Schrödinger group on Zhidkov spaces

Clément Gallo UMR de Mathématiques, Bat. 425 Université Paris-Sud 91405 Orsay, France.

October 8, 2003

Résumé On considère le problème de Cauchy pour l’équation de Schrödinger non linéaire sur ℝ^{n}, dans des espaces de fonctions non (nécessairement) nulles à l’infini. Ce problème se pose par exemple dans l’étude de la stabilité des “dark solitons”. On montre que l’opérateur de Schrödinger génère un groupe sur les espaces de Zhidkov X^{k}(ℝ^{n}) pour k > n/2, puis que le problème de Cauchy pour SNL est localement bien posé dans ces mêmes espaces de Zhidkov. On montre aussi la conservation des invariants classiques, et, dans certains cas, on en déduit que le problème de Cauchy est globalement bien posé.

Abstract. We consider the Cauchy problem for nonlinear Schrödinger equations on ℝ^{n} with non zero boundary condition at infinity, a situation which occurs in stability studies of dark solitons. We prove that the Schrödinger operator generates a group on Zhidkov spaces X^{k}(ℝ^{n}) for k > n/2, and that the Cauchy problem for NLS is locally well-posed on the same Zhidkov spaces. We justify the conservation of classical invariants which implies in some cases the global well-posedness of the Cauchy problem.

Key words. Nonlinear Schrödinger equations, Dark solitons, Cauchy problem

AMS subject classifications. 35Q55, 35A07

1 Introduction

This paper is devoted to the Cauchy problem for the Nonlinear Schrödinger equation (NLS)

\{ \begin{matrix}iu_{t} + Δu + f(|u|^{2})u = 0, (t, x) ∈ ℝ × ℝ^{n} \\ u(0) = u_{0} \end{matrix}  (1)

with non zero boundary condition at infinity. Such boundary conditions occur in the “defocusing” case (e.g. f(|u|^{2}) = 1 − |u|^{2}), and are pertinent to many physical contexts. In nonlinear optics, the so-called dark soliton (see [KLD]) is a solution of (1) of the form u(x, t) = u_{v}(x − vt). For instance, for n = 1 and f(r) = 1 − r, we can compute that, for v ∈ (−\sqrt{2}, \sqrt{2}), u_{v}(x − vt) solves (1), where u_{v} is given by:

u_{v}(x) = \sqrt{1 − \frac{v^{2}}{2}} tanh (\sqrt{1 − \frac{v^{2}}{2}} \frac{x}{\sqrt{2}}) + i\frac{v}{\sqrt{2}}  (2)

The Gross-Pitaevskii equation (see [BeS1], [BeS2] and references therein)

iu_{t} + Δu + (1 − |u|^{2})u = 0

with the boundary condition u → 1 as |x| → ∞ is a model for Superfluid Helium II at a temperature near zero and for Bose-Einstein condensation. More generally, NLS with the boundary condition |u| → ρ_{0} as |x| → ∞ where ρ_{0} is a positive constant such that f(ρ_{0}^{2}) = 0 occurs in several physical contexts (see [BGMP]), an especially interesting particular case being the cubic-quintic “ψ^{3} − ψ^{5}” NLS.

These nonlinear Schrödinger equations possess solitons or solitary waves (see [BeS1], [dB], [M], [KLD]) and it is natural to study the Cauchy problem in spaces the solitary waves belongs to. Of course they can not be the usual Sobolev spaces H^{s}(ℝ^{n}) because of the boundary condition at infinity. A possibility would be to work in the affine space 1 + H^{s}(ℝ^{n}), when u → 1 as |x| → ∞, and this actually was done in [BeS1]. This approach obviously fails when only |u| (but not u) tends to a constant at infinity, as it is the case for the dark soliton (2). Also, the solitary wave ϕ could be only slowly decaying at infinity, implying that ϕ − 1 ∉ L^{2}(ℝ^{n}) (see [G] for the travelling wave of Gross-Pitaevskii equation in the 2-dimensional case).

In [Z0] Zhidkov introduces in the one-dimensional case the spaces X^{k} (with k a natural number), which consist of functions defined on ℝ, bounded and uniformly continuous, with derivatives up to order k in L^{2}, and proved that the Cauchy problem for NLS is locally well-posed in X^{k}.

Our aim here is to complete and generalize Zhidkov results. We introduce the Zhidkov spaces X^{k}(ℝ^{n}) in higher dimensions and prove that the linear Schrödinger equation defines a strongly continuous group on X^{k}(ℝ^{n}) if and only if k > n/2, and consequently that the Cauchy problem for NLS is locally well-posed in X^{k}(ℝ^{n}) if k > n/2. We also justify rigorously the conservation of natural invariants of the NLS yielding some global well-posedness in X^{k}(ℝ^{n}) for some defocusing NLS. A byproduct is the complete justification of the result of Zhiwu Lin in [ZL] that gives a criterion of stability for dark solitons of a class of NLS equations.

This paper is organized as follows. In section 2, we define the Zhidkov spaces X^{k}(ℝ^{n}) and state some useful properties of these spaces. In section 3, we prove that the linear Schrödinger equation is well posed in X^{k}(ℝ^{n}) if and only if k > n/2. In section 4, we show that the Cauchy problem for NLS is locally well-posed in X^{k}(ℝ^{n}), if k > n/2. In section 5, we introduce the renormalized energy for (1) and prove its conservation, for n = 1 or 2, under some hypothesis on f and u_{0}, and we show that in dimension 1, it implies the globalness of the solution of (1) in a defocusing case.

Notations Throughout this paper, C denotes a constant that can change from line to line. If j is a positive integer and u is a map of class C^{j} from ℝ^{n} into ℂ, and x, v_{1}, ..., v_{j} ∈ ℝ^{n}, we denote by D^{j}u(x)(v_{1}...v_{j}) the j^{th} differential of u at x applied to (v_{1}, ..., v_{j}). If E is a Banach space, we denote by C_{b}(ℝ, E) the space of continuous bounded functions from ℝ into E.

2 Some properties of Zhidkov spaces

Definition 2.1 Let n, k ∈ ℕ^{∗}. We define the space X^{k}(ℝ^{n}) as the closure for the norm

||u||_{X^{k}(ℝ^{n})} := ||u||_{L^{∞}} + ∑_{1⩽|α|⩽k} ||∂^{α}u||_{L^{2}}

of the space {u ∈ C^{k}(ℝ^{n}), bounded, uniformly continuous, with ∇u ∈ H^{k−1}}.

Note that a function in X^{k} is uniformly continuous.

Proposition 2.1 Let n, k ∈ ℕ^{∗}. Then X^{k+1}(ℝ^{n}) is dense in X^{k}(ℝ^{n}).

Proof. Let u ∈ X^{k}(ℝ^{n}), (ρ_{l})_{l⩾1} a mollifier sequence (i.e. ρ_{l} ∈ C_{c}^{∞}(ℝ^{n}), ∫_{ℝ^{n}} ρ_{l} = 1, ρ_{n} ⩾ 0, Supp ρ_{l} ⊂ B(0, 1/l)). Then ρ_{l} ∗ u is a sequence in X^{k+1} that converges to u in X^{k}. □

We show now a regularity result for functions in X^{k}(ℝ^{n}), for k > n/2.

Proposition 2.2 Let n ∈ ℕ^{∗}, let k = ⌊n/2⌋ + 1, and p ∈ ℝ such that

\{ \begin{matrix}p > n & if n is even \\ p = 2n & if n is odd \end{matrix} .

Then ∇u ∈ L^{p}(ℝ^{n}) for u ∈ X^{k}(ℝ^{n}) and there exists C > 0 such that

|u(x) − u(y)| ⩽ C|x − y|^{1−n/p}||∇u||_{L^{p}}, x, y ∈ ℝ^{n} .  (3)

Proof. By our choice of p and k, Sobolev’s embedding implies that

||∇u||_{L^{p}(ℝ^{n})} ⩽ C||∇u||_{H^{k−1}(ℝ^{n})} ⩽ C||u||_{X^{k}(ℝ^{n})}, u ∈ X^{k}(ℝ^{n}) .  (4)

Following the proof of Morrey’s theorem given in [B], we can show that there exists a constant C > 0 that depends only on n and p such that for any compact set K ⊂ ℝ^{n}, and any cube Q = [−r, r]^{n} containing K,

|u(x) − u(y)| ⩽ C|x − y|^{1−n/p}||∇u||_{L^{p}(Q)}, x, y ∈ K, u ∈ H_{loc}^{k}(ℝ^{n}) .  (5)

Since X^{k}(ℝ^{n}) ⊂ H_{loc}^{k}(ℝ^{n}), (4) and (5) prove the announced result. □

Remark . In our proof, we did not use the fact that the elements of X^{k}(ℝ^{n}) are uniformly continuous. Therefore, for k > n/2, an equivalent definition for X^{k} could be

X^{k}(ℝ^{n}) = {u ∈ L^{∞}(ℝ^{n}), ∇u ∈ H^{k−1}(ℝ^{n})} .

In particular, for k > n/2, H^{k}(ℝ^{n}) ⊂ X^{k}(ℝ^{n}).

3 The Schrödinger group on X^{k}(ℝ^{n})

In this section, we prove that if k > n/2, the Schrödinger operator defines a group on X^{k}(ℝ^{n}). More precisely,

Theorem 3.1 Let n ∈ ℕ^{∗}, k > n/2 and u_{0} ∈ X^{k}(ℝ^{n}). For t ∈ ℝ and x ∈ ℝ^{n}, the quantity

S(t)u_{0}(x) = \{ \begin{matrix}e^{−inπ/4}π^{−n/2}lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}u_{0}(x + 2\sqrt{t}z)dz & if t ⩾ 0 \\ e^{inπ/4}π^{−n/2}lim_{ε→0} ∫_{ℝ^{n}} e^{(−i−ε)|z|^{2}}u_{0}(x + 2\sqrt{−t}z)dz & if t ⩽ 0 \end{matrix} (6)

makes sense, and the family of operators (S(t))_{t∈ℝ} defines a strongly continuous group on X^{k}(ℝ^{n}). Moreover, there exists a constant C > 0 that depends only on n and k, such that for every u_{0} ∈ X^{k}(ℝ^{n}) and t ∈ ℝ,

||S(t)u_{0}||_{X^{k}} ⩽ C(1 + |t|^{ρ})||u_{0}||_{X^{k}}  (7)

where

ρ = \{ \begin{matrix}1/2 & if n is even \\ 1/4 & if n is odd \end{matrix} .

For convenience, we also denote S(t)u_{0}(x) by u(t, x).

All the computations below will be performed with t ⩾ 0. The case t ⩽ 0 is similar. Before starting the proof of the theorem itself, we need to prove some technical lemmas.

Lemma 3.1 Let n, k ⩾ 1, u_{0} ∈ X^{k}(ℝ^{n}), β > 0, x ∈ ℝ^{n} and t > 0. We define g : (β, ∞) → ℂ by

g(r) = ∫_{𝕊^{n−1}} u_{0}(x + 2\sqrt{t}rv)dv  (8)

then g ∈ X^{k}(β, ∞) (the definition of which is clear), and for all j ∈ {1...k}, g^{(j)} ∈ L^{2}(β, ∞, r^{n−1}dr) with

||g^{(j)}||_{L^{2}(β,∞,r^{n−1}dr)} ⩽ (2\sqrt{t})^{j−n/2}|𝕊^{n−1}|^{1/2}||u_{0}||_{X^{k}(ℝ^{n})} .  (9)

Proof. For any r ∈ (β, ∞), |g(r)| ⩽ |𝕊^{n−1}|||u_{0}||_{L^{∞}}, hence g ∈ L^{∞}(β, ∞).

