Relation between hyperbolic Nizhnik-Novikov-Veselov equation and stationary Davey-Stewartson II equation

Relation b et w een h yp erb olic Nizhnik-No vik o v-V eselo v equation and stationary Da v ey-Stew artson I I equation Zi-Xia ng Zhou Sc ho ol of Mathemati cal Sciences, F udan Universit y , Shanghai 2004 3 3, China Emai l : zxzhou@fudan. edu.cn Abstract A Lax system in three v ariables is presente d, t w o e qu a tions of which f o rm the Lax pair of the stationary Da ve y-Stewartson I I equation. With certain nonlinear con- strain ts, the full integrabilit y condition of this Lax system con tains the hyp erb ol ic Nizhnik-No vik o v-V eselo v equation and its standard Lax pair. Th e Darb oux transf o r- mation for the Dav ey-Stew artson I I equation is u sed to so lve the h yp erb olic Nizhnik- No vik o v-V eselo v equation. Using Darb oux transform ation, global n -soliton sol u t ions are obtained. It is pro ved that eac h n -solito n solution approac h es zero uniformly and exp onen tially at sp atial i n finit y and is asymptotic to n 2 lumps of p eaks at temp oral infinity . 1 In tro duct ion The Nizhnik-No vik ov - V eselo v (NNV) equation [16, 17, 1 9 ] is an imp ortan t 2 +1 dimensional in tegrable equation whic h is a natural generalization of the KdV equation to 2 +1 dimensions. It is useful in b oth mech anics and differential geometry [11, 12]. The NNV equation has b een solv ed b y v arious me tho ds s uch as in v erse s cattering [2], biline ar metho d [18], biline ar B¨ ac klund transforma t io n [8], binary Dar b oux t r ans fo r ma t ion [14] and so on [1, 6, 7, 9, 13, 15]. Ho w ev er, one can not construct the usual Darb oux transformation (without in tegration) b ecaus e the principal part of the first equation of its L a x pair is tw o dimens ional wa v e op erator or Laplace op erator. Starting from the idea of nonlinearization [3], many high dimensional integrable systems w ere reduced to lo we r dimensional ones so that interes ting solutions lik e soliton solutions and quasi-p erio dic solutions can b e obtained from low er dimensional systems. Es p ecially , the KP equation [4, 10], the DSI equation and the 2 + 1 dimens ional N -w av e equation [21] w ere related to some 1+1 dimensional AKNS systems. F o llo wing this idea, in this pap er, w e presen t a Lax system of three v a riables , t w o equations o f whic h form the Lax pair of the 1 stationary Da ve y-Stewartson I I (DS I I) equation. With the nonlinear constrain ts (14), the full integrabilit y condition o f this Lax system con tains the hyperb olic NNV equation and its standard Lax pair. The DSI I equation has a Darb oux tra nsfor ma t ion without in tegrat ion. With the relatio ns giv en by (1 4 ), the Da rboux transformatio n for DSI I equation is us ed to solve the hyperb olic NNV equation. This Darb oux transformation without in tegrat io n is mor e suitable for sym- b olic calculation than the kno wn binary Darb oux transformatio n. It is well known that DSI equation has solutions approaching zero exp onen tially at spatial infinit y , but D SI I equation has not. Ho we ver, w e get soliton solution u of the hyperb olic NNV equation f r om t ha t o f the stationar y DSI I eq uatio n so that u approac hes zero exp onen tially at spatial infinity . This is p ossible b ecause the solution u of the h yp erbolic NNV equation is giv en b y i( g − ¯ g ) as in (14) , not f , the solution of the stationary DSI I equation. These soliton solutions are differen t fro m the kno wn one deriv ed by binary Da rboux transformatio n or bilinear metho d etc. and the b eha vior of the solutio ns is more complicated. In Section 2, after reviewing the h yp erbolic NNV equation and the stationary DSI I equa- tion together with their standard Lax pairs, a new Lax system (10 ) is presen ted in whic h an extra equation is added t o the standa r d Lax pair of the stationary DSI I equation. With the nonlinear constraints (14), the integrabilit y condition of this Lax system includes b oth the h yp erbolic NNV equation and its standard Lax pair. T he Darb oux transformation for the new Lax system is giv en in Section 3 and the general expression of m ulti-solito n solutions is presen ted in Section 4. In Section 5, the explicit expressions and b eha vior of single-soliton solutions are discussed. In Section 6, it is prov ed that eac h n - s olito n solutio n approac hes zero uniformly and exp onen tially at spatial infinity . In Section 7, it is pro ve d that eac h n -soliton solution is asymptotic to n 2 lumps of p eaks at temp oral infinity . Finally , some linear algebraic lemmas are presen ted in the Appendix. 2 Hyp erb olic Nizhnik-No vik o v-V eselo v equation and Da v ey-Stew arts on I I equatio n The h yp erb olic NNV eq uatio n is u t = u ξ ξξ + u ηη η + 3( uv ) ξ + 3( uw ) η , v η = u ξ , w ξ = u η , (1) whic h has a Lax pair f ξ η + uf = 0 , f t = f ξ ξξ + f ηη η + 3 v f ξ + 3 w f η . (2) By taking the new co ordinates x = ξ − η , y = ξ + η , the hyperb olic NNV equation (1) b ecome s u t = 2 u y y y + 6 u xxy + 3( u ( v + w )) y + 3( u ( v − w )) x , ( ∂ y − ∂ x ) v = ( ∂ y + ∂ x ) u, ( ∂ y + ∂ x ) w = ( ∂ y − ∂ x ) u, (3) 2 and the Lax pair (2) b ecomes f y y − f xx + uf = 0 , f t = 2 f y y y + 6 u xxy + 3( v + w ) f y + 3( v − w ) f x . (4) On the other hand, the DSI I equation is − i f τ = f xx − f y y − i( g − ¯ g ) f , ( ∂ y − i ∂ x ) g = ( ∂ x − i ∂ y )( | f | 2 ) , (5) whic h has a Lax pair Ψ y = i J Ψ x + P Ψ , Ψ τ = 2 i J Ψ xx + 2 P Ψ x + Q Ψ (6) where J =   1 0 0 − 1   , P =   f − ¯ f   , Q =   g f x − i f y − ¯ f x − i ¯ f y ¯ g   . (7) If ( f , g ) is indep ende nt of τ , (5 ) b ec omes the stat io nary DSI I equation f xx − f y y − i( g − ¯ g ) f = 0 , ( ∂ y − i ∂ x ) g = ( ∂ x − i ∂ y )( | f | 2 ) . (8) T aking Ψ( x, y , τ ) = Φ( x, y )e 