Rend. Sem. Mat. Univ. Politec. Torino
Vol. 69, 1 (2011), 91 – 96
M. H. Mortad
GLOBAL SPACE-TIME LP -ESTIMATES
FOR THE WAVE OPERATOR ON L2
Abstract. We prove and disprove several estimates of the following type:
kuk∗ ≤ akuk2 + bkuk2
∂2
∂t 2
where =
− △x is the usual wave operator defined on L2 (R × Rn ), n ≥ 2 and where
k · k∗ represents different norms yet to be determined.
1. Introduction
Many authors have worked on Strichartz estimates for the wave equation since the
paper of Strichartz [10]. Some of the important papers are [11, 4, 3, 8].
In this work we examine a slightly different type of these estimates. The question asked here is: To what L p -space (or another space) does a function u belong to if u
and u already belong to L2 (R × Rn ) where n ≥ 2? This question actually constitutes
the follow-up of a work done by the author (see [6]). It was proved in [6] (among
others) that the following two inequalities do hold:
(1)
kukL p (R2 ) ≤ akukL2(R2 ) + bkukL2(R2 ) where 2 ≤ p < ∞,
(2)
ess sup ku(·,t)kLr (Rn ) ≤ akukL2(Rn+1 ) + bkukL2(Rn+1 ) ,
t∈R
where 2 < r <
(3)
2n
n−1
while the following one does not hold:
kukL∞(R2 ) ≤ akukL2(R2 ) + bkukL2(R2 ) .
Note that Estimate (1) was a consequence of the following important inequality
(which also appeared in [6]):
kukBMO(R2 ) ≤ akuk2 + bkuk2,
for some constants a and b.
In this paper, we prove that in R × Rn and for n ≥ 2, the estimate
kukL p (Rn ×R) ≤ akukL2(Rn ×R) + bkukL2(Rn ×R)
holds if and only if p ≤
2n+2
n−1 .
91
92
M. H. Mortad
In the end, we answer some questions left open in [6].
It is worth mentioning that the author has a similar work (see [7]) done for the
time dependent Schrödinger operator. See also [1] for some related work for the Airy
operator. For literature on PDEs and Fourier transforms, see [2] and [5].
2. Main results
Here is the first main result in the paper.
T HEOREM 1. Let n ≥ 2 and let p =
such that
2n+2
n−1 .
Then for all a > 0, there exists b > 0
kukL p (R×Rn ) ≤ akukL2(R×Rn ) + bkukL2(R×Rn )
(4)
for all u ∈ L2 (R × Rn) such that u ∈ L2 (R × Rn ).
Proof. Let the x-Fourier transform of u, i.e. û(t, ξ), be supported in the region |ξ| ≤ 2.
Then
k△ukL2 (Rn+1 ) . kukL2 (Rn+1 )
and the right hand side will be equivalent to
k∂t2 ukL2 (Rn+1 ) + kukL2(Rn+1 )
which controls kukLtpL2x (Rn+1 ) thanks to the Sobolev embedding theorem (in the t variable). Hence by the frequency localization,
ku(t, ·)kLxp (Rn ) . ku(t, ·)kL2x (Rn ) .
Now, consider the case where û(t, ξ) is supported in the region |ξ| ≥ 1. Recall
the Strichartz estimate for u = 0, that is
kukL p (Rn+1 ) . k∂t u(0, ·)k
1
H− 2
+ ku(0, ·)k
1
H2
.
One can hence choose any value of s instead of 0 for the initial data time and obviously
restrict the left to a finite interval. The Duhamel’s principle (in the case u 6= 0) then
implies that
kukL p ([0,T ]×Rn ) . kuk
1
. T 2 kuk
1
Lt1 H − 2 ([0,T ]×Rn )
+ inf (k∂t u(s, ·)k
0≤s≤T
1
−1
Lt2 H 2 ([0,T ]×Rn )
+ T − 2 (k∂t u(s)k
1
1
H− 2
−
Lt2 H 2 ([0,T ]×Rn )
+ ku(s, ·)k
+ kuk
1
H2
)
1
Lt2 H 2 ([0,T ]×Rn )
)
Applying a Littlewood–Paley decomposition reduces the work to the case û(t, ξ)
is supported where |ξ| ∈ [λ, 3λ], with λ ≥ 1. Taking T = λ leads to
kukL p ([0,λ]×Rn ) . kukL2([0,λ]×Rn ) + λ−1 k∂t ukL2 ([0,λ]×Rn ) + kukL2([0,λ]×Rn ) .
L p -estimates for the wave operator
93
Summing over a disjoint decomposition of R into intervals of length λ and remembering that p ≥ 2 yield
kukL p (Rn+1 ) . kukL2 (Rn+1 ) + λ−1 k∂t ukL2 (Rn+1 ) + kukL2(Rn+1 ) .
The proof will be complete once we show that for û(t, ξ) supported where |ξ| ∈
[λ, 3λ] we may bound
λ−1 k∂t ukL2 (Rn+1 ) . kukL2(Rn+1 ) + kukL2(Rn+1 ) .
