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A Singular Critical Potential For The Schrödinger operator
Thomas Duyckaerts
To cite this version:
Thomas Duyckaerts. A Singular Critical Potential For The Schrödinger operator. 2003. �hal-
00000760v2�
ccsd-00000760 (version 2) : 20 Oct 2003
THOMAS DUYCKAERTS
Abstract. We construct a potential V on R
d, smooth away from one pole, and a sequence of quasi- modes for the operator −∆+ V , which concentrate on this pole. No smoothing effect, Strichartz estimates nor dispersive inequalities hold for the corresponding Schr¨ odinger equation.
1. Introduction Consider a Schr¨ odinger operator:
(1) P = −∆ + V, ∆ =
X d
j=1
∂ 2
∂x 2 j ,
on R d , d ≥ 1, where V is a real potential, small at infinity, and the initial value problem for the Schr¨odinger equation:
(2)
( i∂ t U − P U = 0 U
↾t=0= U 0 ∈ L 2 ( R d ).
For sufficiently smooth potentials V , the equation (2) implies a local smoothing effect:
(3) ||χU|| L
2(]0,T[,H
1/2(R
d)) ≤ C||U 0 || L
2(
Rd) , χ ∈ C 0
∞( R d ),
where T may be finite, and in some cases infinite, and C may depend on V , χ and T. This well-known estimate (see [2], [6] and references therein) is essential in the study of (2) and related non-linear equations.
In this paper, we investigate the minimal regularity to be assumed on V such that (3) is still true. In [6], A. Ruiz and L. Vega consider the equation (2) as a perturbation of the free Schr¨odinger equation to show an inequality which implies (3). The following potentials fall within this scope:
• V ∈ L p , p ≥ d/2.
• Potentials belonging to some Morey-Campanato spaces, (which are larger than L d/2 ) with a smallness condition in these spaces (see [6] for details). For example, potentials with inverse- square singularities:
(4) V = X
j=1..N
a j
|x − p j | 2 + V 0 , V 0 ∈ L p , p ≥ d/2, where the |a j |’s are small enough.
Let us mention that potentials in the article above may be time-dependent but we shall not develop this point here. In [5], (see also [3]), the author studies potentials of type (4) with regular V 0 , assuming only the following positivity properties on the a j ’s:
a j + (d/2 − 1) 2 > 0, j = 1..N, and shows non-trapping type inequality on the truncated resolvent:
(5) ∀λ ∈ C , Re λ 2 ≥ C, Im λ 2 6= 0, ||χ(P − λ 2 )
−1 χ|| L
2ջL
2≤ C
|λ| , χ ∈ C 0
∞( R d ), One may deduce from (5) estimates of type (3) (cf [2]).
1
2 THOMAS DUYCKAERTS
The preceding works do not study potentials V ∈ L p , p < d/2, and potentials of type (4) but with poles of higher order. In this paper, we show that the space L d/2 and the inverse-square singularities are critical, by constructing an unipolar potential whose pole is of the order of (log r) 2 /r 2 , r = |x|, near 0, and such that (3), (5) and some other usual estimates on solutions of Schr¨odinger equations do not hold.
Theorem 1. Let d ≥ 1, N ≥ 0 be integers. There exist:
• a radial, positive potential V on R d , which has compact support and such that:
V ∈ C
∞( R d \{0}) (6)
| log r| 2
Cr 2 ≤ V (r) ≤ C| log r| 2
r 2 , r ≤ r 0 , (7)
• an increasing sequence (λ n ) n
≥n
0of positive real numbers, diverging to +∞,
• a sequence of radial C
∞functions (u n ) n
≥n
0, whose support is of the following form:
½ c 1
log λ n
λ n
≤ r ≤ c 2
log λ n
λ n
¾
; such that:
(−∆ + V )u n − λ 2 n u n = f n
(8)
||u n || L
1= 1 (9)
∀j ∈ N , || d j
dr j f n || L
∞= O(λ j n
−N ), n ջ +∞.
