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HAL Id: hal-03300023

https://hal.archives-ouvertes.fr/hal-03300023

Preprint submitted on 26 Jul 2021

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Hs BOUNDS FOR THE DERIVATIVE NONLINEAR SCHRÖDINGER EQUATION

Hajer Bahouri, Trevor Leslie, Galina Perelman

To cite this version:

Hajer Bahouri, Trevor Leslie, Galina Perelman. Hs BOUNDS FOR THE DERIVATIVE NONLINEAR SCHRÖDINGER EQUATION. 2021. �hal-03300023�

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HAJER BAHOURI, TREVOR M. LESLIE, AND GALINA PERELMAN

ABSTRACT. We study the derivative nonlinear Schr¨odinger equation on the real line and obtain global-in- time bounds on high order Sobolev norms.

1. INTRODUCTION

We consider the Cauchy problem for the derivative nonlinear Schr¨odinger equation (DNLS) on the real lineR:

(1)

i∂tu+∂x2u=−i∂x(|u|2u), u

t=0 =u0 ∈Hs(R), s≥ 12.

We remark right away that the DNLS isL2 critical, as it is invariant under the scaling (2) u(t, x)7→uµ(t, x) := √

µu(µ2t, µx), µ >0.

The DNLS equation was introduced by Mio-Ogino-Minami-Takeda and Mjølhus [20, 21] as a model for studying magnetohydrodynamics, and it has received a great deal of attention from the mathematics community after being shown to be completely integrable by Kaup-Newell [13]. The infinitely many conserved quantities admitted by the DNLS equation play an important role in the wellposedness theory.

The first three—the mass, momentum, and energy—are as follows.

M(u) :=

Z

R

|u|2dx, (3)

P(u) := Im Z

R

uuxdx+1 2

Z

R

|u|4dx, (4)

E(u) :=

Z

R

|ux|2− 3

2Im(|u|2uux) + 1 2|u|6

dx.

(5)

Before stating our main result, let us give a very brief review of what is known about the wellposedness of the DNLS equation. More detailed overviews can be found, for example, in the introductions of [2]

and [14]. Local wellposedness in Hs(R) for s ≥ 12 was proven by Takaoka [25], improving earlier work [22] by Ozawa. On the other hand, fors < 12, the uniform continuity of the data-to-solution map fails in Hs(R) [3, 26]. One can, however, close the 12-derivative gap between the H12 threshold and the critical space L2(R) by working in more general Fourier-Lebesgue spaces, c.f. Gr¨unrock [6] and references therein.

Date: July 22, 2021.

This material is based upon work supported by the National Science Foundation under Grant No. DMS-1928930 while the authors participated in a program hosted by the Mathematical Sciences Research Institute in Berkeley, California, during the Spring 2021 semester.

1

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A line of results, due to Hayashi-Ozawa [8], Colliander-Keel-Staffilani-Takaoka-Tao [4], Wu [28], and Guo-Wu [7], establishes global well-posedness of the DNLS equation in Hs(R)fors ≥ 12, for ini- tial data having mass less than4π. Another line (Pelinovsky-Saalmann-Shimabukuro [23], Pelinovsky- Shimabukuro [24], and Jenkins-Liu-Perry-Sulem [12,11, 10]) uses inverse scattering techniques to es- tablish global wellposedness under stronger regularity and decay assumptions on the initial data, but without a smallness requirement on the mass.

The first and third authors proved in [2] that the DNLS equation is globally well-posed in Hs(R) fors≥ 12 and that solutions generated fromH12 initial data remain bounded inH12(R)for all time. There have also been some recent works below the aforementioneds = 12 threshold of uniformHs continuity with respect to initial data [3, 26]. Klaus-Schippa [17] gave Hs a priori estimates for 0 < s < 12 in the case of small mass, Killip-Ntekoume-Vis¸an [14] improved the small mass assumption to 4π and furthermore proved a global wellposedness result in Hs(R), 16 ≤ s < 12, for initial data with mass less than4π. Very recently, Harrop-Griffiths, Killip, and Vis¸an [9] have removed the small mass assumption both from theirHsa priori bounds,0< s < 12, as well as from their global wellposedness result inHs(R) with 16 ≤s < 12.

In this paper, we are concerned with the global-in-time boundedness of solutions to the DNLS equation inHsspaces. We prove that a uniform-in-time bound inHs(R)holds for alls≥ 12.

Theorem 1.1. Supposeuis a solution to the DNLS equation with initial datau0 ∈Hs(R),s ≥ 12. There exists a finite positive constantC =C(s,ku0kHs(R)), such that

sup

t∈R

ku(t)kHs(R) ≤C(s,ku0kHs(R)).

The main idea is to build off of theHsbounds with0< s < 12 from [9] and to take advantage of the complete integrability of the equation. As in [2], [9], the present work relies heavily on the conservation of the transmission coefficient for the spectral problem associated to the DNLS equation. This property has already been used in many other works; of particular relevance to us are the papers of G´erard [5], Killip-Vis¸an-Zhang [16], Killip-Vis¸an [15], and Koch-Tataru [18], on the cubic NLS and KdV equations.

