Introduction to scattering for radial 3D NLKG below energy norm

J. Differential Equations 248 (2010) 893–923

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Journal of Differential Equations www.elsevier.com/locate/jde

Introduction to scattering for radial 3D NLKG below energy norm Tristan Roy Department of Mathematics, Box 951555, University of California, Los Angeles, CA 90095-1555, USA

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 27 September 2008 Revised 10 August 2009 Available online 1 December 2009 Keywords: Global existence Scattering Almost Morawetz–Strauss estimates Strichartz estimates

We prove scattering for some radial 3D semilinear Klein–Gordon equations with rough data. First we prove Strichartz-type estimates in mixed norm spaces. Then by using these decays we establish some local bounds. By combining these results to a Morawetz-type estimate and a radial Sobolev inequality we control the variation of an almost conserved quantity on arbitrary large intervals. Once we have showed that this quantity is controlled, we prove that some of these local bounds can be upgraded to global bounds. This is enough to establish scattering. All the estimates involved require a delicate analysis due to the nature of the nonlinearity and the lack of scaling. © 2009 Elsevier Inc. All rights reserved.

1. Introduction In this paper we consider the p-defocusing Klein–Gordon equation on R3

∂tt u − u + u = −|u | p −1 u

(1)

with data u (0) = u 0 , ∂t u (0) = u 1 lying in H s , H s−1 respectively. Here H s is the standard inhomogeneous Sobolev space, i.e. H s is the completion of the Schwartz space S (R3 ) with respect to the norm

   f  H s :=  D s f L 2 (R3 ) E-mail address: [email protected]. 0022-0396/$ – see front matter doi:10.1016/j.jde.2009.11.002

© 2009 Elsevier Inc.

All rights reserved.

(2)

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T. Roy / J. Differential Equations 248 (2010) 893–923

where  D  is the operator defined by

 s  D s f (ξ ) := 1 + |ξ | ˆf (ξ )

(3)

and ˆf denotes the Fourier transform

ˆf (ξ ) :=



f (x)e −ix·ξ dx

(4)

R3

We are interested in the strong solutions of the p-defocusing Klein–Gordon equation on some interval [0, T ], i.e. maps u, ∂t u that lie in C ([0, T ], H s (R3 )), C ([0, T ], H s−1 (R3 )) respectively and that satisfy



 sin(t  D ) u (t ) = cos t  D  u 0 + u1 − D

t

sin((t − t  ) D ) 

D

0

    |u | p −1 t  u t  dt 

(5)

The p-defocusing Klein–Gordon equation is closely related to the p-defocusing wave equation, i.e.

∂tt v −  v = −| v | p −1 v

(6)

with data v (0) = v 0 , ∂t v (0) = v 1 . (6) enjoys the following scaling property

v (t , x) →

2

1

1

1

2 . p −1

2

p −1 +1

,

x



λ λ  

λ p −1

λ −

t

u0

2

v 1 (x) → 3 2

u

λ p −1

v 0 (x) →

We define the critical exponent sc :=



1

u1

x

λ   x

(7)

λ

˙ sc × H˙ sc −1 norm of (u 0 , u 1 ) One can check that the H

is invariant under the transformation (7). (6) was demonstrated to be locally well-posed by Lindblad 2 , p > 3 by using an iterative argument. In fact their results and Sogge [7] in H s × H s−1 , s > 32 − p − 1

extend immediately to (1).2 ˙ 1 critical. If 3 < p < 5 If p = 5 then sc = 1 and this is why we say that the nonlinearity |u | p −1 u is H ˙ 1 subcritical. then sc < 1 and the regime is H It is well known that smooth solutions to (1) have a conserved energy





E u (t ) :=

1



2

  ∂t u (t , x)2 dx + 1 2

R3

+

1 2



1 p+1



  ∇ u (t , x)2 dx + 1



2

R3

  u (t , x)2 dx

R3

  u (t , x) p +1 dx

R3

˙ m denotes the standard homogeneous Sobolev space endowed with the norm  f  ˙ m :=  D m f  L 2 (R3 ) . Here H H By rewriting for example (1) in the “wave” form ∂tt u − u = −|u | p −1 u − u.

(8)

T. Roy / J. Differential Equations 248 (2010) 893–923

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In fact by standard limit arguments the energy conservation law remains true for solutions (u , ∂t u ) ∈ H s × H s−1 , s  1. Since the lifespan of the local solution depends only on the H s × H s−1 norm of the initial data (u 0 , u 1 ) (see [7]) then it suffices to find an a priori pointwise in time bound in H s × H s−1 of the solution (u , ∂t u ) to establish global well-posedness. The energy captures the evolution in time of the H 1 × L 2 norm of the solution. Since it is conserved we have global existence of (1). The scattering theory (namely, the existence of the bijective wave operators) in the energy space3 for (1) has been extensively studied for a large range of exponents p. In particular Brenner [1,2] was able to prove that if 73 < p < 5, then every solution scatters as T goes to infinity. In fact he showed

˙ 1 subcritical and L 2 supercritical,4 scattering for all dimension n, n  3 and for all exponent p that is H 4 i.e. 1 + n4 < p < 1 + n− . Later Nakanishi [10,11] was able to extend these results to n = 1 and 2. 2 In this paper we are interested in proving scattering results for data below the energy norm, i.e. for s < 1. We will assume that (1) has radial data. The main result of this paper is the following one.

Theorem 1. The p-radial defocusing Klein–Gordon equation on R3 is globally well-posed in H s × H s−1 , 1 > s > s( p ) and there exists a scattering state (u +,0 , u +,1 ) ∈ H s × H s−1 such that







lim  u ( T ), ∂t u ( T ) − K (u +,0 , u +,1 ) H s × H s−1 = 0

T →∞

(9)

with

 K (t ) :=

sin (t  D ) D

cos (t  D ) − D  sin t  D 

cos (t  D )

 (10)

3 < p < 5 and

s p :=

⎧ ⎨1 −

(5− p )( p −3) , 2( p −1)( p −2)

3
⎩1 −

(5 − p ) , 2( p −1)(6− p )

4 p<5

2

(11)

Throughout the paper ∇ denotes the gradient operator. Let sc , θ1 , . . . , θ3 denote the following numbers

sc :=

3 2

−

2

(12)

p−1

(2s−1)(4− p )

, 3
s( p −1)( p −2) (4s−1)( p −4) , s( p −1)(6− p )

θ1 :=

( p +2)( p −3)

( p −1)( p −2) , ( p +2)(5− p ) (6− p )( p −1) ,

θ2 :=

4 p<5 3< p4 4 p<5

(13)

(14)

and

θ3 :=

3 4

I.e. with data (u 0 , u 1 ) ∈ H 1 × L 2 . Since if p > 1 + n4 then sc > 0.

4− p , s( p −1)( p −2) p −4 , s( p −1)(6− p )

3< p4 4 p<5

(15)

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T. Roy / J. Differential Equations 248 (2010) 893–923

We write F ( v ) for the following function

F ( v ) := | v | p −1 v

(16)

I f (ξ ) := m(ξ ) ˆf (ξ )

(17)

Let I be the following multiplier

ξ

where m(ξ ) := η( N ),

η is a smooth, radial, nonincreasing in |ξ | such that 

|ξ |  1

1,

η(ξ ) := ( 1 )1−s , |ξ |  2 |ξ |

(18)

and N  1 is a dyadic number playing the role of a parameter to be chosen. We shall abuse the notation and write m(|ξ |) for m(ξ ), thus for instance m( N ) = 1. Some estimates that we establish throughout the paper require a Paley–Littlewood decomposition. We set it up now. Let φ(ξ ) be a real, radial, nonincreasing function that is equal to 1 on the unit ball {ξ ∈ R3 : |ξ |  1} and that that is supported on {ξ ∈ R3 : |ξ |  2}. Let ψ denote the function

ψ(ξ ) := φ(ξ ) − φ(2ξ )

(19)

If ( M , M 1 , M 2 ) ∈ 2Z are dyadic numbers such that M 2 > M 1 we define the Paley–Littlewood operators in the Fourier domain by

