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MathGroup Archive 2004

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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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