(D13) CSM$USERS:[WHEREMAN.PROGRAMS.NPAINLEVE.SINGLE]P_SG.OUT;4 (C14) batch("np_exec.max")$ (C15) /* ************************************************************************* */ /* Batch file NP_EXEC.MAX */ /* ************************************************************************* */ exec_painleve (eq, alpha, do_resonances, max_resonance, do_simplification)$ ---------------------------------------------------------------- 3 PAINLEVE ANALYSIS OF EQUATION, - 2 F F + 2 F F - F + F = 0 T X T X ---------------------------------------------------------------- ALPHA SUBSTITUTE U G FOR f IN ORIGINAL EQUATION. 0 MINIMUM POWERS OF g ARE [2 ALPHA - 2, 3 ALPHA, ALPHA] 2 ALPHA - 2 2 * COEFFICIENT OF G IS - 2 U ALPHA G G 0 T X 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 . 3 ALPHA 3 * COEFFICIENT OF G IS - U 0 ALPHA * COEFFICIENT OF G IS U 0 ---------------------------------------------------------------- FOR EXPONENTS ( 2 ALPHA - 2 ) AND ( 3 ALPHA ) OF g, WITH alpha = - 2 , POWER OF g is - 6 ----> SOLVE FOR U 0 2 1 TERM U (4 G G - U ) -- IS DOMINANT 0 T X 0 6 G IN EQUATION. ---------------------------------------------------------------- 1 ) WITH U = 4 G G ----> FIND RESONANCES 0 T X ALPHA R + ALPHA SUBSTITUTE U G + U G FOR f IN EQUATION 0 R 2 2 R - 6 TERM ( 8 (G ) (G ) (R - 2) (R + 1) ) U G IS DOMINANT T X R IN EQUATION. THE ONLY NON-NEGATIVE INTEGRAL ROOT IS [R = 2] WITH MAXIMUM RESONANCE = 2 ----> CHECK RESONANCES. 2 ==== \ K - 2 SUBSTITUTE POWER SERIES > G U FOR f IN EQUATION. / K ==== K = 0 WITH U = 4 G G 0 T X 1 2 2 * COEFFICIENT OF -- IS - 16 (G ) (4 G + U ) (G ) 5 T T X 1 X G U = - 4 G 1 T X 1 * COEFFICIENT OF -- IS 0 4 G U IS ARBITRARY ! 2 COMPATIBILITY CONDITION IS SATISFIED ! ---------------------------------------------------------------- FOR EXPONENTS ( 2 ALPHA - 2 ) AND ( ALPHA ) OF g, WITH alpha = 2 , POWER OF g is 2 ----> SOLVE FOR U 0 2 TERM - U (4 U G G - 1) G IS DOMINANT 0 0 T X IN EQUATION. ---------------------------------------------------------------- 1 1 ) WITH U = ------- ----> FIND RESONANCES 0 4 G G T X ALPHA R + ALPHA SUBSTITUTE U G + U G FOR f IN EQUATION 0 R R + 2 TERM ( (R - 2) (R + 1) ) U G IS DOMINANT R IN EQUATION. THE ONLY NON-NEGATIVE INTEGRAL ROOT IS [R = 2] WITH MAXIMUM RESONANCE = 2 ----> CHECK RESONANCES. 2 ==== \ K + 2 SUBSTITUTE POWER SERIES > G U FOR f IN EQUATION. / K ==== K = 0 1 WITH U = ------- 0 4 G G T X 2 2 4 U (G ) (G ) - G 3 1 T X T X * COEFFICIENT OF G IS - ----------------------- 2 2 4 (G ) (G ) T X G T X U = ------------- 1 2 2 4 (G ) (G ) T X 4 * COEFFICIENT OF G IS 0 U IS ARBITRARY ! 2 COMPATIBILITY CONDITION IS SATISFIED ! ---------------------------------------------------------------- FOR EXPONENTS ( 3 ALPHA ) AND ( ALPHA ) OF g, POWER OF g IS NOT MINIMAL -- SKIP THIS VALUE OF ALPHA. ---------------------------------------------------------------- (C16) output()$ ---------------------------------------------------------------- AT THE END OF THE COMPUTATIONS THE FOLLOWING ARE AVAILABLE: * U VALUE(S) (type uval[j,k,l] where 1 <= j <= 2 and 0 <= k <= [2, 2] and 1 <= l <= [1, 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 <= 2 ) j stands for j_th alpha * COMPATIBILITY CONDITION(S) (type compcond[j,k] where 1 <= j <= 2 and 1 <= k <= [1, 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 <= 2 and 1 <= k <= [1, 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() > ---------------------------------------------------------------- (C17) /* ************************** END of NP_EXEC.MAX ************************** */ (C18) closefile();