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arXiv:math/0602571v1 [math.AP] 27 Feb 2006
Scattering and small data completeness for the critical nonlinear
Schr¨odinger equation
Hans Lindblad∗and Avy Soffer†
University of California at San Diego and Rutgers University
February 2, 2008
Abstract
We prove Asymptotic Completeness of one dimensional NLS with long range nonlinearities. We
also prove existence and expansion of asymptotic solutions with large data at infinity.
1 Introduction
We consider the problem of scattering for the critical nonlinear Schr¨odinger equation in one space
dimension:
(1.1) i∂tv+∂2
xv−β|v|2v−γ|v|4v= 0
For related results in higher dimensions see e.g.[D1, G-V1, G-V2, G-V3, G-V4, GO] and cited references.
In one dimension the scattering problem for NLS and/or Hartree long range type were studied before
in [HKN,ST,Oz]. There are many other works in this direction, most are cited in the above references.
The work closest to ours, as far as the results are concerned is [HN]. In this paper the asymptotic
completeness is proved, and the L∞decay of the solution is shown. We use a different method, much
simpler, and we get an explicit construction of the phase function and the asymptotic form of the
solution as well. We also prove by the same method the existence theory of wave operators for large
data in the repulsive case, and small data in the general nonlinear case. In the work of [G-V4], the
Hartree equation in 3 or more dimensions is considered; the analysis uses, among other things the
representation of the solution in hyperbolic coordinates which we also use. But the approach used in
this paper is very different and much more involved than the work presented here. See also [Nak].
Recall first that a solution of the linear Schr¨odinger, i.e. β=γ= 0, with fast decaying smooth
initial data satisfies
(1.2) u(t, x)∼t−1/2eix2/4tbu(0, x/t)
where bu(t, ξ) = Ru(t, x)e−ixξ dx denotes the Fourier transform with respect to xonly.
∗Part of this work was done while H.L. was a Member of the Institute for Advanced Study, Princeton, supported by
the NSF grant DMS-0111298 to the Institute. H.L. was also partially supported by the NSF Grant DMS-0200226.
†Also a member of the Institute of Advanced Study, Princeton.Supported in part by NSF grant DMS-0100490.
1
We make the following ansatz for the solution of the nonlinear problem
(1.3) v(t, x) = t−1/2eix2/4tV(t, y), s =t, y =x/t
Plugging this into (1.1) gives, since (i∂t+∂2
x)t−1/2eix2/4t= 0,
(1.4) (i∂t+∂2
x)v(t, x) = (i∂t+∂2
x)t−1/2eix2/4tV(s, y)=t−1/2eix2/4ti∂t+∂2
x+i(x/t)∂x)V(s, y)
=t−1/2eix2/4ti∂s+s−2∂2
y)V(s, y)
Hence (1.1) becomes
(1.5) Ψ(V) = i∂sV−βs−1|V|2V−γs−2|V|4V+s−2∂2
yV= 0
It is easy to check that the general solution to the ODE
L(g) = id
dsg−β
s|g|2g−γ
s2|g|4g= 0
is of the form
g=aeiφ,where φ=−βa2ln |s|+γa4
s+b
for some constants aand b.
It is therefore natural with the following ansatz for the solution of the nonlinear problem
(1.6) V(s, y)∼V0(s, y) = a(y)eiφ(s,y), φ(s, y) = −βa(y)2ln |s|+b(y)
where a(y) and b(y) are any smooth sufficiently fast decaying functions of y=x/t.
First we show scattering, i.e. given any a(y) and b(y) as above we show that there is a solution V
as above.
Theorem 1.1. Suppose that a(y)and b(y)are polynomially decaying smooth real valued functions and
let v0(t, x) = t−1/2eix2
/4tV0(t, x/t), where V0is given by (1.6). Then if β≥0or βis small (1.1) has a
smooth solution v∼v0as t→ ∞, satisfying
(1.7) k(v−v0)(t, ·)kL∞+k(v−v0)(t, ·)kL2≤C(1 + ln(1 + t))2(1 + t)−1
We then show asymptotic completeness for small initial data, i.e. that there is a an asymptotic
expansion of the form (1.6).
