\section{Regular Singular Point Analysis}
In \S 5 we demonstrated the existence of symmetric monopoles for arbitrary
$\lambda \ge 0$ and $\epsilon > 0$. In this section we explore their
form near the regular singular point of the equations \eqref{ode},
namely $r=0$.
\begin{theorem}
Let $(\gamma, \varphi)$ be a bounded finite-action
solution to the ODE system \eqref{ode}
for some fixed $\epsilon >0$ and $\lambda \ge 0$. Then there exist
constants $a_1$ and $b_2$ such that
\begin{equation}
\begin{split}
\varphi(r) = & a_1 r + O(r^3); \\
\gamma(r) = & b_2 r^2 + O(r^4); \\
\gamma'(r) = & 2 b_2 r + O(r^3);
\end{split}
\end{equation}
near $r=0$. In particular, $\varphi(0)=\gamma'(0)=0$.
\end{theorem}
\begin{proof}
We begin with the equation for $\varphi$, which we write as
\begin{equation}
\label{phieq}
\varphi'' + \frac{2\varphi'}{r} - \frac{2 \varphi}{r^2}
= \frac{8 \varphi}{r^2} (\gamma + \gamma^2) + 2 \lambda \varphi (\varphi^2-1).
\end{equation}
Since $\varphi$ is bounded and $\gamma(0)=0$, the terms on the right hand side
are less singular than those on the left hand side, and to leading order
$\varphi$ resembles the solution to the homogeneous equation
\begin{equation}
\varphi'' + \frac{2\varphi'}{r} - \frac{2 \varphi}{r^2} = 0.
\end{equation}
The general solution to this equation is $\varphi = a_1 r + a_{-2}r^{-2}$.
However, $\varphi$ is bounded by assumption, so $a_{-2}=0$.
Thus the solution to \eqref{phieq}
is, to leading order, $a_1 r$.
Next we turn to the equation for $\gamma$, namely
\begin{equation}
\label{gammaeq}
\gamma'' - \frac{2 \gamma}{r^2} = \frac{2}{\epsilon} \varphi^2 (1+2\gamma)
+ \frac{2\gamma}{r^2}(2 \gamma^2 + 3\gamma).
\end{equation}
Again, since $\varphi$ is bounded and $\gamma(0)=0$, this may be viewed as
a perturbation of the homogeneous linear equation
\begin{equation}
\gamma'' - \frac{2 \gamma}{r^2} = 0,
\end{equation}
whose solution is $\gamma = b_{-1} r^{-1} + b_2 r^2$. Since $\gamma$ is
bounded, $b_{-1}=0$. Thus our solution to \eqref{gammaeq} is, to leading
order, $b_2 r^2$.
With these basic results, we can estimate the right hand sides of
\eqref{phieq} and \eqref{gammaeq}. The right hand side of \eqref{phieq} is
$O(r)$, which gives an $O(r)$ correction to $\varphi''$, hence an $O(r^3)$
correction to $\varphi$. The right hand side of \eqref{gammaeq} is
$O(r^2)$, thus giving an $O(r^3)$ correction to $\gamma'$ and an
$O(r^4)$ correction to $\gamma$.
\end{proof}
One can do an expansion for $\varphi$ and $\gamma$ in powers of $r$.
Indeed, only odd powers contribute to $\varphi$ and only even powers
contribute to $\gamma$. This is seen by induction. By Theorem 6.1,
$\varphi$ is odd and $\gamma$ is even through order $r^2$. However, if
$\varphi$ is odd and $\gamma$ is even through order $r^k$, then the
right hand sides of \eqref{phieq} and \eqref{gammaeq} are odd and even,
respectively, to order $r^k$, and so $\varphi$ and $\gamma$ are odd
and even, respectively, to order $r^{k+2}$. Thus $\varphi$ and $\gamma$
are odd and even to all orders in $r$.
