Energy solution to Schrödinger-Poisson system in the two-dimensional whole space

0.18 MB·English

by  Satoshi Masaki

ENERGY SOLUTION TO SCHRO¨DINGER-POISSON
SYSTEM IN THE TWO-DIMENSIONAL WHOLE SPACE
SATOSHIMASAKI
Abstract. We consider the Cauchy problem of the two-dimensional
Schr¨odinger-Poisson system in the energy class. Though the Newto-
0
1 nian potential diverges at the spatial infinity in the logarithmic order,
0 global well-posedness is proven in both defocusing and focusing cases.
2 The key is a decomposition of the nonlinearity into a sum of the lin-
ear logarithmic potential and a good remainder, which enables us to
n
apply the perturbation method. Our argument can be adapted to the
a
J one-dimensional problem.

5
2

] 1. Introduction
P
A
This paper is devoted to the study of the Scho¨dinger-Poisson system
.
h 1
i∂ u+ ∆u= λPu, (t,x) R1+2,
t t
a 2 ∈
m (1.1)  ∆P = u2,
 − | |
[ 
u(0,x) = u0(x),

1 where λ is a real constant. We suppose P is the Newtonian potential
v 
8 1
0 (1.2) P = (log x u2)
−2π | |∗| |
3
4 where denotes the convolution. For a suitable u, this is the unique strong
∗
. solution of ∆P = u2 under the condition
1
− | |
0
0 P 0 as x , P L∞(R2), P(0) = (log y )u(y)2dy
1 |∇ | → | | → ∞ ∇ ∈ R2 | | | |
Z
:
v (see [11]). When the dimensions are larger than two, the Schro¨dinger-
i PoissonsystemisaspecialcaseoftheHartreeequationandoneofthetypical
X
example of the nonlinear Schro¨dinger equation with a nonlocal nonlinearity,
r
a and there is large amount of literature (see [6] and references therein). On
the other hand, the two-dimensional case is less studied. In [1, 18], (1.1) is
considered with some restrictive assumptions such as a neutrality condition
which confirms that the Newtonian potential (1.2) does not diverge at the
spatial infinity and in particular belongs to L2 space. The Poisson equation
is sometimes posed with a background (or doping profile):
∆P = u 2 b,
− | | −
wherebisagivenpositivefunction. Then,theneutralityconditionis u2
| | −
bdx = 0 or equivalently (u2 b)(0) = 0. When we consider the problem
F | | − R
in dimensions less than three, this condition is useful to control P. Notice
that this condition excludes all nontrivial solutions when b 0, and that
≡
we need to remove this condition for the study of (1.1). In [11], the above
1
2 S.MASAKI
assumptions are removed and the existence of a unique local solution is
proven for data in the usual Sobolev space Hs(R2) (s > 2) despite the fact
that the nonlinear potential diverges at the spatial infinity. Since (1.2) is
not necessarily defined for u Hs (s >2) we introduced a new formula
∈
1 x y
P = log | − | u(y)2dy
−2π R2 y | |
Z (cid:18) | | (cid:19)
which makes sense merely if u2 Lp(R2) (p (1,2)). We underline that
| | ∈ ∈
the local solutions given there do not have finite energy (the energy is given
in (1.5) below). Our aim in this paper is to prove that there exists a time-
global solution if initial data has finite energy.
For our analysis, the following reduction is crucial: We guess that the
Newtonian potential (1.2) may behave like 1 u 2 log x at the spatial
−2π k kL2 | |
infinity, which will be the bad part of the nonlinearity, and decompose the
nonlinearity as
λ λ x y
λPu= u 2 (log x )u u log | − | u(y)2dy,
−2π k kL2 h i − 2π R2 x | |
Z (cid:18) h i (cid:19)
where x = (1+ x 2)1/2. We then obtain
h i | |
1 λ λ x y
i∂ u+ ∆u+ u 2 (log x )u= u log | − | u(y)2dy.
t 2 2π k kL2 h i −2π R2 x | |
Z (cid:18) h i (cid:19)
ItwillturnoutthatthebadpartofP iscorrectlyextractedfromtheoriginal
nonlinearity and therefore the behavior of the “new nonlinearity” becomes
better. Notice that one can also expect that u is conserved because λ
k kL2
is a real number. Hence, putting
λ
m := u 2 ,
−2π k 0kL2
we reach to the equation
1 λ x y
i∂ u+ ∆ mlog x u = u log | − | u(y)2dy,
t
(1.3) 2 − h i −2π R2 x | |
 (cid:18) (cid:19) Z (cid:18) h i (cid:19)
u(0,x) = u0(x).
