Klein-Gordon Equations on Modulation Spaces

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Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 947642, 15 pages http://dx.doi.org/10.1155/2014/947642

Research Article Klein-Gordon Equations on Modulation Spaces

Guoping Zhao,1 Jiecheng Chen,2 and Weichao Guo3

1 Department of Mathematics, Zhejiang University, Hangzhou 310027, China 2 Department of Mathematics, Zhejiang Normal University, Jinhua 321004, China 3 Department of Mathematics, Xiamen University, Xiamen 361005, China

Correspondence should be addressed to Weichao Guo; [email protected]

Received 16 January 2014; Accepted 26 April 2014; Published 20 May 2014

Academic Editor: Simeon Reich

Copyright © 2014 Guoping Zhao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

𝑎 We consider the Cauchy problem for a family of Klein-Gordon equations with initial data in modulation spaces 𝑀𝑝,1.Wedevelop the well-posedness, blowup criterion, stability of regularity, scattering theory, and stability theory.

1. Introduction assumption is also necessary in this paper; in fact, this is an open problem that if In this paper, we consider the Cauchy problem for the 󵄩 󵄩 R × 󵄩 𝜆 󵄩 𝜆 following nonlinear Klein-Gordon equation in the space 󵄩|𝑢| 𝑢󵄩 ≤𝐶‖𝑢‖𝑀 ‖𝑢‖𝑀 𝑛 𝑛 󵄩 󵄩𝑀𝑝,1 𝑝,1 𝑝,1 (2) R = R𝑡 × R𝑥: holds for any positive real constant 𝜆. We recall that 𝑠𝑘 = 𝑛/2 − 2/𝑘 is the critical index for 𝐻𝑠 𝑢𝑡𝑡 + (𝐼−Δ) 𝑢+𝑓(𝑢) =0, (1). Up to now, we cannot solve (1)in for the case that 𝑠<𝑠𝑘 (the sup-critical case). On the other hand, we notice 𝑢(𝑡0,𝑥)=𝑢0, (1) that the modulation space 𝑀𝑝,1 has low regularity property. More precisely, for sufficiently large 𝑠,wehavethefollowing 𝑢𝑡 (𝑡0,𝑥)=𝑢1, embedding: 𝑠 𝐻 ⊂𝑀𝑝,1. (3) 𝑢(𝑡, 𝑥) 𝐼×R𝑛 where is a complex-valued function in for some In other words, the modulation space 𝑀𝑝,𝑞 has lower regu- 𝐼 𝑡 (𝑢 ,𝑢 ) 𝑠 time interval containing 0,theinitialdata 0 1 lies in larity than 𝐻 for large 𝑠.So,forlarge𝑠𝑘 (high dimension 𝑀𝑠 ×𝑀𝑠−1 1≤𝑝≤ 𝑠 the product of modulation spaces 𝑝,1 𝑝,1 ( for instance), one can solve (1)in𝑀𝑝,1 which contains sup- 𝑘+1 𝑠 ∞, 𝑠≥0), and the nonlinear term 𝑓(𝑢) = 𝜋(𝑢 ) is any critical initial dates in 𝐻 for 𝑠<𝑠𝑘. 𝑠 (𝑘 + 1)-time product of 𝑢 and 𝑢, 𝑘∈N. To understand this The local well-posedness of(1)in𝑀𝑝,1 (1≤𝑝≤∞, research problem and its historical developments, the reader 𝑠≥0)isaresultofBenyí and Okoudjou [4]; see also mayseeRuzhanskyetal.[1]forabriefsurveyofnonlinear Wang [2]foraglobalresultwithsmallinitialdata.These 𝑠 𝑠−1 evolution equations on the modulation spaces. Concerning results say that, for (𝑢0,𝑢1)∈𝑀𝑝,1 ×𝑀𝑝,1 , there exists the well-posedness of solution to the Schrodinger¨ equation apositive𝑇 = 𝑇(‖𝑢0‖𝑀𝑠 ,‖𝑢1‖𝑀𝑠−1 )≤∞such that (1) in the modulation space, readers can refer to [2, 3]. 𝑝,1 𝑝,1 𝑢(𝑡, 𝑥) ∈ 𝐶([0,𝑠 𝑇],𝑀 ) We give some remarks about our results. The known has a unique solution 𝑝,1 .Moreover, study of the Klein-Gordan equations (or other dispersive the lifetime of the solution can be proved to be bounded equations) on modulation spaces must be based on the below by a decreasing positive function 𝑔 depending on 𝑓(𝑢) ‖𝑢0‖ 𝑠 ,‖𝑢1‖ 𝑠−1 𝑇(𝑢0,𝑢1) ≥ 𝑔(‖𝑢0‖ 𝑠 ,‖𝑢1‖ 𝑠−1 ) assumption that the nonlinear term is a polynomial. This 𝑀𝑝,1 𝑀𝑝,1 ;thatis, 𝑀𝑝,1 𝑀𝑝,1 .It 2 Abstract and Applied Analysis

𝑠 𝑠 also asserts that if a strong 𝑀𝑝,1 solution keeps its 𝑀𝑝,1 norm From Lemma 19, we can verify the condition (8)by bounded in a bounded interval, it can be extended beyond choosing 𝐼 sufficiently small. So this theorem already gives ‖𝑢0‖ 𝑠 ,‖𝑢1‖ 𝑠−1 the endpoint. Hence, the following blowup criterion holds: local existence for large 𝑀𝑝,1 𝑀𝑝,1 data. On the other hand, by inequality (63), we have the following global result 𝑇<∞implies lim ‖𝑢(𝑡)‖𝑀𝑠 =∞. 𝑡→𝑇 𝑝,1 (4) as an application.

In this paper, we will develop a stronger blowup criterion Corollary 2 (global well-posedness for small fine data). Let which says that a blowup solution cannot blow up too slowly 2𝜎(𝑝) = (𝑛 + 2)(1/2 −1/𝑝) (𝑢 ,𝑢 )∈𝑀𝑠+2𝜎(𝑝) × . Assume that 0 1 𝑝󸀠,1 (see Corollary 8 and Remark 9). We also study the regularity 𝑀𝑠+2𝜎(𝑝)−1 of solutions and show that the regularity is stable along 𝑝󸀠,1 satisfies the lifetime. As an application, the global existence of low 󵄩 󵄩 󵄩 󵄩 󵄩𝑢0󵄩 𝑠+2𝜎(𝑝) + 󵄩𝑢1󵄩 𝑠+2𝜎(𝑝)−1 ≤𝛿0, regularity ensures the global existence of high regularity. 𝑀 󸀠 𝑀 󸀠 (10) 𝑠 𝑝 ,1 𝑝 ,1 Compared with 𝐶(𝐼,𝑝,1 𝑀 ) (used in [4]), the space 𝑟 𝑠 for some small constant 𝛿0 >0. Then, there exists a unique 𝐿𝑡(𝐼,𝑝,1 𝑀 ) seems more suitable for applying continuity argu- 𝑟 𝑠 global solution 𝑢∈𝐿(R,𝑀 ).Similarly,uisalsoastrong ment, which is the key point for obtaining the perturbation 𝑡 𝑝,1 𝑢∈𝐶(R,𝑀𝑠 )∩𝐶1(R,𝑀𝑠−1) theorem, especially the long-time version. So we choose solution; that is, 𝑝,1 𝑝,1 .Onealsohas 𝑟 𝑠 𝐿𝑡(𝐼,𝑝,1 𝑀 ) asourworkspaceandestablishthenonlinear the bound 󵄩 󵄩 󵄩 󵄩 estimate associated with this work space in Section 2. ‖𝑢‖𝐿𝑟 (R,𝑀𝑠 ) ≲ 󵄩𝑢0󵄩 𝑠+2𝜎(𝑝) + 󵄩𝑢1󵄩 𝑠+2𝜎(𝑝)−1 . 𝑡 𝑝,1 󵄩 󵄩𝑀 󵄩 󵄩𝑀 (11) In Section 3, we will establish the local theory. Wefirst use 𝑝󸀠,1 𝑝󸀠,1 the fixed point theorem to construct a local-in-time solution More precisely, we have the following global well- 𝑢∈𝐿𝑟(𝐼,𝑠 𝑀 ) 𝑡 𝑝,1 to (1). Then, we verify that such solution is posedness result which gives the decay rate of solutions. 𝑠 𝑠 astrong𝑀𝑝,1 solution in the sense that 𝑢∈𝐶(𝐼,𝑀𝑝,1)∩ 1 𝑠−1 Corollary 3 (another form of global well-posedness for small 𝐶 (𝐼,𝑝,1 𝑀 ) andisuniqueinthecategoryofstrongsolution. (𝑢 ,𝑢 )∈𝑀𝑠+2𝜎(𝑝) ×𝑀𝑠+2𝜎(𝑝)−1 Finally, we study the regularity of solutions and deduce a fine data). Assume that 0 1 𝑝󸀠,1 𝑝󸀠,1 stronger blowup criterion which implies the high rate of satisfies 󵄩 󵄩 󵄩 󵄩 blowup. We will develop the scattering results in Section 4. 󵄩𝑢 󵄩 + 󵄩𝑢 󵄩 <𝛿 󵄩 0󵄩𝑀𝑠+2𝜎(𝑝) 󵄩 1󵄩𝑀𝑠+2𝜎(𝑝)−1 In Section 5,weestablishastabilitytheoryfor(1)andobtain 𝑝󸀠,1 𝑝󸀠,1 (12) the continuous dependence as a corollary. 𝛿>0 Denote the operator 𝜔=Id −Δ,and for some small constant . Then, there exists a unique global solution 𝑢: −1 2 𝜑 (𝜔) := F 𝜑((1 + |𝜉|) ) F (5) 𝑛(1/2−1/𝑝) 𝑢∈𝑋:={𝑢: (1+|𝑡|) ‖𝑢‖ 𝑠 <∞}. sup 𝑀𝑝,1 (13) for any function 𝜑. Using this notation, we define 𝑡∈R

1/2 In the proof of Theorem 1, uniqueness is an immediate sin (𝑡𝜔 ) 𝐾 (𝑡) := (6) conclusion by the fixed point theorem. But, in fact, we have 𝜔1/2 the following stronger result. and the Klein-Gordon semigroup: 𝑠 Theorem 4 (unconditional uniqueness in 𝑀𝑝,1). Let I be a 𝑖𝑡𝜔1/2 time interval containing 𝑡0, 𝑝∈[1,∞], 𝑘∈N,andlet𝑢, V ∈ 𝐺 (𝑡) := 𝑒 . (7) 𝑠 𝐶(𝐼,𝑝,1 𝑀 ) be two strong solutions to (1) in the sense of (45) 𝑢(𝑡 )=V(𝑡 ) 𝑢 (𝑡 )=V (𝑡 ) We now state our main results. Unless otherwise speci- with the same initial data 0 0 , 𝑡 0 𝑡 0 ;then 𝑢=V 𝑀𝑠 𝑡∈𝐼 fied, we assume that the letters 𝑘, 𝑛 are integers such that in 𝑝,1 for all . 1/(𝑘 + 2) + 1/𝑛(𝑘 + 1) <1/2 and (𝑝, 𝑟) is 𝑘-admissible (Definition 25). First, we have the following local theorem. By combining the above uniqueness result with the local theorem, one can define the maximal interval 𝐼 of the Theorem 1 (local well-posedness). Let I be a compact time strong solution; thus, we have the following standard blowup 𝑠 𝑠−1 interval that contains 𝑡0.Let(𝑢0,𝑢1)∈𝑀𝑝,1 ×𝑀𝑝,1 satisfy criterion. 𝑠 𝑠−1 󵄩 󸀠 󵄩 Theorem 5 𝑢 ∈𝑀 𝑢 ∈𝑀 󵄩𝐾 (𝑡 − 𝑡 )𝑢 +𝐾(𝑡−𝑡 )𝑢 󵄩 ≤𝜂 (blowup criterion). Let 0 𝑝,1, 1 𝑝,1 , 󵄩 0 0 0 1󵄩𝐿𝑟 (𝐼,𝑀𝑠 ) (8) 𝑡 𝑝,1 and let 𝐼=(𝑇min,𝑇max) be the maximal interval. If 𝑇max < ∞(𝑇min >−∞), then one has for some 0<𝜂≤𝜂0,where𝜂0 is a small constant (depending 𝑢∈ only on n, k). Then, there exists a unique solution ‖𝑢‖ 𝑟 𝑠 =∞(‖𝑢‖ 𝑟 𝑠 =∞). 𝑟 𝑠 𝑠 𝐿𝑡([𝑡0,𝑇max],𝑀𝑝,1) 𝐿𝑡([𝑇min,𝑡0],𝑀𝑝,1) (14) 𝐿𝑡(𝐼,𝑝,1 𝑀 ) to (1).Moreover,uisa𝑀𝑝,1 strong solution to (1) 𝑠 1 𝑠−1 in the sense that 𝑢∈𝐶(𝐼,𝑀𝑝,1)∩𝐶(𝐼,𝑝,1 𝑀 ),andonealso The above blowup criterion will be improved soon as an has application of Lemmas 31 and 32.Forcompleteness,wealso give a proof for this weak version. Then, we give a regularity ‖𝑢‖ 𝑟 𝑠 ≤2𝜂. 𝐿𝑡(𝐼,𝑀𝑝,1) (9) result. Abstract and Applied Analysis 3

Theorem 6 (persistence of regularity). Let 1≤𝑝1,𝑝2 ≤∞, to (1), with 𝑒 and 𝑢−𝑢̃, 𝑢𝑡 − 𝑢̃𝑡 small in a suitable space, is 𝑠2 0≤𝑠1,𝑠2 <∞,and𝑘∈N,andlet𝑢 be a strong 𝑀𝑝 ,1 solution it possible to show that the genuine solution 𝑢 to (1)stays 2 𝑠 𝑠 𝐼 𝑢(𝑡 )∈𝑀1 very close to 𝑢̃ in some sense (for instance, in the 𝑀𝑝,1)? to (1) with its maximal existence interval .If 0 𝑝1,1 and 𝑠1−1 𝑠1 Note that the question of continuous dependence of the data 𝑢𝑡(𝑡0)∈𝑀𝑝 ,1 for some 𝑡0 ∈𝐼,then𝑢 is also a strong 𝑀𝑝 ,1 1 1 corresponds to the case 𝑒=0and the uniqueness theory to solution with the same maximal interval. the case 𝑒=0, 𝑢(𝑡0)=𝑢(𝑡̃ 0).Wehavethefollowingshort- time perturbations and long-time perturbations. Remark 7. Combining the above theorem with global result 𝑀 ×𝑀−1 in 2,1 2,1 in [2], one can easily get the global well- Theorem 11 (short-time perturbations). Let I be a compact 𝑀𝑠 ×𝑀𝑠−1 𝑝∈[1,2] posedness in 𝑝,1 𝑝,1 with small initial data for time interval, and let 𝑢̃ be an approximate solution to (1) in the 𝑠 and 𝑠≥0. sense of (19). Assume that 𝑢̃ has a uniform 𝑀𝑝,1 bound: 󵄩 󵄩 Thus, it is not possible to develop a singularity which 𝑢̃ ∞ 𝑠 ≤𝑀, 󵄩𝑢̃ 󵄩 ≤ 𝑀̃ 𝑠 𝑠 −1 𝑠 ‖ ‖𝐿 (𝐼,𝑀 ) 󵄩 𝑡󵄩𝐿∞(𝐼,𝑀𝑠−1) (20) 𝑀 1 ×𝑀1 𝑀 2 × 𝑡 𝑝,1 𝑡 𝑝,1 causes the 𝑝1,1 𝑝1,1 norm to blow up while the 𝑝2,1 𝑠2−1 𝑀𝑝 ,1 norm remains bounded. We also see that the regularity ̃ 𝑠 2 for some constant 𝑀, 𝑀>0.Let𝑡0 ∈𝐼and let 𝑢(𝑡0)∈𝑀𝑝,1, 𝑠2 is stable, because if a solution lies in 𝐶(𝐼,𝑝 𝑀 ,1) and is not 𝑠−1 2 𝑢𝑡(𝑡0)∈𝑀𝑝,1 be close to 𝑢(𝑡̃ 0), 𝑢̃𝑡(𝑡0),respectively,inthesense 𝑠1 𝑠1−1 𝑀 ×𝑀 𝑡0 in 𝑝1,1 𝑝1,1 at some initial time ,itneverbelongsto that 𝑠1 𝑠1−1 𝑀𝑝 ,1 ×𝑀𝑝 ,1 at any later (or earlier) time. As an application 1 1 󵄩 󵄩 󸀠 󵄩 󵄩 ̃󸀠 󵄩𝑢(𝑡0)−𝑢(𝑡̃ 0)󵄩 𝑠 ≤𝑀, 󵄩𝑢𝑡(𝑡0)−𝑢̃𝑡(𝑡0)󵄩 𝑠−1 ≤ 𝑀 (see also Lemmas 31 and 32), we have the following stronger 󵄩 󵄩𝑀𝑝,1 󵄩 󵄩𝑀𝑝,1 version of blowup criterion. (21)

