A New Expression of Soliton Solution to the Ultradiscrete Toda Equation

A New Expression of Soliton Solution to the Ultradiscre te T o da Equation Hidetomo Nagai Ma jor in Pure and Applied Mathematics, Gradua te Sc ho ol of F undamental Science and Engineering, W aseda Universit y , 3-4 -1, Okubo, Shinjuku-ku, T okyo 169-8 55, Japan E-mail: n1a9g 8a1i@t oki.waseda.jp Abstract. A new type of mult i-soliton solution to the ultradiscr ete T oda equation is prop osed. The solutio n can be transformed into a nother expressio n of solution in a per turbation form. A direct pro of of the solution is also given. 1. In tro duction Ultradiscretization is a limiting pr o cedure to transfor m a discrete equation with con tin uo us dependen t v aria bles into an ultradiscrete equation with discrete dep enden t v ariables[1]. Ultradiscrete system is applied to the traffic flo w[2 , 3] and the sorting algorithm[4] for example. An impor tan t formula o f the pro cedure is a simple limiting form ula, lim ǫ → +0 ǫ log( e a/ǫ + e b/ǫ ) = ma x ( a, b ) . (1) F or example, a ssume a discrete equation, X n +1 = 1 + X n X n − 1 . (2) If w e use a transformation of v aria ble X n = e x n /ǫ and tak e a limit ǫ → +0, w e obtain an ultradiscrete equation, x n +1 = max(0 , x n ) − x n − 1 . (3) If initial v alues, for example, x 0 and x 1 are all integer, x n for a n y n is also. The dep enden t v a r ia ble X is considered to b e discretized in this meaning. If w e apply this procedure to a multi-soliton solution to a discre te soliton equation, w e obta in an ultradiscrete solution to an ultradiscrete soliton equation consisten tly . F or example, a set of equation a nd solution in an ultr a discrete v ersion is automatically deriv ed b y ultradiscretiz ing the disc rete Kortew eg–de V ries equation and its solutio n where the e quation is transformed in to a soliton ce llular automaton called ‘b ox and ball system’ with a binary state v alue[1, 5]. A New Expr e s s ion of Soliton Solution to the Ultr adiscr ete T o da Equation 2 Ho w eve r, there is a difficulty called ‘negativ e problem’ in the ultradiscretizing pro cedure. If ‘+’ in the left side o f (1) is replaced b y ‘ − ’, the limit b y ǫ → +0 is not well-define d. The refore, a multi-soliton solution expressed by a determinan t can not b e ultradiscretized though that in a p erturbation form ‡ can. A pro of of discrete solution to discre te solito n equation is o ften realized b y using an iden tit y of determinan ts. Henc e, it is impor t an t to find a ‘determinan t‘ form of ultradiscrete solution corresp onding to determinan t solutions. Ab out this problem, T ak ahashi and Hirota sho w ed a new expression of solution to the ultra discrete KdV equation recen tly[6 ]. The solution is expresse d by a form o f ultradiscretized