hahn-Banach an Riesz theorems (see [3]). One obtains weak solutions for which it is not possible to apply the a priori energy estimate. Thus, in the thir step, one introuces a tangential mollifier, regularizes the weak 1

2 solution, applies the a priori estimate to the regularize sequence, an passes to the limit. This proceure was introuce in [11]. The fourth step is to take into account the initial atum U 0, an it was first achieve in [18]. In all the above mentionne results, it is crucial that the first step yiels an energy estimate without loss of erivatives. (We shall say that such problems are stable problems). Such an estimate hols either because the bounary conitions are maximally issipative (or strictly issipative, which is even better), either because the uniform Lopatinskii conition is satisfie. However, it is known that this stability conition is not met by some physically interesting problems. Examples of situations where the uniform Lopatinskii conition breaks own are provie by elastoynamics (with the well-known Rayleigh waves [22, 20]), shock waves or contact iscontinuities in compressible flui mechanics, see e.g. [12, 16]. For such nonstable problems, there is no L 2 estimate, but in some weakly stable situations, one can prove a priori energy estimates with a loss of one tangential erivative from the source terms to the solution. Without entering etails, these problems are those for which the so-calle Lopatinskii eterminant vanishes at orer 1 in the hyperbolic region of the cotangent of the bounary T R t,y R R. For noncharacteristic problems, such energy estimates with loss of one erivative have been erive by the author in [5], an for uniformly characteristic problems, similar energy estimates have been erive by P. Secchi an the author in [6]. (Note that for the Rayleigh waves problem, the Lopatinskii eterminant vanishes in the elliptic region of the cotangent of the bounary, an the situation is slightly better, as shown in [20]). In this paper, we show how to solve the IBVP for such weakly stable problems where losses of erivatives occur. More precisely, we show how to construct solutions of (1), with U 0 = 0, provie that we have an a priori estimate with a loss of one tangential erivative, both for the initial problem (1) an for a ual problem. The construction of a weak solution is quite classical, but still, it requires some attention. Then, we shall regularize our weak solution by using a tangential mollifier. Unlike in the case of stable problems, where any tangential mollifier is suitable, we shall show here that the choice of the mollifier is crucial in our context. Our result is that weak solutions are what we shall call semi-strong solutions. In the en, we shall prove a well-poseness result (in the sense of Haamar) for the IBVP (1), when U 0 = 0. The case of general initial ata is aresse in our final remarks. The paper is organize as follows. In view of possible applications to nonlinear problems, we have chosen to work with low regularity coefficients. Of course, this choice will introuce technical ifficulties, an we have foun it appropriate to give in section 2 all the notations an results on paraifferential calculus that will be use throughout this paper. In section 3, we state precisely our weak stability assumption, an give our main result. In section 4, we prove that (1) amits weak solutions, an that these weak solutions are semi-strong solutions. Up to a few technical etails, this ensures well-poseness for zero initial ata. In section 5, we give some extensions of our results, an make a few comments. 2 Paraifferential calculus with a parameter In this section, we collect some efinitions an results on paraifferential calculus. We refer to the original works by Bony an Meyer [1, 15] an also to [14, 17] for the introuction of the parameter. The reaer will fin etaile proofs in these references. We first introuce some norms on the usual Sobolev spaces. For all γ 1, an for all s R, we equip the space H s (R ) with the following norm: u 2 s,γ := 1 (2π) R λ 2s,γ (ξ) û(ξ) 2 ξ, λ s,γ (ξ) := (γ 2 + ξ 2 ) s/2. We shall write 0 rather than 0,γ for the (usual) L 2 norm. 2

3 The classification of paraifferential symbols (with a parameter) is the following: Definition 2.1. A paraifferential symbol of egree m R an regularity k (k N) is a function a(x, ξ, γ) : R R [1, + [ C q q such that a is C with respect to ξ an for all α N, there exists a constant C α verifying (ξ, γ), α ξ a(, ξ, γ) W k, (R ) C α λ m α,γ (ξ). The set of paraifferential symbols of egree m an regularity k is enote by Γ m k (R ). It is equippe with the obvious semi-norms. We enote by Σ m k (R ) the subset of paraifferential symbols a Γ m k (R ) such that for a suitable ε ]0, 1[ the partial Fourier transform of a satisfies (ξ, γ), Supp F x a(, ξ, γ) {ζ R / ζ ε (γ 2 + ξ 2 ) 1/2 }. Of course, the symbols in Σ m k (R ) are C functions with respect to both variables x an ξ, an for all a Σ m k (R ), we have the estimates (x, ξ, γ), β x α ξ a (x, ξ, γ) C α,β λ m α + β,γ (ξ). Thus any symbol a Σ m k (R ) belongs to Hörmaner s class S1,1 m [9] an efines an operator Op γ (a) on the Schwartz class S(R ) by the usual formula u S(R ), x R, Op γ (a)u(x) := 1 (2π) e ix ξ a(x, ξ, γ) û(ξ) ξ. R We shall use the following terminology: Definition 2.2. A family of operators {P γ } efine for γ 1 will be sai of orer m (m R) if the operators P γ are uniformly boune from H s+m (R ) to H s (R ) for all s, inepenently of γ: γ 1, u H s+m (R ), P γ u s,γ C s u s+m,γ. The following Theorem is crucial: Theorem 2.1. If a Σ m k (R ), k N an m R, the family {Op γ (a)} is of orer m. More precisely, for all s R, there exists a positive constant C such that γ 1, u H s+m (R ), Op γ (a)u s,γ C u s+m,γ. The constant C only epens on s, m, on the confinement parameter ε ]0, 1[, an on a finite number N of semi-norms of a (N only epens on s an m). The regularization of symbols in the class Γ m k (R ) is achieve by a convolution with amissible cut-off functions: Definition 2.3. Let ψ : R R [1, + [ [0, + [ be a C function such that the following estimates hol for all α, β N : (ζ, ξ, γ), α ζ β ξ ψ (ζ, ξ, γ) C α,β λ α β,γ (ξ). We shall say that ψ is an amissible cut-off function if there exist real numbers 0 < ε 1 < ε 2 < 1 satisfying ψ(ζ, ξ, γ) = 1 if ζ ε 1 (γ 2 + ξ 2 ) 1/2, ψ(ζ, ξ, γ) = 0 if ζ ε 2 (γ 2 + ξ 2 ) 1/2. 3

