(D17)       CSM$USERS:[WHEREMAN.PROGRAMS.NPAINLEVE.SINGLE]P_KS.OUT;1


(C18) batch("np_exec.max")$


(C19) /* ************************************************************************* */
/*                           Batch file NP_EXEC.MAX                          */
/* ************************************************************************* */

exec_painleve (eq, alpha, do_resonances, max_resonance, do_simplification)$

You are using the simplification suggested by KRUSKAL

You selected G(T,X,...) = X - H(T,...)

----------------------------------------------------------------

PAINLEVE ANALYSIS OF EQUATION,  F        + F    + F F  + F   = 0
                                 X X X X    X X      X    T

----------------------------------------------------------------

                  ALPHA
  SUBSTITUTE  U  G       FOR f IN ORIGINAL EQUATION.
               0

  MINIMUM POWERS OF g ARE  [2 ALPHA - 1, ALPHA - 4]

                       2 ALPHA - 1       2
    * COEFFICIENT OF  G             IS  U  ALPHA
                                         0

     NOTE : THIS TERM VANISHES FOR  ALPHA = 0 ,

     VERIFY IF  ALPHA = 0  LEADS TO DOMINANT BEHAVIOR,

     IF IT DOES THEN RUN THE PROGRAM AGAIN WITH THIS USER 

     SUPPLIED VALUE OF ALPHA.

     HENCE, PUT BETA =  0 .

                       ALPHA - 4
    * COEFFICIENT OF  G           IS  U  (ALPHA - 3) (ALPHA - 2) (ALPHA - 1)
                                       0

                                                                           ALPHA

    NOTE : THIS TERM VANISHES FOR   [ALPHA = 0, ALPHA = 1, ALPHA = 2, ALPHA = 3]

                                                                               ,

    VERIFY IF ANY OF THESE VALUES FOR ALPHA LEADS TO DOMINANT

    BEHAVIOR,

    IF IT DOES THEN RUN THE PROGRAM AGAIN WITH THIS VALUE

    AS USER SUPPLIED ALPHA, CALLED BETA.

----------------------------------------------------------------

  FOR EXPONENTS ( 2 ALPHA - 1 ) AND ( ALPHA - 4 ) OF g,

    WITH alpha =  - 3 , POWER OF g is  - 7  ----> SOLVE FOR  U
                                                              0

                               1
       TERM  - 3 (U  - 120) U  --  IS DOMINANT
                   0         0  7
                               G

       IN EQUATION.

----------------------------------------------------------------

1 )  WITH  U  = 120  ----> FIND RESONANCES
            0

                    ALPHA         R + ALPHA
    SUBSTITUTE  U  G       +  U  G           FOR f IN EQUATION
                 0             R

                                2                    R - 7
       TERM ( (R - 6) (R + 1) (R  - 13 R + 60) ) U  G       IS DOMINANT
                                                  R

       IN EQUATION.

       THE ONLY NON-NEGATIVE INTEGRAL ROOT IS  [R = 6]

 

    WITH MAXIMUM RESONANCE =  6  ----> CHECK RESONANCES.

                              6
                             ====
                             \      K - 3
    SUBSTITUTE POWER SERIES   >    G      U   FOR f IN EQUATION.
                             /             K
                             ====
                             K = 0

    WITH U  = 120
          0

                         1
        * COEFFICIENT OF -- IS  - 480 U
                          6            1
                         G

        U  = 0
         1

                         1
        * COEFFICIENT OF -- IS  - 24 (19 U  - 60)
                          5               2
                         G

             60
        U  = --
         2   19

                         1
        * COEFFICIENT OF -- IS  360 (H  - U )
                          4           T    3
                         G

        U  = H
         3    T

                                  120 (722 U  + 11)
                         1                  4
        * COEFFICIENT OF -- IS  - -----------------
                          3              361
                         G

               11
        U  = - ---
         4     722

                         1
        * COEFFICIENT OF -- IS  - 120 U
                          2            5
                         G

        U  = 0
         5

                         1
        * COEFFICIENT OF - IS  0
                         G

        U  IS ARBITRARY !
         6

       COMPATIBILITY CONDITION IS SATISFIED !

----------------------------------------------------------------


(C20) output()$

----------------------------------------------------------------

AT THE END OF THE COMPUTATIONS THE FOLLOWING ARE AVAILABLE:

* U VALUE(S)

  (type uval[j,k,l] where 1 <= j <=  1 and 0 <= k <=  [6]

  and 1 <= l <=  [1] )

  j stands for j_th alpha,k stands for u[k],l stands for

  l_th solution set for u[0] 

* ALPHA VALUE(S)

  (type alpha[j] where 1 <= j <=  1 )

  j stands for j_th alpha

* COMPATIBILITY CONDITION(S)

  (type compcond[j,k] where 1 <= j <=  1 and 1 <= k <=  [1] )

  j stands for j_th alpha,k stands for k_th solution set for u[0]

* RESONANCE(S)

  (type res[j,k] where 1 <= j <=  1 and 1 <= k <=  [1] )

  j stands for j_th alpha,k stands for k_th solution set for u[0]

----------------------------------------------------------------

TO SEE THIS MENU AGAIN JUST TYPE < output() >

----------------------------------------------------------------



(C21) /* ************************** END of NP_EXEC.MAX ************************** */



(C22) closefile()$