(10)
Corollary 1. Let N be any integer greater than 1, P = −∆ +V , where V is the potential of the preceding theorem, and χ a function in C 0
∞( R d ) which does not vanish in 0. Then:
• (lack of local Strichartz estimates)
(11) ∀q, q 0 ∈ [1, +∞], q > q 0 , ∀T > 0, ∀C > 0, ∃U 0 ∈ C 0
∞, ||χU(t)|| L
1(]0,T[,L
q(
Rd)) > C||U 0 || L
q0(
Rd) ;
• (lack of local smoothing effect)
(12) ∀σ > 0 ∀T > 0, ∀C > 0, ∃U 0 ∈ C 0
∞, ||χU(t)|| L
1(]0,T[,H
σ(
Rd)) > C ||U 0 || L
2(
Rd) ;
• (lack of local dispersion)
(13) ∀q ∈]1, +∞], ∀T > 0, ∀C > 0, ∃U 0 ∈ C 0
∞, ||χU(T )|| L
q(
Rd) > C ||U 0 || L
q′(
Rd) ;
• (lack of Strichartz estimates with loss of derivative) (14) ∀q ∈ [1, +∞], ∀σ ∈ [0, 1], 1
2 − 1 q > σ
d ,
∀T > 0, ∀C > 0, ∃U 0 ∈ C 0
∞, ||χU (t)|| L
1(]0,T[,L
q(R
d)) > C||U 0 || D(P
σ/2) . In the preceding statements, we wrote U (t) the solution of the equation (2) with initial value U 0 and q
′the conjugate exponent of q which is defined by: 1/q + 1/q
′= 1.
Remarks. • If d > 2, the hypothesis (6) and (7) implie:
V ∈ \
p<d/2
L p .
• Of course, the sequence (f n ) n
≥n
0invalidates (5) when N ≥ 2.
• It will be clear in the proof of the theorem than one may construct quasi-modes of infinite order (i.e. such that(10) holds with O(λ
−∞n ) instead of O(λ j n
−N )) by taking a singularity just a bit stronger:
| log r| 2+ε
Cr 2 ≤ V (r) ≤ C| log r| 2+ε
r 2 , ε > 0.
With a still stronger singularity, one may force f n to be exponentially decreasing in −λ n .
• Classical Strichartz estimates:
||u|| L
p(]0,T[,L
q(
Rd)) ≤ C||u 0 || L
2(
Rd) , p > 2, 2 p + d
q = d 2 ,
are invalidated by (11), with the exception of the case q = 2, p = +∞ which is always true by the L 2 conservation law. Likewise, (14) shows that Strichartz estimates with loss of derivative (see [2]) of the form:
||u|| L
p(]0,T[,L
q(
Rd)) ≤ C||u 0 || D(P
σ/2) , 1 2 − 1
q > σ
d , σ ∈ [0, 1],
are false. In this inequality, the limit case σ/d = 1/2 − 1/q is an immediate consequence of the Sobolev inclusion
L q ( R d ) ⊂ H σ ( R d ) ⊂ D(P σ/2 ),
and of the conservation of the norm D(P σ/2 ) of any solution of (2). Notice that this last inclusion implies that (14) is still true in usual Sobolev spaces, i.e when D(P σ/2 ) is replaced by H σ ( R n ).
• The potential V and quasi-modes u n being of compact support, as close to 0 as desired, the pre- ceding counter-example is still valid in any regular domain, for a Laplace operator with Dirichlet or Neumann Dirichlet boundary condition.
• Examples of linear Schr¨odinger equations which do not admit the classical dispersive inequality or Strichartz estimates are given in [1] and [4]. In these two cases, the particular behaviour of the equation arises from the metric which defines the Laplace operator. Let us mention that the idea to use quasi-modes in relation with Strichartz inequalities is owed to M. Zworski (see [8]).
