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

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The adjoint IBVP 115

a duality formula holds. Of course, L



will be the adjoint operator L

∗

found in

Chapter 1,

∗

= −∂

−



)

∂

Assume that u, v ∈ D (Ω × R)

(for the sake of simplicity, the time variable runs

through the whole line) and let us compute



Ω

((v, f) − (u, F ))dx dt =



Ω

((v, Lu) −(u, L

∗

v))dx dt



Ω

(∂

(u, v)+



∂

u, v))dx dt (4.4.18)

= −



∂Ω

u, v)dy dt,

from Green’s formula. The left-hand side is a bilinear form in the variables (u, f)

and (v, F), respectively. Similarly, we search a decomposition of A

in such a

way that the right-hand side be bilinear in (u, g)and(v, h). It will be suﬃcient

that A

reads C

N + M

B, for suitable matrices M and N, for then

u, v)=(Nu,Cv)+(Bu,Mv)=(Nu,h)+(g,Mv).

Moreover, in order that the (backward) adjoint IBVP be well-posed, we ask in

particular that it be normal, which at least requires that C is q × n,whereq is the

number of negative eigenvalues of (A

)

,thatisofA

. In the non-characteristic

case, this means q = n − p, while in general it only implies q ≤ n − p.

Let us begin with the non-characteristic case. We choose any (n − p) × n

matrix X that is onto, such that





∈ GL

(R).

Such a matrix exists since B is onto. Let us write the inverse blockwise as (Y,D),

where D is n × (n − p). From the identity YB+ DX = I

,wemaychooseC :=

, M =(A

Y )

and N = X.

We notice that R(D) ⊂ kerB,sinceBD =0

p×(n−p)

. However, both spaces

have the same dimension n − p,sinceB is onto and D is one-to-one. Hence

R(D)=kerB and

kerC =(R(A

D))

⊥

=(A

R(D))

⊥

=(A

kerB)

⊥

. (4.4.19)

Identity (4.4.19) shows that kerC, which is the meaningful object in a boundary

condition, does not depend on the choice of the complement X. It actually does

not even depend on the procedure; we leave the reader to verify that the duality

identity (4.4.18) only needs kerC =(A

kerB)

⊥

The characteristic case can be treated in the same spirit. However, the matrix

built above is (n − p) × n instead of being q × n. Hence, we must ﬁrst

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

reduce the matrix A

to the block-diagonal form



0 a



∈ GL

n−m

(R).

Because of kerA

⊂ kerB,wethenhaveB =(0,B

), where B

is p × (n − m)

and is onto. Then we proceed as before, replacing B by B

and n by n − m.We

thus ﬁnd a matrix C

, such that a

= M

+ C

and kerC

=(a

kerB

)

⊥

Then the following matrices work:

C =





,M=(0,M

),N=(0,N

Finally, we obtain the result:

Proposition 4.5 Let a pair (L, B) be given, where L is hyperbolic, B ∈

p×n

(R), where p is the number of incoming characteristics, and kerA

⊂ kerB.

Let L

∗

be the adjoint of L, with q the number of its incoming characteristics,

that is the number of negative eigenvalues of A

. Then there exists a matrix

C ∈ M

q×n

(R) and matrices M,N such that A

= C

N + M

B. The matrix

C is characterized uniquely, up to left multiplication, by the identity kerC =

(kerB))

⊥

When u, v are smooth ﬁelds on Ω × R, decaying fast enough at inﬁnity, the

following identity holds



Ω

((Lu, v) − (L

∗

v, u))dx dt +



∂Ω

((Nu,Cv)+(Bu,Mv))dy dt =0.

(4.4.20)

As a corollary, we obtain

Theorem 4.1 Given a normal hyperbolic IBVP, deﬁned by (L, B), there exists

an adjoint IBVP, deﬁned by the pair (L

∗

,C), which is normal hyperbolic (back-

ward in time) and satisﬁes the identity (4.4.20). One has

∗

= −∂

−



)

∂

= C

N + M

where C ∈ M

q×n

(R) satisﬁes kerC =(A

(kerB))

⊥

. Moreover,

kerA

=kerN ∩ kerB. (4.4.21)

Finally, the former IBVP is the adjoint of the latter.

Proof It remains to prove that the adjoint problem is backward normal and

to check the validity of (4.4.21).