P  M f (ξ ) := φ



 P M f (ξ ) := ψ



ξ



M

ξ

ˆf (ξ )



M

ˆf (ξ )

ˆ  P > M f (ξ ) := f (ξ ) − P  M f (ξ )  P M f (ξ ) := P  M f (ξ ) 128  P  M f (ξ ) := P > M f (ξ ) 128

P M 1 <. M 2 f := P  M 2 f − P < M 1 f Since



M ∈2 Z

(20)

ψ( Mξ ) = 1 we have f =



PM f

(21)

M ∈2Z

Notice also that

f = P M f + P M f It T is a multiplier with nonnegative symbol m then T 1 

2 f (ξ ) = ψ 2 ( M ) ˆf (ξ ). instance P M 1

ξ

1 2

(22) 1

denotes then multiplier with symbol m 2 . For

T. Roy / J. Differential Equations 248 (2010) 893–923

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Throughout this paper we constantly use Strichartz-type estimates. Notice that some Strichartz estimates for the Klein–Gordon equation already exist in Besov spaces [5]. Here we have chosen to q work in the L t L rx spaces in order to avoid too many technicalities. The following proposition is proved in Section 7. q

Proposition 2 (Strichartz estimates for Klein–Gordon equations in L t L rx spaces). Assume that u satisfies the following Klein–Gordon equation on Rd , d  3

⎧ ⎨ ∂tt u − u + u = Q u (0, x) = u 0 (x) ⎩ ∂t u (0, x) = u 1 (x).

(23)

Let T  0. Then

  u Lq ([0,T ]) Lrx + ∂t  D −1 u Lq ([0,T ]) Lq + u Lt∞ ([0,T ], H m ) + ∂t u Lt∞ ([0,T ], H m−1 ) t

x

t

 u 0  H m + u 1  H m−1 +  Q Lq˜ ([0,T ]) Lr˜

(24)

x

t

under the following assumptions

• (q, r ) is m-wave admissible, i.e. (q, r ) lies in the set W of wave-admissible points

  1 d−1 d−1 W := (q, r ): (q, r ) ∈ (2, ∞] × [2, ∞), +  q

2r

4

(25)

it obeys the following constraint

1 q

+

d r

=

d 2

−m

(26)

and

  2(d − 1) (q, r ) = 2, d−3

(27)

  1 1 1 1  := (˜q, r˜): + = 1, + = 1 W q˜ q r˜ r

(28)

 of W , i.e. • (˜q, r˜) lies in the dual set W

and it satisfies the following inequality

1 q˜

+

d r˜

−2=

1 q

+

d r

(29)

Remark 3. Notice that the constraints that (q, r , q˜ , r˜ ) must satisfy are essentially the same to those in the Strichartz estimates for the wave equation [7]. These similarities are not that surprising. Indeed the relevant operator is e it  D  , e it D for the Klein–Gordon, wave equations respectively.5 They are similar to each other on high frequencies.

5

f (ξ ) := |ξ | ˆf (ξ ). With D multiplier defined by D

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T. Roy / J. Differential Equations 248 (2010) 893–923

Now we explain the main ideas of this paper. Our first objective is to establish global well-posedness of (1) for data in H s × H s−1 , 1 > s > s( p ). Unfortunately since the solution lies in H s × H s−1 pointwise in time the energy (8) is infinite. Therefore we introduce the following mollified energy





E Iu (t ) :=

1



2

  ∂t Iu (t , x)2 dx + 1



2

R3

+

1 p+1



   D Iu (t , x)2 dx + 1



2

R3

   Iu (t , x)2 dx

R3

   Iu (t , x) p +1 dx

(30)

R3

This is the I -method originally designed by J. Colliander, M. Keel, G. Staffilani, H. Takaoka and T. Tao [4] to study global existence for rough solutions of semilinear Schrödinger equations. Since the multiplier gets closer to the identity operator as the parameter N goes to infinity6 we expect the variation of the smoothed energy to approach zero as N grows. However it is not equal to zero and it needs to be controlled on an arbitrary large interval. The semilinear Schrödinger and Wave equations have a scaling property. In [4,16] the authors were able after scaling to make the mollified energy at time zero smaller than one. Then by using the Strichartz estimates they locally bounded some numbers that allowed them to find an upper bound of its local variation. Iterating the process they managed to yield an upper bound7 of its total variation. Choosing appropriately the parameter N they bounded it by a constant. Unfortunately the p-defocusing Klein–Gordon equation does not have any scaling symmetry. We need to control the variation of (30) by a fixed quantity. A natural choice is a constant C > 1 multiplied by the mollified energy E ( Iu 0 ) := E ( Iu (0)) at time zero. It occurs that this is possible if E ( Iu 0 ) is bounded by a constant depending on the parameter N: see (48) and (49). But Proposition 4 shows that E ( Iu 0 ) is bounded by a power of N. Therefore we can choose N to control the mollified energy as long as s > s( p ). Since the pointwise in time H s × H s−1 norm of the solution is bounded by the mollified energy (see (59)) we have global well-posedness. Now we are interested in proving asymptotic completeness by using the I -method. Notice that this method has already been used in [16] to prove scattering below the energy norm for semilinear Schrödinger equations with a power type nonlinearity. We would like to establish (9). Notice first that if this result is true then it implies that the pointwise in time H s × H s−1 bound of the solution is bounded by a function that does not depend on time. Therefore in view of the previous paragraph, the variation of the smoothed energy should not depend on time T . To this end we use some tools. Recall that this variation is estimated by using local bounds of some quantities, namely some Z m,s p +2 p +2

(see Proposition 5). We divide the whole interval [0, T ] into subintervals where the L t L x of Iu is small and we control these numbers on them by the Strichartz estimates and a continuity argument. Notice that in this process we are not allowed to create powers of time T 8 since it will eventually p +2 p +2 norm of the solution force us to choose N as a function of T . We also need to control the L t L x on [0, T ]. Morawetz and Strauss [8,9] proved a weighted long time estimate (see (118)) depending on the energy. Combining this result to a radial Sobolev inequality (see (53))9 we can control the p +2 p +2 Lt L x norm of u by some power of the energy. Of course since the solution lies in H s × H s−1 , s < 1 we cannot use this inequality as such. Instead we prove an almost Morawetz–Strauss estimate (see Proposition 9 and Proposition 8) by substituting u for Iu in the establishment of (118). This approach was already used in [15]. Notice here that the upper bound of (54) does not depend on T either. The almost conservation law (see Proposition 6) is proved in Section 3 by performing a low– high frequency decomposition and using the smoothness of F 10 when we estimate the low frequency

6 7 8 9 10

Formally speaking. Depending on N, the time and the initial data. By using Hölder locally in time. This is the only place where we rely crucially on the assumption of spherical symmetry. Namely F is C 1 if p > 3.

T. Roy / J. Differential Equations 248 (2010) 893–923

899

part of the variation. Combining all these tools we are able to iterate and globally bound the mollified p +2 p +2 energy and the L t L x norm of u by a function of N and the data. These global results allow us to update a local control of the Z m,s to a global one. It occurs that scattering holds if some integrals are finite. By using the global control of the Z m,s in the Cauchy criterion we prove these facts. This is enough to establish scattering. 2. Proof of Theorem 1 In this section we prove Theorem 1 assuming that the following propositions are true. Proposition 4 (Mollified energy at time 0 is bounded by N 2(1−s) ). Assume that sc < s < 1. Then



p +1 

E ( Iu 0 )  N 2(1−s) u 0 2H s + u 1 2H s−1 + u 0  H s

(31)

Proposition 5 (Local Boundedness). Assume that u satisfies (1). Let M = [0, s] ∪ {1−}. There exists N = N (u 0  H s , u 1  H s−1 )  1 such that if J , time interval, satisfies





sup E Iu (t )  3E ( Iu 0 )

(32)

t∈ J

and

 Iu L p+2 ( J ) L p+2  x

t

1 N + ( E ( Iu 0 ))

(33)

1−θ2 2θ2

then 1

Z ( J , u )  E 2 ( Iu 0 )

(34)

Z ( J , v ) := sup Z m,s ( J , v )

(35)

where, given a function v,

m∈M

and

Z m,s ( J , v ) :=

sup

  ∂t  D −m I v 

(q,r )−m wave adm

q

L t ( J ) L rx

  +  D 1−m I v Lq ( J ) Lr t

x

(36)

We recall that, if a and M are two real number, then M a+ := M a+α and M a− := M a−α for α 1. Proposition 6 (Almost Conservation Law). Assume that u satisfies (1). Let J = [a, b] be a time interval. Let 3p −5 3  p < 5 and s  2p . Then

     Z p +1 ( J , u )  sup E Iu (t ) − E Iu (a)   5− p t∈ J

N

2

−

(37)

Remark 7. Notice that if p = 3 then the upper bound is O ( N11− ) modulo Z p +1 ( J , u ). This result has already been established in [15] for a slighly different problem, i.e. the defocusing cubic wave equation by using a multilinear analysis.

