Penalization approach for mixed
hyperbolic systems with constant
coefficients satisfying a Uniform
Lopatinski Condition.
B. Fornet
October 24, 2018
Abstract
In this paper, we describe a new, systematic and explicit way of
approximating solutions of mixed hyperbolic systems with constant
coefficients satisfying a Uniform Lopatinski Condition via different Pe-
nalization approaches.
LATP, Université de Provence,39 rue Joliot-Curie, 13453 Marseille cedex 13, France.
LMRS, Université de Rouen,Avenue de l’Université, BP.12, 76801 Saint Etienne du
Rouvray, France.
http://arxiv.org/abs/0704.0497v1
1 Introduction.
In this paper, we describe a new, systematic and explicit way of approximat-
ing solutions of mixed hyperbolic systems with constant coefficients satisfy-
ing a Uniform Lopatinski Condition via different Penalization approaches.
In applied Mathematics like, for instance, in the study of fluids dynamics,
the method of penalization is used to treat boundary conditions in the case
of complex geometries. By replacing the boundary condition by a singular
perturbation of the PDE extended to a larger domain, this method allows
the construction of an approximate, often more easily computable, solution.
We consider mixed boundary value problems for hyperbolic systems:
Aj∂j ,
on {xd ≥ 0}, with boundary conditions on {xd = 0}. The n× n real valued
matrices Aj are assumed constant. Of course, we assume the coefficients to
be constant as a first approach, aiming to generalize the results obtained here
in future works. We assume that the boundary {xd = 0} is noncharacteristic,
which means that detAd 6= 0. We denote by
y := (x1, . . . , xd−1) and x := xd. The problem writes:
(1.1)
Hu = f, {x > 0},
Γu|x=0 = Γg,
u|t<0 = 0 ,
where the unknown u(t, x) ∈ Rn, Γ : Rn → Rp is linear and such that
rg Γ = p; which implies that Γ can be viewed as a p×n real valued constant
matrix. Let us fix T > 0 once and for all for this paper. Let Ω+
denotes
the set [0, T ] × Rd+ and ΥT denote the set [0, T ] × R
d−1. f is a function in
Hk(Ω+
), g is a function in Hk(ΥT ), where k ≥ 3 or k = ∞, such that:
f |t<0 = 0 and g|t<0 = 0. We make moreover the following Hyperbolicity
assumption on H :
Assumption 1.1. For all (η, ξ) ∈ Rd−1 × R− {0}, the eigenvalues of
ηjAj + ξAd
are real, semi-simple and of constant multiplicity.
Let us introduce now the frequency variable ζ := (γ, τ, η), where iτ + γ,
with γ ≥ 0, and τ ∈ R stands for the frequency variable dual to t and
η = (η1, . . . , ηd−1) where ηj ∈ R is the frequency variable dual to xj. We
note:
A(ζ) := − (Ad)
(iτ + γ)Id+
iηjAj
Denote by M a N ×N, complex valued, matrix; E−(M)[resp E+(M)] is the
linear subspace generated by the generalized eigenvectors associated to the
eigenvalues of M with negative [resp positive] real part. If F and G denote
two linear subspaces of CN such that dimF+dimG = N, det(F,G) denotes
the determinant obtained by taking orthonormal bases in each space. Up to
the sign, the result is independent of the choice of the bases. We shall now
explicit the Uniform Lopatinski Condition assumption:
Assumption 1.2. (H,Γ) satisfies the Uniform Lopatinski Condition i.e for
all ζ such that γ > 0, there holds:
(1.2) |det(E−(A), ker Γ)| ≥ C > 0.
The mixed hyperbolic system (1.1) has a unique solution in
Hk(Ω+
), and, since H is hyperbolic with constant multiplicity, for all γ
positive, the eigenvalues of A stay away from the imaginary axis. More-
over, as emphasized for instance by Chazarain and Piriou in [3] and Mé-
tivier in [8], there is a continuous extension of the linear subspace E−(A)
to {γ = 0, (τ, η) 6= 0
} that we will denote by Ẽ−(A). Ẽ+(A) extends as
well continuously to {γ = 0, (τ, η) 6= 0
} and we will denote Ẽ+(A) this
extension. Moreover, there holds:
Ẽ−(A)
Ẽ+(A) = C
We can refer the reader to [3], [6], [7], or [8] for detailed estimates concern-
ing mixed hyperbolic problems satisfying a Uniform Lopatinski Condition.
Moreover, we can refer to [10] for the proof of the continuous extension of the
linear subspaces mentioned above in the hyperbolic-parabolic framework.
Remark 1.3. As a consequence of the uniform Lopatinski condition, there
holds, for all ζ 6= 0 :
rg Γ = p = dim Ẽ−(A(ζ)).
1.1 A Kreiss Symmetrizer Approach.
We will now describe a penalization method involving a Kreiss Symmetrizer
and a matrix constructed by Rauch in [12], in the construction of our singular
perturbation. Note well that we have some freedom in both the choice of the
Kreiss Symmetrizer and of Rauch’s matrix. Let us denote respectively by û,
f̂ , and ĝ the tangential Fourier-Laplace transform of u, f, and g. Since the
Uniform Lopatinski Condition is holding for the mixed hyperbolic system
(1.1), there is, see [9] a Kreiss symmetrizer S for the problem:
(1.3)
∂xû = Aû+ f̂ , {x > 0},
Γû|x=0 = Γĝ,
That is to say there exists a matrix S(ζ), homogeneous of order zero in ζ,
C∞ in R+ × Rd − {0
} and there are λ > 0, δ > 0 and C1 such that:
• S is hermitian symmetric.
• ℜ (SA) ≥ λId.
• S ≥ δId− C1Γ
An algebraic result proved by Rauch in [12] can be reformulated as follow,
and a proof is recalled in section 2.2:
Lemma 1.4. There is a hermitian symmetric, uniformly definite positive,
N ×N matrix B such that:
ker Γ = E+((S)
−1B).
Moreover B depends smoothly of ζ.
Remark 1.5. This result is proved by constructing explicit matrices satisfy-
ing the desired properties. Thus, it is not merely an existence result and we
can use the explicitly known matrix B in our construction of a penalization
operator.
Let us denote by R := B
2 and SR := R
−1SR−1. We will denote by P−
the projector on E−(SR) parallel to E+(SR) and by P
+ the projector on
E+(SR) parallel to E−(SR); P
− and P+ denoting the associated Fourier
multiplier. We recall that, denoting by F the tangential Fourier transform,
the Fourier multiplier P−(∂t, ∂y, γ) [resp P
+(∂t, ∂y, γ)] is then defined, for all
w ∈ Hk(Rd+1), and γ > 0, by:
−(∂t, ∂y, γ)w
= P−(ζ)F(w),
[resp
+(∂t, ∂y, γ)w
= P+(ζ)F(w)],
in the future we will rather write:
±(∂t, ∂y, γ)w
= P±(ζ)F(w).
We fix, once and for all, γ > 0 big enough. Let us consider then the solution
uε of the well-posed Cauchy problem on the whole space (1.4):
(1.4)
Huε +
Muε1x<0 = f1x>0 +
θ1x<0, {x ∈ R},
uε|t<0 = 0,
where
M := −eγtAdS
−1RP−Re−γt,
θ := −eγtAdS
−1RP−Γg̃,
and S(∂t, ∂y) [resp R(∂t, ∂y)] denotes the Fourier multiplier associated to
S(ζ) [resp R(ζ)]. Let us define g̃ by:
g̃ := e−x
In what follows, ĝ will denote the Fourier-Laplace transform of g̃. Let us
denote by
ũ := u−1x<0 + u1x≥0 = u
1x≤0 + u1x>0.
u denotes the solution of (1.1), and thus belongs to Hk(Ω+
). u− is a function
belonging to Hk(Ω−
) and such that u−|x=0 = u|x=0. More precisely, u
− can
be computed by: eγtF−1
R−1(v̂− + P−Γĝ)
, where v̂− is the solution of the
problem:
SR∂xv̂
− − P+SRARv̂
− = P+SRARP
−Γĝ, {x < 0},
v̂−|x=0 = P
+Rû|x=0,
and û denotes the Fourier-Laplace transform of the solution u of (1.1).
Theorem 1.6. For all k ≥ 3, if f ∈ Hk(Ω+
) and g ∈ Hk(ΥT ), then there
holds:
‖uε − u−‖
Hk−3(Ω
+ ‖uε − u‖
Hk−3(Ω
= O(ε),
where uε denotes the solution of the Cauchy problem (1.4) and u denotes
the solution of the mixed hyperbolic problem (1.1). If g = 0 then:
‖uε − u−‖
+ ‖uε − u‖
= O(ε).
Of course, since uε is defined for all {x ∈ R}, its limit as ε → 0+, ũ is
can be viewed as an ”extension” of u on the fictive domain {x < 0}. The
”extension” resulting from our method of penalization gives a continuous ũ
across {x = 0}, while the method used in [2] gave simply: ũ|x<0 = 0. We
have the following Corollaries:
Corollary 1.7. Assume for example that f ∈ H∞(Ω+
) and
g ∈ H∞(ΥT ) then
‖uε − u‖
Hs(Ω+
= O(ε); ∀s > 0.
Corollary 1.8. If f belongs to L2(Ω+
) and g = 0 then:
‖uε − ũ‖L2(ΩT ) = 0.
One of the interest of this first approach lies in the rate of convergence
of uε towards u. Indeed, in general, a boundary layer will form near the
boundary in this kind of singular perturbation problem. For example in the
paper by Bardos and Rauch [2], as confirmed by Droniou [4], a boundary
layer forms. It is also the case in [11], as analyzed in our Appendix. There
are also boundary layers phenomena in the parabolic context: see the ap-
proach proposed by Angot, Bruneau and Fabrie [1] for instance. However,
surprisingly, and like in the penalization method proposed by Fornet and
Guès in [5], our method allows the convergence to occur without formation
of any boundary layer on the boundary. As a result, this leads to the kind
of sharp stability estimate given in Theorem 1.6. These results concern the
case where f and g are sufficiently regular. The reason is that we construct
an approximate solution. In the case of g only in L2(ΥT ), such a simple
treatment does not work. However, let δ > 0 be given. If we approximate
f and g by smooth functions fν ∈ H
) and gν ∈ H
∞(ΥT ) such that
‖f − fν‖L2(Ω+
< δ and ‖g − gν‖L2(ΥT ) < δ, by the uniform Lopatinski
condition, we get:
‖uν − u‖L2(Ω+
< Cν,
where uν is the solution of the mixed hyperbolic problem (1.1) with data fν
and gν . We can now apply Corollary 1.8 to uν , and obtain by penalization
a sequence uεν in L
2(ΩT ) such that: limε→0+ u
ν = uν in L
). Finally, by
choosing, ε sufficiently small, we get ‖u − uεν‖L2(Ω+
< 2Cδ. By choosing ε
and ν as functions of δ, and noting u(δ) = u
, we have:
(1.5) lim
‖u(δ) − u‖L2(Ω+
1.2 A second Approach.
In the first approach we have just introduced, it is necessary to compute
a Kreiss’s Symmetrizer and a Rauch’s matrix. In view of future numerical
applications, we will now introduce another method preventing the compu-
tation of these matrices. The price to pay is that we need the preliminary
computation of v, which is by definition the solution of the Cauchy problem
on the free space:
(1.6)
Hv = f, (t, y, x) ∈ ΩT ,
v|t<0 = 0 ∀(y, x) ∈ R
Let us denote P−(ζ) the spectral projector on Ẽ−(A(ζ)) parallel to
Ẽ+(A(ζ)), andP
+(ζ) the spectral projector on Ẽ+(A(ζ)) parallel to Ẽ−(A(ζ)).
Let us introduce P±(∂t, ∂y, γ), the Fourier multiplier associated to P
±(ζ).
Let us denote by Π the projector on Ẽ−(A(ζ)) parallel to KerΓ, which has
a sense because of the Uniform Lopatinski Condition and denote Π the
associated Fourier multiplier. We define then h̃ by:
h̃ := e−x
−(e−γtv|x=0) +Πe
−γt(g − v|x=0)
where g denotes the function involved in the boundary condition of the
mixed hyperbolic problem (1.1). Now, let us consider the following singularly
perturbed Cauchy problem on the whole space:
(1.7)
Huε +
−e−γtuε1x<0 = f1x>0 +
γth̃1x<0,
uε|t<0 = 0 .
Let us denote by
ũ := u−1x<0 + u1x≥0 = u
1x≤0 + u1x>0.
u denotes the solution of (1.1) thus belonging toHk(Ω+
) and u− is a function
belonging to Hk(Ω−
) and such that u−|x=0 = u|x=0. More precisely, u
− can
be computed by: eγtF−1(F(h̃)+v̂−), where v̂− is the solution of the problem:
+v̂−)−A(P+v̂−) = 0, {x < 0},
+v̂−|x=0 = P
+û|x=0.
and û denotes the Fourier-Laplace transform of the solution u of (1.1). The
problem (1.7) is well-posed and, for all ε > 0, there exists a unique
uε ∈ Hk(ΩT ) solution. We will fix γ adequately big beforehand. We observe
then the following result:
Theorem 1.9. For all k ≥ 3, if f ∈ Hk(Ω+
) and g ∈ Hk(ΥT ), then there
holds:
‖uε − u−‖
Hk−3(Ω
+ ‖uε − u‖
Hk−3(Ω
= O(ε),
where uε denotes the solution of the Cauchy problem (1.7) and u denotes
the solution of the mixed hyperbolic problem (1.1).
The singular perturbation involved in the definition of uε does not de-
pend either of Kreiss’s Symmetrizer or Rauch’s matrix. As a result, for
this method of penalization far less computations are necessary in order to
obtain our singular perturbation. Note well that the proof of the energy
estimates in Theorem 1.9 is completely different from the proof of the en-
ergy estimates in Theorem 1.6. Indeed, for our first approach our singularly
perturbed problem was treated as a Cauchy problem, contrary to our second
approach where it was interpreted as a transmission problem.
Corollary 1.10. Assume for example that f ∈ H∞(Ω+
) and
g ∈ H∞(ΥT ) then
‖uε − u‖
= O(ε); ∀s > 0.
Of course, we see that the same problem of regularity arises in Theorem
1.9 and Theorem 1.6. However, by a simple density argument, we can also
prove here the exact analogous of (1.5).
Remark 1.11. In the case where f = 0, then the solution v of (1.6) is
v = 0 and thus, the perturbed cauchy problem (1.7) rewrites:
Huε +
−e−γtuε1x<0 =
γte−x
Πe−γtg
1x<0, {x ∈ R},
uε|t<0 = 0 .
2 Underlying approach leading to the proof of
Theorem 1.6.
2.1 Some preliminaries.
Since the Uniform Lopatinski Condition holds, there is S, homogeneous of
order zero in ζ, and such that there are λ > 0, δ > 0 and C1 and there holds:
• S is hermitian symmetric.
• ℜ (SA) ≥ λId.
• S ≥ δId− C1Γ
S is then called a Kreiss Symmetrizer for the problem:
(2.1)
∂xû = Aû+ f̂ , {x > 0},
Γû|x=0 = Γĝ,
where f̂ and ĝ denotes respectively the Fourier-Laplace transforms of f and
g̃; and û denotes the Fourier-Laplace transform of the solution u of the well-
posed mixed hyperbolic problem (1.1). û is also solution, for all fixed ζ 6= 0
of the following equation:
(2.2)
S∂xû = SAû+ S(Ad)
−1f̂ , {x > 0},
Γû|x=0 = Γĝ,
Remark 2.1. Following our current assumptions, Γ is independent of ζ 6= 0,
however, more general boundary conditions, of the form:
Γ(ζ)û|x=0 = Γ(ζ)ĝ,
can be treated. It would imply taking as boundary condition for (1.1):
Γγu|x=0 = Γγg,
with for γ big enough,
Γγ := Γ(∂t, ∂y)e
where, Γ(∂t, ∂y) denotes the Fourier multiplier associated to Γ(ζ), that is to
say is defined by:
F(Γ(∂t, ∂y)u) = Γ(ζ)F(u).
Referring for example to [3] and [7], Kreiss has proved that the exis-
tence of a Kreiss symmetrizer for the symbolic equation is sufficient to prove
the well-posedness of the associated pseudodifferential equation (here (1.1)).
Indeed, multiplying by û and integrating by parts the equation:
S∂xû = SAû+ S(Ad)
leads to the desired a priori estimates. For all ζ 6= 0, S(ζ) is hermitian
symmetric and definite positive on ker Γ. Let us sum up the properties crucial
in the proof of the well-posedness of our problem:
Proposition 2.2. For all ζ = (τ, γ, η) such that τ2 + γ2 +
j=1 η
j = 1,
there holds:
• S(ζ) is hermitian symmetric.
• ℜ (SA) (ζ) := 1
(SA+ (SA)∗)(ζ) is positive definite.
• −S(ζ) is definite negative on ker Γ and ker Γ is of same dimension as
the number of negative eigenvalues in −S(ζ).
Note that, by homogeneity of S, it is equivalent for the properties in Propo-
sition 2.2 to hold for |ζ| = 1 or for |ζ| > 0. As a consequence of the first
point and third point of Proposition 2.2, the Lemma 1.4 applies and gives a
matrix B such that: ker Γ = E+(S
−1B). In the sequel, such a matrix B is
fixed once for all.
The following chapter contains a proof of Lemma 1.4 assorted of a detailed
construction of B.
2.2 Detailed proof of Lemma 1.4: Construction of the ma-
trices B solving Lemma 1.4.
As we will emphasize in next chapter, Lemma 1.4 is a crucial feature in our
first method of Penalization. The aim of this chapter is to give a more com-
plete proof rather than simply recalling Rauch’s result and, in the process,
to precise how the matrices B solving Lemma 1.4 are constructed. For all
ζ 6= 0, S(ζ) is hermitian symmetric, uniformly definite positive on Ẽ+(A(ζ)),
and uniformly definite negative on Ẽ−(A(ζ)); as a consequence, S(ζ) keeps
exactly p positive eigenvalues and N − p negative eigenvalues for all ζ 6= 0.
Basically, knowing that S is uniformly definite positive on ker Γ; we search
to express ker Γ in a way involving S. Consider q ∈ ker Γ, since, for all ζ 6= 0,
E−(S(ζ))
E+(S(ζ)) = C
N , we can split q in:
q := q+ + q−
with q+ ∈ E+(S(ζ)) and q
− ∈ E−(S(ζ)).
