Textos de

Metodos Matematicos

30Blow up and scattering in the nonlinear

Schrodinger equation

Thierry Cazenave

Universidade Federal do Rio de Janeiro

Centro de Ciencias Matematicas e da Natureza

INSTITUTO DE MATEMATICA

Blow up and scattering in the nonlinear Schrodinger equation

Thierry Cazenave

Analyse Numerique URA CNRS 189

Universite Pierre et Marie Curie

4, place Jussieu

75252 Paris Cedex 05 France

Second Edition

Instituto de Matematica–UFRJ

Rio de Janeiro, RJ

1996

This monograph contains the material of second course on nonlinear Schrodinger

equations given by Dr. Cazenave at the Instituto de Matematica da Universidade Fed-

eral do Rio de Janeiro (IM-UFRJ) during the period of July-August 1993. The first

course appeared as An introduction to nonlinear Schrodinger equations, Textos de Metodos

Matematicos n. 22 (first edition 1989) and n. 26 (second edition 1993). We are very grate-

ful to Dr. Cazenave for these fruitful lectures and we hope that these publications could

help the researchers working on nonlinear Schrodinger equations.

Rio de Janeiro, August 1994.

Eduardo Siqueira

Elaine Machtyngier

Rolci Cipolatti

Contents:

1. Introduction and notation . . . . . . . . . . . . . . . . . . . . . . . . 2

2. Some general results . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3. The pseudo-conformally invariant case. Sharp existence and blow up results . . 18

4. Some applications of the pseudo-conformal transformation . . . . . . . . . . 38

5. A global existence and decay result in the non-invariant case . . . . . . . . . 52

6. Some results on the scattering theory in the non-invariant case . . . . . . . . 59

7. The Cauchy problem for a nonautonomous Schrodinger equation . . . . . . . 74

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

1

1. Introduction and notation. In these notes, we describe some results concerning

the blow up phenomenon and the scattering theory for the nonlinear Schrodinger equation.

For simplicity, we consider the model caseiut +4u+ λ|u|αu = 0, (t, x) ∈ [0, T )×RN ,

u(0, x) = ϕ(x), x ∈ RN ,(1.1)

where α > 0 and λ ∈ R, and ϕ is a given initial value.

It is well-known that if λ ≤ 0 or if α < 4/N , then all solutions of (1.1) are global and

bounded in the energy space. Then a natural question is the asymptotic behavior of the

solutions. Since the solutions of the linear equation converge locally to 0 as t→ ±∞, one

expects that the “small” solutions of (1.1) behave asymptotically like solutions of the linear

equation, at least for α large enough. In the repulsive case λ ≤ 0, one expects the same

property to hold also for “large” solutions. Indeed, a scattering theory was developped

for (1.1), for both low energy and large energy. Of course, the possible values of α for

which the scattering operator can be constructed depend on the space in which one works.

The most natural spaces are H1, L2 and H1 ∩ L2(|x|2dx). Some satisfactory results are

available, though it seems that some of them are not optimal with respect to the range of

α’s.

On the other hand, if α ≥ 4/N and λ > 0, then some solutions of (1.1) blow up in

finite time. Several questions naturally arise. What norms of u actually blow up? What

is the blow up rate? What is the asymptotic profile of u? For the heat equation, there

are satisfactory answers to these questions (see [12,13,14,15,16,22,23,24,25,33,42,43,44,45,

49,50,51]), while for the Schrodinger equation, they are essentially open. More precisely,

besides theH1 norm, there is a family of Lp norms which blows up. However, the optimality

is in general unknown, since there does not exist any pointwise upper bound on any Lp

norm for p > 2. As for the blow up rate, there is a lower bound which is a trivial

consequence of the local existence theory, but there is no pointwise upper bound. As a

matter of fact, in the so-called pseudo-conformal case, there exist explicit solutions that

blow up twice as fast as the lower bound. This shows at least that the lower bound is

not always optimal. The asymptotic profile is completely unknown, except for the explicit

example mentioned above and some related examples.

In Section 2, we recall some basic properties of the equation (1.1), and we present the

2

general results concerning the blow up: the blow up theorems, the lower estimate and the

blow up of a family of Lp norms.

In Section 3, we consider the so-called “critical” or “pseudo-conformal” case α = 4/N .

We first establish sharp existence results concerning the initial value problem in H1(RN )

and L2(RN ). Next, we describe some properties of the blowing up solutions that are

known only for the critical nonlinearity.

In Section 4, we still consider the “pseudo-conformal” case α = 4/N , and we introduce

the pseudo-conformal transformation. We deduce some results concerning the blow up of

solutions and the scattering theory. These results may appear as miracles, in that they are

obtained by merely applying the pseudo-conformal transformation. (Note, however, that

some of these results are recovered in the following section by apparently more natural

methods.) The fact that one obtains by the same method both results on the blow up

phenomenon and results on the asymptotic behavior of global solutions may look surprising.

This is, however, natural since the pseudo-conformal transformation closely relates the

asymptotic behavior of the global solutions and the behavior near the blow up time of the

nonglobal solutions.

In Section 5, we extend certain of the results of Section 4 to the non-conformal case

α 6= 4/N . They are obtained by a fixed point argument based on certain Strichartz type

inequalities.

In Section 6, we present some results concerning the scattering theory in the non-

conformal case (i.e. α 6= 4/N). Even though the equation is no longer invariant by

the pseudo-conformal transformation, we still apply the transformation. This leads to a

nonautonomous nonlinear Schrodinger equation. The local (in time) study of that equa-

tion corresponds, via the transformation, to the study of the asymptotic behavior of the

solutions of (1.1). Finally, Section 7 is devoted to the study of the initial value problem

for the nonautonomous equation cited above.

We conclude this section by introducing the notation that will be used throughout

these notes. For m ∈ Z, we denote by Hm(RN ) the Sobolev space Hm(RN ,C), equipped

with its usual norm ‖ · ‖Hm . For 1 ≤ p ≤ ∞, we denote by Lp(RN ) the Lebesgue space

Lp(RN ,C), equipped with its usual norm ‖ ·‖Lp, and for m ∈ N, we denote by Wm,p(RN )

the Sobolev space space Wm,p(RN ,C), equipped with its usual norm ‖ · ‖W m,p . Given an

3

interval I ⊂ R, a Banach space X and 1 ≤ p ≤ ∞, we denote by Lp(I,X) the Banach

space of measurable functions u : I → X such that the function t 7→ ‖u(t)‖X belongs to

Lp(I). Its norm is denoted by ‖ · ‖Lp(I,X), or ‖ · ‖Lp(I), or even ‖ · ‖Lp when there is no

risk of confusion. Given 1 ≤ p ≤ ∞, the conjugate exponent p′ of p is defined by

1

p+

1

p′= 1.

We define the space Y → L2(RN ) by Y = u ∈ L2(RN ); | · |u(·) ∈ L2(RN ) endowed

with the norm ‖u‖Y = ‖u‖L2 + ‖xu‖L2 , and the space X → H1(RN ) by

X = H1(RN ) ∩ Y,

endowed with the norm ‖u‖X = ‖u‖H1 + ‖u‖Y .

We say that a pair (q, r) of real numbers is admissible if 2 ≤ r <2N

N − 2(2 ≤ r ≤ ∞

if N = 1) and if2

q= N

(1

2− 1

r

).

Note that in particular 2 < q ≤ ∞.

We denote by (T (t))t∈R the group of isometries generated by i4 in L2(RN ), i.e.

T (t) = eit4.

Finally, we recall Strichartz’ estimates. Let I 3 0 be any open interval of R (bounded

or not) and let J = I, let (γ, ρ) be an admissible pair, let ϕ ∈ L2(RN ) and let f ∈Lγ′(I, Lρ′(RN )). Then, for any admissible pair (q, r), the function

u(t) = T (t)ϕ+

∫ t

0

T (t− s)f(s) ds

belongs to C(J, L2(RN )) ∩ Lq(I, Lr(RN )), and

‖u‖Lq(I,Lr) ≤ C‖ϕ‖L2 + C‖f‖Lγ′ (I,Lρ′ ), (1.2)

for some constant C independent of I. (See [5], Theorem 3.2.5.)

4

2. Some general results. We recall that under fairly general assumptions, and in

particular for all the solutions that we consider in these notes, u is a solution of (1.1) on

some interval I 3 0 if and only if

u(t) = T (t)ϕ+ iλ

∫ t

0

T (t− s)|u(s)|αu(s) ds, (2.1)

for all t ∈ I. See for example [5], Remark 2.5.1. We first recall the existence results.

Theorem 2.1. Assume λ ∈ R and 0 ≤ α <4

N − 2(0 ≤ α < ∞ if N = 1). For

every ϕ ∈ H1(RN ), there exists a unique solution u of (1.1) defined on a maximal interval

(−T∗, T ∗) with T∗, T∗ > 0, u ∈ C((−T∗, T ∗), H1(RN )) ∩ C1((−T∗, T ∗), H−1(RN )).

In addition, we have the following blow up alternative. Either T ∗ = ∞ (respec-

tively, T∗ = ∞), or else T ∗ < ∞ and limt↑T ∗

‖u(t)‖H1 = +∞ (respectively, T∗ < ∞ and

limt↓−T ∗

‖u(t)‖H1 = +∞).

Moreover, there is conservation of charge and energy, i.e.

‖u(t)‖L2 = ‖ϕ‖L2 , (2.2)

and

E(u(t)) =:

∫

RN

(1

2|∇u(t, x)|2 − λ

α+ 2|u(t, x)|α+2

)dx = E(ϕ), (2.3)

for all t ∈ (−T∗, T ∗).Finally, if ϕ ∈ X, then u ∈ C((−T∗, T ∗), X) and we have the pseudo-conformal

conservation law

‖(x+ 2it∇)u(t)‖2L2 − 8λt2

α+ 2‖u(t)‖α+2

Lα+2

= ‖xϕ‖2L2 − 4λ

4−Nα

α+ 2

∫ t

0

s

∫

RN

|u(s, x)|α+2 dx ds, (2.4)

for all t ∈ (−T∗, T ∗).

See [5], Theorem 4.3.1 and Remark 4.3.2 for the existence and the conservation of

charge and energy; Lemma 6.4.3 for the X regularity; Theorem 7.1.1 for the pseudo-

conformal conservation law.

Remark 2.2. The solution u has some further properties, which we recall below.

5

(i) For any admissible pair (q, r), we have u ∈ Lqloc((−T1, T2),W

1,r(RN )) for −T∗ <

−T1 < T2 < T ∗.

(ii) The solution u depends continuously on the initial value ϕ in the following sense. The

mappings ϕ 7→ T∗ and ϕ 7→ T ∗ are lower-semicontinuous H1(RN ) → R. Moreover, if

ϕn −→n→∞

ϕ in H1(RN ) and if un is the corresponding solution of (1.1), then un −→n→∞

u

in C([−T1, T2], H1(RN )) for any bounded interval [−T1, T2] ⊂ (−T∗, T ∗).

(iii) Moreover, if ϕn −→n→∞

ϕ in X, then un −→n→∞

u in C([−T1, T2], X).

(iv) If ϕ ∈ H2(RN ), then u ∈ C((−T∗, T ∗), H2(RN )) ∩ C1((−T∗, T ∗), L2(RN )).

See [5], Theorem 5.3.1 and Remark 5.3.5 for the W 1,r smoothing effect; Theorem 4.3.1

for the continuous dependence in H1(RN ); Corollary 6.4.4 for the continuous dependence

in X; Theorem 5.2.1 and Remark 5.2.9 for the H2 regularity.

Remark 2.3. If α < 4/N , then the Cauchy problem for (1.1) is globally well posed

in L2(RN ) (for any λ ∈ R). More precisely, we have the following result. (See [5],

Theorem 6.3.1.)

For every ϕ ∈ L2(RN ), there exists a unique solution u of (1.1), u ∈ C(R, L2(RN ))∩Lθ

loc(R, Lα+2(RN )) with θ =

4(α+ 2)

Nα. Moreover, ut ∈ Lθ

loc(R, H−2(RN )). In addition,

u ∈ Lqloc(R, L

r(RN )) for every admissible pair (q, r).

There is conservation of charge, i.e. (2.2) holds for all t ∈ R.

The solution u depends continuously on the initial value ϕ in the sense that if ϕn −→n→∞

ϕ

in L2(RN ), then un −→n→∞

u in C([−T, T ], L2(RN )) for any T <∞, where un is the solution

of (1.1) with initial value ϕn.

Remark 2.4. Here are some sufficient conditions for global existence.

(i) It follows immediately from the blow up alternative and the conservations of charge

and energy that if λ ≤ 0, then for every ϕ ∈ H1(RN ) the solution is global and

bounded, i.e. supt∈R

‖u(t)‖H1 <∞.

(ii) The same properties hold for λ > 0 provided α < 4/N . Indeed, it follows from

Gagliardo-Nirenberg’s inequality that

‖u‖α+2Lα+2 ≤ C‖∇u‖

Nα2

L2 ‖u‖α+2−Nα2

L2 .

6

Therefore, by conservation of charge and energy

‖u(t)‖2H1 ≤ C + C‖u(t)‖

Nα2

H1 ‖ϕ‖α+2−Nα2

L2 ,

and the conclusion follows, sinceNα

2< 2.

(iii) By applying the same argument in the case λ > 0 and α = 4/N , one obtains easily

that there exists ε > 0 such that if ‖ϕ‖L2 ≤ ε, then u is global and bounded. (See

Section 3.)

(iv) If λ > 0 and α > 4/N , then there exists ε > 0 such that if ‖ϕ‖H1 ≤ ε, then u is global

and bounded. See [5], Theorem 6.2.1.

If λ > 0 and α ≥ 4/N , then some solutions of (1.1) blow up in finite time. We recall

the first result in this direction. (See Glassey [21].)

Theorem 2.5. Assume λ > 0 and4

N≤ α <

4

N − 2(4/N ≤ α <∞ if N = 1). If ϕ ∈ X

is such that E(ϕ) < 0, then T∗ <∞ and T ∗ <∞, i.e. the solution u of (1.1) blows up for

both t > 0 and t < 0.

Proof. We recall the idea of the proof. (See for example [5], Theorem 6.4.7 for the

complete proof.) Since ϕ ∈ X, it follows that u ∈ C((−T∗, T ∗), X), and in particular

xu ∈ C((−T∗, T ∗), L2(RN )). One shows, by an approximation argument, that the mapping

t 7→ ‖xu(t, x)‖2L2 belongs to C2(−T∗, T ∗), and that

d2

dt2

∫

RN

|x|2|u(t, x)|2 dx = 4NαE(ϕ)− 2(Nα− 4)

∫

RN

|∇u(t, x)|2 dx ≤ 4NαE(ϕ). (2.5)

In particular, the nonnegative function ‖xu(t, x)‖2L2 has a second derivative which is

bounded by a negative constant, which implies that T∗ <∞ and T ∗ <∞.

Remark 2.6. The assumption E(ϕ) < 0 means in some sense that ϕ is “large”. Indeed,

if ϕ is small enough, then the solution is global; and if ϕ ∈ X is not identically zero, then

E(kϕ) < 0 for |k| large enough.

The above proof of blow up is interesting. The condition for blow up is E(ϕ) < 0.

However, the argument is based on the study of ‖xu(t, x)‖2L2, and this quantity is defined

7

for ϕ ∈ X, but not for a general ϕ ∈ H1(RN ). The question as to whether or not

negative energy implies blow up for general H1 solutions is open. There is, however, a

recent progress in this direction. Ogawa and Tsutsumi [34] have shown (under the extra

assumptions N ≥ 2 and α ≤ 4) that if ϕ ∈ H1(RN ) is spherically symmetric and if

E(ϕ) < 0, then u blows up in finite time. The result is the following.

Theorem 2.7. Assume N ≥ 2, λ > 0 and4

N≤ α <

4

N − 2(2 ≤ α ≤ 4 if N = 2). If

ϕ ∈ H1(RN ) is such that E(ϕ) < 0 and if ϕ is spherically symmetric, then T∗ < ∞ and

T ∗ <∞, i.e. the solution u of (1.1) blows up for both t > 0 and t < 0.

The proof is in some way an adaptation of the proof of Theorem 2.5. Roughly speaking,

instead of calculating ‖xu(t, x)‖2L2, we evaluate ‖M(x)u(t, x)‖2

L2 where M : RN → R is a

function such that M(x) = |x| for |x| ≤ R and M is constant for |x| large. Then, we use

the decay properties of the spherically symmetric functions of H1(RN ) to estimate certain

integrals for |x| large that appear in the calculation of ‖M(x)u(t, x)‖2L2. Note that, as

opposed to the case ϕ ∈ X, the appropriate function M(x) depends on the initial value ϕ.

The proof makes use of the following lemma.

Lemma 2.8. Let N ≥ 1 and let k ∈ C1([0,∞)) be a nonnegative function such that

r−(N−1)k(r) ∈ L∞(0,∞) and r−(N−1)(k′(r))− ∈ L∞(0,∞). There exists a constant C

such that

‖k 12u‖L∞(RN ) ≤ C‖u‖

12

L2(RN )

(‖r−(N−1)kur‖

12

L2(RN )+ ‖r−(N−1)(k′)−‖

12

L∞‖u‖12

L2(RN )

),

for all spherically symmetric functions u ∈ H1(RN ).

Proof. By density, we may assume that u ∈ D(RN). For s > 0, we have

k(s)|u(s)|2 = −∫ ∞

s

d

ds

(k(s)|u(s)|2

)ds

= −∫ ∞

s

k′(s)|u(s)|2 ds− 2

∫ ∞

s

k(s)Re (u(s)ur(s)) ds

≤∫ ∞

s

(k′(s))−|u(s)|2 ds+ 2

∫ ∞

s

k(s)|u(s)| |ur(s)| ds

≤ C‖r−(N−1)(k′)−‖L∞‖u‖2L2(RN ) + C‖u‖L2(RN )‖r−(N−1)kur‖L2(RN ).

Hence the result.

8

Proof of Theorem 2.7. By scaling, we may assume λ = 1. Let u be as in the statement

of the theorem. Consider a function Ψ ∈W 4,∞(RN ), and set

V (t) =1

2

∫

RN

Ψ(x)|u(t, x)|2 dx,

for all t ∈ (−T∗, T ∗). We claim that

d2

dt2V (t) = 2

∫

RN

(H(Ψ)∇u,∇u) dx− α

α+ 2

∫

RN

4Ψ|u|α+2 dx− 1

2

∫

RN

42Ψ|u|2 dx, (2.6)

for all t ∈ (−T∗, T ∗), where the Hessian matrix H(Ψ) is given by H(Ψ) = (∂j∂kΨ)1≤j,k≤N .

If ϕ ∈ H2(RN ), then u is an H2 solution (see Remark 2.2 (iv)) and (2.6) follows from

elementary calculations (see Kavian [27]). The general case follows by approximating ϕ

in H1(RN ) by a sequence (ϕn)n≥0 ⊂ H2(RN ) and using the continuous dependance (see

Remark 2.2 (ii)). Next, we rewrite (2.6) as follows.

d2

dt2V (t) = 2NαE(u(t))− 2

∫

RN

Nα

2|∇u|2 − (H(Ψ)∇u,∇u)

dx

+α

α+ 2

∫

RN

(2N −4Ψ)|u|α+2 dx− 1

2

∫

RN

42Ψ|u|2 dx. (2.7)

Let now ρ ∈ D(R) be such that ρ(x) ≡ ρ(4 − x), ρ ≥ 0,

∫

R

ρ = 1, supp(ρ) ⊂ (1, 3) and

ρ′ ≥ 0 on (−∞, 2). We define the function θ by

θ(r) = r −∫ r

0

(r − σ)ρ(σ) dσ,

for r ≥ 0. We consider ε ∈ (0, 1), to be specified later, and we set

Ψ(x) ≡ Ψ(r) =1

εθ(εr2),

and

γ(x) ≡ γ(r) = 1− θ′(εr2)− 2εr2θ′′(εr2) =

∫ εr2

0

ρ(s) ds+ 2εr2ρ(εr2),

for x ∈ RN and r = |x|. Elementary calculations show that

(H(Ψ)∇u,∇u) = 2(1− γ(r))|ur|2,

4Ψ = 2N(1− γ(r)) + 4(1−N)εr2θ′′(εr2),(2.8)

and that

42Ψ = ε(4N(N + 2)θ′′(εr2) + 16(N + 2)εr2θ′′′(εr2) + 16(εr2)2θ′′′′(εr2)

).

9

In particular, there exists a constant a such that

‖42Ψ‖L∞ ≤ 2aε. (2.9)

It now follows from (2.7), (2.8) and (2.9) that

d2

dt2V (t) ≤ 2NαE(ϕ)− 4

∫γ(r)|ur|2 +

2Nα

α+ 2

∫γ(r)|u|α+2 + aε‖u‖2

L2 , (2.10)

where we have used the above relations and the propertiesNα

4≥ 1 and θ′′ ≤ 0. We claim

that there exists b and c such that

2Nα

α+ 2

∫γ(r)|u|α+2 ≤ bε

(N−1)α

4 ‖u‖α+4

2

L2

(∫γ|ur|2

)α4

+ cεNα4 ‖u‖α+2

L2 . (2.11)

Indeed, we first observe that γ(r) ≤ 1 + 2 sups≥0 sρ(s), so that

∫γ(r)|u|α+2 ≤ C‖γ 1

4u‖αL∞‖u‖2L2

≤ C‖u‖α+4

2

L2 ‖r−(N−1)γ12ur‖

α2

L2 + C‖u‖α+2L2 ‖r−(N−1)γ−

12 (γ′)−‖

α2

L∞ .

The first inequality follows from the property α ≤ 4 and the second from Lemma 2.8

(one verifies easily that γ12 ∈ C1([0,∞))). Observe that γ(r) ≡ 0 for r ≤ ε−

12 , so that

‖r−(N−1)γ12ur‖L2 ≤ ε

N−12 ‖γ 1

2ur‖L2 . Next, note that γ ≥ 1/2 for εr2 ≥ 2. Furthermore,

γ′(r) = 6εrρ(εr2) + 4ε2r3ρ′(εr2),

so that γ′(r) ≥ 0 for εr2 ≤ 2 and

γ′(r) ≥ −4ε2r3|ρ′(εr2)| ≥ −4ε12 ‖s 3

2 ρ′(s)‖L∞(0,∞).