Noticethat mlog x isnowcompletely independentofuandthatitthere-
− h i
fore can be regarded as a linear potential. In what follows, we work with
this equation. Observe that if there exists a solution to (1.3) conserving
u , then it is also a solution of (1.1).
k kL2
Now, thelinear partoftheequation isnoti∂ +(1/2)∆ buti∂ +(1/2)∆
t t
−
mlog x . Thus, a natural choice of the function space on which we shall
work ihs niot the Sobolev space H1(R2) any more, but the following one:
:= u H1(R2); log x u L2 ,
H { ∈ h i ∈ }
(1.4)
u := u p+ log u .
k kH k kH1(R2) h·i L2(R2)
(cid:13)p (cid:13)
If m > 0, that is, if λ < 0, then the a(cid:13)bove space(cid:13)coincides with the form
(cid:13) (cid:13)
domain of the positive operator 1∆ +mlog x . Our main result is the
−2 h i
following:
ENERGY SOLUTION TO 2D-SP SYSTEM 3
Theorem 1.1. The problem (1.3) is globally well-posed in . Moreover,
H
the solution conserves u(t) and the energy
k kL2
1 λ
(1.5) E(t) = u(t) 2 (log x y )u(t,x)2 u(t,y)2dxdy.
2k∇ kL2 − 4π R2 | − | | | | |
Z
Corollary 1.2. The Problem (1.1) is globally well-posed in .
H
Remark 1.3. Let u C(R; ) be a solution of (1.3) (and of (1.1)) given in
∈ H
Theorem 1.1. Then, v := uexp( i λ t log u(s) 2 ds) solves
− 2π 0 k |·| kL2
1 λ R p x y
i∂ v+ ∆v = v log | − | v(y)2dy,
t
(1.6) 2 −2π R2 y | |
 Z (cid:18) | | (cid:19)
v(0,x) = u0(x).
Notice that the nonlinearity of (1.6) makes sense without the momentum
condition log v L2. This observation explains why existence of a
time-local soluti|o·n|ca∈n be proven by assuming only u Hs(R2) (s > 1) in
p 0 ∈
[11].
1.1. Consequent results. Our argument is also applicable to (1.1) involv-
ing a power type nonlinearity:
1
i∂ u+ ∆u= λPu+η up−1u, (t,x) R1+2,
t
2 | | ∈
(1.7)  ∆P = u2,
 − | |

u(0,x) = u (x),
0

where η is a real number and p >2.
Theorem 1.4. The problem (1.7) is globally well-posed in if either one
H
of the following conditions is satisfied:
(1) η > 0, λ R and p > 2;
(2) η < 0, λ ∈ R, and 2 6 p < 3;
∈
(3) η < 0, λ > 0, p = 3, and u is small;
k 0kH
(4) η < 0, λ < 0, p > 3, and u is small.
k 0kH
Moreover, the solution conserves u(t) and the energy
k kL2
1 λ
E (t):= u(t) 2 (log x y )u(t,x)2 u(t,y)2dxdy
(1.8) p 2k∇ kL2 − 4π ZR2 | − | | | | |
η
p+1
+ u(t) .
p+1k kLp+1
The proof is done with a straight-forward modification (see Section 4).
The case where p = 3 is known as the L2-critical case. Since the -norm
H
contains derivative, it seems difficult to treat the case 1 < p < 2. Never-
theless, we can show global well-posed in a slightly smaller function space
1,2 := u H1(R2);ulog x L2 .
