Acta Mathematica Sinica, English Series July, 2002, Vol.18, No.3, pp. 549–564

The Smoothness of Scattering Operators for Sinh-Gordon and Nonlinear Schr¨ odinger Equations

Bao Xiang WANG Department of Mathematics, Hebei University, Baoding 071002, P. R. China E-mail: [email protected]

Abstract We show that the scattering operator carries a band in H s (Rn ) × H s−1 (Rn ) into H s (Rn ) × H s−1 (Rn) for the sinh-Gordon equation and an analogous result also holds true for the nonlinear Schr¨ odinger equation with an exponential nonlinearity, where s ≥ n/2 is arbitrary and n ≥ 2. Therefore, the scattering operators are infinitely smooth for the above two equations. Keywords Sinh-Gordon equation, Nonlinear Schr¨ odinger equation, The smoothness of scattering operator MR(2000) Subject Classification 35L15, 35Q55, 35Q53, 35P25

1

Introduction

This paper is devoted to the study of the existence and smoothness of scattering operators for the following equations: utt − u + (eu − e−u )/2 = 0, 2

iut + u + µ(eλ|u| − 1)u = 0,

(1.1) (1.2)

where u is a complex-valued function of (t, x) ∈ R × Rn (or a real-valued function in (1.1)),  is the Laplace operator on Rn , λ, µ ∈ R. (1.1) is known as the sinh-Gordon equation which arises from the soliton theory, and (1.2) is derived in the theory of nonlinear optics. There are several studies on (1.1) and (1.2) in recent years (cf. [1, 2] and references therein). In the present paper we discuss the existence and smoothness of the scattering operators. The definition of the scattering operator can be found in [2]. Let k ∈ R, Ω ⊂ H k . For Equation (1.2), if a scattering operator is such that S : Ω → H k , then we say that S is k-order smooth. Moreover, if S : Ω ∩ H s → H s is s-order smooth for all s ≥ k, then S is said to be an infinitely smooth scattering operator. For Equation (1.1), we can give a similar definition for the infinitely smooth scattering operator. For convenience, we also introduce the following notion: Received July 2, 1999, Accepted April 27, 2001 Supported by the National Natural Science Foundation of China, Grant 19901007.

Wang B. X.

550

Definition 1.1

Let 0 ≤ r ≤ s < ∞. Bδr,s = {(ϕ, ψ) ∈ H s × H s−1 : ϕ H r + ψ H r−1 ≤ δ}

is said to be a band in H s × H s−1 . One sees that Bδr,s is a whole neighbourhood of (0, 0) in H s × H s−1 if r = s. If r < s, the

norm of (ϕ, ψ) in H s × H s−1 can be arbitrarily large whenever (ϕ, ψ) ∈ Bδr,s has a sufficiently small norm in H r × H r−1 . Now we state our main results. Theorem 1.2

Let n/2 ≤ s < ∞, n ≥ 3. Then there exists δ > 0 such that the scattering n/2,s

into H s × H s−1 . Moreover, S is an infinitely smooth

operator S of (1.1) carries the band Bδ scattering operator. Theorem 1.3

Let n ≥ 2, n/2 ≤ s < ∞. Then there exists δ > 0 such that the scattering n/2,s

operator S of (1.2) carries the band Bδ

= {u ∈ H s : u H n/2 ≤ δ} into H s . Moreover, S is

an infinitely smooth scattering operator. Remark

Theorem 1.2 also holds true for two space dimensions. But the proof presented in

this paper cannot be developed for the case n = 2. Remark

The existence of the scattering operator S of (1.2) was first shown by Nakamura

and Ozawa in [1]. More precisely, they proved that S : N → H n/2 for some neighbourhood N of 0 in H n/2 . Putting s = n/2 in Theorem 1.3, we can derive their result as a corollary. The proof of Theorem 1.3 relies upon some nonlinear mapping estimates in Besov spaces which are different from that of [1]. Remark

As indicated in [1], H n/2 is a critical space which cannot be embedded into L∞ . It

follows that the L∞ norm of the element in Bδ

n/2,s

2

can also be independent of δ > 0.

Some Inequalities in Besov and Sobolev Spaces

In this section we collect some embedding inequalities in Besov and Sobolev spaces. If we do not emphasize the dependence of embedding constants on the inequalities, then these embedding relations are well-known. But the embedding constant is of importance for the proof of nonlinear estimates, so, it seems to be necessary to give exact versions of these inequalities and their proofs will only be sketched. Let ψ ∈ C0∞ (Rn ) satisfy supp ψ ⊂ {x : 1/2 ≤ |x| ≤ 2} and ψ(x) > 0 if 1/2 < |x| < 2.

−k x))−1 , j = 0, ±1, ±2, . . . and ϕ0 (x) = 1 − ∞ Let ϕj (x) = ψ(2−j x)( ∞ k=−∞ ψ(2 j=1 ϕj (x). For
−k −1 ψ(2 x)) . For any s ∈ R convenience, we write ϕj ≡ ϕj , j = 1, 2, . . ., ϕ(x) = ψ(x)( ∞ k=−∞ s s (B˙ p,q ) and and for any p, q with 1 ≤ p, q < ∞, the norms on (homogeneous) Besov spaces Bp,q s s ˙ (homogeneous) Triebel spaces Fp,q (Fp,q ) are defined by (cf. [3, 4])  1q  q1    ∞ ∞ q q sk −1 k sk −1 s u Bp,q = 2 F ϕ Fu Lp , u B˙ p,q = 2 F ϕk Fu Lp , s k=0

 ∞  1q    sk −1 k  q   s = u Fp,q (2 F ϕ Fu)   k=0

k=−∞

 ∞  q1    sk −1  q   , u F˙ s =  (2 F ϕk Fu)  p,q p

L

k=−∞

Lp

.

The Smoothness of Scattering Operators for Sinh-Gordon and Nonlinear Schr¨ odinger Equations

551

The (homogeneous) Sobolev spaces H s,p (H˙ s,p ) are defined by H s,p = (I − )−s/2 Lp , H˙ s,p = (−)−s/2 Lp , H s,2 = H s and H˙ s,2 = H˙ s . We denote by S = S(Rn ) the Schwartz spaces and ormander multiplier spaces, by Mp the H¨ ρ Mp =

sup f ∈S, f Lp =1

F −1 ρFu Lp .

