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Smoothing effect for the regularized Schrödinger

equation with non controlled orbits

Lassaad Aloui, Moez Khenissi, Georgi Vodev

To cite this version:

Lassaad Aloui, Moez Khenissi, Georgi Vodev. Smoothing effect for the regularized Schrödinger equa-tion with non controlled orbits. Communicaequa-tions in Partial Differential Equaequa-tions, Taylor & Francis, 2013, 38 (2), pp.265–275. �hal-00587257v2�

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Smoothing effect for the regularized

Schrdinger equation with non controlled

orbits

Lassaad Aloui

Dpartement de Mathmatiques, Facult des Sciences de Bizerte, Tunisie Email: lassaad.aloui@fsg.rnu.tn

Moez Khenissi

Dpartement de Mathmatiques, cole Suprieure des Sciences et de Technologie de Hammam Sousse, Rue Lamine El Abbessi, 4011 Hammam Sousse, Tunisie

Email: moez.khenissi@fsg.rnu.tn

Georgi Vodev

Universit de Nantes, Dpartement de Mathmatiques, UMR 6629 du CNRS, 2, rue de la Houssinire, BP 92208, 44332 Nantes Cedex 03, France

e-mail: georgi.vodev@math.univ-nantes.fr

Abstract

We prove that the geometric control condition is not necessary to obtain the smooth-ing effect and the uniform stabilization for the strongly dissipative Schr¨odinger equation.

Key words: Smoothing effect, Resolvent estimates, Stabilization and Geometric Control.

1 Introduction and statement of results

It is well known that the Schr¨odinger equation enjoys some smoothing prop-erties. One of them says that if u0 ∈ L2(Rd) with compact support, then the

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solution of the Schr¨odinger equation      i∂tu− ∆u = 0 in R × Rd u(0, .) = u0 in Rd, (1) satisfies u∈ C∞(R\ {0} × Rd).

We say that the Schr¨odinger propagator has an infinite speed. Another type of gain of regularity for system (1) is the Kato-1/2 smoothing effect (see [8], [15], [16]), namely any solution of (1) satisfies

Z R Z |x|<R|(1 − ∆) 1 4u(t, x)|2dxdt≤ C Rku0k2L2(Rd). (2)

In particular, this result implies that for a.e. t∈ R, u(t, .) is locally smoother than u0 and this happens despite the fact that (1) conserves the global L2 norm. The Kato-effect has been extended to variable coefficients operators with non trapping metric by Doi ([9], [10])) and to non trapping exterior domains by Burq, Gerard and Tzvetkov [4]. On the other hand, Burq [3] proved that the nontrapping assumption is necessary for the H1/2 smoothing effect. Moreover, using Ikawa’s result [11], he showed, in the case of several convex bodies satisfying certain assumptions, that the smoothing effect with an ε > 0 loss still holds.

Recently, the first author [1,2] has introduced the forced smoothing effect for Schr¨odinger equation. The idea is inspired from the stabilization problem and it consists of acting on the equation in order to produce some smoothing effects. More precisely, in [2] the following regularized Schr¨odinger equation on a bounded domain Ω⊂ Rd is considered:

            

i∂tu− ∆Du + ia(x)(−∆D)12a(x)u = 0 in R× Ω,

u(0, .) = f in Ω, u|R×∂Ω = 0,

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where ∆D denotes the Dirichlet realization of the Laplace operator on Ω and a(x) is a smooth real-valued function. Under the geometric control condition (GCC) on the set w = {a 6= 0}, it is proved in [2] that any solution with initial data in Hs

D(Ω) belongs to L2loc((0,∞), HDs+1(Ω)), where s ∈ [−s0, s0] and s0 ≥ 1 depends on the behavior of a(x) near the boundary. When the function a is constant near each component of the boundary, we have s0 =∞. Then by iteration of the last result, a C∞-smoothing effect is proved in [2]. Note that these smoothing effects hold away from t = 0 and they seem strong compared with the Kato effect for which the GCC is necessary. Therefore the case when w ={a 6= 0} does not control geometrically Ω is very interesting.