4 An example of cut-off function is the following: we choose a nonnegative C function χ 0 on R R such that ξ γ 2 1 ξ γ 2 2 = χ 0 (ξ 1, γ 1 ) χ 0 (ξ 2, γ 2 ), { χ 0 (ξ, γ) = 1 if ( γ 2 + ξ 2) 1/2 1/2, χ 0 (ξ, γ) = 0 if ( γ 2 + ξ 2) 1/2 1. We efine a function ϕ 0 (ξ, γ) := χ 0 (ξ/2, γ/2) χ 0 (ξ, γ). Then the function ψ 0 efine by ψ 0 (ζ, ξ, γ) := p 0 χ 0 (2 2 p ζ, 0) ϕ 0 (2 p ξ, 2 p γ) (2) is an amissible cut-off function (one can take ε 1 = 1/16 an ε 2 = 1/2). If ψ is an amissible cut-off function, the inverse Fourier transform K ψ of ψ(, ξ, γ) satisfies (ξ, γ), α ξ Kψ (, ξ, γ) L 1 (R ) C α λ α,γ (ξ). These L 1 bouns for the erivatives α ξ Kψ yiel the following result: Proposition 2.1. Let ψ be an amissible cut-off function. The mapping a σa ψ (x, ξ, γ) := K ψ (x y, ξ, γ) a(y, ξ, γ) y R is continuous from Γ m k (R ) to Σ m k (R ) for all m (the confinement parameter of σa ψ is ε 2 ). If a Γ m 1 (R ), then a σa ψ Γ m 1 0 (R ). In particular, if ψ 1 an ψ 2 are two amissible cut-off functions an a Γ m 1 (R ), then σ ψ 1 a σ ψ 2 a Σ m 1 0 (R ). Fixing an amissible cut-off function ψ, we efine the paraifferential operator T ψ,γ a formula T ψ,γ a := Op γ (σ ψ a ). by the If ψ 1 an ψ 2 are two amissible cut-off functions an a Γ m 1 (R ), then Proposition 2.1 an Theorem 2.1 show that the family {T ψ 1,γ a T ψ 2,γ a } is of orer (m 1). The symbolic calculus is base on the following Theorem: Theorem 2.2. Let a Γ m 1 (R ) an b Γ m 1 (R ). Then ab Γ m+m 1 (R ) an the family {T ψ,γ a T ψ,γ b T ψ,γ ab } γ 1 is of orer m + m 1 for all amissible cut-off function ψ. Let a Γ m 1 (R ). Then the family {(T ψ,γ a ) T ψ,γ a } γ 1 is of orer m 1 for all amissible cut-off function ψ. Let a Γ m 2 (R ) an b Γ m 2 (R ). Then ab Γ m+m 2 (R ) an the family {T ψ,γ a T ψ,γ b T ψ,γ ab T ψ,γ i j ξ j a xj b } γ 1 is of orer m + m 2 for all amissible cut-off function ψ. 4

5 Let a Γ m 2 (R ). Then the family {(Ta ψ,γ ) T ψ,γ a T ψ,γ i j ξ j xj a } γ 1 is of orer m 2 for all amissible cut-off function ψ. An easy consequence of Theorem 2.2 is that, for any symbols a Γ m 2 (R ) an b Γ m 2 (R ) that commute, the remainer T γ a T γ b T γ b T γ a T γ i{a,b} = [T γ a, T γ b ] T γ i{a,b} is of orer m + m 2. Here above, the notation {a, b} stans for the Poisson bracket of a an b: {a, b} := ξj a xj b xj a ξj b. j We now stuy the case of paraproucts: they are efine by the particular choice of ψ 0 as cut-off function, where ψ 0 is efine by (2). We shall write Ta γ instea of T ψ 0,γ a for the associate paraifferential operators. We have the following important result: Theorem 2.3. Let a W 1, (R ), u L 2 (R ) an γ 1. Then we have a u T γ a u 0 C γ a W 1, (R ) u 0, a xj u T γ a xj u 0 C a W 1, (R ) u 0, a u T γ a u 1,γ C a W 1, (R ) u 0, for a suitable constant C that is inepenent of (a, u, γ). If in aition a W 2, (R ), we have a u T γ a u 1,γ C γ a W 2, (R ) u 0, a xj u T γ a xj u 1,γ C a W 2, (R ) u 0, for a suitable constant C that is inepenent of (a, u, γ). We can exten the paraifferential calculus to symbols efine on a half-space in the following way: let Ω enote the half-space R ]0, + [= R The space L2 x (Ht,y) s is equippe with the norm u 2 s,γ := + 0 u(, x ) 2 s,γ x. Again, we shall write 0 rather than 0,γ when s = 0 (that is, for the usual norm in L 2 (Ω)). We enote by Γ m k (Ω) the set of symbols a(x 0,..., x, ξ, γ) efine on Ω R [1, + [ such that the mapping x a(, x, ) is boune into Γ m k (R ). We efine the paraifferential operator Ta γ by the formula u C 0 (Ω), x 0, (T γ a u)(, x ) := T γ a(x ) u(, x ). Using Theorem 2.3 an integrating with respect to x, we obtain for all symbol a W 1, (Ω) an all u L 2 (Ω) the estimates: a u T γ a u 0 C γ a W 1, (Ω) u 0, a xj u T γ a xj u 0 C a W 1, (Ω) u 0, j = 0,..., 1. When a W 2, (Ω), one obtains an estimate with a gain of two tangential erivatives: a u T γ a u 1,γ C γ a W 2, (Ω) u 0, a xj u T γ a xj u 1,γ C a W 2, (Ω) u 0, j = 0,..., 1. 5

6 3 Statement of the result Recall that Ω enotes the half-space R t,y R + x. We first make the following assumption on the coefficients of (1): Assumption 1. The A j s are efine on Ω an belong to W 2, (Ω). There exists δ > 0 such that for all (t, x) Ω one has et A (t, x) δ. The matrix B is efine on R an belongs to W 2, (R ). It has maximal rank p, where p equals the number of positive eigenvalues of A (that is, the number of incoming characteristics). The system is symmetric hyperbolic, that is, there exists a (real) matrix value mapping S W 2, (Ω) verifying (t, x) Ω, S(t, x) = S(t, x) T, S(t, x) δ I, S(t, x) A j (t, x) = A j (t, x) T S(t, x). We now make our first weak stability assumption on system (1): Assumption 2. For any D 1 W 1, (Ω), an for any symbol D 2 Γ 0 1 (Ω), there exists a constant C (that epens only on δ, A j W 2, (Ω), D 1 W 1, (Ω), B W 2, (R ) an on a finite number of seminorms of the symbol D 2 ) an there exists a constant γ 0 1 such that for all U C0 (Ω) an for all γ γ 0 one has γ U U x =0 2 0 C γ 3 f 2 1,γ + 1 ) γ 2 g 2 1,γ, where f := A 1 γu + x0 U + A j xj U + D 1 U + T γ D 2 U, g := B U. x =0 Before stating our last assumption, we make a couple of remarks. In the erivation of energy estimates, one usually replaces the linear operator U A 1 γu + x0 U + A j xj U + D 1 U + T γ D 2 U by its paraifferential version U T γ (γ+iξ 0 )A 1 1 U + T γ iξ j A 1 A j U + x U + T γ U + T γ A 1 D D 1 2 U, an treats the errors as source terms 1. These errors have the following form: ( ) γ A 1 U T γ U, or A 1 A 1 A j xj U T γ U, or A 1 iξ j A 1 A D 1 U T γ U. j A 1 D 1 To absorb these errors in an estimate with a loss of one tangential erivative, one nees the regularity state in assumptions 1 an 2 for the coefficients (see Theorem 2.3 in the preceeing section). 1 Recall that we use a tangential symbolic calculus for which x is seen as a parameter. 6

