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

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Main results in the non-characteristic case 125

Testing (4.5.27) against functions L

∗

v where v ∈ D(Ω × R), we obtain Lu = f in

the distributional sense and γ

Bu − g, Mv = 0. However, the equality kerC =

kerB)

⊥

and the fact that B is onto imply that M :kerC → R

is onto, and

therefore we may replace Mz by any test function. It follows that γ

Bu = g.

With Lemma 4.7 and Corollary 4.2, we conclude with the well-posedness of

the Boundary Value Problem:

Lemma 4.8 Given f(x, t) and g(y,t) in the classes L

, there exists a unique u

in L

, solution of the Boundary Value Problem Lu = f and γ

Bu = g, relative

to the half-space Ω × R.

Moreover, this solution satisﬁes (4.5.26).

Applying this result to the adjoint problem, we obtain

Corollary 4.4 The space X

deﬁned above coincides with L

−γ

4.5.4 Improved estimates

In order to pass from the well-posedness of the BVP to that of the IBVP, we

need to improve (4.5.26) in the following way.

Lemma 4.9 With the above assumptions, every smooth and compactly sup-

ported function u satisﬁes for every T ∈ R:

−2γT



Ω

u(T )

dx + γ



−∞



Ω×R

−2γt

u

dx dt



−∞



∂Ω×R

−2γt

γ

u

dy dt (4.5.31)

≤ C





−∞



Ω×R

−2γt

Lu

dx dt +



−∞



∂Ω×R

−2γt

γ

Bu

dy dt



where the constant C does not depend on γ, T or u.

Proof The ﬁrst step is to replace the integrals over R by integrals over (−∞,T).

This immediately follows from Corollary 4.3 and from the existence part. Let

be deﬁned by

f = Lu if t>T and

f =0 if t<T. Deﬁne in a similar way ˜g.

Then let ˜u ∈ L

be the solution associated to

f and ˜g.Then˜u vanishes for t<T

(Corollary 4.3). The estimate (4.5.26) for v := u − ˜u precisely reads



−∞



Ω×R

−2γt

u

dx dt +



−∞



∂Ω×R

−2γt

γ

u

dy dt

≤ C





−∞



Ω×R

−2γt

Lu

dx dt +



−∞



∂Ω×R

−2γt

γ

Bu

dy dt



126 Initial boundary value problem in a half-space with constant coeﬃcients

We turn to the estimate of the L

-norm of u(T ). The proof below, which we

borrow from Rauch’s PhD thesis [161], covers the physically signiﬁcant case of a

Friedrichs-symmetric operator. For a proof in full generality, we refer to [162].

Thus let us assume that L is Friedrichs symmetric. Integrating the identity

∂



−2γt

|u|





∂



−2γt

∗



=2e

−2γt



Lu, u−γ|u|



on Ω × (−∞,T), we obtain

−2γT

u(T )

+2γ



−∞

−2γt

u(t)



−∞

−2γt

(u(t),Lu(t))

dt +



−∞

−2γt

u(t),γ

u(t))

dt.

The Cauchy–Schwarz inequality yields

−2γT

u(T )

≤

2γ



−∞

−2γt

Lu(t)

dt + C



−∞

−2γt

γ

u(t)

dt.

We conclude with the help of (4.5.26). 

Working now as in Lemma 4.7, we obtain

Proposition 4.6 Let u, Lu and γ

Bu be of class L

.Thenu is continuous in

time, with values in L

(Ω), and satisﬁes 4.5.31.

4.5.5 Existence for the initial boundary value problem

Existence Given three functions f ∈ L

(Ω × R

), g ∈ L

(∂Ω × R

)andu

∈

(Ω), we deﬁne a linear form on L

−γ

(see Corollary 4.4 above) by



∗

v):=



+∞



Ω

(v, f)dx dt +



Ω

· v(0) dx +



+∞



∂Ω

(g, γ

Mv)dy dt.

We may think that the time integrals run over R

, and that we have extended f

and g by zero to negative times.