Since dim kerΓ = dimE+(S(ζ)) = p, these two linear subspaces are in bijec-
tion. Let us give the two main ideas behind this proof: one idea is to detail
the bijection between q ∈ kerΓ and q+ ∈ E+(S(ζ)) as it satisfies some con-
straints, the other is to come down to the model case where the eigenvalues
of S are either 1 or −1. Let us denote:
S̃−1 =
−IdN−p 0
0 Idp
In a first step, we will prove the following result:
Proposition 2.3. If we assume that V is a linear subspace of CN of dimen-
sion p, and that there is C > 0 such that, for all q ∈ V, there holds:
〈S̃−1q, q〉 ≥ C〈q, q〉,
then the two following equivalent properties hold:
• There is a hermitian symmetric, positive definite matrix R̃, such that:
[q ∈ V] ⇔
q ∈ E+(R̃S̃R̃)
which is equivalent to:
V = E+(R̃
• There is a hermitian symmetric, positive matrix R̃, such that:
[q ∈ V] ⇔
R̃q ∈ E+(R̃S̃
−1R̃)
which is equivalent to:
V = E+(S̃
−1R̃2).
Moreover, we can link the two properties by taking:
R̃2 = S̃R̃
Proof. In this proof, we will show how to construct some matrices R̃
satisfying the required properties. There is a (N − p)× p matrix ℵ of rank
N − p such that ‖ℵ‖ ≤ 1 and:
V = {q ∈ CN , q− = ℵq+},
where q+[resp q−] denotes the projector on E+((S̃)
−1) [resp E−((S̃)
−1)]parallel
to E−((S̃)
−1) [resp E+((S̃)
−1)]. Indeed, dimV = p = dimE+((S̃)
−1), and
N = E−((S̃)
E+((S̃)
−1). Moreover, there ic C > 0 such that, for all
q ∈ V, there holds:
〈(S̃)−1q, q〉 = −〈q−, q−〉+ 〈q+, q+〉 ≥ C〈q, q〉.
and thus
|q+|2 − |ℵq+|2 ≥ C|q|2,
which implies that ‖ℵ‖ < 1. We will show now that, for R̃ constructed as
follow:
IdN−p −ℵ
−ℵ∗ Idp
there holds:
[q ∈ V] ⇔
R̃q ∈ E+(R̃S̃
−1R̃)
First,we see that the constructed R̃ is trivially hermitian symmetric and
positive definite since ‖ℵ‖ < 1. First, we have:
R̃S̃−1R̃ =
−IdN−p +NN
0 Idp −N
R̃q =
q− − ℵq+
−ℵ∗q− + q+
Thus, since ‖ℵ‖ < 1, there holds:
R̃q ∈ E+(R̃S̃
−1R̃)
q− − ℵq+ = 0
⇔ [q ∈ V] .
We will now prove that we have:
(R̃)−1E+(R̃S̃
−1R̃) = E+(S̃
−1R̃2).
Since R̃S̃−1R̃ is hermitian symmetric, the linear subspace E+(R̃S̃
−1R̃) is
generated by the eigenvectors of R̃S̃−1R̃ associated to positive eigenvalues.
A basis of (R̃)−1E+(R̃S̃
−1R̃) is thus given by ((R̃)−1vj)j where vj denotes
an eigenvector of R̃S̃−1R̃ associated to a positive eigenvalue λj . We have:
R̃S̃−1R̃vj = λjvj.
Let us denote wj = (R̃)
−1vj, we have then:
R̃S̃−1R̃2wj = λjR̃wj ⇔ S̃
−1R̃2wj = λjwj .
As a result, wj is an eigenvector of S
−1R̃2 associated to the eigenvalue λj
hence we obtain that:
(R̃)−1E+(R̃S̃
−1R̃) = E+(S̃
−1R̃2).
We can also prove, the same way, that:
R̃E+(R̃S̃R̃) = E+(R̃
Now, taking
R̃2 = S̃R̃
we can check that:
E+(S̃
−1R̃2) = E+(R̃
which concludes the proof. ✷
Lemma (??) is a Corollary of the following Proposition:
Proposition 2.4. If S−1 denotes a smooth in ζ 6= 0, matrix-valued function
in the space of hermitian symmetric matrices with p positive eigenvalues and
N −p negative eigenvalues and ker Γ denotes a linear subspace of dimension
p and there is C > 0 such that, for all q ∈ ker Γ, there holds:
〈S−1q, q〉 ≥ C〈q, q〉,
then the two following equivalent properties hold:
• There is a smooth in ζ 6= 0, matrix-valued function R, in the space of
hermitian symmetric, positive matrices such that:
[q ∈ KerΓ] ⇔
∀ζ 6= 0, R−1(ζ)q ∈ E+(R(ζ)S(ζ)R(ζ))
which is equivalent to:
∀ζ 6= 0, KerΓ = E+(R
2(ζ)S(ζ)).
• There is a smooth in ζ 6= 0, matrix-valued function R, in the space of
hermitian symmetric, positive matrices such that:
[q ∈ KerΓ] ⇔
∀ζ 6= 0, R(ζ)q ∈ E+(R(ζ)S
−1(ζ)R(ζ))
which is equivalent to:
∀ζ 6= 0, KerΓ = E+(S
−1(ζ)R2(ζ)).
Moreover, for all ζ 6= 0, these two properties can be linked by taking:
(R(ζ))2 = S(ζ)(R(ζ))2S(ζ).
Proof. We will show here that Proposition 2.4 can be deduced from Propo-
sition 2.3. For all ζ 6= 0, S(ζ) is a hermitian symmetric matrix, moreover S
depends smoothly of ζ. As a consequence S−1 is also a hermitian symmetric
matrix depending smoothly of ζ, and as such, there is a nonsingular matrix
V such that:
S̃−1 = V ∗
Let us denote Λ the diagonalized version of S−1 with eigenvalues sorted by
increasing order, then there is Z depending smoothly of ζ such that, for all
ζ 6= 0, we have:
Z∗(ζ) = Z−1(ζ),
Λ(ζ) = Z∗(ζ)
(ζ)Z(ζ).
As a consequence, V depends smoothly of ζ since, for all ζ 6= 0:
V (ζ) = (Λ(ζ))−
2Z(ζ),
where Λ is the diagonal matrix obtained by taking the absolute value of each
eigenvalue of Λ. For the sake of simplicity, let us omit the dependence in ζ.
Now, for all q ∈ V −1 ker Γ, there is C > 0, such that:
〈S̃−1q, q〉 = 〈V ∗S−1V q, q〉 = 〈S−1(V q), (V q)〉 ≥ C〈(V q), (V q)〉.
Moreover V is nonsingular, thus there is C ′ > 0, such that, for all
q ∈ V −1 ker Γ, there holds:
〈S̃−1q, q〉 ≥ C ′〈q, q〉.
Moreover dimV −1 ker Γ = p, using Proposition 2.3, for all fixed ζ 6= 0, there
is a hermitian symmetric, positive definite matrix R̃(ζ), such that:
V −1(ζ) ker Γ = E+((R̃(ζ))
2S̃(ζ)) = R̃(ζ)E+(R̃(ζ)S̃(ζ)R̃)(ζ).
We will now prove that we can construct R̃ depending smoothly of ζ. First
there is a (N − p) × p matrix ℵ of rank N − p, depending smoothly of ζ,
such that fore all ζ 6= 0 ‖ℵ(ζ)‖ ≤ 1 and:
V −1(ζ) ker Γ = {q ∈ CN , q− = ℵ(ζ)q+},
where q+[resp q−] denotes the projector on E+((S̃)
−1) [resp E−((S̃)
−1)]parallel
to E−((S̃)
−1) [resp E+((S̃)
−1)]. R̃ is given, for all ζ 6= 0, by:
R̃(ζ) =
S̃−1(ζ)R̃2(ζ)S̃−1(ζ),
with R̃ given by:
R̃(ζ) =
IdN−p −ℵ(ζ)
−ℵ∗(ζ) Idp
Since S̃−1 = V ∗
V, there holds: S̃ = V ∗SV, and, as a consequence:
(V R̃)−1 ker Γ = E+(R̃V
∗SV R̃).
As R̃V ∗SV R̃ is hermitian symmetric, a basis of the linear subspace E+(R̃V
∗SV R̃)
is given by the eigenvectors of R̃V ∗SV R̃ associated to positive eigenvalues.
This leads us to consider vj = (V R̃)
−1uj satisfying:
R̃V ∗SV R̃vj = λjvj.
We have:
R̃V ∗SV R̃(V R̃)−1uj = λj(V R̃)
−1uj.
hence:
(V R̃)R̃V ∗Suj = λjuj .
Since (V R̃)R̃V ∗ = (R̃V ∗)∗(R̃V ∗) is hermitian symmetric and positive defi-
nite, we can then define its square root. We define R by:
(R̃V ∗)∗(R̃V ∗).
Since both R̃ and V depends smoothly of ζ, so does R. Moreover, there
holds:
R2Suj = λjuj,
which gives:
ker Γ = V R̃E+(R̃V
∗SV R̃) = E+(R
We have thus proved there is a smooth in ζ 6= 0, matrix-valued function R,
in the space of hermitian symmetric, positive matrices such that:
[q ∈ KerΓ] ⇔
∀ζ 6= 0, R−1(ζ)q ∈ E+(R(ζ)S(ζ)R(ζ))
which is equivalent to:
∀ζ 6= 0, KerΓ = E+(R
2(ζ)S(ζ)).
Now consider R defined, for all ζ 6= 0, by:
R(ζ) =
S(ζ)(R(ζ))2S(ζ),
R(ζ) =
(R̃(ζ)V ∗(ζ)S(ζ))∗(R̃(ζ)V ∗(ζ)S(ζ)).
ζ 7→ R(ζ) is smooth and, for all ζ, R(ζ) is a hermitian symmetric, positive
definite matrix. Moreover, there holds:
[q ∈ KerΓ] ⇔
∀ζ 6= 0, R(ζ)q ∈ E+(R(ζ)S
−1(ζ)R(ζ))
which is equivalent to:
∀ζ 6= 0, KerΓ = E+(S
−1(ζ)R2(ζ)).
Let us detail the computation of R(ζ).
R(ζ) =
S(ζ)V (ζ)R̃
(ζ)V ∗(ζ)S(ζ).
Moreover
(ζ) = S̃−1(ζ)R̃2(ζ)S̃−1(ζ),
we have thus:
R(ζ) =
R̃(ζ)S̃−1(ζ)V ∗(ζ)S(ζ)
R̃(ζ)S̃−1(ζ)V ∗(ζ)S(ζ)
which gives:
B(ζ) =
R̃(ζ)S̃−1(ζ)V ∗(ζ)S(ζ)
R̃(ζ)S̃−1(ζ)V ∗(ζ)S(ζ)
We recall that R̃ is given, for all ζ 6= 0, by:
R̃(ζ) =
IdN−p −ℵ(ζ)
−ℵ∗(ζ) Idp
and that for all ζ 6= 0, V (ζ) is given by:
V (ζ) = (Λ(ζ))−
2Z(ζ),
where
Λ(ζ) = Z∗(ζ)
(ζ)Z(ζ)
with Λ is a diagonal matrix with real coefficients: (λ1, . . . , λN ), and Λ de-
notes the diagonal matrix with diagonal coefficients (|λ1|, . . . , |λN |).
Remark 2.5. In the construction of B the only freedom we have resides in
the choice of ℵ.
2.3 A change of dependent variables.
Let us denote by R := B
2 and v̂ := Rû. v̂ is hence solution of (2.3):
(2.3)
R−1SR−1∂xv̂ = R
−1SAR−1v̂ +R−1S(Ad)
−1f̂ , {x > 0},
ΓR−1v̂|x=0 = Γĝ,
We will adopt the following notations: SR := R
−1SR−1, AR := RAR
and ΓR := ΓR
−1. We first observe that:
ker ΓR = R ker Γ = RE+((S)
−1R2).
but S−1
= RS−1R thus
ker ΓR = RE+(R
−1SRR) = E+(SR).
This is where Lemma 1.4 is used in a crucial manner. Let us denote by
− the projector on E−(SR) parallel to E+(SR) and by by P
+ the projector
on E+(SR) parallel to E−(SR); P
− and P+ denoting the associated Fourier
multiplier. Since SR is hermitian symmetric, P
− is in fact the orthogonal
projector on E−(SR). The problem (2.3) can then be written:
SR∂xv̂ = SRARv̂ +R
−1S(Ad)
−1f̂ , {x > 0},
−v̂|x=0 = P
−Γĝ,
This problem is well-posed because, as a direct Corollary of Proposition 2.2,
we have:
Proposition 2.6. For all ζ such that τ2 + γ2 + |η|2 = 1, there holds:
• SR(ζ) is hermitian symmetric.
• ℜ (SRAR) (ζ) is positive definite.
• −SR(ζ) is definite negative on ker ΓR and the dimension of ker ΓR is
the same as the number of negative eigenvalues of −SR(ζ).
Proof. For the sake of simplicity, let us omit the dependence in ζ in our
notations.
• SR := R
−1SR−1, and both S and R are hermitian thus SR is hermi-
tian.
• SRAR = R
−1SAR−1, thus for all q ∈ CN , there holds:
2〈ℜ(SRAR)q, q〉 = 〈SRARq, q〉+〈q, SRARq〉 = 〈R
−1SAR−1q, q〉+〈q,R−1SAR−1q〉,
since R−1 is hermitian, we have then:
= 〈SAR−1q,R−1q〉+ 〈R−1q, SAR−1q〉 = 2〈ℜ(SA)R−1q,R−1q〉.
Since ℜ(SA) is positive definite and R is invertible, ℜ (SRAR) is thus
positive definite.
• By construction of R, it satisfies ker ΓR = E+(SR), with SR hermitian.
As a consequence −SR is definite negative on ker ΓR and the dimension
of ker ΓR is the same as the number of negative eigenvalues of −SR.
✷ Let us mention that, since R and S remains uniformly bounded in
ζ 6= 0, f̂ and R−1S(Ad)
−1f̂ belongs to the same space. In a same spirit as
[5], this suggests the following singular perturbation of (2.3):
SR∂xv̂
−v̂ε1x<0 = SRARv̂
−Γĝ1x<0 +R
−1S(Ad)
−1f̂ , {x ∈ R},
This is equivalent to perturb (2.2) as follow:
S∂xû
RP−Rûε1x<0 = SAû
RP−Γĝ1x<0 + S(Ad)
−1f̂ , {x ∈ R},
Finally, this induces the following perturbation for (1.1):
(2.4)
Huε +
Muε1x<0 = f1x>0 +
θ1x<0, {x ∈ R},
uε|t<0 = 0,
where
M := −eγtAdS
−1RP−Re−γt,
θ = −eγtAdS
−1RP−Γg̃,
and S(∂t, ∂y) [resp R(∂t, ∂y)] denotes the Fourier multiplier associated to
S(ζ) [resp R(ζ)].
3 Proof of Theorem 1.6.
First, we construct an approximate solution of equation (2.4) (which is also
equation (1.4)), then prove suitable energy estimates that ensures uε and its
approximate solution both converges towards the same limit as
ε → 0+.
3.1 Construction of the approximate solution.
uε is the solution of the well-posed Cauchy problem:
Huε +
Muε1x<0 = f1x>0 +
θ1x<0, {x ∈ R},
uε|t<0 = 0.
uε is moreover the solution of the well-posed Cauchy problem:
Huε +
Muε1x<0 = SA
f1x>0 +
θ1x<0, {x ∈ R},
uε|t<0 = 0.
The associated equation after tangential Fourier-Laplace transform writes :
S∂xû
RP−Rûε1x<0 − SAû
ε = −
RP−Γĝ1x<0 + S(Ad)
−1f̂1x>0, {x ∈ R}.
or alternatively:
ûε = R−1v̂ε
SR∂xv̂
−v̂ε1x<0 = SRARv̂
−Γĝ1x<0 +R
−1S(Ad)
−1f̂ , {x ∈ R},
We will use the following formulation as a transmission problem in our con-
struction of an approximate solution:
SR∂xv̂
ε+ = SRARv̂
ε+ +R−1S(Ad)
−1f̂ , {x > 0},
SR∂xv̂
−v̂ε− = SRARv̂
−Γĝ, {x < 0},
v̂ε+|x=0+ = v̂
ε−|x=0− .
For Ω an open regular subset of Rd+1, and ρ ∈ N, let us introduce the
weighted spaces H
γ (Ω) defined by:
H̺γ (Ω) = {̟ ∈ e
γtL2(Ω), ‖̟‖H̺γ (Ω) < ∞};
where
γ (Ω)
α,|α|≤̺
γρ−|α|‖e−γt∂α̟‖2
L2(Ω).
We will construct an approximate solution uεapp of u
ε. uεapp will be con-
structed as follow:
uεapp = u
app1x>0 + u
app1x<0,
where uε±app is an approximate solution of u
ε± satisfying the following ansatz:
uε±app =
U±j (ζ, x)ε
where the profiles U±j belong toH
), where Ω±
stands for [0, T ]×Rd±.
Denote
v̂εapp = RF(e
−γtuεapp) := v̂
app1x>0 + v̂
app1x<0.
v̂ε±app is then an approximate solution of v
ε± and is of the form:
v̂ε±app =
j (ζ, x)ε
where
= RF(e−γtU±
and conversely
U±j = e
γtF−1
R−1V ±j
The profiles U±j can be constructed inductively at any order. Let us show
how the first profiles are constructed: Identifying the terms in ε−1 gives:
−V −0 = P
−Γĝ.
Hence, P+V −0 remains to be computed in order to obtain the profile
0 = e
γtF−1
Identifying the terms in ε0 gives then that V +0 is solution of the well-posed
problem:
(3.1)
SR∂xV
0 = SRARV
−1S(Ad)
−1f̂ , {x > 0},
−V +0 |x=0 = P
−Γĝ.
The associated profile
U+0 = e
γtF−1
R−1V +0
belongs then toHkγ (Ω
).Moreover, the problem (3.1) is Kreiss-Symmetrizable
and thus the trace of the profile U+0 , see [3] for instance, satisfies:
U+0 |x=0 ∈ H
γ (ΥT ).
