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 265

admits a unique solution u ∈ L

(

Ω × [0,T]). Furthermore, the trace u

∂Ω×[0,T ]

belongs to L

(∂Ω × [0,T]), and there exist γ

and c>0 depending only on ω so

that for all γ ≥ γ

γ e

−γt

u

(Ω×(0,T ))

+ e

−γt

|∂Ω×[0,T ]



(∂Ω×(0,T ))

≤ c



e

−γt

f

(Ω×(0,T ))

+ e

−γt

g

(∂Ω×(0,T ))



(9.2.43)

Furthermore, if f ∈ H

(

Ω × [0,T]) and g ∈ H

(∂Ω × [0,T]) with k ≥ 1 and

∂

f =0, ∂

g =0at t =0for all j ∈{0,...,k− 1}, then u belongs to H

(

Ω ×

[0,T]) and satisﬁes ∂

u =0at t =0for all j ∈{0,...,k− 1}.

Remark 9.12

i) According to Theorem 9.8, the equality in (9.2.42) is meaningful at least

in H

−1/2

(Ω) for any square-integrable u satisfying (9.2.40) with f also

square-integrable.

ii) It will appear in the proof of Theorem 9.12 that solving the homogeneous

IBVP (9.2.40)–(9.2.42) for f ∈ L

(

Ω × [0,T]) and g ∈ L

(∂Ω × [0,T]) is

equivalent to solving







Lu =



f on Ω × (−∞,T) ,

Bu = g on ∂Ω × (−∞,T) ,

u =0 onΩ× (−∞, 0) ,

for



f ∈ L

(

Ω × (−∞,T]) and g ∈ L

(∂Ω × (−∞,T]) vanishing on (−∞, 0).

Proof To prove the existence of a solution u,weﬁrstextendf and g by zero

for t<0andt>T. The resulting functions, say

f and ˘g, obviously belong to

γt

(

Ω × R)ande

γt

(∂Ω × R), respectively, for any real number γ. Therefore,

we may apply Theorem 9.9 to

f and ˘g as source term and boundary term,

respectively. This yields a solution ˘u ∈ e

(

Ω × R) of the boundary value

problem

L˘u =

f on Ω × R ,B˘u =˘g on ∂Ω × R . (9.2.44)

By construction u := ˘u

|t∈[0,T ]

clearly satisﬁes (9.2.40) and (9.2.41). However, the

fact that u vanishes at t = 0 needs some veriﬁcation.

Similarly as in the proof of Theorem 9.8, we denote by 1 the character-

istic function of R

, and for any function v, v

(x, t)=v(x, t) 1(t). By that

theorem, where Ω × R is replaced by {(x, t); t>0 } (whose boundary is non-

characteristic because of the hyperbolicity of L in the direction of t), we know

that ˘u admits a trace ˘u

|t=0

∈ H

−1/2

) and we have the jump formula

L(˘u

)=(L˘u)

+˘u

|t=0

⊗ δ

{t=0}

266 Variable-coeﬃcients initial boundary value problems

So we will be able to conclude that ˘u

|t=0

= 0 if we can show that L(˘u

)=(L˘u)

which will be the case if ˘u

=˘u. The latter equality equivalently means that

˘u

t<0

= 0, which follows from the following support theorem.

Theorem 9.13 In the framework of Theorem 9.9, if f ∈ L

(

Ω × R) and g ∈

(∂Ω × R) both vanish for t<t

, then the solution u of the boundary value

problem

Lu = f on Ω × R ,Bu= g on ∂Ω × R

also vanishes for t<t

We postpone the proof of this result, and complete the proof of Theorem

9.12.

For the uniqueness of the solution u, we must show that the only solution in

of the homogeneous problem

Lu =0on Ω× (0,T) ,

Bu =0on ∂Ω × (0,T) ,

|t=0

=0on Ω,

is the trivial solution 0. So assume u is a solution. Using the same notation as

before and again the jump formula we see u

solves the problem

Lu =0on Ω× (−∞,T) ,

Bu =0on ∂Ω × (−∞,T) .

