Re: Help on mathieu equation needed

*Subject*: [mg2197] Re: [mg2141] Help on mathieu equation needed*From*: Ralheid at aol.com*Date*: Fri, 13 Oct 1995 06:25:44 GMT*Approved*: usenet@wri.com*Distribution*: local*Newsgroups*: wri.mathgroup*Organization*: Wolfram Research, Inc.*Sender*: daemon at wri.com ( )

-ernst wallenborn.writes >I have a 2nd order nonlinear diff. equation, which looks like d^2w > ----- + (a - 2q cos 3v) w = 0 ,(0<=v<=2 Pi) dv^2 > i.e. very similar to the mathieu equation (e.g. > Abramowitz&Stegun 20.1.1) > except that cos2v is replaced by cos3v. How can i solve this in > Mathematica? You can look at the following Eqn = w''[v] + w (a - 2q Cos[3 v] ) == 0; Eqn = w''[v]/w[v] == - (a - 2q Cos[3 v] ); Homogenous Solution: eqn0 = w0''[v]/w0[v] == 0; Flatten[ DSolve[eqn0,w0[v],v] ] w0[v_] = w0[v] /. % C[1] + v*C[2] Constant Solution: eqn1 = w1''[v]/w1[v] == - a Flatten[ DSolve[%,w1[v],v] ] w1[v_] = w1[v] /. % E^(-I*a^(1/2)*v)*C[1] + E^(I*a^(1/2)*v)*C[2] Fourier Solution: eqn2 = w2''[v]/w2[v] == c[n] E^(I n v) Flatten[ DSolve[eqn2,w2[v],v] ] w2[v_,n_] = w2[v] /. % BesselJ[0, 2*(n^(-2))^(1/2)*(E^(I*n*v)*c[n])^ (1/2)]*C[1] + 2*BesselK[0,2*(-n^(-2))^(1/2)*(E^(I*n*v)*c[n])^ (1/2)]*C[2] Only two terms of the fourier solution show up the n=-3 and n=3 terms w2[v,-3], and w2[v, 3]. Full Solution solution[v_]= w0[v] + w1[v] + w2[v,-3] + w2[v,3] Taking out dummy constants C[1] etc. and replacing c[3] & c[-3] with q, the constants of the 3rd term in the fourier series we have: test[v_]= c1 + E^(-I*a^(1/2)*v)*c2 + v*c3 + E^( I*a^(1/2)*v)*c4 + BesselJ[0, (2*(E^(-3*I*v)*q)^ (1/2))/3]*c5 + BesselJ[0, (2*(E^( 3*I*v)*q)^ (1/2))/3]*c6 + 2*BesselK[0, (2*I)/3*(E^(-3*I*v)*q)^ (1/2)]*c7 + 2*BesselK[0, (2*I)/3*(E^( 3*I*v)*q)^ (1/2)]*c8 I cannot prove that the above is a correct solution. or a solution at all. There are obvious questions of the validity of these superpositions, and the use of the -3 term in the evaluation of the fourier solution may cause some problems. Further work is possible to extract REAL terms from the above.