2 i λ 2 τ in (6), w e g e t the Lax pair for (8) as Φ y = i J Φ x + P Φ , 2 i λ 2 Φ = 2 i J Φ xx + 2 P Φ x + Q Φ (9) The first equation of (4) a nd the first equation of (8) are similar, and t he second equation of (4) is of order 3 . Hence w e introduce an extra equation to the La x pair (9) so that the whole system b ecome s Φ y = M ( ∂ )Φ ≡ i J Φ x + P Φ , 2 i λ 2 Φ = L ( ∂ ) Φ ≡ 2 i J Φ xx + 2 P Φ x + Q Φ , Φ t = N ( ∂ )Φ ≡ 1 6 i J Φ xxx + 16 P Φ xx + R Φ x + S Φ (10) where J , P , Q ar e giv en b y (7) , R = 4   3 g + i | f | 2 4 f x − 2 i f y − 4 ¯ f x − 2 i ¯ f y 3 ¯ g − i | f | 2   , S = 2   3 g x + 2 i ¯ f f x + ¯ f f y − f ¯ f y 6 f xx − 2 i f xy − i( g − ¯ g ) f + 2 | f | 2 f − 6 ¯ f xx − 2 i ¯ f xy + i( g − ¯ g ) ¯ f − 2 | f | 2 ¯ f 3 ¯ g x − 2 i f ¯ f x + f ¯ f y − ¯ f f y   , (11) 3 and L ( ∂ ), M ( ∂ ) and N ( ∂ ) refer to differen tial op erators with respect to x whose co efficien t s are 2 × 2 matrices, ∂ = ∂ x . The in tegr a bilit y conditions of (10) include the follo wing equations: f y y − f xx + uf = 0 , f t = 2 f y y y + 6 f xxy + 3( v + w ) f y + 3( v − w ) f x , (12) ( ∂ y − i ∂ x ) g = ( ∂ x − i ∂ y )( | f | 2 ) , i 2 g t = − 2 g xxx + 2 ¯ f f xxy + 2 f ¯ f xxy + 4 i ¯ f f xxx + 4 i f ¯ f xxx +2( ¯ f x − i ¯ f y ) f xy + 2( f x − i f y ) ¯ f xy + 2( i ¯ f x + 2 ¯ f y ) f xx + 2( i f x + 2 f y ) ¯ f xx +(2 | f | 2 − i( g − ¯ g ))( | f | 2 ) y + (6 i | f | 2 + ( g − ¯ g ))( | f | 2 ) x − 2 | f | 2 ¯ g x + 6 i g g x , (13) where u = i( g − ¯ g ) , v = 2 | f | 2 + ( g + ¯ g ) , w = 2 | f | 2 − ( g + ¯ g ) . (14) Note that (12) is exactly the same as the origina l Lax pair ( 4 ) of the hyperb olic NNV equation. By dir ect calculation, w e kno w that ( u, v , w ) satisfies the h yp erb olic NNV equation (3) prov ided that f and g satisfy (12)–(14). Therefore, explicit solutions o f the h yp e rb olic NNV equation can b e obtained from tho s e of (1 2 )–(1 4). Clearly , the solutions of ( 1 2 )–(1 4) are only part of those of the h yp erb olic NNV equation. Ho w ev er, they include some in teresting ones whic h will b e show n in the rest o f this paper. 3 Darb oux transformation The binary D arboux tra ns for ma t io n fo r the h yp erb olic NNV equation is w ell-know n [14]. In tegrations are needed in constructing explicit solutions. Ho w ev er, for DSI I equation, usual Darb oux transformation without in tegration is kno wn. This Darb oux transformation is simpler than the binary Darb oux tr ans fo r ma t io n for the h yp erb olic NNV equation, and can b e easily used to t he stationary DSI I equation so that explicit solutions of the hyperb olic NNV equation can b e c onstructed. Note that the co efficien ts of L ( ∂ ), M ( ∂ ), N ( ∂ ) s atisfy i J, P , Q, R, S ∈ Σ (15) where Σ = { A is a 2 × 2 m at rix | K AK − 1 = ¯ A } =      a b − ¯ b ¯ a      a, b ∈ C    , (16) K = − 1 1 ! . That is, L ( ∂ ), M ( ∂ ), N ( ∂ ) satisfy K L ( ∂ ) K − 1 = ¯ L ( ∂ ) , K M ( ∂ ) K − 1 = ¯ M ( ∂ ) , K N ( ∂ ) K − 1 = ¯ N ( ∂ ) . (17) 4 Hence, if Φ = ξ η ! is a solutio n of (10) with λ = λ 0 , then K ¯ Φ = − ¯ η ¯ ξ ! is a solutio n of (10) with λ = ± i ¯ λ 0 . The Darb oux transformation of arbitrary order is constructed as fo llows [5, 20]. Supp ose G ( ∂ ) = ∂ n + G 1 ( x, y , t ) ∂ n − 1 + · · · + G n ( x, y , t ) (18) is a D arboux op erator for (10), i.e., there exist L ′ ( ∂ ), M ′ ( ∂ ), N ′ ( ∂ ) whic h ha ve the same form as L ( ∂ ), M ( ∂ ), N ( ∂ ) with f and g replaced by certain f ′ and g ′ , suc h that Φ ′ = G ( ∂ ) Φ satisfies λ Φ ′ = L ′ ( ∂ )Φ ′ , Φ ′ y = M ′ ( ∂ )Φ ′ , Φ ′ t = N ′ ( ∂ )Φ ′ . (19) If s o, G ( ∂ ) satisfie s L ′ ( ∂ ) G ( ∂ ) = G ( ∂ ) L ( ∂ ) , M ′ ( ∂ ) G ( ∂ ) = G ( ∂ ) M ( ∂ ) + G y ( ∂ ) , N ′ ( ∂ ) G ( ∂ ) = G ( ∂ ) N ( ∂ ) + G t ( ∂ ) . (20) Since L ( ∂ ), M ( ∂ ), N ( ∂ ) satisfy the relations ( 1 7), and L ′ ( ∂ ), M ′ ( ∂ ), N ′ ( ∂ ) satisfy the similar relations K L ′ ( ∂ ) K − 1 = ¯ L ′ ( ∂ ) , K M ′ ( ∂ ) K − 1 = ¯ M ′ ( ∂ ) , K N ′ ( ∂ ) K − 1 = ¯ N ′ ( ∂ ) , (21) w e w an t that G ( ∂ ) satisfies K G ( ∂ ) K − 1 = ¯ G ( ∂ ). W rite G j =   a j b j − ¯ b j ¯ a j   . (22) Denote L ′ ( ∂ ) = 2 i J ∂ 2 + 2 P ′ ∂ + Q ′ , the n the first equation of (20) leads to (2 i J ∂ 2 + 2 P ′ ∂ + Q ′ )( ∂ n + G 1 ∂ n − 1 + · · · + G n ) = ( ∂ n + G 1 ∂ n − 1 + · · · + G n )(2 i J ∂ 2 + 2 P ∂ + Q ) , (23) in w hich the coefficien ts of ∂ n +1 and ∂ n giv e P ′ = P − i[ J, G 1 ] , Q ′ = Q − 2 i[ J, G 2 ] − 2[ P , G 1 ] + 2 i [ J , G 1 ] G 1 + 2 nP x − 4 i J G 1 ,x . (24) Hence, a fter t he action of Darb oux transformation, f ′ = f − 2 i b 1 , g ′ = g − 4 i a 1 ,x − 2( ¯ f b 1 − f ¯ b 1 ) − 4 i | b 1 | 2 , u ′ = u + 8 | b 1 | 2 − 4 i( ¯ f b 1 − f ¯ b 1 ) + 4( a 1 + ¯ a 1 ) x , v ′ = v + 8 | b 1 | 2 − 4 i( ¯ f b 1 − f ¯ b 1 ) − 4 i( a 1 − ¯ a 1 ) x , w ′ = w + 8 | b 1 | 2 − 4 i( ¯ f b 1 − f ¯ b 1 ) + 4 i( a 1 − ¯ a 1 ) x . (25) 5 No w tak e n distinct complex num b ers λ 1 , · · · , λ n with λ j = µ j + i ν j ( µ j ’s a nd ν j ’s a re real). Let Φ j =   ξ j η j   b e a column solution of (10) with λ = λ j , then Φ n + j ≡ K ¯ Φ j =   − ¯ η j ¯ ξ j   is a s olutio n of (10) with λ = ± i ¯ λ j ( j = 1 · · · , n ). The Darb oux tr a ns for ma t io n is determine d b y the system of linear algebraic equations G ( ∂ )Φ j = 0 ( j = 1 , · · · , 2 n ) (26) if it has a unique solution [20]. Denote ξ =     ξ 1 . . . ξ n     , η =     η 1 . . . η n     , a =     a 1 . . . a n     , b =     b 1 . . . b n     , (27) then (2 6) b ec omes T   a b   = −   ∂ n ξ − ∂ n ¯ η   (28) where T =   A B − ¯ B ¯ A   , (29) A =  ∂ n − 1 ξ · · · ξ  , B =  ∂ n − 1 η · · · η  . (30) (26) ha s a unique solution if and only if det T 6 = 0. 