The double pendulum is a famous example of a chaotic system. It consists of one pendulum hanging from the end of another pendulum, which in turn hangs from a fixed point. In the traditional version, both arms are perfectly rigid.
Here's an easy way to generalize the model so that the arms are not perfectly rigid. Let $\mathbf{x}$ and $\mathbf{y}$ be the variable locations of the ends of the two arms, where masses $m_x$ and $m_y$ are located. (The arms themselves are massless.) Take the system's equations of motion to be \begin{align} \newcommand{\bfx}{\mathbf{x}} \newcommand{\bfy}{\mathbf{y}} \newcommand{\bfg}{\mathbf{g}} m_x\ddot\bfx + \nabla_\bfx V(\bfx,\bfy) &= 0 \\ m_y\ddot\bfy + \nabla_\bfy V(\bfx,\bfy) &= 0 \end{align} with potential energy \begin{align} V(\bfx,\bfy) = &- m_x\bfg\cdot\bfx + f\big(k_x,L_x,|\bfx|\big) \\ &- m_y\bfg\cdot\bfy + f\big(k_y,L_y,|\bfy-\bfx|\big) \end{align} where $\bfg$ is the acceleration of gravity (a downward-pointing vector) and the function $f$ is defined by $$ f(k.L,x) = k(L^2-x^2)^2. $$ The $L$s are the nominal arm-lengths, and the $k$s are the degree of rigidity of the arms. For finite $k$s, the potential energy $V$ is a smooth function of $\bfx$ and $\bfy$. The perfectly rigid version corresponds to $k_x,k_y\to\infty$: in that limit, any deviation from the nominal lengths $L_x,L_y$ costs infinite energy.
How chaotic is the system with finite-but-large values of $k_x$ and $k_y$, if chaos is quantified in the standard way(s)? Intuitively, the closer $k_x,k_y$ are to zero, the less chaotic the system should be, because taking the limit $k_x,k_y\to 0$ gives a pair of freely-falling masses that don't interact with each other at all. But can we determine how the degree of chaos (quantified in a standard way) scales with $k_x,k_y$, at least roughly?
