Benzoni-Gavage S., Serre D. Multi-dimensional Hyperbolic Partial Differential Equations: First-order Systems and Applications

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Energy estimates 235

Clearly, (L

) is an open condition. In other words, the constant C is locally

uniform in

. Furthermore, by homogeneity degree 0 of E

−

, the condition (L

)

is equivalent (with the same constant C)to(L

), where X

∈

is deﬁned by

=(y, 0,t,η

,τ

), (η

,τ

)=(τ,η)/(τ,η) if X =(y, 0,t,η,τ).

Remark 9.5 As already observed in Remark 9.4 above, the stable sub-

space E

−

(y, 0,t,0, 1) of A(y, 0,t,0, 1) coincides with the stable subspace

(A(y, 0,t,ν(y, 0))). Therefore, the condition (L

)atX =(y, 0,t,0, 1) requires

in particular that the intersection of kerB(y, t)andE

(A(y, 0,t,ν(y, 0)) be

zero. If we also know that the rank of B(y,t) equals the dimension of

(A(y, 0,t,ν(y, 0)), the normality condition in (9.0.1) is just a reformulation of

(y,0,t,0,1)

): this is why there will be no need to mention (9.0.1) in the assump-

tions of the main theorems ((9.0.1) will be a consequence of those assumptions).

One may observe additionally that (L

(y,0,t,0,1)

) amounts to a one-dimensional

stability condition, in which no transversal modes (in e

iηy

) are considered.

The uniform Kreiss–Lopatinski˘ı condition

The extension of (L

)topointsX with Re τ = 0 looks the same but with

a careful deﬁnition of E

−

(X), no longer the stable subspace of A(X). For,

the matrix A(y, 0,t,η,τ) with Re τ = 0 is no longer hyperbolic in general.

This is where the so-called neutral modes come into play: by deﬁnition, the

time behaviour of neutral modes is e

τt

with Re τ = 0; but in fact only those

modes with amplitude in E

−

(X) are to be considered, with E

−

(X) deﬁned by

continuous extension of the projector P

−

(X)asE

−

(X)=Im(P

−

(X)).

Lemma 9.1 Assume the operator L is constantly hyperbolic, for all X

∈ X

Then there exists a projector P

−

(X) of rank p such that

−

(X) = lim

→X

−

(X) .

This innocent-looking result is highly non-trivial: observe indeed that in gen-

eral some eigenvalues of A(y, x

,t,η,τ) cross the contour Γ as Re τ approaches

zero. The proof for a constant-coeﬃcients operator L is given in Chapter 5.

A careful look at the arguments shows that they remain valid for variable

coeﬃcients. The details are left to the reader.

Remark 9.6 In general, for X =(y, 0,t,η,τ) ∈ X

with Re τ = 0, the actual

stable subspace of A(X) is strictly embedded in E

−

(X), which also contains a

part of the center subspace of A(X).

Once the subspace E

−

(X) is properly deﬁned at all points X ∈ X

,the

Lopatinski˘ı condition at those points still reads as (L

Now, what we call the uniform Kreiss–Lopatinski˘ı condition is merely the

following.

(UKL) The condition (L

) is satisﬁed for all X ∈ X

236 Variable-coeﬃcients initial boundary value problems

Remark 9.7 The term uniform attached to Kreiss–Lopatinski˘ı refers to the

fact that the constant C in (L

) may be chosen to be uniform in X

. Indeed,

for X =(y, 0,t,η,τ) ∈ X

, the constant C in (L

) depends continuously on

(y, t, η, τ), is independent of (y, t) outside a compact subset of R

(provided that

L and B have constant coeﬃcients outside a compact set) and (τ, η) lies in the

compact set P

Practical veriﬁcation of (UKL)

The usual way to check whether (UKL) is satisﬁed consists in ﬁnding a basis

of E

−

(X), say (e

(X),...,e

(X)) (a classical argument of Kato [95], p. 99–101

implies that this basis can be chosen to depend smoothly on X) and in looking

for the zeroes of the so-called Lopatinski˘ı determinant

∆(X):=det(B(y, t)e

(X),...,B(y, t)e

(X)) .

As a matter of fact, the mapping B(y, t)

−

(X)

is an isomorphism if and only if

∆(X) = 0 and therefore (UKL) is clearly equivalent to

∆(X) =0 ∀X ∈ X

In practice, the Lopatinski˘ı determinant ∆(X) is usually derived for X ∈

ﬁrst, and then extended by continuity to points X =(y, 0,t,η,τ) such that

Re τ = 0. However, we warn the reader that this limiting procedure requires

some care to avoid introducing fake zeroes. In any case, we claim the search for

zeroes of ∆ is mostly algebraic. See Section 4.6 for more details (and also Chapter

13.2.18 for an actual example).