7 The crucial point in assumption 2 is that the energy estimate is inepenent of the lower orer term in the interior equation. More precisely, if the energy estimate hols with D 1 = D 2 = 0, it is not clear whether it also hols for arbitrary D 1 an D 2. This is a major ifference with the stable case where there is no loss of erivative (an therefore, one can treat lower orer terms as source terms in energy estimates). In the framework of weakly stable problems, the lower orer terms in the interior equations can not be neglecte an one nees to pay special attention. One way to rephrase assumption 2 is the following: energy estimates with loss of one tangential erivative hol, inepenently of the lower orer terms, an inepenently of their nature (meaning classical, or paraifferential, or a linear combination of the two). We now turn to our last assumption, that is the analogue of assumption 2 for a ual problem. First recall the following efinition: Definition 3.1. A ual problem for (1) is a linear problem that reas: { t V + AT j x j V + D V = f (t, x), t ]0, T [, x R +, M (t, y) V = g x =0 (t, y), t ]0, T [, y R 1, where M is a (n p) n matrix of maximal rank such that (t, y) R, B (t, y) T B(t, y) + M (t, y) T M(t, y) = A (t, y, 0), (3) for suitable p n an (n p) n matrices B an M, an such that B, M, M belong to W 2, (R ). Our final assumption is that the energy estimate with loss of one tangential erivative is also satisfie by one ual problem 2, when the parameter γ is change into γ: Assumption 3. There exists a ual problem (that is, a matrix M satisfying (3)) such that for any D 1 W 1, (Ω), an for any symbol D 2 Γ 0 1 (Ω), there exists a constant C (that epens only on δ, A j W 2, (Ω), D 1 W 1, (Ω), M W 2, (R ) an on a finite number of seminorms of the symbol D 2 ) an there exists a constant γ 0 1 such that for all V C0 (Ω) an for all γ γ 0 one has γ V V x =0 2 0 C γ 3 f 2 1,γ + 1 ) γ 2 g 2 1,γ, where f := (A T ) 1 γv x0 V A T j xj V + D 1 V + T γ D 2 V, g := M V. x =0 In terms of the Lopatinskii conition, assumption 3 means that for one ual problem, the backwar Lopatinskii conition egenerates at orer 1 in the hyperbolic region of the cotangent of the bounary. In practice, one can usually compute explicit ual bounary conitions for which the Lopatinskii eterminant equals that of the original problem (1). Thus the erivation of energy estimates for a ual problem is usually a irect consequence of energy estimates for the original problem. In all what follows, we always make assumptions 1, 2 an 3. The result is the following: Theorem 3.1. Let D W 1, (Ω), an let T > 0. Then, for all functions f(t, x) an g(t, y) verifying: f, t f, x1 f,, x 1 f L 2 (Ω T ), Ω T :=], T [ R +, g H 1 (ω T ), ω T :=], T [ R 1, 2 Note that, with our efinition, the ual problem is not uniquely efine. 7

8 an such that f an g vanish for t < 0, there exists a unique U L 2 (], T [ R +), whose trace on {x = 0} belongs to L 2 (], T [ R 1 ), that vanishes for t < 0, an that is a solution to { t U + A j(t, x) xj U + D(t, x)u = f(t, x), t ], T [, x R +, B(t, y) U = g(t, y), x =0 t ], T [, y R 1. In aition, U C([0, T ]; L 2 (R +)) an the following estimate hols for all t [0, T ] an all real number γ γ 0 : e 2γt U(t) 2 L 2 (R + ) + γ e γs U 2 L 2 (Ω + t) e γs U x =0 2 L 2 (ω t) C γ e γs f 2 L 2 (Ω + 1 t) γ 3 e γs t,y f 2 L 2 (Ω + t) e γs g 2 L 2 (ω + 1 ) t) γ 2 e γs g 2 L 2 (ω t). The constant C an the parameter γ 0 only epen on δ, A j W 2, (Ω), D W 1, (Ω), B W 2, (R ) an M W 2, (R ). 4 Proof of the main result In this section, we first show existence an uniqueness of solutions for the Bounary Value Problem, with source terms (f, g) in weighte spaces. For γ 1, we efine the spaces L 2 γ(ω) := exp(γt)l 2 (Ω), H 1 γ(ω) := exp(γt)h 1 (Ω). We also efine the spaces H(Ω) := {v L 2 (Ω) s.t. t v, x1 v,..., x 1 v L 2 (Ω)} = L 2 (R + x ; H 1 (R t,y)), H γ (Ω) := exp(γt)h(ω) = {v D (Ω) s.t. exp( γt)v H(Ω)}, H(Ω) := {v H(Ω) s.t. t v, x1 v,..., x v H(Ω)}. (4) The spaces L 2 γ(r ) an H 1 γ(r ) are efine in a similar way. The space L 2 γ(ω) is equippe with the obvious norm: v L 2 γ (Ω) := exp( γt)v 0, an the space H γ (Ω) is equippe with the norm v Hγ(Ω) := ṽ 1,γ with ṽ := exp( γt)v. Similarly, the space H 1 γ(r ) is equippe with the norm w H 1 γ (R ) := w 1,γ with w := exp( γt)w. Some elementary, though useful, properties of the spaces H(Ω) an H(Ω) are collecte in appenix A at the en of this paper. In particular, we show that elements of H(Ω) amit a trace in H 3/2 (R ) (though they o not necesarily belong to H 2 (Ω), since the efinition (4) oes not require x 2 f L 2 (Ω)). We consier a zero orer coefficient D W 1, (Ω), that we fix once an for all, an we wish to prove a well-poseness result for the following Bounary Value Problem: { LU := t U + A j(t, x) xj U + D(t, x)u = f(t, x), (t, x) Ω, B(t, y) U = g(t, y), (t, y) (5) x =0 R, when the source terms f an g belong to H γ (Ω) an Hγ(R 1 ), an γ is large. In view of assumption 2, we expect to obtain a unique solution U in L 2 γ(ω) whose trace on the bounary {x = 0} belongs to L 2 γ(r ). In the en, we shall localize this result on a finite time interval. 8

9 For later use, we efine the norm of the coefficients: N := A j W 2, (Ω) + D W 1, (Ω) + B W 2, (R ) + M W 2, (R ), (6) where M is given by assumption 3, an represents the ual bounary conitions. 4.1 Preliminary estimates We first show that the original problem, as well as the ual problem, satisfy an energy estimate in L 2 (H 1 ) when the source terms are in L 2 (that is, we can shift the inices of regularity). More precisely, we have the following result: Lemma 4.1. Let D 1 W 1, (Ω), an let D 2 Γ 0 1 (Ω). There exists a constant C (that epens only on δ, A j W 2, (Ω), D 1 W 1, (Ω), B W 2, (R ) an on a finite number of seminorms of the symbol D 2 ) an there exists a constant γ 1 1 such that for all U C0 (Ω) an for all γ γ 1 one has γ U U x =0 2 0 C γ 3 f 1 2 1,γ + 1 ) γ 2 g 1 2 1,γ, where f 1 := T γ (γ+iξ 0 )A 1 1 U + T γ iξ j A 1 A j U + x U + T γ A 1 D 1+D 2 U, g 1 := T γ B U x =0, an one also has γ U 2 1,γ + U x =0 2 1,γ C γ 3 f ) γ 2 g 2 2 0, where f 2 := A 1 γu + x0 U + A j xj U + D 1 U, g 2 := B U. x =0 (7) Proof. The first inequality is easily prove using the estimates given in Theorem 2.3: A 1 B U x =0 T γ B U x =0 1,γ C B W 1, (R ) U x =0 0, (γu + x 0 U) T γ (γ+iξ 0 )A 1 A 1 A j xj U T γ iξ j A 1 A j U 1,γ C A 1 W 2, (Ω) U 0, U 1,γ C A 1 A j W 2, (Ω) U 0, thanks to assumption 1. Consequently, using assumption 2, the triangle inequality an choosing γ large enough, one can absorb the error terms in the left han sie of the inequality. We now turn to the secon estimate. Let U C0 (Ω), an efine It is clear that we have f := T γ (γ+iξ 0 )A 1 g := T γ B U x =0, 1 U + T γ iξ j A 1 A j W := T γ (γ 2 + ξ 2 ) 1/2 U = T γ λ 1,γ U. U + x U + T γ U, A 1 D 1 W 0 = U 1,γ, an W x =0 0 = U x =0 1,γ, 9

