Analytic controllability of the wave equation over a cylinder

URL:http://www.emath.fr/cocv/

ANALYTIC CONTROLLABILITY OF THE WAVE EQUATION OVER A CYLINDER

Brice Allibert

Abstract. We analyze the controllability of the wave equation on a cylinder when the control acts on the boundary, that does not satisfy the classical geometric control condition. We obtain precise esti- mates on the analyticity of reachable functions. As the control time increases, the degree of analyticity that is required for a function to be reachable decreases as an inverse power of time. We conclude that any analytic function can be reached if that control time is large enough. In theC^∞ class, a precise description of all reachable functions is given.

R´esum´e. On donne des estimations sur l’analycit´e n´ecessaire pour qu’une fonction soit contrˆolable en temps fini sur un cylindre pour l’´equation des ondes. Cette valeur d´ecroit polynomialement avecT. On en d´eduit que toute fonction analytique peut ˆetre contrˆol´ee en un temps assez grand. On donne de plus une description pr´ecise des fonctionsC^∞ qui sont contrˆolables de cette fa¸con.

AMS Subject Classification. 93B03, 93B05, 35L05, 35L20, 35B37.

Received September 8, 1997. Revised April 30, 1998.

1. Introduction

In this paper, we consider the control problem for the wave equation over a cylindrical surface. We denote this surfaceCand we suppose, for the sake of simplicity, that its radius is 1 and its length π. We can parameterize CinR³byx²+y²= 1 andz∈(0, π). We will also denotex= cosθandy= sinθso that (z, θ)∈(0, π)×S¹is a set of coordinates overC.

The controlled part of the boundary will be Γ =∂C∩ {z= 0}. Thus the uniqueness time for this problem (i.e. the time that is needed to guarantee that a solution of the wave equation vanishing on Γ with its normal derivative, vanishes everywhere) is Tu = 2π. As usual, we shall denote E0 = H₀¹(C)⊕L²(C) and E₋1 = L²(C)⊕H⁻¹(C).

The goal of this paper is to give results about the spaceFT of controlled functions in timeT > Tu,e.g. the set of all functionsuin E₋1 for which there exists a controlg(θ, t) inL²(Γ×(0, T)) such that the solution of problem

2u= 0 overC×(0, T) (u, ∂tu)|^t=0=u u|^∂C×(0,T)=g1Γ×(0,T)

(1.1)

Keywords and phrases:Wave equation, attainable set, analyticity, boundary control.

1CMAT, ´Ecole Polytechnique, UMR 7640 du CNRS, F-91128 Palaiseau, France; e-mail:[email protected] c EDP Sciences, SMAI 1999

satisfies

(u, ∂tu)|^t=T = 0.

Let us introduce a few more notations in order to state our theorems. For any positive real numberα, we will denote

G0,α=



u(z, θ) =X

n,k

αn,ke^−α|n|sinkze^inθ|(αn,k)n,k ∈l²



 G₋1,α =



u(z, θ) =X

n,k

αn,ke^−α|n|sinkze^inθ|

αn,k

√n²+k²

n,k

∈l²



 Gα=G0,α×G₋1,α.

It is easily seen that ifα⁰ ≥αthenGα⁰ ⊂Gαand that G0=E₋1.

Let us remark that all the elements ofGα are holomorphic on the complex band|=m θ| < α. Conversely, if a functionu(z, θ) = P

n,kαn,ksinkz e^inθ ofL²((0, π)×S¹) is holomorphic on a band|=m θ| < α+for a positive, then we can prove that it belongs toG0,α andG₋1,α.

Proof. Indeed, the function v(x, θ) 7→ u(x, θ+iα) belongs to L²((0, π)×S¹). So the sequence (βn,k) of its Fourier coefficients belongs to l²(Z×N). Now by analyticity and periodicity,

βn,k = Z

v(x, θ)e^−inθsinkz dθ dz

= Z

v(x, θ−iα)e^{−in(θ−iα)}sinkz dθ dz

= e^−nα Z

u(x, θ)e^−inθsinkz dθ dz

= e^−nααn,k.

So the sequence (e^−nααn,k) belongs to l². For symmetric reasons, the sequence (e^+nααn,k) also does. So (e^|n|ααn,k) belongs tol².

Hence

u=X

n,k

γn,ke^−|n|αsinkz e^inθ,

with (γn,k)∈l²(Z×N). SoubelongsG0,α. It is easy to see that it also belongs toG₋1,α. In order to give quantitative results aboutFT, we introduce the value

αC(T) = inf{α∈R⁺ such thatGα⊂FT}·

We know (see [1] and [2]) that for any timeT > Tu,

αC(T)≤π.

