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

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How energy estimates imply well-posedness 255

9.2 How energy estimates imply well-posedness

As for the Cauchy problem, energy estimates can be used in a duality argument

to show the well-posedness of Initial Boundary Value Problems (IBVP). However,

this is far from being straightforward, as already seen in Chapter 4 in the

constant-coeﬃcients case. The ﬁrst step is to show the well-posedness of the

Boundary Value Problem, posed for t ∈ R in weighted spaces, with boundary

data for x ∈ ∂Ω. This is basically where the energy estimates/duality argument

are used. The second step deals with the well-posedness of the special, IBVP with

zero initial data, which we also call homogeneous IBVP (the term homogeneous

referring by convention to the initial data and not to the boundary data, contrary

to the terminology of Chapter 7). The ﬁnal step concerns the general IBVP, with

compatibility conditions needed for the regularity of solutions.

Sections 9.2.1, 9.2.2 and 9.2.3 describe these steps successively, for smooth

coeﬃcients. Section 9.2.4 will be devoted to coeﬃcients with poorer regularity.

9.2.1 The Boundary Value Problem

The resolution of the Boundary Value Problem (BVP) relies on a duality

argument, which requires the deﬁnition of an adjoint BVP. We proceed as in

Section 4.4. We ﬁrst observe that for smooth enough functions u and v,



Ω×R

( v

Lu − u

∗

v)=



Ω×R

∂

u)+



∂



∂Ω



(x, t) A(x, t, ν(x)) u(x, t)dµ(x)dt,

(where µ denotes the measure on ∂Ω) after integration by parts. We thus need

to decompose the matrix A(x, t, ν(x)) according to the boundary matrix B(x, t)

in order to formulate an adjoint BVP. This is the purpose of the following

abstract result, which can be applied to W := ∂Ω × R and, with a slight abuse

of notation, A(x, t)=A(x, t, ν(x)) (invertible if and only if our BVP is non-

characteristic).

Lemma 9.4 Given a smooth manifold W , assume that A ∈ C

∞

(W ; GL

(R))

and B ∈ C

∞

(W ; M

p×n

(R)).IfB is everywhere of maximal rank p and if kerB

admits a smooth basis, there exists N ∈ C

∞

(W ; M

(n−p)×n

(R)) such that

=kerB ⊕ kerN

everywhere on W . Furthermore, there exist

C ∈ C

∞

(W ; M

(n−p)×n

(R)) and M ∈ C

∞

(W ; M

p×n

(R))

such that

=kerC ⊕ kerM, A = M

B + C

N, kerC =(A kerB)

⊥

everywhere on W .

256 Variable-coeﬃcients initial boundary value problems

Proof of (9.1.27) The row vectors (of length n)ofB, b

,...,b

say are,

by assumption, C

∞

functions on W . By assumption, there exists also a family

of independent vector-valued smooth functions (e

p+1

,...,e

) spanning kerB

everywhere on W . Hence we have

kerB = Span(e

p+1

,...,e

) = ( Span(b

∗

,...,b

∗

))

⊥

and (b

∗

,...,b

∗

p+1

,...,e

) is a basis of R

. The linear mapping

→ R

n−p

which associates to any vector u ∈ R

its (n − p) last components in that basis

is obviously one-to-one on kerB, and its matrix (in the ‘canonical’ bases of R

and R

n−p

)is

N =



0 I

n−p



−1

with P :=



∗

...b

∗

p+1

...e



By construction, the mapping w ∈ W → N(w) ∈ M

(n−p)×n

(R) is of class C

∞

Therefore, the square matrix

B :=





is invertible everywhere on W , and its inverse B

−1

is a C

∞

function on W .

Deﬁning D and Y as the (n × p)and(n × (n − p)) blocks in B

−1

,insuchaway

that

















= I

we obtain M and C as

M =(AY)

and C =(AD)

By construction, M and C are C

∞

on W . The remaining algebraic details are

left to the reader. 

