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HAL Id: hal-01936113

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Scattering in weighted L -space for a 2D nonlinear Schr’́odinger equation with inhomogeneous exponential

nonlinearity

Abdelwahab Bensouilah, van Duong Dinh, Mohamed Majdoub

To cite this version:

Abdelwahab Bensouilah, van Duong Dinh, Mohamed Majdoub. Scattering in weighted L

2

-space

for a 2D nonlinear Schr’́odinger equation with inhomogeneous exponential nonlinearity. 2018. �hal-

01936113�

(2)

SCHR ¨ ODINGER EQUATION WITH INHOMOGENEOUS EXPONENTIAL

2

NONLINEARITY

3

ABDELWAHAB BENSOUILAH, VAN DUONG DINH, AND MOHAMED MAJDOUB

4

Abstract. We investigate the defocusing inhomogeneous nonlinear Schr¨odinger equation i∂tu+ ∆u=|x|−b

eα|u|2−1−α|u|2

u, u(0) =u0, x∈R2,

with 0< b <1 andα= 2π(2−b). First we show the decay of global solutions by assuming that the initial data u0 belongs to the weighted space Σ(R2) = {u ∈ H1(R2) : |x|u ∈ L2(R2)}. Then we combine the local theory with the decay estimate to obtain scattering in Σ when the Hamiltonian is below the value (1+b)(2−b)2 .

1. Introduction and main result

5

This paper is concerned with the scattering theory for the following initial value problem ( i∂

t

u + ∆u = |x|

−b

e

α|u|2

− 1 − α|u|

2

u,

u(0) = u

0

, (1.1)

where u = u(t, x) is a complex-valued function in space-time R × R

2

, 0 < b < 1 and α =

6

2π(2 − b).

7

The classical nonlinear Schr¨ odinger equation (b = 0) with pure power or exponential

8

nonlinearities arises in various physical contexts, as for example the self trapped beams in

9

plasma, the propagation of a laser beam, water waves at the free surface of an ideal fluid and

10

plasma waves (see [21]).

11

From the mathematical point of view, the classical NLS equation, i.e., problem (1.1) with

12

b = 0, has attracted considerable attention in the mathematical community and the well-

13

posedness theory as well as the scattering has been extensively studied, see for instance

14

[2, 3, 7, 9, 19, 22]. We refer the reader to [8, 33] and references therein for more properties

15

and information on nonlinear Schr¨ odinger equations.

16

In particular, in [9] a notion of criticality was proposed and the authors established in

17

both subcritical and critical regimes the existence of global solutions in the functional space

18

C( R , H

1

( R

2

)) ∩ L

4loc

( R , W

1,4

( R

2

)). Later on in [19], the scattering in the energy space was

19

Date: October 25, 2018.

2000Mathematics Subject Classification. 35-xx, 35L70, 35Q55, 35B40, 35B33, 37K05, 37L50.

Key words and phrases. Inhomogeneous nonlinear Schr¨odinger equation; Decay solutions; Virial identity;

Scattering; WeightedL2-space; Exponential nonlinearity, Singular Moser-Trudinger inequality.

1

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obtained in the subcritical case. Note that the critical case was investigated in [3] where the

1

scattering is proved in the radial framework.

2

The situation in the case b > 0 is less understood. Recently, in [5] the authors established

3

the global well-posedness in the energy space for 0 < b < 1. A natural question to ask then

4

is the long time behavior of global solutions, that is the scattering. This means that every

5

global solution of (1.1) approaches solutions to the associated free equation

6

i∂

t

v + ∆v = 0, (1.2)

in the energy space H

1

as t → ±∞. The main difficulty is how to obtain the interaction

7

Morawetz inequality? Recall that the interaction Morawetz inequality is nothing but the

8

convolution of the classical one with the mass density. This in particular leads to a priori

9

global bound of the solution in L

4t

(L

8x

) which is the main tool for the scattering in the energy

10

space (see for instance [3, 19, 24]). Note that the interaction Morawetz inequalities were first

11

established for the NLS with power-type nonlinearity, and the proof depends heavily on the

12

form of nonlinearity. Of course the proof can be easily adapted to more general homogeneous

13

nonlinearities. More precisely, for linear combination of powers it suffices that all the powers

14

are quadratic or higher with positive coefficients. The problem with singular weight (or for

15

non-homogeneous nonlinearity) is much more difficult and should be investigated separately.

