Interactive Inverted Pendulum Solver

Horizontal

\[ x = x_p(t) + l \sin (\theta) = A \cos(\omega t - \delta) + l \sin (\theta), \] \[ y = l \cos (\theta) \] The velocity components of the pendulum bob are: \[ \dot{x} = -A \omega \sin(\omega t - \delta) + l \dot{\theta} \cos (\theta), \] \[ \dot{y} = -l \dot{\theta} \sin (\theta) \] The total kinetic energy $ T $ is therefore: \[ T = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) \] Expanding this expression, we get: \[ T = \frac{1}{2} m \left[ \left(-A \omega \sin(\omega t - \delta) + l \dot{\theta} \cos (\theta)\right)^2 + \left(l \dot{\theta} \sin (\theta)\right)^2 \right] \] Expanding further and simplifying, we obtain: \[ T = \frac{1}{2} m \left( A^2 \omega^2 \sin^2(\omega t - \delta) + l^2 \dot{\theta}^2 - 2 A l \omega \dot{\theta} \cos (\theta) \sin(\omega t - \delta) \right) \] The potential energy $ V $ is due to gravity and is given by: \[ V = -m g y = -m g l \cos (\theta) \] The Lagrangian $ L $ is given by: \[ L = T - V = \frac{1}{2} m \left( A^2 \omega^2 \sin^2(\omega t - \delta) + l^2 \dot{\theta}^2 - 2 A l \omega \dot{\theta} \cos (\theta) \sin(\omega t - \delta) \right) + m g l \cos (\theta) \] Using the Euler-Lagrange equation: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0, \] \[ \frac{\partial L}{\partial \dot{\theta}} = m l^2 \dot{\theta} - m A l \omega \cos (\theta) \sin(\omega t - \delta). \] \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) = m l^2 \ddot{\theta} - m A l \omega \left( -\dot{\theta} \sin (\theta) \sin(\omega t - \delta) + \cos (\theta) \omega \cos(\omega t - \delta) \right) \] \[ \frac{\partial L}{\partial \theta} = m l \left( A \omega \dot{\theta} \sin(\omega t - \delta) \sin (\theta) - g \sin (\theta) \right) \] \[ m l^2 \ddot{\theta} - m A l \omega \left( -\dot{\theta} \sin (\theta) \sin(\omega t - \delta) + \cos (\theta) \omega \cos(\omega t - \delta) \right) - m l \left( A \omega \dot{\theta} \sin(\omega t - \delta) \sin (\theta) - g \sin (\theta) \right) = 0 \] \[ l \ddot{\theta} - A \omega \left( -\dot{\theta} \sin (\theta) \sin(\omega t - \delta) + \cos (\theta) \omega \cos(\omega t - \delta) \right) - \left( A \omega \dot{\theta} \sin(\omega t - \delta) \sin (\theta) - g \sin (\theta) \right) = 0 \] \[ l \ddot{\theta} - A \omega \left( -\dot{\theta} \sin (\theta) \sin(\omega t - \delta) + \cos (\theta) \omega \cos(\omega t - \delta) + \dot{\theta} \sin(\omega t - \delta) \sin (\theta) \right) + g \sin (\theta) = 0 \] \[ \ddot{\theta} - \frac{ A \omega}{l} \left( -\dot{\theta} \sin (\theta) \sin(\omega t - \delta) + \cos (\theta) \omega \cos(\omega t - \delta) + \dot{\theta} \sin(\omega t - \delta) \sin (\theta) \right) + \frac{g}{l} \sin (\theta) = 0 \] \[ \ddot{\theta} + \frac{g}{l} \sin (\theta) = \frac{ A \omega}{l} \left( -\dot{\theta} \sin (\theta) \sin(\omega t - \delta) + \cos (\theta) \omega \cos(\omega t - \delta) + \dot{\theta} \sin(\omega t - \delta) \sin (\theta) \right) \] \[ \ddot{\theta} + \frac{g}{l} \sin (\theta) = \frac{ A \omega}{l} \left( \cos (\theta) \omega \cos(\omega t - \delta) \right) \] \begin{equation} \boxed{\ddot{\theta} + \frac{g}{l} \sin (\theta) = \frac{ A \omega^2}{l} \left( \cos (\theta) \cos(\omega t - \delta) \right)} \label{eq:horizontal_original_no_damping} \tag{1} \end{equation} Expanding about $\theta = 0$ yields to first order $\sin(\theta) \sim 0 + \cos(0) (\theta-0) = \theta$, $\cos(\theta) \sim 1$, yielding \[ \boxed{\ddot{\theta} + \frac{g}{l}\theta = \frac{ A \omega^2}{l}\cos(\omega t - \delta)} \] with solution \[\theta{\left(t \right)} = \frac{A \omega^{2} \cos{\left(\delta - \omega t \right)}}{g - l \omega^{2}} + C_{1} \sin{\left(\frac{\sqrt{g} t}{\sqrt{l}} \right)} + C_{2} \cos{\left(\frac{\sqrt{g} t}{\sqrt{l}} \right)}\] To second order, $\sin(\theta) \sim \theta$, $\cos(\theta) \sim 1 - \dfrac{\theta^2}{2}$, yielding \begin{equation} \boxed{\ddot{\theta} + \frac{g}{l}\theta = \frac{ A \omega^2}{l}\left(1 - \dfrac{\theta^2}{2}\right)\cos(\omega t - \delta)} \label{eq:hoizontal_second_order_no_damping} \tag{2} \end{equation} Considering linear damping simply modifies equation \ref{eq:horizontal_original_no_damping} to be: \begin{equation} \boxed{\ddot{\theta} + \frac{g}{l} \sin (\theta) = \frac{ A \omega^2}{l} \left( \cos (\theta) \cos(\omega t - \delta) \right) -\lambda \dot{\theta}}\label{eq:horizontal_original_linear_damping}\tag{3} \end{equation}