Once more, 

is well-deﬁned on some subspace of L

−ρ

and continuous, for

every ρ larger than γ. This gives the existence of a u

in L

, with the property

that 

(F )=(u

,F)everyF in L

−ρ

.SinceL

−ρ

∩ L

−γ

is dense in L

−γ

, we see

that u

= u

almost everywhere. In other words, the common value u belongs to

for every ρ larger than γ.

Testing 

against elements of X

given by the vsinD(Ω × R

∗

), we obtain

the diﬀerential equation

Lu = f, (t =0)

Main results in the non-characteristic case 127

in the distributional sense. This allows us to apply Green’s identity



Ω×(S,T )

((Lu, v) − (u, L

∗

v))dx dt + γ

Bu,γ

Mv

∂Ω×(S,T )

+ γ

Nu,γ

Cv

∂Ω×(S,T )

= u(T

−

),v(T )−u(S

),v(S), (4.5.32)

both in (−∞, 0) and (0, +∞). This gives



Ω×R

∗

(v · Lu − u · L

∗

v)dx dt = γ

Bu,γ

Mv

∂Ω×R

+ γ

Nu,γ

Cv

∂Ω×R

−[u]

t=0

,v(0)

Ω

where [u]

t=0

denotes the diﬀerence of the traces of u at t =0

and t =0

−

.We

thus obtain

γ

Bu,Mv

∂Ω×R



+∞



∂Ω

(g, Mv)dy dt, (4.5.33)

[u]

t=0

,v(0)

Ω



Ω

· v(0) dx. (4.5.34)

The argument that we developed in the previous section applies to (4.5.33)

and shows that γ

Bu = gχ

t>0

. In particular, Lu and γ

Bu vanish for t<0.

Given >0, let φ ∈ C

∞

be a function of time, satisfying φ ≡ 1fort<−,and

φ ≡ 0fort>−/2. Then u



:= φu is in L

,aswellasLu



and



.Since



and γ



vanish for t<− (where they coincide with Lu and γ

Bu), and

since u



, Lu



, γ



belong to L

for every ρ>γ, Corollary 4.3 tells that u



vanishes for t<−.Since is arbitrary, we deduce that u vanishes on t<0.

In particular, the trace of u at t =0

−

is zero, and (4.5.34) amounts to saying

that u(t =0

)=u

. Hence there exists a solution of the full IBVP, which lies in

(Ω × R

). We shall see in a moment that this is unique. Because of the bound

|

∗

v)|≤C



−1

f

+ γ

−1/2

u



+ γ

−1/2

g



L

∗

v

−γ

given by (4.5.28), this solution satisﬁes the estimate

γu

≤ C



f

+ u



+ g



, (4.5.35)

for every γ>γ

,whereC does not depend on γ.

Uniqueness Let u ∈ L

(Ω × R

) be such that Lu = 0. Hence γ

u and u(0)

make sense. Assume that u(0) ≡ 0andγ

Bu ≡ 0. Green’s Formula yields



Ω×R

(u, L

∗

v)dx dt = γ

Nu,γ

Cv

∂Ω×R

Note that Lu itself is not in L

(Ω × R).

128 Initial boundary value problem in a half-space with constant coeﬃcients

for every v ∈ D (

Ω × R). Extending u to negative times by zero, we obtain



Ω×R

(u, L

∗

v)dx dt = γ

Nu,γ

Cv

∂Ω×R

In particular, Lu =0onΩ× R and γ

Bu =0on∂Ω × R. Using Corollary 4.2,

we deduce u ≡ 0.

Improved estimates Apply (4.5.32) with (S, T )=(0, +∞):



Ω×(0,+∞)

(u, L

∗

v)dx dt −γ

Nu,γ

Cv =



Ω×(0,+∞)

(f,v)dx dt + g, γ

Mv

+ u

,v(0).

According to (4.5.28), the left-hand side is bounded above by



−1

f

+ γ

−1/2

u



+ γ

−1/2

g



(L

∗

v

−γ

+ γ

Cv

−γ

) .

