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

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Construction of a Kreiss symmetrizer under (UKL) 145

non-characteristic (det A

=0). Assume ﬁnally that the IBVP satisﬁes the uni-

form Kreiss–Lopatinskiˇı condition.

Then there exists a matrix-valued C

∞

-map (τ, η) → K(τ, η),onRe τ ≥ 0,

η ∈ R

d−1

, |τ| + |η| =0, such that

i) the matrix Σ(τ,η):=K(τ, η)A

is Hermitian,

ii) there exists a number c>0 such that, for every (τ,η) and every x ∈ kerB,

the inequality x

∗

Σ(τ,η)x ≤−cx

holds,

iii) there exists a number c

> 0 such that, for every (τ,η), the inequality

Re M ≥ c

(Re τ)I

holds in the sense of symmetric matrices, where

M = M (τ,η):=−(ΣA)(τ, η)=K(τ,η)(τI

+ iA(η))

and Re M denotes the Hermitian matrix

(M + M

∗

If instead, the matrices A

and B are parametrized (for instance, if L has

variable coeﬃcients) with regularity C

with respect to the parameters z, then

such a symmetrizer Σ can be chosen with the same regularity: derivatives ∂

τ,η

∂

are continuous whenever l ≤ k.

Comments

Since A

is invertible, it is equivalent to search for a K or for a Σ. In

the following, we shall always work in terms of Σ. The situation with a

characteristic boundary (det A

= 0) raises signiﬁcant new diﬃculties. It

will be treated in Chapter 6.

The matrix K is called a Kreiss symmetrizer or a dissipative symmetrizer,

or simply a symmetrizer. It plays, in the present framework, the role that

the identity I

played for symmetric operators with a strictly dissipative

boundary condition. As a matter of fact, if L is symmetric, then K ≡ I

satisﬁes trivially point i), while Re M =(Reτ)I

. Finally, ii) is simply the

dissipation assumption.

When L is symmetric, it may happen that the IBVP satisﬁes the (UKL)

condition though the boundary condition is not dissipative. In such a case,

the symmetrizer K provided by the theorem diﬀers from I

All the arguments hold true when the data depend on parameters, exactly

as developed in the proof. Hence we shall present them for a single data,

depending only on (τ,η).

The proof of Theorem 5.1 is long and technical, though interesting in its own

way. We shall split it into several steps. From Step 1 to Step 18, we detail the

construction in the case of a strictly hyperbolic operator. We examine afterwards

which steps need a further study for a constantly hyperbolic operator and how

to adapt the proof to this more general framework.

146 Construction of a symmetrizer under (UKL)

Step 1 We shall actually build a K that is homogeneous with degree zero, with

respect to (τ,η). Therefore, it will be enough to build K or Σ for (τ,η)inthe

unit hemisphere, Re τ ≥ 0, η ∈ R

d−1

, |τ|

+ |η|

=1.

Step 2 Since the hemisphere is compact, and since the properties to ﬁll deﬁne

convex cones, it will be enough to localize the construction, namely to build

aΣ

loc

in the vicinity of every point, then to cover the hemisphere by a ﬁnite

set of such neighbourhoods, and at last to deﬁne a Σ from the corresponding

loc

s, using a partition of unity with real non-negative coeﬃcients. From now

on, we thus give ourselves a point (τ

,η

) and we look for a solution Σ near

this point.

Step 3 The case of interior points, namely Re τ

> 0 is easier, since, choosing

Σ independently of (τ,η), it is enough to ﬁnd a Σ in H

satisfying Σ|

ker

< 0

and Re M>0 at the sole point (τ

,η

). Hence, the construction needs only to

be done pointwisely instead of locally, at interior points.

Step 4 Given an interior point (τ,η), we build a symmetizer. We shall use two

lemmas.

Lemma 5.3 Let A be a hyperbolic matrix, and let E

−

, E

be its stable and

unstable invariant subspaces. Let us deﬁne

−

:= {H ∈ H

; E

−

⊂ kerH}.

Then the map T : H → Re (HA) is an automorphism of X

−

Moreover, if T (H) ≥ 0 with E

−

=kerT (H), then H ≥ 0 with E

−

=kerH.

Proof We ﬁrst prove that T has a right-inverse. For that, let X ∈ X

−

be given.

If x = x

−

+ x

with x

∈ E

, we deﬁne

∗

Hx :=



−∞

y(s)

∗

Xy(s)ds,

where y is the solution of dy/ds = Ay, y(0) = x

. The integral converges since y

decays exponentially fast at −∞. The above equality deﬁnes a unique Hermitian

matrix H ∈ X

−

Given h ∈ R and x(h):=x

−

+ y(h), we have

x(h)

∗

Hx(h):=



−∞

y(s)

∗

Xy(s)ds.

