Finding Solution to Laplace Equation

[Nick R]

Usually, we use the technique of "separation of variables" as follows:

In a "separable coordinate system", we assume a separable solution

\Phi=A(a)B(b)C(c)

Then we obtain 3 ODEs for A(a), B(b), C(c)

We note that there are actually entire families of solutions to each ODE, that happen to be complete/orthonormal. So we can write A(a), B(b), C(c) as linear combinations of these guys:

A(a)=\sum E_n A_n(a)
B(b)=\sum F_n B_n(b)
C(c)=\sum G_n C_n(c)

Then we use the boundary conditions and exploit the orthogonality of A_n, B_n, C_n to find the coefficents E_n, F_n, G_n. This involves integrating at the boundaries.

So we get

\Phi=\sum_i\sum_j\sum_k D_{ijk} A_i (a)B_j (b)C_k (c)

Where

D_{ijk} = E_i F_j G_k

OK. Easy.

NOW HERE COMES THE QUESTION

WHY can't I expand \Phi in some other set of functions?

In general, I can expand any 3 variable function like so

\sum_n \sum_m \sum_l A_{nml} \zeta_n (x) \nu_m (y) \mu_l (z)

Where the coefficients are

\int \int \int \Phi \zeta_n^* (x) \nu_m^* (y) \mu_l^* (z) dx dy dz

Well, I suppose I answered my own question... You have to KNOW \Phi to find the coefficients in this general case: but the technique of separation of variables allows us to find the coefficients without knowing \Phi in the first place...

And the other way to do it would be to find a set of orthogonal functions that are eigenfunctions of the Laplace operator... Then we don't need to know \Phi apriori to find the coefficients either - and this works for finding solutions to Poisson's equation as well (which is good because we can't use separation of variables there).

Because I spent so long thinking about this, I am going to press "submit" for the heck of it and see if I get any interesting comments. This was motivated by wondering WHY separation of variables actually works (without a doubt) and the answer seems to be, because it is really equivalent to expanding \Phi in a complete set of functions (a very specific set of complete functions - so that we can actually find the coefficients in the linear combination).

 

[Nick R]

OK so yesterday I somehow lost sight of how you actually do separation of variables when I typed out the above.

Yesterday I said we assume

\Phi = \Phi=A(a)B(b)C(c)

Where

A(a)=\sum E_n A_n(a)
B(b)=\sum F_m B_m(b)
C(c)=\sum G_l C_l(c)

And then find coefficients using boundary conditions/orthogonality then write the answer as a triple sum.

No!

Actually, we find

\phi_{nm} = A_n(a)B_m(b)C_{nm}(c)

and represent \Phi as a double sum of these two index "separable solutions"

\Phi = \sum_n\sum_m D_{nm}\phi_{nm}

And use orthogonality/boundary conditions to find the coefficients D_{nm}

This is distinctly different from the general expansion for a function of three variables. I.e. doing separation of variables and doing an expansion of a function in orthogonal functions are totally different things.

So apparently, the only way to know that separation of variables works is, if you find a solution that satisfies the boundaries, and that obeys \nabla^2\Phi = 0 in the region under consideration, then that's the only possible solution...

Gosh what a waste of time.