Note that by continuity of the flow, and the preservation of the Schwartz class under the flow, we lose nothing by restricting attention to the Schwartz class; we will thus work exclusively with Schwartz functions for the remainder of the manuscript. We will also suppress the time dependence when it does not play a role.

One can easily prove Theorem 1.1 in the special case s = 1, using the conserved quantity E(u).

Indeed, simply rearranging (5) yields

kuk2H˙1(R)=E(u)−1

2kuk6L6(R)+3 2

Z

R

Im(|u|2uux) dx.

Clearly, the last term can be bounded above in absolute value by 12kuk2H˙1(

R) +Ckuk6L6(R), whence the desired bound follows by Sobolev embedding.

The higher-order Sobolev norms of integer order can be dealt with similarly, once we have a formula for the corresponding higher-order conserved quantities. We will show that for any nonnegative integer`, one of the conserved quantities is equal to a constant multiple ofkuk2H˙`(

R), plus terms which are of lower order. For nonintegers, we will use a sort of ‘generalized energy’, comparable tokuk2H˙s(

R), that will be defined in terms of the transmission coefficient of the DNLS spectral problem. We sketch presently the background necessary to define these objects precisely; for more details, see, for example, [1,12,11,10, 13,19,24,27].

The DNLS equation can be obtained as a compatibility condition of the following system [13]:

xψ =U(λ)ψ,

tψ = Υ(λ)ψ.

(6)

(4)

Here λ ∈ C is a spectral parameter, independent of t and x, and ψ = ψ(t, x, λ) is C2-valued. The operatorsU(λ)andΥ(λ)are defined by

U(λ) = −iσ32+iλU), Υ(λ) = −i(2λ4−λ2|u|23+

0 2λ3u−λ|u|2u+iλux

−2λ3u+λ|u|2u+iλux 0

(7) ,

where

σ3 =

1 0 0 −1

, U =

0 u u 0

.

To be more specific about the sense in which the DNLS is a compatibility condition, we note that u satisfies the DNLS equation if and only ifU andΥsatisfy the so-called ‘zero-curvature’ representation

∂U

∂t − ∂Υ

∂x + [U,Υ] = 0.

The first equation of (6) can be written in the form

(8) Lu(λ)ψ := (iσ3x−λ2−iλU)ψ = 0, which defines the scattering transform associated to the DNLS. Let us denote

+:={λ∈C: Imλ2 >0}.

Then givenu∈ S(R)andλ∈Ω+, there are unique solutions to (8) (the “J¨ost solutions”) exhibiting the following behavior at±∞:

ψ1(x, λ) =e−iλ2x 1

0

+o(1)

, asx→ −∞, ψ2+(x, λ) =e2x

0 1

+o(1)

, asx→+∞.

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Finally, we denote byau(λ)the Wronskian of the J¨ost solutions defined above:1 (10) au(λ) = det(ψ1(x, λ), ψ+2(x, λ)).

Using the second equation in (6), it can be shown that au(λ) is time-independent if u is a solution of (1). Furthermore, au is a holomorphic function of λ in Ω+, and one may determine the behavior ofau at infinity by transforming (8) into a Zakharov-Shabat spectral problem, linear with respect to the spectral parameter, c.f. [13], [23]. The equivalence between the two problems allows us to write

(11) lim

|λ|→∞, λ∈Ω+

au(λ) =e

i 2kuk2

L2(R). For fixedu, we can thus define the logarithm so that

(12) lim

|λ|→∞,λ∈Ω+

lnau(λ) =−i

2kuk2L2(R). Moreover,lnau(λ)admits an asymptotic expansion of the following form:

(13) lnau(λ) =

X

j=0

Ej(u)

λ2j as|λ| → ∞, λ∈Ω+.

Since au(λ) is time-independent, the quantities Ej(u) are conservation laws. They are all polynomial in u, u, and their derivatives. Furthermore, the Ej(u)’s inherit scaling properties fromau(λ). That is, for µ > 0, the fact thatauµ(λ) = au(λµ)implies that Ej(uµ) = µjEj(u), for each j ∈ N. The first

1Thetransmission coefficientmentioned earlier is the inverse ofau(λ).

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several of the Ej(u)’s are (up to multiplicative constants) the conserved quantities (3)–(5) mentioned earlier:

E0(u) = −i

2kuk2L2(R)=−i

2M(u), E1(u) = i

4P(u), E2(u) = −i 8E(u).

For each ` ∈ N, the quantity E2`(u) can be used to control kuk2H˙`(

R). Let us define, for ρ positive sufficiently large andL∈N,

(14) ϕL(u, ρ) = Im

"

lnau(p iρ)−

2L+1

X

j=0

Ej(u) (iρ)j

# .

If u is a solution of the DNLS equation, then ϕL(u, ρ) is time-independent, being a sum of time- independent quantities.

