(D17)      CSM$USERS:[WHEREMAN.PROGRAMS.NPAINLEVE.SINGLE]P_BOUS.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,...)

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

                                F
                                 X X X X                    2
PAINLEVE ANALYSIS OF EQUATION,  -------- + 2 F F    + 2 (F )  + F     = 0
                                   3            X X       X      T T

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

                  ALPHA
  SUBSTITUTE  U  G       FOR f IN ORIGINAL EQUATION.
               0

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

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

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

    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.

                       ALPHA - 4
    * COEFFICIENT OF  G           IS  

                                    U  (ALPHA - 3) (ALPHA - 2) (ALPHA - 1) ALPHA
                                     0
                                    --------------------------------------------
                                                         3

    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 - 2 ) AND ( ALPHA - 4 ) OF g,

    WITH alpha =  - 2 , POWER OF g is  - 6  ----> SOLVE FOR  U
                                                              0

                            1
       TERM  20 U  (U  + 2) --  IS DOMINANT
                 0   0       6
                            G

       IN EQUATION.

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

1 )  WITH  U  = - 2  ----> FIND RESONANCES
            0

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

                                                    R - 6
       TERM ( (R - 6) (R - 5) (R - 4) (R + 1) ) U  G       IS DOMINANT
                                                 R

       IN EQUATION.

       THE 3 NON-NEGATIVE INTEGRAL ROOTS ARE  [R = 4, R = 5, R = 6]

 

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

                              6
                             ====
                             \      K - 2
    SUBSTITUTE POWER SERIES   >    G      U   FOR f IN EQUATION.
                             /             K
                             ====
                             K = 0

    WITH U  = - 2
          0

                         1
        * COEFFICIENT OF -- IS  - 40 U
                          5           1
                         G

        U  = 0
         1

                         1                2
        * COEFFICIENT OF -- IS  - 12 ((H )  + 2 U )
                          4             T        2
                         G

                   2
               (H )
                 T
        U  = - -----
         2       2

                         1
        * COEFFICIENT OF -- IS  - 4 (H    + 2 U )
                          3           T T      3
                         G

               H
                T T
        U  = - ----
         3      2

                         1
        * COEFFICIENT OF -- IS  0
                          2
                         G

        U  IS ARBITRARY !
         4

       COMPATIBILITY CONDITION IS SATISFIED !

                         1
        * COEFFICIENT OF - IS  0
                         G

        U  IS ARBITRARY !
         5

       COMPATIBILITY CONDITION IS SATISFIED !

        * COEFFICIENT OF 1 IS  0

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


(D21)                                 DONE


(C22) closefile();