Smoothing effect for the regularized Schrödinger equation with non controlled orbits

0.15 MB·English

by  Lassaad Aloui

Smoothing effect for the regularized
Schrdinger equation with non controlled
orbits
2
1
Lassaad Aloui
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2

n Dpartement de Mathmatiques, Facult des Sciences de Bizerte, Tunisie
a
Email: [email protected]
J

8
1
Moez Khenissi

]
P
A Dpartement de Mathmatiques, cole Suprieure des Sciences et de Technologie de
. Hammam Sousse, Rue Lamine El Abbessi, 4011 Hammam Sousse, Tunisie
h
Email: [email protected]
t
a
m
[ Georgi Vodev

1
v Universit de Nantes, Dpartement de Mathmatiques, UMR 6629 du CNRS, 2, rue
1
de la Houssinire, BP 92208, 44332 Nantes Cedex 03, France
1
7 e-mail: [email protected]
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.
1
0
2
1
Abstract
:
v
i
X Weprovethatthegeometriccontrolconditionisnotnecessarytoobtainthesmooth-
r ing effect and the uniform stabilization for the strongly dissipative Schro¨dinger
a
equation.
Key words: Smoothing effect, Resolvent estimates, Stabilization and Geometric
Control.
1 Introduction and statement of results
It is well known that the Schro¨dinger equation enjoys some smoothing prop-
erties. One of them says that if u L2(Rd) with compact support, then the
0
∈
Preprint submitted to Elsevier 19 January 2012
solution of the Schro¨dinger equation
i∂ u ∆u = 0 in R Rd
 t − × (1)
u(0,.) = u in Rd,
0

