On the uniqueness of solutions to the Gross-Pitaevskii hierarchy

0.17 MB·English

by  Sergiu Klainerman

ON THE UNIQUENESS OF SOLUTIONS TO THE
GROSS-PITAEVSKII HIERARCHY
SERGIU KLAINERMAN AND MATEI MACHEDON
7
0
0
2 Abstract. The purpose of this note is to give a new proof of
uniqueness of the Gross- Pitaevskii hierarchy, first established in
n
[1], in a different space, based on space-time estimates similar in
a
J spirit to those of [2].

3

1
v
6
0
0
1 1. Introduction
0
7
0 TheGross-Pitaevskiihierarchyreferstoasequenceoffunctionsγ(k)(t,x ,x′),
/ k k
h k = 1,2,···,wheret ∈ R,x = (x ,x ,··· ,x ) ∈ R3k,x′ = (x′,x′,··· ,x′) ∈
k 1 2 k k 1 2 k
p R3k which are symmetric, in the sense that
-
h
t
a γ(k)(t,x ,x′) = γ(k)(t,x′,x )
m k k k k
:
v and
i
X
γ(k)(t,x ,···x ,x′ ,···x′ ) = γ(k)(t,x ,···x ,x′,···x′) (1)
r σ(1) σ(k) σ(1) σ(k) 1 k 1 k
a
for any permutation σ, and satisfy the Gross-Pitaevskii infinite linear
hierarchy of equations,
k
i∂t +∆xk −∆x′k γ(k) = Bj,k+1(γ(k+1)). (2)
j=1
(cid:0) (cid:1) X
with prescribed initial conditions
γ(k)(0,x ,x′) = γ(k)(x ,x′).
k k 0 k k
Here ∆xk, ∆x′k refer to the standard Laplace operators with respect to
the variables x ,x′ ∈ R3k and the operators B = B1 −B2
k k j,k+1 j,k+1 j,k+1
1
2 SERGIUKLAINERMANANDMATEI MACHEDON
are defined according to,
B1 (γ(k+1))(t,x ,x′)
j,k+1 k k
= δ(x −x )δ(x −x′ )γ(k+1)(t,x ,x′ )dx dx′
j k+1 j k+1 k+1 k+1 k+1 k+1
Z Z
B2 (γ(k+1))(t,x ,x′)
j,k+1 k k
= δ(x′ −x )δ(x′ −x′ )γ(k+1)(t,x ,x′ )dx dx′ .
j k+1 j k+1 k+1 k+1 k+1 k+1
Z Z
In other words B1 , resp. B2 , acts on γ(k+1)(t,x ,x′ ) replac-
j,k+1 j,k+1 k+1 k+1
ing both variables x and x′ by x , resp x′. We shall also make
k+1 k+1 j j
use of the operators,
Bk+1 = B
j,k+1
1≤j≤k
X
One can easily verify that a particular solution to (2) is given by,
k
γ(k)(t,x ,x′) = φ(t,x )φ(t,x′) (3)
k k j j
j=1
Y
where each φ satisfies the non-linear Schr¨odinger equation in 3+1 di-
mensions
(i∂ +∆)φ = φ|φ|2, φ(0,x) = φ(x) (4)
t
In [1] L. Erd¨os, B. Schlein and H-T Yau provide a rigorous derivation
of the cubic non-linear Schr¨odinger equation (4) from the quantum
dynamics of many body systems. An important step in their program
is to prove uniqueness to solutions of (2) corresponding to the special
initial conditions
k
γ(k)(0,x ,x′) = γ(k)(x ,x′) = φ(x )φ(x′) (5)
k k 0 k k j j
j=1
Y
with φ ∈ H1(R3). To stateprecisely the uniqueness result of [1], denote
Sj = (1−∆xj)1/2, Sj′ = (1−∆x′j)1/2 and S(k) = kj=1Sj · kj=1Sj′. If
the operator given by the integral kernel γ(k)(x ,x′) is positive (as an
k Qk Q
operator), then so is S(k)γ(k)(x ,x′), and the trace norm of S(k)γ(k) is
k k
kγ(k)k = S(k)γ(k)(x ,x′) dx .