Therefore, ‖r−(N−1)γ−12 (γ′)−‖L∞ ≤ Cε

N2 , and (2.11) follows. It now follows from (2.10),

(2.11) and conservation of charge that

d2

dt2V (t) ≤ 2NαE(ϕ)− 4

∫γ(r)|ur|2 + bε

(N−1)α

4 ‖ϕ‖α+4

2

L2

(∫γ|ur|2

)α4

+cεNα4 ‖ϕ‖α+2

L2 + aε‖ϕ‖2L2 .

Finally, since α ≤ 4, we may apply the inequality xα4 ≤ x+ 1 to obtain

d2

dt2V (t) ≤ 2NαE(ϕ)− (4− bε

(N−1)α

4 ‖ϕ‖α+4

2

L2 )

∫γ(r)|ur|2

+bε(N−1)α

4 ‖ϕ‖α+4

2

L2 + cεNα2 ‖ϕ‖α+2

L2 + aε‖ϕ‖2L2 . (2.12)

10

We note that the constants a, b and c do not depend on ϕ and ε. Since E(ϕ) ≤ 0 and

α ≤ 4, it follows immediately from (2.12) that one can choose ε > 0 depending only on ϕ

through ‖ϕ‖L2 and E(ϕ) such that

d2

dt2V (t) ≤ NαE(ϕ), (2.13)

for all t ∈ (−T∗, T ∗). Since V (t) ≥ 0, (2.13) implies that T∗ <∞ and T ∗ <∞.

Remark 2.9. There are two limitations in the above proof. The first one is α ≤ 4. If

α > 4, there appear powers of ‖γ 12ur‖L2 larger than 2 with positive coefficients in (2.12).

This is due to the homogeneity in Lemma 2.8. The other limitation is N ≥ 2, since if

N = 1 the power of ε in the second and third terms of the right hand side of (2.12) vanish.

This is due to the fact that the radially symmetric functions in dimension 1 do not have

any decay property.

However, in the critical case N = 1, α = 4, Ogawa and Tsutsumi [35] have proved that

all negative energy H1 solutions blow up in finite time without any symmetry assumption.

Their method is a more sophisticated version of the above argument.

We now give a lower estimate for blowing up solutions (see [9]).

Theorem 2.10. Assume λ > 0 and4

N≤ α <

4

N − 2(4/N ≤ α < ∞ if N = 1). If

ϕ ∈ H1(RN ) is such that T ∗ <∞, then there exists δ > 0 such that

‖∇u(t)‖L2 ≥ δ

(T ∗ − t)1α−N−2

4

, (2.14)

for 0 ≤ t < T ∗. A similar estimate holds near −T∗ if T∗ <∞.

Proof. Generally speaking, every time a local existence result is obtained for an evolution

equation, the proof also gives a lower estimate of blow up. Here, we do not go through the

entire local existence argument, but instead we give a direct proof. Set r = α+2 and let q

be such that (q, r) is an admissible pair. Let ϕ be as above, and let u be the corresponding

solution of (1.1). It follows from Holder’s inequality that

‖∇(|u|αu)‖Lr′ ≤ C‖ |u|α|∇u| ‖Lr′ ≤ C‖u‖αLr‖∇u‖Lr . (2.15)

11

By conservation of energy, we have

λ‖u‖rLr = −rE(ϕ) +

r

2‖∇u‖2L2 .

It follows that

λ‖u‖αLr ≤ C(1 + ‖∇u‖2

L2)αr ≤ C(1 + ‖∇u‖L2)

2αr .

It follows from (2.15) and the above inequality that for any 0 < t < τ < T ∗,

‖∇(|u|αu)‖Lq′((t,τ),Lr′) ≤ C(1 + ‖∇u‖L∞((t,τ),L2))2αr ‖∇u‖Lq′((t,τ),Lr)

≤ C(τ − t)q−q′

qq′ (1 + ‖∇u‖L∞((t,τ),L2))2αr ‖∇u‖Lq((t,τ),Lr).

Set now

ft(τ) = 1 + ‖∇u‖L∞((t,τ),L2) + ‖∇u‖Lq((t,τ),Lr),

so that by the above inequality,

‖∇(|u|αu)‖Lq′((t,τ),Lr′ ) ≤ C(τ − t)q−q′

qq′ ft(τ)1+ 2α

r . (2.16)

On the other hand, it follows from the equation (2.1) and the Strichartz’ estimates (1.2)

that

‖∇u‖L∞((t,τ),L2) + ‖∇u‖Lq((t,τ),Lr) ≤ C‖∇u(t)‖L2 + C‖∇(|u|αu)‖Lq′((t,τ),Lr′),

for 0 < t < τ < T ∗. By (2.16), this implies that

ft(τ) ≤ C(1 + ‖∇u(t)‖L2) + C(τ − t)q−q′

qq′ ft(τ)1+ 2α

r , (2.17)

for 0 < t < τ < T ∗.

Consider now t ∈ (0, T ∗). Note that if T ∗ <∞, it follows from the blow up alternative

that ft(τ) → ∞ as τ ↑ T ∗. Note also that ft is continuous and nondecreasing on (t, T ∗)

and that

ft(τ)−→τ↓t

1 + ‖∇u(t)‖L2 .

Therefore, there exists τ0 ∈ (t, T ∗) such that ft(τ0) = (C + 1)(1 + ‖∇u(t)‖L2), where C is

the constant in (2.17). Choosing τ = τ0 in (2.17) yields

1 + ‖∇u(t)‖L2 ≤ C(1 + C)1+2αr (τ − t)

q−q′

qq′ (1 + ‖∇u(t)‖L2)1+2αr

≤ (1 + C)2+2αr (T ∗ − t)

q−q′

qq′ (1 + ‖∇u(t)‖L2)1+2αr ;

12

and so,

1 + ‖∇u(t)‖L2 ≥ 1

(1 + C)1+rα (T ∗ − t)

r(q−q′)

2αqq′

.

Hence the result, since t ∈ [0, T ∗) is arbitrary andr(q − q′)

2αqq′=

1

α− N − 2

4.

Before proceeding further, we establish an immediate consequence of the above result

concerning the blowing up of certain Lp norms of the solution.

Corollary 2.11. Assume λ > 0 and4

N≤ α <

4

N − 2(4/N ≤ α < ∞ if N = 1). If

ϕ ∈ H1(RN ) is such that T ∗ <∞, then ‖u(t)‖Lp −→t↑T ∗

∞ for all p >Nα

2. Moreover,

‖u(t)‖Lp ≥ δ

(T ∗ − t)1α− N

2p

, (2.18)

for 0 < t < T ∗ ifNα

2< p ≤ α+ 2 and

‖u(t)‖Lp ≥ δ

(T ∗ − t)4−(N−2)α

α2 ( 12−

1p ), (2.19)

for 0 < t < T ∗ if p ≥ α+ 2. A similar estimate holds near −T∗ if T∗ <∞.

Remark. Note that if N ≥ 3 and p >2N

N − 2or if N = 2 and p = ∞, then it may happen

that ‖u(t)‖Lp = ∞ for some (or all) t ∈ (−T∗, T ∗). Clearly, this does not contradict the

above estimates. Note, however, that u ∈ Lqloc((−T∗, T ∗),W 1,r(RN )) for every admissible

pair (q, r), so that by Sobolev’s embedding theorem, ‖u(t)‖Lp < ∞ for almost all t ∈(−T∗, T ∗) provided N ≤ 3 or N ≥ 4 and p <

2N

N − 4.

Proof of Corollary 2.11. Suppose firstNα

2< p ≤ α + 2. It follows from Gagliardo-

Nirenberg’s inequality that

‖u‖α+2Lα+2 ≤ C‖∇u‖2−µ

L2 ‖u‖α+µLp ,

with µ =4p− 2Nα

2N − (N − 2)p. By conservation of energy and the above inequality, we obtain

‖∇u(t)‖2L2 ≤ 2E(ϕ) +

2λ

α+ 2‖u(t)‖α+2

Lα+2

≤ C + C‖∇u(t)‖2−µL2 ‖u(t)‖α+µ

Lp ,

13

for all 0 < t < T ∗. Since ‖∇u(t)‖L2 −→t↑T ∗

∞, it follows that

‖∇u(t)‖µL2 ≤ C‖u(t)‖α+µ

Lp .

By Theorem 2.10, this implies

‖u(t)‖Lp ≥ ε

(T ∗ − t)µ

α+µ ( 1α−N−2

4 ).

(2.18) follows, sinceµ

α+ µ

(1

α− N − 2

4

)=

1

α− N

2p.

Suppose now p ≥ α+ 2. It follows from Holder’s inequality that

‖u‖α+2Lα+2 ≤ ‖u‖

αpp−2

Lp ‖u‖2(p−(α+2))

p−2

L2 .

Therefore, by conservation of charge and energy,

‖∇u(t)‖2L2 ≤ 2E(ϕ) +

2λ

α+ 2‖u(t)‖α+2

Lα+2

≤ C + C‖u(t)‖αp

p−2

Lp ‖ϕ‖2(p−(α+2))

p−2

L2 ,

for all 0 < t < T ∗. (2.19) now follows from Theorem 2.10 and the above inequality.

Remark 2.12. Theorem 2.10 and Corollary 2.11 give lower estimates of ‖∇u‖L2 and

‖u‖Lp near blow up. They do not give any upper estimate. It is interesting to compare

these results to the corresponding ones for the heat equation. If one considers the equation

ut − 4u = |u|p−1u with Dirichlet boundary condition, then a simple argument (even

simpler than the proof of Theorem 2.10) gives the lower estimate ‖u‖L∞ ≥ (T ∗ − t)− 1α . If

α <4

N − 2, then it is known that this is the actual blow up rate of the solutions (see [49,

15, 22, 44]). However for larger α’s, some solutions blow up faster (see [25]). A lower

estimate is obtained as well for ‖u‖Lp , p >Nα

2. In some cases it is known that ‖u‖Lp also

blows up for p =Nα

2(see [51]) and that ‖u‖Lp remains bounded for p <

Nα

2(see [12]).

Remark 2.13. In the case α > 4/N , one does not know the exact blow up rate of

any blowing up solution. In addition, one does not know whether or not ‖u‖Lp blows up

for 2 < p ≤ Nα

2. On the other hand, there is an upper estimate of integral form (see

Merle [28]). More precisely, for ϕ ∈ X it follows from (2.5) that

d2

dt2‖xu(t)‖2

L2 ≤ a− b‖∇u(t)‖2L2,

14

for some constants a, b > 0. Since ‖xu(t)‖2L2 ≥ 0, this implies that

∫ T∗

0

∫ t

0

‖∇u(s)‖2L2 ds dt <∞.

Since ∫ T∗

0

∫ t

0

‖∇u(s)‖2L2 ds dt =

∫ T∗

0

(T ∗ − t)‖∇u(t)‖2L2 dt,

it follows immediately from Holder’s inequality that

∫ T∗

0

‖∇u(t)‖µL2 dt <∞,

for 0 ≤ µ < 1. If ϕ ∈ H1(RN ) and ϕ is spherically symmetric, then one obtains the same

estimate. Indeed, by using the fact that α > 4/N one can improve (2.13) as follows

d2

dt2V (t) ≤ NαE(ϕ)− (Nα− 4)‖∇u‖2

L2 ,

and the conclusion is the same.

Remark 2.14. In the case α = 4/N , then (2.14) becomes

‖∇u(t)‖L2 ≥ δ

(T ∗ − t)12

, (2.20)

and (2.18) and (2.19) become

‖u(t)‖Lp ≥ δ

(T ∗ − t)N2 ( 1

2−1p), (2.21)

for p > 2. In particular, ‖u‖Lp blows up for p > 2. Since ‖u‖L2 is constant, estimate (2.21)

is optimal with respect to p. On the other hand, it is known that the blow up rate given

by (2.20) and (2.21) is not always optimal, since some solutions blow up twice as fast (see

Section 4, and in particular Remark 4.3). Whether or not there exist solutions that blow

up at rates different from the rate of these explicit solutions is an open question. Note also

that (2.14) gives a lower estimate of blow up in the case α < 4/N , but in this case there

is no blow up.

Remark 2.15. In the proof of Theorem 2.5, one evaluatesd2

dt2‖xu(t)‖2

L2 in order to

show that if T ∗ were infinite, then ‖xu(t)‖2L2 would become negative in finite time. Note,

15

however that this does not imply that ‖xu(t)‖2L2 → 0 as t ↑ T ∗. This is sometimes the case

(see Section 4 and in particular Remark 4.3), but not always. First, observe that by the

invariance of the equation under space translation, one constructs easily a solution such

that ‖xu(t)‖2L2 6→ 0 as t ↑ T ∗. Indeed, it follows from the conservation of momentum (i.e.∫

u∇u =

∫ϕ∇ϕ) that, given x0 ∈ RN ,

∫|x− x0|2|u|2 =

∫|x|2|u|2 + |x0|2

∫|ϕ|2 + 2

∫x · x0|ϕ|2 + 4tIm

∫ϕx0 · ∇ϕ,

so that ‖(x− x0)u‖L2 will not converge to 0 for x0 large enough. Moreover, in general

inf‖(x− x0)u(t)‖L2 ; t ∈ [0, T ∗), x0 ∈ RN > 0.

To see this, we follow an argument of Merle [31]. Assume α = 4/N and consider a

real valued, spherically symmetric function ϕ ∈ H1(RN ) such that xϕ(x) 6∈ L2(RN )

and E(ϕ) < 0. Let (ϕn)n≥0 ⊂ X be a sequence of real valued, spherically symmetric

functions such that ϕn −→n→∞

ϕ inH1(RN ) and let un be the corresponding solutions of (1.1).

In particular, E(ϕn) −→n→∞

E(ϕ) and ‖ϕn‖L2 −→n→∞

‖ϕ‖L2 . Therefore, it follows from the

proof of Theorem 2.7 (see in particular formula (2.13)) that there exists a function Ψ ∈W 4,∞(RN ), Ψ ≥ 0 such that

d2

dt2

∫Ψ|un|2 ≤ 2E(ϕ) < 0,

for 0 ≤ t < T ∗(ϕn). On the other hand, since ϕn is real valued, one verifies easily that

d

dt

∫Ψ|un|2

∣∣∣∣t=0

= 0,

so that ∫Ψ|un|2 ≤ 2

∫Ψ|ϕ|2 + t2E(ϕ),

for 0 ≤ t < T ∗(ϕn) and for n large enough. This implies that there exists T 0 < ∞ such

that

T ∗(ϕn) ≤ T 0, (2.22)

for n large enough. On the other hand, for every x0 ∈ RN , we have (see the proof of

Theorem 2.5)

‖(x− x0)un(t)‖2L2 = ‖(x− x0)ϕn‖2L2 + 8E(ϕn)t2,

16

for 0 ≤ t < T ∗(ϕn). In particular, for n large enough,

‖(x− x0)un(t)‖2L2 ≥ ‖(x− x0)ϕn‖2L2 + 16E(ϕ)t2,

for 0 ≤ t < T ∗(ϕn). Since infx0∈RN

‖(x− x0)ϕn‖2L2 −→n→∞

∞, it now follows from (2.22) that

inf‖(x− x0)un(t)‖L2 ; 0 ≤ t < T ∗(ϕn), x0 ∈ RN −→n→∞

∞,

which proves the claim.

Remark 2.16. There is an abundant literature devoted to the determination of the blow

up rate by means of numerical computations. See [39] for a survey of some of this literature.

These experiments lead to the utmost confusion. For a given equation, the computed blow

up rates vary according to the authors, and not by an epsilon. For example, in the case

α = N = 2, the rates (as of 1993) range from (Tm − t)− 12 to (Tm − t)−

32 (see in particular

the fascinating Table 1 of [39], p. 409). Even more confusing, for a given author, the rate

may vary from one paper to another. On the top of this, it seems that the only blow up

rate which is known to be achieved (by the pseudo-conformally self-similar solutions in the

critical case α = 4/N) was never observed, or at least never reported. This last point does

not seem to worry too much the authors, the usual interpretation being that this blow up

rate is “exceptional”.

There might be several reasons to explain this situation. First of all when computing

numerically a blowing up solution, no matter how fine is the scheme and how close is

the computed solution to the actual solution, one has to stop the computation at some

point. Without any a priori information on the blow up rate, one cannot be sure that the

asymptotics of the solution will not change drastically between the time when one stops

the computation and the actual blow up time. Another major difficulty is the following.

It seems that most of the numerical computations were made by truncating the solution

for |x| large. However the solution, and in particular the blow up time, is quite sensitive

to small variations of the phase of the initial value for |x| large (see Lemma 4.7 and

Theorem 5.1). An interesting experiment was made to compare several methods for the

linear and nonlinear Schrodinger equations, in some cases where an explicit solution exists.

In particular, it is seen that the method of truncation of the domain can lead to large errors.

It seems that a spectral method that takes into account the values of the initial datum for

|x| large is much more accurate. (See Ahjaou [1], Ahjaou and Kavian [2].)

17

3. The pseudo-conformally invariant case. Sharp existence and blow up re-

sults. In this section, we assume α = 4/N . We recall that if λ ≤ 0, then all the

solutions of (1.1) with initial value ϕ ∈ H1(RN ) are global and bounded in H1(RN ) (see

Remark 2.4 (i)). If λ > 0, then the solutions are global provided ϕ ∈ H1(RN ) verifies

‖ϕ‖L2 < δ, for some δ > 0 (see Remark 2.4 (iii)). In fact, one can determine the optimal δ.

Let R be the (unique) spherically symmetric, positive ground state of the elliptic equation

−4R+ R = |R| 4N R, (3.1)

in RN (see for example [5], Definition 8.1.14 and Theorems 8.1.5 , 8.1.6 and 8.1.7). Note

that any ground state of (3.1) is of the form eiθR(x− y) for some θ ∈ R and y ∈ RN . We

have the following result of Weinstein [46].

Theorem 3.1. Assume λ > 0. If ϕ ∈ H1(RN ) is such that λ1α ‖ϕ‖L2 < ‖R‖L2 , then the

solution u of (1.1) is global and supt∈R

‖u(t)‖H1 <∞.

Remark 3.2. The condition ‖ϕ‖L2 < λ−1α ‖R‖L2 is sharp, in the sense that for any

ρ > λ−1α ‖R‖L2 (in fact, even for ρ = λ−

1α ‖R‖L2 , see Remark 4.3) there exists ϕ ∈ H1(RN )

such that ‖ϕ‖L2 = ρ and such that u blows up in finite time for both t < 0 and t > 0.

Indeed, let ψ(x) = R(√λx), so that ‖ψ‖L2 = λ−

1α ‖R‖L2 and ψ is a solution of

−4ψ + λψ = λ|ψ|αψ.

It follows that E(ψ) = 0 (see [5], formula (8.1.20)). Let ρ > λ−1α ‖R‖L2, set γ =

λ1α

ρ

‖R‖L2

> 1, and consider ϕρ = γψ. We have ‖ϕρ‖L2 = ρ and

E(ϕρ) = γα+2E(ψ)− γα+2 − γ2

2‖∇ψ‖2

L2 < 0;

and so, the corresponding solution uρ of (1.1) blows up in finite time (see Theorem 2.5).

Remark 3.3. In dimension 1, an elementary calculation shows that

R(x) =3

14

√cosh(2x)

.

In particular,

‖R‖2L2 = π

√3.

18

Therefore, it follows from Theorem 3.1 that if ϕ ∈ H1(RN ) is such that λ1α ‖ϕ‖L2 < 3

14√π,

then the solution u of (1.1) is global and supt∈R

‖u(t)‖H1 <∞.

The proof of Theorem 3.1 relies on the following lemma.

Lemma 3.4. The best constant in the Gagliardo-Nirenberg inequality

1

α+ 2‖ψ‖α+2

Lα+2 ≤C

2‖∇ψ‖2

L2‖ψ‖αL2,

is C = ‖R‖−αL2 .

Proof. We follow the argument of [46]. We have to show that

infu∈H1,u6=0

J(u) =2‖R‖α

L2

α+ 2, (3.2)

where

J(u) =‖∇u‖2L2‖u‖α

L2

‖u‖α+2Lα+2

.

We set

σ = infu∈H1,u6=0

J(u),

and we consider a minimizing sequence (un)n≥0. We observe that by Gagliardo-Nirenberg’s

inequality, σ > 0. We consider vn defined by vn(x) = µnun(λnx) with

λn =‖un‖L2

‖∇un‖L2

and µn =‖un‖

N−22

L2

‖∇un‖N2

L2

,

so that ‖vn‖L2 = ‖∇vn‖L2 = 1 and

‖vn‖−(α+2)Lα+2 = J(vn) = J(un) −→

n→∞σ > 0.

By symmetrization, we may assume that vn is spherically symmetric; and so, there exists

a subsequence, which we still denote by (vn)n≥0 and v ∈ H1(RN ) such that vn −→n→∞

v in

H1(RN ) weak and in Lα+2(RN ) strong. Since ‖v‖Lα+2 = limn→∞

‖vn‖Lα+2 = σ−1

α+2 > 0, it

follows that v 6= 0. This implies that

J(v) = σ, and ‖v‖L2 = ‖∇v‖L2 = 1. (3.3)

19

In particular,d

dtJ(v + tw)|t=0 = 0 for all w ∈ H1(RN ), and taking in account (3.3), we

obtain

−4v +α

2v = σ

α+ 2

2|v|αv.

Let now u be defined by v(x) = au(bx) with a =

(α

σ(α+ 2)

) 1α

and b =(α

2

) 12

, so that u

is a solution of (3.1) and

J(u) = J(v) = σ.

Since u satisfies equation (3.1), it follows from Pohozaev’s identity that

1

2‖∇u‖2L2 =

λ

α+ 2‖u‖α+2

Lα+2 ,

and that 2‖∇u‖2L2 = N‖u‖2

L2 (see [5], formulas (8.1.20) and (8.1.21)); and so

J(u) =N

N + 2‖u‖

4N

L2 =2

α+ 2‖u‖α

L2 . (3.4)

Since R also satisfies equation (3.1), it satisfies the same identity, and since u minimizes

J , we must have J(R) ≥ J(u), which implies that ‖u‖L2 ≤ ‖R‖L2. On the other hand, R

being the ground state of (3.1), it is also the solution of (3.1) of minimal L2 norm, so that

‖R‖L2 ≤ ‖u‖L2 . Therefore, ‖R‖L2 = ‖u‖L2 and the result now follows from (3.4).