Thanks to this lemma we have the identity



Ω×R

( v

Lu − u

∗

v)+



∂Ω×R

((Mv)

(Bu)+(Nu)

(Cv))

= 0 for all u, v. (9.2.33)

Furthermore, the (uniform) Lopatinski condition for the BVP (9.1.5) is equiva-

lent to the backward (uniform) Lopatinski condition for the adjoint BVP

∗

u)(x, t)=f(x, t),x∈ Ω ,t∈ R , (9.2.34)

(Cu)(x, t)=g(x, t),x∈ ∂Ω ,t∈ R . (9.2.35)

Indeed, Theorem 4.2 in Chapter 4, is valid pointwisely in (x, t).

How energy estimates imply well-posedness 257

We use below the following slight abuse of notation. For any (Hilbert) space H

of functions u of (x, t), e

γt

H stands for the space of functions u such that (x, t) →

u

(x, t)=e

−γt

u(x, t) belongs to H. This will be used for H = H

), H =

; H

)) and H = H

× R

Our aim is to solve (9.1.5)(9.1.6) in e

γt

; H

)) for s ≥ 0. For ‘small’

values of s (i. e. s ≤ d/2 + 1) the boundary condition (9.1.6) is to be understood

in the sense of traces, thanks to the following result attributed to Friedrichs.

Theorem 9.8 Assuming ∂Ω is non-characteristic for the operator L (whose

coeﬃcients are supposed to be C

∞

, or even Lipschitz, functions of (x, t) that are

constant outside a compact subset of Ω × R), we consider the subspace

E = {u ∈ L

(Ω × R); Lu ∈ L

(Ω × R) }

equipped with the graph norm. Then D (

Ω × R) is dense in E and the map

Ω × R) →D(∂Ω × R)

ϕ → ϕ

|∂Ω×R

admits a unique continuous extension from E to H

−1/2

(∂Ω × R).

Proof For general domains Ω, the question reduces, through co-ordinate charts,

to the case of a hyperplane, see [31] for more details. For simplicity, we directly

assume that

Ω={x ; x

> 0}.

Furthermore, multiplying L by (A

)

−1

, we may assume without loss of generality

that the coeﬃcient of ∂

in L is the identity matrix – by assumption (A

)

−1

smooth and uniformly bounded.

We ﬁrst show the existence of a constant C so that for any ϕ ∈ D(

Ω × R),

ϕ

|∂Ω×R



−1/2

(∂Ω×R)

≤ C ( ϕ

(Ω×R)

+ Lϕ

(Ω×R)

) .

Indeed, if 1 denotes the characteristic function of R

, we may associate to any

function u the function u

deﬁned by u

(x, t)=u(x, t) 1(x

). Then, by the

so-called jump formula

L(ϕ

)=(Lϕ)

+ ϕ

|{x

=0}

⊗ δ

=0}

Now, an easy calculation (using Fourier transform) shows that the H

−1/2

norm

of ϕ

|{x

=0}

is proportional to the H

−1

norm of ϕ

|{x

=0}

⊗ δ

=0}

. Therefore,

there exists C>0sothat

ϕ

|{x

=0}



−1/2

≤ C ( L(ϕ

)

−1

+ (Lϕ)



−1

)

≤ C ( C



ϕ



+ (Lϕ)



)=C ( C



ϕ

+ Lϕ

)

where C



is the norm of L as an operator L

→ H

−1

258 Variable-coeﬃcients initial boundary value problems

To show the density of D(Ω × R)inE, we can restrict ourselves – by

standard cut-oﬀ – to the approximations of functions u ∈ E that are compactly

supported. Take a smooth kernel ρ ∈ C

∞

d+1

; R

) compactly supported in

≤ 0} such that



d+1

ρ = 1 and consider the associated molliﬁer ρ

(x)=

−d−1

ρ(x/ε, t/ε). We know that the convolution operator by ρ

,sayR

, tends to

the identity in L

when ε goes to 0. In particular, for any u ∈ E, R

) tends to

in L

d+1

) and thus ϕ

:= R

)

|{x

>0}

tends to u in L

(Ω × R). Since ρ

is supported in {x

≤ 0}, the smooth function ϕ

is also compactly supported

if u is so. It remains to show that Lϕ

tends to Lu.But

Lϕ

=(R

◦ L)(u

)

|{x

>0}

+[L, R

](u

)

|{x

>0}

By Theorem C.14 (which extends a classical lemma of Friedrichs [63] to Lipschitz

coeﬃcients), the last term is known to tend to 0 in L

. Furthermore, since

L(u

) − (Lu)

is supported in {x

=0} and ρ

is supported in {x

≤ 0},the

ﬁrst term equivalently reads

◦ L)(u

)

|{x

>0}

= R

((Lu)

)

|{x

>0}

which does tend to Lu in L

(Ω × R). 