By backward normal, we mean that the IBVP obtained from (L

∗

,C)bythe

time reversion t →−t is normal. This amounts to saying that C ∈ M

q×n

(R),

The adjoint IBVP 117

where q is the number of negative eigenvalues of (A

)

(or of A

as well), and

that R

=kerC ⊕E

(−(A

)

). In other words, it remains to check that

=(kerC)

⊥

+(E

) ⊕ kerA

). (4.4.22)

However, our construction satisﬁes (kerC)

⊥

= R(C

)=A

(kerB). Therefore,

(4.4.22) follows from the sequence

= R(A

)+kerA

= A

(kerB ⊕ E

)) + kerA

⊂ A

(kerB)

)) + kerA

Last, kerN ∩ kerB ⊂ kerA

is trivial, and the converse follows from the facts

that kerA

⊂ kerB and that C is onto. 

The method of duality will need the following important fact.

Theorem 4.2 Let (L, B) deﬁne a normal hyperbolic IBVP on Ω={x

> 0},

and let (L

∗

,C) deﬁne its dual IBVP. Let (τ,η) ∈ C × R

d−1

be given, such that

Re τ>0. It holds that

= E

−

(τ,η) ⊕ kerB) ⇐⇒ (C

= E

∗

−

(−¯τ,−η) ⊕ kerC),

where E

∗

−

(z,σ) denotes the stable invariant subspace for the diﬀerential-algebraic

equation

)



+(zI

+ iA(σ)

)V =0.

Consequently, (L, B) satisﬁes the Kreiss–Lopatinski˘ı property if and only if

∗

,C) satisﬁes the Kreiss–Lopatinski˘ı property, backward in time. And similarly,

(L, B) satisﬁes the uniform Kreiss–Lopatinski˘ı property if and only if (L

∗

,C)

satisﬁes the uniform Kreiss–Lopatinski˘ı property, backward in time.

Proof We ﬁrst note that, whenever V solves (A

)



− (¯τI

+ iA(η)

)V =

0andU solves A



+(τI

+ iA(η))U =0, then (V

∗



= 0 and therefore

U is constant. If, moreover, U and V decay as x → +∞, this constant is

zero. In other words, we have

Lemma 4.6 Assume that Re τ>0 and η ∈ R

d−1

.Ifu ∈ E

−

(τ,η) and v ∈

∗

−

(−¯τ,−η), then v

∗

u =0.

By continuity, the equality also holds true if Re τ ≥ 0.

Let us assume that C

= E

∗

−

(−¯τ,−η) ⊕ kerC,thatisC(E

∗

−

)=C(C

)=C

Then let u belong to kerB ∩ E

−

(τ,η). From Lemma 4.6 and the decomposition

of A

, we have, for all v in E

∗

−

(−¯τ,−η),

0=v

∗

u =(Cv)

∗

Nu,

which amounts to saying that Nu = 0. From (4.4.21), we obtain u ∈ kerA

Proposition 4.3 then gives u = 0, proving that C

= E

−

(τ,η) ⊕ kerB.

The converse follows after the exchange of (L, B)and(L

∗

,C).

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

We now prove the uniform part of the theorem. Let us assume that the dual

IBVP satisﬁes (UKL) condition, meaning that there exists a positive constant c

such that, for every complex number τ with Re τ>0, for every η ∈ R

d−1

and

every vector v ∈ E

∗

−

(−¯τ,−η), it holds that

|(A

)

v|≤c

|Cv|.

From Theorem 4.1, there exists a positive number c

such that the following

inequality holds

u|≤c

(|Nu| + |Bu|).

Using the Kreiss–Lopatinski˘ı condition for the dual problem, and then its uni-

formity, we may write

|Nu| =sup



(Cv,Nu)

|Cv|

; v ∈ E

∗

−

(−¯τ,−η)



≤ c

sup



(Cv,Nu)

|(A

)

; v ∈ E

∗

−

(−¯τ,−η)



Since the kernel of M contains that of (A

)

, there exists another constant c

such that |Mv|≤c

|(A

)

v|. Using now the decomposition of A

, we conclude

that

|Nu|≤c

|Bu| + c

sup



(v, A

|(A

)

; v ∈ E

∗

−

(−¯τ,−η)



The use of Lemma 4.6 now gives the expected result: if u ∈ E

−

(τ,η), then

u|≤c

(1 + c

)|Bu|,

where the constants do not depend on the pair (τ,η). 

4.5 Main results in the non-characteristic case

The main statement of this chapter is the following strong well-posedness result

in L

. It completes the study of the strongly dissipative symmetric case (see

Chapter 3).

Theorem 4.3 Let Ω be the half-plane {x ∈ R

; x

> 0}.Let

L = ∂



α=1

∂

+ R

be a constantly hyperbolic ﬁrst-order operator, with constant coeﬃcients. Assume

that the boundary is non-characteristic (det A

=0), and let p be the number of

positive eigenvalues of A

(the number of incoming characteristics). Let B ∈

p×n

(R) have rank p.