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T. Roy / J. Differential Equations 248 (2010) 893–923

Proposition 8 (Estimate of integrals). Let J be a time interval. Let v be a function. Then for i = 1, 2 we have p +1   ( J , v)  R i ( J , v )  Z

N

(38)

5− p 2 −

with

  R 1 ( J , v ) := J R3

 ∇ I v (t , x).x  F ( I v ) − I F ( v ) dx dt |x|

(39)

and

  R 2 ( J , v ) :=

I v (t , x) 

|x|

R3

J



F ( I v ) − I F ( v ) dx dt

(40)

Proposition 9 (Almost Morawetz–Strauss Estimate). Let u be a solution of (1) and let T  0. Then

T  0 R3

      | Iu (t , x)| p +1 dx dt  sup E Iu (t ) + R 1 [0, T ], u + R 2 [0, T ], u |x| t ∈[0, T ]

(41)

These propositions will be proved in the next sections. The proof of Theorem 1 is made of four steps

• Boundedness of the mollified energy and the quantity  Iu  L p+2 L p+2 . x

t

p +2 p +2 Lx

We will prove that we can control the mollified energy E ( Iu ) and the L t arbitrary large intervals [0, T ], T  0. More precisely let

F T :=

⎧ ⎨ ⎩

norm of Iu on

⎫

supt ∈[0, T  ] E ( Iu (t ))  2E ( Iu 0 ) ⎬

T  ∈ [0, T ]:

 Iu 

p +2

p +2

Lt

3

p +2

([0, T  ]) L x

 C E 2 ( Iu 0 ) ⎭

(42)

We claim that F T = [0, T ] for some universal constant C  0 and N = N (u 0  H s , u 1  H s−1 )  1 to be chosen later. Indeed

◦ F T = ∅ since 0 ∈ F T ◦ F T is closed by continuity ◦ F T is open. Let T ∈ F T . By continuity there exists δ > 0 such that for all T  ∈ ( T − δ, T + δ) ∩ [0, T ] we have





sup E Iu (t )  3E ( Iu 0 )

t ∈[0, T  ]

(43)

and

 Iu 

p +2

p +2

Lt

3

p +2

([0, T  ]) L x

 2C E 2 ( Iu 0 )

(44)

T. Roy / J. Differential Equations 248 (2010) 893–923

901

Let P = ( J j )1 j l be a partition of [0, T  ] such that  Iu  L p+2 ( J ) L p+2 = j x t 1, . . . , l − 1 and  Iu  L p+2 ( J ) L p+2  t

l

x

1−θ2 2θ 2

lE

1−θ2 2θ 2

for all j = ( Iu 0 )

with N + defined in Proposition 5. Then by (44)

1

N+ E

1

N+ E

( Iu 0 ) ( p +2)(1−θ2 ) 3 +2 2θ2

( Iu 0 ) N +

(45)

By Proposition 5 and 6 we get after iteration



 E sup E Iu (t ) − E ( Iu 0 ) 

( p +2)(1−θ2 ) 3 p +1 +2+ 2 2θ2

t ∈[0, T ]

N

( Iu 0 )

(46)

5− p 2 −

We recall that, if A and B are two real numbers such that A  B, then the constant determined by  in A  B is the smallest constant among the K such that A  K B. Let C 1 be the constant determined by  in (46). If we can choose N  1 such that

C1

E

( p +2)(1−θ2 ) 3 p +1 +2+ 2 2θ2

N

( Iu 0 )

5− p 2 −

 E ( Iu 0 )

(47)

then supt ∈[0, T  ] E ( Iu (t ))  2E ( Iu 0 ). The constraint (47) is equivalent to (5− p )( p −3)

E ( Iu 0 ) 

N ( p−1)( p−2)

−

(48)

2( p −3) ( p −1)( p −2)

C1 if 3 < p  4 and

(5− p )2

E ( Iu 0 ) 

N (6− p)( p−1)

−

(49)

2(5− p ) (6− p )( p −1)

C1

if 4  p < 5 after plugging (14) into (47). By Proposition 4 it suffices to prove that there exists N = N (u 0  H s , u 1  H s−1 )  1 such that



p +1 



p +1 

N 2(1−s) max u 0  H s , u 1  H s−1 , u 0  H s

(5− p )( p −3)

−

(5− p )2

−

 N ( p−1)( p−2)

(50)

in order to satisfy (48) and

N 2(1−s) max u 0  H s , u 1  H s−1 , u 0  H s

 N (6− p)( p−1)

(51)

in order to satisfy (49). Such a choice is possible if and only if s > s( p ). By Proposition 9, Proposition 8 and (43) we get

T   0 R3

| Iu (t , x)| p +1 E dx dt  E ( Iu 0 ) + |x|

( p +2)(1−θ2 ) 3 p +1 +2+ 2 2θ2

N

5− p − 2

( Iu 0 )

 E ( Iu 0 )

(52)

902

T. Roy / J. Differential Equations 248 (2010) 893–923

Combining (52) to the well-known pointwise radial Sobolev inequality

   Iu (t , x)   Iu (t , .) H 1 |x|

(53)

we have

 Iu 

p +2

3

p +2

Lt

p +2

([0, T  ]) L x

 E 2 ( Iu 0 )

(54)

and we assign to C the constant determined by  in (54).

• Global existence We have just proved that





sup E Iu (t )  2E ( Iu 0 )

t ∈[0, T ]

(55)

and

 Iu 

p +2

p +2

Lt

3

p +2

([0, T ])x

 C E 2 ( Iu 0 )

(56)

for some well-chosen N = N (u 0  H s , u 1  H s−1 )  1 and 1 > s > s( p ). Therefore by Proposition 4





sup E Iu (t ) u 1  s−1 ,u 0  H s 1 H

t ∈[0, T ]

(57)

and

 Iu 

p +2

p +2

Lt

p +2

([0, T ]) L x

u 1  H s−1 ,u 0  H s 1

(58)

Here A u 1  H s−1 ,u 0  H s B means that there exists a constant K := K (u 0  H s , u 1  H s−1 ) such that A  K B. Now by Plancherel and (57)

   u ( T ), ∂t u ( T ) 

H s × H s −1

   E Iu ( T ) u 1  H s−1 ,u 0  H s 1

(59)

This proves global well-posedness of (1) with data (u 0 , u 1 ) ∈ H s × H s−1 , 1 > s > s( p ). Moreover by continuity we have





sup E Iu (t ) u 1  t ∈R

H s −1

,u 0  H s 1

(60)

and

 Iu 

p +2

p +2

Lt

p +2

(R) L x

u 1  H s−1 ,u 0  H s 1

(61)

T. Roy / J. Differential Equations 248 (2010) 893–923

903

• Global estimates J j = [a j , b j ])1 j ˜l be a partition of [0, ∞) such that Let P := (

 Iu L p+2 ( J t

p +2 j )L x



1 N + ( E ( Iu 0 ))

(62)

1−θ2 2θ2

with N + defined in Proposition 5. Notice that from Proposition 4 and (61) the number of interval ˜l satisfies