Vertical

\[ x = l \sin (\theta), \] \[ y = y_p(t) + l \cos (\theta) = A \cos(\omega t - \delta) + l \cos (\theta) \] The velocity components of the pendulum bob are: \[ \dot{x} = l \dot{\theta} \cos (\theta), \] \[ \dot{y} = -A \omega \sin(\omega t - \delta) - l \dot{\theta} \sin (\theta) \] The total kinetic energy $ T $ is therefore: \[ T = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) \] Expanding this expression, we get: \[ T = \frac{1}{2} m \left[ \left(l \dot{\theta} \cos (\theta)\right)^2 + \left(-A \omega \sin(\omega t - \delta) - l \dot{\theta} \sin (\theta)\right)^2 \right] \] Expanding further and simplifying, we obtain: \[ T = \frac{1}{2} m \left( A^2 \omega^2 \sin^2(\omega t - \delta) + l^2 \dot{\theta}^2 + 2 A l \omega \dot{\theta} \sin (\theta) \sin(\omega t - \delta) \right) \] The potential energy $ V $ is due to gravity and is given by: \[ V = -m g y = -m g \left(A \cos (\omega t - \delta) + l \cos (\theta)\right) \] The Lagrangian $ L $ is given by: \[ L = T - V = \frac{1}{2} m \left( A^2 \omega^2 \sin^2(\omega t - \delta) + l^2 \dot{\theta}^2 + 2 A l \omega \dot{\theta} \sin (\theta) \sin(\omega t - \delta) \right) + m g \left(A \cos (\omega t - \delta) + l \cos (\theta)\right) \] Using the Euler-Lagrange equation: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0, \] \[ \frac{\partial L}{\partial \dot{\theta}} = m l^2 \dot{\theta} + m A l \omega \sin (\theta) \sin(\omega t - \delta). \] \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) = m l^2 \ddot{\theta} + m A l \omega \left( \dot{\theta} \cos (\theta) \sin(\omega t - \delta) + \sin (\theta) \omega \cos(\omega t - \delta) \right) \] \[ \frac{\partial L}{\partial \theta} = m l \left( A \omega \dot{\theta} \sin(\omega t - \delta) \cos (\theta) - g \sin (\theta) \right) \] \[ m l^2 \ddot{\theta} + m A l \omega \left( \dot{\theta} \cos (\theta) \sin(\omega t - \delta) + \sin (\theta) \omega \cos(\omega t - \delta) \right) - m l \left( A \omega \dot{\theta} \sin(\omega t - \delta) \cos (\theta) - g \sin (\theta) \right) = 0 \] \[ l \ddot{\theta} + A \omega \left( \dot{\theta} \cos (\theta) \sin(\omega t - \delta) + \sin (\theta) \omega \cos(\omega t - \delta) \right) - \left( A \omega \dot{\theta} \sin(\omega t - \delta) \cos (\theta) - g \sin (\theta) \right) = 0 \] \[ l \ddot{\theta} + A \omega \left( \dot{\theta} \cos (\theta) \sin(\omega t - \delta) + \sin (\theta) \omega \cos(\omega t - \delta) - \dot{\theta} \sin(\omega t - \delta) \cos (\theta) \right) + g \sin (\theta) = 0 \] \[ \ddot{\theta} + \frac{A \omega}{l} \left( \dot{\theta} \cos (\theta) \sin(\omega t - \delta) + \sin (\theta) \omega \cos(\omega t - \delta) - \dot{\theta} \sin(\omega t - \delta) \cos (\theta) \right) + \frac{g}{l} \sin (\theta) = 0 \] \[ \ddot{\theta} + \frac{g}{l} \sin (\theta) = -\frac{A \omega}{l} \left( \sin (\theta) \omega \cos(\omega t - \delta) \right) \] \begin{equation} \boxed{\ddot{\theta} + \frac{g}{l} \sin (\theta) = -\frac{A \omega^2}{l} \left( \sin (\theta) \cos(\omega t - \delta) \right)} \label{eq:vertical_original_no_damping} \tag{6} \end{equation} Expanding about $\theta = 0$ yields to first order $\sin(\theta) \sim 0 + \cos(0) (\theta-0) = \theta$, $\cos(\theta) \sim 1$, yielding: \[ \boxed{\ddot{\theta} + \frac{g}{l} \theta = -\frac{A \omega^2}{l} \left( \theta \cos(\omega t - \delta) \right)} \] To second order, $\sin(\theta) \sim \theta$, $\cos(\theta) \sim 1 - \dfrac{\theta^2}{2}$, yielding \begin{equation} \boxed{\ddot{\theta} + \frac{g}{l} \theta = -\frac{A \omega^2}{l} \left( \theta \cos(\omega t - \delta) \right)} \label{eq:vertical_second_order_no_damping} \tag{7} \end{equation} Considering linear damping simply modifies equation \ref{eq:vertical_original_no_damping} to be: \begin{equation} \boxed{\ddot{\theta} + \frac{g}{l} \sin (\theta) = -\frac{A \omega^2}{l} \left( \sin (\theta) \cos(\omega t - \delta) \right) -\lambda \dot{\theta}}\label{eq:vertical_original_linear_damping} \tag{8} \end{equation}