Main results in the non-characteristic case 119

Then the IBVP

Lu = f, in Ω × (0,T),

Bu = g, on ∂Ω × (0,T),

u = u

, in Ω ×{0},

is strongly well-posed in L

if, and only if, the uniform Kreiss–Lopatinski˘ı

condition holds. In other words:

On the one hand, if (3.2.19) holds for smooth solutions, then the IBVP

satisﬁes (UKL).

On the other hand, assuming that (UKL) holds, we have the following

existence and uniqueness property: for all data f ∈ L

(Ω × (0,T)), g ∈

(∂Ω × (0,T)) and u

∈ L

(Ω), there exists a unique u ∈ L

(Ω × (0,T))

with the following properties:

– It satisﬁes Lu = f in Ω ×(0,T),

– Its trace on ∂Ω × (0,T) (which is known to belong to H

−1/2

because of

Lu ∈ L

(Ω × (0,T))) is square-integrable, and satisﬁes Bu = g,

– It belongs to C ([0,T]; L

(Ω)), and satisﬁes u(t =0)=u

– Finally, (3.2.19) holds for every γ>γ

. Here, γ

and the constant C

depend only on A

,B,R, but not on T,f,g or u

;whenR =0, one may

take γ

=0.

A related result holds when the boundary is characteristic. However, we

shall establish it for a smaller class of admissible operators. We postpone the

corresponding analysis to Chapter 6.

4.5.1 Kreiss’ symmetrizers

The proof that (UKL) implies well-posedness follows the lines of Chapter 2, but

displays new ideas, coming partly from Chapter 3.

We begin by the analysis of the BVP, that is a boundary value problem, posed

for all time t ∈ R, thus without initial condition. We ﬁrst prove the analogue of

(3.2.19), but without ﬁnal and initial states:



−2γt

(γu(t)

+ γ

u(t)

)dt

≤ C



−2γt



(Lu)(t)

+ γ

Bu(t)



dt. (4.5.23)

It readily implies that the solution is unique. Since the adjoint BVP satisﬁes

the same assumptions as the direct one, it enjoys the same estimate. Through

a duality argument, using Hahn–Banach and Riesz theorems, we obtain the

existence of the solution of the BVP. The resolution stands of course in the

space associated to the norms present in (4.5.23), say in L

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

The passage from the well-posedness of the BVP to that of the IBVP is made

in three steps, and is due to Rauch. The ﬁrst one is causality; we prove that if

the source f and the boundary data vanish for negative times, then so does

the solution. This allows us to solve the IBVP when the initial data vanishes

identically. The next step consists in a new estimate, namely that of u(·,T)in

,whenu(·, 0) ≡ 0. The last one is again a duality argument.

The proof of the estimate (4.5.23) mimics that of the Friedrichs-symmetric

case with strict dissipation, considered in Chapter 3. However, a dissipative

symmetrizer has not been given a priori and we have to build it. A main technical

diﬃculty is that this symmetrizer, called a Kreiss symmetrizer, is symbolic, thus

it depends on the frequencies (τ, η). Its construction is lengthy and is postponed

to Chapter 5. Theorem 5.1 tells us that there exists a map (τ,η) → K(τ, η)

(the Kreiss symmetrizer), deﬁned for η ∈ R

and Re τ>0, with the following

properties:

i) (τ,η) → K is bounded and homogeneous of degree zero,

ii) Σ(τ,η):=K(τ,η)A

is Hermitian,

iii) There exists a positive constant c

, independent of (τ,η), such that

∗

Σ(τ,η)w ≤−c

w

, ∀w ∈ kerB (4.5.24)

iv) There exists a positive constant, say again c

, independent of (τ,η), such

that

Re (v

∗

M(τ,η)v) ≥ c

(Re τ)v

, ∀v ∈ C

, (4.5.25)

where

M(τ,η):=K(τ,η)(τI

+ iA(η)).

Note that in the Friedrichs-symmetric, strictly dissipative case, one can simply

choose K ≡ I

, which is classical instead of symbolic. Point ii) is the symmetry

property, while point iii) is the strict dissipation.

Remark Estimate (4.5.23) can be used the same way as (3.2.19) to show

the necessity of (UKL). We leave the reader to adapt the calculations of

Section 4.3.1.

4.5.2 Fundamental estimates

Recall that the construction of the Kreiss symmetrizer is postponed to the next

chapter, under appropriate assumptions. We thus suppose that L is constantly

hyperbolic, that the boundary is non-characteristic and that the boundary

condition satisﬁes (UKL), and we admit in the remainder of the present chap-

ter that these properties ensure the existence of a dissipative symmetrizer K

(Theorem 5.1).