This means that any initial condition that can be continued as an holomorphic function with respect toθover the band|=m θ| ≤π+ can be controlled in any timeT > Tu.

In this paper, we will improve this result by proving the following two theorems

Theorem 1.1. For any positive numberδ, there is a constantCδ such that for any timeT > Tu, αC(T)≤ Cδ

T^1−δ·

Theorem 1.2. There exists a constantc such that for any timeT > Tu, αC(T)≥ c

T²·

Notice that Theorem 1.1 implies that any analytic initial condition belongs to some FT for T large enough.

Namely, if we denoteF_∞=[

T

FT,

C^ω×C^ω⊂F_∞. (1.2)

Melrose and Sj¨ostrand have proved that the analytic wave front of a solution of the wave equation propagates at the speed of light along the geodesics of the surface. (A simple definition of the analytic wave front set is given in Sect. 3.1.) In our case, it means that the analytic wave front of a solution of problem (1.1) will propagate until it reaches the boundary (at least).

The orthogonal sections of the cylindrical surface are geodesics that never hit the boundary. So if the analytic wave front set of the initial data of problem (1.1) contains a point on one of these geodesics, it will propagate, without ever hitting the boundary. So at any timeT,u|^t=T or∂tu|^t=T will have at least one point in its analytic wave front. Therefore, it will never be 0. Hence, no such function can belong toFT, even ifT is very big. This means that these functions do not belong toF_∞.

We will prove in this article that up to a periodisation of the problem that is needed to define the wave front at the boundary, these are the only functions of C^∞ that do not belong to F_∞, which gives us a geometric description of this space.

To be more precise, for any distributionu(z, θ) in H⁻¹((0, π)×S¹), let us denote P u the distribution in D⁰(R×S¹) obtained by putting first, forz∈(0, π),u(π+z) =−u(π−z) and then by putting for any integer k, u(z+ 2kπ) =u(z). Considering P u allows us to see the singularities ofuon the boundary of the cylinder.

Then we have the following theorem:

Theorem 1.3. Ifubelongs toE₋1, ifP ubelongs to C^∞×C^∞ and has no analytic wave front in the direction of the captive geodesics, thenubelongs toF_∞

Notice that this includes (1.2) because ifuis analytic, thenP uhas no wave front at all!

In the first section, we will prove Theorem 1.1; in the second one, we will show that a little improvement of this proof lets us prove Theorem 1.3. In the last section, we will prove Theorem 1.2 and give an explicit value for the constantc.

2. Proof of Theorem 1.1

In order to prove this theorem, we will study the following observation problem, that is adjoint to (1.1)

2u= 0 overC×R^t u|^∂C×Rt= 0

(u, ∂tu)|^t=0=u∈E0.

(2.1) Let us denote

Ku= ∂u

∂n|^Γ×Rt.

We will use the HUM method to turn observation estimates for problem (2.1) into control properties for problem (1.1).

In problem (2.1), we can separate the variablesθ andz, by using Fourier series. Let us give a few definitions with this respect

Definition 2.1. Letu=



X

n∈^Z

k∈N^∗

α^j_n,ksinkz e^inθ





j=1,2

be an initial data inE0. We shall denote

u∈E₀ⁿ if m6=n⇒α^j_m,k= 0, u∈E₀⁽¹⁾ if |k|>|n| ⇒α^j_n,k = 0, u∈E₀⁽²⁾ if |k| ≤ |n| ⇒α^j_n,k = 0, u∈E₀^i,n⇔u∈E₀⁽ⁱ⁾∩E₀ⁿ.

The spaceE0 can be split into









E0= M

n

E₀^1,n

!L M

n

E₀^2,n

!

u=u¹+u²=X

n

(u^1,n+u^2,n)

We will callu¹ the “low frequency term” andu² the “high frequency term”.

These two sequences of vectors form an Hilbert basis forE0

e¹_n,k= r2

π

sinkz e^inθ

√1 +k²+n²,0

!

; e²_n,k = 0, r2

πsinkz e^inθ

! .

For anyn, we will restrict the initial data to functions ofE₀ⁿ and estimate the constants that appears in the usual observation inequalities

||u||²E0 ≤C(n, T)||Ku||²L²(Γ×(0,T)),

(see for instance [4]). As the Bardos-Lebeau-Rauch geometric control hypothesis does not hold, we do not expect these constantsC(n, T) to be bounded inn. We will see that the way they go to the infinity is closely related to the size of the space of controlled data, through the HUM method.