satisfies
u C∞(R 0 Rd).
∈ \{ }×
We say that the Schro¨dinger 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
ZRZ|x|<R|(1−∆)41u(t,x)|2dxdt ≤ CRku0k2L2(Rd). (2)
In particular, this result implies that for a.e. t R, u(t,.) is locally smoother
∈
than u and this happens despite the fact that (1) conserves the global L2
0
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
Schro¨dinger 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 Schro¨dinger equation
on a bounded domain Ω Rd is considered:
⊂
i∂tu−∆Du+ia(x)(−∆D)21a(x)u = 0 in R×Ω,
u(0,.) = f in Ω, (3)
u|R×∂Ω = 0,
where ∆ denotes the Dirichlet realization of the Laplace operator on Ω and
D
a(x) is a smooth real-valued function. Under the geometric control condition
(GCC) on the set w = a = 0 , it is proved in [2] that any solution with
{ 6 }
initial data in Hs(Ω) belongs to L2 ((0, ),Hs+1(Ω)), where s [ s ,s ]
D loc ∞ D ∈ − 0 0
and s 1 depends on the behavior of a(x) near the boundary. When the
0
≥
function a is constant near each component of the boundary, we have s = .
0
∞
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 = 0 does not control geometrically Ω is very interesting.
{ 6 }
2
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 O Rd be the union of a finite number of bounded strictly
∪i=1 i ⊂
convex bodies, O , satisfying the conditions of [11], namely:
i
For any 1 i,j,k N, i = j, j = k, k = i, one has
• ≤ ≤ 6 6 6
Convex Hull(O O ) O = . (4)
i j k
∪ ∩ ∅
Denote by κ the infimum of the principal curvatures of the boundaries of
•
thebodies O , and Lthe infimum ofthe distances between two bodies. Then
i
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 Ω = Oc B is connected, where Oc = Rd O. In the present paper we
0
∩ \
will consider the regularized Schro¨dinger equation (3) in Ω . For a bounded
0
domain Ω of Rd and any s R, we denote by Hs(Ω) the Hilbert space
∈ D
Hs(Ω) = u = a e , γ2s a 2 < ,
D { X j j X j | j| ∞}
j j
where γ2 are the eigenvalues of ∆ and e is the corresponding or-
{ j} − D { j}
thonormal basis of L2(Ω). We have the following interpolation inequalities:
g g st g 1−st for all g Ht (Ω ), 0 s t. (5)
k kHDs(Ω0) ≤ k kHDt (Ω0)k kL2(Ω0) ∈ D 0 ≤ ≤
Clearly, Hs(Ω) and H−s(Ω) are in duality and Hs(Ω) is the domain of
D D D
(−∆D)2s. Remark also that HDs(Ω) = Hs(Ω) is the usual Sobolev space for
1
0 s < and Hs(Ω) = u Hs(Ω), ∆ju = 0,2j s 1 for
≤ 2 D { ∈ |∂Ω ≤ − 2}
1
s . Throughout this paper a C∞(Ω ) will be a real-valued function
0
≥ 2 ∈
such that suppa x Ω : dist(x,∂B) 2ε and a = Const = 0 on
0 0
⊂ { ∈ ≤ } 6
x Ω : dist(x,∂B) ε , where 0 < ε 1 is a constant. Under this
0 0 0
{ ∈ ≤ } ≪
assumption the following properties hold for all s R and n N:
∈ ∈
(Ps) the multiplication by a maps HDs(Ω0) into itself,
(Qs,n) the commutator [a,(−∆D)n] maps HDs(Ω0) into HDs−2n+1(Ω0).
SetBa = a(x)( ∆D)12a(x)anddefinetheoperatorAa = ∆D+iBa onL2(Ω0)
− −
with domain
D(A ) = f L2(Ω );A f L2(Ω ), f = 0 on ∂Ω .
a 0 a 0 0
{ ∈ ∈ }
Since the properties ( s) and ( s,n) hold for all s R and n N, the problem
P Q ∈ ∈
( 3) is well posed in Hs(Ω ) for all s R. Moreover the operator A generates
D 0 ∈ a
a semi-group, U(t), such that for f Hs(Ω ), U(t)f C([0,+ [,Hs(Ω ))
∈ D 0 ∈ ∞ D 0
3
is the unique solution of (3). It is easy to see that the spectrum, sp(A ), of
a
A consists of complex numbers, τ , satisfying τ . Furthermore, since
a j j
| | → ∞
a(x) is not identically zero, we have
sp(A ) τ C,Imτ > 0 .
a
⊂ { ∈ }
The resolvent
(A τ)−1 : L2(Ω ) L2(Ω )
a 0 0
− →
isholomorphicon Imτ < 0 andcanbeextended toameromorphicoperator
{ }
on C. Our main result is the following
Theorem 1 If the function a is as above, there exist positive constants σ
0
and C such that for Imτ < σ we have
0
| |
(Aa τ)−1 C τ −21 log2 τ , (6)
(cid:13) − (cid:13)L2(Ω0)→L2(Ω0) ≤ h i h i
(cid:13) (cid:13)
(cid:13) (cid:13)
where τ = 1+ τ 2.
h i q | |
A similar bound has been recently proved in [5] for the Laplace operator in
Ω with strong dissipative boundary conditions on ∂B, provided B is strictly
0
convex (viewed from the exterior). Note also that a better bound (with log
insteadoflog2)wasobtainedin[6],[7]inthecaseofthedampedwaveequation
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 Schro¨dinger 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
t
u(t) = ei(t−τ)Aaf(τ)dτ
Z
0
satisfies
kukL2THDs+1−ε(Ω0) ≤ CkfkL2THDs(Ω0) (7)
for all T > 0 and f L2Hs(Ω ).
∈ T D 0
(ii)If v Hs(Ω ), then
0 ∈ D 0
v C∞((0,+ ) Ω ) (8)
0
∈ ∞ ×
where v is the solution of (3) with initial data v .
0
4
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 u in L2(Ω ), we have
0 0
u ce−αt u , t > 1.
k kL2(Ω0) ≤ k 0kL2(Ω0) ∀
ThisresultshowsthatwecanstabilizetheSchro¨dingerequationbya(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 (A
a
−
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)21a(x))u = v in Ω0, (9)
u = 0,
|∂Ω0

with v L2(Ω ). Clearly, it suffices to prove the following
0
∈
Proposition 4 Under the assumptions of Theorem 1, there exists λ > 0
0
such that for every λ > λ and every solution u of (9) we have
0
log2λ
u . v . (10)
k kL2(Ω0) λ k kL2(Ω0)
Proof. Let χ C∞(B) be such that χ = 0 on x B : dist(x,∂B) ε /3 ,
∈ 0 { ∈ ≤ 0 }
χ = 1 on x B : dist(x,∂B) ε /2 . Clearly we have
0
{ ∈ ≥ }
(1 χ)u . au . (11)
k − kL2(Ω0) k kL2(Ω0)
On the other hand, the function χu satisfies the equation
(∆D +λ2)χu = χv +iχa(x)(−∆D)21a(x)u+[∆D,χ]u in Rd \O, (12)
u = 0.
∂O
|

