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) 155

The problem clearly decouples. On the one hand, we wish to ﬁnd, for each j,

a negative-deﬁnite matrix

of the form above. This can be found easily when

> 1, since then there is no constraint on the entries s

,....Whenm

=1,

there are two cases. Either e

is trivial, and then

=(s

), where the constraint

is s

< 0, obviously compatible with our task. Or e

is non-trivial, then

void and there is nothing to prove. Hence the problem concerning the

scan

always be solved.

On the other hand, we have to ﬁnd a Hermitian matrix σ

, such that

Re (iσ

) is positive-deﬁnite, and the restriction of σ

to the space deﬁned

by x

= P

is negative-deﬁnite. Noting that, from the Kreiss–Lopatinski˘ı

condition, this subspace is transverse to the stable subspace of −iβ

,thisis

exactly the same problem as the one solved in Step 4.

Step 19 We now turn to the case of a constantly hyperbolic operator. Only

Steps 6, 9, 11, 13, 14, 16 and 18 need some adaptation.

In Step 6, the description of β

is given by Proposition 5.1. The blocks β

have distinct eigenvalues. Each one is a regular Jordan block J(µ; q, N).

Step 20 The adaptation of Lemma 5.4 in Step 9 is subtle. Actually, the

dimension of the space of symmetric matrices S such that Sβ

(ρ, η)=

(ρ, η)S,

though constant, will not be equal to m, but to another number, see below.

Let us ﬁrst note that we can consider instead the complex dimension of the

set of complex symmetric matrices with this property

. Now, any conjugation

β → P

−1

βP induces the transformation S →

PSP on the solutions, thus pre-

serving the dimension. We use this argument in two ways. First, we may assume

that

(ρ, η) = diag(...,β

(ρ, η),...),

where β

are smooth functions of their arguments, and the β

(ρ

,η

)s are the

regular Jordan blocks J(µ

; q

), with distinct real eigenvalues, described in

Step 19. Then the fact recalled in Step 10 tells us that solutions S must be block-

diagonal too, say diag(...,S

,...), since the β

s keep disjoint spectra. Next, we

may assume that each β

(ρ, η) has a Jordan form (here, P need not depend

smoothly on (ρ, η)). Let us note that the arguments in Section 5.1 adapt to every

eigenvalue of β(ρ, η) (not only the pure imaginary ones): to each eigenvalue, there

corresponds a unique Jordan block, a regular one. More precisely, β

(ρ, η)isa

collection of blocks J(ω; q

), with ω in some ﬁnite set Ω

. The geometric

multiplicity q

is the same for all elements ω, because geometric multiplicity is

upper semicontinuous, and because all ωs are roots of the same factor P

in the

characteristic polynomial (see Proposition 1.7).

We prefer considering solutions in Sym

, because the latter is not a

complex vector space.

156 Construction of a symmetrizer under (UKL)

Again, the diagonal block S

must be of the form diag(...,S

,...), where

J(ω; q

As in step 13, we ﬁnd that S

must have the form (5.2.9), where now the terms

,... must be q

× q

symmetric matrices. Such matrices from a vector space of

dimension N

+1)/2. Summing over Ω

, we see that the set of solutions S

has dimension N

+1)/2. Finally, the set of solutions S has dimension



+1),

obviously a constant number in a neighbourhood of (ρ

,η

Step 21 In Step 11, Lemma 5.6 adapts straightforwardly to the context of

regular Jordan block J = J(µ; q; N). The existence of a skew-symmetric real

matrix t such that

Jt − tJ + y is positive-deﬁnite is equivalent to the positivity

of the upper-left diagonal q × q block y

. The proof goes by induction on N.

Therefore, condition (5.2.8) is unchanged up to the fact that y

is now a q × q

block, instead of a scalar.

Then, y

= s

,wheres

∈ Sym

is to be chosen. The lower-left

block a

may be written, as in Step 14, as LR,whereL = 



Q(A

)

−1

, R = Q

−1



and















, and 



=(...,0

, 0

,...).