16

For instance, it was noticed in [11] that the interaction Morawetz inequality for the NLS

17

with singular nonlinearity N (x, u) = |x|

−b

|u|

α

u may not hold due to the lack of momentum

18

conservation law.

19

This is why we restrict ourselves to initial data belonging to the weighted L

2

-space Σ :=

20

H

1

∩ L

2

(|x|

2

dx). Note that the scattering in Σ for the NLS with N (x, u) = |x|

−b

|u|

α

u was

21

considered by the second author in [10].

22

The scattering in the energy space will be investigated in a forthcoming paper, and we

23

believe that some ideas developed in [3] will be helpful.

24

Remark 1.1. We stress that the two-dimensional nonlinear Klein-Gordon equation with pure

25

exponential nonlinearity was studied in [16, 18, 17], and a similar trichotomy based on the

26

energy was defined. Recently, M. Struwe [30, 31] was able to construct global smooth solution

27

for smooth initial data and prove the scattering [29].

28

Before stating our main result, let us recall that solutions of (1.1) satisfy the conservation

29

of mass and Hamiltonian

30

M(u(t)) := ku(t)k

L2

, (1.3)

31

H(u(t)) :=

Z

|∇u(t, x)|

2

+ 1 α

Z

e

α|u(t,x)|2

− 1 − α|u(t, x)|

2

− α

2

2 |u(t, x)|

4

dx

|x|

b

. (1.4) Our main result is the following.

32

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Theorem 1.2. Let u

0

∈ Σ be such that H(u

0

) <

(1+b)(2−b)2

. Then the corresponding global

1

solution u of (1.1) satisfies u ∈ L

4

( R , C

1/2

) and there exist u

±0

∈ Σ such that

2

t→±∞

lim ke

−it∆

u(t) − u

±0

k

Σ

= 0.

Let us make some comments. First, we see that

(1+b)(2−b)2

→ 1 as b → 0. Thus our

3

result extends the one in [19] for initial data in Σ. Second, the condition H(u) <

(1+b)(2−b)2

4

illustrates the interaction between the wave function u and the potential |x|

−b

. More precisely,

5

a sufficient condition for scattering is when the energy of the wave is less than a fixed amount

6

depending on the sole parameter b that characterizes the weight function involved in the

7

Hamiltonian of (1.1). Finally, a natural question that one could raise is the following: is the

8

value

(1+b)(2−b)2

critical for scattering, in the sense that if the energy of the wave exceeds the

9

latter quantity, would one get scattering?

10

Remark 1.3. For all 0 < b < 1,

89

(1+b)(2−b)2

< 1.

11

The proof of Theorem 1.2 follows a standard strategy for the classical NLS equation. We first derive a decaying property for global solutions by using the pseudo-conformation law. We then show two types of global bounds for the solution u and its weighted variant (x + 2it∇)u.

More precisely, we will show that

kuk

S1(R)

< ∞, k(x + 2it∇)uk

S0(R)

< ∞, (1.5) where

12

kuk

S1(R)

:= kuk

L(R,H1)

+ kuk

L4(R,W1,4)

, kuk

S0(R)

:= kuk

L(R,L2)

+ kuk

L4(R,L4)

. The proof of these global bounds relies on the decaying property, the singular Moser-Trudinger

13

inequality and the Log estimate. The main difficulty comes from the singular weight |x|

−b

14

which does not belong to any Lebesgue space. To overcome this problem, we will take the

15

advantage of Lorentz spaces. Note that |x|

−b

∈ L

2b,∞

( R

2

), where L

p,∞

is the Lorentz space.

16

Once these global bounds are established, the scattering in weighted L

2

space Σ follows easily.

