(D21) CSM$USERS:[WHEREMAN.PROGRAMS.NPAINLEVE.SINGLE]P_MKDV.OUT;1 (C22) batch("np_exec.max"); (C23) /* ************************************************************************* */ /* Batch file NP_EXEC.MAX */ /* ************************************************************************* */ exec_painleve (eq, alpha, do_resonances, max_resonance, do_simplification)$ You are using the simplification suggested by KRUSKAL You selected G(T,X,...) = X - H(T,...) ---------------------------------------------------------------- 2 2 PAINLEVE ANALYSIS OF EQUATION, 2 A F - 3 F F + F = 0 X X X X T ---------------------------------------------------------------- ALPHA SUBSTITUTE U G FOR f IN ORIGINAL EQUATION. 0 MINIMUM POWERS OF g ARE [3 ALPHA - 1, ALPHA - 3] 3 ALPHA - 1 3 * COEFFICIENT OF G IS - 3 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 . ALPHA - 3 2 * COEFFICIENT OF G IS 2 U A (ALPHA - 2) (ALPHA - 1) ALPHA 0 NOTE : THIS TERM VANISHES FOR [ALPHA = 0, ALPHA = 1, ALPHA = 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. ---------------------------------------------------------------- FOR EXPONENTS ( 3 ALPHA - 1 ) AND ( ALPHA - 3 ) OF g, WITH alpha = - 1 , POWER OF g is - 4 ----> SOLVE FOR U 0 1 TERM - 3 U (2 A - U ) (2 A + U ) -- IS DOMINANT 0 0 0 4 G IN EQUATION. ---------------------------------------------------------------- 1 ) WITH U = 2 A ---> FIND RESONANCES 0 ALPHA R + ALPHA SUBSTITUTE U G + U G FOR f IN EQUATION 0 R 2 R - 4 TERM ( 2 A (R - 4) (R - 3) (R + 1) ) U G IS DOMINANT R IN EQUATION. THE 2 NON-NEGATIVE INTEGRAL ROOTS ARE [R = 3, R = 4] WITH MAXIMUM RESONANCE = 4 ----> CHECK RESONANCES. 4 ==== \ K - 1 SUBSTITUTE POWER SERIES > G U FOR f IN EQUATION. / K ==== K = 0 WITH U = 2 A 0 1 2 * COEFFICIENT OF -- IS 24 U A 3 1 G U = 0 1 1 * COEFFICIENT OF -- IS 2 A (H + 6 U A) 2 T 2 G H T U = - --- 2 6 A 1 * COEFFICIENT OF - IS 0 G U IS ARBITRARY ! 3 COMPATIBILITY CONDITION IS SATISFIED ! * COEFFICIENT OF 1 IS 0 U IS ARBITRARY ! 4 COMPATIBILITY CONDITION IS SATISFIED ! ---------------------------------------------------------------- 2 ) WITH U = - 2 A ---> FIND RESONANCES 0 ALPHA R + ALPHA SUBSTITUTE U G + U G FOR f IN EQUATION 0 R 2 R - 4 TERM ( 2 A (R - 4) (R - 3) (R + 1) ) U G IS DOMINANT R IN EQUATION. THE 2 NON-NEGATIVE INTEGRAL ROOTS ARE [R = 3, R = 4] WITH MAXIMUM RESONANCE = 4 ----> CHECK RESONANCES. 4 ==== \ K - 1 SUBSTITUTE POWER SERIES > G U FOR f IN EQUATION. / K ==== K = 0 WITH U = - 2 A 0 1 2 * COEFFICIENT OF -- IS 24 U A 3 1 G U = 0 1 1 * COEFFICIENT OF -- IS - 2 A (H - 6 U A) 2 T 2 G H T U = --- 2 6 A 1 * COEFFICIENT OF - IS 0 G U IS ARBITRARY ! 3 COMPATIBILITY CONDITION IS SATISFIED ! * COEFFICIENT OF 1 IS 0 U IS ARBITRARY ! 4 COMPATIBILITY CONDITION IS SATISFIED ! ---------------------------------------------------------------- (C24) 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 <= [4] and 1 <= l <= [2] ) 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 <= [2] ) 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 <= [2] ) j stands for j_th alpha,k stands for k_th solution set for u[0] ---------------------------------------------------------------- TO SEE THIS MENU AGAIN JUST TYPE < output() > ---------------------------------------------------------------- (C25) /* ************************** END of NP_EXEC.MAX ************************** */ (D25) DONE (C26) closefile();