Let ε > 0. Since u_{0} is uniformly continuous, there exists some δ > 0 such that |y−z| ⩽ δ implies |u_{0}(y)−u_{0}(z)| ⩽ ε. Let r_{1}, r_{2} be such that |r_{1}−r_{2}| ⩽ δ/(2\sqrt{t}). Then we get: |g(r_{1}) − g(r_{2})| ⩽ |𝕊^{n−1}|ε and g is uniformly continuous.

For j ∈ {1...k}, the Cauchy-Schwarz inequality and a change of variables yield:

∫_{β}^{∞} |g^{(j)}(r)|^{2}r^{n−1}dr

⩽ (2\sqrt{t})^{2j}|𝕊^{n−1}| ∫_{β}^{∞} ∫_{𝕊^{n−1}} |D^{j}u_{0}(x + 2\sqrt{t}rv)(v...v)|^{2}r^{n−1}dvdr
= (2\sqrt{t})^{2j}|𝕊^{n−1}| ∫_{2\sqrt{t}β}^{∞} ∫_{𝕊^{n−1}} |D^{j}u_{0}(x + rv)(v...v)|^{2} \frac{r^{n−1}dvdr}{(2\sqrt{t})^{n}}
⩽ (2\sqrt{t})^{2j−n}|𝕊^{n−1}|||u_{0}||_{X^{k}(ℝ^{n})}^{2}

which is the announced inequality. □

We state next two elementary lemmas which are straightforward consequences of the Leibniz formula

Lemma 3.2 Let k ∈ ℕ. There exists constants (b_{l,k})_{0⩽l⩽k} such that

(\frac{d}{dr} (\frac{1}{r}.))^{k} = ∑_{l=0}^{k} b_{l,k} \frac{1}{r^{2k−l}} \frac{d^{l}}{dr^{l}} .  (10)

Lemma 3.3 There exists constants (a_{k,j})_{0⩽j⩽k} such that

(\frac{d}{dr} (\frac{1}{r}.))^{k} (r^{n−1}.) = ∑_{j=0}^{k} a_{k,j} \frac{1}{r^{2k−n+1−j}} \frac{d^{j}}{dr^{j}} .  (11)

We are now ready to prove theorem 3.1. We fix n ∈ ℕ^{∗}, and we introduce a function χ ∈ C^{∞}(ℝ^{n}) such that:

• χ is radial

• χ increases along any half-line issued at 0

• χ ≡ 0 on {x, |x| ⩽ 1}

• χ ≡ 1 on {x, |x| ⩾ 2}

For any β > 0, we define χ_{β} = χ(./β). For convenience, we will also use the notation χ_{β}(|.|) = χ_{β}(.).

Proof of theorem 3.1. We assume first that k = ⌊n/2⌋ + 1. The first step of the proof is to show that the limit in formula (6) is well defined. Let us fix t > 0, β > 0 and ε > 0. We split the integral in (6) into two parts:

∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}u_{0}(x + 2\sqrt{t}z)dz = ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}(1 − χ_{β}(z))u_{0}(x + 2\sqrt{t}z)dz_{(I)}

+∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}χ_{β}(z)u_{0}(x + 2\sqrt{t}z)dz_{(II)}

We first consider the term (I):

|e^{(i−ε)|z|^{2}}(1 − χ_{β}(z))u_{0}(x + 2\sqrt{t}z)| ⩽ ||u_{0}||_{L^{∞}}1_{B(0,2β)}(z)

Then by Lebesgue’s theorem, the limit as ε → 0 of (I) exists and

|lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}(1 − χ_{β}(z))u_{0}(x + 2\sqrt{t}z)dz| ⩽ ||u_{0}||_{L^{∞}}|B(0, 1)|(2β)^{n} .(12)

We now consider the term (II). We compute it by using polar coordinates, and with the notations of lemma 3.1, we integrate by parts (which is justified by the regularity results on g proved in lemma 3.1) using lemma 3.3:

∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}χ_{β}(z)u_{0}(x + 2\sqrt{t}z)dz

= ∫_{0}^{∞} e^{(i−ε)r^{2}}χ_{β}(r)r^{n−1} (∫_{𝕊^{n−1}} u_{0}(x + 2\sqrt{t}rv)dv) dr

= ∫_{β}^{∞} (\frac{1}{2(i − ε)r} \frac{d}{dr} .)^{k} (e^{(i−ε)r^{2}}) χ_{β}(r)r^{n−1} (∫_{𝕊^{n−1}} u_{0}(x + 2\sqrt{t}rv)dv) dr

= (\frac{−1}{2(i − ε)})^{k} ∫_{β}^{∞} e^{(i−ε)r^{2}} ∑_{j=0}^{k} a_{k,j} \frac{1}{r^{2k−n+1−j}} \frac{d^{j}}{dr^{j}} [χ_{β}(r)g(r)] dr

= (\frac{−1}{2(i − ε)})^{k} ∑_{j=0}^{k} a_{k,j} ∑_{l=0}^{j} \binom{j}{l} ∫_{β}^{∞} e^{(i−ε)r^{2}} \frac{1}{r^{2k−j}} g^{(l)}(r)χ_{β}^{(j−l)}(r)r^{n−1}dr(13)

where we have used the Leibniz formula in the last equality. We will now apply Lebesgue’s theorem to each term of this sum.

For l = 0, j ∈ {0...k}, we have:

|e^{(i−ε)r^{2}} \frac{1}{r^{2k−j}} g(r)χ_{β}^{(j)}(r)r^{n−1}|

⩽ \frac{|𝕊^{n−1}|||u_{0}||_{L^{∞}}}{r^{2k−n+1}} \{ \begin{matrix}||χ||_{L^{∞}} & if j = 0 \\ \frac{r^{j}}{β^{j}} ||χ^{(j)}||_{L^{∞}} & if j ∈ \{1...k\} and r ⩽ 2β \\ 0 & if j ∈ \{1...k\} and r > 2β \end{matrix}

⩽ \frac{|𝕊^{n−1}|||u_{0}||_{L^{∞}}}{r^{2k−n+1}} 2^{j}||χ^{(j)}||_{L^{∞}}

Since k > n/2, ∫_{β}^{∞} dr/r^{2k−n+1} = β^{n−2k}/(2k − n) < ∞ and we can pass to the limit as ε → 0 by Lebesgue’s theorem. We obtain

|lim_{ε→0} ∫_{β}^{∞} e^{(i−ε)r^{2}} \frac{1}{r^{2k−j}} g(r)χ_{β}^{(j)}(r)r^{n−1}dr| ⩽ \frac{|𝕊^{n−1}|||u_{0}||_{L^{∞}}2^{j}||χ^{(j)}||_{L^{∞}}}{(2k − n)β^{2k−n}} . (14)

For l ⩾ 1, j ∈ {l...k}, we have:

|e^{(i−ε)r^{2}} \frac{1}{r^{2k−j}} g^{(l)}(r)χ_{β}^{(j−l)}(r)r^{n−1}| ⩽ ||χ^{(j−l)}||_{L^{∞}} \frac{1}{β^{j−l}} \frac{1}{r^{2k−j}} |g^{(l)}(r)|r^{n−1} .

Notice that 2(2k − j) − (n − 1) > 1, so that r → 1/r^{2k−j} ∈ L^{2}(β, ∞, r^{n−1}dr). The Cauchy-Schwarz inequality together with (9) lead to

∫_{β}^{∞} \frac{1}{r^{2k−j}} |g^{(l)}(r)|r^{n−1}dr

⩽ \frac{1}{(4k − 2j − n)^{1/2}β^{2k−j−n/2}} (2\sqrt{t})^{l−n/2}|𝕊^{n−1}|^{1/2}||u_{0}||_{X^{k}} .

So Lebesgue’s theorem can be applied, and

|lim_{ε→0} ∫_{β}^{∞} e^{(i−ε)r^{2}} \frac{1}{r^{2k−j}} g^{(l)}(r)χ_{β}^{(j−l)}(r)r^{n−1}dr|

⩽ \frac{|𝕊^{n−1}|^{1/2}||χ^{(j−l)}||_{L^{∞}}}{(4k − 2j − n)^{1/2}} ||u_{0}||_{X^{k}} \frac{(2\sqrt{t})^{l−n/2}}{β^{2k−l−n/2}}  (15)

We fix now β = 1. For j ∈ \{1...k\}, the quantities (2\sqrt{t})^{j−n/2} are majorized by C(t^{(1−n/2)/2} + t^{(k−n/2)/2}) where C is a positive constant. Therefore, using (12),(14) and (15), there exists a positive constant C such that for all u_{0} ∈ X^{k}, for all t > 0,

||u(t)||_{L^{∞}(ℝ^{n})} ⩽ C(1 + t^{(1−n/2)/2} + t^{(k−n/2)/2})||u_{0}||_{X^{k}(ℝ^{n})}  (16)

The second step of the proof consists in proving that u(t) → u_{0} in L^{∞} as t → 0. We first introduce some definitions.

Definition 3.1 For l ∈ {0...k} and h ∈ L^{2}(β, ∞, r^{n−1}dr), we define

T_{h}^{l} := ∫_{β}^{∞} e^{(i−ε)r^{2}}h(r)g^{(l)}(r)r^{n−1}dr

and we will say that such a quantity is “of type T^{(l)}”. If h can be written h(r) = χ_{β}^{(p)}(r)/r^{q} with 2q − n > 0, we will say that the “order” of T_{h}^{l} in β is p + q − n/2. If h is given as a linear combination of such terms, we define the “order” of T_{h}^{l} as the lowest order of non-zero monomials in the expression of h (remark that our “definition” of the order of T_{h}^{l} depends on the decomposition of h).

Let α ∈ (0, 1/(2n+1)), and m > 0 such that m ⩾ (n/2−1)/α. The following technical lemma gives a new expression of the term (II). It consists in doing as much integration by parts as necessary, in order to express (II) as a sum of terms of orders ⩾ m, which we are able to estimate in an appropriate way (see estimate (19) below.

Lemma 3.4 (II) can be written as a linear combination of terms of type T^{(l)} with l ∈ {0...k}, such that for l ∈ {1...k − 1}, the order of the terms of type T^{(l)} in this linear combination is ⩾ m.

Proof. Let l ∈ {1...k − 1}, p, q ∈ ℕ such that 2q − n > 0 and h(r) = χ_{β}^{(p)}(r)r^{q}. We transform T_{h}^{l} by integrations by parts, as in (13). After some computations, we get (this is actually (13) where we have replaced k by k − l, g by g^{(l)} and χ_{β} by χ_{β}^{(p)}(r)/r^{q}):

T_{h}^{l} =(\frac{−1}{2(i − ε)})^{k−l} ∑_{j=0}^{k−l} a_{k−l,j} ∑_{c=0}^{j} \binom{j}{c} ∑_{b=0}^{j−c} \binom{j − c}{b}(−q)...(−q − (j − c − b − 1))

× ∫_{β}^{∞} \frac{e^{(i−ε)r^{2}}χ_{β}^{(p+b)}(r)g^{(l+c)}(r)}{r^{2(k−l)+q−c−b}}r^{n−1}dr . (17)

Since 2(2(k − l) + q − c − b) − n ⩾ 2(k − l) + 2q − n, k ⩾ l and 2q − n > 0, we have written T_{h}^{l} as a linear combination of terms of types T^{(l)}, T^{(l+1)}...T^{(k)}. Moreover, the order of the terms of type T^{(l)} (that correspond to c = 0) in this sum is:

p + b + 2(k − l) + q − b − n/2 = \underbrace{p + q − n/2}_{order of T_{h}^{l}} +2 \underbrace{(k − l)}_{⩾1} .