17

This paper is organized as follows. In Section 2, we recall some useful tools needed in our

18

problem. The pseudo-conformal law is derived in Section 3. The decaying property of global

19

solutions in Lebesgue spaces is showed in Section 4. Sections 5 and 6 are devoted to the

20

proofs of global bounds (1.5). We shall give the proof of our main result in Theorem 1.2 in

21

Section 7.

22

2. Useful Tools

23

In this section, we collect some known and useful tools.

24

Proposition 2.1 (Moser-Trudinger inequality [1]).

25

Let α ∈ [0, 4π). A constant c

α

exists such that

26

k exp(α|u|

2

) − 1k

L1(R2)

≤ c

α

kuk

2L2(R2)

, (2.1)

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for all u in H

1

( R

2

) such that k∇uk

L2(R2)

≤ 1. Moreover, if α ≥ 4π, then (2.1) is false.

1

Remark 2.2. We point out that α = 4π becomes admissible in (2.1) if we require kuk

H1(R2)

2

1 rather than k∇uk

L2(R2)

≤ 1. Precisely, we have

3

sup

kukH1≤1

k exp(4π|u|

2

) − 1k

L1(R2)

< ∞, (2.2) and this is false for α > 4π. See [25] for more details.

4

Theorem 2.3. [26] Let 0 < b < 2 and 0 < α < 2π(2 − b). Then, there exists a positive

5

constant C = C(b, α) such that

6

Z

R2

e

α|u(x)|2

− 1

|x|

b

dx 6 C Z

R2

|u(x)|

2

|x|

b

dx, (2.3)

for all u ∈ H

1

( R

2

) with k∇uk

L2(R2)

6 1.

7

We point out that α = 2π(2 − b) becomes admissible in (2.3) if we require kuk

H1(R2)

≤ 1

8

instead of k∇uk

L2(R2)

≤ 1. More precisely, we have

9

Theorem 2.4. [27] Let 0 < b < 2. We have

10

sup

kukH1(

R2)61

Z

R2

e

α|u(x)|2

− 1

|x|

b

dx < ∞ if and only if α ≤ 2π(2 − b). (2.4) The following lemma will be very useful.

11

Lemma 2.5. Let 0 < b < 2 and γ ≥ 2. Then, there exists a positive constant C = C(b, γ) > 0

12

such that

13

Z

R2

|u(x)|

γ

|x|

b

dx ≤ Ckuk

γH1(R2)

, (2.5) for all u ∈ H

1

( R

2

).

14

Proof. Note that

15

k|x|

−b

k

Lr(B)

< ∞ if b < 2

r , k|x|

−b

k

Lr(Bc)

< ∞ if b > 2

r , (2.6)

where B = B(0, 1) is the unit ball in R

2

and B

c

= R

2

\B . Write

16

Z

R2

|x|

−b

|u(x)|

γ

dx = Z

B

|x|

−b

|u(x)|

γ

dx + Z

Bc

|x|

−b

|u(x)|

γ

dx.

We have from the Sobolev embedding H

1

( R

2

) ⊂ L

q

( R

2

) for any q ∈ [2, ∞) that

17

Z

Bc

|x|

−b

|u(x)|

γ

dx ≤ kuk

γLγ(

R2)

. kuk

γH1(

R2)

.

The first term is estimated as follows. Since 0 < b < 2, there exists ε > 0 small such that

18

b <

1+ε2

. We apply (2.6) with r = 1 + ε and get

19

Z

B

|x|

−b

|u(x)|

γ

dx ≤ k|x|

−b

k

L1+ε(B)

k|u|

γ

k

L1+εε (R2)

. kuk

γ

L(1+ε)γε (R2)

. kuk

γH1(

R2)

.

Combining the two terms, we prove the desired estimate.

20

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Remark 2.6. The inequality (2.5) fails for b > 2. Indeed, let u ∈ D( R

2

) (the space of smooth compactly supported functions) be a radial function such that u(x) ≡ 1 for |x| 6 1. Then, u ∈ H

1

(R

2

) and

Z

R2

|u(x)|

γ

|x|

b

dx > 2π Z

1

0

rdr

r

b

= +∞.

We also recall the so-called Gagliardo-Nirenberg inequalities and Sobolev embedding.