Now, we introduce a smooth cut-oﬀ function θ such that θ ≡ 1on(−∞,τ],

τ<T,andθ ≡ 0on[T,+∞). Then both θu

and L(θu

) belong to L

(

Ω × R),

and the trace of Bθu

belongs to L

(∂Ω × R). Furthermore, L(θu

) ≡ 0and

Bθu

≡ 0fort<τ. Hence by Theorem 9.13 again, θu

= 0 for t<τ.Since

this is true for all τ<Tand θu

= u for t ∈ [0,τ], we infer that u ≡ 0(a.e)in

[0,T), as expected.

So the unique solution of (9.2.40)–(9.2.42) is necessarily the one constructed

in the ﬁrst step, i.e. u =˘u

|t∈[0,T ]

. Therefore, the (weighted) localized L

estimate

(9.2.43) for the homogeneous IBVP is a consequence of the weighted L

estimate

(9.2.36) for the BVP, applied to ˘u.

Finally, the assumptions on the partial derivatives of f and g if f ∈ H

(

Ω ×

[0,T]) and g ∈ H

(∂Ω × [0,T]) show that they admit extensions as functions in

(

Ω × R)andH

(∂Ω × R), respectively, which vanish for t<0. By Theorem

9.9, the solution of the corresponding BVP belongs to H

(

Ω × R), and by

Theorem 9.13, its restriction to (−∞,T]) depends only on f and g (and not

on their extensions for t>T), so it coincides with the restriction of ˘u (solution

of (9.2.44) above) to (−∞,T]): this implies ˘u

|t<T

∈ H

(

Ω × (−∞,T]), hence

u =˘u

|t∈[0,T ]

∈ H

(

Ω × [0,T]), and the fact that ˘u

|t<0

= 0 does imply that

∂

u =0att = 0 for all j ∈{0,...,k− 1}. 

How energy estimates imply well-posedness 267

Proof of Theorem 9.13 Without loss of generality, we may assume t

=0.

By assumption, f belongs to e

γt

(

Ω × R)andg belongs to e

γt

(∂Ω × R)

for all γ>0. So, by Theorem 9.9, for all n ∈ N, there exists a unique u

∈

(Ω × R) with γ

= γ

+ n such that

= f on Ω × R ,Bu

= g on ∂Ω × R ,

and by the energy estimate (9.2.36),

e

−γ





e

−γ

f

+ e

−γ

g

=: C

Furthermore, we claim u

is independent of n for large enough n. Indeed, consider

a C

∞

function

θ : R → [0, 1]

t ≤ 0 → 1 ,

t>1 → e

−t

Then, for all n, v

(x, t):=θ(t)(u

n+1

− u

)(x, t) deﬁnes a function in

(∂Ω × R) such that

(L − θ



(t)/θ(t)) v

=0 on Ω×R ,Bv

=0 on ∂Ω × R .

Therefore, by the uniqueness part of Theorem 9.9 applied to the operator

(L − θ



(t)/θ(t)) (the additional smooth, zero-order term being harmless), we ﬁnd

that v

=0(forn large enough so that γ

is larger than the γ

corresponding

to the modiﬁed operator).

Then the fact that u

= u vanishes for t<0, is an easy consequence of the

uniform bound

e

−γ

u

≤ C.

Indeed, take ε>0andϕ ∈ D(Ω × (−∞, −ε)). Then

|u, ϕ|≤e

−γ

u

e

ϕ

≤ C e

− γ

ϕ

which goes to zero when n → +∞. Therefore, u

|t∈(−∞,−ε)

≡ 0 (a.e.) for

all ε>0. 2

9.2.3 The general IBVP (smooth coeﬃcients)

As already pointed out in Chapter 4 for constant-coeﬃcients problems, the

resolution of the Initial Boundary Value Problem (9.0.2)–(9.0.4) with non-zero

initial data u

is not a direct consequence of previous results.