10 an we are thus le to erive an energy estimate of W in L 2. Thanks to assumption 1 an to Theorem 2.2, we compute: T γ (γ+iξ 0 )A 1 1 W + T γ iξ j A 1 A j W + x W + T γ W A 1 D 1 = T γ f + T γ λ U + 1 T γ U + 1,γ i{(γ+iξ 0 )A 1,λ 1,γ } {ξ j A 1 A j,λ 1,γ } Rγ 2 U = T γ f + T γ λ W + 1 T γ W + 1,γ i{(γ+iξ 0 )A 1,λ 1,γ }λ 1,γ {ξ j A 1 A j,λ 1,γ }λ 1,γ Rγ 1 W, where R γ 2 is a family of orer 2, an Rγ 1 is a family of orer 1. In a similar way, we also compute T γ B W = T γ x g + R γ =0 λ 1,γ 1 W, x =0 where, once again, R γ with the symbol is a family of orer 1. Now, we apply the first estimate of Lemma We get D 2 := i{(γ + iξ 0 )A 1 γ W W x =0 2 0 C, 1 λ 1,γ }λ 1,γ {ξ j A 1 A j, λ 1,γ }λ 1,γ Γ 0 1(Ω). γ 3 T γ f + R γ λ 1,γ 1 W 2 1,γ + 1 ) γ 2 T γ g + R γ λ 1,γ 1 W x =0 2 1,γ. Going back to the efinition of W, an choosing γ large enough, we have alreay obtaine the L 2 (H 1 ) estimate for the paraifferential problem, namely: γ U 2 1,γ + U x =0 2 1,γ C γ 3 f ) γ 2 g 2 0. The result now follows from Lemma 4.2 that we give just below. This Lemma enables us to control the istance (in L 2 ) between f an f 2, an the istance between g an g 2 (f 2 an g 2 are efine by (7)). Lemma 4.2. Let γ 1, a W 1, (R ) an v H 1 (R ). Then one has (a T γ a ) v 0 C a W 1, (R ) v 1,γ, for a suitable constant C that oes not epen on γ, a, v. If, in aition, a W 2, (R ), one has (a T γ a ) v 0 C γ a W 2, (R ) v 1,γ, (a T γ a ) xj v 0 C a W 2, (R ) v 1,γ. Proof. We prove Lemma 4.2 for v in the Schwartz class S(R ). The conclusion follows from a ensity/continuity argument. Decompose v as v = v + j xj v j, with v 0 γ v 1,γ, v j 0 v 1,γ. 10

11 This ecomposition hols with We easily get v (ξ) := γ 2 γ 2 + ξ 2 v(ξ), v j(ξ) := iξ j γ 2 + ξ 2 v(ξ). a v T γ a v 0 (a T γ a )v 0 + j (a T γ a ) xj v j 0 C a W 1, (R ) v 1,γ, thanks to Theorem 2.3. The secon part of Lemma 4.2 is prove in the same way, an we omit the etails. When ealing with symbols efine on a half-space, one simply integrates the estimates of Lemma 4.2. The result is a gain of one or two tangential erivatives, epening on the regularity of the multiplicator. Then one can en the proof of Lemma 4.1. The etails are left to the reaer. Of course, similar a priori estimates hol true for the ual problem (with γ change into γ), since the assumptions an the regularity of the coefficients are exactly the same. We therefore have: Lemma 4.3. Let D W 1, (Ω). There exists a constant C (that epens only on δ, A j W 2, (Ω), D W 1, (Ω) an M W 2, (R )) an there exists a constant γ 1 1 such that for all V C 0 (Ω) an for all γ γ 1 one has γ V 2 1,γ + V x =0 2 1,γ C γ 3 f ) γ 2 g 2 0, where f := (A T ) 1 γv x0 V A T j xj V + D V, g := M V. x =0 Recall that M represents the bounary conitions for the ual problem. With the help of our L 2 (H 1 ) estimate, we are going to construct weak solutions of (5). 4.2 Existence of weak solutions This paragraph is evote to the proof of the following result: Proposition 4.1. There exists γ 2 (N, δ) 1 such that for γ γ 2, f H γ (Ω), an g H 1 γ(r ), there exists U L 2 γ(ω) satisfying U x =0 of istributions). H 1/2 γ (R ) an U is a solution to (5) (in the sense Proof. We first commute (5) with the weight exp( γt), an we are le to search a function Ũ L 2 (Ω) that is a solution to { L γ Ũ := γũ + LŨ = f(t, x), (t, x) Ω, B(t, y) Ũ = g(t, y), (t, y) (8) x =0 R, with ( f, g) := exp( γt)(f, g) H(Ω) H 1 (R ). The formal ajoint (L γ ) of the operator L γ is efine by (L γ ) V := γv t V A T j xj V + D T xj A T j V. 11

12 Using relation (3) an formally integrating by parts, (8) reas V C 0 (Ω), Ũ, (Lγ ) V L 2 (Ω) = f, V L 2 (Ω) + g, B V x =0 L 2 (R ) + MŨ x =0, M V x =0 L 2 (R ), where the scalar proucts are enote as follows: U 1, U 2 L 2 (Ω) := U 1 (x) U 2 (x) x, Ω U 1, U 2 L 2 (R ) := U 1 (t, y) U 2 (t, y) ty. R We efine a set of appropriate test functions: } F := {V C0 (Ω) s.t. M V = 0 x=0 C0 (Ω). Thanks to assumption 3 (with D 2 = 0 an D 1 = D T xj A T j ), we observe that the operator (L γ ) is one-to-one in the vector space F. Consequently, we may efine a linear form l with the following formula: V F, l[(l γ ) V ] := f, V L 2 (Ω) + g, B V x =0 L 2 (R ). (9) The following estimate is now a consequence of Lemma 4.3: l[(l γ ) V ] f 1,γ V 1,γ + C g 1,γ V x =0 1,γ ( ) C V 1,γ + V x =0 1,γ C γ 3/2 (Lγ ) V 0, provie that γ is large enough, say γ γ 2 (N, δ). Now we apply Hahn-Banach theorem an we can thus exten l as a (continuous) linear form over the whole space L 2 (Ω). Thanks to Riesz theorem, we conclue that there exists a function Ũ L2 (Ω) verifying: V F, l[(l γ ) V ] = Ũ, (Lγ ) V L 2 (Ω). In particular, it is clear that L γ Ũ = f in the sense of istributions. Using that A is invertible, the trace of Ũ on the bounary {x = 0} is well-efine an belongs to H 1/2 (R ), see [3, chapter 7] 3. Moreover, the following Green s formula hols: V C 0 (Ω), Ũ, (Lγ ) V L 2 (Ω) = f, V L 2 (Ω) + A Ũ x =0, V x =0 H 1/2 (R ),H 1/2 (R ). Combining with the efinition of l, see (9), we obtain: Using (3), we observe that the matrix V F, g BŨ x =0, B V x =0 H 1/2 (R ),H 1/2 (R ) = 0. ( B M ) W 2, (R ) is invertible. We can therefore conclue that BŨ x = g. This completes the proof of Proposition = The proof in [3] is one with C boune coefficients, but it extens to Lipschitzean coefficients by using paraifferential techniques to estimate commutators, see e.g. [4] an appenix B for such estimates. 12

