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

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Strictly dissipative symmetric IBVPs 95

3.2.1 The a priori estimate

We assume that u is a smooth solution of the full IBVP, with smooth decay as

|x| goes to inﬁnity. Multiplying (3.1.1) by u

∗

and taking the real part, we obtain

∂

|u|



α=1

∂

∗

u)=2Re(f,u).

Integrating on Ω, we obtain



Ω

|u(t)|

dx =2Re



Ω

(f(t),u(t)) dx +



∂Ω

∗

u dy. (3.2.18)

Using Lemma 3.3, we obtain



Ω

|u(t)|

dx + ε



∂Ω

|u(t)|

dy ≤ 2Re



Ω

(f(t),u(t)) dx + C



∂Ω

|g(t)|

dy.

Let γ be a positive number. We apply the latter estimates to v := exp(−γt)u,

for which Lv = −γv +exp(−γt)f =: F , v(t =0)=u

,andBv =exp(−γt)g on

the boundary. By Young’s inequality, we have

2Re (F, v) ≤ exp(−2γt)



|f|

− γ|u|



We thus obtain

−2γt

u(t)

+ γe

−2γt

u(t)

+ εe

−2γt

γ

u(t)

≤ e

−2γt



f(t)

+ Cg(t)



where ·

denotes the norm in either L

(Ω) or L

(∂Ω). We now integrate

from 0 to T>0 and obtain

−2γT

u(T )



−2γt



γu(t)

+ εγ

u(t)



≤u





−2γt



f(t)

+ Cg(t)



dt.

Notation For positive γ,T, we deﬁne a ‘norm’ ·

γ,T

u

γ,T



∂Ω

−2γt

|γ

u(y, t)|

dy dt + γ



Ω

−2γt

|u(x, t)|

dx dt.

We warn the reader that this expression does not deﬁne a functional space, and

therefore cannot be considered as a genuine norm, since the corresponding closure

of D(

Ω) is isomorphic to L

(Ω) × L

(∂Ω), so that the function on the boundary

is no longer a trace of the interior function.

We summarize this weighted estimate in the following

96 Friedrichs-symmetric dissipative IBVPs

Proposition 3.1 For the symmetric, strictly dissipative IBVP in Ω={x

0}, there holds the following a priori estimate for every positive numbers γ,T,

−2γT

u(T )

+ u

γ,T

(3.2.19)

≤ C



u(0)



−2γt



(Lu)(t)

+ γ

Bu(t)





where the constant C>0 depends neither upon f,g,u

,u, nor on γ, T .

The characteristic case The deﬁnition of strict dissipation given in Deﬁnition

3.3 was appropriate when the boundary is not characteristic. Of course, one says

that the boundary is characteristic if the normal matrix A(ν) is singular. In the

general case, we ask that the quadratic form v → (A

v, v), deﬁned on kerB,be

non-positive, and vanishes only on kerA

(in particular, kerA

⊂ kerB, as quoted

before). This assumption yields the fact that the quadratic form w → ε|A

w, w) − C|Bw|

is non-positive and vanishes only on kerA

(compare with

Lemma 3.3). The norm ·

γ,T

in the a priori estimate (3.2.19) is then weakened

into

u

γ,T



∂Ω

−2γt

|γ

u(y, t)|

dy dt + γ



Ω

−2γt

|u(x, t)|

dx dt.

3.2.2 Construction of ˆu and u

The existence of the solution of the homogeneous boundary value problem (that

is, with g ≡ 0) was obtained in Section 3.1. Since strict dissipativeness implies

dissipativeness, we may use this construction to solve our problem when g ≡ 0.

Following the details of this construction, we verify that the estimate (3.2.19)

holds true (without the g term of course). Since our problem is linear, it remains

to treat the pure boundary value problem. Hence we shall assume that u

≡ 0

and f ≡ 0. Since u

≡ 0, we may extend the solution and the data g by zero to

negative times. We therefore have to solve a problem in the domain deﬁned by

x ∈ Ω=R

d−1

× (0, ∞),t∈ R,

and we may assume that g is supported in a slab ∂Ω × [0,T]. This problem is

the occasion to describe for the ﬁrst time, but in a simple context, the use of

the Fourier–Laplace transform (see Appendix B.3). We warn the reader that we

apply the Fourier transformation to the variable y only, since x

does not vary

on the whole line but only

on (0, +∞).

We now face the pure boundary value problem. Taking the Fourier–Laplace

transform (in variables (y,t)) of the PDEs and of the boundary condition, this

One might apply the Laplace transform to the variable x

, but this would not help anyway.