In order to establish bounds on the Hsnorm ofu, fors ≥ 12, we will show thatR

R ρ2s−1ϕ[s](u, ρ)dρ withR > 0large enough controls theH˙sseminorm ofu, in a sense to be made precise later. Here and below, we use[s]to denote the integer part of a real numbers.

Our proof of Theorem1.1relies on a good understanding of the structure of the remainder associated to the expansion (13). Note that whenλ2 = iρ, the imaginary part of this remainder (which is what we really use) is simplyϕL(u, ρ). In Section2, we will introduce a determinant characterization ofau(λ);

we use this characterization to formulate a technical statement (Lemma 2.1 below) on the size of the remainder. Assuming the result of Lemma2.1, we will prove Theorem1.1at the end of Section2. Then, in Section 3, we will prove our technical Lemma, completing the circle of ideas. Most of the work is contained in this last section.

Before moving on, let us establish a few notational conventions that we wish to add to the ones intro- duced above. First of all, we use the following normalization for the Fourier transform:

f(ζ) =b 1

√2π Z

R

eixζf(x) dx.

The symbolNwill denote the nonnegative integers, andN = N\{0}. We will use k · k2 to denote the Hilbert-Schmidt norm, andk · kwill denote the operator norm onL2(R). And we will use the following shorthand for derivatives:

D=−i∂x, L0 =iσ3x.

Whenever 2 ≤ p < ∞, we will uses(p)to denote the Sobolev exponents(p) = 121p such that the embeddingHs(p)(R),→Lp(R)holds.

Finally, we set notation for the following subset ofΩ+:

Γδ={λ∈Ω+:δ <arg(λ2)< π−δ}.

This notation will be useful in some of the intermediate steps we use to prove Theorem 1.1, as our estimates will frequently depend on Im|λ|λ22 (which is≤C(δ)onΓδ). However, the value ofδ > 0will be inconsequential for our final steps, where we will takeλ2to be pure imaginary. Therefore, for simplicity of presentation, we willfixδ >0once and for all and suppress dependence onδin all bounds below.

2. PROOF OF THE MAINRESULT

2.1. The Determinant Characterization of au(λ). An important property of au(λ) is the fact that it can be realized as a perturbation determinant:

(15) au(λ)2 = det(I −Tu(λ)2),

where

Tu(λ) =iλ(L0−λ2)−1U, λ ∈Ω+.

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The operatorTu(λ)is Hilbert-Schmidt, with

(16) kTu(λ)k22 = |λ|2

Im(λ2)kuk2L2(R). As a consequence of (15), we may write2

(17) lnau(λ) =−

X

k=1

Tr(Tu(λ)2k)

2k , ifkTu(λ)k<1.

This series will converge wheneverλ ∈ Γδ has large enough modulus; indeed, using the explicit kernel of(L0−λ2)−1, it can easily be shown that for anyp > 2, we have

(18) kTu(λ)k. |λ|kukLp(R) Im(λ2)1−1p

, λ∈Ω+, u∈Lp(R).

In particular, we can findR0 =R0(kuk

H13(R))such thatkTu(λ)k ≤ 12 for allλ∈Γδsatisfying|λ|2 ≥R0. We will fix the notationR0 for use below.

As we shall see later, each term of the series (17) can be expanded in powers ofλ−2:

(19) − Tr(Tu(λ)2k)

2k =

X

j=k−1

µj,k(u) λ2j . According to (13) and (17), theEj(u)’s should then satisfy

(20) Ej(u) =

j+1

X

k=1

µj,k(u).

We will use the following notation for the remainders after truncation of the expansions (17) and (19):

(21) lnau(λ) =−

2L+2

X

k=1

Tr(Tu(λ)2k)

2k +τL(u, λ), L∈N;

(22) −Tr(Tu(λ)2k)

2k =

2L+1

X

j=k−1

µj,k(u)

λ2jLk(u, λ), k ∈ {1, . . . ,2L+ 2}, L∈N.

The primary difficulty of the proof of Theorem1.1—and indeed, the subject of Lemma2.1—is the un- derstanding of the size and structure of the remainder termsτLk(u, λ), and to a lesser extent, theµj,k(u)’s.

On the other hand, forλ∈Γδwith large enough modulus, it is easy to bound theτL(u, λ)’s. For example, ifkTu(λ)k ≤ 12, then

L(u, λ)|=

lnau(λ) +

2L+2

X

k=1

TrTu2k(λ) 2k

X

k=2L+3

kTu(λ)k2k−2kTu(λ)k22

.kTu(λ)k4L+4kTu(λ)k22 .

kuk4L+4Hs(p)(R)kuk2L2(R)

|λ|(4L+4)(1−2p) , 2< p <∞, s(p) = 1 2 −1

p. (23)

The following table summarizes the various relationships among the quantities introduced above and will be helpful to keep track of the numerology. More precise information about the µj,k(u)’s andτLk(u, λ)’s will be provided below.

