Vol. 4, No. 2 (2007) 1–25 c
World Scientific Publishing Company 3
ON THE GLOBAL EXISTENCE OF ROUGH SOLUTIONS OF THE
5
CUBIC DEFOCUSING SCHR ¨ODINGER EQUATION IN R2+1
Y. F. FANG
7
Department of Mathematics, Cheng Kung University Tainan, 701 Taiwan
MANOUSSOS G. GRILLAKIS
9
Department of Mathematics, University of Maryland College Park, Md 20742, USA mng@math.umd.edu 11 Received 23 May 2006 Accepted 10 Oct. 2006 13 Communicated by P. G. LeFloch
Abstract. We consider the cubic defocusing Schr¨odinger equation in two space
dimen-15
sions and prove that if the initial data are inH1/2, then there exists a global solution in time. The proof combines the argument from [5] with some new correlation estimates
17
for the Schr¨odinger equation.
Keywords: I-method; Strichartz estimates; correlation estimates; conservation laws. 19
Mathematics Subject Classification 2000: 35L40, 76N10
1. Introduction
21
The purpose of the present paper is to demonstrate global existence of rough solutions for the cubic defocusing Schr¨odinger equation in 2 + 1 dimensions. To make the statement more precise, we would like to consider the following evolution equation
i∂tψ− ∆ψ + |ψ|2ψ = 0, (1.1a)
ψ(0, x) = ψ0(x); (t, x)∈ R × R2, (1.1b) i.e. x∈ R2are the space variables and t denotes the time variable. For the evolution
Eq. (1.1), we would like to assume that ψ0∈ H1/2(R2) and show that there exists
23
a global in time solution with ψ(t) ∈ H1/2(R2). Certain conserved quantities are
crucial to the behaviour of the solutions. The conservation of energy means that
25 the integral E(t)def= R2 ∇ψ2 +1 2|ψ| 4 dx (1.2a) 27 1
is independent of time. On the other hand, the evolution equation preserves the
1
total mass which means that the integral
M (t)def=
R2
ψ2dx (1.2b)
3
is also independent of time. It is rather easy to demonstrate, using the two conserved quantities above, that a solution of (1.1) exists for all time provided that the initial data are sufficiently smooth, namely in H1. The reason for assuming ψ
0 ∈ H1 is
that one would like to take advantage of the conservation of energy, see the intergal in (1.2a), which is positive definite and which implies that the H1 norm of the solution is bounded as long as the solution exists. In order to explain the type of result that we would like to demonstrate in the present work, we have to explain first why the Eq. (1.1) admit local weak solutions if we only assume that the initial data are in L2. To see why this is so, we have to use the Strichartz estimates, more
precisely consider the linear Schr¨odinger equation with a forcing term f ,
iψt− ∆ψ = f, (1.3a)
ψ(0, x) = ψ0(x). (1.3b)
The solution of (1.3) can be written in integral form as a combination of two terms, namely ψ(t) = ψL(t) + φ(t) where
5
ψL(t)def= e−i∆tψ0; φ(t)def= t
0
e−i∆(t−s)f (s)ds. (1.4) The local in time construction of solutions is based on the well known Strichartz
7
estimates
ψLL4(R2+1)≤ C1ψ0L2(R2); φL4(R2+1)≤ C2fL4/3(R2+1). (1.5)
9
Pick a time interval, say [0, T ], with the final time T to be chosen and set up an iteration scheme as follows
11
iψt(k+1)− ∆ψ(k+1)=−χ[0,T ](t)|ψ(k)|2ψ(k); ψ(k+1)(0, x) = ψ0(x) (1.6) with ψ(0) satisfying i∂tψ(0)− ∆ψ(0)= 0, ψ(0)(0) = ψ
0and χ[0,T ] the characteristic 13
function of the interval [0, T ]. The Strichartz estimate in (1.5) ψLL4(R2+1) ≤
C1ψ0L2(R2) implies that there exists some time interval, say [0, T ], such that
15
ψLL4(R2×[0,2T ])≤ , (1.7)
where is a fixed small number which will depend only on C2, see (1.5). Let us define a local in time norm,
Xk=defψ(k)
L4(R2×[0,T ]).
Applying the estimates (1.5), we obtain an inequality of the form Xk+1≤ + CX3
k,
1
from which it follows that Xk is a bounded sequence. By considering the evolution equation for differences ψ(k+1)− ψ(k), one can show that the iteration is in fact a
3
contraction with respect to the L4norm, thus constructing a local solution of (1.1)
in the time interval [0, T ]. A crucial observation is the fact that T is defined via (1.7)
5
and we do not have any control of its size. Repeating the construction starting with data at time T , we can construct a solution on successive time intervals, but there is
7
no guarantee that the intervals will not shrink rapidly thus constructing a solution only on a finite time interval. On the other hand, if we use the conservation of energy
9
this cannot happen, since the energy estimate controls the H1norm of the solution, but we need to assume that ψ0∈ H1. It is an interesting question to ask whether 11
equation (1.1) has global in time solutions if one assumes that ψ0∈ Hsfor 0≤ s < 1. This question was first posed by Bourgain, see [1, 2], in which he demonstrated
13
global existence for s > 3/5. The range of s was subsequently improved in [5, 6] to s > 4/7 using the crucial idea of an almost conservation law for the energy. In
15
the present work we would like to demonstrate global existence for s = 1/2. There are two key ingredients needed in the proof, the first is the almost conservation law
17
and the second is a correlation estimate. The paper is divided as follows, in Sec. 2, we present the main argument leading to global existence, we essentially follow the
19
argument in [6]. In Secs. 3 and 4, we derive a bilinear estimate due to Bourgain and the almost conservation law developed in [5, 6]. Section 5 presents a method
21
for deriving correlation estimates. Recently Coliander et al. showed existence for
s > 1/2 by improving the almost conservation law. 23
2. Main Argument
Presently we would like to give the main argument of the proof of global existence for the following problem
i∂tψ− ∆ψ + |ψ|2ψ = 0, (2.1a)
ψ(0, x) := ψ0(x)∈ H1/2(R2). (2.1b) We will follow the technique used in [5, 6] which is based on the multiplier I and
25
the idea of an almost conservation law. Before we start let us specify some notation conventions. We would like to define the Fourier multiplier ms,N(ξ) to be a smooth