\[ x = x_p(t) + l \sin (\theta) \] \[ y = l \cos (\theta) \] The velocity components of the pendulum bob are: \[ \dot{x} = \dot{x}_p(t) + l \dot{\theta} \cos (\theta), \] \[ \dot{y} = -l \dot{\theta} \sin (\theta) \] The total kinetic energy $ T $ is therefore: \[ T = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) \] Expanding this expression, we get: \[ T = \frac{1}{2} m \left[ \left(\dot{x}_p(t) + l \dot{\theta} \cos (\theta)\right)^2 + \left(l \dot{\theta} \sin (\theta)\right)^2 \right] \] Expanding further and simplifying, we obtain: \[ T = \frac{1}{2} m \left( \dot{x_p}^2(t) + l^2 \dot{\theta}^2 + 2 l \dot{x}_p(t) \dot{\theta} \cos (\theta) \right) \] The potential energy $ V $ is due to gravity and is given by: \[ V = -m g y = -m g l \cos (\theta) \] The Lagrangian $ L $ is given by: \[ L = T - V = \frac{1}{2} m \left( \dot{x_p}^2(t) + l^2 \dot{\theta}^2 + 2 l \dot{x}_p(t) \dot{\theta} \cos (\theta) \right) + m g l \cos (\theta) \] Using the Euler-Lagrange equation: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0, \] \[ \frac{\partial L}{\partial \dot{\theta}} = m l^2 \dot{\theta} + m l \dot{x}_p(t) \cos (\theta) \] \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) = m l^2 \ddot{\theta} + m l \left( \ddot{x}_p(t)\cos(\theta) - \dot{x}_p(t)\dot{\theta}\sin(\theta) \right) \] \[ \frac{\partial L}{\partial \theta} = -m l \left( \dot{x}_p(t)\dot{\theta}\sin(\theta) + g \sin (\theta) \right) \] \[ m l^2 \ddot{\theta} + m l \left( \ddot{x}_p(t)\cos(\theta) - \dot{x}_p(t)\dot{\theta}\sin(\theta) \right) + m l \left( \dot{x}_p(t)\dot{\theta}\sin(\theta) + g \sin (\theta) \right) = 0 \] \[ l \ddot{\theta} + \left( \ddot{x}_p(t)\cos(\theta) - \dot{x}_p(t)\dot{\theta}\sin(\theta) \right) + \left( \dot{x}_p(t)\dot{\theta}\sin(\theta) + g \sin (\theta) \right) = 0 \] \[ l \ddot{\theta} + \ddot{x}_p(t)\cos(\theta) + g \sin (\theta) = 0 \]

$\underline{\textbf{Steps to learn } \mathbf{x_p(t)}}$

Generate a function $x_p(t)$
Numerically integrate $\theta(t)$ from $t = 0$ to $t_\text{final}$ with $\theta(0) = \theta_0$, $\dot{\theta}(0) = 0$
Measure the loss as $\mathcal{L} \sim \max(\vec{\theta} - \theta_0)$
Optimize the parameters in $x_p(t)$ to minimize $\mathcal{L}$
Repeat steps 1-4 until convergence.

