PDE solver problems

*To*: mathgroup at smc.vnet.net*Subject*: [mg19711] PDE solver problems*From*: Arnold Gregory Civ AFRL/SNAT <Gregory.Arnold at sn.wpafb.af.mil>*Date*: Sat, 11 Sep 1999 16:36:12 -0400*Sender*: owner-wri-mathgroup at wolfram.com

I'm having some problems with the PDE solver in Mathematica & I wanted to know if anybody else has experienced the same difficulties and/or worked around them. For the equation: eq=(-(u2*Y1*Y2) + u1*(X2^2 + Y2^2))*D[Phi[u1,u2],u2] + Y1*(-(u2*Y1) + u1*Y2)* D[Phi[u1, u2],u1] Mathematica (v4.0.0), Dsolve[eq,Phi[u1,u2],{u1,u2}] returns: {{Phi[u1, u2] -> C[1][1/(u1^2*(X2^2 + u2^2*Y1^2 - 2*u2*Y1*Y2 + Y2^2))]}} This solution is close, but it does not satisfy the PDE when I check it. The solution I know is: Phi[u1, u2] -> C[1][(u2 - X2)*(u2 + X2)*Y1^2 - 2*u1*u2*Y1*Y2 + u1^2*(X2^2 + Y2^2)] Of course, I'm trying to do similar problems for which I don't know the answer & now I'm worried that Mathematica will not be able to help me out. Anybody have a suggestion for what I might be doing wrong or how to work around? I've also noticed that Mathematica will occasionally add extraneous constants to solutions of PDE's (or unnecessarily invert them as in the example above). For example, if the 'simplest' solution is Phi[u,v]->C[1][u^2+v^2], it might instead return Phi[u,v]->C[1][c*(u^2+v^2)], where "c" is a constant in my original PDE. I'll have to dig out some true examples of where this occurred... has anybody else noticed this and perhaps developed a workaround? Finally, Mathematica seems to have trouble solving the following form of equations: mf = Phi[x1, y1, x2, y2]; eq = -y1*D[mf, x1] + x1*D[mf, y1] - y2*D[mf, x2] + x2*D[mf, y2]; Dsolve[ eq,mf,Apply[List,mf]] It complains about "Inconsistent or Redundant Transcendental equations". But I know the solution is of the form: Phi[x1, y1, x2, y2] -> C[1][x1^2 + y1^2, -x2*y1+x1*y2, x1*x2+y1*y2] (x2^2+y2^2 is obviously a valid solution, too, but it can be derived from these solutions) Anybody have suggestions for solutions of this form? Thanks for your time! Greg