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

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Facts about the characteristic case 165

When τ → 0, these roots tend to those of Q(0; η, i·). Because of (6.1.10), these

limits have a non-zero real part (notice that η = 0 here.) To obtain an expression

of Q(0,η,iµ), we use again Schur’s formula, obtaining

P (X; ξ)= det(XI

n−m

− a

(η) − ξ

)det(XI

− a

(η)

×(XI

n−m

− a

(η) − ξ

))

−1

(η)).

Expanding

(XI

n−m

− a

(η) − ξ

)

−1

= −



k≥0

(η)+ξ

)

−k−1

and using (6.1.5), we ﬁnd

det(XI

−a

(XI

n−m

−a

−ξ

))

−1

)=X

det(I

+ a

+ ξ

)

−2

)

+ O(X

m+1

Hence

Q(X; ξ)=det(XI

n−m

− a

(η) − ξ

)det(I

+ a

(η)(a

(η)+ξ

)

−2

(η))

+ O(X).

Setting X =0andξ

= iµ, we ﬁnd the limit equation. 

Comments

i) It happens frequently that a

(η)a

(η) =0

n−m

(we have seen that this

is a necessary condition for stabilizability.) For instance, if L is Friedrichs

symmetric, it happens whenever a

(η) = 0, a natural fact when the IBVP

does not decouple between a trivial ODE v

= f

andanIBVPofsmaller

size. If a

(η)a

(η) =0

n−m

, then the matrix A

(η, τ) is unbounded as

τ → 0, though its eigenvalues have ﬁnite limits. Hence it does not remain

uniformly diagonalizable: Eigenvectors associated to distinct eigenvalues

tend to become parallel. However, eigenvalues do not merge in general (a

counterintuitive fact, but A

does not have a limit as τ → 0); for instance,

if m = n − 2, then p =1 and Q(τ,iη,·) has exactly one root of positive

real part and one of negative real part, this dichotomy persists as τ → 0,

since the roots of Q(0; iη, ·) may not belong to iR. Hence the roots remain

distinct.

ii) In the Friedrichs-symmetric case, (6.1.6) tells us exactly that the range

R(a

(η)) is an isotropic subspace for the quadratic form q

deﬁned on

n−m

= R

by (a

)

−1

.Sinceq

is non-degenerate with p positive and

p negative eigenvalues, its maximal isotropic subspaces have dimension

p. Hence rk a

(η) ≤ p. Let us examine two examples. The ﬁrst one is

Maxwell’s system of electromagnetism, for which (d, n, m, p)=(3, 6, 2, 2).

Easy calculations show that rk a

(η)=2=p for every η =0. Thus our

inequality is sharp. Its optimality is reinforced by the following observation.

166 The characteristic IBVP

Since p =(n − m)/2 = 2, the set I

)ofq

-isotropic planes is a manifold

of dimension one

. This dimension ﬁts with that of the projective space on

which the map η → R(a

(η)) is deﬁned. One checks easily that the range

of this map is precisely a connected component of I

The second example is the linearized isentropic gas dynamics (or

acoustics). With the normalization λ ≡ 0, the ground velocity is zero. In

suitable units, the operator reads







+divu

+ ∇ρ



Here, (n, m, p)=(d +1,d− 1, 1). Again, the equality rk a

(η)=1=p

holds. However, the set of isotropic lines is discrete (it consists in two

elements). Thus the map η → R(a

(η)) is constant, although it is deﬁned

on a projective space of dimension d − 2. This constancy could be checked

by direct calculation.

Why should λ ≡ 0 hold?

Our ﬁrst motivation is an observation made by Majda and Osher in [127]. As

mentioned in Proposition 6.2, the eigenvalues of A

(τ,η) have ﬁnite limits as

τ tends to zero, the singularity of A

. It turns out that when the eigenvalue,

responsible for the characteristic nature of the boundary, is not a linear function

of the frequency, then A

(τ,η) admit eigenvalues that tend to inﬁnity as τ →

0. Majda and Osher showed that such a behaviour, although compatible with

strong L

-estimates, is responsible for a loss in the estimates of the derivatives.