Failure of (UKL)

If a zero of ∆ is found in

, the well-posedness of the BVP (9.1.5)(9.1.6) is

hopeless: as was shown in Chapter 4 the existence of non-trivial normal modes

with Re τ>0 is responsible for a Hadamard instability.

If ∆ does not vanish on

but has a zero (y,0,t,η,τ) with Re τ =0,wesay

the BVP is weakly stable. The non-trivial kernel of B(y, t)

−

(y,0,t,η,τ)

implies the

existence of non-trivial neutral modes, which oscillate as e

τt

and are bounded

but not necessarily square-integrable transversally to the boundary: in this case

(UKL) fails but it is still possible to construct (weaker) Kreiss’ symmetrizers and

obtain energy estimates with (a limited) loss of derivatives; this goes beyond the

scope of this book and we refer to [41,170] for more details.

Construction of Kreiss’ symmetrizers

The hard part of the job is the construction of Kreiss’ symmetrizers for constant-

coeﬃcients problems: In general, it involves an intricate piece of matrix analysis

and algebraic geometry, see Chapter 5; It is easier in the speciﬁc case of gas

dynamics, see Chapter 13. Using the constant-coeﬃcients construction we can

indeed construct, for ﬁxed (y,x

,t) ∈ Y, Hermitian matrices r(y, x

,t,η,τ)

Energy estimates 237

and invertible matrices T (y, x

,t,η,τ) satisfying the inequalities in (9.1.16) and

(9.1.17) for all (τ,η) ∈ P

The main thing to do here is to allow variations of (y, x

,t) and check the

smoothness of r and T with respect to (x, t). In fact, the careful construction

made in Chapter 5 is robust under perturbation of coeﬃcients. We will not go

into details in all cases. For clarity, we just give the explicit construction in the

easiest cases, namely the ﬁrst and second one in the following list.

i) At points of

, Kreiss’ symmetrizers are easily found by a block reduction

of the matrix A(X) and the Lyapunov matrix theorem.

ii) At points X

∈ X

where the matrix A is smoothly diagonalizable –

which means there exist invertible matrices T (X) depending smoothly on

X in some neighbourhood of X

such that T (X)

−1

A(X)T (X) is diagonal –

Kreiss’ symmetrizers exist in diagonal form.

iii) At ‘generic’ points of X

, the construction of Kreiss’ symmetrizers is

undoubtedly cumbersome.

On the one hand, the smooth diagonalizability condition in ii) implies in

particular that the distinct eigenvalues of A(X) are smooth functions of X,

analytic in (η, τ), in some neighbourhood of X

. On the other hand, iii) comprises

the so-called case of glancing points, where two (or more) eigenvalues of A(X)

collide, and is much more involved. It requires a speciﬁc assumption, named

by Majda the block-structure condition, which is now known to hold true for

any constantly hyperbolic system [134]. For the case of systems with variable

multiplicities, we refer to the recent work of M´etivier and Zumbrun [135].

Remark 9.8 In gas dynamics, A(X) is smoothly diagonalizable at all points

except the glancing points, in such a way that case iii) reduces to those special,

glancing points, where the construction of a Kreiss’ symmetrizer is still rather

easy (see Chapter 13).

Construction of Kreiss symmetrizers in case i) Let us ﬁx X

∈

and a

neighbourhood V of X

where the projector P

−

(X)ontoE

−

(X) is well-

deﬁned, while P

(X):=I

− P

−

(X) is a projector onto the unstable subspace

(X)ofA(X). Using Kato’s argument [95], we can ﬁnd smooth bases of the

ranges of P

−

(X)andP

(X). (Here and in what follows, smooth always means

∞

in (y, x

,t) and analytic in (τ,η).) Then the matrix T (X)composedofthe

corresponding column vectors is invertible, depends smoothly on X ∈ V ,and

reduces A(X) to a block-diagonal form:

a(X):=T (X)

−1

A(X) T (X)=



−

(X) 0

0 A

(X)



where A

−

only has a spectrum of negative real part and A

only has a spectrum

of positive real part. We now invoke the following result, which is a variant of

the Lyapunov matrix theorem.