Let 1 ≤ p ≤ ∞. Then there exists a continuous function ω : R → R+

Lemma 2.1 (Cf. [3])

such that for any s ∈ R, k = 0, 1, 2, . . .,    (1 + |ξ|2 )s/2 k   ϕ   ≤ ω(s),  2ks Mp

(2.1)

where ω(·) is independent of p but may depend on n. Proof

Since ρ(t·) Mp = ρ(·) Mp for any t = 0, it is easy to see that −k

the same norm in Mp as (2 −k

D (2 α

2 s/2

+ |ξ| )

2 s/2

+ |ξ| )

(1+|ξ|2 )s/2 ϕ(2−k ξ) 2ks

has

ϕ(ξ). It follows from supp ϕ ⊂ {ξ : 1/2 ≤ |ξ| ≤ 2} that

ϕ(ξ) L2 ≤ ω(s) ∈ C(R, R+ ),

|α| ≤ [n/2] + 1.

In view of Bernstein’s multiplier theorem we get the desired results. Similarly to (2.1) we have  s    |ξ|    2ks ϕk  ≤ ω(s), Mp

k = 0, ±1, ±2, . . . .

(2.2)

s 0 s 0 → Fp,q and (I − )s/2 : Bp,q → Bp,q are all It follows from (2.1) that (I − )s/2 : Fp,q

isomorphic mappings and s s , 0 ≤ (I − )s/2 f Fp,q ≤ ω2 (s) f Fp,q ω1 (s) f Fp,q

σ Fp,q

(2.3)

σ Bp,q .

by Analogous results hold for homogeneous Besov which is also true if we replace s s ˙ ˙ . spaces Bp,q and homogeneous Triebel spaces Fp,q In the following we denote by Cp a constant that depends only on p and n. Proposition 2.2

Let 1 < p < ∞. There exists a constant Cp > 0 such that for any q, s, r

with p ≤ q < ∞, s ∈ R and 1 ≤ r < ∞, 

u B˙ s ≤ Cp q 1/p u B˙ s+n/p−n/q . q,r

Proof

p,r

(2.4)

In [5], it has been shown that there exists Cp > 0 such that 

u Lq ≤ Cp q 1/p u H˙ n/p−n/q,p ,

(2.5)

for all q ≥ p. It follows that 

F −1 ϕk Fu Lq ≤ Cp q 1/p F −1 ϕk Fu H˙ n/p−n/q,p 

1

1

= Cp q 1/p F −1 ϕk |ξ|n( p − q ) Fu Lp ≤ Cp q 1/p



1 

1

1

F −1 ϕk+j |ξ|n( p − q ) F(F −1 ϕk Fu) Lp .

(2.6)

j=−1

In view of (2.2) one sees that there exists a constant C0 = sup0≤s≤n/p ω(s) that is independent 1 1 1 1 of q and k verifying ϕk+j |ξ|n( p − q ) Mp ≤ C0 2(k+j)n( p − q ) . So, we have F −1 ϕk Fu Lq ≤ C0 Cp q 1/p



1 

1

1

2(k+j)n( p − q ) F −1 ϕk Fu Lp

j=−1

≤ Cp q

1 − q1 )k 1/p n( p

2

F −1 ϕk Fu Lp .

(2.7)

Wang B. X.

552

(2.7) is multiplied by 2sk and equipped with the lr -norm, and we get the desired results. Clearly, from (2.2) we see that 

u H˙ s,q ≤ Cp q 1/p u H˙ s+n/p−n/q,p . Remark on Proposition 2.2

(2.8)

In (2.4) and (2.8), if p ranges over a compact interval [p1 , p2 ],

p1 > 1, then sup{Cp : p ∈ [p1 , p2 ]} < ∞ (see [5]). Let 1 < p < ∞. Then there exist a constant Cp > 0 and a continuous

Proposition 2.3

function ω : R+ → R+ such that for any s1 , s2 ∈ R, s1 ≤ s2 , u H s1 ,p ≤ Cp ω(s2 − s1 ) u H s2 ,p . Proof

(2.9)

For any s1 , s2 ∈ R and s1 ≤ s2 , we have u H s1 ,p = F −1 (1 + |ξ|2 )s1 −s2 F(I − )s2 /2 u Lp .

Since sup(1 + |ξ|2 )s1 −s2 = 1 and sup 1≤|α|≤[n/2]+1

|ξ||α| |Dα (1 + |ξ|2 )s1 −s2 | ≤ ω(s2 − s1 ),

ω ∈ C(R+ , R+ ),

in view of the Mihlin multiplier theorem we see that there exists a constant Cp > 0 such that (1 + |ξ|2 )s1 −s2 Mp ≤ Cp ω(s2 − s1 ). Proposition 2.4 Let 1 < p < ∞. Then there exist a constant Cp > 0 and a continuous function ω : R+ → R+ such that for any s ≥ 0, u H˙ s,p ≤ Cp ω(s) u H s,p . Proof

(2.10)

For any s > 0, we have

  u H˙ s,p = F −1

 |ξ|s s2  F(I − ) u  p. L (1 + |ξ|2 )s/2 In view of the Mihlin multiplier theorem, we see that         |ξ|s |ξ|s |α|  α   ≤ Cp ω(s),  ≤ C sup |ξ| D p   (1 + |ξ|2 )s/2  2 s/2 (1 + |ξ| )  0≤|α|≤[n/2]+1 Mp where ω(·) ∈ C(R+ , R+ ). From this the result is deduced. 0 0 = Lp (cf. [4]), i.e. there exist Cp1 , Cp2 > 0 such that Cp1 u Fp,2 ≤ For given 1 < p < ∞, Fp,2

0 . It follows from (2.3) that u Lp ≤ Cp2 u Fp,2

s s . ≤ u H s,p ≤ Cp2 ω2 (s) u Fp,2 Cp1 ω1 (s) u Fp,2

Also, noticing that

s Bp,min(p,q)

⊂

s Fp,q

⊂

s Bp,max(p,q)

(2.11)

with the embedding constant 1 (cf. [4, p.

47]), it follows that if s is taken over a closed interval [s1 , s2 ], then s , u H s,p ≤ Cp u Bp,2

2≤p<∞

(2.12)

and sup{Cp : 2 ≤ p ≤ N } < ∞ for some fixed N < ∞. s (s > 0, p > 1): It is convenient to use an equivalent norm on B˙ p,2  ∞  dt 1/2 t−2(s−[s]) sup h Dα u 2Lp . u oB˙ s = p,2 t |h|≤t 0

(2.13)

|α|=[s]

s Similarly, u Lp + u oB˙ s is an equivalent norm on Besov spaces Bp,2 . In the present paper we p,2

use only the cases p ∈ [2, ρ] and s ∈ (0, s0 ] for some ρ, s0 < ∞. On the basis of Theorem 6.2.5 in [3] we have C1 u B˙ s ≤ u 0B˙ s ≤ (C2 /(s−[s]) ) u B˙ s , p,2

p,2

p,2

(2.13a)

The Smoothness of Scattering Operators for Sinh-Gordon and Nonlinear Schr¨ odinger Equations

553

where C1 and C2 depend only on s0 . 