Since V +0 has just be computed, V
0 |x=0 is given by: V
0 |x=0 − V
0 |x=0 = 0
and thus, there holds:
−V +0 |x=0 = P
−V −0 |x=0.
Moreover
SR∂xV
0 − P
−V −1 = SRARV
0 , {x < 0}.
Projecting this equation on E+(SR) collinearly to E−(SR) gives then:
SR∂xP
0 − P
+SRARV
0 = 0, {x < 0},
Since
+SRARV
0 = P
+SRARP
+V −0 + P
+SRARP
−Γĝ,
we have then:
SR∂x(P
+V −0 )− P
+SRAR(P
+V −0 ) = P
+SRARP
−Γĝ, {x < 0},
and as a consequence, P+V −0 is solution of the following problem:
(3.2)
SR∂x(P
0 )− P
+SRAR(P
0 ) = P
+SRARP
−Γĝ {x < 0},
+V −0 |x=0 = P
+V +0 |x=0.
Let us precise how (3.2) has to be interpreted: we denote w = P+V −0 . w is
then totally polarized on E+(SR), and satisfies the problem:
(3.3)
+w = w
SR∂xw − P
+SRARw = P
+SRARP
−Γĝ {x < 0},
w|x=0 = P
+V +0 |x=0.
As we will see, the problem (3.3) is Kreiss-Symmetrizable and thus well-
posed. Indeed, for all ζ such that τ2 + γ2 + |η|2 = 1, we have, omitting the
dependencies in ζ in our notations:
• For all q ∈ CN , there holds:
〈SRq, q〉 = 〈q, SRq〉.
• Since Re(SRAR) is positive definite and P
+ is an orthogonal projector,
there is C > 0 such that, for all q ∈ E+(SR), there holds:
〈P+SRARP
+q, q〉+ 〈q,P+SRARP
+q〉 ≥ C〈q, q〉.
Indeed, for all q ∈ E+(SR), there holds:
〈P+SRARP
+q, q〉 = 〈P+SRARP
+q,P+q〉 = 〈SRARP
+q,P+q〉.
• −SR is definite negative on kerP
+ that is to say, that there is c > 0
such that, for all q ∈ kerP+, there holds:
〈−SRq, q〉 ≤ −c〈q, q〉.
Moreover kerP+ has the same dimension as the number of negative
eigenvalues in −SR.
The profile U−0 can then be computed by:
U−0 := e
γtF−1
R−1(w + P−Γĝ)
belongs to Hkγ (Ω
), moreover its trace U−0 |x=0 belongs
to Hkγ (ΥT ). Consider now the equation:
1 = SR∂xV
0 − SRARV
0 , {x < 0}.
Since P−V −1 |x=0 = P
−V +1 |x=0, V
1 is solution of the well-posed problem:
SR∂xV
1 = SRARV
1 , {x > 0},
−V +1 |x=0 = SR∂xV
0 |x=0 − SRARV
0 |x=0.
Due to the loss of regularity in the boundary condition, the associated profile
U+1 = e
γtF−1
R−1V +1
belongs to H
), moreover its trace U+1 |x=0 belongs
γ (ΥT ). Moreover, applying P
+ to the equation:
−V −2 + SRARV
1 = SR∂xV
1 , {x < 0},
we obtain:
SR∂xP
+V −1 = P
+SRARP
+V −1 + P
+SRARP
−V −1 , {x < 0},
+V −1 |x=0 = P
+V +1 |x=0.
As before, let us take P+V −1 as the unknown of the well-posed problem:
SR∂x(P
+V −1 )− P
+SRAR(P
+V −1 ) = P
+SRAR
SR∂xV
0 − SRARV
, {x < 0},
(P+V −1 )|x=0 = P
1 |x=0.
This problem is Kreiss-Symmetrizable since, for all ζ such that
τ2 + γ2 + |η|2 = 1, there holds:
• For all q ∈ CN , there holds:
〈SRq, q〉 = 〈q, SRq〉.
• There is C > 0 such that for all q ∈ E+(SR), there holds:
〈P+SRARP
+q, q〉+ 〈q,P+SRARP
+q〉 ≥ C〈q, q〉.
• −SR is definite negative on kerP
+ that is to say, that there is c > 0
such that, for all q ∈ kerP+, there holds:
〈−SRq, q〉 ≤ −c〈q, q〉.
Moreover kerP+ has the same dimension as the number of negative
eigenvalues in −SR.
However, due to a loss of regularity in both the source term and the boundary
condition, the associated profile
U−1 = e
γtF−1
+V −1 + SR∂xV
0 − SRARV
belongs to H
). The construction of the following profiles can be
pursued at any order the same way. In practice, we take:
uε+app = U
uε−app = U
0 + εU
As a result, the approximate solution writes uεapp := u
app1x>0 + u
app1x<0;
where uε+app belongs to H
) and uε−app belongs to H
). uεapp is then
solution of a well-posed problem of the form:
(3.4)
Huεapp +
Muεapp1x<0 = f1x>0 +
θ1x<0 + εr
ε, {x ∈ R},
uεapp|t<0 = 0 .
Where rε := rε+1x>0 + r
ε−1x<0, with r
ε+ ∈ H
) and
rε− ∈ Hk−3γ (Ω
Remark 3.1. In the case where g = 0, the loss of regularity in the profiles
is delayed by one step. As a result, in this case we obtain:
uε+app ∈ H
uε−app ∈ H
rε+ ∈ Hkγ (Ω
rε− ∈ H
3.2 Stability estimates
We will begin by proving energy estimates on the following equation:
(3.5) SRARê
ε − SR∂xê
−êε1x<0 = εr̂
ε, {x ∈ R},
where êε = R
F(e−γtuε)−F(e−γtuεapp)
:= ŵε; with wε = uε − uεapp.
Refering to (3.4), wε is the solution of the Cauchy problem:
(3.6)
Hwε +
Mwε1x<0 = εr
wε|t<0 = 0 .
For a fixed positive ε, the perturbation is nonsingular and thus the principal
part of the pseudodifferential operator H+ 1
M is the same as the principal
part of H. Hence, there is a unique solution of the Cauchy problem (3.6):
wε which belongs to Hk−3γ (ΩT ). In order to simplify the notations, in this
chapter we shall denote by L2 and H
γ the spaces: L
2(ΩT ) and H
γ (ΩT ).
We recall the definition of the weighted spaces: H
γ (ΩT ) for ρ ∈ N.
H̺γ (ΩT ) = {̟ ∈ e
γtL2(ΩT ), ‖̟‖H̺γ (ΩT ) < ∞};
where
γ (ΩT )
α,|α|≤̺
γρ−|α|‖e−γt∂α̟‖2
L2(ΩT )
For fixed positive ε, there holds:
∂x〈SRê
ε, êε〉L2 dx = 0.
2Re〈SR∂xê
ε, êε〉L2 dx = 0.
Using the equation, we have then:
Re〈SRARê
−êε − εr̂ε, êε〉L2 dx = 0.
which is equivalent to:
Re〈SRARê
ε, êε〉L2 dx =
Re〈P−êεεr̂ε, êε〉L2 dx
Re〈r̂ε, êε〉L2) dx.
But Re〈SRARê
ε, êε〉 = 〈Re (SRAR) ê
ε, êε〉 and Re (SRAR) is positive defi-
nite, hence there is C > 0, independent of ε such that:
Cγ‖êε‖2
Re〈P−êε, êε〉 ≤
Re〈εr̂ε, êε〉L2 dx.
Thus, because P− is an orthogonal projector, for all positive λ, there holds:
Cγ‖êε‖2
‖P−êε‖2
‖êε‖2
‖εr̂ε‖2
Choosing λ big enough we have C− ε
> 0 and the following energy estimate:
γ‖êε‖2
‖P−êε‖2
‖r̂ε‖2
This shows that êε converges towards zero in L2 when ε tends to zero, with
a rate in O(ε). We recall that êε is given by:
êε := RF
e−γt(uεapp − u
and r̂ε is given by:
r̂ε := RF
e−γtrε
Moreover, since R and P− are two uniformly bounded, uniformly definite
positive matrices, there are two positive real numbers α and β such that,
for all ζ 6= 0 and x ∈ R, there holds:
• α‖F
e−γt(uεapp − u
≤ ‖êε‖2
• α‖P−F
e−γt(uεapp − u
≤ ‖P−êε‖2
• ‖r̂ε‖2
≤ β‖F
e−γtrε
Applying then Plancherel’s equality we obtain then:
γ‖uεapp − u
eγtL2
uεapp − u
eγtL2
‖rε‖2
eγtL2
We have thus proved there are two positive constants c and C such that:
cγ‖uεapp − u
eγtL2
uεapp − u
eγtL2
‖rε‖2
eγtL2
Let us denote by ‖.‖∗
+ ‖.‖2
. More generally, when
rε ∈ H̺, there is two positive constants cρ and Cρ such that:
cργ‖u
app − u
uεapp − u
‖rε‖∗2
As we have seen during the construction of the profiles, ̺ = k− 3 in general
and ρ = k − 3
in the case where g = 0.
3.3 End of the proof of Theorem 1.6.
As a consequence of our stability estimate, there holds:
‖uεapp − u
Hk−3(Ω
+ ‖uεapp − u
Hk−3(Ω
= O(ε2).
Moreover, by construction of uεapp, there holds:
‖uεapp − u
Hk−3(Ω
+ ‖uεapp − u‖
Hk−3(Ω
= O(ε2).
Hence, we have proved that:
‖uε − u−‖2
Hk−3(Ω
+ ‖uε − u‖2
Hk−3(Ω
= O(ε2).
By the same arguments, if g = 0, there holds:
‖uε − u−‖2
+ ‖uε − u‖2
= O(ε2).
This completes the proof of Theorem 1.6.
4 Proof of Theorem 1.9.
Like in the proof of Theorem 1.6, we begin by constructing formally an
approximate solution of equation (1.7). We prove then suitable energy esti-
mates that ensures both uε and its approximate solution converges towards
ũ as ε → 0+.
4.1 Construction of an approximate solution.
The goal of this Lemma is to replace the boundary condition Γu|x=0 = Γg of
problem (1.1) by a condition of the form P−(e−γtu)|x=0 = h with a suitable
h ∈ Hk(ΥT ).
Lemma 4.1. Let u denote the unique solution in Hk(Ω+
) of the mixed
hyperbolic problem (1.1), P+(∂t, ∂y, γ)
e−γtu
does not depend of the choice
of the boundary operator Γ and of g. Let us introduce the function h of
Hk(ΥT ) defined by:
e−γtv|x=0
e−γt(g − v|x=0)
The solution u of the mixed hyperbolic problem (1.1) is the unique solution
of the following well-posed mixed hyperbolic problem (4.1):
(4.1)
Hu = f, {x > 0},
−(∂t, ∂y, γ)
e−γtu|x=0
u|t<0 = 0 .
In addition, the mapping (f, g) → h is linear continuous from
Hk(Ω+
)×Hk(ΥT ) to H
k(ΥT ).
Proof. Let v denote a solution in Hk(ΩT ) of the equation:
Hv = f, (t, y, x) ∈ ΩT ,
v|t<0 = 0 .
We introduce then U which is, by definition, the solution of the following
mixed hyperbolic problem:
HU = 0, {x > 0},
Γ(∂t, ∂y, γ)U|x=0 = Γ(∂t, ∂y, γ)g − Γ(∂t, ∂y, γ)v|x=0,
U|t<0 = 0 .
The right hand side of the boundary condition is, a priori,
in Hk−
2 (ΥT ). Hence the solution U belongs to H
2 (Ω+
). By construction
we have:
(4.2) u = U+ v.
Hence, since u ∈ Hk(Ω+
) and v ∈ Hk(Ω+
), in fact we have:
U ∈ Hk(Ω+
Let Û denote the Fourier-Laplace transform in (t, y) of U (Fourier-Laplace
transform tangential to the boundary) given by: F(e−γtU). It satisfies the
following symbolic equation:
∂xÛ = A(ζ)Û, {x > 0},
Γ(ζ)Û|x=0 = Γ(ζ)ĝ − Γ(ζ)v̂|x=0,
where ĝ and v̂ denotes respectively the tangential Fourier-Laplace transform
of g and v. Since A(ζ) is independent of x, projecting the above equation on
E+(A(ζ)) gives then:
Û = A(ζ)P+Û.
Moreover P+Û|x=0 ∈ E−(A(ζ))
E+(A(ζ)) since limx→∞P
Û = 0. Hence,
there holds:
Û = 0,
and thus
Û = P−Û.
The boundary condition:
Γ(ζ)Û|x=0 = Γ(ζ)ĝ − Γ(ζ)v̂|x=0
is equivalent to:
Û|x=0 ∈ ĝ − v̂|x=0 +KerΓ.
We have thus:
Û|x=0 ∈ ĝ − v̂|x=0 + ker Γ.
Let us denote by Π the projector on Ẽ−(A) parallel to ker Γ, which has a
sense because the Uniform Lopatinski Condition holds.
Since Û|x=0 ∈ Ẽ−(A), and of the Uniform Lopatinski Condition, the above
boundary condition is equivalent to:
Û|x=0 = Π(ĝ − v̂|x=0),
and thus, because P−Û|x=0 belongs to E−(A), we have:
Û|x=0 = Π(ĝ − v̂|x=0).
As a consequence, we obtain:
−û|x=0 = P
−v̂|x=0 +Π(ĝ − v̂|x=0).
Hence, there holds:
e−γtu|x=0
e−γtv|x=0
e−γt(g − v|x=0)
:= h.
P+(∂t, ∂y, γ)
e−γtu
= P+(∂t, ∂y, γ)
e−γtv
, thus it does not depend of the
choice of the boundary operator Γ and of g.Moreover, since u|x=0 ∈ H
k(ΥT ),
it follows that
g ∈ Hk(ΥT ). Now, since the Uniform Lopatinski Condition holds, u satisfies
the following energy estimate:
eγtL2(Ω
+ ‖u|x=0‖
eγtL2(ΥT )
≤ γ‖f‖2
eγtL2(Ω
+ ‖g‖eγtL2(ΥT ),
More generally, we have:
Hkγ (Ω
+ ‖u|x=0‖
Hkγ (ΥT )
≤ γ‖f‖2
Hkγ (Ω
+ ‖g‖Hkγ (ΥT ).
where ‖̟‖2
|α|=0 γ
k−|α|‖∂α̟‖2
eγtL2
h = P−(e−γtu|x=0) hence
L2(ΥT )
≤ C‖e−γtu|x=0‖
L2(ΥT )
= C‖u|x=0‖
eγtL2(ΥT )
and for 0 ≤ j ≤ d− 1, there holds:
‖∂jh‖
L2(ΥT )
≤ cj‖ηjP
−F(e−γtu)|x=0‖ ≤ c
j‖u|x=0‖H1γ(ΥT ).
More generally, we have:
Hkγ (ΥT )
≤ Ckγ‖f‖
Hkγ (Ω
+ Ck‖g‖
Hkγ (ΥT )
But γ is a positive real number fixed once and for all at the beginning of the
paper, hence this proves that the mapping (f, g) → h is continuous from
Hk(Ω+
)×Hk(ΥT ) to H
k(ΥT ). ✷
As we will see, Lemma 4.1 is central in our construction of an approximate
solution. We will construct an approximate solution
uεapp := u
app1x>0 + u
app1x<0,
along the following ansatz:
uε+app :=
εju+j (t, y, x),
with u+j ∈ H
), u+j |x=0 ∈ H
γ (ΥT ); and
uε−app :=
j (t, y, x),
with u−j ∈ H
), u−j |x=0 ∈ H
γ (ΥT ). As usual, we will refer to the
terms u±
as profiles. We will rather work on the reformulation of
problem (1.7) as the transmission problem (4.3):
(4.3)


Huε+ = f, {x > 0},
Huε− +
−e−γtuε− =
γth̃, {x < 0},
uε+|x=0 − u
ε−|x=0 = 0,
uε±|t<0 = 0 .
Plugging uε+app and u
app in (4.3) and identifying the terms with same power
in ε, we obtain the following profiles equations:
• Identification of the terms of order ε−1 :
(4.4) Ade
−e−γtu
0 = Ade
γth̃, {x < 0}.
• Identification of the terms of order ε0 :
(4.5) Hu−0 +Ade
−e−γtu−1 = 0, {x < 0}.
(4.6) Hu+0 = f, {x > 0},
• Identification of the terms of order εj for j ≥ 1 :
(4.7) Hu
j +Ade
−e−γtu
j+1 = 0, {x < 0}.
(4.8) Hu
j = 0, {x > 0},
• Translation of the continuity condition over the boundary on the pro-
files:
For all 1 ≤ j ≤ M, there holds:
(4.9) u
j |x=0 − u
j |x=0 = 0.
Denote by û±j := F(e
−γtu±j ) . We have then:
j := e
γtF−1(û±j ).
We will now give the equations satisfied by the Fourier-Laplace transform
of the profiles: û±j . First, equation (4.4) is equivalent to:
−û−0 = F(h̃), {x < 0}.
We deduce from this equation that there holds:
−û−0 |x=0 = ĥ.
Then, using (4.9) for j = 0, and (4.6) gives that, for γ big enough,
0 = F(e
−γtû
where û+0 is the solution of the well-posed first order ODE in x:
0 −Aû
0 = F(e
−γt(Ad)
−1f), {x > 0},
−û+0 |x=0 = h,
Thus u+0 is solution of:
Hu+0 = f, {x > 0},
eγtP−e−γtu+0 |x=0 = h.
Thanks to Lemma 4.1, we recognize u+0 as the solution of our starting mixed
hyperbolic problem (1.1). Once u+0 is known, so is û
0 and thus û
0 |x=0 is
given by:
û−0 |x=0 = û
0 |x=0.
Moreover,
u+0 |x=0 = u
0 |x=0 ∈ H
γ (ΥT ).
By (4.5), there holds:
0 −Aû
−û−1 = 0, {x < 0}.
As a consequence, P+û−0 is given by the well-posed ODE:
0 )−A(P
0 ) = 0, {x < 0},
+û−0 |x=0 = P
+û+0 |x=0.
Indeed, since kerP+(ζ) = E−(A(ζ)), this problem satisfies the Uniform
Lopatinski Condition: for all ζ 6= 0, there holds:
E−(A(ζ))
E+(A(ζ)) = C
For γ big enough, by linearity of the inverse Fourier transform, u−0 can then
be computed by:
u−0 := e
γtF−1(P−û−0 ) + e
γtF−1(P+û−0 ).
