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