Rows of L and columns of R span the left and right kernels of ρI

+ A(η, µ)

(notations of Lemma 5.2). In particular, a

∈ GL

(R). When N (which plays the

rˆole of m

) is larger than 1, 



holds, hence LA

R =0

, which is consistent

with identity (5.1.1) and the fact that dρ/dµ vanishes at this point. Then, we are

free to compose L and R with two independent invertible q × q matrices (this is

reﬂected in a modiﬁcation of the change of basis Q). This allows us to prescribe

any arbitrary invertible value to a

, for instance ±I

. Consequently, y

= ±2s

and we may choose any positive- or negative-deﬁnite s

On the other hand, when N =1,LA

R = I

holds, hence from (5.1.1)

= −



dρ

dµ



−1

= −2



dρ

dµ



−1

In conclusion, s

must be chosen positive-deﬁnite if E

is non-trivial, but

negative-deﬁnite if E

is trivial.

Construction of a Kreiss symmetrizer under (UKL) 157

Step 22 In the requirements of Step 16, the ﬁrst item must be read ±s

∈

SPD

for every j (instead of s

= 0). Next, each diagonal block R

must be

chosen in the block-diagonal form

= diag(...,

,I

,

−1

,

−3

,...).

The rest of the step works in exactly the same way.

Finally, there is even more room for the choice of

in Step 18. It may again

be chosen so as to be negative-deﬁnite.

The proof is complete. 2

Comment The construction of the dissipative symmetrizer with some uniform

estimates reduces quite easily (Steps 1 and 2) to a local problem. At interior

points, the construction can even be done pointwise (Step 3), with a Lyapunov

stability argument (Step 4). Thus, most of the proof is devoted to the con-

struction in the neighbourhood of boundary points (iρ, η). At such points, we

separate the hyperbolic and the central part of B(ρ, η) (Steps 5 and 6). The

symmetrizer is block-diagonal in terms of the invariants subspaces of B(ρ, η)

(Step 7). The treatment of the hyperbolic part is similar to the case of interior

points (Step 18). Thus there remains the central part. What is hidden in the proof

above is that the construction is rather simple when the central part of B(ρ, η)is

semisimple, that is diagonalizable. Suppose that its eigenvalues are simple, or that

their multiplicities are locally constant. Then the symmetrizer is block-diagonal

and certain parts of the analysis (Steps 13 to 18) are essentially trivial. Thus

the deeper part of the proof concerns the so-called glancing points, which are

boundary points (iρ, η) at which two (or more) real eigenvalues of B(ρ, η)cross

each other. This is the only place where we make use of the assumption of strict

or constant hyperbolicity. Finally, Steps 19 to 22 only adapt the construction

from the strictly to the constantly hyperbolic case.

THE CHARACTERISTIC IBVP

6.1 Facts about the characteristic case

In this section, we place ourselves in the characteristic case: A

is singular. We

denote by m the dimension of its kernel. Using a linear transformation, we may

assume that



0 a



with a

∈ GL

n−m

(R). Since we may not hope for a better result than in the non-

characteristic case, we also assume that the operator L is constantly hyperbolic.

From Theorem 1.7, the upper-left block in the block decomposition of A(η)is

of the form l(η)I

,wherel is a linear form on the space of frequencies

.In

other words, l is a vector in the physical space, whose last component vanishes.

Without loss of generality, the change of variable (x, t) → (x − tl, t) preserves the

physical space {x

> 0} and leads to the situation where

A(η)=



(η)

(η) a

(η)



We recall that the boundary matrix B satisﬁes

kerA

⊂ kerB, which means here

that B has the form B =(0

p×m

In the following, we often use block decomposition. For instance, a generic

vector u ∈ C

will split as (v, w)

, with v ∈ C

.ThusthekernelofA

is given

by the equation w = 0. From Proposition 4.3, we have E

−

(τ,η) ∩ kerA

= {0},

hence E

−

(τ,η) is isomorphic to its projection e

−

(τ,η)onthew-component. The

subspace e

−

(τ,η) can be deﬁned equivalently as the stable subspace of the Schur

complement

(τ,η):=−(a

)

−1



τI

n−m

+ ia

(η)+

(η)a

(η)



The reciprocal map from e

−

(τ,η)toE

−

(τ,η) is obviously

w →



−

(η)w, w



The matrix A

has been shown to be hyperbolic (Lemma 4.3).