Strictly dissipative symmetric IBVPs 97

problem is formally equivalent to the following:

(τI

+ iA(η))ˆu + A

∂ˆu

∂x

=0,x

> 0, (3.2.20)

Bˆu(η,0,τ)=ˆg(η, τ). (3.2.21)

The problem (3.2.20) and (3.2.21) is clearly a set of uncoupled diﬀerential-

algebraic problems, parametrized by (η,τ). We may solve separately each of

them (or merely almost all of them). Doing so, the partial derivative will of

course be seen as an ordinary derivative.

The solutions of (3.2.20) are smooth functions (see their description in

Section 4.2.3, when the boundary is characteristic). Since ˆu(·, ·,γ+ i·) is expected

to be square-integrable with respect to dη dx

dσ,ˆu(η, ·,τ)mustbesquare-

integrable on R

. This means here that it decays exponentially fast as x

→ +∞.

In other words, the integrability property translates into

ˆu(η, 0,τ) ∈ E

−

(η, τ). (3.2.22)

This, together with the boundary condition (3.2.21), determines in a unique way

ˆu(η, 0,τ) (adapt Lemma 3.2), and hence ˆu(η, x

,τ).

Once we have determined the function ˆu,itremainstoshowthatitisthe

Laplace–Fourier transform of some function u, which satisﬁes the IBVP. We begin

by noticing that, since the coeﬃcients of (3.2.20) are holomorphic in τ,themap

τ → E

−

(η, τ) is holomorphic too. And since g itself is τ-holomorphic, the trace

ˆu(η, 0,τ) is holomorphic, as the solution of a Cramer system with holomorphic

coeﬃcients. Hence, ˆu itself is τ -holomorphic.

We now turn to the estimate of ˆu. Since it decays exponentially fast as x

→

+∞, we may multiply (3.2.20) by ˆu

∗

, take the real part, and integrate with

respect to x

on R

. Because of the symmetry of A(η)andA

, we obtain

(Re τ)



+∞

|ˆu|

ˆu(0)

∗

ˆu(0).

Let now ε>0andC>0 be such that the quadratic form w → ε|A

w, w) − C|Bw|

is non-positive. We deduce

(Re τ)



+∞

|ˆu|

+ ε|A

ˆu(0)|

≤ C|ˆg|

This inequality implies that if ˆg is square-integrable in (η,σ) for some value of

γ, then the left-hand side enjoys the same property. An integration yields the

inequality



+∞



d−1

|ˆu|

dη dx

dσ + ε



ˆu|

dη dσ ≤ C



|ˆg|

dη dσ.

(3.2.23)

98 Friedrichs-symmetric dissipative IBVPs

In particular, the trace at x

=0 of A

ˆu is square-integrable. Assuming, for

instance, that g belongs to L

(∂Ω × R

) (actually, some growth is allowed as t

increases), the right-hand side of (3.2.23) is bounded. Thus, ˆu(·, ·,γ+ i·)

an O(1/γ). Since ˆu is τ-holomorphic, the theorem of Paley–Wiener (see Rudin

[169], chapter 19) implies that there exists a function u :Ω× (0, +∞) with the

following properties:

For every γ>0, the function e

−γt

u is square-integrable,

The function ˆu is the Fourier–Laplace transform of u, with respect to the

variables (y, t).

The inverse Fourier–Laplace transform of (3.2.20) shows that u satisﬁes the

system Lu = 0. In particular, Lu is square-integrable and A

u must admit a

trace on x

= 0, of class H

−1/2

, which we denote abusively γ

u. Since kerB

contains kerA

, the trace of Bu makes sense as well.

The following estimate follows from (3.2.23) and the Parseval formula:



+∞

−2γt

u(t)

(Ω)

dt ≤ C



+∞

−2γt

g(t)

(∂Ω)

dt. (3.2.24)

At last, arguing as in Section 3.1.1, we see that e

−γt

u is square-integrable

(rather than of class H

−1/2

), and that the following estimate holds



+∞

−2γt

u(t)

(Ω)

dt + ε



+∞

−2γt

γ

u(t)

(∂Ω)

≤ C



+∞

−2γt

g(t)

(∂Ω)

dt. (3.2.25)

We summarize our result in the following statement.

Theorem 3.3 Let L be a Friedrichs-symmetric (hence hyperbolic) operator. Let

Ω=R

d−1

× (0, +∞) be the spatial domain. Finally, let B be a strictly dissipative

boundary matrix.