1

Proposition 2.7 (Gagliardo-Nirenberg inequalities [12, 23]).

2

We have

3

kuk

Lm+1

. kuk

1−θLq+1

k∇uk

θLp

, (2.7) where

θ = pN (m − q)

(m + 1)[N (p − q − 1) + p(q + 1)] , 0 ≤ q < σ − 1, q < m < σ,

4

σ =

(

(p−1)N+p

N−p

if p < N

∞ if p ≥ N

In particular, for N = 2, we obtain

5

kuk

Lq

. kuk

2/qL2

k∇uk

1−2/qL2

, 2 ≤ q < ∞. (2.8) Proposition 2.8 (Sobolev embeddings).

6

We have

7

W

s,p

( R

N

) , → L

q

( R

N

), 1 ≤ p < ∞, 0 ≤ s < N p , 1

p − s N ≤ 1

q ≤ 1

p . (2.9)

8

W

1,p

( R

N

) , → C

1−Np

( R

N

), p > N. (2.10) The following estimate is an L

logarithmic inequality which enables us to establish

9

the link between ke

4π|u|2

− 1k

L1

T(L2(R2))

and dispersion properties of solutions of the linear

10

Schr¨ odinger equation.

11

Proposition 2.9 (Log estimate [15]).

12

Let 0 < β < 1. For any λ >

2πβ1

and any 0 < µ ≤ 1, a constant C

λ

> 0 exists such that, for

13

any function u ∈ H

1

( R

2

) ∩ C

β

( R

2

), we have

14

kuk

2L

≤ λkuk

2µ

log

C

λ

+ 8

β

µ

−β

kuk

Cβ

kuk

µ

, (2.11)

where

15

kuk

2µ

:= k∇uk

2L2

+ µ

2

kuk

2L2

. (2.12) Recall that C

β

(R

2

) denotes the space of β-H¨ older continuous functions endowed with the norm

kuk

Cβ(R2)

:= kuk

L(R2)

+ sup

x6=y

|u(x) − u(y)|

|x − y|

β

.

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We refer to [15] for the proof of this proposition and more details. We just point out that

1

the condition λ >

2πβ1

in (2.11) is optimal.

2

We also recall the so-called Strichartz estimates. We say that (q, r) is an L

2

-admissible

3

pair if

4

0 ≤ 2

q = 1 − 2

r < 1. (2.13)

In particular, note that (

1−2σ2

,

σ1

) is an admissible pair for any 0 < σ < 1/2 and W

1,σ1

( R

2

) , → C

1−2σ

( R

2

).

Proposition 2.10 (Strichartz estimates [8]).

5

Let I ⊂ R be a time interval and let t

0

∈ I. Then, for any admissible pairs (q, r) and (˜ q, r), ˜

6

we have

7

kvk

Lq(I,W1,r(R2))

. kv(t

0

)k

H1(R2)

+ ki∂

t

v + ∆vk

Lq˜0

(I,W1,˜r0(R2))

. (2.14) The following continuity argument (or bootstrap argument) will be useful for our purpose.

8

Theorem 2.11 (Continuity argument).

Let X : [0, T ] → R be a nonnegative continuous function, such that, for every 0 6 t 6 T , X(t) 6 a + bX(t)

θ

,

where a, b > 0 and θ > 1 are constants such that a <

1 − 1

θ

1

(θb)

1/(θ−1)

and X(0) 6 1 (θb)

1/(θ−1)

. Then, for every 0 6 t 6 T , we have

X(t) 6 θ θ − 1 a.

Proof. We sketch the proof for reader’s convenience. The function f : x 7−→ bx

θ

− x + a is

9

decreasing on [0, (θb)

1/(1−θ)

] and increasing on [(θb)

1/(1−θ)

, ∞). The assumptions on a and

10

X(0) imply that f ((θb)

1/(1−θ)

) < 0. As f (X(t)) > 0, f (0) > 0 and X(0) 6

(θb)1/(θ−1)1

, we

11

deduce the desired result.

