= Solution
Assume $\mu^2,\lambda>0$. At zero source, the <vacuum manifold> is the sphere of radius $v_0=\mu/\sqrt\lambda$. The source tilts the <scalar potential> by $-\epsilon\phi_1$. Its force should be small compared with the radial restoring-force scale $\mu^2v_0$, so a useful quantitative condition is
$$
\boxed{|\epsilon|\ll\mu^2v_0=\frac{\mu^3}{\sqrt\lambda}.}
$$
For nonzero $\epsilon$, the exact <global symmetry in field theory> of the <Lagrangian> is $O(n-1)$, rotating $\phi_2,\ldots,\phi_n$. The global <vacuum> aligns with the source:
$$
\langle\phi\rangle=(\operatorname{sgn}(\epsilon)w,0,\ldots,0),\qquad
\lambda w^3-\mu^2w=|\epsilon|,\qquad
w=v_0+\frac{|\epsilon|}{2\mu^2}+O\!\left(\frac{\epsilon^2}{\mu^4v_0}\right).
$$
This <vacuum> preserves the entire remaining $O(n-1)$ <symmetry>. There is therefore no exact <Goldstone boson> for $\epsilon\ne0$. Expanding the <scalar potential> to quadratic order gives
$$
\boxed{m_{\perp}^2=-\mu^2+\lambda w^2=\frac{|\epsilon|}{w},\qquad
m_{\parallel}^2=-\mu^2+3\lambda w^2=2\mu^2+\frac{3|\epsilon|}{w}.}
$$
Thus there are $n-1$ degenerate light <pseudo-Goldstone bosons> and one heavy radial <scalar>. The small-source condition makes their squared <masses> parametrically smaller than the radial squared <mass>. Sending $\epsilon$ to zero restores the original massless modes.
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