p ermanen t whic h is defined by a signature-free determinan t. It su ggests that this t yp e of solution is an ultradiscrete corres p ondence to a determinan t one to a discrete soliton equation. In this pap er, the author prop o ses a new ty p e of m ulti-soliton solution to the ultradiscrete T o da equation with a similar structure of expression. T he discrete T o da equation introduced b y Hirota[7] is log(1 + V m n +1 ) − 2 log(1 + V m n ) + log(1 + V m n − 1 ) = log (1 + δ 2 V m +1 n ) − 2 log(1 + δ 2 V m n ) + log(1 + δ 2 V m − 1 n ) . (5) Using transformations including parameters ǫ and L ( L > 0 ), V m n = e u m n /ǫ , δ = e − L/ 2 ǫ , (6) and taking the limit ǫ → +0, we obtain the ultra discrete T o da equation[8, 9] u m n +1 − 2 u m n + u m n − 1 = max (0 , u m +1 n − L ) − 2 max(0 , u m n − L ) + max(0 , u m − 1 n − L ) . (7) Its bilinear equation can also b e ultradiscretized and giv en b y f m n +1 + f m n − 1 = max(2 f m n , f m +1 n + f m − 1 n − L ) , (8) where u m n is related to f m n as u m n = f m +1 n − 2 f m n + f m − 1 n . (9) A multi-soliton solution in a p erturbation form to (8) is r ep orted in [9] and is deriv ed b y ultradiscretizing that to the discrete T o da equation. It follows a form, f m n = max µ i =0 , 1  N X i =1 µ i s i ( m, n ) + X 1 ≤ i 0 (16) without loss o f generality . Under this assumption, ( 1 1) reduces to f m n = 1 2 max σ i = ± 1  N X i =1 σ i s i + N X i =1  1 + σ i 2  ( N − i ) p i + q i ε i N X j = i +1 ε j  + σ i i − 1 X j =1 1 − σ j 2 ( p i + ε i ε j q i )  , (17) A New Expr e s s ion of Soliton Solution to the Ultr adiscr ete T o da Equation 4 where s i denotes s i ( m, n ) for abbreviation. The pro of of equiv alence b etw een (11) a nd (17) is shown in App endix A. W e obtain fro m (17), f m n ≎ 1 2 max σ i = ± 1  N X i =1 (1 + σ i ) s i + N X i =1  1 + σ i 2  ( N − i ) p i + q i ε i N X j = i +1 ε j  + σ i i − 1 X j =1 1 − σ j 2 ( p i + ε i ε j q i )  , (18) where f m n ≎ g m n means that f m n and g m n giv e the same u m n through the righ t side of transformation (9). Moreo v er, (1 8) reduces to f m n ≎ max µ i =0 , 1  N X i =1 µ i  s i + ( N − i ) p i 2 + q i ε i N X j = i +1 1 2 ε j + i − 1 X j =1 ( p i + ε i ε j q i ) + 1 2 N X j = i +1 ( p j + ε i ε j q j )  − X 1 ≤ i 0 (25) without loss o f generality and then (22 ) is expressed by f m n = max µ i =0 , 1  N X i =1 µ i s i − X 1 ≤ i 0) − 1 ( x < 0) . (27) Substituting ( 2 6) in to the left side of (24), w e obta in f m n +1 + f m n − 1 = max µ i =0 , 1  X 1 ≤ i ≤ N µ i ( s i − ε i q i ) − X 1 ≤ i 0. Under this a ssumption, q i is giv en b y q i = − min (0 , − p i + L ) (35) and 0 ≤ q i ≤ p i . Moreo v er, a ij = p j + ε i ε j q j ≥ 0. Hence, w e give the following prop osition to prov e. Prop osition 2 F unctions g l ( σ i ) , g r 1 ( σ i ) and g r 2 ( σ i ) define d by g l ( σ i ) = − N X i =1 σ i ε i q i − 1 2 X 1 ≤ i 0) . Define maxim um v alues of g l ( σ i ), g r 1 ( σ i ) and g r 2 ( σ i ) b y ¯ g l , ¯ g r 1 and ¯ g r 2 resp ectiv ely . They are giv en if σ i satisfies the fo llo wing conditions, ¯ g l : σ i = ( − 1) i i Y l =1 ε l ( i : odd) , ( − 1) i i − 1 Y l =1 ε l ( i : ev en) , ¯ g r 1 : σ i = i Y l =1 ε l ( i : odd) , − i − 1 Y l =1 ε l ( i : ev en) , ¯ g r 2 : σ i = 1 ( i = 1) , ( − 1) i ε 1 i Y l =1 ε l ( i : odd, i ≥ 3) , ( − 1) i ε 1 i − 1 Y l =1 ε l ( i : ev en) . (38) A New Expr e s s ion of Soliton Solution to the Ultr adiscr ete T o da Equation 7 Substituting t hese conditions, we can deriv e ¯ g l = N X i =1 1 + ( − 1) i 4 p i + q 1 − N X i =2  1 + ( − 1) i 4 i Y l =1 ε l − 1 + ( − 1) i − 1 4  1 + i − 1 Y l =1 ε l  q i , (39) ¯ g r 1 = N X i =1 1 + ( − 1) i 4 p i + N X i =2  1 + ( − 1) i 4 i Y l =1 ε l + 1 + ( − 1) i − 1 4  1 − i − 1 Y l =1 ε l  q i , (40) ¯ g r 2 = p 1 + N X i =1 1 + ( − 1) i 4 p i − N X i =2  1 + ( − 1) i 4 i Y l =1 ε l − 1 + ( − 1) i − 1 4  1 + i − 1 Y l =1 ε l  q i . (41) Though the whole pro of on the ab ov e relations (39) ∼ (41) is long and omitted, an imp ortant form ula for the pro of is sho wn in App endix B. If p 1 ≥ L , then we obtain ¯ g r 1 − ¯ g l = − q 1 + N X i =2  1 + ( − 1) i 2 i Y l =1 ε l − 1 + ( − 1) i − 1 2 i − 1 Y l =1 ε l  q i ≤ − q 1 + q 2 − q 3 + q 4 − . . . + ( − 1) N q N ≤ 0 , (42) and ¯ g r 2 − ¯ g l − L = p 1 − q 1 − L = 0 . (43) If p 1 < L , then q i = 0 for a ny 1 ≤ i ≤ N and ¯ g r 1 − ¯ g l = 0 , ¯ g r 2 − ¯ g l − L = p 1 − L ≤ 0 . (44) Th us a r elat io n max( ¯ g r 1 − ¯ g l , ¯ g r 2 − ¯ g l − L ) = 0 , (45) holds and (37) is satisfied. 4. Conclusion A new express ion of mu lti-soliton solution to the ultradiscrete T o da equation is prop osed. The solution is equiv alen t to a solution in a p erturbation for m. A direct pro of of solution is also giv en. The fo r mer solution expressed b y max in (11) can b e deriv ed by ultradiscretization. F or example, let us consider F = X π i ψ π 1 ( m, n ) ψ π 2 ( m + 1 , n − ε π 1 ) ψ π 3 ( m + 2 , n − ε π 1 − ε π 2 ) · · · ψ π N ( m + N − 1 , n − ε π 1 − ε π 2 − · · · − ε π N − 1 ) . (46) If w e use a transforma t io n ψ i ( m, n ) = cosh( s i ( m, n ) / 2 ε ) , (47) then w e giv e the right side of (11) by lim ǫ → +0 ǫ log F . (48) A New Expr e s s ion of Soliton Solution to the Ultr adiscr ete T o da Equation 8 It suggests that a discrete solution as a corr esp o ndence of ultradiscrete solution can exists. Ho w ev er, suc h a solution has not y et b een found. Moreo v er, pro of o f discrete solution to discrete soliton equation is oft en r ealized by using an iden tity o f determinan ts. T o find the discrete corresp ondence of solution a nd to prov e an ultradiscrete