Corollary 8 (stronger blowup criterion). Let 𝑝∈[1,∞], 𝑘∈ 󸀠 ̃󸀠 for some 𝑀 , 𝑀 >0. Moreover, assume the following smallness N, and a strong solution of Cauchy problem (1) blowsupina conditions: finite time 𝑇max <∞(𝑇min >−∞)if and only if

‖𝑢̃‖ 𝑟 𝑠 ≤𝜀, 𝐿𝑡(𝐼,𝑀𝑝,1) 𝑐 ‖𝑢(𝑡)‖𝐿𝑘([𝑡 ,𝑇 ],𝑀0 ) =∞(‖𝑢(𝑡)‖𝐿𝑘([𝑇 ,𝑡 ],𝑀0 ) =∞). 𝑡 0 max ∞,1 𝑡 min 0 ∞,1 󵄩 󵄩 󸀠 ̃ (15) 󵄩𝐾 (𝑡 −0 𝑡 )(𝑢(𝑡0)−𝑢(𝑡0)) 󵄩 (22) Remark 9. From another point of view, the above blowup +𝐾 (𝑡0 −𝑡 )(𝑢𝑡 (𝑡0)−𝑢̃𝑡 (𝑡0))󵄩 𝑟 𝑠 ≤𝜀, 󵄩𝐿𝑡(𝐼,𝑀𝑝,1) ‖𝑢(𝑡)‖ 𝑠 criterion implies that 𝑀𝑝,1 cannot blow up too slowly 𝑡 𝑇 when tends to a finite blowup time ;thatis, ‖𝑒‖𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ) ≤𝜀, 𝑡 𝑝󸀠,1

‖𝑢(𝑡)‖ 𝑠 𝑀𝑝,1 lim =+∞ (16) for some 0<𝜀<𝜀𝑐,where𝜀𝑐 is a small constant. 𝑡→𝑇|𝑇 − 𝑡|−1/𝑘+𝜀 Then, there exists a solution 𝑢 to (1) with initial values 𝑢(𝑡 ), 𝑢 (𝑡 ) 𝑡 for every 𝜀>0. 0 𝑡 0 at time 0 satisfying 󵄩 󵄩 󵄩(𝜕2 +𝐼−Δ)(𝑢−𝑢)̃ + 𝑒󵄩 ≲𝜀, We also obtain a scattering theorem for these equations 󵄩 𝑡 󵄩𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ) (23) 𝑟 𝑠 𝑡 𝑝󸀠,1 provided a bounded 𝐿𝑡(R,𝑀𝑝,1) norm. ‖𝑢−𝑢̃‖𝐿𝑟 (𝐼,𝑀𝑠 ) ≲𝜀, 𝑟 𝑠 𝑡 𝑝,1 (24) Theorem 10 (𝐿𝑡(R,𝑀𝑝,1) bounds imply scattering). Let 𝑢0 ∈ 𝑠 𝑠−1 𝑠 󸀠 ̃󸀠 𝑀 ,𝑢 ∈𝑀 𝑢 𝑀 ‖𝑢−𝑢̃‖ ∞ 𝑠 ≲𝐶(|𝐼|) (𝑀 + 𝑀 +𝜀), 𝑝,1 1 𝑝,1 ,andlet be a global strong 𝑝,1 solution to 𝐿𝑡 (𝐼,𝑀𝑝,1) (25) (1) such that 󵄩 ̃ 󵄩 󸀠 ̃󸀠 󵄩𝑢𝑡 − 𝑢𝑡󵄩𝐿∞(𝐼,𝑀𝑠−1) ≲𝐶(|𝐼|) (𝑀 + 𝑀 +𝜀), (26) ‖𝑢‖ 𝑟 𝑠 ≤𝑀 𝑡 𝑝,1 𝐿𝑡(R,𝑀𝑝,1) (17) 󵄩 󵄩 󸀠 ± ± 𝑠 󵄩 󵄩 ̃󸀠 󵄩𝑢𝑡󵄩𝐿∞(𝐼,𝑀𝑠 ) ≲𝐶(|𝐼|) (𝑀 + 𝑀 + 𝑀 +𝜀), (27) for some constant 𝑀>0.Then,thereexistV1 , V2 ∈𝑀𝑝,1 such 𝑡 𝑝,1 V± ≜ 𝐺(𝑡)V±+𝐺(𝑡)V± that 1 2 aresolutionstothefreeKlein-Gordon 󵄩 󵄩 ̃ 󸀠 ̃󸀠 󵄩𝑢𝑡󵄩 ∞ 𝑠−1 ≲𝐶(|𝐼|) (𝑀+𝑀 + 𝑀 +𝜀). equation 𝑢𝑡𝑡 +(𝐼−Δ)𝑢=0,and 𝐿𝑡 (𝐼,𝑀𝑝,1 ) (28) 󵄩 ±󵄩 󵄩𝑢(𝑡) − V 󵄩 𝑠 󳨀→ 0 Theorem 12 󵄩 󵄩𝑀𝑝,1 (18) (long-time perturbations). Let I be a compact time interval, and let 𝑢̃ be an approximate solution to (1) in 𝑡→±∞ as . the sense of (19) for some function e. Assume that Finally, we will discuss the stability theory. The stability ‖𝑢̃‖𝐿𝑟 (𝐼,𝑀𝑠 ) ≤𝐿, (29) theory for (1) means that given an approximate solution 𝑡 𝑝,1 󵄩 󵄩 ̃ 𝑢̃ + (𝐼−Δ) 𝑢=𝑓̃ (𝑢̃) +𝑒 ‖𝑢̃‖𝐿∞(𝐼,𝑀𝑠 ) ≤𝑀, 󵄩𝑢̃𝑡󵄩 ∞ 𝑠−1 ≤ 𝑀 𝑡𝑡 (19) 𝑡 𝑝,1 󵄩 󵄩𝐿𝑡 (𝐼,𝑀𝑝,1 ) (30) 4 Abstract and Applied Analysis

̃ for some constants 𝑀, 𝑀,and𝐿.Let𝑡0 ∈𝐼and let 𝑢(𝑡0)∈ with 𝐶 = 𝐶(|𝐼|,𝐿 ‖𝑢 𝑘+1(𝐼,𝑀𝑠 ));thatis,𝑢𝑛 →𝑢0 in 𝑠 𝑠−1 𝑡 𝑝,1 𝑀 , 𝑢𝑡(𝑡0)∈𝑀 be close to 𝑢(𝑡̃ 0), 𝑢̃𝑡(𝑡0),respectively,inthe 𝑠 1 𝑠−1 𝑝,1 𝑝,1 𝐶(𝐼,𝑝,1 𝑀 )∩𝐶 (𝐼,𝑝,1 𝑀 ). sense that 󵄩 ̃ 󵄩 󸀠 Also, one can deduce continuous dependence for 𝑝∈ 󵄩𝑢(𝑡0)−𝑢(𝑡0)󵄩𝑀𝑠 ≤𝑀, 𝑝,1 [1, ∞], 𝑘 ∈ N directly without using perturbation theorem, (31) 󵄩 󵄩 ̃󸀠 and the proof is not difficult, so we omit the details. 󵄩𝑢𝑡(𝑡0)−𝑢̃𝑡(𝑡0)󵄩 𝑠−1 ≤ 𝑀 󵄩 󵄩𝑀𝑝,1 Corollary 14. Assume that (𝑝, 𝑟) is a k-admissible pair. Denote 𝑀󸀠, 𝑀̃󸀠 >0 𝐴 𝑀𝑠+2𝜎(𝑝) ×𝑀𝑠+2𝜎(𝑝)−1 for some . Moreover, assume the following smallness by the subset of 𝑝󸀠,1 𝑝󸀠,1 ,suchthat,forevery conditions: (𝑢0,𝑢1)∈𝐴,theCauchyproblem(1) with initial data 𝑢0,𝑢1 𝑠 󵄩 󸀠 󵄩 ̃ has a global strong 𝑀𝑝,1 solution 𝑢 on R and ‖𝑢‖𝐿𝑟 (𝐼,𝑀𝑠 ) <∞. 󵄩𝐾 (𝑡 −0 𝑡 )(𝑢(𝑡0)−𝑢(𝑡0)) 𝑡 𝑝,1 𝐴 𝑀𝑠+2𝜎(𝑝) ×𝑀𝑠+2𝜎(𝑝)−1 󵄩 Then, the set is open in 󸀠 󸀠 . 󵄩 𝑝 ,1 𝑝 ,1 +𝐾(𝑡−𝑡0)(𝑢𝑡(𝑡0)−𝑢̃𝑡(𝑡0))󵄩 𝑟 𝑠 ≤𝜀, 󵄩𝐿𝑡(𝐼,𝑀𝑝,1) (32) 2. Preliminaries ‖𝑒‖𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ) ≤𝜀, 𝑡 𝑝󸀠,1 If 𝑋 and 𝑌 are two quantities (typically nonnegative), we will for some 0<𝜀<𝜀1,where𝜀1 =𝜀1(𝐿) is a small constant. Then, often use the notation 𝑋≲𝑌to denote the statement that there exists a solution 𝑢 to (1) with initial values 𝑢(𝑡0), 𝑢𝑡(𝑡0) at 𝑋≤𝐶𝑌for some absolute constant 𝐶>0,where 𝐶 can time 𝑡0 satisfying depend on 𝑛, 𝑘, 𝑝,𝑟, but it might be different from line to 𝑛 󵄩 󵄩 line. Given 𝑘=(𝑘1,𝑘2,...,𝑘𝑛)∈Z ,wewrite|𝑘| := |𝑘1|+ 󵄩(𝜕2 +𝐼−Δ)(𝑢−𝑢̃) +𝑒󵄩 ≲𝐶(𝐿) 𝜀, 󵄩 𝑡 󵄩𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ) (33) |𝑘2|+⋅⋅⋅+|𝑘𝑛|, |𝑘|∞ := sup𝑖=1,2,...,𝑛|𝑘𝑖|,and⟨𝑘⟩ := 1 + |𝑘|.Let 𝑡 𝑝󸀠,1 𝑟 𝑛 𝑛 𝐿𝑥(R ) denote the Banach space of functions 𝑓:R → C ‖𝑢−𝑢̃‖ 𝑟 𝑠 ≲𝐶(𝐿) 𝜀, whose norm 𝐿𝑡(𝐼,𝑀𝑝,1) (34) 1/𝑟 󵄩 󵄩 󵄨 󵄨𝑟 󸀠 ̃󸀠 󵄩𝑓󵄩 := (∫ 󵄨𝑓(𝑥)󵄨 𝑑𝑥) <∞, 1≤𝑟<∞. ‖𝑢−𝑢̃‖𝐿∞(𝐼,𝑀𝑠 ) ≲𝐶(|𝐼| ,𝐿) (𝑀 + 𝑀 +𝜀), 󵄩 󵄩𝐿𝑟 (R𝑛) 󵄨 󵄨 (41) 𝑡 𝑝,1 (35) 𝑥 R𝑛

󵄩 󵄩 󸀠 󵄩 󵄩 ̃󸀠 The norm ‖𝑓‖ ∞ 𝑛 is defined with the usual modification. 󵄩𝑢𝑡 − 𝑢̃𝑡󵄩 ∞ 𝑠−1 ≲𝐶(|𝐼| ,𝐿) (𝑀 + 𝑀 +𝜀), 𝐿𝑥 (R ) 󵄩 󵄩𝐿𝑡 (𝐼,𝑀𝑝,1 ) (36) We also abbreviate ‖⋅‖𝐿𝑟 (R𝑛) for ‖⋅‖𝐿𝑟 ,or‖⋅‖𝑟, when there is 𝑟 𝑥 󸀠 ̃󸀠 𝐿 (𝐼, 𝑋) ‖𝑢‖ ∞ 𝑠 ≲𝐶(|𝐼| ,𝐿) (𝑀+𝑀 + 𝑀 +𝜀) , no confusion. Weuse 𝑡 to denote the space-time norm: 𝐿𝑡 (𝐼,𝑀𝑝,1) (37) 1/𝑟 󵄩 󵄩 󸀠 𝑟 󵄩 󵄩 ̃ ̃󸀠 ‖𝑢‖ 𝑟 =(∫ ‖𝑢‖ 𝑑𝑡) 󵄩𝑢𝑡󵄩𝐿∞(𝐼,𝑀𝑠−1) ≲𝐶(|𝐼| ,𝐿) (𝑀+𝑀 + 𝑀 +𝜀). (38) 𝐿 (𝐼,𝑋) 𝑋 (42) 𝑡 𝑝,1 𝑡 𝐼 As applications of the above stability theorems, we have with the usual modifications when 𝑝, 𝑞,or𝑟 is infinite. For the following corollaries. the operator 𝜔=Id −Δ,theoperator