H { ∈ h i∈ }
Theorem 1.5. Suppose 1 < p < 2. For η,λ R The problem (1.7) is glob-
∈
ally well-posed in the space 1,2. Moreover, the solution conserves u(t)
H k kL2
and the energy E (t) given in (1.8).
p
4 S.MASAKI
We can also handle the one-dimensional problem
1 λ
i∂ u+ ∂ u= (x u2)u+η up−1u, (t,x) R1+1,
t xx
(1.9) 2 −2 | |∗| | | | ∈

u(0,x) = u (x),
 0
where λ,η R and p > 2. The one dimensional problem was studied in
∈
[7, 14, 15]. The global well-posedness of (1.9) was shown in the space f
{ ∈
H1(R); x f L2(R) in [15], and in the space f H1(R); x f L2(R)
| | ∈ } { ∈ | | ∈ }
with a presence of background in [7], provided λ > 0 and data is small
p
relative to the background. We can prove the global well-posedness result
of (1.9) including these results.
Theorem1.6. Theproblem (1.9)isgloballywell-posedin f H1(R); x f
L2(R) if λ R and either one of the following condition{s is∈satisfied: | | ∈
} ∈ p
(1) η > 0, λ R, and p > 2;
(2) η < 0, λ ∈ R, and 2 6 p < 5;
∈
(3) η < 0, λ > 0, p = 5, and u + u is small;
k 0kH1 k |·| 0kL2
(4) η < 0, λ < 0, p > 5, and u + u is small.
k 0kH1 kp|·| 0kL2
The solution conserves u and the energy
k kL2 p
(1.10)
1 λ η
E(t) := ∂ u 2 x y u(x)2 u(y)2dxdy+ u p+1 .
2 k x kL2(R)− 2 R2| − || | | | p+1k kLp+1(R)
ZZ
e Theone-dimensionalversionofTheorem1.5isasfollows,whichreproduce
the same result in [15, Theorem 2.1] when η < 0 and λ > 0.
Theorem 1.7. Suppose 1 < p < 2. For η,λ R The problem (1.9) is
globally well-posed in the space Σ := u H1(R∈2); x u L2 . Moreover,
{ ∈ | | ∈ }
the solution conserves u(t) and the energy E(t) given in (1.10).
k kL2
As in the two dimensional case, the key is a “reduction” of (1.9) to
e
1 λ u 2 λ
i∂ u+ ∂ u+ k 0kL2 x u = u (x y x ) u(y)2dy+η up−1u,
t xx
 2 2 | | −2 R | − |−| | | | | |
Z
u(0,x) = u0(x).
We briefly mention about other related works. Oh considered in [12]

the Cauchy problem of the nonlinear Schro¨dinger equation with general
potentialandL2-subcriticalpower-typenonlinearity, andprovedglobalwell-
posedness in the form domain of 1∆+V, provided the potential V > 0
−2
satisfies ∂αV L∞ for α > 2 (see also [6]). In particular, the case where
∈ | |
thepotentialV isaquadraticpolynomialisextensivelystudied. Inthiscase,
we have several special properties such as explicit representations of linear
solutions,calledMehler’sformula,and/oroftheHeisenbergobservables. We
refer the reader to [2, 3, 4, 10, 19] for H1-subcritical and H1-critical power-
type nonlinearity and to [5] for H1-subcritical Hartree type nonlinearity. In
[16], the ground states of (1.1) is treated.
The rest of the paper is organized as follows: We collect some basic es-
timates in Section 2, and, in Section 3 we prove Theorem 1.1. Section 4 is
devoted to the study of (1.7).
ENERGY SOLUTION TO 2D-SP SYSTEM 5
2. Preliminaries
2.1. Strichartz estimate. We first summarize the properties on the oper-
ator
1
(2.1) A:= ∆ mlog x ,
2 − h i
where m = 0 is a real constant. For any m, A is essentially self-adjoint
on C∞(R26 ) (see [13]). Since our potential is sub-quadratic, that is, since
0
∂αlog x 0 as x for α = 2 and ∂αlog x L∞ for α > 3, the
| h i|→ | | → ∞ | | h i ∈ | |
following estimate is established in [17]: For any T > 0,
eitAϕ 6C t −1 ϕ
L∞ | | k kL1
for t [ T,T], where C(cid:13)depen(cid:13)ds on T (see also [8]). Once we know this
∈ − (cid:13) (cid:13)
type of estimate, the Strichartz estimate follows by interpolation. We say
that a pair (q,r) is admissible if 26 r < and 2/q = δ(r) := 1 2/r.
