Journal of Applied Mathematics and Physics
Vol.02 No.13(2014), Article ID:52539,108 pages
10.4236/jamp.2014.213143
On the Cauchy Problem for Von Neumann-Landau Wave Equation
Chuangye Liu1, Minmin Liu2
1Laboratory of Nonlinear Analysis, Department of Mathematics, Central China Normal University, Wuhan, China
2School of Science, Wuhan Institute of Technology, Wuhan, China
Email: chuangyeliu1130@126.com, ocbmml@126.com
Copyright © 2014 by authors and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/
Received 21 October 2014; revised 16 November 2014; accepted 11 December 2014
ABSTRACT
In present paper we prove the local well-posedness for Von Neumann-Landau wave equation by the T. Kato’s method.
Keywords:
Von Neumann-Landau Wave Equation, Strichartz Estimate, Cauchy Problem
1. Introduction
For the stationary Von Neumann-Landau wave equation, Chen investigated the Dirichlet problems [1] , where the generalized solution is studied by Function-analytic method. The present paper is related to the Cauchy problem: the Von Neumann-Landau wave equation
, (1)
where for
is an unknown complex valued function on
and
is a nonlinear complex valued function.
If the plus “+” is replaced by the minus “−” on right hand in Equation (1), then the resulted equation is the Schrödinger equation. For the Schrödinger equation, the well-posedness problem is investigated for various nonlinear terms. In terms of the nonlinear terms
, the problem (1) can be divided into the subcritical case and the critical case for
solutions. We are concerned with the subcritical case and obtain a local well- posedness result by the T. Kato’s method.
The paper is organized as follows. Section 2 contains the list of assumptions on the interaction term and the main result is presented. Section 3 is concerned with the Strichartz estimates. Finally, in Section 4, the main result is proved.
2. Statement of the Main Result
In this section we list the assumptions on the interaction term and state the main result. Firstly, we recall that the definition of admissible pair [2] .
Definition 2.1. Fix,
. We say that a pair
of exponents is admissible if
, (2)
and
(3)
Remark 2.1. The pairs is always admissible, so is the
if
The two pairs are called the endpoint cases.
Secondly, let satisfy
, (4)
and
, (5)
for all such that
with
(6)
where is a constant independent of
Set
, (7)
for all measurable function and a.e.
.
Finally, let us make the notion of solution more precise.
Definition 2.2. Let be an interval such that
We say that
is a strong
-solution of (1) on
if
satisfies the integral equation
, (8)
for all where
The main result is the following theorem:
Theorem 1. Suppose Let
satisfy (4)-(6). If
(considered as a function
) is of class
, then the Cauchy problem (1) is locally well posed in
More specially, the following properties hold:
(i) For any there exists a time
and constant
such that for each
in the ball
there exists a unique strong
-solution
to the Equation (1) in
such that
(9)
where and
is an admissible pair.
(ii) The map is continuous from
to
(iii) For every the unique solution
is defined on a maximal interval
with
and
(iv) There is the blowup alternative: If then
as
(respectively, if
then
as
).
Remark 2.2. It follows from Strichartz estimates that
,
for any admissible pair
Remark 2.3. For the Schrödinger equations, the similar results hold [2] . It implies a fact that the ellipticity of the operator is not the key point in the local well-posedness problem.
3. Strichartz Estimates
In this subsection, we recall that the Strichartz estimates. Let denote a general Fourier variable in
Let
then by Fourier transform(denoting by
or
) we have
, (10)
for any It is easy to verify that the
is a self-adjoint unbounded operator on
with the domain
Then, by Stone theorem we see that
is an unitary group on
. Moreover,
can be expressed explicitly by Fourier transform.
, (11)
for any By the direct compute, we conclude
(12)
The following result is the fundamental estimate for
Lemma 1. If and
then
maps
continuously to
and
(13)
where is the dual exponent of
defined by the formula
Proof. For the proof please see [3] or [4] . □
The following estimates, known as Strichartz estimates, are key points in the method introduced by T. Kato [5] .