Theorem 1.2. Suppose that f∈C∞
0. Then if ε > 0is sufficiently small (1.1) has a global solution
with data v(0, x) = εf(x). Moreover there are functions a(y)and b(y)such that with v0(t, x) =
t−1/2eix2
/4tV0(t, x/t), where V0is given by (1.6),v∼v0as t→ ∞;
(1.8) k(v−v0)(t, ·)kL∞≤C(1 + t)−3/2+Cε
2
2 The first order asymptotics and small data existence at infinity
The ansatz we use is an approximate solution of the form
v0(t, x) = s−1/2eix2
/4tV0(s, y),where V0(s, y) = a(y)eiφ(s,y), φ(s, y) = −βa(y)2ln |s|+b(y)
where a(y) and b(y) are any smooth sufficiently fast decaying functions of y=x/t and s=t.
(2.1) i ∂t+∂2
x−β|v0|2−γ|v0|4v0=t−1/2eix2/4ti ∂s+s−2∂2
y−βs−1|V0|2−γs−2|V0|4V0
=s−5/2eix2/4t−γ|V0|4V0+∂2
yV0=F0.
Assuming that a(y) and b(y) decay polynomially we have
|∂i
s∂j
yv0| ≤ CN(1 + βln |s|)j
s(1 + |y|)1/2(1 + |y|)N
and
(2.2) |∂i
s∂j
yF0| ≤ CN(1 + βln |s|)2+j
s(1 + |y|)5/2(1 + |y|)N
for any N. It follows that
|∂i
t∂j
xv0| ≤ CN
t+|x|1/2(1 + |x/t|)N
and
(2.3) |∂i
t∂j
xF0| ≤ CN(1 + βln |t|)2
t+|x|5/2(1 + |x/t|)N
for any N.
We now consider
w=v−v0
(i ∂t+∂2
x)w=G(v0, w) + F0
where
G(v0, w) = β|v0+w|2(v0+w)− |v0|2v0+γ|v0+w|4(v0+w)− |v0|4v0
The solution of the PDE
(2.4) (i ∂t+∂2
x)w=F
with vanishing final data at infinity is given by
w(t, x) = Z∞
tZE(t−s, x −y)F(s, y)dyds
where Eis the forward fundamental solution of i∂t+∂2
x.
3
Lemma 2.1. Suppose that
(2.5) i∂tw+∂2
xw=F
Then
(2.6) kw(t, ·)kL2≤ kw(t0,·)kL2+Zt
t0
kF(s, ·)kL2ds
Proof. The energy identity for this equation is
(2.7) d
dt Z|w(t, x)|2dx = 2 Zℑ(Fw)(t, x)dx
where ℑis the imaginary part.
The energy estimate is therefore
(2.8) kw(t, ·)kL2≤Z∞
t
kF(s, ·)kL2ds
From differentiating the equation it also follows that
(2.9) X
|α|≤1
k∂αw(t, ·)kL2≤Z∞
tX
|α|≤1
k∂αF(s, ·)kL2ds
We will now use the above inhomogeneous estimate together with an iterative procedure to get existence
for the equation in the previous section, of a solution wdecaying at infinity to zero fast, in a sense
having vanishing data at infinity. We therefore put up an iteration
(i∂t+∂2
x)w0=F0,(i∂t+∂2
x)wk+1 =G(v0, wk) + F0
where the solutions are defined as convolution with the fundamental solution that vanishes at infinity
(more precise later on). We must now first find the right estimates for w0and thereafter make an
assumption that the other iterates have similar bounds. It follows from (2.3) that
X
|α|≤1
k∂αF0(t, x)kL2≤C(1 + βln |1 + t|)2
t2
and hence Z∞
tX
|α|≤1
k∂αF0(t, ·)kL2dt ≤K(1 + βln |1 + t|)2
t
for some fixed constant K. We therefore make the inductive assumption that
(2.10) X
|α|≤1
k∂αwk(t, ·)kL2≤2K(1 + βln |1 + t|)2
t
4
Lemma 2.2.
(2.11) kw(t, ·)k2
L∞≤ kw(t, ·)kL2k∂xw(t, ·)kL2
Proof. Follows by H¨older’s inequality w2≤2R|w||wx|dx ≤2kwkL2k∂wkL2.