Whenever m ≥ 2, this is in contradiction with Assumption 1.1 of [127], where the upper-left

block was required to have simple eigenvalues.

This property is called reﬂexivity in [150, 151].

Facts about the characteristic case 159

As mentioned in Section 5.1, (τ,η) → e

−

(τ,η), and (τ,η) → E

−

(τ,η)also,

admit continuous extensions at boundary points, except perhaps at τ =0.Itis

unclear whether the property E

−

(τ,η) ∩ kerA

= {0} extends to such points,

though it will be a necessary condition for the uniform Kreiss–Lopatinski˘ı

condition, because of kerA

⊂ kerB.

6.1.1 A necessary condition for strong well-posedness

We now present a new restriction that the strong well-posedness imposes to the

IBVP. Amazingly enough, this restriction concerns the operator L only, but not

the boundary matrix B. It is thus of a very diﬀerent nature from the uniform

Lopatinski˘ı condition. Of course, this condition is trivially satisﬁed when the

boundary is non-characteristic.

We start from the estimate in Deﬁnition 4.6. Considering a trivial initial data

≡ 0, we may extend u by zero to negative times and then take the Laplace–

Fourier transform in (t, y). The estimate implies



+∞

γ|ˆu(γ + iσ,η,x

dσdηdx



ˆu(γ + iσ, η, 0)|

dσdη

≤ C





+∞

Lu(γ + iσ,η,x

dσdηdx



|Bˆu(γ + iσ, η, 0)|

dσdη



Since

Lu(τ,η,x

)=(τ + iA(η))ˆu + A

ˆu



decouples with respect to (τ, η), this

estimate is equivalent to

(Re τ)



+∞

|ˆu(τ,η,x

+ |A

ˆu(τ,η,0)|

≤ C



Re τ



+∞

Lu(τ,η,x

+ |Bˆu(τ,η,0)|



for every pair (τ,η) with Re τ>0. The constant C, being the same as above,

does not depend on (τ, η).

We now specialize to the solutions of the diﬀerential-algebraic system (τI

iA(η))ˆu + iA

ˆu



= 0 that decay at +∞. These solutions take values in E

−

(τ,η).

Using Bˆu = B

w with ˆu =: (v, w)

, the above estimate amounts to

(Re τ)



v

+ w



+ |w(0)|

≤ C|B

. (6.1.1)

Since τv + ia

(η)w = 0, (6.1.1) reduces to the inequality

(Re τ)



|τ|

−2

a

(η)w

+ w



+ |w(0)|

≤ C|B

w(0)|

. (6.1.2)

160 The characteristic IBVP

Estimate (6.1.2) splits into three inequalities, among which two are familiar to

us. For instance, one of them is the uniform Kreiss–Lopatinski˘ı condition



Re τ>0,η∈ R

d−1

,w(0) ∈ e

−

(τ,η)



=⇒ (|w(0)|≤C|B

w(0)|).

The new fact is the estimate

a

(η)w

≤

C|τ|

√

Re τ

w(0)|.

However, since we shall require the (UKL) condition, the only new information

is the coarser estimate

a

(η)w

≤

C|τ|

√

Re τ

|w(0)|, ∀w(0) ∈ e

−

(τ,η), (6.1.3)

where w(x

):=exp(x

(τ,η))w(0).

We emphasize that (6.1.3) does not depend on the particular choice of a

boundary condition. It is a property of e

−

(τ,η) itself, which is necessary in order

that some boundary condition exists for which the IBVP is strongly well-posed.

