Inverted pendulum

Example

Take a cart of mass $M > 0$ that moves on a horizontal line, with position $x(t)$. A massless rigid rod of length $\ell > 0$ is hinged on the cart and carries a point mass $m > 0$ at its free end. Let $\theta(t)$ be the angle of the rod from the upward vertical, and let $g > 0$ be the acceleration of gravity. The only input is a horizontal force $u(t)$ on the cart. The Lagrangian $L = \frac{1}{2} M \dot x^2 + \frac{1}{2} m \left( \dot x^2 + 2 \ell \dot x \dot\theta \cos\theta + \ell^2 \dot\theta^2 \right) - m g \ell \cos\theta$ gives the equations of motion $$(M + m)\, \ddot x + m \ell \left( \ddot\theta \cos\theta - \dot\theta^2 \sin\theta \right) = u,$$ $$\ell\, \ddot\theta + \ddot x \cos\theta - g \sin\theta = 0.$$ There are two degrees of freedom, $x$ and $\theta$, and one input, so the cart-pole is an underactuated system.

The upright position $\theta = 0$, $\dot\theta = 0$, $\dot x = 0$ with $u = 0$ is an equilibrium. The linearisation at it is $(M + m) \ddot x + m \ell \ddot\theta = u$ and $\ell \ddot\theta + \ddot x = g \theta$. Eliminating $\ddot x$ gives $$\ddot\theta = a\, \theta - b\, u, \qquad a = \frac{(M + m) g}{M \ell}, \quad b = \frac{1}{M \ell}.$$ With $u = 0$ the solutions include $\theta(t) = \theta_0 e^{\sqrt{a}\, t}$, so the upright position is not Lyapunov stable. With the feedback $u = k_1 \theta + k_2 \dot\theta$, the angle obeys $\ddot\theta + b k_2 \dot\theta + (b k_1 - a) \theta = 0$. If $k_2 > 0$ and $k_1 > a / b = (M + m) g$, both coefficients are positive, so both roots of $s^2 + b k_2 s + (b k_1 - a)$ have negative real part and $\theta(t) \to 0$.