Solving algebraic equations with some conditions
- To: mathgroup at smc.vnet.net
- Subject: [mg92007] Solving algebraic equations with some conditions
- From: Kay-Michael Voit <kay at voits.net>
- Date: Tue, 16 Sep 2008 19:26:36 -0400 (EDT)
Hello,
I am trying to solve a system of 3 PDEs using a Fourier ansatz:
nL[x_, t_] =
Sum[cn[p, l] Exp[I (p k x + l \[Omega] t)], {p, -1, 1}, {l, -1, 1}];
EscL[x_, t_] =
Sum[cE[p, l] Exp[I (p k x + l \[Omega] t)], {p, -1, 1}, {l, -1, 1}];
iL[t_] = Sum[ci[l] Exp[I l \[Omega] t], {l, -1, 1}];
So I have a system of up to 55 algebraic equations (Not every Exp[I (p k
x + l \[Omega] t)] occures after inserting the ansatz, thus some will be
trivial) for the 55 Fourier coefficient. The equations are not linear,
but linear in every coefficient. I was told that this would take some
time but should be possible.
Additionally, I know that the functions are all real, thus I get more
equations:
K[p,l]=Conjugate[K[p,l]]
Unfortunately, I have the impression that Mathematica cannot handle
these in Solve[], as simple examples produce things like Conjugate^-1[I].
A possibility I considered is constructing the coefficients with real
and imaginary part:
iL[t_]
= Sum[(ci[l, re] + ci[l, im] I) Exp[I l \[Omega] t], {l, -1, 1}];
Then I had to give Mathematica assumtions that these are all real so
that an expression like
-H Wges+(I cn[0,0,im]+cn[0,0,re])/\[Tau]==0
will be correktly solved to
{cn[0,0,im] -> 0, cn[0,0,re]->\[Tau] H Wges}
(also assuming that \[Tau], H, Wges are real)
I am not sure how to achieve that.
Indeed I have another condition for the coefficients:
K[p,l][\[omega]]=K[p,-l]][-\[omega]]
In this case, I have completely no idea how to model this in Mathematica.
Can anybody give me a hint (on the problem or on literature) how to
solve this problem?
Thanks in advance
Kay-Michael Voit