Diﬀerentiating at the origin, we obtain

∗

H + HA)x = x

∗

Xx,

which means T (H)=X (note that only the component of x



(0) along E

important in this calculation.)

The integral thus deﬁnes a right-inverse, which proves that T is an automor-

phism. If, moreover, X = T (H) is non-negative, the integral formula shows that

Construction of a Kreiss symmetrizer under (UKL) 147

H is non-negative. Assume additionally that E

−

=kerX.Ifx ∈ kerH,wehave



−∞

y(s)

∗

Xy(s)ds =0,

which implies that y ∈ E

−

,thatisy ≡ 0 and hence x ∈ E

−

. 

We apply Lemma 5.3 to A(τ,η). Let us choose a non-negative element K

−

,whosekernelisE

−

(τ,η), and let us deﬁne H

= T

−1

). From Lemma 5.3,

it is non-negative and E

−

(τ,η)=kerH

. Similarly, choosing a non-positive

Hermitian matrix K

−

such that E

(τ,η)=kerK

−

, the equation Re (HA)=K

−

admits a unique non-negative Hermitian solution H

−

whose kernel is E

(τ,η).

From the Kreiss–Lopatinski˘ı condition C

= E

−

(τ,η) ⊕ kerB, there exists a

matrix P such that, for x

∈ E

−

+ x

∈ kerB) ⇐⇒ (x

−

= Px

Since the restriction of H

to E

(τ,η) is positive-deﬁnite, the restriction of

− P

∗

−

P to E

(τ,η) is positive-deﬁnite for a large enough positive number

b. We now deﬁne Σ := −bH

+ H

−

. Because of the choice of b, it satisﬁes point

ii). Moreover, Re M = bK

− K

−

. As a sum of two non-negative Hermitian

matrices, it is non-negative and its kernel is contained in the intersection of both

kernels, which is E

−

(τ,η) ∩ E

(τ,η)={0}. This matrix is therefore positive-

deﬁnite.

Remark When A(τ,η) is not hyperbolic, a fact that happens at some boundary

points, at least when |τ| >> |η|, there cannot exist an Hermitian Σ such that

Re (ΣA) is negative-deﬁnite. Actually, if x is an eigenvector associated with a

pure imaginary eigenvalue µ of A,thenRex

∗

ΣAx =(Reµ)x

∗

Σx =0.

Summary After Step 4, it remains to construct a Σ in a neighbourhood of any

given point (iρ

,η

) of the boundary.

Step 5 We recall our notation τ = γ + iρ.LetQ ∈ GL

(R) be given. Making

the change of unknowns v = Qu amounts to replacing the matrices A

, B(ρ, η)

and B by

:= QA

−1

,β(ρ, η):=QB(ρ, η)Q

−1

,b:= BQ

−1

Then, a local solution Σ(τ, η) of our problem yields a local solution σ(τ, η)of

the problem associated to the matrices a

, β and b, through the correspondence

Σ=:

QσQ. Similarly, M is replaced by m, deﬁned by M =:

QmQ.Onemay

even allow Q to depend smoothly on (ρ, η).

Step 6 Let (iρ

,η

) be a non-zero boundary point. Using a change of basis as

above, we may assume that

B(ρ

,η

−1

=: β

= diag(β

,β

where ‘c’ and ‘h’ stand for the central and hyperbolic parts of A(iρ

,η

). Namely,

has only real eigenvalues, while β

has only non-real eigenvalues. One may

148 Construction of a symmetrizer under (UKL)

even assume that β

has a canonical Jordan form. From the strict hyperbolicity

of L, we know that eigenspaces of B(ρ

,η

), when associated to real eigenvalues,

are lines because

ker(B(ρ, η) − µI

)=ker(ρI

+ A(η, −µ)).

Therefore, distinct Jordan blocks of β

have distinct eigenvalues. We shall denote

by m the size of β

and by β

(1 ≤ j ≤ J) its Jordan blocks. The size and the

eigenvalue of β

are denoted by m

and µ

Next, it is well-known

that an analytical function Q(ρ, η) may be found in

a neighbourhood of (ρ

,η

), so that Q(ρ

,η

)=Q

and

Q(ρ, η)B(ρ, η)Q(ρ, η)

−1

=: β(ρ, η) = diag(β

(ρ, η),β

(ρ, η)).

We point out that, by a continuity argument, β

still has non-real eigenvalues,

though the eigenvalues of β

need not remain real. From upper semicontinuity

of geometric multiplicities, the eigenspaces of β

are lines.

