Time= 40 msecs (D2) CSM$USERS:[WHEREMAN.HIROTA]H_KP.OUT;1 (C3) batchload("hir_sing.max")$ Batching the file hir_sing.max Batchload done. Time= 4430 msecs (C4) n:3$ Time= 0 msecs (C5) B(f,g):= Dxt[1,1](f,g) + Dx[4](f,g) + 3*Dy[2](f,g)$ Time= 0 msecs (C6) name:Kadomtsev_Petviashvili$ Time= 0 msecs (C7) hirota(b,name,n,1,1,true,false)$ /*********************************************************/ /* WELCOME TO THE MACSYMA PROGRAM HIR_SING.MAX */ /* BY WILLY HEREMAN AND WUNING ZHUANG */ /* FOR THE CALCULATION OF SOLITONS */ /* OF THE KADOMTSEV_PETVIASHVILI EQUATION */ /* WITH HIROTA'S METHOD */ /* Version 1.0, released on May 29, 1995 */ /* Copyright 1995 */ /*********************************************************/ The equation in f corresponding to the given bilinear operator is 2 4 3 2 2 d F dF 2 d F dF d F d F 2 dF dF d F 3 F --- - 3 (--) + F --- - 4 -- --- + 3 (---) - -- -- + F ----- = 0 2 dY 4 dX 3 2 dT dX dT dX dY dX dX dX 2 4 For this equation the polynomial P(K,-OMEGA,L) = - K OMEGA + 3 L + K The equation has at least a one- and two-soliton solution. For the KADOMTSEV_PETVIASHVILI equation, We use the dispersion relation 2 4 3 L + K I I OMEGA[I] = --------- K I In the expansion of f we use THETA = K X - OMEGA T + L Y + CST. Starting the random test(s) for the existence of a 3 soliton solution. Wavenumbers k[i] selected for the random number test(s): K = 4 1 K = 11 2 K = 6 3 Wavenumbers l[i] selected for the random number test(s): L = 17 1 L = 13 2 L = 10 3 The equation passed the random number test(s) for the existence of a 3 soliton solution. Starting the symbolic test for the existence of a 3 soliton solution. The equation passed the symbolic test for the existence of a 3 soliton solution. Starting the random test(s) for the existence of a 4 soliton solution. Wavenumbers k[i] selected for the random number test(s): K = 2 1 K = 7 2 K = 12 3 K = 15 4 Wavenumbers l[i] selected for the random number test(s): L = 2 1 L = 8 2 L = 12 3 L = 14 4 The equation passed the random number test(s) for the existence of a 4 soliton solution. Starting the construction of the three-soliton solution. The coefficient a[i,j] is calculated via the polynomial form. 2 4 The polynomial is P(K,-OMEGA,L) = - K OMEGA + 3 L + K The coefficient a[i,j] = 2 2 2 2 (K L - K K - L K + K K ) (K L + K K - L K - K K ) I J I J I J I J I J I J I J I J --------------------------------------------------------------- 2 2 2 2 (K L - K K - L K - K K ) (K L + K K - L K + K K ) I J I J I J I J I J I J I J I J The coefficient b[1,2,3] is calculated via the polynomial form. 2 2 The coefficient b[1,2,3] = (K L - K K - L K + K K ) 1 2 1 2 1 2 1 2 2 2 2 2 (K L + K K - L K - K K ) (K L - K K - L K + K K ) 1 2 1 2 1 2 1 2 1 3 1 3 1 3 1 3 2 2 2 2 (K L + K K - L K - K K ) (K L - K K - L K + K K ) 1 3 1 3 1 3 1 3 2 3 2 3 2 3 2 3 2 2 2 2 (K L + K K - L K - K K )/((K L - K K - L K - K K ) 2 3 2 3 2 3 2 3 1 2 1 2 1 2 1 2 2 2 2 2 (K L + K K - L K + K K ) (K L - K K - L K - K K ) 1 2 1 2 1 2 1 2 1 3 1 3 1 3 1 3 2 2 2 2 (K L + K K - L K + K K ) (K L - K K - L K - K K ) 1 3 1 3 1 3 1 3 2 3 2 3 2 3 2 3 2 2 (K L + K K - L K + K K )) 2 3 2 3 2 3 2 3 THETA + THETA + THETA THETA + THETA 3 2 1 3 2 The function f = B %E + A %E 1, 2, 3 2, 3 THETA + THETA THETA THETA + THETA THETA 3 1 3 2 1 2 + A %E + %E + A %E + %E 1, 3 1, 2 THETA 1 + %E + 1 At the end of the computations the form of the function f and the coefficients a[i,j] and b[1,2,3] are available. The explicit factored form of a[1,2] and b[1,2,3] can be obtained by entering factor(a[1,2]); and factor(b[1,2,3]); The explicit forms of theta[i] and omega[i] are also available. The form of f can be obtained by typing f; . The explicit form of f can be obtained by typing expression(f); . Time= 51680 msecs (C8) kill(all)$ Time= 20 msecs (C1) closefile();