\begin{align*} \sin(\theta) &\approx \sin(\theta_0) + \cos(\theta_0)(\theta - \theta_0) \\ &= \sin(\theta_0) + \cos(\theta_0)\theta - \cos(\theta_0)\theta_0 \\ &= \underbrace{\sin(\theta_0) - \cos(\theta_0)\theta_0}_{\alpha} + \underbrace{\cos(\theta_0)}_\beta \theta \\ &\equiv \alpha + \beta \theta \end{align*} \[ \ddot{\theta} + \frac{g}{l}\sin(\theta) = -\frac{A \omega^2}{l}\cos(\omega t - \delta)\sin(\theta) \] Define: \[ F(t) = -\frac{A \omega^2}{l}\cos(\omega t - \delta) \] Then: \[ \ddot{\theta} + \frac{g}{l} (\alpha + \beta\theta) = F(t)(\alpha + \beta\theta) \] \[ \ddot{\theta} = -\frac{g}{l} (\alpha + \beta\theta) - \frac{A \omega^2}{l}\cos(\omega t - \delta)(\alpha + \beta\theta) \] \[ \ddot{\theta} = -\frac{\left(\alpha+\beta\theta\right)}{l} \left(g + A \omega^2\cos(\omega t - \delta)\right) \] \[ \ddot{\theta} + \frac{g}{l} (\alpha + \beta\theta) = - \frac{A \omega^2}{l}\cos(\omega t - \delta)(\alpha + \beta\theta) \] \[ \ddot{\theta} + \frac{g}{l} \left(\sin(\theta_0) + \cos(\theta_0)(\theta - \theta_0)\right) = - \frac{A \omega^2}{l}\cos(\omega t - \delta)\left(\sin(\theta_0) + \cos(\theta_0)(\theta - \theta_0)\right) \] \[ \ddot{\theta} + \frac{g}{l}\beta \left(\frac{\alpha}{\beta} + \theta\right) = F(t)\beta \left(\frac{\alpha}{\beta} + \theta\right) \] Let: \[ \varphi = \frac{\alpha}{\beta} + \theta, \quad \dot{\varphi} = \dot{\theta}, \quad \ddot{\varphi} = \ddot{\theta} \] \[ \ddot{\varphi} + \frac{g}{l}\beta \varphi = F(t)\beta \varphi \] \[ \ddot{\varphi} + \frac{g}{l}\beta \varphi - F(t)\beta \varphi = 0 \] \[ \ddot{\varphi} + \frac{g}{l}\beta \left(1 - \frac{l}{g}F(t)\right) \varphi = 0 \] \[ \ddot{\varphi} + \frac{g}{l} \beta \left(1 + \frac{l}{g} \frac{A \omega^2}{l}\cos(\omega t - \delta)\right)\varphi = 0 \qquad \texttt{//Setting $\delta = 0$} \] \[ \rightarrow \ddot{\varphi} + \frac{g}{l} \beta \left(1 + \frac{A \omega^2}{g}\cos(\omega t)\right)\varphi = 0 \] Let: \[ \tau = \omega t \rightarrow \frac{\partial^2 \varphi}{\partial \tau^2} = \frac{1}{\omega^2} \frac{\partial^2 \varphi}{\partial t^2} \] \[ \frac{\partial^2 \varphi}{\partial \tau^2} + \frac{g \beta \omega^2}{l} \left( 1 + \frac{A \omega^2}{g} \cos(\tau) \right) \varphi = 0 \] \[ \frac{\partial^2 \varphi}{\partial \tau^2} + \frac{g \beta \omega^2}{l}\varphi + \frac{\beta A \omega^4}{l} \cos(\tau) \varphi = 0 \] \[ \frac{\partial^2 \varphi}{\partial \tau^2} + \left(\frac{g \beta \omega^2}{l} - \frac{-\beta A \omega^4}{l} \cos(\tau)\right) \varphi = 0 \] \[ \frac{\partial^2 \varphi}{\partial \tau^2} + \left(k - m \cos(\tau) \right) \varphi = 0 \] where \[ k = \left( \frac{g\omega^2}{l } \right) \cos(\theta_0), \quad m = -\frac{A \omega^4}{l} \cos(\theta_0), \quad \tau = \omega t \] \[ \varphi = \frac{\sin(\theta_0) - \cos(\theta_0) \theta_0}{\cos\theta_0} + \theta = \tan(\theta_0) - \theta_0 + \theta \] \[ k > 0 \quad \text{for} \quad \theta_0 \in \left( 2\pi n - \frac{\pi}{2}, 2\pi n + \frac{\pi}{2} \right) \] \[ k < 0 \quad \text{for} \quad \theta_0 \in \left( 2\pi n + \frac{\pi}{2}, 2\pi n + \frac{3\pi}{2} \right) \] \[ \text{where } \: n \in \mathbb{Z} \] \[ \mathbf{k < 0} \quad \Rightarrow \quad 2 \sqrt{\frac{k(k-1)(k-4)}{3k-8}} < m(k) < \frac{1}{4} \left(\sqrt{(9 - 4k)(13 - 20k)} - (9 - 4k)\right) \] $\mathbf{k > 0} \:\Rightarrow$ the following: \[ \quad 0 < m(k) < \left(\frac{1}{4} \sqrt{(9 - 4k)(13 - 20k)} - (9 - 4k)\right) \] \[ \qquad \left(0 < k < \frac{1}{4}\right) \] \[ 0 < m(k) < \frac{1}{4} \left( 9 - 4k \mp \sqrt{(9 - 4k)(13 - 20k)} \right) \] \[ \left(\frac{1}{4} < k < \frac{13}{20}\right) \] \[ 0 < m(k) < \sqrt{\frac{2(k-1)(k-4)(k-9)}{k-5}} \] \[ \left(\frac{13}{20} < k < 1\right) \] \[ 0 < m(k) < 2 \sqrt{\frac{k(k-1)(k-4)}{3k-8}} \] \[ (k > 1) \]

Interactive Inverted Pendulum Viewer

Horizontal

Vertical

\(\underline{\textbf{Steps to learn } \mathbf{x_p(t)}}\)