Typically, boundary data of class H

are needed in order that the trace of the

solution be of class H

The following analysis, due to Ohkubo [150], is related to the former com-

ment and gives a partial explanation for the choice of a constant (or a linear)

eigenvalue. Let us assume that the IBVP is well-posed in L

,inthesenseof

Deﬁnition 4.6. We wish to show that this well-posedness extends to the class

. Thus we give ourselves data f = Lu, g = γ

Bw and a ≡ 0thatbelongto

in their respective domains. By assumption, we know that there exists a

unique solution u =(v, w) that is at least square-integrable on (0,T) ×Ω, and

such that w has a square-integrable trace along the boundary. Thus we seek for

the integrability of the ﬁrst-order derivatives. First of all, we diﬀerentiate the

equation Lu = f in directions parallel to the boundary (tangential derivatives).

We see that ∂

u solves the IBVP

= ∂

f, γ

= ∂

g, X

(0) = 0.

Since the data are square-integrable, we obtain that ∂

u ∈ L

and that ∂

w has

a well-deﬁned trace that is square-integrable too. The same procedure may be

It may be identiﬁed to the set of straight lines contained in a one-sheeted hyperboloid. In

particular, it has two connected components.

Facts about the characteristic case 167

applied to the time derivative X

= ∂

f, γ

= ∂

g, X

(0) = f (0).

Hence ∂

u and ∂

w have the required properties.

We now turn towards the normal derivative ∂

u. Using the second part of

the system, we have

∂

w =(a

)

−1

− ∂

w − a

(∇

)w − a

(∇

)v) . (6.1.11)

Equation (6.1.11) immediately tells us that ∂

w is square-integrable and admits a

square-integrable trace

. However, it remains to study ∂

v. To handle this term,

we diﬀerentiate the ﬁrst part of the system and we use (6.1.11). We obtain an

ODE

∂



∂

v − a

(∇

)(a

)

−1



= ∂

− a

(∇

)(a

)

−1

+ a

(∇

)(a

)

−1

(∇

+ a

(∇

)(a

)

−1

(∇

)v.

Assuming Property (6.1.6) and also

(η)(a

)

−1

(η) ≡ 0, (6.1.12)

which together imply (6.1.5), the right-hand side reduces to ∂

−

(∇

)(a

)

−1

, a square-integrable data. Thus ∂

v − a

(∇

)(a

)

−1

w, hence

∂

v, is square-integrable (notice that we do not need a boundary condition for

∂

v.)

Ohkubo’s conclusion is that a suﬃcient condition for the IBVP be well-posed

in H

is that both (6.1.6) and (6.1.12) hold true. Though it is not clear that these

conditions are necessary, it looks reasonable to restrict ourselves to the class of

such operators. This is the choice that we shall adopt in Section 6.2. Using a series

expansion, we see that these conditions imply (6.1.5), which means that kerA(ξ)

has dimension m whenever det(a

(η)+ρa

) = 0. By a continuity argument, we

see that zero is an eigenvalue of A(ξ) of multiplicity larger than or equal to m,

for every direction ξ, whence λ ≡ 0.

6.1.3 Facts in two space dimensions

From now on, we assume that our operator L is symmetric, with a zero eigenvalue

of constant multiplicity m ≥ 1. We denote by p =(n − m)/2thenumberof

positive (negative) eigenspeeds in every direction ξ =0.

We shall need to consider the isotropic cone of A(ξ):

Γ(ξ):={u ∈ R

; u

A(ξ)u =0}.

Notice that the analysis would be complete at this stage if the boundary was not characteristic.

This explains why only one condition (UKL) governs the well-posedness of non-characteristic IVBPs

in every Sobolev space, while we need an extra condition in the characteristic case, to pass from the

well-posedness in L

to that in H

The way we used (6.1.11) does not seem optimal. We expect that a cleverer analysis could yield

the weaker condition (6.1.5), instead of (6.1.6) and (6.1.12).