4 Expressio n of so liton solu tions F o r zero seed solution u = v = w = f = g = 0, (10) b ecomes Φ xx = λ 2 J Φ , Φ y = i J Φ x , Φ t = 16 i J Φ xxx (31) with Φ = ( ξ , η ) T . Hence tak e ξ j = κ (1) j (e ρ (1) j + i σ (1) j + e − ρ (1) j − i σ (1) j ) , η j = κ (2) j (e ρ (2) j + i σ (2) j + e − ρ (2) j − i σ (2) j ) (32) where ρ (1) j = Re ( λ j x + i λ j y + 16 i λ 3 j t ) + ρ (1) j 0 = µ j x − ν j y + 16( ν 3 j − 3 µ 2 j ν j ) t + ρ (1) j 0 , ρ (2) j = Re ( i λ j x + λ j y − 16 λ 3 j t ) + ρ (2) j 0 = − ν j x + µ j y − 16( µ 3 j − 3 µ j ν 2 j ) t + ρ (2) j 0 , σ (1) j = Im ( λ j x + i λ j y + 16 i λ 3 j t ) + σ (1) j 0 = ν j x + µ j y + 16( µ 3 j − 3 µ j ν 2 j ) t + σ (1) j 0 , σ (2) j = Im ( i λ j x + λ j y − 16 λ 3 j t ) + σ (2) j 0 = µ j x + ν j y + 16( ν 3 j − 3 µ 2 j ν j ) t + σ (2) j 0 , (33) 6 κ (1) j , κ (2) j are non-zero constan ts, ρ (1) j 0 , ρ (2) j 0 , σ (1) j 0 , σ (2) j 0 are real constan ts. By solving a j ’s and b j ’s fro m (2 8), the D a rboux transformation (25) giv es the n -soliton solution f = − 2 i b 1 , g = − 4 i a 1 ,x − 4 i | b 1 | 2 , u = 8 | b 1 | 2 + 4( a 1 + ¯ a 1 ) x , v = 8 | b 1 | 2 − 4 i( a 1 − ¯ a 1 ) x , w = 8 | b 1 | 2 + 4 i( a 1 − ¯ a 1 ) x . (34) Hereafter w e omit the primes on f , g , u, v , w for those obtained by t he a c tio n of Da r b oux transformation. Let K n =   − I n I n   . Denote ζ = ξ − ¯ η ! , R j ··· k =  ∂ j ζ ∂ j − 1 ζ , · · · , ∂ k ζ  (35) for j ≥ k , then T =  R n − 1 ··· 0 K n ¯ R n − 1 ··· 0  . (36) Let Π =   ∂ n ζ ∂ n − 1 ζ ∂ n − 2 ζ R n − 3 ··· 0 K n ¯ R n − 1 ··· 0 0 0 ∂ n +1 ζ ∂ n ζ ∂ n − 1 ζ 0 0 R n − 2 ··· 0 K n ¯ R n − 1 ··· 0   . (37) Theorem 1 When det T 6 = 0 , the multi-soliton solution u of the hyp erb olic NNV e quation given by (34) c an b e written as u = − 8 Re det Π (det T ) 2 . ( 38) Pro of. Solved from (28) by Cramer r ule, a 1 = − (det T ) − 1    ∂ n ζ R n − 2 ··· 0 K n ¯ R n − 1 ··· 0    , (39) b 1 = − (det T ) − 1    R n − 1 ··· 0 ∂ n ζ K n ¯ R n − 2 ··· 0    . (40) a 1 + ¯ a 1 = − (det T ) − 1     ∂ n ζ R n − 2 ··· 0 K n ¯ R n − 1 ··· 0    +    ∂ n ζ R n − 2 ··· 0 K n ¯ R n − 1 ··· 0     = − (det T ) − 1     ∂ n ζ R n − 2 ··· 0 K n ¯ R n − 1 ··· 0    +    R n − 1 ··· 0 K n ∂ n ¯ ζ K n ¯ R n − 2 ··· 0     = − (det T ) − 1 (det T ) x = − tr( T − 1 T x ) , (41) a 1 ,x + ¯ a 1 ,x = − tr( T − 1 T xx ) + tr(( T − 1 T x ) 2 ) . (42) 7 Denote e I k = I k × k 0 ( n − k ) × k ! . Let h = e a e b ! b e the solution of T h = − ∂ n +1 ζ where e a = ( e a 1 , · · · , e a n ) T , e b = ( e b 1 , · · · , e b n ) T , then a 1 ,x + ¯ a 1 ,x = − tr   − e a − a e I n − 2 ¯ e b ¯ b 0 − e b − b 0 − ¯ e a − ¯ a e I n − 2   + tr   − a e I n − 1 ¯ b 0 − b 0 − ¯ a e I n − 1   2 = a 2 1 + ¯ a 2 1 + e a 1 + ¯ e a 1 − a 2 − ¯ a 2 − 2 | b 1 | 2 . (43) According to (34), u = 8 Re ( a 2 1 + e a 1 − a 2 ) . (44) Let d = − (det T ) 2 ( a 2 1 + e a 1 − a 2 ), the n b y C ramer rule, d = −    ∂ n ζ R n − 2 ··· 0 K n ¯ R n − 1 ··· 0    2 + det T    ∂ n +1 ζ R n − 2 ··· 0 K n ¯ R n − 1 ··· 0    + det T    ∂ n ζ ∂ n − 1 ζ R n − 3 ··· 0 K n ¯ R n − 1 ··· 0    . (45) Using La plac e expansion of det Π, d = det Π. Hence u = − 8 Re d (det T ) 2 = − 8 Re det Π (det T ) 2 . (46) The the or em is prov ed. Remark 1 A c c or ding to L emma 2 of A, det T ≥ 0 holds everywher e. Howev er, det T > 0 may not hold everywher e when the p ar ameters ρ ( k ) j 0 and σ ( k ) j 0 take some sp e cial values, as wil l shown in the next se ction fo r sing le so lit on solution. On the other h and, det T > 0 ho lds everywher e in generic c ase, which wil l b e shown her e. The Darb oux op er ator G ( ∂ ) of or der n c an b e c onstructe d by c omp osing n Darb oux op- er a tors of or der one as fol lows. F or given λ 1 , · · · , λ n and Φ 1 , · · · , Φ n as ab ove, let H j = (Φ j , Φ n + j ) j (= 1 , · · · , n ) . I f det H 1 6 = 0 , then ∆ 1 ( ∂ ) = ∂ − H 1 ,x H − 1 1 is a D arb oux op er- ator of or der one. It tr ansforms ( u , v , w, f , g ) to ( u (1) , v (1) , w (1) , f (1) , g (1) ) and tr ansforms H j to H (1) j = ∆ 1 ( ∂ ) H j = H j,x − H 1 ,x H − 1 1 H j ( j = 2 , 3 , · · · , n ) . A gain, if det H (1) 2 6 = 0 , then ∆ 2 ( ∂ ) = ∂ − H (1) 2 ,x ( H (1) 2 ) − 1 is a Darb oux op er ator of or d er one for the L ax p air w it h ( u (1) , v (1) , w (1) , f (1) , g (1) ) . It tr ansforms ( u (1) , v (1) , w (1) , f (1) , g (1) ) to ( u (2) , v (2) , w (2) , f (2) , g (2) ) and tr ansforms H (1) j to H (2) j = ∆ 2 ( ∂ ) H (1) j = H (1) j,x − H (1) 2 ,x ( H (1) 2 ) − 1 H (1) j ( j = 3 , 4 , · · · , n ) . Con- tinuing this pr o c ess, we get H ( k ) j ( k = 1 , · · · , n − 1; j = k + 1 , · · · , n ) an d ∆ j ( ∂ ) ( j = 1 , · · · , n ) . A c c or din g to [20], G ( ∂ ) = ∆ n ( ∂ )∆ n − 1 ( ∂ ) · · · ∆ 1 ( ∂ ) , det T = det( H ( n − 1) n ) det( H ( n − 2) n − 1 ) · · · det( H (1) 2 ) det( H 1 ) . (47) 8 Henc e det T 6 = 0 if al l det H ( j − 1) j 6 = 0 . Supp ose H ( j − 1) j =   ξ ( j − 1) j − ¯ η ( j − 1) j η ( j − 1) j ¯ ξ ( j − 1) j   , then det H ( j − 1) j = | ξ ( j − 1) j | 2 + | η ( j − 1) j | 2 = 0 if and only if ξ ( j − 1) j = 0 and η ( j − 1) j = 0 hold simultane ously. F or fixe d j , this gives a system of four r e al e quations Re ξ ( j − 1) j = 0 , Im ξ ( j − 1) j = 0 , Re η ( j − 1) j = 0 , Im η ( j − 1) j = 0 (48) for thr e e r e al varia b les x, y , t . It has no solution unless the p ar ameters ρ ( k ) j 0 and σ ( k ) j 0 ( j = 1 , · · · , n ; k = 1 , 2) take sp e cia l values. This sh o ws that det T > 0 holds everywher e fo r generic ρ ( k ) j 0 and σ ( k ) j 0 . Ther efor e , the multi-soliton solution u is glob al for generic ρ ( k ) j 0 and σ ( k ) j 0 . 