Let u be given in D(

Ω × R

), meaning that u is extendable to R

× R as a

∞

function with compact support. For the sake of clarity, we deﬁne f = Lu

Main results in the non-characteristic case 121

and g = γ

Bu. Let us deﬁne the Laplace transform in time, Fourier transform

in y

h →Lh(τ,η):=



d−1

×R

−τt−iη·y

h(y, t)dy dtη∈ R

d−1

, Re τ>0.

Since we shall deal with smooth and compactly supported functions, we shall

never discuss the convergence of the integral. We need the following auxiliary

functions:

U(·, ·,x

):=L[u(·,x

, ·)],F(·, ·,x

):=L[f(·,x

, ·)],G:= Lg.

Integration by parts yields the identities

L[∂

h]=iη

Lh, α =1,...,d− 1, L[∂

h]=∂

Lh, L[∂

h]=τLh.

Therefore, we have

(τI

+ iA(η)+R)U + A



= F,

wheretheprimedenotesthex

-derivative. Multiplying on the left by U

∗

K(τ,η),

we have

∗

(M(τ,η)+K(τ,η)R)U + U

∗

Σ(τ,η)U



= U

∗

K(τ,η)F.

Let us take the real part in this identity. Since Σ is Hermitian, we have

Re (U

∗

Σ(τ,η)U



∗

Σ(τ,η)U)



Using points i) and iv) above, and integrating in x

over R

,wederivean

inequality

Re τ − C

)



∞

U

≤

U(0)

∗

Σ(τ,η)U(0) + C



∞

UF dx

We now appeal to Lemma 3.3: There exist positive constants ε and C,such

that the Hermitian form w → εw

+ w

∗

Σ(τ,η)w − CBw

is non-positive.

Checking the proof of the lemma, we easily see that the constants may be chosen

independently of (τ,η). We may write c

instead of ε. We deduce therefore the

bound

Re τ − C

)



∞

U

U(0)

≤ C



∞

UF dx

+ CBU(0)

where the argument 0 stands for the x

-variable. If C

> 0, we take a threshold

larger than C

, to obtain



∞

U

U(0)

≤ C



∞

UF dx

+ CBU(0)

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

for γ =Reτ>γ

. Using now the Cauchy–Schwarz inequality, we obtain



∞

U

+ U(0)

≤ C





∞

F 

+ BU(0)



Integrating in η, we obtain

γU(τ )

+ U(τ, 0)

≤ C



F (τ)

+ BU(τ,0)



where the L

-norm is taken in terms of η and, for U and F ,intermsofx

too.

Integrating then with respect to the imaginary part of τ and using Plancherel

formula, we obtain the weighted estimate



Ω×R

−2γt

u

dx dt +



∂Ω×R

−2γt

γ

u

dy dt (4.5.26)

≤ C





Ω×R

−2γt

Lu

dx dt +



∂Ω×R

−2γt

γ

Bu

dy dt



We consider now less-regular functions. For this purpose, we deﬁne the

weighted spaces L

of measurable functions u such that (x, t) → e

−γt

u(x, t)is

square-integrable. These are Hilbert spaces with the obvious norms

u

:= e

−γt

u

Such spaces may concern functions deﬁned either on Ω × R

,oron∂Ω × R

.We

deﬁne in a similar way weighted Sobolev spaces H

. We note that, given u in L

such that Lu ∈ L

, the trace γ

u is well-deﬁned in a class H

−1/2

, weighted by

γt

, which we denote H

−1/2

Lemma 4.7 Assume that u, Lu and γ

Bu are in the class L

. Then

i) The trace γ

u belongs to the class L

ii) The function u satisﬁes (4.5.26).

Proof Using a cut-oﬀ function, we may restrict ourselves to the case of a

compactly supported functions.

We begin with the easy case where u belongs to H

(Ω × R) and has a compact

support. Then point i) is obvious. On the other hand, u is the limit in H

of functions u



∈ D (

Ω × R). Hence Lu and γ

Bu are the limits in L

of the

sequences Lu



and γ



. The latter satisfy (4.5.26), which remains valid in the

limit.