The conclusion of this computation is that passing from the canonical expression of T_{h}^{l} to its new expression, the order of type T^{(l)} terms increases at least by 2.

We now use the above calculation to show the result by induction on l ∈ {0...k − 1}. Let us consider, for l ∈ {0...k − 1}, the induction hypothesis H_{l}: “(II) can be written as a sum of terms of type T^{(γ)}, 0 ⩽ γ ⩽ k, so that if 1 ⩽ γ ⩽ l, the term of type T^{(γ)} in this sum has order ⩾ m .” Formula (13) implies that

(II) = ∑_{l=0}^{k} T_{h^{(l)}}^{l}, whith h^{(l)}(r) = ∑_{j=l}^{k} a_{kj}\binom{j}{l}\frac{χ_{β}^{(j−l)}(r)}{r^{2k−j}} .

We have 2(2k − j) − n ⩾ 2k − n > 0, so that H_{0} is true thanks to (13). Let us now take l ∈ {1...k −1} and suppose H_{l−1}. The induction hypothesis implies that (II) can be written under the form: (II) = ∑_{γ}^{k} λ_{γ}T_{h_{γ}}^{γ} where λ_{γ} ∈ ℂ,

h_{γ} is a linear combination of χ_{β}^{(p)}(r)/r^{q}, and for γ ∈ {1...l − 1}, T_{h_{γ}}^{γ} ’s order is at least m. Applying the former calculation to the term T_{h_{l}}^{l}, we get a new expression of (II) where the terms of type T^{(γ)} with γ ⩽ l − 1 are unchanged (in particular, the new expression still verifies H_{l−1}), the order of the term of type T^{(l)} has increased by 2(k − l) ⩾ 2. We can start this process again, as long as it is necessary (but with a finite number of steps) to ensure that the term of type T^{(l)} is of order ⩾ m, and hence H_{l} is true. So we have proved that H_{k−1} is true, which is the result of the lemma. □

We give now an upper bound on |T_{h}^{l}|, for l ∈ {1...k}, with h(r) = χ_{β}^{(p)}(r)/r^{q} and 2q − n > 0:

|T_{h}^{l}| ⩽ \frac{||χ^{(p)}||_{∞}}{β^{p}} ∫_{β}^{∞} \frac{1}{r^{q}} |g^{(l)}(r)|r^{n−1}dr

⩽ \underbrace{\frac{||χ^{(p)}||_{∞}}{\sqrt{2q − n}} ||u_{0}||_{X^{k}}|𝕊^{n−1}|^{1/2}}_{=:C} \frac{(2\sqrt{t})^{l−n/2}}{β^{p+q−n/2}}  (18)

where we have used the Cauchy-Schwarz inequality and lemma 3.1. If we assume that β ⩾ 1 and T_{h}^{l}’s order is at least m, then

|T_{h}^{l}| ⩽ C \frac{(2\sqrt{t})^{l−n/2}}{β^{m}} .  (19)

We write β on the form β = β̃/(2\sqrt{t})^{α}, for t ⩽ 1 and β̃ > 2^{α}. Let us fix δ > 0. The choice of m and the fact that l ⩾ 1 ensure that l − n/2 + αm ⩾ 0. We choose β̃ > 2^{α} large enough such that for all t ⩽ 1,

\frac{(2\sqrt{t})^{l−n/2+αm}}{β̃^{m}} ⩽ δ , l ∈ \{1...k\}

and

\frac{|𝕊^{n−1}|||u_{0}||_{∞}2^{j}||χ^{(j)}||_{L^{∞}}}{(2k − n)β̃^{2k−n}} (2\sqrt{t})^{α(2k−n)} ⩽ δ , j ∈ \{0...k\} .

The property H_{k−1} proven in Lemma 3.4, (19) and (14) imply that there exists a constant C > 0 (which does not depend on δ) such that

∀ε > 0, |∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}χ_{β̃/(2\sqrt{t})^{α}}(z)u_{0}(x + 2\sqrt{t}z)dz| ⩽ Cδ .  (20)

Proposition 2.2 with p = 2n yields

|lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}(1 − χ_{β}(z))u_{0}(x + 2\sqrt{t}z)dz

−lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}(1 − χ_{β}(z))u_{0}(x)dz|

⩽ ∫_{ℝ^{n}}(1 − χ_{β}(z))|u_{0}(x + 2\sqrt{t}z) − u_{0}(x)|dz
⩽ ∫_{|z|⩽2β} C(2\sqrt{t}2β)^{1/2}||∇u_{0}||_{L^{p}}dz
⩽ C(2\sqrt{t})^{1/2}β^{n+1/2} = C(2\sqrt{t})^{1/2−α(n+1/2)}β̃^{n+1/2} .  (21)

By the choice of α, 1/2 − α(n + 1/2) > 0. So we can take t_{0} < 1 such that

∀t ⩽ t_{0}, C(2\sqrt{t})^{1/2−α(n+1/2)}β̃^{n+1/2} ⩽ δ  (22)

Therefore, for t ⩽ t_{0},

|lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}u_{0}(x + 2\sqrt{t}z)dz − lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}u_{0}(x)dz|

⩽ |lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}χ_{β}(z)u_{0}(x + 2\sqrt{t}z)dz| + |lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}χ_{β}(z)u_{0}(x)dz|

+ |lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}(1 − χ_{β}(z))(u_{0}(x + 2\sqrt{t}z) − u_{0}(x))dz|

⩽ Cδ + Cδ + δ  (23)

by (20), (20) applied to z → u_{0}(x) instead of u_{0} and (22). Inequality (23) and the well-known identity

lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}dz = π^{n/2}e^{inπ/4}

imply that u(t) → u_{0} = u(0) in L^{∞} as t → 0, t > 0. We could prove in the same way the left continuity at 0, and this, combined with the group property S(t+s) = S(t)S(s), t, s ∈ ℝ (which is easy to verify) shows that for all u_{0} ∈ X^{k}, S(.)u_{0} ∈ C(ℝ, L^{∞}(ℝ^{n})).

We consider now the general case k > n/2. For any multi-index α such that 1 ⩽ |α| ⩽ k, for any t > 0 (we have a similar formula for t < 0),

∂^{α}u(t) = \frac{e^{−inπ/4}}{π^{n/2}} lim_{ε→0} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}∂^{α}u_{0}(x + 2\sqrt{t}z)dz .

Here, ∂^{α}u_{0} ∈ L^{2}. Since (6) defines classically a unitary group on L^{2}, the map t → ∂^{α}u(t) belongs to C(ℝ, L^{2}(ℝ^{n})), with ||∂^{α}u(t)||_{L^{2}} = ||∂^{α}u_{0}||_{L^{2}}, t ∈ ℝ. Therefore we can conclude that for all u_{0} ∈ X^{k}, t → S(t)u_{0} ∈ C(ℝ, X^{k}(ℝ^{n})).

Moreover, since for all u_{0} ∈ X^{k}, S(.)u_{0} is bounded on [−1, 1], the Banach-Steinhaus theorem implies that ||S(t)||_{L(X^{k})} is bounded on [−1, 1]. Combining this with (16), we get (7). □

We give now the infinitesimal generator of S(t).

Theorem 3.2 Let k > n/2. The generator of the group (S(t))_{t∈ℝ} on X^{k}(ℝ^{n}) defined in theorem 3.1 is iΔ, its domain is X^{k+2}(ℝ^{n}).

Proof. We denote by A = S′(0) the generator of (S(t))_{t}. We split the proof into three steps. In the first step, we show that if u_{0} ∈

X^{k+4}(ℝ^{n}), (S(t)u_{0}−u_{0})/t ⟶_{t→0} iΔu_{0} in X^{k}(ℝ^{n}), and therefore (X^{k+4}(ℝ^{n}), iΔ) ⊂

(D(A), A). In the second step, we prove that (X^{k+2}(ℝ^{n}), iΔ) ⊂ (D(A), A). We conclude in the third step.

1^{st} step. Let u_{0} ∈ X^{k+4}(ℝ^{n}) ⊂ C^{4}(ℝ^{n}). We want to show that

\frac{S(t)u_{0} − u_{0}}{t} ⟶_{t→0} iΔu_{0} in X^{k} .  (24)

We will prove (24) for t → 0, t > 0. The proof is similar in the case t → 0, t < 0. By Taylor’s formula

u_{0}(x+2\sqrt{t}z)−u_{0}(x) = ∇u_{0}(x).2\sqrt{t}z+∫_{0}^{1}(1−s)D^{2}u_{0}(x+2\sqrt{t}z)(2\sqrt{t}z, 2\sqrt{t}z)ds ,

the fact that

∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}∇u_{0}(x).2\sqrt{t}zdz = 0 ,

and

iΔu_{0}(x) = \frac{1}{2} lim_{ε→0} \frac{e^{−inπ/4}}{π^{n/2}} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}D^{2}u_{0}(x)(2\sqrt{t}z, 2\sqrt{t}z)dz ,

we obtain

(\frac{S(t)u_{0}− u_{0}}{t} − iΔu_{0})(x)

= 4\frac{e^{−inπ/4}}{π^{n/2}} lim_{ε→0} ∫_{0}^{1}(1 − s)∫_{ℝ^{n}}e^{(i−ε)|z|^{2}}(D^{2}u_{0}(x + 2s\sqrt{t}z) − D^{2}u_{0}(x))(z,z)dzds .

Next,

∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}(D^{2}u_{0}(x + 2s\sqrt{t}z) − D^{2}u_{0}(x))(z, z)dz
= (\frac{−1}{2(i − ε)})^{2} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}Δ^{2}u_{0}(x + 2s\sqrt{t}z)(2s\sqrt{t})^{2}dz

−\frac{1}{2(i − ε)} ∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}(Δu_{0}(x + 2s\sqrt{t}z) − Δu_{0}(x))dz ,

hence

(\frac{S(t)u_{0} − u_{0}}{t} − iΔu_{0}) (x) = −4 [t ∫_{0}^{1}(1 − s)s^{2}S(ts^{2})Δ^{2}u_{0}(x)ds
+\frac{1}{2i} ∫_{0}^{1}(1 − s)(S(ts^{2})Δu_{0} − Δu_{0})(x)ds] .(25)

Since k > n/2, Δ^{2}u_{0} ∈ H^{k}(ℝ^{n}) ⊂ X^{k}(ℝ^{n}). Therefore (S(ts^{2})Δ^{2}u_{0})_{t,s∈[0,1]} is bounded in X^{k} and the first term in the right hand side of (25) tends to 0 in

X^{k} as t → 0. Since Δu_{0} ∈ H^{k+2}(ℝ^{n}) ⊂ X^{k}(ℝ^{n}), S(ts^{2})Δu_{0} − Δu_{0} ⟶_{t→0} 0 in

X^{k}, uniformly in s ∈ [0, 1]. Thus (24) has been proven. Therefore, if u_{0} ∈ X^{k+4}(ℝ^{n}), u_{0} ∈ D(A) and Au_{0} = iΔu_{0}.

2^{nd} step. Let u_{0} ∈ X^{k+2}(ℝ^{n}). Thanks to Proposition 2.1, there exists a sequence (v_{0}^{l})_{l∈ℕ} ⊂ X^{k+4}(ℝ^{n}) such that v_{0}^{l} → u_{0} in X^{k+2} ⊂ X^{k}. Hence iΔv_{0}^{l} → iΔu_{0} in H^{k} ⊂ X^{k}. Therefore (u_{0}, iΔu_{0}) belongs to the closure of {(v_{0}, iΔv_{0}), v_{0} ∈ X^{k+4}(ℝ^{n})} in X^{k}×X^{k}, and since the infinitesimal generator of a strongly continuous semigroup is a closed operator (see corollary 2.5 in [P] p5), this implies that u_{0} ∈ D(A) and Au_{0} = iΔu_{0}, i.e. (X^{k+2}(ℝ^{n}), iΔ) ⊂ (D(A), A).