168 The characteristic IBVP

The intersection of these cones as ξ runs through R

is denoted Γ. Obviously,

kerA(ξ) is a subset of Γ(ξ). Actually, we proved in Theorem 1.7 that kerA(ξ



) ⊂

Γ(ξ) for every ξ,ξ



with ξ



= 0. This implies not only

kerA(ξ



) ⊂ Γ, (6.1.13)

but also

kerA(ξ)+kerA(ξ



) ⊂ Γ(ξ) ∩ Γ(ξ



), ∀ξ =0,ξ



=0. (6.1.14)

We now assume that d = 2 and denote by H the sum of the kernels kerA(ξ)as

ξ runs the unit circle. Then Γ = Γ(ξ) ∩ Γ(ξ



) whenever ξ ∧ ξ



= 0. Hence (6.1.13),

(6.1.14) and the fact that each Γ(ξ) is a quadratic cone give us:

Proposition 6.3 Assume that d =2, the operator L is symmetric and the zero

eigenvalue has constant multiplicity m ≥ 1. Then it holds that



ξ=0

kerA(ξ)=:H ⊂ Γ:=

Γ(ξ). (6.1.15)

Comments

It is worth pointing out that the sum (of vector spaces of dimension m)

H can be a rather big subspace. However, Propositions 6.2 and 6.3 tell us

that, as any isotropic subspace Γ(ξ), its dimension is less than or equal

to m + p. What is even more striking is that in the examples of physical

interest

, the dimension of this sum is actually equal to m + p. Hence all

the quadratic forms A(ξ) have a maximal isotropic subspace in common!

As we shall see in the following, this property plays a crucial role, so that

we shall take it as an assumption for the planar restrictions of L.

Property (6.1.15) would be false in higher space dimension, for there is

no reason why u

A(ξ





would vanish for u ∈ kerA(ξ)andu



∈ kerA(ξ



)

when ξ,ξ



,ξ



are linearly independent.

In the next result, we assume that H is of maximal dimension (namely m + p)

and we use the canonical form for the symbol A(ξ):

A(ξ)=



(ξ)

(ξ) a

(ξ)



∈ Sym

:= a

Since H contains kerA

= R

×{0}, it splits as R

× H

where dim H

= p.

Lemma 6.1 Assume that d =2 and dim H = m + p (maximal dimension).

Then H

is its own a

-orthogonal space. If ξ =0, then H is its own A(ξ)-

orthogonal subspace.

Proof Since a

is non-degenerate, dim H

⊥

=2p − dim H

= p.SinceH

⊥

contains H

, it thus equals H

. Similarly, H is self-orthogonal with respect

For example, acoustics, linearized gas dynamics, Maxwell’s equations, linear elasticity.

Facts about the characteristic case 169

to A(ξ). Since it contains kerA(ξ), its A(ξ)-orthogonal must be of dimension

m + n − dim H =dimH, whence the result. 

Corollary 6.1 Assume that d =2, the operator L is Friedrichs symmetric and

that the subspace H :=



ξ=0

kerA(ξ) is of maximal dimension, namely m + p.

Then it is possible to choose the u-co-ordinates in such a way that A(ξ) reads

A(ξ)=



m+p

β(ξ)

δ(ξ)



, (6.1.16)

where δ(ξ) ∈ Sym

, while β(ξ) ∈ M

(m+p)×p

is injective for ξ =0.

Conversely, if A(ξ) reads as above and β(ξ) is injective for ξ =0, then the

operator L = ∂

+ A(∇

) is Friedrichs symmetric, and zero is an eigenspeed of

multiplicity m in every direction of the plane.

Proof Assume that H is of maximal dimension m + p. One chooses orthogonal

co-ordinates in which H = R

m+p

×{0}. Since the co-ordinates are orthogonal,

the symbol remains symmetric. Because H is isotropic for every A(ξ), the symbol

has the form described in (6.1.16), with δ(ξ) symmetric. Let q belong to kerβ(ξ)

and ξ =0.Then

A(ξ)







δ(ξ)q



∈ H

⊥

From Lemma 6.1, we deduce that (0,q)

∈ H, which means q = 0. Hence β(ξ)

is injective.

Conversely, assuming that β(ξ) is injective, we see that kerA(ξ)=kerβ(ξ)

{0}. Since the rank of β

equals that of β,namelyp, we ﬁnd that kerβ(ξ)

is of

dimension m + p − p = m. 

