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Re: [ff3d-users] trivial solution

From: Stephane Del Pino
Subject: Re: [ff3d-users] trivial solution
Date: Mon, 7 Mar 2005 22:42:14 +0100
User-agent: KMail/1.7.2

Le Lundi 7 Mars 2005 20:38, Neil Christensen a écrit :
> Hi,
>     I am trying solve the pde:
>     div(grad(phi))  + phi = 0
> with the boundary condition
>     phi = 0 on <1,0,0>.
> As you know the analytic solution to this is
>     phi = sum_i  a_i cos(k_i . r) + b_i sin(k_i . r)
> where a_i, b_i and k_i are chosen such that the boundary condition is
> satisfied.  In my first attempt, the boundary I used was a square.  The
> problem is that when I run ff3d on it, I only get the trivial solution
> phi = 0.  How can I get other solutions?
I just think you can't: if I am right
      div(grad(phi))  + phi = f in omega
      phi = g on the border
is a hilled posed problem in the sense that their is an infinite number of 
solutions to this problem. For instance, let us assume that 
 f = 1, g = 1 and omega = ]0,1[
Let us denote phi_1 a solution of this problem. Now
 phi_0 = sum_i a_i sin(i*pi*x)
satisfies for all reals numbers a_i
        div(grad(phi_0))  + phi_0 = 0 in ]0,1[,
   with phi_0 = 0 at 0 and 1
So, by linearity phi_0+phi_1 is also solution of the first problem.

I think you cannot solve this kind of problem ... uniqueness is often 
mandatory to solve problems.

If you meant solving
     -div(grad(phi))  + phi = 0
it is different, since then, the problem is well posed: it has a unique 
Since 0 is a solution, it is *the* solution.

I hope this answers your question,
Best regards,

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