Proof of Theorem 3.1. Let ϕ ∈ H1(RN ) verify λ1α ‖ϕ‖L2 < ‖R‖L2 and let u be the

maximal solution of (1.1). By conservation of charge and energy and by Lemma 3.4, we

have for all t ∈ (−T∗, T ∗),

1

2‖∇u(t)‖2

L2 ≤ E(ϕ) +λ

α+ 2‖u(t)‖α+2

Lα+2

≤ E(ϕ) +λ

2‖R‖αL2

‖∇u(t)‖2L2‖u(t)‖α

L2

≤ E(ϕ) +λ‖ϕ‖α

L2

2‖R‖αL2

‖∇u(t)‖2L2 ;

and so,1

2

(1− λ‖ϕ‖α

L2

‖R‖αL2

)‖∇u(t)‖2

L2 ≤ E(ϕ).

Hence the result.

20

The case α = 4/N is critical for the L2 theory. However, we have the following result.

(See [8,9].)

Theorem 3.5. Assume α = 4/N . For any ϕ ∈ L2(RN ), there exists a unique maximal

solution u ∈ C((−T∗, T ∗), L2(RN )) ∩ Lα+2loc ((−T∗, T ∗), Lα+2(RN )) of (1.1).

We have the following blow up alternative. Either T ∗ = ∞ (respectively, T∗ = ∞),

or else T ∗ < ∞ and ‖u‖Lq((0,T ∗),Lr) = ∞ for every admissible pair (q, r) with r ≥ α + 2

(respectively, T∗ < ∞ and ‖u‖Lq((−T∗,0),Lr) = ∞ for every admissible pair (q, r) with

r ≥ α+ 2).

Moreover u ∈ Lqloc((−T∗, T ∗), Lr(RN )), for every admissible pair (q, r).

If in addition ϕ ∈ H1(RN ), then u ∈ C((−T∗, T ∗), H1(RN )).

There is conservation of charge, i.e. (2.2) holds for all t ∈ (−T∗, T ∗).Finally, the solution u depends continuously on the initial value ϕ in the following

sense. The mappings ϕ 7→ T∗ and ϕ 7→ T ∗ are lower-semicontinuous L2(RN ) → R.

Moreover, if ϕn −→n→∞

ϕ in L2(RN ) and if un is the corresponding solution of (1.1), then

un −→n→∞

u in C([−T1, T2], L2(RN )) ∩ Lα+2([−T1, T2], L

α+2(RN )) for any bounded interval

[−T1, T2] ⊂ (−T∗, T ∗).

Proof. We argue only for t ≥ 0, the argument for t ≤ 0 being the same. (Note that if

u(t) satisfies (1.1), then u(−t) also satisfies (1.1).) We proceed in five steps.

Step 1. We first establish several estimates. Consider 0 < T ≤ ∞, let I = (0, T ) and set

J = I. Given u, v ∈ Lα+2(I, Lα+2(RN )), set f = |u|αu− |v|αv. Then

‖f‖L

α+2α+1 (I,L

α+2α+1 )

≤ (α+ 1)(‖u‖α

Lα+2(I,Lα+2) + ‖v‖αLα+2(I,Lα+2)

)

‖u− v‖Lα+2(I,Lα+2). (3.5)

This follows from the estimate | |u|αu − |v|αv| ≤ (α + 1)(|u|α + |v|α)|u − v| and Holder’s

inequality.

Given u ∈ Lα+2(I, Lα+2(RN )), set

F(u)(t) =

∫ t

0

T (t− s)|u(s)|αu(s) ds.

It follows from (3.5) and Strichartz’ estimate (1.2) that

F(u) ∈ C(J, L2(RN )) ∩ Lq(I, Lr(RN )),

21

for every admissible pair (q, r) and that

‖F(u)‖Lq(I,Lr) ≤ C‖u‖α+1Lα+2(I,Lα+2)

, (3.6)

for some constant C independent of T . In addition, if v ∈ Lα+2(I, Lα+2(RN )), then

‖F(u)−F(v)‖Lq(I,Lr) ≤ C(‖u‖α

Lα+2(I,Lα+2) + ‖v‖αLα+2(I,Lα+2)

)

‖u− v‖Lα+2(I,Lα+2). (3.7)

Step 2. Let 0 < T ≤ ∞, let I = (0, T ) and set J = I. We show that there exists δ > 0

such that if ϕ ∈ L2(RN ) verifies

‖T (·)ϕ‖Lα+2(I,Lα+2) < δ, (3.8)

then there exists a unique solution u ∈ C(J, L2(RN )) ∩ Lα+2(I, Lα+2(RN )) of (1.1). In

addition, u ∈ Lq(I, Lr(RN )) for any admissible pair (q, r). Moreover, u depends contin-

uously on ϕ, in the sense that if ϕ, ψ ∈ L2(RN ) both verify (3.8) and if u, v denote the

corresponding solutions of (1.1), then

‖u− v‖L∞(I,L2) + ‖u− v‖Lα+2(I,Lα+2) ≤ K‖ϕ− ψ‖L2 , (3.9)

for some constant K independent of T , u, v.

Let δ > 0, to be chosen later, and let ϕ and T be as above. Consider the set

E = u ∈ Lα+2(I, Lα+2(RN )); ‖u‖Lα+2(I,Lα+2) ≤ 2δ.

For u ∈ E, set

H(u)(t) = T (t)ϕ+ iλ

∫ t

0

T (t− s)|u(s)|αu(s) ds,

for t ∈ I. It follows easily from (3.6), (3.7) and (3.8) that if δ is small enough (independently

of ϕ and T ), then H is a strict contraction on E. Thus H has a fixed point u, which is the

unique solution of (1.1) in E. Applying (3.6) and (1.2), we see that u ∈ C(J, L2(RN )) ∩Lq(I, Lr(RN )), for every admissible pair (q, r). (3.9) follows easily from (3.7) and (1.2).

Step 3. u being as in Step 2, we show that if ϕ ∈ H1(RN ), then u ∈ C(J,H1(RN )).

It follows from Theorem 2.1 that (1.1) has a solution v ∈ C([0, T ∗), H1(RN )). v is

in particular an L2 solution, so that by uniqueness u and v coincide as long as they are

22

both defined. Therefore, we need only show that T ∗ > T . Assume to the contrary that

T ∗ ≤ T . Since the equation (1.1) is invariant by space translations and since the gradient

is the limit of the finite differences quotient, it follows easily from (3.9) that

‖∇v‖L∞((0,T ∗),L2) ≤ C‖∇ϕ‖L2 ,

which contradicts the blow up alternative for the H1 solutions.

Step 4. u being as in Step 2, we show that there is conservation of charge.

Indeed, let ϕn → ϕ in L2(RN ), with ϕn ∈ H1(RN ). It follows from (1.2) that for n

large enough, ϕn verifies (3.8), so that by (3.9), we have un → u in C(J, L2(RN )), where

un is the solution associated to ϕn. On the other hand, it follows from Step 3 that un is

an H1 solution, so that ‖un(t)‖L2 = ‖ϕn‖L2 for all t ∈ I. Passing to the limit as n→∞,

we obtain ‖u(t)‖L2 = ‖ϕ‖L2 for all t ∈ I.

Step 5. Conclusion. Let ϕ ∈ L2(RN ). Since by (1.2), T (·)ϕ ∈ Lα+2(R, Lα+2(RN )),

it follows that ‖T (·)ϕ‖Lα+2((0,T ),Lα+2) → 0 as T ↓ 0. Therefore, (3.8) is satisfied for

T small enough, and we can apply Step 2 to construct a unique local solution. Let

u ∈ C([0, T ∗), L2(RN )) be the maximal solution. It remains to establish the blow up

alternative and the continuous dependence. We show the blow up alternative by contra-

diction. Suppose that T ∗ <∞ and that ‖u‖Lα+2((0,T ∗),Lα+2) <∞. Let 0 ≤ t ≤ t+s < T ∗.

It follows that

T (s)u(t) = u(t+ s)− iλ

∫ s

0

T (s− σ)|u(t+ σ)|αu(t+ σ) dσ.

By (3.6), there exists C such that

‖T (·)u(t)‖Lα+2((0,T ∗−t),Lα+2) ≤ ‖u‖Lα+2((t,T ∗),Lα+2) + C‖u‖α+1Lα+2((t,T ∗),Lα+2)

.

Therefore, for t close enough to T ∗, we have

‖T (·)u(t)‖Lα+2((0,T ∗−t),Lα+2) < δ/2.

It follows from Step 2 that u can be extended after T ∗, which is a contradiction. This

shows that

‖u‖Lα+2((0,T ∗),Lα+2) = ∞.

23

Let now (q, r) be an admissible pair such that r > α+2. It follows from Holder’s inequality

that for any T < T ∗,

‖u‖Lα+2((0,T ),Lα+2) ≤ ‖u‖µL∞((0,T ),L2)‖u‖

1−µLq((0,T ),Lr) ≤ ‖ϕ‖µ

L2‖u‖1−µLq((0,T ),Lr), (3.10)

with µ =2(r − α− 2)

(α+ 2)(r − 2). Letting T ↑ T ∗, we obtain ‖u‖Lq((0,T ∗),Lr) = ∞.

Finally, we show the continuous dependence. Consider T < T ∗. We have u ∈C([0, T ], L2(RN )), so that it follows from (1.2) and an obvious compactness argument

that there exists τ > 0 such that

‖T (·)u(t)‖Lα+2((0,τ),Lα+2) ≤ δ/2,

for all t ∈ [0, T ], where δ is as in (3.8). Consider an integer n such that T ≤ nτ , let K ≥ 1

be the constant in (3.9) and let M be such that ‖T (·)v‖Lα+2(R,Lα+2) ≤M‖v‖L2 (see (1.2)).

Let ε > 0 be small enough so that MKn−1ε < δ/2. We claim that if ‖ϕ− ψ‖L2 ≤ ε, then

T ∗(ψ) > T and ‖u − v‖C([0,T ],L2) + ‖u − v‖Lα+2((0,T ),Lα+2) ≤ nKn‖ϕ − ψ‖L2, where v is

the solution corresponding to the initial value ψ. Indeed, if ‖ϕ− ψ‖L2 ≤ ε, then

‖T (·)ψ‖Lα+2((0,T/n),Lα+2) ≤ ‖T (·)ϕ‖Lα+2((0,T/n),Lα+2)

+ ‖T (·)(ϕ− ψ)‖Lα+2((0,T/n),Lα+2)

≤ δ/2 +Mε < δ.

Therefore, it follows from Step 2 that T ∗(ψ) > T/n and that

‖u− v‖C([0,T/n],L2) + ‖u− v‖Lα+2((0,T/n),Lα+2) ≤ K‖ϕ− ψ‖L2 .

In particular, ‖u(T/n) − v(T/n)‖L2 ≤ Kε. The claim follows by iterating this argument

n times. This completes the proof.

Remark 3.6. The blow up alternative in Theorem 3.5 is not very handy. It does not

concern the L2 norm of u. Indeed, despite of the conservation of charge, T∗ and T ∗ can

be finite in some cases. For example, assume λ > 0 and let ϕ ∈ X be such that E(ϕ) < 0.

It follows from Theorem 2.5 that u blows up in H1, for both t > 0 and t < 0. Therefore,

it follows from Theorem 3.5 that T ∗ <∞ and T∗ <∞.

24

Remark 3.7. We conjecture that if λ < 0, then T∗ = T ∗ = ∞ for all ϕ ∈ L2(RN ).

However, we only have the following partial result. Assume λ < 0, and suppose that

ϕ ∈ L2(RN ) is such that xϕ(x) ∈ L2(RN ). Then T ∗ = T∗ = ∞ and in addition u ∈Lq(R, Lr(RN )) for every admissible pair (q, r). Indeed, consider a sequence (ϕn)n≥0 ∈ X,

with ϕn −→n→∞

ϕ in L2(RN ) and xϕn(x) bounded in L2(RN ). The corresponding solutions

satisfy un ∈ C(R, X) (see Remark 2.4), and from the pseudo-conformal conservation

law (2.4) we see that ‖un(t)‖α+2Lα+2 ≤ Ct−2 for all t ∈ R. By continuous dependence, this

implies that ‖u(t)‖α+2Lα+2 ≤ Ct−2 for almost all t ∈ (−T∗, T ∗). In particular, we see that

if T ∗ < ∞, then u ∈ Lα+2((0, T ∗), Lα+2(RN )), which contradicts the blow up alternative

(note that u ∈ Lα+2((0, T ), Lα+2(RN )) for all 0 < T < T ∗). We see as well that T∗ = +∞.

In addition, it is clear that the above estimate implies that u ∈ Lα+2(R, Lα+2(RN )). The

estimate for an arbitrary admissible pair follows from inequality (3.10).

Remark 3.8. There exists η > 0 such that if

‖T (·)ϕ‖Lα+2(R,Lα+2) < η, (3.11)

then T∗ = T ∗ = ∞. Moreover, we have u ∈ Lq(R, Lr(RN )) for every admissible pair (q, r).

This follows easily from Step 2 of the proof of Theorem 3.5 (see in particular (3.8)). How-

ever, this conclusion does not hold in general for large data. Indeed, if λ > 0, there exist

nontrivial solutions (standing waves) of the form u(t, x) = eiωtφ(x), with φ ∈ H1(RN ),

φ 6= 0 (see [5], Section 8.1). These solutions obviously do not belong to Lq(R, Lr(RN ))

if q < ∞. On the other hand, it follows from Strichartz’ estimates (1.2) that (3.11) is

satisfied if

‖ϕ‖L2 < µ, (3.12)

for µ small enough.

Remark 3.9. Assume λ > 0 and let d be the supremum of the µ’s such that (3.12) implies

global existence. Then it is clear that d ≤ λ−1α ‖R‖L2, where R is as in Theorem 3.1 (see

Remark 3.2). Whether or not d = λ−1α ‖R‖L2 is an open question. However, we can

show that if ϕ ∈ L2(RN ) verifies ‖ϕ‖L2 < λ−1α ‖R‖L2, and if in addition xϕ ∈ L2(RN ),

then T∗ = T ∗ = +∞ and u ∈ Lq(R, Lr(RN )) for every admissible pair (q, r). Indeed,

25

consider a sequence (ϕn)n≥0 ⊂ X, with ϕn −→n→∞

ϕ in L2(RN ) and xϕn(x) bounded in

L2(RN ). The corresponding solutions satisfy un ∈ C(R, X) (see Theorem 3.1), and from

the pseudo-conformal conservation law (2.4) we see that (see [5], formula (7.1.6))

8t2E(vn(t)) = ‖xϕn‖L2

for all t ∈ R, where vn(t) = e−i|x|2

4t un(t). In particular, ‖vn(t)‖L2 = ‖un(t)‖L2 = ‖ϕn‖L2 ,

so that there exists ε > 0 such that ‖vn(t)‖L2 ≤ λ−1α ‖R‖L2 − ε for n large enough. It

follows that there exists C such that ‖∇vn(t)‖2L2 ≤ CE(vn(t)) for all t ∈ R (see the proof

of Theorem 3.1). By Lemma 3.4, this implies that

‖vn(t)‖α+2Lα+2 ≤ CE(vn(t))‖ϕn‖

4N

L2 ≤C

t2,

for all t ∈ R. We conclude as in Remark 3.7 above.

Theorem 3.5 has an immediate application to the study of the blowing up solutions.

Theorem 3.10. Let ϕ ∈ L2(RN ) be such that T ∗ <∞. If (tn)n≥0 is any sequence such

that tn ↑ T ∗, then u(tn) does not have any strong limit in L2(RN ). A similar statement

holds for T∗.

Proof. Assume that u(tn) −→n→∞

w in L2(RN ). By continuous dependence, we have

T ∗(u(tn)) ≥ 1

2T ∗(w) > 0 for n large enough. This implies that T ∗(ϕ) ≥ tn +

1

2T ∗(w)

for n large. This is absurd, since tn → T ∗.

In fact, one can prove a stronger result which implies the above theorem (see [10]).

Theorem 3.11. There exists ρ > 0 with the following property. Let ϕ ∈ L2(RN )

be such that T ∗ < ∞. If L is the set of weak L2 limit points of u(t) as t ↑ T ∗, then

‖w‖2L2 ≤ ‖ϕ‖2

L2 − ρ2, for all w ∈ L. A similar statement holds for T∗.

Proof. It follows from Step 2 of the proof of Theorem 3.5 (see in particular (3.8)) that

there exists δ > 0 such that if

‖T (·)φ‖Lα+2((0,τ),Lα+2) < δ,

26

then T ∗(φ) > τ . Taking φ = u(t), we deduce that

‖T (·)u(t)‖Lα+2((0,T ∗−t),Lα+2) ≥ δ,

for all t ∈ [0, T ∗). Therefore, given any ψ ∈ L2(RN ) and any t ∈ [0, T ∗), we have

δ ≤ ‖T (·)(u(t)− ψ)‖Lα+2((0,T ∗−t),Lα+2) + ‖T (·)ψ‖Lα+2((0,T ∗−t),Lα+2)

≤ c‖u(t)− ψ‖L2 + ‖T (·)ψ‖Lα+2((0,T ∗−t),Lα+2),

where c is the constant in the corresponding Strichartz’ inequality (1.2). Since

‖T (·)ψ‖Lα+2((0,T ∗−t),Lα+2) −→t↑T ∗

0,

it follows that

lim inft↑T ∗

‖u(t)− ψ‖L2 ≥ δ

c.

Therefore, if u(tn) ψ for some sequence tn ↑ T ∗, then

δ2

c2≤ lim inf

n→∞‖u(tn)− ψ‖2

L2 = lim infn→∞

(u(tn)− ψ, u(tn)− ψ)L2

= lim infn→∞

(‖u(tn)‖2L2 + ‖ψ‖2

L2 − 2(u(tn), ψ)L2

)= ‖u(tn)‖2L2 − ‖ψ‖2

L2

= ‖ϕ‖2L2 − ‖ψ‖2

L2,

and the result follows.

For H1 spherically symmetric blowing up solutions in dimension N ≥ 2, there is a

minimal amount of concentration of the L2 norm at the origin, as shows the following

result (see Merle and Tsutsumi [32] and Weinstein [48]).

Theorem 3.12. Assume N ≥ 2 and λ > 0, and let R be the spherically symmetric,

positive ground state of equation (3.1). Let γ : (0,∞) → (0,∞) be such that γ(s)−→s↓0

∞

and s12 γ(s)−→

s↓00. If ϕ ∈ H1(RN ) is spherically symmetric and such that T ∗ <∞, then

lim inft↑T ∗

‖u(t)‖L2(Ωt) ≥ λ−1α ‖R‖L2 ,

where Ωt = x ∈ RN ; |x| < |T ∗ − t| 12 γ(T ∗ − t). A similar statement holds for T∗.

As a consequence of Theorem 3.12, we have the following result.

27

Corollary 3.13. Assume N ≥ 2 and λ > 0, and let R be the spherically symmetric,

positive ground state of equation (3.1). Let ϕ ∈ H1(RN ) be spherically symmetric and

such that T ∗ <∞. If L is the set of weak L2 limit points of u(t) as t ↑ T ∗, then

‖w‖2L2 ≤ ‖ϕ‖2

L2 − λ−2α ‖R‖2

L2 ,

for all w ∈ L. A similar statement holds for T∗.

Remark 3.14. Note that the minimal loss of L2 norm given by Corollary 3.13 is optimal.

Indeed, there exist solutions that blow up in finite time, and for which ‖ϕ‖2L2 = λ−

2α ‖R‖2

L2

(see Remark 4.1). Corollary 3.13 improves the conclusion of Theorem 3.11 for H1 spher-

ically symmetric solutions, in the sense that it gives the optimal value of ρ.

Proof of Corollary 3.13. Assume tn ↑ T ∗ and u(tn) w in L2(RN ). Given ε > 0, we

have u(tn) w in L2(|x| > ε); and so,

‖w‖2L2(|x|>ε) ≤ lim inf

n→∞‖u(tn)‖2L2(|x|>ε).

On the other hand,

‖u(tn)‖2L2(|x|>ε) = ‖u(tn)‖2L2 − ‖u(tn)‖2L2(|x|<ε)

= ‖ϕ‖2L2 − ‖u(tn)‖2L2(|x|<ε)

≤ ‖ϕ‖2L2 − ‖u(tn)‖2L2(Ωt)

,

and the result follows from Theorem 3.12.

Proof of Theorem 3.12. Set ω(t) = ‖∇u(t)‖−1L2 , so that ω(t)−→

t↑T ∗0. We show that

lim inft↑T ∗

‖u(t)‖L2(|x|<ω(t)γ(T ∗−t)) ≥ λ−1α ‖R‖L2 . (3.13)

The theorem follows from (3.13) and (2.20), since γ is arbitrary. To prove (3.13), we argue

by contradiction and we assume that there exists tn ↑ T ∗ such that

limn→∞

‖u(tn)‖L2(|x|<ω(tn)γ(T ∗−tn)) < λ−1α ‖R‖L2 . (3.14)

We set

vn(x) = ω(tn)N2 u(tn, ω(tn)x),

28

so that

‖vn‖L2 = ‖u(tn)‖L2 = ‖ϕ‖L2 ,

‖∇vn‖L2 = 1,

E(vn) = ω(tn)2E(u(tn)) = ω(tn)2E(ϕ) −→n→∞

0.

(3.15)

It follows in particular that

E(vn) =1

2− λ

α+ 2‖vn‖α+2

Lα+2 ,

so that

‖vn‖α+2Lα+2 −→

n→∞

α+ 2

2λ6= 0. (3.16)

By (3.15), (vn)n≥0 is a bounded sequence in H1(RN ), so that there exists a subsequence,

which we still denote by (vn)n≥0, and w ∈ H1(RN ) such that vn w weakly inH1(RN ) as

n→∞. Since the vn’s are spherically symmetric, it follows that vn −→n→∞

w in Lα+2(RN ).