˜l  E

( p +2)(1−θ2 ) 2θ2

( Iu 0 )

u 1  H s−1 ,u 0  H s 1

(63)

Moreover by slightly modifying the steps between (89) and (95) and by (60) we have 1





θ ( p −1)+1

Z s,s ( J j , u )  E 2 Iu (a j ) + C 1 Z s,3s

θ ( p −1)+1

1 2

 E ( Iu 0 ) + C 1 Z s,3s

θ ( p −1)+1

( J j , u ) + C 2 Z s,s

θ ( p −1)+1

( J j , u ) + C 2 Z s,s

( J j , u)

( J j , u)

(64)

with C 1 , C 2 , and θ defined in (96), (97) and (93) respectively. Even if it means increasing the value of N = N (u 0  H s , u 1  H s−1 )  1 in (50) and (51) we can assume that (98) and (99) hold. Therefore by Lemma 10 and Proposition 4 we have 1

Z s,s ( J j , u )  E 2 ( Iu 0 )

u 1  H s−1 ,u 0  H s 1

(65)

By (65) and (63) we have

Z s,s (R, u ) u 1 

H s −1

,u 0  H s 1

(66)

• Scattering Let

 v (t ) :=

 v 0 :=

u (t ) ∂t u (t ) u0 u1

 (67)



(68)

and

 t unl (t ) =

t 0

0

sin ((t −t  ) D ) (|u | p −1 (t  )u (t  )) dt  D

cos ((t − t  ) D )(|u | p −1 (t  )u (t  )) dt 

 (69)

Then we get from (5)

v(t ) = K(t )v0 − unl (t )

(70)

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T. Roy / J. Differential Equations 248 (2010) 893–923

Recall that the solution u scatters in H s × H s−1 if there exists

 v+,0 :=

u 0,+ u 1,+

 (71)

such that

  v(t ) − K(t )v+,0 

(72)

H s × H s −1

has a limit as t → ∞ and the limit is equal to 0. In other words since K(t ) is bounded on H s × H s−1 it suffices to prove that the quantity

 −1  K (t )v(t ) − v+,0 

(73)

H s × H s −1

has a limit as t → ∞ and the limit is equal to 0. A computation shows that

K−1 (t ) =



cos (t  D )

− sin(Dt D )

 (74)

 D  sin (t  D ) cos (t  D )

But

K−1 (t )v(t ) = v0 − K−1 (t )unl (t )

(75)

By Proposition 2

 −1  K (t 1 )unl (t 1 ) − K−1 (t 2 )unl (t 2 )

H s × H s −1

   unl (t 1 ) − unl (t 2 ) H s × H s−1    |u | p −1 u  2 2 L t1+s ([t 1 ,t 2 ]) L x2−s

     D 1−s I |u | p −1 u 

2

2

L t1+s ([t 1 ,t 2 ]) L x2−s

(76)

If we let J := [t 1 , t 2 ] in (89) and follow the same steps up to (95) we get from (55)

 1−s  p −1   D  I | u | u 

2 L t1+s

2 ([t 1 ,t 2 ]) L x2−s

θ ( p −1)+1 

 C 1 Z s,3s

[t 1 , t 2 ], u

θ ( p −1)+1 

+ C 2 Z s,s



[t 1 , t 2 ], u



(77)

By (66), (76) and (77)





lim K−1 (t 1 )unl (t 1 ) − K−1 (t 2 )unl (t 2 ) H s × H s−1 = 0

t 1 →∞

(78)

uniformly in t 2 . This proves that K−1 (t ) v (t ) has a limit in H s × H s−1 as t goes to infinity. Moreover





lim v(t ) − K(t )v+,0 ( H s , H s−1 ) = 0

t →∞

with v+,0 defined in (71)

(79)

T. Roy / J. Differential Equations 248 (2010) 893–923

∞

sin (t   D ) 

u +,0 := u 0 +

    |u | p −1 t  u t  dt 

(80)

      cos t   D  |u | p −1 t  u t  dt 

(81)

D

0

905

and

∞ u +,1 := u 1 − 0

3. Proof of “Mollified energy at time 0 is bounded by N 2(1−s) ” In this section we aim at proving Proposition 4. By Plancherel we have

 |ξ |2N

N



  u 1 (ξ )2 dξ +

 Iu 1 2L 2 

N 2 (1 − s ) 

|ξ |2N

2 (1 − s )

|ξ |2(1−s)

 u 1 (ξ )2 dξ

u 1 2H s−1

(82)

Similarly



 2 |ξ |2 u 0 (t , ξ ) dξ +

∇ Iu 0 2L 2  |ξ |2N

N

 |ξ |2

|ξ |2N

2 (1 − s )

N 2 (1 − s ) 

|ξ |2(1−s)

u 0 2H s

 u 0 (ξ )2 dξ (83)

Moreover by the assumption s > sc p +1

p +1

p +1

 u 0  L p +1   P N u 0  L p +1 +  P  N u 0  L p +1 N

1) ( p +1)( 32(( pp − +1) −s)

p +1

u 0  H s

p +1

 N 2 (1 − s )  u 0  H s

(84)

4. Proof of “Local Boundedness” Before attacking the proof of Proposition 5 let us prove a short lemma. Lemma 10. Let x(t ) be a nonnegative continuous function of time t such that x(0) = 0. Let X be a positive constant and let αi , C i , i ∈ {1, . . . , m} be nonnegative constants such that

C i X α i −1 1

(85)

and

x(t )  X +

m  i =1

C i xαi (t )

(86)

906

T. Roy / J. Differential Equations 248 (2010) 893–923

Then

x(t )  X

If we let x(t ) :=

x(t ) X

(87)

then we have m 

x(t )  1 +

C i X αi −1 xαi (t )

(88)

i =1

and x(0) = 0. Applying a continuity argument to x we have x(t )  1. This implies (87). Plugging  D 1−m I into (24) we have







Z m,s ( J , u )  E 2 ( Iu 0 ) +  D 1−m I |u | p −1 u  1

2

(89)

2

L t1+m ( J ) L x2−m

There are three cases

• m = s. By (89), the fractional Leibnitz rule and Hölder inequality





Z s,s ( J , u )  E 2 ( Iu 0 ) +  D 1−s Iu  1

2 L ts

1

 E 2 ( Iu 0 ) + Z s,s ( J , u )u 

2 ( J ) L x1−s

 p −1  | u | 

p −1

2( p −1)

Lt

2( p −1)

( J )L x

 1 p −1  E 2 ( Iu 0 ) + Z s,s ( J , u )  P N u  2( p−1) Lt

We are interested in estimating  P N u 

L t2 ( J ) L 2x

p −1

2( p −1) 2( p −1) Lx

Lt

2( p −1)

( J )L x

+  P N u 



p −1

2( p −1)

Lt

(90)

2( p −1)

( J )L x

. There are two cases

◦ 3  p  4. By interpolation and (33) we have

 P N u

p −1

2( p −1)

Lt

θ3 ( p −1)

 P N u

2 L ts

2

( J ) L x1−s

θ1 ( p −1) θ ( p −1 )  P N u  2p+2 p +2 L t∞ ( J ) L 2x Lt ( J ) L x

  P N u

2( p −1)

( J )L x

 1−s θ3 ( p −1) θ1 ( p −1) θ ( p −1 )   Iu  2p+2 Iu  2 2 p +2  D  L t∞ ( J ) L 2x Lt ( J ) L x L ts ( J ) L x1−s

  Iu 

 

E

(θ1 +θ2 −1)( p −1) 2

( Iu 0 )

N+ E

(−θ3 )( p −1) 2

( Iu 0 )

N+

θ ( p −1 )

Z s,3s

θ ( p −1 )

Z s,3s

( J , u)

( J , u)

(91)

◦ p > 4. By interpolation, Sobolev inequality and (33) we have

 P N u

p −1

2( p −1) Lt (

2( p −1) J )L x

 θ ( p −1 ) θ ( p −1 )   P N u  1∞ 6  P N u  2p+2 Lt ( J ) L x

Lt

 θ ( p −1 ) θ ( p −1 )  ∇ Iu  1∞ 2  Iu  2p+2 Lt ( J ) L x

Lt

(

p +2 J )L x



θ3 ( p −1)