The fundamental tools used in this paper are time-space Lp −Lp estimates which are usually named the Strichartz inequalities (cf. [6, 7] and their references). Using the ideas of [7, 8, 9] and [10], we can get a generalized Strichartz inequality which is of importance in the proof of our main results. For any 1 ≤ r < ∞, we denote by r the dual number of r (1/r + 1/r = 1) and 2 1 1 2σ(r) = = − . n+1 γ(r)(n − 1) 2 r

t Let U (t) = exp(it(I − )1/2 ) and A := 0 U (t − τ ) · dτ. We have

(2.14)

U (t)ϕ Lγ(ρ) (R,B s−σ(ρ) ) ≤ C ϕ H s ,

(2.15)

Af Lγ(q) (R,B s−σ(q) ) ≤ C f Lγ(ρ) (R,B s+σ(ρ) )

(2.16)

ρ,2

ρ ,2

q,2

if ρ, q ∈ [2, ∞) satisfy γ(ρ),γ(q) ∈ (2, ∞]. Moreover, if ρ, q ∈ [2, 2(n−1) n−3 − ε] (ρ, q ∈ [2, 1/ε] if n = 3, ε > 0 can be arbitrarily small), then the constant C in (2.15) and (2.16) can depend only on ε > 0. Let S(t) = exp(it∆). Cazenave and Weissler in [6] showed the following time-space Lp − Lp



estimates in Lebesgue-Besov spaces: S(t)ϕ Lq (R,B˙ s ) ≤ C ϕ H˙ s , r,2   t    S(t − τ )f (τ )dτ  q ≤ C f Lγ  (R,B˙ s  ˙s L (R,Br,2 )

0

ρ ,2

(2.17) ),

(2.18)

 (q, r) is said to be an admissible pair if 2/q =  2n/(n − 2), n > 2  n(1/2 − 1/r) and 2 ≤ r < 2∗ = . ∞, n≤2

where (q, r) and (γ, ρ) are admissible pairs

3

Nonlinear Estimates

t 2 Let f (u) = (eλ|u| − 1)u and Ls f = s S(t − τ )f (u(τ ))dτ . It is easy to see that f (u) =
∞ k 2k k=1 λ |u| u/k!. Now we indicate some ideas in the nonlinear estimates which are different from that of [1]. Let (q, r) be an admissible pair and (qk , rk ) be a sequence of admissible pairs. Considering the estimates of Ls (f (u)), in view of (2.18) we have ∞  |λ|k |u|2k u qk . Ls f (u) Lq (R,B˙ s ) ≤ C L (R,B˙ s ) r,2 k! r ,2 k=1

k

So, it seems necessary to choose a suitable sequence of admissible pairs (qk , rk ) to carry out the (see below, Lemma 3.4). nonlinear estimates |u|2k u qk ˙s L

(R,B

r ,2 k

)

For the sinh-Gordon equation, the ideas in the nonlinear estimates are basically the same as the above. Since the semi-group exp(it(I −∆)1/2 ) lacks smoothness, it follows that the nonlinear estimate of sinh-Gordon equations is more complicated than that of nonlinear Schr¨ odinger equations. Therefore, we give only the details of the nonlinear estimates of the sinh-Gordon equation.

Wang B. X.

554

Lemma 3.1 (Cf. [11])

Let α be a multi-index with |α| ≥ 1. Suppose that f (u) ∈ C |α| . Then

we have Dα f (u) =

|α|  

C(α1 , . . . , αq )f (q) (u)

q=1 Λqα

q

Dαi u,

(3.1)

i=1

|α| q   |f (q) (u) − f (q) (v)| |Dαi v| |D (f (u) − f (v))| ≤ C α

q=1 Λqα

+ |f (q) (u)|

q  i−1 

  

i=1

i=1 q

Dαj u

j=1

j=i+1

  Dαj vDαi (u − v) ,

where C and C(α1 , . . . , αq ) denote constants depending only on α1 , . . . , αq .  α + ... + α = α q 1 q q a ≡ 1 are provided and Λ = . α j=q+1 j 1 ≤ |α1 | ≤ . . . ≤ |αq |

0 j=1

(3.2) aj ≡ 1 and

Lemma 3.2 (Convexity H¨older Inequality) Let 1 < pi , qi < ∞, 0 ≤ θi ≤ 1, σi , σ ∈ R (i =
N
N
N
N 1, . . . , N ). Suppose that i=1 θi = 1, σ = i=1 θi σi , 1/p = i=1 θi /pi , and 1/q = i=1 θi /qi . Then we have ∩N B˙ σi ⊂ B˙ σ and for any v ∈ ∩N B˙ σi , p,q

pi ,qi

i=1

i=1

v B˙ pσ ,q ≤ i

Lemma 3.3 ρ=

i

N i=1

pi ,qi

v θB˙iσi .

(3.3)

pi ,qi

Let p be a natural number, s(p) = n/2 − 2/p ≥ 1/2 and 1/2 ≤ s < ∞. Let

2(n−1)(2+p) (n−1)(2+p)−4 ,

n ≥ 3. Then there exists a constant C > 0 which is independent of p and ρ

such that 

up+1 B˙ s−1+σ(ρ) ≤ p[s]+p/ρ C p+1 u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . Bρ,2

ρ ,2

(3.4)

ρ,2

If s − 1 + σ(ρ) is an integer, then (3.4) also holds true if we replace the homogeneous Besov σ spaces B˙ r,2 by the homogeneous Sobolev spaces H˙ σ,r . Proof

We only consider the case where s and s − 1 + σ(ρ) are not integers. It is easy to see

that σ(ρ) ≤ 1/2, s(p) + σ(ρ) ≥ 1, 0 ≤ 1 − 2σ(ρ) ≤ s(p) − σ(ρ). Moreover, one can easily deduce that