We turn to the general case of an L

-function with compact support. Let



be a molliﬁer in the variables (y,t) (the tangential variables.) A convolution

by ρ



yields a function u



, compactly supported and C

∞

with respect to y and

t. Besides, Lu



= ρ



∗ (Lu)andγ



= ρ



∗ (γ

Bu) are smooth in (y, t). The

sequences (u



)



,(Lu



)



and (γ



)



converge, respectively, to u, Lu and γ

in L

. In particular, they are Cauchy in L

Main results in the non-characteristic case 123

When 0 ≤ α<d,wehave∂



=(∂



) ∗ u, with ∂

:= ∂

. Hence ∂



square-integrable. Next, the identity

∂



=(A

)

−1

(Lu



− ∂



−

d−1



α=0

∂



)

shows that ∂



is square-integrable

too. Hence u



belongs to H

(Ω × R), and

we are allowed to use (4.5.26).

This estimate shows that the sequences (u



)



and (γ



)



are Cauchy, hence

converge, in L

. This proves that γ

u is actually an L

-function. Similarly, u



Cauchy in C (I; L

(Ω)), for every bounded interval I. Hence, it converges in this

space and that proves the continuity of u with respect to time, with L

-values.

Finally, the estimate passes to the limit. 

Lemma 4.7 immediately gives the following results:

Corollary 4.2 Let f(x, t) and g(y,t) be given in the classes L

. Then the solu-

tion of the boundary value problem Lu = f (x ∈ Ω, t ∈ R), γ

Bu = g (y ∈ ∂Ω,

t ∈ R), if it exists, must be unique in the class L

Corollary 4.3 Assume that u, Lu and γ

Bu are of class L

for every γ larger

than some ﬁnite threshold. Assume also that Lu and Bu vanish identically for

t<T. Then u vanishes for t<T.

Proof Because of translational invariance, we may take T = 0, meaning that

Lu ≡ 0andγ

Bu ≡ 0fort<0. Then the right-hand side of (4.5.26) is an o(e

γ

)

as γ → +∞, for every positive . Hence the left-hand side has the same property,

which implies that u ≡ 0fort<0. 

Corollary 4.3 is a principle of causality.

4.5.3 Existence and uniqueness for the boundary value problem in L

We proceed by duality, using the fact that the dual space to L

is L

−γ

,when

using the standard product

(u, v):=



Ω×R

u(x, t) · v(x, t)dx dt.

In particular, we have

u

=sup

|(u, v)|

v

−γ

Assume that u and Lu belong to L

. Then the well-deﬁned boundary trace γ

belongs to H

−1/2

, and we have a Green’s formula: For every v ∈ H

,itholds

We point out the importance that the boundary be non-characteristic in this argument.

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

that

(Lu, v) − (u, L

∗

v)+γ

Bu,γ

Mv + γ

Nu,γ

Cv =0, (4.5.27)

where the adjoint operator L

∗

is deﬁned by

∗

:= −∂

−



)

∂

and we recall that A

= M

B + C

N. The boundary terms in (4.5.27) are

duality products between H

−1/2

and H

1/2

−γ

Since A(ξ)andA(ξ)

are similar, L

∗

is constantly hyperbolic too. We have

seen (Theorem 4.2) that the backward adjoint IBVP satisﬁes (UKL). Therefore,

the latter admits a dissipative symmetrizer

and we may use an estimate similar

to (4.5.26): If v, L

∗

v and γ

Cv are of class L

−γ

,thenγ

v is of class L

−γ

,andv

satisﬁes



Ω×R

2γt

v

dx dt +



∂Ω×R

2γt

γ

v

dy dt

≤ C





Ω×R

2γt

L

∗

v

dx dt +



∂Ω×R

2γt

γ

Cv

dy dt



, (4.5.28)

at least for γ>γ

Deﬁne the subspace X

of L

−γ

, whose elements are the functions of the form

∗

v,wherev and L

∗

v are in L

−γ

, such that γ

Cv ≡ 0. From (4.5.28), the map

∗

v → v is well-deﬁned and continuous from X

into L

−γ

Given functions f(x, t)andg(y, t) of class L

, we may deﬁne a linear form 

on X

∗

v → (L

∗

v):=



Ω×R

(v, f)dx dt +



∂Ω×R

(g, Mv)dy dt,

where we recall that A

= M

B + C

N. Estimate (4.5.28) and the Cauchy–

Schwarz inequality show that  is continuous, with

|(L

∗

v)|≤v

−γ

f

+ Cγ

v

−γ

g

≤ C



−1

f

+ γ

−1/2

g



L

∗

v

−γ

Thanks to the Hahn–Banach and Riesz Theorems, there exists a function u(x, t),

of class L

, satisfying

(L

∗

v)=(u, L

∗

v), (4.5.29)

and

γu

≤ C



−1

f

+ g



. (4.5.30)

A dissipative symmetrizer for a backward IBVP obeys the same requirements as for the direct

IBVP, except that an inequality has to be reversed. We leave the reader to write the details.