In particular, we have E(w) ≤ 0; and by (3.16), w 6= 0. By applying Lemma 3.4, we now

obtain

λ1α ‖w‖L2 ≥ ‖R‖L2. (3.17)

Let M > 0. We have in particular

‖w‖L2(|x|<M) = limn→∞

‖vn‖L2(|x|<M)

= limn→∞

‖u(tn)‖L2(|x|<Mω(tn))

≤ lim infn→∞

‖u(tn)‖L2(|x|<ω(tn)γ(T ∗−tn)),

since γ(s) →∞ as s ↓ 0. Since M is arbitrary, we find by applying (3.17)

lim infn→∞

‖u(tn)‖L2(|x|<ω(tn)γ(T ∗−tn)) ≥ ‖w‖L2 ≥ λ−1α ‖R‖L2,

which contradicts (3.14). This completes the proof.

In fact, Corollary 3.13 can be generalized to nonradial solutions (and also to the

dimension 1). More precisely, we have the following result.

Theorem 3.15. Let R be the spherically symmetric, positive ground state of equa-

tion (3.1). Let ϕ ∈ H1(RN ) be such that T ∗ <∞. If L is the set of weak L2 limit points

of u(t) as t ↑ T ∗, then

‖w‖2L2 ≤ ‖ϕ‖2

L2 − λ−2α ‖R‖2

L2 ,

29

for all w ∈ L. A similar statement holds for T∗.

Theorem 3.15 is an immediate consequence of the following proposition.

Proposition 3.16. Let (un)n≥0 ⊂ H1(RN ) \ 0 and u ∈ L2(RN ) be such that un u

in L2(RN ) as n→∞. If furthermore ‖∇un‖L2 −→n→∞

∞ and

lim supn→∞

E(un)

‖∇un‖2L2

≤ 0,

then ‖u‖2L2 ≤ lim inf

n→∞‖un‖2L2 − λ−

2α ‖R‖2

L2.

For the proof of Proposition 3.16, we use the concentration-compactness techniques.

Given a bounded sequence (un)n≥0 ⊂ L2(RN ), we set

ρn(t) = supy∈RN

∫

|x−y|≤t

|un(x)|2 dx,

so that ρn is a nondecreasing function of t for all n ≥ 0, and

ρ(t) = lim infn→∞

ρn(t),

so that ρ is also a nondecreasing function of t. Finally, we define

µ ((un)n≥0) = limt→∞

ρ(t).

With this notation, we have the following lemma.

Lemma 3.17. If (un)n≥0 ⊂ H1(RN ) is such that

(i) ‖un‖2L2 = a > 0;

(ii) 0 < infn≥0

‖∇un‖L2 ≤ supn≥0

‖∇un‖L2 <∞;

(iii) lim supn→∞

E(un) ≤ 0;

then µ ((un)n≥0) ≥ λ−2α ‖R‖2

L2.

Proof of Proposition 3.16 (assuming Lemma 3.17). Let (un)n≥0 be as in the

statement of Proposition 3.16. Set a = lim infn→∞

‖un‖2L2 . By considering a subsequence, we

may assume that

‖un‖2L2 −→n→∞

a.

30

Set ωn = ‖∇un‖−1L2 and define vn(x) = ω

2Nn un(ωnx). It follows that ‖vn‖2L2 = ‖un‖2L2 ,

‖∇vn‖2L2 = 1 and

lim supn→∞

E(vn) = lim supn→∞

E(un)

‖∇un‖2L2

≤ 0.

We first show that a > 0. Indeed, it follows from Lemma 3.4 that

E(vn) ≥ 1

2

(1− λ

‖vn‖αL2

‖R‖αL2

)‖∇vn‖2L2 ,

which implies that λ‖vn‖αL2 ≥ ‖R‖α

L2; and so, a ≥ λ−2α ‖R‖2

L2 . We now set

wn =

√a

‖vn‖L2

vn,

so that (wn)n≥0 verifies the assumptions of Lemma 3.17. Therefore,

µ ((wn)n≥0) ≥ λ−2α ‖R‖2

L2 .

Let ε > 0. It follows from the above inequality that there exists T such that

ρ(T ) ≥ λ−2α ‖R‖2

L2 − ε

4;

and so, for n large enough,

ρn(T ) ≥ λ−2α ‖R‖2

L2 − ε

2.

Therefore, there exists a sequence (yn)n≥0 ⊂ RN such that

∫

|x−yn|≤T

|wn(x)|2 dx ≥ λ−2α ‖R‖2

L2 − ε,

for n large, which means that

a

‖un‖2L2

∫

|x−zn|≤tn

|un(x)|2 dx ≥ λ−2α ‖R‖2

L2 − ε,

with zn = ωnyn and tn = ωnT . Note that tn −→n→∞

0. By possibly extracting a subsequence,

we may assume that either |zn| −→n→∞

∞ or else zn −→n→∞

z for some z ∈ RN . In the first

case, consider M > 0. Since un u in L2(|x| ≤M), we have

‖u‖2L2(|x|≤M) ≤ lim infn→∞

‖un‖2L2(|x|≤M) = lim infn→∞

‖un‖2L2 − ‖un‖2L2(|x|≥M)

= a− lim supn→∞

‖un‖2L2(|x|≥M) ≤ a− λ−2α ‖R‖2

L2 + ε.

31

The result follows by letting M ↑ ∞ then ε ↓ 0. In the second case, consider δ > 0. Since

un u in L2(|x| ≥ δ), we have

‖u‖2L2(|x−z|≥δ) ≤ lim infn→∞

‖un‖2L2(|x−z|≥δ) = lim infn→∞

‖un‖2L2 − ‖un‖2L2(|x−z|≤δ)

= a− lim supn→∞

‖un‖2L2(|x−z|≤δ) ≤ a− λ−2α ‖R‖2

L2 + ε.

The result follows by letting δ ↓ 0 then ε ↓ 0. This completes the proof.

Proof of Lemma 3.17. We claim that there exists δ > 0, depending only on N and λ

with the following property. If (un)n≥0 ⊂ H1(RN ) is such that

‖un‖2L2 = a > 0; (3.18)

0 < infn≥0

‖∇un‖L2 ≤ supn≥0

‖∇un‖2L2 <∞; (3.19)

lim supn→∞

E(un) ≤ 0; (3.20)

µ ((un)n≥0) < λ−2α ‖R‖2

L2; (3.21)

then there exists a sequence (un)n≥0 ⊂ H1(RN ) which satisfies (3.19), (3.20) and (3.21),

and such that ‖un‖2L2 = a − β, for some β ≥ δ. The result follows, since if (un)n≥0 ⊂H1(RN ) is as in the statement of the lemma, and if µ ((un)n≥0) < λ−

2α ‖R‖2

L2, then one

can apply the claim k times with k large enough so that a− kδ < 0 in order to obtain a

sequence (un)n≥0 ⊂ H1(RN ) such that ‖un‖2L2 < 0, which is absurd. Therefore, we need

only prove the claim, and we consider (un)n≥0 ⊂ H1(RN ) satisfying (3.18)—(3.21). We

proceed in several steps.

Step 1. There is a constant K such that

‖w‖α+2Lα+2 ≤ K

(sup

y∈RN

∫

|x−y|≤t

|w(x)|2 dx)α

2 (∫

RN

|∇w|2 +1

t2

∫

RN

|w|2), (3.22)

for all w ∈ H1(RN ) and all t > 0. Indeed, it follows from [5], Lemma 8.3.7 that there is a

constant K such that

‖w‖α+2Lα+2 ≤ K

(sup

y∈RN

∫

|x−y|≤1

|w(x)|2 dx)α

2

‖w‖2H1 ,

for all w ∈ H1(RN ). By changing w(x) to w(tx), we obtain (3.22).

32

We now set

δ =

(α+ 2

2K

) 2α

> 0, (3.23)

we set

µ = µ ((un)n≥0) ,

and we consider the functions ρn(t) introduced for the definition of µ.

Step 2. There exists a sequence nk →∞ and tnk→∞ such that µ = lim

k→∞ρnk

(tnk), as

follows immediately from the definition of µ. Therefore,

µ = limn→∞

ρn(tn). (3.24)

with tn →∞, after renaming the sequence (un)n≥0.

Step 3. There exists a subsequence (ρnk)k≥0 and a nondecreasing function γ such that

ρnk(t) −→

k→∞γ(t), uniformly on bounded intervals.

Indeed, fix n ≥ 0 and t > 0. We first show that for all t > 0 there exists yn(t) ∈ RN

such that

ρn(t) =

∫

|x−yn(t)|≤t

|un(x)|2 dx. (3.25)

Indeed, fix t > 0. There is a sequence (ynk )k≥0 ⊂ RN such that

ρn(t) = limk→∞

∫

|x−ynk|≤t

|un(x)|2 dx > 0.

We claim that the sequence (ynk )k≥0 is bounded. Otherwise, there exists a subsequence

(ynkj

)j≥0 such that |x− ynkj| ≤ t ∩ |x− yn

k`| ≤ t = ∅ for j 6= `; and so,

∫

RN

|un|2 ≥∑

j≥0

∫

|x−ynkj|≤t

|un(x)|2 dx = +∞,

which is absurd. Therefore, (ynk )k≥0 has a convergent subsequence, and its limit yn(t)

verifies (3.25). We now claim that the sequence (ρn)n≥0 is uniformly Holder continuous

on bounded intervals. Indeed, let r > 2 be such that H1(RN ) → Lr(RN ) and consider

0 ≤ s ≤ t <∞. We have

|ρn(t)− ρn(s)| = ρn(t)− ρn(s) =

∫

|x−yn(t)|≤t

|un|2 −∫

|x−yn(s)|≤s

|un|2

=

∫

s≤|x−yn(t)|≤t

|un|2 +

∫

|x−yn(t)|≤s

|un|2 −∫

|x−yn(s)|≤s

|un|2

≤∫

s≤|x−yn(t)|≤t

|un|2,

33

by (3.25) and the definition of ρn(s). Therefore,

|ρn(t)− ρn(s)| ≤ C‖un‖2H1 |s ≤ |x− yn(t)| ≤ t| r−2r ,

and the result follows, since un is bounded in H1(RN ). By Ascoli’s theorem, there exists

a subsequence, which we still denote by (ρn)n≥0 and a nondecreasing function γ such that

ρn(t) −→n→∞

γ(t), (3.26)

uniformly on bounded intervals. Note that by passing to a subsequence, we do not lose

property (3.24).

Step 4. We have

µ = limt→∞

γ(t) ≥ δ, (3.27)

where δ is defined by (3.23). Moreover,

µ = limn→∞

ρn(tn/2), (3.28)

where (tn)n≥0 is as in (3.24).

Indeed, note first that by definition of µ,

µ ≤ lim supt→∞

lim infn→∞

ρn(t) = lim supt→∞

γ(t) = limt→∞

γ(t).

On the other hand, it follows from (3.24) and the nondecreasing character of ρn that for

all t > 0,

µ = limn→∞

ρn(tn) ≥ limn→∞

ρn(t) = γ(t).

By letting t ↑ ∞, we obtain µ ≥ limt→∞

γ(t), so that µ = limt→∞

γ(t). Finally, it follows

from (3.22) and (3.23) that

E(un) ≥ 1

2

(1−

(ρn(tn)

δ

)α2

)∫|∇un|2 −

K

(α+ 2)t2nρn(tn)

α2 .

If µ < δ, then we obtain by letting n→∞ and applying (3.24) and (3.19)

lim supn→∞

E(un) ≥ 1

2

(1−

(µδ

)α2

)infn≥0

∫|∇un|2 > 0,

which is absurd. This proves (3.27).

34

Next, since ρn(tn/2) ≤ ρn(tn), it follows from (3.24) that lim supn→∞

ρn(tn/2) ≤ µ. Fi-

nally, given t > 0, we have tn/2 ≥ t for n large, so that

ρn(tn/2) ≥ ρn(t) −→n→∞

γ(t).

Therefore, lim infn→∞

ρn(tn/2) ≥ γ(t). (3.28) follows, by letting t→∞.

Step 5. Construction of the sequence (un)n≥0. Let θ ∈ C∞([0,∞)) be such that

θ(t) ≡ 1 for 0 ≤ t ≤ 1/2, θ(t) ≡ 0 for t ≥ 3/4 and 0 ≤ θ ≤ 1. Let ϕ ∈ C∞([0,∞)) be such

that ϕ(t) ≡ 0 for 0 ≤ t ≤ 3/4, ϕ(t) ≡ 1 for t ≥ 1 and 0 ≤ ϕ ≤ 1. Finally, set ψ = 1−θ−ϕ.

We define

θn(t) = θ

(t

tn

), ϕn(t) = ϕ

(t

tn

), ψn(t) = ψ

(t

tn

),

where tn is as in (3.24). It follows in particular that θn + ϕn + ψn = 1 and that

‖∇θn‖L∞ + ‖∇ϕn‖L∞ + ‖∇ψn‖L∞ ≤ C

tn. (3.29)

Finally, we setvn(x) = θn(|x− yn(tn/2)|)un(x),

un(x) = ϕn(|x− yn(tn/2)|)un(x),

wn(x) = ψn(|x− yn(tn/2)|)un(x),

where yn(t) is as in (3.25).

Step 6. Conclusion. We have

ρn(tn/2) =

∫

|x−yn(tn/2)|≤tn/2

|un|2 ≤∫

RN

|vn|2

≤∫

|x−yn(tn/2)|≤tn

|un|2 ≤∫

|x−yn(tn)|≤tn

|un|2 ≤ ρn(tn),

so that

‖vn‖2L2 −→n→∞

µ, (3.30)

by (3.24) and (3.28). Furthermore,∫

RN

|wn|2 ≤∫

tn/2≤|x−yn(tn/2)|≤tn

|un|2

=

∫

|x−yn(tn/2)|≤tn

|un|2 −∫

|x−yn(tn/2)|≤tn/2

|un|2

≤∫

|x−yn(tn)|≤tn

|un|2 −∫

|x−yn(tn/2)|≤tn/2

|un|2

= ρn(tn)− ρn(tn/2),

35

so that

‖wn‖2L2 −→n→∞

0. (3.31)

Since un = vn +wn + un and since vn and un have disjoint supports, it follows from (3.30),

(3.31) and (3.18) that

‖un‖2L2 −→n→∞

a− µ. (3.32)

In addition, since |un| ≤ |un|, it follows immediately from (3.21) that

µ ((un)n≥0) ≤ µ ((un)n≥0) < λ−2α ‖R‖2

L2 . (3.33)

On the other hand, it follows from Lemma 3.4, (3.21), (3.30) and (3.31) that there exists

σ > 0 such that

E(vn) ≥ σ‖∇vn‖2L2 and E(wn) ≥ σ‖∇wn‖2L2 , (3.34)

for n large. On the other hand, one verifies easily (since vnun ≡ 0) that

∣∣ |un|α+2 − |vn|α+2 − |wn|α+2 − |un|α+2∣∣ ≤ C|un|α+1|wn|,

so that ∣∣∣∣∫|un|α+2 −

∫|vn|α+2 −

∫|wn|α+2 −

∫|un|α+2

∣∣∣∣ −→n→∞0. (3.35)

(Note that un is bounded in H1(RN ) and that wn is bounded in H1(RN ) and converges

to 0 in L2(RN ), hence in Lα+2(RN ).) Furthermore, it follows from an easy calculation

that

|∇un|2 − |∇vn|2 − |∇wn|2 − |∇un|2 = 2(∇ψn · (∇θn +∇ϕn)|un|2

+ψn(θn + ϕn)|∇un|2 +∇(ψn(θn + ϕn))Re(un∇un))

≥ −Ct2n|un|2 −

C

tn|un| |∇un|,

by (3.29), so that

lim infn→∞

∫|∇un|2 −

∫|∇vn|2 −

∫|∇wn|2 −

∫|∇un|2

≥ 0.

This, together with (3.35), implies that

lim infn→∞

E(un)− E(vn)− E(wn)− E(un) ≥ 0. (3.36)

36

It now follows from (3.20), (3.34) and (3.36) that

lim supn→∞

E(un) ≤ 0. (3.37)

Next, it follows immediately from the definition of un that

‖∇un‖2L2 ≤ C‖un‖2H1 . (3.38)

Finally, we show that

lim infn→∞

‖∇un‖L2 > 0. (3.39)

To prove this, we argue by contradiction and we assume that there exists a subsequence,

which we still denote by (un)n≥0 such that ‖∇un‖L2 −→n→∞

0. It then follows from (3.32) and

Lemma 3.4 that E(un) −→n→∞

0. (3.20), (3.36) and (3.34) now imply that ‖∇vn‖L2 −→n→∞

0

and ‖∇wn‖L2 −→n→∞

0. Therefore, ‖∇un‖L2 −→n→∞

0, which contradicts (3.19); and so we have

proved (3.39).

It now follows from (3.33), (3.37), (3.38) and (3.39) that the sequence (un)n≥0 veri-

fies (3.19), (3.20) and (3.21). Changing un to

√a− µ

‖un‖L2

un and applying (3.32), we see that

the new sequence (un)n≥0 verifies (3.18)—(3.21). This completes the proof.

37

4. Some applications of the pseudo-conformal transformation. In this section,

we still assume α = 4/N . In this case, the pseudo-conformal conservation law (2.4),

introduced by Ginibre and Velo [20], becomes an exact conservation law. This conservation

law is associated to a group of transformations which leaves invariant the set of solutions

of (1.1) (see Ginibre and Velo [19]). We describe below this group of transformations (the

pseudo-conformal transformation).

Let b ∈ R. Given (t, x) ∈ R×RN , we define the conjugate variables (s, y) ∈ R×RN

by

s =t

1 + bt, y =

x

1 + bt,

or equivalently

t =s

1− bs, x =

y

1− bs.

Given u defined on (−S1, S2)×RN with 0 ≤ S1, S2,≤ ∞, we set

T1 =

∞ if bS1 ≤ −1,

S1

1 + bS1if bS1 > −1,

T2 =

∞ if bS2 ≥ 1,

S2

1− bS2if bS2 < 1,

and we define ub on (−T1, T2) by

ub(t, x) = (1 + bt)−N2 ei

b|x|2

4(1+bt) u

(t

1 + bt,

x

1 + bt

),

or equivalently

ub(t, x) = (1− bs)N2 ei

b|y|2

4(1−bs) u(s, y). (4.1)

Note that

‖ub(t)‖L2 = ‖u(s)‖L2 , (4.2)

and more generally,

‖ub(t)‖Lβ+2 = (1− bs)Nβ

2(β+2) ‖u(s)‖Lβ+2 , (4.3)

if β ≥ 0. In particular,

‖ub(t)‖Lα+2 = (1− bs)2

α+2 ‖u(s)‖Lα+2 , (4.4)

so that if bs1 > −1 and bs2 < 1, then

‖ub‖Lα+2((−t1,t2)Lα+2) = ‖u‖Lα+2((−s1,s2)Lα+2), (4.5)

38

with t1 =s1

1 + bs1and t2 =

s21− bs2

. Next, if u ∈ C((−S1, S2), X), then it is clear that

ub ∈ C((−T1, T2), X). In addition,

‖xub(t)‖L2 = (1− bs)−1‖yu(s)‖L2, (4.6)

‖∇ub(t)‖L2 =1

2‖(y + 2i(1− bs)∇)u(s)‖L2, (4.7)

‖∇u(s)‖L2 =1

2‖(x+ 2i(1 + bt)∇)ub(t)‖L2. (4.8)

The interest of the above transformation lies in the following result.

Theorem 4.1. If u ∈ C((−S1, S2), L2(RN )) ∩ Lα+2

loc ((−S1, S2), Lα+2(RN )) is a solution

of (1.1), then ub ∈ C((−T1, T2), L2(RN )) ∩ Lα+2

loc ((−T1, T2), Lα+2(RN )) is also a solution

of (1.1).

Moreover, if u ∈ C((−S1, S2), X), then ub ∈ C((−T1, T2), X).

Proof. It is clear that ub ∈ C((−T1, T2), L2(RN )). In addition, it follows from (4.5) that

ub ∈ Lα+2loc ((−T1, T2), L

α+2(RN )).

Furthermore, one shows that if 0 ≤ S1, S2 < ∞ and if bS1 > −1 and bS2 < 1, then

the mapping u 7→ ub is continous C([−S1, S2], L2(RN )) ∩ Lα+2((−S1, S2), L

α+2(RN )) →C([−T1, T2], L

2(RN )) ∩ Lα+2((−T1, T2), Lα+2(RN )).

Consider now a solution u ∈ C([−S1, S2], L2(RN )) ∩ Lα+2((−S1, S2), L

α+2(RN ))

of (1.1), with S1 and S2 as above. Let ϕ = u(0). We have in particular T∗(ϕ) > S1

and T ∗(ϕ) > S2. Consider (ϕn)n≥0 ⊂ H2(RN ) such that ϕn → ϕ in L2(RN ). By con-

tinuous dependence, we have T∗(ϕn) > S1 and T ∗(ϕn) > S2 for n large enough. We

denote by un the corresponding solutions of (1.1). Note that un is an H2 solution. A

tedious, but straightforward calculation shows that (un)b satisfies (1.1) almost everywhere

on (−T1, T2)×RN . The conclusion follows from the continuity property mentioned above

and the continous dependence.

Remark 4.2. Note that the pseudo-conformal transformation preserves both the space

L2(RN ) and the space X. On the other hand, it does not preserve the space H1(RN ).

Remark 4.3. The pseudo-conformal transformation has a simple application, which

yields interesting information on the blow up (see Weinstein [47]). Assume for simplicity

39

that λ = 1 and let ψ be a nontrivial solution of (3.1). (Note that ψ(x) has exponential de-

cay as |x| → ∞, see for example [5], Theorem 8.1.1.) It follows that u(t, x) = eitψ(x) is the

solution of (1.1) with ϕ = ψ, and that T ∗(ϕ) = T∗(ϕ) = +∞. We set v(t, x) = u−1(t, x),

i.e.

v(t, x) = (1− t)−N2 e−i

|x|2

4(1−t) ei t1−tψ

(x

1− t

), (4.9)

for x ∈ RN and t < 1. It follows that for any 1 ≤ p ≤ ∞,

‖v(t)‖Lp = (1− t)−N(p−2)

2p ‖ψ‖Lp , (4.10)

for all t < 1. Therefore, ‖v‖Lα+2((0,1),Lα+2) = +∞, so that T ∗ = 1. Furthermore, it follows

from (4.7) that

‖∇v(t)‖2L2 =

1

4

∫

RN

∣∣∣∣(x+

2i

1− t∇)ψ(x)

∣∣∣∣2

dx,

so that

(1− t)‖∇v(t)‖L2 −→t↑1

‖∇ψ‖L2 . (4.11)

It follows in particular from (4.10) and (4.11) that v blows up twice as fast as the lower

estimates (2.20) and (2.21). This implies that, at least in the case α = 4/N , the lower

estimates (2.20) and (2.21) are not optimal for all the blowing up solutions.