We can therefore write an asymptotic expansion
\begin{equation} \label{powers}
\begin{split}
\varphi(r) & \sim \sum_{n\hbox{\small{} odd}} a_n r^n, \\
\gamma(r) & \sim \sum_{n\hbox{\small{} even}} b_n r^n.
\end{split}
\end{equation}
Plugging this expansion into equations \eqref{phieq} and \eqref{gammaeq}
and equating coefficients of $r^{n-2}$ gives recursion relations of the form
\begin{equation}\label{recursion}
\begin{split}
(2k)(2k+3) a_{2k+1} = & \hbox{ algebraic expression involving }
a_1, b_2, \ldots,b_{2k}, \\
(2k+1)(2k-2) b_{2k} = & \hbox{ algebraic expression involving }
a_1, b_2, \ldots,b_{2k-2}.
\end{split}
\end{equation}
These relations do not constrain $a_1$ or $b_2$.
However, once we have $a_1$ and $b_2$, the remaining coefficients
are determined. The first few are
\begin{equation}
\begin{split}
a_3 = & (4 a_1 b_2 - \lambda a_1)/5; \\
b_4 = & \left ( 3 b_2^2 + \epsilon^{-1} a_1^2 \right )/5; \\
a_5 = & (4 a_1 b_4 + 4 a_3 b_2 + 4 a_1 b_2^2 + \lambda(a_1^3 - a_3))/14; \\
b_6 = & \left ( b_2^3 + 3 b_2 b_4 + \epsilon^{-1}
(a_1 a_3 + a_1^2 b_2)
\right ) / 7; \\
a_7 = & [4( a_1 b_6 + a_3 b_4 + a_5 b_2 + a_3 b_2^2 + 2 a_1 b_2 b_4)
+ \lambda(3 a_1^2 a_3 - a_5)]/27; \\
b_8 = & \left [ 3(2 b_2^2b_4 + b_4^2 + 2 b_2 b_6) + \epsilon^{-1}
(a_3^2 + 2 a_1 a_5
+ 2 a_1^2 b_4 + 4 a_1 a_3 b_2 ) \right] / 27; \\
a_9 = & \big [4(a_1 b_8 + a_3 b_6 + a_5 b_4 + a_7 b_2 + a_5 b_2^2
+ 2 a_3 b_2 b_4 + a_1 b_4^2 + 2 a_1 b_2 b_6) \\
& + \lambda(3 a_1^2 a_5 + 3 a_1 a_3^2 - a_7)\big ]/44; \\
b_{10} = & \big [ 3(b_2^2 b_6 + b_2 b_4^2 + b_2 b_8 + b_4 b_6) \\
& + \epsilon^{-1} (a_1 a_7 + a_3 a_5 + a_1^2 b_6 + a_3^2 b_2
+ 2 a_1a_3b_4 + 2 a_1a_5b_2)
\big ] / 22.
\end{split}
\end{equation}
\section{Symmetries and Stability}
%{\rm This section probably does not belong in the paper. However, I
%thought I'd include these results for our own reference -- Lorenzo}
The action functional and the resulting Euler-Lagrange equations
are invariant under two natural symmetries.
\begin{eqnarray}
\label{phisym}
\varphi(r) \to - \varphi(r) & \qquad & \gamma(r) \to + \gamma(r); \\
\label{gammasym}
\varphi(r) \to \varphi(r) & \qquad & \gamma(r) \to -1 - \gamma(r).
\end{eqnarray}
The first symmetry \eqref{phisym}
comes from the isometry $\vec x \to - \vec x$ of $B^3$,
which of course respects rotational symmetry.
Since $\vec \sigma \cdot \vec x$ is odd and $\vec \sigma \cdot (\vec x \times
\vec {dx})$ is even, pulling the pair $(A,\phi)$ back by this isometry flips
the sign of $\varphi$ while preserving $\gamma$. Using this symmetry, we
can fix the sign of $\varphi(1)$, which we henceforth take to be positive.
The second symmetry \eqref{gammasym}
is a gauge transformation by $(\vec\sigma \cdot \vec x)/r$.