13 We have constructe a solution Ũ L2 (Ω) of (8), whose trace belongs to H 1/2 (R ) (so that the bounary conitions have a clear meaning). We point out that the L 2 (H 1 ) estimate given by Lemma 4.3 is crucial in orer to obtain Ũ L2 (Ω). If we ha only use the L 2 estimate given by assumption 3, we woul have obtaine a solution Ũ L2 (H 1 ). In view of assumption 2, we expect the function Ũ to amit a trace in L2 an to satisfy an appropriate energy estimate, namely: γ Ũ Ũ x =0 2 0 C γ 3 f 2 1,γ + 1 ) γ 2 g 2 1,γ. In the next paragraph, we show that this property hols. In particular, there exists a unique solution of (5) in L 2 γ(ω), an its trace belongs to L 2 γ(r ) when γ is large. Of course, such an existence-uniqueness result will hol inepenently of the zero orer term D. Before showing this result, we first state an analogue of Proposition 4.1 when the source terms are in L 2 γ. As a matter of fact, we want to solve the BVP for source terms with tangential erivatives in L 2 γ, but in the analysis, we shall see that we also nee to solve BVPs with source terms that are only in L 2 γ. Proposition 4.2. There exists γ 2 (N, δ) 1 such that for γ γ 2, f L 2 γ(ω), an g L 2 γ(r ), there exists U L 2 (R + ; Hγ 1 (R )) satisfying U x =0 H 3/2 γ (R ) an U is a solution to (5) (in the sense of istributions). Proof. Most of the proof is similar to the proof of Proposition 4.1. Keeping the same notations for the linear form l, an for the vector space F, an using assumption 2, we easily obtain the existence of Ũ L2 (R + ; H 1 (R )) such that V F, l[(l γ ) V ] = Ũ, (Lγ ) V L 2 (H 1 ),L 2 (H 1 ). In particular, one has L γ Ũ = f in the sense of istributions. The problem is now to give a meaning to the bounary conitions. This is solve by a trace lemma, which we state in appenix C at the en of this paper. Using this result, we can conclue that the trace of Ũ on {x = 0} is well-efine an belongs to H 3/2 (R ). Moreover, the following Green s formula hols: V C 0 (Ω), Ũ, (Lγ ) V L 2 (H 1 ),L 2 (H 1 ) = f, V L 2 (Ω)+ A Ũ x =0, V x =0 H 3/2 (R ),H 3/2 (R ). Using a continuity/ensity argument, the equality hols for all functions V H 2 (Ω) such that M V = 0 on the bounary. As was one in the proof of Proposition 4.1, we obtain: g BŨ x =0, B V x =0 H 3/2 (R ),H 3/2 (R ) = 0, provie that V H 2 (Ω) an M V = 0 on the bounary. Because B an M belong to W 2, (R ), for all function µ H 3/2 (R ), there exists V H 2 (Ω) such that µ = B V, an x =0 M V x =0 = 0. We can therefore conclue that g = BŨ x = Weak=semi-strong The result is the following: Theorem 4.1. Let ( f, g) H(Ω) H 1 (R ), an let Ũ L2 (Ω) be a solution to (8) 4, for γ sufficiently large. Then there exist a sequence (U ν ) in H(Ω), a boune sequence ( ν ) in the set of symbols Γ 0 1 (Ω), an a boune sequence (bν ) in the set of symbols Γ 1 1 (R ), that satisfy the following properties: 4 Recall that the trace of Ũ is automatically in H 1/2 (R ) an the bounary conitions make sense. 13

14 U ν Ũ in L2 (Ω), U ν x =0 Ũ x =0 in H 1/2 (R ), L γ U ν + A T γ ν U ν f in H(Ω), B U ν x =0 + T γ b ν U ν x =0 g in H1 (R ). In particular, Ũ x =0 belongs to L2 (R ) an the following energy estimate hols: γ Ũ Ũ x =0 2 0 C γ 3 f 2 1,γ + 1 ) γ 2 g 2 1,γ. (10) Recall that the space H(Ω) is efine by (4) 5. There is a similar result for solutions of (5) with source terms in L 2 γ. One simply nees to shift the inices (the regularize sequence belongs to H γ (Ω) an so on). The a priori estimate in L 2 (Hγ 1 ) is the inequality (7) in Lemma 4.1. We omit the proof in this case, an focus on Theorem 4.1. As etaile in the introuction, we are going to introuce a tangential mollifier in orer to regularize Ũ. For all ε ]0, 1], we efine the following symbol ϑ ε: (ξ, γ) R [1, + [, ϑ ε (ξ, γ) := as well as the corresponing Fourier multiplier: Θ γ ε := T γ ϑ ε = (γ 2 ε t,y ) 1. 1 γ 2 + ε ξ 2, With slight abuse of notations, we let Θ γ ε act on functions efine over R an on functions efine over the half-space Ω (where we use symbolic calculus with respect to the tangential coorinates, an the Fourier transform has to be unerstoo as a partial Fourier transform). This mollifier is exactly the one use in [7] (after introucing the parameter γ). As we shall see later on, it has some particularly nice commutation properties with the operator L γ (these properties are expresse by relation (2.11) in [7]). Elementary properties of the mollifier Θ γ ε are liste below: Lemma 4.4. Let γ 1, ε ]0, 1] an s R. Then for all v L 2 (R + ; H s (R )), one has: Θ γ ε v s,γ 1 γ 2 v s,γ, Θ γ ε v s+1,γ 1 γ ε v s,γ, Θ γ ε v s+2,γ 1 ε v s,γ. If v L 2 (R + ; H s+2 (R )), one has Θ γ ε v v/γ 2 s,γ ε γ 4 v s+2,γ. In particular, one has Θ γ ε v v/γ 2 s,γ 0 when ε 0, for all v L 2 (R + ; H s (R )). The proof of Theorem 4.1 is base on several estimates of commutators. Before starting the proof, we recall a lemma by Frierichs: 5 Recall also that the trace on {x = 0} of any element v H(Ω) is well-efine an belongs to H 3/2 (R ). In particular, it belongs to H 1 (R ) an the thir point of the Theorem makes sense, see Theorem A.1 in appenix A. 14