The estimates will have to take into account both the high frequency and the low frequency term. For the former, the eigenfrequencies are well separated, and a usual Ingham technique will provide us the required estimate. The other term is more complicated, as the gap between eigenfrequencies goes to zero whenngoes to the infinity. This part of the spectrum will require a more sophisticated proof, based upon the technique of biorthogonal sequences.

At first, we will give the estimates and show how we can prove the theorem out of them. Then, we will prove those estimates.

Here is the main proposition upon which our proof will be based.

Proposition 2.2. (low frequency estimate). For any positive and δ, there exist a timeT1(, δ) smaller than

Cδ

^1+δ and a positive constantC,δ such that for any integern and any initial conditionuin E₀ⁿ, the solution u of problem (2.1) satisfies

||u¹||²E₀ ≤C,δ e^2|n| Z

S¹

Z T₁(,δ)

−T₁(,δ)|Ku(θ, t)|² dt dθ.

We will show this proposition later on. Let us first see how we can prove Theorem 1.1 out of them. To do this we will need a few more ingredients. First, a little lemma that we can show right now

Lemma 2.3. (High frequency estimate). For any time T > 2√

2, there is a constant CT such that for any integernand any initial condition uinE₀^2,n, the solutionuof problem (2.1) satisfies

||u||²E₀ ≤CT||Ku||²L²(Γ×(0,T)).

Proof. It is a mere application of the classical Ingham technique. Asubelongs toE₀⁽²⁾,u(z, θ, t) can be written as

u(z, θ, t) = X

k>|n|

α^±_n,ksinkz e^inθe^±it

√n²+k²

. Now, asT >2π√

2> ^2π

inf_k>|n|(√

n²+(k+1)²−√ n²+k²), Z

S¹

Z T 0

∂u

∂z(z= 0, θ, t)

² dt dθ= 4π² Z T

0

X

k>|n|

kα^±_n,ke^±it

√n²+k²

2

dt, ≥C X

k>|n|

|kα^±_n,k|²,

(see [8] p. 222 for a detailed proof of the Ingham estimate for series with gaps).

So Z

S¹

Z T 0

∂u

∂z(z= 0, θ, t)

² dt dθ≥C X

k>|n|

|(1 +|n|+k)α^±_n,k|²≥C||u||²E0. Hence,

||Ku||²L²(Γ×(0,T)) ≥C||u||²E₀.

We will also need the following caracterisation ofFT by the HUM method

Lemma 2.4. An initial datav ofE₋1 belongs toFT if and only if there exists a constantCv such that for any initial datauinE0, the solutionuof problem (2.1) satisfies

| hv, uiE₋₁,E0| ≤Cv||Ku||L²(Γ×(0,T)). The proof of this lemma can be found in [6] or [4].

Now let us prove the theorem. Pick a positive value forδand, and takev inG. We can put v(z, θ) =X

n

e^−|n|Vⁿ(z)e^inθ,

with

||Vⁿ(z)e^inθ||^E−1

n ∈l²(Z).

TakeT(, δ) = sup(T1(, δ),2π√

2). AsT1(, δ)≤ ^1+δC , for small, we also haveT(, δ)≤ ^1+δC .

For anyuofE0, we have

| hv, uiE₋₁,E0|=

X

n

e^−|n|

Vⁿ(z)e^inθ, uⁿ

E₋₁,E₀

. So

| hv, uiE₋₁,E0| ≤X

n

e^−|n|||Vⁿ(z)e^inθ||^E−1||uⁿ||^E0. (2.2) Now for any integernand anyuinE₀ⁿ, we have

||u||²E₀ =||u¹||²E₀+||u²||²E₀. So, through Proposition 2.2 and Lemma 2.3,

||u||²E0 ≤C,δ e^2|n| Z

S¹

Z T₁(,δ)

−T1(.δ)

|Ku(θ, t)|² dt dθ + C Z

S¹

Z T(,δ) 0

Ku²(θ, t)² dt dθ.

So

||u||²E₀ ≤ Z

S¹C,δ⁰ e^2|n| Z T(,δ)

−T(,δ)|Ku(t)|² dt dθ+C Z

S¹

Z T(,δ)

0 |Ku(θ, t)|² dt dθ +C

Z

S¹

Z T(,δ) 0

Ku¹(θ, t)² dt dθ.