1/2 Corollary 13 (continuous dependence). Assume that 1/(𝑘 + sin (𝑡𝜔 ) 𝐾 (𝑡) := (43) 2) ≤ 1/𝑝 < 1/2 − 1/𝑛(𝑘 +1),and𝑢(𝑡) is a strong solution to 1/2 𝑠 𝑠−1 𝜔 (1) with initial data 𝑢(𝑡0)=𝑢0 ∈𝑀𝑝,1, 𝑢𝑡(𝑡0)=𝑢1 ∈𝑀𝑝,1 .If 𝑠 𝑠−1 and the Klein-Gordon semigroup (𝑢0,𝑛,𝑢1,𝑛)→(𝑢0,𝑢1) in 𝑀𝑝,1 ×𝑀𝑝,1 ,then 1/2 𝐺 (𝑡) := 𝑒𝑖𝑡𝜔 (44) lim 𝑇max (𝑢0,𝑛,𝑢1,𝑛)≥𝑇max (𝑢0,𝑢1), 𝑛→∞ (39) have been defined in Section 1.Thus,wemayrecallDuhamel’s 𝑇 (𝑢 ,𝑢 )≤𝑇 (𝑢 ,𝑢 ). 𝑛→∞lim min 0,𝑛 1,𝑛 min 0 1 formula: 󸀠 𝑢 (𝑡) =𝐾 (𝑡 −0 𝑡 )𝑢0 +𝐾(𝑡−𝑡0)𝑢1 For every compact interval 𝐼⊂(𝑇min(𝑢0,𝑢1), 𝑇max(𝑢0,𝑢1)),let 𝑢𝑛 be the solution to (1) on 𝐼 with initial 𝑢0,𝑛, 𝑢1,𝑛,andthenwe 𝑡 (45) have − ∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏. 𝑡 󵄩 󵄩 0 󵄩𝑢𝑛 −𝑢󵄩𝐿∞(𝐼,𝑀𝑠 ) 𝑡 𝑝,1 Also, we recall the integral form of Gronwall’s inequality. 󵄩 󵄩 󵄩 󵄩 ≲𝐶(󵄩𝑢 −𝑢 󵄩 + 󵄩𝑢 −𝑢 󵄩 ), 󵄩 0,𝑛 0󵄩𝑀𝑠 󵄩 1,𝑛 1󵄩𝑀𝑠−1 Lemma 15 (Gronwall inequality and integral form [5]). Let 𝑝,1 𝑝,1 + (40) 𝐴:[𝑡0,𝑡1]→R be continuous and nonnegative, and 󵄩 󵄩 󵄩(𝑢𝑛)𝑡 −𝑢𝑡󵄩 ∞ 𝑠 suppose that A obeys the integral inequality 󵄩 󵄩𝐿𝑡 (𝐼,𝑀𝑝,1) 𝑡 󵄩 󵄩 󵄩 󵄩 ≲𝐶(󵄩𝑢 −𝑢 󵄩 + 󵄩𝑢 −𝑢 󵄩 ) 𝐴 (𝑡) ≤𝐶+∫ 𝐵 (𝑠) 𝐴 (𝑠) 𝑑𝑠 󵄩 0,𝑛 0󵄩𝑀𝑠 󵄩 1,𝑛 1󵄩𝑀𝑠−1 (46) 𝑝,1 𝑝,1 𝑡0 Abstract and Applied Analysis 5

+ for all 𝑡∈[𝑡0,𝑡1],where𝐶≥0and 𝐵:[𝑡0,𝑡1]→R is Below, we list some basic properties for the space 𝑠 𝑛 continuous and nonnegative. Then, we have 𝑀𝑝,𝑞(R ). 𝑡 𝐴 (𝑡) ≤𝐶exp (∫ 𝐵 (𝑠) 𝑑𝑠) (47) Lemma 17 (embedding). Let 𝑠1,𝑠2 ∈ R and 0< 𝑡 0 𝑝1,𝑝2,𝑞1,𝑞2 ≤∞. 𝑡∈[𝑡,𝑡 ] 𝑠 𝑠 for all 0 1 . 𝑠 ≤𝑠 𝑝 ≤𝑝 𝑞 ≤𝑞 𝑀 1 ⊂𝑀2 (1) If 2 1, 1 2,and 1 2,then 𝑝1,𝑞1 𝑝2,𝑞2 . 𝑠 𝑠 S := S(R𝑛) S󸀠 := S󸀠(R𝑛) 𝑞 <𝑞 𝑠 −𝑠 >𝑛/𝑞 −𝑛/𝑞 𝑀 1 ⊂𝑀2 Let be the Schwartz space and (2) If 2 1 and 1 2 2 1,then 𝑝,𝑞1 𝑝,𝑞2 . the tempered distribution space. We introduce the definition of modulation space, which was introduced by Feichtinger [6] Proposition 18 (isomorphism). Let 0<𝑝,𝑞≤∞, 𝑠, 𝜎 ∈ 𝜎/2 𝑠 𝑠−𝜎 in 1983 by short-time Fourier transform. We will also display R.Then,𝐽𝜎 =(𝐼−Δ) :𝑀𝑝,𝑞 →𝑀𝑝,𝑞 is an isomorphic somebasicpropertiesofthisfunctionspace. mapping. Applying the frequency-uniform localization techniques, one can get an equivalent definition of modulation spaces (see The proof of Proposition 18 can be found in [6]forthe [7] for details) as follows. Let 𝑄𝑘 be the unit cube with the cases 1≤𝑝,𝑞≤∞and [7, 8]forthecases0<𝑝,𝑞<1.We 𝑛 center at 𝑘,so{𝑄𝑘}𝑘∈Z𝑛 constitutes a decomposition of R . denote 𝐽=𝐽1. 𝑛 First, we construct a smooth cut-off function. Let 𝜌∈S(R ) 𝑛 𝑠 and let 𝜌:R → [0, 1] be a smooth function satisfying Lemma 19 (for uniform boundedness of 𝐺(𝑡) in 𝑀𝑝,𝑞,see 𝜌(𝜉) =1 for |𝜉|∞ ≤1/2and 𝜌(𝜉) =0 for |𝜉| ≥.Let 1 𝜌𝑘 [9]). Let 𝑠∈R, 1≤𝑝≤∞,and0<𝑞<∞.Onehas 𝜌 be a translation of , 󵄩 󵄩 𝑛|1/2−1/𝑝|󵄩 󵄩 𝑛 󵄩𝐺(𝑡)𝑓󵄩𝑀𝑠 ≤𝐶(1+|𝑡|) 󵄩𝑓󵄩𝑀𝑠 , (54) 𝜌𝑘 (𝜉) =𝜌(𝜉−𝑘) ,𝑘∈Z . (48) 𝑝,𝑞 𝑝,𝑞 𝑛 We see that 𝜌𝑘(𝜉) = 1 in 𝑄𝑘,so∑𝑘∈Z𝑛 𝜌𝑘(𝜉) ≥ 1 for all 𝜉∈R . where the constant C depends only on p, q, s, and n. Denote −1 Onecanalsofindtheseestimatesin[10]andamore 𝑖𝑡(𝑚2𝐼+|Δ|)1/2 𝑛 general estimate on 𝑒 with 𝑚 =0̸ , 𝑡≥1in Chen 𝜎𝑘 (𝜉) =𝜌𝑘 (𝜉) (∑𝜌𝑙(𝜉)) ,𝑘∈Z . (49) 𝑙∈Z𝑛 and Fan [11]. Chen and Fan also showed that the exponent |1/𝑝 − 1/2| |1/𝑝−1/2| isthebestpossibleinthefactor𝑡 if 𝑛 equals Then, {𝜎𝑘(𝜉)}𝑘∈Z𝑛 satisfies the following properties: 1[11]. 󵄨 󵄨 󵄨𝜎𝑘 (𝜉)󵄨 ≥𝑐, ∀𝜉∈𝑄𝑘, Lemma 20 (for truncated decay estimate of 𝐺(𝑡),see 󵄨 󵄨 󸀠 supp (𝜎𝑘) ⊂ {𝜉 : 󵄨𝜉−𝑘󵄨∞ ≤1}, Proposition 4.2 in [2]). Let 𝑠∈R,2≤𝑝<∞, 1/𝑝+1/𝑝 =1, and 0<𝑞<∞, 𝜃∈[0,1]; 𝑛 (50) ∑ 𝜎𝑘 (𝜉) ≡1, ∀𝜉∈R , 𝑛 1 1 𝑘∈Z 2𝜎 (𝑝) = (𝑛+2) ( − ). 2 𝑝 (55) 󵄨 𝛼 󵄨 𝑛 + 𝑛 󵄨𝐷 𝜎𝑘 (𝜉)󵄨 ≤𝐶|𝛼|,∀𝜉∈R ,𝛼∈(Z ∪ {0}) . 𝑛 One has In fact, {𝜎𝑘(𝜉)}𝑘∈Z𝑛 constitutes a smooth decomposition of R 󵄩 󵄩 −𝑛𝜃(1/2−1/𝑝)󵄩 󵄩 𝜎 (𝜉) = 𝜎(𝜉 −𝑘) 󵄩𝐺(𝑡)𝑓󵄩 ≤ 𝐶(1 + |𝑡|) 󵄩𝑓󵄩 𝑠+𝜃2𝜎(𝑝) , and 𝑘 ,inwhich 󵄩 󵄩𝑀𝑠 󵄩 󵄩𝑀 (56) 𝑝,𝑞 𝑝󸀠,𝑞 −1 where the constant C depends only on p, q, s, and n. 𝜎 (𝜉) =𝜌(𝜉) ( ∑ 𝜌𝑙(𝜉)) . (51) 𝑛 𝑙∈Z + Lemma 21 (algebra property [4]). Let 𝑚∈Z ,𝑠 ≥. 0 The frequency-uniform decomposition operators can be Assume that 1/𝑝1 +⋅⋅⋅+1/𝑝𝑚 =1/𝑝0, 1/𝑞1 +⋅⋅⋅+1/𝑞𝑚 = exactly defined by 𝑚−1+1/𝑞0 with 0<𝑝𝑖 ≤∞,1≤𝑞𝑖 ≤∞for 1≤𝑖≤𝑚. −1 ◻ := F 𝜎 F (52) One has 𝑘 𝑘 󵄩 󵄩 𝑛 󵄩 𝑚 󵄩 𝑚 for 𝑘∈Z . 󵄩 󵄩 󵄩 󵄩 󵄩∏𝑢𝑖󵄩 ≤𝐶∏󵄩𝑢𝑖󵄩𝑀𝑠 , 󵄩 󵄩 𝑝𝑖,𝑞𝑖 (57) 󵄩 𝑖=1 󵄩𝑀𝑠 𝑖=1 Definition 16 (modulation space). Let 𝑠∈R,0<𝑝,𝑞≤∞, 𝑝0,𝑞0 and one defines the modulation space where C is independent of 𝑢𝑖. { 𝑠 𝑛 󸀠 Lemma 22 (Leibniz rule for modulation space [3]). Let 𝑚∈ 𝑀𝑝,𝑞 (R )={𝑓∈S : + Z ,𝑠≥0and 1/𝑝1 +⋅⋅⋅+1/𝑝𝑚 =1/𝑝0 ,1/𝑞1 +⋅⋅⋅+1/𝑞𝑚 = { 𝑚−1+1/𝑞0 ,and0<𝑝𝑖 ≤∞,1≤𝑞𝑖 ≤∞for 1≤𝑖≤𝑚. 1/𝑞 } One has 󵄩 󵄩 𝑠𝑞󵄩 󵄩𝑞 󵄩 󵄩 󵄩𝑓󵄩𝑀𝑠 := ( ∑ ⟨𝑘⟩ 󵄩𝑘𝑓󵄩𝑝) <∞ . 󵄩 𝑚 󵄩 𝑚 𝑚 𝑝,𝑞 } 󵄩 󵄩 󵄩 󵄩 𝑘∈Z𝑛 󵄩∏𝑢 󵄩 ≤𝐶∑ ∏󵄩𝑢 󵄩 , } 󵄩 𝑖󵄩 󵄩 𝑖󵄩𝑀𝛿(𝑖,𝑗)𝑠 󵄩 󵄩 𝑝𝑖,𝑞𝑖 (58) 󵄩 𝑖=1 󵄩𝑀𝑠 𝑗=1 𝑖=1 (53) 𝑝0,𝑞0 6 Abstract and Applied Analysis

where C is independent of 𝑢𝑖,and𝛿(𝑖, 𝑗) =1 if 𝑖=𝑗and Proposition 29 (nonlinear estimate). Let 𝑡0 ∈𝐼, 𝐹(𝑢) = 𝑢 𝑢 𝑢 𝑘+1 vanishes otherwise. Particularly, if we choose 𝑖 to be or , ∏𝑖=1 𝑢𝑖.Forany𝑘-admissible pair (𝑝, 𝑟),onehas 𝑝𝑖 =∞and 𝑞𝑖 =1for 𝑖=0,1,...,𝑚.Wehave 󵄩 󵄩 󵄩 𝑡 󵄩 󵄩 𝑘+1 󵄩 𝑘 󵄩∫ 𝐾 (𝑡−𝜏) 𝐹 (𝑢 (𝜏)) 𝑑𝜏󵄩 󵄩𝜋(𝑢 )󵄩 ≤𝐶‖𝑢‖ 𝑠 ‖𝑢‖ . 󵄩 󵄩 󵄩 󵄩 𝑠 𝑀 𝑀0 (59) 󵄩 󵄩 󵄩 󵄩𝑀∞,1 ∞,1 ∞,1 𝑡0 𝑟 𝑠 󵄩 󵄩𝐿𝑡(𝐼,𝑀𝑝,1)

Lemma 23 ≲ ‖𝐹(𝑢)‖ 𝑟/(𝑘+1) (for Bernstein multiplier theorem, see Proposition 𝐿 (𝐼,𝑀𝑠 ) (64) 𝐿∈Z,𝐿 > 𝑛/2 𝜕𝛼 𝜌∈𝐿2,𝑖 = 𝑡 𝑝,1 1.11 in [10]). Let , 𝑥𝑖 1,2,...,𝑛, 0 ≤ 𝛼 ≤𝐿 𝜌 𝐿 1≤ 𝑘+1 .Then, is a multiplier on 𝑝, 󵄩 󵄩 𝑝≤∞ 𝐶 ≲ ∏󵄩𝑢𝑖󵄩 𝑟 𝑠 . .Moreover,thereexistsaconstant such that 󵄩 󵄩𝐿𝑡(𝐼,𝑀𝑝,1) 𝑖=1