We are now in a position to prove the following, which covers both the

frameworks of Theorem 9.4 and Theorem 9.5.

Theorem 9.9 Assume that Ω is either (globally) diﬀeomorphic to a half-

space or a relatively compact domain with C

∞

boundary. We also make the

usual assumptions (CH) (constant hyperbolicity of the operator L), (NC) (non-

characteristicity of the boundary ∂Ω with respect to L), (N) (normality of the

boundary data B) with the additional fact that kerB admits a smooth basis on

∂Ω × R and (UKL) (uniform Kreiss–Lopatinski˘ı condition for (L, B) in Ω).

Then there exists γ

≥ 1 such that for all γ ≥ γ

, for all f ∈ e

γt

(Ω × R)

and all g ∈ e

γt

(∂Ω × R), there is one and only one solution u ∈ e

γt

(Ω × R)

of the Boundary Value Problem (9.1.5)(9.1.6). Furthermore, the trace of u on

∂Ω × R belongs to e

γt

(∂Ω × R),andu

−γt

u enjoys an estimate

γ u



(Ω×R)

+ (u

)

|∂Ω×R



(∂Ω×R)

≤ c







(Ω×R)

+ g



(∂Ω×R)



(9.2.36)

for some constant c>0 depending only on γ

. Furthermore, for all k ∈ N, there

exists γ

≥ γ

such that, for all γ ≥ γ

, for all f ∈ e

γt

(Ω × R) and all g ∈

γt

(∂Ω × R), the solution u of (9.1.5)(9.1.6) belongs to e

γt

(Ω × R) and

its trace belongs to e

γt

(∂Ω × R).

Note: The ﬁrst part of the theorem is the extension to variable coeﬃcients of

Lemma 4.8 in Chapter 4. The last part speciﬁes the degree of regularity of the

solution for more regular data.

How energy estimates imply well-posedness 259

Proof As for the Cauchy problem, a technical diﬃculty lies in the fact that

the energy estimate is known a priori for solutions smoother than L

. More pre-

cisely, either Theorem 9.4 or Theorem 9.5 gives (9.1.21) for u ∈ D (

Ω × R). This

implies (9.1.21) a priori only for u ∈ e

γt

(Ω × R): indeed, we can approach

u ∈ e

γt

(Ω × R)bysomeu

∈ D (Ω × R) in such a way that u

goes to u in

γt

(Ω × R), (u

)

|∂Ω×R

tends to the trace of u in e

γt

(∂Ω × R), but also Lu

tends to Lu in e

γt

(Ω × R). In fact, it will turn out that (9.1.21) is met as

soon as u belongs to L

(Ω × R): this fact may be viewed as a consequence of the

regularity part of the theorem, which we admit for the moment. 

Uniqueness It is straightforward if we use the regularity part. For, by linearity

it suﬃces to prove that the only e

γt

solution of the homogeneous problem

Lu =0 onΩ× R ,Bu=0on∂Ω × R (9.2.37)

is 0. But if u ∈ e

γt

(∂Ω × R) solves (9.2.37), then u belongs to e

(∂Ω × R)

(because 0 belongs to H

!) and therefore satisﬁes the energy estimate (9.2.36)

with γ ≥ γ

and f ≡ 0, g ≡ 0, which of course implies u = 0 almost everywhere.

Existence It is shown by duality as in the constant-coeﬃcients case (also see

Theorem 2.6 in Chapter 2 for the Cauchy problem). The details are slightly

diﬀerent here because we have not proved yet the a priori estimate for L

data.