Hk k k x′k=xk k
Z
(cid:0) (cid:1)(cid:12)
(cid:12)
ON THE UNIQUENESS OF SOLUTIONS TO THE GROSS-PITAEVSKII HIERARCHY3
The authors of [1] prove uniqueness of solutions to (2) in the set of
symmetric, positive operators γ satisfying, for some C > 0,
k
sup kγ(k)(t,·,·)k ≤ Ck (6)
Hk
0≤t≤T
In that work, the equations (2) are obtained as a limit of the BBGKY
hierarchy (see [1]), and it is proved that solutions to BBGKY with
initial conditions (5) converge, in a weak sense, to a solution of (2) in
the space (6).
The purpose of this note is to give a new proof of uniqueness of the
Gross-Pitaevskii hierarchy (2), inadifferent space, motivated, inpart,
by space-time type estimates, similar in spirit to those of [2].
Our norms will be
kR(k)γ(k)(t,·,·)k (7)
L2(R3k×R3k)
Here, Rj = (−∆xj)1/2, Rj′ = (−∆x′j)1/2 and R(k) = kj=1Rj · kj=1Rj′.
Notice that for a symmetric, smooth kernel γ, for which the associated
Q Q
linear operator is positive we have
kR(k)γ(k)(t,·,·)k
L2(R3k×R3k)
≤ kS(k)γ(k)(t,·,·)k
L2(R3k×R3k)
≤ kγ(k)(t,·,·)k
Hk
since |S(k)γ(k)(x,x′)|2 ≤ S(k)γ(k)(x,x)S(k)γ(k)(x′,x′). This is similar to
the condition a a −|a |2 ≥ 0 which is satisfied by all n×n positive
ii jj ij
semi-definite Hermitian matrices.
Our main result is the following:
Theorem 1.1 (Main Theorem). Consider solutions γ(k)(t,x ,x′) of
k k
the Gross-Pitaevskii hierarchy (2), with zero initial conditions, which
verify the estimates,
T
kR(k)B γ(k)(t,·,·)k dt ≤ Ck (8)
j,k L2(R3k×R3k)
Z0
for some C > 0 and all 1 ≤ j < k. Then kR(k)γ(k)(t,·,·)k =
L2(R3k×R3k)
0 for all k and all t.
4 SERGIUKLAINERMANANDMATEI MACHEDON
Therefore,solutionsto(2)verifyingtheinitialconditions(5),areunique
in the space-time norm (8). We plan to address the connection with so-
lutions of BBGKY in a future paper. The following remark is however
reassuring.
Remark 1.2. The sequence γ(k), given by (3) with φ an arbitrary solu-
tion of (4) with H1 data, verifies (8) for every T > 0 sufficiently small.
Moreover, if the H1 norm of the initial data is sufficiently small then
(8) is verified for all values of T > 0.
Proof. Observe thatR(k)B γ(k)(t,x ,...,x ;x′ ...,x′)canbewritten
1,k 1 k 1 k
in the form,
R(k)B γ(k)(t,·,·) = R |φ(t,x )|2φ(t,x ) R (φ(t,x )···R (φ(t,x )
1,k 1 1 1 2 2 k k
· R′(φ(t,x′)···R′(φ(t,x′)
1(cid:0) 1 k (cid:1) k
Therefore, in [0,T]×R3k ×R3k we derive
kR(k)Bj,kγ(k)kL1tL2 ≤ k|R1(|φ|2φ)kL1tL2x ·kR2φkL∞t L2x···kRkφkL∞t L2x
· kR1′φkL∞t L2xkR2′φkL∞t L2x ···kRk′φkL∞t L2x
≤ Ck∇(|φ|2φ)k ×k∇φk2k−1
L1tL2x L∞t L2x
where the norm on the left is in [0,T]×R3k×R3k and all norms on the
right hand side are taken relative to the space-time domain [0,T]×R3.