2This series expansion oflnau(λ)is consistent with the definition (12).

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TrTu2(λ)

2 TrTu4(λ)

4 TrTu6(λ)

6 · · · TrTu4L+2(λ)

4L+ 2 TrTu4L+4(λ) 4L+ 4

lnau(λ) = µ0,1(u) E0(u)

+ µ1,1(u)

λ2 + µ1,2(u) λ2

E1(u) λ2 + µ2,1(u)

λ4 + µ2,2(u)

λ4 + µ2,3(u) λ4

E2(u) λ4 ..

.

.. .

..

. . .. ...

+ µ2L,1(u)

λ4L + µ2L,2(u)

λ4L + µ2L,3(u)

λ4L + · · · + µ2L,2L+1(u) λ4L

E2L(u) λ4L + µ2L+1,1(u)

λ4L+2 + µ2L+1,2(u)

λ4L+2 + µ2L+1,3(u)

λ4L+2 + · · · + µ2L+1,2L+1(u)

λ4L+2 + µ2L+1,2L+2(u) λ4L+2

E2L+1(u) λ4L+2

+ τL1(u, λ) + τL2(u, λ) + τL3(u, λ) + · · · + τL2L+1(u, λ) + τL2L+2(u, λ) + τL(u, λ)

2.2. Structure of the Traces. In this section, we record all the information about the traces that we need in order to prove our main result. We deal first with the easy case of TrTu(λ)2, about which we need more explicit information. A straightforward computation gives us

(24) TrTu2(λ) = 2iλ2

Z

R

|bu(ζ)|2 ζ+ 2λ2dζ.

We determine the expansion ofTrTu2(λ)by simply substituting into (24) the identity 2λ2

ζ+ 2λ2 =

2L+1

X

j=0

− ζ 2λ2

j

+ ζ

ζ+ 2λ2 ζ

2 2L+1

, L∈N,

to obtain

(25) − TrTu2(λ)

2 =

2L+1

X

j=0

1

λ2j · i (−2)j+1

Z

R

ζj|u(ζ)|b 2

| {z }

=:µj,1(u)

− i

4L+1λ4L+2 Z

R

ζ2L+2|u(ζ)|b 2 ζ+ 2λ2

| {z }

=:τL1(u,λ)

, L∈N.

Now we state our main Lemma, which describes the structure of the otherµj,k(u)’s andτLk(u, λ)’s.

Lemma 2.1. For anyk ∈N, L∈N, the traces TrTu2k(λ)admit the decomposition(22). Theµj,k(u)’s andτLk(u, λ)’s satisfy the properties below, where for anyn∈Nwe denoteσ(n) = max{n,13}.

• Each µj,k(u) is a homogeneous polynomial of degree 2k in u, u, and their derivatives; it is homogeneous with respect to the natural scaling. We have

2`,2(u)|.kuk3Hσ(`−1)(R)kukH`(R), `∈N,

2`,k(u)|.kuk2kHσ(`−1)(R), `∈N, k ∈ {3, . . . ,2`+ 1},

2`+1,k(u)|.kuk2kH`(R), `∈N, k ∈ {2, . . . ,2`+ 2}.

(26)

(8)

• For|λ|2 > R0,λ∈Γδ, we have the following bounds:

L2(u, λ)|.α

kuk3Hσ(L)(

R)kukHL+α(R)

|λ|4L+2+2α , L∈N, 0≤α <1;

(27)

Lk(u, λ)|. kuk2kHL(R)

|λ|4L+4 , L∈N, k∈ {3, . . . ,2L+ 2}.

(28)

We postpone the proof of the Lemma until Section3.

2.3. Proof of Theorem1.1. In this section, we will prove Theorem1.1, assuming the result of Lemma2.1.

Fors ∈ N, the conclusion follows easily from Lemma 2.1, together with (20), (25), and an induction argument; we provide the details presently. Actually, the case s = 1 was already proved in the Intro- duction. Therefore, let us turn to our inductive hypothesis. For k = 1, . . . , `−1, we assume that the following bound holds.

(29) sup

t∈R

ku(t)kHk(R) ≤C(k,ku0kHk(R)).

We will prove that the same bound holds withk=`≥2.

First of all, for any integer`≥2, and any timet, we have ku(t)k2H˙`(

R) =C(`)µ2`,1(u(t)) by (25)

=C(`)

"

E2`(u(t))−

2`+1

X

k=2

µ2`,k(u(t))

#

by (20)

≤C(`)E2`(u0) + 1

2ku(t)k2H˙`(

R)+C(`,ku0kH`−1(R)).

To pass to the last line, we used time-independence of E2`(u(t)), the bounds (26), and our inductive hypothesis (29) (withk =`−1). Finally, using that

E2`(u0) =

2`+1

X

k=1

µ2`,k(u0)≤C(`,ku0kH`(R)), we get

sup

t∈R

ku(t)kH˙`(R)≤C(`,ku0kH`(R)),

which finishes the induction argument, and thus the proof of Theorem1.1fors ∈N.