As the problem is well posed,

||u||²E₀ ≤ Z

S¹

C_,δ⁰ e^2|n| Z T(,δ)

−T(,δ)|Ku(t)|² dt dθ+C⁰ ||u¹||²E₀. So by Proposition 2.2,

||u||²E0≤Z

S¹C_,δ⁰ e^2|n| Z T(,δ)

−T(,δ)

|Ku(t)|² dt dθ + C⁰C,δ e^2|n| Z

S¹

Z T1(,δ)

−T₁(,δ)

|Ku(θ, t)|² dt dθ.

Thus

||u||²E₀ ≤Z

S¹C_,δ⁰ e^2|n| Z T(,δ)

−T(,δ)

|Ku(t)|² dt dθ.

If we put this into (2.2), we get

| hv, uⁿiE₋₁,E0| ≤ C,δ

X

n

||Vⁿ(z)e^inθ||^E−1

sZ

S¹

Z 2T(,δ) 0

|Kuⁿ(θ, t)|² dt dθ

≤ C,δ

sX

n

||Vⁿ(z)e^inθ||²E₋₁

vu utX

n

Z

S¹

Z 2T(,δ)

0 |Kuⁿ(θ, t)|² dt dθ

≤ C,δCv||Ku||^L2(Γ×(0,2T(,δ))).

So by Lemma 2.4,vbelongs toFT. This proves Theorem 1.1.

2.1. Proof of Proposition 2.2 (low frequencies)

As the gap between frequencies goes to zero in the low part of the spectrum, Ingham techniques cannot be used in the proof of this proposition. Instead, we will use a biorthogonal sequence method. The idea is to build a sequence of compactly supported functionsh^k,n(t) such that for any j 6= k, h^k,n(t) is orthogonal to the eigenfunctione^t√

n²+j²

, and whose scalar product withe^t√n2+k2, on the contrary, is not too small. The sequence of functionsh^k,n(t) is said to be biorthogonal to the sequence e^t√n2+k2. The observation constants will appear to be related to theL² norms of the functionsh^k,n(t). We will be able to decrease these norms if we allow the support of the functions to grow.

Let us state precisely the lemma we will prove about the biorthogonal sequence

Lemma 2.5. For any odd integerqand any positive real number, there is a timeT1(q, )smaller thanCq^q+11−q such that for any couple(n, k0)inN^∗2, we can find anL² functionh^k_,q^0,n for which the following properties hold

(i) h^k_,q^0,n is supported by [−T1(q, ), T1(q, )].

(ii) ||h_,q^k0,n||²L² ≤C e²ⁿ. (iii) Ifk6=k0,

Z

h^k_,q^0,n(t)e^±it√n2+k2 dt= 0.

(iv) If(n, k0)∈I={(k, n)∈N^∗2|k≤n}, Z

h^k_,q^0,n(t)e^±it√

n²+k²₀

dt ≥ _nNq^c · The constants depend only on andq.

Moreover, h^k_,q^0,n can be chosen as even or odd. They will be denotedhek₀,n

,q andhok₀,n ,q .

The reader may keep in mind thatq is a technical parameter, designed to get better polynomial estimates for the timeT. Let us first show how Proposition 2.2 can be proved out of this lemma.

Letube an element ofE₀ⁿ, and putu=X

j

α¹_n,je¹_n,j+α²_n,je²_n,j . For any (n, k0) in I, any integerM and any θin S¹,

Z hek0,n

,q (t)K



X^M

j=1

α¹_n,je¹_n,j+α²_n,je²_n,j

(t, θ)dt= Z

hek0,n ,q (t)

XM j=1

α¹_n,jKe¹_n,j(t, θ) +α²_n,jKe²_n,j(t, θ) dt

= XM j=1

α¹_n,j

Z hek₀,n

,q (t)Ke¹_n,j(t, θ)dt+α²_n,j Z

hek₀,n

,q (t)Ke²_n,j(t, θ)dt

= 1

√2πe^inθ XM j=1

"

α¹_n,j j p1 +j²+n²

Z hek0,n

,q (t) e^it√

j²+n²

+e^−it√

j²+n² dt

+α²n,j

j ip

n²+j² Z

hek0,n ,q (t)

e^it

√j²+n²

−e^−it

√j²+n² dt

# .

Now, asche k0,n

,q is even andcho k0,n ,q is odd, Z

hek0,n ,q (t)K



X^M

j=1

α¹_n,je¹_n,j+α²_n,je²_n,j

(t, θ)dt= r2

πe^inθ XM j=1

α¹_n,j j p1 +j²+n²

Z hek0,n

,q (t)e^it√

j²+n²

dt.