12

3. Pseudo-conformal law

13

In this section, we show a decaying property of global solutions to (1.1). Note that the

14

conservation laws of mass and Hamiltonian give the boundedness of the L

2

and the H

1

norms

15

but are insufficient to provide a decay estimate in (more general) Lebesgue spaces. To obtain

16

such a decay we will take advantage of the pseudo-conformal law.

17

More precisely, we define the following quantities

18

V(t) :=

Z

|x|

2

|u(t, x)|

2

dx, (3.1) M(t) := 2

Z

I (¯ u(t, x)x · ∇u(t, x)) dx, (3.2)

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K(t) := k(x + 2it∇)u(t)k

2L2

+ 4t

2

α

Z

e

α|u(t,x)|2

− 1 − α|u(t, x)|

2

− α

2

2 |u(t, x)|

4

dx

|x|

b

, (3.3) G(t) := 4(2 − b)

α

Z

e

α|u(t,x)|2

− 1 − α|u(t, x)|

2

− α

2 |u(t, x)|

4

dx

|x|

b

− 8 α

Z

e

α|u(t,x)|2

(α|u(t, x)|

2

− 1) + 1 − α

2

2 |u(t, x)|

4

dx

|x|

b

=:

Z

g(|u(t, x)|

2

) dx

|x|

b

,

(3.4)

where

g(τ ) = 4(2 − b) α

e

ατ

− 1 − ατ − α

2

2 τ

2

− 8 α

e

ατ

(ατ − 1) + 1 − α

2

2 τ

2

. (3.5) Proposition 3.1. Let u

0

∈ Σ and u the corresponding global solution to (1.1). Then

1

dV(t)

dt = 2M(t), (3.6)

d

2

V(t)

dt

2

= 8H(u

0

) − G(t), (3.7)

dK(t)

dt = tG(t), (3.8)

G(t) ≤ 0, ∀ t ∈ R . (3.9)

Proof. A straightforward computation gives (3.6). Let N (x, u) := |x|

−b

e

α|u|2

− 1 − α|u|

2

u.

Following [32] for instance, we find that d

2

V(t)

dt

2

= 8 Z

|∇u|

2

dx + 4 Z

x · {N (x, u), u}

p

dx, where {f, g}

p

= R f∇¯ g − g∇ f ¯

is the momentum bracket.

2

Now compute the momentum bracket {N (x, u), u}

p

. Expand N (x, u) in a formal series N (x, u) = |x|

−b

X

k=2

α

k

k! |u|

2k

u.

Using the fact

3

{|x|

−b

|u|

β

u, u}

p

= − β

β + 2 ∇(|x|

−b

|u|

β+2

) − 2

β + 2 ∇(|x|

−b

)|u|

β+2

, one gets

{N (x, u), u}

p

=

X

k=2

α

k

k! {|x|

−b

|u|

2k

u, u}

p

= −

X

k=2

k α

k

(k + 1)! ∇(|x|

−b

|u|

2k+2

) −

X

k=2

α

k

(k + 1)! ∇(|x|

−b

)|u|

2k+2

.

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An integration by parts leads Z

x · {N (x, u), u}

p

= 2

Z

X

k=2

k α

k

(k + 1)! |u|

2k+2

! dx

|x|

b

+ b

Z

X

k=2

α

k

(k + 1)! |u|

2k+2

! dx

|x|

b

= 2 α

Z

e

α|u|2

(α|u|

2

− 1) + 1 − α

2

2 |u|

4

dx

|x|

b

+ b

α Z

e

α|u|2

− 1 − α|u|

2

− α

2

2 |u|

4

dx

|x|

b

, where we have used

X

k=2

k α

k

(k + 1)! |u|

2k+2

= 1 α

e

α|u|2

(α|u|

2

− 1) + 1 − α

2

2 |u|

4

,

X

k=2

α

k

(k + 1)! |u|

2k+2

= 1 α

e

α|u|2

− 1 − α|u|

2

− α

2

2 |u|

4

.

Therefore,

d

2

V(t)

dt

2

= 8k∇u(t)k

2L2

+ 8 α

Z

e

α|u|2

(α|u|

2

− 1) + 1 − α

2

2 |u|

4

dx

|x|

b

+ 4b

α Z

e

α|u|2

− 1 − α|u|

2

− α

2

2 |u|

4

dx

|x|

b

.