27
function in frequency variables which is less or equal than one and has the following properties 29 ms,N(ξ)def= 1, if |ξ| ≤ N, (N/|ξ|)1−s, if |ξ| ≥ 2N, (2.2)
for 0 < s < 1, however in the present work we will choose s = 1/2. We will use the multiplier above to define Iψ, which whenever convenient we will also denote by
ψI, as follows
ψI(t, x) = Iψ(t, x) =defF−1m
We will also use the following spaces introduced by Bourgain, see [1, 2]. 1 ψXs,α def= ξs Sαψ L2, (2.4a)
where we adopt the notation convention
3
ξdef= |ξ| + 1; S def= τ− |ξ|2; Sdef= | S| + 1. (2.4b) When we consider localization in time, we will use the notation Xs,αI or simply write
5
Xs,αloc, where I stands for some time interval [T1, T2]. This means that we consider
ψXI s,α def = inf e ψ ψXs,α : ψ(t) = ψ(t), t∈ I. (2.5) 7
The s-derivative of a function is defined in the standard way
Dsψ =F−1
|ξ|sf. (2.6)
9
The energy of the function ψI is defined as follows, see the definition of energy in (1.2a), 11 E(Iψ)(t)def= R2 t ∇ψ I2+1 2ψI 4 dx. (2.7)
Finally, we will employ the Strichartz estimates t 0 e−i(t−s)∆f (s)ds Lp x,Lrt ≤ C1 f Lp x,Lrt , (2.8a) e−it∆ψ0 Lp x,Lrt ≤ C0 ψ0 L2 (R2), (2.8b)
where the exponents p and r satisfy the admissibility condition, see [5, 6],
13 1 p+ 1 r = 1 2; r > 2. (2.8c)
Let us list here the ingredients that are essential for the demonstration of global
15
existence. The first observation is the fact that if ψ(t, x) is a solution of (2.1), we can scale it and obtain a new solution, namely the scaled function,
17 ψ(λ)(t, x)def= 1 λψ t λ2, x λ (2.9)
satisfies the same equation with initial data ψ(λ)0 = (1/λ)ψ0(x/λ). This scaling
19
preserves the L2 norm of ψ(t) as well as the L4 space-time norm. Now let us scale
the solution ψ→ ψ(λ) so that E(Iψ0(λ))≤ 1/4, more precisely under the scaling in
21 (2.9), we have ∇Iψ(λ) 0 2L2≈ N λψ0 2 H1/2 and Iψ0(λ)4L4≈ 1 λ2ψ0 4 L4. (2.10) 23
Notice that from the Sobolev inequality we haveψL4 ≤ CψH1/2. With all the
1
above in mind, we can choose λ in the following manner
λ = CNψ02H1/2, (2.11)
3
where C some fixed large positive number so that the scaled function ψ0(λ)satisfies,
E(Iψ0(λ)) < 1/4. (2.12)
5
We will choose N at the end of the argument to be sufficiently large.
The almost conservation law is the next essential ingredient, see [5, 6] where this
7
idea was introduced. The idea is that the energy of ψI is almost conserved in the following sense 9 E(Iψ)(T )−E(Iψ)(0) ≤ 1 N(3−δ)/2 C1Iψ4 X1,(1+δ)/2[0,T ] + C2Iψ6X1,(1+δ)/2[0,T ] . (2.13) The third crucial ingredient is an a priori estimate which reads as follows. A solution
11
of (2.1) satisfies the a priori estimate
ψ4 L4(R2×[T1,T2])≤ ψ02L2 sup [T1,T2] ψ(t)2 H1/2 T2− T1. (2.14) 13
This type of estimate for three space dimensions was derived first in [6, 7]. A general method for developing this type of apriori estimates will be given in Sec. 5. In order
15
to present the proof of global existence, we will need a series of technical lemmata. The first lemma that we will need is the following, see [3].
17
Lemma 2.1. Assume that q(ξ1, ξ2), with ξ1,2∈ R2, is a smooth function satisfying |∂αq| ≤ C(α)(1 + |ξ|)−|α|; ξ = (ξ
1, ξ2) (2.15)
19
for every multiindex α and define a quadratic expression as follows Q[f, g](x) =F−1 q(ξ− η, η) f (ξ− η)g(η)dη . (2.16) 21
The following estimate holds
Q[f, g]Lp≤ CfLp1gLp2, where 1 p1 + 1 p2 = 1 p. (2.17) 23
This lemma appears in [3] and we will omit its proof. We will use Lemma 2.1 above to claim that as far as the Lpnorm is concerned we can think that DI(ψ3)∼ 25
(ψ2)(DIψ). In order to apply Lemma 2.1 we need to assume that the multiplier I has smooth symbol which can be achieved by considering a smooth approximation
27
Theorem 2.2. A pair of exponents is called admissible if it satisfies 1 1 p+ 1 r = 1 2; r > 2 (2.18a)
and with this choice of exponents the following Strichartz type estimates holds true 3
ψLp
x,Lrt ≤ C(δ)ψX0,(1+δ)/2. (2.18b)
These estimates are well known, see Bourgain [1, 2], also see next section. Next
5
we will need a lemma that says that if the space-time L4 norm of ψ is small on
some interval [0, T ], we can control all higher order norms.
7
Lemma 2.3 (Local existence withHs(R2) data, wheres ≥ 0). Let us define
the quantity, 9 µ([0, t])def= D[0,t] ψ4 dxdt. (2.19a)
If µ([0, T ])≤ µ0, where µ0 is some universal constant then the following estimates are true, Dsψ Lp xLrt(D[0,T ])≤ C Dsψ0 L2, (2.19b) DIψ L4(D[0,T ])≤ C DIψ0 L2. (2.19c)
Proof. The proof imitates the argument for local existence given in the
introduc-11
tion, the function Dsψ satisfies the equation,
i∂tDsψ− ∆Dsψ=−Dsχ[0,T ](t)|ψ|2ψ (2.20)
13
in the interval [0, T ], where χ[0,T ] denotes the characteristic function of the time interval [0, T ]. Let us take p = r = 4 for simplicity. Let us write D := R2× [0, T ] for
15
simplicity. Applying Strichartz estimates as stated in (2.8a) and (2.8b), we obtain Dsψ L4 (D) ≤ C1 D sψ 0 L2+ C2 Ds |ψ|2ψ L4/3(D). (2.21) 17
Because of Lemma 2.1, we can write Ds(|ψ|2ψ)∼ ψ2(Dsψ) and using Holder on
the right-hand side of (2.21) with exponents 4 and 4/3, we obtain
19 Dsψ L4(D)≤ C1 Dsψ0 L2+ C2 µ1/2([0, T ]) Dsψ L4(D). (2.22)
We need to assume C22µ([0, T ])≤ 1/4 in order to complete the proof of the lemma. 21
The proof of (2.19b) is similar so we will omit it. Lemma 2.4. Assume D1/2ψ
0 L2 ≤ B where B is some fixed constant, and
23
µ([0, T ]) ≤ µ1(B) where µ1(B) is a constant that depends only on B, then we
have the estimate 25
Iψ X[0,T ]
1,(1+δ)/2 ≤ C
DIψ0 L2
Proof. Let us consider the evolution equation below
1
i∂tDIψ1− ∆DIψ1=−DIχ[0,T ](t)|ψ|2ψ, (2.24) so that DIψ(t) = DIψ1(t) in the interval [0, T ]. Again we will denote D := R2×
[0, T ]. For Eq. (2.24) the following estimate holds true for any δ > 0, see [1, 2], S(1+δ)/2DIψ1 L2
(D)≤ C1 DIψ0 L2+ C(δ) S
−(1−δ)/2DIχ
[0,T ]|ψ|2ψ L2(D),
(2.25) where C(δ)∼ δ−1. Employing Lemma 2.1 again, we have that DI(ψ3)∼ ψ2(DIψ).