𝑛 𝑛/2𝐿 󵄩 󵄩 󵄩 󵄩1−𝑛/2𝐿 󵄩 𝐿 󵄩 𝐾(𝑡)𝜔1/2 = (𝐺(𝑡) − 𝐺(−𝑡))/2𝑖 󵄩𝜌󵄩 ≤𝐶󵄩𝜌󵄩 2 (∑󵄩𝜕 𝜌󵄩 ) . (60) Proof. Observe that .For 󵄩 󵄩𝑀𝑝 󵄩 󵄩𝐿 󵄩 𝑥𝑖 󵄩𝐿2 𝑖=1 any 𝑘-admissible pair (𝑝, 𝑟), using the general Minkowski inequality, Proposition 18,andLemma20,wehave Lemma 24. Let 𝑠∈R, 0<𝑝,𝑞<∞,andletΩ be a compact 𝑛 Ω ̂ 󵄩 𝑡 󵄩 subset of R .Then,S ={𝑓:𝑓∈S and Supp𝑓⊂Ω}is 󵄩 󵄩 𝑠 󵄩∫ 𝐾(𝑡 − 𝜏)𝐹(𝑢(𝜏))𝑑𝜏󵄩 dense in 𝑀 . 󵄩 󵄩 𝑝,𝑞 󵄩 𝑡0 󵄩 𝑟 𝑠 󵄩 󵄩𝐿𝑡(𝐼,𝑀𝑝,1) 𝑘 𝑎𝑑𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒 𝑝𝑎𝑖𝑟 󵄩 𝑡 󵄩 Definition 25 ( - ). One calls the exponent 󵄩 󵄩 (𝑝, 𝑟) 𝑘 𝛾 ≲ 󵄩∫ ‖𝐾 (𝑡−𝜏) 𝐹 (𝑢 (𝜏))‖𝑀𝑠 𝑑𝜏󵄩 -admissible if there exists another exponent such 󵄩 𝑝,1 󵄩 󵄩 𝑡0 󵄩𝐿𝑟 (𝐼) that 𝑡 󵄩 󵄩 󵄩 𝑡 󵄩 1 𝑘 󵄩 −𝑛𝜃(1/2−1/𝑝) 󵄩 + =1, ≲ 󵄩∫ (1+|𝑡−𝜏|) ‖𝐹(𝑢)‖ 𝑠+𝜃2𝜎(𝑝)−1 𝑑𝜏󵄩 󵄩 𝑀 󸀠 󵄩 󵄩 𝑡0 𝑝 ,1 󵄩 𝑟 𝛾 𝑟 󵄩 󵄩𝐿𝑡(R) 󵄩 󵄩 1 1 𝑛 1 1 󵄩 −𝑛𝜃(1/2 −1/𝑝) 󵄩 ≲ 󵄩∫ (1+|𝑡−𝜏|) ‖𝐹(𝑢)‖ 𝑠+𝜃2𝜎(𝑝)−1 𝜒 (𝜏)𝑑𝜏󵄩 ≤ ≤ ∧𝑛( − ), (61) 󵄩 𝑀 𝐼 󵄩 𝑘+1 𝛾 𝑛+2 2 𝑝 󵄩 R 𝑝󸀠,1 󵄩 𝑟 𝐿𝑡(R) 1 1 1 1 (65) ≤ < − . 𝑘+2 𝑝 2 𝑛 (𝑘+1) for any 𝜃∈[0,1]. If 1/𝛾 = 𝑛/(𝑛 + 2) ∧ 𝑛(1/2 −1/𝑝),then1/𝛾 < 1 and there Remark 26. From Definition 25,if(𝑝, 𝑟) is 𝑘-admissible, we exists 𝜃∈(0, 1] such that can easily verify that 1 1 1 𝑛 1 1 1 1 =𝜃𝑛( − )= ∧𝑛( − ). 𝑘 + 1 ≤ 𝑟 < ∞, 𝑟𝑛( − )>1. 𝛾 2 𝑝 𝑛+2 2 𝑝 (66) 2 𝑝 (62) 𝜃 𝜃2𝜎(𝑝) − 1≤0 Moreover, we have the following inequality: With this ,wehave .ObservingDefinition25 and Remark 27,wehave 󵄩 󵄩 󵄩𝐺(𝑡0 −𝑡 )𝑢0󵄩𝐿𝑟 (R,𝑀𝑠 ) 𝑡 𝑝,1 1 𝑘+1 1 − 𝑛𝜃 (1/2 − 1/𝑝) = − 󵄩 󵄩 (67) 󵄩 󵄩 󵄩 󵄩 𝑟 𝑟 1 ≲ 󵄩(1 + |𝑡 − 𝑡 |)−𝑛(1/2−1/𝑝)󵄩𝑢 󵄩 󵄩 󵄩 0 󵄩 0󵄩𝑀𝑠 󵄩 (63) 󵄩 𝑝󸀠,1 󵄩𝐿𝑟 (R) 𝑡 and 𝑟/(𝑘 + 1) >1.Sointhiscasewechoose𝜃 satisfying (66) 󵄩 󵄩 and exploit Lemma 17 and the Hardy-Littlewood-Sobolev ≲ 󵄩𝑢0󵄩𝑀𝑠 . 𝑝󸀠,1 inequality to have

1/(𝑘 + 2) + 1/𝑛(𝑘 + 1) <1/2 𝑘 󵄩 𝑡 󵄩 Remark 27. If , there exist - 󵄩 󵄩 1/(𝑘 + 2) + 1/𝑛(𝑘 + 1) <1/2 󵄩∫ 𝐾(𝑡 − 𝜏)𝐹(𝑢(𝜏))𝑑𝜏󵄩 admissible pairs. The condition 󵄩 󵄩 󵄩 𝑡0 󵄩𝐿𝑟 (𝐼,𝑀𝑠 ) canensurethat2<𝑘𝑛. 𝑡 𝑝,1 󵄩 𝑠 󵄩 −𝑛𝜃(1/2 − 1/𝑝) Definition 28. One defines the strong 𝑀𝑝,1 solution to (1) ≲ 󵄩∫ (1 + |𝑡 − 𝜏|) 󵄩 𝑢∈𝐶(𝐼,𝑀𝑠 )∩𝐶1(𝐼,𝑠−1 𝑀 ) R as follows: the distribution 𝑝,1 𝑝,1 is 󵄩 󵄩 the solution to (1)inthesenseof(45) with the initial data × ‖𝐹(𝑢)‖ 𝑠 𝜒 (𝜏)𝑑𝜏󵄩 𝑠 𝑠−1 𝑀 𝐼 󵄩 (𝑢(𝑡 ), 𝑢 (𝑡 )) = (𝑢 ,𝑢 )∈𝑀 ×𝑀 𝑝󸀠,1 󵄩 𝑟 0 𝑡 0 0 1 𝑝,1 𝑝,1 . 𝐿𝑡(R) 󵄩 󵄩 We establish the following nonlinear estimate. = 󵄩(‖𝐹 (𝑢)‖𝑀𝑠 𝜒𝐼 (𝑡)) 󵄩 𝑝󸀠,1 Abstract and Applied Analysis 7

󵄩 𝑘+1 −𝑛𝜃(1/2−1/𝑝)󵄩 Corollary 30. 𝑘∈N 𝑓(𝑢) = 𝜋(𝑢 ) ∗(1 + |𝑡|) 󵄩 Let , ,andthenonehas 󵄩‖ 𝑟 𝐿𝑡(R) 󵄩 󵄩 󵄩 𝑡 󵄩 󵄩 󵄩 󵄩∫ 𝐾(𝑡 − 𝜏)𝑓(𝑢(𝜏))𝑑𝜏󵄩 󵄩 󵄩 󵄩 󵄩 ≲ 󵄩‖𝐹(𝑢)‖ 𝑠 𝜒 (𝜏)󵄩 󵄩 𝑡0 󵄩 𝑟 𝑠 󵄩 𝑀 𝐼 󵄩 󵄩 󵄩𝐿𝑡(𝐼,𝑀𝑝,1) 󵄩 𝑝󸀠,1 󵄩 𝑟/(𝑘+1) 𝐿𝑡 󵄩 󵄩 (73) ≲ 󵄩𝑓 (𝑢)󵄩𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ) = ‖𝐹 (𝑢)‖𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ). 𝑡 𝑝,1 𝑡 𝑝󸀠,1 𝑘+1 ≲ ‖𝑢‖ 𝑟 𝑠 . (68) 𝐿𝑡(𝐼,𝑀𝑝,1) If (1/𝛾) < 𝑛/(𝑛 + 2) ∧ 𝑛(1/2 −1/𝑝), then there exists 𝜃∈ [0, 1] such that 3. Local Well-Posedness 1 1 1 𝑛 1 1 In this section, we establish the local theory for the Cauchy <𝜃𝑛( − )≤ ∧𝑛( − ). 𝑘, 𝑛 𝛾 2 𝑝 𝑛+2 2 𝑝 (69) problem (1).Intherestofthispaper,weassumethat satisfy 1/(𝑘+2)+1/𝑛(𝑘+1)<, 1/2 so there exist 𝑘-admissible With this 𝜃,wehave𝛾𝜃𝑛(1/2 − 1/𝑝) >1 and 𝜃2𝜎(𝑝) − exponents by Remark 27. 1≤0. Taking advantage of Young’s inequality and Holder’s¨ inequality, we obtain 3.1. Proof of Existence Part of Theorem 1. We use the fixed pointargumenttoconstructalocalsolution.Let𝐵𝑎 ={𝑢: 󵄩 𝑡 󵄩 󵄩 󵄩 ‖𝑢‖𝐿𝑟 (𝐼,𝑀𝑠 ) ≤𝑎}, and define a map J on 𝐵𝑎: 󵄩 󵄩 𝑡 𝑝,1 󵄩∫ 𝐾(𝑡 − 𝜏)𝐹(𝑢(𝜏))𝑑𝜏󵄩 󵄩 𝑡0 󵄩 𝑟 𝑠 󵄩 󵄩𝐿𝑡(𝐼,𝑀𝑝,1) 󸀠 J (V) =𝐾 (𝑡 −0 𝑡 )𝑢0 +𝐾(𝑡−𝑡0)𝑢1 󵄩 −𝑛𝜃(1/2−1/𝑝)󵄩 ≲ 󵄩(1 + |𝑡|) 󵄩 𝑡 󵄩 󵄩𝐿𝛾 (74) 𝑡 (70) − ∫ 𝐾 (𝑡−𝜏) 𝑓 (V) 𝑑𝜏. 󵄩 󵄩 𝑡0 󵄩 󵄩 × 󵄩‖𝐹(𝑢)‖𝑀𝑠 𝜒𝐼(𝜏)󵄩 󵄩 𝑝󸀠,1 󵄩𝐿𝑟/(𝑘+1) 𝑡 We want to choose suitable 𝑎 and 𝜂 so that J :𝐵𝑎 →𝐵𝑎 is a contraction. By corollary of Proposition 29,wehave ≲ ‖𝐹(𝑢)‖𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ). 𝑡 𝑝󸀠,1 𝑘+1 ‖J(V)‖ 𝑟 𝑠 ≤𝜂+𝐶𝑎 . 𝐿𝑡(𝐼,𝑀𝑝,1) (75) In general, we get the first result. Now, by Lemma 21 and 𝑘 Holder’s¨ inequality, we have Let 𝜂1 satisfy 𝐶(2𝜂1) =1/2and choose 𝜂≤𝜂1,andthenwe have J :𝐵2𝜂 →𝐵2𝜂. Indeed, 󵄩 𝑡 󵄩 󵄩 󵄩 󵄩∫ 𝐾(𝑡 − 𝜏)𝐹(𝑢(𝜏))𝑑𝜏󵄩 𝑘+1 󵄩 󵄩 ‖J(V)‖ 𝑟 𝑠 ≤𝜂+𝐶(2𝜂) 󵄩 𝑡0 󵄩 𝑟 𝑠 𝐿 (𝐼,𝑀 ) 󵄩 󵄩𝐿𝑡(𝐼,𝑀𝑝,1) 𝑡 𝑝,1 (76) 𝑘 ≲ ‖𝐹(𝑢)‖𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ) ≤ 𝜂 + 𝐶(2𝜂1) (2𝜂) ≤ 2𝜂. 𝑡 𝑝󸀠,1

𝑟/(𝑘+1) (𝑘+1)/𝑟 We also have 󵄩𝑘+1 󵄩 󵄩 󵄩 󵄩 󵄩 ‖J(𝑢) − J(V)‖𝐿𝑟 (𝐼,𝑀𝑠 ) ≲(∫ 󵄩∏𝑢𝑖 (𝜏)󵄩 𝑑𝜏) 𝑡 𝑝,1 𝐼 󵄩 󵄩 󵄩 𝑖=1 󵄩𝑀𝑠 (71) 𝑝󸀠,1 󵄩 󵄩 ≤𝐶󵄩𝑓(𝑢) −V 𝑓( )󵄩𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ) (𝑘+1)/𝑟 𝑡 𝑝󸀠,1 (77) 𝑘+1 󵄩 󵄩𝑟/(𝑘+1) ≲(∫ ∏󵄩𝑢 (𝜏)󵄩 𝑑𝜏) 𝑘 𝑖 𝑠 𝑟 𝑠 󵄩 󵄩𝑀 󸀠 ≤𝐶(𝑘+1) 2𝜂 ‖𝑢−V‖𝐿 (𝐼,𝑀 ), 𝐼 𝑖=1 (𝑘+1)𝑝 ,1 𝑡 𝑝,1

𝑘+1 if ‖𝑢‖𝐿𝑟 (𝐼,𝑀𝑠 ),‖V‖𝐿𝑟 (𝐼,𝑀𝑠 ) ∈𝐵2𝜂. 󵄩 󵄩 𝑡 𝑝,1 𝑡 𝑝,1 󵄩 󵄩 𝑘 ≲ ∏󵄩𝑢𝑖 (𝑡)󵄩 𝑟 𝑠 . 𝜂 𝜂 𝐶(𝑘 + 1)2𝜂 ≤1/2 𝐿𝑡(𝐼,𝑀 󸀠 ) So we can shrink 1 to 2 so that 2 𝑖=1 (𝑘+1)𝑝 ,1 and find a possible smaller constant 𝜂0 ≤𝜂2.Then,when 𝑖(𝑡−𝑡0)Δ 𝑝 ≤ 𝑘+2 ‖𝑒 𝑢0‖ 𝑟 𝑠 ≤𝜂≤𝜂0 J :𝐵2𝜂 →𝐵2𝜂 By the fact and the embedding theorem 𝐿𝑡(𝐼,𝑀𝑝,1) ,wehave and (Lemma 17), we obtain it is nothing but a contraction map. We now obtain 𝑢∈𝐵2𝜂 which is the fixed point of J that solves Cauchy problem (1) 𝑘+1 𝑘+1 󵄩 󵄩 󵄩 󵄩 ‖𝑢‖ 𝑟 𝑠 ≤2𝜂 󵄩 󵄩 󵄩 󵄩 in the sense of integral form. Of course, 𝐿𝑡(𝐼,𝑀𝑝,1) . ∏󵄩𝑢𝑖 (𝑡)󵄩 𝑟 𝑠 ≲ ∏󵄩𝑢𝑖(𝑡)󵄩 𝑟 𝑠 . (72) 󵄩 󵄩𝐿𝑡(𝐼,𝑀 󸀠 ) 󵄩 󵄩𝐿𝑡(𝐼,𝑀𝑝,1) 𝑖=1 (𝑘+1)𝑝 ,1 𝑖=1 3.2.ProofofStrongSolutionPartofTheorem1. In this sub- 𝑟 𝑠 section, we will verify that the local solution 𝑢∈𝐿𝑡(𝐼,𝑝,1 𝑀 ) 𝑀𝑠 𝑢∈𝐶(𝐼,𝑀𝑠 )∩ From Proposition 29,wehaveanimmediatecorollaryas is a 𝑝,1 strong solution in the sense that 𝑝,1 1 𝑠−1 follows. 𝐶 (𝐼,𝑝,1 𝑀 ). 8 Abstract and Applied Analysis