Finally, we would like to point out that for the potential V introduced in the theorem, one may define the operator P without ambiguity. The operator −∆ + V has a natural meaning on C 0
∞( R d \{0}), and is essentially self-adjoint on this space, under a positivity asumption implied by (7). We call P its unique self-adjoint extension, which is of course positive. We refer to Reed and Simon [7, Th X.30] for any precision.
The author would like to thank Nicolas Burq, its advisor and Clotilde Fermanian Kammerer for their precious help.
Section 2 of the paper is devoted to the proof of the corollary, and section 3 to that of the theorem.
2. Proof of the corollary.
According to H¨ older’s inequality, for any function v on R d , with a finite volume support:
(15) ||v|| L
q0≤ B 1/q
0−1/q ||v|| L
q, q > q 0 , where B is the volume of the support. Let U n (t) = e
−iλ
2nt u n . Then:
i∂ t U n − P U n = −e
−iλ
2nt f n , U n↾t=0 = u n
U n (t) = e
−itP u n + ie
−iλ
2nt Z t
0
e i(s
−t)P f n ds.
(16)
Assume that (11) is false. According to (16):
Z T 0
||χU n (t)|| L
q(R
d) dt ≤ Z T
0
||χe
−itP u n || L
q(R
d) dt + Z T
0
Z t 0
||χe i(s
−t)P f n || L
q(R
d) dsdt.
Since the support of u n concentrates in 0, and χ does not vanish in 0, the left term of this inequality is, for n large enough, greater than:
1 C
Z T 0
||u n || L
q(
Rd) dt ≥ 1 C
µ log λ n
λ n
¶ d(1/q
−1/q
0)
||u n || L
q0(
Rd) .
4 THOMAS DUYCKAERTS
Using the negation of (11), the right term is dominated by:
||u n || L
q0(
Rd) + ||f n || L
q0(
Rd) . Hence we obtain, using (10):
µ λ n
log λ n
¶ d(1/q
0−1/q)
||u n || L
q0= O ¡
||u n || L
q0+ λ
−n N ¢ ,
which leads to the announced contradiction since N ≥ 1 and q > q 0 , the norm of u n in L q
0being greater than 1 by (9).
By Sobolev inequalities, (11) implies (12), taking in (11) q 0 = 2 and q > 2 close enough to 2. So (12) holds.
Let us assume that (13) is not true. Then we get, by (16):
||χu n (T )|| L
q(
Rd) ≤ C||u n || L
q′(
Rd) + Z T
0
||χe i(s
−T) f n || L
1(
Rd) ds
≤ C ³
||u n || L
q′(
Rd) + ||f n || L
2(R
d) ´ .
To obtain the second inequality we have bounded, up to a multiplicative constant, the L q
′norm on the support of χ by the L 2 norm. The contradiction is again a simple consequence of (15) and the fact that q
′< q.
We shall now prove (14). Otherwise, we would have by (16):
(17)
Z T 0
||u n (t)|| L
q(R
d) dt ≤ C||u n || D(P
σ/2) + Z T
0
Z t 0
||e i(s
−t)P f n || D(P
σ/2) ds dt.
Of course, e i(s
−t)P maps isometrically the domain of P σ/2 onto itself. We have:
||f n || 2 D(P
1/2) = Z
|∇f n (x)| 2 dx + Z
(1 + V (x))|f n (x)| 2 dx.
On the support of f n , as on that of u n , |x| ≥ log Cλ λ
nn. Using the superior bound of V in (7), we get, on the support of f n :
|V (r)| ≤ C λ 2 n
(log λ n ) 2 (log log λ n − log λ n ) 2 ≤ Cλ 2 n . It follows using (10) and N ≥ 2 that:
(18) ||f n || D(P
1/2) ≤ C (||∇f n || L
2+ λ n ||f n || L
2) ≤ Cλ
−n 1 .
Furthermore, the equation P u n − λ 2 n u n = f n implies, using once again the bounds (9) and (10) on the norms of u n and f n :
||u n || 2 D(P
1/2) ≤ ||f n || L
2||u n || L
2+ λ 2 n ||u n || 2 L
2≤ Cλ 2 n ||u n || 2 L
2.
