Re: Need your help to solve a PDE.

*To*: mathgroup at smc.vnet.net*Subject*: [mg53134] Re: Need your help to solve a PDE.*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Fri, 24 Dec 2004 05:59:40 -0500 (EST)*Organization*: The University of Western Australia*References*: <cpmhbv$o8j$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

In article <cpmhbv$o8j$1 at smc.vnet.net>, "Satya Das" <satyaranjandas77 at yahoo.com> wrote: > Hi All, > I need to solve an equation. It would be a great help if you solve this. > The problem is mentioned at > http://www.geocities.com/satyaranjandas77/PDE.pdf. I had to go this way > because I was not able to send an attachment. You could have appended a (small) Notebook to your message. > I had made many attempts myself, but since it is more than 5 years I am out > of college I need to relearn how to solve PDE. :( > Is it possible to solve this equation using mathmatica? Not using built-in functions. However, if you can solve the equation by hand then, of course you can use Mathematica to expedite the solution. > Even the power series solution will work just fine for me. > > Thanks in advance, > satya > > --- Below are some more related info ---- > > I did try Phi(r, theta) = F(r)G(theta), but this did not work. A superposition of separable solutions will wokr. > I had tried the followings too: > Phi(r, theta) = (1/r) Psi(r, theta) -> to reduce the equation to look more > like solvable. > Phi(r, theta) = Psi(r sin(theta)/r_0, theta) -> to make the equation > dimensionless, and from the equation it seems like r sin(theta) is more > natural. > Phi(r, theta) = Psi(r_0 sin(theta)/r, theta) -> here r and r_0 have changed > their position. > > I did not try power series because the equation is in two variables. > Boundary condition is that Phi vanishes as r->oo. > Other condition on Phi is that it is symmetric about theta = pi/2 => Phi(r, > theta) = Phi(r, pi - theta). > > I even tried a trial function: > Phi(r, theta) = (r /(r_0^2 sin^2 theta)) (1 - exp(-r^2/(r_0^2 sin^2 > theta))), but this does not satisfy the equation. Nor does it satisfy your boundary condition as r -> Infinity. > You may ignore my attempts because I am not too sure if I was doing the > right thing. > I realized that my mathematical capabilities have been deteriorated during > last five years. Changing variables to polar coordinates Phi[r, Theta] -> Psi[x, y] where {x,y}=={r Cos[Theta], r Sin[Theta]} looks promising: ((r0^2 + y^2) Derivative[0,1][Psi][x, y] + y (y^2 - r0^2) * (Derivative[0,2][Psi][x,y]+Derivative[2,0][Psi][x,y]))/(r0^2 y) == 0 This is separable as Psi[x, y] -> X[x] Y[y] with separation constant c^2 > 0. The solution for X[x], valid for large positive x, is straightforward. X[x] = Exp[-c x] (really x is Abs[x] and Abs[Cos[Theta]] satisfies your symmetry condition). The solution for Y[y] is more complicated but for large y (y >> r0) is asymptotically like BesselJ[0, c y] (assuming a finite solution for y->0). Hope this helps. Cheers, Paul -- Paul Abbott Phone: +61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) 35 Stirling Highway Crawley WA 6009 mailto:paul at physics.uwa.edu.au AUSTRALIA http://physics.uwa.edu.au/~paul

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