3^{rd} step. Let u_{0} ∈ D(A). We want to show that Au_{0} = iΔu_{0} in D′. Since X^{k+2}(ℝ^{n}) is dense in X^{k}(ℝ^{n}), there exists a sequence (v_{0}^{l})_{l∈ℕ} ∈ X^{k+2}(ℝ^{n}) that tends to u_{0} in X^{k}(ℝ^{n}), and then

Av_{0}^{l} = iΔv_{0}^{l} ⟶_{l→∞} iΔu_{0} in D′ .

Let us show that Av_{0}^{l} → Au_{0} in D′. Let ϕ ∈ C_{c}^{∞}(ℝ^{n}). We have

< Av_{0}^{l}, ϕ >_{D′,D}=< ϕ, Av_{0}^{l} >_{(X^{k})′,X^{k}}=< A^{∗}ϕ, v_{0}^{l} >_{(X^{k})′,X^{k}}

→ < A^{∗}ϕ, u_{0} >_{(X^{k})′,X^{k}}=< ϕ, Au_{0} >_{(X^{k})′,X^{k}}=< Au_{0}, ϕ >_{D′,D} . (26)

To justify this calculation, we need to prove that ϕ ∈ D(A^{∗}). Since the Schrödinger group is continuous in the Schwartz space S, one has for any u ∈ D(A),

< Au, ϕ >_{D′,D} = lim_{t→0} ⟨\frac{S(t) − id}{t}u, ϕ⟩_{D′,D}
= lim_{t→0} ∫_{ℝ^{n}} u(y) (\frac{S(−t) − id}{t}\overline{ϕ}) (y)dy = − < iΔϕ, u >_{D′,D}

and then

| < S′(0)u, ϕ >_{D′,D} | ⩽ ||Δϕ||_{L^{1}}||u||_{X^{k}}, u ∈ D(A) .

It follows that ϕ ∈ D(A^{∗}) and (26) is justified. Therefore, Au_{0} = iΔu_{0} in D′. Now, u_{0} ∈ X^{k}, Au_{0} = iΔu_{0} ∈ X^{k}, hence Δ∇u_{0} ∈ (H^{k−1})^{n} and since ∇u_{0} ∈ (H^{k−1})^{n}, this yields ∇u_{0} ∈ (H^{k+1})^{n} and we conclude that u_{0} ∈ X^{k+2}. □

Finally, we show that the assumption k > n/2 we made in theorem 3.1 is sharp. More precisely, we have:

Proposition 3.1 Let us take n ∈ ℕ^{∗}. For x ∈ ℝ^{n}, we define

u_{0}(x) = \frac{e^{−i|x|^{2}}}{(1 + |x|^{2})^{n/2}log\sqrt{2 + |x|^{2}}} .

Then u_{0} ∈ X^{⌊n/2⌋}(ℝ^{n}), but if we define u(t, x) by formula (6), u(1/4, 0) = ∞.

Proof. It is clear that u_{0} ∈ L^{∞}(ℝ^{n}) and is uniformly continuous (because ∇u_{0} is bounded). Let α be a multi-index with 1 ⩽ |α| ⩽ ⌊n/2⌋. The “worst term” in ∂^{α}u_{0} is the one obtained when deriving e^{−i|x|^{2}} |α| times. This term is (−2ix)^{α}e^{−i|x|^{2}}(1 + |x|^{2})^{−n/2}/log\sqrt{2 + |x|^{2}}. Plainly,

∫_{ℝ^{n}} |\frac{(−2ix)^{α}e^{−i|x|^{2}}}{(1 + |x|^{2})^{n/2}log\sqrt{2 + |x|^{2}}} |^{2} dx ⩽ 2^{|α|} ∫_{ℝ^{n}} \frac{(|x|^{2})^{|α|}}{(1 + |x|^{2})^{n}log^{2}\sqrt{2 + |x|^{2}}} dx

= 2^{|α|}|𝕊^{n−1}| ∫_{0}^{∞} \frac{r^{2|α|+n−1}dr}{(1 + r^{2})^{n}log^{2}\sqrt{2 + r^{2}}} .

The right hand side term in the above inequality is finite, because 2n − (2|α| + n − 1) ⩾ 1 and ∫^{∞} \frac{dr}{rlog^{2}r} < ∞. Therefore u_{0} ∈ X^{⌊n/2⌋}(ℝ^{n}). Moreover, since u_{0} ∈ L^{2}, u_{0} ∈ H^{⌊n/2⌋}(ℝ^{n}), and u ∈ C(ℝ, H^{⌊n/2⌋}) is a solution of the linear Schrödinger equation.

For t = 1/4, x = 0,

∫_{ℝ^{n}} e^{(i−ε)|z|^{2}}u_{0}(x + 2\sqrt{t}z)dz = ∫_{ℝ^{n}} \frac{e^{−ε|y|^{2}}dy}{(1 + |y|^{2})^{n/2}log\sqrt{2 + |y|^{2}}}

= |𝕊^{n−1}| ∫_{0}^{∞} \frac{e^{−εr^{2}}r^{n−1}dr}{(1 + r^{2})^{n/2}log\sqrt{2 + r^{2}}}

which tends to +∞ as ε → 0, by the monotone convergence theorem and since ∫^{∞} dr/(rlogr) = ∞. An amusing fact is that for (t, x) ≠ (1/4, 0), u(t, x) is well defined by formula (6) and then u is a continuous function of (t, x) ∈ ℝ × ℝ^{n}\(1/4, 0). □

Remark The “ill-posedness” result in proposition 3.1 pertains to the dispersive blow-up phenomena described in [BS1] for KdV type equations and in [BS2] for general dispersive equations.

4 Local existence for NLS in X^{k}(ℝ^{n})

We consider now the Cauchy problem (1) with a more general nonlinearity:

\{ \begin{matrix}iu_{t} + Δu + F (u) = 0 \\ u(0) = u_{0} ∈ X^{k}(ℝ^{n}) \end{matrix}  (27)

where k > n/2 and F : X^{k}(ℝ^{n}) → X^{k}(ℝ^{n}) is of class C^{1}. We have shown in section 3 that the Schrödinger group S(t) defines a strongly continuous group on X^{k}(ℝ^{n}). By a classical fixed point argument one obtains the local wellposedness of the Cauchy problem (27). Namely, we have:

Theorem 4.1 Let M > 0. Then there exists T_{+}(M) > 0 and T_{−}(M) < 0 such that for all u_{0} ∈ X^{k}(ℝ^{n}) with ||u_{0}||_{X^{k}} ⩽ M, there exists an unique mild solution u ∈ C([T_{−}(M), T_{+}(M)], X^{k}) of (27). We recall that a mild solution of (27) is a solution of the integral equation

u(t) = S(t)u_{0} + i ∫_{0}^{t} S(t − s)F (u(s))ds , t ∈ [T_{−}, T_{+}] .  (28)

We also recall (see [P] or [CH]) that if u solves (28), it is a solution of (27) in the space C(ℝ, H_{loc}^{k−2}), and that if u_{0} ∈ D(S′(0)) = X^{k+2}, u ∈ C^{1}([T_{−}, T_{+}], X^{k}) ∩ C([T_{−}, T_{+}], X^{k+2}) is the classical solution of (27).

Proof. It is a direct application of proposition 4.3.3 in [CH]. □

We have furthermore

Theorem 4.2 For every u_{0} ∈ X^{k}(ℝ^{n}), there exists T_{∗}(u_{0}) ∈ [−∞, 0) and T^{∗}(u_{0}) ∈ (0, +∞] such that

• there exists a maximal solution u ∈ C(T_{∗}(u_{0}), T^{∗}(u_{0}), X^{k}) which is the unique solution of (28), for all T_{±}, T_{∗}(u_{0}) < T_{−} < 0 < T_{+} < T^{∗}(u_{0}),

• either T^{∗}(u_{0}) = +∞ or ||u(t)||_{X^{k}} ⟶_{t↑T^{∗}(u_{0})} +∞,

• either T_{∗}(u_{0}) = −∞ or ||u(t)||_{X^{k}} ⟶_{t↓T_{∗}(u_{0})} +∞.

Proof. It is a direct application of theorem 4.3.4 in [CH]. □

Theorem 4.3 If u_{0} ∈ X^{k+2}(ℝ^{n}), the solution u ∈ C(T_{∗}(u_{0}), T^{∗}(u_{0}), X^{k}) of (28) given by theorems 4.1 and 4.2 is a classical solution of (27) in X^{k}, which means that u∈C(T_{∗}(u_{0}), T^{∗}(u_{0}), X^{k+2}) ∩C^{1}(T_{∗}(u_{0}), T^{∗}(u_{0}), X^{k}).

Proof. It is a direct application of theorem 1.5 in [P], because F : X^{k} → X^{k} is of class C^{1}, and because the domain of iΔ (which is the generator of the group S(t)) on X^{k} is X^{k+2}. □

Here is a typical example where Theorem 4.1 applies.

Proposition 4.1 Let f : ℝ_{+} → ℝ be of class C^{k+1}. Then F : X^{k}(ℝ^{n}) → X^{k}(ℝ^{n}) defined by F (u) = f(|u|^{2})u is of class C^{1}.

Proof. Let u ∈ X^{k}(ℝ^{n}). It is clear that F (u) ∈ L^{∞}.

For any multi-index α such that 1 ⩽ |α| ⩽ k, ∂^{α}F (u) can be written as a linear combination of terms of type u^{a}\overline{u}^{b}f^{(l)}(|u|^{2})∂^{α_{1}}u...∂^{α_{r}}u∂^{β_{1}} \overline{u}...∂^{β_{s}} \overline{u}, where a, b, r , s, l are integers and α_{i}, β_{j} are multi-indices such that l ⩽ |α| ⩽ k and ∑ |α_{i}| + ∑ |β_{j}| ⩽ |α| ⩽ k.

By Sobolev’s embeddings and generalized Hölder’s inequality, it can be shown easily that ∂^{α}F (u) ∈ L^{2}, and hence F (u) ∈ X^{k}.

The proof that for u ∈ X^{k}, F′(u) maps X^{k} into X^{k} and that F′ is continuous is similar. □

We end up this paragraph by proving the persistency of higher regularity.

Proposition 4.2 Let n ∈ ℕ^{∗}, k > n/2 + 1, and u_{0} ∈ X^{k}(ℝ^{n}). We suppose f ∈ C^{k+1}(ℝ_{+}). For l ∈ {⌊n/2⌋ + 1...k}, we denote by u ∈ C(T_{∗}(l), T^{∗}(l), X^{l}) the solution of (28) with F (u) = f(|u|^{2})u, where ]T_{∗}(l), T^{∗}(l)[ is the maximal existence interval of u in X^{l}(ℝ^{n}). Then

T^{∗} := T^{∗}(⌊n/2⌋ + 1) = ... = T^{∗}(k)

and similarly, T_{∗} := T_{∗}(⌊n/2⌋ + 1) = ... = T_{∗}(k).