So by (iii), ifM≥k0, Z

hek0,n ,q (t)K



X^M

j=1

α¹_n,je¹_n,j+α²_n,je²_n,j

(t, θ)dt= r2

πe^inθα¹_n,k0 k0

1 +k₀²+n² Z

hek0,n ,q (t)e^it√

k²₀+n²

dt.

So as (n, k0)∈I, by (iv), Z

hek0,n ,q (t)K



X^M

j=1

α¹_n,je¹_n,j+α²_n,je²_n,j

(t, θ)dt

≥ |α¹_n,k0| c n^Nq· Therefore, ifM goes to the infinity,

Z hek₀,n

,q (t)Ku(t, θ)dt

≥ |α¹_n,k0| c

n^Nq · (2.3)

Similarly

Z hok₀,n

,q (t)Ku(t, θ)dt

≥ |α²_n,k0| · c n^Nq· Now

||u¹||²E₀ =X

k≤n

|α¹n,k|²+|α²n,k|². So by (2.3), for anyθ∈S¹,

||u¹||²E₀ ≤CX

k≤n

n^2Nq Z

hek0,n

,q (t)Ku(t, θ)dt

² + same withho. So by (i),

||u¹||²E0 ≤CX

k≤n

n^2Nq Z

|h^k_,q^0,n(t)|² dt

Z T1(q,)

−T1(q,)

|Ku(t, θ)|²dt.

So by (iii),

||u¹||²E0 ≤CX

k≤n

n^2Nq e²ⁿ

Z T1(q,)

−T1(q,)

|Ku(t, θ)|² dt.

Thus

||u¹||²E₀ ≤C n^2Nq0 e²ⁿ Z

S¹

Z T₁(q,)

−T₁(q,)|Ku(t, θ)|² dt dθ, with T1(q, )≤Cq^q+11−q =^C1+δ^δ andδ→0⁺ whenq→+∞.

As we can shift slightlyto eliminate the polynomial inn, we have proved Proposition 2.2.

We still have to prove Lemma 2.5. We will do this in three steps. At first, we build a sequence of functions f^k,n for which (i), (iii) and (iv) hold. This construction is explicit, and it is equivalent with the usual way of building biorthogonal sequences, by taking infinite products. The problem with these functions is that theirL² norm behaves likee^nπ, which is much too big for (ii). Though, we will see that, on the Fourier side, most of this norm is concentrated in [−n, n].

Then, by the stationary phase formula, we will build a sequence of functionsgⁿ that are exponentially small on [−n, n] (in frequency), and reasonably bounded outside.

At last, we will puth=f∗gand prove how the properties of each functions compensate in such a way that hsatisfies (i) to (iv).

2.1.1. Definition off andg

Let us define the sequence of functionsf^k,n. For any (n, k) in Z^∗×N^∗ we put

f^k,n(t) =F⁻¹

sinπ√ τ²−n²

√τ²−n²

1 τ²−(n²+k²)

·

(F⁻¹ is the reciprocal of the Fourier transform.) The following properties hold for f^k,n:

(f-i) fb^k,n ∈ O(C), fb^k,n ∈ L²(R) and ∀τ ∈ C , |fb^k,n(τ)| ≤ Cn,k e^{π|=m τ|}. So, by Paley-Wiener, f^k,n belongs toL²(−π, π).

(f-iii)∀k∈N^∗\{k0},fb^k0,n(±√

n²+k²) = 0.

(f-iv)∃N ∈N,∀(n, k)∈I , fb^k,n(±√

n²+k²)≥ n^cN·

However,||f^k,n||L² ≥C eⁿ, therefore (ii) doesn’t hold. Instead, we have the following estimate for any (n, k0) inZ^∗×N^∗ and anyτ in [−n, n]

|fb^k0,n(τ)| ≤C e^nπ√₁

−|^τn|². (2.4)

For any odd integerq, lethq(x) be the solution ofy⁰= 1 +y^q−1 such thaty(0) = 0. It is defined over (−xq, xq) for a positivexq. It is odd, strictly increasing and analytic. Moreover,hq(x) =x+αqx^q+o(x^q) whenxis close to 0, with a positiveαqand hq(x)→+∞whenx→xq.

Its reciprocal function will be denotedHq. It is defined overR, odd, bounded byxq and satisfies Hq(x) = x−αqx^q+o(x^q) whenxis close to 0.