Introduce the space

E := {v ∈ e

−γt

(Ω × R); Cv

|∂Ω×R

=0}

equipped with the norm

v

:= e

γt

v

(Ω×R)

Similarly, to simplify the writing we denote here

|v|

:= |e

γt

(∂Ω×R)

The energy estimate for the backward and homogeneous boundary value problem

associated with L

∗

reads

γ v

+ |v|

≤

L

∗

v

for some c>0 independent of both v and γ. In particular, it shows that L

∗

restricted to E is one-to-one and thus enables us to deﬁne a linear form  on L

∗

(L

∗

v)=



Ω×R

f +



∂Ω×R

(Mv)

260 Variable-coeﬃcients initial boundary value problems

satisfying the estimate

|(L

∗

v)|≤f

−γ

v

+ M

∞

|g|

−γ

|v|





f

−γ

1/2

|g|

−γ



L

∗

v

Therefore, by the Hahn–Banach theorem  extends to a continuous form on the

weighted space e

−γt

(Ω × R), and by the Riesz theorem, there exists u in the

dual space e

γt

(Ω × R) such that

(L

∗

v)=



Ω×R

∗

Thanks to (9.2.33), this yields



Ω×R

Lu +



∂Ω×R

(Mv)

Bu −



Ω×R

f −



∂Ω×R

(Mv)

g =0

for all v ∈ E . In particular, for v ∈ D (Ω × R), the boundary terms vanish and

we infer that Lu = f in the sense of distribution. Consequently, we have



∂Ω×R

(Mϕ)

Bu −



∂Ω×R

(Mϕ)

g =0

for all compactly supported C

∞

function ϕ on the boundary ∂Ω × R such that

Cϕ = 0. Since at each point, M

|KerC

:KerC→ C

is onto, and M has C

∞

coeﬃcients, this implies



∂Ω×R

Bu −



∂Ω×R

g =0

for all ψ ∈ D (∂Ω × R), hence the boundary condition Bu = g in the sense of

distributions.

Regularity A key ingredient is the following tricky result.

Theorem 9.10 (Chazarain–Piriou) Assume that the operator L and the bound-

ary operator B have C

∞

coeﬃcients, constant outside a compact subset of Ω × R,

that the boundary ∂Ω is non-characteristic for the operator L, and that the pair

(L, B) is endowed with the a priori estimate

γ ϕ

(Ω×R)

+ ϕ

|∂Ω



(∂Ω×R)

≤ c



L

ϕ

(Ω×R)

+ Bϕ

(∂Ω×R)



(9.2.38)

for some c>0 independent of γ ≥ γ

and ϕ ∈ H

(Ω × R). Then for all integer

m ≥−1, the four conditions

v ∈ H

(Ω × R) ∩ L

(Ω × R) ,L

v ∈ H

m+1

(Ω × R) for all γ ≥ γ

|∂Ω×R

∈ H

(∂Ω × R) ,Bv

|∂Ω×R

∈ H

m+1

(∂Ω × R) ,

How energy estimates imply well-posedness 261

imply

v ∈ H

m+1

(Ω × R) and v

|∂Ω×R

∈ H

m+1

(∂Ω × R) .

Note: in this statement, v

|∂Ω×R

is to be understood as the trace of v on ∂Ω × R,

as deﬁned in Theorem 9.8, and belongs at least to H

−1/2

(∂Ω × R).

Admitting temporarily Theorem 9.10, we can easily complete the proof

of the regularity part in Theorem 9.9. For, if u ∈ e

γt

(Ω × R) is such that

Lu = f belongs to e

γt

(Ω × R)andBu

|∂Ω×R

= g belongs to e

γt

(∂Ω × R),

Theorem 9.10 applied to m = −1andv = u

readily implies u

|∂Ω×R

belongs

to e

γt

(∂Ω × R), as claimed. Now, if additionally f ∈ e

γt

(Ω × R)andg ∈

γt

(∂Ω × R), we can apply Theorem 9.10 to m =0 and v = u

, and thus

obtain that u belongs to e

γt

(Ω × R) and its trace to e

γt

(∂Ω × R). By

induction, we thus show that u and its trace inherit exactly the Sobolev index

of f and g.

Energy estimate To complete the proof of Theorem 9.9 it remains to show

that the energy estimate in (9.2.36) is valid as soon as u belongs to e

γt

.To

prove this fact we can regularize the data f and g and use the regularity part of

Theorem 9.9 together with the uniqueness of solutions. For any f ∈ e

γt

(Ω ×

R)andg ∈ e

γt

(∂Ω × R), there exist f

∈ D (Ω × R)andg

∈ D (∂Ω × R)such

that

−γt

(Ω×R)

−−−−−−→

ε→0

−γt

f, e

−γt

(∂Ω×R)

−−−−−−−→

ε→0

−γt

For all ε, there exists u

∈ e

γt

(Ω × R) with γ ≥ γ

solving the regularized

problem

= f

on Ω × R ,Bu

= g

on ∂Ω × R .