It remains to consider the situation wheres /∈N. We start by recording the characterization ofϕL(u, ρ) in terms of the remainders τLk(u,√

iρ), and we also set notation for the quadratic part ofϕL(u, ρ). We also note that the caseL= 0is included in the definition.

ϕL(u, ρ) = Im

"

lnau(p iρ)−

2L+1

X

j=0

Ej(u) (iρ)j

#

= Im

"2L+2 X

k=1

τLk(u,p

iρ) +τL(u,p iρ)

#

, L∈N, (30)

ϕL,0(u, ρ) = ImτL1(u,p

iρ) = (−1)L 22L+1ρ2L

Z

R

ζ2L+2|bu(ζ)|2

ζ2+ 4ρ2 dζ, L∈N.

(31)

The conclusion of Theorem1.1for nonintegers≥ 12 will be deduced from the following two Lemmas.

Lemma 2.2. Supposeu∈ S(R),s >0,s /∈N, andR >0. Then the following comparison holds.

(32)

Z

R+

ρ2s−1[s],0(u, ρ)|dρ.skuk2H˙s(R) .s

Z R

ρ2s−1[s],0(u, ρ)|dρ+R2(s−[s])kuk2H˙[s](R).

(9)

Proof. Let us define the functionfν :R →R, for0< ν < 1, byfν(z) = |z|1+z2ν−12 . Note thatfν ∈L1(R) for this range ofν.

We make a direct substitution of the formula (31) for ϕ[s],0(u, ρ) into the left side of (32), then we switch the order of integration. Continuing the computation yields

Z R

ρ2s−1[s],0(u, ρ)|dρ= 1 22[s]+1

Z

R

ζ2[s]+2|u(ζ)|b 2 Z

R

ρ2(s−[s])−1 ζ2+ 4ρ2 dρdζ

= 1

22s+1 Z

R

|ζ|2s|bu(ζ)|2 Z

2R

|ζ|

fs−[s](z) dzdζ

= 1

4s+1kfs−[s]kL1(R)kuk2H˙s(R)− 1 2

Z

R

ζ 2

2s

|bu(ζ)|2 Z 2R|ζ|

0

fs−[s](z) dzdζ.

We estimate the second term on the right by means of the trivial replacement 1+z1 2 ≤1:

1 2

Z

R

ζ 2

2s

|bu(ζ)|2 Z 2R|ζ|

0

fs−[s](z) dzdζ ≤ 1 2

Z

R

ζ 2

2s

|bu(ζ)|2 Z 2R|ζ|

0

z2(s−[s])−1dzdζ = R2(s−[s])

s−[s] · kuk2H˙[s](

R)

4[s]+1 .

The comparison (32) follows.

Lemma 2.3. Supposeu∈ S(R),s >0,s /∈N. Denotingβ = max{[s], s+[s]+14([s]+1), 13}, we have

(33) |ϕ[s](u, ρ)−ϕ[s],0(u, ρ)| ≤ C(s,kukHβ(R))

ρs+[s]+1 (kukHs(R)+ 1), ∀ρ≥R0.

Proof. Choosep > 2to solve2([s] + 1)(1− 2p) =s+ [s] + 1. (Note thats(p) = s+[s]+14([s]+1) for this choice ofp.) Then forρ > R0, we have

[s](u, ρ)−ϕ[s],0(u, ρ)| ≤

2[s]+2

X

k=2

[s]k(u,p

iρ)|+|τ[s] (u,p

iρ)| by (30),(31)

≤C(s)

kuk3Hβ(R)kukHs(R) ρs+[s]+1 +

2[s]+2

X

k=3

kuk2kH[s](R)

ρ2[s]+2 +

kuk4[s]+4Hs(p)(R)kuk2L2(R)

ρ(2[s]+2)(1−p2)

by (27),(28),(23)

≤ C(s,kukHβ(R))

ρs+[s]+1 (kukHs(R)+ 1).

In the second line, we understand the sum overkto be empty if[s] = 0.

The conclusion of Theorem 1.1 for noninteger s ≥ 12 follows from Lemmas2.2 and 2.3, the time- independence of the quantityϕ[s](u, ρ)for solutions of the DNLS equation, and the bound

(34) sup

t∈R

ku(t)kHβ(R) ≤C(β,ku0kHβ(R)),

whereβ = max{[s],4([s]+1)s+[s]+1,13}is as in the statement of Lemma2.3. The bound (34) follows from our induction argument ifs >1and from the result of Harrop-Griffiths, Killip, and Vis¸an [9] if 12 ≤s <1.