Proof. We make the proof for T^{∗}, it is similar for T_{∗}. A priori, for l > ⌊n/2⌋ + 1, T^{∗} ⩾ T^{∗}(l). We suppose by contradiction that T^{∗} > T^{∗}(l). For t ∈]T_{∗}, T^{∗}[,

u(t) = S(t)u_{0} + i ∫_{0}^{t} S(t − s) (f(|u(s)|^{2})u(s)) ds

hence for t ∈]T_{∗}(l), T^{∗}(l)[,

||u(t)||_{X^{l}} ⩽ ||S(t)||_{L(X^{l},X^{l})}||u_{0}||_{X^{l}} + ∫_{0}^{t} ||S(t − s)||_{L(X^{l},X^{l})}||f(|u(s)|^{2})u(s)||_{X^{l}}ds

We know by (7) that ||S(t)||_{L(X^{l},X^{l})} is bounded on [0, T^{∗}(l)], and that u is continuous from [0, T^{∗}(l)] into X^{⌊n/2⌋+1}(ℝ^{n}), so there exists M > 0 such that for all t ∈ [0, T^{∗}(l)], ||u(t)||_{X^{⌊n/2⌋+1}} ⩽ M . In particular, ||u(t)||_{L^{∞}} ⩽ M, which implies that there exists M̃ > 0 such that ||f(|u(s)|^{2})u(s)||_{X^{l}} ⩽ M̃||u(s)||_{X^{l}}. Finally,

||u(t)||_{X^{l}} ⩽ C||u_{0}||_{X^{l}} + ∫_{0}^{t} CM̃||u(s)||_{X^{l}}ds

and Gronwall’s lemma concludes as usual that ||u(t)||_{X^{l}} can not blow up at T^{∗}(l), which is a contradiction with the definition of T^{∗}(l). □

5 Conserved quantities and global well-posedness of NLS in X^{k}(ℝ^{n})

We first show the conservation of the renormalized energy for (1), for a “regular” initial data in X^{k}(ℝ^{n}), in the case n = 1 or 2.

Proposition 5.1 Let n = k = 1, 2, u_{0} ∈ X^{k+2}(ℝ^{n}), f ∈ C^{k+1}(ℝ_{+}). We denote by u ∈ C(T_{∗}, T^{∗}, X^{k+2}(ℝ^{n})) ∩ C^{1}(T_{∗}, T^{∗}, X^{k}(ℝ^{n})) the solution to the Cauchy problem (1) obtained in theorem 4.3, where (T_{∗}, T^{∗}) is its maximal existence interval. Let V (r) := − ∫^{r} f(s)ds. We assume that ∫_{ℝ^{n}} V (|u_{0}(x)|^{2})dx converges (in a sense we will precise in the proof). Then for all t ∈ (T_{∗}, T^{∗}), ∫_{ℝ^{n}} V (|u(t, x)|^{2})dx converges (in the same sense), and the energy is conserved:

∫_{ℝ^{n}} [|∇u(t, x)|^{2} + V (|u(t, x)|^{2})] dx = ∫_{ℝ^{n}} [|∇u_{0}(x)|^{2} + V (|u_{0}(x)|^{2})] dx (29)

proof. Let us multiply (1) by u_{t}^{−} and take the real part:

2Re(Δuu_{t}^{−}) = ∂_{t}V (|u|^{2}) .

We fix t ∈ (T_{∗}, T^{∗}), x ∈ ℝ^{n} and integrate over [0, t]:

2Re ∫_{0}^{t} Δu(s, x)u_{t}^{−}(s, x)ds = V (|u(t, x)|^{2}) − V (|u_{0}(x)|^{2})  (30)

We choose a non-increasing function θ ∈ C_{c}^{∞}(ℝ) such that:

θ(x) = \{ \begin{matrix}1 & if & x ⩽ 1 \\ 0 & if & x ⩾ 2 \end{matrix} .

For R > 0, we define θ_{R}(x) := θ(|x|/R), x ∈ ℝ^{n}. We have then:

||∇θ_{R}||_{L^{2}(ℝ^{n})} = R^{n/2−1} (∫_{ℝ^{n}} |θ′(|y|)|^{2}dy)^{1/2} .

The last integral is finite because θ′(|.|) ∈ C_{c}^{∞}(ℝ^{n}). In particular, {∇θ_{R}}_{R⩾1} is bounded in L^{2}(ℝ^{n}).

We now multiply (30) by θ_{R}(x) and integrate over ℝ^{n}:

2Re ∫_{ℝ^{n}} ∫_{0}^{t}Δu(s, x)u_{t}^{−}(s, x)dsθ_{R}(x)dx

= ∫_{ℝ^{n}} V (|u(t, x)|^{2})θ_{R}(x)dx − ∫_{ℝ^{n}} V (|u_{0}(x)|^{2})θ_{R}(x)dx .  (31)

Using Fubini’s theorem and an integration by parts, we calculate the left hand side of (31):

2Re ∫_{ℝ^{n}} ∫_{0}^{t} Δu(s, x)u_{t}^{−}(s, x)dsθ_{R}(x)dx

= − ∫_{ℝ^{n}} (|∇u(t, x)|^{2} − |∇u_{0}(x)|^{2}) θ_{R}(x)dx

−2Re ∫_{0}^{t} ∫_{ℝ^{n}} ∇u(s, x)∇θ_{R}(x)u_{t}^{−}(s, x)dxds

For convenience, we assume t > 0. Since ∇θ_{R} is supported in {x, |x| ⩾ R}, the Cauchy Schwarz inequality implies

|∫_{0}^{t} ∫_{ℝ^{n}} ∇u(s, x)∇θ_{R}(x)u_{t}^{−}(s, x)dxds|

⩽ sup_{s∈[0,t]}||u_{t}(s)||_{L^{∞}} (∫_{0}^{t} ||∇u(s)||_{L^{2}(|x|⩾R)}ds) ||∇θ_{R}||_{L^{2}(ℝ^{n})} ,

and this last quantity converges to 0 as R → ∞, because ||∇θ_{R}||_{L^{2}} is bounded. Moreover, ∫_{ℝ^{n}} (|∇u(t, x)|^{2} − |∇u_{0}(x)|^{2}) θ_{R}(x)dx clearly converges to ||∇u(t)||_{L^{2}}^{2}− ||∇u_{0}||_{L^{2}}^{2} as R → ∞. Hence, if we assume that

∫_{ℝ^{n}} V (|u_{0}(x)|^{2})dx := lim_{R→∞} ∫_{ℝ^{n}} V (|u_{0}(x)|^{2})θ_{R}(x)dx

exists (it a priori depends on the choice of θ), taking the limit in (31), for all t ∈ (T_{∗}, T^{∗}), we obtain that

∫_{ℝ^{n}} V (|u(t, x)|^{2})dx := lim_{R→∞} ∫_{ℝ^{n}} V (|u(t, x)|^{2})θ_{R}(x)dx

exists, and the energy is conserved. □

Remark This proof does not seem to work for n ⩾ 3, because we have used the existence of {θ_{R}}_{R⩾1} ⊂ C_{c}^{∞}(ℝ^{n}), with θ_{R}(x) ≡ 1 for |x| ⩽ R and ∇θ_{R} bounded in L^{2}(ℝ^{n}), and it can be shown that such a sequence does not exist for n ⩾ 3.

Next, we give a variant of theorem 5.1 in dimension 1, that will be useful later.

Proposition 5.2 Let ϕ, ϕ_{0} : ℝ → ℝ non-increasing functions of class C^{∞}, such that

ϕ(x) = ϕ_{0}(x) ≡ \{ \begin{matrix}1 & if & x ⩽ 0 \\ 0 & if & x ⩾ 1 \end{matrix} .

For R > 0, we define ϕ_{R}^{+}(x) = ϕ(x − R)ϕ_{0}(−x) and ϕ_{R}^{−}(x) = ϕ_{R}^{+}(−x).

Then if lim_{R→∞} ∫_{ℝ} V (|u_{0}(x)|^{2})ϕ_{R}^{±}(x)dx exists, so does lim_{R→∞} ∫_{ℝ} V (|u(t, x)|^{2})ϕ_{R}^{±}(x)dx

.

Proof. It suffices to replace θ_{R} by ϕ_{R}^{±} in the proof of theorem 5.1. We obtain in the same way that

∫_{ℝ} [V (|u(t, x)|^{2}) − V (|u_{0}(x)|^{2}) + (|∇u(t, x)|^{2} − |∇u_{0}(x)|^{2})] ϕ_{R}^{±}(x)dx

= ∓2Re∫_{0}^{t}∫_{ℝ}∇u(s, x)(∇ϕ(±x − R)ϕ_{0}(∓x) − ϕ(±x − R)∇ϕ_{0}(∓x))\overline{u_{t}}(s, x)dxds.

Passing to the limit as R → ∞, the fact that (∇ϕ(±x − R))_{R⩾0} is bounded in L^{2} and the assumption that ∫_{ℝ} V (|u_{0}(x)|^{2})ϕ_{R}^{±}(x)dx converges as R → ∞ imply

that lim_{R→∞} ∫_{ℝ} V (|u(t, x)|^{2})ϕ_{R}^{±}(x)dx exists, with

lim_{R→∞} ∫_{ℝ} V (|u(t, x)|^{2})ϕ_{R}^{±}(x)dx − lim_{R→∞} ∫_{ℝ} V (|u_{0}(x)|^{2})ϕ_{R}^{±}(x)dx

= − ∫_{ℝ} (|∇u(t, x)|^{2} − |∇u_{0}(x)|^{2}) ϕ_{0}(∓x)dx
±2Re ∫_{0}^{t} ∫_{ℝ} ∇u(s, x)∇ϕ_{0}(∓x)\overline{u_{t}}(s, x)dxds .

Remark that the limit lim_{R→∞} ∫_{ℝ} V (|u(t, x)|^{2})ϕ_{R}^{±}(x)dx depends on ϕ_{0} but not on

ϕ if lim_{R→∞} ∫_{ℝ} V (|u_{0}(x)|^{2})ϕ_{R}^{±}(x)dx does not. □

We want now to improve Proposition 5.1, to make it work for u_{0} ∈ X^{k}(ℝ^{n}), with k > n/2, and not only k > n/2+2. The price to pay is an extra assumption on the nonlinearity f:

Theorem 5.1 Let n = k = 1 or 2. We suppose that f ∈ C^{k+1}(ℝ_{+}) and that there exists some ρ_{0} > 0 such that f(ρ_{0}) = 0 and f′(ρ_{0}) < 0. We define

V (r) := − ∫_{ρ_{0}}^{r} f(s)ds .

If n = 1, we assume that {r, V (r) = 0} is discrete, and if n = 2, we assume that V is non-negative on ℝ_{+}. Let 0 < C_{1} < 1 < C_{2}, and δ_{0} > 0 (δ_{0} < ρ_{0}) such that

|r − ρ_{0}| ⩽ δ_{0} ⇒ C_{1} \frac{V′′(ρ_{0})}{2}(r − ρ_{0})^{2} ⩽ V (r) ⩽ C_{2} \frac{V′′(ρ_{0})}{2}(r − ρ_{0})^{2}

(the assumptions on f ensure that such a δ_{0} does exist). Finally we assume that u_{0} ∈ X^{k}(ℝ^{n}), x → V (|u_{0}(x)|^{2}) ∈ L^{1}(ℝ^{n}) and there exists 0 < δ_{1} < δ_{0} and A > 0 such that |x| ⩾ A implies ||u_{0}(x)|^{2} − ρ_{0}| ⩽ δ_{1}. Then, denoting (T_{∗}, T^{∗}) the maximal existence interval of the solution of (1) associated to u_{0}, the energy

E(u(t)) := ∫_{ℝ^{n}} [|∇u(t, x)|^{2} + V (|u(t, x)|^{2})] dx  (32)

is finite and conserved for all t ∈ (T_{∗}, T^{∗}).