The energy estimate (9.2.36) is valid for the (smooth) diﬀerence (u

− u



)

and thus shows that both (u

)

ε>0

and ((u

)

|∂Ω×R

)

ε>0

are Cauchy sequences,

in e

γt

(Ω × R)ande

γt

(∂Ω × R) respectively. Therefore, there exist u ∈

γt

(Ω × R)andu

∈ e

γt

(∂Ω × R) such that

e

−γt

− u)

(Ω×R)

→ 0ande

−γt

((u

)

|∂Ω×R

− u

)

(Ω×R)

→ 0

as ε goes to 0. By construction, Lu

= f

converges to f = Lu in e

γt

(Ω ×

R). Consequently, by Theorem 9.8 the trace u

|∂Ω×R

converges to u

|∂Ω×R

γt

−1/2

(∂Ω × R), hence by uniqueness of limits in the sense of distributions,

|∂Ω×R

= u

. In the limit, we thus get Bu

|∂Ω×R

= g,andu solves the same

BVP as u. Therefore u

= u and by passing to the limit in the energy estimate

(9.2.36) for u

, we obtain (9.2.36) for u. 2

Remark 9.11 The previous regularization method and Corollary 9.1 applied

to u

, show (at least when Ω is a half-space) that there exists C

> 0sothat

for f ∈ H

(Ω × R)andg ∈ H

(∂Ω × R) with γ ≥ γ

, the solution u of the

262 Variable-coeﬃcients initial boundary value problems

Boundary Value Problem (9.1.5)(9.1.6) satisﬁes

γ u

(Ω×R)

+ u



(∂Ω×R)

≤C



f

(Ω×R)

+ g

(∂Ω×R)



(9.2.39)

Proof of Theorem 9.10 (Note: In the case m = −1, the only thing to show

is that the trace of v belongs to L

(∂Ω × R).)

For general domains Ω, the proof should naturally involve co-ordinate charts,

and to some extent this would hide the heart of the matter. We prefer to give the

proof in the simpler case Ω = {x ; x

> 0 }, Ω × R being identiﬁed with R

= {(y,t,x

); x

≥ 0 },and∂Ω × R with R

= {(y,t) }. The outline follows

the proof given by Chazarain and Piriou [31] in the case of smooth compact Ω,

save for co-ordinate charts.

A ﬁrst useful remark is that it suﬃces to show v actually belongs to

; H

m+1

)). Indeed, Proposition 2.3 applied with x

instead of t shows

that v ∈ L

; H

m+1

)) and L

v ∈ H

× R

)) imply v ∈ H

m+1

Next, we are to use the following observation on Sobolev spaces.

Proposition 9.2 Take s ∈ R.Ifv ∈ H

) and if there exists C>0 so that

v

s,θ



(1 + ξ

)

s+1

1+ θξ

|v(ξ)|

dξ ≤ C

for all θ ∈ (0, 1], then v belongs to H

s+1

(This is a consequence of Fatou’s Lemma: take the lim inf of the inequality when

θ goes to 0.) So, showing additional regularity for v ∈ H

amounts to ﬁnding a

uniform bound of v

s,θ

. In this respect, one may try to use the left part of the

two-sided estimate given by the following technical result.

Proposition 9.3 Take s ∈ R, r>s+1 and ρ ∈ D (R

) such that

ρ(ξ)=O(ξ

)

in the neighbourhood of 0 and that ρ does not vanish identically on any ray

{tξ; t ∈ R

}, ξ =0.Forε>0, we deﬁne

: x → ε

−d

ρ(x/ε) ,

and consider R

the convolution operator by ρ

. Then there exist C and C



> 0

such that

C v

s,θ

≤v

)

+ n

s,θ

(v) ≤ C



v

s,θ

(v):=



R

v

)

−2(s+1)





−1

dε

for all θ ∈ (0, 1] and all v ∈ H

How energy estimates imply well-posedness 263

(The proof is technical but elementary: it can be found in [31], Chapter 2. There

do exist kernels ρ satisfying all the requirements: take say σ ∈ D (R

) such that



σ = 0 and deﬁne, for instance, ρ =∆

r/2

σ.) Another important preliminary

result is the following.