It follows that
kwk(t, ·)kL∞≤2K(1 + βln |1 + t|)2
t
Also using the estimates for v0,X
|α|≤1
|∂αv0| ≤ C0/t1/2
we get X
|α|≤1
k∂αG(v0, wk)kL2≤C1β
tX
|α|≤1
k∂αwkkL2,if t≥t0
for some number t0=t0(β)<∞.t0is chosen such that the r.h.s. of equation (2.10) is smaller than
1. Hence by the energy inequality and the inductive assumption we get for t≥t′
0(β);
(2.12) X
|α|≤1
k∂αwk+1(t, ·)kL2≤Z∞
t
C1β
sX
|α|≤1
k∂αwk(s, ·)kL2ds +K(1 + ln |1 + t|)2
t,
≤Z∞
t
C1β2K(1 + βln |1 + s|)2ds
s2+K(1 + ln |1 + t|)2
t≤(C2β+ 1)K(1 + ln |1 + t|)2
t
Hence (2.10) follows also for kreplaced by k+1, if βis so small that C2β≤1. This proves the theorem
for small β.
3 Global existence and decay for the initial value problem
Here we show that (1.1) has a global solution for small initial data and that the solution decays like
t−1/2. Let us suppose we are given initial data when s=t= 1,say in C∞
0.
Lemma 3.1. Suppose that gis real valued and
(3.1) i∂sV−gV =F.
Then
(3.2) |V(s)| ≤ |V(s0)|+Zs
s0
|F(σ)|dσ.
Proof. Multiplying with the integrating factor eiG(s), where G=Rg ds gives ∂sV eiG=−i F eiG and
the lemma follows from integrating this.
5
It now follows that if (1.5) holds then
(3.3) |V(s, y)| ≤ |V(1, y)|+Zs
1
|∂2
yV(σ, y)|dσ
σ2.
Hence the desired bound for Vwould follow if we can prove that for some fixed δ < 1;
(3.4) |∂2
yV(s, y)| ≤ Cε(1 + s)δ.
We will now derive this bound from energy bounds. We will assume that
(3.5) |V| ≤ C0ε
Writing (1.5) in the form
(3.6) i∂sV−s−2∂2
yV=F=βs−1|V|2V+γs−2|V|4V
and differentiating the above equation with respect to ygives
(3.7) i∂s−s−2∂2
yV(k)=F(k), V (k)=∂k
yV, F (k)=∂k
yF
We claim that
(3.8) kF(k)(s, ·)kL2≤Cs−11 + s−1kV(s, ·)k2
L∞kV(s, ·)k2
L∞kV(k)(s, ·)kL2, k = 0,1,2,3.
In fact
|F(0)| ≤ C s−1(1 + s−1|V|2)|V|3
(3.9)
|F(1)| ≤ C s−1(1 + s−1|V|2)|V|2|∂yV|(3.10)
|F(2)| ≤ C s−1(1 + s−1|V|2)|V|2|∂2
yV|+|V| |∂yV|2
(3.11)
|F(3)| ≤ C s−1(1 + s−1|V|2)|V|2|∂3
yV|+|V| |∂yV| |∂2
yV|+|∂yV|3
(3.12)
For k= 0,1 this is obvious and for k≥2 this follows from interpolation (proved by just integrating by
parts):
Lemma 3.2.
(3.13) k∂j
yVkk/j
L2k/j ≤CkVkk/j−1
L∞k∂k
yVkL2
For k= 2 we have
(3.14) k∂yV(t, ·)k2
L4≤CkV(t, ·)kL∞k∂2
yV(t, ·)kL2,
Similarly for k= 3 we have
k∂yV(t, ·)k3
L6≤CkV(t, ·)k2
L∞k∂3
yV(t, ·)kL2,(3.15)
k∂2
yV(t, ·)k3/2
L3≤CkV(t, ·)k2
L∞k∂3
yV(t, ·)kL2.(3.16)
6
Lemma 3.3. Suppose that
(3.17) i∂sW−s−2∂2
yW=F.
Then
(3.18) kW(s, ·)kL2≤ kW(s0,·)kL2+Zs
s0
kF(σ, ·)kL2dσ.
