(D1) CSM$USERS:[WHEREMAN.PROGRAMS.NPAINLEVE.SINGLE]P_CYLKP.OUT;2 (C2) batch("p_cylkp.dat"); (C3) /* This is data file for the cylindrical KP equation */ /* in book Ablowitz and Clarkson on p 64 */ do_simplification: true; (D3) TRUE (C4) prefer_variable:x; (D4) X (C5) /* giving_uzero : false ; solving_uzero : false ; */ eq : ftxy[1,1,0](t,x,y)+6*f*ftxy[0,2,0](t,x,y) + 6*(ftxy[0,1,0](t,x,y))^2 + ftxy[0,4,0](t,x,y)+a(t)*ftxy[0,1,0](t,x,y) + b(t)*ftxy[0,0,2](t,x,y); 2 (D5) F B(T) + F A(T) + F + 6 F F + 6 (F ) + F Y Y X X X X X X X X T X (C6) /* where a(t) and b(t) are arbitrary functions */ (D6) DONE (C7) batch("np_exec.max"); (C8) /* ************************************************************************* */ /* Batch file NP_EXEC.MAX */ /* ************************************************************************* */ exec_painleve (eq, alpha, do_resonances, max_resonance, do_simplification)$ You are using the simplification suggested by KRUSKAL You selected G(T,X,...) = X - H(T,...) ---------------------------------------------------------------- PAINLEVE ANALYSIS OF EQUATION, F B(T) + F A(T) + F + 6 F F Y Y X X X X X X X 2 + 6 (F ) + F = 0 X T X ---------------------------------------------------------------- ALPHA SUBSTITUTE U G FOR f IN ORIGINAL EQUATION. 0 MINIMUM POWERS OF g ARE [2 ALPHA - 2, ALPHA - 4] 2 ALPHA - 2 2 * COEFFICIENT OF G IS 6 U ALPHA (2 ALPHA - 1) 0 1 NOTE : THIS TERM VANISHES FOR [ALPHA = -, 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. ALPHA - 4 * COEFFICIENT OF G IS U (ALPHA - 3) (ALPHA - 2) (ALPHA - 1) 0 ALPHA NOTE : THIS TERM VANISHES FOR [ALPHA = 0, ALPHA = 1, ALPHA = 2, ALPHA = 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. ---------------------------------------------------------------- FOR EXPONENTS ( 2 ALPHA - 2 ) AND ( ALPHA - 4 ) OF g, WITH alpha = - 2 , POWER OF g is - 6 ----> SOLVE FOR U 0 1 TERM 60 U (U + 2) -- IS DOMINANT 0 0 6 G IN EQUATION. ---------------------------------------------------------------- 1 ) WITH U = - 2 ----> FIND RESONANCES 0 ALPHA R + ALPHA SUBSTITUTE U G + U G FOR f IN EQUATION 0 R R - 6 TERM ( (R - 6) (R - 5) (R - 4) (R + 1) ) U G IS DOMINANT R IN EQUATION. THE 3 NON-NEGATIVE INTEGRAL ROOTS ARE [R = 4, R = 5, R = 6] WITH MAXIMUM RESONANCE = 6 ----> CHECK RESONANCES. 6 ==== \ K - 2 SUBSTITUTE POWER SERIES > G U FOR f IN EQUATION. / K ==== K = 0 WITH U = - 2 0 1 * COEFFICIENT OF -- IS - 120 U 5 1 G U = 0 1 1 2 * COEFFICIENT OF -- IS - 12 ((H ) B(T) - H + 6 U ) 4 Y T 2 G 2 (H ) B(T) - H Y T U = - --------------- 2 6 1 * COEFFICIENT OF -- IS - 4 (H B(T) - A(T) + 6 U ) 3 Y Y 3 G H B(T) - A(T) Y Y U = - ---------------- 3 6 1 * COEFFICIENT OF -- IS 0 2 G U IS ARBITRARY ! 4 COMPATIBILITY CONDITION IS SATISFIED ! 1 * COEFFICIENT OF - IS 0 G U IS ARBITRARY ! 5 COMPATIBILITY CONDITION IS SATISFIED ! 2 H B(T) - A(T) + 4 H A(T) B(T) - 2 A (T) Y Y T T Y Y * COEFFICIENT OF 1 IS - ----------------------------------------------- 6 U IS ARBITRARY ? 6 COMPATIBILITY CONDITION: 2 H B(T) - A(T) + 4 H A(T) B(T) - 2 A (T) Y Y T T Y Y - ----------------------------------------------- = 0, 6 *** CONDITION IS NOT SATISFIED. *** *** CHECK FOR FREE PARAMETERS OR PRESENCE OF U . *** 0 ---------------------------------------------------------------- (C9) 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 <= [6] 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() > ---------------------------------------------------------------- (C10) /* ************************** END of NP_EXEC.MAX ************************** */ (D10) DONE (C11) closefile();