238 Variable-coeﬃcients initial boundary value problems

Lemma 9.2 If A is a matrix whose spectrum entirely lies in the open half-plane

{z ;Rez>0 }

there exists a unique positive-deﬁnite Hermitian matrix H such that

Re (HA)=I.

Furthermore, H admits the explicit representation

H =



−∞

∗

dt.

As a consequence, if A depends smoothly on a parameter, so does H.

Applying Lemma 9.2 to A

(X)and−A

−

(X) we obtain two positive-deﬁnite

Hermitian matrices H

(X)andH

−

(X), depending smoothly on X ∈ V such

that

Re (H

(X) A

(X)) = ±I.

Let us now consider a matrix r(X) of the form

r(X)=



−H

−

(X) 0

0 µH

(X)



with µ>0 to be determined. By construction of H

Re (r(X) a(X)) =



0 µI



So (9.1.16) is satisﬁed as soon as µ ≥ 1 (recall that for X ∈ X

, the last com-

ponent has a real part γ ≤ 1). This is all that we have to do if X

is not in

.IfX ∈

, the actual choice of µ will come from the fulﬁllment of (9.1.17).

By (UKL), up to diminishing V to a compact neighbourhood of X

, we ﬁnd a

constant C such that

P

−

(X) V 

≤ C



P

(X) V 

+ B(y, t) V 



for all V ∈ C

for all X ∈ V .

Because of the block structure of r(X), we have for all v =



−



∈ C

∗

r(X) v = −v

∗

−

(X) v

−

+ µv

∗

(X) v

≥v

−



+ bµv



−(1 + c) v

−



with b>0 such that H

≥ bI and c>0 such that H

−

≤ cI. Now, denoting

V = T (X) v we observe that

−

(X) V = T (X)



−



(X) V = T (X)





Energy estimates 239

and thus

T (X)

−1

P

(X) V ≤v

≤T (X)

−1

P

(X) V  .

Therefore

∗

r(X) v ≥T (X)

−2

( P

−

(X) V 

+ bµP

(X) V 

)

−(1 + c) T (X)

−1



P

−

(X) V 

≥T (X)

−2



P

−

(X)V 

+(bµ − C(1 + c)T (X)

−1



)P

(X)V 



−C (1 + c) T (X)

−1



B(y, t) V 

So if µ is chosen greater than

:= 2 C (1 + c)max

X∈V

T (X)

−1



/b ,

we have

∗

r(X) v ≥ α v

− β B(y, t) T (X) v 

for

α =

min

X∈V

T (X)

−2

min



bµ



and β =

bµ

This proves the estimate in (9.1.17).

Construction of Kreiss symmetrizers in case ii) It basically works as in

the previous case, by passing to the limit in the projection operators. Thus we

still have an estimate of the form

P

−

(X) V 

≤C



P

(X) V 

+ B(y, t) V 



for all V ∈ C

and X ∈ V ,

where V is a neighbourhood of X

in X. Reducing A(X) to diagonal form

a(X)=



−

(X) 0

0 A

(X)



with A

(X) diagonal and

±Re A

(X) ≥ CγI,

we get (9.1.16), and also (9.1.17) for µ large enough, merely by setting

r(X)=



−I

0 µI



240 Variable-coeﬃcients initial boundary value problems

Construction of Kreiss symmetrizers in case iii) We refer to Chapter 5.

See also [31].

In conclusion, we have the following.

Theorem 9.2 Under the following assumptions:

The operator L is constantly hyperbolic, that is, the matrices A(x, t, ξ) are

diagonalizable with real eigenvalues of constant multiplicities on Ω × R ×

\{0});

The boundary ∂Ω of Ω={x ∈ R

; x

> 0} is non-characteristic, in that

the matrix A

(x, t) is non-singular along ∂Ω × R,

The boundary matrix B is of constant, maximal rank equal to the number

of incoming characteristics along ∂Ω × R, which are deﬁned as the positive

eigenvalues of A

(counted with multiplicity),

The uniform Kreiss–Lopatinski˘ı condition (UKL) is satisﬁed;

there exists a Kreiss’ symmetrizer (according to Deﬁnition 9.3) at any point of

As a consequence of this theorem together with Proposition 9.1 and Theorem

9.1 we have the following.