3
Interpolating between the Strichartz type estimate ψ L4 t,x ≤ C(δ) ψ X0,(1+δ)/2 (2.26) 5
and the Plancherel identity,ψL2= ψL2, we obtain the estimate
ψ Lp t,x ≤ C(δ) ψ X 0,(1−δ)/2, where p = 4 1 + θ; θ = 2δ 1 + δ. (2.27) 7
The dual to the estimate above reads
S−(1−δ)/2ψ L2 ≤ C(δ)ψLp; p=
4
3− θ (2.28)
9
and applying (2.28) on the last term of (2.25), we obtain
S−(1−δ)/2DIχ[0,T ]|ψ|2ψ L2 ≤ C(δ) χ[0,T ]ψ2(DIψ) Lp. (2.29)
11
Now an application of Holder’s inequality with exponents, 1 3− θ+ 2(1− θ) 3− θ + θ 3− θ = 1, (2.30) 13
where the first exponents is applied to DIψ, the second to|ψ|8(1−θ)/(3−θ) and the
third to|ψ|8θ/(3−θ) gives
15
ψ2(DIψ) Lp ≤ C(δ)ψ(1−θ)/2L4(D) ψ2θL8(D)DIψL4(D). (2.31)
The Strichartz estimate, see Lemma 2.3, gives
17
D1/2ψ L8/3
x L8t(D)≤ C
D1/2ψ0 L2 (2.32)
provided that µ[0, T ]≤ µ0 while the Sobolev embedding W1/2,8/3⊂ L8 combined
with (2.32) in (2.25) gives Iψ X[0,T ] 1,(1+δ)/2 ≤ C1 DIψ0 L2+ C(δ)D1/2ψ02θL2 µ(1−θ)/2([0, T ])IψX[0,T ] 1,(1+δ)/2. (2.33) Let us choose δ small but fixed, for example δ = 2−10. From the inequality (2.33),
19
we conclude that
IψX[0,T ]
1,(1+δ)/2 ≤ CDIψ0L2 (2.34a)
provided that
1
C(δ)D1/2ψ02θL2
µ(1−θ)/2([0, T ])< 1/2, (2.34b) which means that we would like to have µ([0, T ]) smaller than some fixed number,
3
say µ1(B) since we assumed that D1/2ψ0L2 < B. This concludes the proof of
Lemma 2.4.
5
Finally the almost conservation law, see (2.13) combined with Lemma 2.4 and the scaling gives an estimate for the energy of Iψ(λ). First recall that with the 7
scaling in (2.11), we have
D1/2ψ(λ)0 L2 = (1/√λ) D1/2ψ0 L2 ∼ (1/CN1/2) < 1 (2.35a)
9
and λ was chosen so that E(Iψ(λ)0 ) < 1/4, so we have E(Iψ(λ))(T )− E(Iψ(λ))(0) ≤ C N(3−δ)/2DIψ (λ) 0 4L2 ≤ C N(3−δ)/2, (2.35b) 11
provided that µ[0, T ] < µ1(B). We are now in a position to demonstrate the global in time existence of (2.1), again we will follow the argument in [6].
13
Proof of global existence. Start with data ψ0 ∈ Cc∞ so that ψ(t, x) is a global solution of our problem. Now scale the solution ψ(λ) so that E(Iψ(λ)
0 ) ≤ 1/4, i.e. 15
the scaling parameter λ is chosen according to (2.11). The multiplier I depends on
N , see (2.2), and we will choose N at the very end of the argument to be sufficiently 17
large. For the time being let us pick T0arbitrarily large denote D[0, t] := R2× [0, t]
and define the following set
19
Sdef= t : 0 < t≤ T0 andψ(λ)L4(D[0,t])≤ At1/8 (2.36)
with A a constant to be chosen as follows
21
Adef= Kψ03/4L2 + 1
, (2.37)
where K is some large number to be chosen later.
23
Claim 2.5. We claim that S = [0, T0], i.e. the L4 space-time norm on the slice R2× [0, t] is bounded by some constant times t1/8.
25
Proof of Claim 2.5: Assume not, sinceψ(λ)
L4(D[0,t]) is a continuous function,
there exist some T ∈ [0, T0] with the properties, D[0, T ] := R2× [0, T ], 27
ψ(λ)
From the correlation estimate (2.14), we know that 1 ψ(λ) L4(D[0,T )])≤ ψ01/2L2 (R2) sup [0,T ] ψ(λ)(t)1/2 H1/2(R2) T1/8, (2.39)
and our next goal is to estimate, using the almost conservation law, the quantity
3 below sup [0,T ] ψ(λ)(t)1/2 H1/2(R2). (2.40) 5
Let us decompose ψ(λ) in high and low frequencies, i.e. we write
ψ(λ)= P≤Nψ(λ)+ P≥Nψ(λ). (2.41)
7
This is a decomposition of ψ(λ) on frequencies |ξ| ≤ N and |ξ| ≥ N respectively.
