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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
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