A natural (or naive) approach to the general IBVP is to consider and solve

the Cauchy problem

Lv =0,v

|t=0

= u

268 Variable-coeﬃcients initial boundary value problems

(with u

extended by zero outside Ω) and then look for u = w + v with

Lw = f in Ω ,Bw= g − Bv on ∂Ω ,w

|t=0

=0. (9.2.45)

Unfortunately, for L

data f and u

, v is square-integrable in Ω but its trace

on ∂Ω has no reason to be also square-integrable (recall that by Theorem

9.8 it is a priori H

−1/2

), which obviously hinders the resolution of (9.2.45).

An alternative is to consider as a preliminary problem not a Cauchy

problem but an Initial Boundary Value Problem of the form

Lv =0 inΩ,B

v =0 on∂Ω ,v

|t=0

= u

, (9.2.46)

no matter what B

is, provided it yields a solution v with L

trace on ∂Ω:

if we are able to solve (9.2.46) in such a way that v

|∂Ω

is in L

,thenwe

will be able to solve (9.2.45) in L

(thanks to Theorem 9.12), hence also

the general IBVP (by taking u = w + v). In other words, the resolution

of the general IBVP relies on the resolution of a particular IBVP, with a

possibly diﬀerent boundary matrix, with non-zero initial data u

but with

zero source term f and zero boundary data g.

For the resolution of (9.2.46), a natural idea is to proceed as for the Cauchy

problem, by duality. In this respect, we need a pointwise estimate of the form

v(T )

(Ω)

≤ e

γT

v(0)

(Ω)

, (9.2.47)

for γ large enough, independent of v ∈ C

([0,T]; L

(Ω)) ∩ C ([0,T]; H

(Ω)) such

that

Lv =0 inΩ,B

v =0 on∂Ω .

Deriving such an estimate was the main purpose of the paper by Rauch [162],

following his PhD thesis [161]. We shall not go into (thankless) details regarding

the derivation of (9.2.47) in general, for which we refer to [162]. But in the case

of a Friedrichs-symmetrizable operator L (as in [161]), the task is much easier:

proving (9.2.47) is a matter of integrations by parts and of suitable choice of the

boundary matrix B

, as we explain now.

Let us assume that L admits a Friedrichs symmetrizer S

, with

σI

≤ S

≤ σ

−1

for some positive σ.Wetakev ∈ C

([0,T]; L

(Ω)) ∩ C ([0,T]; H

(Ω)) to ensure

that our computations are valid. If Lv =0then



Ω

∗

v = −2



Ω



∗

∂

v +



Ω

∗

(∂

) v

= −



∂Ω



∗

v +



Ω

∗

(∂



∂

)) v.

How energy estimates imply well-posedness 269

If ∂Ω is non-characteristic, the matrix

(x, t) A(x, t, ν(x))

is hyperbolic along ∂Ω, which allows us to deﬁne B

(x, t) as the spectral

projection onto its stable subspace. If B

v =0 on∂Ωthen

v(x, t)

∗

(x, t) A(x, t, ν(x)) v(x, t) ≥ 0

hence



Ω

∗

v ≤



Ω

∗

(∂



∂

)) v.

This implies, after integration,

v(t)

≤ e

γt

v(0)

with

γ =

2σ

∂



∂

)

∞

(Ω×(0,t))

Theorem 9.14 (Rauch) Under the assumptions of Theorem 9.2, there exists

≥ 1 and C

> 0 such that for all γ ≥ γ

, for all T>0 and all u ∈ D (R

d−1

× [0,T]),

−2γT

u

|t=T



(Ω)

+ γ e

−γt

u

(Ω×(0,T ))

+ e

−γt

|∂Ω



(∂Ω×(0,T ))

≤ C



u

|t=0



(Ω)

e

−γt

Lu

(Ω×(0,T ))

+e

−γt



(∂Ω×(0,T ))



Proof hints If (additionally) L is Friedrichs symmetrizable, a duality argument

and the counterpart of the energy estimate (9.2.43) for the adjoint homogeneous

IBVP easily yield the reﬁned energy estimate as stated in Theorem 9.14 (for

details, see the proof of Theorem 9.19 below, which holds true for Lipschitz

coeﬃcients). Without Friedrichs symmetrizability, the proof of Theorem 9.14 is

much more technical (and valid only for smooth coeﬃcients). Rauch [162] ﬁrst

derives a H

d−1

estimate, by using a method due to G˚arding and Leray, then

an estimate of negative index for the adjoint problem, and ﬁnally shows how to

‘raise’ this estimate up to an L

one.