Step 7 From the last two steps, we are led to the construction of a local solution

σ(τ,η) in a neighbourhood V of (iρ

,η

), associated to the matrices a

, β and b.

Here, all three matrices depend smoothly on (ρ, η), but do not depend on γ.

We shall specialize σ as follows:

σ(γ + iρ, η) = diag(S + iγT, σ

where S = S(ρ, η), T = T (ρ, η), σ

are such that

S ∈ Sym

,T∈ Alt

,σ

∈ H

n−m

In particular, S and T have real entries

and this ensures that σ is Hermitian.

From m := σ(γ(a

)

−1

+ iβ), we obtain

m =



iSβ

+ γ(S(a

)

−1

− Tβ

)+O(γ

) O(γ)

O(γ) iσ

+ O(γ)



From the Remark in Step 4, there is no hope that the term Sβ

would help in

the positivity of Re m. Therefore, we shall ask that

Sβ

∈ Sym

, (5.2.2)

for every (ρ, η) in a neighbourhood W of (ρ

,η

). Denoting

Y (ρ, η):=S(a

)

−1

− Tβ

(S(a

)

−1

− Tβ

)

= S(a

)

−1

+(a

)

−t

S − Tβ

See, for instance, the procedure used in the proof of Theorem 2.3.

We recall that, unless another ground ﬁeld is speciﬁed, elements of Sym

and Alt

have real

entries.

Construction of a Kreiss symmetrizer under (UKL) 149

we obtain

Re m =



Y (ρ, η)+O(γ

) O(γ)

O(γ)Re(iσ

)+O(γ)



. (5.2.3)

Step 8 We wish now to replace the local construction by a pointwise one. Let

us ﬁrst assume that Y (ρ, η) and Re (iσ

) be positive-deﬁnite at (ρ

,η

). By

continuity, they remain so, uniformly in some neighbourhood of (ρ

,η

)thatwe

still call W. Then, decomposing vectors of C

into their ‘c’ and ‘h’ components,

we have for every (ρ, η)inW

Re (X

∗

mX) ≥ cγ|X

+ c|X

+ O(γ)|X

||X

where c is some positive constant. Using the Cauchy–Schwarz inequality, we see

that m satisﬁes point iii) near (iρ

,η

In other words, the point iii) will be fulﬁlled locally whenever the following

properties hold

Y (ρ

,η

) ∈ SPD

(R), (5.2.4)

Re (iσ

) ∈ HPD

n−m

. (5.2.5)

Step 9 We ﬁnish the work begun in Step 8. To do so, we study condition

(5.2.2).

Lemma 5.4 The equation Sβ

(ρ, η)=

(ρ, η)S deﬁnes a subspace of dimen-

sion m in Sym

, for every (ρ, η) in a small neighbourhood of (ρ

,η

). This

subspace varies smoothly with (ρ, η).

Proof We only have to prove that the map Λ

S → Sβ

−

Sym

→ Alt

is onto. Then the analytic dependence of kerΛ follows from that of Λ and from

the constancy of the dimension.

Recall that the eigenspaces of β

are (not necessarily real) lines for every

value of (ρ, η)inW. Since the property to prove is equivalent to the injectivity

of the transpose of Λ:

s → β

s − s

Alt

→ Sym

the Lemma will be a direct consequence of the following one, together with the

fact that

is similar to its transpose. 

Lemma 5.5 Let two matrices s, D ∈ M

eigenspaces of D are lines, that s is alternate, and that

Ds −sD =0.Then

s =0.

150 Construction of a symmetrizer under (UKL)

Proof The fact that the eigenspaces of D are lines means that the minimal

polynomial of D equals its characteristic one. This amounts to saying that there

exists a vector x such that C

equals the Krylov subspace Span{x, Dx, D

x,...}

(see Exercise 16, Chapter 2 in [187]). In other words, as a C[D]-module, C

has

dimension one.

Denoting by a the alternate form deﬁned by s,wehaveassumedthat

a(Dz, y)=a(z,Dy) for every z,y ∈ C

. Applying recursively this equality, we

obtain for every k, l ∈ N

a(D

x, D

x)=a(D

x, D

x)=−a(D

x, D

x),

so that a(D

x, D

x) = 0. Since the vectors D

x span C

, we conclude that a =0,

that is s =0. 

Assume now that a matrix S

in Sym

satisﬁes S

∈ Sym

. Then, from

Lemma 5.4, it is possible to ﬁnd an analytic function (ρ, η) → S from W to

Sym

that satisﬁes (5.2.2) in W and S(ρ

,η

)=S

Hence, in order that σ satisfy all requirements near a point (ρ

,η

), it is

enough that (5.2.4) and (5.2.5) hold and

∈ Sym

. (5.2.6)

In other words, we have replaced the local construction of σ by a pointwise one.