For this reason, we use the following notion.

Deﬁnition 6.1 Let L be a hyperbolic operator and λ(ξ) be an eigenvalue of A(ξ),

of constant multiplicity m in a neighbourhood of e

, with λ(e

)=0. Without loss

of generality, we may assume also that dλ(e

)=0;henceA

and A(η) have the

structure described above. We then say that L is stabilizable

if (6.1.3) holds

true, with a constant C independent of (η, τ) in the unit hemisphere.

The previous analysis shows that:

Proposition 6.1 With the notations above, assume that there exists a boundary

matrix β with βu := β

w, such that the IBVP associated to (L, β) in the half-

plane {x

> 0} is strongly well-posed. Then L is stabilizable.

We shall discuss two examples, taken from [127]. In the ﬁrst one, the property

does not hold, while in the second, we show that it does for every Friedrichs-

symmetrizable system. The latter might be seen as a consequence of Proposi-

tion 6.1, since it is rather easy to ﬁnd a strictly dissipative boundary condition

for a symmetric operator, and such a boundary condition automatically satisﬁes

(UKL).

First example We consider the case where a

≡ 0. Then the operator L

decouples, since w obeys a diﬀerential system L

w = f

, which does not involve

Remember that in a characteristic IBVP, we expect a boundary estimate of only A

u,thatisof

the w component.

One should say stabilizable in direction e

Facts about the characteristic case 161

the component v at all. Since the boundary condition involves w only, the IBVP

itself decouples between a sub-IBVP

w = f

,w|

t=0

= w

= g

and an ODE for v,wherew enters as a source term:

= f

− a

(∇

)w.

Let us assume that L

is constantly hyperbolic and that the sub-IBVP satisﬁes

the (UKL) condition. Then, for L

data, the solution w is uniquely deﬁned (see

Theorem 4.3) and is L

. The fact that the source term a

(∇

)w is only of class

−1

is not necessarily a cause of trouble; after all, the pure Cauchy problem is

well-posed in L

since L is constantly hyperbolic

, though it decouples as well.

In the present case, e

−

(τ,η) is precisely the stable subspace at frequency

(τ,η) for the operator L

. From Lemma 4.5, it extends continuously at boundary

points, and in particular at (0,η), provided η is non-zero. Since A

(τ,η)doesnot

display any singularity, we may pass to the limit in (6.1.3) if the full IBVP

is well-posed, obtaining a

(η)w = 0 almost everywhere when w(0) ∈ e

−

(0,η).

Since w is continuous (as the solution of the diﬀerential equation w



= A

(0,η)w),

this implies a

(η)w(0) = 0. In other words, the well-posedness of the full IBVP

requires that

−

(0,η) ⊂ kera

(η), ∀η ∈ R

d−1

. (6.1.4)

In most cases, (6.1.4) implies that a

≡ 0. For instance, if n − m =2 and L

has one positive velocity in each direction, then e

−

(0,η) is a line, spanned

by some eigenvector r of A

(0,η). Let λ be the corresponding eigenvalue:

(λa

+ ia

(η))r = 0. This vector cannot be real, since otherwise λ would be

purely imaginary, λ = iµ,andL

would have a zero velocity in the direction

(η, µ), contradicting the assumption of constant hyperbolicity for L.Sincea

real valued, (6.1.4) implies that both Re r and Im r belong to its kernel. Hence

the kernel has dimension at least two, which means a

≡ 0.

In conclusion, if a

≡ 0andifL

is constantly hyperbolic with non-vanishing

velocities, so that L is itself constantly hyperbolic, then most of the choices of

the non-zero matrix a

(η) violate the stability condition (6.1.3).

We ﬁnish this section by giving an example of such an operator L

(n − m =

2, strict hyperbolicity, with a negative and a positive velocity in each direction):

= ∂





∂



0 −1



∂

We point out that the above analysis works under the weaker assumption

(η)a

(η) ≡ 0.