Theorem 9.11 If R

is a smoothing operator as in Proposition 9.3, if P

(resp., B) is a diﬀerential operator of order 1 (resp., 0), with C

∞

coeﬃcients

that are constant outside a compact subset of R

, there exists C



> 0 such that



[ P,R

] v

)

−2(s+1)





−1

dε

≤ C



v

s,θ



[ B,R

] v

)

−2(s+2)





−1

dε

≤ C



v

s,θ

for all θ ∈ (0, 1],allv ∈ H

(These are specials cases of a general result on pseudo-diﬀerential operators,

the proof of which involves commutator decompositions and repeated use of

Proposition 9.3; see [31], Chapter 4.)

In particular, if (A

)

−1

is smooth and uniformly bounded, we may apply

Theorem 9.11 to

P (x

)=−(A

)

−1



∂

d−1



j=1

∂



which depends continuously on x

(viewed here as a parameter): this will give

an estimate for the operator L

:= (A

)

−1

L since [ L

]=[−P,R

]. And

applying Theorem 9.11 to B =(A

)

−1

will eventually give an estimate for the

operator

:= (A

)

−1

= ∂

− P (x

) − γ (A

)

−1

uniformly in γ.

Now, the characterization of H

functions that are actually in H

s+1

given by Proposition 9.2, the two-sided inequality provided by Proposition 9.3,

the commutator’s estimates given by Theorem 9.11, and the energy estimate

(9.2.38), altogether enable us to complete the proof of Theorem 9.10. Consider

v ∈ H

× R

) ∩ L

× R

) such that L

v ∈ H

m+1

× R

), v

∈

)andBv

∈ H

m+1

)form ≥−1, and deﬁne ϕ

:= R

(v), where

is a smoothing operator as in Proposition 9.3, in the variables (y, t) ∈ R

.By

construction, the function ϕ

belongs to L

; H

+∞

)), and

= R

ϕ +[L

] ϕ

belongs to L

× R

): for the ﬁrst term we use the smoothness of (A

)

−1

that L

ϕ is at least L

and that R

is a bounded operator on L

; for the

264 Variable-coeﬃcients initial boundary value problems

commutator, we use that there are no derivatives with respect to x

and apply

Friedrichs Lemma ( [63], or Theorem C.14) in the (y, t) variables. Therefore,

applying again Proposition 2.3, with x

playing the role of t, we infer that ϕ

belongs to H

× R

), and we can apply the L

estimate (9.2.38) to ϕ

,which

reads

γ ϕ



×R

)

+ (ϕ

)



)

≤ c



L



×R

)

+ B(ϕ

)



)



Multiplying by ε

−2m−3





−1

, integrating in ε ∈ (0, 1], and using Theo-

rem 9.11 (with s = m and s = m − 1) we get a constant C>0 such that



+∞

v

m,θ

+ v



m,θ

≤ C





+∞

v

m,θ

+ γ



+∞

v

m−1,θ



+∞

m,θ

v(x

)) dx

+ n

m,θ

(Bv

)+v



m−1,θ



For γ large enough the ﬁrst term in the right-hand side can be absorbed in the

left-hand side. Therefore, applying Propositions 9.2 and 9.3, we get



+∞

v

m,θ

+ v



m,θ

≤ C



L

v

m+1

))

+ Bv



m+1

)

+ γ v

))

+ v



)



The right-hand side being independent of θ, Proposition 9.2 thus shows that v

belongs to L

; H

m+1

)) and v

belongs to H

m+1

). This completes

the proof of Theorem 9.10. 2

9.2.2 The homogeneous IBVP

Theorem 9.12 In the framework of Theorem 9.9, if f ∈ L

(

Ω × [0,T]) and

g ∈ L

(∂Ω × [0,T]) the problem

Lu = f on Ω × (0,T) , (9.2.40)

Bu = g on ∂Ω × (0,T) , (9.2.41)

|t=0

=0 on Ω , (9.2.42)