It also follows from (4.10) that ‖v(t)‖Lp −→t↑1

0 if 1 ≤ p < 2, so that

v(t) 0, (4.12)

in L2(RN ) as t ↑ 1. In particular, the loss of L2 norm at blow up is equal to ‖R‖L2 if

ψ is a ground state of (3.1), but it is larger if ψ is an excited state. (Note that excited

states exist if N ≥ 2, see [4].) Therefore, the loss of L2 norm given by Theorem 3.15 is not

always optimal.

Note also that by (4.6),

‖xv(t)‖L2 = (1− t)‖xψ‖L2 −→t↑1

0. (4.13)

(c.f. Remark 2.15.) In particular, v(t)−→t↑1

0 in L2(|x| ≥ ε) for any ε > 0. Furthermore,

one verifies easily that v(t)−→t↑1

0 in H1(|x| ≥ ε) and in L∞(|x| ≥ ε) (this last point

because ψ has exponential decay). Therefore, v(t) blows up only at x = 0. Furthermore,

40

it follows from an easy calculation that |v(t)|2−→t↑1

‖ψ‖2L2δ in D′(RN ), where δ is the Dirac

measure at x = 0.

Finally, we observe that formula (4.9) also makes sense for t > 1, and that v given

by (4.9) is also a solution of (1.1) for t > 1. As a matter of fact, the properties of v as t ↑ 1

and as t ↓ 1 are similar. Formula (4.9) gives (formally) an extension of the solution v beyond

the blow up time T ∗ = 1. We know that v satisfies (1.1) on (−∞, 1) and on (1,∞), and we

investigate in what sense v may be a solution near t = 1. Note first that v ∈ C((−∞, 1)∪(1,∞), L2(RN )) and that ‖v(t)‖L2 = ‖ψ‖L2 for t 6= 1, so that by property (4.12) v is

discontinuous in L2(RN ) at t = 1. On the other hand, v ∈ L∞(R, L2(RN )), so that 4v ∈L∞(R, H−2(RN )). Furthermore, it follows from (4.10) that ‖ |v(t)|αv(t)‖L1 = c|1− t|N−4

2 .

Therefore, if we assume N ≥ 3, then |v|αv ∈ L1loc(R, L

1(RN )). If m is an integer such

that L1(RN ) → H−m(RN ) (so that in particular m ≥ 2), then it follows that |v|αv ∈L1

loc(R, H−m(RN )). Therefore, ut ∈ L1

loc(R, H−m(RN )), so that u ∈ C(R, H−m(RN )).

It follows that v(t) → 0 in H−m(RN ) as t→ 1. This implies easily that v satisfies (1.1) in

D′(R, H−m(RN )). Therefore, we see that v can be extended in a reasonable sense beyond

the blow up time T ∗ = 1. However, the meaning of this extension is not quite clear.

Indeed, if we define

v(t) =

v(t) if t < 1,

0, if t ≥ 1,

then the above argument shows that v is also an extension of v beyond T ∗ = 1, which sat-

isfies (1.1) in D′(R, H−m(RN )). As a matter of fact, one can define many such extensions.

For example, since equation (1.1) is invariant by space translation and by multiplication

by a constant of modulus 1, it follows easily that for any y ∈ RN and ω ∈ R,

v(t) =

v(t) if t < 1,

eiωv(t, · − y), if t ≥ 1,

is also an extension of v beyond T ∗ = 1, which satisfies (1.1) in D′(R, H−m(RN )). It is

not clear whether or not there is such an extension which is, in some way, more natural

than the others.

Remark 4.4. In dimension 1, the solutions considered in the above remark are com-

pletely explicit. Indeed,

R(x) =3

14

√cosh(2x)

,

41

(see Remark 3.3), so that

u(t, x) = eitR(x)

is a solution of the Schrodinger equation iut + uxx + |u|4u = 0; and so,

v(t, x) =1√

1− te−i x2

4(1−t) ei t1−t

314

√cosh( 2x

1−t )

is also a solution of the equation, that blows up at t = 1.

Remark 4.5. Let v(t) be as in Remark 4.3. Given y ∈ RN , set vy(t) = v(t, · − y), so

that vy is a solution of (1.1) for which T ∗ = 1, and that blows up at the point y ∈ RN .

Given (y`)1≤`≤k with yj 6= y` for j 6= `,

w(t) =k∑

`=1

vy`(t),

is a function that blows up at t = 1, and only at the points y`. On the other hand,

since (1.1) is nonlinear, w is not a solution of (1.1). However, Merle [29] shows that there

exists a solution u of (1.1) on [0, 1) for which T ∗ = 1 and which is asymptotic as t ↑ 1 to

w. This shows in a way the stability of the type of blow up displayed by v.

Remark 4.6. Assume for simplicity λ = 1 and let R be the positive, spherically sym-

metric ground state of (3.1). It follows that for any γ ∈ R, µ > 0 and y ∈ RN ,

v(t, x) = eiγeiµ2tR(µ(x − y)) is a solution of (1.1). For any b < 0 and x1 ∈ RN ,

u(t, x) = vb(t, x− x1) is therefore also a solution. An easy calculation shows that

u(t, x) =

(ω

T ∗ − t

)N2

eiθ−i

|x−x1|2

4(T∗−t)+i ω2

T∗−tR

(ω

T ∗ − t((x− x1)− (T ∗ − t)x0)

), (4.14)

with T ∗ = −1

b, ω = µT ∗, x0 =

y

T ∗, θ = γ − µ2T ∗.

Let now ϕ ∈ H1(RN ) be such that ‖ϕ‖L2 = ‖R‖L2 and such that T ∗ <∞. It follows

from Merle [30] that there exist θ ∈ R, ω > 0, x0, x1 ∈ RN such that u is given by (4.14).

Similarly, if T∗ <∞, then there exist θ ∈ R, ω > 0, x0, x1 ∈ RN such that u is given by

u(t, x) =

(ω

T∗ + t

)N2

eiθ+i|x−x1|

2

4(T∗+t)−i ω2

T∗+tR

(ω

T∗ + t((x− x1)− (T∗ + t)x0)

).

42

In other words, the only solutions that blow up on the critical sphere are those obtained

from the ground state by the pseudo-conformal transformation. Note in particular that if

u is a solution on the critical L2 sphere, then T ∗ and T∗ cannot both be finite.

Consider now an initial value ϕ ∈ L2(RN ), and let u ∈ C((−T∗, T ∗), L2(RN )) ∩Lα+2

loc ((−T∗, T ∗), Lα+2(RN )) be the maximal solution of (1.1). Given b ∈ R, set

ϕb(x) = eib|x|2

4 ϕ(x),

and let ub ∈ C((−T b∗ , T

∗b ), L2(RN ))∩Lα+2

loc ((−T b∗ , T

∗b ), Lα+2(RN )) be the maximal solution

of (1.1) corresponding to the initial value ϕb. By uniqueness and Theorem 4.1, it follows

that ub coincides with ub given by (4.1), as long as ub is defined. In particular,

T ∗b = ∞ if bT ∗ ≥ 1, T ∗b ≥T ∗

1− bT ∗if bT ∗ < 1,

and

T b∗ = ∞ if bT∗ ≤ −1, T b

∗ ≥T ∗

1 + bT∗if bT∗ > −1.

Before proceeding further, we introduce some notation. We set

B+ = ϕ ∈ L2(RN ); T ∗ <∞,

N+ = ϕ ∈ L2(RN ); T ∗ = ∞ and ‖u‖Lα+2((0,∞),Lα+2) = ∞,

R+ = ϕ ∈ L2(RN ); T ∗ = ∞ and ‖u‖Lα+2((0,∞),Lα+2) <∞.

Similarly, we set

B− = ϕ ∈ L2(RN ); T∗ <∞,

N− = ϕ ∈ L2(RN ); T∗ = ∞ and ‖u‖Lα+2((−∞,0),Lα+2) = ∞,

R− = ϕ ∈ L2(RN ); T∗ = ∞ and ‖u‖Lα+2((−∞,0),Lα+2) <∞.

We observe that if u satisfies (1.1) on [0, T ], then v(t) = u(−t) satisfies (1.1) on [−T, 0]

with the initial value v(0) = ϕ. This implies that

B− = B+ = ϕ ∈ L2(RN ); ϕ ∈ B+,

N− = N+ = ϕ ∈ L2(RN ); ϕ ∈ N+,

R− = R+ = ϕ ∈ L2(RN ); ϕ ∈ R+.

43

The following result is an immetiate consequence of (4.5) and of the blow up alternative

of Theorem 3.5.

Lemma 4.7. The following properties hold.

(i) If ϕ ∈ B+, then ϕb ∈ B+ and T ∗b =T ∗

1− bT ∗for all b <

1

T ∗; ϕ 1

T∗∈ N+; ϕb ∈ R+ for

all b >1

T ∗.

(ii) If ϕ ∈ N+, then ϕb ∈ B+ and T ∗b = −1

bfor all b < 0; ϕb ∈ R+ for all b > 0.

(iii) If ϕ ∈ R+, then T ∗b > −1

bfor all b < 0; ϕb ∈ R+ for all b ≥ 0.

Similar properties hold for t < 0. More precisely,

(iv) If ϕ ∈ B−, then ϕb ∈ B− and T b∗ =

T∗1 + bT∗

for all b > − 1

T∗; ϕ− 1

T∗∈ N−; ϕb ∈ R−

for all b < − 1

T∗.

(v) If ϕ ∈ N−, then ϕb ∈ B− and T b∗ =

1

bfor all b > 0; ϕb ∈ R− for all b < 0.

(vi) If ϕ ∈ R−, then T b∗ >

1

bfor all b > 0; ϕb ∈ R− for all b ≤ 0.

Corollary 4.8. Let ϕ ∈ L2(RN ). Then either T ∗(ϕb) = ∞ for all b ∈ R, or else there

exists b0 ∈ R such that T ∗(ϕb) < ∞ for all b < b0 and T ∗(ϕb) = ∞ for all b ≥ b0. In the

first case, we have ϕb ∈ R+ for all b ∈ R. In the latter case ϕb0 ∈ N+ and ϕb ∈ R+ for all

b > b0.

Similarly, either T∗(ϕb) = ∞ for all b ∈ R, or else there exists b1 ∈ R such that

T∗(ϕb) < ∞ for all b > b1 and T ∗(ϕb) = ∞ for all b ≤ b1. In the first case, we have

ϕb ∈ R− for all b ∈ R. In the latter case ϕb1 ∈ N− and ϕb ∈ R− for all b < b1.

Proof. We begin with the remark that (ϕb)c = ϕb+c and (ub)c = ub+c for all b, c ∈ R.

Suppose first that T ∗(ϕb) = ∞ for all b ∈ R. If there exists b ∈ R such that

ϕb ∈ N+, then ‖ub‖Lα+2((0,∞),Lα+2) = ∞. By property (ii) of Lemma 4.7, this implies that

ϕb−1 = (ϕb)−1 ∈ B+, which is absurd; and so, ϕb ∈ R+ for all b ∈ R.

Suppose now that there exists c ∈ R such that ϕc 6∈ R+. Then ϕc ∈ N+∪B+. Setting

b0 =1

T ∗(ϕc), it follows from properties (i) and (ii) of Lemma 4.7 that ϕb ∈ B+ for b < b0,

ϕb0 ∈ N+ and ϕb ∈ R+ for b > b0. The properties for t < 0 are proved similarly.

44

Corollary 4.9. R+ and R− are open subsets of L2(RN ). R+ ∩X and R− ∩X are open

subsets of X.

Proof. Let ϕ ∈ R+. Set ψ = ϕ−1. It follows from property (iii) of Lemma 4.7 that

T ∗(ψ) > 1. By continuous dependence (see Theorem 3.5), it follows that there exists ε > 0

such that if ‖ψ − φ‖L2 ≤ ε, then T ∗(φ) > 1. This implies that φ1 ∈ R+. Setting γ = φ1,

we see that if ‖ϕ− γ‖L2 ≤ ε, then γ ∈ R+. The result for R− is proved similarly, and the

results for R+ ∩X and R− ∩X follow, since X → L2(RN ).

Remark 4.10. The above results call for several comments.

(i) Assume λ < 0. In this case, we know that if ϕ ∈ Y , with

Y = u ∈ L2(RN ); xu(x) ∈ L2(RN ),

then T ∗(ϕ) = T∗(ϕ) = +∞ (see Remark 3.7). This implies, in view of Corollary 4.8,

that Y ⊂ R+ ∩R−. The question as to whether or not L2(RN ) = R+ = R− is open

(see Remark 3.7).

(ii) Assume λ > 0. In this case, we know that some solutions blow up in finite time,

for both t > 0 and t < 0. However, it follows from Lemma 4.7 that, given any

ϕ ∈ L2(RN ), ϕb ∈ R+ for b large enough. This means that blow up is not related

to the size of the initial value only, since changing ϕ to ϕb is only a modification

of the phase. Note also that the multiplier eib|x|2

4 which leads to global existence is

independent of the initial value, except for the parameter b. This property will be

generalized to the case α 6= 4/N in the next section.

(iii) Still assume λ > 0. It follows from the above observation that R+ is an unbounded

subset of L2(RN ). As well, R− is an unbounded subset of L2(RN ). In addition,

it follows from Remark 3.8 and Corollary 4.8 that there exists µ > 0 such that if

‖ϕ‖L2 < µ, then ϕb ∈ R+ ∩ R− for all b ∈ R. In particular, the open set R+ ∩ R−

contains 0 in its interior. We do not know whether or not R+ ∩ R− is unbounded.

(iv) Still assume λ > 0. If N ≥ 2, then equation (3.1) has an unbounded sequence

of solutions (see Berestycki and Lions [4]). On the other hand, if ϕ is a solution

45

of (3.1), then u(t, x) = eitϕ(x) is a solution of (1.1), for which T ∗ = T∗ = ∞ and

‖u‖Lα+2(−∞,0),Lα+2) = ‖u‖Lα+2(0,∞),Lα+2) = ∞. Therefore, ϕ ∈ N+ ∩ N−; and so,

N+ ∩ N− is an unbounded subset of L2(RN ).

(v) Still assume λ > 0. It follows from Remark 3.9 that if ϕ ∈ Y verifies ‖ϕ‖L2 <

λ−1α ‖R‖L2 , then ϕ ∈ R+ ∩R−. We do not know whether or not the same conclusion

holds without the assumption ϕ ∈ Y .

(vi) Assume for simplicity λ = 1. Set

E =a

N2 eiγeiµ|x−y1|

2

R(a(x− y1)− y0); a > 0, γ, µ ∈ R, y0, y1 ∈ RN.

If ϕ ∈ E , then

u(t, x) =

(a

1 + 4µt

)N2

eiγeiµ|x−y1|

2

1+4µt ei a2t(1+4µt)R

(a

1 + 4µt(x− y1)− y0

),

for −∞ < t < − 1

4µif µ < 0, for − 1

4µ< t <∞ if µ > 0, and for −∞ < t <∞ if µ = 0

(see Remark 4.6). On the other hand, if ϕ ∈ X, ϕ 6∈ E , then ϕ ∈ R+ ∩ R−. Indeed,

if ϕ 6∈ E , then eib|x|2

4 ϕ 6∈ E for all b ∈ R. Therefore, it follows from Remark 4.6

that T ∗ = T∗ = ∞. As well, we have T ∗(ϕb) = T∗(ϕb) = ∞ for all b ∈ R. By

Corollary 4.8, this implies that ϕ ∈ R+ ∩ R−. If we assume only ϕ ∈ H1(RN ), then

it follows as well from Remark 4.6 that T ∗ = T∗ = ∞, but we do not know whether

or not T ∗(ϕb) = T∗(ϕb) = ∞ for all b ∈ R, since ϕb need not be in H1(RN ) for b 6= 0.

In particular, we do not know if ϕ ∈ R+ ∩ R−. If we assume only ϕ ∈ L2(RN ), then

we cannot apply Remark 4.6, and we do not even know whether or not T ∗ = T∗ = ∞.

Remark 4.11. For the nonlinear Klein-Gordon equation, and under appropriate assump-

tions on the nonlinearity, the set of initial values for which the solution is global, is a closed

subset of the energy space (see [6]). The same property holds for a class of nonlinear heat

equations (see [7]). For the nonlinear Schrodinger equation, the situation is quite different.

In fact, if λ > 0, then the set of initial values for which the solution is global (for t > 0, for

t < 0, or for both t > 0 and t < 0) is not closed, in any of the spaces L2(RN ), H1(RN )

or X. Indeed, assume for simplicity λ = 1, and let ϕ = R−1, so that ‖ϕ‖L2 = ‖R‖L2 and

T ∗ = 1. For 0 < ε < 1, set ϕε = (1 − ε)ϕ. We have ‖ϕε‖L2 = (1 − ε)‖R‖L2 < ‖R‖L2;

46

and so ϕε ∈ R+ ∩R− (see Remark 4.10 (v)). On the other hand, we have ϕε−→ε↓0

ϕ in X.

Therefore, a limit point (in X, hence also in H1(RN ) and in L2(RN )) of a sequence of

R+ ∩R− can belong to B+. By considering ϕ = R1, we see as well that a limit point of a

sequence of R+ ∩ R− can belong to B−.

Remark 4.12. In the case λ > 0, the pseudo-conformal transformation relates the be-

havior at blow up of blowing up solutions to the behavior at infinity of solutions with

initial values in N+ or N−. To see this, consider ϕ ∈ X such that T ∗(ϕ) <∞, and let u be

the corresponding solution of (1.1). Set ψ = ϕb with b = 1/T ∗, and let v be the solution

of (1.1) with the initial value ψ. It follows from Lemma 4.7 that ψ ∈ N+. On the other

hand, we have (see (4.4))

‖u(t)‖α+2Lα+2 =

(T ∗

T ∗ − t

)2 ∣∣∣∣∣∣∣∣v(

T ∗t

T ∗ − t

) ∣∣∣∣∣∣∣∣α+2

Lα+2

,

for all 0 ≤ t < T ∗. By conservation of energy, ‖u(t)‖α+2Lα+2 ≈ ‖∇u(t)‖2

L2 near T ∗. Therefore,

we see on the above identity that a lower estimate (respectively, an upper estimate) of

‖∇u(t)‖L2 at blow up corrseponds to a lower estimate (respectively, an upper estimate) of

‖v(t)‖Lα+2 at infinity. In particular, the lower estimate (2.20) implies that

lim inft→∞

t‖u(t)‖α+2Lα+2 > 0,

for all solutions u of (1.1) with ϕ ∈ X ∩ N+, ϕ 6= 0. On the other hand, if all solutions

of (1.1) with initial data in X∩N+ were bounded from above in Lα+2(RN ), then we would

obtain the upper estimate

‖∇u(t)‖L2 ≤ C

T ∗ − t,

at blow up, and if all solutions of (1.1) with initial data in X ∩ N+ were bounded from

below in Lα+2(RN ), then we would obtain the lower estimate

‖∇u(t)‖L2 ≥ c

T ∗ − t,

at blow up.

We now give some applications to the scattering theory. Some of these results will be

generalized to the case α 6= 4/N in Section 6. The reader is referred to the beginning of

47

Section 6 for the definitions of the scattering states u±, of the wave operators Ω± and of

the scattering operator S.

Before stating the results, we introduce some notation. We denote by F the Fourier

transform

Fu(ξ) =

∫

RN

e−2πix·ξu(x) dx,

and we denote by F−1 its inverse. We define the dilation Dβ , β > 0, by

Dβu(x) = βN2 u(βx),

and the multiplier Mb, b ∈ R by

Mbu(x) = eib|x|2

4 u(x).

With this notation, and using the explicit kernel

T (t)u =1

(4πit)N2

∫

RN

ei|x−y|2

4t u(y) dy,

elementary calculations show that

T (t)Dβ = DβT (β2t), (4.15)

and that

T (t)Mb = M b1+bt

D 11+bt

T

(t

1 + bt

), (4.16)

if 1 + bt > 0 and

T

(−1

b

)Mb = i−

N2 M−bD− b

4πF . (4.17)

We have the following result (see [10]).

Theorem 4.13. The following properties hold.

(i) Let ϕ ∈ L2(RN ) with T ∗ = ∞, and let u be the corresponding solution of (1.1). The

limit

u+ = limt→∞

T (−t)u(t) (4.18)

exists in L2(RN ) if and only if ϕ ∈ R+. The limit (4.18) exists in X if and only if

ϕ ∈ R+ ∩X. In addition,

u+ = iN2 F−1D4πM−1v(1), (4.19)

48

where v is the solution of (1.1) with the initial value ϕ−1. (Note that since ϕ ∈ R+,

we have T ∗(ϕ−1) > 1.) In addition, the mapping U+ : u 7→ u+ is a bicontinuous

bijection R+ → F−1(R−) for the L2 topology and R+ ∩X → F−1(R− ∩X) for the

X topology.

(ii) Let ϕ ∈ L2(RN ) with T∗ = ∞, and let u be the corresponding solution of (1.1). The

limit

u− = limt→−∞

T (−t)u(t) (4.20)

exists in L2(RN ) if and only if ϕ ∈ R−. The limit (4.20) exists in X if and only if

ϕ ∈ R− ∩X. In addition,

u− = i−N2 FD4πM1w(−1), (4.21)

where w is the solution of (1.1) with the initial value ϕ1. (Note that since ϕ ∈ R−, we

have T∗(ϕ1) > 1.) In addition, the mapping U− : u 7→ u− is a bicontinuous bijection

R− → F(R+) for the L2 topology and R− ∩X → F(R+ ∩X) for the X topology.

(iii) The wave operators Ω± are bicontinuous bijections R± → F∓1(R∓) for the L2 topol-

ogy and R± ∩X → F∓1(R∓ ∩X) for the X topology.