(10)

Let us give the remaining details of the proof of Theorem1.1presently. For anyt ∈R, we have ku(t)k2H˙s(R).s

Z R0

ρ2s−1[s],0(u(t), ρ)|dρ+R2(s−[s])0 ku(t)k2H[s](R)

≤C(s) Z

R0

ρ2s−1[s](u(t), ρ)|dρ+C(s, R0,ku(t)kHβ(R))(ku(t)kHs(R)+ 1)

≤C(s) Z

R0

ρ2s−1[s](u0, ρ)|dρ+C(s,ku0kHs(R))(ku(t)kHs(R)+ 1)

≤ 1

2ku(t)k2Hs(R)+C(s,ku0kHs(R)),

which establishes the desired conclusion. Note that the first line in the calculation above is simply the upper bound in Lemma2.2. To pass from the first line to the second, we use Lemma2.3, followed by the lower bound of Lemma2.2. We use (34) and the time independence ofϕ[s](u(t), ρ)to pass to the third line. Finally, we justify the last line by noting that

Z R0

ρ2s−1[s](u0, ρ)|dρ.sC(s,ku0kHs(R)),

which follows from an application of Lemma2.3, followed by the lower bound in Lemma2.2.

3. PROOF OFLEMMA 2.1

3.1. Outline of the Proof. In this section, we expand eachTr(Tu2k(λ))in powers ofλ−2, up to a spec- ified order, and we establish bounds on the remainders, in order to prove our key Lemma 2.1. In Sec- tion3.2, we consider the caseL = 0, which is easy to treat explicitly but does not fit naturally into our argument for the other cases. WhenL≥1, we follow the strategy of [5], deducing the expansions of the traces from the expansion of the resolvent Lu(λ)−1. The relationship between Tu(λ) andLu(λ)is the following:

(35) Lu(λ) = (L0−λ2)(I−Tu(λ)).

Therefore,

(36) Lu(λ)−1 = (I −Tu(λ))−1(L0−λ2)−1 =

X

n=0

Tu(λ)n(L0 −λ2)−1

| {z }

=:Rn

, kTu(λ)k<1.

The point is that

(37) Tu2k(λ) = iλR2k−1U.

Thus, the part of L−1u (λ)that is of relevance to us is R2k−1, i.e., the term in the expansion (36) that is homogeneous of degree2k−1inu, u. In particular, we seek an expansion ofλR2k−1in powers ofλ−2, up to orderλ4L+2for a givenL∈N, and a good understanding of the remainder term.

Our strategy will be to examine the symbolR(x, ζ)of the pseudodifferential operatorLu(λ)−1. In Sec- tion3.3, we will expand the diagonal and antidiagonal partsRd(x, ζ)andRa(x, ζ)ofR(x, ζ)in powers of λ−2, determining recursively the form of each term of the expansion. Homogeneity considerations will then give us the desired expansion of λR2k−1 (and thus of TrTu2k(λ)) in powers of λ−2. In Sec- tion3.4, we identify theµj,k(u)’s from (22) and separate them from the remainder term. In Section3.5 we estimate the remainder term, finishing the proof of the Lemma. The final Section3.6consists of the proof by induction of a technical result stated in Section3.3.1, on the form of the terms of the expansions forRdandRa.

(11)

3.2. CaseL= 0. Let us note first of all that the desired decomposition in the caseL= 0reads lnau(λ) = [µ0,1(u) +λ−2µ1,1(u) +τ01(u, λ)

| {z }

=−12TrTu2(λ)

] + [λ−2µ1,2(u) +τ02(u, λ)

| {z }

=−14TrTu4(λ)

] +τ0(u, λ).

(See the table in Section2.1.) The only term which we have not already understood isτ02(u, λ); in order to treat it, we decomposeTu4(λ)explicitly as follows. A computation (the details of which are contained, for instance, in [2]) tells us that

TrTu4(λ) =i(2λ2)2 Z

R

u(x) (D+ 2λ2)−1u(x)2

(D−2λ2)−1u(x) dx.

Then, making a few simple manipulations, we can bring the right side of the equation above into the following form.

TrTu4(λ) = i

−2λ2 Z

R

u(x)

u(x)−(D+ 2λ2)−1Du(x) 2

u(x)−(D−2λ2)−1Du(x) dx

=− i 2λ2

Z

R

|u(x)|4dx− Z

R

|u|2u(x)(D−2λ2)−1Du(x) dx−2 Z

R

|u|2u(x)(D+ 2λ2)−1Du(x) dx

+ 2 Z

R

|u(x)|2(D+ 2λ2)−1Du(x)(D−2λ2)−1Du(x) dx+ Z

R

((D+ 2λ2)−1Du(x))2u(x)2dx

− Z

R

u(x)((D+ 2λ2)−1Du(x))2((D−2λ2)−1Du(x)) dx

=− 4 λ2

i

8kuk4L4(R)

| {z }

1,2(u)

−4τ02(u, λ).

To estimateτ02(u, λ), we use the following simple Lemma, the proof of which we omit.

Lemma 3.1. The following estimates hold, forλ∈Γδ.