Using Theorem 9.14 we can prove the L

-well-posedness of the general IBVP,

as stated below.

Theorem 9.15 In the framework of Theorem 9.9, for all f ∈ L

(

Ω × [0,T]),

g ∈ L

(∂Ω × [0,T]), u

∈ L

(

Ω), the problem







Lu = f on Ω × (0,T) ,

Bu = g on ∂Ω × (0,T) ,

|t=0

= u

on Ω ,

270 Variable-coeﬃcients initial boundary value problems

admits a unique solution u ∈ L

(

Ω × [0,T]), which is such that u

∂Ω×[0,T ]

∈

(∂Ω × [0,T]). Furthermore, u belongs to C ([0,T]; L

(

Ω)) and satisﬁes an

estimate of the form

u(T )

(Ω)

u

(Ω×(0,T ))

+ u

|∂Ω×[0,T ]



(∂Ω×(0,T ))

≤ c



u



(Ω)

+ T f

(Ω×(0,T ))

+ g

(∂Ω×(0,T ))



with c independent of u

, f, g, u and T .

Proof Once we have the energy estimate of Theorem 9.14, the proof of L

-well-

posedness follows a standard procedure, and works even for rough (Lipschitz)

coeﬃcients. To avoid too much repetition, we refer to the proof of Theorem 9.19

below. 

Additional regularity and compatibility conditions

We recall that for the IBVP with zero initial data, Theorem 9.12 says solutions

are H

for H

source term f and H

boundary data g, provided that

∂



f =0 and ∂



g =0 at t =0, for all  ∈{0,...,k− 1}.

In fact, these conditions are not optimal: weaker conditions are suﬃcient to

ensure the H

regularity of solutions, even for the general IBVP. These are

compabilitity conditions between the source term f , the boundary data g and the

initial data u

, looking rather complicated but easy to understand. For, if Lu = f

with u of class C

with respect to time and f of class C

p−1

, a straightforward

proof by induction shows that

∂

u =

q−1



=0



q − 1







∂

q−1−

u + ∂

q−1

where P

:= ∂

+ L = −



(x, t) ∂

and P

:= −



(∂

)(x, t) ∂

for all

integer p ≥ 1; now if Bu = g on ∂Ω, Leibniz’ rule shows that

∂



g =





q=0







(∂

−q

B) ∂

Consequently, the compatibility conditions (necessary for such a u to exist) are

(CC

) the functions u

: x ∈

Ω → u

(x) deﬁned inductively by

(x)=

q−1



=0



q − 1







q−1−

(x)+∂

q−1

f(x, 0) for all q ∈{1,...,p}

How energy estimates imply well-posedness 271

are such that

∂



g(x, 0)





q=0







(∂

−q

B)(x, 0) u

(x) for all x ∈ ∂Ω, and all  ∈{1,...,p}.

Here above we have used the notation



:= −



(∂



)(x, 0) ∂

Theorem 9.16 (Rauch–Massey) In the framework of Theorem 9.15, if, more-

over, the initial data belongs to H

(

Ω), the source term f belongs to H

(

Ω ×

[0,T]), and the boundary data belongs to H

k+1/2

(∂Ω × [0,T]) for some integer

k ≥ 1,andifu

, f, g satisfy the compatibility conditions in (CC

) for all

p ∈{0,...,k− 1}, then the solution u belongs to C

([0,T]; H

k−r

(

Ω)) for all

r ∈{0,...,k}.

We omit the proof here; see [165].