We thus obtain the estimate

γu

+ γ

Nu

≤ C



f

+ u



+ g



However, since kerN ∩ kerB =kerA

= {0}, this really means

γu

+ γ

u

≤ C



f

+ u



+ g



. (4.5.36)

If u

vanished, the argument employed for uniqueness would show that, after

extension by zero to negative times, Lu still belongs to L

. Then (4.5.32) would

be valid. We are now going to prove that it remains valid for general data u

. For that purpose, it is enough to assume f ≡ 0andg ≡ 0. By density, we

may also assume that u

belongs to D(Ω).

From the uniqueness result above, tangential derivatives ∂

u (α =1,...,

d − 1), being the solutions of the IBVP corresponding to f

=0, g

=0 and

0α

:= ∂

, belong to L

. Similarly, ∂

u is the solution corresponding to f

=0,

= 0 and the initial data −



∂

, and thus belongs to L

. Hence

∂

u = −∂

u −



∂

u is L

. Hence u is H

(γ) and we may integrate the

energy identity on the slab Ω × (0,T), as in the proof of Lemma 4.7.

4.5.6 Proof of Theorem 4.3

It remains to treat the case of a time interval (0,T). Let f ∈ L

(Ω × (0,T)),

g ∈ L

(∂Ω × (0,T)) and u

∈ L

(Ω) be given. The extensions of f and g by

zero, for times t>T, belong to L

for every γ. We thus obtain a unique solution

u of the IBVP in Ω × R

. Its restriction to times t ∈ (0,T) furnishes a solution

of the IBVP in the slab, with the required estimate.

We now prove uniqueness. Assume that f, g and u

vanish identically. and

u ∈ L

(Ω × (0,T)) satisﬁes the IBVP. If  ∈ (0,T), choose φ in D(R) such that

Main results in the non-characteristic case 129

φ(t)=1 if t<T−  and φ(t)=0 if t>T− /2. Extending u by zero to t ∈

(0,T), we obtain that φu and L[φu]areinL

for every γ.SinceL[φu]and

B(φu) vanish for t<T− , Corollary 4.3 tells that u vanish for T − .Since

is arbitrary, we conclude that u =0.

4.5.7 Summary

We summarize below the strategy that we followed for proving the existence and

uniqueness, and establishing estimates for the IBVP.

With the help of the Kreiss’ symmetrizer, we establish an a priori estimate

in L

,whenu ∈ D(Ω × R

By truncation and convolution, we extend this estimate to the case where

u, Lu and γ

Bu are of class L

This implies uniqueness for the BVP.

This implies also a causality property: If Lu ≡ 0andγ

Bu ≡ 0 in the past

(say for t<T), then u ≡ 0 in the past.

Since (UKL) and constant hyperbolicity pass to the adjoint BVP, we

also have an estimate for the latter when its data and solution are of

class L

−γ

By a duality argument, which uses Hahn–Banach and Riesz Theorems, the

BVP is solvable in the class L

, for data in L

Thanks to the existence result and to the causality, one may replace R

(−∞,T) in the estimates. Also, the IBVP with a zero initial data admits

a solution.

Thanks to the energy estimate (when the operator is Friedrichs symmetric),

we also have an estimate of u(T )inL

(Ω).

We deduce that, for data in L

, the solution is continuous with values in

(Ω).

Thanks to the causality property and to the time-pointwise estimate, and

using a duality argument, the IBVP is solvable in the class L

As above, the IBVP has a causality property and estimates on Ω × (0,T)

insteadofΩ× (0, +∞).

4.5.8 Comments

BVP vs IBVP We emphasize that the existence and the estimates for the

Boundary Value Problem do not automatically imply the corresponding results

for the Initial Boundary Value Problem. Let us consider an abstract diﬀerential

equation

= Ax + f, (4.5.37)

130 Initial boundary value problem in a half-space with constant coeﬃcients

which is a model for the homogeneous BVP (Bu(0,t) ≡ 0). Let us assume that

(4.5.37) has an existence property with an estimate



−2γt

u(t)

dt ≤



−2γt

f(t)

dt, ∀γ>0, (4.5.38)

where the norm is taken in some Hilbert space. Letting γ → +∞, we ﬁnd as

usual that if f vanished in the past, then u does too.

