(d17) /usr/people/whereman/programs/npainleve/npainsing/p_5pkdv.out (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,...) ---------------------------------------------------------------- 2 PAINLEVE ANALYSIS OF EQUATION, f + cc f f + bb f f + aa f f x x x x x x x x x x x x + f = 0 t ---------------------------------------------------------------- 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 u aa 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 * COEFFICIENT OF g IS 2 u (alpha - 1) alpha (alpha cc - 2 cc + alpha bb) 0 2 cc NOTE : THIS TERM VANISHES FOR [alpha = -------, alpha = 0, alpha = 1] , cc + bb 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 - 2 u (12 u cc + 6 u bb + u aa + 360) -- IS DOMINANT 0 0 0 0 7 g IN EQUATION. ---------------------------------------------------------------- 2 2 3 sqrt(4 cc + 4 bb cc + bb - 40 aa) + 6 cc + 3 bb 1 ) WITH u = - --------------------------------------------------- 0 aa ----> FIND RESONANCES alpha r + alpha SUBSTITUTE u g + u g FOR f IN EQUATION 0 r 3 2 TERM ( (r - 6) (r + 1) (aa r - 15 aa r 2 2 2 - 3 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) r - 6 cc r - 3 bb cc r + 86 aa r 2 2 + 12 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) 2 2 2 2 + 6 bb sqrt(4 cc + 4 bb cc + bb - 40 aa) + 24 cc + 24 bb cc + 6 bb r - 7 - 240 aa) ) u g IS DOMINANT r IN EQUATION. At end of domcoefr(powU0), this is coefr: 3 2 2 2 (r - 6) (r + 1) (aa r - 15 aa r - 3 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) r 2 2 2 - 6 cc r - 3 bb cc r + 86 aa r + 12 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) 2 2 2 2 + 6 bb sqrt(4 cc + 4 bb cc + bb - 40 aa) + 24 cc + 24 bb cc + 6 bb - 240 aa) This coefr is going into the analyser: (r - 6) (r + 1) 3 2 2 2 2 (aa r - 15 aa r - 3 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) r - 6 cc r 2 2 - 3 bb cc r + 86 aa r + 12 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) 2 2 2 2 + 6 bb sqrt(4 cc + 4 bb cc + bb - 40 aa) + 24 cc + 24 bb cc + 6 bb - 240 aa) This expr is at the start of the analyser, before going into stripper: 3 2 2 2 (r - 6) (r + 1) (aa r - 15 aa r - 3 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) r 2 2 2 - 6 cc r - 3 bb cc r + 86 aa r + 12 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) 2 2 2 2 + 6 bb sqrt(4 cc + 4 bb cc + bb - 40 aa) + 24 cc + 24 bb cc + 6 bb - 240 aa) THE ONLY NON-NEGATIVE INTEGRAL ROOT IS [r = 6] . CONCLUSION: APART FROM R = -1, THE EQUATION HAS ONLY 1 NON-NEGATIVE ROOT(S). THERE SHOULD BE 3 MORE NON-NEGATIVE INTEGER SINGLE ROOT(S). THE EQUATION FAILS THE TEST. THE REMAINING UNSOLVED POLYNOMIAL IS: 3 2 2 2 aa r - 15 aa r - 3 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) r 2 2 2 - 6 cc r - 3 bb cc r + 86 aa r + 12 cc sqrt(4 cc + 4 bb cc + bb - 40 aa) 2 2 2 2 + 6 bb sqrt(4 cc + 4 bb cc + bb - 40 aa) + 24 cc + 24 bb cc + 6 bb - 240 aa . ANALYSIS OF THE POLYNOMIAL IN R WAS COMPLETE. WITH MAXIMUM RESONANCE = 6 ----> CHECK RESONANCES. 6 ==== \ k - 2 SUBSTITUTE POWER SERIES > g u FOR f IN EQUATION. / k ==== k = 0 *** FATAL ERROR : NON-ZERO LEADING COEFFICIENT REMAINS AFTER SUBSTITUTING 2 2 3 sqrt(4 cc + 4 bb cc + bb - 40 aa) + 6 cc + 3 bb u = - --------------------------------------------------- . *** 0 aa This is reso at the end of function find_resonances: [6] PART fell off end. Returned to Macsyma Toplevel. (c20) QUIT();