Hence, according to Proposition 4.8 of [3], it follows that
logλ
χu . ( v + au ). (13)
k kL2(Ω0) λ k kL2(Ω0) k kH1(Ω0)
5
By (11) and (13),
logλ logλ
u . v + au + au . (14)
k kL2(Ω0) λ k kL2(Ω0) λ k kH1(Ω0) k kL2(Ω0)
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 ). Then we have, for λ 1,
0
∈ ∈ ≫
ψu . λ ψu + v +λ1/2 u , (15)
Hs+1 Hs L2 L2
k k k k k k k k
ψu . λ−1 ψu +λ−1 v +λ−1/2 u . (16)
Hs Hs+1 L2 L2
k k k k k k k k
Proof. The function w = ( ∆ )s/2ψu := P u satisfies the equation
D s
−
( ∆ λ2 +iB )w = P v +[∆ ,P ]u+i[B ,P ]u in Ω ,
 − D − a s D s a s 0 (17)
w = 0.
∂O
|

Multiplying equation (17) by w, integrating by parts and taking the real part,
we obtain
( ∆ )1/2w 2 λ2 w 2 = Re P v+[∆ ,P ]u+i[B ,P ]u,w . (18)
k − D kL2(Ω0)− k kL2(Ω0) h s D s a s i
Using that [∆ ,P ] = ( ∆ )s/2[∆ ,ψ], we deduce from (18)
D s D D
−
( ∆ )1/2w 2 . λ2 w 2 + ψv P u + u w . (19)
k − D kL2 k kL2 k kL2k 2s kL2 k kHs+1k kL2
This implies
ψu 2
k kHs+1
. λ2 ψu 2 + v ψu + u ψu
k kHs k kL2k kH2s k kHs+1k kHs
. λ2 ψu 2 + v 2 +ε ψu 2 + u ψu
k kHs k kL2 k kH2s k kHs+1k kHs
. λ2 ψu 2 + v 2 +ε ψu 2 +ε ψu 2 +λ−2 u 2 +λ2 ψu 2 .
k kHs k kL2 k kHs+1 k kHs k kHs+1 k kHs
(20)
Choosing ε small enough, we obtain
ψu 2 . λ2 ψu 2 + v 2 +λ−2 u 2 . (21)
k kHs+1 k kHs k kL2 k kHs+1
Taking ψ = 1 in the previous estimate, we get for λ 1
≫
u 2 . λ2 u 2 + v 2 . (22)
k kHs+1 k kHs k kL2
Inserting (22) in (21), we obtain
ψu 2 . λ2 ψu 2 + v 2 + u 2 . (23)
k kHs+1 k kHs k kL2 k kHs
6
On the other hand, since s [0,1], by interpolation we have
∈
u 2 u u λ−1 u 2 +λ u 2 . (24)
k kHs ≤ k kH1k kL2 ≤ k kH1 k kL2
Choosing s = 0 in (22), we get
u 2 . λ2 u 2 + v 2 . (25)
k kH1 k kL2 k kL2
Combining (24) and (25), we obtain
u 2 . λ u 2 +λ−1 v 2 . (26)
k kHs k kL2 k kL2
Inserting this estimate in (23) we get
ψu 2 . λ2 ψu 2 + v 2 +λ u 2 . (27)
k kHs+1 k kHs k kL2 k kL2
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∞(Ω ). Then we have, for
0
∈
λ 1,
≫
ψu . λ−1/2 ψu +λ−1 v +λ−1/2 u , (28)
k kL2 k kH1/2 k kL2 k kL2
ψu . λ1/2 ψu +λ−1/2 v + u . (29)
k kH1 k kH1/2 k kL2 k kL2
Proof. Using Lemma 5 together with an interpolation argument, we get
ψu 2 . λ−1 ψu 2 +λ ψu 2
k kH1 k kH3/2 k kH1/2
. λ−1(λ2 ψu 2 + v 2 +λ u 2 )+λ ψu 2 (30)
k kH1/2 k kL2 k kL2 k kH1/2
. λ ψu 2 +λ−1 v 2 + u 2 ,
k kH1/2 k kL2 k kL2
which proves (29). Using (18) and (29) we obtain
ψu 2 . λ−2 ψu 2 +λ−2 v 2 +λ−1 u 2
k kL2 k kH1 k kL2 k kL2 (31)
. λ−1 ψu 2 +λ−2 v 2 +λ−1 u 2 ,
k kH1/2 k kL2 k kL2
which proves (28).
We now return to the proof of Proposition 4. Using Lemma 6 and the estimate
(14), we get
logλ logλ
u . v + au for λ 1. (32)
k kL2 λ k kL2 √λ k kH1/2 ≫
7
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)21au,aui = Imhv,ui ≤ kvkL2kukL2. (33)
By (32) and (33), we get
logλ logλ
u . v + v 1/2 u 1/2. (34)
k kL2 λ k kL2 √λ k kL2 k kL2
This implies
logλ logλ logλ √λ
u . v + ( v +ε u ), (35)
k kL2 λ k kL2 √λ √λ k kL2 logλk kL2
for any ε > 0. Choosing ε small enough, we get for large λ
log2λ
u . v , (36)
L2 L2
k k λ k k
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 σ such that
0
k(Aa −τ)−1kHDs(Ω0)→HDs+1−ε(Ω0) ≤ C (37)
holds for Imτ < σ .
0
| |
Proof. Let u and f satisfy the equation
( ∆ τ +iB )u = f in Ω ,
 − D − a 0 (38)
u = 0 on ∂Ω .
0

