Solving Poisson-Boltzmann equation in Cylindrical and Spherical Coordinates

[thepopasmurf]

Homework Statement

I don't have a specific problem in mind, it's more that I forgot how to solve the particular equation from first principles.

$\nabla^{2} \Phi = k^{2}\Phi$

Places I've looked so far have just quoted the results but I would like the complete method or the appropriate substitution.

Homework Equations

The relevant equations would be the definition of the Laplacian operator in cylindrical or spherical coordinates.

The Attempt at a Solution

My attempt was to sub in

$\Phi = r^a \times exp(br)$

and solve for a and b, but I got both a and b equal to plus and minus 1 without a way to eliminate the postitive a solution.

 

[Mute]

Well, the usual 'first principles' way would be to assume a solution of the form $\Phi(\mathbf{r}) = A(\rho)B(\phi)C(z)$ in cylindrical coordinates or $\Phi(\mathbf{r}) = R(r)\Theta(\theta)G(\phi)$ in spherical coordinates. If you then divide the entire equation by $\Phi$ again, you can rearrange terms until one side of the equation depends only on one of the three variables, implying it must be a constant, which the other side of the equation is then also equal to.

For a very simple example, say you had the 2d problem $\nabla^2 \Phi(x,y) = 0$, and you let $\Phi(x,y) = X(x)Y(y)$. Plugging this in and dividing by $\Phi$ again gives $\partial_x X(x)/X(x) + \partial_y Y(y)/Y(y) = 0$, or $\partial_x X(x)/X(x) = -\partial_y Y(y)/Y(y)$, and hence it must be the case that $\partial_x X(x)/X(x) = k$ and $-\partial_y Y(y)/Y(y) = k$, where k is some constant (not your equation's k, in this example!).

Also, keep in mind you will need some boundary conditions for the problem. Based on your boundary conditions, you may be able to argue that the solution should be spherically or cylindrically symmetric, so you can through out the angular or height dependences, but if you want an absolutely generally solution, you have to keep those terms as you could have boundaries conditions which vary with angle or height.