Theorem 9.3 Under the assumptions of Theorems 9.1 and 9.2, for all s ∈ R,

there exists γ

≥ 1 and C

> 0 such that for all γ ≥ γ

and u ∈D(R

d−1

×R

×R),

γ u



))

+ u

γ | x



)

≤ C



L

u



))

+ Bu

γ | x



)



where u

(x, t)=e

−γt

u(x, t) and v

)

= Λ

s,γ

v

)

Remark 9.9 It is also possible to derive estimates in weighted Sobolev spaces

d−1

× R

× R):={u =e

γt

u

; u

∈ H

d−1

× R

× R) }

for m ∈ N. (Sometimes, we will use the (abuse of) notation e

γt

instead of

.) Recalling that for functions of (y, t) ∈ R

v

)





|α|≤m

2(m−|α|)

∂

v

)

(the sign  standing for two-sided inequalities with constants independent of γ

and v, see Remark C.1), hence also

e

−γt

u 

)





|α|≤m

2(m−|α|)

e

−γt

∂

u

)

Energy estimates 241

it is natural to equip H

(

O), for any open domain O of R

, with the norm

deﬁned by

u

(



|α|≤m

2(m−|α|)

e

−γt

∂

u

(O)

The result of Theorem 9.3 may be viewed as a L

; H

)) estimate:

γ u

))

+ u



)

≤ C



Lu

))

+ Bu



)



In fact, this also implies a H

d−1

× R

× R) estimate, as shown in the

following.

Corollary 9.1 Under the assumptions of Theorems 9.1 and 9.2, for all m ∈ N,

there exists γ

≥ 1 and C

> 0 such that for all γ ≥ γ

and u ∈ D (R

d−1

× R),

γ u

d−1

×R

×R)

+ u



)

≤ C



Lu

d−1

×R

×R)

+ Bu



)



Proof Since

u

d−1

×R

×R)

= e

−γt

u

d−1

×R

×R)

the case m = 0 is just a reformulation of the estimate in Theorem 9.3 (with

s =0).

For m ≥ 1, recall that

u

d−1

×R

×R)



|α|≤m

2(m−|α|)

e

−γt

∂

u

d−1

×R

×R)

We already have an estimate for the terms of this sum that do not involve

derivatives with respect to x

, for which α

= 0. It remains to bound the terms

for which α

≥ 1. This is done by using the equality

−γt

∂

u =(A

)

−1

( L

u

− γ u

− ∂

u

−

d−1



j=1

∂

u

) , (9.1.20)

which already implies the H

estimate. Indeed, we have

e

−γt

∂

u

d−1

×R

×R)

≤ C



L

u



d−1

×R

×R)

+ C



u



))

where C



> 0 depends only on (A

)

−1



∞

and A



∞

.Since

u

d−1

×R

×R)

= u

))

+ e

−γt

∂

u

d−1

×R

×R)

242 Variable-coeﬃcients initial boundary value problems

we get by using the L

; H

)) estimate,

γ u

d−1

×R

×R)

+ u



)

≤

Lu

))

+ C Bu



)



e

−γt

Lu

d−1

×R

×R)

+2C



γ u

))

hence, using again the L

; H

)) estimate,

γ u

d−1

×R

×R)

+ u



)

≤ max (C(1 + 2C



), 2C



)

Lu

d−1

×R

×R)

+ C(1 + 2C



) Bu



)

More generally, the H

estimates for k ≤ m can be proved by induction, dif-

ferentiating the equality (9.1.20). Details are omitted (see the proof of Theorem

9.7, which even shows H

estimates for less-smooth coeﬃcients). 

9.1.4 Non-planar boundaries

Theorem 9.3/Corollary 9.1 have natural extensions to more general domains Ω.

At ﬁrst, we may consider a domain Ω diﬀeomorphic to a half-space. Since a

change of variables preserves the constant hyperbolicity of the operator (see, for

instance, the proof of Theorem 2.10 in Chapter 2) and the non-characteristicness

of the boundary, we get an extended version of Theorem 9.3 (for s =0) by

changing the other assumptions accordingly. In particular, the uniform Kreiss–

Lopatinski˘ı condition has to do with points in the set

:= {X =(x, t, ξ, τ); (x, ξ) ∈ T

∗

∂Ω ,τ∈ C

, |τ |

+ ξ

=1},

where T

∗

∂Ω is the cotangent bundle of ∂Ω. (We recall that τ ∈ C

means Re τ ≥

0.) As done before, we denote by ν the exterior unit normal vector ﬁeld on ∂Ω.

Then, for all X ∈ X

we denote by E

−

(X) the stable subspace of

A(X):=A(x, t, ν(x))

−1

( τI

+ iA(x, t, ξ)).

In this generalized framework, the condition (L

) reads as before:

) There exists C>0 so that for all V ∈ E

−

(X), V ≤C B(y, t) V .