Using the conservation law (1.4), we conclude the proof of (3.7). To prove (3.8), we first remark that

K(t) = V(t) − t dV(t)

dt + 4t

2

H(u

0

).

Hence

dK(t)

dt = −t d

2

V(t)

dt

2

+ 8tH(u

0

),

and the conclusion follows. Finally, for the sign of G, a simple computation shows that (for

1

all τ ≥ 0)

2

g

0

(τ ) = −8(αxe

αx

− e

αx

+ 1) − 4b(e

αx

− αx − 1) ≤ 0.

Since g(0) = 0, we get (3.9).

3

As a consequence of Proposition 3.1, we have

4

Corollary 3.2. Let u

0

∈ Σ and u the corresponding global solution to (1.1). Then k(x + 2it∇)u(t)k

2L2

+ 4t

2

α Z

e

α|u(t,x)|2

− 1 − α|u(t, x)|

2

− α

2

2 |u(t, x)|

4

dx

|x|

b

= kxu

0

k

2L2

+ Z

t

0

τ G(τ ) dτ.

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4. Decay estimate

1

Theorem 4.1. Let u

0

∈ Σ and u the corresponding global solution to (1.1). Then, for all

2

t 6= 0 and 2 ≤ q < ∞,

3

ku(t)k

Lq

≤ C

q

ku

0

k

Σ

|t|

−(1−2q)

, where C

q

> 0 is a constant depending only on q.

4

Proof. Set v(t, x) := e

−i|x|

2

4t

u(t, x). We see that k(x + 2it∇)u(t)k

2L2

= 4t

2

k∇v(t)k

2L2

. Hence,

5

by Corollary 3.2,

6

4t

2

H(v(t)) = kxu

0

k

2L2

+ Z

t

0

τ G(τ ) dτ.

Using (3.9), we get

7

4t

2

k∇v(t)k

2L2

≤ kxu

0

k

2L2

, or equivalently

8

k∇v(t)k

L2

. |t|

−1

.

The conservation of mass, the fact that |u| = |v| and the Gagliardo-Nirenberg inequality

9

(2.8), yield, for all 2 ≤ q < ∞,

10

ku(t)k

Lq

= kv(t)k

Lq

. |t|

−(1−2q)

.

The proof is complete.

11

A natural and useful consequence from the previous theorem is the following bound esti-

12

mate.

13

Corollary 4.2. Let u

0

∈ Σ and u the corresponding global solution to (1.1). Let 1 ≤ p <

14

∞, 2 ≤ q < ∞ be such that

15

p

1 − 2 q

> 1. (4.1)

Then, for all T > 0, we have

kuk

Lp([T ,∞);Lq)

. T

1p+2q−1

p

1 −

2q

− 1

1/p

< ∞.

For bounded time intervals, the local theory allows us to remove the assumption (4.1) to

16

obtain

17

Corollary 4.3. Let u

0

∈ Σ and u the corresponding global solution to (1.1). Let 1 ≤ p <

∞, 2 ≤ q < ∞ and 0 < T < S < ∞. Then

kuk

Lp([T ,S];Lq)

≤ C, where C > 0 depends only on p, q, T, S, ku

0

k

Σ

.

18

(11)

Another important consequence that will be used to obtain global bounds asserts that one

1

can decompose any time interval (T, ∞) with T > 0 into a finite number of intervals on which

2

the L

pt

(L

qx

) norm is sufficiently small for every (p, q) satisfying (4.1). More precisely, we have

3

Corollary 4.4. Let u

0

∈ Σ and u the corresponding global solution to (1.1). Let 1 ≤ p <

4

∞, 2 ≤ q < ∞, ε > 0 and T > 0. Assume that the condition (4.1) is fulfilled. Then there

5

exists L ≥ 1 not depending on u and time intervals I

1

, I

2

, · · · , I

L

such that

L

[

`=1

I

`

= [T, ∞)

6

and

7

kuk

Lp(I`;Lq)

≤ ε, ∀ ` = 1, 2, · · · , L. (4.2) Proof. From Corollary 4.2, one can choose S > T sufficiently large (not depending on u) such that kuk

Lp([S,∞);Lq)

≤ ε. Define

T

`

= T + ` S − T

m , ` = 0, 1, · · · , m,

where m ≥ 1 to be chosen later. Using H¨ older’s inequality in time, we obtain that

8

kuk

Lp(T`,T`+1];Lq)

S − T m

2p1

kuk

L2p([T,S];Lq)

.