In the following, we shall consider symmetric operators in any space dimen-

sion d ≥ 2. Given a vector η ∈ R

d−1

= R

d−1

×{0}, we shall denote H(η)thesum

of the kernels kerA(ξ)asξ = 0 runs over the plane spanned by η and e

.From

Proposition 6.3, H(η) is isotropic for both A

and A(η). Since H(η) contains

kerA

= R

×{0}, it splits as R

× H

(η), where H

(η) is isotropic for both

and a

(η). As mentioned above, we shall assume that H(η) is of maximal

dimension, namely m + p, meaning that H

(η) has dimension p. In particular,

we have the following identity:

)

−1

(η)=H

(η)

⊥

. (6.1.17)

6.1.4 The space E

−

(0,η)

We now characterize the limit E

−

(0,η) of the stable subpace E

−

(τ,η)when

τ → 0 with Re τ>0. Recall that the latter is associated to the diﬀerential-

algebraic system

(τI

+ iA(η))u + A



=0.

170 The characteristic IBVP

Following the analysis of the previous section, we assume that L is Friedrichs

symmetric, that zero is an eigenvalue of multiplicity m ≥ 1, that d ≥ 2 and that

dim H

(η)=p for every η ∈ R

d−1

\{0}. As above, we use the block decomposi-

tion u =(v, w)

with v ∈ C

, w ∈ C

n−m

= C

and

A(η + ξ



(η)

(η) a

(η)+ξ



If w ∈ H

(η)andv ∈ R

,then(v, w)

∈ H(η), hence (v, w)

∈ Γ(η), mean-

ing that 2v

(η)w + w

(η)w = 0. Since this is valid for every v, we conclude

that a

(η)w = 0, whence

(η) ⊂ kera

(η). (6.1.18)

Introducing H

(η):=H

(η)

⊥

(with respect to the standard Euclidian product),

we derive that

R(a

(η)) ⊂ H

(η). (6.1.19)

The diﬀerential system rewrites

τv + ia

(η)w =0,ia

(η)v +(τ + ia

(η))w + a



=0,

from which we may eliminate v:



+(τ + ia

(η))w +

(η)a

(η)w =0. (6.1.20)

We shall denote e

−

(τ,η)thew-projection of E

−

(τ,η), that is the stable subspace

of the system (6.1.20).

In the decomposition C

= H

(η) ⊕

⊥

(η), we denote by π = π(η)the

orthogonal projection onto H

(η). With q

= πw and q

= τ

−1

(1 − π)w ∈ H

(η),

we have w = q

+ τq

and the system reads



+ ia

(η)q

+ a

(η)a

(η)q

+ τ{a



+ q

+ ia

(η)q

+ τq

} =0. (6.1.21)

We write an equivalent system by projecting onto H

(η)andH

(η) separately.

The important point is that the ﬁrst three terms in (6.1.21) belong to H

(η)

(use the fact that H

(η) is isotropic simultaneously for a

and for a

(η), and use

(6.1.19)). Hence the system becomes



+ ia

(η)q

+ a

(η)a

(η)q

= −τ(1 − π){a



+ ia

(η)q

+ τq

πa



+ q

+ iπa

(η)q

=0.

This can be written in a compact form as





= Z(τ,η)





Since we proceeded by changes of variables, Z(τ,η) is conjugated to A

(τ,η),

hence it is of hyperbolic type. Its stable subspace is of dimension p. What is even

Facts about the characteristic case 171

more important is that Z(τ, η) has a limit Z(0,η) that is hyperbolic too. As a

matter of fact, the formal limit of the above system, as τ → 0, writes as



+ ia

(η)q

+ a

(η)a

(η)q

=0, (6.1.22)

πa



+ q

+ iπa

(η)q

=0. (6.1.23)

Since a

is invertible, the ﬁrst line determines q



in terms of q

, q

. On the other

hand, if πa

q =0,thena

q ∈ H

(η). From (6.1.17), we conclude that q ∈ H

(η).

Hence the second line determines q



, showing that the system can be written in

the form q



= Z(0,η)q and that Z(0,η) is the limit of Z(τ,η).

Proposition 6.4 The matrix Z(0,η) is hyperbolic: It does not have pure imag-

inary eigenvalues.

Proof Let q satisfy Z(0,η)q = iαq with α ∈ R. Denoting by ir

+ r

the

decomposition of q,wehave

−(αa

+ a

(η))r

+ a

(η)a

(η)r

=0,

π(αa

+ a

(η))r

+ r

=0.

One may assume that r

and r

are real. Multiplying the ﬁrst line by r

,the

second one by r

and summing, we obtain the identity

+ |a

(η)r

=0.