Thanks to this simpliﬁcation, we shall drop the index ‘0’ from now on.

Summary After Step 9, it remains to construct three matrices S ∈ Sym

T ∈ Alt

and σ

∈ H

n−m

with the properties that at a given point (ρ, η)inthe

sphere S

d−1

, (5.2.2), (5.2.4) and (5.2.5) hold together with

σ|

ker

< 0, (5.2.7)

where σ := diag(S, σ

Step 10 Since T occurs only in (5.2.4), we analyse ﬁrst this property, assuming

that S is known. Then Y = Y

− Tβ

T ,whereY

∈ Sym

is given in

terms of S. Let us remember that β = diag(β

,...), where the β

s are Jordan

blocks with distinct real eigenvalues µ

. We decompose T =: (T

)

j,k

block-wise

accordingly. We note that the block T

occurs only in the (j, k)-block of Y ,

through

− T

. Therefore, using the well-known fact that X → NX −



is an automorphism of M

(K) whenever N and N



have disjoint spectra,

we may choose uniquely the oﬀdiagonal blocks T

in such a way that Y be block-

diagonal. We emphasize that, Y

being symmetric, this choice is consistent with

the skew-symmetric form of T that we ask for: it holds that T

= −

Step 11 We continue Step 10 by determining the (skew-symmetric) diagonal

blocks T

. For that purpose, we use the following lemma.

Lemma 5.6 Let y belong to Sym

,andletβ be an upper Jordan block of size

m. Then the following properties are equivalent,

Construction of a Kreiss symmetrizer under (UKL) 151

there exists a skew-symmetric t, such that

βt − tβ + y is positive-deﬁnite.

> 0.

Proof The direct implication is trivial, because one always has (tβ)

=0.

To prove the converse, we proceed by induction. There is nothing to prove if

m =1.Ifm ≥ 2, we assume that the lemma is true at order m − 1. We use hats

for (m − 1) × (m − 1) upper-left blocks. Since y

> 0, the induction hypothesis

ensures the existence of an (m − 1) × (m − 1) skew-symmetric matrix

t,such

that ss

m−1

t −

β +ˆy>0. We now deﬁne

t :=



−

X 0



,X:=



m−2



A straightforward computation gives

βt − tβ + y =



m−1

· 2u + y



where the oﬀdiagonal terms do not depend on u. Choosing u>0 large enough,

the resulting matrix is positive-deﬁnite. 

Summary Thanks to Steps 10 and 11, we have reduced our task to the

construction of S ∈ Sym

and σ

∈ H

n−m

with the properties that at a given

point (ρ, η) in the sphere S

d−1

, (5.2.2), (5.2.5) and (5.2.7) hold together with

> 0, 1 ≤ j ≤ J, (5.2.8)

where y

stands for the upper-left coeﬃcient of the jth diagonal block Y

of Y

Step 12 We decompose as above S blockwise. Then (5.2.2), together with

S ∈ Sym

, give

= S

. Using the fact mentioned in Step 10, we ﬁnd

that S

=0whenj = k. Hence, S has to be block-diagonal. We shall denote by

its diagonal blocks.

Step 13 Hence, Sβ

= diag(S

,...) and (5.2.2) reduces to S

∈ Sym

Since S

∈ Sym

as well, we ﬁnd by inspection that the S

s have the general

form







O .







, (5.2.9)

with arbitrary numbers s

,...,s

Step 14 Since the block Y

equals S

,wherea

is the jth diagonal

block of (a

)

−1

, we obtain y

=2s

,wheres

stands for the s

-coeﬃcient of

and a

stands for the last coeﬃcient of the ﬁrst column of a

. In other words,

152 Construction of a symmetrizer under (UKL)

=(a

)

. At this point, it is essential to know the sign of a

. For this, we

note that a

equals 



)

−1



,where















, and 



=(...,0, 1, 0,...)

are the right and left eigenvectors of β, corresponding to the block β

. Deﬁning

 := 



Q and r = Q

−1



, we obtain a

= (A

)

−1

r,wherer,  are right and

left eigenvectors of B. With the notations of Lemma 5.2, r = R and  = LA

(this again shows (5.1.1) when m

≥ 2), hence a

= LR. In particular, a

is non-zero.