Recall that s = 1/2 in (2.2) and (2.3). For the low frequencies, we have, interpolating between L2 and H1
P≤N[ψ(λ)(t) H1/2 ≤ P≤N[ψ(λ)(t) 1/2L2 P≤N[ψ(λ)(t) 1/2H1 (2.42a)
≤ ψ01/2L2 ∇Iψ(λ)(t) 1/2L2. (2.42b)
For the high frequencies, again with s = 1/2, we can write P≥N[ψ(λ)(t) H1/2≤ 1
N1/2 ∇Iψ (λ)(t)
L2. (2.42c)
9
Combining the above inequalities (2.42a) and (2.42b) in (2.39), we have the overall estimate ψ(λ) L4(D[0,T ]) ≤ T1/8ψ03/4L2 sup [0,T ] ∇Iψ(λ)(t) 1/4 L2 + T1/8ψ0 1/2 L2 N1/2 sup [0,T ] ∇ Iψ(λ)(t) 1/2L2 . (2.43) Since we know from (2.38) that, ψ(λ)4
L4(D[0,T ]) ≤ (2A)4T1/2, we can split any
interval [0, T ], where T ≤ T0, in subintervals, say Jk, with k = 1, 2, . . . , L, write
Dk:= R2× J
k and for each slice Dk, we have
ψ(λ)4
L4(Dk)≤ µ1(B),
(2.44) where µ1 := min{µ0, µ1}, see Lemmata 2.3 and 2.4. The number of slices, which
we will call L, is at most like
11
L∼(2A) 4T1/2
Because of (2.44) we can apply the almost conservation law (2.35) for each of the
1
slices Dk. We need to controlD1/2ψ(λ)(T
k)L2 by some constant. Observe that
|ξ|| ψ(ξ)|2dξ≤ψL2+ 1E1/2(Iψ) (2.46)
3
and since the L2norm is conserved, we can boundD1/2ψ(λ)(Tk)L2 by a constand
B, see Lemma (2.4) as long as we know that E(Iψ(λ)(T
k)) < 1/2. Applying the
5
estimate in (2.13) L times on the successive slices, we obtain sup
[0,T ]
EIψ(λ)(t)≤ EIψ0(λ)+ L
N3/2−δ, (2.47)
7
where we know from scaling, see (2.12), that EIψ0(λ) ≤ 14. Thus in order to guarantee that the energy of scaled function Iψ(λ)(t) satisfies E(Iψ(λ)(t)) < 1/2 for 9
all t∈ [0, T0], we would like to choose
L N3/2−δ ≤ 1 4 or N 3/2−δ∼ 4(2A)4T 1/2 0 µ1(B) , (2.48) 11
so that we have the bound sup
[0,T ]
EIψ(λ)(t)≤ 1/2 (2.49)
13
and substituting (2.49) back to (2.43), we obtain
ψ(λ)
L4(D[0,T ])≤ 2ψ03/4L2T1/8, (2.50)
15
which contradicts (2.38) if K 2. This concludes the proof of the claim and notice that the choice of N is given by (2.48).
17
Finally recall that we assumed that ψ0 ∈ Cc∞ hence if ψ0 ∈ H1/2 we can
approximate it by ψ0 ∈ Cc∞ so that ψ0− ψ0H1/2 → 0. Since we considered the
19
solution in an arbitrarily large interval of time [0, T0] and the proof of Claim 2.5 involves only the H1/2 norm of the solution we obtain existence of solutions for 21
arbitrarily large time. This concludes the proof of global existence.
3. Bourgain’s Quadratic Estimate
23
Let us start by explaining a quadratic estimate derived by Bourgain [1, 2], which estimate is crucial in deriving the almost conservation law in the next section. The
25
main estimate is stated in (3.11) and rephrased in (3.14) in a manner which is convenient for our purpose. Consider two functions ψ1(t, x) and ψ2(t, x) and their
27
corresponding Fourier transforms, say ψ1(τ, ξ) and ψ2(τ, ξ) respectively. Let us write the Fourier transform of the product ψ1ψ2
29 ψ1ψ2(τ, ξ) = R×R2 ψ1(τ− σ, ξ − η) ψ2(σ, η)dηdσ. (3.1)
The idea is to rewrite (3.1) in parabolic variables, let us define first the new variables,
1
udef= τ− σ − |ξ − η|2; vdef= σ− |η|2 (3.2)
and write ψ2(v, η) and ψ1(u, ξ− η) for the functions ψ2and ψ1respectively. At this
3
point it is convenient to define
pdef= |ξ − η|2+|η|2; η
1def= η− ξ/2, (3.3)
5
so that we can write
|η1|2=
2p− |ξ|2/4. (3.4)
7
Let us call ρ := |η1| the length of the vector η1 so that we can write η1 = ρeiα, where α is the angle of the vector η1 and express ρ as a function of p and ξ via the
9 formula ρ(p, ξ) = 1 2 2p− |ξ|2. (3.5) 11
Notice that dη = dη1= ρdρdα = (1/4)dpdα, moreover we have that p = τ− u − v, so that we can write ρ as a function of (τ, u, v, ξ)
13
ρ(τ, u, v, ξ) = 1
2
2(τ− u − v) − |ξ|2. (3.6)
Combining all these we can write
15
ψ1ψ2(t, x) =F−1{Φ(τ, ξ)} , (3.7a)
where Φ(τ, ξ) is given by the following integral
17
Φ(τ, ξ)def=
S1×R×R
ψ1(u, ξ/2− ρeiα)ψ2(v, ξ/2 + ρeiα) dαdudv (3.7b) and ρ(u, v, ξ) is given by the formula (3.6). Now let us consider a dyadic
decompo-19
sition of ψ1and ψ2 as follows
u∼ 2l1; v∼ 2l2; |ξ| ∼ 2j1; |η| ∼ 2j2 (3.8)
21
and assume without loss of generality that 2j1 2j2. The crucial observation is
that since
23
ξ/2− ρeiα∼ 2j1; ξ/2 + ρeiα∼ 2j2; 2j1 2j2, (3.9)
we have that the angle α of the vector η1 is constrained in an interval [−α0, α0],
25
where the size of the interval is restricted by
α0∼2
j2
2j1. (3.10)
27
Plancherel in (3.7a) and (3.7b) gives the estimate ψ1ψ2 2
L2≤ C2j2−j12l12l2ψ12L2ψ22L2. (3.11)
The estimate above can be expressed in a form which is more convenient for our purpose. Let us define first,
sdef= |s| + 1; Sdef= τ − |ξ|2 (3.12a) b(ξ1, ξ2)def= min ξ 1 ξ2 1/2 , ξ 2 ξ1 1/2 . (3.12b)
Using the function b we can define the quadratic expression, for δ > 0, arbitrarily
1 small b(−1+δ)ψ1ψ2def= F−1 b(−1+δ)(ξ− η, η) ψ1(τ− σ, ξ − η) ψ2(σ, η)dσdη , (3.13) 3
and the estimate in (3.11) implies the following estimate for the quadratic expression in (3.13)
5
b(−1+δ)(ψ1ψ2)
L2 ≤ C(δ) S(1+δ)/2ψ1 L2 S(1+δ)/2ψ2 L2; δ > 0. (3.14)
Notice that from the definition in (3.12b), we have that b−1+δ≥ 1.