In view of the standard energy identity for the nonlinear equation (4)
we have apriori bounds for supt∈[0,T]k∇φ(t)kL2(R3). Therefore we only
need to provide a uniform bound for the norm k∇(|φ|2φ)k . We
L1L2
t x
shall show below that this is possible for all values of T > 0 provided
that the H1 norm of φ(0) is sufficiently small. The case of arbitrary
size for kφ(0)k and sufficiently small T is easier and can be proved
H1
in a similar manner.
WeshallrelyonthefollowingStrichartzestimate(see[3])forthelinear,
inhomogeneous, Schr¨odinger equation i∂ φ+∆φ = f in [0,T]×R3,
t
kφkL2tL6x ≤ C kfkL1tL2x +kφkL∞t L2x (9)
(cid:0) (cid:1)
We start by using H¨older inequality, in [0,T]×R3,
k∇(|φ|2φ)k ≤ Ck∇φk kφ2k ≤ Ck∇φk kφk2
L1tL2x L2tL6x L2tL3x L2tL6x L4tL6x
ON THE UNIQUENESS OF SOLUTIONS TO THE GROSS-PITAEVSKII HIERARCHY5
Using (9) for f = |φ|2φ we derive,
k∇φkL2tL6x ≤ C k∇(|φ|2φ)kL1tL2x +k∇φkL∞t L2x
(cid:0) (cid:1)
Denoting,
A(T) = k|φ|2φk
L1L2([0,T]×R3)
t x
B(T) = k∇(|φ|2φ)k
L1L2([0,T]×R3)
t x
we derive,
B(T) ≤ C B(T)+k∇φ(0)k kφk2
L2 L4L6
t x
≤ C(cid:0)B(T)+k∇φ(0)kL2(cid:1)kφkL∞t L6xkφkL2tL6x
≤ C A(T)+kφ(0)k B(T)+k∇φ(0)k k∇φ(0)k
(cid:0) L2 (cid:1) L2 L2
x
On the other han(cid:0)d, using (9) again(cid:1)(cid:0), (cid:1)
A(T) ≤ C kφ3k +kφ(0)k
L1L2 L2
t x
≤ C(A(T)3 +kφ(0)k
(cid:0) L2 (cid:1)
Observe thislast inequality implies that, forsuffic(cid:1)iently small kφ(0)k ,
L2
A(T)remainsuniformlyboundedforallvaluesofT. Thus, forallvalues
of T, with another value of C,
B(T) ≤ C B(T)+k∇φ(0)k k∇φ(0)k
L2 L2
x
from which we get a un(cid:0)iform bound for B(T(cid:1)) provided that k∇φ(0)k
L2
is also sufficiently small.
(cid:3)
TheproofofTheorem (1.1)isbased ontwo ingredients. Oneis express-
ingγ(k) intermsofthefuture iteratesγ(k+1)···, γ(k+n) usingDuhamel’s
formula. Since B(k+1) = k B is a sum of k terms, the iterated
j=1 j,k+1
Duhamel’s formula involves k(k+1)···(k+n−1) terms. These have
P
to be grouped into much fewer O(Cn) sets of terms. This part of our
paper follows in the spirit of the Feynman path combinatorial argu-
ments of [1]. The second ingredient is the main novelty of our work.
We derive a space-time estimate, reminiscent of the bilinear estimates
of [2].
