Time= 30 msecs

(D2)              CSM$USERS:[WHEREMAN.HIROTA]H_M5KDV.OUT;1

(C3) batchload("hir_sing.max")$

Batching the file hir_sing.max
Batchload done.

Time= 4280 msecs

(C4) n:3$

Time= 0 msecs

(C5) B(f,g):= Dxt[3,1](f,g) + Dx[6](f,g)$

Time= 10 msecs

(C6) name:modified_5th_KdV$

Time= 10 msecs

(C7) hirota(b,name,n,1,3,true,true)$

/*********************************************************/

/*    WELCOME TO THE MACSYMA PROGRAM HIR_SING.MAX        */

/*            BY WILLY HEREMAN AND WUNING ZHUANG         */

/*              FOR THE CALCULATION OF SOLITONS          */

/*             OF THE  MODIFIED_5TH_KDV  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 


   6          5        2   4         3          3        2    2        3
  d F     dF d F      d F d F       d F 2   dF d F      d F  d F      d F   dF
F --- - 6 -- --- + 15 --- --- - 10 (---)  - -- --- + 3 ----- --- - 3 ------ --
    6     dX   5        2   4         3     dT   3     dT dX   2          2 dX
  dX         dX       dX  dX        dX         dX            dX      dT dX

                                                                       4
                                                                      d F
                                                                 + F ------  = 0
                                                                          3
                                                                     dT dX


                                                   6    3
For this equation the polynomial P(K,-OMEGA,L) =  K  - K  OMEGA

The equation has at least a one- and two-soliton solution.

For the  MODIFIED_5TH_KDV  equation, 

We use the dispersion relation 

              3
 OMEGA[I] =  K
              I

In the expansion of f we use THETA = K X - OMEGA T + 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   =  10
 1

K   =  13
 2

K   =  3
 3

The equation did not pass the random number test(s) for 

the existence of a 3 soliton solution.

The test with random numbers lead to the value 

90783226912842067968000000

which is so large that it may not be reliable on some computers

(due to storage problems with large integers).

Set the random-number test(s) equal to zero, and run

the symbolic test(s). 

Starting the construction of the two-soliton solution.

The coefficient a[i,j] is calculated via the polynomial form.

                                    6    3
The polynomial is P(K,-OMEGA,L) =  K  - K  OMEGA

                                   4
                          (K  - K )
                            J    I
The coefficient a[i,j] =  ----------
                                   4
                          (K  + K )
                            J    I

                          THETA  + THETA      THETA      THETA
                               2        1          2          1
The function f =  A     %E                + %E       + %E       + 1
                   1, 2

At the end of the computations the form of the function f

and the coefficient a[1,2] are explicitly available.

The explicit factored form of a[1,2] can be obtained

by typing factor(a[1,2]); 

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= 18610 msecs

(C8) kill(all)$

Time= 30 msecs

(C1) closefile();