Following up with that process of construction, we can go on with the con-
struction of the profiles at any order. Indeed, assume that all the profiles
(u+j , u
j ) up to order j have been computed. Then consider the equation
obtained through identification:
j+1 = −∂xû
j +Aû
j , {x < 0}.
We see there is a loss of regularity between û−j+1 and û
Let us say that u±j ∈ H
). Considering the traces, we have then:
j |x=0 ∈ H
γ (ΥT ). We will show in this part how the Sobolev regularity
of the profiles u±j+1, which is by definition mj+1, can be computed know-
ing mj . To begin with P
j+1 belongs to H
). P−u+j+1|x=0, which
belongs to H
γ (ΥT ), is known by P
j+1|x=0 = e
γtF−1(P−û+j+1|x=0),
with:
j+1|x=0 = P
j+1|x=0.
Hence, û+j+1 := F(e
−γtu+j+1) is the solution of the first order ODE in x :
j+1 −Aû
j+1 = 0, {x > 0},
−û+j+1|x=0 = P
−û−j+1|x=0.
Since kerP−(ζ) = E+(A(ζ)), this problem satisfies the Uniform Lopatinski
Condition: for all ζ 6= 0, there holds:
E−(A(ζ))
E+(A(ζ)) = C
As a consequence, this problem is well-posed and, u+j+1 ∈ H
).More-
over, there holds:
u+j+1|x=0 = u
j+1|x=0 ∈ H
γ (ΥT ).
Indeed, P+û+j+1 ∈ H
∞(Rd+1+ ) hence P
+u+j+1|x=0 ∈ H
γ (ΥT ) and thus
u+j+1|x=0 ∈ H
γ (ΥT ). Furthermore, we have:
u−j+1|x=0 = u
j+1|x=0.
Applying P+ on the following equation:
−û−j+2 = −∂xû
j+1 +Aû
j+1, {x < 0};
we obtain then the equation:
j+1)−AP
j+1 = 0, {x < 0}.
Remark 4.2.
−û−j+2 = −∂xû
j+1 +Aû
j+1, {x < 0}.
shows that the ”Fourier profile” û−j+1 must be so that −∂xû
j+1 + Aû
j+1 is
polarized on E−(A). It is indeed the case because we search for û
j+1 satisfy-
+û−j+1)−AP
+û−j+1 = 0, {x < 0}.
j+1 is given by:
j+1 := e
γtF−1(P−û−j+1) + e
γtF−1(P+û−j+1).
with P+u−j+1 = e
γtF−1(P+û−j+1) belongs to H
) and is the unique
solution of the well-posed first order ODE:
+û−j+1)−A(P
+û−j+1) = 0, {x < 0},
+û−j+1|x=0 = P
+û+j+1|x=0.
The profile u−j+1 belongs to H
). This achieves to show that the
knowledge of (u+j , u
j ), allows us to compute (u
j+1, u
j+1).
Moreover mj+1 = mj −
, that is to say that a construction of each supple-
mentary profile consummate 3
of Sobolev regularity. In practice, we take:
uε+app = u
uε−app = u
0 + εu
As a result, the approximate solution writes uεapp := u
app1x>0 + u
app1x<0;
where uε+app belongs to H
) and uε−app belongs to H
). The so de-
fined uεapp is solution of a well-posed problem of the form:
(4.10)
Huεapp +
−e−γtuεapp1x<0 = f1x>0 +
γth̃1x<0 + εr
uεapp|t<0 = 0 .
Where rε := rε+1x>0 + r
ε−1x<0, with r
ε+ ∈ H
) and
rε− ∈ Hk−3γ (Ω
4.2 Asymptotic Stability of the problem as ε tends towards
zero.
Denote by vε = uεapp − u
ε. By construction of uεapp, v
ε is solution of the
following Cauchy problem:
(4.11)
Hvε +
−e−γtvε1x<0 = εr
vε|t<0 = 0 .
For all positive ε, this problem is well-posed. In order to investigate the
stability of this problem as ε goes to zero, we will reformulate it as a trans-
mission problem. The restrictions of vε to {x > 0} and {x < 0}, respectively
denoted by vε+ and vε− are solution the following transmission problem:
(4.12)


Hvε+ = εrε+, {x > 0},
Hvε− +
−e−γtvε− = εrε−, {x < 0},
vε+|x=0 − v
ε−|x=0 = 0,
vε±|t<0 = 0 .
Let us denote by V ε the function, valued in R2N , defined for all {x > 0}
and (t, y) ∈ [0, T ]× Rd−1 by:
V ε(t, y, x) =
V ε+(t, y, x)
V ε−(t, y,−x)
vε is solution of the Cauchy problem (4.11) iff V ε is solution of the mixed
hyperbolic problem on a half space (4.13) given below:
(4.13)
H̃V ε +BεV ε = εRε, {x > 0},
Γ̃V ε|x=0 = 0,
V ε|t<0 = 0 ,
where
H̃ = ∂t +
0 −Ad
γtP−e−γt
Rε(t, y, x) =
rε+(t, y, x)
rε−(t, y,−x)
Id −Id
Returning to the construction of our approximate solution, we have
Rε ∈ H
)×Hk−3γ (Ω
) and is such that Rε|t<0 = 0.
In fact Rε ∈ Hk
) with k′ = k−3. For all positive ε, there exists a unique
solution V ε in Hkγ (Ω
) to the above problem. We will prove here that this
solution converges, uniformly in ε, towards 0 in Hk
), as ε vanishes. As
in the proof of Kreiss Theorem, see [3] for instance, existence of solution
for mixed hyperbolic systems like (1.7) or (4.13), are obtained through the
proof of both direct and ”dual”a priori estimates on an adjoint problem. This
estimates results in the constant coefficient case of estimates on the Fourier-
Laplace transform of the solution. Additionally, if this ”Fourier” estimate
can be proved, both direct and ”dual” energy estimates are deduced from
it. In a first step, let us recall formally how to conduct the Fourier-Laplace
transform of a mixed hyperbolic problem:
Hu = f, {x > 0},
Γu|x=0 = g,
u|t<0 = 0 ,
Denote by u∗ := e
−γtu, u∗ is in particular a solution of the following problem:
Hu∗ + γu∗ = e
−γtf, {x > 0},
Γu∗|x=0 = g .
We take then the tangential (with respect to (t,y)) Fourier transform of this
equation, which gives:
Ad∂xû∗ + (γ + iτ)û∗ + iηj
Aj û∗ = F
e−γtf
, {x > 0},
Γû∗|x=0 = ĝ .
Multiplying this equation by A−1
, we obtain that u∗ is solution of the fol-
lowing ODE in x:
∂xû∗ −Aû∗ = (Ad)
e−γtf
, {x > 0},
Γû∗|x=0 = ĝ .
Note that, û∗ and u can be freely deduced from each other through the
formulas:
û∗ = F(e
−γtu)
u = eγtF−1(û∗).
We shall now introduce a rescaled solution V ε of the solution V ε of
(4.13) defined as follows: V ε(t, y, x) := V ε(t, y, εx), and the rescaled remain-
der: Rε(t, y, x) := Rε(t, y, εx). Denoting by V̂
= F(e−γtV ), the associated
equation writes then:
− εÃV̂
= ε2R̂ε, {x > 0},
|x=0 = 0 .
where
M(ζ) =
0 P−(ζ)
We remark that
εÃ(ζ) = Ã(εζ) = Ã(ζ̂),
with ζ̂ = (τ̂ , γ̂, η̂) := εζ. Moreover P− is homogeneous of order zero in ζ.
Let us denote R̃ε(ζ̂ , x) := R̂ε(ζ, x) and Ṽ
(ζ̂ , x) := V̂
(ζ, x). Hence Ṽ
solution of the following problem:
−Ã(ζ̂) +M(ζ̂)
= ε2R̃ε(ζ̂ , x), {x > 0},
|x=0 = 0 .
As a consequence, the Uniform Lopatinski Condition for problem (4.13)
writes: For all γ̂ > 0,
|det(E−(Ã(ζ̂)−M(ζ̂), ker Γ)| ≥ C > 0.
In view of the proof of the Proposition (4.3), we recall that the spaces E±(A)
have to be considered in the extended sense defined above.
Proposition 4.3. Since H satisfies the hyperbolicity Assumption in As-
sumption 1.1, the Uniform Lopatinski Condition is satisfied for our present
problem; that is to say that, for all ζ̂ such that γ̂ > 0 there holds:
|det(E−(Ã(ζ̂)−M(ζ̂), ker Γ)| ≥ C > 0.
Proof. We will begin to show that the Uniform Lopatinski Condition writes
as well that for all ζ̂ 6= 0 there holds:
(4.14) E+(A(ζ̂)−P
−(ζ̂))
E−(A(ζ̂)) = {0} .
This notation keeps a sense for ζ̂ such that γ̂ = 0 because we will prove
a posteriori that the involved linear subspaces continuously extends from
{ζ̂ , γ̂ > 0} to {ζ̂ , γ̂ = 0}. Then we will prove that, for all ζ̂, the property
4.14 holds true. The Uniform Lopatinski Condition writes actually, for all
ζ̂ 6= 0 :
E−(Ã(ζ̂)−M(ζ̂))
ker Γ̃ = {0}.
and thus, since we have:
E−(Ã(ζ̂)−M(ζ̂)) = E−(A(ζ̂))× E+(A(ζ̂)−P
−(ζ̂)),
and by definition of Γ̃, the Uniform Lopatinski Condition writes then that,
for all ζ̂ 6= 0, there holds:
E+(A(ζ̂)−P
−(ζ̂))
E−(A(ζ̂)) = {0}.
Lemma 4.4.
A(ζ̂)−P−(ζ̂)
A(ζ̂)
A(ζ̂)−P−(ζ̂)
A(ζ̂)
Proof. For all ζ̂ 6= 0, there is an invertible N × N matrix with complex
coefficients P (ζ̂) such that: P−1AP is trigonal and the diagonal coefficients
are sorted by increasing order of their real parts. Let us denote by ν the
dimension of E− (A) . The above matrix P traduces the change of basis from
the canonical basis of CN into (v1, . . . , vν , vν+1, . . . , vN ), where
Span ((vk)1≤k≤ν) = E− (A) ,
Span ((vk)ν+1≤k≤N ) = E+ (A) .
Moreover, there holds
P−1P−P = D
where D is the diagonal matrix whose ν first diagonal terms are equal to 1
and the N − ν last diagonal terms are null.
P−1(A−P−)P = P−1AP −D.
P−1AP−D is also trigonal, with the same eigenvalues with positive real part
as P−1AP and the same eigenvalues with negative real part as P−1AP −Id.
As a consequence, for all ζ̂ 6= 0, there holds:
A(ζ̂)−P−(ζ̂)
A(ζ̂)
A(ζ̂)−P−(ζ̂)
A(ζ̂)
As a consequence of Lemma 4.4, the rescaled Uniform Lopatinski Con-
dition for ε > 0, ε → 0 happens to be exactly the same as the one written
for bigger positive ε. Indeed, it writes, for all ζ̂ 6= 0 :
E+(A(ζ̂))
E−(A(ζ̂)) = {0}.
✷ The Lopatinski condition is satisfied, and, as a result, the following,
uniform in ε, energy estimate holds for V ε, for all γ ≥ γk′ > 0 :
γ‖V ε‖2
+ ‖V ε|2x=0‖Hk′γ (ΥT )
‖εRε‖2
which is equivalent to:
(4.15) γ‖V ε‖2
+ ‖V ε|x=0‖
γ (ΥT )
‖εRε‖2
This proves the convergence of V ε towards zero in Hk
). The weight
γ is fixed beforehand thus, in fact, the solution of (4.13) tends to zero in
) at a rate at least in O(ε).
5 End of proof of Theorem 1.9.
Let us consider V ε defined by:
V ε(t, y, x) :=
uε+app(t, y, x)− u
ε+(t, y, x)
uε−app(t, y,−x)− u
ε−(t, y,−x)
This notation is perfectly fine because the so-defined function is solution of
an equation of the form (4.13). Moreover, thanks to the stability estimate
(4.15), there is γk positive such that, for all γ > γk, we have:
γ‖uεapp−u
+γ‖uεapp−u
+‖uεapp−u
γ (ΥT )
‖εRε+‖2
Hence, it follows that:
‖uεapp − u
Hk−3(Ω
+ ‖uεapp − u
Hk−3(Ω
= O(ε2).
Moreover, by construction of uεapp, we have:
‖uεapp − u‖
Hk−3(Ω
+ ‖uεapp − u
Hk−3(Ω
= O(ε2).
As a result, we obtain that there holds:
‖uε − u‖2
Hk−3(Ω
+ ‖uε − u−‖2
Hk−3(Ω
= O(ε2).
This concludes the proof of Theorem 1.9.
6 Appendix: answer to a question asked in [11].
In this chapter, we will show that the loss of convergence observed numeri-
cally in [11] in a neighborhood of the boundary is due to a boundary layer
phenomenon. We consider the 1-D wave equation:
(6.1)
∂ttU − c
2∂xxU = 0, (x, t) ∈]0, π[×R
U |x=0 = U |x=π = 0,
U |t=0(x) = sin(x),
∂tU |t=0 = 0.
As in [11], we define then U ε = U ε+1x>0 + U
ε−1x<0 by:
(6.2)


ε+ − c2∂xxU
ε+ = 0, (x, t) ∈]0, π[×R+,
ε− − c2∂xxU
U ε− = 0, (x, t) ∈]−∞, 0[×R+,
U ε+|x=0 − U
ε−|x=0 = 0
ε+|x=0 − ∂xU
ε−|x=0 = 0
U ε+|x=π = 0.
U ε±|t=0(x) = sin(x), {±x > 0}.
ε±|t=0 = 0, {±x > 0}.
We will now construct formally an approximate solution U ε±app of U
ε± satis-
fying the following ansatz:
U ε+app =
U+j (t, x)ε
U ε−app =
t, x,
where the profiles U−j (t, x, z) := U
j (t, x) + U
j (t, z), with
e−αzU∗−j = 0,
for some α > 0. Since the stability estimates are trivial here, we will only
focus on the construction of
U εapp := U
app1x>0 + U
app1x<0.
Plugging U ε±app into problem (6.2) and identifying the terms with same power
of ε, we obtain the following equations:
U−0 = 0,
Moreover, U∗−0 = 0 as it is the only solution of the problem:
U∗−0 − c
2∂zzU
0 = 0, {z < 0},
0 |z=0 = 0,
0 = 0.
U ε+app converges towards U
0 as ε → 0
+. As awaited U+0 is the solution of the
well-posed 1-D wave equation:


0 − c
2∂xxU
0 = 0, (x, t) ∈]0, π[×R
U+0 |x=0 = U
0 |x=0 + U
0 |z=0 = 0.
U+0 |x=π = 0.
0 |t=0(x) = sin(x), {x > 0}.
0 |t=0 = 0, {x > 0}.
Let us write the following profiles equations: First, we can see that, for all
j ≥ 1, there holds:
where U∗−1 is the solution of the well-posed profile equation:
1 − c
2∂zzU
1 = −∂ttU
0 = 0, {z < 0},
1 |z=0 = ∂xU
0 |x=0,
U∗−1 = 0.
Hence U∗−1 is given by:
U∗−1 = c∂xU
0 |x=0e
We will show now that the profiles can be computed as any order. Assume
that U∗−j has been computed, U
j is solution of the well-posed 1-D wave
equation: 


− c2∂xxU
= 0, (x, t) ∈]0, π[×R+,
j |x=0 = U
j |z=0.
U+j |x=π = 0.
U+j |t=0(x) = 0, {x > 0}.
0 |t=0 = 0, {x > 0}.
U∗−j+1 is then solution of the well-posed profile equation:
U∗−j+1 − c
2∂zzU
j+1 = −∂ttU
j , {z < 0},
j+1|z=0 = ∂xU
|x=0,
U∗−j+1 = 0.
Let us answer the question asked in [11]: U ε− is bound to present boundary
layer behavior in {x = 0−}, indeed its approximate solution is composed
exclusively of boundary layer profiles, which describes quick transitions at
the boundary using a fast scale in ε. As a result of the loss in convergence
induced by the boundary layer, the following estimate holds:
‖U ε − U‖L2(]−∞,π[×R+) = O(ε
In [11], their small parameter is µ = ε2, as a result, adopting the same
notations as them, our estimate writes:
‖Uµ − U‖L2(]−∞,π[×R+) = O(µ
which is in agreement with the estimates given in [11]. Like in the penaliza-
tion approach proposed by Bardos and Rauch [2] and underlined by Droniou
in [4], the boundary layer only forms on one side of the boundary. The ap-
proximation U ε+ of U, is computed by taking U ε+|x=0 = U
ε−|x=0, thus, in
numerical applications, the boundary layer phenomenon also affects the rate
of convergence of U ε+ towards U,
as ε → 0+.
References
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into account obstacles in viscous flows. Numerische Mathematik 1999;
81:497-520.
[2] C. Bardos, J. Rauch, Maximal positive boundary value problems as limits
of singular perturbation problems, Trans. Amer. Math. Soc.,270 (1982),
pp 377-408.
[3] J. Chazarain, A. Piriou, Introduction to the theory of linear partial dif-
ferential equations. translated from the french , Studies in Mathematics
and its Applications, 14 , North Holland Publishing Co., Amsterdam-
New York,1982.
[4] J. Droniou, Perturbation Singulière par Pénalisation d’un Système Hy-
perbolique, Rapport de stage (1997).
[5] B. Fornet, O.Guès, Penalization approach of semi-linear symmetric hy-
perbolic problems with dissipative boundary conditions, preprint (2007).
[6] O. Guès, G. Métivier, M. Williams, K.Zumbrun Uniform stability esti-
mates for constant-coefficient symmetric hyperbolic boundary value prob-
lems. preprint (2005).
[7] H.O. Kreiss, Initial boundary value problems for hyperbolic systems,
Comm. Pure Appl. math 13 (1970), 277-298.
[8] G. Métivier, Small Viscosity and Boundary Layer Methods : Theory,
Stability Analysis, and Applications, Birkhauser (2003).
[9] G. Métivier, K. Zumbrun Viscous Boundary Layers for Noncharacteris-
tic Nonlinear Hyperbolic Problems, Preprint.