Here, the velocities of L

may not vanish, in order that L be constantly hyperbolic.

162 The characteristic IBVP

Friedrichs-symmetrizable operators We have the following result:

Theorem 6.1 Let L be Friedrichs symmetrizable. Then (6.1.3) holds true.

Proof Multiplying by the Friedrichs symmetrizer, we may assume that L has

the form

L = S

∂



∂

with symmetric matrices S

and S

, the latter being positive-deﬁnite. Let u be

a decaying solution of

(τS

+ iS(η))u + S



=0.

Multiplying on the left by u

∗

, taking the real part and integrating, we obtain

(Re τ)u

≤

∗

(0)S

u(0) =

∗

(0)s

w(0).

This contains in particular Inequality (6.1.3), since the v component of u is

−iτ

−1

(η)w.



6.1.2 The case of a linear eigenvalue

The purpose of this section is to identify a natural class of operators L for

which a symbolic dissipative symmetrizer could be constructed. Our motivations

are twofold. First, we wish to admit some of the operators that we encounter

frequently in physics. Second, the L

-well-posedness necessitates a few additional

properties, one of them being stabilizability.

Let ξ → λ(ξ) be the eigenvalue of A(ξ), responsible for the characteristicity

of the boundary. Thus λ(e

) = 0. Because of the constant hyperbolicity, λ

is diﬀerentiable for ξ = 0. Since it is homogeneous of degree one, we deduce

dλ(e

= 0. On the other hand, we do not alter the nature of the IBVP by

chosing a moving frame that travels at a constant speed, parallel to the boundary.

This amounts to changing the variable as (x, t) → (x



,t), with x



= x − tl and

= 0 (see above). Choosing l =dλ(e

), we are led to the case where dλ(e

)=0.

There are realistic cases where this reduction yields the property λ ≡ 0. For

instance, λ could have been linear in the initial setting, a fact that happens,

for instance, when its multiplicity m is strictly larger than n/2 (Corollary 1.1).

This is the case in hydrodynamics when the ﬂuid velocity is tangential along

the boundary (no mass transfer across the boundary). A second possibility

is that rotational invariance makes the spectrum depend only on |ξ|.Then

λ ≡ 0 follows directly from λ(e

) = 0. This happens for Maxwell’s equations in

electromagnetism.

We summarize in the following proposition the properties displayed by the

matrices A(ξ)whenλ ≡ 0 is an eigenvalue of constant multiplicity m ≥ 1. In par-

ticular, the eigenvalues of A

remain bounded as τ goes to zero. This fundamental

Facts about the characteristic case 163

property is crucial when estimating the derivatives of the solution of the IBVP

with compatible data. It has been shown in [127], through counterexamples,

that the failure of this property is responsible for a loss of order in the estimates

of derivatives: L

-norms of derivatives of order r ≥ 1 require the L

-norms of

derivatives of the boundary data, at some order (typically 2r) strictly larger

than r. See also a discussion at the end of the section.

Proposition 6.2 Assume that d ≥ 2 and L is constantly hyperbolic, with λ =0

being an eigenvalue of A(ξ), of multiplicity m ≥ 1 (for ξ =0). Without loss of

generality, assume that A

has the form



0 a



Then:

Given p, the number of positive eigenvalues of A

, the matrix A(ξ) has

precisely p positive eigenvalues and p negative eigenvalues for ξ =0.In

particular, n − m =2p is even.

One has

A(η, ρ)=



(η)

(η) a

(η)+ρa



,η∈ R

d−1

,ρ∈ R.

It holds that

(η)(a

(η)+ρa

)

−1

(η)=0

, (6.1.5)

whenever ξ =(η, ρ) ∈ C

satisﬁes det(a

(η)+ρa

) =0. In particular, it

holds that

(η)(a

)

−1

(η)=0

,η∈ C

d−1

. (6.1.6)

More generally, we have

(η)(a

)

−1



(η)(a

)

−1



(η)=0

,η∈ C

d−1

,k∈ N. (6.1.7)

If ξ =(η, ξ

) ∈ C

satisﬁes det(a

(η)+ξ

) =0, then dim(kerA(ξ)) = m.