Then, for every data u

∈ L

(Ω), g ∈ L

((0,T) × ∂Ω) and f ∈ L

((0,T) ×

Ω), there exists a unique solution of the Initial Boundary Value Problem (3.1.1)–

(3.1.3) in the class u ∈ L

((0,T) × Ω). In addition, A

u admits a trace on the

boundary, of class L

((0,T) × ∂Ω). Finally, we have u ∈C([0,T]; L

(Ω)),and

the estimate (3.2.19).

INITIAL BOUNDARY VALUE PROBLEM IN A HALF-SPACE

WITH CONSTANT COEFFICIENTS

In this chapter, we drop the assumption of Friedrichs-symmetry and therefore

dissipativeness, while keeping the other features of Chapter 3: L is a hyperbolic

operator with constant coeﬃcients, and the spatial domain is a half-space. We

impose linear boundary conditions. Of course, most physical problems yield a

Friedrichs-symmetric operator, but we wish to consider boundary conditions that

are not dissipative in the sense of Deﬁnitions 3.1, 3.2 and 3.3.

Our main purpose is the well-posedness of such an Initial Boundary Value

Problem (IBVP). We postpone the study of more natural linear problems (gen-

eral smooth domain, variable coeﬃcients) to Chapter 9, which uses the results

displayed in the present one, together with those of Chapter 2.

4.1 Position of the problem

Let

L := ∂



α=1

∂

be a hyperbolic operator (with A

∈ M

(R)), and let B be a constant real-valued

q × n matrix. Let Ω be the half-space in R

, deﬁned by the inequality x

> 0.

Denote the tangential variables (x

,...,x

d−1

)byy.Wehave

Ω={(y, x

); y ∈ R

d−1

> 0}.

The general problem that we have in mind is still

(Lu)(x, t)=f(x, t),x

,t> 0,y∈ R

d−1

, (4.1.1)

Bu(y, 0,t)=g(y,t),t>0,y∈ R

d−1

, (4.1.2)

u(x, 0) = u

(x),x

> 0,y∈ R

d−1

. (4.1.3)

Here above, the source term f(x, t), the boundary data g(y, t) and the initial

data u

(x) are given in suitable functional spaces.

Since practical applications involve systems with variable coeﬃcients, the

present chapter investigates robust existence and stability results. As for the

Cauchy problem (see Chapter 1), or the symmetric strictly dissipative case

(Chapter 3), the terminology ‘robust results’ refers to strong well-posedness

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

theorems, where the solution not only exists and is unique, but is estimated

in the same norms as the data.

4.1.1 The number of scalar boundary conditions

Before going further, we observe that we may always choose the matrix B with

full rank q, since otherwise g would have to satisfy compatibility conditions,

and we could reduce the set of boundary conditions by extracting r independent

lines, with r = rank B. Therefore, we shall always assume that q =rankB. In

particular, we have q ≤ n. The most signiﬁcant object is kerB, rather than B

itself, since a multiplication on the left by a regular q × q matrix G transforms

our boundary condition into an equivalent one, whose matrix is B



:= GB.

As we shall see in Sections 4.2 and 4.3, the matrix B in the boundary condition

(4.1.2) must satisfy several requirements of algebraic type, for the IBVP to be

well-posed. In this section, we concentrate on its rank q, that is the number of

scalar boundary data that we need.

With this purpose in mind, let us consider data that do not depend on the

tangential variable y:wehavef = f(x

,t), g = g(t)andu

= u

,t). Assuming

that the IBVP is well-posed (which means at least existence and uniqueness)

in some appropriate functional space, the corresponding solution must have the

same translational invariance : u = u(x

,t). This means that the reduced system

(∂

+ A

∂

)u = f, Bu(0,t)=g(t),u(x

, 0) = u

) (4.1.4)

is well-posed in the quarter-plane {x

> 0,t>0}. By assumption (hyperbolicity)

is diagonalizable. Up to a linear tranformation of dependent variables, we may

assume that A

is already diagonal, A

= diag(a

,...,a

), with a

≥···≥a

Each co-ordinate u

must obey the transport equation

(∂

+ a

∂

= f

. (4.1.5)

Denote by p the number of positive eigenvalues: a

> 0 ≥ a

p+1

. We say that

p is the number of incoming characteristics

. Split the unknown into an incoming

and an outgoing part,













−







p+1







Integrating (4.1.5), we observe that u

−

is uniquely determined by its value at

initial time:

,t)=u

− a

t)+



− a

(t − s),s)ds, j ≥ p +1.

In a half-space deﬁned by the reverse inequality x

< 0, the incoming characteristics correspond

to a

< 0.