6 SERGIUKLAINERMANANDMATEI MACHEDON
Theorem 1.3. Let γ(k+1)(t,x ,x′ ) verify the homogeneous equa-
k+1 k+1
tion,
i∂t +∆k±+1 γ(k+1) = 0, ∆(±k+1) = ∆xk+1 −∆x′k+1 (10)
(cid:0)γ(k+1)(0,x (cid:1) ,x′ ) = γ(k+1)(x ,x′ ).
k+1 k+1 0 k+1 k+1
Then there exists a constant C, independent of j,k, such that
kR(k)B (γ(k+1))k (11)
j,k+1 L2(R×R3k×R3k)
≤ CkR(k+1)γ(k+1)k
0 L2(R3(k+1)×R3(k+1))
2. Proof of the estimate
Without loss of generality, we may take j = 1 in B . It also suffices
j,k+1
to estimate the term in B1 , the term in B2 can be treated in the
j,k+1 j,k+1
same manner. Let γ(k+1) be as in (10). Then the Fourier transform of
γ(k+1) with respect to the variables (t,x ,x′) is given by the formula,
k k
δ(τ +|ξ |2 −|ξ′|2)γˆ(ξ,ξ′)
k k
where τ corresponds to the time t and ξ = (ξ ,ξ ,...,ξ ), ξ′ =
1 2 k
k k
(ξ′,ξ′,...,ξ′) correspond to the space variables x = (x ,x ,...,x )
1 2 k k 1 2 k
and x′ = (x′,x′,...,x′). We also write ξ = (ξ ,ξ ), ξ′ =
k 1 2 k k+1 k k+1 k+1
(ξ′,ξ′ ) and,
k k+1
|ξ |2 = |ξ |2 +...+|ξ |2 +|ξ |2 = |ξ |2 +|ξ |2
1 k k+1 k+1
k+1 k
|ξ′ |2 = |ξ′|2 +...+|ξ′|2 +|ξ′ |2
k+1 1 k k+1
The Fourier transform of B1 (γ(k+1)), with respect to the same vari-
1,k+1
ables (t,x ,x′), is given by,
k k
δ(···)γˆ(ξ −ξ −ξ′ ,ξ ,··· ,ξ ,ξ′ )dξ dξ′ (12)
1 k+1 k+1 2 k+1 k+1 k+1 k+1
Z Z
where,
δ(···) = δ(τ +|ξ −ξ −ξ′ |2 +|ξ |2 −|ξ |2 −|ξ′ |2)
1 k+1 k+1 k+1 1 k+1
and γ denotes the initial condition γ(k+1). By Plancherel’s theorem,
0
estimate (11) is equivalent to the following estimate,
kI [f]k ≤ Ckfˆk , (13)
k L2(R×Rk×Rk) L2(Rk+1×Rk+1)
ON THE UNIQUENESS OF SOLUTIONS TO THE GROSS-PITAEVSKII HIERARCHY7
applied to f = R(k+1)γ, where,
|ξ |fˆ(ξ −ξ −ξ′ ,ξ ,··· ,ξ ,ξ′ )
I [f](τ,ξ ,ξ′) = δ(...) 1 1 k+1 k+1 2 k+1 k+1 dξ dξ′ .