(iv) The scattering operator S = Ω−1+ Ω− = U+Ω− is a bicontinuous bijection U−(R+ ∩

R−) → U+(R+∩R−) for the L2 topology and U−(R+∩R−∩X) → U+(R+∩R−∩X)

for the X topology.

Proof. Let v be the solution of (1.1) with the initial value ϕ−1. Since T ∗ = ∞, we have

T ∗(ϕ−1) ≥ 1. On [0, 1), we have v = u−1, i.e.

v(t) = M− 11−t

D 11−t

u

(t

1− t

).

Therefore, it follows from (4.15) and (4.16) that

T (−t)v(t) = M−1T

(− t

1− t

)u

(t

1− t

), (4.22)

for all t ∈ [0, 1). Assume that the limit (4.18) exists in L2(RN ). It follows from (4.22) that

v(t) has a limit in L2(RN ) as t ↑ 1. Thus T ∗(ϕ−1) > 1; and so ϕ ∈ R+. If the limit (4.18)

exists in X, then we obtain as well ϕ ∈ R+ ∩X.

49

Conversely, if ϕ ∈ R+, then v has a limit in L2(RN ) as t ↑ 1, and it follows from (4.22)

that the limit (4.18) exists in L2(RN ). If ϕ ∈ R+ ∩ X, then v has a limit in X as t ↑ 1,

and it follows from (4.22) that the limit (4.18) exists in X. This proves the first part of (i).

Suppose now that the limit (4.18) exists in L2(RN ), so that v(1) = limt↑1

v(t) exists in

L2(RN ). It follows from (4.22) that

T (−1)v(1) = M−1u+.

Therefore, by (4.17),

v(1) = T (1)M−1u+ = i−N2 M−1D 1

4πFu+,

from which (4.19) follows.

We now show that U+(R+) = F−1(R−). Given ϕ ∈ R+, we have by (4.19)

U+ϕ = iN2 F−1D4πM−1v(1),

and we show that D4πM−1v(1) ∈ R−. Indeed, since v is defined on [0, 1], we have

T∗(v(1)) > 1; and so, M−1v(1) ∈ R−. Furthermore, since the equation (1.1) is invari-

ant by the scaling

u(t, x) → βN2 u(β2t, βx),

for any β > 0, it is clear that Dβ(R−) = R−, so that D4πM−1v(1) ∈ R−; and so,

U+(R+) ⊂ F−1(R−).

Conversely, let η ∈ F−1(R−). We have Fη ∈ R−, and it follows from the observation

just above that D 14πFη ∈ R−. Therefore, T∗(M1D 1

4πFη) > 1. Let z(t) be the solution

of (1.1) with the initial valueM1D 14πFη. Since z is defined on [−1, 0], we have T ∗(z(−1)) >

1; and so, M1z(−1) ∈ R+, thus i−N2 M1z(−1) ∈ R+. We set ϕ = i−

N2 M1z(−1). We have

ϕ−1 = i−N2 z(−1), so that the solution v of (1.1) with the initial value ϕ−1 is i−

N2 z(t− 1).

Therefore, by (4.19),

U+ϕ = iN2 F−1D4πM−1v(1) = i

N2 F−1D4πM−1i

−N2 z(0)

= iN2 F−1D4πM−1i

−N2 M1D 1

4πFη = η.

Therefore, η ∈ U+(R+); and so U+(R+) = F−1(R−).

50

The continuity properties of U+ and its inverse follow from the continuous dependence

of the solutions of (1.1) (applied to the solution v). This completes the proof of (i).

(ii) is proved by the same argument, or simply by observing that if u(t) satisfies (1.1)

on [0,∞), then u(−t) satisfies (1.1) on (−∞, 0] with the initial value ϕ. Properties (iii)

and (iv) are immediate consequences of (i) and (ii).

Remark 4.14. Here are some comments about Theorem 4.13.

(i) Assume λ < 0. As observed before (see Remark 4.10), we do not know whether or not

R+ = R− = L2(RN ). However, we know that R+ ∩X = R− ∩X = X. This implies

that the scattering operator S, as well as the operators U± and Ω± are bicontinuous

bijections X → X. In addition, it follows from Corollary 4.9 (see also Remark 3.8)

that these operators are defined on L2 neighborhoods of 0. In addition, with the

notation of Remark 4.10, we have Y ⊂ R+ ∩ R−. Since F−1Y = H1(RN ), this

implies that H1(RN ) ⊂ U+(R+) ∩ U−(R−). We do not know whether or not S is

defined H1(RN ) → H1(RN ).

(ii) Assume λ > 0. It follows from Remark 4.10 that, if B denotes the ball or L2(RN ) of

radius λ−1α ‖R‖L2 , X∩B ⊂ R+∩R−. Since the operators S, U± and Ω± are isometric

in L2(RN ), it follows easily that they all are bicontinuous bijections X ∩B → X ∩Bfor the X norm. In addition, it follows from Corollary 4.9 (see also Remark 3.8) that

these operators are defined on open subsets of L2(RN ) containing 0. However, as

observed before (see Remark 4.10), we do not know whether or not they are defined

on B.

51

5. A global existence and decay result in the non-invariant case. In this sec-

tion, we extend some of the porperties of Lemma 4.7 to the case α 6= 4/N (see also

Remark 4.10 (ii)). We define

α0 =2−N +

√N2 + 12N + 4

2N, (5.1)

and we observe that2

N≤ 4

N + 2< α0 <

4

N<

4

N − 2,

if N ≥ 2 and2

N< α0 <

4

N,

if N = 1. The main result of this section is the following (see [11]).

Theorem 5.1. Assume α0 < α <4

N − 2, where α0 is defined by (5.1), and let ϕ ∈ X.

Given b ∈ R, set ϕb(x) = eib|x|2

4 and let ub be the solution of (1.1) with the initial value

ϕb. There exists b0 < ∞ such that if b ≥ b0, then T ∗(ϕb) = ∞. Moreover, we have

ub ∈ Lq((0,∞),W 1,r(RN ) for any admissible pair (q, r). Finally, if

a =2α(α+ 2)

4− α(N − 2), (5.2)

then ub ∈ La((0,∞), Lα+2(RN )).

Remark 5.2. Here are some comments about Theorem 5.1.

(i) Let β =4(α+ 2)

Nα, so that (β, α+ 2) is an admissible pair. One verifies easily that if

α > 4/N , then β < a, where a is given by (5.2), and that if α < 4/N , then β > a. Next,

if u ∈ Lq((0,∞),W 1,r(RN )) for every admissible pair (q, r), we have in particular u ∈Lβ((0,∞), Lα+2(RN )), and also u ∈ L∞((0,∞), Lα+2(RN )) by Sobolev’s embedding

theorem. Therefore, we have also u ∈ La((0,∞), Lα+2(RN )) if α ≥ 4/N . On the

other hand, if α < 4/N , then the property u ∈ La((0,∞), Lα+2(RN )) expresses a

better decay at infinity.

(ii) If λ < 0, then all solutions of (1.1) are global. Therefore, Theorem 5.1 means that all

the solutions ub have a certain decay as t→∞ for b large enough. In fact, we will see

in the next section that the conclusions of Theorem 5.1 hold with b0 = 0.

52

(iii) If λ > 0 and α < 4/N , then all solutions of (1.1) are global. Therefore, Theorem 5.1

means that ub has a certain decay as t → ∞ if b is large enough. Note that certain

solutions do not decay, in particular the standing waves, i.e. the solutions of the form

eiωtϕ(x) (see [5], Chapter 8).

(iv) If λ > 0 and α ≥ 4/N , then (1.1) posesses solutions that blow up in finite time.

Theorem 5.1 means that for any ϕ ∈ X, the initial value ϕb gives rise to a solution

which is global and which decays as t→∞ provided b is large enough.

(v) Suppose λ > 0 and α ≥ 4/N . If ϕ ∈ X is such that E(ϕ) < 0, then the maximal

solution u of (1.1) blows up in finite time, for both t > 0 and t < 0. Theorem 5.1

implies that if b is large enough, then the maximal solution ub of (1.1) with initial

value ϕb is positively global and decays as t→∞. Of course, E(ϕb) ≥ 0 for such b’s,

and one may wonder if ub still blows up at a finite negative time. The answer is yes,

as the following argument shows. Changing ϕb to ϕb (which changes ub(t) to ub(−t)),it suffices to show that if E(ϕ) < 0, then for all b > 0 the solution u of (1.1) with

initial value ϕ(x)e−ib|x|2

4 blows up at a positive finite time. Let T ∗ be the maximal

existence time of u, and let f(t) = ‖ | · |u(t, ·)‖2L2. We have (see Section 2)

f ′(t) = 4Im

∫

RN

xu∇u dx,

and

f ′′(t) = 16E(u(0))− λ4(Nα− 4)

α+ 2

∫

RN

|u|α+2.

Therefore,

f(t) = f(0) + tf ′(0) + 8E(u(0))t2 − λ4(Nα− 4)

α+ 2

∫ t

0

∫ s

0

∫

RN

|u|α+2 dx dσ ds,

for all 0 ≤ t < T ∗. It follows that

f(t) ≤ f(0) + tf ′(0) + 8E(u(0))t2, (5.3)

for all 0 ≤ t < T ∗. Setting P (t) = f(0) + tf ′(0) + 8E(u(0))t2 for all t ≥ 0, a

straightforward calculation shows that

P (t) = ‖xϕ‖2L2 + 4t

(F (ϕ)− b

2‖xϕ‖2

L2

)+ 8t2

(E(ϕ) +

b2

8‖xϕ‖2

L2 − b

2F (ϕ)

),

53

with

F (ϕ) = Im

∫

RN

xϕ∇ϕdx.

In particular,

P

(1

b

)=

8

b2E(ϕ) < 0;

and it follows easily from (5.3) that T ∗ < 1/b. Hence the result.

(vi) We do not know whether or not Theorem 5.1 has any extension to the case α ≤α0. Note, however, that in the case N 6= 2, α = α0 and λ < 0, it follows from

Corollary 6.9 and Theorem 6.15 below that for any nontrivial solution u of (1.1), we

have (1 + t)‖u(t)‖aLα+2 → ` > 0 as t → ±∞. In particular, no nontrivial of (1.1)

belongs to La((0,∞), Lα+2(RN )).

Theorem 5.1 is based on the following Strichartz inequality, which involves non-

admissible pairs.

Lemma 5.3. Let (q, r) be an admissible pair with r > 2. Fix a >q

2and define a by

1

a+

1

a=

2

q. (5.4)

If T > 0 and f ∈ La′((0, T ), Lr′(RN )), then u defined by

u(t) =

∫ t

0

T (t− s)f(s) ds,

belongs to La((0, T ), Lr(RN )). Furthermore, there exists a constant C, depending only on

N, r and a such that

‖u‖La((0,T ),Lr) ≤ C‖f‖La′((0,T ),Lr′), (5.5)

for every f ∈ La′((0, T ), Lr′(RN )).

Proof. By density, we need only prove estimate (5.5) for f ∈ C([0, T ],S(RN)). It is well

known (see [5], Proposition 3.2.1) that

‖T (t)‖L(Lr′ ,Lr) ≤ |4πt|−N( 12−

1r ),

whenever 2 ≤ r ≤ ∞. Therefore,

‖u(t)‖Lr ≤∫ t

0

(4π(t− s))−2q ‖f(s)‖Lr′ ,

54

and so (5.5) is an immediate consequence of the Riesz potential inequalities (Stein [36],

Theorem 1, p. 119).

Corollary 5.4. Let r = α+ 2, let (q, r) be the corresponding admissible pair, and let a

be given by (5.2). Then a > q/2 if and only if α > α0, with α0 defined by (5.1). For such

values of α and a, and for 0 < T ≤ ∞, we have the following estimates for G defined by

G(f)(t) =

∫ t

0

T (t− s)f(s) ds,

for 0 ≤ t < T .

(i) If u ∈ La((0, T ), Lr(RN )), then G(|u|αu) ∈ La((0, T ), Lr(RN )). Furthermore, there

exists C depending only on N and α such that

‖G(|u|αu)‖La((0,T ),Lr) ≤ C‖u‖α+1La((0,T ),Lr), (5.6)

for every u ∈ La((0, T ), Lr(RN )).

(ii) If u ∈ La((0, T ), Lr(RN )) ∩ Lq((0, T ),W 1,r(RN )) and if (γ, ρ) is any admissible pair,

then G(|u|αu) ∈ Lγ((0, T ),W 1,ρ(RN )). Furthermore, there exists C depending only

on N,α and ρ such that

‖G(|u|αu)‖Lγ((0,T ),W 1,ρ) ≤ C‖u‖αLa((0,T ),Lr)‖u‖Lq((0,T ),W 1,r), (5.7)

for every u ∈ La((0, T ), Lr(RN )) ∩ Lq((0, T ),W 1,r(RN )).

Proof. The first part of the lemma is a simple calculation, which we omit. For asser-

tions (i) and (ii) consider a defined by (5.4). Since (α+ 1)r′ = r, (α+ 1)a′ = a and

1

q′=

1

q+α

a,

we see that

‖ |u|αu‖La′((0,T ),Lr′) = ‖u‖α+1La((0,T ),Lr)

and (applying Holder’s inequality twice) that

‖ |u|αu‖Lq′((0,T ),W 1,r′) ≤ C‖u‖αLa((0,T ),Lr)‖u‖Lq((0,T ),W 1,r).

55

The results now follow from (5.5) and (1.2) respectively.

By applying Corollary 5.4, we obtain the following estimate for the solutions of (1.1).

Proposition 5.5. Assume that α0 < α <4

N − 2, where α0 is defined by (5.1). If

u ∈ C([0,∞), H1(RN )) is a solution of (1.1), and if u ∈ La((0,∞), Lα+2(RN )) with a

defined by (5.2), then u ∈ Lγ((0,∞),W 1,ρ(RN )) for every admissible pair (γ, ρ).

Proof. We already know (see Remark 2.2) that u ∈ Lγ((0, T ),W 1,ρ(RN )), for every

T <∞ and every admissible pair (γ, ρ). For a fixed T ≥ 0, we set v(t) = u(t+ T ); and so

v(t) = S(t)u(T ) + iλ

∫ t

0

T (t− s)|v|αv(s) ds.

As before, set r = α + 2, and let q be such that (q, r) is an admissible pair. Let (γ, ρ) be

any admissible pair, and let a be defined by (5.2). It follows from (1.2) and (5.7) that

‖v‖Lγ((0,τ),W 1,ρ) ≤ C1‖u(T )‖H1 + C2‖u‖αLa((T,T+τ),Lr)‖v‖Lq((0,τ),W 1,r), (5.8)

for every τ > 0. Choosing T large enough so that C2‖u‖αLa((T,∞),Lr) ≤ 1/2 and letting

(γ, ρ) = (q, r), we see that

‖v‖Lq((0,τ),W 1,r) ≤ 2C1‖u(T )‖H1 ,

for every τ > 0. Hence u ∈ Lq((0,∞),W 1,r(RN )). That u ∈ Lγ((0,∞),W 1,ρ(RN )) for

any admissible pair (γ, ρ) now follows from (5.8).

We are now in a position to prove the following global existence result.

Theorem 5.6. Assume α0 < α <4

N − 2, where α0 is defined by (5.1), and let a be given

by (5.2). There exists ε > 0 such that if ϕ ∈ H1(RN ) and ‖T (·)ϕ‖La((0,∞),Lα+2) ≤ ε, then

the maximal solution u of (1.1) is positively global. Moreover, u ∈ La((0,∞), Lα+2(RN )),

and u ∈ Lγ((0,∞),W 1,ρ(RN )) for every admissible pair (γ, ρ).

Proof. Let ε > 0 and set r = α+ 2; let ϕ ∈ H1(RN ) be such that

‖T (·)ϕ‖La(0,∞;Lr) ≤ ε;

56

and let u be the maximal solution of (1.1) defined on [0, T ∗), with 0 < T ∗ ≤ ∞. Consider

q such that (q, r) is an admissible pair. It follows from (5.6), (1.2) and (5.7) that there

exists K independent of ϕ such that

‖u‖La((0,T ),Lr) ≤ ε+K‖u‖α+1La((0,T ),Lr), (5.9)

and

‖u‖Lq((0,T ),W 1,r) ≤ K‖ϕ‖H1 +K‖u‖αLa((0,T ),Lr)‖u‖Lq((0,T ),W 1,r), (5.10)

for every T < T ∗. Assume that ε satisfies

2α+1Kεα < 1. (5.11)

Since ‖u‖La(0,T ;Lr) depends continuously on T , it follows from (5.9) and (5.11) that

‖u‖La(0,T ∗;Lr) ≤ 2ε. (5.12)

Applying (5.10) and (5.12), we obtain

‖u‖Lq(0,T ∗;W 1,r) ≤ 2K‖ϕ‖H1 . (5.13)

Applying now (1.2) and (5.7) with (γ, ρ) = (∞, 2), and using (5.12) and (5.13), we see that

‖u‖L∞(0,T ∗;H1) <∞.

Therefore T ∗ = ∞, and the result follows from (5.12) and Proposition 5.5.

Proof of Theorem 5.1. Since ϕ ∈ X, we have ϕb ∈ X, for every b ∈ R. Let (q, r) be

the admissible pair such that r = α+2 and let a be defined by (5.3). Using formulas (4.15)

and (4.16), one verifies easily that

‖T (·)ϕb‖aLa((0,∞),Lr) =

∫ 1/b

0

(1− bτ)2(a−q)

q ‖T (τ)ϕ‖aLr dτ.

Since ‖T (τ)ϕ‖Lr ≤ C‖T (τ)ϕ‖H1 ≤ C‖ϕ‖H1 and2(a− q)

q> −1, it follows that

limb↑∞

‖T (·)ϕb‖La(0,∞;Lr) = 0.

57

The result now follows from Theorem 5.6.

Remark 5.7. Assume that α0 < α <4

N − 2, where α0 is defined by (5.1), and let

ϕ ∈ X. There exists s0 < ∞ such that for every s ≥ s0, the maximal solution of (1.1)

with initial value ψs = T (s)ϕ is positively global and verifies u ∈ La((0,∞), Lα+2(RN )),

and u ∈ Lγ((0,∞),W 1,ρ(RN )) for every admissible pair (γ, ρ). Indeed, since ϕ ∈ X,

‖T (·)ϕ‖La(0,∞;Lr) is finite (See [5], Corollary 3.3.4); and so, the result follows from Theo-

rem 5.6.

58

6. Some results on the scattering theory in the non-invariant case. We first

recall the basic notions of the scattering theory. For convenience, we consider the scattering

inX, though in principle the same notions can be introduced inH1(RN ) or even in L2(RN )

if α ≤ 4/N .

Let ϕ ∈ X be such that the solution of (1.1) is defined for all t ≥ 0, i.e. T ∗ = ∞. If

the limit

u+ = limt→∞

T (−t)u(t), (6.1)

exists in X, we say that u+ is the scattering state of ϕ (at +∞). As well, if ϕ ∈ X is such

that the solution of (1.1) is defined for all t ≤ 0, i.e. T∗ = ∞, and if the limit

u− = limt→−∞

T (−t)u(t), (6.2)

exists in X, we say that u− is the scattering state of ϕ at −∞. We set

R+ = ϕ ∈ X; T ∗ = ∞ and the limit (6.1) exists,

and

R− = ϕ ∈ X; T∗ = ∞ and the limit (6.2) exists.

Since changing t to −t in equation (1.1) corresponds to changing u to u, we see that

R− = R+ = ϕ ∈ X; ϕ ∈ R+.

We define the operators

U± :R± → X

ϕ 7→ u±

and we set

U± = U±(R±).

We also have

U− = U+ = v ∈ X; v ∈ U+.

If the mappings U± are injective, we set

Ω± = U−1± : U± →R±.

59

The mappings Ω± are called the wave operators. Finally, we set

O± = U±(R+ ∩R−), (6.3)

so that

O− = O+ = v ∈ X; v ∈ O+.

Finally, the scattering operator S is the mapping

S = U+Ω− : O− → O+. (6.4)

In other words, u+ = Su− if and only if there exists ϕ ∈ X such that T ∗ = T∗ = ∞ and

such that T (−t)u(t) → u± as t→ ±∞.

These notions are relevant for conservative equations with dispersion, in order to

describe the asymptotic behavior of the solutions. For example, for the linear Schrodinger

equation, i.e. if λ = 0, then u(t) does not have any strong limit in X if ϕ 6= 0. Indeed,

u(t) 0 in L2(RN ) as t→ ±∞ and ‖u(t)‖L2 ≡ ‖ϕ‖L2 . On the other hand, the operators

introduced above are well-defined. In fact, in the linear case, Ω± and S all coincide with

the identity in X. Note, however, that in general these operators are nonlinear.

A scattering theory can be developped in H1(RN ) for α > 4/N and N ≥ 3. If λ < 0,

we have a global theory (i.e. S is a bijection H1(RN ) → H1(RN )), and if λ > 0, we have

low energy scattering (i.e. S and S−1 are defined on a neighborhood of 0 in H1(RN )).

These results are due to Ginibre and Velo [18] (see also [5], Chapter 7).

It seems that a scattering theory in L2(RN ) exists only for α = 4/N , and that the

only available results are those of the preceding section (i.e. low energy scattering).

The scattering theory in X was first developped by Ginibre and Velo [20] for α ≥ 4/N

by using the pseudo-conformal conservation law, and then improved by Tsutsumi [40] for

α > α0.

We now apply the pseudo-conformal transformation (4.1) with b < 0, and we suppose

for convenience that b = −1. Moreover, throughout this section we systematically consider

the variables (s, y) ∈ R×RN defined by

s =t

1− t, y =

x

1− t,

60

or equivalently,

t =s

1 + s, x =

y

1 + s.