By interpolation on the norms of the left hand side, and since 0 ≤ σ ≤ 1, we get:
(19) ||u n || D(P
σ/2) ≤ Cλ σ n ||u n || L
2. According to (18), (19) and the inequality (17):
||u n || L
q≤ Cλ σ n ||u n || L
2+ o(1).
Hence, with (15):
µ λ n
log λ n
¶ d/2
−d/q
||u n || L
2≤ Cλ σ n ||u n || L
2+ o(1),
which is absurd since d 2 − d q > σ.
3. D´ emonstration du th´ eor` eme.
Denote by r the euclidian norm of x and by
′the radial derivative dr d . We would like to find radial functions V , f n , u n such that:
(20) f n (r) = −u
′′n (r) − d − 1
r u
′n (r) + V (r)u n (r) − λ 2 n u n (r),
with f n small and λ n diverging to infinity. We shall first change functions to get rid of the first-order derivative in this equation. Set:
(21) u n = r
−d−21v n , f n = r
−d−21g n , W = V + d 2 − 4d + 3 4r 2 . Thus (20) becomes:
(20’) g n (r) = −v n
′′+ W v n − λ 2 n v n . Let:
y(s) = e
−√s
2+1 , b(s) = − 1
(s 2 + 1) 3/2 + s 2 s 2 + 1 , which are C
∞solutions of the equation on R :
−y
′′(s) + b(s)y(s) = 0 (22)
y(s) > 0, ∀j ∈ N , |y (j) (s)| ≤ C j e
−|s
|(23)
|b(s)| ≤ 1.
(24)
We shall write:
y ω,a (r) = y(ω(r − a)), b ω,a (r) = ω 2 b(ω(r − a)), where a and ω are two real parameters. We have:
(22’) − y
′′ω,a + b ω,a y ω,a = 0;
Let q(λ) be a strictly increasing positive function defined in a neighbourhood of +∞ such that:
λ
ջlim +
∞q(λ) = +∞
(25)
λ
ջlim +
∞q(λ) λ = 0.
(26)
Let (λ n ) n
≥n
0be the sequence defined by the equations:
10 n = q(λ n ), so that λ n diverges faster to +∞ than 10 n .
Choose a cutoff function χ: χ ∈ C 0
∞(]−1, 1[) , χ = 1 on
·
− 1 2 , 1
2
¸ . Set:
ψ n (r) = χ(10 n r) − χ(10 n+1 r), χ n (r) = χ
µ 10 n+1
µ r − 3
10 n+1
¶¶
,
6 THOMAS DUYCKAERTS
-
5 10
n+35 10
n+21 10
n+12 10
n+12.5 10
n+13 10
n+13.5 10
n+14 10
n+15 10
n+11 10
n1 10
n−1r
z }| {
supp χ n
z }| {
supp ψ n
−1
z }| { supp ψ n+1
z }| { χ n = 1
| {z }
ψ n = 1
| {z }
supp ψ n
Figure 1. Cutoff functions near 3/10 n+1
The ψ n ’s form a partition of unity near 0. The support of each χ n is included in that of ψ n , away from those of the ψ j ’s, j 6= n:
supp ψ n ⊂
· 5 10 n+2 , 1
10 n
¸
1
10 n+1 ≤ r ≤ 5
10 n+1 ⇒ ψ n (r) = 1 X
n
≥n
0ψ n (r) = χ(10 n
0r) (27)
supp χ n ⊂
· 2 10 n+1 , 4
10 n+1
¸
⊂ {ψ n = 1}
(28)
25
10 n+2 ≤ r ≤ 35
10 n+2 ⇒ χ n (r) = 1.