solution b y some ultradiscretized identit y are considered to b e imp or t a n t future problems. Ac kno wledgmen ts The author is grateful to Prof. Daisuk e T ak ahashi and Prof. Ry o go Hirota for man y fruitful discussions and helpful advices. App endix A . Equiv alence b et w een (11) and (17) In this a pp endix, we pro v e the fo llowing prop osition. Prop osition 3 max( | s 1 ( m, n ) | , | s 2 ( m, n ) | , . . . , | s N ( m, n ) | ) = max σ i = ± 1  N X i =1 σ i s i + N X i =1  1 + σ i 2  ( N − i ) p i + q i ε i N X j = i +1 ε j  + σ i i − 1 X j =1 1 − σ j 2 ( p i + ε i ε j q i )  (A.1) wher e s i ( m, n ) = p i m − ε i q i n + c i , − 1 ≤ ε i ≤ 1 , (A.2) p i ≥ 0 , q i ≥ 0 , p 1 − q 1 ≥ p 2 − q 2 ≥ . . . ≥ p N − q N ≥ 0 . (A.3) If p i and q i satisfy (15) a nd (16), conditions of (A.3) is all satisfied. Therefore, if the ab ov e prop osition is pro v ed, equiv alence b etw een (11) and (17) is sho wn. Note that the k ey formula sho wn in [6] is a sp ecial case of (A.1) with ε i = ε j for any 1 ≤ i, j ≤ N . Pro of. The left side of (A.1) is equal to max π i N X i =1    s π i + ( i − 1) p π i + q π i ε π i i − 1 X j =1 ε π j    , (A.4) where ( π 1 , π 2 , . . . , π N ) is a p erm utation o f natural n um b ers 1 ∼ N . Noting | x | = max( x, − x ), the ab ov e expression is rewritten by max π i  max σ π i = ± 1 N X i =1 σ π i  s π i + ( i − 1) p π i + q π i ε π i i − 1 X j =1 ε π j  = max σ i = ± 1  N X i =1 σ i s i + max π i N X i =1 σ π i  ( i − 1 ) p π i + q π i ε π i i − 1 X j =1 ε π j  . (A.5) A New Expr e s s ion of Soliton Solution to the Ultr adiscr ete T o da Equation 9 Th us, what to prov e is the follo wing. max π i N X i =1 σ π i  ( i − 1) p π i + q π i ε π i i − 1 X j =1 ε π j  = N X i =1  1 + σ i 2  ( N − i ) p i + q i ε i N X j = i +1 ε j  + σ i i − 1 X j =1 1 − σ j 2 ( p i + ε i ε j q i )  . (A.6) W e use a mathematical induction on N . The claim is t r ivial fo r N = 1. Then, let us assume (A.6) holds for a certain N and define L ( π 1 , π 2 , . . . , π N +1 ) b y L ( π 1 , π 2 , . . . , π N +1 ) = N +1 X i =1 σ π i  ( i − 1) p π i + q π i ε π i i − 1 X j =1 ε π j  . (A.7) F or N + 1, assum e a certain p erm utation ( π 1 , π 2 , . . . , π N +1 ) and let α b e a num b er satisfying 1 ≤ α ≤ N + 1 and π α = 1. In the case of σ 1 = 1, w e hav e L ( π 1 , . . . , π α − 1 , π α +1 , . . . , π N +1 , π α ) − L ( π 1 , π 2 , . . . , π N +1 ) = ( N − α + 1) p 1 + q 1 ε 1 N +1 X j = α +1 ε π j − N +1 X i = α +1 σ π i ( p π i + ε 1 ε π i q π i ) = N +1 X i = α +1 ( p 1 − σ π i p π i + ε 1 ε π i ( q 1 − σ π i q π i )) ≥ 0 . (A.8) In the case of σ 1 = − 1, w e hav e L ( π α , π 1 , . . . , π α − 1 , π α +1 , . . . , π N +1 ) − L ( π 1 , π 2 , . . . , π N +1 ) = α − 1 X i =1 ( p 1 + ε 1 ε π i q 1 + σ π i ( p π i + ε 1 ε π i q π i )) ≥ 0 . (A.9) A