[10] G. Métivier, K. Zumbrun Symmetrizers and Continuity of Stable Sub-
spaces for Parabolic-Hyperbolic Boundary Value Problems, Preprint.
[11] A. Paccou, G. Chiavassa, J. Liandrat and K. Schneider A penalization
method applied to the wave equation. C. R. Acad. Sci. Paris Serie II,
(2003)
[12] J. Rauch, Boundary value problems as limits of problems in all space.,
In Séminaire Goulaouic-Schwartz (1978/1979), pages Exp. No. 3, 17.
École Polytech., Palaiseau, 1979.
	Introduction.
	A Kreiss Symmetrizer Approach.
	A second Approach.
	Underlying approach leading to the proof ofTheorem ??.
	Some preliminaries.
	Detailed proof of Lemma ??: Construction of the matrices B solving Lemma ??.
	A change of dependent variables.
	Proof of Theorem ??.
	Construction of the approximate solution.
	Stability estimates
	End of the proof of Theorem ??.
	Proof of Theorem ??.
	Construction of an approximate solution.
	Asymptotic Stability of the problem as  tends towards zero.
	End of proof of Theorem ??.
	Appendix: answer to a question asked in paccou.

  In this paper, we describe a new, systematic and explicit way of
approximating solutions of mixed hyperbolic systems with constant coefficients
satisfying a Uniform Lopatinski Condition via different Penalization
approaches.

Introduction.
In this paper, we describe a new, systematic and explicit way of approximat-
ing solutions of mixed hyperbolic systems with constant coefficients satisfy-
ing a Uniform Lopatinski Condition via different Penalization approaches.
In applied Mathematics like, for instance, in the study of fluids dynamics,
the method of penalization is used to treat boundary conditions in the case
of complex geometries. By replacing the boundary condition by a singular
perturbation of the PDE extended to a larger domain, this method allows
the construction of an approximate, often more easily computable, solution.
We consider mixed boundary value problems for hyperbolic systems:
Aj∂j ,
on {xd ≥ 0}, with boundary conditions on {xd = 0}. The n× n real valued
matrices Aj are assumed constant. Of course, we assume the coefficients to
be constant as a first approach, aiming to generalize the results obtained here
in future works. We assume that the boundary {xd = 0} is noncharacteristic,
which means that detAd 6= 0. We denote by
y := (x1, . . . , xd−1) and x := xd. The problem writes:
(1.1)
Hu = f, {x > 0},
Γu|x=0 = Γg,
u|t<0 = 0 ,
where the unknown u(t, x) ∈ Rn, Γ : Rn → Rp is linear and such that
rg Γ = p; which implies that Γ can be viewed as a p×n real valued constant
matrix. Let us fix T > 0 once and for all for this paper. Let Ω+
denotes
the set [0, T ] × Rd+ and ΥT denote the set [0, T ] × R
d−1. f is a function in
Hk(Ω+
), g is a function in Hk(ΥT ), where k ≥ 3 or k = ∞, such that:
f |t<0 = 0 and g|t<0 = 0. We make moreover the following Hyperbolicity
assumption on H :
Assumption 1.1. For all (η, ξ) ∈ Rd−1 × R− {0}, the eigenvalues of
ηjAj + ξAd
are real, semi-simple and of constant multiplicity.
Let us introduce now the frequency variable ζ := (γ, τ, η), where iτ + γ,
with γ ≥ 0, and τ ∈ R stands for the frequency variable dual to t and
η = (η1, . . . , ηd−1) where ηj ∈ R is the frequency variable dual to xj. We
note:
A(ζ) := − (Ad)
(iτ + γ)Id+
iηjAj
Denote by M a N ×N, complex valued, matrix; E−(M)[resp E+(M)] is the
linear subspace generated by the generalized eigenvectors associated to the
eigenvalues of M with negative [resp positive] real part. If F and G denote
two linear subspaces of CN such that dimF+dimG = N, det(F,G) denotes
the determinant obtained by taking orthonormal bases in each space. Up to
the sign, the result is independent of the choice of the bases. We shall now
explicit the Uniform Lopatinski Condition assumption:
Assumption 1.2. (H,Γ) satisfies the Uniform Lopatinski Condition i.e for
all ζ such that γ > 0, there holds:
(1.2) |det(E−(A), ker Γ)| ≥ C > 0.
The mixed hyperbolic system (1.1) has a unique solution in
Hk(Ω+
), and, since H is hyperbolic with constant multiplicity, for all γ
positive, the eigenvalues of A stay away from the imaginary axis. More-
over, as emphasized for instance by Chazarain and Piriou in [3] and Mé-
tivier in [8], there is a continuous extension of the linear subspace E−(A)
to {γ = 0, (τ, η) 6= 0
} that we will denote by Ẽ−(A). Ẽ+(A) extends as
well continuously to {γ = 0, (τ, η) 6= 0
} and we will denote Ẽ+(A) this
extension. Moreover, there holds:
Ẽ−(A)
Ẽ+(A) = C
We can refer the reader to [3], [6], [7], or [8] for detailed estimates concern-
ing mixed hyperbolic problems satisfying a Uniform Lopatinski Condition.
Moreover, we can refer to [10] for the proof of the continuous extension of the
linear subspaces mentioned above in the hyperbolic-parabolic framework.
Remark 1.3. As a consequence of the uniform Lopatinski condition, there
holds, for all ζ 6= 0 :
rg Γ = p = dim Ẽ−(A(ζ)).
1.1 A Kreiss Symmetrizer Approach.
We will now describe a penalization method involving a Kreiss Symmetrizer
and a matrix constructed by Rauch in [12], in the construction of our singular
perturbation. Note well that we have some freedom in both the choice of the
Kreiss Symmetrizer and of Rauch’s matrix. Let us denote respectively by û,
f̂ , and ĝ the tangential Fourier-Laplace transform of u, f, and g. Since the
Uniform Lopatinski Condition is holding for the mixed hyperbolic system
(1.1), there is, see [9] a Kreiss symmetrizer S for the problem:
(1.3)
∂xû = Aû+ f̂ , {x > 0},
Γû|x=0 = Γĝ,
That is to say there exists a matrix S(ζ), homogeneous of order zero in ζ,
C∞ in R+ × Rd − {0
} and there are λ > 0, δ > 0 and C1 such that:
• S is hermitian symmetric.
• ℜ (SA) ≥ λId.
• S ≥ δId− C1Γ
An algebraic result proved by Rauch in [12] can be reformulated as follow,
and a proof is recalled in section 2.2:
Lemma 1.4. There is a hermitian symmetric, uniformly definite positive,
N ×N matrix B such that:
ker Γ = E+((S)
−1B).
Moreover B depends smoothly of ζ.
Remark 1.5. This result is proved by constructing explicit matrices satisfy-
ing the desired properties. Thus, it is not merely an existence result and we
can use the explicitly known matrix B in our construction of a penalization
operator.
Let us denote by R := B
2 and SR := R
−1SR−1. We will denote by P−
the projector on E−(SR) parallel to E+(SR) and by P
+ the projector on
E+(SR) parallel to E−(SR); P
− and P+ denoting the associated Fourier
multiplier. We recall that, denoting by F the tangential Fourier transform,
the Fourier multiplier P−(∂t, ∂y, γ) [resp P
+(∂t, ∂y, γ)] is then defined, for all
w ∈ Hk(Rd+1), and γ > 0, by:
−(∂t, ∂y, γ)w
= P−(ζ)F(w),
[resp
+(∂t, ∂y, γ)w
= P+(ζ)F(w)],
in the future we will rather write:
±(∂t, ∂y, γ)w
= P±(ζ)F(w).
We fix, once and for all, γ > 0 big enough. Let us consider then the solution
uε of the well-posed Cauchy problem on the whole space (1.4):
(1.4)
Huε +
Muε1x<0 = f1x>0 +
θ1x<0, {x ∈ R},
uε|t<0 = 0,
where
M := −eγtAdS
−1RP−Re−γt,
θ := −eγtAdS
−1RP−Γg̃,
and S(∂t, ∂y) [resp R(∂t, ∂y)] denotes the Fourier multiplier associated to
S(ζ) [resp R(ζ)]. Let us define g̃ by:
g̃ := e−x
In what follows, ĝ will denote the Fourier-Laplace transform of g̃. Let us
denote by
ũ := u−1x<0 + u1x≥0 = u
1x≤0 + u1x>0.
u denotes the solution of (1.1), and thus belongs to Hk(Ω+
). u− is a function
belonging to Hk(Ω−
) and such that u−|x=0 = u|x=0. More precisely, u
− can
be computed by: eγtF−1
R−1(v̂− + P−Γĝ)
, where v̂− is the solution of the
problem:
SR∂xv̂
− − P+SRARv̂
− = P+SRARP
−Γĝ, {x < 0},
v̂−|x=0 = P
+Rû|x=0,
and û denotes the Fourier-Laplace transform of the solution u of (1.1).
Theorem 1.6. For all k ≥ 3, if f ∈ Hk(Ω+
) and g ∈ Hk(ΥT ), then there
holds:
‖uε − u−‖
Hk−3(Ω
+ ‖uε − u‖
Hk−3(Ω
= O(ε),
where uε denotes the solution of the Cauchy problem (1.4) and u denotes
the solution of the mixed hyperbolic problem (1.1). If g = 0 then:
‖uε − u−‖
+ ‖uε − u‖
= O(ε).
Of course, since uε is defined for all {x ∈ R}, its limit as ε → 0+, ũ is
can be viewed as an ”extension” of u on the fictive domain {x < 0}. The
”extension” resulting from our method of penalization gives a continuous ũ
across {x = 0}, while the method used in [2] gave simply: ũ|x<0 = 0. We
have the following Corollaries:
Corollary 1.7. Assume for example that f ∈ H∞(Ω+
) and
g ∈ H∞(ΥT ) then
‖uε − u‖
Hs(Ω+
= O(ε); ∀s > 0.
Corollary 1.8. If f belongs to L2(Ω+
) and g = 0 then:
‖uε − ũ‖L2(ΩT ) = 0.
One of the interest of this first approach lies in the rate of convergence
of uε towards u. Indeed, in general, a boundary layer will form near the
boundary in this kind of singular perturbation problem. For example in the
paper by Bardos and Rauch [2], as confirmed by Droniou [4], a boundary
layer forms. It is also the case in [11], as analyzed in our Appendix. There
are also boundary layers phenomena in the parabolic context: see the ap-
proach proposed by Angot, Bruneau and Fabrie [1] for instance. However,
surprisingly, and like in the penalization method proposed by Fornet and
Guès in [5], our method allows the convergence to occur without formation
of any boundary layer on the boundary. As a result, this leads to the kind
of sharp stability estimate given in Theorem 1.6. These results concern the
case where f and g are sufficiently regular. The reason is that we construct
an approximate solution. In the case of g only in L2(ΥT ), such a simple
treatment does not work. However, let δ > 0 be given. If we approximate
f and g by smooth functions fν ∈ H
) and gν ∈ H
∞(ΥT ) such that
‖f − fν‖L2(Ω+
< δ and ‖g − gν‖L2(ΥT ) < δ, by the uniform Lopatinski
condition, we get:
‖uν − u‖L2(Ω+
< Cν,
where uν is the solution of the mixed hyperbolic problem (1.1) with data fν
and gν . We can now apply Corollary 1.8 to uν , and obtain by penalization
a sequence uεν in L
2(ΩT ) such that: limε→0+ u
ν = uν in L
). Finally, by
choosing, ε sufficiently small, we get ‖u − uεν‖L2(Ω+
< 2Cδ. By choosing ε
and ν as functions of δ, and noting u(δ) = u
, we have:
(1.5) lim
‖u(δ) − u‖L2(Ω+
1.2 A second Approach.
In the first approach we have just introduced, it is necessary to compute
a Kreiss’s Symmetrizer and a Rauch’s matrix. In view of future numerical
applications, we will now introduce another method preventing the compu-
tation of these matrices. The price to pay is that we need the preliminary
computation of v, which is by definition the solution of the Cauchy problem
on the free space:
(1.6)
Hv = f, (t, y, x) ∈ ΩT ,
v|t<0 = 0 ∀(y, x) ∈ R
Let us denote P−(ζ) the spectral projector on Ẽ−(A(ζ)) parallel to
Ẽ+(A(ζ)), andP
+(ζ) the spectral projector on Ẽ+(A(ζ)) parallel to Ẽ−(A(ζ)).
Let us introduce P±(∂t, ∂y, γ), the Fourier multiplier associated to P
±(ζ).
Let us denote by Π the projector on Ẽ−(A(ζ)) parallel to KerΓ, which has
a sense because of the Uniform Lopatinski Condition and denote Π the
associated Fourier multiplier. We define then h̃ by:
h̃ := e−x
−(e−γtv|x=0) +Πe
−γt(g − v|x=0)
where g denotes the function involved in the boundary condition of the
mixed hyperbolic problem (1.1). Now, let us consider the following singularly
perturbed Cauchy problem on the whole space:
(1.7)
Huε +
−e−γtuε1x<0 = f1x>0 +
γth̃1x<0,
uε|t<0 = 0 .
Let us denote by
ũ := u−1x<0 + u1x≥0 = u
1x≤0 + u1x>0.
u denotes the solution of (1.1) thus belonging toHk(Ω+
) and u− is a function
belonging to Hk(Ω−
) and such that u−|x=0 = u|x=0. More precisely, u
− can
be computed by: eγtF−1(F(h̃)+v̂−), where v̂− is the solution of the problem:
+v̂−)−A(P+v̂−) = 0, {x < 0},
+v̂−|x=0 = P
+û|x=0.
and û denotes the Fourier-Laplace transform of the solution u of (1.1). The
problem (1.7) is well-posed and, for all ε > 0, there exists a unique
uε ∈ Hk(ΩT ) solution. We will fix γ adequately big beforehand. We observe
then the following result:
Theorem 1.9. For all k ≥ 3, if f ∈ Hk(Ω+
) and g ∈ Hk(ΥT ), then there
holds:
‖uε − u−‖
Hk−3(Ω
+ ‖uε − u‖
Hk−3(Ω
= O(ε),
where uε denotes the solution of the Cauchy problem (1.7) and u denotes
the solution of the mixed hyperbolic problem (1.1).
The singular perturbation involved in the definition of uε does not de-
pend either of Kreiss’s Symmetrizer or Rauch’s matrix. As a result, for
this method of penalization far less computations are necessary in order to
obtain our singular perturbation. Note well that the proof of the energy
estimates in Theorem 1.9 is completely different from the proof of the en-
ergy estimates in Theorem 1.6. Indeed, for our first approach our singularly
perturbed problem was treated as a Cauchy problem, contrary to our second
approach where it was interpreted as a transmission problem.
Corollary 1.10. Assume for example that f ∈ H∞(Ω+
) and
g ∈ H∞(ΥT ) then
‖uε − u‖
= O(ε); ∀s > 0.
Of course, we see that the same problem of regularity arises in Theorem
1.9 and Theorem 1.6. However, by a simple density argument, we can also
prove here the exact analogous of (1.5).
Remark 1.11. In the case where f = 0, then the solution v of (1.6) is
v = 0 and thus, the perturbed cauchy problem (1.7) rewrites:
Huε +
−e−γtuε1x<0 =
γte−x
Πe−γtg
1x<0, {x ∈ R},
uε|t<0 = 0 .
2 Underlying approach leading to the proof of
Theorem 1.6.
2.1 Some preliminaries.
Since the Uniform Lopatinski Condition holds, there is S, homogeneous of
order zero in ζ, and such that there are λ > 0, δ > 0 and C1 and there holds:
• S is hermitian symmetric.
• ℜ (SA) ≥ λId.
• S ≥ δId− C1Γ
S is then called a Kreiss Symmetrizer for the problem:
(2.1)
∂xû = Aû+ f̂ , {x > 0},
Γû|x=0 = Γĝ,
where f̂ and ĝ denotes respectively the Fourier-Laplace transforms of f and
g̃; and û denotes the Fourier-Laplace transform of the solution u of the well-
posed mixed hyperbolic problem (1.1). û is also solution, for all fixed ζ 6= 0
of the following equation:
(2.2)
S∂xû = SAû+ S(Ad)
−1f̂ , {x > 0},
Γû|x=0 = Γĝ,
Remark 2.1. Following our current assumptions, Γ is independent of ζ 6= 0,
however, more general boundary conditions, of the form:
Γ(ζ)û|x=0 = Γ(ζ)ĝ,
can be treated. It would imply taking as boundary condition for (1.1):
Γγu|x=0 = Γγg,
with for γ big enough,
Γγ := Γ(∂t, ∂y)e
where, Γ(∂t, ∂y) denotes the Fourier multiplier associated to Γ(ζ), that is to
say is defined by:
F(Γ(∂t, ∂y)u) = Γ(ζ)F(u).
Referring for example to [3] and [7], Kreiss has proved that the exis-
tence of a Kreiss symmetrizer for the symbolic equation is sufficient to prove
the well-posedness of the associated pseudodifferential equation (here (1.1)).
Indeed, multiplying by û and integrating by parts the equation:
S∂xû = SAû+ S(Ad)
leads to the desired a priori estimates. For all ζ 6= 0, S(ζ) is hermitian
symmetric and definite positive on ker Γ. Let us sum up the properties crucial
in the proof of the well-posedness of our problem:
Proposition 2.2. For all ζ = (τ, γ, η) such that τ2 + γ2 +
j=1 η
j = 1,
there holds:
• S(ζ) is hermitian symmetric.
• ℜ (SA) (ζ) := 1
(SA+ (SA)∗)(ζ) is positive definite.
• −S(ζ) is definite negative on ker Γ and ker Γ is of same dimension as
the number of negative eigenvalues in −S(ζ).
Note that, by homogeneity of S, it is equivalent for the properties in Propo-
sition 2.2 to hold for |ζ| = 1 or for |ζ| > 0. As a consequence of the first
point and third point of Proposition 2.2, the Lemma 1.4 applies and gives a
matrix B such that: ker Γ = E+(S
−1B). In the sequel, such a matrix B is
fixed once for all.
The following chapter contains a proof of Lemma 1.4 assorted of a detailed
construction of B.
2.2 Detailed proof of Lemma 1.4: Construction of the ma-
trices B solving Lemma 1.4.
As we will emphasize in next chapter, Lemma 1.4 is a crucial feature in our
first method of Penalization. The aim of this chapter is to give a more com-
plete proof rather than simply recalling Rauch’s result and, in the process,
to precise how the matrices B solving Lemma 1.4 are constructed. For all
ζ 6= 0, S(ζ) is hermitian symmetric, uniformly definite positive on Ẽ+(A(ζ)),
and uniformly definite negative on Ẽ−(A(ζ)); as a consequence, S(ζ) keeps
exactly p positive eigenvalues and N − p negative eigenvalues for all ζ 6= 0.
Basically, knowing that S is uniformly definite positive on ker Γ; we search
to express ker Γ in a way involving S. Consider q ∈ ker Γ, since, for all ζ 6= 0,
E−(S(ζ))
E+(S(ζ)) = C
N , we can split q in:
q := q+ + q−
with q+ ∈ E+(S(ζ)) and q
− ∈ E−(S(ζ)).
Since dim kerΓ = dimE+(S(ζ)) = p, these two linear subspaces are in bijec-
tion. Let us give the two main ideas behind this proof: one idea is to detail
the bijection between q ∈ kerΓ and q+ ∈ E+(S(ζ)) as it satisfies some con-
straints, the other is to come down to the model case where the eigenvalues
of S are either 1 or −1. Let us denote:
S̃−1 =
−IdN−p 0
0 Idp
In a first step, we will prove the following result:
Proposition 2.3. If we assume that V is a linear subspace of CN of dimen-
sion p, and that there is C > 0 such that, for all q ∈ V, there holds:
〈S̃−1q, q〉 ≥ C〈q, q〉,
then the two following equivalent properties hold:
• There is a hermitian symmetric, positive definite matrix R̃, such that:
[q ∈ V] ⇔
q ∈ E+(R̃S̃R̃)
which is equivalent to:
V = E+(R̃
• There is a hermitian symmetric, positive matrix R̃, such that:
[q ∈ V] ⇔
R̃q ∈ E+(R̃S̃
−1R̃)
which is equivalent to:
V = E+(S̃
−1R̃2).
Moreover, we can link the two properties by taking:
R̃2 = S̃R̃
Proof. In this proof, we will show how to construct some matrices R̃
satisfying the required properties. There is a (N − p)× p matrix ℵ of rank
N − p such that ‖ℵ‖ ≤ 1 and:
V = {q ∈ CN , q− = ℵq+},
where q+[resp q−] denotes the projector on E+((S̃)
−1) [resp E−((S̃)
−1)]parallel
to E−((S̃)
−1) [resp E+((S̃)
−1)]. Indeed, dimV = p = dimE+((S̃)
−1), and
N = E−((S̃)
E+((S̃)
−1). Moreover, there ic C > 0 such that, for all
q ∈ V, there holds:
〈(S̃)−1q, q〉 = −〈q−, q−〉+ 〈q+, q+〉 ≥ C〈q, q〉.
and thus
|q+|2 − |ℵq+|2 ≥ C|q|2,
which implies that ‖ℵ‖ < 1. We will show now that, for R̃ constructed as
follow:
IdN−p −ℵ
−ℵ∗ Idp
there holds:
[q ∈ V] ⇔
R̃q ∈ E+(R̃S̃
−1R̃)
First,we see that the constructed R̃ is trivially hermitian symmetric and
positive definite since ‖ℵ‖ < 1. First, we have:
R̃S̃−1R̃ =
−IdN−p +NN
0 Idp −N
R̃q =
q− − ℵq+
−ℵ∗q− + q+
Thus, since ‖ℵ‖ < 1, there holds:
R̃q ∈ E+(R̃S̃
−1R̃)
q− − ℵq+ = 0
⇔ [q ∈ V] .
We will now prove that we have:
(R̃)−1E+(R̃S̃
−1R̃) = E+(S̃
−1R̃2).
Since R̃S̃−1R̃ is hermitian symmetric, the linear subspace E+(R̃S̃
−1R̃) is
generated by the eigenvectors of R̃S̃−1R̃ associated to positive eigenvalues.
A basis of (R̃)−1E+(R̃S̃
−1R̃) is thus given by ((R̃)−1vj)j where vj denotes
an eigenvector of R̃S̃−1R̃ associated to a positive eigenvalue λj . We have:
R̃S̃−1R̃vj = λjvj.
Let us denote wj = (R̃)
−1vj, we have then:
R̃S̃−1R̃2wj = λjR̃wj ⇔ S̃
−1R̃2wj = λjwj .
As a result, wj is an eigenvector of S
−1R̃2 associated to the eigenvalue λj
hence we obtain that:
(R̃)−1E+(R̃S̃
−1R̃) = E+(S̃
−1R̃2).
We can also prove, the same way, that:
R̃E+(R̃S̃R̃) = E+(R̃
Now, taking
R̃2 = S̃R̃
we can check that:
E+(S̃
−1R̃2) = E+(R̃
which concludes the proof. ✷
Lemma (??) is a Corollary of the following Proposition:
Proposition 2.4. If S−1 denotes a smooth in ζ 6= 0, matrix-valued function
in the space of hermitian symmetric matrices with p positive eigenvalues and
N −p negative eigenvalues and ker Γ denotes a linear subspace of dimension
p and there is C > 0 such that, for all q ∈ ker Γ, there holds:
〈S−1q, q〉 ≥ C〈q, q〉,
then the two following equivalent properties hold:
• There is a smooth in ζ 6= 0, matrix-valued function R, in the space of
hermitian symmetric, positive matrices such that:
[q ∈ KerΓ] ⇔
∀ζ 6= 0, R−1(ζ)q ∈ E+(R(ζ)S(ζ)R(ζ))
which is equivalent to:
∀ζ 6= 0, KerΓ = E+(R
2(ζ)S(ζ)).
• There is a smooth in ζ 6= 0, matrix-valued function R, in the space of
hermitian symmetric, positive matrices such that:
[q ∈ KerΓ] ⇔
∀ζ 6= 0, R(ζ)q ∈ E+(R(ζ)S
−1(ζ)R(ζ))
which is equivalent to:
∀ζ 6= 0, KerΓ = E+(S
−1(ζ)R2(ζ)).
Moreover, for all ζ 6= 0, these two properties can be linked by taking:
(R(ζ))2 = S(ζ)(R(ζ))2S(ζ).
Proof. We will show here that Proposition 2.4 can be deduced from Propo-
sition 2.3. For all ζ 6= 0, S(ζ) is a hermitian symmetric matrix, moreover S
depends smoothly of ζ. As a consequence S−1 is also a hermitian symmetric
matrix depending smoothly of ζ, and as such, there is a nonsingular matrix
V such that:
S̃−1 = V ∗
Let us denote Λ the diagonalized version of S−1 with eigenvalues sorted by
increasing order, then there is Z depending smoothly of ζ such that, for all
ζ 6= 0, we have:
Z∗(ζ) = Z−1(ζ),
Λ(ζ) = Z∗(ζ)
(ζ)Z(ζ).
As a consequence, V depends smoothly of ζ since, for all ζ 6= 0:
V (ζ) = (Λ(ζ))−
2Z(ζ),
where Λ is the diagonal matrix obtained by taking the absolute value of each
eigenvalue of Λ. For the sake of simplicity, let us omit the dependence in ζ.
Now, for all q ∈ V −1 ker Γ, there is C > 0, such that:
〈S̃−1q, q〉 = 〈V ∗S−1V q, q〉 = 〈S−1(V q), (V q)〉 ≥ C〈(V q), (V q)〉.
Moreover V is nonsingular, thus there is C ′ > 0, such that, for all
q ∈ V −1 ker Γ, there holds:
〈S̃−1q, q〉 ≥ C ′〈q, q〉.
Moreover dimV −1 ker Γ = p, using Proposition 2.3, for all fixed ζ 6= 0, there
is a hermitian symmetric, positive definite matrix R̃(ζ), such that:
V −1(ζ) ker Γ = E+((R̃(ζ))
2S̃(ζ)) = R̃(ζ)E+(R̃(ζ)S̃(ζ)R̃)(ζ).
We will now prove that we can construct R̃ depending smoothly of ζ. First
there is a (N − p) × p matrix ℵ of rank N − p, depending smoothly of ζ,
such that fore all ζ 6= 0 ‖ℵ(ζ)‖ ≤ 1 and:
V −1(ζ) ker Γ = {q ∈ CN , q− = ℵ(ζ)q+},
where q+[resp q−] denotes the projector on E+((S̃)
−1) [resp E−((S̃)
−1)]parallel
to E−((S̃)
−1) [resp E+((S̃)
−1)]. R̃ is given, for all ζ 6= 0, by:
R̃(ζ) =
S̃−1(ζ)R̃2(ζ)S̃−1(ζ),
with R̃ given by:
R̃(ζ) =
IdN−p −ℵ(ζ)
−ℵ∗(ζ) Idp
Since S̃−1 = V ∗
V, there holds: S̃ = V ∗SV, and, as a consequence:
(V R̃)−1 ker Γ = E+(R̃V
∗SV R̃).
As R̃V ∗SV R̃ is hermitian symmetric, a basis of the linear subspace E+(R̃V
∗SV R̃)
is given by the eigenvectors of R̃V ∗SV R̃ associated to positive eigenvalues.
This leads us to consider vj = (V R̃)
−1uj satisfying:
R̃V ∗SV R̃vj = λjvj.
We have:
R̃V ∗SV R̃(V R̃)−1uj = λj(V R̃)
−1uj.
hence:
(V R̃)R̃V ∗Suj = λjuj .
Since (V R̃)R̃V ∗ = (R̃V ∗)∗(R̃V ∗) is hermitian symmetric and positive defi-
nite, we can then define its square root. We define R by:
(R̃V ∗)∗(R̃V ∗).
Since both R̃ and V depends smoothly of ζ, so does R. Moreover, there
holds:
R2Suj = λjuj,
which gives:
ker Γ = V R̃E+(R̃V
∗SV R̃) = E+(R
We have thus proved there is a smooth in ζ 6= 0, matrix-valued function R,
in the space of hermitian symmetric, positive matrices such that:
[q ∈ KerΓ] ⇔
∀ζ 6= 0, R−1(ζ)q ∈ E+(R(ζ)S(ζ)R(ζ))
which is equivalent to:
∀ζ 6= 0, KerΓ = E+(R
2(ζ)S(ζ)).
Now consider R defined, for all ζ 6= 0, by:
R(ζ) =
S(ζ)(R(ζ))2S(ζ),
R(ζ) =
(R̃(ζ)V ∗(ζ)S(ζ))∗(R̃(ζ)V ∗(ζ)S(ζ)).
ζ 7→ R(ζ) is smooth and, for all ζ, R(ζ) is a hermitian symmetric, positive
definite matrix. Moreover, there holds:
[q ∈ KerΓ] ⇔
∀ζ 6= 0, R(ζ)q ∈ E+(R(ζ)S
−1(ζ)R(ζ))
which is equivalent to:
∀ζ 6= 0, KerΓ = E+(S
−1(ζ)R2(ζ)).
Let us detail the computation of R(ζ).
R(ζ) =
S(ζ)V (ζ)R̃
(ζ)V ∗(ζ)S(ζ).
Moreover
(ζ) = S̃−1(ζ)R̃2(ζ)S̃−1(ζ),
we have thus:
R(ζ) =
R̃(ζ)S̃−1(ζ)V ∗(ζ)S(ζ)
R̃(ζ)S̃−1(ζ)V ∗(ζ)S(ζ)
which gives:
B(ζ) =
R̃(ζ)S̃−1(ζ)V ∗(ζ)S(ζ)
R̃(ζ)S̃−1(ζ)V ∗(ζ)S(ζ)
We recall that R̃ is given, for all ζ 6= 0, by:
R̃(ζ) =
IdN−p −ℵ(ζ)
−ℵ∗(ζ) Idp
and that for all ζ 6= 0, V (ζ) is given by:
V (ζ) = (Λ(ζ))−
2Z(ζ),
where
Λ(ζ) = Z∗(ζ)
(ζ)Z(ζ)
with Λ is a diagonal matrix with real coefficients: (λ1, . . . , λN ), and Λ de-
notes the diagonal matrix with diagonal coefficients (|λ1|, . . . , |λN |).
Remark 2.5. In the construction of B the only freedom we have resides in
the choice of ℵ.
2.3 A change of dependent variables.
Let us denote by R := B
2 and v̂ := Rû. v̂ is hence solution of (2.3):
(2.3)
R−1SR−1∂xv̂ = R
−1SAR−1v̂ +R−1S(Ad)
−1f̂ , {x > 0},
ΓR−1v̂|x=0 = Γĝ,
We will adopt the following notations: SR := R
−1SR−1, AR := RAR
and ΓR := ΓR
−1. We first observe that:
ker ΓR = R ker Γ = RE+((S)
−1R2).
but S−1
= RS−1R thus
ker ΓR = RE+(R
−1SRR) = E+(SR).
This is where Lemma 1.4 is used in a crucial manner. Let us denote by
− the projector on E−(SR) parallel to E+(SR) and by by P
+ the projector
on E+(SR) parallel to E−(SR); P
− and P+ denoting the associated Fourier
multiplier. Since SR is hermitian symmetric, P
− is in fact the orthogonal
projector on E−(SR). The problem (2.3) can then be written:
SR∂xv̂ = SRARv̂ +R
−1S(Ad)
−1f̂ , {x > 0},
−v̂|x=0 = P
−Γĝ,
This problem is well-posed because, as a direct Corollary of Proposition 2.2,
we have:
Proposition 2.6. For all ζ such that τ2 + γ2 + |η|2 = 1, there holds:
• SR(ζ) is hermitian symmetric.
• ℜ (SRAR) (ζ) is positive definite.
• −SR(ζ) is definite negative on ker ΓR and the dimension of ker ΓR is
the same as the number of negative eigenvalues of −SR(ζ).
Proof. For the sake of simplicity, let us omit the dependence in ζ in our
notations.
• SR := R
−1SR−1, and both S and R are hermitian thus SR is hermi-
tian.
• SRAR = R
−1SAR−1, thus for all q ∈ CN , there holds:
2〈ℜ(SRAR)q, q〉 = 〈SRARq, q〉+〈q, SRARq〉 = 〈R
−1SAR−1q, q〉+〈q,R−1SAR−1q〉,
since R−1 is hermitian, we have then:
= 〈SAR−1q,R−1q〉+ 〈R−1q, SAR−1q〉 = 2〈ℜ(SA)R−1q,R−1q〉.
Since ℜ(SA) is positive definite and R is invertible, ℜ (SRAR) is thus
positive definite.
• By construction of R, it satisfies ker ΓR = E+(SR), with SR hermitian.
As a consequence −SR is definite negative on ker ΓR and the dimension
of ker ΓR is the same as the number of negative eigenvalues of −SR.
✷ Let us mention that, since R and S remains uniformly bounded in
ζ 6= 0, f̂ and R−1S(Ad)
−1f̂ belongs to the same space. In a same spirit as
[5], this suggests the following singular perturbation of (2.3):
SR∂xv̂
−v̂ε1x<0 = SRARv̂
−Γĝ1x<0 +R
−1S(Ad)
−1f̂ , {x ∈ R},
This is equivalent to perturb (2.2) as follow:
S∂xû
RP−Rûε1x<0 = SAû
RP−Γĝ1x<0 + S(Ad)
−1f̂ , {x ∈ R},
Finally, this induces the following perturbation for (1.1):
(2.4)
Huε +
Muε1x<0 = f1x>0 +
θ1x<0, {x ∈ R},
uε|t<0 = 0,
where
M := −eγtAdS
−1RP−Re−γt,
θ = −eγtAdS
−1RP−Γg̃,
and S(∂t, ∂y) [resp R(∂t, ∂y)] denotes the Fourier multiplier associated to
S(ζ) [resp R(ζ)].
3 Proof of Theorem 1.6.
First, we construct an approximate solution of equation (2.4) (which is also
equation (1.4)), then prove suitable energy estimates that ensures uε and its
approximate solution both converges towards the same limit as
ε → 0+.
3.1 Construction of the approximate solution.
uε is the solution of the well-posed Cauchy problem:
Huε +
Muε1x<0 = f1x>0 +
θ1x<0, {x ∈ R},
uε|t<0 = 0.
uε is moreover the solution of the well-posed Cauchy problem:
Huε +
Muε1x<0 = SA
f1x>0 +
θ1x<0, {x ∈ R},
uε|t<0 = 0.
The associated equation after tangential Fourier-Laplace transform writes :
S∂xû
RP−Rûε1x<0 − SAû
ε = −
RP−Γĝ1x<0 + S(Ad)
−1f̂1x>0, {x ∈ R}.
or alternatively:
ûε = R−1v̂ε
SR∂xv̂
−v̂ε1x<0 = SRARv̂
−Γĝ1x<0 +R
−1S(Ad)
−1f̂ , {x ∈ R},
We will use the following formulation as a transmission problem in our con-
struction of an approximate solution:
SR∂xv̂
ε+ = SRARv̂
ε+ +R−1S(Ad)
−1f̂ , {x > 0},
SR∂xv̂
−v̂ε− = SRARv̂
−Γĝ, {x < 0},
v̂ε+|x=0+ = v̂
ε−|x=0− .
For Ω an open regular subset of Rd+1, and ρ ∈ N, let us introduce the
weighted spaces H
γ (Ω) defined by:
H̺γ (Ω) = {̟ ∈ e
γtL2(Ω), ‖̟‖H̺γ (Ω) < ∞};
where
γ (Ω)
α,|α|≤̺
γρ−|α|‖e−γt∂α̟‖2
L2(Ω).
We will construct an approximate solution uεapp of u
ε. uεapp will be con-
structed as follow:
uεapp = u
app1x>0 + u
app1x<0,
where uε±app is an approximate solution of u
ε± satisfying the following ansatz:
uε±app =
U±j (ζ, x)ε
where the profiles U±j belong toH
), where Ω±
stands for [0, T ]×Rd±.
Denote
v̂εapp = RF(e
−γtuεapp) := v̂
app1x>0 + v̂
app1x<0.
v̂ε±app is then an approximate solution of v
ε± and is of the form:
v̂ε±app =
j (ζ, x)ε
where
= RF(e−γtU±
and conversely
U±j = e
γtF−1
R−1V ±j
The profiles U±j can be constructed inductively at any order. Let us show
how the first profiles are constructed: Identifying the terms in ε−1 gives:
−V −0 = P
−Γĝ.