Examples. We give some examples for which Theorem 5.1 is valid

• the cubic defocusing NLS equation:

iu_{t} + Δu + (ρ_{0} − |u|^{2})u = 0, x ∈ ℝ^{n}, n = 1 or 2.  (33)

In this case, f(r) = ρ_{0} − r, V (r) = (ρ_{0} − r)^{2}/2 ⩾ 0, and the assumptions of theorem 5.1 are verified. In fact, here, the assumption of the existence of δ_{1} and A can be relaxed, because it is a consequence of V (|u_{0}|^{2}) ∈ L^{1}.

• more generally, the “pure power case”:

iu_{t} + Δu + α(ρ_{0}^{p} − |u|^{2p})u = 0, x ∈ ℝ^{n}, n = 1 or 2.  (34)

where α > 0 and p ⩾ 1/2 if n = 1, p ⩾ 1 if n = 2. Here, f(r) = α(ρ_{0}^{p} −r^{p}), V (r) ⩾ 0 and V (r) = 0 if and only if r = 0. The assumption f ∈ C^{k+1}(ℝ^{+}) in Theorem 5.1 is verified only if p = 1 or p ⩾ 2 in the one-dimensional case, and if p = 1, 2 or p ⩾ 3 in the case n = 2. However, in other cases, X^{k} ∋ u → f(|u|^{2})u ∈ X^{k} is of class C^{1}, and it suffices for the conclusion of Theorem 5.1.

• the cubic-quintic NLS equation:

iu_{t} + Δu − α_{1}u + α_{3}u|u|^{2} − α_{5}u|u|^{4} = 0, x ∈ ℝ  (35)

where α_{1}, α_{3}, α_{5} are positive constant such that 3/16 < α_{1}α_{5}/α_{3}^{2} < 1/4. In this case, using some scale transformations (see [BGMP]), (35) can be rewritten as

iu_{t} + Δu + (|u|^{2} − ρ_{0})(2A + ρ_{0} − 3|u|^{2})u = 0  (36)

with 0 < A < ρ_{0}. Here, f(r) = (r − ρ_{0})(2A + ρ_{0} − 3r) and V (r) = (r − ρ_{0})^{2}(r − A), and the assumptions of theorem 5.1 are verified. They are also verified for other values of the parameters α_{1}, α_{3}, α_{5} (for instance A ⩽ 0), but this seems to be less interesting from a physical point of view (see [BGMP]).

Proof of theorem 5.1. Let us take a mollifier sequence (ρ_{l})_{l⩾1} (with ∫ ρ_{l} = 1, Supp ρ_{l} ⊂ B(0, 1/l) and ρ_{l} ⩾ 0).

In a first step, we will control |ρ_{0} − |ρ_{l} ∗ u_{0}(x)|^{2}| for l large and |x| ⩾ A.

|ρ_{0} − |ρ_{l} ∗ u_{0}(x)|^{2}|

⩽ (\sqrt{ρ_{0}} + ||u_{0}||_{∞}) \{ \begin{matrix}|ρ_{l} ∗ u_{0}(x)| − ρ_{0} & if |ρ_{l} ∗ u_{0}(x)| ⩾ \sqrt{ρ_{0}} & (case 1) \\ ρ_{0} − |ρ_{l} ∗ u_{0}(x)| & if |ρ_{l} ∗ u_{0}(x)| < \sqrt{ρ_{0}} & (case 2) \end{matrix}

In case 1, we have:

0 ⩽ |ρ_{l} ∗ u_{0}(x)| − \sqrt{ρ_{0}} ⩽ ∫_{ℝ^{n}} ρ_{l}(x − y)(|u_{0}(y)| − \sqrt{ρ_{0}})dy
⩽ ∫_{ℝ^{n}} ρ_{l}(x − y)||u_{0}(y)| − \sqrt{ρ_{0}}|dy .  (37)

In case 2, note that for |x| ⩾ A,

|u_{0}(x)|^{2} ⩾ ρ_{0} − |ρ_{0} − |u_{0}(x)|^{2}| ⩾ ρ_{0} − δ_{1} > 0 .  (38)

Let α := \sqrt{ρ_{0} − δ_{1}} and v ∈ ℂ such that |v| = 1 and v\overline{u_{0}(x)} ∈ iℝ. For y ∈ ℝ^{n}, the decomposition of u_{0}(y) on the ℝ-basis (u_{0}(x), v) of ℂ can be written as

u_{0}(y) = Re[u_{0}(y)\overline{u_{0}(x)}]\frac{u_{0}(x)}{|u_{0}(x)|^{2}} + P (u_{0}(y))v .

Then, because of Pythagoras’ theorem,

0 ⩽ \sqrt{ρ_{0}} − |ρ_{l} ∗ u_{0}(x)| ⩽ \sqrt{ρ_{0}} − |(∫_{ℝ^{n}} ρ_{l}(x − y)Re[u_{0}(y)\overline{u_{0}(x)}]dy) \frac{u_{0}(x)}{|u_{0}(x)|^{2}} |

We choose p as in Proposition 2.2, and l_{0} ∈ ℕ^{∗} such that

|x − y| ⩽ 1/l_{0} implies |u_{0}(x) − u_{0}(y)| ⩽ C|x − y|^{1−n/p}||∇u_{0}||_{L^{p}} ⩽ α/2 .

So for l ⩾ l_{0} and y ∈ B(x, 1/l), we have

Re[u_{0}(y)\overline{u_{0}(x)}] = |u_{0}(x)|^{2} + Re[(u_{0}(y) − u_{0}(x))\overline{u_{0}(x)}]
⩾ |u_{0}(x)|^{2} − |u_{0}(y) − u_{0}(x)||u_{0}(x)| ⩾ α^{2}/2 > 0 ,

and therefore, for l ⩾ l_{0},

0 ⩽ \sqrt{ρ_{0}} − |ρ_{l} ∗ u_{0}(x)|
⩽ \sqrt{ρ_{0}} − ∫_{ℝ^{n}} ρ_{l}(x − y)Re[u_{0}(y)\overline{u_{0}(x)}]dy \frac{1}{|u_{0}(x)|}
= \sqrt{ρ_{0}} − |u_{0}(x)| + \frac{1}{|u_{0}(x)|} ∫_{ℝ^{n}} ρ_{l}(x − y)Re[(u_{0}(x) − u_{0}(y))\overline{u_{0}(x)}]dy
⩽ |\sqrt{ρ_{0}} − |u_{0}(x)|| + ∫_{ℝ^{n}} ρ_{l}(x − y)|u_{0}(x) − u_{0}(y)|dy .  (39)

Let ε > 0. Since ρ_{l} ∗ u_{0} → u_{0} in L^{∞}, there exists l_{1} ⩾ l_{0} such that l ⩾ l_{1} implies |||ρ_{l} ∗u_{0}|^{2} −|u_{0}|^{2}||_{L^{∞}} ⩽ δ_{0} −δ_{1} and then for |x| ⩾ A, ||ρ_{l} ∗u_{0}(x)|^{2} −ρ_{0}| ⩽ δ_{0}. Therefore, for |x| ⩾ A and l ⩾ l_{1}, we have:

0 ⩽ V (|ρ_{l} ∗ u_{0}(x)|^{2}) ⩽ C_{2} \frac{V′′(ρ_{0})}{2}||ρ_{l} ∗ u_{0}(x)|^{2} − ρ_{0}|^{2} .  (40)

Let B ⩾ A + 1. (37), (39) and (40) imply, for l ⩾ l_{1}:

∫_{|x|⩾B} V (|ρ_{l} ∗ u_{0}(x)|^{2})dx

⩽ C_{2} \frac{V′′(ρ_{0})}{2}(\sqrt{ρ_{0}} + ||u_{0}||_{∞})^{2}
× ∫_{|x|⩾B} [(∫_{ℝ^{n}} ρ_{l}(x − y)||u_{0}(y)| − \sqrt{ρ_{0}}|dy)^{2} 1_{|ρ_{l}∗u_{0}(x)|−\sqrt{ρ_{0}}⩾0}

+ (|\sqrt{ρ_{0}} − |u_{0}(x)|| + ∫_{ℝ^{n}} ρ_{l}(x − y)|u_{0}(y) − u_{0}(x)|dy)^{2} 1_{|ρ_{l}∗u_{0}(x)|−\sqrt{ρ_{0}}⩽0}] dx

⩽ C_{2} \frac{V′′(ρ_{0})}{2}(\sqrt{ρ_{0}} + ||u_{0}||_{∞})^{2} ∫_{|x|⩾B} [∫_{ℝ^{n}} ρ_{l}(x − y)||u_{0}(y)| − \sqrt{ρ_{0}}|^{2}dy

+ 2||u_{0}(x)| − \sqrt{ρ_{0}}|^{2} + 2 (∫_{ℝ^{n}} ρ_{l}(x − y)|u_{0}(y) − u_{0}(x)|dy)^{2}] dx . (41)

We will now control each integral in the right hand side of (41). We begin by the first one:

∫_{|x|⩾B} ∫_{ℝ^{n}} ρ_{l}(x − y)||u_{0}(y)| − \sqrt{ρ_{0}}|^{2}dydx
⩽ ∫_{|x|⩾B} ∫_{|y|⩾B−1/l} ρ_{l}(x − y)\frac{||u_{0}(y)|^{2} − ρ_{0}|^{2}}{ρ_{0}} dydx
⩽ \frac{1}{ρ_{0}} ∫_{|y|⩾B−1/l} ||u_{0}(y)|^{2} − ρ_{0}|^{2}dy
⩽ \frac{2}{C_{1}ρ_{0}V′′(ρ_{0})} ∫_{|y|⩾B−1/l} V (|u_{0}|^{2})dy .  (42)

Next,

∫_{|x|⩾B} 2|\sqrt{ρ_{0}} − |u_{0}(x)||^{2}dx ⩽ \frac{2}{ρ_{0}} ∫_{|x|⩾B} |ρ_{0} − |u_{0}(x)|^{2}|^{2}dx
⩽ \frac{4}{C_{1}ρ_{0}V′′(ρ_{0})} ∫_{|x|⩾B} V (|u_{0}|^{2})dx . (43)

It is a bit more difficult to find an upper bound to the third one. We provisionally admit the following lemma:

Lemma 5.1 Let n ∈ ℕ^{∗}, k > n/2 and u_{0} ∈ X^{k}(ℝ^{n}). Then for all (x, y), the function f : [0, 1] → ℂ, t → u_{0}(x + t(y − x)) is absolutely continuous.