 P N u

2 L ts

6

( J ) L x1−s

 1−s θ3 ( p −1)   Iu  2 p +2  D 

( J )L x

2

L ts ( J ) L x1−s

T. Roy / J. Differential Equations 248 (2010) 893–923

E



(θ1 +θ2 −1)( p −1) 2

( Iu 0 )

θ ( p −1 )

Z s,3s

N+ E



(−θ3 )( p −1) 2

( Iu 0 )

θ ( p −1 )

Z s,3s

N+

907

( J , u)

( J , u)

(92)

p −1

Now we estimate  P  N u  2( p−1) 2( p −1) . Let L ( J )L x

t

1

θ :=

(93)

s ( p − 1)

By interpolation we have

 P N u 

p −1

2( p −1)

Lt

2( p −1)

( J )L x

θ ( p −1 )

  P N u 

2 L ts

(1−θ )( p −1)

2 ( J ) L x1−s

 P N u 

θ ( p −1 )

 D 1−s Iu  

2

2

L ts ( J ) L x1−s

N (1−s)θ ( p −1)

E

(1−θ )( p −1) 2

(1−θ )( p −1) L t∞ ( J ) L 2x ( p − 1 )( 1 −θ )(1−s) N + N

 D  Iu 

θ ( p −1 )

( Iu 0 )

2(s( p −1)−1) 2s−1

L t∞ ( J ) L x

Z s,s

( J , u)

(94)

N (1−s)( p −1) N +

1 since s > sc  p − . Therefore we get from (89), (91), (92) and (94) 1

θ ( p −1)+1

1

Z m,s ( J , u )  E 2 ( Iu 0 ) + C 1 Z s,3s

θ ( p −1)+1

( J , u ) + C 2 Z s,s

( J , u)

(95)

with

C 1 :=

E

−θ3 ( p −1)

( Iu 0 )

2

N+

(96)

and

C 2 :=

E

(1−θ )( p −1) 2

( Iu 0 )

N (1−s)( p −1) N +

(97)

Notice that by Proposition 4

C1 E

θ3 ( p −1) 2

( Iu 0 ) 

1 N+

1

(98)

and

C2 E

θ ( p −1) 2

( Iu 0 ) 

1 N+

1

(99)

908

T. Roy / J. Differential Equations 248 (2010) 893–923

if we choose N = N (u 0  H s , u 1  H s−1 )  1. Applying Lemma 10, we get 1

Z s,s ( J , u )  E 2 ( Iu 0 )

(100)

• m < s. Notice that by (91), (92), (94), (98), (99) and (100)

u 

p −1

2( p −1)

Lt

2( p −1)

( J )L x



1 N+

1

(101)

Moreover







Z m,s ( J , u )  E 2 ( Iu 0 ) +  D 1−m I |u | p −1 u  1

   E ( Iu 0 ) +  D 1−m Iu  1 2

2

2

L tm ( J ) L x1−m

1

 E 2 ( Iu 0 ) + Z m,s ( J , u )u 

2

L t1+m

2( p −1)

Lx

u 

p −1 Lt

2 ( J ) 2− m

p −1

2( p −1)

Lt

2( p −1)

( J )L x

(102)

2( p −1)

( J )L x

By (101) and (102) and Lemma 10, we get (34).

• m = 1− = 1 − α with α small. We have







Z m,s ( J , u )  E 2 ( Iu 0 ) +  D 1−(1−) I |u | p −1 u  L 1+ ( J ) L 2− 1

x

t

   E ( Iu 0 ) + N + |u | p −1 u L 1+ ( J ) L 2− 1 2

x

t

     E ( Iu 0 ) + N + | P N u | p −1 P N u  L 1+ ( J ) L 2− + N + | P N u | p −1 P  N u  L 1+ ( J ) L 2− x x t t     + p −1 + p −1   + N | P N u | P N u L 1+ ( J ) L 2− + N | P  N u | P  N u L 1+ ( J ) L 2− (103) 1 2

x

t

x

t

But by (101) we have





N + | P N u | p −1 P N u  L 1+ ( J ) L 2−  N +  Iu  L 2+ ( J ) L ∞−  Iu  t

x

x

t

p −1

2( p −1)

Lt

2( p −1)

( J )L x

  p −1  N +  D 1−(1−) Iu  L 2+ ( J ) L ∞− u  2( p−1) x

t

 N + u  

p −1

2( p −1)

Lt

2( p −1)

( J )L x

Lt

2( p −1)

( J )L x

Z 1−,s ( J , u )

Z 1−,s ( J , u )

(104)

N+

Similarly

 +   N | P  N u | p −1 P N u 

L t1+ ( J ) L 2x −



Z 1−,s ( J , u ) N+

(105)

T. Roy / J. Differential Equations 248 (2010) 893–923

Moreover since s >



909

p −3 2



N + | P N u | p −1 P  N u  L 1+ L 2−  N +  P N u 

p −1

x

t

( p −1)+

Lt p −1

 N + Z 1−,s ( J , u )

6( p −1) − p −3

 P N u 

( J )L x

6 − 6− p

L t∞ ( J ) L x

 D  Iu L ∞ ( J ) L 2x N

t 5− p − 2

1



E 2 ( Iu 0 ) N

5− p 2 −

p −1

Z 1−,s ( J , u )

(106)

By Proposition 4 we have for N = N (u 0  H s , u 1  H s−1 )  1 2

  N + | P  N u | p −1 P  N u  L 1+ ( J ) L 2−  N + t

 D 1−( p −) Iu  N

p

Z2

p −,s

N

p+

Lt

x



p 2p − p −2

( J )L x

5− p 2

( J , u)

5− p − 2

p



E 2 ( Iu 0 ) N

5− p − 2

1

 E 2 ( Iu 0 ) since s >

2 p

(107)

> sc . Now by (103), (105), (106) and (107)

1

Z 1−,s ( J , u )  E 2 ( Iu 0 ) +

Z 1−,s ( J , u ) N+

1

+

E 2 ( Iu 0 ) N

5− p 2 −

p −1

Z 1−,s ( J , u )

(108)

1

Let C 3 :=

E 2 ( Iu 0 ) N

5− p − 2

. Then by Proposition 4

C3 E

p −2 2

( Iu 0 ) 1

(109)

and

1 N+

1

(110)

if N = N (u 0  H s , u 1  H s−1 )  1. From Lemma 10, (101), (108) (109) and (110) we get (34). 5. Proof of “Almost Morawetz–Strauss estimate” In this section we prove Proposition 9. First we recall the proof of the Morawetz–Strauss estimate based upon the important equality [8,9,14]

910

T. Roy / J. Differential Equations 248 (2010) 893–923

   ∇u · x u  + ∂tt u − u + u + |u | p −1 u |x| |x|     ∇u · x u = ∂t  + ∂t u |x| |x|      | u | p +1 x | u |2 x ∇u · x |∂t u |2 · x |u |2 · x |∇ u |2 u − + ∇ + + −  u + + div − 2|x| |x| |x| 2|x| ( p + 1)|x| 2 |x|3  p +1 2 |∇ u · x| p − 1 |u | 1 + |∇ u |2 − (111) + p + 1 |x| |x| |x|

 

Integrating (111) with respect to space and time we have

T 

p−1 p+1

0 R3



=−

|u | p +1 (t , x) dx dt + 2π |x| 



R3



R3

|u |2 (0, t ) dt 0

 ∇ u ( T , x) · x u ( T , x) + ∂t u ( T , x) dx |x| |x|





+

T

 ∇ u (0, x) · x u (0, x) + ∂t u (0, x) dx |x| |x|

(112)

if u satisfies (1). By Cauchy–Schwartz

           ∇ u · x + u ∂t u ( T , x) dx  E 12 (u )   |x| |x| R3

R3

2  1  2  ∇ u ( T , x). · x u   dx  +   |x| |x|

(113)