∞ −
Lemma 2.1 (Strichartz’sestimate). For any T > 0, the following properties
hold:
Suppose ϕ L2(R2). For any admissible pair (q,r), there exists a
• ∈
constant C = C(T,q,r) such that
eitAϕ 6 C ϕ .
Lq((−T,T);Lr) k kL2
(cid:13) (cid:13)
Let I ( T,T(cid:13)) be a(cid:13)n interval and t0 I. For any admissible pairs
• ⊂ − ∈
(q,r) and (γ,ρ), there exists a constant C = C(t,q,r,γ,ρ) such that
t
ei(t−s)AF(s)ds 6 CkFkLγ′(I;Lρ′)
(cid:13)Zt0 (cid:13)Lq(I;Lr)
(cid:13) (cid:13)
for ever(cid:13)(cid:13)y F Lγ′(I;Lρ′). (cid:13)(cid:13)
∈
2.2. Some estiamtes.
Lemma 2.2. Let W be an arbitrary weight function such that W, ∆W
L∞(R2). It holds for all T > 0, admissible pair (q,r), and ϕ ∇ that ∈
∈ H
[ ,eitA]ϕ 6 C T ϕ ,
∇ Lq((−T,T);Lr) | |k k2
(cid:13)[W,eitA]ϕ(cid:13) 6 C T (1+ )ϕ .
(cid:13) (cid:13)Lq((−T,T);Lr) | |k ∇ k2
(cid:13) (cid:13)
(cid:13) (cid:13)
Proof. Since v = eitAϕ solves i∂ v+Av = 0, an explicit calculation shows
t
t mx
[ ,eitA]ϕ = i ei(t−s)A eisAϕds
∇ − 1+x2
Z0
and
t 1
[W,eitA]ϕ = i ei(t−s)A W + ∆W eisAϕds.
∇ ·∇ 2
Z0 (cid:18) (cid:19)
The Strichartz estimate therefore gives the desired estimates. (cid:3)
The following is useful for estimates of the nonlinearity in (1.3).
6 S.MASAKI
Lemma 2.3. Set a function
|x−y|
log
hxi
K(x,y)=
1+log y
h i
of x,y R2. For any p [1, ) and ε > 0, there exist a function W(x,y) >
∈ ∈ ∞
0 with kWkL∞y Lpx 6 ε and a constant C0 such that
K(x,y) 6 C +W(x,y)
0
| |
holds for all (x,y) R2+2.
∈
Proof. Take η (0,1] and set W(x,y) = K(x,y)1 (x,y). If η is
|x−y|6η
∈ | |
sufficiently small then
log x +log y +η 1
W(,y) 6 k | |kLp(|x|6η) h| | ik kLp(|x|6η) 6 ε
k · kLp 1+log y
h i
since log x belongs to Lp (R2) for all p < . Moreover, by (2.12) of [11],
| | loc ∞
√3
sup K(x,y) 6 1+log
η
|x−y|>η
for any η 6 1, which completes the proof. (cid:3)
Remark 2.4. In 1D case, the corresponding estimate is
x y x
| − |−| | 6 1.
1+ y
(cid:13) | | (cid:13)L∞x,y(R2)
(cid:13) (cid:13)
(cid:13) (cid:13)
3.(cid:13)Proof of t(cid:13)he theorem
3.1. Local well-posedness.
Lemma 3.1. Let (q ,r ) be an admissible pair with r > 2. For any u ,
0 0 0 0
∈ H
there exist an existence time T = T( u ) and a unique solution u
k 0kH ∈
C(( T,T); ) Lq0(( T,T);Lr0) C1(( T,T); ∗). The solution conserves
− H ∩ − ∩ − H
u(t) andtheenergy (1.5). Moreover, themapu uiscontinuousfrom
k kL2 0 7→
to C(( T,T); ).
H − H
Proof. We write Lp(( T,T);X) = LpX, for short. Define a Banach space
− T
:= f L∞(( T,T); ); f 6 M
HT,M { ∈ − H k kHT }
with norm
kfkHT := kfkL∞T H+kfkLqT0W1,r0 + loghxif Lq0Lr0 .