1 ρ

−

s(p)−σ(ρ) n

=

p+4 np(2+p)



p

and 1 s(p) − σ(ρ) − ρ n

Denoting f (u) = up+1 , we have f (u) B˙ s−1+σ(ρ) = ρ ,2

 0

∞ −2v

t

 +

1 1 − 2σ(ρ) 1 − = . ρ n ρ



|α|=[s−1+σ(ρ)]

sup h Dα f (u) 2Lρ

|h|≤t

(3.5)

dt t

1/2 ,

(3.6)

where v = s + σ(ρ) − [s + σ(ρ)]. It follows from |f (q+1) (u)| ≤ δ(q)Aqp+1 |u|p−q that |f (q) (uh ) − f (q) (u)| ≤ δ(q)Aqp+1 (|uh |p−q + |u|p−q )|uh − u|, where δ(q) =

 1, q ≤ p 0, q > p

(3.7)

. When s − 1 + σ(ρ) < 1, (3.3) is easily deduced and so, we can

The Smoothness of Scattering Operators for Sinh-Gordon and Nonlinear Schr¨ odinger Equations

555

assume that [s − 1 + σ(ρ)] ≥ 1. From (3.2) and (3.7) one easily sees that h D f (u) α

Lρ

≤C

|α|   q=1 Λqα

+

δ(q)Aqp+1

q     p−q p−q + |u| )|uh − u| Dαi u (|uh | i=1

q  q i−1     Dαj uh Dαj uDαi (uh − u) |uh |p+1−q i=1

j=1



Lρ

j=i+1

|α| q     δ(q)Aqp+1 Iq Lρ + IIqi Lρ . = q=1 Λqα

Lρ

(3.8)

i=1

For brevity we denote  Aq :=

Bq :=

q 

∞

t−2v sup Iq 2Lρ |h|≤t

0

q  

Bqi :=

i=1

∞

0

i=1

dt t

1/2 ,

t−2v sup IIqi 2Lρ |h|≤t

dt t

1/2 .

First, we consider the case s ∈ [1/2, s(p) + 1]. From s ≤ s(p) + 1 one sees that s(p) − σ(ρ) ≥ v. Put  1 s(p) − σ(ρ) 1 1 s(p) − σ(ρ) − v 1 − = − , = (p − q) , r0 ρ n λ ρ n 1 1 (s − σ(ρ)) − |α1 | 1 (s(p) − σ(ρ)) − |αi | 1 , , i = 2, ..., q. = − = − r1 ρ n ri ρ n We easily see that 0 ≤ s −σ(ρ) −|α1 | ≤ s(p) −σ(ρ) and if q ≥ 2, then |αi | ≤ s −1+σ(ρ) −|α1 | ≤
q s(p) − σ(ρ), i = 2, ..., q. It follows that 1/λ, 1/ri > 0 and 1/λ + i=0 1/ri = 1/ρ . In view of H¨ older’s inequality we have Iq Lρ ≤ |u|

p−q



Lr0

uh − u Lλ

q

Dαi u Lri .

(3.9)

i=1

Hence, by (2.4) and (2.8) we have 

Aq ≤ pp/ρ Cρp+1 u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . Bρ,2

ρ,2

(3.10)

Since ρ is taken over a bounded interval [2, 6], from the Remark of Proposition 2.2 we see that Cρ can be independent of ρ and p. To estimate Bq is easier than Aq and the details are omitted. So, we have shown that (3.3) holds for any s ∈ [1/2, s(p) + 1]. It suffices to consider the case s ∈ [s(p) + 1, ∞) and the proof will be split up into the following two steps: Step 1

We estimate Aq . First, we consider the case q = 1. Let   1 1 1 s(p) − σ(ρ) 1 s(p) − σ(ρ) − , = − , = (p − q) r0 ρ n λ ρ n 1 1 1 − 2σ(ρ) 1 s − σ(ρ) − v − |αq | = − . = − r1 ρ n ρ n

Wang B. X.

556 1 λ

We see that 1/λ, 1/r0 , 1/r1 > 0 and I1 Lρ

+

1 r0

+

1 r1

=

1 ρ .

It follows from (2.8) and (2.12) that

≤

|u|p−1 Lr0 uh − u Lλ Dα1 u Lr1

≤

p(p−1)/ρ C p u p−1 ˙ s−σ(ρ)−v,ρ ˙ s(p)−σ(ρ),ρ uh − u Lλ u H H

≤

p(p−1)/ρ C p u p−1 uh − u Lλ u B˙ s−σ(ρ)−v , ˙ s(p)−σ(ρ)

 

Bρ,2

ρ,2

(3.11)

where C is independent of p and ρ. By (3.11) and (2.4) we have 

u B˙ v u B˙ s−σ(ρ)−v A1 ≤ p(p−1)/ρ C p u p−1 ˙ s(p)−σ(ρ) Bρ,2

≤p



p/ρ

C

p+1

λ

ρ,2

u p−1 s(p)−σ(ρ) u B ˙ s(p)−σ(ρ)+v u B˙ s−σ(ρ)−v . B˙ ρ,2 ρ,2

(3.12)

ρ,2

Noticing that s(p) − σ(ρ) + v ≤ s − σ(ρ) and s − σ(ρ) − v ≥ s(p) − σ(ρ), in view of the convexity H¨older inequality,  A1 ≤ pp/ρ C p+1 u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . (3.13) Bρ,2

ρ,2

Next, we consider the case q ≥ 2. The estimation of Iq Lρ proceeds in the following three cases: (I) |αq | ≥ s − s(p) − v. In this case one can easily deduce that |αi | < s(p) − σ(ρ) for all i = 1, . . . , q − 1. Indeed, assume for the contrary that |αq−1 | > s(p) − σ(ρ). It follows that |αq | + |αq−1 | > s − σ(ρ) − v ≥ [s − 1 + σ(ρ)]. This is a contradiction. Put   1 1 s(p) − σ(ρ) 1 1 s(p) − σ(ρ) − , = − , = (p − q) r0 ρ n λ ρ n 1 (s(p) − σ(ρ)) − |αi | 1 , i = 1, ..., q − 1, = − ri ρ n 1 (s − σ(ρ) − v) − |αq | 1 . = − rq ρ n From |αq | > s − s(p) − v we see that 0 ≤ s − σ(ρ) − v − |αq | ≤ s(p) − σ(ρ). It follows that
q 1/rq > 0. Similarly, one easily sees that 1/λ, 1/ri > 0 (i = 0, . . . , q − 1) and λ1 + i=0 r1i = 1/ρ . In view of the H¨ older inequality we have 