15 Lemma 4.5 (Frierichs). Let a W 1, (Ω). There exists a constant C that epens only on a W 1, (Ω) such that, for all γ 1, for all ε ]0, 1], an for all v L 2 (Ω), one has [a, Θ γ ε ] v 1,γ C v 0. Furthermore, one has [a, Θ γ ε ] v 1,γ 0 when ε 0, for all v L 2 (Ω). Let a W 2, (Ω) an j {0,..., 1}. There exists a constant C that epens only on a W 2, (Ω) such that, for all γ 1, for all ε ]0, 1], an for all v L 2 (Ω), one has [a xj T γ a xj, Θ γ ε ] v 1,γ C v 0. Furthermore, one has [a xj T γ a xj, Θ γ ε ] v 1,γ 0 when ε 0, for all v L 2 (Ω). We postpone the proof of Lemma 4.5 to appenix B (the first part is well-known), an we now give the proof of Theorem 4.1. Proof. Define A irect computation yiels U ε := Θ γ ε Ũ L2 (R + x ; H 2 (R t,y)). A 1 Lγ U ε = Θ γ ε (A 1 f) + γ [A 1 1, Θγ ε ] Ũ + [A 1D, Θγ ε ] Ũ + j=0 [A 1 A j xj, Θ γ ε ] Ũ, where L γ is efine by (8), an where we use the convention A 0 = I. Thanks to Lemma 4.4 an to Lemma 4.5, we alreay have Θ γ ε (A 1 f) + γ [A 1, Θγ ε ] Ũ + [A 1D, Θγ ε ] Ũ = 1 γ 2 A 1 f + r ε, r ε 1,γ 0. This is because A 1 ecomposition W 2, (Ω), D W 1, (Ω), an A 1 f L 2 (R + ; H 1 (R )). Using the [A 1 A j xj, Θ γ ε ] Ũ = [A 1A j xj T γ ia 1 A jξ j, Θ γ ε ] Ũ + [T γ an using Lemma 4.5 (recall that A 1 A j W 2, (Ω)), we obtain A 1 Lγ U ε = 1 γ 2 A 1 1 f + r ε + j=0 ia 1 A jξ j, Θ γ ε ] Ũ, [T γ ia 1 A jξ j, Θ γ ε ] Ũ, (11) where r ε 1,γ tens to 0. The remaining commutators are zero orer terms in Ũ, uniformly with respect to ε. Therefore, one cannot neglect them an treat these commutators as source terms. What saves the ay is that these commutators can be ecompose in the following way 6 : [T γ ia 1 A jξ j, Θ γ ε ] Ũ = T γ j,ε U ε + r ε, where j,ε is a symbol in Γ 0 1 (Ω) that is uniformly boune with respect to ε. Inee, we use Theorem 2.2 to compute [T γ ia 1 A jξ j, Θ γ ε ] Ũ = T γ {A 1 A jξ j,ϑ ε} Ũ + R γ ε Ũ = T γ j,ε ϑ ε Ũ + R γ ε Ũ = T γ j,ε U ε + R γ ε Ũ, (12) 6 If ϑ ε ha compact support in ξ, such a ecomposition woul not hol. The choice of the mollifier ϑ ε is therefore crucial. 15

16 where the symbol j,ε is given by j,ε = 1 k=0 an where R γ ε is of orer 1, uniformly with respect to ε: 2ε ξ j ξ k γ 2 + ε ξ 2 x k (A 1 A j) Γ 0 1(Ω), (13) R γ ε V 1,γ C V 0, V L 2 (Ω), ε ]0, 1]. The symbols j,ε efine by (13) are uniformly boune in Γ 0 1 (Ω) with respect to ε. We observe that ϑ ε = γ 2 +εσ ε, with σ ε boune in Γ 2 1 (Ω). We also observe that j,ε = εα j,ε, with α j,ε boune in Γ 2 1 (Ω). Consequently, the remainer Rγ ε in (12) satisfies lim ε 0 Rγ ε V 1,γ = 0, V C0 (Ω). Thanks to the uniform boun on R γ ε, we may conclue that R γ ε Ũ 1,γ tens to zero as ε tens to zero. Using (12) in equation (11), an efining 1 ε := j,ε Γ 0 1(Ω), j=0 we obtain A 1 Lγ U ε + T γ U ε = 1 ε γ 2 A 1 f + r ε, r ε 1,γ 0, (14) an the symbols ε are boune in Γ 0 1 (Ω). Note that (14) also reas x U ε = A 1 1 γu ε + x0 U ε + A j xj U ε + DU ε T γ U ε + 1 ε γ 2 A 1 f + r ε H(Ω), j=0 an we thus have U ε H(Ω), with H(Ω) efine by (4). For the bounary conitions, one procees in an entirely similar way, an gets BU ε x =0 + T γ b ε U ε x =0 = 1 γ 2 g + r ε, r ε 1,γ 0, (15) with symbols b ε boune in Γ 1 1 (R ). Using (14) an (15), the first part of the Theorem is prove, provie that we efine (with slight abuse of notations): U ν := γ 2 U εν H(Ω), ν := εν, b ν := b εν, ε ν := 1 ν + 1. Now we show that Ũ x =0 L2 (R ) an that (10) hols. First note that the operators V A 1 Lγ V + T γ ν V an V BV x =0 are continuous from the space H(Ω) into H(Ω) an from H(Ω) into H 1 (R ) (use Theorem A.1 in appenix A for the bounary operator). Since C0 (Ω) is ense in H(Ω) (see Proposition A.1 in appenix A), it is clear that the a priori energy estimate given by assumption 2 still hols when U H(Ω) (an not only when U C0 (Ω)). 16

17 Thanks to assumption 2, an to the bouneness of ν in Γ 0 1 (Ω), we know that there exists a constant C = C(N, δ) an a positive number γ 3 (N, δ) such that ) γ U ν U ν x =0 2 0 C for all ν N, an for all γ γ 3. Decomposing γ 3 A 1 Lγ U ν + T γ U ν 2 ν 1,γ + 1 γ 2 BU ν 2 x =0 1,γ BU ν x =0 = (BU ν x =0 + T γ b ν U ν x =0 ) T γ b ν U ν x =0, an using (14)-(15), we get (for γ large enough): where γ U ν U ν x =0 2 0 C γ 3 A 1 f + rν int 2 1,γ + 1 γ 2 g + rb ν 2 1,γ r int ν 1,γ 0, r b ν 1,γ 0., ), (16) The sequence (U ν ) is thus boune in L2 x (R ), an therefore, up to extracting a subsequence, =0 it converges weakly in L 2 (R ) towar some function u L 2 (R ). Since the whole sequence (U ν ) converges towar Ũ in x =0 x =0 H 1/2 (R ), this implies Ũ x =0 L2 (R ). Moreover, we know that the sequence (U ν ) converges strongly towar Ũ in L2 (Ω), an (16) yiels γ Ũ Ũ x =0 2 0 C lim inf γ 3 A 1 f + rν int 2 1,γ + 1 ) γ 2 g + rb ν 2 1,γ, C γ 3 f 2 1,γ + 1 ) γ 2 g 2 1,γ. This completes the proof. We summarize Proposition 4.1 an Theorem 4.1 by the following well-poseness result for the Bounary Value Problem (5): Theorem 4.2. Let D W 1, (Ω). There exists γ 3 (N, δ) such that for γ γ 3, f H γ (Ω) an g H 1 γ(r ), there exists a unique solution U L 2 γ(ω) of the following system: { LU = t U + A j(t, x) xj U + D(t, x)u = f(t, x), (t, x) Ω, B(t, y) U x =0 = g(t, y), (t, y) R, This solution satisfies U x =0 L2 γ(r ) an the following estimate hols: γ U 2 L 2 γ (Ω) + U x =0 2 L 2 γ (R ) C γ 3 f 2 H + 1 ) γ(ω) γ 2 g 2 Hγ 1(R ). In aition, there exists a sequence (U ν ) in H 1 γ(ω) that satisfies the following properties: U ν U in L 2 γ(ω), U ν x =0 U x =0 in L 2 γ(r ), LU ν f in L 2 γ(ω), B U ν x =0 g in L2 γ(r ). The last part of the Theorem is prove in [17], using Frierichs lemma (lemma 4.5). We make the following important comments: even though the source terms f an g have tangential erivatives in L 2 γ, there is no hope to prove, for instance, that LU ν converges towar f in H γ (Ω), see Theorem 4.1. Eventually, the convergence of the traces in L 2 γ can be obtaine because we alreay know (thanks to Theorem 4.1) that the trace of U belongs to L 2 γ(r ). When the source terms are only in L 2 γ, there is an analogous result: 17