This is of the form of (4.2), with $\theta = \pi/2$. From (4.4) it is
clear that this transformation sends $\gamma$ to $-1-\gamma$ without generating
a $\beta$ term or changing $\varphi$. Applying this to a connection with
$\gamma(0)=0$ yields a new connection with $\gamma(0)=-1$. This connection
has finite action but is singular at the origin, reflecting the singular
gauge transformation that generated it.
We now consider stability properties of the ODE system \eqref{ode}.
These ODEs have several fixed points, namely
\begin{equation}
(\gamma,\varphi) = \hbox{ $( -\frac{1}{2}, 1 )$, $(-\frac{1}{2}, -1)$,
$(-\frac{1}{2}, 0 )$, $(0,0)$ or $(-1,0)$}.
\end{equation}
The point $\left (-\frac{1}{2}, -1 \right )$ is related to
$\left ( -\frac{1}{2}, 1 \right)$ by the symmetry \eqref{phisym}, while
$(-1,0)$ is related to $(0,0)$ by \eqref{gammasym}, so we do not need to
study these. What remains is $\left ( -\frac{1}{2}, 1 \right)$,
$\left (-\frac{1}{2}, 0 \right)$, and (0,0).
For the $\gamma = -1/2$ fixed points, we define $\delta = \gamma + 1/2$,
and the equation for $\gamma$ becomes
\begin{equation}
\delta'' = 4 \delta \left ( \frac{\varphi^2}{\epsilon}
- \frac{1}{4 r^2} + \frac{\delta^2}{r^2} \right ).
\end{equation}
The fixed point $\gamma = -1/2$ is stable for $\varphi=1$ when $r^2 <
\epsilon/4$, but is unstable if $r^2 > \epsilon/4$. This defines a
natural length scale to the problem, namely $\sqrt{\epsilon}/2$. We
should expect our solutions to behave qualitatively differently for
$r$ less than or greater than this length scale. Of course, if
$\epsilon>4$, then all radii $r$ are less than this length scale. In
the case of $\varphi=0$, the value $\gamma=1/2$ is always stable,
regardless of $\epsilon$ or $r$.
Near $\gamma=-1/2$, the equation for $\varphi$ becomes
\begin{equation}
\varphi'' + 2 \frac{\varphi'}{r} = 2 \varphi \left (
\frac{4 \delta^2}{r^2} + \lambda(\varphi^2-1) \right ).
\end{equation}
The behavior of this fixed point depends on the value of $\varphi$.
Near $\varphi=0$ we have
$$ (r\varphi)'' = -2\lambda r\varphi + O(\delta^2) + O(\varphi^2),$$
which is stable for all positive values of $r$.
Near $\varphi=1$, however, we write $\varphi = 1 + \zeta$ and have
$$ (r\zeta)'' = 4 \lambda (r \zeta) + O(\zeta^2)
+ O(\delta^2).
$$
This is unstable as long as $\lambda >0$, and has natural length
scale $1/\sqrt{4\lambda}$.
To summarize, the fixed point $(-1/2,0)$ is stable, while the fixed point
$(-1/2,1)$ is unstable. If $r^2 < \epsilon/4$, then there is one unstable
mode, corresponding to growth of $\varphi-1$. If $r^2 > \epsilon/4$, then
there are two unstable modes, one for $\varphi$ and one for $\gamma$.
Finally, there is the fixed point (0,0). Near (0,0) we have
\begin{equation}
\begin{split}
\gamma'' = & 2 \gamma/r^2 + \hbox{ higher order}, \\
\varphi'' + 2 \varphi'/r = & 2
\varphi (r^{-2} - \lambda) +\hbox{ higher order.}
\end{split}
\end{equation}
This fixed point is always unstable, with $\gamma$ growing rapidly. $\varphi$
will grow exponentially if $r < 1/\sqrt{\lambda}$, and will oscillate if
$r > 1/\sqrt{\lambda}$.