5 Single sol iton solu tion By ta king n = 1, the single soliton can be obtained as u = 16 A B 2 (49) where B = ( κ (1) 1 ) 2 cosh(2 ρ (1) 1 ) + ( κ (2) 1 ) 2 cosh(2 ρ (2) 1 ) + ( κ (1) 1 ) 2 cos(2 σ (1) 1 ) + ( κ (2) 1 ) 2 cos(2 σ (2) 1 ) , A = − ( µ 2 1 − ν 2 1 )( κ (1) 1 ) 4 cosh(2 ρ (1) 1 ) cos(2 σ (1) 1 ) − 2 µ 1 ν 1 ( κ (1) 1 ) 4 sinh(2 ρ (1) 1 ) sin(2 σ (1) 1 ) − ( µ 2 1 + ν 2 1 )( κ (1) 1 ) 2 ( κ (2) 1 ) 2 sinh(2 ρ (1) 1 ) sin(2 σ (2) 1 ) +( µ 2 1 − ν 2 1 )( κ (2) 1 ) 4 cosh(2 ρ (2) 1 ) cos(2 σ (2) 1 ) + 2 µ 1 ν 1 ( κ (2) 1 ) 4 sinh(2 ρ (2) 1 ) sin(2 σ (2) 1 ) +( µ 2 1 + ν 2 1 )( κ (1) 1 ) 2 ( κ (2) 1 ) 2 sinh(2 ρ (2) 1 ) sin(2 σ (1) 1 ) +( µ 2 1 − ν 2 1 )(( κ (2) 1 ) 4 − ( κ (1) 1 ) 4 ) . (50) The solution is singular if B = 0, i.e. | ξ 1 | 2 + | η 1 | 2 = 0. This is equiv alen t to ρ (1) 1 = ρ (2) 1 = 0, 2 σ (1) 1 = j π + π / 2, 2 σ (2) 1 = k π + π / 2 f or certain in tegers j and k . In con trast, the solution is global if and only if | ξ 1 | 2 + | η 1 | 2 6 = 0 ev erywhere, i.e . the parameters s atisfy µ 1 ( ρ (1) 10 − σ (2) 10 + k π + π / 2) + ν 1 ( ρ (2) 10 + σ (1) 10 − j π − π / 2) 6 = 0 . (51) W e alw a ys suppose (51) is satisfied, whic h is e quiv alen t to d et T 6 = 0. When µ 2 1 6 = ν 2 1 , the solution u approac hes zero exp onen tially at spatial infinit y , and the p eaks app ear when neither ρ (1) 1 nor ρ (2) 1 is large. Hence the cen ter of the lump of p eaks lo cates near ρ (1) 1 = 0 and ρ (2) 1 = 0, i.e., x = 64 µ 1 ν 1 t − µ 1 ρ (1) 10 + ν 1 ρ (2) 10 µ 2 1 − ν 2 1 , y = 1 6( µ 2 1 + ν 2 1 ) t − ν 1 ρ (1) 10 + µ 1 ρ (2) 10 µ 2 1 − ν 2 1 . (52) 9 Figure 1: Single soliton solution u : λ 1 = 2 + 0 . 5 i, κ (1) 1 = 1, κ (2) 1 = 1 . 2, ρ (1) 10 = 0, ρ (2) 10 = 1, σ (1) 10 = σ (2) 10 = 0, t = 1. The solutions ar e sho wn in F igure 1 and Figure 2 for differen t parameters. The fig ur e of the solution contains a lum p of p eaks rather than a s ingle p eak, and the shap e depends on the angle arctan µ 2 1 − ν 2 1 2 µ 1 ν 1 b et w een the straig h t lines ρ (1) 1 = 0 and ρ (2) 1 = 0. Nev ertheless, we still call it single soliton solution b ecause it is generated from the zero solution b y Darb oux transformation, and the peaks in the solution nev er separate. Note that although u is lo calized, v and w ar e not. If ν 1 = µ 1 6 = 0, the solution is in v arian t when ( x, y ) is c hanged to  x + mπ 2 µ 1 , y + mπ 2 µ 1  for an y in teger m . Hence the solution is p erio dic. Moreo v er, ρ (1) 1 + ρ (2) 1 = ρ (1) 10 + ρ (2) 10 . The p eaks app ear when neither ρ (1) 1 nor ρ (2) 1 is large. Hence the p eaks lie near the straigh t line x − y − 3 2 µ 2 1 t + ρ (1) 10 − ρ (2) 10 2 µ 1 = 0. The solution is sho wn in Figure 3. Similarly , the s olutio n is also p eriodic if ν 1 = − µ 1 6 = 0. 6 Lo calization of the solut ions In this section, we will prov e that the m ulti-solito n solutions approach zero uniformly a nd exp o ne ntially at spatial infinit y . In order to get g lo bal solutions, we alwa ys supp ose det T 6 = 0 ev erywhere, whic h is true for generic parameters ρ ( k ) j 0 and σ ( k ) j 0 ( j = 1 , · · · , n ; k = 1 , 2). Note t ha t the solutio n of (28) is in v arian t if b oth ξ j and η j (for fixed j ) ar e multiplie d b y a common function. Let ω j =    ξ j if | ξ j | ≥ | η j | η j if | ξ j | < | η j | , (53) ˚ T = diag ( ω 1 , · · · , ω n , ¯ ω 1 , · · · , ¯ ω n ) . (54) 10 Figure 2: Single soliton solution u : λ 1 = 1 . 1 + 0 . 9 i, κ (1) 1 = 1, κ (2) 1 = 1 . 2, ρ (1) 10 = 0, ρ (2) 10 = 1, σ (1) 10 = σ (2) 10 = 0, t = 1. Figure 3 : P erio dic solution u : λ 1 = 1 + i, κ (1) 1 = 1, κ (2) 1 = 1 . 2 , ρ (1) 10 = 0, ρ (2) 10 = 1, σ (1) 10 = σ (2) 10 = 0, t = 1. 11 Let e T = ˚ T − 1 T , then the nor m of each entry of e T cannot exceed 1. Although e T is not con tin uous, | det T | is con tin uous. Let x = r cos θ , y = r sin θ . Since ρ (1) j ’s and ρ (2) j ’s dep en d on x and y linearly , w e can write, fo r k = 1 , 2, ρ ( k ) j ( r cos θ, r sin θ, t ) = ε ( k ) j ( θ ) α ( k ) j ( θ ) r + β ( k ) j , (55) where ε ( k ) j ( θ ) = ± 1 ( j = 1 , · · · , n ) so that α ( k ) j ( θ ) ≥ 0. Here the v a riable t is omitted in α ( k ) j ( θ ), β ( k ) j and ε ( k ) j ( θ ). Clearly α ( k ) j ’s are con tinuous functions. Not e also that ε ( k ) j ( θ ) is not w ell-defined when α ( k ) j ( θ ) = 0. Theorem 2 Supp ose λ 1 , · · · , λ n ar e distinct non - zer o c omplex numb ers such that ¯ λ j 6 = ± i λ l for al l j, l = 1 , · · · , n . u is the n -soliton solution given by (3 8). Then fo r fixe d t , ther e ar e p o sitive c onstants r 0 , χ and C such that | u ( r cos θ, r sin θ, t ) | ≤ C e − χr (56) for r > r 0 and al l e i θ ∈ S 1 . Henc e u ( r cos θ , r sin θ , t ) → 0 uniformly and exp onential ly as r → + ∞ . Pro of. The pro of is div ided in to four steps. Step 1: Obtain the asymptotic beha vior of ξ j ’s and η j ’s. Let λ j = µ j + i ν j where µ j ’s and ν j ’s ar e real, t hen µ j 6 = ± ν j for a ll j = 1 , · · · , n . Let Z ( ε ) = { e i θ ∈ S 1 | tan θ = ε } for ε = ± 1, Z = Z (+1) ∪ Z ( − 1) . If e i θ ∈ Z ( ε ) , then b y (33 ), α (1) j ( θ ) = α (2) j ( θ ) = 1 √ 2 | µ j − ε ν j | > 0, ε (2) j ( θ ) = εε (1) j ( θ ) and ε (1) j ( θ )( µ j − ε ν j ) cos θ > 0 for a ll j = 1 , · · · , n . If e i θ ∈ S 1 \ Z , then α (1) j ( θ ) = | µ j cos θ − ν j sin θ | , α (2) j ( θ ) = | − ν j cos θ + µ j sin θ | with α (1) j ( θ ) 6 = α (2) j ( θ ). F o r δ ∈ (0 , π / 4), define Ω ( ε ) δ = n e i θ    there exists e i θ 0 ∈ Z ( ε ) suc h that | θ − θ 0 | < δ o ( ε = ± 1) , Ω (0) δ = n e i θ    | θ − θ 0 | > δ / 2 for all e i θ 0 ∈ Z o . (57) Then Ω (+1) δ ∪ Ω ( − 1) δ ∪ Ω (0) δ = S 1 , and there exists δ ∈ (0 , π / 4) and ω > 0 suc h that α (1) j ( θ ) > ω , α (2) j ( θ ) > ω , ε (2) j ( θ ) = εε (1) j ( θ ) , ε (1) j ( θ )( µ j − εν j ) cos θ > 0 if e i θ ∈ Ω ( ε ) δ , | α (1) j ( θ ) − α (2) j ( θ ) | > ω if e i θ ∈ Ω (0) δ . (58) F o r e i θ 0 ∈ Z ( ε ) and e i θ ∈ Ω ( ε ) δ with | θ − θ 0 | < δ , ε (1) j ( θ ) is a constan t, α (1) j ( θ ) − α (2) j ( θ ) = ε (1) j ( θ )( µ j cos θ − ν j sin θ ) − ε (2) j ( θ )( − ν j cos θ + µ j sin θ ) = εε (1) j ( θ )( µ j + εν j ) cos θ ( ε − tan θ ) . (59) 12 Hence, if α (1) j ( θ ) > α (2) j ( θ ) for e i θ ∈ Ω ( ε ) δ \ Z ( ε ) with 0 < θ − θ 0 < δ , then α (1) j ( θ ) < α (2) j ( θ ) for e i θ ∈ Ω ( ε ) δ \ Z ( ε ) with − δ < θ − θ 0 < 0, and v i s e versa . Recall t hat ξ j = κ (1) j (e ε (1) j ( θ ) λ j x + i ε (1) j ( θ ) λ j y + 1 6 i ε (1) j ( θ ) λ 3 j t + e − ε (1) j ( θ ) λ j x − i ε (1) j ( θ ) λ j y − 1 6 i ε (1) j ( θ ) λ 3 j t ) = κ (1) j e α (1) j ( θ ) r + ε (1) j ( θ ) β (1) j + i ε (1) j ( θ ) σ (1) j ( θ, r )  1 + e − 2 α (1) j ( θ ) r − 2 ε (1) j ( θ ) β (1) j − 2 i ε (1) j ( θ ) σ (1) j ( θ, r )  , η j = κ (2) j (e i ε (2) j ( θ ) λ j x + ε (2) j ( θ ) λ j y − 1 6 ε (2) j ( θ ) λ 3 j t + e − i ε (2) j ( θ ) λ j x − ε (2) j ( θ ) λ j y + 1 6 ε (2) j ( θ ) λ 3 j t ) = κ (2) j e α (2) j ( θ ) r + ε (2) j ( θ ) β (2) j + i ε (2) j ( θ ) σ (2) j ( θ, r )  1 + e − 2 α (2) j ( θ ) r − 2 ε (2) j ( θ ) β (2) j − 2 i ε (2) j ( θ ) σ (2) j ( θ, r )  . (60) Let Y j ( θ , r ) = ( κ (1) j ) − 1 κ (2) j e ( α (2) j ( θ ) − α (1) j ( θ )) r + ε (2) j ( θ ) β (2) j − ε (1) j ( θ ) β (1) j + i ε (2) j ( θ ) σ (2) j ( θ, r ) − i ε (1) j ( θ ) σ (1) j ( θ, r ) . (61) When r → + ∞ , the follow ing limits hold uniformly . F o r e i θ ∈ Ω ( ε ) δ with α (1) j ( θ ) ≥ α (2) j ( θ ), ξ − 1 j ∂ k ξ j → ( ε (1) j ( θ ) λ j ) k , ξ − 1 j ∂ k η j − ( i ε (2) j ( θ ) λ j ) k Y j ( θ , r ) → 0 . (62) F o r e i θ ∈ Ω ( ε ) δ with α (1) j ( θ ) ≤ α (2) j ( θ ), η − 1 j ∂ k ξ j − ( ε (1) j ( θ ) λ j ) k Y j ( θ , r ) − 1 → 0 , η − 1 j ∂ k η j → ( i ε (2) j ( θ ) λ j ) k . (63) F o r e i θ ∈ Ω (0) δ with α (1) j ( θ ) > α (2) j ( θ ), ξ − 1 j ∂ k ξ j → ( ε (1) j ( θ ) λ j ) k , ξ − 1 j ∂ k η j → 0 . (64) F o r e i θ ∈ Ω (0) δ with α (1) j ( θ ) < α (2) j ( θ ), η − 1 j ∂ k ξ j → 0 , η − 1 j ∂ k η j → ( i ε (2) j ( θ ) λ j ) k . (65) Step 2: There ex ists r 0 > 0 and c 0 > 0 suc h that det e T > c 0 when r > r 0 . When e i θ ∈ Z ( ε ) ( ε = ± 1), α (1) j ( θ ) = α (2) j ( θ ) for j = 1 , · · · , n . (60) implie s | η j ( θ , r ) | | ξ j ( θ , r ) | → γ j ( θ ) ≡ | κ (2) j | e ε (2) j ( θ ) β (2) j | κ (1) j | e ε (1) j ( θ ) β (1) j (66) as r → + ∞ . By ( 6 2), det e T ( θ , r ) =  Y | γ j ( θ ) | > 1 | γ j ( θ ) |  − 2       A ( θ , r ) B ( θ, r ) − ¯ B ( θ , r ) ¯ A ( θ , r )       + o (1) , (67) where A ( θ , r ) and B ( θ , r ) are n × n matrices, whose en tries are A j k ( θ , r ) = ( ε (1) j ( θ ) λ j ) k , B j k ( θ , r ) = ( i ε (2) j ( θ ) λ j ) k Y j ( θ , r ) = ( i εε (1) j ( θ ) λ j ) k Y j ( θ , r ) (68 ) 13 and o (1) refers to the terms whic h ten d to zero as r → + ∞ . Let Λ = ( ε (1) 1 ( θ ) λ 1 , · · · , ε (1) n ( θ ) λ n ) , Γ = diag( Y 1 ( θ , r ) , · · · , Y n ( θ , r )) . (69) By Lemma 3 of A, there exist r 1 > 0 and c 1 > 0 suc h that det e T > c 1 for e i θ ∈ Z ( ε ) and r > r 1 . When e i θ ∈ Ω ( ε ) δ \ Z ( ε = ± 1) with 0 < θ − θ 0 < δ ( θ 0 ∈ Z ( ε ) ), ε (2) j ( θ ) = εε (1) j ( θ ). Supp ose α (1) j ( θ ) > α (2) j ( θ ) for j = 1 , · · · , m and α (1) j ( θ ) < α (2) j ( θ ) for j = m + 1 , · · · , n . By (62) and (63), det e T ( θ , r ) =             A ( θ ) B ( θ , r ) C ( θ , r ) D ( θ ) − ¯ B ( θ , r ) ¯ A ( θ ) − ¯ D ( θ ) ¯ C ( θ , r )             + o (1) =             A ( θ ) B ( θ , r ) ¯ D ( θ ) − ¯ C ( θ , r ) − ¯ B ( θ , r ) ¯ A ( θ ) C ( θ , r ) D ( θ )             + o (1) (70) where A ( θ ) and B ( θ , r ) are m × n matrices, C ( θ , r ) and D ( θ ) are ( n − m ) × n matrices, whose e ntries are giv en by A j k ( θ ) = ( ε (1) j ( θ ) λ j ) k , B j k ( θ , r ) = ( i εε (1) j ( θ ) λ j ) k Y j ( θ , r ) ( j = 1 , · · · , m ) , ¯ D j k ( θ ) = ( − i ε (2) j ( θ ) ¯ λ j ) k , − ¯ C j k ( θ , r ) = ( εε (2) j ( θ ) ¯ λ j ) k ( − ¯ Y j ( θ , r ) − 1 ) ( j = m + 1 , · · · , n ) (71) and o (1) refers to the terms whic h ten d to zero uniformly as r → + ∞ . Let Λ = ( ε (1) 1 ( θ ) λ 1 , · · · , ε (1) m ( θ ) λ m , − i ε (2) m +1 ( θ ) ¯ λ m +1 , · · · , − i ε (2) n ( θ ) ¯ λ n ) , Γ = diag( Y 1 ( θ , r ) , · · · , Y m ( θ , r ) , − ¯ Y m +1 ( θ , r ) − 1 , · · · , − ¯ Y n ( θ , r ) − 1 ) . (72) By Lemma 3 of A, there exist r 2 > 0 and c 2 > 0 suc h that det e T > c 2 for e i θ ∈ Ω (+1) δ ∪ Ω ( − 1) δ \ Z with 0 < θ − θ 0 < δ and r > r 2 . Similarly , when e i θ ∈ Ω (+1) δ ∪ Ω ( − 1) δ \ Z with − δ < θ − θ 0 < 0 ( θ 0 ∈ Z ), there exist r 3 > 0 and c 3 > 0 suc h that det e T > c 3 for r > r 3 . When e i θ ∈ Ω (0) δ , supp ose α (1) j ( θ ) > α (2) j ( θ ) for j = 1 , · · · , m and α (1) j ( θ ) < α (2) j ( θ ) for j = m + 1 , · · · , n , then, b y (64 ) and (6 5), lim r → + ∞ det e T ( θ , r ) =             A ( θ ) 0 0 D ( θ ) 0 ¯ A ( θ ) − ¯ D ( θ ) 0             =             A ( θ ) 0 ¯ D ( θ ) 0 0 ¯ A ( θ ) 0 D ( θ )             = |       A ( θ ) ¯ D ( θ )       | 2 (73) holds unifo r mly , where A ( θ ) is an m × n matrix, D ( θ ) is an ( n − m ) × n matrix, whose en tries are giv en b y A j k ( θ ) = ( ε (1) j ( θ ) λ j ) k , ( j = 1 , · · · , m ) , ¯ D j k ( θ ) = ( − i ε (2) j ( θ ) ¯ λ j ) k ( j = m + 1 , · · · , n ) . (74) 14 Using the condition ¯ λ j 6 = ± i λ l and the prop ert y of V andermonde determinan t, w e know that there exist r 4 > 0 and c 4 > 0 suc h that det e T > c 4 for a ll e i θ ∈ Ω (0) δ and r > r 4 . Let r 0 = max( r 1 , r 2 , r 3 , r 4 ), c 0 = min( c 1 , c 2 , c 3 , c 4 ), then for an y e i θ ∈ S 1 , det e T > c 0 when r > r 0 . Step 3: Denote e Π = ˚ T ˚ T ! − 1 Π, then lim r → + ∞ Re det e Π = 0 for any fixed e i θ ∈ S 1 . When e i θ ∈ Z ( ε ) , c onsidering (62), (63 ) and ε (2) j ( θ ) = εε (1) j ( θ ), let Λ = diag(Λ 1 , Λ 2 ) with Λ 1 = diag( ε (1) 1 ( θ ) λ 1 , · · · , ε (1) n ( θ ) λ n ) , Λ 2 = diag( − i ε ε (1) 1 ( θ ) ¯ λ 1 , · · · , − i ε ε (1) n ( θ ) ¯ λ n ) , (75) ζ = (1 , · · · , 1 | {z } n , − ¯ ξ − 1 1 ¯ η 1 , · · · , − ¯ ξ − 1 n ¯ η n ) T , (76) then ¯ Λ 2 = i ε Λ 1 , and e Π −  Y | γ j ( θ ) | > 1 | γ j ( θ ) |  − 4 Π Λ → 0 as r → + ∞ where γ j ( θ )’s are defined b y (66) and Π Λ is defined by (122). According to Lemma 4 of A, Re