Hence, P+V −0 remains to be computed in order to obtain the profile
0 = e
γtF−1
Identifying the terms in ε0 gives then that V +0 is solution of the well-posed
problem:
(3.1)
SR∂xV
0 = SRARV
−1S(Ad)
−1f̂ , {x > 0},
−V +0 |x=0 = P
−Γĝ.
The associated profile
U+0 = e
γtF−1
R−1V +0
belongs then toHkγ (Ω
).Moreover, the problem (3.1) is Kreiss-Symmetrizable
and thus the trace of the profile U+0 , see [3] for instance, satisfies:
U+0 |x=0 ∈ H
γ (ΥT ).
Since V +0 has just be computed, V
0 |x=0 is given by: V
0 |x=0 − V
0 |x=0 = 0
and thus, there holds:
−V +0 |x=0 = P
−V −0 |x=0.
Moreover
SR∂xV
0 − P
−V −1 = SRARV
0 , {x < 0}.
Projecting this equation on E+(SR) collinearly to E−(SR) gives then:
SR∂xP
0 − P
+SRARV
0 = 0, {x < 0},
Since
+SRARV
0 = P
+SRARP
+V −0 + P
+SRARP
−Γĝ,
we have then:
SR∂x(P
+V −0 )− P
+SRAR(P
+V −0 ) = P
+SRARP
−Γĝ, {x < 0},
and as a consequence, P+V −0 is solution of the following problem:
(3.2)
SR∂x(P
0 )− P
+SRAR(P
0 ) = P
+SRARP
−Γĝ {x < 0},
+V −0 |x=0 = P
+V +0 |x=0.
Let us precise how (3.2) has to be interpreted: we denote w = P+V −0 . w is
then totally polarized on E+(SR), and satisfies the problem:
(3.3)
+w = w
SR∂xw − P
+SRARw = P
+SRARP
−Γĝ {x < 0},
w|x=0 = P
+V +0 |x=0.
As we will see, the problem (3.3) is Kreiss-Symmetrizable and thus well-
posed. Indeed, for all ζ such that τ2 + γ2 + |η|2 = 1, we have, omitting the
dependencies in ζ in our notations:
• For all q ∈ CN , there holds:
〈SRq, q〉 = 〈q, SRq〉.
• Since Re(SRAR) is positive definite and P
+ is an orthogonal projector,
there is C > 0 such that, for all q ∈ E+(SR), there holds:
〈P+SRARP
+q, q〉+ 〈q,P+SRARP
+q〉 ≥ C〈q, q〉.
Indeed, for all q ∈ E+(SR), there holds:
〈P+SRARP
+q, q〉 = 〈P+SRARP
+q,P+q〉 = 〈SRARP
+q,P+q〉.
• −SR is definite negative on kerP
+ that is to say, that there is c > 0
such that, for all q ∈ kerP+, there holds:
〈−SRq, q〉 ≤ −c〈q, q〉.
Moreover kerP+ has the same dimension as the number of negative
eigenvalues in −SR.
The profile U−0 can then be computed by:
U−0 := e
γtF−1
R−1(w + P−Γĝ)
belongs to Hkγ (Ω
), moreover its trace U−0 |x=0 belongs
to Hkγ (ΥT ). Consider now the equation:
1 = SR∂xV
0 − SRARV
0 , {x < 0}.
Since P−V −1 |x=0 = P
−V +1 |x=0, V
1 is solution of the well-posed problem:
SR∂xV
1 = SRARV
1 , {x > 0},
−V +1 |x=0 = SR∂xV
0 |x=0 − SRARV
0 |x=0.
Due to the loss of regularity in the boundary condition, the associated profile
U+1 = e
γtF−1
R−1V +1
belongs to H
), moreover its trace U+1 |x=0 belongs
γ (ΥT ). Moreover, applying P
+ to the equation:
−V −2 + SRARV
1 = SR∂xV
1 , {x < 0},
we obtain:
SR∂xP
+V −1 = P
+SRARP
+V −1 + P
+SRARP
−V −1 , {x < 0},
+V −1 |x=0 = P
+V +1 |x=0.
As before, let us take P+V −1 as the unknown of the well-posed problem:
SR∂x(P
+V −1 )− P
+SRAR(P
+V −1 ) = P
+SRAR
SR∂xV
0 − SRARV
, {x < 0},
(P+V −1 )|x=0 = P
1 |x=0.
This problem is Kreiss-Symmetrizable since, for all ζ such that
τ2 + γ2 + |η|2 = 1, there holds:
• For all q ∈ CN , there holds:
〈SRq, q〉 = 〈q, SRq〉.
• There is C > 0 such that for all q ∈ E+(SR), there holds:
〈P+SRARP
+q, q〉+ 〈q,P+SRARP
+q〉 ≥ C〈q, q〉.
• −SR is definite negative on kerP
+ that is to say, that there is c > 0
such that, for all q ∈ kerP+, there holds:
〈−SRq, q〉 ≤ −c〈q, q〉.
Moreover kerP+ has the same dimension as the number of negative
eigenvalues in −SR.
However, due to a loss of regularity in both the source term and the boundary
condition, the associated profile
U−1 = e
γtF−1
+V −1 + SR∂xV
0 − SRARV
belongs to H
). The construction of the following profiles can be
pursued at any order the same way. In practice, we take:
uε+app = U
uε−app = U
0 + εU
As a result, the approximate solution writes uεapp := u
app1x>0 + u
app1x<0;
where uε+app belongs to H
) and uε−app belongs to H
). uεapp is then
solution of a well-posed problem of the form:
(3.4)
Huεapp +
Muεapp1x<0 = f1x>0 +
θ1x<0 + εr
ε, {x ∈ R},
uεapp|t<0 = 0 .
Where rε := rε+1x>0 + r
ε−1x<0, with r
ε+ ∈ H
) and
rε− ∈ Hk−3γ (Ω
Remark 3.1. In the case where g = 0, the loss of regularity in the profiles
is delayed by one step. As a result, in this case we obtain:
uε+app ∈ H
uε−app ∈ H
rε+ ∈ Hkγ (Ω
rε− ∈ H
3.2 Stability estimates
We will begin by proving energy estimates on the following equation:
(3.5) SRARê
ε − SR∂xê
−êε1x<0 = εr̂
ε, {x ∈ R},
where êε = R
F(e−γtuε)−F(e−γtuεapp)
:= ŵε; with wε = uε − uεapp.
Refering to (3.4), wε is the solution of the Cauchy problem:
(3.6)
Hwε +
Mwε1x<0 = εr
wε|t<0 = 0 .
For a fixed positive ε, the perturbation is nonsingular and thus the principal
part of the pseudodifferential operator H+ 1
M is the same as the principal
part of H. Hence, there is a unique solution of the Cauchy problem (3.6):
wε which belongs to Hk−3γ (ΩT ). In order to simplify the notations, in this
chapter we shall denote by L2 and H
γ the spaces: L
2(ΩT ) and H
γ (ΩT ).
We recall the definition of the weighted spaces: H
γ (ΩT ) for ρ ∈ N.
H̺γ (ΩT ) = {̟ ∈ e
γtL2(ΩT ), ‖̟‖H̺γ (ΩT ) < ∞};
where
γ (ΩT )
α,|α|≤̺
γρ−|α|‖e−γt∂α̟‖2
L2(ΩT )
For fixed positive ε, there holds:
∂x〈SRê
ε, êε〉L2 dx = 0.
2Re〈SR∂xê
ε, êε〉L2 dx = 0.
Using the equation, we have then:
Re〈SRARê
−êε − εr̂ε, êε〉L2 dx = 0.
which is equivalent to:
Re〈SRARê
ε, êε〉L2 dx =
Re〈P−êεεr̂ε, êε〉L2 dx
Re〈r̂ε, êε〉L2) dx.
But Re〈SRARê
ε, êε〉 = 〈Re (SRAR) ê
ε, êε〉 and Re (SRAR) is positive defi-
nite, hence there is C > 0, independent of ε such that:
Cγ‖êε‖2
Re〈P−êε, êε〉 ≤
Re〈εr̂ε, êε〉L2 dx.
Thus, because P− is an orthogonal projector, for all positive λ, there holds:
Cγ‖êε‖2
‖P−êε‖2
‖êε‖2
‖εr̂ε‖2
Choosing λ big enough we have C− ε
> 0 and the following energy estimate:
γ‖êε‖2
‖P−êε‖2
‖r̂ε‖2
This shows that êε converges towards zero in L2 when ε tends to zero, with
a rate in O(ε). We recall that êε is given by:
êε := RF
e−γt(uεapp − u
and r̂ε is given by:
r̂ε := RF
e−γtrε
Moreover, since R and P− are two uniformly bounded, uniformly definite
positive matrices, there are two positive real numbers α and β such that,
for all ζ 6= 0 and x ∈ R, there holds:
• α‖F
e−γt(uεapp − u
≤ ‖êε‖2
• α‖P−F
e−γt(uεapp − u
≤ ‖P−êε‖2
• ‖r̂ε‖2
≤ β‖F
e−γtrε
Applying then Plancherel’s equality we obtain then:
γ‖uεapp − u
eγtL2
uεapp − u
eγtL2
‖rε‖2
eγtL2
We have thus proved there are two positive constants c and C such that:
cγ‖uεapp − u
eγtL2
uεapp − u
eγtL2
‖rε‖2
eγtL2
Let us denote by ‖.‖∗
+ ‖.‖2
. More generally, when
rε ∈ H̺, there is two positive constants cρ and Cρ such that:
cργ‖u
app − u
uεapp − u
‖rε‖∗2
As we have seen during the construction of the profiles, ̺ = k− 3 in general
and ρ = k − 3
in the case where g = 0.
3.3 End of the proof of Theorem 1.6.
As a consequence of our stability estimate, there holds:
‖uεapp − u
Hk−3(Ω
+ ‖uεapp − u
Hk−3(Ω
= O(ε2).
Moreover, by construction of uεapp, there holds:
‖uεapp − u
Hk−3(Ω
+ ‖uεapp − u‖
Hk−3(Ω
= O(ε2).
Hence, we have proved that:
‖uε − u−‖2
Hk−3(Ω
+ ‖uε − u‖2
Hk−3(Ω
= O(ε2).
By the same arguments, if g = 0, there holds:
‖uε − u−‖2
+ ‖uε − u‖2
= O(ε2).
This completes the proof of Theorem 1.6.
4 Proof of Theorem 1.9.
Like in the proof of Theorem 1.6, we begin by constructing formally an
approximate solution of equation (1.7). We prove then suitable energy esti-
mates that ensures both uε and its approximate solution converges towards
ũ as ε → 0+.
4.1 Construction of an approximate solution.
The goal of this Lemma is to replace the boundary condition Γu|x=0 = Γg of
problem (1.1) by a condition of the form P−(e−γtu)|x=0 = h with a suitable
h ∈ Hk(ΥT ).
Lemma 4.1. Let u denote the unique solution in Hk(Ω+
) of the mixed
hyperbolic problem (1.1), P+(∂t, ∂y, γ)
e−γtu
does not depend of the choice
of the boundary operator Γ and of g. Let us introduce the function h of
Hk(ΥT ) defined by:
e−γtv|x=0
e−γt(g − v|x=0)
The solution u of the mixed hyperbolic problem (1.1) is the unique solution
of the following well-posed mixed hyperbolic problem (4.1):
(4.1)
Hu = f, {x > 0},
−(∂t, ∂y, γ)
e−γtu|x=0
u|t<0 = 0 .
In addition, the mapping (f, g) → h is linear continuous from
Hk(Ω+
)×Hk(ΥT ) to H
k(ΥT ).
Proof. Let v denote a solution in Hk(ΩT ) of the equation:
Hv = f, (t, y, x) ∈ ΩT ,
v|t<0 = 0 .
We introduce then U which is, by definition, the solution of the following
mixed hyperbolic problem:
HU = 0, {x > 0},
Γ(∂t, ∂y, γ)U|x=0 = Γ(∂t, ∂y, γ)g − Γ(∂t, ∂y, γ)v|x=0,
U|t<0 = 0 .
The right hand side of the boundary condition is, a priori,
in Hk−
2 (ΥT ). Hence the solution U belongs to H
2 (Ω+
). By construction
we have:
(4.2) u = U+ v.
Hence, since u ∈ Hk(Ω+
) and v ∈ Hk(Ω+
), in fact we have:
U ∈ Hk(Ω+
Let Û denote the Fourier-Laplace transform in (t, y) of U (Fourier-Laplace
transform tangential to the boundary) given by: F(e−γtU). It satisfies the
following symbolic equation:
∂xÛ = A(ζ)Û, {x > 0},
Γ(ζ)Û|x=0 = Γ(ζ)ĝ − Γ(ζ)v̂|x=0,
where ĝ and v̂ denotes respectively the tangential Fourier-Laplace transform
of g and v. Since A(ζ) is independent of x, projecting the above equation on
E+(A(ζ)) gives then:
Û = A(ζ)P+Û.
Moreover P+Û|x=0 ∈ E−(A(ζ))
E+(A(ζ)) since limx→∞P
Û = 0. Hence,
there holds:
Û = 0,
and thus
Û = P−Û.
The boundary condition:
Γ(ζ)Û|x=0 = Γ(ζ)ĝ − Γ(ζ)v̂|x=0
is equivalent to:
Û|x=0 ∈ ĝ − v̂|x=0 +KerΓ.
We have thus:
Û|x=0 ∈ ĝ − v̂|x=0 + ker Γ.
Let us denote by Π the projector on Ẽ−(A) parallel to ker Γ, which has a
sense because the Uniform Lopatinski Condition holds.
Since Û|x=0 ∈ Ẽ−(A), and of the Uniform Lopatinski Condition, the above
boundary condition is equivalent to:
Û|x=0 = Π(ĝ − v̂|x=0),
and thus, because P−Û|x=0 belongs to E−(A), we have:
Û|x=0 = Π(ĝ − v̂|x=0).
As a consequence, we obtain:
−û|x=0 = P
−v̂|x=0 +Π(ĝ − v̂|x=0).
Hence, there holds:
e−γtu|x=0
e−γtv|x=0
e−γt(g − v|x=0)
:= h.
P+(∂t, ∂y, γ)
e−γtu
= P+(∂t, ∂y, γ)
e−γtv
, thus it does not depend of the
choice of the boundary operator Γ and of g.Moreover, since u|x=0 ∈ H
k(ΥT ),
it follows that
g ∈ Hk(ΥT ). Now, since the Uniform Lopatinski Condition holds, u satisfies
the following energy estimate:
eγtL2(Ω
+ ‖u|x=0‖
eγtL2(ΥT )
≤ γ‖f‖2
eγtL2(Ω
+ ‖g‖eγtL2(ΥT ),
More generally, we have:
Hkγ (Ω
+ ‖u|x=0‖
Hkγ (ΥT )
≤ γ‖f‖2
Hkγ (Ω
+ ‖g‖Hkγ (ΥT ).
where ‖̟‖2
|α|=0 γ
k−|α|‖∂α̟‖2
eγtL2
h = P−(e−γtu|x=0) hence
L2(ΥT )
≤ C‖e−γtu|x=0‖
L2(ΥT )
= C‖u|x=0‖
eγtL2(ΥT )
and for 0 ≤ j ≤ d− 1, there holds:
‖∂jh‖
L2(ΥT )
≤ cj‖ηjP
−F(e−γtu)|x=0‖ ≤ c
j‖u|x=0‖H1γ(ΥT ).
More generally, we have:
Hkγ (ΥT )
≤ Ckγ‖f‖
Hkγ (Ω
+ Ck‖g‖
Hkγ (ΥT )
But γ is a positive real number fixed once and for all at the beginning of the
paper, hence this proves that the mapping (f, g) → h is continuous from
Hk(Ω+
)×Hk(ΥT ) to H
k(ΥT ). ✷
As we will see, Lemma 4.1 is central in our construction of an approximate
solution. We will construct an approximate solution
uεapp := u
app1x>0 + u
app1x<0,
along the following ansatz:
uε+app :=
εju+j (t, y, x),
with u+j ∈ H
), u+j |x=0 ∈ H
γ (ΥT ); and
uε−app :=
j (t, y, x),
with u−j ∈ H
), u−j |x=0 ∈ H
γ (ΥT ). As usual, we will refer to the
terms u±
as profiles. We will rather work on the reformulation of
problem (1.7) as the transmission problem (4.3):
(4.3)


Huε+ = f, {x > 0},
Huε− +
−e−γtuε− =
γth̃, {x < 0},
uε+|x=0 − u
ε−|x=0 = 0,
uε±|t<0 = 0 .
Plugging uε+app and u
app in (4.3) and identifying the terms with same power
in ε, we obtain the following profiles equations:
• Identification of the terms of order ε−1 :
(4.4) Ade
−e−γtu
0 = Ade
γth̃, {x < 0}.
• Identification of the terms of order ε0 :
(4.5) Hu−0 +Ade
−e−γtu−1 = 0, {x < 0}.
(4.6) Hu+0 = f, {x > 0},
• Identification of the terms of order εj for j ≥ 1 :
(4.7) Hu
j +Ade
−e−γtu
j+1 = 0, {x < 0}.
(4.8) Hu
j = 0, {x > 0},
• Translation of the continuity condition over the boundary on the pro-
files:
For all 1 ≤ j ≤ M, there holds:
(4.9) u
j |x=0 − u
j |x=0 = 0.
Denote by û±j := F(e
−γtu±j ) . We have then:
j := e
γtF−1(û±j ).
We will now give the equations satisfied by the Fourier-Laplace transform
of the profiles: û±j . First, equation (4.4) is equivalent to:
−û−0 = F(h̃), {x < 0}.
We deduce from this equation that there holds:
−û−0 |x=0 = ĥ.
Then, using (4.9) for j = 0, and (4.6) gives that, for γ big enough,
0 = F(e
−γtû
where û+0 is the solution of the well-posed first order ODE in x:
0 −Aû
0 = F(e
−γt(Ad)
−1f), {x > 0},
−û+0 |x=0 = h,
Thus u+0 is solution of:
Hu+0 = f, {x > 0},
eγtP−e−γtu+0 |x=0 = h.
Thanks to Lemma 4.1, we recognize u+0 as the solution of our starting mixed
hyperbolic problem (1.1). Once u+0 is known, so is û
0 and thus û
0 |x=0 is
given by:
û−0 |x=0 = û
0 |x=0.