18 Theorem 4.3. Let D W 1, (Ω). There exists γ 3 (N, δ) such that for γ γ 3, f L 2 γ(ω) an g L 2 γ(r ), there exists a unique solution U L 2 (R + x : Hγ 1 (R t,y)) of the following system: { t U + A j(t, x) xj U + D(t, x)u = f(t, x), (t, x) Ω, B(t, y) U = g(t, y), (t, y) x =0 R, This solution satisfies U x =0 H 1 γ (R ) an the following estimate hols: ( ) 1 γ 3 U 2 L 2 (R + ;Hγ 1 (R )) + γ2 U x =0 2 Hγ 1 (R ) C γ f 2 L 2 γ(ω) + g 2 L 2 γ(r ). One shoul also keep in min that the ual problem amits similar well-poseness results. 4.4 Well-poseness with zero initial ata. En of the proof Now, we show that Theorem 4.2 an Theorem 4.3 yiel a well-poseness result for the Initial Bounary Value Problem with zero initial ata. We first prove that the classical support lemma extens to weakly stable problems: Lemma 4.6. There exists γ 4 (N, δ) such that, if γ γ 4, (f, g) H γ (Ω) H 1 γ(r ) vanish for t < T 0, then the solution U L 2 γ(ω) of (5) vanishes for t < T 0. Moreover, if γ γ 4, f L 2 γ(ω), an g L 2 γ(r ), then the solution U L 2 (R + ; H 1 γ (R )) of (5) also vanishes for t < T 0. Proof. We give the proof when the source terms are in H γ (Ω) Hγ(R 1 ), but the proof is similar when the source terms are in L 2 γ. There is no loss of generality in assuming T 0 = 0. (Otherwise, use a translation t t T 0 ). We fix a function χ C (R) such that χ oes not vanish an { 1, if x 0, χ(t) = exp( t), if x 1. The function (t, y, x ) Ω χ (t)/χ(t) belongs to W 1, (Ω). enough, the only solution in L 2 γ(ω) to the linear problem LV χ (t) V = 0, (t, x) Ω, χ(t) B(t, y) V = 0, (t, y) x =0 R, Consequently, for all γ large (17) is the trivial solution V = 0, thanks to Theorem 4.2 (we use the essential fact that Theorem 4.2 hols for any zero orer term in W 1, (Ω)). Consier some ata (f, g) H γ (Ω) Hγ(R 1 ) that vanish for t < 0. Then we have (f, g) H γ+j (Ω) Hγ+j 1 (R ) for all integer j. Thanks to Theorem 4.2, we know that there exists a unique U j L 2 γ+j (Ω) satisfying { LUj = f(t, x), (t, x) Ω, B(t, y) U j x =0 = g(t, y), (t, y) R. The function χ(u j+1 U j ) belongs to L 2 γ+j (Ω) an one checks that it is a solution to (17). Therefore it equals zero, an U j+1 = U j = = U 0. Furthermore, we know that sup j N 1 γ + j f H γ+j (Ω) < + an sup j N 1 γ + j g H 1 γ+j (R ) < +, 18

19 because f an g vanish for t < 0. Thus Theorem 4.2 yiels sup j This implies that U 0 vanishes for t < 0. U j L 2 γ+j (Ω) = sup U 0 L 2 j γ+j (Ω) < +. We introuce a few notations: for T > 0, let Ω T := Ω {t < T } =], T [ R +, an let ω T :=], T [ R 1. The spaces L 2 γ(ω T ), L 2 γ(ω T ), an H γ (Ω T ) are efine similarly as L 2 γ(ω) etc. The efinition of the norms in L 2 γ(ω T ) an L 2 γ(ω T ) is clear. As regars the norm in H γ (Ω T ), it is efine by 1 f 2 H γ(ω T ) := γ2 f 2 L 2 γ(ω T ) + xj f 2 L 2 γ(ω T ). The norm of H 1 γ(ω T ) is efine in a similar way. We are now able to en the proof of Theorem 3.1. Proof. We consier source terms f H(Ω T ), an g H 1 (ω T ), that vanish in the past. We note that f an g belong to H γ (Ω T ) an to H 1 γ(ω T ) for all γ 1. We exten f an g as functions f H(Ω) an g H 1 (R ). Because f an g vanish for t < 0, it is also straightforwar that f H γ (Ω) an g H 1 γ(r ) for all γ 1. Consequently, for γ large enough, there exists a unique U L 2 γ(ω) such that j=0 { LU = f (t, x), (t, x) Ω, B(U ) x =0 = g (t, y), (t, y) R, an U satisfies the corresponing energy estimate, see Theorem 4.2. Furthermore, U vanishes in the past, thanks to Lemma 4.6. We also have γ U 2 L 2 γ(ω T ) + (U ) x =0 2 L 2 γ(ω T ) γ U 2 L 2 γ(ω) + (U ) x =0 2 L 2 γ(r ) C γ 3 f 2 H + 1 ) γ(ω) γ 2 g 2 Hγ 1(R ) C γ 3 f 2 H γ(ω T ) + 1 ) γ 2 g 2 Hγ 1(ω T ). The restriction U of U to Ω T belongs to L 2 (Ω T ) because U vanishes in the past an U L 2 γ(ω T ) when γ is large. We have thus constructe a solution in L 2 (Ω T ) to the localize problem: { LU = f(t, x), (t, x) ΩT, B(t, y) U = g(t, y), (t, y) ω (18) x =0 T. We now show uniqueness of such a solution. Let U L 2 (Ω T ) have a trace in L 2 (ω T ), vanish in the past, an satisfy { LU = 0, (t, x) ΩT, BU x =0 = 0, (t, y) ω T. Let ε > 0, an consier a function χ C (R) such that χ(t) = 1 if t T 2ε, an χ(t) = 0 if t T ε. Define U χ := χu. Then U χ L 2 (Ω) an U χ vanishes in the past, so U χ L 2 γ(ω) for all γ 1. Moreover, we compute: { LUχ = χ (t)u, (t, x) Ω, BU χ x =0 = 0, (t, y) R. 19