7
4. Almost Conservation Law
The almost conservation law was developed in a series of papers by the authors of
9
[5, 6], for the Schrodinger equation, see [5, 6]. Presently we would like to rederive the almost conservation law. The derivation is slightly different but we follow the
11
same reasoning. Start by writing the evolution equation for ψI := Iψ, where Iψ :=
F−1(mψ),
13
i∂tψI− ∆ψI+|ψI|2ψI+I|ψ|2ψ− |ψI|2ψI= 0. (4.1) The energy of ψI is defined to be, see (2.7),
15 EψIdef= R2 ∇ ψI2+1 2ψI 4 dx, (4.2)
in imitation of the conserved energy. The energy in (4.2) is not conserved but in
17
the terminology of [5, 6] it is almost conserved. Let us define two quantities
Pdef= I|ψ|2ψ; Rdef= |ψ
I|2ψI, (4.3)
19
so that from the evolution Eq. (4.1), we can derive an evolution equation for E(Iψ), namely 21 dE dt = P− R ; ψI t L2+ P− R ; ψI t L2. (4.4)
Using the evolution equation, we can write
23
φdef= ∂tψ =−i∆ψ + i|ψ|2ψ, (4.5a) so that we can split φ = φ1+ φ2 where we define
25
Notice that Iφ2= iP so if we substitute back in (4.4), we obtain 1 dE dt = 2Im P− R ; Iφ1L2+ 2ImP− R ; RL2. (4.6) The almost conservation of energy means that we would like to prove the estimate
E(T )− E(0) ≤N−(3−δ)/2C1Iψ4X[0,T ]
1,α + C2Iψ 6 X[0,T ] 1,α ; α = (1 + δ)/2. (4.7) Lemma 4.1. Assume 1/2≤ s < 1. For the quantity defined below
3
J (t)def= Is|ψ|2ψ− |Isψ|2Isψ ; φ, (4.8a)
the following estimate holds true 5 T 0 J (t)dt ≤ C N(3−δ)/2Is1ψ3X[0,T ] 1,α Iβ φX[0,T ] −1,α; α = 1 + δ 2 , (4.8b)
where s1 and β are given by 7 s1= s−1− δ 4 ; β = 3− δ 4 ; s≥ 1 2; δ > 0. (4.8c)
At this point let us notice that applying the lemma above to Iφ1, see (4.5) and the first term on the right-hand side of (4.6), we obtain since s1 < s2 imply
ms1 ≤ ms2, T 0 P− R ; Iφ1dt ≤ C N(3−δ)/2Iψ 3 X[0,T ] 1,α Iφ1X−1,α[0,T ] ≤ C N(3−δ)/2Iψ 4 X[0,T ] 1,α . (4.9) Proof. We will ignore complex conjugates since they are irrelevant here and also ignore the dependence of functions on the time variable. Writing the integral above in Fourier space, we obtain
J = 1− m(ξ2+ ξ3+ ξ4) m(ξ2)m(ξ3)m(ξ4) ψI(ξ2) ψI(ξ3) ψI(ξ4) φ(ξ2+ ξ3+ ξ4) dξ2dξ3dξ4. (4.10a) Let us write for simplicity,
9
ms,N(ξ) =
1, if |ξ| < N,
(|ξ|/N)1−s, if |ξ| ≥ N, (4.10b)
and call ξ1:= ξ2+ ξ3+ ξ4. Rewrite the integral appearing in (4.10a) in the following manner J = 1− m(ξ1) m(ξ2)m(ξ3)m(ξ4) |ξ 1| |ξ2||ξ3||ξ4| × |ξ2| ψI(ξ2)|ξ3| ψI(ξ3)|ξ4| ψI(ξ4)|ξ1|−1φ(ξ 1)dξ2dξ3dξ4. (4.10c)
Employing the notation b2,4 := b(ξ2, ξ4) and b1,3 := b(ξ1, ξ3) for the expression b defined in (3.12a and 3.12b) and keeping in mind the estimate in (3.14), write the
integral J in the following way J = |ξ 1| |ξ2||ξ3||ξ4| − a(ξ1)
a(ξ2)a(ξ3)a(ξ4) b1−δ2,4 (ξ2, ξ4)b1−δ1,3 (ξ1, ξ3) ×|ξ2| ψI(ξ2)|ξ4| ψI(ξ4)b−1+δ2,4 |ξ1|−1φ(ξ 1)|ξ3| ψI(ξ3)b−1+δ1,3 dξ2dξ3dξ4, (4.10d) where a(ξ) stands for the function
1
a(ξ) =|ξ|ms,N(ξ) =
|ξ|, if |ξ| ≤ N,
N1−s|ξ|s, if |ξ| > N.. (4.10e) Give a name to the first part of the integrand in (4.10d),
3 Qdef= |ξ1| |ξ2||ξ3||ξ4|− a(ξ1)
a(ξ2)a(ξ3)a(ξ4)
b1−δ2,4 (ξ2, ξ4)b1−δ1,3 (ξ1, ξ3). (4.11) The desired estimate in (4.8b) will follow from the quadratic estimates in (3.14) if
5
we demonstrate the bound below.
Claim 4.2. The quantity Q defined in (4.11) satisfies the bound
7
|Q| ≤ N−(3−δ)/2(λ
1λ2λ3λ4)(1−δ)/4; δ > 0, (4.12a) where λj are the quantities
9
λ(ξ) =
1, if |ξ| ≤ N,
N/|ξ|, if |ξ| ≥ N; λj= λ(ξj). (4.12b)
Proof of Claim 4.2. We can assume without loss of generality that|ξ4| ≤ |ξ3| ≤
11
|ξ4| and write
ρj=|ξj|; j = 1, 2, 3, 4, (4.13)
13
so that ρ4 ≤ ρ3 ≤ ρ2. First let us observe that if ρ4 ≤ ρ3 ≤ ρ2 ≤ N and ρ1 ≤ N then Q = 0. We will examine the size of the quantity Q in three different cases.
15
Case 1. Assume that ρ4 ≤ ρ3 ≤ N ≤ ρ2. The quantity Q is largest when ρ4 ≤
ρ3 N hence ρ1≈ ρ2. In this case we have,
Q = ρ1 ρ2ρ3ρ4 − ρs1 ρs2ρ3ρ4 ρ4 ρ2 (1−δ)/2ρ 3 ρ1 (1−δ)/2 = 1 + ρ1− ρ2 ρ2 − 1 +ρ1− ρ2 ρ2 s 1 (ρ1ρ2)(1−δ)/2(ρ3ρ4)(1+δ)/2 (4.14a) ≈ ρ1− ρ2 ρ2(ρ1ρ2)(1−δ)/2(ρ 3ρ4) 1+δ 2 ≤ ρ3+ ρ4 (ρ3ρ4)(1+δ)/2 N ρ2 3−δ 2 N ρ1 1−δ 2 N−2+δ,
where we used the fact|ρ1−ρ2| ≤ ρ3+ ρ4 N and (1+x)s−1 ≤ Cx for x ∈ (0, 1).
1
The estimate in (4.14a) can be expressed in the following manner
Q ≤Cλ(3−δ)/22 λ(1−δ)/21 N−2+δ. (4.14b)
3
Case 2. Assume ρ4 ≤ N ≤ ρ3 ≤ ρ2. The quantity Q is largest when ρ3 ≈ N,
ρ4 N, hence ρ1≈ ρ2. In this case we have the estimates
Q = ρ1 ρ2ρ3ρ4 − N1−sρs 1 N2(1−s)(ρ2ρ3)sρ4 ρ4ρ3 ρ2ρ1 (1−δ)/2 ≈ ρ1 ρ2ρ3 s 1 ρ4 ρ1 ρ2ρ3 1−s − 1 N1−s ρ3ρ4 ρ1ρ2 (1−δ)/2 (4.15a) ≤ ρ1 ρ2ρ3 s ρ3ρ4 ρ1ρ2 (1−δ)/2 1 N1−s ≈ N ρ3 s−(1−δ)/2 N2 (ρ1ρ2)1−δ2 N−3+δ2 .