Thanks to this lemma, we can write:

|u_{0}(y) − u_{0}(x)| = | ∫_{0}^{1}(y − x).∇u_{0}(x + t(y − x))dt| ,

and

∫_{|x|⩾B} (∫_{ℝ^{n}} ρ_{l}(x − y)|u_{0}(y) − u_{0}(x)|dy)^{2} dx

⩽ ∫_{|x|⩾B} ∫_{ℝ^{n}} ρ_{l}(x − y)|y − x|^{2} ∫_{0}^{1} |∇u_{0}(x + t(y − x))|^{2}dtdydx

⩽ \frac{1}{l^{2}} ∫_{|x|⩾B} ∫_{0}^{1} ∫_{ℝ^{n}} ρ_{l}(\frac{ỹ − x}{t})|∇u_{0}(ỹ)|^{2} \frac{dỹ}{t^{n}}dxdt

⩽ \frac{1}{l^{2}} ∫_{0}^{1} ∫_{ℝ^{n}} ∫_{ℝ^{n}} ρ_{l}(−x̃)|∇u_{0}(ỹ)|^{2}dx̃dỹdt = \frac{||∇u_{0}||_{L^{2}}}{l^{2}}  (44)

(here we have made successively the changes of variables ỹ = ty + (1 − t)x and x̃ = \frac{x−ỹ}{t}). By possibly enlarging B one may assume that

(\sqrt{ρ_{0}} + ||u_{0}||_{∞})^{2} \frac{C_{2}}{C_{1}ρ_{0}} [∫_{|y|⩾B−1} V (|u_{0}|^{2})dy + 2 ∫_{|x|⩾B} V (|u_{0}|^{2})dx] ⩽ ε/2 .(45)

We also choose l_{2} ⩾ l_{1} such that

2C_{2} \frac{V′′(ρ_{0})}{2}(\sqrt{ρ_{0}} + ||u_{0}||_{∞})^{2} \frac{||∇u_{0}||_{L^{2}}}{l_{2}^{2}} ⩽ ε/2 .  (46)

Combining (41), (42), (43), (44), (45), (46), we obtain finally, for l ⩾ l_{2},

∫_{|x|⩾B} V (|(ρ_{l} ∗ u_{0})(x)|^{2})dx ⩽ ε .

In particular, V (|(ρ_{l}∗u_{0})|^{2}) ∈ L^{1}(ℝ^{n}). By possibly enlarging B one may assume that

∫_{|x|⩾B} V (|u_{0}(x)|^{2})dx ⩽ ε .

Moreover, there exists l_{3} ⩾ l_{2} such that for l ⩾ l_{3},

∫_{|x|⩽B} |V (|(ρ_{l} ∗ u_{0})(x)|^{2}) − V (|u_{0}(x)|^{2})|dx ⩽ ε

because ρ_{l} ∗ u_{0} → u_{0} in L^{∞}. The last three inequalities and the triangular inequality show that for l ⩾ l_{3},

∫_{ℝ^{n}} |V (|(ρ_{l} ∗ u_{0})(x)|^{2}) − V (|u_{0}(x)|^{2})|dx ⩽ 3ε

and therefore V (|ρ_{l} ∗ u_{0}|^{2}) → V (|u_{0}|^{2}) in L^{1}.

For l ∈ ℕ^{∗}, we denote by u_{l}(t) the solution of (1) with initial data u_{l}(0) = ρ_{l} ∗ u_{0}, and we denote by (T_{∗}(l), T^{∗}(l)) its maximal existence interval. Theorem 5.1 ensures that for t ∈ (T_{∗}(l), T^{∗}(l))

∫_{ℝ^{n}} [|∇u_{l}(t)|^{2} + V (|u_{l}(t)|^{2})] dx = ∫_{ℝ^{n}} [|∇ρ_{l} ∗ u_{0}|^{2} + V (|ρ_{l} ∗ u_{0}|^{2})] dx . (47)

Let T_{∗} < T̃_{1} < T̃_{2} < T^{∗}. By continuity with respect to the initial data (see for example [CH]), there exists K > 0, δ > 0 such that ||ρ_{l} ∗u_{0} −u_{0}||_{X^{k}} ⩽ δ implies T^{∗}(l) > T̃_{2}, T_{∗}(l) < T̃_{1}, and

||u_{l}(t) − u(t)||_{X^{k}} ⩽ K||ρ_{l} ∗ u_{0} − u_{0}||_{X^{k}} t ∈ [T̃_{1}, T̃_{2}] .

In particular, since ρ_{l} ∗ u_{0} → u_{0} in X^{k}, we have that ∇u_{l}(t) → ∇u(t) in L^{2}. Moreover, ∇ρ_{l} ∗ u_{0} = ρ_{l} ∗ ∇u_{0} → ∇u_{0} in L^{2} and we have already shown that V (|ρ_{l} ∗ u_{0}|^{2}) → V (|u_{0}|^{2}) in L^{1}. In order to take the limit as l → ∞ in (47), it remains to take care of the term ∫_{ℝ^{n}} V (|u_{l}(t)|^{2})dx. We distinguish the cases n = 2 and n = 1.

In the case n = 2 one has V ⩾ 0 and for l ⩾ l_{3}, x → V (|u_{l}(t)|^{2}) is a L^{1} function and ∫_{ℝ^{n}}V (|u_{l}(t)|^{2})dx has a limit as l → ∞. We apply Fatou’s lemma to the sequence (V (|u_{l}(t, .)|^{2}))_{l∈ℕ}. We already know that it is bounded in L^{1}, and the continuity with respect to the initial data ensures that u_{l}(t) → u(t) in X^{k}. Therefore V (|u_{l}(t)|^{2}) → V (|u(t)|^{2}) in L^{∞}. Thus V (|u(t, .)|^{2}) ∈ L^{1} and

∫ V (|u(t, x)|^{2})dx ⩽ lim inf_{l→∞} ∫ V (|u_{l}(t, x)|^{2}) .

We can now pass to the limit in (47):

∫_{ℝ^{n}} [|∇u(t)|^{2} + V (|u(t)|^{2})] dx ⩽ ∫_{ℝ^{n}} [|∇u_{0}|^{2} + V (|u_{0}|^{2})] dx . (48)

By reversing time, we get the inverse inequality, and the conservation of the energy (29) has been proved.

In the case n = 1, we also show that V (|u_{l}(t, .)|^{2}) ∈ L^{1}. For l ⩾ l_{3}, we know that V (|ρ_{l} ∗ u_{0}|^{2}) ∈ L^{1}(ℝ). Hence for all choice of ϕ as in Proposition 5.2,

lim_{R→∞} ∫_{ℝ} V (|ρ_{l} ∗ u_{0}(x)|^{2})ϕ_{R}^{±}(x)dx = ∫_{ℝ} V (|ρ_{l} ∗ u_{0}(x)|^{2})ϕ_{0}(∓x)dx ,

and Proposition 5.2 ensures that lim_{R→∞} ∫_{ℝ} V (|u_{l}(t, x)|^{2})ϕ_{R}^{±}(x)dx exists and does

not depend on the choice of ϕ. This implies that V (|u_{l}(t, x)|^{2}) → 0 as x → ±∞. Indeed, we have the following lemma (we will prove it later):

Lemma 5.2 Let f : ℝ → ℝ an uniformly continuous function such that for all ϕ as in Proposition 5.2, lim_{R→∞} ∫_{ℝ} f(x)ϕ_{R}^{±}(x)dx exists. Then f(x) →_{x→±∞} 0.

Since {r, V (r) = 0} is discrete and u_{l}(t) ∈ C_{b}(ℝ), there exists r_{±}^{l}(t) such that

V (r_{±}^{l}(t)) = 0 and |u_{l}(t, x)|^{2} →_{x→∞} r_{±}^{l}(t) .

Let us show that r_{±}^{l} is continuous on ]T_{∗}(l), T^{∗}(l)[. Let t ∈]T_{∗}(l), T^{∗}(l)[, h so that t + h ∈]T_{∗}(l), T^{∗}(l)[, ε > 0 and x ∈ ℝ. One has

|r_{±}^{l}(t+h)−r_{±}^{l}(t)| ⩽ |r_{±}^{l}(t+h)−u_{l}(t+h, x)|+|u_{l}(t+h, x)−u_{l}(t, x)|+|u_{l}(t, x)−r_{±}^{l}(t)| .

u_{l} ∈ C(]T_{∗}(l), T^{∗}(l)[, X^{k}) ⊂ C(]T_{∗}(l), T^{∗}(l)[, L^{∞}), thus we can choose h small enough in order that

||u_{l}(t + h) − u_{l}(t)||_{L^{∞}} ⩽ ε/3 .

We also choose |x| large enough such that

|r_{±}^{l}(t + h) − u_{l}(t + h, x)| ⩽ ε/3 and |u_{l}(t, x) − r_{±}^{l}(t)| ⩽ ε/3 .

Hence |r_{±}^{l}(t + h) − r_{±}^{l}(t)| ⩽ ε. Therefore r_{±}^{l} is continuous with value in a discrete set, which means that it is constant. For l ⩾ l_{3}, the fact that |||ρ_{l} ∗ u_{0}|^{2} − |u_{0}|^{2}||_{L^{∞}} ⩽ δ_{0} − δ_{1} and the assumption on u_{0} imply that r_{±}^{l}(0) = ρ_{0}, hence

∀l ⩾ l_{3}, ∀t ∈]T_{∗}(l), T^{∗}(l)[, |u_{l}(t, x)|^{2} →_{x→±∞} ρ_{0} .

We choose l_{4} ⩾ l_{3} such that for l ⩾ l_{4},

|||u_{l}(t)|^{2} − |u(t)|^{2}||_{L^{∞}}

⩽ ||u_{l}(t) − u(t)||_{L^{∞}}(||u_{l}(t) − u(t)||_{L^{∞}} + 2||u(t)||_{L^{∞}})
⩽ K||ρ_{l} ∗ u_{0} − u_{0}||_{X^{k}}(K||ρ_{l} ∗ u_{0} − u_{0}||_{X^{k}} + 2 sup_{t∈[T̃_{1},T̃_{2}]}||u(t)||_{X^{k}})⩽ \frac{δ_{0} − δ_{1}}{2} .

Let D > 0 such that |x| ⩾ D implies ||u_{l_{4}}(t, x)|^{2} − ρ_{0}| ⩽ δ_{1}. Then for l ⩾ l_{4}, |x| ⩾ D,

||u_{l}(t, x)|^{2} − ρ_{0}|

⩽ ||u_{l}(t, x)|^{2} − |u(t, x)|^{2}| + ||u(t, x)|^{2} − |u_{l_{4}}(t, x)|^{2}| + ||u_{l_{4}}(t, x)|^{2} − ρ_{0}|
⩽ 2\frac{δ_{0} − δ_{1}}{2} + δ_{1} = δ_{0}

This implies that V (|u_{l}(t, x)|) is non-negative on {x, |x| ⩾ D}, and now (47) yields:

∫_{|x|⩾D} V (|u_{l}(t, x)|^{2})dx

= ∫_{ℝ}[|∇ρ_{l} ∗ u_{0}|^{2} + V (|ρ_{l} ∗ u_{0}|^{2})] dx −∫_{ℝ}|∇u_{l}(t, x)|^{2}dx −∫_{|x|⩽D}V (|u_{l}(t, x)|^{2})dx
→_{l→∞} ∫_{ℝ} [|∇u_{0}|^{2} + V (|u_{0}|^{2})] dx − ∫_{ℝ} |∇u(t, x)|^{2}dx − ∫_{|x|⩽D} V (|u(t, x)|^{2})dx ,

and (V (|u_{l}(t, .)|^{2})1_{{|x|⩾D}})_{l⩾l_{4}} is bounded in L^{1}. We apply Fatou’s lemma to this sequence, and we can conclude similarly to the case n = 2, V ⩾ 0. □

To complete the proof of theorem 5.1, it just remains to prove lemmas 5.1 and 5.2.

Proof of lemma 5.2. We argue by contradiction. Assume that there exists ε > 0 such that for all A > 0, there exists x > A such that |f(x)| > ε. Since f is uniformly continuous, there exists δ ∈ (0, 2), such that |x − y| < δ implies |f(x) − f(y)| ⩽ ε/2. We may thus construct a sequence (x_{n})_{n∈ℕ} such that x_{n} → ∞ and |f(y)| ⩾ ε/2 as soon as |y − x_{n}| ⩽ δ. We may assume moreover that the intervals ]x_{n} − δ, x_{n} + δ[ are disjoint, and that for instance f(x_{n}) > 0. We choose ϕ as in proposition 5.2, namely

ϕ(x) = \{ \begin{matrix}1 & if & x ⩽ 0 \\ 0 & if & x ⩾ δ/2 \end{matrix} .

