Laplace Equation in Cartesian Coor.

1. Dec 8, 2011

Apteronotus

Solving the Laplace equation in Cartesian Coordinates leads to the 2nd order ODEs:
$\frac{X''}{X}=k_1, \qquad \frac{Y''}{Y}=k_2 \qquad \frac{Z''}{Z}=k_3$
In each case the sign of $k_i$ will determine if the solution (to the particular ODE) is harmonic or not.

Hence, if two people solve the Laplace equation, one may get a solution that is harmonic in $x$, and the other harmonic in $z$.

How could this be?
Does this simply mean that both solutions satisfy the Laplace equation, and hence the solution is not unique?

If we have non-uniqueness then when considering an actual physical situation (such as the potential of an electric field) how can we be sure that the solution we get is the true solution?

Thanks

2. Dec 8, 2011

dextercioby

Because PDEs (or even ODEs) are always accompanied by boundary/limit conditions which restrict (even really heavily) the dimension of the space of solutions. Giving no boundary condition for the solution means a great deal of arbitrary in the shape of the solutions.

3. Dec 17, 2011

Apteronotus

So if I have the BC then how do I know which of the 6 (i think) possible solutions is the actual solution?
Trial and error?

4. Dec 29, 2011

meldraft

Your boundary condition will specify which one you have to choose. Try solving the equation in a 3D domain where you specify the BCs ;)

5. Jan 14, 2012

Apteronotus

Thank you, both!