It is not hard, using the Parseval Identity, to prove that τI −A has a bounded

inverse for every τ of positive real part, with the estimate

(τI − A)

−1

≤



Re τ

. (4.5.39)

However, (4.5.38) is essentially equivalent to (4.5.39), and it is not possible from

there to derive a pointwise estimate for the semigroup generated by A (assuming

that it exists). This prevents us from proving anything about the Cauchy problem

for (4.5.37) without some additional information about A.

Historically, this diﬃculty was encountered by Kreiss [103], who solved only

the (non-homogeneous) BVP. The extension of his results to the full IBVP

was obtained later by Rauch [161, 162]. We shall face this diﬃculty in the

homogeneous IBVP (that is with a zero boundary condition, see Chapter 7)

and in the so-called WR case, see Chapter 8.

4.6 A practical tool

4.6.1 The Lopatinski˘ı determinant

We are going to deﬁne in this section a (not very) practical tool called the

Lopatinski˘ı determinant. This is a function (τ,η) → ∆(τ, η), with the following

properties

i) It is well-deﬁned for η ∈ R

d−1

and Re τ>0,

ii) It is jointly analytic in (τ,η), and therefore holomorphic in τ,

iii) It vanishes precisely at points violating the Lopatinski˘ı condition.

To ﬁll these three properties, it is enough to construct a basis

β(τ,η)={X

(τ,η),...,X

(τ,η)}

of E

−

(τ,η), which satisﬁes the ﬁrst two ones. Then we deﬁne the Lopatinski˘ı

determinant as

∆(τ,η):=det(BX

(τ,η),...,BX

(τ,η)), (4.6.40)

since the vanishing of the determinant is equivalent to the existence of a non-

trivial linear combination

X :=



(τ,η)

such that BX = 0, which amounts to X ∈ kerB ∩ E

−

(τ,η).

A practical tool 131

To construct β, we use the procedure described by Kato in [95], Section

4.2, which goes as follows. Let z → P (z) be an analytic function with values in

projectors, where z ranges on a simply connected domain. From identity P

= P ,

one easily ﬁnds P



=[Q, P ], with Q := [P



,P]. Then the linear Cauchy problem



= QM, M (z

)=I

is globally solvable and yields the formula

M(z)

−1

P (z)M(z) ≡ P (z

Therefore, given a basis β

of the range of P (z

), the set β(z):=M(z)β

is a

basis of the range of P (z), and is analytic in z.

When P depends on several variables, this procedure cannot be done simul-

taneously in general. If Q

:= [∂P/∂z

,P], simultaneity requires the compati-

bility condition ∂Q/∂z

− ∂Q/∂z

=[Q

], though in practice we have oddly

∂Q/∂z

− ∂Q/∂z

=2[Q

]! However, we may apply Kato’s procedure succes-

sively to each of the arguments, provided that at each step, a Cauchy problem

is posed in a simply connected region. Because ODEs with analytic coeﬃcients

propagate analyticity, the resulting matrix M is jointly analytic in its arguments.

The inelegant fact is that the result depends on the order in which we solve ODEs,

because of the lack of compatibility.

Applying these ideas to the projectors π

−

(τ,η), which are jointly analytic,

we now have

Lemma 4.10 For η ∈ R

d−1

and Re τ>0, the space E

−

(τ,η) admits a basis

β(τ,η), which is jointly analytic in (τ,η), and thus holomorphic in τ.

The symmetric case We restrict ourselves here to the non-characteristic case.