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In this work we give an example of geometry where the geometric control condition is not satisfied but the C∞ smoothing effect holds. More precisely, let O = N

i=1Oi ⊂ Rd be the union of a finite number of bounded strictly convex bodies, Oi, satisfying the conditions of [11], namely:

• For any 1 ≤ i, j, k ≤ N, i 6= j, j 6= k, k 6= i, one has

Convex Hull(Oi∪ Oj)∩ Ok =∅. (4) • Denote by κ the infimum of the principal curvatures of the boundaries of the bodies Oi, and L the infimum of the distances between two bodies. Then if N > 2 we assume that κL > N (no assumption if N = 2).

Let B be a bounded domain containing O with smooth boundary and such that Ω0 = Oc∩ B is connected, where Oc = Rd\ O. In the present paper we will consider the regularized Schr¨odinger equation (3) in Ω0. For a bounded domain Ω of Rd and any s∈ R, we denote by Hs

D(Ω) the Hilbert space HDs(Ω) ={u = X j ajej, X j γ2sj |aj|2 <∞}, where {γ2

j} are the eigenvalues of −∆D and {ej} is the corresponding or-thonormal basis of L2(Ω). We have the following interpolation inequalities:

kgkHs D(Ω0)≤ kgk s t Ht D(Ω0)kgk 1−s t L2(Ω 0) for all g ∈ H t D(Ω0), 0≤ s ≤ t. (5) Clearly, Hs

D(Ω) and HD−s(Ω) are in duality and HDs(Ω) is the domain of (−∆D)s2. Remark also that Hs

D(Ω) = Hs(Ω) is the usual Sobolev space for 0 ≤ s < 1

2 and H s

D(Ω) = {u ∈ Hs(Ω), ∆ju|∂Ω = 0, 2j ≤ s − 12} for s 1

2. Throughout this paper a ∈ C

(Ω0) will be a real-valued function such that supp a ⊂ {x ∈ Ω0 : dist(x, ∂B)≤ 2ε0} and a = Const 6= 0 on {x ∈ Ω0 : dist(x, ∂B)≤ ε0}, where 0 < ε0 ≪ 1 is a constant. Under this assumption the following properties hold for all s∈ R and n ∈ N:

(Ps) the multiplication by a maps Hs

D(Ω0) into itself, (Qs,n) the commutator [a, (−∆D)

n] maps Hs

D(Ω0) into HDs−2n+1(Ω0). Set Ba= a(x)(−∆D)

1

2a(x) and define the operator Aa =−∆D+iBaon L2(Ω0)

with domain

D(Aa) ={f ∈ L2(Ω0); Aaf ∈ L2(Ω0), f = 0 on ∂Ω0}.

Since the properties (Ps) and (Qs,n) hold for all s∈ R and n ∈ N, the problem ( 3) is well posed in Hs

D(Ω0) for all s∈ R. Moreover the operator Aagenerates a semi-group, U(t), such that for f ∈ Hs

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is the unique solution of (3). It is easy to see that the spectrum, sp(Aa), of Aa consists of complex numbers, τj, satisfying j| → ∞. Furthermore, since a(x) is not identically zero, we have

sp(Aa)⊂ {τ ∈ C, Im τ > 0}. The resolvent

(Aa− τ)−1 : L2(Ω0)→ L2(Ω0)

is holomorphic on {Im τ < 0} and can be extended to a meromorphic operator on C. Our main result is the following

Theorem 1 If the function a is as above, there exist positive constants σ0 and C such that for |Im τ| < σ0 we have

(Aa− τ)−1 L2(Ω 0)→L2(Ω0)≤ Chτi −1 2 log2hτi, (6) where hτi = q1 +|τ|2.

A similar bound has been recently proved in [5] for the Laplace operator in Ω0 with strong dissipative boundary conditions on ∂B, provided B is strictly convex (viewed from the exterior). Note also that a better bound (with log instead of log2) was obtained in [6], [7] in the case of the damped wave equation on compact manifolds without boundary under the assumption that there is only one closed hyperbolic orbit which does not pass through the support of the dissipative term. This has been recently improved in [14] for a class of compact manifolds with negative curvature, where a strip free of eigenvalues has been obtained under a pressure condition.

As an application of this resolvent estimate we obtain the following smoothing result for the associated Schr¨odinger propagator.