Iq Lρ ≤ p(p−1)/ρ C p u p−1 uh − u Lλ u B˙ s−σ(ρ)−v . ˙ s(p)−σ(ρ) Bρ,2

ρ,2

(3.14)

In the same way as in the case of q = 1 one easily deduces that 

Aq ≤ pp/ρ C p+1 u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . Bρ,2

ρ,2

(II) s(p) − σ(ρ) ≤ |αq | < s − s(p) − v. Put  1 s(p) − σ(ρ) 1 s(p) − σ(ρ) 1 1 − = − , = (p − q) , r0 ρ n λ ρ n 1 (s(p) − σ(ρ) + |αi |) − |αi | 1 , i = 1, ..., q − 1, = − ri ρ n
q−1 1 1 1 − 2σ(ρ) 1 (s − σ(ρ) − v − i=1 |αi |) − |αq | = − . = − rq ρ n ρ n

(3.15)

The Smoothness of Scattering Operators for Sinh-Gordon and Nonlinear Schr¨ odinger Equations

Obviously, we have

1 λ

+


q

1 i=0 ri

=

1 ρ .

557

Thus, in view of the H¨ older inequality we have

Iq Lρ ≤ |u|p−q Lr0 uh − u Lλ

q−1

Dαi u Lri Dαq u Lrq .

(3.16)

i=1

By (3.16), (2.4), (2.8) and (2.12) we have 

Aq≤ p(p−1)/ρ C p u p−q u B˙ v ˙ s(p)−σ(ρ) Bρ,2

λ,2



q−1 i=1

u B˙ s(p)−σ(ρ)+|αi | u B˙ |αq |+1−2σ(ρ) ρ,2

≤ pp/ρ C p+1 u p−q u B˙ s(p)−σ(ρ)+v ˙ s(p)−σ(ρ) Bρ,2

q−1

ρ,2

i=1

ρ,2

u B˙ s(p)−σ(ρ)+|αi | u B˙ |αq |+1−2σ(ρ) . (3.17) ρ,2

ρ,2

In view of s(p) − σ(ρ) ≤ |αq | ≤ s − s(p) − v, we have s(p) − σ(ρ) ≤ s(p) − σ(ρ) + v, s(p) − σ(ρ) + |αi |, |αq | + 1 − 2σ(ρ) ≤ s − σ(ρ). Let θi (i = 0, 1, . . . , q) satisfy s(p) − σ(ρ) + v = θ0 (s(p) − σ(ρ)) + (1 − θ0 )(s − σ(ρ)), s(p) − σ(ρ) + |αi | = θi (s(p) − σ(ρ)) + (1 − θi )(s − σ(ρ)),

i = 1, . . . , q − 1,

|αq | + 1 − 2σ(ρ) = θq (s(p) − σ(ρ)) + (1 − θq )(s − σ(ρ)).
q
q older One easily sees that i=0 θi = q and i=0 (1 − θi ) = 1. In view of the convexity H¨ inequality, from (3.17) one sees that 

Aq ≤ pp/ρ C p+1 u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . Bρ,2

(3.18)

ρ,2

(III) |αq | < s−s(p)−v and |αq | < s(p)−σ(ρ). In this case we easily see that |αi | ≤ s(p)−σ(ρ) for all i = 1, . . . , q. In view of s(p)+σ(ρ) ≥ 1 we have [s−1+σ(ρ)] = s−1+σ(ρ)−v ≥ s−s(p)−v.
q It follows that i=1 |αi | = [s − 1 + σ(ρ)] ≥ s − s(p) − v. Thus, we can take appropriate βi
q satisfying 0 ≤ βi ≤ |αi | and i=1 βi = s − s(p) − v. Put  1 s(p) − σ(ρ) 1 s(p) − σ(ρ) 1 1 − = − , = (p − q) , r0 ρ n λ ρ n 1 (s(p) − σ(ρ) + βi ) − |αi | 1 , i = 1, ..., q. = − ri ρ n One can easily verify that s(p) − σ(ρ) ≤ s(p) − σ(ρ) + βi ≤ s(p) − σ(ρ) + |αq | ≤ s − σ(ρ) − v and 0 ≤ s(p) − σ(ρ) + βi − |αi | ≤ s(p) − σ(ρ). So, 1/λ, 1/ri > 0 (i = 0, 1, . . . , q). Noticing that q  1 s(p) − σ(ρ)
q (β − |α |) 1 s(p) − σ(ρ) 1 1  1 i i + − + − = , =p − i=1 λ i=0 ri ρ n n ρ n ρ in view of H¨older’s inequality, together with (2.8) and (2.12) we have 

Iq Lρ ≤ p(p−1)/ρ C p u p−q uh − u Lλ ˙ s(p)−σ(ρ) Bρ,2

q

u B˙ s(p)−σ(ρ)+βi .

i=1

ρ,2

(3.19)

By (3.19) and (2.4) we have 

Aq ≤ p(p−1)/ρ C p u p−q u B˙ v ˙ s(p)−σ(ρ) Bρ,2

λ,2



q

u B˙ s(p)−σ(ρ)+βi

i=1

≤ pp/ρ C p+1 u p−q u B˙ s(p)−σ(ρ)+v ˙ s(p)−σ(ρ) Bρ,2

ρ,2

ρ,2

q i=1

u B˙ s(p)−σ(ρ)+βi . ρ,2

(3.20)

Wang B. X.

558

Let θi (i = 0, 1, . . . , q) satisfy s(p) − σ(ρ) + v = θ0 (s(p) − σ(ρ)) + (1 − θ0 )(s − σ(ρ)), s(p) − σ(ρ) + βi = θi (s(p) − σ(ρ)) + (1 − θi )(s − σ(ρ)), It is easy to see that 0 ≤ θi ≤ 1,


q

i=0 θi

= q and


q

i=0 (1

i = 1, . . . , q.

− θi ) = 1. So, (3.20) implies that



Aq ≤ pp/ρ C p+1 u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . Bρ,2

(3.21)

ρ,2

Step 2 We estimate Bq . Since the proof of this part is basically parallel to that of Step 1, we only give an outline of the proof. First, we consider the case of q = 1. Let  1 s(p) − σ(ρ) 1 − = p , r0 ρ n 1 1 − 2σ(ρ) 1 s − σ(ρ) − v − |α1 | 1 = − . = − r1 ρ n ρ n In view of the H¨ older inequality we have 

II11 Lρ ≤ pp/ρ C p u p˙ s(p)−σ(ρ) Dα1 (uh − u) Lr1 .

