An inverted pendulum is not stable. Nevertheless, as pointed out by Pyotr Kapitza (1894--1984) in 1951, if you oscillate vertically the suspension point at a sufficientlyfast frequency Ω, you can stabilise it. A similar phenomenon occurs in arotating saddle (see video), and is also used to create radio-frequencyionic traps.

Let us start considering a very familiar one-dimensional system: a planar pendulum made of a massless rod of length ℓ ending with a point mass m. In the familiar swing, the driving occurs in different ways: if you “drive” the swing yourself, you do it by effectively modifyingthe position of your “center-of-mass”, hence the effective length ℓ(t) of the “pendulum”. If you are pushed by someone else, then you have a pendulum with a “periodic external force”. We use the generalize coordinate q = θ that denotes the angle formed with the vertical (θ = 0 being the downward position), and y0(t) denotes the position of its suspension point, we can derive the equations of motion from the Lagrangian formalism. In a short while we will assume that y0(t) = Acos Ωt, where A is the amplitude of the driving and Ω the driving frequency, but for the time being, let us proceed by keeping y0(t to be general. In a system of reference with the y-axis oriented upwards and the x-axis horizontally, the position x(t) and y(t) of the massm is:

\[ \begin{cases} x(t) &= \ell \,\sin\theta (t) , \\ y(t) &= y_0 (t) - \ell\,\cos \theta (t) , \end{cases} \qquad \Longrightarrow \qquad \begin{cases} \dot{x}(t) &= \ell \dot{\theta} \cos\theta , \\ \dot{y}(t) &= \dot{y}_0 + \ell \dot{\theta} \sin\theta . \end{cases} \]
The Lagrangean is given by
\begin{align*} {\cal L}\left( \theta , \dot{\theta} , t \right) &= \frac{m}{2} \left( \dot{x}^2 + \dot{y}^2 \right) - mgy \\ &= \frac{m}{2} \,\ell^2 \dot{\theta}^2 + m\ell \dot{y}_0 \dot{\theta} \sin\theta + mg\ell\,\cos\theta , \end{align*}
where we dropped the term \( \frac{m}{2}\,\dot{y}_0^2 - mgy_0 \) because it would not enter in the Euler--Lagrange equations due to pure dependence on time t. The associated momentum is given by:
\[ p_{\theta} = \frac{\partial {\cal L}}{\partial \theta} = m\ell^2 \dot{\theta} + m\ell \dot{y}_0 \sin\theta \qquad \Longrightarrow \qquad \dot{\theta} = \frac{p_{\theta}}{m\ell^2} - \frac{\dot{y}_0}{\ell} \,\sin\theta . \]
The Hamilton’s equationsread:
\[ \begin{cases} \dot{\theta} &= \frac{\partial H}{\partial p_{\theta}} = \frac{p_{\theta}}{m\ell^2} , \\ \dot{p}_{\theta} &= - \frac{\partial H}{\partial \theta} = -m \left[ g - A \Omega^2 \cos\Omega t \right] \sin \theta . \end{cases} \]
Hence, transforming it into a second-order equation:
\begin{equation} \label{EqI.1} \frac{{\text d}^2 \theta}{{\text d} t^2} + \frac{1}{\ell} \left[ g - A\Omega^2 \cos (\Omega t) \right] \sin \theta = 0 , \end{equation}
where \( \omega_0^2 = g/\ell \) is the square of the frequency of the unperturbed pendulum in the linear regime, Ω is the driving frequency, A is the amplitude of the driving, which we model as y0(t) = Acos(Ωt).