• If0≤α1 ≤α2 ≤1, then

(38)

(D±2λ2)−1Du ˙

Hα1(R) .α2−α1

kukH˙α2(R)

(2 Im(λ2))α2−α1, ∀u∈Hα2(R).

• If2≤p <∞, then

(39)

(D±2λ2)−1Du

Lp(R) .p kukHs(p)(R), ∀u∈Hs(p)(R).

We estimate one of the terms defining τ02(u, λ)explicitly; the others can be dealt with in an entirely similar way.

1 λ2

Z

R

u(x)((D+ 2λ2)−1Du(x))2((D−2λ2)−1Du(x)) dx

≤ 1

|λ|2kukL6(R)k(D+ 2λ2)−1Duk2L6(R)k(D−2λ2)−1DukL2(R) .α

kuk3

H13(R)kukHα(R)

|λ|2+2α . We conclude thatτ02(u, λ)satisfies the required bound, finishing the caseL= 0.

3.3. Expanding the Resolvent.

(12)

3.3.1. Formal Expansion ofRaandRd. As stated above, forL≥1we seek an expansion of the symbol ofL−1u (λ), in powers ofλ−2. That is, we seek to understandR(x, ζ)in the expression

(40) L−1u (λ)f = 1

√2π Z

dζeixζR(x, ζ)fb(ζ).

The identityLu(λ)R(x, D) = Iimplies

(41) iσ3xR(x, ζ)−(ζσ32)R(x, ζ)−iλU(x)R(x, ζ) =I.

Introducing the new variablep= λζ2, this reads

(42) iσ3xR(x, ζ)−λ2(pσ3+ 1)R(x, ζ)−iλU(x)R(x, ζ) =I.

We splitRinto its diagonal and antidiagonal partsRdandRa, respectively, R(x, ζ) =Rd(x, ζ) +Ra(x, ζ),

and we also split equation (42) accordingly:

(43) iσ3xRd(x, ζ)−λ2(pσ3+ 1)Rd(x, ζ)−iλU(x)Ra(x, ζ) =I; (44) iσ3xRa(x, ζ)−λ2(pσ3+ 1)Ra(x, ζ)−iλU(x)Rd(x, ζ) = 0.

Setting the notation

Rd(x, ζ) =X

k≥0

1

λ2+2kRdk(x, p), Ra(x, ζ) =X

k≥0

1

λ3+2kRak(x, p), we rewrite (43) and (44) in expanded form:

I =−(pσ3+ 1)Rd0+

X

k=1

3xRdk−1−(pσ3+ 1)Rkd−iU Rak−1

λ2k ;

(45)

0 = −(pσ3+ 1)Ra0−iU Rd0+

X

k=1

3xRak−1−(pσ3+ 1)Rak−iU Rdk

λ2k .

(46)

We thus obtain the recursive system (47)–(49) below.

(47) Rd0(x, p) =−pσ3−1

p2−1 , Ra0(x, p) =− iU p2−1, (48) Rkd(x, p) = 1

p2−1

−iU Rak−1(x, p) +i∂xRdk−1(x, p)σ3

(pσ3−1), k≥1,

Rka(x, p) = 1 p2−1

iU Rdk(x, p) +i∂xRak−1(x, p)σ3

(pσ3 + 1)

= 1

p2−1

U2Rak−1(x, p)−U ∂xRk−1d (x, p)σ3+i∂xRak−1(x, p)σ3(pσ3+ 1) ,

k ≥1.

(49)

Note that we used the formula for Rdk(x, p)to pass to the second line in the formula for Rak(x, p). We also used several times the fact thatσ3A=−Aσ3 for any antidiagonal matrix.

We use the computations above to clarify the form of the Rdk’s and Rak’s; the precise statement is contained in the following Lemma.

(13)

Lemma 3.2. TheRdk’s andRak’s take the following form:

(50) Rkd(x, p) =

k

X

r=1

Rdk,r(x, p), k≥1,

(51) Rka(x, p) =

k

X

r=0

Rak,r(x, p), k≥0,

where the entries of the Rdk,r’s and Rk,ra ’s are homogeneous polynomials of degrees 2r and 2r + 1, respectively, inu, u, and their derivatives. More specifically, settingQγ = ∂xγ1U· · ·∂xγnU, forγ ∈ Nn, we have

Rdk,r(x, p) = 1 (p2−1)k+1

X

γ∈N2r

|γ|=k−r

Qγ(x)P|γ|(p)(pσ3−1), (52)

Rak,r(x, p) = 1 (p2−1)k+1

X

γ∈N2r+1

|γ|=k−r

Qγ(x)P|γ|(p).

(53)

Here and below we use the notationPnto denote any diagonal matrix whose diagonal entries are poly- nomials inphaving degree at mostn.

We postpone the proof of this Lemma until Section3.6, so as not to interrupt the flow of ideas.

3.3.2. The Truncated Expansion, and a Formula forR2m−1. For a fixedN ∈ N, we set the following notation. (Later we will setN = 2L.)