Pickδ >1, close to 1, that will be fixed later. We define the functionsg+ as follows for any odd integerq, any integernand any real time T

g+n

T,q(t) =1(−T,T)e^{inδxqT hq(}

xq Tt)

. So

c g+n

T,q(τ) = Z T

−T

e^{inδxqT hq(}

xq T t)−iτ t

dt.

Put Ψq(s) =_x^T

qHq

_δx

q

T s , then

c g+ n T,q(τ) =

Z +∞

−∞ e^{ins−iτΨq(s)} Ψ⁰q(s)ds.

If we write θq(s) = _x¹

qHq(δxqs), then c

g+ n T,q(τ) =

Z +∞

−∞ θ⁰q

s T

e^inT

s

T−^τnθ_q(_T^s)

ds=T

Z +∞

−∞ θ⁰q(v)e^inT

v−n^τθ_q(v)

dv.

We will show later on the following lemma about the functionsg+n T,q.

Lemma 2.6. For big enoughT, there are three positive real constantsC_q¹, C_q,T² , c³_q,T,δ and an integern(q)such that

•for any integer n and any real numberτ smaller than ⁿ_δ, cg+n

T,q(τ)≤C_q,T² e^{−T n Cq1min{(1δ−nτ)}

q−1q ,1}.

•For any integerngreater thann(q)and any integerk0 smaller thann, there is a timeTn,k0 in[T, T+ 1]such

that

cg+n T_n,k0,q

q

n²+k₀²

≥c³_q,T,δ

√n ·

Let us see how this lemma allows us build a sequence of functions hthat satisfy Lemma 2.5.

2.1.2. Definition and properties of h

First, let us notice that we can get a lemma that is symmetrical to Lemma 2.6 for functions g₋ⁿ_T,q(t) = 1(−T,T)e^{−inδxqT hq(}

xq Tt)

. (tis replaced by−t).

As g₋ⁿ_T,q=g+n

T,q,Tn,k0,+=Tn,k0,−. So if we put

gpn

T,q(t) =1(−T,T)cos

n T δxq

hq

xq

T t

=<e(g+n

T,q) == 1 2(g+n

T,q(t) +g₋ⁿ_T,q(t)), we get, for|n^τ| ≤ ¹δ

bgpn

T,q(τ)≤Cq,T e^{−T n Cq(1δ−|τn|)}

q−1q

· (2.5)

Forn≥n() andk0≤n, asC_q,T² e^−nTuC1q ≤ ^c₂^{3q,T ,δ√}_n ifnis big enough, cg_±ⁿ_Tn,k

0,q(τ)≤ 1 2

cg_∓ⁿ_Tn,k

0,q(τ) forτ=∓q

n²+k²₀.

As we can increase the constant to deal with the finite number of (n, k) inIwhosenis not big enough, we get for any (n, k0) inIand τ=±p

n²+k²₀, bg_pT

n,k0,q(τ)≥cq,T,δ

√n · (2.6)

So we get forgp the following lemma, that corresponds to Lemma 2.6

Lemma 2.7. For big enoughT, there are three positive real constantsCq¹, C_q,T² , c³_q,T,δ and an integern(q)such that

•for any integer n and any real numberτ in[−ⁿδ,ⁿ_δ], bgpn

T,q(τ)≤C_q,T² e^{−T n Cq1(1δ−|τn|)}

q−1q

.

•For any integerngreater thann(q)and any integerk0 smaller thann, there is a timeTn,k0 in[T, T+ 1]such

that

bgp n T_n,k0,q

±q

n²+k²₀

≥ c³_q,T,δ

√n .

We could get similar results forgin

T,q(t) =1(−T,T)sin n_δx^T

qhq(^x_T^qt)

==m(g+n

T,q). ge is even whereas go is odd.

Now let us define the functionshby a convolution product, as was announced.

Letbe a positive real number. Chooseδ such thatπq 1−(_δ¹

)²=₂ andT such that sup

β∈[0,¹

δ]

πp

1−β²−C_q¹T 1

δ −β _q−1^q !

≤. (2.7)

As the derivative of the function to maximize is √^−πβ

1−β² +_q−^q₁TC_q¹(_δ¹

−β)^q−11 , it is enough to chooseT such that this derivative is 0 for value β such thatπp

1−β²=. We haveδ= 1 + _8π²2 +o(²),β= 1−2π²²+o(²). Thus _δ¹

−β∼cq², hence T∼cq^q+11−q.

Now, let us define a sequence of timesT_{n,k 0}.