Theorem 2 Let s∈ R. Under the assumptions of Theorem 1, we have (i) For each ε > 0 there is a constant C > 0 such that the function

u(t) = Z t 0 e i(t−τ )Aaf (τ )dτ satisfies kukL2 THDs+1−ε(Ω0)≤ C kfkL2THsD(Ω0) (7)

for all T > 0 and f ∈ L2

THDs(Ω0). (ii)If v0 ∈ HDs(Ω0), then

v ∈ C∞((0, +∞) × Ω0) (8) where v is the solution of (3) with initial data v0.

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Theorem 1 also implies the following stabilization result.

Theorem 3 Under the assumptions of Theorem 1, there exist α, c > 0 such that for the solution u of (3) with initial data u0 in L2(Ω0), we have

kukL2(Ω

0) ≤ ce

−αtku0kL

2(Ω

0), ∀ t > 1.

This result shows that we can stabilize the Schr¨odinger equation by a (strongly) dissipative term that does not satisfy the geometric control condition of [12]. In fact, to have the exponential decay above it suffices to have the estimate (6) with a constant in the right-hand side.

2 Resolvent estimates

This section is devoted to the proof of Theorem 1. Since the resolvent (Aa− τ )−1 is meromorphic on C and has no poles on the real axes, it suffices to prove (6) for|τ| ≫ 1. Let u be a solution of the following equation

     (∆D + λ2− ia(x)(−∆D)12a(x))u = v in Ω0, u|∂Ω0 = 0, (9)

with v ∈ L2(Ω0). Clearly, it suffices to prove the following

Proposition 4 Under the assumptions of Theorem 1, there exists λ0 > 0 such that for every λ > λ0 and every solution u of (9) we have

kukL2(Ω 0) .

log2λ

λ kvkL2(Ω0). (10)

Proof. Let χ∈ C∞

0 (B) be such that χ = 0 on {x ∈ B : dist(x, ∂B) ≤ ε0/3}, χ = 1 on {x ∈ B : dist(x, ∂B) ≥ ε0/2}. Clearly we have

k(1 − χ)ukL2(Ω

0).kaukL2(Ω0). (11)

On the other hand, the function χu satisfies the equation

    

(∆D + λ2)χu = χv + iχa(x)(−∆D)12a(x)u + [∆D, χ]u in Rd\ O,

u|∂O = 0. (12)

Hence, according to Proposition 4.8 of [3], it follows that

kχukL2(Ω 0).

log λ

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By (11) and (13), kukL2(Ω 0). log λ λ kvkL2(Ω0)+ log λ λ kaukH1(Ω0)+kaukL2(Ω0). (14)

To estimate the second and the third terms in the right hand side of (14), we need the following

Lemma 5 Let s∈ [0, 1] and ψ ∈ C(Ω

0). Then we have, for λ≫ 1,

kψukHs+1 .λkψukHs +kvkL2 + λ1/2kukL2, (15)

kψukHs .λ−1kψukHs+1 + λ−1kvkL2 + λ−1/2kukL2. (16)

Proof. The function w = (−∆D)s/2ψu := Psu satisfies the equation

    

(−∆D − λ2+ iBa)w = Psv + [∆D, Ps]u + i[Ba, Ps]u in Ω0, w|∂O = 0.

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Multiplying equation (17) by w, integrating by parts and taking the real part, we obtain k(−∆D)1/2wk2L2(Ω 0)−λ 2 kwk2L2(Ω 0)= RehPsv +[∆D, Ps]u+i[Ba, Ps]u, wi. (18)

Using that [∆D, Ps] = (−∆D)s/2[∆D, ψ], we deduce from (18) k(−∆D)1/2wk2 L2 .λ2kwk2L2 +kψvkL2kP2sukL2 +kukHs+1kwkL2. (19) This implies kψuk2 Hs+1 .λ2kψuk2

Hs +kvkL2kψukH2s +kukHs+1kψukHs

.λ2kψuk2

Hs +kvk2L2 + εkψuk2H2s+kukHs+1kψukHs

.λ2kψuk2

Hs +kvk2L2 + εkψuk2Hs+1 + εkψukH2s + λ−2kuk2Hs+1+ λ2kψuk2Hs.