Let’s see that the following estimate holds
kukHD2(Ω0) ≤ Chτi12 log2hτikfkL2(Ω0), (39)
8
for Imτ < σ . Using Proposition 1 we get
0
| |
u = ∆ u
k kHD2(Ω0) k D kL2(Ω0)
= τu iB u+f
k − a kL2(Ω0) (40)
≤ Chτi12 log2hτikfkL2(Ω0) +CkukHD1(Ω0)
≤ Chτi21 log2hτikfkL2(Ω0) +εkukHD2(Ω0) +CεkfkL2(Ω0).
Choosing ε small enough, we obtain (39). Using (39) and (5) with s = 1 ε,
−
t = 2, we obtain
u 2 u 1−ε u 1+ε
k kHD1−ε(Ω0) ≤ k kHD2(Ω0)k kL2(Ω0)
log2 τ
≤ C(hτi12 log2hτi)1−ε( τ h21 i)1+εkfk2L2(Ω0) (41)
h i
C f 2 .
≤ k kL2(Ω0)
Then we get (37) for s = 0, i.e.
k(−∆D −τ +iBa)−1kL2(Ω0)→HD1−ε(Ω0) ≤ C. (42)
We will prove (37) for s = 2N with N N, namely
∈
kukHD2N+1−ε(Ω0) . kfkHD2N(Ω0). (43)
Let f H2N(Ω ) and let ube the corresponding solution of (38). The function
∈ D 0
( ∆ )Nu satisfies
D
−
(∆ +τ iB )(( ∆ )Nu) = ( ∆ )Nf i[B ,( ∆ )N]u. (44)
D a D D a D
− − − − −
Using (42), we obtain
( ∆ )Nu . ( ∆ )Nf + [B ,( ∆ )N]u (45)
k − D kHγ(Ω0) k − D kL2(Ω0) k a − D kL2(Ω0)
where γ = 1 ε. Since
−
u = ( ∆)Nu , (46)
k kHD2N+γ(Ω0) k − kHγ(Ω0)
we obtain
u . ( ∆ )Nf + [B ,( ∆ )N]u . (47)
k kH2N+γ(Ω0) k − D kL2(Ω0) k a − D kL2(Ω0)
On the other hand, we have
[Ba,( ∆D)N] = a( ∆D)12[a,( ∆D)N]+[a,( ∆D)N]( ∆D)21a. (48)
− − − − −
9
Using the properties ( s) and ( s,n) we get
P Q
[B ,( ∆ )N]u . u . (49)
k a − D kL2(Ω0) k kHD2N(Ω0)
Consequently
u . f + u
k kHD2N+γ(Ω0) k kHD2N(Ω0) k kHD2N(Ω0) (50)
. f +ε u +C u .
k kHD2N(Ω0) k kHD2N+γ(Ω0) εk kHDγ(Ω0)
Choosing ε small enough and using (42) we obtain
kukHD2N+γ(Ω0) . kfkHD2N(Ω0) +kukHDγ(Ω0)
. f + f (51)
k kHD2N(Ω0) k kL2(Ω0)
. f ,
k kHD2N(Ω0)
which proves (37) for s = 2N. Using the identity
[a,( ∆ )−N] = ( ∆ )−N[( ∆ )N,a]( ∆ )−N, (52)
D D D D
− − − −
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.Wewillfirstprove(7).Extendf by0fort R [0,T].The
∈ \
Fourier transforms (in t) of u and f are holomorphic in the domain Imz < 0
and satisfy the equation
( z ∆ +iB )u(z, ) = f(z, ). (53)
D a
− − · ·
b
b
Wetakez = λ iσ,λ R,σ > 0,andwelet σ tendtozero. UsingProposition
− ∈
7, we get
kukL2(R;HDs+1−ε(Ω0)) . kfkL2(R;HDs(Ω0)), s ∈ R. (54)
The fact that thebFourier transform ofbany 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,+ [, then the function
∈ 0 ∞
w(t, ) = ϕ(t)v(t, ) satisfies the equation
· ·
i∂ w ∆ w+iB w = iϕ′(t)v in R Ω ,
 t − D a + × 0
w(0,·) = 0 in Ω0, (55)
w|R×∂Ω0 = 0.
10