Given 0 ≤ a < b ≤ ∞ and u defined on (a, b)×RN , we set

v(t, x) = (1− t)−N2 u

(t

1− t,

x

1− t

)e−i

|x|2

4(1−t) , (6.5)

for x ∈ RN anda

1 + a< t <

b

1 + b. In particular, if u is defined on (0,∞), then v is

defined on (0, 1). Transformation (6.5) reads as well, using the variables (s, y),

v(t, x) = (1 + s)N2 u(s, y)e−i

|y|2

4(1+s) . (6.6)

One verifies easily that, given 0 ≤ a < b <∞, u ∈ C([a, b], X)∩C1([a, b], H−1(RN )) if and

only if v ∈ C([ a1+a

, b1+b

], X) ∩ C1([ a1+a

, b1+b

], H−1(RN )). Furthermore, a straightforward

calculation (see Theorem 4.1) shows that u satisfies (1.1) on (a, b) if, and only if v satisfies

the equation

ivt +4v + λ(1− t)Nα−4

2 |v|αv = 0, (6.7)

on the interval(

a1+a ,

b1+b

). Note that the term (1 − t)

Nα−42 is regular, except possibly

at t = 1, where it is singular for α < 4/N . Furthermore, we have the following identities

(see (4.6)–(4.8)):

‖v(t)‖β+2Lβ+2 = (1 + s)

Nβ2 ‖u(s)‖β+2

Lβ+2 , β ≥ 0, (6.8)

‖∇v(t)‖2L2 =

1

4‖(y + 2i(1 + s)∇)u(s)‖2

L2, (6.9)

‖∇u(s)‖2L2 =

1

4‖(x+ 2i(1− t)∇)v(t)‖2

L2. (6.10)

It follows from (6.8) and conservation of charge for (1.1) that

d

dt‖v(t)‖L2 = 0. (6.11)

Moreover, if we set

E1(t) =1

2‖∇v(t)‖2

L2 − (1− t)Nα−4

2λ

α+ 2‖v(t)‖α+2

Lα+2 ,

E2(t) = (1− t)4−Nα

2 E1(t) = (1− t)4−Nα

21

2‖∇v(t)‖2

L2 − λ

α+ 2‖v(t)‖α+2

Lα+2 ,

61

and

E3(t) =1

8‖(x+ 2i(1− t)∇)v(t)‖2

L2 − (1− t)Nα2

λ

α+ 2‖v(t)‖α+2

Lα+2 .

Then,d

dtE1(t) = −(1− t)

Nα−62

4−Nα

2

λ

α+ 2‖v(t)‖α+2

Lα+2 , (6.12)

d

dtE2(t) = (1− t)

2−Nα2

Nα− 4

4‖∇v(t)‖2

L2 , (6.13)

andd

dtE3(t) = 0. (6.14)

Indeed, (6.12) and (6.13) are equivalent, and both are equivalent to the pseudo-conformal

conservation law for u, by (6.8) and (6.9). As well, (6.14) is equivalent to the conservation

of energy for u, by (6.8) and (6.10).

The results that we present in this section are based on the following observation.

Proposition 6.1. Assume that 0 < α <4

N − 2, let u ∈ C([0,∞), X) be a solution

of equation (1.1), and let v ∈ C([0, 1), X) be the corresponding solution of (6.7) defined

by (6.5). Then, T (−s)u(s) has a strong limit in X (respectively, in L2(RN )) as s→∞ if

and only if v(t) has a strong limit in X (respectively, in L2(RN )) as t ↑ 1, in which case

lims→∞

T (−s)u(s) = ei|x|2

4 T (−1)v(1), (6.15)

in X (respectively, in L2(RN )).

Proof. It follows from (4.22) that

T (−s)u(s) = ei|x|2

4 T

(− s

1 + s

)v

(s

1 + s

),

from which the result follows.

The following result implies that if α ≤ 2/N , then no scattering theory can be devel-

opped for the equation (1.1) (see Barab [3], Strauss [37], [38], Tsutsumi and Yajima [41]).

Theorem 6.2. Assume α ≤ 2

N. If N = 1, assume further α ≤ 1. Let ϕ ∈ L2(RN ) and

let u ∈ C(R, L2(RN )) be the corresponding solution of (1.1). If ϕ 6= 0, then T (−t)u(t)

62

does not have any strong limit in L2(RN ), neither as t → ∞ nor as t → −∞. In other

words, no nontrivial solution of (1.1) has scattering states, even in L2(RN ).

Proof. We consider the case t → ∞, the argument for t → −∞ being the same. We

argue by contradiction and we assume T (−t)u(t) −→t→∞

u+ in L2(RN ). In particular, we

have

‖u+‖L2 = ‖u(t)‖L2 = ‖ϕ‖L2 > 0.

On the other hand, it follows from (6.15) that v(t)−→t↑1

w in L2(RN ), with

w = T (1)

(e−i

|x|2

4 u+

)6= 0.

Since α+ 1 ≤ 2, we have |v(t)|αv(t)−→t↑1

|w|αw 6= 0 in L2

α+1 (RN ). Let now θ ∈ D(RN) be

such that

〈i|w|αw, θ〉 = 1. (6.16)

It follows from (6.7) that

d

dt〈v(t), θ〉 = 〈i4v, θ〉+ λ(1− t)

Nα−42 〈i|v|αv, θ〉

= 〈iv,4θ〉+ λ(1− t)Nα−4

2 〈i|v|αv, θ〉.

Therefore, by (6.16), and since v is bounded in L2(RN ),

∣∣∣∣d

dt〈v(t), θ〉

∣∣∣∣ ≥1

2|λ|(1− t)

Nα−42 ,

for 1−ε ≤ t < 1, ε > 0 small enough. SinceNα− 4

2< −1, it follows that |〈v(t), θ〉| −→

t↑1∞,

which is absurd.

Remark 6.3. In the case N = 1 and 1 < α ≤ 2, we have the following result. Let ϕ ∈ Xand let u ∈ C(R, X) be the corresponding solution of (1.1). If ϕ 6= 0, then T (−t)u(t)does not have any strong limit in X, neither as t → ∞ nor as t → −∞. The proof is

similar. One needs only observe that, since v(t) is bounded in X, hence in H1(R), we have

v(t)−→t↑1

w in Lp(R) for every 2 ≤ p ≤ ∞; and so, |v(t)|αv(t)−→t↑1

|w|αw in L2(R).

On the other hand, if α > 2/N and if λ ≤ 0, then every solution of (1.1) in X has

scattering states in L2(RN ), as shows the following result (see Tsutsumi and Yajima [41]).

63

Theorem 6.4. Assume2

N< α <

4

N − 2(2 < α < ∞ if N = 1), and λ < 0. Let ϕ ∈ X

and let u ∈ C(R, X) be the corresponding solution of (1.1). There exist u± ∈ L2(RN )

such that

T (−t)u(t) −→t→±∞

u±,

in L2(RN ).

Remark 6.5. Here are some comments on Theorem 6.4.

(i) If α ≥ α0, then we will show below that u± ∈ X and that T (−t)u(t) −→t→±∞

u± in

X (see Theorems 6.14 and 6.15). The same conclusion holds if α > 4N+2 (α > 2 if

N = 1), and if ‖ϕ‖X is small enough (Theorem 6.7). If α ≤ 4N+2

, or if α < α0 and

‖ϕ‖X is large, then we do not know whether or not u± ∈ X.

(ii) Theorem 6.4 does not apply to the case λ > 0. In fact, if α <4

N + 2there are

arbitrarily small initial values ϕ ∈ X which do not have a scattering state, even in the

sense of L2(RN ). To see this, let ϕ ∈ X be a nontrivial solution of the equation

−4ϕ+ ϕ = λ|ϕ|αϕ.

(See for example [5], Chapter 8.) Given a > 0, set ϕa(x) = a2αϕ(ax). It follows

that −4ϕa + a2ϕa = λ|ϕa|αϕa. Therefore, ua(t, x) = eia2tϕa(x) satisfies (1.1), and

T (−t)ua(t) = eia2tT (−t)ϕa does not have any strong limit as t→∞ in L2(RN ). On

the other hand, one verifies easily that if α <4

N + 2, then ‖ua‖X −→

a↓00. However,

we will see below (Theorem 6.7) that if α > 4N+2

, then small initial values in X have

scattering states in X at ±∞.

Proof of Theorem 6.4. By Proposition 6.1, we need only show that v(t) has a strong

limit in L2(RN ) as t ↑ 1. As observed above (Remark 6.5), we have a better result when

α ≥ α0. Therefore, we may assume that α < α0, and in particular α < 4/N . Therefore, it

follows from (6.11) and (6.13) that

‖v(t)‖L2 ≤ C, (6.17)

‖v(t)‖Lα+2 ≤ C, (6.18)

‖∇v(t)‖L2 ≤ C(1− t)Nα−4

4 , (6.19)

64

for all t ∈ [0, 1). By using the embeddings Lα+2α+1 (RN ) → H−1(RN ) → H−2(RN ), and the

equation (6.7), we obtain

‖vt‖H−2 ≤ ‖4u‖H−2 + C(1− t)Nα−4

2 ‖ |v|αv‖H−2 ≤ C‖v‖L2 + C(1− t)Nα−4

2 ‖v‖α+1Lα+2 .

Therefore,

‖vt‖H−2 ≤ C + C(1− t)Nα−4

2 ,

by (6.17) and (6.18). It follows that vt ∈ L1((0, 1), H−2(RN )). Therefore, there exists

w ∈ H−2(RN ) such that v(t)−→t↑1

w in H−2(RN ). By using again (6.17), we obtain w ∈L2(RN ) and

v(t) w, (6.20)

in L2(RN ) as t ↑ 1. Consider now ψ ∈ H1(RN ), and let 0 ≤ t ≤ τ < 1. We have by (6.7)

(v(τ)− v(t), ψ)L2 =

∫ τ

t

〈vt, ψ〉H−1,H1 ds

=

∫ τ

t

(i∇v,∇ψ)L2 ds+

∫ τ

t

(1− s)Nα−4

2 〈iλ|v|αv, ψ〉L

α+2α+1 ,Lα+2

ds;

and so,

|(v(τ)− v(t), ψ)L2| ≤ C‖∇ψ‖L2

∫ τ

t

‖∇v‖L2 ds+ C‖ψ‖Lα+2

∫ τ

t

(1− s)Nα−4

2 ‖v‖α+1Lα+2 ds

≤ C‖∇ψ‖L2

∫ τ

t

(1− s)Nα−4

4 ds+ C‖ψ‖Lα+2

∫ τ

t

(1− s)Nα−4

2 ds,

by (6.19) and (6.18). Letting τ ↑ 1 and applying (6.20), we obtain

|(w − v(t), ψ)L2| ≤ C‖∇ψ‖L2

∫ 1

t

(1− s)Nα−4

4 ds+ C‖ψ‖Lα+2

∫ 1

t

(1− s)Nα−4

2 ds

≤ C(1− t)Nα4 ‖∇ψ‖L2 + C(1− t)

Nα−22 ‖ψ‖Lα+2 .

We now let ψ = v(t) and we apply again (6.19) and (6.18). It follows that

|(w − v(t), v(t))L2| ≤ C(1− t)Nα4 (1− t)

Nα−44 + C(1− t)

Nα−22

≤ C(1− t)Nα−2

2 −→t↑1

0.(6.21)

Finally,

‖v(t)− w‖2L2 = −(w − v(t), v(t))L2 + (w − v(t), w)L2 −→

t↑10,

65

by (6.21) and (6.20). This completes the proof.

Besides the fact that Theorem 6.4 does not apply to the case λ > 0, it does not

either allow us to construct the wave and scattering operators, since the initial value ϕ

and the scattering states u± do not belong to the same space. We will improve this

result under more restrictive assumptions on α by solving the initial value problem for the

nonautonomous equation (6.7), and by applying Proposition 6.1 which relates the behavior

of u at infinity and the behavior of v at t = 1.

However, we want to solve the Cauchy problem for (6.7) starting from any time

t ∈ [0, 1], including t = 1 where the nonautonomous term might be singular. In order to

do this, we define the function

f(t) =

λ(1− t)

Nα−42 if −∞ < t < 1,

λ if t ≥ 1,(6.22)

and we consider the equation

ivt +4u+ f(t)|v|αv = 0. (6.23)

Under appropriate assumptions on α, the initial value problem for (6.23) can be solved

starting from any time t ∈ R, and we have the following result.

Theorem 6.6. Assume4

N + 2< α <

4

N − 2(2 < α < ∞, if N = 1). Then, for

every t0 ∈ R and ψ ∈ X, there exist Tm(t0;ψ) < t0 < TM (t0;ψ) and a unique, maximal

solution v ∈ C((Tm, TM ), X) ∩ C1((Tm, TM ), H−1(RN )) of equation (6.23). The solution

v is maximal in the sense that if TM < ∞ (respectively Tm > −∞), then ‖u(t)‖H1 → ∞,

as t ↑ TM (respectively t ↓ Tm). In addition, the solution v has the following properties.

(i) If TM = 1, then lim inft↑1

(1 − t)δ‖v(t)‖H1 > 0 with δ =N + 2

4− 1

αif N ≥ 3, δ any

number larger than 1− 1

αif N = 2, and δ =

1

2− 1

αif N = 1.

(ii) The solution v depends continuously on ψ in the following way. The mapping ψ 7→ TM

is lower semicontinuous X → (0,∞], and the mapping ψ 7→ Tm is upper semicontin-

uous X → [−∞, 0). In addition, if ψn −→n→∞

ψ in X and if [S, T ] ∈ (Tm, TM ), then

vn → v in C([S, T ], X), where vn denotes the solution of (6.23) with initial value ψn.

66

Proof. The result follows immediately by applying Theorems 7.1 and 7.2 of Section 7

with h(t) = f(t− t0).

We now give some applications to the scattering theory for (1.1).

Theorem 6.7. Assume4

N + 2< α <

4

N − 2(2 < α < ∞, if N = 1). Then, the

following properties hold.

(i) The setsR± and U± are open subsets ofX containing 0. The operators U± : R± → U±are bicontinuous bijections (for the X topology) and the operators Ω± : U± → R±

are bicontinuous bijections (for the X topology).

(ii) The sets O± are open subsets of X containing 0, and the scattering operator S is a

bicontinuous bijection O− → O+ (for the X topology).

Remark 6.8. Theorem 6.7 implies that there is a low energy scattering theory in X for

the equation (1.1), provided4

N + 2< α <

4

N − 2(2 < α < ∞, if N = 1). As observed

before (see Remark 6.5), if α < 4N+2

and λ > 0, then there is no low energy scattering.

Proof of Theorem 6.7. Let ϕ ∈ X and let u be the corresponding solution of (1.1).

Let v be the solution of (6.7) with the initial value ψ defined by ψ(x) = ϕ(x)e−i|x|2

4

(see Theorem 6.6). It follows that v is defined by (6.6), as long as (6.6) makes sense.

Therefore, it follows from Proposition 6.1 and Theorem 6.6 that ϕ ∈ R+ if and only if

TM (0;ψ) > 1, and that in this case u+ = ei|x|2

4 T (−1)v(1). Therefore, the open character

of R+ and the continuity of the operator U+ follow from the continuous dependence of v

on ψ (property (ii) of Theorem 6.6).

Let now y ∈ X, and set w = T (1)

(e−i

|x|2

4 y

), so that y = ei

|x|2

4 T (−1)w. It follows

from Proposition 6.1 and Theorem 6.6 that y = U+ϕ for some ϕ ∈ X (i.e. y ∈ U+) if and

only if Tm(1, w) < 0. In this case, ϕ = ei|x|2

4 z(0), where z is the solution of (6.7) with

the initial value z(1) = w. Therefore, the open character of U+ and the continuity of the

operator Ω+ = (U+)−1 follow as above from the continuous dependence of z on w.

As observed before, the similar statements for R−, U−, U− and Ω− are equivalent,

by changing t to −t and u(t) to u(−t). Therefore, we have proved part (i) of the theo-

rem. Part (ii) now follows from part (i) and the definitions of O± and S (formulas (6.3)

and (6.4)).

67

Corollary 6.9. Assume4

N + 2< α <

4

N − 2(2 < α <∞, if N = 1). If ϕ ∈ R+, ϕ 6= 0,

then for any p ∈ [2,∞) such that (N − 2)p ≤ 2N there exists ` > 0 such that

(1 + s)N(p−2)

2 ‖u(s)‖pLp −→

s→+∞`.

A similar statement holds if ϕ ∈ R−.

Proof. Let ϕ ∈ R+, ϕ 6= 0, and let u be the solution of (1.1). We have T (−t)u(t) −→t→∞

u+

inX. Since U+(0) = 0 and U+ is a bijection, we have u+ 6= 0. Let now v be defined by (6.6).

It follows from Proposition 6.1 that v(t)−→t↑1

T (1)

(e−i

|x|2

4 u+

)6= 0 in X. By Sobolev’s

embedding, for any p ∈ [2,∞) such that (N − 2)p ≤ 2N , there exists ` > 0 such that

‖v(t)‖Lp → ` as t ↑ 1. The result now follows from (6.8).

Before proceeding fruther, we need the following lemma.

Lemma 6.10. Let α0 be defined by (5.1), and let a be defined by (5.2). Assume α0 <

α <4

N − 2and let v ∈ C([0, 1), X) be a solution of (6.7), and let TM ≥ 1 be the maximal

existence time of v. If (1− t)Nα−4

2α v ∈ La((0, 1), Lα+2(RN )), then TM > 1.

Proof. Suppose (1 − t)Nα−4

2α v ∈ La((0, 1), Lα+2(RN )). Set r = α + 2, and let (q, r) be

the corresponding admissible pair. Given 0 ≤ t0 ≤ t < 1, it follows from equation (6.7)

and estimate (1.2) that

‖v‖L∞((t0,t),H1) + ‖v‖Lq((t0,t),W 1,r) ≤ C‖v(t0)‖H1 + C‖(1− s)Nα−4

2 |v|αv‖Lq′((t0,t),W 1,r′).

Applying Holder’s inequality, we obtain

‖(1− s)Nα−4

2 |v|αv‖Lq′((t0,t),W 1,r′) ≤ C‖(1− s)Nα−4

2 |v|α‖L

qq−2 ((t0,t),L

α+2α )

‖v‖Lq((t0,t),W 1,r).

Sinceαq

q − 2= a, it follows that

‖(1− s)Nα−4

2 |v|αv‖Lq′((t0,t),W 1,r′) ≤ C‖(1− s)Nα−4

2α v‖αLa((t0,t),Lα+2)‖v‖Lq((t0,t),W 1,r);

and so,

‖v‖L∞((t0,t),H1) + ‖v‖Lq((t0,t),W 1,r) ≤ C‖v(t0)‖H1

+ C‖(1− s)Nα−4

2α v‖αLa((t0,t),Lα+2)‖v‖Lq((t0,t),W 1,r).

68

Choosing t0 close enough to 1 so that C‖(1− s)Nα−4

2α v‖αLa((t0,1),Lα+2) ≤ 1/2, we get

‖v‖L∞((t0,t),H1) + ‖v‖Lq((t0,t),W 1,r) ≤ 2C‖v(t0)‖H1 .

It follows that v remains bounded in H1(RN ) as t ↑ 1, and the result follows from Theo-

rem 6.6.

Corollary 6.11. Let α0 be defined by (5.1), and let a be defined by (5.2). Assume

α0 < α <4

N − 2, let ϕ ∈ X be such that T ∗ = ∞ and let u ∈ C([0,∞), X) be the solution

of (1.1). Then ϕ ∈ R+ if and only if u ∈ La((0,∞), Lα+2(RN )).

Proof. Assume ϕ ∈ R+. It follows from Corollary 6.9 that sups≥0

(1+s)Nα2 ‖u(s)‖α+2

Lα+2 <∞.

Therefore,

‖u(s)‖aLα+2 ≤ C(1 + s)−

Nαa2(α+2) .

SinceNαa

2(α+ 2)> 1, we obtain u ∈ La((0,∞), Lα+2(RN )).

Conversely, if u ∈ La((0,∞), Lα+2(RN )) and if v is defined by (6.6), then it follows

from (6.8) that (1 − t)Nα−4

2α v ∈ La((0, 1), Lα+2(RN )). Lemma 6.10 now implies that v(t)

exists beyond t = 1, and it follows from Proposition 6.1 that ϕ ∈ R+.

Corollary 6.12. Let α0 be defined by (5.1), and let ϕ ∈ X. Then eib|x|2

4 ϕ ∈ R+ if b is

large enough, and e−ib|x|2

4 ϕ ∈ R− if b is large enough.

Proof. The result is an immediate consequence of Corollary 6.11 and Theorem 5.1.

We now establish further properties of the wave operators Ω±.

Theorem 6.13. Assume4

N + 2< α <

4

N − 2(2 < α < ∞, if N = 1). Then, the

following properties hold.

(i) If λ < 0, then U± = X. Therefore, the wave operators Ω± are bicontinuous bijections

X → R±.

(ii) If λ > 0 and α < 4/N , then U± = X. Therefore, the wave operators Ω± are bicontin-

uous bijections X → R±.

69

(iii) If λ > 0 and α ≥ 4/N , then U± 6= X.

Proof. Assume λ < 0, or λ > 0 and α < 4/N . Let w ∈ X and let z be the solution

of (6.7) with the initial value z(1) = w. By Theorem 6.6, z is defined on some interval

[1− ε, 1] with ε > 0. Set

φ(y) = eN2 ei

ε|y|2

4 z(1− ε, εy) ∈ X.

Let u be the solution of equation (1.1) with the initial value

u

(1− ε

ε

)= φ. (6.24)

Since λ < 0, or λ > 0 and α < 4/N , it follows that u is global. Therefore, we may define

ϕ = u(0). We claim that ϕ ∈ R+ and that u+ = ei|x|2

4 T (−1)w. Indeed, consider v defined

by (6.6). We see by applying (6.6) with t = 1− ε and (6.24) that

v(1− ε) = z(1− ε),

so that by uniqueness v ≡ z; and so the claim follows from Proposition 6.1. Thus we have

proved (i) and (ii).

Assume now that λ > 0 and α ≥ 4/N . Let Φ ∈ X be such that E(Φ) < 0, so that

T ∗(Φ) < ∞ and T∗(Φ) < ∞. Given b ∈ R, set φ(x) = eib|x|2

4 Φ(x), and let w be the

corresponding solution of (1.1). It follows from Theorem 5.1 and Remark 5.2 (v) that if

b is large enough, we have T∗(φ) < ∞, T ∗(φ) = ∞ and w ∈ La((0,∞), Lα+2(RN )). By

Corollary 6.11, this implies that φ ∈ R+. Let ω = U+(φ). Let now τ be large enough so

that −T∗(φ)+τ > 0, and set u(t) = w(t+τ). It follows that u is defined on (−T∗(φ)+τ,∞)

and that u blows up as t ↓ −T∗(φ) + τ > 0. In particular, u cannot be defined on (0,∞).

