(d19) /usr/people/whereman/programs/upainleve/upainsing/p_tony.out (c20) batch("np_exec.max"); (c21) /* ************************************************************************* */ /* Batch file NP_EXEC.MAX */ /* ************************************************************************* */ exec_painleve (eq, alpha, do_resonances, max_resonance, do_simplification)$ THERE ARE NO PARAMETERS IN THE EQUATION. You are using the simplification suggested by Kruskal. You selected G(T,X,...) = X - H(T,...). ---------------------------------------------------------------- 2 PAINLEVE ANALYSIS OF EQUATION, f + 5 f f + 10 f f + 10 f f x x x x x x x x x x x x + f = 0 t WITHOUT ANY PARAMETERS. ---------------------------------------------------------------- alpha SUBSTITUTE u g FOR f IN ORIGINAL EQUATION. 0 MINIMUM POWERS OF g ARE [3 alpha - 1, 2 alpha - 3, alpha - 5] 3 alpha - 1 3 * COEFFICIENT OF g IS 10 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 . 2 alpha - 3 2 * COEFFICIENT OF g IS 5 u (alpha - 1) alpha (3 alpha - 2) 0 2 NOTE : THIS TERM VANISHES FOR [alpha = -, alpha = 0, alpha = 1] , 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. alpha - 5 * COEFFICIENT OF g IS u (alpha - 4) (alpha - 3) (alpha - 2) 0 (alpha - 1) alpha NOTE : THIS TERM VANISHES FOR [alpha = 0, alpha = 1, alpha = 2, alpha = 3, alpha = 4] , 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 ( 3 alpha - 1 ) AND ( 2 alpha - 3 ) OF g, WITH alpha = - 2 , POWER OF g is - 7 ----> SOLVE FOR u 0 2 1 TERM - 20 u (u + 6) -- IS DOMINANT 0 0 7 g IN EQUATION. ---------------------------------------------------------------- 1 ) WITH u = - 6 ----> FIND RESONANCES 0 alpha r + alpha SUBSTITUTE u g + u g FOR f IN EQUATION 0 r r - 7 TERM ( (r - 8) (r - 7) (r - 6) r (r + 1) ) u g IS DOMINANT r IN EQUATION. THE 4 NON-NEGATIVE INTEGRAL ROOTS ARE [r = 0, r = 6, r = 7, r = 8] . ANALYSIS OF THE POLYNOMIAL IN R WAS COMPLETE. WITH MAXIMUM RESONANCE = 8 ----> CHECK RESONANCES. 8 ==== \ k - 2 SUBSTITUTE POWER SERIES > g u FOR f IN EQUATION. / k ==== k = 0 WITH u = - 6 0 1 * COEFFICIENT OF -- IS - 420 u 6 1 g u = 0 1 1 * COEFFICIENT OF -- IS - 720 u 5 2 g u = 0 2 1 * COEFFICIENT OF -- IS - 720 u 4 3 g u = 0 3 1 * COEFFICIENT OF -- IS - 12 (h + 40 u ) 3 t 4 g h t u = - -- 4 40 1 * COEFFICIENT OF -- IS - 180 u 2 5 g u = 0 5 1 * COEFFICIENT OF - IS 0 g u IS ARBITRARY ! 6 COMPATIBILITY CONDITION IS SATISFIED ! * COEFFICIENT OF 1 IS 0 u IS ARBITRARY ! 7 COMPATIBILITY CONDITION IS SATISFIED ! * COEFFICIENT OF g IS 0 u IS ARBITRARY ! 8 COMPATIBILITY CONDITION IS SATISFIED ! PART fell off end. Returned to Macsyma Toplevel. (c22) QUIT();