(d15) 		    /usr/people/whereman/painsing/p_asg.out

(c16) batch("np_exec.max")$

(c17) /* ************************************************************************* */
/*                           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,...)
----------------------------------------------------------------
						 2		      2	   3
PAINLEVE ANALYSIS OF EQUATION,  2 f f    - 2 (f )  + 2 f f    - 2 (f )  - f  + f
				     x x       x	  t t	    t

									     = 0
----------------------------------------------------------------
		  alpha
  SUBSTITUTE  u  g       FOR f IN ORIGINAL EQUATION.
	       0
  MINIMUM POWERS OF g ARE  [2 alpha - 2, 3 alpha, alpha]
		       2 alpha - 2	     2		  2
    * COEFFICIENT OF  g		    IS  - 2 u  alpha ((h )  + 1)
					     0	        t
     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 .
		       3 alpha	       3
    * COEFFICIENT OF  g	        IS  - u
				       0
		       alpha
    * COEFFICIENT OF  g	      IS  u
				   0
----------------------------------------------------------------
  FOR EXPONENTS ( 2 alpha - 2 ) AND ( 3 alpha ) OF g,
    WITH alpha =  - 2 , POWER OF g is  - 6  ----> SOLVE FOR  u
							      0
	      2	       2	   1
       TERM  u  (4 (h )  - u  + 4) --  IS DOMINANT
	      0	     t	    0	    6
				   g
       IN EQUATION.
----------------------------------------------------------------
		      2
1 )  WITH  u  = 4 (h )  + 4  ----> FIND RESONANCES
	    0	    t
		    alpha	  r + alpha
    SUBSTITUTE  u  g	   +  u  g	     FOR f IN EQUATION
		 0	       r
		     2	   2			   r - 6
       TERM ( 8 ((h )  + 1)  (r - 2) (r + 1) ) u  g       IS DOMINANT
		   t			        r
       IN EQUATION.
       THE ONLY NON-NEGATIVE INTEGRAL ROOT IS  [r = 2]
 
    WITH MAXIMUM RESONANCE =  2  ----> CHECK RESONANCES.
			      2
			     ====
			     \	    k - 2
    SUBSTITUTE POWER SERIES   >	   g	  u   FOR f IN EQUATION.
			     /		   k
			     ====
			     k = 0
		    2
    WITH u  = 4 (h )  + 4
	  0	  t
			 1	        2     2
        * COEFFICIENT OF -- IS  16 ((h )  + 1)  (4 h    - u )
			  5	      t		    t t	   1
			 g
        u  = 4 h
	 1      t t
			 1
        * COEFFICIENT OF -- IS  0
			  4
			 g
        u  IS ARBITRARY !
	 2
       COMPATIBILITY CONDITION IS SATISFIED !
----------------------------------------------------------------
  FOR EXPONENTS ( 2 alpha - 2 ) AND ( alpha ) OF g,
    WITH alpha =  2 , POWER OF g is  2  ----> SOLVE FOR  u
							  0
			    2		   2
       TERM  - u  (4 u  (h )  + 4 u  - 1) g   IS DOMINANT
		0     0	  t	   0
       IN EQUATION.
----------------------------------------------------------------
		     1
1 )  WITH  u  = -----------  ----> FIND RESONANCES
	    0	      2
		4 (h )  + 4
		    t
		    alpha	  r + alpha
    SUBSTITUTE  u  g	   +  u  g	     FOR f IN EQUATION
		 0	       r
				    r + 2
       TERM ( (r - 2) (r + 1) ) u  g	   IS DOMINANT
				 r
       IN EQUATION.
       THE ONLY NON-NEGATIVE INTEGRAL ROOT IS  [r = 2]
 
    WITH MAXIMUM RESONANCE =  2  ----> CHECK RESONANCES.
			      2
			     ====
			     \	    k + 2
    SUBSTITUTE POWER SERIES   >	   g	  u   FOR f IN EQUATION.
			     /		   k
			     ====
			     k = 0
		   1
    WITH u  = -----------
	  0	    2
	      4 (h )  + 4
		  t
						  4	       2
				  h    + 4 u  (h )  + 8 u  (h )  + 4 u
			  3	   t t	    1   t	 1   t	      1
        * COEFFICIENT OF g  IS  - -------------------------------------
						    2	  2
					     4 ((h )  + 1)
						  t
		    h
		     t t
        u  = - --------------
	 1	      2	    2
	       4 ((h )  + 1)
		    t
			  4
        * COEFFICIENT OF g  IS  0
        u  IS ARBITRARY !
	 2
       COMPATIBILITY CONDITION IS SATISFIED !
----------------------------------------------------------------
  FOR EXPONENTS ( 3 alpha ) AND ( alpha ) OF g,
    POWER OF g IS NOT MINIMAL 
						-- SKIP THIS VALUE OF ALPHA.
----------------------------------------------------------------

(c18) output()$
----------------------------------------------------------------
AT THE END OF THE COMPUTATIONS THE FOLLOWING ARE AVAILABLE:
* U VALUE(S)
  (type uval[j,k,l] where 1 <= j <=  2 and 0 <= k <=  [2, 2]
  and 1 <= l <=  [1, 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 <=  2 )
  j stands for j_th alpha
* COMPATIBILITY CONDITION(S)
  (type compcond[j,k] where 1 <= j <=  2 and 1 <= k <=  [1, 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 <=  2 and 1 <= k <=  [1, 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() >
----------------------------------------------------------------

(c19) /* ************************** END of NP_EXEC.MAX ************************** */


(c20) closefile();