(D14)      CSM$USERS:[WHEREMAN.PROGRAMS.NPAINLEVE.SINGLE]P_RES0.OUT;3


(C15) batch("np_exec.max");


(C16) /* ************************************************************************* */
/*                           Batch file NP_EXEC.MAX                          */
/* ************************************************************************* */

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

  SUBSTITUTE  X ----> G + X0

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

                                 2                3
PAINLEVE ANALYSIS OF EQUATION,  F  F      - 3 (F )   = 0
                                    G G G       G

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

                  ALPHA
  SUBSTITUTE  U  G       FOR f IN ORIGINAL EQUATION.
               0

  MINIMUM POWERS OF g ARE  [3 ALPHA - 3]

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

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

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

    USING THE USER-SUPPLIED VALUE alpha = - 2

    WITH alpha =  - 2 , POWER OF g is  - 9  ----> SOLVE FOR  U
                                                              0

               1
       TERM  0 --  IS DOMINANT
                9
               G

       IN EQUATION.

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

               2                          R - 9
       TERM ( U  (R - 10) R (R + 1) ) U  G       IS DOMINANT
               0                       R

       IN EQUATION.

       THE 2 NON-NEGATIVE INTEGRAL ROOTS ARE  [R = 0, R = 10]

 

    WITH MAXIMUM RESONANCE =  10  ----> CHECK RESONANCES.

                              10
                             ====
                             \         K - 2
    SUBSTITUTE POWER SERIES   >    U  G       FOR f IN EQUATION.
                             /      K
                             ====
                             K = 0

    WITH U  ARBITRARY ?
          0

                         1            2
        * COEFFICIENT OF -- IS  - 18 U  U
                          8           0  1
                         G

        U  = 0
         1

                         1            2
        * COEFFICIENT OF -- IS  - 48 U  U
                          7           0  2
                         G

        U  = 0
         2

                         1            2
        * COEFFICIENT OF -- IS  - 84 U  U
                          6           0  3
                         G

        U  = 0
         3

                         1             2
        * COEFFICIENT OF -- IS  - 120 U  U
                          5            0  4
                         G

        U  = 0
         4

                         1             2
        * COEFFICIENT OF -- IS  - 150 U  U
                          4            0  5
                         G

        U  = 0
         5

                         1             2
        * COEFFICIENT OF -- IS  - 168 U  U
                          3            0  6
                         G

        U  = 0
         6

                         1             2
        * COEFFICIENT OF -- IS  - 168 U  U
                          2            0  7
                         G

        U  = 0
         7

                         1            2
        * COEFFICIENT OF - IS  - 144 U  U
                         G            0  8

        U  = 0
         8

                                     2
        * COEFFICIENT OF 1 IS  - 90 U  U
                                     0  9

        U  = 0
         9

        * COEFFICIENT OF G IS  0

        U   IS ARBITRARY !
         10

       COMPATIBILITY CONDITION IS SATISFIED !


(C17) 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 <=  [COUNTERT ]
                                                               1

  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() >

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



(C18) /* ************************** END of NP_EXEC.MAX ************************** */


(D18)                                 DONE


(C19) closefile();