Lemma 2. Let and
be any admissible exponents. Then, we have the homogeneous Strichartz estimate
(14)
the dual homogeneous Strichartz estimate
(15)
and the inhomogeneous Strichartz estimate
, (16)
for any interval and real number
Proof. For the proof please see [3] or [4] in the non-endpoint case. On the other hand, the proof in the endpoint case follows from the theorem 1.2 in [6] and the lemma 1 in the present paper. □
4. The Proof of Theorem
Proof. Let be such that
for
and
for
Setting
one easily verifies that for any
(17)
Set for
Using (17), we deduce from Hölder’s inequality that
(18)
And it follows from Remark 1.3.1 (vii) in [2] that
(19)
We now proceed in four steps.
Step 1. Proof of (i). Fix to be chosen later, and let
,
be such that
is an ad- missible pair, and set
Consider the set
, (20)
equipped with the distance
(21)
We claim that is a complete metric space. Indeed, let
be a Cauchy sequence. Clearly,
is also a Cauchy sequence in
and
In particular, there exists a function
such that
in
and
as
Applying theorem 1.2.5 in [2] twice, we conclude that
,
and that
thus, in
as
Taking up any Since
is continuous
it follows that
is measurable, and we deduce easily that
Similarly, since
is continuous
we see that
Using inequalities (18) and (19) and Remark
1.2.2 (iii) in [2] , We deduce the following:
and
Using the embedding and Hölder’s inequality in time, we deduce from the above estimates that
, (22)
and
(23)
Given. For any
let
be defined by
(24)
It follows from (22) and Strichartz estimates (lemma 2) that
(25)
and
(26)
Also, we deduce from (23) that
(27)
Finally, note that We now proceed as follows. For any
we set
and we let
be the unique positive number so that
(28)
It then follows from (26) and (28) that for any
(29)
Thus, and by (27) we obtain
(30)
In particular, is a strict contraction on
By Banach’s fixed-point theorem,
has a unique fixed
point that is
satisfies (8). By (25),
. By the definition 2.2, we con- clude that
is a strong
-solution of (1) on
Note that
is decreasing on
, then the estimate (9) holds for
by letting
in (29).
For uniqueness, assume that are two strong
-solution of (1) on
with the same initial value
. Then, we have
(31)
For simplicity, we set
,
for and
For any interval
by (18) and Strichartz estimates (16), then we obtain
(32)
Similarly, for we have
(33)
Note that Then, it follows from that
, (34)
where the constant and the constant
is independent of
by above inequalities. Note that
we conclude that
by the lemma 4.2.2 in [2] . So
Step 2. Proof of (ii). Suppose that in
as
By the part (i), we denote
and
by
the unique solution of (1) corresponding to the initial value and
, respectively. We will show that
in
as
Note that
(35)
and the estimate (29) which implies that (27) holds for Note that the choosing of the time
in (28), it follows from (27) with (30) that
(36)
Hence, we have
(37)
Next, we need to estimate Note that
commutes with
, and so
(38)
A similar identity holds for We use the assumption
which implies that
where
is a
real matrix. Therefore, we may write
(39)
Note that and
are also
so that
and from (17) we deduce that
and
for any
and some constant
Therefore, arguing as in Step 1, we obtain the estimate
(40)
By choosing as (28) and noting that
from (40) we obtain that
(41)
There, if we prove that
, (42)
as then we have
, (43)
as which, combined with (37), yields the desired convergence. we prove (42) by contradiction, and we assume that there exists
and a subsequence, which we still denote by
such that
(44)
By using (37) and possibly extracting a subsequence, we may assume that a.e. on
and that there exists
such that
a.e. on
In particular, both
and
converge to 0 a.e. on
Since
and
we obtain from the dominated convergence a contradiction with (44).
Step 3. Proof of (iii). Consider and let
It follows from part (i) there exists a solution
,
of (1).
Step 4. Proof of (iv). Suppose now that and assume that there exist
and a sequence
such that
Let
be such that
By part (i), from the initial data
one can extend
up to
, which contradicts maximality. Therefore,
One shows by the same argument that if then
This completes the proof. □
Acknowledegments
We are grateful to the anonymous referee for many helpful comments and suggestions, which have been incorporated into this version of the paper. C. Liu was supported in part by the NSFC under Grants No. 11101171, 11071095 and the Fundamental Research Funds for the Central Universities. And M. Liu was supported by science research foundation of Wuhan Institute of Technology under grants No. k201422.
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