After expansion we have

     ∇ u ( T , x). · x u ( T , x) 2  dx =  +  |x| |x| 

R3

R3

2      ∇ u ( T , x) · x 2 ∇( |u | (2T ,x) ) · x |u |2 ( T , x)  dx + 2  dx + dx   2 |x| |x| |x|2

    ∇ u ( T , x) · x 2   dx =   |x|

R3

R3

R3

 E (u )

(114)

Here we used the identity



|u |2 ( T , x)x div 2|x|2



∇( |u |

2

=

( T ,x) 2

|x|2

)·x

+

|u |2 ( T , x) 2|x|2

(115)

Combining (112) and (116) we get

    ∇ u ( T , x) · x u ( T , x) 2   dx  E (u ) +  |x| |x| 

R3

(116)

T. Roy / J. Differential Equations 248 (2010) 893–923

911

Similarly

    ∇ u (0, x). · x u (0, x) 2  dx  E (u )  +  |x| |x| 

(117)

R3

We get from (112), (116) and (117) the Morawetz–Strauss estimate

T  0 R3

|u | p +1 (t , x) dx dt  E (u ) |x|

(118)

Now we plug the multiplier I into (111) and we redo the computations. We get (41). 6. Proof of “Almost conservation law” and “Estimate of integrals” The proof of Proposition 6, 8 relies on the following lemma Lemma 11. Let G such that G  L ∞ ( J ) L 2  Z ( J , v ). If s  t

 

x

3p −5 2p

> sc and 3  p < 5 then

p +1    ( J , v) G F ( I v ) − I F ( v )  dx dt  Z

N

J R3

5− p 2 −

(119)

Proof. We have

 

   G F ( I v ) − I F ( v )  dx dt

J R3

     G L ∞ ( J ) L 2x  F ( I v ) − F ( v ) L 1 ( J ) L 2 + G L ∞ ( J ) L 2x  F ( v ) − I F ( v ) L 1 ( J ) L 2 t t x x t t       Z ( J , v )  F ( I v ) − F ( v ) 1 2 +  F ( v ) − I F ( v ) 1 2 Lt ( J ) L x

Lt ( J ) L x

(120)

Let





X 1 :=  F ( I v ) − F ( v ) L 1 ( J ) L 2 t

x

(121)

and





X 2 :=  F ( v ) − I F ( v ) L 1 ( J ) L 2 t

x

(122)

We are interested in estimating X 1 . By the fundamental theorem of calculus we have the pointwise bound

     F ( I v ) − F ( v )  max | I v |, | v | p −1 | I v − v |

(123)

912

T. Roy / J. Differential Equations 248 (2010) 893–923

Plugging this bound into X 1 we get

X 1   P N v 



1 N

5− p 2 −

p −1

4( p −1) + 7− p Lt (

4( p −1) − p −3 J )L x

 P N v 

  1  D 1−(1−) I v  p − 4( p −1) Lt

 1− 3p−5  p 2p I v  +  D  p

+ 7− p

(



4 − p −3 Lt (

4( p −1) − p −3 J )L x

4 + 5− p J )L x

+  P N v 

 1−( p−3 +)   D  2 I v

p −1 p

2p

Lt ( J ) L x

4 − p −3

Lt

 P  N v L p ( J ) L 2p t

x

4 + 5− p

( J )L x

2p Lt ( J ) L x

p −1

 

p

Z 1−,s ( J , v ) Z p−3 +,s ( J , v ) + Z 3p−5 ( J , v ) 2

2p

N

,s

5− p − 2

Z p( J , v) N

(124)

5− p − 2

Now we turn to X 2 . On low frequencies we use the smoothness of F whereas on high frequencies we take advantage of the regularity of u, lying in H s . More precisely by the fundamental theorem of calculus we have

F ( v ) = F ( P N v + P N v )

 1 = F ( P N v ) +

 | P N v + s P N v |

p −1

ds P  N v

0

 1 + 0

P N v + s P N v P N v + s P N v

 | P N v + s P N v |

p −1

ds P  N v

(125)

Therefore





X 2   P  N F ( v ) L 1 ( J ) L 2 t

    P  N F ( P N v )

x

L t1 ( J ) L 2x

    + | P N v | p −1 P  N v  L 1 ( J ) L 2 + | P  N v | p −1 P  N v L 1 ( J ) L 2 x

t

t

 X 2,1 + X 2,2 + X 2,3

(126)

with



x



X 2,1 :=  P  N F ( P N v ) L 1 ( J ) L 2 t

x





X 2,2 := | P N v | p −1 P  N v  L 1 ( J ) L 2 t

and





X 2,3 := | P  N v | p −1 P  N v  L 1 ( J ) L 2 . t

But again by the fundamental theorem of calculus

x

x

T. Roy / J. Differential Equations 248 (2010) 893–923

X 2,1 

913

1 N

 ∇ F ( P N v )

L t1 ( J ) L 2x

   | P N v | p −1 P N v 1 p −1   | P N v | ∇ P N v + ∇ P N v   N P v N

(127) L t1 ( J ) L 2x

Therefore

X 2,1 



1 N

 P N v 

4( p −1) + 7− p

Lt

1 N

p −1

5− p 2 −

4( p −1) − p −3

∇ P N v 

 1−(1−)  p −1  D  I v  4( p−1) + Lt

4 − p −3

Lt

( J )L x

7− p

(

4( p −1) − p −3 J )L x

4 + 5− p

( J )L x

 1−( p−3 +)   D  2 I v

4 − p −3

Lt

4 + 5− p

( J )L x

p −1

 

Z 1−,s ( J , v ) Z p−3 +,s ( J , v ) N

2 5− p − 2

Z p( J , v) N

(128)

5− p 2 −

Moreover

X 2,2 

1 N

5− p − 2

 1−(1−)  p −1  D  I v  4( p−1) + Lt

7− p

(

4( p −1) − p −3 J )L x

 1−( p−3 +)   D  2 I v

4 − p −3

Lt

4 + 5− p

( J )L x

p −1

 

Z 1−,s ( J , v ) Z p−3 +,s ( J , v ) N

2 5− p − 2

Z p( J , v) N

(129)

5− p − 2

As for X 2,3 we have

X 2,3   P  N v 



 D 1−

p p

2p

Lt ( J ) L x

3p −5 2p

N

I v

p p

2p

Lt ( J ) L x

5− p 2

p

 

Z 3p−5 ( J , v ) ,s

2

N

5− p 2

Z p( J , v) N

2

5− p − 2

(130)

Let t  ∈ J = [a, b]. Then if u is a solution to (1) then

                E Iu t − E Iu (a)  =    ∂ Iu F ( Iu ) − I F ( u ) t   [a,t  ] R3

 

 [a,t  ]

R3

   ∂t Iu F ( Iu ) − I F (u ) 

(131)

914

T. Roy / J. Differential Equations 248 (2010) 893–923

Notice that

∂t Iu L ∞ ( J ) L 2x  Z 0,s ( J , u )

(132)

t

Applying Lemma 11 with G := ∂t Iu to (131) we get (37). Notice also that

  ∇Iv · x    |x|  ∞ 2  ∇ I v Lt∞ ( J ) L 2x L ( J )L x t

 Z 0, s ( J , v )

(133)

and that

   Iv     |x|  ∞ 2  ∇ I v Lt∞ ( J ) L 2x Lt ( J ) L x  Z 0, s ( J , v )

(134)

∇ I v (t ,x).x we get (38) from (133) and Lemma 11 for i |x| Similarly (38) holds for i = 2 if we let G (t , x) := I v|(xt|,x) .

by Hardy inequality. Letting G (t , x) :=

= 1.

