(d23) 	     /home/ugoktas/macsyma/painleve/unal/system/sp_kdv.out

(c24) batch("sp_exec.max");

(c25) /* sp_exec.max */

exec_painleve(eqlist,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 THE EQUATION(S),
f       + f  f   + f    = 0
 1	   1  1	    1
  x x x	       x     t
 
------------------------------------------------------------------------
		     alpha
			  1
SUBSTITUTE  u	    g	     FOR f
	     1, [0]		  1
 
 
IN THE ORIGINAL EQUATION(S).
THE MINIMUM POWERS OF g IN EQUATION  1  ARE
[alpha  - 3, 2 alpha  - 1, alpha  - 1]
      1		    1	        1
By balancing any two or more terms, we obtain the solution set:
[[alpha  = - 2]]
       1
Now, we separate the cases above into different groups.
Next, we check the remaining steps with this solution.
THE POWER OF g is  - 5 IN EQUATION  1 .
 
----> SOLVE FOR  u
		  1, [0]
------------------------------------------------------------------------
				 1
TERM  - 2 u	  (u	   + 12) --  IS DOMINANT
	   1, [0]   1, [0]	  5
				 g
IN EQUATION  1 .
------------------------------------------------------------------------
WITH [u	      = - 12] ------> FIND RESONANCES
       1, [0]
 
		     alpha		 r + alpha
			  1			  1
SUBSTITUTE  u	    g	     +  u       g	     FOR  f  
	     1, [0]		 1, [r]			   1

							     IN THE EQUATION(S).
 
THIS IS THE MATRIX FOR RESONANCES:
 
[ (r - 6) (r - 4) (r + 1) ]
 
THIS IS THE EQUATION FOR RESONANCES:
(r - 6) (r - 4) (r + 1)  = 0
 
THESE ARE THE RESONANCES:
[r = - 1, r = 4, r = 6]
 
WITH MAXIMUM RESONANCE =  6  ----> CHECK RESONANCES.
 
			  6
			 ====
			 \	        k - 2
SUBSTITUTE POWER SERIES   >    u       g       FOR f   IN THE EQUATION(S).
			 /      1, [k]		    1
			 ====
			 k = 0
 
------------------------------------------------------------------------
* u	  = - 12
   1, [0]
------------------------------------------------------------------------
* u	  = 0
   1, [1]
------------------------------------------------------------------------
* u	  = h
   1, [2]    t
------------------------------------------------------------------------
* u	  = 0
   1, [3]
------------------------------------------------------------------------
* THE COEFFICIENT MATRIX FOR THE VECTOR
  [u	  ]
    1, [4]
 
  HAS A DETERMINANT EQUAL TO ZERO.
  THE COMPATIBILITY CONDITION IS SATISFIED !
  THERE IS/ARE  1  FREE FUNCTION(S) !
------------------------------------------------------------------------
------------------------------------------------------------------------
	    h
	     t t
* u	  = ----
   1, [5]    6
------------------------------------------------------------------------
* THE COEFFICIENT MATRIX FOR THE VECTOR
  [u	  ]
    1, [6]
 
  HAS A DETERMINANT EQUAL TO ZERO.
  THE COMPATIBILITY CONDITION IS SATISFIED !
  THERE IS/ARE  1  FREE FUNCTION(S) !
------------------------------------------------------------------------
------------------------------------------------------------------------

(c26) /* sp_exec.max */
(d26) 				      done

(c27) closefile();