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Re: solving a PDE

• To: mathgroup at smc.vnet.net
• Subject: [mg40716] Re: [mg40711] solving a PDE
• From: Dr Bob <majort at cox-internet.com>
• Date: Tue, 15 Apr 2003 03:56:35 -0400 (EDT)
• References: <200304140806.EAA06435@smc.vnet.net>
• Reply-to: majort at cox-internet.com
• Sender: owner-wri-mathgroup at wolfram.com

pde = (1 - M^2)D[phi[x, y], x, x] + D[phi[x, y], y, y] == 0;
bc1 = Derivative[0, 1][phi][x, (d/2)] == U k A Cos[kx];
bc2 = Derivative[0, 1][phi][x, (-d/2)] == U k A Cos(kx);
eqns = {pde, bc1, bc2};
DSolve[eqns, phi, {x, y}]

{{phi -> Function[{x, y}, C[1][(Sqrt[-1 + M^2]*x)/
         (1 - M^2) + y] + C[2][-((Sqrt[-1 + M^2]*
          x)/(1 - M^2)) + y]]}}

or

phi[x, y] /. DSolve[eqns, phi, {x, y}] // First

Bobby

On Mon, 14 Apr 2003 04:06:37 -0400 (EDT), C B <cbhat at purdue.edu> wrote:

>
> I am trying to solve the following PDE
> (1-M^2)D[phi[x,y],x,x]+ D[phi[x,y],y,y] = 0
>
> With the following BC's
> BC1: Derivative[0,1][phi][x,(d/2)] = U k A Cos(kx)
> BC2: Derivative[0,1][phi][x,(-d/2)] = U k A Cos(kx)
>
> How do I get Mathematica to do it?
>
>
>
>
>
>



-- 
majort at cox-internet.com
Bobby R. Treat

• References:
  • solving a PDE
    • From: C B <cbhat@purdue.edu>