Consider again the pendulum linearized about its unstable upper position, given by the equation \(\ddot{\varphi}(t)-\varphi(t)=u(t)\), and assume that we use the PD control law \(u(t)=-\alpha \varphi(t)-\beta \dot{\varphi}(t)\) to obtain an asymptotically stable closed-loop system $\ddot{\varphi}(t)+b \dot{\varphi}(t)+a \varphi(t)=0\( (with \)a=\alpha-1>0\( and \)b=\beta>0$ ). Introduce the natural frequency \(\omega:=\sqrt{a}\) and the damping factor \(\zeta:=b /(2 \sqrt{a})\), so that the equation now reads $$ \ddot{\varphi}(t)+2 \zeta \omega \dot{\varphi}(t)+\omega^{2} \varphi(t)=0 . $$ (A) Prove the following facts: 1\. If \(\zeta<1\) (the "underdamped" case), all solutions are decaying oscillations. 2\. If \(\zeta=1\) (the "critically damped" case) or if \(\zeta>1\) ("overdamped"), then all solutions for which \(\varphi(0) \neq 0\) are such that \(\varphi(t)=0\) for at most one \(t>0\). 3\. If \(\zeta \geq 1\), then every solution that starts from rest at a displaced position, that is, \(\varphi(0) \neq 0\) and \(\dot{\varphi}(0)=0\), approaches zero monotonically. (In this case, we say that there is no "overshoot.") 4\. Show rough plots of typical solutions under the three cases $\zeta<1, \zeta=1\(, and \)\zeta>1$.

This is a second-order linear ordinary differential equation (ODE) with constant coefficients. To solve the given ODE, we will try the exponential solution, \(\varphi(t) = e^{rt}\), where \(r\) is a constant. Substituting this into the given ODE, we get: \[ e^{rt}(r^2 + 2 \zeta \omega r + \omega^2) = 0 \] This is a quadratic equation in \(r\), we will find its roots and then use them to get the general solution.

We have a quadratic equation \(r^2 + 2 \zeta \omega r + \omega^2 = 0\). Using the quadratic formula, we can find the roots as: \[ r = \frac{-2 \zeta \omega \pm \sqrt{(2 \zeta \omega)^2 - 4 \omega^2}}{2} \] Simplify the expression to get: \[ r = -\zeta \omega \pm \omega \sqrt{\zeta^2 - 1} \]

Using the roots obtained in the previous step, we will analyze the three cases of damping ratio, \(\zeta\). Case 1: \(\zeta<1\) (underdamped): In this case, \(\zeta^2 - 1 < 0\), so the roots are complex conjugates: \(r = -\zeta \omega \pm j\omega \sqrt{1-\zeta^2}\). The general solution is: \[ \varphi(t) = e^{-\zeta \omega t}(C_1 \cos{\omega_d t} + C_2 \sin{\omega_d t}) \] where \(\omega_d = \omega \sqrt{1-\zeta^2}\) and \(C_1\) and \(C_2\) are constants determined by the initial conditions. All solutions are decaying oscillations as the exponential term decays over time. Case 2: \(\zeta=1\) (critically damped): In this case, \(\zeta^2 - 1 = 0\), so we have repeated real roots: \(r = -\omega\). The general solution is: \[ \varphi(t) = (C_1 + C_2 t)e^{-\omega t} \] All solutions with \(\varphi(0) \neq 0\) are such that \(\varphi(t)=0\) for at most one \(t>0\). Case 3: \(\zeta>1\) (overdamped): In this case, \(\zeta^2 - 1 > 0\), so the roots are real and distinct: \(r_1 = -\zeta \omega - \omega \sqrt{\zeta^2 - 1}\) and \(r_2 = -\zeta \omega + \omega \sqrt{\zeta^2 - 1}\). The general solution is: \[ \varphi(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \] Similar to the critically damped case, all solutions with \(\varphi(0) \neq 0\) are such that \(\varphi(t) = 0\) for at most one \(t>0\). If \(\zeta \geq 1\) and \(\dot{\varphi}(0) = 0\), every solution approaches zero monotonically, as both exponential terms decay over time without oscillations.

For the underdamped case (\(\zeta < 1\)), the plot will show decaying oscillations, starting from a displaced position. The oscillations' amplitude will decay exponentially over time. For the critically damped case (\(\zeta = 1\)), the plot will show an initial displacement and a smooth, non-oscillatory decay to zero without overshooting. For the overdamped case (\(\zeta > 1\)), the plot will show an initial displacement and a smooth, non-oscillatory decay to zero without overshooting, but it will be slower compared to the critically damped case.