We notice that, when m

=1,



holds and therefore 



=1> 0, which

translates into LA

R>0. This ﬁxes the respective orientations of L and R.Inthe

opposite case, one has 



=0,thatisLA

R = 0, which leaves the orientations

of L, R independent of each other. In other words, the sign of a

is well-deﬁned

if m

=1,astheoneof(LR)(LA

R), while it is free (depending on our choice

of the Jordan basis) when m

> 1.

Recalling Proposition 5.1, we know that E

−

(iρ, η) is a direct sum of invariant

subspaces E

(1 ≤ j ≤ J)andE

, corresponding either to the block β

or to β

The dimension of E

is half the size n − m of β

, while that of E

belongs to

− 1/2,m

+1/2].

If m

= 1, then Proposition 5.1 and Lemma 5.2 show that a

is positive if

and only if E

is non-trivial. Therefore a

> 0 if and only if R ∈ E

−

(iρ, η).

If m

> 1, then E

is non-trivial. The sign of a

may be choosen at our

convenience.

Step 15 Let us denote e

−

:= QE

−

(τ,η), so that, by the (UKL) condition, it

holds that e

−

∩ kerb = {0}. From the description given in Proposition 5.1, we

have e

−

= e

⊕ e

,wheree

⊂ C

×{0} and e

⊂{0}×C

n−m

. The component

, which corresponds to the stable subspace of A(iρ, η), is the invariant subspace

of β associated to the eigenvalues (of β

) of negative imaginary parts. The

‘central’ component e

is spanned by some of the m vectors of the Jordan

basis for β

. Within the Jordan basis associated to the jth block β

, the l

ﬁrst

vectors are in e

, while the other ones are not, where |l

− m

/2|≤1/2. This

allows us to write e

= ⊕

. Denoting also by f

the subspace spanned by the

− l

last vectors of this basis, by f

their direct sum and by f

the ‘unstable’

component, namely the invariant subspace associated to the eigenvalues (of β

)

of positive imaginary parts, we obtain C

= e

⊕ e

⊕ f

=: e

−

⊕ f

. Note

that, in spite of the notation, f

is not the limit, as (τ, η) tends to (iρ, η), of

(τ,η).

Construction of a Kreiss symmetrizer under (UKL) 153

Given a vector x in C

, we write its block-decomposition

x =













with obvious notations. Then, in each block x

(1 ≤ j ≤ J), we split into e

and

components:



j−



Finally, we denote by x

= x

+ x

the decomposition in {0}×C

n−m

. Hence,

the decomposition of x into e

−

and f

components reads

x = x

−

+ x









j−

























In the following, we shall identify x

−

, x

with the vectors







j−



















respectively.

Step 16 The assumption that the Kreiss–Lopatinski˘ı property holds at the

boundary point (iρ, η) means exactly that the subspace kerb has an equation of

the form x

−

= Px

,whereP is an appropriate linear map. From the description

of x

above, we write P blockwise

P =





1≤j,k≤J

Let us summarize what we are looking for. We wish to ﬁnd matrices

∈ Sym

of the form (5.2.9), and a Hermitian matrix σ

, with the following

properties:

for every j, s

=0,

when m

=1,thens

is negative if e

is trivial and positive otherwise,

Re (iσ

) is positive-deﬁnite,

the restriction of σ = diag(...,S

,...,σ

)tokerb is negative-deﬁnite.

154 Construction of a symmetrizer under (UKL)

An important point is that we do not require that β

be a standard Jordan block.

Therefore we are free to make a change of variables x → x



= Rx, provided it

alters neither the block-diagonal form of σ, nor the form (5.2.9) of each diagonal

block. We choose a diagonal change of variable

R = R() = diag(...,R

,...,I

n−m

where

= diag(...,

,,

−1

,

−3

,...) =: diag(U

In other words,



j−

= U

j−



= V



= x



= x

It is important to note that this change of variables does not modify the overall

structure (5.2.9). Therefore, we have to solve the same problem as described

above, with the only change that the equation of kerb is replaced by x



−

= P ()x



for some real number  =0,where

P ()=



−1



1≤j,k≤J

Step 17 When  → 0, P () tends to



0 P



1≤j,k≤J

Let us assume that a solution σ



of the above problem has been found, when

kerb is replaced by the subspace deﬁned by x



−

= P



. Then, by continuity, σ



is still a solution of the problem for some small non-zero value of . Hence, going

back to  = 1 through the change of variable, we obtain our matrix σ. This shows

that we need only to solve the problem in Step 16, when kerb is the subspace

deﬁned by

j−

=0 (1≤ j ≤ J),x

= P

Step 18 We study the latter problem. When x

j−

=0andx

= P

,then

∗

σx =



j=1

∗

+ x

∗

where

is one of the following three matrices



