𝑠 󵄩 𝑡𝑛 3.2.1. To Prove 𝑢 ∈ 𝐶(𝐼,𝑀 ). It is equivalent to prove that 󵄩 𝑝,1 + 󵄩∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏 󵄩 0 󵄩 󵄩 󵄩𝑢(𝑡𝑛,⋅)−𝑢(𝑡,⋅)󵄩𝑀𝑠 󳨀→ 0 𝑝,1 (78) 𝑡 󵄩 󵄩 − ∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏󵄩 𝑡 →𝑡 0 󵄩 𝑠 as 𝑛 . We may assume without loss of generality that 󵄩𝑀𝑝,1 𝑡0 =0, ≜ 𝐼+̃ 𝐼𝐼.̃ 󵄩 󵄩 󵄩𝑢(𝑡𝑛,⋅)−𝑢(𝑡,⋅)󵄩𝑀𝑠 𝑝,1 (83) 󵄩 󵄩 ≤ 󵄩𝐾󸀠(𝑡 )𝑢 −𝐾󸀠(𝑡)𝑢 󵄩 󵄩 𝑛 0 0󵄩 𝑠 𝑀𝑝,1 First, by the Minkowski inequality, we have that 󵄩 󵄩 󵄩 󵄩 󵄩 𝑡 󵄩 + 󵄩𝐾(𝑡𝑛)𝑢1 − 𝐾(𝑡)𝑢1󵄩𝑀𝑠 󵄩 𝑛 󵄩 𝑝,1 ̃ 󵄩 󵄩 𝐼=󵄩∫ (𝐾 𝑛(𝑡 −𝜏)−𝐾(𝑡−𝜏))𝑓(𝑢 (𝜏)) 𝑑𝜏󵄩 󵄩 󵄩 𝑠 󵄩 𝑡 󵄩 0 󵄩𝑀 󵄩 𝑛 (79) 𝑝,1 󵄩 (84) + 󵄩∫ 𝐾(𝑡𝑛 −𝜏)𝑓(𝑢 (𝜏)) 𝑑𝜏 󵄩 𝑡𝑛 󵄩 0 󵄩 󵄩 ≤ ∫ 󵄩(𝐾(𝑡𝑛 − 𝜏) − 𝐾(𝑡 − 𝜏))𝑓(𝑢(𝜏))󵄩𝑀𝑠 𝑑𝜏. 𝑡 󵄩 0 𝑝,1 󵄩 − ∫ 𝐾(𝑡 − 𝜏)𝑓(𝑢(𝜏))𝑑𝜏󵄩 0 󵄩𝑀𝑠 𝑝,1 For ‖(𝐾(𝑡𝑛 − 𝜏) − 𝐾(𝑡 − 𝜏))𝑓(𝑢(𝜏))‖𝑀𝑠 ≲ ‖𝑓(𝑢(𝜏))‖𝑀𝑠 ≲ 𝑝,1 𝑝󸀠,1 𝑘+1 𝑘+1 1 ≜ 𝐼 + 𝐼𝐼 + 𝐼𝐼𝐼. ‖𝑢‖ 𝑠 𝑘+1 ≤ 𝑟 ‖𝑢‖ 𝑠 ∈𝐿 𝑀𝑝,1 and ,wehave 𝑀𝑝,1 , 1 −1 𝑠 so ‖(𝐾(𝑡𝑛 − 𝜏) − 𝐾(𝑡 − 𝜏))𝑓(𝑢(𝜏))‖𝑀𝑠 ∈𝐿.Since Recall that 𝑢0,𝐽 𝑢1 ∈𝑀𝑝,1.For𝐼, 𝐼𝐼, by density Lemma 24, 𝑝,1 ‖(𝐾(𝑡𝑛 − 𝜏) − 𝐾(𝑡 − 𝜏))𝑓(𝑢(𝜏))‖𝑀𝑠 →0,as 𝑡𝑛 →𝑡, Lemma 19, triangle inequality, and the definition of 𝐺(𝑡),we 𝑝,1 𝑠 Ω ̃ only need to prove that 𝐺(𝑡)V ∈𝐶(𝐼,𝑀𝑝,1) for V ∈ S . therefore, 𝐼→0,as 𝑡𝑛 →𝑡. Using Hausdorff-Young inequality, we have Secondly, as in the proof of Proposition 29,we get that

󵄩 󵄩 𝑡 󵄩◻𝑘(𝐺(𝑡𝑛)V − 𝐺(𝑡)V)󵄩 𝑝 𝐼𝐼̃ ≲ ∫ (1+|𝑡−𝜏|)−𝑛(1/2 − 1/𝑝) 󵄩 1/2 1/2 󵄩 𝑡 󵄩 𝑖𝑡 (1+|𝜉|2) 𝑖𝑡(1+|𝜉|2) 󵄩 𝑛 󵄩 𝑛 󵄩 ≲ 󵄩𝜎𝑘 (𝑒 −𝑒 ) ̂V(𝜉)󵄩 󵄩 󵄩 󵄩 󵄩𝑝󸀠 (80) 󵄩 󵄩 × 󵄩𝑓 (𝑢 (𝜏))󵄩𝑀𝑠 𝑑𝜏 𝑝󸀠,1 (85) 󵄩 1/2 1/2 󵄩 󵄩 𝑖𝑡 (1+|𝜉|2) 𝑖𝑡(1+|𝜉|2) 󵄩 ≲ 󵄩(𝑒 𝑛 −𝑒 ) ̂V(𝜉)󵄩 . 𝑡 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩𝑝 ≲ ∫ 󵄩𝑓 (𝑢 (𝜏))󵄩 𝑠 𝑑𝜏. 󵄩 󵄩𝑀 󸀠 𝑡𝑛 𝑝 ,1 1/2 1/2 𝑖𝑡 (1+|𝜉|2) 𝑖𝑡(1+|𝜉|2) ̂V ∈ S ‖(𝑒 𝑛 −𝑒 )̂V(𝜉)‖ →0 Since ,so 𝑝 as 𝑘+1 𝑘+1 1 𝑡 →𝑡 Since ‖𝑓(𝑢)‖ 𝑠 ≲‖𝑢‖ 𝑠 , ‖𝑢‖ 𝑠 ∈𝐿.Wehave 𝑛 , by Lebesgue’s dominated convergence theorem. 𝑀 󸀠 𝑀𝑝,1 𝑀𝑝,1 Ω 𝑝 ,1 Since V ∈ S , there exists only the finite number of 𝑘 such that ◻𝑘(𝐺(𝑡𝑛)V − 𝐺(𝑡)V) =0̸ ,sowehave ̃ 𝐼𝐼 󳨀→ 0, as 𝑡𝑛 󳨀→ 𝑡. (86) 󵄩 󵄩 󵄩𝐺(𝑡 )V − 𝐺(𝑡)V󵄩 󳨀→ 0 𝑡 󳨀→ 𝑡. 󵄩 𝑛 󵄩𝑀𝑠 as 𝑛 (81) 𝑝,1 Because 𝐼𝐼𝐼 ≤ 𝐼+̃ 𝐼𝐼̃,itleadsto Thus, 󵄩 󵄩 𝐼𝐼𝐼 󳨀→ 0, as 𝑡𝑛 󳨀→ 𝑡. (87) 𝐼=󵄩𝐾󸀠(𝑡 )𝑢 −𝐾󸀠(𝑡)𝑢 󵄩 󳨀→ 0 𝑡 󳨀→ 𝑡, 󵄩 𝑛 0 0󵄩𝑀𝑠 as 𝑛 𝑝,1 𝑠 (82) Accordingly, (78)holds;thatis,𝑢∈𝐶(𝐼,𝑀𝑝,1). 󵄩 󵄩 𝐼𝐼 = 󵄩𝐾(𝑡𝑛)𝑢1 − 𝐾(𝑡)𝑢1󵄩 𝑠 󳨀→ 0 as 𝑡𝑛 󳨀→ 𝑡. 󵄩 󵄩𝑀𝑝,1 𝑢 (𝑡) 𝑀𝑠−1 𝐼𝐼𝐼 3.2.2. 𝑡 Exists and Is Continuous in 𝑝,1 Sense. First of For , 𝑠−1 all, we consider the existence of derivative in 𝑀𝑝,1 sense. 󵄩 𝑡𝑛 󵄩 𝐼𝐼𝐼 = 󵄩∫ 𝐾(𝑡𝑛 −𝜏)𝑓(𝑢 (𝜏)) 𝑑𝜏 Recall that 󵄩 0 𝑢 (𝑡) =𝐾󸀠 (𝑡 − 𝑡 )𝑢 +𝐾(𝑡−𝑡 )𝑢 𝑡 󵄩 0 0 0 1 󵄩 − ∫ 𝐾(𝑡 − 𝜏)𝑓(𝑢(𝜏))𝑑𝜏󵄩 𝑡 0 󵄩𝑀𝑠 (88) 𝑝,1 − ∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏. 𝑡 󵄩 𝑡 0 󵄩 𝑛 ≤ 󵄩∫ 𝐾(𝑡 −𝜏)𝑓(𝑢 (𝜏)) 𝑑𝜏 󵄩 𝑛 −1 𝑠 󵄩 0 󵄩 For 𝑢0, 𝐽 𝑢1 ∈𝑀𝑝,1, and the definition of 𝐺(𝑡),weshould 𝑠 𝑡𝑛 󵄩 𝐺(𝑡)𝜓(𝑥) 𝜓∈𝑀 󵄩 only deal with the derivatives of for 𝑝,1 and − ∫ 𝐾(𝑡 − 𝜏)𝑓(𝑢(𝜏))𝑑𝜏󵄩 𝑡 0 󵄩𝑀𝑠 ∫ 𝐾(𝑡 − 𝜏)𝑓(𝑢(𝜏))𝑑𝜏. 𝑝,1 𝑡0 Abstract and Applied Analysis 9

By the density Lemma 24,forevery𝜀>0, there exists For the nonlinear part, Ω 𝑠 V ∈ S ∩𝑀𝑝,1,suchthat‖𝜓 − V‖𝑀𝑠 <𝜀. For the derivative 󵄩 𝑡 𝑡 𝑝,1 󵄩∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏 −∫ 3 𝐾(𝑡 −𝜏)𝑓(𝑢 (𝜏)) 𝑑𝜏 𝑠 󵄩 𝑡 𝑡 3 𝐺(𝑡)𝜓(𝑥) 𝑡=𝑡 𝜓∈𝑀 󵄩 0 0 of at 3 for 𝑝,1,wehave 󵄩 󵄩 𝑡−𝑡 󵄩 3 󵄩 󵄩 󵄩 𝑡 󵄩 󵄩𝐺(𝑡)𝜓 − 𝐺(𝑡 )𝜓 1/2 󵄩 3 󵄩 󵄩 3 −𝑖𝜔 𝐺(𝑡 )𝜓󵄩 − ∫ 𝐾󸀠(𝑡 − 𝜏)𝑓(𝑢(𝜏))𝑑𝜏󵄩 󵄩 3 󵄩 3 󵄩 󵄩 𝑡−𝑡3 󵄩 𝑠−1 𝑡 󵄩 𝑀𝑝,1 0 󵄩 𝑠−1 󵄩𝑀𝑝,1 󵄩 󵄩 󵄩 󵄩 󵄩 𝑡3 󵄩𝐺(𝑡)𝜓 3− 𝐺(𝑡 )𝜓 󵄩 󵄩∫ (𝐾 (𝑡−𝜏) −𝐾(𝑡 −𝜏))𝑓(𝑢 (𝜏)) 𝑑𝜏 = 󵄩 −𝑖𝐺(𝑡 )𝜓󵄩 󵄩 𝑡 3 󵄩 1/2 3 󵄩 ≤ 󵄩 0 󵄩 (𝑡 −3 𝑡 )𝜔 󵄩𝑀𝑠 󵄩 𝑝,1 󵄩 𝑡−𝑡 󵄩 3 󵄩 󵄩 󵄩 󵄩𝐺(𝑡)(𝜓 − V)−𝐺(𝑡3)(𝜓 − V)󵄩 󵄩 ≤ 󵄩 󵄩 𝑡3 󵄩 󵄩 1/2 󵄩 󸀠 󵄩 󵄩 (𝑡 −3 𝑡 )𝜔 󵄩𝑀𝑠 (89) − ∫ 𝐾 (𝑡3 − 𝜏)𝑓(𝑢(𝜏))𝑑𝜏󵄩 𝑝,1 󵄩 𝑡0 󵄩 󵄩𝑀𝑠−1 󵄩 󵄩 𝑝,1 󵄩𝐺(𝑡)V −𝐺(𝑡 )V 󵄩 󵄩 3 󵄩 󵄩 𝑡 󵄩 + 󵄩 −𝑖𝐺(𝑡3)V󵄩 󵄩∫ 𝐾(𝑡 − 𝜏)𝑓(𝑢(𝜏))𝑑𝜏󵄩 󵄩 1/2 󵄩 󵄩 𝑡 󵄩 󵄩 (𝑡 −3 𝑡 )𝜔 󵄩 𝑠 󵄩 3 󵄩 󵄩 󵄩𝑀𝑝,1 + 󵄩 󵄩 󵄩 󵄩 󵄩 𝑡−𝑡3 󵄩 󵄩 󵄩 󵄩 󵄩 𝑠−1 󵄩 󵄩 󵄩 󵄩𝑀𝑝,1 + 󵄩𝑖𝐺(𝑡3)(𝜓 − V)󵄩𝑀𝑠−1 𝑝,1 󵄩 𝑡 (93) 󵄩 3 (𝐾 (𝑡−𝜏) −𝐾(𝑡3 −𝜏)) 󸀠 = 󵄩∫ ( −𝐾 (𝑡 −𝜏)) ≜𝐼𝑉+𝑉+𝑉𝐼. 󵄩 𝑡−𝑡 3 󵄩 𝑡0 3 󵄩 󵄩 󵄩 × 𝑓(𝑢(𝜏))𝑑𝜏󵄩 𝑉 󵄩 𝑠−1 For , by the Hausdorff-Young inequality and the Lebesgue 𝑀𝑝,1 dominated convergence theorem, we have 󵄩 𝑡 󵄩 󵄩∫ 𝐾(𝑡 − 𝜏)𝑓(𝑢(𝜏))𝑑𝜏󵄩 󵄩 𝑡 󵄩 + 󵄩 3 󵄩 󵄩 󵄩 󵄩 𝑡−𝑡3 󵄩 󵄩 󵄩 󵄩 󵄩𝑀𝑠−1 󵄩 𝐺 (𝑡) V −𝐺(𝑡3) V 󵄩 𝑝,1 󵄩◻ ( −𝑖𝐺(𝑡 ) V)󵄩 󵄩 𝑘 (𝑡 − 𝑡 )𝜔1/2 3 󵄩 󵄩 󵄩 3 󵄩 𝑡3 󵄩 𝑝 󵄩 (𝐾 (𝑡−𝜏) −𝐾(𝑡3 −𝜏)) 󸀠 ≲ ∫ 󵄩( −𝐾 (𝑡 −𝜏)) 󵄩 󵄩 󵄩 𝑡−𝑡 3 󵄩 𝑖𝑡⟨𝜉⟩ 𝑖𝑡3⟨𝜉⟩ 󵄩 𝑡0 󵄩 3 󵄩 𝑒 −𝑒 𝑖𝑡 ⟨𝜉⟩ 󵄩 (90) 󵄩 3 ̂󵄩 ≲ 󵄩𝜎𝑘 ( −𝑖𝑒 ) V󵄩 󵄩 󵄩 (𝑡 − 𝑡 )⟨𝜉⟩ 󵄩 󸀠 󵄩 󵄩 3 󵄩𝑝 󵄩 ×𝑓(𝑢 (𝜏)) 󵄩 𝑑𝜏 󵄩𝑀𝑠−1 󳨀→ 0 as 𝑡󳨀→𝑡3. 𝑝,1 󵄩 󵄩 + max 󵄩𝐾(𝑡 − 𝜏)𝑓(𝑢(𝜏))󵄩 𝑠−1 . 󵄩 󵄩𝑀𝑝,1 𝜏∈[𝑡,𝑡3]or[𝑡3,𝑡] V ∈ SΩ ∩𝑀𝑠 𝑘 As 𝑝,1, so there is only the finite number of such 𝜔(𝑡, 𝑥) ∈ 𝐶(𝐼,𝑠 𝑀 ) 𝐾(𝑡)𝜔(𝑡, 𝑥)∈ 1/2 If 𝑝,1 ,thenwehave that ◻𝑘((𝐺(𝑡)V −𝐺(𝑡3)V)/(𝑡 −3 𝑡 )𝜔 −𝑖𝐺(𝑡3)V) =0̸ .Thus,we 𝑠−1 1/2 𝐶(𝐼,𝑝,1 𝑀 ). In fact, taking advantage of (92) and the Lebesgue get 𝑉→0as 𝑡→𝑡3;thatis,(𝐺(𝑡)V(𝑥))𝑡 =𝑖𝜔 𝐺(𝑡)V(𝑥) in 𝑠−1 Ω 𝑠 dominated convergence theorem, we can get 𝑀𝑝,1 ,forV ∈ S ∩𝑀𝑝,1. 󵄩 󵄩 𝐼𝑉 󵄩𝐾(𝑡)𝜔(𝑡, 𝑥)3 −𝐾(𝑡 )𝜔(𝑡3,𝑥)󵄩 𝑠−1 For , by the Bernstein multiplier theorem, we have 󵄩 󵄩𝑀𝑝,1 󵄩 󵄩 ≤ 󵄩(𝐾(𝑡) −3 𝐾(𝑡 ))𝜔(𝑡3,𝑥)󵄩 𝑠−1 󵄩 󵄩 󵄩 󵄩𝑀𝑝,1 󵄩 𝐺(𝑡)(𝜓 − V)−𝐺(𝑡 )(𝜓 − V)󵄩 󵄩◻ 3 󵄩 ≲‖𝜓−V ‖ . (94) 󵄩 𝑙 1/2 󵄩 𝑝 (91) 󵄩 󵄩 󵄩 (𝑡 − 𝑡 )𝜔 󵄩 + 󵄩𝐾(𝑡)(𝜔(𝑡, 𝑥)3 −𝜔(𝑡 ,𝑥))󵄩 𝑠−1 󵄩 3 󵄩𝑝 󵄩 󵄩𝑀𝑝,1