S − T m

2p1

≤ ε,

for m ≥ 1 sufficiently large and for all ` = 0, 1, · · · , m− 1. This finishes the proof of Corollary

9

4.4.

10

5. Global bounds 1

11

In this section, we give the proof of the first global bound in (1.5). For a time slab I ⊂ R , we define S

1

(I ) via

kuk

S1(I)

= kuk

L(I,H1)

+ kuk

L4(I,W1,4)

. By the Strichartz estimates, we have

12

kuk

S1(I)

. ku(T )k

H1

+ ki∂

t

u + ∆uk

L

2 1+2δ(I,W1,

1

1−δ)

, (5.1)

for any 0 < δ < 1/2 and T ∈ I . Note that

2

1+2σ

,

1−δ1

is the conjugate pair of the Schr¨ odinger

13

admissible pair

2

1−2σ

,

σ1

.

14

Theorem 5.1. Let u

0

∈ Σ be such that H(u

0

) <

(1+b)(2−b)2

. Then the corresponding global

15

solution u to (1.1) satisfies u ∈ S

1

( R ).

16

(12)

Proof. It suffices to estimate the nonlinear term in some dual Strichartz norm as in (5.1). We

1

have

2

|∇N (x, u)| . |x|

−b

|∇u||u|

2

e

α|u|2

− 1

+ |x|

−b−1

|u|

e

α|u|2

− 1 − α|u|

2

:= A + B.

Let 0 < δ <

12

to be chosen adequately, and let I be a time slab. Let us first estimate the

3

norm kAk

L1+2δ2 (I,L1−δ1 )

. By H¨ older’s inequality,

4

kAk

L1−δ1

. k∇uk

L1−δ2

kuk

2

L4δ

e

α|u|2

− 1

|x|

b

L

2 1−2δ

.

The term

eα|u|2−1

|x|b

L

2 1−2δ

can be estimated using Lorentz spaces. Indeed, by (A.1), we get

e

α|u|2

− 1

|x|

b

L1−2δ2

. ke

α|u|2

− 1k

1−θL1

ke

α|u|2

− 1k

θL

k|x|

−b

k

L2b,∞

. ke

α|u|2

− 1k

θL

,

where θ := δ +

1+b2

. Note that we can choose 0 < δ <

1−b2

so that θ ∈ (0, 1). Here we have used the Moser-Trudinger inequality (2.1) to obtain that ke

α|u|2

− 1k

L1

. 1 since k∇uk

2L2

< H(u

0

) <

(1+b)(2−b)2

< 1. Hence

kAk

L1+2δ2 (I,L1−δ1 )

. k∇uk

L1−δ2

kuk

2

L4δ

ke

α|u|2

− 1k

θL

L1+2δ2 (I)

+ k∇uk

L

2 1−δ

kuk

2

L4δ

ke

α|u|2

− 1k

θL

L

2 1+2δ(J)

,

where I = {t ∈ I/ ku(t)k

L

≤ 1} and J = {t ∈ I/ ku(t)k

L

≥ 1}. The first term in the right hand side can be easily estimated as follows

k∇uk

L

2 1−δ

kuk

2

L4δ

ke

α|u|2

− 1k

θL

L

2

1+2δ(I)

. k∇uk

L2δ(I,L

2 1−δ)

kuk

2

L1+δ4 (I,L4δ)

. kuk

S1(I)

kuk

2

L1+δ4 (I,L4δ)

, (5.2) where the following interpolation inequality is used

k∇uk

L2δ(I,L

2

1−δ)

≤ k∇uk

1−2δL(I,L2)

k∇uk

L4(I,L4)

. (5.3) Let us turn to the second term. For t ∈ J, we obtain using (2.11) with β =

12

δ2

that

ke

α|u|2

− 1k

θL

. 1 +

kuk

C12δ2

kuk

µ

!

