Can a system of cubic anharmonic oscillators have multiple stable equilibria? Consider the Hamiltonian for a System of $N$ anharmonic oscillators
$H= \sum_{i = 1}^N (\frac{p_i^2}{2m_i}+\frac{1}{2}k_iq_i^2)+\sum_{i,j=1}^N b_{ijk}q_iq_jq_k$
with specific constants $k_i,b_{ijk}, m_i$ and the respective Positions $q_i$ and Momenta $p_i$. The equilibria of the system are the Solutions of the equations $\frac{\partial H}{\partial q_i} = 0$. Since there might be multiple Solutions of this equation due to nonlinearity, it might be possible that there exist multiple equilibria.
The interesting Question is whether there exist multiple stable equilibria. In the one-particle case we have the Equilibrium condition $k_1q_1+3b_{111}q_1^2=0$ with Solutions $q_1=0,q_1=-\frac{k_1}{3b_{111}}$. The second derivative $\frac{\partial^2H}{\partial q_i q_j}$ determines whether an Equilibrium Position is stable (when it is positive definite) or not (otherwise). For one-dimensional case we know that $q_1 = 0$ is a stable Equilibrium, but the other Equilibrium Point isn't.
Question: When I have a System of cubic anharmonic oscillators, can I have more than one stable Equilibrium configurations?
My idea: I think yes, because consider only two anharmonic oscillators:
$k_1q_1+3b_{111}q_1^2+(b_{121}+b_{211}+b_{112})q_1q_2+(b_{122}+b_{221}+b_{212})q_2^2=0$ (1)
$k_2q_2+3b_{222}q_2^2+(b_{122}+b_{221}+b_{212})q_1q_2+(b_{112}+b_{121}+b_{211})q_1^2=0$ (2)
and then solve (2) iteratively by fixed-point Iteration
$q_2 = - \frac{3b_{222}q_2^2+(b_{122}+b_{221}+b_{212})q_1q_2+(b_{112}+b_{121}+b_{211})q_1^2}{k_2}$
with initial guess $q_2 = 0$. Then my first Iteration yields: $q_2 \propto q_1^2$. Substituting this Approximation into (1) leads to a cubic equation in $q_1$ which has three Solutions. From Analysis, it is known that there might be two stable Solutions (positive derivative values) in a one-variable cubic equation.
 A: Generically, I would bet on the scenario that there is always only one stable equilibrium and a bunch of unstable ones.
Your Hamiltonian looks like $$H = \frac{1}{2m_1}\, p_1^2 + .. + \frac{1}{2m_n}\, p_n^2 +
U^{(3)}(q_1, ..., q_n)$$ where $U^{(3)}$ is a sum of a homogeneous quadratic plus a homogeneous cubic cubic polynomial with respect to the variables $q_1,...,q_n$.  The equilibrium points are the solutions to the algebraic (no more than inhomogeneous quadratic) equations:
\begin{align}
&\frac{\partial H}{\partial p_1} = \frac{1}{m_1}\, p_1 = 0\\
&...\\
&\frac{\partial H}{\partial p_n} = \frac{1}{m_1}\, p_2 = 0\\
&\frac{\partial H}{\partial q_1} = \frac{\partial U^{(3)}}{\partial q_1}(q_1, ..., q_n)  = 0\\
&...\\
&\frac{\partial H}{\partial q_n} = \frac{\partial U^{(3)}}{\partial q_n}(q_1, ..., q_n)  = 0
\end{align}
Since the first half of the equations yields $p_1 = ... = p_n = 0$, you are left with the quadratic equations from the second half:
\begin{align}
&\frac{\partial H}{\partial q_1} = \frac{\partial U^{(3)}}{\partial q_1}(q_1, ..., q_n)  = 0\\
&...\\
&\frac{\partial H}{\partial q_n} = \frac{\partial U^{(3)}}{\partial q_n}(q_1, ..., q_n)  = 0
\end{align} only for the variables $q_1, ..., q_n$. Thus, since $H$ can serve as a Lyapunov stability function, the local minima of the cubic multi-variable polynomial are the stable (but not asymptotically) equilibria. You can check that since in this special case $U^{(3)}$ is a sum of a homogeneous quadratic and a homogeneous cubic polynomial, the point $q_1 = ... = q_n = 0$ is always a stable equilibrium, because $U^{(3)}$ has a local minimum there, assuming that the parameters $k_i$ are all positive, which should be the case of harmonic oscilators. However, if you pick another equilibrium of $H$, call it equilibrium 1, then noting stops you from drawing a one-dimensional line in the space $q_1,...,q_n$ that connects the zero equilibrium to equilibrium 1. Furthermore, if you restrict the polynomial $U^{(3)}$ onto this line, you obtain a single variable cubic polynomial that has a local minimum at the zero equilibrium. Then the other equilibrium 1, is also a critical point and the only option is for it to be a local maximum for $U^{(3)}$ restricted to the line, which means that equilibrium 1 cannot be a local minimum in the ambient space $q_1,...,q_n$ and hence, cannot be a stable equilibrium.