Position of the problem 101

Let  be a linear form vanishing on the vector space B(R

×{0

n−p

}), and let

L denote B,sothatL

= ···= L

=0( and L are row vectors with q and n

components, respectively). From (4.1.2), we obtain



j=p+1



(−a

t)+



(s − t),s)ds



= g(t). (4.1.6)

When  = 0, (4.1.6) is a non-trivial compatibility condition for the data (u

,g,f).

Such a constraint prevents a general existence result from being obtained.

Therefore, well-posedness requires

B(R

×{0

n−p

})=R

which implies p ≥ q. In other words,

Existence for every reasonnable data requires that the number of boundary

conditions be less than or equal to the number of incoming characteristics.

We now turn to uniqueness, by considering the homogeneous IBVP (f ≡ 0, u

≡

0, g ≡ 0). From above, we already know that u

−

≡ 0. Let R ∈ R

be such that

(R, 0)

∈ kerB. Let us choose a smooth function v of one variable, with v ≡ 0

on [0, +∞), and deﬁne u

by its co-ordinates

,t):=v



− t



,j=1,...,p.

We verify immediately that u is a solution of the IBVP. By uniqueness,

we must have u ≡ 0. This means R = 0. Therefore, well-posedness requires

×{0

n−p

}) ∩ kerB = {0}. Since the dimension of kerB is n − q, this means

in particular p ≤ q. In other words,

Uniqueness requires that the number of boundary conditions be larger than or

equal to the number of incoming characteristics.

Gathering these results, we obtain that for the IBVP (4.1.1)–(4.1.3) to be well-

posed, it is necessary that

=(R

×{0}) ⊕ kerB.

In particular, the number of independent scalar boundary conditions (the rank of

B) must be equal to the number of incoming characteristics,thatis

q = p.

Going back to a general matrix A

, not necessarily diagonal, this reads

Proposition 4.1 For the IBVP (4.1.1)–(4.1.3) to be well-posed, it is necessary

that

= U(A

) ⊕ kerB, (4.1.7)

where we recall that U(A

) is the unstable subspace of A

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

4.1.2 Normal IBVP

The previous analysis concerned only the well-posedness of the one-dimensional

IBVP, where the diﬀerential operator is L

:= ∂

+ A

∂

. Going back to the

general case where A

is diagonalizable (but not necessarily diagonal), we

conclude that in order that this reduced IBVP be well-posed in reasonable spaces

like L

, it is necessary and suﬃcient to have

=(kerB) ⊕ E

), (4.1.8)

where E

) denotes the unstable invariant subspace of A

As we already saw in Section 3.1.1, the need for a correct deﬁnition of the

trace of Bu when u and Lu belong to a weak class, like L

(Ω × (0,T)), requires

that kerA

⊂ kerB. This leads us to the following.

Deﬁnition 4.1 We say that the IBVP (4.1.1)–(4.1.3) is normal if

B ∈ M

p×n

(R), where p is the number of positive eigenvalues of A

kerA

⊂ kerB,

property (4.1.8) holds true.

Of course, dimensionality ensures that p =rankB for a normal IBVP. Using

the characteristics, one easily shows that, in one space dimension, a normal

hyperbolic IBVP is well-posed in L

, in the sense that, for every data

∈ L

),g∈ L

(0,T),f∈ L

× (0,T)),

there exists a unique solution u ∈ L

× (0,T)), which satisﬁes, moreover,

u(0, ·) ∈ L

(0,T), with obvious norm estimates.

4.2 The Kreiss–Lopatinski˘ı condition

We derive in this section a condition that turns out to be necessary for any

kind of very weak well-posedness. The idea is very similar to the one followed in

Section 1.1. In particular, we examine only the semigroup aspects of the IBVP,

that is we shall only consider a homogeneous boundary condition (g ≡ 0). Of

course, as seen in Chapter 3, we do not expect a simultaneous characterization

of the well-posedness of non-homogeneous IBVP.

A Fourier transform with respect to the full space variable x being impossible,

we content ourselves with F

, the Fourier transform with respect to the tangen-

tial variables. As we saw in Chapter 3, it is worth treating the time variable by

a Laplace transform. Hence, we begin by looking for particular solutions of the

form (normal mode analysis)

u(x, t):=e

τt+iη·y

U(x

where η ∈ R

d−1

and τ ∈ C. Since we aim to ﬁnd a necessary condition, we are

only interested in those solutions that could contradict well-posedness, that is

The Kreiss–Lopatinski˘ı condition 103

those that grow rapidly as time increases, while being temperate in space. Thus

we restrict ourselves to complex numbers τ of positive real part.