T
(cid:13)p (cid:13)
We show that if r > 2 then there exist M = M(cid:13) ( u ) a(cid:13)nd T = T( u )
0 (cid:13) k 0kH (cid:13) k 0kH
such that
Q[u](t,x) := (eitAu )(x)
0
i t y
+ ei(t−s)A log |·− | u(s,y)2dy u(s, )ds (x)
2π (cid:18)Z0 (cid:18)ZR2 h·i | | (cid:19) · (cid:19)
becomes a contraction map from to itself, where A is defined in (2.1).
T,M
H
ENERGY SOLUTION TO 2D-SP SYSTEM 7
Set
|x−y|
log
hxi
K(x,y) = .
1+log y
h i
Then, by Lemma 2.3, there exist a nonnegative function W L∞Lr0′ and a
∈ y x
constant C such that
0
K(x,y) 6 C +W(x,y).
0
| |
Recall that r (2, ) and so r′ := r /(r 1) (1,2). We hence see that
0 ∈ ∞ 0 0 0− ∈
Pu= K(x,y)(1+log y )u(y)2u(x)dydx
h i | |
ZZ
satisfies
2
Pu 6 C( u + u ) 1+log x u .
k kL2 k kL2 k kLr0 h i L2
Take L1 norm to yield (cid:13)(cid:13)p (cid:13)(cid:13)
T (cid:13) (cid:13)
(3.1)
kPukL1TL2 6 C(T kukL∞T L2 +T21+r10 kukLqT0Lr0) 1+loghxiu 2L∞L2.
T
(cid:13)p (cid:13)
By the Strichartz estimate, we end up with (cid:13) (cid:13)
(cid:13) (cid:13)
(3.2) kQ[u]kL∞T L2 +kQ[u]kLqT0Lr0 6 Cku0kL2 +C(T +T21+r10)kuk3HT .
We next estimate Q[u]. One easily sees that
∇
t
Q[u] = eitA u i ei(t−s)A (Pu)(s)ds
0
∇ ∇ − ∇
Z0
t
+[ ,eitA]u i [ ,ei(t−s)A](Pu)(s)ds.
0
∇ − ∇
Z0
We deduce from Lemma 2.2 with (q,r) = ( ,2) that
∞
t t
[ ,ei(t−s)A](Pu)(s) ds 6 (t s) Pu(s) ds 6 t Pu .
Z0 (cid:13)∇ (cid:13)L2 Z0 − k kL2 | |k kL1TL2
The r(cid:13)ight hand side is bo(cid:13)unded as in (3.1). [ ,eitA]u is handled similarly.
(cid:13) (cid:13) 0
∇
Mimicking (3.1), we infer that
(3.3)
kP∇ukL1TL2 6 C(T k∇ukL∞T L2 +T12+r10 k∇ukLqT0Lr0) 1+loghxiu 2L∞L2
T
(cid:13)p (cid:13)
Now, let us estimate ( P)u. It writes (cid:13) (cid:13)
(cid:13) (cid:13)
∇
x y x
( P)(x)u(x) = − u(y)2dy u(x),
∇ R2 x y 2 − 1+x2 | |
(cid:18)Z (cid:18)| − | (cid:19) (cid:19)
and so
( P)u 6 C (x −1 u2)+ −1 u 2 u
k ∇ kL2 | | ∗| | h·i k kL2 Lr20r−02 k kLr0
(cid:13) (cid:13)
6 C((cid:13)u 2 + u 2 ) u (cid:13)
(cid:13)k kLr20r−01 k kL2 k kLr0 (cid:13)
6 C( u 2 + u 2 ) u
k kL2 k∇ kL2 k kLr0
8 S.MASAKI
by the Hardy-Littlewood-Sobolev and the Sobolev inequalities. We see that
(3.4) k(∇P)ukL1TL2 6 C(T kuk2L∞T L2 +T21+r10 kuk2LqT0Lr0)kukL∞T L2.
We deduce from the Strichartz estimate that
(3.5) k∇Q[u]kL∞T L2+k∇Q[u]kLqT0Lr0 6 Ck∇u0kH+C(T+T12+r10)kuk3HT .