k k k |ξ −ξ −ξ′ ||ξ ||ξ′ | k+1 k+1
Z Z 1 k+1 k+1 k+1 k+1
ApplyingtheCauchy-Schwarzinequalitywithmeasures, weeasilycheck
that,
|ξ |2
|I [f]|2 ≤ δ(...) 1 dξ dξ′
k |ξ −ξ −ξ′ |2|ξ |2|ξ′ |2 k+1 k+1
Z Z 1 k+1 k+1 k+1 k+1
· δ(...)|fˆ(ξ −ξ −ξ′ ,ξ ,··· ,ξ ,ξ′ )|2dξ dξ′
1 k+1 k+1 2 k+1 k+1 k+1 k+1
Z Z
If we can show that the supremum over τ, ξ ···ξ ,ξ′ ···ξ′ of the first
1 k 1 k
integral above is bounded by a constant C2, we infer that,
kI [f]k2 ≤ C2 δ(...)|fˆ(ξ −ξ −ξ′ ,ξ ,··· ,ξ ,ξ′ )|2dξ dξ′ dτ
k L2 1 k+1 k+1 2 k+1 k+1 k+1 k+1
Z Z Z
≤ C2kfˆk2
L2(Rk+1×Rk+1)
Thus,
kI [f]k2 ≤ C2kfˆk2
k L2(R×Rk×Rk) L2(Rk+1×Rk+1)
as desired. Thus we have reduced matters to the following,
Proposition 2.1. There exists a constant C such that
δ(τ +|ξ −ξ −ξ′ |2 +|ξ |2 −|ξ′ |2)
1 k+1 k+1 k+1 k+1
Z
|ξ |2
1 dξ dξ′ ≤ C
|ξ −ξ −ξ′ |2|ξ |2|ξ′ |2 k+1 k+1
1 k+1 k+1 k+1 k+1
uniformly in τ,ξ .
1
The proof is based on the following lemmas
Lemma 2.2. Let P be a 2 dimensional plane or sphere in R3 with the
usual induced surface measure dS. Let 0 < a < 2, 0 < b < 2, a+b > 2.
Let ξ ∈ R3. Then there exists C independent of ξ and P such that
1 C
dS(η) ≤
|ξ −η|a|η|b |ξ|a+b−2
ZP
8 SERGIUKLAINERMANANDMATEI MACHEDON
Proof. If P = R2 and ξ ∈ R2 this is well known. The same proof
works in our case, by breaking up the integral I ≤ I +I +I over the
1 2 3
overlapping regions:
Region 1: |ξ| < |η| < 2|ξ|. In this region |ξ −η| < 3|ξ| and
2
1 1
I ≤C dS(η)
1
|ξ|b |ξ −η|a
ZP∩{|ξ−η|<3|ξ|}
1
1 1
≤C dS(η)
|ξ|b |ξ −η|a
i=−∞ZP∩{3i−1|ξ|<|ξ−η|<3i|ξ|}
X
1
1 1 C
≤C (3i|ξ|)2 =
|ξ|b |3iξ|a |ξ|a+b−2
i=−∞
X
where we have used the obvious fact that the area of P ∩ {3i−1|ξ| <
|ξ −η| < 3i|ξ|} is ≤ C(3i|ξ|)2.
Region 2: |ξ| < |ξ −η| < 2|ξ|. In this region |η| < 3|ξ| and
2
1 1
I ≤C dS(η)
2 |ξ|a |η|b
ZP∩{|η|<3|ξ|}
C
≤ (14)
|ξ|a+b−2
in complete analogy with region 1.
Region 3: |η| > 2|ξ| or |ξ − η| > 2|ξ|. In this region, |η| > |ξ| and
2|ξ −η| ≥ |ξ −η|+|ξ| ≥ |η|, thus
1
I ≤C dS(η)
3 |η|a+b
ZP∩{|η|>|ξ|}
∞
1
≤C dS(η)
|η|a+b
i=1 ZP∩{2i−1|ξ|<|η|<2i|ξ|}
X
∞
1 C
≤C (2i|ξ|)2 =
|2iξ|a+b |ξ|a+b−2
i=1
X
(15)
(cid:3)
We also have
ON THE UNIQUENESS OF SOLUTIONS TO THE GROSS-PITAEVSKII HIERARCHY9
Lemma 2.3. Let P, ξ as in Lemma (2.2) and let ǫ = 1 . Then
10
1 C
dS(η) ≤
|ξ −η|2−ǫ|ξ −η||η|2−ǫ |ξ|3−2ǫ
ZP 2
Proof. Consider separately the regions |η| > |ξ|, and |ξ −η| > |ξ|, and
2 2
(cid:3)
apply the previous lemma.