It follows that q ∈ H

(η) ∩ kera

(η). Then the equations reduce to (αa

(η))q ∈ H

(η). All this means

A(ξ)





∈{0}×H

(η)=H(η)

⊥

,ξ:= η + αe

Using Lemma 6.1, we conclude that (0,q)

∈ H(η), namely q ∈ H

(η). Since

q ∈ H

(η), this gives q =0. 

Lemma 6.2 There exists a linear map N (0,η) such that the equation of the

stable subspace of Z(0,η) is

= N(0,η)q

Moreover, one has kerN (0,η) ⊂ kera

(η).

Proof Since the dimension of the stable subspace is p,wehavetoprovethat

its intersection with H

(η) is trivial. Thus, let q(x) be a solution of the limit

diﬀerential system, vanishing at +∞. Multiply (6.1.22) by q

∗

and (6.1.23) by q

∗

sum both equations, then take the real part. One obtains

+ |a

(η)q

Re (q

∗

)=0.

172 The characteristic IBVP

Integrate now from 0 to +∞, we obtain



+∞



+ |a

(η)q



dx =Re(q

(0)

∗

(0)).

If q

(0) vanishes, we deduce that q

≡ 0, hence q

(0) = 0. This proves that the

stable subspace is parametrized by q

Assume now that N (0,η)r =0withr ∈ H

(η). The solution of the diﬀerential

system with initial data q

(0) = 0 = N (0,η)r, q

(0) = r, vanishes at +∞. Hence

≡ 0anda

(η)q

≡ 0. Whence a

(η)r =0. 

A standard perturbation argument tells us that the stable subspace of Z(0,η)

is the limit of that of Z(τ,η). Thus there exists a linear map N(τ, η) that depends

smoothly on (τ,η)whenτ is small, such that the latter has an equation q

N(τ,η)q

. In other words, E

−

(τ,η) is given by

−

(τ,η)={(−ia

(η)r, N (τ,η)r + τr); r ∈ H

(η)}.

Passing to the limit as τ → 0, we obtain

{(−ia

(η)r, N (0,η)r); r ∈ H

(η)}⊂E

−

(0,η). (6.1.24)

When the right-hand side in (6.1.24) has dimension p, the embedding is an

equality that gives a complete description of E

−

(0,η). However, one does not see

any good reason why this would hold true for every η ∈ R

d−1

\{0}.

From Lemma 6.2, equality occurs in (6.1.24) unless N(0,η) is singular. We

examine the latter property. If r ∈ kerN(0,η), then there exists (see the proof of

the lemma) a non-trivial q(x) that takes values in H

(η) ∩ kera

(η), and such

that (see (6.1.23))

π(ξ



+ ia

(η)q)=0.

Let x

exp(σx)r

be a leading term in q as x → +∞,thenReσ<0andπ(σa

(η))r

= 0. In other words, the vector U := (0,r

)

satisﬁes

U ∈ H(η)

⊥

,A(η + ξ

)U ∈ H(η)

⊥

, (6.1.25)

for a ξ

= −iσ that is non-real. Conversely, let U = 0 satisfy (6.1.25) with a non-

real ξ

.WehaveU =(0,r)

with r ∈ H

(η). Up to a complex conjugacy, we

may assume that the imaginary part of ξ

is positive. Then q(x):=exp(iξ

is a decaying solution of our diﬀerential system that satisﬁes q

(0) = 0 and

(0) = r = 0. Hence N (0,η) is singular. We summarize our results in the

following statement.

Proposition 6.5 The linear map N(0,η) is singular if and only if there exists

a non-zero vector U ∈ C

that solves the generalized eigenvalue problem (6.1.25).

When N(0,η) is regular, then

−

(0,η)={(−ia

(η)M(0,η)z, z); z ∈ H

(η)}, (6.1.26)

Facts about the characteristic case 173

where M (0,η):=N(0,η)

−1

. In particular, the w-projection of E

−

(0,η), as well

as the limit of e

−

(τ,η) as τ → 0,equalH

(η).

Note that the eigenvalue problem (6.1.25) could not have a non-trivial solution

when ξ

is real, as shown in Proposition 6.4.