αθλkuk2µ

,

for some 0 < µ < 1 and λ >

π(1−δ)1

to be chosen later. Since kuk

2µ

= k∇uk

2L2

+ µ

2

kuk

2L2

<

5

H(u

0

) + µ

2

M(u

0

) =: K

2

(µ), we bound

6

ke

α|u|2

− 1k

θL

. 1 +

kuk

C12δ2

K(µ)

!

αθλK2(µ)

.

(13)

Since K

2

(µ) → H(u

0

) <

(1+b)(2−b)2

as µ → 0, we can choose 0 < µ < 1 sufficiently small so

1

that K

2

(µ) <

(1+b)(2−b)2

. Moreover, as

θK1−δ2(µ)

1+b2

K

2

(µ) <

2−b1

as δ → 0, we choose 0 <

2

δ <

1−b2

sufficiently small such that

θK1−δ2(µ)

<

2−b1

. At final, we choose

π(1−δ)1

< λ <

αθK22(µ) 3

so that αθλK

2

(µ) < 2. It follows that

4

ke

α|u|2

− 1k

θL

. (1 + kuk

C12δ2

)

2

. kuk

2

C12δ 2

, where we have used the fact that ku(t)k

C12δ2

≥ ku(t)k

L

≥ 1 for all t ∈ J. Therefore,

k∇uk

L

2 1−δ

kuk

2

L4δ

ke

α|u|2

− 1k

θL

L

2 1+2δ(J)

.

k∇uk

L

2 1−δ

kuk

2

L4δ

kuk

2

C12δ2

L

2 1+2δ(I)

. k∇uk

L2δ(I,L

2 1−δ)

kuk

2

L2δ(I,L4δ)

kuk

2

L1−δ4 (I,C12δ2)

. kuk

2

L2δ(I,L4δ)

kuk

3S1(I)

. (5.4) The last estimate follows from (5.3) and the fact

kuk

L

4

1−δ(I,C12δ

2)

. kuk

L

4 1−δ(I,W1,

4 1+δ)

. kuk

δL(I,H1)

kuk

1−δL4(I,W1,4)

. kuk

S1(I)

.

Combining inequalities (5.2) and (5.4), we end up with

5

kAk

L

2 1+2δ(I,L

1

1−δ)

. kuk

S1(I)

kuk

2

L

4 1+δ(I,L4δ)

+ kuk

2

L2δ(I,L4δ)

kuk

3S1(I)

. (5.5) Let us now estimate the term kBk

L1+2δ2 (I,L1−δ1 )

. Taking

1−δ1

< p, q < ∞ such that

1p

+

1q

= 1−δ

6

and applying H¨ older’s inequality, we get

7

kBk

L

1 1−δ

.

|x|

−b−1

|u|

e

α|u|2

− 1

L1−δ1

. kuk

Lq

e

α|u|2

− 1

|x|

b+1

Lp

.

Clearly,

8

e

α|u|2

− 1

|x|

b+1

p

Lp

. e

α(p−1)kuk2L

Z e

α|u|2

− 1

|x|

p(b+1)

dx.

Since

1−δ1

<

b+12

for 0 < δ <

1−b2

, we choose

1−δ1

< p <

b+12

. Hence we can apply the singular Moser-Trudinger inequality for the term

Z

eα|u|2−1

|x|p(b+1)

dx to obtain kBk

L

2 1+2δ(I,L

1 1−δ)

.

kuk

Lq

e

α

p−1

p kuk2L

L

2 1+2δ(I)

+

kuk

Lq

e

α

p−1

p kuk2L

L

2 1+2δ(J)

. Note that the choice of p leads to q >

1−b−2δ2

. Therefore,

9

kBk

L

2 1+2δ(I,L

1

1−δ)

. kuk

L

2

1+2δ(I,Lq)

+ kuk

Lγ(I,Lq)

e

α

p−1

p kuk2L

Lρ(J)

,

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