q

7. Strichartz estimates for NLKG in L t L rx spaces The techniques used in the proof of these estimates are, broadly speaking, standard [7,6]. However some subtleties appear because unlike the homogeneous Schrodinger and wave equations the homogeneous defocusing Klein–Gordon equation does not enjoy any scaling property. Now we mention them. Regarding the estimates involving the homogeneous part of the solution we apply, broadly speaking, a “T T ∗ ” argument to the truncated cone operators localized at all the frequencies11 instead of applying it at frequency equal to one and then use a scaling argument for the other frequencies. The inhomogeneous estimates are slightly more complicated to establish. In the first place we try to reduce the estimates (see (172)) localized at all frequencies to the estimate at frequency one (see (180)). This strategy does not totally work because of the lack of scaling. However the remaining estimate (see (184)), after duality is equivalent to a homogeneous estimate on high frequencies (see (186)) that has already been established. Let u be the solution of (23) with data (u 0 , u 1 ). We can substitute [0, T ] for R in (24) without loss of generality. Let ul (t ) := cos (t  D )u 0 +

sin (t  D ) D u1

and unl (t ) := −

t 0

sin (t −t  ) D  D

Q (t  ). We need to show

  ul Lt∞ H m + ∂t ul (t ) L ∞ H m−1  u 0  H m + u 1  H m−1 t   −1   q ul L Lrx + ∂t  D  ul Lq Lr  u 0  H m + u 1  H m−1 t t x   −  unl Lq Lrx + ∂t  D  1 unl Lq Lr   Q Lq˜ Lr˜ t

t

x

t

(135) (136) (137)

x

and

unl Lt∞ H m + ∂t unl Lt∞ H m−1   Q Lq˜ Lr˜ t

x

By Plancherel theorem we have (135). We prove (136), (137) and (138) in the next subsections.

11

I.e. to e it  D  P M , M ∈ 2Z : see (149).

(138)

T. Roy / J. Differential Equations 248 (2010) 893–923

915

7.1. Proof of (136) By decomposition and substitution it suffices to prove

 it  D   e u0 

 u 0  H m

q

L t L rx

(139)

If we could prove for every Schwartz function f

 it  D   e P 1 f 

  f L2

(140)

 M m  f L2

(141)

q

L t L rx

and

 it  D   e PM f 

q

L t L rx

 for M ∈ 2Z , M > 1, then (136) would follow. Indeed let P M := P M  2M and P 1 := P 2 . Applying 2

(141) to f := P M f we have

 it  D   e PM f 

q

L t L rx

 M m  P M f L2   P ˙m M f H

(142)

Similarly plugging f := P 1 f into (140) we have

 it  D   e P 1 f 

q

L t L rx

  P 1 f  L 2   f Hm

(143)

Before moving forward, we recall the fundamental Paley–Littlewood equality [13]: if 1 < p < ∞ and h is Schwartz then

h

Lp

   12    2   ∼ | P M h|  M ∈2Z

(144) Lp

We plug h := P >1 f into (144). Hence by Minkowski inequality and Plancherel theorem

 it  D   e P >1 f 

q

L t L rx

  1    it  D  2 2       e PM f  M 1



  12  it  D  2 e PM f  q r Lt L x

M 1



q

L t L rx

 M 1

  f Hm

2  P M f H ˙m

 12

(145)

916

T. Roy / J. Differential Equations 248 (2010) 893–923

since q  2 and r  2. Combining (143) to (145) we get (139). It remains to prove (140) and (141). Let T 1 ( f ) := e it  D  P 1 f and T M ( f ) := e it  D  P M f , M ∈ 2Z , M > 1. We have



φ(ξ )e it ξ  ˆf (ξ )e i ξ.x dξ

T 1 ( f )(t , x) :=

(146)

R3

and if M ∈ 2Z , M > 1 let



 T M ( f )(t , x) :=

ψ



ξ M

e it ξ  ˆf (ξ )e i ξ.x dξ

(147)

R3

We would like to prove

   T 1 ( f )

q

  f L2

(148)

q

  f Hm

(149)

L t L rx

and

   T M ( f )

L t L rx

By a “T T ∗ ” argument we are reduced showing for every continuous in time Schwartz in space function g

   T 1 T ∗ ( g ) 1

q

L t L rx

 g

q

(150)



L t L rx

and similarly

   T M T ∗ ( g ) M

with

1 q

+

1 q

= 1 and

1 r

+

1 r

q

L t L rx

 M 2m  g 

q



L t L rx

(151)

= 1. But a computation shows that T 1 T 1∗ ( g ) = K 1 ∗ g



=

    K 1 t − t  , . ∗ g t  , . dt 

(152)

and ∗ TMTM (g) = K M ∗ g



=

    K M t − t  , . ∗ g t  , . dt 

(153)

with





K 1 t − t  , x :=

 R3

  φ(ξ )2 e i ξ (t −t  ) e i ξ ·x dξ

(154)

T. Roy / J. Differential Equations 248 (2010) 893–923

917

and

KM

   2  ξ  i ξ (t −t  ) i ξ ·x t − t , x := ψ e e dξ M 







(155)

R3

On one hand by Plancherel equality we have

     K M t − t, . ∗ g t, . 

     g t  , . L2

(156)

        K M t − t  , . L∞  g t  , . L1

(157)

L2

On the other hand

     K M t − t, . ∗ g t, . 

L∞

where  K M (t − t  , .) L ∞ is estimated by the stationary phase method [5, p. 441]

   K M t − t, . 

 L∞

 M min 1,



1

d

( M |t − t  |)

d −1 2

  min 1,

M

 12  (158)

|t − t  |

and

   K1 t − t, . 

 L∞

 min 1,



1

(159)

d

|t − t  | 2

By complex interpolation we have

        K 1 t − t  , . ∗ g t  , .  r  min 1, L

1− 2r

1 d

|t − t  | 2

    g t , . 

Lr

(160)



and

         K M t − t, . ∗ g t, .  r   K M t − t   g t  , . Lr L

(161)

with

   K M (t ) := M d min 1, and r  such that d 2r

1 r

+

1 r



1

( M |t |)

d −1 2

  min 1,

M

 12 1− 2r (162)

|t |

= 1. Observe that if (q, r ) is wave admissible and (q, r ) = (∞, 2) then

1 q

+

<

d 4

d . 4

Therefore there are two cases. First we estimate  T 1 T 1∗  L q L r . There are two cases

<

t

x

• Case 1: r > 2. Then since (q, r ) is wave admissible and (q, r ) = (∞, 2) we also have and by (152), Young’s inequality and (160)

1 q

+

d 2r

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T. Roy / J. Differential Equations 248 (2010) 893–923

  T 1 T ∗ g  1

q L t L rx

  1− 2r    1  q  g  q r   min 1 , d   2 Lt L x Lt |t | 2  g

q

(163)



L t L rx

• Case 2: r = 2. Then q = ∞. Then by (152) and (156) we get (150).       We turn to (151). We write  KM = K , K M ,a + K M ,b + K M ,c in (161) with K M ,a := K M χ M ,b := |t | 1  KMχ 1

 |t | M and K M ,c M

M

:=  K M χ|t | M . We have by Young’s inequality and (26)

       K M ,a t − t   g t  , .  r   Lx

2

q

Lt

 M d(1− r ) χ|t | 1  M

 M 2m  g 

q

q

L t2

g

q



L t L rx

(164)



L t L rx

   To estimate  K M ,b (t − t ) ∗  g (t , .) L r   L q there are two cases t

• Case 1:

1 q

+

d−1 2r

<

d−1 4

By Young’s inequality, (162) and (26) we have

 2         M 2 (1 − r )     g  q r  K M ,b t − t   g t  , .  L r   Lq   χ 1 d −1  q  M t  M L Lx x t ( Mt ) 2 Lt2  M 2m  g Lq Lr 

(165)

x

• Case 2:

1 q

+

d−1 2r

=

d−1 . 4

By (162) we have

     d +1 2    K g t , .  L r   M 2 (1 − r ) M ,b t − t



 g (t  , .)Lr  |t − t  |

x

R

M

d(1− 2r )− 2q



d −1 2 2 (1 − r )

 g (t  , .)Lr  2

|t − t  | q

R

dt 

dt 

(166)

By (27), (26) and Hardy–Littlewood–Sobolev inequality [13]

       K M ,b t − t   g t  , .  r   Lx

q

Lt

 M 2m  g 

q

(167)