Theorem 9.4 We assume that

Ω is diﬀeomorphic to a half-space. Other

assumptions are, the coeﬃcients (in the operator L and in the boundary matrix

B) are constant outside a compact subset of

Ω × R and furthermore:

(CH) The operator L is constantly hyperbolic;

(NC) The boundary ∂Ω is non-characteristic;

Energy estimates 243

(N) The boundary matrix B(x, t) is of constant, maximal rank p =

dim E

(A(x, t, ν(x))) for (x, t) ∈ ∂Ω × R;

(UKL) (L

) holds for all X ∈ X

Then there exists C>0 such that for all u ∈ D (

Ω × R) and all γ ≥ 1,

γ u



(Ω×R)

+ u



(∂Ω×R)

≤ C



L

u



(Ω×R)

+ Bu



(∂Ω×R)



(9.1.21)

where u

(x, t)=e

−γt

u(x, t).

Proof If

Φ:

Ω → D = {x ; x

≥ 0 }

x → x =Φ(x)

is a diﬀeomorphism then under the space–time change of variables Ψ : (x, t) →

(Φ(x),t), the operator L becomes



L such that

(



Lv)(x, t)=(Lv)(x, t)

for all (smooth enough) v = v ◦ Ψ. By the chain rule we easily see that



L = ∂



j=1



∂



where



(x, t):=



k=1

∂

x

(x) A

(x, t) for all j ∈{1,...,d}.

Observing that



A( x, t,



ξ ):=





(x, t)=A(x, t, ξ)

with ξ =



ξ · dΦ(x) , i.e. ξ



j=1



∂

x

(x) for all k ∈{1,...,d}

(so that



ξ ∈ T

∗

x

∂D), we claim that the stable subspace of



A(y, 0,t,η,τ):=−(



(y, 0,t))

−1



τI

+ i

d−1



j=1



(y, 0,t)



is the same as the stable subspace of

A(x, t, ξ, τ)=A(x, t, ν(x))

−1

( τI

+ iA(x, t, ξ))

244 Variable-coeﬃcients initial boundary value problems

for x =Φ

−1

(y, 0) ∈ ∂Ωandξ =(η, 0) · dΦ(x). Indeed, we have



A(y, 0,t,η,τ)=−





k=1

∂

x

(x) A

(x, t)



−1



τI

+ iA(x, t, ξ)



and ∂

x

(x),...,∂

x

(x) are, up to a positive factor, the components of −ν(x).

Therefore, the uniform Kreiss–Lopatinski˘ı condition for the problem



Lu =



Bu = g

in the half-space {(y, z); z>0 } is equivalent to the uniform Kreiss–Lopatinski˘ı

condition stated above for the original problem

Lu = f, Bu = g

in Ω.

So a Boundary Value Problem in the domain Ω is fully equivalent to a Bound-

ary Value Problem in the half-space D, the latter satisfying the assumptions of

Theorem 9.3 if the former satisﬁes the assumptions of Theorem 9.4. And the L

estimates for either one of the BVP are clearly equivalent. 

If Ω is a smooth relatively compact domain instead of being globally diﬀeo-

morphic to a half-space, the same result is true.

Theorem 9.5 Assume Ω is a relatively compact domain with C

∞

boundary.

Then the energy estimate (9.1.21) in Theorem 9.4 is valid under the other

assumptions (CH), (NC), (N) and (UKL).

Proof The ideas are the same as in the proof of Theorem 9.4, except that we

use co-ordinate charts instead of a global diﬀeomorphism. Let (U

)

j∈{0,...,J}

acoveringof

Ω by chart open subsets, with U

⊂ Ω, and consider (ϕ

)

j∈{0,...,J}

an associated partition of unity, that is, ϕ

∈ D (U

) for all j and



j=0

≡ 1 .

In what follows, we use the convenient notation , which means ‘less than or

equal to a (harmless) constant times . . .’

On the one hand, by Theorem 2.13, which applies to the present case because

of our constant hyperbolicity assumption (see Theorems 2.3 and 2.2), there is a

> 0sothatforu ∈ D(Ω) and γ ≥ γ

γ e

−γt

u 

(Ω×R)



e

−γt

L(ϕ

u)

(Ω×R)

Then, since the commutator [ L, ϕ

] is of order 0, this implies

γ e

−γt

u 

(Ω×R)



e

−γt

Lu

(Ω×R)

e

−γt

u

(Ω×R)