Fork0≤n, we take the values given by Lemma 2.7 withT =T. Fork0> n, we putT_{n,k 0} =T.

Then

T_{n,k 0}∈[T, T+ 1], so

c¹_q^q+11−q ≤T_{n,k 0}≤c²_q^q+11−q. Let us define the sequence of functions has follows

che k0,n

,q (τ) =fb^k0,n· gben T_n,k

0,q(τ), cho

k₀,n

,q (τ) =fb^k0,n· gbon T_n,k

0,q(τ).

The indexheorhomeans thathis even or odd (This index will not be written for results valid in both cases).

Now let us check each of the properties of Lemma 2.5 for the functions h.

• (i) This point is easy because h^k_,q^0,n is the convolution product off^k0,n which is supported by [−π, π] and gⁿ_T

n,k0,q, which is supported by [−T_{n,k 0}, T_{n,k 0}].

So if we putT1(q, ) =π+T_{n,k 0},h^k_,q^0,n is supported by [−T1(q, ), T1(q, )] withT1(q, )≤Cq^1−qq+1.

• (ii) This is where the small values of g compensate the big size of f. As over R\[−n, n] the L² norm offbis bounded by a polynomial and|bg|^L∞ is bounded by 2T1(q, ), the problem is concentrated in [−n, n].

We must estimate Z n

−n

bh^k_,q^0,n(τ)² dτ. Now over _n^τ ∈[−1,1], by (2.4),

|fb^k0,n(τ)|²≤C e^2πn√₁

−|^τn|². So if|^τn| ≥ δ¹ then

|bh^k_,q^0,n(τ)|²≤C eⁿ.

Now by Lemma 2.7, for|^τn| ≤ δ¹,

|bgⁿ_T

n,k0,q|²≤C e^{−2Tn,k 0nCq1(δ1−|τn|)}

q−1q

. So from (2.7),

|bh^k_,q^0,n(τ)|²≤C e²ⁿ. Hence

||bh^k_,q^0,n||²L²≤Ce²ⁿ.

• (iii) This point is a direct consequence of property (f-iii). Indeed if k 6= k0, fb^k0,n(±√

n²+k²) = 0, so bh^k,q^0,n(±√

n²+k²) = 0. Which is exactly (iii) on the Fourier side.

•(iv) This is true because (f-iv) is not worsened too much by the product withg.

Indeed for (n, k0)∈I, by (f-iv) and (2.6), bh^k_,q^0,n(±q

n²+k²₀) ≥ C

n^N cq,T,δ

√n ≥Cq,

n^N0· Which is again the Fourier transcription of (iv).

In order to end our proof, we only have to prove Lemma 2.6 left.

2.1.3. Proof of Lemma 2.6 Recall that

c g+ n

T(τ) =T Z +∞

−∞ θ_q⁰(v)e^{inT(v−τnθq(v))} dv.

Put α=nT andβ =_n^τ. We will consider

φ(α, β) = Z

θ_q⁰(v)e^iα

v−βθ_q(v) dv, with αgoing to +∞.

Depending upon the value of β as compared with ¹_δ, the phase in the integral will have stationary points or not. This will allow us to use classical asymptotic estimates in both cases.

Case β≤ ¹δ

In that case, the phase is not stationary. There will be an exponential decrease with respect toα. Let us prove it by shifting slightly in the imaginary direction.

ν−βθ (ν)_δ

ν

Figure 1. Non stationnary phase.

For any real numberv and anyβ smaller than ¹_δ,

=m

v+i−βθq(v+i)

= −β =m θq(v+i)

= −β =m

θq(v+i)−θq(v)

= −β =m Z v+i

v

θ⁰q(z)dz

= −β =m Z v+i

v

δdz 1 +δ^q−1x^qq⁻¹z^q−1

= −βδ <e Z 1

0

du

1 +δ^q−1x^qq⁻¹(v+iu)^q−1· Therefore, ifβ≤0,

=m

v+i−βθq(v+i)

≥ifβ≤0.

Ifβ >0, then

=m

v+i−βθq(v+i)

≥−βδ Z 1

0

du

1 +δ^q−1x^qq⁻¹(v+iu)^q−1 · Now for any real numberv,

Z 1 0

du

1 +δ^q−1x^qq⁻¹(v+iu)^q−1

| {z }

I

≤ 1

1−cq^q−1,

because

either v, thenI≤ 1+v^cq−1 ≤1,

eitherv≤Mqand in that case|v+iu|^q−1≤Cq^q−1⇒

⇒ |1 +δ^q−1x^q_q⁻¹(v+iu)^q−1| ≥1−cq^q−1⇒I≤ 1−cq^1q−1.