20 Observe that χ U L 2 γ(ω) for all γ 1, an χ U vanishes for t < T 2ε. We can thus apply Lemma 4.6 (with source terms in L 2 γ) 7, an conclue that U χ vanishes for t < T 2ε. Consequently, U vanishes for t < T, that is, U = 0. To en the proof of Theorem 3.1, we show the continuity with respect to the time variable. We consier source terms f H(Ω T ) an g H 1 (ω T ) that vanish in the past, an we continue these functions as f H(Ω) an g H 1 (R ). We alreay know that the unique solution U L 2 (Ω T ) of (18) that vanishes in the past, is the restriction to Ω T of the solution U L 2 γ(ω) to the global problem { LU = f, (t, x) Ω, BU x =0 = g, (t, y) R, Moreover, the trace of U on {x = 0} belongs to L 2 (R ), an vanishes in the past. To prove Theorem 3.1, it is sufficient to show the continuity of U with respect to the time variable. Thanks to Theorem 4.2, we know that there exists a sequence (U ν ) in H 1 γ(ω) verifying U ν U in L 2 γ(ω), U ν x =0 U x =0 in L 2 γ(r ), LU ν LU = f in L 2 γ(ω), B U ν x =0 g in L 2 γ(r ). Using the Frierichs symmetrizer S (see assumption 1), the classical energy estimate for symmetric hyperbolic systems in a half-space reas (see e.g. [11]): e 2γt U ν (t) 2 L 2 (R + ) + γ U ν 2 L 2 γ(ω t) C γ LU ν 2 L 2 γ (Ωt) + U ν x =0 2 L 2 γ (ωt) ), an we also have e 2γt U ν (t) U ν (t) 2 L 2 (R + ) + γ U ν U ν 2 L 2 γ(ω t) ( ) 1 C γ LU ν LU ν 2 L 2 γ(ω + U ν t) U ν x =0 x =0 2 L 2 γ(ω t). Passing to the limit, we obtain the continuity of U with respect to the time variable. previous estimates for the trace of U on ω t yiels the estimate state in Theorem Some remarks 5.1 The IBVP with general initial ata Using Theorem 3.1, one woul like to show well-poseness of the IBVP (1) with general initial ata. Assume first that the coefficients A j s an D, as well as the Frierichs symmetrizer S, are C, boune, an with boune erivatives. Exten those coefficients to the whole space R +1, so that the system remains symmetric hyperbolic. Then for all f L 1 (]0, T [; H 2 (R )), an for all U 0 H 2 (R ), one can construct a solution U (1) C 0 ([0, T ]; H 2 (R )) C 1 ([0, T ]; H 1 (R )) to the Cauchy problem { t U (1) + A j(t, x) xj U (1) + D(t, x)u (1) = f(t, x), t ]0, T [, x R, U (1) t=0 = U 0 (x), x R. For the IBVP (1), one seeks the solution uner the form U = U (1) + U (2), with U (2) solution to t U (2) + A j(t, x) xj U (2) + D(t, x)u (2) = 0, t ]0, T [, x R +, B(t, y)u (2) (1) x = g BU =0, t ]0, T [, y R 1 x, =0 U (2) t=0 = 0, x R. 7 It is crucial here to have a well-poseness result for source terms in L 2 γ. The 20

21 Consequently, if the source term g belongs to H 1 (]0, T [ R 1 ), an if the initial ata U 0 H 2 (R +) satisfy the compatibility conition g t=0 = B(0, y)(u 0 ) x =0, then one can solve the IBVP (1) with a source term f L 1 (]0, T [; H 2 (R +)), thanks to Theorem 3.1. However, this strategy harly applies when the A j s are in W 2, (Ω) an D is only in W 1, (Ω). The problem is to solve the Cauchy problem with initial ata, for instance in H 2 (R ), an to obtain a solution on ]0, T [ R, such that its trace on ]0, T [ R 1 belongs to H 1. This oes not seem possible with a zero orer coefficient in W 1,. We therefore prefer not to pursue this issue, which is a little beyon the scope of this paper. However, for C boune coefficients, an with ata that satisfy the above mentione compatibility conition, the techniques of [18] shoul yiel a well-poseness result for the IBVP (1). The result of [19] even suggests that, uner this compatibility conition, the IBVP is well-pose with initial ata in H 1 (R +). When ealing with nonlinear problems, one usually solves the nonlinear equations by a sequence of linearize problems with zero initial ata an source terms that vanish in the past, see e.g. [12, 14, 17]. This is another reason why we o not pursue the stuy of general initial ata. 5.2 Uniformly characteristic IBVP In applications, it often happens that the bounary is characteristic, that is, the eterminant of the matrix A vanishes on the bounary. In many of these cases (at least in many physically relevant problems), the rank of A is constant only on the bounary, an the bounary conitions are maximally issipative. In such situations, the corresponing IBVP has been stuie in great etails by many authors, even at the level of quasilinear equations, see e.g. [8, 21] an the references cite therein. When the bounary conitions satisfy the uniform Lopatinskii conition, an when the rank of A is constant in a neighborhoo of the bounary, the linear IBVP was stuie in [13]. When losses of tangential erivatives occur, an when the bounary is uniformly characteristic (this happens for instance in the stuy of contact iscontinuities, see [6]), one can reprouce the analysis evelope above. More precisely, assume that there exist two invertible matrices Q 1,2 (t, x) W 2, (Ω), such that 0 n0 (t, x) Ω, Q 1 (t, x)a (t, x)q 2 (t, x) = I n+ I n, where I k is the ientity matrix in R k, an n 0, n ± are fixe integers. Then the problem { t U + A j(t, x) xj U + D(t, x)u = f(t, x), (t, x) Ω T, B(t, y) U = g(t, y), x =0 (t, y) ω T, with source terms f H(Ω T ) an g H 1 (ω T ) that vanish in the past, satisfies a well-poseness result that is entirely analogous to Theorem 3.1, provie that the analogues of assumptions 2 an 3 for characteristic problems are satisfie. The only ifference is that we can control only the noncharacteristic part of the trace of the solution U on the bounary. With the help of our analysis, the verification of the well-poseness of the linearize equations for the vortex sheets problem (as stuie in [6]) is thus essentially reuce to the calculation of the Lopatinskii eterminant for a suitable ual problem. 21

22 Acknowlegments I thank Patrick Gérar for bringing the reference [7] to my attention. I also thank Guy Métivier for enlightening numerous technical etails. Research of the author was supporte by the EU finance network HYKE, HPRN-CT A Some properties of anisotropic Sobolev spaces In R +1, a generic point is enote by x = (x 0,..., x ). We use the notation x = (x, x ) with x R an x R. We also use the notation Ω = R +1 + = {x R+1 s.t. x > 0}. We efine the following spaces H(R +1 ) := {f L 2 (R +1 ) s.t. x0 f,..., x 1 f L 2 (R +1 )}, H(R +1 ) := {f H(R +1 ) s.t. x0 f,..., x f H(R +1 )}, H(Ω) := {f L 2 (Ω) s.t. x0 f,..., x 1 f L 2 (Ω)}, H(Ω) := {f H(Ω) s.t. x0 f,..., x f H(Ω)}. The spaces H(R +1 ) an H(R +1 ) are equippe with the following norms 8 : f 2 H(R +1 ) := 1 (2π) +1 (1 + ξ 2 ) f(ξ) 2 ξ, R +1 f 2 H(R +1 ) := 1 (2π) +1 (1 + ξ 2 )(1 + ξ 2 ) f(ξ) 2 ξ, R +1 where we have ecompose ξ = (ξ, ξ ) R R. The spaces H(Ω) an H(Ω) are equippe with the norms: f 2 H(Ω) := f 2 L 2 (Ω) + x 0 f 2 L 2 (Ω) + + x 1 f 2 L 2 (Ω), f 2 H(Ω) := f 2 H(Ω) + x 0 f 2 H(Ω) + + x f 2 H(Ω). The following ensity result is stanar, an is prove by a truncation/regularization argument (see [2] for etails): Proposition A.1. The space C0 (R+1 ) is ense in both H(R +1 ) an H(R +1 ). The space C0 (Ω) is ense in both H(Ω) an H(Ω). The following result is also very classical: Proposition A.2. There exist two continuous linear mappings E : H(Ω) H(R +1 ) an E : H(Ω) H(R +1 ) such that for all u H(Ω) (resp. u H(Ω)), Eu = u (resp. Eu = u) almost everywhere in Ω. Observe that for the extension operator E, it is sufficient to consier a continuation by 0 outsie of Ω (this is because there is no normal erivative x in the efinition of H(Ω)). We en this short appenix with the following result: Theorem A.1. The mapping Γ : u C 0 (Ω) u(x, 0) C 0 (R ) can be continue in a unique way as a continuous linear mapping Γ : H(Ω) H 3/2 (R ). 8 Here we take γ = 1 for the sake of simplicity, but it is clear that introucing the parameter γ in the efinition of the norms oes not change the results state below. 22