A Variation on Laplace's equation (context: Yang-Mills N-Instantons, Rajaraman's book)

Question

Statement of the problem

I need to solve the equation

\begin{align} 0 = \frac{1}{\phi} \partial_{\sigma}\partial_{\sigma} \phi \hspace{20mm} (1) \end{align}

where $\phi$ is a scalar field and we're working in Euclidean 4-space. More specific questions are asked below. Basic advice on this would also be appreciated.

Context

I'm trying to work through 't Hooft's Euclidean Yang-Mills N-instanton solution. The idea is that we make the ansatz for the gauge potentials

\begin{align} A_{\mu} = i \overline{\sigma}_{\mu \nu} \partial^{\nu}(\ln \phi(x)) \end{align}

where $\overline{\sigma}_{\mu \nu}$ can be thought of as a constant 4x4 matrix that is unimportant to my question. We use the self-duality of the field strength tensor that arises from this ansatz to obtain the equation (1) for $\phi$

The Difficulties

One generally good source has been Rajaraman's "Solitons and Instantons." The relevant section is displayed

My difficulties are:

1. I am assuming by "singular" Rajaraman means that the function contains singularities. Why, explicitly, does $\phi$ being non-singular imply that we we need only solve $\Box \phi = 0$? Surely if $\phi$ contains any zeros then $1/\phi$ raises issues?
2. Why does $\Box \phi = 0$ only permit the constant solution and not say $\phi(x) = a_{\mu} x^{\mu} + b$ where $a_{\mu}$ and $b$ are constants? Is this because of the form of the ansatz, or is it something I'm missing?
3. In the case where $\phi$ is allowed to be singular, why does $\phi = 1/|x|^2$ solve the equation (1) for $x \neq 0$? Surely we have

\begin{align} \frac{1}{\phi} \partial_{\sigma}\partial_{\sigma} \phi &= |x|^2 \partial_{\sigma}\partial_{\sigma} \frac{1}{|x|^2}\\ &= |x|^2 \partial_{\sigma}\bigg(-\frac{2 x_{\sigma}}{|x|^4}\bigg)\\ &=|x|^2\partial_{\sigma} (\frac{-2 x_{\sigma}}{|x|^4})\\ &= |x|^2\partial_{\sigma} (\frac{-2 x_{\sigma}}{(\sum_{i} x_i^2)^2})\\ &= |x|^2\frac{-2}{(\sum_i x_i^2)^2} + |x|^2\frac{8 x_{\sigma}^2}{(\sum_i x_i^2)^3}\\ &=\frac{-2}{|x|^2} + \frac{8 x_{\sigma}^2}{|x|^4}\\ &\neq 0 \end{align}

(edit) I see where I went wrong above, the last three lines should be

\begin{align} &= \sum_{\sigma}(|x|^2\frac{-2}{(\sum_i x_i^2)^2}) + |x|^2\frac{8 x_{\sigma}^2}{(\sum_i x_i^2)^3}\\ &=\frac{-8}{|x|^2} + \frac{8 |x|^2}{|x|^4}\\ &= 0 \end{align}

Answer 1

1) If $\phi$ is non-singular, you can safely multiply both sides by $\phi$ and get $\square \phi = \phi*0 = 0 $. If $\phi$ is singular you can't do this because $\phi * 0$ is undefined. Equivalently, where $1/\phi=0$ you can have $\square\phi$ nonzero, as shown in (4.64). The singularities in $\phi$ will map to zeroes in $1/\phi$.

2) I believe only the linear solution you mentioned is allowed, which results in unbounded field values at infinity. I assume that's reason enough to reject it.

3) I believe your math is incorrect. $\partial_\sigma (x_\sigma / |x|^4)$ is zero. In 4 dimensions, you should get two equal terms with opposite signs.