This will be proved once we can show that
λ−2 η2 . (η2 − |ξ|2 )2 + 1, |ξ| ≈ λ ≥ 1
by means of Fourier transforms. But, the previous inequality can easily be verified in
the cases η ≤ 4λ and η ≥ 4λ.
The proof is complete.
Another result which generalizes one which appeared in [6] is the following:
P ROPOSITION 1. For all a > 0, there exists b > 0 such that
∞
∑
k=−∞
ess sup ku(., .,t)k2L2 (Rn ) ≤ akuk2L2(R×Rn ) + bkuk2L2(R×Rn )
k≤t≤k+1
for all u ∈ L2 (R × Rn) such that u ∈ L2 (R × Rn ).
Proof. The proof is based on energy estimates. Let u ∈ L2 (R × Rn ) be such that u ∈
L2 (R × Rn).
For any s ∈ [a, b], we have
ess sup ku(·,t)kL2 (Rn ) ≤
a≤t≤b
Z b
a
ku(·,t)kL2 (Rn ) dt + ku(·, s)kL2 (Rn ) .
Hence
ess sup ku(·,t)kL2 (Rn ) ≤
a≤t≤b
Z b
a
1
ku(·,t)kL2 (Rn ) dt +
b−a
Z b
a
ku(·,t)kL2 (Rn ) dt.
Thus
1
1
ess sup ku(·,t)kL2 (Rn ) ≤ (b − a) 2 kukL2 ([a,b]×Rn ) + (b − a)− 2 kukL2 ([a,b]×Rn ) .
a≤t≤b
Finally, we easily get:
1
1
ess sup ku(·,t)kL2 (Rn ) ≤ c 2 kukL2 ([a,b]×Rn ) + c− 2 kukL2 ([a,b]×Rn ) .
k≤t≤k+1
The square-summation over k yields the desired result. The proof is over.
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M. H. Mortad
3. Counterexamples
The next result tells us that p =
2n+2
n−1
is best possible in Theorem 1.
P ROPOSITION 2. Let n ≥ 2 and let p > 2n+2
n−1 . Then there are no constants
a and b such that the estimate of Theorem 1 holds for all u ∈ L2 (R × Rn ) such that
u ∈ L2 (R × Rn).
Proof. There are two different ways to establish this proposition. One can be found in
[1] and one in [7]. The operators considered in the quoted papers are different from the
wave operator but the method can be adapted to this case.
We finish this paper by answering an open question raised in [6] about the self2
∂2
2
2
adjointness of ∂t∂ 2 − ∂x
2 + V where V is positive and belongs to Lloc (R ). We then
∂
insisted on the positivity of V since it was already known back then that ∂t∂ 2 − ∂x
2 +V
was not essentially self-adjoint for some V which was not positive.
The proof exploits the non essential self-adjointness of the one-dimensional
Laplacian perturbed by some negative potential.
2
∂2
∂t 2
2
P ROPOSITION 3. There exists a positive V belonging to L2loc (R2 ) such that
−
+ V is not essentially self-adjoint on C0∞ (R2 ).
∂2
∂x2
Proof. The operator − dtd 2 − t 4 is not essentially self-adjoint on C0∞ (R) (details can be
found in [9]) meaning that
2
2
d2
d
4
4
− 2 − t f (t) = −i f (t) or
+ t f (t) = i f (t)
dt
dt 2
has a non-zero solution in L2 (R). Now the perturbed one-dimensional Laplacian by x2
has as eigenvalue 21 , i.e. there is a non-zero g ∈ L2 (R) such that
d2
1
2
− 2 + x g(x) = g(x)
dx
2
Now by adding up the last two displayed equations we get
(5)
∂2
∂2
−
∂t 2 ∂x2
1
f (x)g(t).
f (x)g(t) + (t + x ) f (x)g(t) = i +
2
4
2
Take ϕ(x,t) = f (x)g(t). Since f , g are both in L2 (R), ϕ will be in L2 (R2 ) and Equation
5 will have a non-zero solution in L2 (R2 ) and yet V (x,t) = t 4 + x2 is nonnegative and
it belongs to ∈ L2loc (R2 ).
L p -estimates for the wave operator
95
A conjecture
We ask the interested reader the following rather hard question which is true in the case
of the Laplacian (see [9]) but it remains open for general operators (not known for the
Wave and the time-dependent Schrödinger operators at least).
C ONJECTURE . Let P be an essentially self-adjoint partial differential operator
with domain C0∞ (Rn ), n ≥ 2. If there do not exist two positive constants a and b such
that for all C0∞ (Rn )
k f k∞ ≤ akP f k2 + bk f k2,
then P + V is not essentially self-adjoint on C0∞ (Rn ) for real-valued V belonging to
L2 (Rn ).
Acknowledgement. I wish to sincerely thank Professor Hart Smith (Department of
Mathematics, University of Washington) for his help to prove Theorem 1.
References
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96
AMS Subject Classification: 35B45, 35L05
Mohammed Hichem MORTAD
Department of Mathematics, University of Oran (Es-Senia)
B.P. 1524, El Menouar, Oran 31000, ALGERIA
e-mail: mhmortad@gmail.com, mortad@univ-oran.dz
Lavoro pervenuto in redazione il 08.04.2011
M. H. Mortad