Examples of physical interest

In practice, we are interested in examples that come from physics, where conser-

vation laws are likely to hold. We have in mind ﬂuid dynamics, electromagnetism

and elasticity. The systems are known as acoustics, linearized gas dynamics,

Maxwell’s equations and linear isotropic elasticity. All of them have a zero

eigenspeed of constant multiplicity; in the case of linearized gas dynamics, one

must assume a solid boundary, and choose a frame moving with the boundary

velocity.

Except for the elasticity, one has the remarkable feature

(η) ≡ 0

. (6.1.27)

We emphasize that (6.1.27) prevents the eigenvalue problem (6.1.25) from hav-

ing a non-trivial solution, since H(η)isitsownA

-orthogonal. Hence the w-

projection of E

−

(0,η) equals H

(η) in every direction.

Let us examine linear elastodynamics. There is no loss of generality in

assuming that d = 2, since all the above analysis deals within planes spanned

by e

and a tangent vector η. The system reads as

∂

F + ∇z =0,∂

z +divT =0,

with F (x, t) ∈ M

(R), z(x, t) ∈ R

and

T = λ(F + F

)+µ(TrF )I

The vector ﬁeld z represents the opposite of the material velocity, while the stress

tensor T is an isotropic function of the inﬁnitesimal deformation tensor. Actually,

the skew-symmetric part F

− F

decouples from the rest and we may restrict

ourselves to the system that governs the evolution of z and the symmetric part

of F . Since the system admits a quadratic energy

|z|

|F + F

(TrF )

it is Friedrichs symmetrizable. Given c

√

2λ + µ (the velocity of ‘pressure

waves’), the choice of variables

u := (2



λ(λ + µ) F

√

λ (F

+ F

),c

+ µF

)

puts the system in the symmetric form Lu =0, where A(ξ) has the canonical

form

A(ξ)=



(η)

(η) a

(η)+ξ



,ξ=(η,ξ

174 The characteristic IBVP

We have

(η)+ξ



b(ξ)



,b(ξ)=



√

λη

√

ηµ/c



(6.1.28)

and a

(η)=(0, 0, 2η



λ(λ + µ)/c

, 0)

In these variables, one checks that H(e

)=R

×{0}.Sincen =5, m =1

and p =2,H(e

) is of dimension m + p and the previous analysis applies. Let us

consider now the eigenvalue problem (6.1.25). If (0,r)

is a solution, then r has

the form (0,q)

with q ∈ C

and q ∈ kerb(ξ) ∩ kera

(η). Here, ξ =(η, ξ

) with

a non-real ξ

.Since

det b(ξ)=(ξ

− µ

)

√

cannot vanish (otherwise ξ

would be real), we must have q =0.Thus,thereis

no non-trivial solution. We conclude that N(0, e

) is non-singular.

More generally, in any space dimension and for every tangent vector η =0,

the space H(η) is of dimension m + p and the limit E

−

(0,η) is parametrized as

in (6.1.26).

6.1.5 Conclusion

Let us summarize the results obtained in this section. It is diﬃcult to characterize

hyperbolic operators L admitting an eigenvalue λ of constant multiplicity m ≥ 1,

such that λ(e

) = 0. Hence, we have simpliﬁed the class of admissible operators

by using as natural as possible arguments.

First, we explained why a linear λ is likely to occur. Without loss of generality,

this allows us to restrict ourselves to λ ≡ 0. A nice consequence of this assumption

is the boundedness of the spectrum of A

(τ,η)asτ → 0, a property needed for

an acceptable estimate of derivatives of the solution of the IBVP. Another one

is that n − m =: 2p is even whenever d ≥ 2.

Next, the concept of stabilization led us to a slightly more restricted class. We

showed that symmetric operators are stabilizable, but that far-from-symmetric

ones (say a

≡ 0

n−m

) are not. Since the examples encountered in the natural

sciences are endowed with a quadratic ‘entropy’ (being often an energy) and thus

are Friedrichs symmetrizable, we restrict ourselves to symmetric symbols A(ξ).

Without loss of generality, one may write

A(ξ)=



(η)

(η)+ξ



,ξ= η + ξ

For every non-zero tangent vector, we introduce the sum H(η)ofthekernels

kerA(ξ)asξ runs over Rη ⊕ Re

. We showed that it is an isotropic subspace

for all these A(ξ) simultaneously. In particular, its dimension is not larger than

m + p. Examples from physics suggest we can assume that

i) the dimension of H(η)ismaximal, namely it equals m + p for every tangent

η =0,