(29)
We shall denote by y n and b n the following functions:
y n (r) = y
λn 2,
310n+1
(r) = y µ λ n
2 µ
r − 3 10 n+1
¶¶
(30)
b n (r) = b
λn2
,
10n+13(r) = λ 2 n 4 b
µ λ n
2 µ
r − 3 10 n+1
¶¶
. (31)
Each of the function −b n may be seen as a potential well which concentrates (by equation (22’)) the energy of y n near 3/10 n+1 . We shall construct W so that the equation g n (r) = 0 is exactly −v
′′n + b n v n = 0 on a small interval (including the support of v n ) around the point 3/10 n+1 . The size of this interval will be of the same order as 10
−n , smaller than the equation parameter ω = λ n /2. We shall choose v n as a cutoff of y n . The exponential decay of y, combined with the difference of order between these two scales will induced small enough error terms for (9) and (10) to hold.
Set (see the figure):
W (r) = X
n
≥n
0ψ n (r) ¡
b n (r) + λ 2 n ¢ (32)
v n (r) = α n y n (r)χ n (r), (33)
where the α n ’s are strictly positive constants yet to be determined. The functions g n , u n , f n and the
potential V are thus defined by (20’) and (21).
- 6
r W (r)
3 10
n+23 10
n+1λ 2 n+1
λ 2 n
Figure 2. The potential W Lemma 1. The following inequalities hold:
∀j ∈ N , ∃C > 0, || d j f n
dr j || L
∞(
Rd) ≤ Cα n λ 1+j n (q(λ n ))
d+12e
−20q(λnλn)(34)
||u n || L
1(
Rd) ≥ α n
C λ
−n 1 (q(λ n ))
−d−12. (35)
Proof. According to (28), on the support of χ n , W is equal to b n + λ 2 n . Hence:
(36) g n (r) = −v
′′n (r) + b n (r)v n (r).
Thus, using (22’):
f n (r) = r
−d−12g n (r) = r
−d−12α n (y n (r)χ
′′n (r) + 2y
′n (r)χ
′n (r)) . The derivative of f n of order j is then of the form:
(37) d j f n
dr j = X
j
1+j
2+j
3=j+1
β j
1,j
2,j
3d j
1dr j
1y n
d j
2dr j
2χ
′n d j
3dr j
3r
−d−12. On the support of χ
′n :
¯ ¯
¯ ¯ r − 3 10 n+1
¯ ¯
¯ ¯ > 1
10 n+1 = 1
10q(λ n ) .
8 THOMAS DUYCKAERTS
So according to the bounds (23) of y and its derivatives:
|y n (j
1) (r)| ≤ Cλ (j n
1) e
−λn2| r
−10n+13|
≤ λ j n
1e
−20q(λnλn)Furthermore,
|χ (j n
2+1) | ≤ C (q(λ n )) j
2+1 ≤Cλ j n
2q(λ n )
¯ ¯
¯ ¯
¯ µ d
dr
¶ j
3³
r
−d−21´ ¯
¯ ¯
¯ ¯
≤ C (q(λ n ))
d−1
2
+j
3≤Cλ j n
3(q(λ n ))
d−1 2
,
for on the support of χ n , r ≥ 10
n+11 . These three inequalities, together with (37), implies (34).
We shall now prove (35). By the definition of y n : λ n
2
¯ ¯
¯ ¯ r − 3 10 n
¯ ¯
¯ ¯ ≤ 1
2 ⇒ y n (r) ≥ m = sup
|
s
|≤12|y(s)|.
Furthermore, if r is as above, and n big enough, then χ n (r) = 1 and so:
u n (r) = r
−d−21v n (r) = r
−d−21α n y n (r) Hence:
||u n || L
1≥ mα n
Z
λ
n| r
−103n|
≤1
r
−d−21r d
−1 dr
≥ α n
C
¡ 10
−n ¢
d−12λ
−n 1 .
¤ Choose M > 1 and set:
(38) q(λ) = λ
M log λ .
The positive function q is strictly increasing for big λ’s and satisfies (25) and (26). In addition, lemma 1 implies, bounding from above q(λ n ) by λ n :
|| d j f n
dr j || L
∞≤ Cα n λ
d+3 2
+j
−M20n
||u n || L
1≥ α n
C λ
d+1