Moreover,
u+0 |x=0 = u
0 |x=0 ∈ H
γ (ΥT ).
By (4.5), there holds:
0 −Aû
−û−1 = 0, {x < 0}.
As a consequence, P+û−0 is given by the well-posed ODE:
0 )−A(P
0 ) = 0, {x < 0},
+û−0 |x=0 = P
+û+0 |x=0.
Indeed, since kerP+(ζ) = E−(A(ζ)), this problem satisfies the Uniform
Lopatinski Condition: for all ζ 6= 0, there holds:
E−(A(ζ))
E+(A(ζ)) = C
For γ big enough, by linearity of the inverse Fourier transform, u−0 can then
be computed by:
u−0 := e
γtF−1(P−û−0 ) + e
γtF−1(P+û−0 ).
Following up with that process of construction, we can go on with the con-
struction of the profiles at any order. Indeed, assume that all the profiles
(u+j , u
j ) up to order j have been computed. Then consider the equation
obtained through identification:
j+1 = −∂xû
j +Aû
j , {x < 0}.
We see there is a loss of regularity between û−j+1 and û
Let us say that u±j ∈ H
). Considering the traces, we have then:
j |x=0 ∈ H
γ (ΥT ). We will show in this part how the Sobolev regularity
of the profiles u±j+1, which is by definition mj+1, can be computed know-
ing mj . To begin with P
j+1 belongs to H
). P−u+j+1|x=0, which
belongs to H
γ (ΥT ), is known by P
j+1|x=0 = e
γtF−1(P−û+j+1|x=0),
with:
j+1|x=0 = P
j+1|x=0.
Hence, û+j+1 := F(e
−γtu+j+1) is the solution of the first order ODE in x :
j+1 −Aû
j+1 = 0, {x > 0},
−û+j+1|x=0 = P
−û−j+1|x=0.
Since kerP−(ζ) = E+(A(ζ)), this problem satisfies the Uniform Lopatinski
Condition: for all ζ 6= 0, there holds:
E−(A(ζ))
E+(A(ζ)) = C
As a consequence, this problem is well-posed and, u+j+1 ∈ H
).More-
over, there holds:
u+j+1|x=0 = u
j+1|x=0 ∈ H
γ (ΥT ).
Indeed, P+û+j+1 ∈ H
∞(Rd+1+ ) hence P
+u+j+1|x=0 ∈ H
γ (ΥT ) and thus
u+j+1|x=0 ∈ H
γ (ΥT ). Furthermore, we have:
u−j+1|x=0 = u
j+1|x=0.
Applying P+ on the following equation:
−û−j+2 = −∂xû
j+1 +Aû
j+1, {x < 0};
we obtain then the equation:
j+1)−AP
j+1 = 0, {x < 0}.
Remark 4.2.
−û−j+2 = −∂xû
j+1 +Aû
j+1, {x < 0}.
shows that the ”Fourier profile” û−j+1 must be so that −∂xû
j+1 + Aû
j+1 is
polarized on E−(A). It is indeed the case because we search for û
j+1 satisfy-
+û−j+1)−AP
+û−j+1 = 0, {x < 0}.
j+1 is given by:
j+1 := e
γtF−1(P−û−j+1) + e
γtF−1(P+û−j+1).
with P+u−j+1 = e
γtF−1(P+û−j+1) belongs to H
) and is the unique
solution of the well-posed first order ODE:
+û−j+1)−A(P
+û−j+1) = 0, {x < 0},
+û−j+1|x=0 = P
+û+j+1|x=0.
The profile u−j+1 belongs to H
). This achieves to show that the
knowledge of (u+j , u
j ), allows us to compute (u
j+1, u
j+1).
Moreover mj+1 = mj −
, that is to say that a construction of each supple-
mentary profile consummate 3
of Sobolev regularity. In practice, we take:
uε+app = u
uε−app = u
0 + εu
As a result, the approximate solution writes uεapp := u
app1x>0 + u
app1x<0;
where uε+app belongs to H
) and uε−app belongs to H
). The so de-
fined uεapp is solution of a well-posed problem of the form:
(4.10)
Huεapp +
−e−γtuεapp1x<0 = f1x>0 +
γth̃1x<0 + εr
uεapp|t<0 = 0 .
Where rε := rε+1x>0 + r
ε−1x<0, with r
ε+ ∈ H
) and
rε− ∈ Hk−3γ (Ω
4.2 Asymptotic Stability of the problem as ε tends towards
zero.
Denote by vε = uεapp − u
ε. By construction of uεapp, v
ε is solution of the
following Cauchy problem:
(4.11)
Hvε +
−e−γtvε1x<0 = εr
vε|t<0 = 0 .
For all positive ε, this problem is well-posed. In order to investigate the
stability of this problem as ε goes to zero, we will reformulate it as a trans-
mission problem. The restrictions of vε to {x > 0} and {x < 0}, respectively
denoted by vε+ and vε− are solution the following transmission problem:
(4.12)


Hvε+ = εrε+, {x > 0},
Hvε− +
−e−γtvε− = εrε−, {x < 0},
vε+|x=0 − v
ε−|x=0 = 0,
vε±|t<0 = 0 .
Let us denote by V ε the function, valued in R2N , defined for all {x > 0}
and (t, y) ∈ [0, T ]× Rd−1 by:
V ε(t, y, x) =
V ε+(t, y, x)
V ε−(t, y,−x)
vε is solution of the Cauchy problem (4.11) iff V ε is solution of the mixed
hyperbolic problem on a half space (4.13) given below:
(4.13)
H̃V ε +BεV ε = εRε, {x > 0},
Γ̃V ε|x=0 = 0,
V ε|t<0 = 0 ,
where
H̃ = ∂t +
0 −Ad
γtP−e−γt
Rε(t, y, x) =
rε+(t, y, x)
rε−(t, y,−x)
Id −Id
Returning to the construction of our approximate solution, we have
Rε ∈ H
)×Hk−3γ (Ω
) and is such that Rε|t<0 = 0.
In fact Rε ∈ Hk
) with k′ = k−3. For all positive ε, there exists a unique
solution V ε in Hkγ (Ω
) to the above problem. We will prove here that this
solution converges, uniformly in ε, towards 0 in Hk
), as ε vanishes. As
in the proof of Kreiss Theorem, see [3] for instance, existence of solution
for mixed hyperbolic systems like (1.7) or (4.13), are obtained through the
proof of both direct and ”dual”a priori estimates on an adjoint problem. This
estimates results in the constant coefficient case of estimates on the Fourier-
Laplace transform of the solution. Additionally, if this ”Fourier” estimate
can be proved, both direct and ”dual” energy estimates are deduced from
it. In a first step, let us recall formally how to conduct the Fourier-Laplace
transform of a mixed hyperbolic problem:
Hu = f, {x > 0},
Γu|x=0 = g,
u|t<0 = 0 ,
Denote by u∗ := e
−γtu, u∗ is in particular a solution of the following problem:
Hu∗ + γu∗ = e
−γtf, {x > 0},
Γu∗|x=0 = g .
We take then the tangential (with respect to (t,y)) Fourier transform of this
equation, which gives:
Ad∂xû∗ + (γ + iτ)û∗ + iηj
Aj û∗ = F
e−γtf
, {x > 0},
Γû∗|x=0 = ĝ .
Multiplying this equation by A−1
, we obtain that u∗ is solution of the fol-
lowing ODE in x:
∂xû∗ −Aû∗ = (Ad)
e−γtf
, {x > 0},
Γû∗|x=0 = ĝ .
Note that, û∗ and u can be freely deduced from each other through the
formulas:
û∗ = F(e
−γtu)
u = eγtF−1(û∗).
We shall now introduce a rescaled solution V ε of the solution V ε of
(4.13) defined as follows: V ε(t, y, x) := V ε(t, y, εx), and the rescaled remain-
der: Rε(t, y, x) := Rε(t, y, εx). Denoting by V̂
= F(e−γtV ), the associated
equation writes then:
− εÃV̂
= ε2R̂ε, {x > 0},
|x=0 = 0 .
where
M(ζ) =
0 P−(ζ)
We remark that
εÃ(ζ) = Ã(εζ) = Ã(ζ̂),
with ζ̂ = (τ̂ , γ̂, η̂) := εζ. Moreover P− is homogeneous of order zero in ζ.
Let us denote R̃ε(ζ̂ , x) := R̂ε(ζ, x) and Ṽ
(ζ̂ , x) := V̂
(ζ, x). Hence Ṽ
solution of the following problem:
−Ã(ζ̂) +M(ζ̂)
= ε2R̃ε(ζ̂ , x), {x > 0},
|x=0 = 0 .
As a consequence, the Uniform Lopatinski Condition for problem (4.13)
writes: For all γ̂ > 0,
|det(E−(Ã(ζ̂)−M(ζ̂), ker Γ)| ≥ C > 0.
In view of the proof of the Proposition (4.3), we recall that the spaces E±(A)
have to be considered in the extended sense defined above.
Proposition 4.3. Since H satisfies the hyperbolicity Assumption in As-
sumption 1.1, the Uniform Lopatinski Condition is satisfied for our present
problem; that is to say that, for all ζ̂ such that γ̂ > 0 there holds:
|det(E−(Ã(ζ̂)−M(ζ̂), ker Γ)| ≥ C > 0.
Proof. We will begin to show that the Uniform Lopatinski Condition writes
as well that for all ζ̂ 6= 0 there holds:
(4.14) E+(A(ζ̂)−P
−(ζ̂))
E−(A(ζ̂)) = {0} .
This notation keeps a sense for ζ̂ such that γ̂ = 0 because we will prove
a posteriori that the involved linear subspaces continuously extends from
{ζ̂ , γ̂ > 0} to {ζ̂ , γ̂ = 0}. Then we will prove that, for all ζ̂, the property
4.14 holds true. The Uniform Lopatinski Condition writes actually, for all
ζ̂ 6= 0 :
E−(Ã(ζ̂)−M(ζ̂))
ker Γ̃ = {0}.
and thus, since we have:
E−(Ã(ζ̂)−M(ζ̂)) = E−(A(ζ̂))× E+(A(ζ̂)−P
−(ζ̂)),
and by definition of Γ̃, the Uniform Lopatinski Condition writes then that,
for all ζ̂ 6= 0, there holds:
E+(A(ζ̂)−P
−(ζ̂))
E−(A(ζ̂)) = {0}.
Lemma 4.4.
A(ζ̂)−P−(ζ̂)
A(ζ̂)
A(ζ̂)−P−(ζ̂)
A(ζ̂)
Proof. For all ζ̂ 6= 0, there is an invertible N × N matrix with complex
coefficients P (ζ̂) such that: P−1AP is trigonal and the diagonal coefficients
are sorted by increasing order of their real parts. Let us denote by ν the
dimension of E− (A) . The above matrix P traduces the change of basis from
the canonical basis of CN into (v1, . . . , vν , vν+1, . . . , vN ), where
Span ((vk)1≤k≤ν) = E− (A) ,
Span ((vk)ν+1≤k≤N ) = E+ (A) .
Moreover, there holds
P−1P−P = D
where D is the diagonal matrix whose ν first diagonal terms are equal to 1
and the N − ν last diagonal terms are null.
P−1(A−P−)P = P−1AP −D.
P−1AP−D is also trigonal, with the same eigenvalues with positive real part
as P−1AP and the same eigenvalues with negative real part as P−1AP −Id.
As a consequence, for all ζ̂ 6= 0, there holds:
A(ζ̂)−P−(ζ̂)
A(ζ̂)
A(ζ̂)−P−(ζ̂)
A(ζ̂)
As a consequence of Lemma 4.4, the rescaled Uniform Lopatinski Con-
dition for ε > 0, ε → 0 happens to be exactly the same as the one written
for bigger positive ε. Indeed, it writes, for all ζ̂ 6= 0 :
E+(A(ζ̂))
E−(A(ζ̂)) = {0}.
✷ The Lopatinski condition is satisfied, and, as a result, the following,
uniform in ε, energy estimate holds for V ε, for all γ ≥ γk′ > 0 :
γ‖V ε‖2
+ ‖V ε|2x=0‖Hk′γ (ΥT )
‖εRε‖2
which is equivalent to:
(4.15) γ‖V ε‖2
+ ‖V ε|x=0‖
γ (ΥT )
‖εRε‖2
This proves the convergence of V ε towards zero in Hk
). The weight
γ is fixed beforehand thus, in fact, the solution of (4.13) tends to zero in
) at a rate at least in O(ε).
5 End of proof of Theorem 1.9.
Let us consider V ε defined by:
V ε(t, y, x) :=
uε+app(t, y, x)− u
ε+(t, y, x)
uε−app(t, y,−x)− u
ε−(t, y,−x)
This notation is perfectly fine because the so-defined function is solution of
an equation of the form (4.13). Moreover, thanks to the stability estimate
(4.15), there is γk positive such that, for all γ > γk, we have:
γ‖uεapp−u
+γ‖uεapp−u
+‖uεapp−u
γ (ΥT )
‖εRε+‖2
Hence, it follows that:
‖uεapp − u
Hk−3(Ω
+ ‖uεapp − u
Hk−3(Ω
= O(ε2).
Moreover, by construction of uεapp, we have:
‖uεapp − u‖
Hk−3(Ω
+ ‖uεapp − u
Hk−3(Ω
= O(ε2).
As a result, we obtain that there holds:
‖uε − u‖2
Hk−3(Ω
+ ‖uε − u−‖2
Hk−3(Ω
= O(ε2).
This concludes the proof of Theorem 1.9.
6 Appendix: answer to a question asked in [11].
In this chapter, we will show that the loss of convergence observed numeri-
cally in [11] in a neighborhood of the boundary is due to a boundary layer
phenomenon. We consider the 1-D wave equation:
(6.1)
∂ttU − c
2∂xxU = 0, (x, t) ∈]0, π[×R
U |x=0 = U |x=π = 0,
U |t=0(x) = sin(x),
∂tU |t=0 = 0.
As in [11], we define then U ε = U ε+1x>0 + U
ε−1x<0 by:
(6.2)


ε+ − c2∂xxU
ε+ = 0, (x, t) ∈]0, π[×R+,
ε− − c2∂xxU
U ε− = 0, (x, t) ∈]−∞, 0[×R+,
U ε+|x=0 − U
ε−|x=0 = 0
ε+|x=0 − ∂xU
ε−|x=0 = 0
U ε+|x=π = 0.
U ε±|t=0(x) = sin(x), {±x > 0}.
ε±|t=0 = 0, {±x > 0}.
We will now construct formally an approximate solution U ε±app of U
ε± satis-
fying the following ansatz:
U ε+app =
U+j (t, x)ε
U ε−app =
t, x,
where the profiles U−j (t, x, z) := U
j (t, x) + U
j (t, z), with
e−αzU∗−j = 0,
for some α > 0. Since the stability estimates are trivial here, we will only
focus on the construction of
U εapp := U
app1x>0 + U
app1x<0.
Plugging U ε±app into problem (6.2) and identifying the terms with same power
of ε, we obtain the following equations:
U−0 = 0,
Moreover, U∗−0 = 0 as it is the only solution of the problem:
U∗−0 − c
2∂zzU
0 = 0, {z < 0},
0 |z=0 = 0,
0 = 0.
U ε+app converges towards U
0 as ε → 0
+. As awaited U+0 is the solution of the
well-posed 1-D wave equation:


0 − c
2∂xxU
0 = 0, (x, t) ∈]0, π[×R
U+0 |x=0 = U
0 |x=0 + U
0 |z=0 = 0.
U+0 |x=π = 0.
0 |t=0(x) = sin(x), {x > 0}.
0 |t=0 = 0, {x > 0}.
Let us write the following profiles equations: First, we can see that, for all
j ≥ 1, there holds:
where U∗−1 is the solution of the well-posed profile equation:
1 − c
2∂zzU
1 = −∂ttU
0 = 0, {z < 0},
1 |z=0 = ∂xU
0 |x=0,
U∗−1 = 0.
Hence U∗−1 is given by:
U∗−1 = c∂xU
0 |x=0e
We will show now that the profiles can be computed as any order. Assume
that U∗−j has been computed, U
j is solution of the well-posed 1-D wave
equation: 


− c2∂xxU
= 0, (x, t) ∈]0, π[×R+,
j |x=0 = U
j |z=0.
U+j |x=π = 0.
U+j |t=0(x) = 0, {x > 0}.
0 |t=0 = 0, {x > 0}.
U∗−j+1 is then solution of the well-posed profile equation:
U∗−j+1 − c
2∂zzU
j+1 = −∂ttU
j , {z < 0},
j+1|z=0 = ∂xU
|x=0,
U∗−j+1 = 0.
Let us answer the question asked in [11]: U ε− is bound to present boundary
layer behavior in {x = 0−}, indeed its approximate solution is composed
exclusively of boundary layer profiles, which describes quick transitions at
the boundary using a fast scale in ε. As a result of the loss in convergence
induced by the boundary layer, the following estimate holds:
‖U ε − U‖L2(]−∞,π[×R+) = O(ε
In [11], their small parameter is µ = ε2, as a result, adopting the same
notations as them, our estimate writes:
‖Uµ − U‖L2(]−∞,π[×R+) = O(µ
which is in agreement with the estimates given in [11]. Like in the penaliza-
tion approach proposed by Bardos and Rauch [2] and underlined by Droniou
in [4], the boundary layer only forms on one side of the boundary. The ap-
proximation U ε+ of U, is computed by taking U ε+|x=0 = U
ε−|x=0, thus, in
numerical applications, the boundary layer phenomenon also affects the rate
of convergence of U ε+ towards U,
as ε → 0+.
References
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pp 377-408.
[3] J. Chazarain, A. Piriou, Introduction to the theory of linear partial dif-
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	Introduction.
	A Kreiss Symmetrizer Approach.
	A second Approach.
	Underlying approach leading to the proof ofTheorem ??.
	Some preliminaries.
	Detailed proof of Lemma ??: Construction of the matrices B solving Lemma ??.
	A change of dependent variables.
	Proof of Theorem ??.
	Construction of the approximate solution.
	Stability estimates
	End of the proof of Theorem ??.
	Proof of Theorem ??.
	Construction of an approximate solution.
	Asymptotic Stability of the problem as  tends towards zero.
	End of proof of Theorem ??.
	Appendix: answer to a question asked in paccou.