On the other hand, we have

T (−t)u(t) = T (τ)T (−(t+ τ)w(t+ τ)) −→t→∞

T (τ)ω.

We claim that T (τ)ω 6∈ U+. Indeed, if T (τ)ω were in U+, then there would exist a solu-

tion z(t) of (1.1) such that T (−t)z(t) −→t→∞

T (τ)ω. By Proposition 6.1 and the uniqueness

property of Theorem 6.6, we would have z = u. This is absurd, since u cannot be defined

on (0,∞). Therefore U+ 6= X, and U− = U+ 6= X. This completes the proof.

70

We now study the asymptotic completeness.

Theorem 6.14. Let α0 be defined by (5.1), and assume α0 < α <4

N − 2

(i) If λ < 0, then R+ = R− = X. In particular, U±, Ω± and S are bicontinuous bijections

X → X.

(ii) If λ > 0, then R± 6= X. Moreover, R± are unbounded subsets of L2(RN ).

Proof. Assume first λ < 0. Let ϕ ∈ X, let u be the solution of (1.1) and let v be defined

by (6.6). If α ≥ 4/N , then it follows from (6.12) that E1(t) is nonincreasing, which implies

that ‖∇v(t)‖L2 is bounded as t ↑ 1. Since ‖v(t)‖L2 is also bounded by (6.11), it follows that

‖v(t)‖H1 is bounded as t ↑ 1. By Theorem 6.6, this implies that v(t) has a limit as t ↑ 1;

and so ϕ ∈ R+ by Proposition 6.1. If α < 4/N , then it follows from (6.13) that ‖v(t)‖Lα+2

remains bounded as t ↑ 1. This implies that (1 − t)Nα−4

2α v ∈ La((0, 1), Lα+2(RN )); and

so, by Lemma 6.10, v(t) has a limit as t ↑ 1. One concludes as above. This shows that

R± = X. Property (i) now follows from Theorem 6.13.

Assume now that λ > 0. The unbounded character of R± follows from Corollary 6.12.

Consider now a standing wave, i.e. a solution of the form u(t, x) = eiωtφ(x), with φ ∈ X,

φ 6= 0 (see [5], Section 8.1). It is clear that φ 6∈ R+ and φ 6∈ R−; and so, R+ 6= X and

R− 6= X.

Finally, we extend the asymptotic completeness result to the case α = α0 (see [11]).

Theorem 6.15. Assume N = 1 or N ≥ 3. Let α0 be defined by (5.1), and assume

α = α0. If λ < 0, then R+ = R− = X. In particular, U±, Ω± and S are bicontinuous

bijections X → X.

Proof. Since R− = R+, we need only show that R+ = X. Let ϕ ∈ X, let u ∈ C(R, X)

be the solution of (1.1) and let v be defined by (6.6). Note that, since u is defined on

[0,∞), v is defined on [0, 1). By Proposition 6.1, ϕ ∈ R+ if v(t) has a limit in X as t ↑ 1.

Therefore, in view of Theorem 6.6, we need only show that

supt∈[0,1)

‖v(t)‖H1 <∞. (6.25)

71

We argue by contradiction and we assume that (6.25) does not hold, i.e.

lim supt↑1

‖v(t)‖H1 = ∞. (6.26)

We consider separately the cases N ≥ 3 and N = 1.

Case N ≥ 3. By (6.26) and property (i) of Theorem 6.6, we have

‖∇v(t)‖2H1 ≥ a

(1− t)N+2

2 − 2α

,

for some constant a > 0 and all t ∈ [0, 1). By applying (6.13), we obtain

d

dtE2(t) ≤ − b

(1− t)Nα−2

2 + N+22 − 2

α

,

for some constant b > 0. Since α = α0, the above inequality means

d

dtE2(t) ≤ − b

(1− t),

which implies that E2(t)−→t↑1

−∞. This is absurd, since E2(t) ≥ 0. This completes the

proof in the case N ≥ 3.

Case N = 1. The argument is the same as above, except that we first need to improve

the lower estimate of blow up given by property (i) of Theorem 6.6. We claim that

‖v(t)‖H1 ≥ a

(1− t)(α−2)(α+4)

4α

, (6.27)

for some constant a > 0 and all t ∈ [0, 1). Indeed, note first that it follows from (6.13)

thatd

dtE2(t) ≤ 0;

and so,

supt∈[0,1)

‖v(t)‖Lα+2 <∞.

Fix t0 ∈ [0, 1). It follows from the equation (6.23) and the Strichartz estimate (1.2) that

‖v‖L∞((t0,t),H1) ≤ C‖v(t0)‖H1 + C‖f |v|αv‖L1((t0,t),H1),

for all t ∈ (t0, 1). On the other hand,

‖ |v|αv‖H1 ≤ C‖v‖αL∞‖v‖H1 ,

72

and, by Gagliardo-Nirenberg’s inequality,

‖v‖L∞ ≤ C‖v‖2

α+4

H1 ‖v‖α+2α+4

Lα+2 .

Therefore, it follows from the four above inequalities that there exists a constant K inde-

pendent of t0 and t such that

‖v‖L∞((t0,t),H1) ≤ K‖v(t0)‖H1 +K‖f‖L1(t0,t)‖v‖3α+4α+4

L∞((t0,t),H1). (6.28)

Now by (6.26), there exists t1 ∈ (t0, 1) such that ‖v‖L∞((t0,t−1),H1) = (K + 1)‖v(t0)‖H1 .

Taking t = t1 in (6.28), we obtain

‖v(t0)‖H1 ≤ K ((K + 1)‖v(t0)‖H1)3α+4α+4 ‖f‖L1(t0,t),

hence

1 ≤ K ((K + 1))3α+4α+4 ‖v(t0)‖

2αα+4

H1 ‖f‖L1(t0,t1).

Since ‖f‖L1(t0,t1) ≤ ‖f‖L1(t0,1) ≤ C(1− t) α−22 , we obtain (6.27). We now conclude exactly

as in the case N ≥ 3. This completes the proof.

Remark 6.16. Here some comments concerning Theorem 6.15.

(i) We do not know whether or not the conclusion of Theorem 6.15 holds in the case N =

2. The argument that we used in the case N = 1 does not seem to be applicable, due

to the particular homogeneity of the Gagliardo-Nirenberg’s inequality in dimension 2.

(ii) If4

N + 2< α < α0, then we do not know whether or not R+ = R− = X. Showing

this property amounts to showing that no solution of (6.7) can blow up at t = 1.

Remark 6.17. Ginibre and Velo [17] extended the construction of the wave operators Ω±

to a wider range of α’s, by working in the space Hρ(RN )∩F(Hρ(RN )), where 0 < ρ < 2.

The lower bound on α for this method is given by

α > max

ρ− 1,

4

N + 2ρ,

2

N

If N = 3, one obtains the lower bound α > 2/N by taking ρ = 3/2. If N ≥ 4, there is still

a gap between the admissible values of α and the lower bound α > 2/N for the scattering

theory given by Theorem 6.2.

73

7. The Cauchy problem for a nonautonomous Schrodinger equation. In this

section, we study the Cauchy problem for equation (6.7) starting from any point t ∈ [0, 1].

In fact, we consider the more general Cauchy problem (see [11])

ivt +4v + h(t)|v|αv = 0,

v(0) = ψ,(7.1)

where h ∈ L1loc(R,R). We will study equation (7.1) under the equivalent form

v(t) = T (t)ψ + i

∫ t

0

T (t− s)h(s)|v(s)|αv(s) ds. (7.2)

We have the following existence and uniqueness result.

Theorem 7.1. Assume 0 < α <4

N − 2(0 < α < ∞ if N = 1). Let θ =

4

4− α(N − 2)(θ = 1, if N = 1; θ > 1 and (2 − α)θ ≤ 1, if N = 2), and consider a real valued

function h ∈ Lθloc(R,R). Then, for every ψ ∈ H1(RN ), there exist T ∗, T∗ > 0 and

a unique, maximal solution v ∈ C((−T∗, T ∗), H1(RN )) ∩ W 1,θloc ((−T∗, T ∗), H−1(RN )) of

equation (7.2). The solution v is maximal in the sense that if T ∗ < ∞ (respectively

T∗ < ∞), then ‖u(t)‖H1 → ∞, as t ↑ T ∗ (respectively t ↓ −T∗). In addition, the solution

v has the following properties.

(i) If T ∗ <∞, then lim inft↑T ∗

‖v(t)‖αH1‖h‖Lθ(t,T ∗) > 0.

(ii) If T∗ <∞, then lim inft↓−T∗

‖v(t)‖αH1‖h‖Lθ(−T∗,t) > 0.

(iii) u ∈ Lqloc((−T∗, T ∗),W 1,r(RN )), for every admissible pair (q, r).

(iv) There exists δ > 0, depending only on N,α and θ such that if

‖ψ‖αθH1

∫ τ

−τ

|h(s)|θ ds ≤ δ,

then [−τ, τ ] ⊂ (−T∗, T ∗) and ‖v‖Lq((−τ,τ),W 1,r) ≤ K‖ψ‖H1 for every admissible pair

(q, r), where K depends only on N,α, θ and q. In addition, if ψ′ is another initial

value satisfying the above condition and if v′ is the corresponding solution of (7.2),

then ‖v − v′‖L∞((−τ,τ),L2) ≤ K‖ψ − ψ′‖L2 .

(v) If ψ ∈ X, then v ∈ C((−T∗, T ∗), X). In addition,

‖xv(t)‖2L2 = ‖xψ‖2

L2 + 4Im

∫ t

0

∫

RN

v(s)(x · ∇v(s)) dx ds,

74

for all t ∈ (−T∗, T ∗).

Proof. For technical reasons, we suppose first that N ≥ 3. Afterwards, we will indicate

the modifications needed to handle the cases N = 2 and N = 1. Let 2∗ =2N

N − 2, and

define r by

1− 2

r=

α

2∗. (7.3)

Since (N − 2)α < 4, it follows that 2 < r < 2∗. Therefore, there exists q such that (q, r) is

an admissible pair. A simple calculation shows that

1

q′=

1

θ+

1

q. (7.4)

By (1.2), there exists K such that

‖T (·)ψ‖L∞(R,H1) + ‖T (·)ψ‖Lq(R,W 1,r) ≤ K‖ψ‖H1 ,

for every ψ ∈ H1(RN ). Given M > 0 and 0 ≤ T1, T2 such that T1 + T2 > 0, let

E = v ∈ C([−T1, T2], H1(RN )) ∩ Lq((−T1, T2),W

1,r(RN ));

‖v‖L∞((−T1,T2),H1) + ‖v‖Lq((−T1,T2),W 1,r) ≤ (K + 1)M.

Endowed with the metric d(u, v) = ‖v − u‖Lq((−T1,T2),Lr), (E, d) is a complete metric

space. Given v ∈ E, it follows from (7.3), (7.4), Sobolev’s and Holder’s inequalities that

h|v|αv ∈ Lq′((−T1, T2),W1,r′(RN )) and that

‖h|v|αv‖Lq′((−T1,T2),W 1,r′) ≤ C‖h‖Lθ(−T1,T2)‖v‖αL∞((−T1,T2),L2∗)‖v‖Lq((−T1,T2),W 1,r)

≤ C1‖h‖Lθ(−T1,T2)(K + 1)α+1Mα+1.(7.5)

Furthermore, given u, v ∈ E, one has as well

‖h(|v|αv − |u|αu)‖Lq′((−T1,T2),Lr′ ) ≤ C‖h‖Lθ(−T1,T2)(‖v‖αL∞((−T1,T2),H1)

+‖u‖αL∞((−T1,T2),H1))‖v − u‖Lq((−T1,T2),Lr); (7.6)

and so,

‖h(|v|αv − |u|αu)‖Lq′((−T1,T2),Lr′ ) ≤ C2‖h‖Lθ(−T1,T2)(K + 1)αMαd(u, v). (7.7)

75

Given v ∈ E and ψ ∈ H1(RN ) such that ‖ψ‖H1 ≤M , let G(v) be defined by

G(v)(t) = T (t)ψ + i

∫ t

0

T (t− s)h(s)|v|αv(s) ds,

for t ∈ (−T1, T2). It follows from (1.2) and (7.5) that

G(v) ∈ C([−T1, T2], H1(RN )) ∩ Lq((−T1, T2),W

1,r(RN )),

and that

‖G(v)‖L∞((−T1,T2),H1)+‖G(v)‖Lq((−T1,T2),W 1,r) ≤ KM+C3(K+1)α+1Mα+1‖h‖Lθ(−T1,T2).

Therefore, if T1 + T2 is small enough so that

C3(K + 1)α+1Mα‖h‖Lθ(−T1,T2) ≤ 1,

then G(v) ∈ E. Furthermore, (1.2) and (7.7) imply that

d(G(v),G(u)) ≤ C4(K + 1)αMα‖h‖Lθ(−T1,T2)d(u, v).

Consequently, if T2 − T1 is small enough so that

K1Mα‖h‖Lθ(T1,T2) ≤ 1/2, (7.8)

where K1 = (K+1)α+1maxC3, C4, then G is Lipschitz continuous E → E with Lipschitz

constant 1/2. Therefore, G has a unique fixed point v ∈ E, which satisfies equation (7.2).

In addition, the first part of property (iv) follows from (7.8), (7.5) and (1.2), and the second

part from (7.6) and (1.2). Uniqueness in the class C([−T1, T2];H1(RN )) follows from (7.6)

and (1.2). (Note that uniqueness is a local property and needs only to be established for

T1 + T2 small enough.) Now, by uniqueness, v can be extended to a maximal interval

(−T∗, T ∗), and property (iii) follows from property (iv). Suppose that T ∗ <∞. Applying

the above local existence result to v(t), t < T ∗ with T1 = 0, we see from (7.8) that if

K1‖v(t)‖αH1‖h‖Lθ(t,T ∗) ≤ 1/2,

then v can be continued up to and beyond T ∗, which is a contradiction. Therefore, we

have

K1‖v(t)‖αH1‖h‖Lθ(t,T ∗) > 1/2,

76

which proves property (i). Property (ii) is proved by the same argument. Finally, since v

satisfies the equation (7.1) in Lθloc((−T∗, T ∗), H−1(RN ) and h is real valued, property (v) is

proved by standard arguments. For example, multiply the above equation by |x|2e−ε|x|2v,

take the imaginary part and integrate over RN , then let ε ↓ 0 (see Proposition 6.4.2 p.107

in [5] for a similar argument).

If N = 2, the proof is the same as in the case N ≥ 3, except that we set r = 2θ and

use the embedding H1(R2) → Lp(R2) with p =αθ

θ − 1.

If N = 1, the argument is slightly simpler. We let

E = v ∈ C([−T1, T2], H1(R)); ‖v‖L∞((−T1,T2),H1) ≤ 2M,

equipped with the metric d(u, v) = ‖v− u‖L∞(T1,T2;L2), and use the embedding H1(R) →L∞(R).

We now study the continuous dependence of the solutions on the initial value. The

result is the following.

Theorem 7.2. Under the assumptions of Theorem 7.1, suppose there exists θ1 > θ such

that h ∈ Lθ1

loc(R). The solution v of (7.2) given by Theorem 7.1 depends continuously on

ψ in the following way.

(i) The mappings ψ 7→ T ∗ and ψ 7→ T∗ are lower semicontinuous H1(RN ) → (0,∞].

(ii) If ψn −→n→∞

ψ in H1(RN ) and if vn denotes the solution of (7.2) with initial value ψn,

then vn → v in C([−T1, T2], H1(RN )) for any interval [−T1, T2] ∈ (−T∗, T ∗). If in

addition ψn → ψ in X, then vn → v in C([T1, T2], X).

Proof. We proceed in two steps.

Step 1. We show that for every M > 0, there exists τ > 0 such that if ψ ∈ H1(RN ) ver-

ifies ‖ψ‖H1 < M , then [−τ, τ ] ⊂ (−T∗, T ∗), and v has the following continuity properties.

(a) If ‖ψ‖H1 < M , ψn −→n→∞

ψ in H1(RN ), and if vn denotes the solution of (7.2) with

initial value ψn, then vn → v in C([−τ, τ ], H1(RN )).

(b) If ‖ψ‖H1 < M , ψn −→n→∞

ψ in X, and if vn denotes the solution of (7.2) with initial

value ψn, then vn → v in C([−τ, τ ], X).

77

We base our proof on arguments of Kato [26] used in the autonomous case. We

only prove the result in the case N ≥ 3 (see the proof of Theorem 7.1 for the necessary

modifications in the cases N = 1, 2). Given M > 0, we choose τ so that the inequality in

property (iv) of Theorem 7.1 is met whenever ‖ψ‖H1 < M . In particular, if ‖ψ‖H1 < M ,

then [−τ, τ ] ⊂ (−T∗, T ∗). Next, observe that1

θ>

4− αN

4; and so we may assume without

loss of generality that1

θ1>

4− αN

4. Therefore, if we define σ by

1

σ=

2

N

(1− 1

θ1

),

then 2 < ασ <2N

N − 2. Let now ρ be defined by

1− 2

ρ=

1

σ.

Since1

σ<

2

N, it follows that 2 < ρ <

2N

N − 2. Finally, let γ be such that (γ, ρ) is an

admissible pair. It follows easily from Holder’s inequality that for every −∞ < a < b <∞,

‖hwz‖Lγ′ ((a,b),Lρ′ ) ≤(∫ b

a

|h(s)|θ1‖w(s)‖θ1

Lσ

)1/θ1

‖z‖Lγ((a,b),Lρ). (7.9)

Consider now ψ such that ‖ψ‖H1 ≤ M , and let ψn be as in (a). Let v, vn be the corre-

sponding solutions of (7.2). It follows from (1.2) that there exists C, depending only on γ,

such that

‖v − vn‖Lγ((−τ,τ),W 1,ρ) + ‖v − vn‖L∞((−τ,τ),H1) ≤ C‖ψ − ψn‖H1

+ ‖h(|v|αv − |vn|αvn)‖Lγ′ ((−τ,τ),W 1,ρ′).(7.10)

On the other hand, a straightforward calculation shows that

|∇(|v|αv − |vn|αvn)| ≤ C|vn|α|∇v −∇vn|+ φ(v, vn)|∇v|, (7.11)

where C depends on α, and the function φ(x, y) is bounded by C(|x|α + |y|α) and verifies

φ(x, y)−→y→x

0. Therefore, applying (7.9), (7.10) and (7.11), we get

‖v − vn‖Lγ((−τ,τ),W 1,ρ) + ‖v − vn‖L∞((−τ,τ),H1) ≤ C‖ψ − ψn‖H1

+ ‖h‖Lθ1(−τ,τ)‖vn‖L∞((−τ,τ),Lασ)‖v − vn‖Lγ((−τ,τ),W 1,ρ)

+ C

(∫ b

a

|h(s)|θ1‖φ(v, vn)‖θ1

Lσ

)1/θ1

‖v‖Lγ((−τ,τ),W 1,ρ).

(7.12)

78

Note that by property (iv) of Theorem 7.1, vn is bounded in H1(RN ), hence in Lασ(RN ),

with the bound, for t ∈ [−τ, τ ], depending only on ‖ψn‖H1 , hence (for large values of n)

only on M . As well, the bound on ‖v‖Lγ((−τ,τ),W 1,ρ) depends only on M . Therefore, it

follows from (7.12) that

‖v − vn‖Lγ((−τ,τ),W 1,ρ) + ‖v − vn‖L∞((−τ,τ),H1) ≤ C‖ψ − ψn‖H1

+ C‖h‖Lθ1(−τ,τ)‖v − vn‖Lγ((−τ,τ),W 1,ρ) + C

(∫ b

a

|h(s)|θ1‖φ(v, vn)‖θ1

Lσ

)1/θ1

,

where the constant C depends only on M . Therefore, if we consider τ possibly smaller so

that C‖h‖Lθ1(−τ,τ) ≤ 1/2 (note that τ still depends on M), it follows that

‖v − vn‖Lγ((−τ,τ),W 1,ρ) + ‖v − vn‖L∞((−τ,τ),H1) ≤C‖ψ − ψn‖H1

+ C

(∫ b

a

|h(s)|θ1‖φ(v, vn)‖θ1

Lσ

)1/θ1

.

Therefore, property (a) follows, provided we show that

(∫ b

a

|h(s)|θ1‖φ(v, vn)‖θ1

Lσ

)−→n→∞

0.

By the dominated convergence theorem, it suffices to verify that

‖φ(v, vn)‖Lσ −→n→∞

0,

for all t ∈ [−τ, τ ]. To see this, we argue by contradiction. We assume that there exists t

and a subsequence, which we still denote by vn(t), such that ‖φ(v(t), vn(t))‖Lσ ≥ µ > 0.

Note that vn(t) → v(t) in L2(RN ) and vn(t) is bounded in H1(RN ) by property (iv) of

Theorem 7.1. Therefore, by Sobolev’s and Holder’s inequalities, vn(t) → v(t) in Lασ(RN ).

It follows that there exists a subsequence, which we still denote by vn(t), and a func-

tion f ∈ Lασ(RN ) such that vn(t) → v(t) almost everywhere in RN and |vn(t)| ≤ f

almost everywhere in RN . Applying the dominated convergence theorem, it follows that

‖φ(v(t), vn(t))‖Lσ → 0, which is a contradiction. Hence property (a).

Property (b) follows from property (a) and Theorem 7.1 (v). Briefly, use Holder’s

inequality on the formula in Theorem 7.1 (v) to obtain a uniform bound in X on the

solutions vn. The integral term then converges along subsequences where xvn converges

weakly in L2(RN ) to xv, and the rest of the proof is standard.

79

Step 2. Conclusion. Let ψ ∈ H1(RN ), let v be the maximal solution of (7.2) given by

Theorem 7.1 and let [−T1, T2] ⊂ (−T∗, T ∗). Set

M =1

2sup

−T1≤t≤T2

‖v(t)‖H1 ,

and consider τ > 0 given by Step 1. By applying Step 1 m times, where (m − 1)τ <

T1 + T2 ≤ mτ , we see that if ‖ψ − ψ‖H1 is small enough, then the solution of (7.2) with

initial value ψ exists on [−T1, T2]. Hence property (i). Property (ii) follows easily from the

same argument.

80

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85