R(N)(x, p) =

N

X

k=0

Rdk(x, p) λ2+2k

| {z }

=:R(N)d (x,p)

+

N−1

X

k=0

Rak(x, p) λ3+2k

| {z }

=:R(N)a (x,p)

= Rd0(x, p) λ2

| {z }

=:R(N)d,0(x,p)

+

N

X

r=1 N

X

k=r

Rdk,r(x, p) λ2+2k

| {z }

=:R(N)d,r(x,p)

+

N−1

X

r=0 N−1

X

k=r

Rak,r(x, p) λ3+2k

| {z }

=:R(N)a,r(x,p)

. (54)

The symbol R(N)(x, p)is a truncated expansion of R(x, p) in inverse powers of λ, having diagonal and antidiagonal partsR(N)d ,R(N)a , respectively. The point of this definition is that, using Lemma3.2, we know thatR(N)d,r is homogeneous of degree2rinu,u, and their derivatives, whileR(Na,r)is homogeneous of degree 2r + 1 in these quantities. Expanding R(N) according to (54) and applying the recursive identities (45)–(46), we see thatR(N)(x, p)satisfies

[iσ3x−λ2(pσ3+ 1)−iλU(x)]R(N)(x, p) =I+Y(N)(x, p), whereY(N)(x, p) = Yd(N)(x, p) +Ya(N)(x, p),

Yd(N)(x, p) = 1

λ2+2N3(∂xRdN)(x, p), Ya(N)(x, p) =− 1

λ1+2NRaN(x, p)(pσ3−1).

This implies

(55) L−1u (λ) = R(N)(x, λ−2D)−L−1u (λ)Y(N)(x, λ−2D).

(14)

Recall thatR2m−1is the term in the expansion (36) which is homogeneous of order2m−1inu,u, and their derivatives. On the other hand, the portion ofR(N)which is of this homogeneity is preciselyR(N)a,m−1. Combining these considerations with (55), we see thatR2m−1is the difference betweenR(N)a,m−1 and the part ofL−1u (λ)Y(N)(x, λ−2D)that is homogeneous of degree2m−1inu,u, and their derivatives. Using the expansion (36) to isolate this part, we obtain:

R2m−1 =Ra,m−1(N) (x, λ−2D)− 1 λ2+2N

X

k+r0=m−1 k≥0,1≤r0≤N

Tu(λ)2k+1(L0−λ2)−1(L0RdN,r0)(x, λ−2D)

− 1 λ1+2N

X

k+r0=m−1 k≥0,0≤r0≤N

Tu(λ)2k(L0−λ2)−1RaN,r0(x, λ−2D)(λ−2L0+ 1).

(56)

3.4. Extracting theµj,m(u)’s. Combining (56) with (37), (54), and pulling out inverse powers ofλ, we easily find the following formula forTrTu2m(λ)withm≥2, truncated atN = 2L.

Tr(Tu2m(λ)) =

2L−1

X

j=m−1

1

λ2j+2 Tr[iU Raj,m−1(x, λ−2D)]

+ X

k+r=m−1 k≥0,1≤r≤2L

(−1)k λ4L+4+2kTr

(U(λ−2L0−1)−1)2k+2(L0Rd2L,r)(x, λ−2D)

+ X

k+r=m−1 k≥0,0≤r≤2L

i(−1)k+1 λ4L+2+2kTr

(U(λ−2L0−1)−1)2k+1R2L,ra (x, λ−2D)(λ−2L0+ 1) . (57)

We will refer to the three sums above asI,II, andIII, respectively.

We now identify the coefficients µj,m(u)’s and verify that they satisfy the properties claimed in Lemma 2.1. The claimed homogeneity properties will be clear from the formulas that we derive be- low; we will just need to verify the bounds (26). The latter are also straightforward to verify but will require us to use the structure of theRdk,r’s andRak,r’s from (52)–(53).

The first sum has the form

−2m

2L−1

X

j=m−1

µj,m(u) λ2j with

µj,m(u) = − i

2mλ2 Tr

U Rj,m−1a (x, λ−2D)

=− i 4mπ

X

γ∈N2m−1

|γ|=j−(m−1)

Tr

"

Z

U(x)Qγ(x) dx

Z P|γ|(λζ2) ((λζ2)2−1)j+1

dζ λ2

# (58) .

Note that ‘Tr’ denotes an operator trace in the first line, whereas it refers to the2×2matrix trace in the second and third lines. We will use the notation ‘Tr’ similarly in what follows without further comment.

Sinceλis presumed to lie inΓδ, a comparison of the degrees in the numerator and denominator ensures that the integrals overζare finite and their values are independent ofλ. The total number of derivatives in the x-integrals isj −(m−1), and we distribute these so that the highest order of the derivatives that fall on a single U is as small as possible. We list the bounds onµj,m(u) according tomand the parity ofj. In each case below,`is a strictly positive integer.

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