󳨀→ 0 as 𝑡󳨀→𝑡3. Using the almost orthogonality of modulation space, we 𝑠 Recall that 𝑓(𝑢) ∈ 𝐶(𝐼,𝑝,1 𝑀 ) and apply (92) and the Lebesgue obtain 𝐼𝑉≲𝜓−V𝑀𝑠 <𝜀. 𝑝,1 dominated convergence theorem to the first term of (93); we For 𝑉𝐼,byLemma19,wehave𝑉𝐼 = ‖𝑖𝐺(𝑡3)(𝜓 − V)‖𝑀𝑠 ≲ 𝑝,1 have ‖𝜓 − V‖𝑀𝑠 <𝜀. 󸀠󵄨 𝑝,1 𝑡 󵄨 𝑠 󵄨 Accordingly, for 𝜓∈𝑀𝑝,1, (∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏) 󵄨 𝑡 󵄨 0 𝑡󵄨𝑡=𝑡 3 (95) 𝑡3 1/2 𝑠−1 = ∫ 𝐾󸀠 (𝑡 −𝜏)𝑓(𝑢 (𝜏)) 𝑑𝜏 𝑀𝑠−1. (𝐺 (𝑡) 𝜓)𝑡 =𝑖𝜔 𝐺 (𝑡) 𝜓 in 𝑀𝑝,1 . (92) 3 in 𝑝,1 𝑡0 10 Abstract and Applied Analysis

Consequently, and that 𝑢 (𝑡) =−𝐽2𝐾(𝑡−𝑡 )𝑢 +𝐾󸀠 (𝑡 − 𝑡 )𝑢 𝑡 𝑡 0 0 0 1 ∫ ⟨𝑡 − 𝜏⟩−𝑛(1/2−1/𝑝)⟨𝜏⟩−𝑛(1/2−1/𝑝)(𝑘+1)𝑑𝜏 𝑡 𝑡/2 󸀠 𝑠−1 (96) − ∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏 in 𝑀𝑝,1 . 𝑡/2 𝑡 −𝑛(1/2−1/𝑝) −𝑛(1/2−1/𝑝)(𝑘+1) 0 = ∫ ⟨𝜏⟩ ⟨𝑡 − 𝜏⟩ 𝑑𝜏 0 Next, the proof of time continuity of 𝑢𝑡 is similar to 𝑢. (102) 𝑡/2 It only needs to take care of the difference of smoothness −𝑛(1/2−1/𝑝) −𝑘𝑛(1/2−1/𝑝) −(1/2−1/𝑝) and the action of the Bessel potential. Finally, we obtain 𝑢∈ ≲⟨𝑡⟩ ⟨𝑡⟩ ∫ ⟨𝜏⟩ 𝑑𝜏 𝑠 1 𝑠−1 𝑠 0 𝐶(𝐼,𝑝,1 𝑀 )∩𝐶 (𝐼,𝑝,1 𝑀 );thatis,itisthestrong𝑀𝑝,1 solution. ≲⟨𝑡⟩−𝑛(1/2−1/𝑝). 3.3. Global Well-Posedness for Small Fine Data. Let us construct a time decaying norm as Combining the above four inequalities, we have 󵄩 󵄩 󵄩 󵄩 𝑛(1/2−1/𝑝) Γ T𝑢 ≲ 󵄩𝑢 󵄩 + 󵄩𝑢 󵄩 + Γ(𝑢)𝑘+1. Γ (𝑢) = (1 + |𝑡|) ‖𝑢‖ 𝑠 . ( ) 󵄩 0󵄩 𝑠+2𝜎(𝑝) 󵄩 1󵄩 𝑠+2𝜎(𝑝)−1 sup 𝑀 𝑀 󸀠 𝑀 󸀠 (103) 𝑡∈R 𝑝,1 (97) 𝑝 ,1 𝑝 ,1 ThisideafortheNLScanbetracedbacktotheworkofStrauss Hence, if we choose 𝛿≲𝑀/2and 𝑀 small enough, [12]andWangandHudzik[2]. We consider the following then using the standard contraction mapping argument the mapping: Cauchy problem (1) has a unique solution satisfying (13). T :𝑢(𝑡) 󳨅→ 𝐾 󸀠 (𝑡) 𝑢 +𝐾(𝑡) 𝑢 0 1 3.4. Uniqueness. In order to prove the uniqueness in a 𝑡 (98) category of strong solution, we only need to verify it locally; − ∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏 that is, to prove that the set {𝑡∈𝐼:𝑢=V} is open. If 0 𝑢(𝑡1)=V(𝑡1) for 𝑡1 ∈𝐼, we can choose an interval 𝐼1 sufficient D ={𝑢:Γ(𝑢)≤𝑀} 𝑑(𝑢, V)=Γ(𝑢− in the metric space with small so that 𝑡1 ∈𝐼1 ⊂𝐼and V).Forany𝑢∈D,inviewofLemma20,wegetthat 󵄩 󵄩 󵄩 󵄩 ‖𝑢−V‖ ∞ 𝑠 󵄩 󵄩 󵄩 󵄩 𝐿𝑡 (𝐼1,𝑀𝑝,1) Γ (T𝑢) ≲ 󵄩𝑢0󵄩 𝑠+2𝜎(𝑝) + 󵄩𝑢1󵄩 𝑠+2𝜎(𝑝)−1 𝑀 󸀠 𝑀 󸀠 𝑝 ,1 𝑝 ,1 󵄨 󵄨 ≤𝐶(󵄨𝐼 󵄨 ,𝑘) {‖𝑢‖ ∞ 𝑠 , 𝑡 (99) 󵄨 1󵄨 max 𝐿𝑡 (𝐼1,𝑀𝑝,1) +Γ(∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏) 0 ‖V‖ ∞ 𝑠 } 𝐿𝑡 (𝐼1,𝑀𝑝,1) (104) and(recallthenotationthat⟨𝑡⟩ = 1 + |𝑡|) 𝑡 × ‖𝑢−V‖ ∞ 𝑠 𝐿 (𝐼1,𝑀 ) Γ(∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏) 𝑡 𝑝,1 0 1 ≤ ‖𝑢−V‖𝐿∞(𝐼 ,𝑀𝑠 ). 𝑛(1/2−1/𝑝) 2 𝑡 1 𝑝,1 ≲ sup⟨𝑡⟩ 𝑡∈R So 𝑢=V on 𝐼1, which concludes that 𝑡1 is an interior point. 𝑡 So 𝐼 is open. × ∫ ⟨𝑡 − 𝜏⟩−𝑛(1/2−1/𝑝) 0 ‖𝑢‖ 𝑟 𝑠 <∞ 3.5. Blowup Criterion. We assume that 𝐿𝑡([𝑡0,𝑇],𝑀𝑝,1) , × ‖𝑢‖𝑀𝑠 𝑑𝜏 (100) 𝑝,1 write the integral equation on [𝑡𝑚,𝑇], and use nonlinear 𝑘+1 𝑛(1/2−1/𝑝) estimate to deduce that ≲ (Γ (𝑢)) sup ⟨𝑡⟩ 󵄩 󵄩 𝑡∈R 󵄩 󸀠 󵄩 󵄩𝐾 (𝑡 −𝑚 𝑡 )𝑢(𝑡𝑚)+𝐾(𝑡−𝑡𝑚)𝑢𝑡 (𝑡𝑚)󵄩 𝑟 𝑠 󵄩 󵄩𝐿𝑡([𝑡𝑚,𝑇],𝑀𝑝,1) 𝑡 −𝑛(1/2−1/𝑝) (105) × ∫ ⟨𝑡 − 𝜏⟩ 𝑘+1 ≲ 𝑢 𝑟 𝑠 + 𝑢 . 0 ‖ ‖𝐿 ([𝑡 ,𝑇],𝑀 ) ‖ ‖ 𝑟 𝑠 𝑡 𝑚 𝑝,1 𝐿𝑡([𝑡𝑚,𝑇],𝑀𝑝,1) ×⟨𝜏⟩−𝑛(1/2−1/𝑝)(𝑘+1)𝑑𝜏. Then, let 𝑡𝑚 close to 𝑇 sufficiently enough so that 󸀠 ‖𝐾 (𝑡 −𝑚 𝑡 )𝑢(𝑡𝑚)+𝐾(𝑡−𝑡𝑚)𝑢𝑡(𝑡𝑚)‖𝐿𝑟 ([𝑡 ,𝑇],𝑀𝑠 ) < (1/2)𝜂0, Owing to 𝑛(1/2 − 1/𝑝)(𝑘 + 1)>1,itfollowsthat 𝑡 𝑚 𝑝,1 where 𝜂0 istheonewechoseinTheorem1.Then,∃𝛿 > 0,such 𝑡/2 󸀠 −𝑛(1/2−1/𝑝) −𝑛(1/2−1/𝑝)(𝑘+1) ‖𝐾 (𝑡 −𝑚 𝑡 )𝑢(𝑡𝑚)+𝐾(𝑡−𝑡𝑚)𝑢𝑡(𝑡𝑚)‖ 𝑟 𝑠 ≤𝜂0 ∫ ⟨𝑡−𝜏⟩ ⟨𝜏⟩ 𝑑𝜏 that 𝐿𝑡([𝑡𝑚,𝑇+𝛿],𝑀𝑝,1) , 0 so we can extend 𝑢 on [𝑡0,𝑇 + 𝛿] by local existence and 𝑡/2 uniqueness, a contradiction. ≲ ⟨𝑡⟩−𝑛(1/2−1/𝑝) ∫ ⟨𝜏⟩−𝑛(1/2−1/𝑝)(𝑘+1)𝑑𝜏 (101) 0 3.6. Persistence of Regularity. We want the norm which can −𝑛(1/2−1/𝑝) ≲ ⟨𝑡⟩ hold the continuation of solution to be as low regularity Abstract and Applied Analysis 11

as possible. It is interesting to make an assertion that the Lemma 32. Let 𝐼 be a bounded time interval containing 𝑡0, 𝑠 𝑠 boundedness of low regularity norm ‖𝑢(𝑡)‖𝑀𝑠2 suffices to 𝑠≥0,andlet𝑢∈𝐶(𝐼,𝑀∞,1) be a strong 𝑀∞,1 solution to (1). 𝑝2,1 𝑠 ‖𝑢‖ 𝑘 0 ‖𝑢(𝑡)‖ 1 If the quantity 𝐿𝑡 (𝐼,𝑀∞,1) is finite, then one has ensure the boundedness of high regularity norm 𝑀𝑝 ,1 𝑠 1 𝑀 1 andthusensurethecontinuationofstrong 𝑝1,1 solution. By the improved version of product lemma (Lemma 22)andthe ‖𝑢 (𝑡)‖𝐿∞(𝐼,𝑀𝑠 ) embedding theorems of modulation spaces, we can establish 𝑡 ∞,1 Lemmas 31 and 32 to achieve our goal. 󵄩 󵄩 󵄩 󵄩 𝑝 ∞ ≲𝐶(|𝐼|) (󵄩𝑢(𝑡0)󵄩𝑀𝑠 + 󵄩𝑢𝑡 (𝑡0)󵄩𝑀𝑠−1 ) Firstly, we “reduce” the regularity index from to . ∞,1 ∞,1 (109)

Lemma 31. 𝐼 𝑡 𝑘 Let be a bounded time interval containing 0,let × (∫ 𝐶 (|𝐼|) ‖𝑢 (𝜏)‖ 0 𝑑𝜏) , 𝑠 𝑠 exp 𝑀∞,1 𝑝∈[1,∞],andlet𝑢∈𝐶(𝐼,𝑀𝑝,1) be a strong 𝑀𝑝,1 solution to 𝐼 ‖𝑢‖ 𝑘 𝑠 (1).Ifthequantity 𝐿𝑡 (𝐼,𝑀∞,1) is finite, then one has where 𝐶(|𝐼|) is some positive constant associated with the ‖𝑢(𝑡)‖ ∞ 𝑠 𝐼 𝐿𝑡 (𝐼,𝑀𝑝,1) length of . 󵄩 󵄩 󵄩 󵄩 𝑀𝑠 𝑢 ≲𝐶(|𝐼|) (󵄩𝑢(𝑡0)󵄩𝑀𝑠 + 󵄩𝑢𝑡 (𝑡0)󵄩𝑀𝑠−1 ) So, for any strong 𝑝,1 solution in a bounded interval 𝑝,1 𝑝,1 (106) 𝑘 0 𝑠 𝐼,ifitkeeps𝐿𝑡(𝐼,∞,1 𝑀 ) norm bounded, it will keep 𝑀𝑝,1 𝑘 ‖𝑢(𝑡)‖ 𝑠 × (∫ 𝐶 (|𝐼|) ‖𝑢(𝜏)‖ 𝑠 𝑑𝜏) , norm bounded in the same interval. Conversely, if 𝑀𝑝,1 exp 𝑀∞.1 𝐼 is bounded in a bounded time interval 𝐼, by embedding 𝑀𝑠 ⊂𝑀0 𝑢(𝑡) 𝐿𝑘(𝐼,0 𝑀 ) 𝐶(|𝐼|) 𝑝,1 ∞,1, will be in 𝑡 ∞,1 .Thus,onecan where is some positive constant associated with the 𝑀𝑠 length of 𝐼. extend a strong solution in 𝑝,1 if and only if the quantity ‖𝑢‖ 𝑘 0 𝐿𝑡 (𝐼,𝑀∞,1) remains bounded. This implies that the blowup Proof. Without loss of generality, we may assume 𝑡0 = inf 𝐼 phenomenon is independent of exact 𝑝 and 𝑠,soifaninitial 𝑠 𝑠 −1 𝑠 𝑠 −1 and 𝐼=[𝑡0,𝑇]for some 𝑇>𝑡0. Writing the integral equation, (𝑢(𝑡 ), 𝑢 (𝑡 )) 𝑀 1 ×𝑀 1 𝑀 2 ×𝑀 2 data 0 𝑡 0 lies in both 𝑝1,1 𝑝1,1 and 𝑝2,1 𝑝2,1 using Lemma 19, and embedding, we have (1≤𝑝1,𝑝2 ≤∞, 0<𝑠1,𝑠2 <∞), they have the same maximal existence interval. 󵄩 󵄩 ‖𝑢(𝑡)‖𝑀𝑠 ≲𝐶(|𝐼|) (󵄩𝑢(𝑡0)󵄩 𝑠 𝑝,1 󵄩 󵄩𝑀𝑝,1

