Goal 1: Verify that the Laplacian is invariant under rotation and translation (we will do this for \mathbb{R}^2).

Let f: \mathbb{R}^2 \rightarrow \mathbb{R} be arbitrary. We know show that:

\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}, where

u(x, y) = xcos(\theta) - ysin(\theta) and

v(x, y) = xsin(\theta) + xcos(\theta).

We use the chain rule. It follows that:

\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}

= \frac{\partial}{\partial x}(\frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}) + \frac{\partial}{\partial y}(\frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y})

= \frac{\partial}{\partial x}(\frac{\partial f}{\partial u}cos(\theta) + \frac{\partial f}{\partial v}sin(\theta)) + \frac{\partial}{\partial y}(-\frac{\partial f}{\partial u}sin(\theta) + \frac{\partial f}{\partial v}cos(\theta))

= \frac{\partial (\frac{\partial f}{\partial u}cos(\theta) + \frac{\partial f}{\partial v}sin(\theta))}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial(\frac{\partial f}{\partial u}cos(\theta) + \frac{\partial f}{\partial v}sin(\theta))}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial(-\frac{\partial f}{\partial u}sin(\theta) + \frac{\partial f}{\partial v}cos(\theta))}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial((-\frac{\partial f}{\partial u}sin(\theta) + \frac{\partial f}{\partial v}cos(\theta))}{\partial v}\frac{\partial v}{\partial y}

= \frac{\partial^2 f}{\partial u^2}(cos^2(\theta) + sin^2(\theta)) + \frac{\partial^2 f}{\partial v \partial u}cos(\theta)sin(\theta) + \frac{\partial^2 f}{\partial u \partial v}cos(\theta)sin(\theta) + \frac{\partial^2 f}{\partial v^2}(cos^2(\theta) + sin^2(\theta)) - (\frac{\partial^2 f}{\partial v \partial u}cos(\theta)sin(\theta) + \frac{\partial^2 f}{\partial u \partial v}cos(\theta)sin(\theta))

= \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}, which is what we want.

We can use similar method to show that the Laplacian is invariant under translation (I’m too lazy to write the rest out).

Goal 2: What do the spherically symmetric harmonic functions look like in \mathbb{R}^2, \mathbb{R}^3?

We know that in \mathbb{R}^2, f(r) = log(r) is a spherically symmetric harmonic function and in \mathbb{R}^3, f(r) = \frac{1}{r} is a spherically symmetric harmonic functions. Are there others?

We will solve the Laplacian equation in polar coordinates to figure out.

In \mathbb{R}^2, a spherically symmetric harmonic function would satisfy:

f_{\theta} = 0 and

f_{rr} + \frac{1}{r}f_{r} + \frac{1}{r^2}f_{\theta\theta} = 0.

This reduces to:

f_{rr} + \frac{1}{r}f_{r} = 0 (a second-order ODE).

Solving this gives:

f(r) = C_1log(r) + C_2, where C_1, C_2 \in \mathbb{R}. This is what spherically symmetric harmonics in \mathbb{R}^2 look like.

What about in \mathbb{R}^3?

They would need to satisfy:

f_{\phi} = f_{\theta} = 0 and

\frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial f}{\partial r}) + \frac{1}{r^2sin^2(\phi)}\frac{\partial^2 f}{\partial \theta^2} + \frac{1}{r^2 sin(\phi)}\frac{\partial}{\partial \phi}(sin(\phi)\frac{\partial}{\partial \phi}) = 0.

This simplifies to:

\frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial f}{\partial r}) = 0

\Leftrightarrow

r^2f'' + 2rf' = 0 (this becomes a 2nd-order ODE).

\Leftrightarrow

rF' + 2F = 0 (where F = f')

\Leftrightarrow

\frac{1}{F}dF = -\frac{2}{r}dr

\Leftrightarrow

log(F) = -2log(r) + C_1

\Leftrightarrow

F = C_1\frac{1}{r^2}

\Leftrightarrow

f'(r) = C_1\frac{1}{r^2}

\Leftrightarrow

f(r) = C_1\frac{1}{r} + C_2 (this is what all spherically symmetric harmonic functions in \mathbb{R}^3 look like).

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