(3.22)

Bρ,2

It follows from (3.22) that 



B1 ≤ pp/ρ C p u p˙ s(p)−σ(ρ) u B˙ |α1 |+v ≤ pp/ρ C p+1 u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . Bρ,2

Bρ,2

r1 ,2

ρ,2

(3.23)

Next, we consider the case of q ≥ 2. In the same way as in Step 1, we divide the estimation into the following three cases: (I) |αq | > s − s(p) − v. Let  1 s(p) − σ(ρ) 1 − = (p + 1 − q) , r0 ρ n 1 s(p) − σ(ρ) − |αi | 1 , i = 1, . . . , q − 1, = − ri ρ n 1 s − σ(ρ) − v − |αq | 1 . = − rq ρ n Similarly to Step 1, we see that for i = 1, . . . , q − 1, 

IIqi Lρ ≤ pp/ρ C p u p−1 Dαi (uh − u) Lri u B˙ s−σ(ρ)−v , ˙ s(p)−σ(ρ) Bρ,2

and

(3.24)

ρ,2



IIqq Lρ ≤ pp/ρ C p u p˙ s(p)−σ(ρ) Dαq (uh − u) Lrq .

(3.25)

Bρ,2

It follows that for i = 1, . . . , q − 1, 



u B˙ |αi |+v u B˙ s−σ(ρ)−v ≤ pp/ρ C p u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) , Bqi ≤ pp/ρ C p u p−1 ˙ s(p)−σ(ρ) Bρ,2

Bρ,2

ρ,2

ri ,2

ρ,2

(3.26)

and 



Bqq ≤ pp/ρ C p u p˙ s(p)−σ(ρ) u B˙ |αq |+v ≤ pp/ρ C p+1 u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . Bρ,2

rq ,2

Bρ,2

ρ,2

(3.27)

The Smoothness of Scattering Operators for Sinh-Gordon and Nonlinear Schr¨ odinger Equations

559

(II) s(p) − σ(ρ) ≤ |αq | ≤ s − s(p) − v. For any i = 1, . . . , q, let  1 s(p) − σ(ρ) 1 − , = (p + 1 − q) r0 ρ n 1 (s(p) − σ(ρ) + |αi |) − |αi | 1 , = − ri ρ n 1 (|αq | + 1 − 2σ(ρ)) − |αq | 1 . = − rq ρ n

i = 1, . . . , q − 1,

Similarly to Step 1, we see that for i = 1, ..., q − 1,  αi ri IIqi ≤ p(p−1)/ρ C p u p+1−q ∆ D u u ˙ s(p)−σ(ρ)+|αj | u B˙ |αq |+1−2σ(ρ) , (3.28) h L s(p)−σ(ρ) ˙ Bρ,2

j=i j≤q−1

and 

IIqq Lρ ≤ pp/ρ C p u p+1−q ˙ s(p)−σ(ρ) Bρ,2

q−1

Bρ,2

ρ,2

u ˙ s(p)−σ(ρ)+|αj | Dαq (uh − u) Lrq .

j=1

Bρ,2

(3.29)

Notice that |αi | + s(p) − σ(ρ) ≤ s − σ(ρ) and |αq | + 1 − 2σ(ρ) + v ≤ s − σ(ρ). Using the convexity H¨ older inequality, we have 

Bq ≤ pp/ρ C p u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . Bρ,2

(3.30)

ρ,2

(III) |αq | ≤ s − s(p) − v and |αq | ≤ s(p) − σ(ρ). Put  1 s(p) − σ(ρ) 1 − = (p + 1 − q) , r0 ρ n 1 s(p) − σ(ρ) + βi − |αi | 1 , i = 1, . . . , q, = − ri ρ n where βi (i = 1, . . . , q) are the same as in Step 1. Noticing that s(p) − σ(ρ) + βi + v ≤ s − σ(ρ) older inequality we get the estimates of and s(p) − σ(ρ) + |αq | + v ≤ s − s(p), in view of the H¨ older inequality we obtain that IIqi Lρ . Moreover, using the convexity H¨ 

Bq ≤ pp/ρ C p u p˙ s(p)−σ(ρ) u B˙ s−σ(ρ) . Bρ,2

(3.31)

ρ,2

Noticing that Aqp+1 ≤ Cp[s] , by (3.6), (3.8), (3.13), (3.15), (3.18), (3.21), (3.23), (3.26), (3.27), (3.30) and (3.31), we get the conclusions. Remark on Lemma 3.3 If s is an integer, then we have v = s − 1 + σ(ρ) − [s − 1 + σ(ρ)] = σ(ρ) = (n + 1)/(n − 1)(2 + P ). Using (2.13a), C2 /v ≤ Cp, we see that Lemma 3.3 also holds in the case that s is an integer. For convenience, we write Lr (R, X) = Lr (X). Lemma 3.4 Let p be an even integer, s(p) = n/2 − 2/p ≥ 0 and 0 ≤ s < ∞. Let r = 2 + 4/n, 2n(2+p) , γ = 2 + p, n ≥ 2. Then there exists a constant C > 0 which is independent of p ρ = n(2+p)−4 and ρ such that |u|p u Lγ  (B˙ s

ρ ,2



)

≤ C p+1 p2[s]+p/ρ u p r

s(p) ˙ s(p) ) L (B˙ r,2 )∩L∞ (H

u Lr (B˙ s

∞ ˙ s r,2 )∩L (H )

σ Moreover, if s is an integer, then (3.32) holds if we replace B˙ q,2 by H˙ σ,q .

.

(3.32)

Wang B. X.

560

Proof It seems necessary to point out that the present ρ is different from that in Lemma 3.3. We have  1 s(p) 1 1 − p (3.33) + = . ρ n ρ ρ Denote f (u) = |u|p u, we see that f (u) B˙ s



ρ ,2



∞ −2v

= 0

t

|α|=[s]

sup h Dα f (u) 2Lρ

|h|≤t

dt 1/2 , t

(3.34)

where v = s − [s]. In view of (3.7) one sees that (3.8) holds also for |α| = [s]. For convenience, we shall use the same notations as in Lemma 3.3. Step 1 Assume that the following two conditions are satisfied: (i) s > 1 is not an integer, i.e. v = s − [s] > 0; (ii) p is sufficiently large, for instance, p > max(8, (8s + 2)/7, 2/(1 − v)). First, we consider the case s ≤ s(p) + 1. From condition (ii) one sees that s(p) > v. We estimate Aq . Denote v 1 v 1 1 1 , . = + = − ρ1 ρ 4n(p − q) ρ2 ρ 4n Also, condition (ii) implies that ρi ∈ [2, 2 + 4/n]. Put 1 1 s(p) s(p) − v 1 1 = , = (p − q) − − , r0 ρ1 n λ ρ2 n 1 1 s − |α1 | 1 s(p) − |αi | 1 , , i = 2, ..., q. = − = − r1 ρ n ri ρ n It is easy to see that 0 ≤ s − |α1 | ≤ s(p) and if q ≥ 2, then |αi | ≤ s − |α1 | ≤ s(p). It follows
q that 1/λ, 1/ri > 0 and by (3.33) we have 1/λ + i=0 1/ri = 1/ρ , whence, Iq Lρ ≤ |u|p−q Lr0 uh − u Lλ

q

Dαi u Lri .