Aﬁeldu deﬁned as above solves Lu = 0 if and only if

(τI

+ iA(η))U + A

=0. (4.2.9)

Up to now, we did not assume anything about A

but hyperbolicity. Since this

allows A

to be singular, Equation (4.2.9) need not be an ODE. We recall and

generalize a notion introduced in Chapter 3:

Deﬁnition 4.2 We consider the hyperbolic IBVP (4.1.1)–(4.1.3). The boundary

=0} is said to be characteristic if the matrix A

is singular (that is det A

0). When the IBVP is posed in a more general spatial domain Ω with a smooth

boundary, ∂Ω is said to be characteristic if the matrix A(ν) is singular, ν being

the outward normal vector ﬁeld. It is non-characteristic otherwise.

This notion is local in nature. The boundary can be characteristic on a part

of the boundary only, this part being either a set of full dimension d − 1in∂Ω,

or a submanifold. However, the theory of the IBVP is essentially open when the

rank of A(ν) varies along a connected component of ∂Ω.

4.2.1 The non-characteristic case

In order to make the exposition as clear as possible, we ﬁrst suppose that the

boundary is non-characteristic. Then (4.2.9) may be recast as an ODE in C

= A(τ,η)U, A(τ,η):=−(A

)

−1

(τI

+ iA(η)). (4.2.10)

The following lemma is fundamental in the theory.

Lemma 4.1 (Hersh) Under the hyperbolicity assumption, and for η ∈ R

d−1

and Re τ>0, the matrix A(τ,η) does not have any pure imaginary eigenvalue.

The number of stable eigenvalues (see the introductory paragraph ‘Notations’),

counted with multiplicities, equals p, the number of positive eigenvalues of A

Proof Let ω be a pure imaginary root of the characteristic polynomial of

A(τ,η),

P (X; τ, η):=det(XI

−A(τ,η)).

Thus ω satisﬁes

det(τI

+ iA(η)+ωA

)=0.

Then, hyperbolicity implies τ ∈ iR, which contradicts the assumption. This

proves that A(τ,η) may not have a pure imaginary eigenvalue. Since P depends

continuously on (τ, η) and has a constant degree, we infer that the number of

roots with positive real part (counted with multiplicities) may not vary locally.

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

Since {Re τ>0}×R

d−1

is a connected set, this number must be constant. We

evaluate it by choosing η =0,τ =1:Wehave

A(1, 0) = −(A

)

−1

whose eigenvalues are the −1/a

s. 

Using this lemma, we decompose the space C

as the direct sum of the

stable and unstable spaces of A(τ,η). Recall that the stable (respectively,

unstable) subspace is the sum of generalized eigenspaces of A(τ, η) correspond-

ing to eigenvalues of negative (respectively, positive) real parts. We denote

−

(τ,η)=E

(A(τ,η)) (respectively, E

(τ,η)=E

(A(τ,η)).) These are the

spaces of incoming (respectively, outgoing) modes. As mentioned in the section

‘Notations’, these spaces can be characterized by means of contour integrals.

Choosing a large enough loop γ in the half-space {Re ω>0}, enclosing the

unstable eigenvalues (namely the ones with positive real parts) of A(τ,η), the

projector onto E

(τ,η), along E

−

(τ,η), is given by the formula

(τ,η)=

2iπ



(zI

−A(τ,η))

−1

dz. (4.2.11)

A similar formula holds for the projector π

−

= I

− π

. Since we may vary

slightly the arguments (τ, η) without changing the contour (because of the

continuity of the roots of a polynomial), we infer from (4.2.11) that the maps

(τ,η) → π

(τ,η) are holomorphic in τ, analytic in η, which amounts to saying:

Lemma 4.2 The stable and unstable subspaces E

(τ,η) depend holomorphically

on τ, analytically on η. In particular, their dimensions do not depend on (τ, η)

as long as η ∈ R

d−1

and Re τ>0.

Fix now η ∈ R

d−1

and Re τ>0. Given an initial datum U(0), the Cauchy

problem for (4.2.10) admits a unique solution U. Decomposing U (0) into a stable

part U

0−

:= π

−

U(0) and an unstable one U

, the solution reads

U(x

)=U

−

)+U

),U

):=exp(x

A(τ,η))U

0±

Because of Lemma 4.1, the matrix

exp



A(τ,η)|

−



decays exponentially fast as x

tends to +∞. Similarly, the inverse of

exp



A(τ,η)|



decays exponentially fast. This shows that U

−

decays exponentially fast, while

is not polynomially bounded, except in the case U

= 0 (that is U (0) ∈

−

(τ,η)). Therefore, in order that U be a tempered distribution on R

,itis

necessary and suﬃcient that U (0) ∈ E

−

(τ,η). In that case, U actually decays

exponentially fast, and is therefore square-integrable.