Assuming the bound
(3.19) kV(s, ·)kL∞≤Kε =δ
with a constant independent of swe have hence proven, using the above lemma and equations (3.7)-
(3.16) that
(3.20) kV(k)(t, ·)kL2≤ kV(k)(1,·)kL2+Zt
1
Dk(δ+δ2)kV(k)(τ, ·)kL2τ−1dτ,
from which it follows that
(3.21) kV(k)(t, ·)kL2≤Ckε(1 + t)Dk(δ+δ2), k = 0,1,2,3.
and by Lemma 2.2
(3.22) kV(k)(t, ·)kL∞≤Ck+1ε(1 + t)Dk(δ+δ2), k = 0,1,2
where Ckis a constant such that Pj≤kkV(j)(1,·)kL2≤Ckε. Now suppose that ε > 0 is so small that
(3.23) D34C3ε+ (4C3ε)2≤1/4.
It then follows from (3.3) that
(3.24) kV(s, ·)kL∞≤4C3εand k∂2
yV(s, ·)kL∞≤C3ε(1 + t)1/2.
This is more than needed in (3.4).
4 The completeness
Lemma 4.1. Suppose that gis real valued and
(4.1) i∂sV−gV =F
Then with G(s, y) = Rs
s0g(τ, y)dτ
(4.2) ∂s|V|+∂s(V eiG)≤2|F|
Proof. Multiplying with Vgives
(4.3) i∂s|V|2=ℑFV
and it follows that |∂s|V|| ≤ |F|. Multiplying with the integrating factor eiG(s), where G=Rg ds gives
∂sV eiG=−i F eiG and the lemma follows.
7
In the application g=−βs−1|V|2−γs−2|V|4and F=s−2∂2
yV. We already have proven that
(4.4) |F(s, y)| ≤ Cεs−2+Cε2
in the previous section. It therefore follows from the above lemma that the limit exists
(4.5) |V(s, y)| − a(y)≤Cεs−1+C ε2,where a(y) = lim
s→∞ |V(s, y)|
It therefore also follows from the lemma that
(4.6) |G(s, y)−φ(s, y)| ≤ C εs−1+C ε2,where φ(s, y) = a(y)2βln |s|+b(y)
and −b(y) is defined as the limit of G(s, y)−a(y)2βln |s|as s→ ∞. Hence
(4.7) V(s, y)−a(y)eiφ(s,y)≤C εs−1+Cε2
5 Higher order asymptotics and large data existence at infinity
We now want to construct a higher order asymptotic expansion at infinity. Therefore, we want to
linearize the operator
L(g) = id
dsg−β
sG1(g)−γ
s2G2(g), G1(g) = |g|2g, G2(g) = |g|4g
We have Gi(V0+W) = Gi(V0) + G′
i(V0)W+O(|W|2), where
G′
1(V0)W= 2|V0|2W+V2
0W , G ′
2(V0)W= 3|V0|4W+ 2|V0|2V2
0W
Here,
V0=a(y)eiφ(s,y)φ(s, y) = −βa(y)2ln |s|+b(y)
Hence the linearized operator is
L0W=L′(V0)W=id
dsW−β
sG′
1(V0)W−γ
s2G′
2(V0)W
Observe that L0is not complex linear. If Zis constant it therefore follows that (k≥1)
L0eiφ ln j|s|
skZ=eiφ ln j|s|
sk+1 (−2βa2−ik)Z−βa2Z+i j eiφ ln j−1|s|
sk+1 Z+eiφ ln j|s|
sk+2 −3γa2Z−2γa4Z
The inverse of
(−2βa2−ik)Z−βa2Z=Y
is given by
Z=1
k2+ 3β2a4−2βa2+i kY+1
k2+ 3β2a4βa2Y
8
and hence
(5.1)
L0eiφ ln j|s|
sk(k2+ 3β2a4)−2βa2+i kY+βa2Y=eiφ ln j|s|
sk+1 Y+i j eiφ lnj−1|s|
sk+1(k2+ 3β2a4)−2βa2+i kY+βa2Y
+eiφ ln j|s|
sk+2(k2+ 3β2a4)γa2(2βa2−i k)[3Y+ 2a2Y] + βa2(2a2Y−3Y)
It follows that
Lemma 5.1. Let Skdenote a finite sum of the form (k≥1)
(5.2) X
k′≥k, j ≥0
ck′j(y)eiφ ln j|s|
sk′, φ =−βa(y)2ln |s|+b(y),
with coefficients decaying polynomially in y. More precisely |∂αcjk ′(y)| ≤ CN(1 + |y|)−N, for any N
and cjk ′= 0 for k′,jsufficiently large. Here ln0|s|= 1.