Using the notation from (4.12b) we can write (4.15a) as follows
Q ≤Cλ1λ2(1−δ)/2λs−(1−δ)/23 N−(3−δ)/2. (4.15b)
5
Case 3. Assume N ≤ ρ4≤ ρ3≤ ρ2. The quantity Q is largest when ρ4≤ ρ3 ρ2, hence ρ1≈ ρ2. Now we can estimate
7 Q = ρ1 ρ2ρ3ρ4 − ρs1 N2(1−s)(ρ 2ρ3ρ4)s ρ3ρ4 ρ1ρ2 (1−δ)/2 = ρ1 ρ2ρ3ρ4 s ρ 1 ρ2ρ3ρ4 1−s − 1 N2(1−s) ρ3ρ4 ρ1ρ2 (1−δ)/2 ≤ ρ1 ρ2ρ3ρ4 s 1 N2(1−s) ρ3ρ4 ρ1ρ2 (1−δ)/2 ≤ ρ1 ρ2 sN ρ3 s−(1−δ)/2N ρ2 1/2N ρ1 1/2N ρ4 s−(1−δ)/2 N−2, (4.16a)
which can be written, using the notation in (4.12b)
9
Q ≤ λ3λ4s−(1−δ)/2λ1λ2(1−δ)/2N−2. (4.16b) The three cases demonstrate the inequality in (4.12a) i.e. we see that for s≥ 1/2,
11
we have that the quantity Q is bounded by
Q ≤CN−(3−δ)/2(λ1λ2λ3λ4)(1−δ)/4, (4.17)
13
where λj := λ(ξj) are given by (4.12b). This concludes the proof of the claim. Hence this completes the proof of Lemma 4.1.
Let us now apply the estimate in Lemma 4.1 for the expression below, see the
1
second term on the right-hand side of (4.6) T 0 P− R; Rdt ≤ C N(3−δ)/2Iψ 3 X1,α[0,T ]Iβ RX[0,T ] −1,α, (4.18a) 3
so that we would like to estimate the expression Sα/ξIβR L2; R =|ψI| 2ψ I. (4.18b) 5
For simplicity we can write
S1def= τ2+ τ3+ τ4−ξ2+ ξ3+ ξ42; Sk def= τk−ξk2, k = 2, 3, 4
(4.19a) and, denoting α := (1 + δ)/2, we have the following inequality
S1α≤ S2α+ Sα3 + S4α+|ξ2||ξ3|α+|ξ3||ξ4|α+|ξ2||ξ4|α. (4.19b)
7
With these observations in mind we have that as far as the L2 norm is concerned,
we can write 9 mβ(ξ) ξ Sα|ψI|2ψI ≈ mξβ(ξ)ψ2 I∗ SαψI+mξβ(ξ)ψI∗|ξ|αψI∗|ξ|αψI. (4.20) Combining the Strichartz type estimate in (2.18a) and (2.18b) with the Sobolev embedding, we obtain the following inequalities
DψIL2 xL∞t + ψI L∞xLr t ≤ C ψI X1,α; 2 < r, (4.21a) SαψI Lq xL2t ≤ C ψI X 1,α; q≥ 2, (4.21b) DαψI Lp xL4t ≤ C ψI X 1,α; p≤ 4/δ. (4.21c)
Finally employing the inequalities in (4.21a)–(4.21c), we have for each of the terms appearing on the right-hand side of (4.20) and for any p > 1 and (1/r) + (1/r) = 1
ξ−1Fψ2ISαψI L2 ≤ C ψI 2L2pr x L∞t SαψI Lpr x L2t ≤ C ψI 3X 1,α, (4.22a) ξ−1FψIDαψI2 L2 ≤ C ψI Lpr x L∞t DαψI 2L2pr x L4t ≤ C ψI 3X 1,α. (4.22b)
Combining (4.22a) and (4.22b) we obtain
11 T 0 P− R ; Rdt ≤ C N(3−δ)/2Iψ 6 X1,α[0,T ] , (4.23)
and this estimate together with (4.9) imply the almost concervation law in (4.7).
5. Correlation Estimates
1
We would like to describe a general method for deriving correlation estimates for Schr¨odinger type equations. The estimate which is relevant here is stated in (5.40).
3
The idea is to view the evolution equation as describing the evolution of a compress-ible dispersive fluid whose pressure is a function of the density. For simplicity we will
5
consider a cubic equation, but the method works for more general nonlinearities. Let us start by considering the Nonlinear Schr¨odinger equation with a defocusing
7
cubic nonlinearity, i.e. the field ψ(t, x) satisfies the equation
iψt− ∆ψ + |ψ|2ψ = 0; ψ : R× Rn→ C, (5.1a)
9
where the space dimension is n = 2, 3 and C is the complex plane. The cubic nonlinearity serves as an example, one can consider more general nonlinear terms
11
as long as they are of defocusing type. Let us adopt the conventions
D[T1, T2]def= Rn× [T1, T2]; Rnt def= (t, x) : x∈ Rn (5.1b)
13
to denote a space-time slab and a time slice for t fixed. A central observation is that the Eq. (5.1a) above conserves energy which means that the integral
15 Edef= Rn t |∇ψ|2+1 2|ψ| 2 dx (5.1c)
is a constant independent of time. Equation (5.1a) has more conservation laws certain of which we wish to exploit in order to obtain apriori estimates. Let us define certain quantities that will be central in our investigation,
ρdef= 1 2|ψ| 2, (5.2a) pj def= 1 2i ψ∇jψ− ψ∇jψ; j = 1, 2, . . . , n, (5.2b) p0def= 1 2i ψψt− ψψt, (5.2c) σjk def= 1 2 ∇jψ∇kψ +∇jψ∇kψ; j, k = 1, 2, . . . , n. (5.2d) What I defined above is a density function ρ(t, x), a momentum vector pj(t, x), with
p0 the time component, and a stress tensor σjk(t, x). The Schr¨odinger evolution gives conservation laws for the density and the space components of the momenta, they are
∂t{ρ} − ∇j{pj} = 0, (5.3a)
∂t{pj} − ∇k{µkj} = 0, (5.3b)
where µkj(t, x) is the tensor
17
The tensor µkj consists of three parts, the term−∆ρ is dispersive while the term
1
2ρ2 describes the pressure as a function of density. Some remarks are in order, Eq. (5.1a) is the evolution equation of the Lagrangian
3 Rn×R − p0+ σ + 2ρ2 dxdt, (5.4a)
where σ = tr(σjk). The integral of the quantity p0is not preserved by the flow, but
5 the integral Rn t − p0+ 2ρ2 dx 7
is, indeed, constant independent of time. This observation follows from a structure equation that (5.1a) satisfies, namely
9
−p0− ∆ρ + σ + 4ρ2= 0. (5.4b)
Notice that the system (5.3a) and (5.3b) describes the conservation laws of an
11
irrotational compressible and dispersive fluid.