When the operator L is symmetric, that is A(ξ)

= A(ξ) for every ξ in R

,an

alternative construction can be done, with the help of the following

Lemma 4.11 In the symmetric case with a non-characteristic boundary, one

has for every η ∈ R

d−1

and Re τ>0

) ∩ E

(τ,η)={0}, (4.6.41)

where E

) stands for the unstable invariant subspace of A

To our knowledge, the validity of property (4.6.41) under the assumption of

hyperbolicity, instead of symmetry, remains an open question.

Proof Let u

belong to E

(τ,η). Then the unique solution



+(τ + iA(η))u =0,u(0) = u

decays exponentially fast at −∞. Multiplying the ODE by u

∗

and integrating,

we obtain

∗

= −2(Re τ )



−∞

|u|

≤ 0.

132 Initial boundary value problem in a half-space with constant coeﬃcients

If, moreover, u

∈ E

), the unique solution of



= A

v, v(0) = u

decays exponentially fast at −∞. Multiplying the ODE by v

∗

and integrating,

we obtain

∗



−∞

≥ 0.

We conclude that u

∗

= 0, which readily implies that u ≡ 0. Thus

=0. 

Thanks to the lemma, the map π

−

(τ,η):E

) → E

−

(τ,η) is injective, thus

bijective since both spaces have dimension p. Now, given a basis γ

of E

we obtain a basis γ(τ, η):=π

−

(τ,η)γ

of E

−

(τ,η), which is jointly analytic.

An abstract deﬁnition of ∆ We use here the exterior algebra Λ(C

). For a

construction of this object, we refer, for instance, to Harris’ book [81]. The (non-

commutative) algebra Λ(C

) is spanned by C

under the associative product

(exterior product) ∧. The exterior product basically satisﬁes X ∧Y = −Y ∧ X

for every X, Y in C

. This rule deﬁnes a graded algebra,

Λ(C

)=Λ

) ⊕···⊕Λ

with

)=C, Λ

)=C

and

dim Λ





Elements of Λ

) are called k-vectors. Given a basis {e

,...,e

} of C

,they

are linear combinations of e

∧···∧e

,wherej

< ···<j

.WhenF is a k-

dimensional subspace of C

, we may deﬁne a k-vector X

by X

:= X

∧···∧

,where{X

,...,X

} is a given basis of F . One veriﬁes that diﬀerent bases

of F give the same X

, up to a non-zero scalar factor. This procedure deﬁnes a

unique one-dimensional subspace in Λ

). The map F → CX

is one-to-one,

but not onto, because not all k-vectors are simple exterior products.

Let M ∈ M

m×n

(k)

) → Λ

)by

∧···∧e

→ Me

∧ e

∧···∧e

+ ···+ e

∧···∧e

k−1

∧ Me

and linearity. It satisﬁes the identity

(k)

∧···∧X

)=MX

∧ X

∧···∧X

+ ···+ X

∧···∧X

k−1

∧ MX

for all X

,...,X

in C

. Let us assume m = n and let (λ

,...,λ

) be the eigen-

values of M, counted with multiplicity. One easily veriﬁes that the eigenvalues

A practical tool 133

of M

(k)

, counted with multiplicities, are the sums

+ ···+ λ

< ···<j

When applying this observation to the matrix A(τ,η), we ﬁnd that its ‘pth

sum’ admits a unique eigenvalue µ(τ,η) of minimal real part, namely the sum

of eigenvalues of A(τ,η) with negative real part. Moreover, µ(τ,η)isasimple

eigenvalue, whose eigenvector is X

,forF := E

−

(τ,η). Using Kato’s argument,

we may construct a jointly analytic choice X(τ, η)ofX

. For instance

X(τ,η)=X

(τ,η) ∧···∧X

(τ,η)

works. Then we may deﬁne



∆(τ,η):=B

(p)

X(τ,η).

This expression belongs to Λ

(p)

), a one-dimensional vector space. The link

between both deﬁnitions is



∆=∆e

∧···∧e

where {e

,...,e

} is any basis of C

with determinant one. We shall not

distinguish ∆ from



∆ in the following.

