A system of non-linear equations with a small parameter Is there any way to solve analytically the following system of equations to the leading order in $\epsilon$:
$$\left\{
\begin{array}{rcl}
  \mu^2 \phi_1 + \lambda \phi_1 (\phi_1^2 + \phi_2^2) + \epsilon =0, \\
  \mu^2 \phi_2 + \lambda \phi_2 (\phi_1^2 + \phi_2^2)= 0. \\
\end{array}
\right.$$
where $\mu^2<0$, $\lambda>0$, $\epsilon^{1/3} \ll |\mu|, \sqrt{-\mu^2/\lambda}$.
 A: Put
$${-\mu^2\over \lambda}=:c^2>0\ .$$
Then we have to solve the system
$$\eqalign{x(c^2-x^2-y^2)&={\epsilon\over \lambda} \cr
y(c^2-x^2-y^2)&=0 \ .\cr}$$
Note that $\epsilon$ does not appear in the second equation. Therefore any solution will have to satisfy one of
$$c^2-x^2-y^2=0,\qquad y=0\ .$$
The first of these enforces $\epsilon=0$, whence is of no use. The second leads to
$$x^3-c^2 x+{\epsilon \over\lambda}=0\ .$$
When $\epsilon=0$ this is solved when $x=0$ and $x=\pm c$. Therefore we shall obtain three different analytic solutions in terms of $\epsilon$. Setting up a power series
$$x_0(\epsilon)=a_1\epsilon+a_2\epsilon^2+a_3\epsilon^3+?\epsilon^4$$
and comparing coefficients gives
$$x_0(\epsilon)={\epsilon\over \lambda c^2}+{\epsilon^3 \over \lambda^3 c^8}+?\epsilon^4\ .$$
Similarly we obtain
$$x_c(\epsilon)=c-{\epsilon \over 2\lambda c^2}-{3\epsilon^2 \over 8\lambda^2 c^5} -{\epsilon^3 \over 2\lambda^3 c^8}+?\epsilon^4\ ,$$
$$x_{-c}(\epsilon)=-c-{\epsilon \over 2\lambda c^2}+{3\epsilon^2 \over 8\lambda^2 c^5} -{\epsilon^3 \over 2\lambda^3 c^8}+?\epsilon^4\ .$$
(I have used Mathematica for the computations.)