Let us proceed to the estimate of log x Q[u]. It holds that
h i
p t
1+log x Q[u]= eitA 1+log x u i ei(t−s)A 1+log x Pu(s)ds
0
h i h i − h i
Z0
p +Rp, p
where
t
R = [ 1+log x ,eitA]u i [ 1+log x ,ei(t−s)A]Pu(s)ds.
0
h i − h i
Z0
p p
A use of Lemma 2.2 with W = 1+log x yields
h i
kRkL∞T L2 +kRkLqT0Lr0 6pCT ku0kH+CT k(1+∇)(Pu)kL1TL2
6 CT ku0kH+CT(T +T21+r10)kuk3HT
where we have used (3.1), (3.3), and (3.4). As in (3.1), it holds that
kP(Wu)kL1TL2 6 C(T kWukL∞T L2 +T21+r10 kWukLqT0Lr0)kWuk2L∞T L2
6 C(T +T21+r10) u 3 ,
k kHT
where W = 1+log x . We conclude from the Strichartz estimate, (3.2),
h i
and (3.5) that
p
kQ[u]kHT 6 C1ku0kH+C2(T +T21+r10)kuk3HT .
A similar argument shows
kQ[u1]−Q[u2]kHT 6 C3(T +T12+r10)(ku1kHT +ku2kHT)2ku1−u2kHT .
Thus, if we take M > 2C u then there exists T = T(M) such that Q
1k 0kH
is a contraction map from to itself.
T,M
H
The conservations of u(t) is shown by multiplying (1.3) by u and
k kL2
integrating the imaginary part. To prove the energy conservation, we need
a regularizing argument. Note that (1.3) can be solved also in the space
f H2(R2) : log x f L2 , which is one of dense subsets of , in an
{ ∈ h i ∈ } H
essentially same way. We omit details. (cid:3)
3.2. Global existence. We first give a useful blow-up criteria.
Lemma 3.2. Suppose u . Let u C(( T ,T ); ) be a unique
0 min max
∈ H ∈ − H
maximal solution given by Lemma 3.1. If T < (resp. T < ), then
max min
∞ ∞
u(t) as t T (resp. t T ).
k∇ kL2 → ∞ ↑ max ↓ − min
ENERGY SOLUTION TO 2D-SP SYSTEM 9
Proof. We only consider positive time. Suppose T < . Then, u(t)
max ∞ k kH
has to diverge as t T . Otherwise, we can extend the solution beyond
max
↑
T by Lemma 3.1. Recall that u(t) = u . Since
max k kL2 k 0kL2
d 2
log x u(t) = 2Re (log x )∂ u(t)u(t)dx
t
dt h i L2 h i
(cid:13)p (cid:13) Z
(cid:13) (cid:13)
(cid:13) (cid:13) = Im (log x )∆u(t)u(t)dx
− h i
Z
x
= Im u(t)u(t)dx,
1+x2 ·∇
Z
it holds that
2 2
log x u(t ) 6 log x u(t ) + t t u u
h i 2 L2 h i 1 L2 | 2− 1|k∇ kL∞((t1,t2);L2)k 0kL2
f(cid:13)(cid:13)opr all Tmin <(cid:13)(cid:13)t1 < t(cid:13)(cid:13)2p< Tmax. This(cid:13)(cid:13)implies that if we assume
(cid:13) − (cid:13) (cid:13) (cid:13)
limsup u(t) <
k∇ kL2 ∞
t↑Tmax
then u(t) never blows up. We hence obtain the lemma. (cid:3)
k kH
Remark 3.3. As in [9], the solution breaks down with concentration at a
point if log x u(t) = 0. However, this does not occur when u(t)
L2
k h i k k∇ k
is bounded above. Indeed, since
p
1 1
kukL2(|x|<r) 6 k1kL4(|x|<r)kukL4 6 Cr2 k∇ukL22
for any r > 0 and since
log x u
u = u u > u k h i kL2,
k kL2(|x|<r) k 0kL2 −k kL2(|x|>r) k 0kL2 − (log r )1/2
p
h i
by letting r = log x u , we obtain
L2
k h i k
1
p 2
log x u −12 6 C k loghxiukL2 +C u 12 ,
k h i kL2   k∇ kL2
log p log x u L2
p k h i k
 D p E
which implies log x u is strictly positive if u < .