(3.35)

u B˙ s(p) u q−1 u B˙ s . Aq ≤ pp/ρ1 C p+1 u p−q ˙ s(p) ˙ s(p)

(3.36)

i=1

In virtue of (2.4) and (2.8) we have 

Bρ

Bρ,2

ρ2 ,2

1 ,2

ρ,2

For the same reason as in Lemma 3.3, C is independent of ρ and p. Let θi ∈ (0, 1) satisfy older 1/ρi = θi /r + (1 − θi )/2, i = 1, 2 and 1/ρ = θ0 /r + (1 − θ0 )/2. Using the convexity H¨ inequality, we get θ (p−q)

(1−θ )(p−q)

Br,2

B2,2

≤ u ˙1s(p) u ˙ s(p)1 u p−q ˙ s(p) Bρ

1 ,2

θ (q−1)

(1−θ )(q−1)

Br,2

B2,2

u q−1 ≤ u ˙0s(p) u ˙ s(p)0 ˙ s(p) Bρ,2

(1−θ )

u B˙ s(p) ≤ u θ˙2s(p) u ˙ s(p)2 ,

,

ρ2 ,2

,

Br,2

B2,2

(1−θ0 )

θ0 u B˙ s ≤ u B ˙ s u B˙ s ρ,2

r,2

2,2

.

Recall that θ1 (p − q) + θ2 + θ0 q = r(p + 1)/(p + 2), i.e. 1/γ  = (θ1 (p − q) + θ2 + θ0 q)/r. Applying H¨older’s inequality about the time variable, we obtain that 

Aq Lγ  ≤ C p+1 pp/ρ u p r

s(p)

˙ s(p) ) L (B˙ r,2 )∩L∞ (H

u Lr (B˙ s

The estimation of Bq is analogous to that of Aq and we omit it.

∞ ˙ s r,2 )∩L (H )

.

(3.37)

The Smoothness of Scattering Operators for Sinh-Gordon and Nonlinear Schr¨ odinger Equations

561

In the following we consider the case s ∈ [s(p) + 1, ∞). Since Bq can be estimated similarly to Aq , we only give the detailed estimation of Aq . Case I

q = 1. Let 1 s(p) 1 = (p − 1) − , r0 ρ1 n

1 s(p) − v 1 = , − λ ρ2 n

1 v 1 = − . r1 ρ n

Using the above interpolation ideas, we can get the estimate of A1 . Case II q ≥ 2. The estimation of Aq proceeds in the following three cases: (I) |αq | ≥ s − s(p). One easily deduces that |αi | < s(p) for all i = 1, . . . , q − 1. Put 1 1 1 s(p) s(p) − v 1 = , = (p − q) − − , r0 ρ1 n λ ρ2 n 1 1 s(p) − |αi | 1 s − |αq | 1 , i = 1, ..., q − 1, . = − = − ri ρ n rq ρ n Analogously to (3.36) we have 

Aq ≤ pp/ρ1 Cρp+1 u p−q u B˙ s(p) u q−1 u B˙ s . ˙ s(p) ˙ s(p) Bρ

Bρ,2

ρ2 ,2

1 ,2

(3.38)

ρ,2

Repeating the above interpolation argument, by (3.38) we get the desired result. (II) s(p) ≤ |αq | ≤ s − s(p). Put 1 1 s(p) s(p) − v 1 1 = , = (p − q) − − , r0 ρ1 n λ ρ2 n 1 (s(p) + |αi |) − |αi | 1 , i = 1, ..., q − 1, = − ri ρ n 1 (|αq | + v) − |αq | 1 . = − rq ρ n Similarly to (I) we can conclude that Aq ≤ p

(p+q)/ρ

C

p+1

u p−q s(p) u B ˙ s(p) B˙ ρ ,2 2

ρ1 ,2

q−1 i=1

u B˙ s(p)+|αi | u B˙ |αq |+v . ρ,2

ρ,2

In the same way as in the estimation of (3.17) we have q−1 i=1

Hence, we have

u B˙ s(p)+|αi | u B˙ |αq |+v ≤ u q−1 u B˙ s . ˙ s(p) ρ,2

Bρ,2

ρ,2

ρ,2



Aq ≤ p(p+q)/ρ C p+1 u p−q u B˙ s(p) u q−1 u B˙ s , ˙ s(p) ˙ s(p) Bρ

1 ,2

ρ2 ,2

Bρ,2

ρ,2

whence, the result is deduced.
q (III) |αq | ≤ min(s−s(p), s(p)). In this case one easily sees that |αi | ≤ s(p). Since i=1 |αi |+
q s(p) > s, we can take βi < |αi | satisfying i=1 βi + s(p) = s. Put 1 1 s(p) s(p) − v 1 1 = , = (p − q) − − , r0 ρ1 n λ ρ2 n 1 1 (s(p) + βi ) − |αi | , i = 1, ..., q. = − ri ρ n

Wang B. X.

562

We have 

Aq ≤ pp/ρ C p+1 u p−q u B˙ s(p) ˙ s(p) Bρ

ρ2 ,2

1 ,2

q

u B˙ s(p)+βi . ρ,2

i=1

By an interpolation we obtain that 

Aq ≤ pp/ρ C p+1 u p−q u B˙ s(p) u q−1 u B˙ s , ˙ s(p) ˙ s(p) Bρ

Bρ,2

ρ2 ,2

1 ,2

ρ,2

which implies the conclusions. Step 2 If condition (i) is not satisfied, the result is easily obtained. So, we only consider the case p ≤ N for some fixed N < ∞. In fact, similarly to Lemma 3.3 we can show that 

≤ p[s] (p(2 + p))p/ρ C0p+1 u p˙ s(p) u B˙ s .

|u|p u B˙ s

Bρ,2

ρ ,2

ρ,2

Taking C = C0 N (N + 2), by an interpolation we can get the conclusions. 4

Proof of Main Results

Let K(t) = (I − )−1/2 sin t(I − )1/2 , K  (t) = cos t(I − ) and As := u− (t) be the solution of the free Klein-Gordon equation u(0) = ϕ− ,

utt + (I − )u = 0,

t s

K(t − τ ) · dτ . Let

ut (0) = ψ − .