Then if k≥1and ψk+1 ∈ Sk+1 there is φk∈ Skand ψk+2 ∈ Sk+2 such that
(5.3) L0φk=ψk+1 +ψk+2
Recall that Ψ(V) = L(V) + s−2∂2
yVand that L0=L′(V0). We have
Lemma 5.2. Let Ψn= Ψ ′(Vn)and suppose that Vn−V0∈ S1. Then if k≥1and ψk+1 ∈ Sk+1 there
is φk∈ Skand ψk+2 ∈ Sk+2 such that
(5.4) Ψnφk=ψk+1 +ψk+2
Proof. First, let φk∈ Skbe as in the previous lemma. Then (Ψ0−L0)φk=s−2∂2
yφk∈ Sk+2.
Furthermore Ψn−Ψ0=s−1βG ′(Vn)−s−1β G ′(V0) = s−1O(Vn−V0)∈ S2so Ψn−Ψ0φk∈ Sk+2.
By the results of previous sections, Ψ(V0)∈ S2. See e.g. equation (2.1). We will now inductively,
for n≥1 construct Vnsuch that Vn−V0∈ S1and Ψ(Vn)∈ Sn+2. Assume that this is true for n≤k.
Then by the above lemma ( with ψk+1 = Ψ(Vk)∈ Sk+2) we can find Vk+1 such that
(5.5) Ψ(Vk) + Ψ′(Vk)(Vk+1 −Vk)∈ Sk+3, Vk+1 −Vk∈ Sk+1.
Furthermore, there are bilinear forms in (X, Z ); G′′
i(U, V )(X, Z ) such that
(5.6) Gi(U) = Gi(V) + G′
i(V)(U−V) + G′′
i(U, V )(U−V , U −V)
Then
(5.7) Ψ(U) = Ψ(V) + Ψ ′(V)(U−V)−β
sG′′
1(U, V )(U−V , U −V)−γ
s2G′′
2(U, V )(U−V , U −V).
Hence
(5.8) Ψ(Vk+1) = Ψ(Vk) + Ψ ′(Vk)(Vk+1 −Vk)
−β
sG′′
1(Vk+1, Vk)(Vk+1 −Vk, Vk+1 −Vk)−γ
s2G′′
2(Vk+1, Vk)(Vk+1 −Vk, Vk+1 −Vk)∈ Sk+3.
9
Let
(5.9) vk(t, x) = t−1/2eix2/4tVk(t, y), s =t, y =x/t
Then (see equation (2.1))
(5.10) i∂tvk+∂2
xvk−β|vk|2vk−γ|vk|4vk=t−1/2eix2/4tΨ(Vk) = Fk
It follows that
(5.11) |∂αFk| ≤ Ck
(t+|x|)2+k
and hence X
|α|≤1
k∂αFN(t, ·)kL2≤KN
tN
for some constant KN. We then define w0= 0 and for k≥1:
(5.12) (i∂t+∂2
x)wk+1 =βG(vN, wk)wk+FN, k ≥0.
We will inductively assume that
(5.13) k∂wk(t, ·)kL2+kwk(t, ·)kL2≤4KN
NtN
Since by H¨older’s inequality
w2≤2Z|w||wx|dx ≤2kwkL2k∂wkL2≤ k∂wk2
L2+kwk2
L2
we also get
kwk(t, ·)kL∞≤4KN
N tN
Since also
|V0|=|a(y)| ≤ C0
it follows that
kvN(t, ·)kL∞=t−1/2kVN(t, ·)k ≤ t−1/2[kV0k+CNt−1]≤2C0
t1/2, t ≥tN=CN
C0
since by construction, VN−V0∈ S1. So C0is independent of N, if tNis sufficiently large. It follows
that
(5.14) kG(v1, wk)(t, ·)kL∞≤8C0
t, t ≥t′
N
Hence by the energy inequality (2.9), (5.12-14)
k∂wk+1(t, ·)kL2+kwk+1(t, ·)kL2≤Z∞
t
β8C0
s
4KN
NsNds+KN
sN+1 ds =32β C0
N+1KN
N tN≤2KN
NtN, t ≥t′′
N
if β > 0 is sufficiently small and t′′
Nis sufficiently large. Hence (5.13) follows also for k+ 1.
10
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