Then

|∫_{ℝ} f(y)ϕ_{x_{n}−δ}(y)dy − ∫_{ℝ} f(y)ϕ_{x_{n}+δ/2}(y)dy|
= ∫_{ℝ} f(y)(ϕ_{x_{n}+δ/2}(y) − ϕ_{x_{n}−δ}(y))dy
⩾ ∫_{x_{n}−δ/2}^{x_{n}+δ/2} f(y)dy ⩾ δε/2

and this is a contradiction. □

Proof of lemma 5.1. Let us fix x, y ∈ ℝ^{n} with x ≠ y. It suffices to show that f : t ∈ [0, 1] → u_{0}(x + t(y − x)) is absolutely continuous. By definition of X^{k}, u_{0} is the limit in X^{k} of a sequence (v_{l})_{l∈ℕ} of functions of class C^{k}. Therefore the functions f_{l} : t ∈ [0, 1] → v_{l}(x + t(y − x)) are absolutely continuous.

We are reduced to prove that the moduli of absolute continuity of f_{l}’s are uniformly bounded. Indeed, if we assume this fact, for all ε > 0, there exists δ > 0 such that for all positive integer m, and for all choice of (α_{j}, β_{j})_{1⩽j⩽m} with 0 ⩽ α_{1} < β_{1} < ... < α_{m} < β_{m} ⩽ 1 and ∑_{j=1}^{m}(β_{j} − α_{j}) ⩽ δ, we have for all l,

∑_{j=1}^{m} |f_{l}(β_{j}) − f_{l}(α_{j})| ⩽ ε/2 .

Then

∑_{j=1}^{m} |f(β_{j})−f(α_{j})| ⩽ ∑_{j=1}^{m} |f_{l}(β_{j})−f_{l}(α_{j})|+∑_{j=1}^{m} [|f(β_{j}) − f_{l}(β_{j})| + |f(α_{j}) − f_{l}(α_{j})|]

choosing l large enough, ||f − f_{l}||_{∞} ⩽ \frac{ε}{4m}. We infer that

∑_{j=1}^{m} |f(β_{j}) − f(α_{j})| ⩽ ε ,

and therefore f is absolutely continuous.

We now prove that the moduli of absolute continuity of f_{l}’s are uniformly bounded. Since v_{l} → u_{0} in X^{k}, v_{l} is bounded in X^{k}. We choose m, α_{j}, β_{j} as above. For all l, we have:

∑_{j=1}^{m} |f_{l}(β_{j}) − f_{l}(α_{j})| = ∑_{j=1}^{m} |∫_{α_{j}}^{β_{j}} f′_{l}(s)ds|

⩽ ∫_{∪^{n}_{j=1}]α_{j},β_{j}[} |f′_{l}(s)|ds

⩽ (∫_{∪^{n}_{j=1}]α_{j},β_{j}[} ds)^{1/2} (∫_{∪^{n}_{j=1}]α_{j},β_{j}[} |f′_{l}(s)|^{2}ds)^{1/2}

⩽ (∑_{j=1}^{m}(β_{j} − α_{j}))^{1/2} (∫_{0}^{1} |f′_{l}(s)|^{2}ds)^{1/2} . (49)

Observe that

∫_{0}^{1} |f′_{l}(s)|^{2}ds = ∫_{0}^{1} |(y − x).∇v_{l}(x + s(y − x))|^{2}ds
⩽ |y − x| ∫_{0}^{|y−x|} |∇v_{l} (x + s\frac{y − x}{|y − x|})|^{2} ds .

Since ∇u_{0} ∈ H^{k−1}(ℝ^{n}), the trace theorem ensures that the mapping that sends a function of H^{k−1}(ℝ^{n}) on its trace on a line D of ℝ^{n} is continuous from H^{k−1}(ℝ^{n}) to H^{(k−1)−(n−1)/2}(D) ⊂ L^{2}(D) because k − 1 − (n − 1)/2 ⩾ ⌊n/2⌋ − (n − 1)/2 ⩾ 0. Therefore, there exists a constant C > 0 such that

(∫_{0}^{1} |f′_{l}(s)|^{2}ds)^{1/2} ⩽ |y − x|^{1/2}||∇v_{l |\{x+t \frac{y−x}{|y−x|} ,t∈ℝ\}} ||_{L^{2}(\{x+t \frac{y−x}{|y−x|} ,t∈ℝ\})}
⩽ C|y − x|^{1/2}||∇v_{l}||_{H^{k−1}(ℝ^{n})}
⩽ C|y − x|^{1/2}sup_{l}||v_{l}||_{X^{k}(ℝ^{n})} .  (50)

Finally, (49) and (50) ensure that f_{l}’s absolute continuity moduli are uniformly bounded. □

Remark. The only reason for which Theorem 5.1 in the case V ⩾ 0 is not valid for n ⩾ 3 is that we did not prove Proposition 5.1 for n ⩾ 3. The whole proof of Theorem 5.1 with V ⩾ 0 would be valid for n ⩾ 3 if Proposition 5.1 was.

We justify next the conservation of the momentum, in the one dimensional case.

Theorem 5.2 Let n = k = 1. The assumptions on the nonlinearity f are as in Theorem 5.1. Let u_{0} ∈ X^{1}(ℝ) such that V (|u_{0}|^{2}) ∈ L^{1} and |u_{0}| ⩾ a > 0. Let 0 ∈ (T̃_{∗}, T̃^{∗}) ⊂ (T_{∗}, T^{∗}) be the maximal interval on which |u(t, x)| > 0, t ∈ (T̃_{∗}, T̃^{∗}), x ∈ ℝ. Then ∀t ∈ (T̃_{∗}, T̃^{∗}), P (u(t)) = P (u_{0}), where the renormalized momentum is given by:

P (u) = Im ∫_{−∞}^{∞} \frac{\overline{u}_{x}u}{|u|^{2}}(|u|^{2} − ρ_{0})  (51)

Proof. It follows from the proof of Theorem 5.1 that for t ∈ (T̃_{∗}, T̃^{∗}), |u(t)|^{2} − ρ_{0} ∈ L^{2} and that |u(t)| ⩾ a(t), where a(t) ∈ ℝ_{+}^{∗} (because |u(t, x)|^{2} ⟶_{x→∞} ρ_{0} and

|u(t, x)| > 0). Moreover, u_{x}(t, .) ∈ L^{2}. Hence P (u(t)) is well defined.

If u_{0} ∈ X^{3}(ℝ), the formal proof of the conservation of P on (T̃_{∗}, T̃^{∗}) (see [ZL], [GSS]) is valid, since u ∈ C(T̃_{∗}, T̃^{∗}, X^{3}(ℝ)) ∩ C^{1}(T̃_{∗}, T̃^{∗}, X^{1}(ℝ)).

Proceeding as in the proof of theorem 5.1, we approach u_{0} ∈ X^{1} by ρ_{l} ∗ u_{0} ∈ X^{3}. Let us fix t ∈ (T̃_{∗}, T̃^{∗}). Since u_{l}(t) → u(t) in X^{1}, for l large enough we have |u_{l}(t)| ⩾ a(t)/2 > 0, and then (T̃_{∗}, T̃^{∗}) ⊂ (T̃_{∗}(l), T̃^{∗}(l)). It follows from the proof of Theorem 5.1 that (|u_{l}(t)|^{2}−ρ_{0}) → (|u(t)|^{2}−ρ_{0}) in L^{2} and ∂_{x}u_{l}(t) → ∂_{x}u(t) in L^{2}. Hence P (u_{l}(t)) → P (u(t)), and the conservation of the momentum, which is true for u_{l}, is also true for u. □

Remark. The above analysis fills a gap in the proof of Theorem 1.1 in [ZL]. Namely, Zhiwu Lin proves a criterion of stability for the traveling bubbles solution of NLS in the one dimensional case, for a nonlinearity f verifying:

(1) f(ρ_{0}) = 0, η_{0} = sup{η, 0 < η < ρ_{0}, V (η) = 0} exists,
0 < η_{0} < ρ_{0}, f(η_{0}) < 0
(2) f′(ρ_{0}) < 0

This proof consists in applying Theorem 3 in [GSS] to the hydrodynamical problem corresponding to (1) (i.e. with the complex unknown u replaced by (r, v) = (ρ_{0} − |u|^{2}, ∂_{x} arg u)). Theorem 5.1 and 5.2 ensure that Assumption 1 of Theorem 3 in [GSS] is verified. Namely, in a neighbourhood of the soliton, our results imply the following facts which were not discussed in [ZL]:

• the local existence for the hydrodynamical Cauchy problem with (r, v) ∈ H^{1} × L^{2} (and not only X^{1} × L^{2}) if the energy is finite at initial time

• the conservation of energy and momentum.

In dimension 1, we prove next that the conservation of energy implies a global existence result for a solution of (1) in X^{1}.

Theorem 5.3 Let n = k = 1. The assumptions on f and u_{0} are as in theorem 5.1, and we assume moreover that there exists some C > 0 such that V (r) ⩾ C(ρ_{0} − r)^{2}. Then u ∈ C_{b}(ℝ, X^{k}).

Proof. We define the energy at initial time

E_{0} = ∫_{ℝ^{n}} [|∇u_{0}(x)|^{2} + V (|u_{0}(x)|^{2})] dx .

We know by Theorem 5.1 that the energy is conserved for t ∈ (T_{∗}, T^{∗}) and that if T^{∗} is finite, ||u(t)||_{X^{k}} → ∞ as t → T^{∗} (and we have a similar result by replacing T^{∗} by T_{∗}) (see [CH]). The conservation of the energy and the fact that V ⩾ 0 imply that for all t, ∫_{ℝ} |∂_{x}u(t)|^{2}dx ⩽ E_{0}. To prove that ||u(t)||_{X^{k}} can not blow up in finite time, it suffices then to show that ||u(t)||_{L^{∞}} can not blow up in finite time. By the Sobolev embedding H^{1}(ℝ) ⊂ L^{∞}(ℝ),

||u(t)||_{L^{∞}}^{2} ⩽ ρ_{0} + |||u(t)|^{2} − ρ_{0}||_{L^{∞}}
⩽ ρ_{0} + C|||u(t)|^{2} − ρ_{0}||_{H^{1}}
= ρ_{0} + C\sqrt{|||u(t)|^{2} − ρ_{0}||_{L^{2}}^{2} + ||∂_{x}(|u(t)|^{2} − ρ_{0})||_{L^{2}}^{2}}

We now use the additional assumption that V (r) ⩾ C(1 − r)^{2}:

||u(t)||_{L^{∞}}^{2} ⩽ ρ_{0} + C\sqrt{E_{0} + 4||u(t)||_{L^{∞}}^{2} ∫ |∂_{x}u(t)|^{2}dx}
⩽ ρ_{0} + C\sqrt{E_{0}} + 2C||u(t)||_{L^{∞}} \sqrt{E_{0}} .

Therefore ||u(t)||_{L^{∞}} can not blow up in finite time, thus (u(t))_{t∈ℝ} is global. Moreover, ||u(t)||_{L^{∞}} is bounded on ℝ and therefore (u(t))_{t∈ℝ} is bounded in X^{1}(ℝ). □

Example. In NLS with a pure defocusing power (34), with n = 1 and p ⩾ 1, we have V (r) ⩾ αρ_{0}^{p−1}(ρ_{0} − r)^{2}/2, hence the assumptions of Theorem 5.3 are verified.

Acknowledgements. The author is grateful to Anne de Bouard and Jean-Claude Saut for their precious help.

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