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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.
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