Let us assume that we have two different solutions of (5.1a) for which we have
13
the corresponding densities ρa and momenta pa where a = 1, 2. Now write the conservation law for the momenta again emphasizing the space and time dependence
15 ∂tp1j(t, x)− ∇k µ1,kj (t, x) = 0. (5.5)
Multiplying the equation above with ρ2(t, y) and using the conservation of mass
equation we obtain an equation, of the form
∂tp1j(t, x)ρ2(t, y)− ∇y,kp1j(t, x)p2,k(t, y)− ∇x,k µ1,kj (t, x)ρ2(t, y) = 0. (5.6) In Eq. (5.6) above, we used the convention∇x,∇yto keep track of which variable
17
is differentiated. Change variables as follows x→ x + y
2 ; y→ x− y
2 (5.7)
19
and symmetrize in the a = 1, 2 indices, i.e. switch the roles of 1 and 2 and add them so that Eq. (5.6) can be written as
21 ∂t Q(12)j (t, x; y) − ∇y,k M(12),kj (t, x; y) − ∇x,k N(12),kj (t, x; y) = 0, (5.8) where we use the convention (12) to denote the symmetrized quantities and the quantities Q(12)j , Mkj(12) and Nkj(12) are defined below,
Q(12)j (t, x; y)def= p(1j (t, x + y)ρ2)(t, x− y), (5.9a)
Mkj(12)(t, x; y)def= µ(1jk(t, x + y)ρ2)(t, x− y) − p(1j (t, x + y)p2)k(t, x− y), (5.9b)
Nkj(12)(t, x; y)def= µ(1jk(t, x + y)ρ2)(t, x− y) + p(1j (t, x + y)p2)k(t, x− y). (5.9c) In what follows we will occasionally skip the indices a = 1, 2 for the two solutions
23
confusion. If we switch y→ −y, we obtain another equation like (5.8) with the sign
1
of the second term switched. Subtracting the two equations we obtain a symmetrized version of (5.8), which reads
3
∂t{QA,j} − ∇y,kMS,jk − ∇x,kNA,jk = 0, (5.10) where the relevant symmetrized or anti-symmetrized quantities are defined below
QA,j(t, x; y)def= Qj(t, x; y)− Qj(t, x;−y), (5.11a)
MS,kj(t, x; y)def= Mkj(t, x; y) + Mkj(t, x;−y), (5.11b)
NA,kj(t, x; y)def= Nkj(t, x; y)− Nkj(t, x;−y). (5.11c) Notice that a similar argument, starting from the conservation of mass Eq. (5.3a)
5
gives the equation
∂t{D(t, x; y} − ∇y,kQkA(t, x; y)− ∇x,kQkS(t, x; y)= 0, (5.12a)
7
where D(t, x; y) and QS,k(t, x; y) are defined as follows
D(12)(t, x; y) = ρ(1(t, x + y)ρ2)(t, x− y), (5.12b)
Q(12)S,k(t, x; y) = p(1k(t, x + y)ρ2)(t, x− y) + p(1k(t, x− y)ρ2)(t, x + y). (5.12c) Symmetrizing Eq. (5.8) gives an equation similar to (5.10)
∂t{QS,j} − ∇y,kMA,jk − ∇x,kNS,jk = 0, (5.13)
9
where MA,jk and NS,jk stand for the anti-symmetrized and symmetrized versions of Mjk and Njkrespectively but we are not going to make use of (5.13) here.
11
A crucial observation is the fact that we can express the stress tensor σkjin terms of the density function ρ and the momenta pj, namely an elementary calculation,
13
using the fact that
∇jρ = 1
2
ψ∇jψ + ψ∇jψ 15
shows that we can write
σkj = 1
2ρ[∇kρ∇jρ + pkpj] . (5.14)
17
The equation above suggests that we split σkj in two parts, a potential part say πkj and a kinetic part, say κkj, in the following manner
19
πkj def= 1
2ρ∇kρ∇jρ; κkj
def= 1
2ρpkpj. (5.15)
The trace of the potential part of the stress tensor tr(πkj) contains part of the potential energy of the field ψ due to compressibility, while the trace of the kinetic part tr(κkj) contains the kinetic energy of the field. Our next step is to examine carefully the tensor MS,kj. We would like to suppress the dependence on t since it
is not relevant in what follows. If we substitute (5.15) in the expression of µkj, we can write a long formula for MS,kj as follows
MS,kj(12)=−p(1j (x + y)pk2)(x− y) − p(1j (x− y)p2)k (x + y) + 2κ(1kj(x + y)ρ2)(x− y) + 2κ(1kj(x− y)ρ2)(x + y) + 2π(1kj(x + y)ρ2)(x− y) + 2πkj(1(x− y)ρ2)(x + y)
− δkj[∆ρ(1(x + y)ρ2)(x− y) + ∆ρ(1(x− y)ρ2)(x + y)]
+ δkj[2(ρ(1)2(x + y)ρ2)(x− y) + 2(ρ(1)2(x− y)ρ2)(x + y)]. (5.16) Let us define two vectors, using the momenta and density
1 Jj1,2(t, x; y)def= p 1 j(x + y) ρ1(x + y) ρ2(x− y) − p 2 j(x− y) ρ2(x− y) ρ1(x + y) (5.17)
and similarly Jj2,1by switching the roles of 1, 2 so that the first four terms in (5.14) can be expressed as a tensor product of the vector Jja,b, i.e. we can write
− p(1
j (x + y)p2)k(x− y) − p(1j (x− y)p2)k (x + y) + 2κ(1kj(x + y)ρ2)(x− y) + 2κkj(1(x− y)ρ2)(x + y) = Jk1,2Jj1,2+ Jk2,1Jj2,1. (5.18) It is worthwhile to notice here that QA,j can be expressed using D and Jj , namely
3 Q(12)A,j =√D1,2J1,2 j + √ D2,1J2,1 j , (5.19)
where D1,2 := ρ1(x + y)ρ2(x− y) and similarly for D2,1. Taking into account the 5
observation in (5.18), we can write the tensors MS,kj as a sum of four terms,
MS,kj(12)= Jk1,2Jj1,2+ Jk2,1Jj2,1+ Φkj+ δkj[W + P ] , (5.20)
7
where the tensor Φkj and the scalar potentials W and P are
Φkj def= 2π(1kj(x + y)ρ2)(x− y) + 2πkj(1(x− y)ρ2)(x + y), (5.21a)
W def= −[∆ρ(1(x + y)ρ2)(x− y) + ∆ρ(1(x− y)ρ2)(x + y)], (5.21b) P def= 2ρ(12(x + y)ρ2)(x− y) + 2ρ(1)2(x− y)ρ2)(x + y). (5.21c) At this point we are ready to derive apriori estimates. For simplicity we will assume that ψ1= ψ2, thus ρ1= ρ2etc. in what follows, the assumption makes the 9