New Expr e s s ion of Soliton Solution to the Ultr adiscr ete T o da Equation 10 Th us, we obtain max π i L ( π 1 , π 2 , . . . , π N +1 ) = 1 + σ 1 2 max π i L ( π 1 + 1 , . . . , π N + 1 , 1) + 1 − σ 1 2 max π i L (1 , π 1 + 1 , . . . , π N + 1) = 1 + σ 1 2  N p 1 + q 1 ε 1 N +1 X j =2 ε j + max π i N X i =1 σ π i +1  ( i − 1) p π i +1 + q π i +1 ε π i +1 i − 1 X j =1 ε π j +1  + 1 − σ 1 2  max π i N X i =1 σ π i +1  ( i − 1) p π i +1 + q π i +1 ε π i +1 i − 1 X j =1 ε π j +1  + N X i =1 σ π i +1 ( p π i +1 + ε 1 ε π i +1 q π i +1 )  = max π i N X i =1 σ π i +1  ( i − 1) p π i +1 + q π i +1 ε π i +1 i − 1 X j =1 ε π j +1  + 1 + σ 1 2  N p 1 + q 1 ε 1 N +1 X j =2 ε j  + 1 − σ 1 2 N X i =1 σ π i +1 ( p π i +1 + ε 1 ε π i +1 q π i +1 ) . (A.10) Using the a ssumption of t he induction, (A.10) reduces to N +1 X i =2  1 + σ i 2  ( N + 1 − i ) p i + q i ε i N X j = i +1 ε j  + σ i i − 1 X j =2 1 − σ j 2  p i + ε i ε j q i  + 1 + σ 1 2  N p 1 + q 1 ε 1 N +1 X j =2 ε j  + 1 − σ 1 2  N +1 X i =2 σ i ( p i + ε 1 ε i q i )  = N +1 X i =1  1 + σ i 2  ( N − i ) p i + q i ε i N X j = i +1 ε j  + σ i i − 1 X j =1 1 − σ j 2  p i + ε i ε j q i  . (A.11) This is equal to the right side of (A.6 ) f o r N + 1 a nd the induction holds. Th us this completes the pro of of Prop.3.  App endix B . Prop osition for pro of on (39) ∼ (41) T o pro v e (39) ∼ (41), the following pro p osition is imp orta n t. The pro of on the prop osition is sho wn in this a pp endix. Prop osition 4 Define g ( σ 1 , σ 2 , . . . , σ N ) by g ( σ 1 , σ 2 , . . . , σ N ) = − X 1 ≤ i 0 . (B.2) Then, g ( σ 1 , σ 2 , . . . , σ N − n − 1 , ¯ σ N − n , ¯ σ N − n +1 , . . . , ¯ σ N ) ≥ g ( σ 1 , σ 2 , . . . , σ N ) (B.3) for any 0 ≤ n ≤ N − 2 , wher e ¯ σ i is given by ¯ σ i = ( 1 ( P N − n − 1 l =1 σ l + P i − 1 l = N − n ¯ σ l ≤ 0) − 1 ( P N − n − 1 l =1 σ l + P i − 1 l = N − n ¯ σ l > 0) (B.4) Pro of. W e use a mathematical induction on n . F or n = 0, we ha v e g ( σ 1 , . . . , σ N − 1 , ¯ σ N ) − g ( σ 1 , . . . , σ N − 1 , − ¯ σ N ) = − 2 ¯ σ N p N N − 1 X i =1 σ i ≥ 0 , (B.5) and (B.3) holds. Assume (B.3) f o r a certain n = k (0 ≤ k ≤ N − 2), that is, g ( σ 1 , . . . , σ N − k − 1 , ¯ σ N − k , ¯ σ N − k +1 , . . . , ¯ σ N ) ≥ g ( σ 1 , . . . , σ N ) . (B.6) Then, what to prov e for induction in the case of n = k + 1 is g ( σ 1 , . . . , σ N − k − 2 , ¯ σ N − k − 1 , ¯ σ N − k , . . . , ¯ σ N ) − g ( σ 1 , . . . , σ N − k − 2 , − ¯ σ N − k − 1 , ˆ σ N − k , . . . , ˆ σ N ) ≥ 0 , (B.7) where ¯ σ N − k − 1 = ( 1 ( P N − k − 2 l =1 σ l ≤ 0) − 1 ( P N − k − 2 l =1 σ l > 0) , ¯ σ i = ( 1 ( P N − k − 2 l =1 σ l + ¯ σ N − k − 1 + P i − 1 l = N − k ¯ σ l ≤ 0) − 1 ( P N − k − 2 l =1 σ l + ¯ σ N − k − 1 + P i − 1 l = N − k ¯ σ l > 0) , ˆ σ i = ( 1 ( P N − k − 2 l =1 σ l − ¯ σ N − k − 1 + P i − 1 l = N − k ˆ σ l ≤ 0) − 1 ( P N − k − 2 l =1 σ l − ¯ σ N − k − 1 + P i − 1 l = N − k ˆ σ l > 0) . (B.8) The left side of (B.7 ) reduces t o −  2 N − k − 2 X i =1 σ i ¯ σ N − k − 1 p N − k − 1 + N − k − 2 X i =1 N X j = N − k σ i ( ¯ σ j − ˆ σ j ) p j + ¯ σ N − k − 1 N X j = N − k ( ¯ σ j + ˆ σ j ) p j + N X i = N − k N X j = i +1 ( ¯ σ i ¯ σ j − ˆ σ i ˆ σ j ) p j  . (B.9) Let us define S by P N − k − 2 i =1 σ i and consider t w o cases of S > 0 or S ≤ 0. In the case of S > 0, ¯ σ i = − 1 for N − k − 1 ≤ i ≤ N − k + S − 1 since S + P i − 1 l = N − k − 1 ¯ σ l > 0, and ˆ σ i = − 1 for N − k ≤ i ≤ N − k + S + 1. Therefore ¯ σ i and ˆ σ i are giv en by ( ¯ σ N − k − 1 , ¯ σ N − k , . . . , ¯ σ N ) = ( − 1 , − 1 , . . . , − 1 | {z } S , 1 , − 1 , 1 , − 1 , 1 , . . . ) , ( − ¯ σ N − k − 1 , ˆ σ N − k , . . . , ˆ σ N ) = (1 , − 1 , − 1 , . . . , − 1 | {z } S , − 1 , 1 , − 1 , 1 , . . . ) . (B.10) A New Expr e s s ion of Soliton Solution to the Ultr adiscr ete T o da Equation 12 Hence, ˆ σ i = ( − ¯ σ i ( i = N − k + S − 1) ¯ σ i (otherwise) , (B.11) and (B.9) reduces to −  − 2 S p N − k − 1 + 2 S p N − k + S − 1 − 2 N − k + S − 2 X j = N − k ¯ σ j p j − 2 N X j = N − k + S ¯ σ j p j + 2 N X j = N − k + S ¯ σ j p j + 2 p N − k + S − 1 N − k + S − 2 X i = N − k ¯ σ i  =2  S p N − k − 1 − S p N − k + S − 1 − N − k + S − 2 X j = N − k p j + ( S − 1) p N − k + S − 1  ≥ 2  S p N − k − 1 − S p N − k + S − 1 − ( S − 1) p N − k + ( S − 1) p N − k + S − 1  =2( p N − k − p N − k + S − 1 ) ≥ 0 . (B.12) Similarly , in t he case of S ≤ 0, ¯ σ i and ˆ σ i are give n by ( ¯ σ N − k − 1 , ¯ σ N − k , . . . , ¯ σ N ) = (1 , 1 , . . . , 1 | {z } − S , 1 , − 1 , 1 , − 1 , 1 , . . . ) , ( − ¯ σ N − k − 1 , ˆ σ N − k , . . . , ˆ σ N ) = ( − 1 , 1 , 1 , . . . , 1 | {z } − S , 1 , 1 , − 1 , 1 , . . . ) . (B.13) Hence, ˆ σ i = ( − ¯ σ i ( i = N − k − S ) ¯ σ i (otherwise) , (B.14) and (B.9) reduces to −  2 S p N − k − 1 − 2 S p N − k − S + 2 N − k − S − 1 X j = N − k ¯ σ j p j + 2 N X j = N − k − S +1 ¯ σ j p j − 2 N X j = N − k − S +1 ¯ σ j p j − 2 N − k − S − 1 X i = N − k ¯ σ i p N − k − S  =2  − S p N − k − 1 + S p N − k − S − N − k − S − 1 X j = N − k p j − S p N − k − S  ≥ 2 S  − p N − k − 1 + p N − k − S + p N − k − p N − k − S  ≥ 0 . (B.15) As a result, (B.7) holds and this completes the pro of of Prop.4.  References [1] T T okihiro, D T ak aha shi, J Matsuk idaira a nd J Satsuma, Phys. R ev. L ett. 7 6 (19 96) 32 47–3 250. [2] K Nishina ri, D T ak aha shi, J . Phys. A: Math. Gen. 31 (1998 ) 5439 –5450 A New Expr e s s ion of Soliton Solution to the Ultr adiscr ete T o da Equation 13 [3] K Nishina ri, D T ak aha shi, J . Phys. A: Math. Gen. 32 (1999 ) 93–1 04 [4] A Nagai, D T ak ahas hi and T T okihiro, Phys. L ett. A 25 5 (19 99) 26 5–27 1. 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