4.6.2 ‘Algebraicity’ of the Lopatinski˘ı determinant

We show in this section that the Lopatinski˘ı locus, that is the set of zeroes of the

Lopatinski˘ı determinant ∆, is a subset of an algebraic manifold of codimension

one. In general, this subset is strict, although it has the same codimension. In

other words, there exists a single polynomial Lop(X, η) such that ∆(τ,η)=0

implies Lop(iτ, η) = 0. Obviously, Lop is a homogeneous polynomial, so that

its zero set may be viewed as a projective variety. More importantly, it has real

coeﬃcients, so that the zeroes (iρ, η) of ∆ on the boundary Re τ = 0 belong to

a real algebraic variety.

Recall ﬁrst that, in the non-characteristic case, ∆ is deﬁned as the deter-

minant in C

of vectors Br

(τ,η),...,Br

(τ,η), where the r

s span the stable

subspace of A(τ, η). When A is diagonalizable, a generic property, r

may be

taken as an eigenvector associated to µ

(τ,η), one of the stable eigenvalues. Using

the polynomial P (X, ξ):=det(XI

+ A(ξ)), the eigenvalues are constrained by

P (τ,iη,µ

) = 0, or equivalently P (−iτ, η, −iµ

)=0.

Since r

solves (τI

+ A(iη, µ

))r

= 0, a choice of r

can be made polyno-

mially in (τ,η,µ

). For instance, we may choose the ﬁrst column of M (τ,η,µ

where M(τ,η,µ) is the transpose of the matrix of cofactors of τI

+ A(iη, µ),

since

(τI

+ A(iη, µ))M(τ,η,µ) = (det(τI

+ A(iη, µ)))I

134 Initial boundary value problem in a half-space with constant coeﬃcients

and the right-hand side vanishes whenever µ is an eigenvalue of A. Note that the

columns of M are non-trivial; for instance, they do depend on τ.Fromnowon,

let us denote by R(τ,η,µ) this choice.

The zeroes of the Lopatinski˘ı determinant therefore satisfy the following list

of polynomial equations:

det(BR(τ,η,µ

),...,BR(τ, η,µ

)) = 0,

P (−iτ, η, −iµ

)=0,

P (−iτ, η, −iµ

)=0.

Using the resultant, we may eliminate µ

, considered as a dummy variable,

between the ﬁrst two equations. These are thus replaced by a polynomial equation

in τ, η,µ

,...,µ

. Using again the resultant, we eliminate successively µ

,...,µ

and end with a single polynomial equation

Lop(iτ, η)=0.

Practical aspects

The procedure described above may be the simplest one, in the sense that it

gives the simplest result Lop.ThisseemstobetruewhenP is irreducible.

However, in practical situations, symmetry properties are responsible for

the presence of multiple eigenvalues of A(ξ), which yield a splitting of P

(see Proposition 1.7, for instance). When P does split, our procedure must

be reconsidered and gives rise to a polynomial of lower degree. See Chapter

13 for convincing examples within gas dynamics.

AﬂawofLop is that it has been built without any consideration about

the sign of the real part of the µ

s. Therefore, its zero set also contains the

zeroes of fake Lopatinski˘ı determinants, where the µ

s are chosen arbitrarily

in the spectrum of A. The manifold deﬁned by Lop(iτ, η) = 0 thus contains

irrelevant parts, which have to be removed on a case-by-case analysis. This

diﬃculty is always encountered when one wishes to check the Lopatinski˘ı

condition, or (UKL).

Another diﬃculty arises when A displays a Jordan block. In this case, the

columns of M do not span the corresponding generalized eigenspace of A,

since it only spans the classical eigenspace. Therefore, it may happen that

Lop admits spurious zeroes, which do not correspond to zeroes of ∆. This

has been the cause of a misunderstanding in [166], where some Lax shocks

in an ideal gas ﬂow were unduly claimed to be unstable. See [42] for a

detailed explanation.

Similarly, there are points (τ,η) where the ﬁrst column of M vanishes,

therefore it does not span an eigenspace. At such points, Lop vanishes