will be identity (5.10), which if we contract with a vector field Xj(y) that depends
1
only on the y variables, we obtain an equation of the form
∂tQA,jXj− ∇y,kMS,jk Xj− ∇x,kNA,jk Xj+ R = 0, (5.22)
3
where the remainder R is a sum of four terms
R = R0+ R1+ R2+ R3 (5.23)
5
given by the following expressions
R0def= (∇kXj) JkJj; R1def= (∇kXj) Φkj, (5.24a)
R2def= (divX) W ; R3def= (divX) P. (5.24b) Let us choose the vector field Xj(y) to be of the particular form, we would like to choose
7
Xj(y)def= b(|y|/N)uj; uj def= yj/|y|, (5.25) where b(s) is an increasing bounded function and N is a large parameter to be chosen appropriately later. Adopt the convention r =|y| and compute
∇kXj = b(r/N ) r [δkj− ukuj] + 1 Nb (r/N )u kuj, (5.26a) divX = 1 r(b(r/N ) + (r/N )b (r/N )) . (5.26b)
Observe that because b(s) is increasing the terms R0, R1 and R3 are positive. We
9
will look more carefully at the R2term. First let us examine the three dimensional, case, i.e. n = 3 with the simple choice b = 1. Let us observe that in three space
11
dimensions we have, see (5.21) for the expression of W
R2= R3×R3×R W |y| dxdydt = C R3×R ρ1(t, x)ρ2(t, x) dxdt. (5.27) 13
Recall the notation in (5.1b) and let us define a correlation function R[T1, T2] for the density and another correlation, say L(t), between the momentum and the density in the following manner
R[T1, T2]def= D[T1,T2] ρ1(t, x)ρ2(t, x) dxdt, (5.28a) L(t)def= Rn t×Rnt QA,jXj dxdy. (5.28b)
With these conventions, after we integrate (5.22), we obtain the estimate
R[0, T ]≤ C [L(0) − L(T )] , (5.29)
15
and one can show, as in [6], that
|L(t)| ≤ Cψ1(t)2
L2ψ2(t)2H1/2+ψ2(t)2L2ψ1(t)2H1/2. (5.30)
It is interesting to recast L(t) in a different light, define a potential function Ua(t, x)
1
and a vector field Pja(t, x) by solving Laplace’s equation
−∆Ua = ρa; −∆Pa
j = paj; a = 1, 2, (5.31)
3
now we can write, omiting the indices a = 1, 2 for simplicity,
L(t) = R3 t ∆U Pj− U∆Pjxjdx. (5.32) 5
Consider the two dimensional case, i.e. n = 2. Assume for simplicity that the two states are equal and make the choice, see (5.25)
7
b(s)def= s log(√e/s), if s≤ 1/√e, (5.33) otherwise we require that b(s) is smooth increasing and bounded by one for s >
9
1/√e. With this choice, we have, with r =|y|
divX = 2
N log(N/r), if r < N/ √
e, (5.34)
11
so that the Laplacian of divX can be written
−∆ (divX) = 2 Nδ(y) +
c(r/N )
N3 , (5.35)
13
where c(r/N ) is a bounded function supported in the region{r > N/√e}.
Integrat-ing (5.22) we obtain the identity below
15 L(T2)− L(T1) = T2 T1 R2 t×R2t R dxdydt = 0. (5.36)
Observe that because Xj is bounded, we can obtain a bound,
17 R2 T×R2T QA,jXj dxdy≤ ψ(T )2L2ψ(T )2H1/2, (5.37)
which estimate follows from the observation
19 R2 t Xj(y)pj(t, x + y)dx ≤ Xj∇jψ H−1/2ψH1/2. (5.38)
Let us make the choice N = (T2− T1)12 so that we have a bound
21 T2 T1 R2 t×R2t c(|y|/N) N3 ρ(t, x + y)ρ(t, x− y) dxdydt ≤ ψ 2 L2/(T2− T1)1/2, (5.39)
which follows from the fact that R2
tρ(t, x) dx =ψ0
2
L2. Finally from the integral
23
of the term we called R2, see (5.24b), we obtain a decay estimate D[T1,T2] ρ2(t, x) dxdt≤ C sup t ψ(t)2 L2ψ(t)2H1/2 (T1− T2)12. (5.40) 25
A remark is in order concerning the name correlation for this type of estimate.
1
Let d(y) be the function with the property∇jd(y) = Xj(y). Multiply (5.12a) with
d and differentiate with respect to time, and use (5.10) to obtain the equation 3
∂t2{dD} + ∇y,kMS,jk Xj− d∂tQkA+∇x,kPA,jk Xj− d∂tQkS= R. (5.41) Integrating with respect to the space variables, we obtain
5 d2 dt2 R2 t×R2t {dD} dxdy = R2 t×R2t {R} dxdy. (5.42)
Notice that the integral
7 C[t]def= R2 t×R2t {dD} dxdy = R2 t×R2t
d(y)ρ(1(t, x + y)ρ2)(t, x− y) dxdy, (5.43) expresses a correlation between the density functions ρa(t, x). Since C[t] is positive
9
and its derivative is integrable in the case where the space dimension is three we obtain that C[t]∼ t for large time. The integral C[t] can be written, after a change of
11 variables, C[t] = 2 R2 t×R2t |x − y|ρ1(t, x)ρ2(t, y)dxdy. 13
We can derive a true space-time estimate for the two dimensional case. Let d(t, y) and Xj(t, y) to be chosen. The conservation laws that we would like to use are
∂t{D} − ∇y,j QjA − ∇x,j QjS = 0, (5.44a)
∂t{QA,j} − ∇y,kMS,jk − ∇x,kNA,jk = 0. (5.44b) Multiply the first equation with d(t, y) and contract the second equation with
Xj(t, y) and add them to obtain,
∂tdD + XjQA,j− ∇y,kdQkA+ MS,jk Xj− ∇x,kdQkS+ NA,jk Xj
+
−∂tdD +∇jd− ∂tXjQjA+∇kXjMS,jk
= 0. (5.45)
Try the vector field of the form
Xj= b |y| tα uj (5.46) 15
with α to be chosen, so that we have
∇kXj= b(|y|/t α) |y| [δkj − ukuj] + 1 tαb (|y|/tα)u kuj (5.47a) divX = 1 |y|b(|y|/tα) + 1 tαb (|y|/tα). (5.47b)
We assumed that b(s) = s log√e/sfor s≤ 1/√e and using this we can compute
divX = 2
tαlog
tα/|y| for|y| ≤ tα/√e, (5.48)