L t L rx

   We estimate  K M ,c (t − t ) g (t , .) L r   L q by applying Young inequality, (26) and (25), i.e. t

       K M ,c t − t   g t  , .  r   Lx

q Lt

 d 2   M ( 2 +1)(1− r )  χt  M 2

t2

r

q

L t2

2

 M q +1 − r  g   M 2m  g  By (161), (164), (165), (167) and (168) we get (151).

   d (1 − 2 )  1

q



L t L rx

q



L t L rx

(168)

T. Roy / J. Differential Equations 248 (2010) 893–923

919

7.2. Proof of (137) By decomposition and substitution it suffices to prove

       e i (t −t  ) D  Q t  dt     t 
    D  Q Lq˜ Lr˜

q

L t L rx

(169)

x

t

By Christ–Kisilev lemma [12]12 it suffices in fact to prove

       e i (t −t  ) D  Q t  dt    

q

    D  Q  Lq˜ Lr˜

(170)

  Q Lq˜ Lr˜

(171)

x

t

L t L rx

If we could prove

       e i (t −t  ) D  P 1 Q t  dt    

q

L t L rx

x

t

and

       e i (t −t  ) D  P M Q t  dt    

q

L t L rx

 M  Q Lq˜ Lr˜

(172)

x

t

 then (135) would follow. Indeed introducing P M as in the previous subsection we have 1 and P

       e i (t −t  ) D  P 1 Q t  dt    

q

L t L rx

  P 1 Q  L q˜ L r˜

x

t

    D  Q 

(173)

q˜

L t L rx˜

and

       e i (t −t  ) D  P M Q t  dt    

q

L t L rx

 M  P M Q  L q˜ L r˜ t

     P M D Q

x

(174)

q˜

L t L rx˜

Therefore we have

       e i (t −t  ) D  P >1 Q t  dt    

q

L t L rx

     

 M ∈2Z , M >1

 M ∈2Z , M >1

12

An original proof of this lemma can be found in [3].

  1     2 2   e i (t −t  ) D  P M Q t  dt      

   12    2  e i (t −t  ) D  P M Q t  dt    q r Lt L x

q

L t L rx

920

T. Roy / J. Differential Equations 248 (2010) 893–923







L t L rx˜

M ∈2Z , M >1

   

 12

   P M  D  Q 2q˜



   P M  D  Q 2

M ∈2Z , M >1

 12    

q˜

L t L rx˜

    D  Q Lq˜ Lr˜

(175)

x

t

Now we establish (171). It is not difficult to see from the proof of (140) and (141) that we also have

 it  D  12  e P 4 f  Lq L r   f  L 2

(176)

 it  D  12  e P 1 f  Lq L r   f  L 2

(177)

x

t

and

x

t

for every Schwartz function f . A dual statement of (177) is

  1      e −it   D  P 2 Q t  dt   1  

  Q Lq˜ Lr˜ t

L2

(178)

x

Composing (176) with (178) we get (171). We turn to (172). We need to prove

         e i (t −t  ) D  ψ ξ Q

t  , ξ dt  e i ξ ·x dξ    M

q

L t L rx

 M  Q Lq˜ Lr˜ t

(179)

x

By the change of variable (ξ, t  ) → ( M , Mt  ) we are reduced showing ξ

    1  1    2  e i ( Mt −t )(|ξ | + M 2 ) 2 ψ(ξ ) Q t , . (ξ ) dt  e iMx·ξ dξ    M M

q

L t L rx

 M 2  Q Lq˜ Lr˜ t

(180)

x

If we could prove that for every Schwartz function G

 S M G Lq Lrx  G Lq˜ Lr˜ t

t

(181)

x

with

 S M G :=

e

i (t −t  )(|ξ |2 +

then (180) would hold. Indeed by (29) we have

1 M2

1

)2

  ψ(ξ ) G t  , ξ dt  e i ξ ·x dξ

(182)

T. Roy / J. Differential Equations 248 (2010) 893–923

    1  1    2  e i ( Mt −t )(|ξ | + M 2 ) 2 ψ(ξ ) Q t , . (ξ ) dt  e iMx·ξ dξ    M M

q

L t L rx

921

       . .  = S Q , Mt , Mx ) ( M   M M

1

 M q˜

+ dr˜ − 1q − dr

 Q Lq˜ Lr˜ t

x

2

 M  Q Lq˜ Lr˜ t

q

L t L rx

(183)

x

By duality and composition with (176) it suffices to show

 it ( D 2 + 1 ) 12 12  e M2 P 1 f Lq Lr   f L 2

(184)

x

t

Again it is not difficult to see from the proof of (149) that

 it  D  12  e P f M

  f Hm

q

L t L rx

(185)

But after performing the change of variable ξ → M ξ we have by (26) and (185)

    1    it ( D 2 + 1 ) 12 12  t 1 x ξ  iM (|ξ |2 +1) 2 iM ·ξ  e 2 M2 P 1 f Lq Lr =  e ψ f ( M .)(ξ ) e d ξ   t x M    1  it  D  12   t x  2   = e P P f ( M .) , M M  M M 

q

L t L rx

q

L t L rx

1  1 d 2   M q + r  P M f ( M .) H m 1

d

d

 M q + r − 2 +m  f L 2   f L2

(186)

7.3. Proof of (138) By decomposition, substitution and Christ–Kisilev lemma [12] it suffices to prove

     e i (t −t  ) D  Q   

L t∞ L 2x

    D 1−m Q  Lq˜ Lr˜

(187)

x

t

If we could prove

       e i (t −t  ) D  P 1 Q t  dt    

L t∞ L 2x

  Q Lq˜ Lr˜ t

(188)

x

and

       e i (t −t  ) D  P M Q t  dt    

L∞ L2 t

then (187) would follow. Indeed

x

 M 1−m  Q Lq˜ Lr˜ t

x

(189)

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T. Roy / J. Differential Equations 248 (2010) 893–923

       e i (t −t  ) D  P 1 Q t  dt    

L t∞ L 2x

  P 1 Q  L q˜ L r˜

x

t

    D 1−m Q 

(190)

q˜

L t L rx˜

and

       e i (t −t  ) D  P M Q t  dt    

L t∞ L 2x

 M 1−m  P M Q  L q˜ L r˜ t

  1−m    P Q M D

x

(191)

q˜

L t L rx˜

Therefore following the same steps to those in (175) we get (187). 1

2 (188) follows from the composition of the trivial inequality e it  D  P  1 f  L ∞ L 2   f  L 2 and (178). t

We turn to (189). We need to prove

         e i (t −t  ) D  ψ ξ Q

t  , ξ dt  e i ξ ·x dξ    M

L t∞ L 2x

x

 M 1−m  Q Lq˜ Lr˜ t

(192)

x

Again by the change of variable (ξ, t  ) → ( M , Mt  ) it suffices to show ξ

    1  1    2  e i ( Mt −t )(|ξ | + M 2 ) 2 ψ(ξ ) Q t , . (ξ ) dt  e iMx·ξ dξ    M M

L t∞ L 2x

 M 2−m  Q Lq˜ Lr˜

(193)

x

t

If we could prove for any Schwartz function

 S M G L ∞ L 2x  G Lq˜ Lr˜ t

t

(194)

x

with S M defined in (182) then substituting q, r for ∞, 2 respectively in (183) we have

    1  1    2  e i ( Mt −t )(|ξ | + M 2 ) 2 ψ(ξ ) Q t , . (ξ ) dt  e iMx·ξ dξ    M M

1

L t∞ L 2x

 M q˜

+ dr˜ − d2

 Q Lq˜ Lr˜ t

 M 2−m  Q Lq˜ Lr˜ t

x

(195)

x

where in the last inequality we used (26) and (29). It remains to prove (194). By duality and composi1

2 tion with the trivial inequality e it  D  P  4 f  L ∞ L 2   f  L 2 it suffices to show (184), which has already

been established.

t

x

Acknowledgment The author would like to thank Terence Tao for suggesting him this problem.

T. Roy / J. Differential Equations 248 (2010) 893–923

923

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