If ξ =(η, ρ) ∈ R

satisﬁes det(a(η)+ρa

) =0, then

(−∞, 0) ∩ Sp



(η)(a

(η)+ρa

)

−2

(η)



= ∅. (6.1.8)

The eigenvalues of A

(τ,η) admit ﬁnite limits as τ → 0 with η =0.The

real part of these limits does not vanish. These limits are the roots µ of

det(a

(η)+iµa

)det(I

+ a

(η)(a

(η)+iµa

)

−2

(η)) = 0,

which is a polynomial equation.

Proof Since λ = 0 has a constant multiplicity and A(ξ) has only real eigen-

values, and since the unit sphere of R

is connected, the number of positive

eigenvalues of A(ξ) is constant, thus equal to p by taking ξ = e

. Since the

164 The characteristic IBVP

spectrum of A(ξ) is the opposite of that of A(−ξ), the number of negative

eigenvalues is p also. The total number of eigenvalues is n on the one hand,

m +2p on the other hand.

The second point is a direct consequence of Theorem 1.7, since π

A(η)π

the upper-left block of A(η)whenξ = e

Assume ξ =(η, ρ) ∈ R

, with det(a

(η)+ρa

) = 0. By assumption,

kerA(η, ρ) has dimension m. One easily ﬁnds

kerA(η, ρ)={(v, −(a

(η)+ρa

)

−1

(η)v) |v ∈ kera

(η)(a

(η)

+ ρa

)

−1

(η)}.

This implies dim kera

(η)(a

(η)+ρa

)

−1

(η) ≥ m. Since this kernel is a sub-

space of R

, it must therefore equal R

. Hence the matrix vanishes. In particular,

we ﬁnd

kerA(η, ρ)={(v, −(a

(η)+ρa

)

−1

(η)v) |v ∈ R

}. (6.1.9)

Analyticity ensures that (6.1.5) holds true even for complex values of (η, ρ).

Hence, formula (6.1.9) remains valid (replacing R

by C

). This shows that

dim kerA(ξ)=m whenever det(a

(η)+ξ

) = 0. Expanding (6.1.5) in terms of

1/ρ as ρ tends to inﬁnity, we obtain (6.1.6) and (6.1.7).

When ξ = 0 is real, diagonalizability tells us that kerA(ξ)

=kerA(ξ). This

amounts to

−1 ∈ Sp



(η)(a

(η)+ρa

)

−2

(η)



By homogeneity, this is equivalent to (6.1.8) (notice that (6.1.8) is obvious if L

is symmetric).

By assumption, the characteristic polynomial P (X; ξ):=det(XI

− A(ξ))

factorizes as X

Q(X; ξ), where Q is itself a polynomial in X.SinceX

unitary, the quotient Q is itself polynomial in ξ. As a polynomial in (X, ξ), Q

is homogeneous of degree m. Since the multiplicity of the null root of P (·; ξ)is

exactly m when ξ = 0 is real, we have

Q(0; ξ) =0,ξ∈ R

\{0}. (6.1.10)

Since P contains the monomial X

n−m

det a

and since det a

= 0, the degree

of Q in ξ

is exactly n − m. Using Schur’s Formula, we have

Q(X; ξ)=det(XI

n−m

− a

(η) − ξ

− X

−1

(η)a

(η)).

Substituting X = iτ and ξ

= iµ, we obtain

Q(iτ; η, iµ) = (det(−ia

)) det(A

(η, τ) − µI

n−m

Hence the eigenvalues of A

(η, τ) are the roots of Q(iτ; η, i·). Since the degree

of Q with respect to its last argument equals its total degree, these roots are

continuous functions of (η, τ) everywhere, and especially at the origin.