Remark 9.13 If u

≡ 0andif ∂



f(x, 0) = 0 for all  ∈{1,...,k− 1},theset

of compatibility conditions (CC

) for p ∈{0,...,k− 1} reduce to

∂



g(x, 0) = 0 for all  ∈{1,...,k− 1},

as expected.

9.2.4 Rough coeﬃcients

The L

-well-posedness of the BVP is true as soon as the coeﬃcients of both (the

principal part of) the operator and the boundary conditions are Lipschitz. More

precisely, we have the following result.

Theorem 9.17 In the framework and with the assumptions of Theorem 9.6,

with the additional fact that W

is contractible, there exists γ

= γ

(ω) ≥ 1 such

that for all γ ≥ γ

, for all v ∈ V

, for all f ∈ e

γt

d−1

× R

× R) and all

g ∈ e

γt

d−1

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

γt

d−1

× R)

of the Boundary Value Problem

u = f on {x

> 0},B

u = g on {x

=0}. (9.2.48)

Furthermore, the trace of u at x

=0 belongs to e

γt

d−1

× R),andu

−γt

u enjoys an estimate

γ u



d−1

×R

×R)

+ u



d−1

×R)

(9.2.49)

≤ c







d−1

×R

×R)

+ g



d−1

×R)



for some constant c>0 depending only on ω.

272 Variable-coeﬃcients initial boundary value problems

Proof As in Theorem 9.9 the existence part relies on a duality argument

using the adjoint BVP, which can still be deﬁned thanks to Lemma 9.4: indeed,

substituting W

to Ω we get C

∞

mappings C, N,andM on W

with values in

(n−p)×n

(R)andM

p×n

(R), respectively, such that R

=kerC(w) ⊕ kerM(w),

(w)=M(w)

B(w)+C(w)

N(w)andkerC(w)=(A

(w)kerB(w))

⊥

for all w ∈ W

; hence the following identity



( z

u − u

)

∗

z)+



((M

u +(N

z)=0

(9.2.50)

for all Lipschitz-continuous v such that v

maps R

d−1

× R into W

(and

consistently with the notation B

, C

:= C ◦ v, M

:= M ◦ v and N

:= N ◦v)

and for all u ∈ e

γt

d−1

× R

× R)andz ∈ e

−γt

d−1

× R

× R).

Note: Thanks to Theorem 9.8, the identity (9.2.50) still holds true when the

regularity of u is relaxed to

u ∈ e

γt

d−1

× R

× R)and L

u ∈ e

γt

d−1

× R

× R)

(or symmetrically, if z ∈ e

−γt

d−1

× R

× R)and(L

)

∗

z ∈ e

γt

d−1

× R), but u ∈ e

γt

d−1

× R

× R)). For completeness, we now explain

the duality argument (which merely parallels what has been done in the proof

of Theorem 9.9). Introduce the subspace of e

−γt

d−1

× R

× R)

E := {z ∈ D (R

d−1

× R

× R); C

=0}

and denote for simplicity

z

:= e

γt

z

d−1

×R

×R)

|z|

:= |e

γt

d−1

×R)

The energy estimate

γ z

+ |z|

≤

(L

)

∗

z

(a consequence of Theorem 9.6 for the backward and homogeneous BVP associ-

ated with (L

)

∗

) enables us to deﬁne a linear form  on (L

)

∗

E by

((L

)

∗

z)=



f +



such that

|((L

)

∗

z)|≤f

−γ

z

+ M



∞

|g|

−γ

|z|





f

−γ

1/2

|g|

−γ



(L

)

∗

z

How energy estimates imply well-posedness 273

By the Hahn–Banach Theorem  thus extends to a continuous form on the

weighted space e

−γt

d−1

× R

× R), which shows by the Riesz Theorem the

existence of u in the dual space e

γt

d−1

× R

× R) such that

((L

)

∗

z)=



)

∗

for all z ∈ E . In particular, this implies by deﬁnition of  on (L

)

∗

E that for all

z ∈ D(R

d−1

× (0, +∞) × R),



( f

z − u

)