Thus

=m

v+i−βθq(v+i)

≥ − βδ

1−cq^q−1

≥ (1−δβ)−c⁰_qβ^q

≥

1 δ−β

−c⁰_qβ^q. Now

min 1

δ−β

−cqβ^q=c⁰_q 1

δ−β _q−1^q

β^1−q1 ≥c⁰⁰_q 1

δ −β _q−1^q

. We can pick two very small real numbersandcq such that for any real numberv,





=m

v+i−βθq(v+i)

≥c^teq 1

δ −β_q−^q₁

ifβ ∈]0,¹_δ],

=m

v+i−βθq(v+i)

≥c^teq ifβ≤0.

So if we shift the integration contour forφcan be shifted fromRtoR+i, we get φ(α, β) =

Z

θq⁰(v+i)e^iα

v+i−βθq(v+i) dv.

Finally, asθ⁰_q(v+i) = _1+(δx ^δ

q(v+i))^q−1, we have|θ⁰_q(v+i)| ≤ 1+v^Cq−1q . Hence, for any real numberαand any β≤ ¹δ,

|φ(α, β)| ≤Z Cq

1 +v^q−1 e^{−α cqmin{(1δ−β)}

q−1q ,1}dv≤Cqe^{−α cqmin{(1δ−β)}

q−1q ,1}.

So if we go back to the original notation, for _n^τ ≤¹δ, cg+n

T,q(τ)≤Cq T e^{−nT cqmin{(1δ−nτ)}

q

q−1,1}. (2.8)

Which is the first part of Lemma 2.6.

ν−βθ (ν)_δ

ν (β,δ)₀

p0(β,δ) ν

Figure 2. Stationnary phase.

Caseβ >1.

In that case, the phase is stationary for two opposite values. We can use the stationary phase formula (see for instance [7] p. 431), and get

φ(α, β) = 1 pHβ,δ

cosαp0(β, δ)

! 

θ_q⁰(v0(β, δ))

√α +

XN j=1

aj(β, δ) α^j√

α



+rβ,δ(α),

whererβ,δ(α)≤ α^CN+1β andα≥Aβ,δ. Hβ,δ denotes the Hessian at the critical points.

Here,CandAare continuous with respect toβandδ, andaj(β, δ) depends only on the first 2j+1 derivatives ofv7→θq(v) atv=v0(β, δ) and onHβ,δ.

Let us computep0(β, δ).

∂

∂v

v−βθq(v)

= 0 ⇔ 1− βδ

1 +δ^q−1x^qq⁻¹v^q−1 = 0

⇒ 1 +δ^q−1x^q_q⁻¹v^q₀⁻¹(β, δ) =βδ

⇒ v0(β, δ) = 1 δxq

(δβ−1)^q−11 . Ifβ takes the values ^√n2_n^+k2 with (n, k)∈I, which implies that√

2≥β≥1, we get C≥v0(β, δ),|p0(β, δ)|,|Hβ,δ| ≥cδ,

so

1≥θ_q⁰(v0(β, δ))≥cq. Moreoveraj(β, δ)≤Cj,δ.

LetT be a positive time. As|p0(β, δ)| ≥cδ, for any (n, k0) in I, we can pick a time Tn,k₀ in [T, T + 1] such that cos

nTn,k₀p0(

√n²+k²₀ n , δ)

≥c⁰_δ.

So ifT is greater thanTu, for any integern≥n(q) ifα=T nandβ =

√n²+k₀²

n ,

θ⁰_q(v0(β, δ))

√α +

XN j=1

aj(β, δ) α^j√

α

≥ |θ⁰_q(v0(β, δ))| 2√

α ,

|rβ,δ(α)| ≤c⁰_δ|Hβ,δ|θ⁰_q(v0(β, δ)) 4√

α ·

So for any integer n≥n(q) any timeT > Tu, and any integerk0≤n, we have found a timeTn,k₀ in [T, T+ 1]

such that

φ nTn,k₀,

pn²+k²₀ n

!≥

c⁰_δ|H|θ⁰_q

v0

√

n²+k²₀ n , δ

4√

np Tn,k0

≥ c

√n·

Which means that

cg+ n T_n,k0,q

q

n²+k₀²

≥CT,q,δ

√n ·

This is the second part of Lemma 2.6.

This completes the proof of Theorem 1.1.