k h i kL2 k∇ kL2 ∞
Proof of Theorepm 1.1. Let us establish a priori estimate of u(t) .
k∇ kL2
We first consider the case λ < 0. Since log x > 0 for x > 1,
| | | |
λ
log x y u(x)2 u(y)2dxdy
− 4π R2+2 | − || | | |
ZZ
λ
> | | log x y u(x)2 u(y)2dxdy
−4π | | − ||| | | |
ZZ|x−y|<1
λ
> | | log x 2 u 2 u 2
−4π k | |kL2(|x|61)k kL4k kL2
By the L2-conservation and the Sobolev embedding, we have
(3.6) u(t) 2 6 2E +C u(t) .
k∇ kL2 0 k∇ kL2
Therefore,thereexistsaconstantM independentoftsuchthat u(t) 6
k∇ kL2
M.
10 S.MASAKI
We now suppose λ > 0. By Lemma 2.3, for any ε > 0 there exists a
constant C such that the following estimate holds:
0
λ
log x y u(x)2 u(y)2dxdy
4π R2+2 | − || | | |
ZZ
λ x y λ 2
6 log | − | u(x)2 u(y)2dxdy+ u 2 log x u
4π R2+2 x | | | | 4π k 0kL2 h i L2
λ ZZ h i 2 λ (cid:13)(cid:13)p (cid:13)(cid:13) 2
6 (C u 2 +ε u 2 ) 1+log x u + u(cid:13)2 log(cid:13)x u
4π 0k 0kL2 k kL4 h i L2 4π k 0kL2 h i L2
λC λ(C +1) (cid:13)p (cid:13) 2 (cid:13)p (cid:13)
6 4π0 ku0k4L2 + 40π (cid:13)(cid:13)ku0k2L2 loghx(cid:13)(cid:13)iu L2 +Cεku0k(cid:13)(cid:13)3L2k∇ukL2 (cid:13)(cid:13)
(cid:13)2p (cid:13)
+Cε u u log x u(cid:13) (cid:13)
k 0kL2k∇ kL2 h i (cid:13)L2 (cid:13)
6 C1+C2(ε+|t|) sup(cid:13)(cid:13)(cid:13)kp∇u(s)kL2(cid:13)(cid:13)(cid:13)+C3ε|t| sup k∇u(s)k2L2,
s∈[0,t] s∈[0,t]
where C (i = 1,2,3) depends only on λ, C , u , and ε. Fix T > 0.
i 0 k 0kH
Taking ε < (8C T)−1, we deduce from the conservation of E(t) that
3
2
(3.7) sup u(s) 6 4E(0)+4C +4C (ε+2T) sup u(s)
k∇ kL2 1 2 k∇ kL2
s∈[0,t] ! s∈[0,t]
for 0 6 t 6 2T. This implies that
sup u(t) 6 C( u ,T) < .
k∇ kL2 k 0kH ∞
t∈[0,2T]
Since T is arbitrary, we obtain the global existence. (cid:3)
4. Remarks on the problem with power nonlinearity
WegivearoughsketchoftheproofsofTheorem1.4and1.5inthissection.
Proof of Theorem 1.4. Thelocal well-posedness partholds if p > 2 as in the
proof of Lemma 3.1. The restriction p > 2 is required when we estimate
(u p−1u u p−1u ) 6 C (u p−2+ u p−2)( u + u )u u
1 1 2 2 p 1 2 1 2 1 2
|∇ | | −| | | | | | | |∇ | |∇ | | − |
+C (u p−1+ u p−1) (u u ).
p 1 2 1 2
| | | | |∇ − |
By exactly the same argument as in Lemma 3.2, the problem of global
existence boils down to obtaining an a priori bound of u(t) . Recall
k∇ kL2
that the conserved energy is
1 λ
E (t):= u(t) 2 (log x y )u(t,x)2 u(t,y)2dxdy
p 2k∇ kL2 − 4π R2 | − | | | | |
Z
η
p+1
+ u(t) .
p+1k kLp+1
The case η > 0. We have
λ
u(t) 2 6 E (t)+ (log x y )u(t,x)2 u(t,y)2dxdy.
k∇ kL2 p 4π R2 | − | | | | |
Z
By the same argument as in the case η = 0, we prove global existence.