∗

z )=0,

hence L

u = f in the sense of distributions. Therefore, L

u ∈ e

γt

d−1

× R) and we may apply the identity (9.2.50). This yields



( B

u − g )=0

for all z ∈ E , hence also



ϕ)

( B

u − g )=0

for all compactly supported C

∞

function ϕ on the boundary R

d−1

× R such

that C

ϕ ≡ 0. Since at each point w ∈ W

, the linear mapping M(w)

|KerC(w)

Ker C(w) → C

is onto, and v

: R

d−1

× R → W

is Lipschitz-continuous,

this implies

ψ, B

u − g  =0

for all ψ ∈ e

−γt

1/2

d−1

× R), hence the boundary condition B

u = g holds

true in e

γt

−1/2

d−1

× R).

In this way we have obtained a weak solution u ∈ e

γt

d−1

× R

× R)of

our BVP. The next step is a so-called weak=strong argument, showing that all

γt

solutions are in fact limits of smooth solutions of regularized problems.

This will imply in particular the validity of the estimate of Theorem 9.6 for any

γt

solution, hence the uniqueness.

We take a standard molliﬁer in the (y,t) variables ρ

,andR

the associated

convolution operator, and deﬁne

= R

−γt

u, F

= R

)

−1

−γt

f, g

= R

−γt

For all ε>0, g

belongs to H

+∞

× R) and goes to g

−γt

g in L

× R)

as ε goes to zero, and F

belongs to L

; H

+∞

× R)) and goes to



)

−1

−γt

f in L

; L

× R)). Similarly, u

belongs to L

; H

+∞

R)) and goes to u

−γt

u in L

; L

× R)), while (u

)

belongs to

+∞

× R) and goes to (u

)

−γt

in H

−1/2

× R).

274 Variable-coeﬃcients initial boundary value problems

If we can show additional regularity of u

in the x

variable, we will be

allowed to apply the energy estimate of Theorem 9.6 to u

and pass to the limit

to obtain the estimate for u

. This will be possible thanks to the following result

on commutators, which is a straightforward consequence of Theorem C.14 and

Remark C.6, stated here as a lemma only for convenience.

Lemma 9.5 With the notations deﬁned above and, as in the proof of

Theorem 9.6,

:= −(A

)

−1



∂

+ γ +

d−1



j=1

(v(x, t)) ∂



we have

lim

ε→0

[P

−γt

u 

=0 and lim

ε→0

[B

−γt



=0.

Going on with the proof of Theorem 9.17, we claim that u

belongs to H

× R). Indeed, we have

∂

= F

+ P

− [P

−γt

where the ﬁrst two terms belong to L

; H

+∞

× R)) and the third one

is bounded in L

; L

× R)) (as a byproduct of the ﬁrst limit in Lemma

9.5), so that ∂

is in L

, as well as the other derivatives ∂

. Consequently,

by density of D (R

× R

× R)inH

× R

× R), this allows us to apply the

inequality in (9.1.22) to u

− u



, hence

γ u

− u





d−1

×R

×R)

+ (u

− u



)



d−1

×R)



(∂

− P

)(u

− u



)

d−1

×R

×R)

+ B

− u



)



d−1

×R)



F

− F





d−1

×R

×R)

[P

− R



−γt

u

d−1

×R

×R)

+ g

− g





d−1

×R)

+ [B

− R



−γt



d−1

×R)

By Lemma 9.5 here above, the commutator terms tend to zero as ε and ε



go to

zero, and the other terms tend to zero as well, as noticed at the beginning. This

shows that (u

)

ε>0

and ((u

)

ε>0

are Cauchy sequences in L

× R

× R)

and L

× R), respectively, which we already knew for the former but not for

the latter: the convergence of traces was a priori known in H

−1/2

× R); by

uniqueness of limits in the sense of distributions, this shows that (u

)

the limit of ((u

)

ε>0

also in L

× R). So by passing to the limit in the

inequality in (9.1.22) for u

, which reads

γ u



d−1

×R

×R)

+ (u

)



d−1

×R)



(∂

− P



d−1

×R

×R)

+ B

)



d−1

×R)