(4.1)

Now we define a mapping T : u(t) −→ u− (t) − A−∞ (sinhu − u). Let ρk =

2(n−1)(k+1) (n−1)(k+1)−2 .

One sees that 2 ≤ ρk ≤ ρ1 . Put

 D=

∞

(4.2)

u ∈ L (H ) ∩ L s

γ(ρ1 )

s−σ(ρ ) (Bρ1 ,2 1 )

u L∞ (H s )∩Lγ(ρ1 ) (B s−σ(ρ1 ) ) ≤ M ρ1 ,2

:

u L∞ (H n/2 )∩Lγ(ρ1 ) (B n/2−σ(ρ1 ) ) ≤ δ

 ,

(4.3)

ρ1 ,2

and for any u, v ∈ D, d(u, v) = u − v L∞ (H 1/2 )∩Lγ(ρ1 ) (B 1/2−σ(ρ1 ) ) .

(4.4)

ρ1 ,2

Denoting θk =

1/2−1/ρk 1/2−1/ρ1 ,

θk 1 − θk 1 = + , γ(ρk ) γ(ρ1 ) ∞

we easily see that 1 θ k 1 − θk , = + ρk ρ1 2 δ−σ(ρ1 )

It follows that L∞ (H δ ) ∩ Lγ(ρ1 ) (Bρ1 ,2 inequality we have

δ − σ(ρk ) = θk (δ − σ(ρ1 )) + (1 − θk )δ, δ−σ(ρk )

) ⊂ Lγ(ρk ) (Bρk ,2

u Lγ(ρk ) (B δ−σ(ρk ) ) ≤ u θkγ(ρ L

ρk ,2

1 ) (B

δ−σ(ρ1 ) ) ρ1 ,2

δ ∈ R.

) and by the convexity H¨ older

k u 1−θ . L∞ (H δ )

(4.5)

Now, we show that T : (D, d) → (D, d) is a contraction mapping. Noticing that sinhu − u =
∞ u2k+1 k=1 (2k+1)! , it follows from (2.16) that 2k+1

A−∞ (sinhu − u) L∞ (H s )∩Lγ(ρ1 ) (B s−σ(ρ1 ) ) ≤ C ρ1 ,2

≤C

∞ A−∞ u  k=1 ∞  k=1

L∞ (H s )∩Lγ(ρ1 ) (B s−σ(ρ1 ) ) ρ1 ,2

(2k + 1)! 1 u2k+1 Lγ(ρk ) (B s−1+σ(ρk ) ) . (2k + 1)! ρ ,2 k

The Smoothness of Scattering Operators for Sinh-Gordon and Nonlinear Schr¨ odinger Equations

563

From Lemma 3.3 and s(2k) < n/2 one concludes that 

u2k+1 B s−1+σ(ρk ) ≤ (2k)[s]+2k/ρk C 2k+1 u 2kn/2−σ(ρk ) u B s−σ(ρk ) . Bρ

ρ ,2 k

Since

1 γ(ρk )

=

2k+1 γ(ρk ) ,

ρk ,2

k ,2

in view of the H¨ older inequality and (4.5) for any u ∈ D, we have 

u2k+1 Lγ(ρk ) (B s−1+σ(ρk ) ) ≤ (2k)[s]+2k/ρk C 2k+1 δ 2k M. ρ ,2 k

Hence, A−∞ (sinhu − u) L∞ (H s )∩Lγ(ρ1 ) (B s−σ(ρ1 ) ) ≤ C ρ1 ,2

 ∞  (2k)[s]+2k/ρk (Cδ)2k

(2k + 1)!

k=1

M,

(4.6)

and similarly, A−∞ (sinhu − u) L∞ (H n/2 )∩Lγ(ρ1 ) (B n/2−σ(ρ1 ) ) ≤ C ρ1 ,2

 ∞  (2k)[s]+2k/ρk (Cδ)2k

k=1

(2k + 1)!

δ.

(4.7)

Since u− (t) = K(t)ϕ− + K  (t)ψ − , by (2.15) we have u− (t) L∞ (H σ )∩Lγ(ρ1 ) (B σ−s(ρ1 ) ) ≤ ( ϕ− H σ + ψ − H σ−1 ), ρ1 ,2

[s]+2k/ρ 2k
∞ k for σ = s, n/2. Take M = 2C( ϕ− H s + ψ − H s−1 ) and let δ satisfy k=1 (2k) (2k+1)!(Cδ) ≤ 1/2. If C( ϕ− H n/2 + ψ − H n/2−1 ) ≤ δ/2, then one can easily verify that T : D → D. Moreover, 2k+1 − v 2k+1 Lγ(ρk ) (B −1/2 ) ∞ u  ρ ,2 k . T u − T v d(u,v) ≤ C (2k + 1)!

k=1

Since L

τk

⊂

−1/2 Bρ ,2 , k

1/τk =

T u − T v d(u,v) ≤ C

1/ρk

+ 1/(2n), applying H¨ older’s inequality and (4.5) we have

 ∞  (2k)2k/ρk (Cδ)2k

k=1

(2k)!

u − v Lγ(ρk ) (B 1/2−σ(ρk ) ) ≤ u − v d(u,v) /2. ρk ,2

Hence, there exists a solution u ∈ D of the integral equation u(t) = u− (t) − A−∞ (sinhu − u).

(4.8)

Moreover, in a standard way we can show that u(t) − u− (t) L∞ (−∞,T ;H s ) → 0, Denote and u+ (0) =

T → −∞.

u+ (t) = u(t) − A∞ (sinhu − u) ϕ+ , u+ t (0)

(4.9)

= ψ + . Then we show in the same way as in the above that u(t) − u− (t) L∞ (T,∞;H s ) → 0,

T → ∞.

Thus, the scattering operator S : (ϕ− , ψ − ) → (ϕ+ , ψ + ) is well defined. The proof of Theorem 1.3 proceeds in the same way as that of Theorem 1.2 and we omit it.

564

Wang B. X.

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