𝑡 4. Scattering Theorems 󵄩 󵄩 󵄩 󵄩 + 󵄩𝑢𝑡 (𝑡0)󵄩𝑀𝑠−1 + ∫ 󵄩𝑓 (𝑢 (𝜏))󵄩𝑀𝑠 𝑑𝜏) 𝑝,1 𝑝,1 The goal of this section is to derive scattering results. 𝑡0

󵄩 󵄩 󵄩 󵄩 ≲𝐶(|𝐼|) (󵄩𝑢(𝑡0)󵄩 𝑠 + 󵄩𝑢𝑡 (𝑡0)󵄩 𝑠−1 󵄩 󵄩𝑀𝑝,1 󵄩 󵄩𝑀𝑝,1 4.1. Proof of Scattering Theorem. Without loss of generality, we assume 𝑡0 =0,andlet 𝑡 𝑘 + ∫ ‖(𝑢(𝜏))‖𝑀𝑠 ‖(𝑢(𝜏))‖𝑀𝑠 𝑑𝜏) 𝑡 ∞,1 𝑝,1 0 𝑢 𝑡 𝐺 (−𝜏) 𝑓 (𝑢 (𝜏)) (107) 2V (𝑡) =𝑢 + 1 − ∫ 𝑑𝜏, 1 0 1/2 1/2 𝑖𝜔 0 𝑖𝜔 𝑘 𝑡 (110) for 𝑡∈[𝑡0,𝑇].Let𝐴(𝑡) = ‖𝑢(𝑡)‖𝑀𝑠 and 𝐵(𝑡) = ‖(𝑢(𝑡))‖𝑀𝑠 , 𝑢1 𝐺 (𝜏) 𝑓 (𝑢 (𝜏)) 𝑝,1 ∞,1 2V (𝑡) =𝑢 − + ∫ 𝑑𝜏. 2 0 1/2 1/2 and both 𝐴 and 𝐵 are continuous and nonnegative on [𝑡0,𝑇]. 𝑖𝜔 0 𝑖𝜔 Using Gronwall inequality, we have 0<𝑠<𝑡 󵄩 󵄩 󵄩 󵄩 For , ‖𝑢(𝑡)‖𝑀𝑠 ≲𝐶(|𝐼|) (󵄩𝑢(𝑡0)󵄩 𝑠 + 󵄩𝑢𝑡 (𝑡0)󵄩 𝑠−1 ) 𝑝,1 󵄩 󵄩𝑀𝑝,1 󵄩 󵄩𝑀𝑝,1

𝑡 (108) 𝑡 𝑘 𝐺 (−𝜏) 𝑓 (𝑢 (𝜏)) × exp (∫ 𝐶 (|𝐼|) ‖𝑢(𝜏)‖𝑀𝑠 𝑑𝜏) V (𝑡) − V (𝑠) =−∫ 𝑑𝜏. ∞.1 1 1 1/2 (111) 𝑡0 𝑠 𝑖𝜔 for 𝑡∈[𝑡0,𝑇]and obtain the conclusion immediately. BythefactweusedinSection3.2.1 that is from Definition 25 for 𝑘-admissible pair (𝑝, 𝑟), there exists 𝛾≥1̃ such that Secondly, we reduce another regularity index from 𝑠 to 0. By the Leibniz rules for modulation space Lemma 22,wecan 1 𝑘+1 1 1 deduce the following lemma. The proof of this lemma is very + =1, 𝛾𝑛̃ ( − )>1. similar to the proof of Lemma 31,soweomitthedetails. 𝛾̃ 𝑟 2 𝑝 (112) 12 Abstract and Applied Analysis

󵄩 󵄩1/𝛾̃ Take advantage of the nonlinear estimate and Holder¨ inequal- ≲ 󵄩(1 + |𝑡|)−𝛾𝑛(1/2−1/𝑝)̃ 󵄩 󵄩 󵄩𝐿1 ity: 𝑡 × ‖𝑢‖𝑘+1 󳨀→ 0 󵄩 󵄩 𝑟 𝑠 󵄩 󵄩 𝐿𝑡([𝑡,∞],𝑀𝑝,1) 󵄩V1 (𝑡) − V1 (𝑠)󵄩 𝑠 󵄩 󵄩𝑀𝑝,1 (117) 𝑡 ≲ ∫ (1+|𝜏|)−𝑛(1/2 − 1/𝑝) − 𝑠 as 𝑡→+∞.SoisV , respectively. In fact, in our proof, we 󵄩 󵄩 + 𝑠 󵄩 󵄩 also have V1 ∈𝑀𝑝,1. × 󵄩𝑓 (𝑢 (𝜏))󵄩𝑀𝑠 𝑑𝜏 𝑝󸀠,1 𝑠 One may ask whether there exists a global 𝑀𝑝,1 solution 𝑡 −𝑛(1/2 − 1/𝑝) to Cauchy problem (1) which is scattering associated with ≲ ∫ (1+|𝜏|) V+, V+ ∈𝑀𝑠 𝑠 1 2 𝑝,1. The following remark gives us a partial answer. 𝑘+1 V+, V+ ∈𝑀𝑠 𝐺(𝑡)V+ +𝐺(−𝑡)V+ ∈ × ‖𝑢‖𝑀𝑠 𝑑𝜏 Remark 33. Let 1 2 𝑝,1 and 1 2 𝑝,1 𝐿𝑟( 𝑇 , ∞), 𝑀𝑠 ) 𝑢 󵄩 󵄩 (113) 𝑡 [ 1 𝑝,1 , and then there exists a solution of (1)on 󵄩 −𝑛(1/2 − 1/𝑝)󵄩 ≲ 󵄩(1+|𝑡|) 󵄩 𝛾̃ 𝐼=[𝑇, ∞) for some 𝑇>𝑇1 such that 󵄩 󵄩𝐿𝑡 󵄩 󵄩 󵄩 𝑘+1 󵄩 󵄩 + + 󵄩 × 󵄩‖𝑢‖𝑀𝑠 󵄩 󵄩 󵄩 󵄩 𝑝,1 󵄩 𝑟/(𝑘+1) 󵄩𝑢(𝑡) − (𝐺 (𝑡) V1 +𝐺(−𝑡) V2 )󵄩 𝑠 󳨀→ 0 𝐿𝑡 ([𝑠,𝑡]) 󵄩 󵄩𝑀𝑝,1 (118) 󵄩 󵄩1/𝛾̃ ≲ 󵄩(1+|𝑡|)−𝛾𝑛(1/2̃ − 1/𝑝) 󵄩 󵄩 󵄩 1 𝑡→+∞ 󵄩 󵄩𝐿𝑡 as . 𝑘+1 × ‖𝑢‖ 𝑟 𝑠 To construct 𝑢,weonlyneedtosolvetheCauchyproblem 𝐿𝑡([𝑠,𝑡],𝑀𝑝,1) from 𝑡=+∞to some finite time 𝑡=𝑇>𝑇1 with initial data 𝑘+1 𝑡=+∞ ≲ ‖𝑢‖ 𝑟 𝑠 . at , that is, to solve the following equation: 𝐿𝑡([𝑠,𝑡],𝑀𝑝,1)

𝑢 (𝑡) =𝐺(𝑡) V+ +𝐺(−𝑡) V+ ‖𝑢‖ 𝑟 𝑠 ≤𝑀 1 2 Since 𝐿𝑡(R,𝑀𝑝,1) ,wehave +∞ 𝐺 (𝑡−𝜏) +𝐺(𝜏−𝑡) (119) + ∫ 𝑓 (𝑢) 𝑑𝜏 󵄩 󵄩 𝑘+1 1/2 󵄩V1 (𝑡) − V1 (𝑠)󵄩𝑀𝑠 ≲ ‖𝑢‖ 𝑟 𝑠 󳨀→ 0 𝑡 2𝑖𝜔 𝑝,1 𝐿𝑡([𝑠,𝑡],𝑀𝑝,1) (114)

𝑡,𝑠→∞ V(𝑡) 𝑀𝑠 𝑡󳨀→ on 𝐼=[𝑇, ∞).Infact,wehave as . This implies that is Cauchy in 𝑝,1 as + + + ‖𝐺(𝑡)V1 +𝐺(−𝑡)V2 ‖𝐿𝑟 ( 𝑇,∞),𝑀𝑠 ) <𝜂0.Ifwechoose𝑇>𝑇1 ∞.DenoteV1 to be the limit: 𝑡 [ 𝑝,1 large enough, where 𝜂0 is as in Theorem 1,thenwecansolve + this equation by the same technique used in the proof of 2V1 = lim 2V1 (𝑡) 𝑡→+∞ Theorem 1 and deduce that there exists a solution to (1)on +∞ 𝐼= 𝑇, ∞) ‖𝑢(𝑡)‖ 𝑟 𝑠 <∞ 𝑡 𝑢 𝐺 (−𝜏) 𝑓 (𝑢 (𝜏)) [ and 𝐿𝑡([𝑇1,∞),𝑀𝑝,1) .So,if is large =𝑢 + 1 − ∫ 𝑑𝜏, 0 1/2 1/2 𝑖𝜔 0 𝑖𝜔 enough, we have − (115) 2V = 2V (𝑡) 󵄩 + + 󵄩 1 𝑡→−∞lim 1 󵄩 󵄩 󵄩𝑢 (𝑡) −(𝐺(𝑡) V1 +𝐺(−𝑡) V2 )󵄩 𝑠 󵄩 󵄩𝑀𝑝,1 𝑢 −∞ 𝐺 (𝜏) 𝑓 (𝑢 (𝜏)) 󵄩 󵄩 =𝑢 − 1 + ∫ 𝑑𝜏. 󵄩 ∞ 󵄩 0 1/2 1/2 󵄩 󵄩 𝑖𝜔 0 𝑖𝜔 = 󵄩∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏󵄩 󵄩 𝑡 󵄩 𝑠 (120) 󵄩 󵄩𝑀𝑝,1

In a similar way, we obtain 𝑘+1 ≲ ‖𝑢‖ 𝑟 𝑠 󳨀→ 0 𝐿𝑡([𝑡,∞],𝑀𝑝,1) V+ = V (𝑡) , V− = V (𝑡) . 2 𝑡→+∞lim 2 2 𝑡→−∞lim 2 (116) 𝑡→∞ 𝑡→ + + + as . A similar argument applies in the situation Recall that V = 𝐺(𝑡)V1 +𝐺(−𝑡)V2 ; so taking advantage of the −∞. nonlinear estimates, we get

󵄩 +󵄩 5. Stability Theory 󵄩𝑢 (𝑡) − V 󵄩 𝑠 󵄩 󵄩𝑀𝑝,1 󵄩 +∞ 󵄩 In this section, we will derive the stability theory including 󵄩 󵄩 = 󵄩∫ 𝐾 (𝑡−𝜏) 𝑓 (𝑢 (𝜏)) 𝑑𝜏󵄩 short-time result and long-time result and deduce some 󵄩 𝑡 󵄩 𝑠 󵄩 󵄩𝑀𝑝,1 corollaries. Abstract and Applied Analysis 13

5.1.ProofofShort-TimePerturbations. By time symmetry, we A standard continuity argument then shows that if we may assume that 𝑡0 = inf 𝐼.LetV =𝑢−𝑢̃.Then,V satisfies the shrink 𝜀𝑐 sufficiently small, then following initial value problem: 𝑆 (𝑇) ≲𝜀 (127) (𝜕 +𝐼−Δ)V =𝑓(𝑢+̃ V) −𝑓(𝑢̃) −𝑒, 𝑡𝑡 for any 𝑇∈𝐼, which implies (23). Using (124)and(127), one V (𝑡 ,𝑥)=𝑢(𝑡 ,𝑥)−𝑢(𝑡̃ ,𝑥), can easily obtain (24). By Lemma 19,wehave 0 0 0 (121) 󵄩 󵄩 󵄩𝐾󸀠(𝑡 − 𝑡 )V(𝑡 )+𝐾(𝑡−𝑡 )V (𝑡 )󵄩 󵄩 0 0 0 𝑡 0 󵄩𝐿∞(𝐼,𝑀𝑠 ) V𝑡 (𝑡0,𝑥)=𝑢𝑡 (𝑡0,𝑥)−𝑢̃𝑡 (𝑡0,𝑥). 𝑡 𝑝,1 󵄩 󵄩 The integral equation is ≲𝐶(𝐼) (󵄩𝑢(𝑡0)−𝑢(𝑡̃ 0)󵄩 𝑠 󵄩 󵄩𝑀𝑝,1 󸀠 V (𝑡) =𝐾 (𝑡 −0 𝑡 ) V (𝑡0)+𝐾(𝑡−𝑡0) V𝑡 (𝑡0) 󵄩 󵄩 + 󵄩𝑢𝑡 (𝑡0)−𝑢̃𝑡 (𝑡0)󵄩 𝑠−1 ) 󵄩 󵄩𝑀𝑝,1 𝑡 (122) (128) ̃ ̃ − ∫ 𝐾 (𝑡−𝜏) (𝑓 (𝑢+V) −𝑓(𝑢) −𝑒)𝑑𝜏. ≲𝐶(𝐼) (𝑀󸀠 + 𝑀̃󸀠), 𝑡0 󵄩 󵄩 𝑇∈𝐼 󵄩 2 󸀠 󵄩 For , define 󵄩−𝐽 𝐾(𝑡0 −𝑡 )V(𝑡0)+𝐾(𝑡 −0 𝑡 )V𝑡(𝑡0)󵄩 ∞ 𝑠−1 󵄩 󵄩𝐿𝑡 (𝐼,𝑀𝑝,1 ) 󵄩 󵄩 𝑆 (𝑇) = 󵄩(𝜕𝑡𝑡 +𝐼−Δ)V +𝑒󵄩 𝑟/(𝑘+1) 𝑠 . 󵄩 󵄩𝐿𝑡 ([𝑡0,𝑇],𝑀 󸀠 ) (123) 󸀠 󸀠 𝑝 ,1 ≲𝐶(𝐼) (𝑀 + 𝑀̃).

