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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
solution.
Since 0 is a solution, it is *the* solution.
I hope this answers your question,
Best regards,
Stephane.