det Π Λ ≡ 0, whic h leads to lim r → + ∞ Re det e Π = 0. When e i θ ∈ S 1 \ Z , supp ose α (1) j ( θ ) > α (2) j ( θ ) fo r j = 1 , · · · , m a nd α (1) j ( θ ) < α (2) j ( θ ) fo r j = m + 1 , · · · , n . By (64) and (65), ξ − 1 j  ∂ k ξ j − ( ε (1) j ( θ ) λ j ) k ξ j  → 0 , ξ − 1 j  ∂ k η j − ( i ε (1) j ( θ ) λ j ) k η j  → 0 ( j = 1 , · · · , m ) , η − 1 j  ∂ k ξ j − ( ε (2) j ( θ ) λ j ) k ξ j  → 0 , η − 1 j  ∂ k η j − ( i ε (2) j ( θ ) λ j ) k η j  → 0 ( j = m + 1 , · · · , n ) (77) as r → + ∞ s ince ξ − 1 j η j → 0 fo r j = 1 , · · · , m and η − 1 j ξ j → 0 for j = m + 1 , · · · , n . Let Λ = diag(Λ 1 , Λ 2 ) w ith Λ 1 = diag( ε (1) 1 ( θ ) λ 1 , · · · , ε (1) m ( θ ) λ m , ε (2) m +1 ( θ ) λ m +1 , · · · , ε (2) n ( θ ) λ n ) , Λ 2 = diag( − i ε (1) 1 ( θ ) ¯ λ 1 , · · · , − i ε (1) m ( θ ) ¯ λ m , − i ε (2) m +1 ( θ ) ¯ λ m +1 , · · · , − i ε (2) n ( θ ) ¯ λ n ) , (78) ζ = (1 , · · · , 1 | {z } m , 0 , · · · , 0 | {z } n − m , 0 , · · · , 0 | {z } m , − 1 , · · · , − 1 | {z } n − m ) T , (79) then ¯ Λ 2 = iΛ 1 , and e Π → Π Λ as r → + ∞ where Π Λ is defined by (1 22 ). According to Lemma 4 in A, w e ha v e lim r → + ∞ Re det e Π = Re det Π Λ = 0. Till now , w e hav e prov ed that det e T has a uniform p ositiv e lo we r b ound for all θ , and lim r → + ∞ Re det Π (det T ) 2 = 0 for an y fixed θ . Step 4: lim r → + ∞ Re det Π (det T ) 2 = 0 uniformly for all e i θ ∈ S 1 as r → + ∞ . Note that Re det Π (det T ) 2 is of form f ( θ , r ) g ( θ , r ) where f ( θ , r ) = m 1 X j =1 e e α j ( θ ) r + e γ j ( θ ) , g ( θ , r ) = m 2 X j =1 e e β j ( θ ) r + e δ j ( θ ) (80) 15 are real-v alued functions of ( θ , r ), e α j ( θ ), e β j ( θ ), e γ j ( θ ), e δ j ( θ ) are (complex v alued) con tinuous functions of θ . Let e a ( θ ) = max 1 ≤ j ≤ m 1 Re e α j ( θ ), e b ( θ ) = max 1 ≤ j ≤ m 2 Re e β j ( θ ). Since lim r → + ∞ f ( θ , r ) g ( θ , r ) = 0 for an y fixed θ , the real con tin uous function e a ( θ ) − e b ( θ ) < 0 ac hiev es its maxim um − χ < 0 on the compact set S 1 . W e ha v e kno wn that det e T has a uniform p ositiv e low er b ound as r ≥ r 0 ( r 0 is indep enden t of θ ), s o has m 2 X j =1 e ( e β j ( θ ) − e b ( θ )) r + e δ j ( θ ) . (81) Hence      f ( θ , r ) g ( θ , r )      ≤ e − χr       m 1 X j =1 e ( e α j ( θ ) − e α ( θ ) ) r + e γ j ( θ )       m 2 X j =1 e ( e β j ( θ ) − e b ( θ )) r + e δ j ( θ ) ≤ C e − χr (82) as r ≥ r 0 where C is a constan t independen t of θ . The theorem is pro v ed. 7 Asymptotic b eha vior of t he sol utions as t → ∞ In this section, the asymptotic b eha vior of the n -solito n solutions as t → ∞ will b e discussed . In order to do so, w e consider the problem in a mo ving fra me . Let x = x 0 + θ 1 t , y = y 0 + θ 2 t where ( θ 1 , θ 2 ) is the v elo cit y of the mov ing f rame , and ( x 0 , y 0 ) is the co ordinate in the mov ing frame w ith this ve lo cit y . Then ρ (1) j ( x 0 + θ 1 t, y 0 + θ 2 t, t ) = ε (1) j α (1) j t + β (1) j , ρ (2) j ( x 0 + θ 1 t, y 0 + θ 2 t, t ) = ε (2) j α (2) j t + β (2) j (83) where ε ( k ) j = ± 1 ( j = 1 , · · · , n ; k = 1 , 2) so that α ( k ) j ≥ 0. W rite λ j = µ j + i ν j ( j = 1 , · · · , n ) where µ j ’s and ν j ’s ar e real, then according to (33), ε (1) j α (1) j = µ j θ 1 − ν j θ 2 + 16( ν 3 j − 3 µ 2 j ν j ) , ε (2) j α (2) j = − ν j θ 1 + µ j θ 2 − 16( µ 3 j − 3 µ j ν 2 j ) , (84) and ε (1) j α (1) j − ε (2) j α (2) j = ( µ j + ν j )( θ 1 − θ 2 + 16( µ 2 j − 4 µ j ν j + ν 2 j )) , ε (1) j α (1) j + ε (2) j α (2) j = ( µ j − ν j )( θ 1 + θ 2 − 16( µ 2 j + 4 µ j ν j + ν 2 j )) . (85) Theorem 3 Supp ose λ j = µ j + i ν j 6 = 0 ( j = 1 , · · · , n ) ar e d ist in c t c omplex numb ers wher e µ j ’s and ν j ’s ar e r e al numb ers , such that µ j 6 = ± ν j for al l j , µ 2 j + 4 µ j ν j + ν 2 j 6 = µ 2 l + 4 µ l ν l + ν 2 l for al l j 6 = l , µ 2 j − 4 µ j ν j + ν 2 j 6 = µ 2 l − 4 µ l ν l + ν 2 l for al l j 6 = l . (86) 16 u is the n -soliton solution giv en by (3 8 ). Then for b ounde d ( x 0 , y 0 ) , lim t →∞ u ( x 0 + θ 1 t, y 0 + θ 2 t, t ) = 0 exc ept when θ 1 = 8( µ 2 l + 4 µ l ν l + ν 2 l − µ 2 j + 4 µ j ν j − ν 2 j ) , θ 2 = 8( µ 2 l + 4 µ l ν l + ν 2 l + µ 2 j − 4 µ j ν j + ν 2 j ) , ( j, l = 1 , 2 , · · · , n ) . (87) Ther efor e, as t → ∞ , u ha s at most n × n lu mp s of p e aks which move in the ab ove velo cities ( θ 1 , θ 2 ) r e s p e ctively. Pro of. W e will alw a ys suppo se that ( θ 1 , θ 2 ) does not satisfy (87). Then, b y (85), α (1) j 6 = 0 whenev er α (1) j = α (2) j . M oreov er, w e only consider the limit t → + ∞ . T he conclusion is t he same for t → −∞ . The proof is divided in to thre e steps. Step 1: Obtain the asymptotic beha vior of ξ j ’s and η j ’s. Supp ose α (1) j > α (2) j for j = 1 , · · · , m ; α (1) j < α (2) j for j = m + 1 , · · · , p ; α (1) j = α (2) j 6 = 0 for j = p + 1 , · · · , n . Then ξ − 1 j  ∂ k ξ j − ( ε (1) j λ j ) k ξ j  → 0 , ξ − 1 j  ∂ k η j − ( i ε (2) j s j λ j ) k η j  → 0 ( j = 1 , · · · , m ) , η − 1 j  ∂ k ξ j − ( ε (1) j s j λ j ) k ξ j  → 0 , η − 1 j  ∂ k η j − ( i ε (2) j λ j ) k η j  → 0 ( j = m + 1 , · · · , p ) , ξ − 1 j  ∂ k ξ j − ( ε (1) j λ j ) k ξ j  → 0 , ξ − 1 j  ∂ k η j − ( i ε (2) j λ j ) k η j  → 0 ( j = p + 1 , · · · , n ) (88) as r → + ∞ where s 1 , · · · , s p are an y constants, since ξ − 1 j η j → 0 for j = 1 , · · · , m and η − 1 j ξ j → 0 for j = m + 1 , · · · , p . No w w e prov e that p can only ta ke n − 1 or n . If p ≤ n − 2, then ε (1) j ε (2) j = ± ε (1) l ε (2) l m ust hold for a n y j 6 = l with p + 1 ≤ j, l ≤ n since b oth sides equal ± 1. If ε (1) j ε (2) j = ε (1) l ε (2) l , then ε (2) j α (2) j = εε (1) j α (1) j , ε (2) l α (2) l = εε (1) l α (1) l hold sim ultaneously where ε = ε (1) j ε (2) j . This con tradicts condition (86). If ε (1) j ε (2) j = − ε (1) l ε (2) l , then ε (2) j α (2) j − εε (1) j α (1) j = 0, ε (2) l α (2) l + εε (1) l α (1) l = 0 hold simu lta ne ously where ε = ε (1) j ε (2) j . This con tradicts the assumption that ( θ 1 , θ 2 ) does not satisfy (87). Hence only p = n or p = n − 1 is p ossible. Step 2: There ex ists t 0 > 0 and c > 0 s uch that det e T > c for t ≥ t 0 . When p = n − 1, | ξ n ( θ , r ) | max( | ξ n ( θ , r ) | , | η n ( θ , r ) | ) → γ 0 ≡ | κ (1) n | e ε (1) n β (1) n max( | κ (1) n | e ε (1) n β (1) n , | κ (2) n | e ε (2) n β (2) n ) . (89) Denote V ( λ j , · · · , λ l ) =      λ n − 1 j · · · 1 . . . . . . λ n − 1 l · · · 1      ( l − j +1) × n (90) 17 for j ≤ l , then, fo r p = n − 1, (88 ) leads to det e T = γ 2 0                   V ( ε (1) 1 λ 1 , · · · , ε (1) m λ m ) 0 0 V ( i ε (2) m +1 λ m +1 , · · · , i ε (2) n − 1 λ n − 1 ) V ( ε (1) n λ n ) ξ − 1 n η n V ( i ε (2) n λ n ) 0 V ( ε (1) 1 ¯ λ 1 , · · · , ε (1) m ¯ λ m ) − V ( − i ε (2) m +1 ¯ λ m +1 , · · · , − i ε (2) n − 1 ¯ λ n − 1 ) 0 − ¯ ξ − 1 n ¯ η n V ( − i ε (2) n ¯ λ n ) V ( ε (1) n ¯ λ n )                   + o (1) = γ 2 0                   V ( ε (1) 1 λ 1 , · · · , ε (1) m λ m ) 0 V ( − i ε (2) m +1 ¯ λ m +1 , · · · , − i ε (2) n − 1 ¯ λ n − 1 ) 0 V ( ε (1) n λ n ) ξ − 1 n η n V ( i ε (2) n λ n ) 0 V ( ε (1) 1 ¯ λ 1 , · · · , ε (1) m ¯ λ m ) 0 V ( i ε (2) m +1 λ m +1 , · · · , i ε (2) n − 1 λ n − 1 ) − ¯ ξ − 1 n ¯ η n V ( − i ε (2) n ¯ λ n ) V ( ε (1) n ¯ λ n )                   + o (1) (91) as t → + ∞ . Let Λ = diag( ε (1) 1 λ 1 , · · · , ε (1) m λ m , − i ε (2) m +1 ¯ λ m +1 , · · · , − i ε (2) n − 1 ¯ λ n − 1 , ε (1) n λ n ) , Γ = diag(0 , · · · , 0 , ξ − 1 n η n ) , ε = ε (1) n ε (2) n , (92) then we get lim inf t → + ∞ det e T > 0 b y Lemm a 3 of A. Similarly , when p = n , lim t → + ∞ det e T =             V ( ε (1) 1 λ 1 , · · · , ε (1) m λ m ) 0 V ( − i ε (2) m +1 ¯ λ m +1 , · · · , − i ε (2) n ¯ λ n ) 0 0 V ( ε (1) 1 ¯ λ 1 , · · · , ε (1) m ¯ λ m ) 0 V ( i ε (2) m +1 λ m +1 , · · · , i ε (2) n λ n )             . =    det V ( ε (1) 1 λ 1 , · · · , ε (1) m λ m , − i ε (2) m +1 ¯ λ m +1 , · · · , − i ε (2) n ¯ λ n )    2 . (93) Using the condition ¯ λ j 6 = ± i λ l , we g e t lim inf t → + ∞ det e T > 0 since the V andermonde determinan t is non- z ero. Step 3: D enote e Π = ˚ T ˚ T ! − 1 Π. If the velocity ( θ 1 , θ 2 ) do es not satisfy (87), then lim t → + ∞ Re det e Π = 0. F ro m (8 8), let Λ = diag(Λ 1 , Λ 2 ) with Λ 1 = diag( ε (1) 1 λ 1 , · · · , ε (1) m λ m , ε (1) m +1 s m +1 λ m +1 , · · · , ε (1) p s p λ p , ε (1) p +1 λ p +1 , · · · , ε (1) n λ n ) , Λ 2 = diag( − i ε (2) 1 ¯ s 1 ¯ λ 1 , · · · , − i ε (2) m ¯ s m ¯ λ m , − i ε (2) m +1 ¯ λ m +1 , · · · , − i ε (2) p ¯ λ p , − i ε (2) p +1 ¯ λ p +1 , · · · , − i ε (2) n ¯ λ n ) , (94) 18 Figure 4: 3 × 3 soliton solution u : λ 1 = 2 + 0 . 5 i, λ 2 = 2 . 5 + 0 . 4 i, λ 3 = 3 + 0 . 3 i, κ (1) j = 1, κ (2) j = 1 . 2, ρ (1) j 0 = 0, ρ (2) j 0 = 0, σ (1) j 0 = 0, σ (2) j 0 = 0 ( j = 1 , 2 , 3), t = 1. ζ = (1 , · · · , 1 | {z } m , 0 , · · · , 0 | {z } p − m , 1 , · · · , 1 | {z } n − p , 0 , · · · , 0 | {z } m , − 1 , · · · , − 1 | {z } p − m , − ¯ ξ − 1 p +1 ¯ η p +1 , · · · , − ¯ ξ − 1 n ¯ η n ) T , (95) then e Π − γ 4 0 Π Λ → 0 for p = n − 1 and e Π − Π Λ → 0 for p = n as t → + ∞ where γ 0 is defined b y (89) and Π Λ is define d by (122). ¯ Λ 2 = i ε Λ 1 holds for ε = ± 1 if and only if s j = εε (1) j ε (2) j for j = 1 , · · · , p , and ε (2) j = εε (1) j for j = p + 1 , · · · , n . Since s j ( j = 1 , · · · , p ) can b e arbitra ry , ε can b e ta k en as ± 1 a rbitrarily , and p = n or n − 1, w e hav e ¯ Λ 2 = i ε Λ 1 b y taking ε = ε (1) n ε (2) n for p = n − 1, and either ε = 1 or ε = − 1 f or p = n . According to Lemma 4 of A, Re det Π Λ ≡ 0. Hence lim r → + ∞ Re det e Π = 0 if ( θ 1 , θ 2 ) do es not satisfy (87). The theorem is pro v ed. Remark 2 If ( θ 1 , θ 2 ) satisfies (87 ), then α (1) j = α (2) j , α (1) l = α (2) l , ε (2) j = ε (1) j , ε (2) l = − ε (1) l for j 6 = l with p + 1 ≤ j, l ≤ n . Henc e ther e is no c om m on ε = ± 1 such that ε (2) i = εε (1) i holds for al l i = p + 1 , · · · , n , which c ontr adicts the c ondition ¯ Λ 2 = i ε Λ 1 in L emma 4 of A. In fac t, the solution do es not tend t o ze r o in this c ase, which c an b e se e n in the fo l lowing example. As an example, a 3 × 3 soliton is sho wn in F ig ure 4 , in whic h there are 9 lumps of p eaks. The lo cal b ehavior of eac h lump of p eaks is still complicated and one of whic h is sho wn in Figure 5 . A Some li near algebr a ic lemmas Lemma 1 Supp ose X and Y ar e 2 n × r and 2 n × (2 n − r ) matric es r esp e ctively, then    X K n ¯ Y    =    Y K n ¯ X    (96) 19 Figure 5: Lo cal b eha vior o f one lump of p eaks in the 3 × 3 soliton solution u . wher e K n = 0 − I n I n 0 ! . Pro of. K n  X K n ¯ Y  K − 1 n =  K n ¯ X − Y    0 I n − I n 0   =  Y K n ¯ X  . (97) The le mma is obtained b y t a k ing the determinants o n b oth sides . Lemma 2 Supp ose T =   A B − ¯ B ¯ A   wher e A and B ar e n × n matric es, then det T ≥ 0 . Pro of. By Lemma 1, det T is real. First suppose b oth A and B are in vertible, then det T = det A det( ¯ A + ¯ B A − 1 B ) = | det A | 2 det( I + A − 1 B A − 1 B ) . (98) Let N = A − 1 B . Supp ose λ is an eigenv alue of ¯ N N , v ∈ R 2 n is a ve ctor in the corresp onding ro ot space R λ , i.e . ( ¯ N N − λI ) m v = 0 for certain p ositiv e in teger m . Then ( ¯ N N − ¯ λI ) m ( ¯ N ¯ v ) = ¯ N ( ¯ N N − λI ) m v = 0 . (99) Hence ¯ N ¯ v ∈ R ¯ λ . If λ is a non-real eigen v alue of ¯ N N of m ultiplicit y k , the m ultiplicit y of ¯ λ is also k . No w supp ose λ < 0 is an eigen v alue of ¯ N N , R λ = V 1 ⊕ · · · ⊕ V m where V 1 , · · · , V m are irre- ducible inv arian t s ubspaces. Su pp ose V 1 = span { ζ , ( ¯ N N − λI ) ζ , · · · , ( ¯ N N − λI ) m − 1 ζ } with ( ¯ N N − λI ) m ζ = 0 . Then ζ 6∈ Image( ¯ N N − λI ), and ( ¯ N N − λI ) m ¯ N ¯ ζ = ¯ N ( ¯ N N − λI ) m ζ = 0, ( ¯ N N − λI ) m − 1 ¯ N ¯ ζ = ¯ N ( ¯ N N − λI ) m − 1 ζ 6 = 0 since det N 6 = 0 . W e will prov e that ζ , ( ¯ N N − λI ) ζ , · · · , ( ¯ N N − λI ) m − 1 ζ , ¯ N ¯ ζ , ( ¯ N N − λI ) ¯ N ¯ ζ , · · · , ( ¯ N N − λI ) m − 1 ¯ N ¯ ζ (100 ) 20 are line arly independent. Supp ose m X j =1 α j ( ¯ N N − λI ) j − 1 ζ + m X j =1 β j ( ¯ N N − λI ) j − 1 ¯ N ¯ ζ = 0 (101) where α 1 , · · · , α m , β 1 , · · · , β m are c omplex n umbers. Acting ( ¯ N N − λI ) m − 1 on both sides of (101), w e get ( ¯ N N − λI ) m − 1 ( α 1 ζ + β 1 ¯ N ¯ ζ ) = 0 . (102) Then ( | α 1 | 2 − λ | β 1 | 2 )( ¯ N N − λI ) m − 1 ζ = − ¯ α 1 β 1 ( ¯ N N − λI ) m − 1 ¯ N ¯ ζ − λ | β 1 | 2 ( ¯ N N − λI ) m − 1 ζ = − ¯ α 1 β 1 ¯ N ( ¯ N N − λI ) m − 1 ζ − λ | β 1 | 2 ( ¯ N N − λI ) m − 1 ζ = β 1 ¯ N β 1 ( ¯ N N − λI ) m − 1 ¯ N ¯ ζ − λ | β 1 | 2 ( ¯ N N − λI ) m − 1 ζ = | β 1 | 2 ¯ N N ( ¯ N N − λI ) m − 1 ζ − λ | β 1 | 2 ( ¯ N N − λI ) m − 1 ζ = | β 1 | 2 ( ¯ N N − λI ) m ζ = 0 . (103) Since λ < 0 and ( ¯ N N − λI ) m − 1 ζ 6 = 0, w e hav e α 1 = β 1 = 0. Con tin uing this process b y acting ( ¯ N N − λI ) m − 2 , · · · , ( ¯ N N − λI ) 0 on b oth sides o f (101) respectiv ely , w e get α 1 = · · · = α n = β 1 = · · · = β n = 0. This prov es the linear indep ende nce of the v ectors in (100). Let e V 1 = span { ¯ N ¯ ζ , ( ¯ N N − λI ) ¯ N ¯ ζ , · · · , ( ¯ N N − λI ) m − 1 ¯ N ¯ ζ } . If ¯ N ¯ ζ = ( ¯ N N − λI ) ζ ′ ∈ Image( ¯ N N − λI ), then ζ = ( ¯ N N − λ I ) N − 1 ¯ ζ ′ ∈ Imag e ( ¯ N N − λ I ), whic h con tradicts the c hoice of ζ . Hence ¯ N ¯ ζ 6∈ Image( ¯ N N − λI ). Moreov er, e V 1 is in v arian t and irreducible under the action of ¯ N N − λI . Hence it mu st b e one of V j with 2 ≤ j ≤ m , which means that m is eve n ( m = 2 k ) and R λ = ( W 1 ⊕ f W 1 ) ⊕ · · · ⊕ ( W k ⊕ f W k ) w here ( W 1 , · · · , W k , f W 1 , · · · , f W k ) is a p erm utation of ( V 1 , · · · , V 2 k ). Therefore, the m ultiplicity of eac h negativ e eigenv alue of ¯ N N mus t b e ev en. Th us, if the eigenv alues of ¯ N N are λ 1 , · · · , λ 2 n (m ultiple eigen v alues are listed rep eatedly), then det T = | det A | 2 det( I + ¯ N N ) = | det A | 2  Y Re λ j 6 =0 | 1 + λ j | 2 Y λ j < 0 (1 + λ j ) 2  1 / 2 Y λ j ≥ 0 (1 + λ j ) ≥ 0 . (104) If A or B is not in v ertible, it is a limit of in v ertible matrices, and the conclusion is also true. The le mma is prov ed. Lemma 3 Supp ose Λ = diag( λ 1 , · · · , λ n ) wher e λ 1 , · · · , λ n ar e distinct c omplex numb ers such that ¯ λ j 6 = ± i λ l for al l j, l = 1 , · · · , n , Γ = diag( γ 1 , γ 2 , · · · , γ n ) . Den ot e V =      λ n − 1 1 λ n − 2 1 · · · 1 . . . . . . . . . λ n − 1 n λ n − 2 n · · · 1      , E = diag  ( i ε ) n − 1 , ( i ε ) n − 2 , · · · , 1  (105) 21 wher e ε = ± 1 . L et T =   V Γ V E − ¯ Γ ¯ V ¯ E ¯ V   , (106) then ther e is a p ositive numb er C d e p ending on Λ only, such that det T > C . Pro of. Denote S ( ˆ m ) k = X j 1 < ··· 0 since ¯ r j + r j 6 = 0 for all j . W rite M = 1 ¯ r j + r k ! n × n =   M 11 M 12 M ∗ 12 M 22   , M + ¯ Γ M Γ = 1 + ¯ γ j γ k ¯ r j + r k ! n × n =   N 11 N 12 N ∗ 12 N 22   (115) where M 11 and N 11 , M 12 and N 12 , M 22 and N 22 are m × m , m × ( n − m ), ( n − m ) × ( n − m ) matrices resp ectiv ely . Then M 11 and N 11 are negative definite Hermitian matrices, so ar e M − 1 11 and N − 1 11 ; M 22 and N 22 are positiv e definite He rmitia n matrices. Let Γ 1 = diag( γ 1 , · · · , γ m ) , Γ 2 = diag( γ m +1 , · · · , γ n ) , (116) then det( − N 11 ) = det( − M 11 + Γ ∗ 1 ( − M 11 )Γ 1 ) ≥ det( − M 11 ) , det( N 22 ) = det( M 22 + Γ ∗ 2 M 22 Γ 2 ) ≥ det( M 22 ) . (117) Hence ( − 1) m det( M + ¯ Γ M Γ) = det ( − N 11 ) det( N 22 + N ∗ 12 ( − N 11 ) − 1 N 12 ) ≥ | det M 11 | · | det M 22 | . (118) On the other hand, ( − 1) m det M = det( − M 11 ) det( M 22 + M ∗ 12 ( − M 11 ) − 1 M 12 ) ≥ | det M 11 | · | det M 22 | > 0 , (119) hence det T ≥ | det V | 2 | det M 11 | · | det M 22 | | det M | . (120) The le mma is prov ed. Lemma 4 L et Λ = diag(Λ 1 , Λ 2 ) whe r e Λ 1 and Λ 2 ar e n × n diagon a l matric es satisfying ¯ Λ 2 = i ε Λ 1 wher e ε = ± 1 . L et ζ b e a 2 n dimensiona l c olumn ve ctor. F or j ≥ k , denote R Λ j ··· k =  Λ j ζ Λ j − 1 ζ , · · · , Λ k ζ  . (121) L et Π Λ =   Λ n ζ Λ n − 1 ζ Λ n − 2 ζ R Λ n − 3 ··· 0 K n ¯ R Λ n − 1 ··· 0 0 0 Λ n +1 ζ Λ n ζ Λ n − 1 ζ 0 0 R Λ n − 2 ··· 0 K n ¯ R Λ n − 1 ··· 0   , (122) wher e K n =   − I n I n   , then det Π Λ is pur ely im aginary. 23 Pro of. ¯ Λ 2 = i ε Λ 1 is eq uiv alen t t o Λ K n = − i εK n ¯ Λ. Denote d Λ = det ∆ Λ . Multiplying ro w j of d Λ b y − λ j and adding it to ro w 2 n + j ( j = 1 , · · · , 2 n ), w e get d Λ =       Λ n ζ Λ n − 1 ζ Λ n − 2 ζ R Λ n − 3 ··· 0 K n ¯ R Λ n − 1 ··· 0 0 0 0 0 0 − R Λ n − 2 ··· 1 i εK n ¯ R Λ n ··· 0 R Λ n − 2 ··· 0 K n ¯ R Λ n − 1 ··· 0       (123) b y using Λ K n = − i εK n ¯ Λ. Adding columns 2 n + 2 , · · · , 3 n − 1 to columns 4 , · · · , n + 1, m ultiplying columns 3 n + 1 , · · · , 4 n − 1 b y − i ε and adding them to columns n + 3 , · · · , 2 n + 1, w e get d Λ =       R Λ n ··· 0 K n ¯ Λ n − 1 ¯ ζ K n ¯ R Λ n − 2 ··· 0 0 0 0 i εK n ¯ Λ n ¯ ζ 0 R Λ n − 2 ··· 0 K n ¯ R Λ n − 1 ··· 0       , (124) By moving the columns, d Λ =       R Λ n ··· 0 K n ¯ R Λ n − 2 ··· 0 0 K n ¯ Λ n − 1 ¯ ζ 0 0 0 R Λ n − 2 ··· 0 i εK n ¯ Λ n ¯ ζ K n ¯ R Λ n − 1 ··· 0       = i ε    R Λ n ··· 0 K n ¯ R Λ n − 2 ··· 0    ·    R Λ n − 2 ··· 0 K n ¯ R Λ n ··· 0    , (125) whic h is purely imaginary according to Lemma 1. The lemma is p rov ed. Ac kno wled gemen ts This w ork w as supp orted by Natio nal Basic R esearch Progra m of China (20 07CB814800) and ST CSM (06JC14005). References References [1] Athorne C and F ordy A P 1987 In tegrable equations in (2+1) dimensionals asso ciated with symmetric and homogeneous spaces J. Math. 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