(20) Choosing ε small enough, we obtain

kψuk2Hs+1 .λ2kψuk2Hs +kvk2L2 + λ−2kuk2Hs+1. (21)

Taking ψ = 1 in the previous estimate, we get for λ≫ 1

kuk2Hs+1 .λ2kukH2s +kvk2L2. (22)

Inserting (22) in (21), we obtain

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On the other hand, since s∈ [0, 1], by interpolation we have kuk2

Hs ≤ kukH1kukL2 ≤ λ−1kuk2

H1 + λkuk2L2. (24)

Choosing s = 0 in (22), we get

kuk2

H1 .λ2kuk2L2 +kvk2L2. (25)

Combining (24) and (25), we obtain

kuk2Hs .λkuk2L2+ λ−1kvk2L2. (26)

Inserting this estimate in (23) we get

kψuk2Hs+1 .λ2kψuk2Hs +kvk2L2+ λkuk2L2. (27)

This completes the proof of (15). Clearly, (16) can be proved in the same way.

We deduce from Lemma 5 the following

Lemma 6 Let u be a solution of (9) and ψ ∈ C(Ω0). Then we have, for λ≫ 1,

kψukL2 .λ−1/2kψukH1/2+ λ−1kvkL2 + λ−1/2kukL2, (28)

kψukH1 .λ1/2kψukH1/2 + λ−1/2kvkL2 +kukL2. (29)

Proof. Using Lemma 5 together with an interpolation argument, we get

kψuk2 H1 .λ−1kψuk2 H3/2 + λkψuk2H1/2 .λ−1(λ2kψuk2 H1/2+kvk2L2 + λkuk2L2) + λkψuk2 H1/2 .λkψuk2 H1/2+ λ −1 kvk2 L2 +kuk2L2, (30)

which proves (29). Using (18) and (29) we obtain

kψuk2 L2 . λ −2 kψuk2 H1 + λ −2 kvk2 L2 + λ −1 kuk2 L2 . λ−1kψuk2 H1/2 + λ −2 kvk2 L2 + λ−1kuk2L2, (31) which proves (28).

We now return to the proof of Proposition 4. Using Lemma 6 and the estimate (14), we get kukL2 . log λ λ kvkL2+ log λ √ λ kaukH1/2 for λ≫ 1. (32)

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Let’s now estimate the H1/2 term. Multiplying equation (9) by u, integrating by parts and taking the imaginary part, we obtain

kauk2H1/2 =h(−∆D) 1 2au, aui = Im hv, ui ≤ kvkL2kukL2. (33) By (32) and (33), we get kukL2 . log λ λ kvkL2 + log λ √ λ kvk 1/2 L2 kuk 1/2 L2 . (34) This implies kukL2 . log λ λ kvkL2 + log λ √ λ ( log λ √ λ kvkL2 + ε √ λ log λkukL2), (35) for any ε > 0. Choosing ε small enough, we get for large λ

kukL2 .

log2λ

λ kvkL2, (36) which is the desired result.

3 Smoothing effect

We will first prove the following

Proposition 7 If a(x) is as in the introduction, then for every s∈ R, ε > 0 there exist positive constants C and σ0 such that

k(Aa− τ)−1kH

s

D(Ω0)→HDs+1−ε(Ω0) ≤ C (37)

holds for |Im τ| < σ0.

Proof. Let u and f satisfy the equation

     (−∆D − τ + iBa)u = f in Ω0, u = 0 on ∂Ω0. (38)

Let’s see that the following estimate holds

kukH2 D(Ω0) ≤ Chτi 1 2 log2hτikfk L2(Ω 0), (39)

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for |Im τ| < σ0. Using Proposition 1 we get kukH2

D(Ω0) = k∆DukL2(Ω0)

= kτu − iBau + fkL2(Ω 0)

≤ Chτi12 log2hτikfkL2(Ω

0)+ CkukHD1(Ω0)

≤ Chτi12 log2hτikfkL2(Ω

0)+ εkukH2D(Ω0)+ CεkfkL2(Ω0).