By nonlinear estimate and the smallness condition, for every By Lemma 20 and Holder¨ inequality, we have that for every 𝑇∈𝐼, 𝑡∈𝐼 󵄩 𝑡 󵄩 ‖V(𝑡)‖𝐿𝑟 ([𝑡 ,𝑇],𝑀𝑠 ) 󵄩 󵄩 𝑡 0 𝑝,1 󵄩∫ 𝐾 (𝑡−𝜏) (𝑓 (𝑢+̃ V) −𝑓(𝑢̃))𝑑𝜏󵄩 󵄩 󵄩 󵄩 𝑡0 󵄩𝑀𝑠 󵄩 𝑡 𝑝,1 󵄩 ≲𝜀+󵄩∫ (𝐾 (𝑡−𝜏) (𝜕𝑡𝑡 +𝐼−Δ)V 󵄩 −𝑛(1/2−1/𝑝) 󵄩 𝑡0 ≲ ∫ (1 + |𝑡−𝜏|) 󵄩 𝐼 󵄩 󵄩 󵄩 +𝑒) 𝑑𝜏󵄩 × 󵄩𝑓 (𝑢+̃ V) −𝑓(𝑢̃)󵄩 𝑑𝜏 󵄩 󵄩 󵄩𝑀𝑠 󵄩 𝑟 𝑠 𝑝󸀠,1 𝐿𝑡([𝑡0,𝑇],𝑀𝑝,1) (129) 󵄩 𝑡 󵄩 󵄩 󵄩 󵄩 󵄩 (124) ≲ ∫ 󵄩𝑓 (𝑢+̃ V) −𝑓(𝑢̃)󵄩 𝑑𝜏 + 󵄩∫ 𝐾 (𝑡−𝜏) 𝑒𝑑𝜏󵄩 󵄩 󵄩𝑀𝑠 󵄩 󵄩 𝐼 𝑝󸀠,1 󵄩 𝑡0 󵄩 𝑟 𝑠 󵄩 󵄩𝐿𝑡([𝑡0,𝑇],𝑀𝑝,1) 𝛽󵄩 󵄩 ≲ |𝐼| 󵄩𝑓 (𝑢+̃ V) −𝑓(𝑢̃)󵄩 𝑟/(𝑘+1) 𝑠 󵄩 󵄩 𝐿𝑡 (𝐼,𝑀 󸀠 ) ≲𝜀+󵄩(𝜕𝑡𝑡 +𝐼−Δ)V +𝑒󵄩𝐿𝑟/(𝑘+1)([𝑡 ,𝑇],𝑀𝑠 ) 𝑝 ,1 𝑡 0 𝑝󸀠,1 ≲𝐶(𝐼) 𝜀. + ‖𝑒‖𝐿𝑟/(𝑘+1)([𝑡 ,𝑇],𝑀𝑠 ) 𝑡 0 𝑝󸀠,1 Similarly, we have ≲𝜀+𝑆(𝑇) . 󵄩 𝑡 󵄩 󵄩 󸀠 󵄩 󵄩∫ 𝐾 (𝑡−𝜏) (𝑓 (𝑢+̃ V) −𝑓(𝑢̃))𝑑𝜏󵄩 ≲𝐶(𝐼) 𝜀, 󵄩 󵄩 󵄩 𝑡0 󵄩𝑀𝑠−1 By Product Lemma 21,wededuce 𝑝,1 󵄩 󵄩 󵄩 󵄩 󵄩 𝑡 󵄩 𝑆 (𝑇) = 󵄩𝑓(𝑢+̃ V)−𝑓(𝑢)̃ 󵄩 𝑟/(𝑘+1) 𝑠 󵄩 󵄩 󵄩 󵄩𝐿 ([𝑡0,𝑇],𝑀 ) 󵄩∫ 𝐾 (𝑡−𝜏) 𝑒𝑑𝜏󵄩 ≲𝐶(𝐼) 𝜀, 𝑡 𝑝󸀠,1 󵄩 󵄩 (130) 󵄩 𝑡0 󵄩 ∞ 𝑠 󵄩 󵄩𝐿𝑡 (𝐼,𝑀𝑝,1) 𝑘+1 𝑘 ≲ ‖V(𝑡)‖𝐿𝑟 ([𝑡 ,𝑇],𝑀𝑠 ) + ‖V(𝑡)‖𝐿𝑟 ([𝑡 ,𝑇],𝑀𝑠 ) 󵄩 󵄩 𝑡 0 𝑝,1 𝑡 0 𝑝,1 󵄩 𝑡 󵄩 󵄩 󸀠 󵄩 󵄩∫ 𝐾 (𝑡−𝜏) 𝑒𝑑𝜏󵄩 ≲𝐶(𝐼) 𝜀, × ‖𝑢(𝑡)̃ ‖ 𝑟 𝑠 (125) 󵄩 𝑡0 󵄩 ∞ 𝑠−1 𝐿𝑡([𝑡0,𝑇],𝑀𝑝,1) 󵄩 󵄩𝐿𝑡 (𝐼,𝑀𝑝,1 )

+⋅⋅⋅+‖V(𝑡)‖ 𝑟 𝑠 andthenweusetheintegralequationtoconclude(25)and 𝐿𝑡([𝑡0,𝑇],𝑀𝑝,1) 𝑘 (26). Finally, (27)and(28) are easy conclusions from (25), × ‖𝑢(𝑡)̃ ‖ 𝑟 𝑠 . (26), and (20). 𝐿𝑡([𝑡0,𝑇],𝑀𝑝,1) After proof of short-time perturbations, we will derive ‖𝑢(𝑡)‖̃ 𝑟 𝑠 ≤𝜀 long-time perturbations by an iterative procedure. In long- Noticing that 𝐿𝑡([𝑡0,𝑇],𝑀𝑝,1) 𝑐,wehave ‖𝑢‖̃ 𝑟 𝑠 ≤𝜀 time perturbations, the smallness condition 𝐿𝑡(𝐼,𝑀𝑝,1) 𝑐 𝑘+1 𝑘 𝑆 (𝑇) ≲(𝜀+𝑆(𝑇)) + (𝜀 + 𝑆(𝑇)) 𝜀 ‖𝑢‖̃ 𝑟 𝑠 ≤𝐿 𝑐 will be replaced by a bound condition 𝐿𝑡(𝐼,𝑀 ) .Inour (126) 𝑝,1 𝑘 proof, we need to divide interval 𝐼 into several subintervals 𝐼𝑗 +⋅⋅⋅+(𝜀+𝑆(𝑇)) 𝜀𝑐 ‖𝑢‖̃ 𝑟 𝑠 𝑘 to gain a small size of 𝐿𝑡(𝐼𝑗,𝑀𝑝,1).The -admissible condition for every 𝑇∈𝐼. which ensures 𝑟<∞allows us to do so. 14 Abstract and Applied Analysis

5.2.ProofofLong-TimePerturbations. For convenience, we Using integral equation also assume that 𝑡0 = inf 𝐼. We subdivide 𝐼 into 𝑁 V (𝑡) =𝐾󸀠 (𝑡 − 𝑡 ) V (𝑡 )+𝐾(𝑡−𝑡 ) V (𝑡 ) subintervals 𝐼𝑗 =[𝑡𝑗,𝑡𝑗+1], 𝑗=0,...,𝑁−1such that 0 0 0 𝑡 0 𝑡 (141) ‖𝑢̃‖ 𝑟 𝑠 ≤𝜀, 𝐿𝑡(𝐼𝑗,𝑀𝑝,1) 𝑐 (131) − ∫ 𝐾 (𝑡−𝜏) (𝑓 (𝑢+̃ V) −𝑓(𝑢̃) −𝑒)𝑑𝜏 𝑡0 where 𝜀𝑐 is as in Theorem 11.Choose𝜀1 ≤𝜀𝑐 in Theorem 12 and Holder¨ inequality, we conclude that and apply short-time perturbations on 𝐼0 to obtain 󵄩 󵄩 ‖V(𝑡)‖ ∞ 𝑠 󵄩 󵄩 𝐿𝑡 (𝐼,𝑀𝑝,1) 󵄩(𝜕𝑡𝑡 +𝐼−Δ)(𝑢−𝑢)̃ +󵄩 𝑒 𝑟/(𝑘+1) 𝑠 ≲𝜀, 󵄩 󵄩𝐿𝑡 (𝐼0,𝑀 󸀠 ) 𝑝 ,1 󵄩 󵄩 󵄩 󵄩 (132) ≲𝐶( 𝐼 ) (󵄩V(𝑡 )󵄩 + 󵄩V (𝑡 )󵄩 ) ‖𝑢−𝑢̃‖ 𝑟 𝑠 ≲𝜀. | | 󵄩 0 󵄩𝑀𝑠 󵄩 𝑡 0 󵄩𝑀𝑠−1 𝐿𝑡(𝐼0,𝑀𝑝,1) 𝑝,1 𝑝,1

Using the integral equation on 𝐼1,wehave 󵄩 󵄩 +𝐶(|𝐼|) (󵄩𝑓 (𝑢+̃ V) −𝑓(𝑢)̃ 󵄩 𝑟/(𝑘+1) 𝑠 󵄩 󵄩𝐿𝑡 (𝐼,𝑀 󸀠 ) (142) 󵄩 󵄩 𝑝 ,1 󵄩 󸀠 󵄩 󵄩𝐾 (𝑡 −1 𝑡 ) V (𝑡1)+𝐾(𝑡−𝑡1) V𝑡 (𝑡1)󵄩 𝑟 𝑠 󵄩 󵄩𝐿𝑡(𝐼1,𝑀𝑝,1) 󵄩 󸀠 󵄩 + ‖𝑒‖ 𝑟/(𝑘+1) 𝑠 ) 󵄩 󵄩 𝐿𝑡 (𝐼,𝑀 󸀠 ) ≲ 󵄩𝐾 (𝑡 −0 𝑡 )V(𝑡1)−𝐾(𝑡−𝑡0)V𝑡(𝑡1)󵄩 𝑟 𝑠 𝑝 ,1 󵄩 󵄩𝐿𝑡(𝐼1,𝑀𝑝,1) 󵄩 󵄩 󸀠 ̃󸀠 + 󵄩(𝜕𝑡𝑡 +𝐼−Δ)V +𝑒󵄩𝐿𝑟/(𝑘+1)([𝑡 ,𝑡 ],𝑀𝑠 ) (133) ≲𝐶(|𝐼| ,𝐿) (𝑀 + 𝑀 +𝜀). 𝑡 0 1 𝑝󸀠,1

+ ‖𝑒‖𝐿𝑟/(𝑘+1)([𝑡 ,𝑡 ],𝑀𝑠 ) In the same way, we have 𝑡 0 1 𝑝󸀠,1 ≲𝜀+𝜀+𝜀. 󵄩 󵄩 ̃󸀠 󸀠 󵄩V𝑡(𝑡)󵄩 ∞ 𝑠−1 ≲𝐶(|𝐼| ,𝐿) (𝑀 +𝑀 +𝜀). 󵄩 󵄩𝐿𝑡 (𝐼,𝑀𝑝,1 ) (143) 𝜀 So we can shrink 1 such that Finally, we can use (29)todeduce(37)and(38). 󵄩 󵄩 󵄩𝐾󸀠(𝑡 − 𝑡 )V(𝑡 )+𝐾(𝑡−𝑡 )V (𝑡 )󵄩 ≤𝜀. 󵄩 1 1 1 𝑡 1 󵄩𝐿𝑟 (𝐼,𝑀𝑠 ) 𝑐 (134) 𝑡 𝑝,1 5.3.ProofofContinuousDependence. Since 1/(𝑘+2) ≤ 1/𝑝 < 1/2 − 1/𝑛(𝑘 + 1), (𝑝,𝑘+1)is a 𝑘-admissible pair. For every We also have compact interval 𝐼⊂(𝑇min(𝑢0), 𝑇max(𝑢0)), noticing that ‖𝑒‖ 𝑟/(𝑘+1) ≤ ‖𝑒‖ 𝑟/(𝑘+1) ≤𝜀≤𝜀. 𝐿 (𝐼 ,𝑀𝑠 ) 𝐿 (𝐼,𝑀𝑠 ) 𝑐 󵄩 󸀠 󵄩 𝑡 1 𝑝󸀠,1 𝑡 𝑝󸀠,1 (135) 󵄩 󵄩 󵄩𝐾 (𝑡 −0 𝑡 )(𝑢0,𝑛 −𝑢0)+𝐾(𝑡−𝑡0)(𝑢1,𝑛 −𝑢1)󵄩 𝑘+1 𝑠 󵄩 󵄩𝐿𝑡 (𝐼,𝑀𝑝,1) So we can use the short-time perturbations to obtain 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≲𝐶(|𝐼|) (󵄩𝑢0,𝑛 −𝑢0󵄩 𝑠 + 󵄩𝑢1,𝑛 −𝑢1󵄩 𝑠−1 ), 󵄩 󵄩 𝑀𝑝,1 𝑀𝑝,1 󵄩(𝜕𝑡𝑡 + 𝐼 − Δ)(𝑢 − 𝑢)̃ + 𝑒󵄩 𝑟/(𝑘+1) 𝑠 ≲𝐶(1) 𝜀, 󵄩 󵄩𝐿𝑡 (𝐼1,𝑀 󸀠 ) 𝑝 ,1 (136) (144) ‖𝑢−𝑢̃‖𝐿𝑟 (𝐼 ,𝑀𝑠 ) ≲𝐶(1) 𝜀. 𝑡 1 𝑝,1 we apply long-time perturbations with 𝑟=𝑘+1and 𝑒=0to conclude that there exists a small constant 𝜀2 = For 𝐼2, repeat the process above to shrink 𝜀1 to continue 𝜀2(|𝐼|, ‖𝑢‖𝐿𝑘+1(𝐼,𝑀𝑠 )) such that if the inductive argument. Accordingly, after a finite number of 𝑡 𝑝,1 𝜀 𝑚∈[1,𝑁−1] 󵄩 󵄩 󵄩 󵄩 “shrinks” of 1,wehave,forall , 󵄩𝑢 −𝑢 󵄩 + 󵄩𝑢 −𝑢 󵄩 ≤𝜀, 󵄩 0,𝑛 0󵄩𝑀𝑠 󵄩 1,𝑛 1󵄩𝑀𝑠−1 2 󵄩 󵄩 𝑝,1 𝑝,1 (145) 󵄩𝐾󸀠(𝑡 − 𝑡 )V(𝑡 )+𝐾(𝑡−𝑡 )V (𝑡 )󵄩 <𝜀 󵄩 𝑚 𝑚 𝑚 𝑡 𝑚 󵄩𝐿𝑟 (𝐼,𝑀𝑠 ) (137) 𝑡 𝑝,1 then (1)hasastrongsolution𝑢𝑛 on 𝐼 with the initial data 𝑢𝑛(𝑡0)=𝑢0,𝑛 and and then, for some 0<𝜀<𝜀1 and 0≤𝑗≤𝐽,wehave 󵄩 󵄩 󵄩 󵄩 󵄩𝑢 −𝑢󵄩 ≲𝐶(|𝐼| , ‖𝑢‖ 𝑘+1 𝑠 ) 𝜀 . 󵄩(𝜕 + 𝐼 − Δ)(𝑢 − 𝑢)̃ + 𝑒󵄩 𝑟/(𝑘+1) ≲ 𝐶 (𝑗) 𝜀, 󵄩 𝑛 󵄩 ∞ 𝑠 𝐿 (𝐼,𝑀 ) 2 󵄩 𝑡𝑡 󵄩 𝑠 󵄩 󵄩𝐿𝑡 (𝐼,𝑀𝑝,1) 𝑡 𝑝,1 (146) 󵄩 󵄩𝐿𝑡 (𝐼𝑗,𝑀 󸀠 ) 𝑝 ,1 (138) ‖𝑢−𝑢̃‖ 𝑟 𝑠 ≲ 𝐶 (𝑗) 𝜀, So is ‖𝑢𝑛 −𝑢‖𝐿∞(𝐼,𝑀𝑠−1). 𝐿𝑡(𝐼𝑗,𝑀𝑝,1) 𝑡 𝑝,1 The proof of Corollary 14 is similar, and we omit it here. where 𝐶(𝑗) only depends on 𝑗.Sowehave 󵄩 ̃ 󵄩 󵄩(𝜕𝑡𝑡 +𝐼−Δ)(𝑢−𝑢) + 𝑒󵄩𝐿𝑟/(𝑘+1)(𝐼,𝑀𝑠 ) Conflict of Interests 𝑡 𝑝󸀠,1 The authors declare that there is no conflict of interests 𝑁−1 (139) ≲ ∑ 𝐶 (𝑗) 𝜀≲𝐶 (𝐿) 𝜀. regarding the publication of this paper. 𝑗=0 Acknowledgments By smallness assumption, we also have This work is partially supported by the NSF of China (Grant ‖𝑢−𝑢̃‖ 𝑟 𝑠 ≲𝐶(𝐿) 𝜀. 𝐿𝑡(𝐼,𝑀𝑝,1) (140) no. 11271330) and PSF of Zhejiang Province (BSH 1302046). Abstract and Applied Analysis 15

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