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Re: Help on mathieu equation needed

  • To: mathgroup at smc.vnet.net
  • Subject: [mg2197] Re: [mg2141] Help on mathieu equation needed
  • From: Ralheid at aol.com
  • Date: Fri, 13 Oct 1995 02:25:44 -0400

-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. 



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