(40)

Choosing ε small enough, we obtain (39). Using (39) and (5) with s = 1− ε, t = 2, we obtain kuk2 HD1−ε(Ω0) ≤ kuk 1−ε H2 D(Ω0)kuk 1+ε L2(Ω 0)

≤ C(hτi12 log2hτi)1−ε(log

2hτi hτi12 )1+εkfk2L2(Ω 0) ≤ Ckfk2 L2(Ω 0). (41)

Then we get (37) for s = 0, i.e.

k(−∆D− τ + iBa)−1kL

2(Ω

0)→HD1−ε(Ω0)≤ C. (42)

We will prove (37) for s = 2N with N ∈ N, namely kukH2N+1−ε

D (Ω0) .kfkHD2N(Ω0). (43)

Let f ∈ H2N

D (Ω0) and let u be the corresponding solution of (38). The function (−∆D)Nu satisfies

(∆D+ τ − iBa)((−∆D)Nu) = (−∆D)Nf − i[Ba, (−∆D)N]u. (44) Using (42), we obtain k(−∆D)NukHγ(Ω0).k(−∆D)NfkL2(Ω 0)+k[Ba, (−∆D) N]ukL2 (Ω0) (45) where γ = 1− ε. Since kukH2N+γ D (Ω0) =k(−∆) Nu kHγ(Ω0), (46) we obtain kukH2N+γ(Ω0).k(−∆D)NfkL2(Ω 0)+k[Ba, (−∆D) N]u kL2(Ω 0). (47)

On the other hand, we have

[Ba, (−∆D)N] = a(−∆D) 1 2[a, (−∆ D)N] + [a, (−∆D)N](−∆D) 1 2a. (48)

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Using the properties (Ps) and (Qs,n) we get k[Ba, (−∆D)N]ukL2(Ω 0).kukHD2N(Ω0). (49) Consequently kukHD2N+γ(Ω0) . kfkHD2N(Ω0)+kukHD2N(Ω0) . kfkH2N D (Ω0)+ εkukHD2N+γ(Ω0)+ CεkukH γ D(Ω0). (50)

Choosing ε small enough and using (42) we obtain

kukH2N+γ D (Ω0) . kfkH 2N D (Ω0)+kukHDγ(Ω0) . kfkH2N D (Ω0)+kfkL2(Ω0) . kfkH2N D (Ω0), (51)

which proves (37) for s = 2N. Using the identity

[a, (−∆D)−N] = (−∆D)−N[(−∆D)N, a](−∆D)−N, (52) we can prove (37) for s =−2N in the same way as in the case s = 2N. Finally, by an interpolation argument we obtain the result for s ∈ R. This completes the proof of Proposition 7.

Proof of Theorem 2. We will first prove (7). Extend f by 0 for t∈ R\[0, T ]. The Fourier transforms (in t) of u and f are holomorphic in the domain Im z < 0 and satisfy the equation

(−z − ∆D+ iBa)u(z,b ·) =f (z,b ·). (53) We take z = λ−iσ, λ ∈ R , σ > 0, and we let σ tend to zero. Using Proposition 7, we get kubkL2(R;Hs+1−ε D (Ω0)).k b fkL2(R;Hs D(Ω0)), s∈ R. (54)

The fact that the Fourier transform of any function from R to a Hilbert space H defines an isometry on L2(R; H) completes the proof of (7).

Now we turn to the proof of (8). Let ϕ ∈ C∞

0 ]0, +∞[, then the function w(t,·) = ϕ(t)v(t, ·) satisfies the equation

            

i∂tw− ∆Dw + iBaw = iϕ′(t)v in R+× Ω0, w(0,·) = 0 in Ω0, w|R×∂Ω0 = 0.

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Using (7) with ε = 1/2 we obtain kwkL2(R +,HDs+1/2(Ω0)) . kϕ ′vkL 2(R +,HDs(Ω0)) . kv0kHs D(Ω0), (56) which implies v ∈ L2loc((0,∞), HDs+1/2(Ω0)). (57) By iteration we obtain v ∈ L2loc((0,∞), HDs+k(Ω0)), ∀k ∈ N. (58) Using the equation satisfied by v, we deduce that

v ∈ Hlock ((0,∞), HDs+k(Ω0)), ∀k ∈ N. (59) This implies that v ∈ C∞((0,∞) × Ω0) and the proof of Theorem 2 is com-pleted.

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