We are ready to prove the main estimate of Proposition (2.1)
Proof. Changing k +1 to 2, we have to show
|ξ |2
I = δ(τ +|ξ −ξ −ξ′|2 +|ξ |2 −|ξ′|2) 1 dξ dξ′ ≤ C
1 2 2 2 2 |ξ −ξ −ξ′|2|ξ |2|ξ′|2 2 2
Z 1 2 2 2 2
The integral is symmetric in ξ − ξ − ξ′ and ξ so we can integrate,
1 2 2 2
without loss of generality, over |ξ −ξ −ξ′| > |ξ |.
1 2 2 2
Case 1. Consider the integral I restricted to the region |ξ′| > |ξ | and
1 2 2
integrate ξ′ first.
2
Notice
δ(τ +|ξ −ξ −ξ′|2 +|ξ |2 −|ξ′|2)dξ′
1 2 2 2 2 2
= δ(τ +|ξ −ξ |2 −2(ξ −ξ )·ξ′ +|ξ |2)dξ′
1 2 1 2 2 2 2
dS(ξ′)
= 2 (16)
2|ξ −ξ |
1 2
where dS is surface measure on a plane P in R3, i.e. the plane ξ′·ω = λ
with ω ∈ S2 and λ = τ+|ξ1−ξ2|2+|ξ2|2.
2|ξ1−ξ2|
In this region
10 SERGIUKLAINERMANANDMATEI MACHEDON
dξ dS(ξ′)
I ≤ |ξ |2 2 2
1 1 |ξ |2|ξ −ξ | |ξ −ξ −ξ′|2|ξ′|2
ZR3 2 1 2 ZP 1 2 2 2
dξ dS(ξ′)
≤ |ξ |2 2 2
1 |ξ |2+2ǫ|ξ −ξ | |ξ −ξ −ξ′|2−ǫ|ξ′|2−ǫ
ZR3 2 1 2 ZP 1 2 2 2
dξ
≤ C|ξ |2 2
1 |ξ |2+2ǫ|ξ −ξ |3−2ǫ
R3 2 1 2
Z
≤ C (17)
Case 2. Consider the integral I restricted to the region |ξ′| < |ξ | and
2 2 2
integrate ξ first.
2
Notice
δ(τ +|ξ −ξ −ξ′|2 +|ξ |2 −|ξ′|2)dξ
1 2 2 2 2 2
ξ −ξ′ ξ −ξ′ ξ −ξ′ ξ −ξ′
= δ τ +| 1 2 −(ξ − 1 2)|2 +|(ξ − 1 2)+ 1 2|2 −|ξ′|2 dξ
2 2 2 2 2 2 2 2
(cid:18) (cid:19)
|ξ −ξ′|2 ξ −ξ′
= δ τ + 1 2 +2|ξ − 1 2|2 −|ξ′|2 dξ (18)
2 2 2 2 2
(cid:18) (cid:19)
dS(ξ )
2
=
4|ξ − ξ1−ξ2′|
2 2
wheredS issurfacemeasureonasphereP,i.e. thesphereinξ centered
2
at 1(ξ −ξ′) and radius 1 |ξ′|2 −τ − |ξ1−ξ2′|2 .
2 1 2 2 2 2
(cid:0) (cid:1)
Thus
dξ′ dS(ξ )
I ≤ |ξ |2 2 2
2 1 ZR3 |ξ2′|2 ZP |ξ2|2|ξ2− ξ1−2ξ2′||ξ1 −ξ2 −ξ2′|2
dξ′ dS(ξ )
≤ |ξ |2 2 2
1 ZR3 |ξ2′|2+2ǫ ZP |ξ2|2−ǫ|ξ2− ξ1−2ξ2′||ξ1 −ξ2 −ξ2′|2−ǫ
dξ′
≤ |ξ |2 2 ≤ C
1 |ξ′|2+2ǫ|ξ −ξ′|3−2ǫ
ZR3 2 1 2
(cid:3)