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 $\mu^2<0$, $\lambda>0$, $\epsilon^{1/3} \ll |\mu|, \sqrt{-\mu^2/\lambda}$.
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.)