Question

By introducing suitable dimensionless variables, the system of nonlinear equations for the damped pendulum [Frqs. (8) of Section 9.3] can be written as $$ d x / d t=y, \quad d y / d t=-y-\sin x \text { . } $$ (a) Show that the origin is a critical point. (b) Show that while \(V(x, y)=x^{2}+y^{2}\) is positive definite, \(f(x, y)\) takes on both positive and negative values in any domain containing the origin, so that \(V\) is not a Liapunov function. Hint: \(x-\sin x>0\) for \(x>0\) and \(x-\sin x<0\) for \(x<0 .\) Consider these cases with \(y\) positive but \(y\) so small that \(y^{2}\) can be ignored compared to \(y .\) (c) Using the energy function \(V(x, y)=\frac{1}{2} y^{2}+(1-\cos x)\) mentioned in Problem \(6(b),\) show that the origin is a stable critical point. Note, however, that even though there is damping and we can epect that the origin is asymptotically stable, it is not possible to draw this conclusion using this Liapunov function. (d) To show asymptotic stability it is necessary to construct a better Liapunov function than the one used in part (c). Show that \(V(x, y)=\frac{1}{2}(x+y)^{2}+x^{2}+\frac{1}{2} y^{2}\) is such a Liapunov function, and conclude that the origin is an asymptotically stable critical point. Hint: From Taylor's formula with a remainder it follows that \(\sin x=x-\alpha x^{3} / 3 !,\) where \(\alpha\) depends on \(x\) but \(0<\alpha<1\) for \(-\pi / 2

Short answer

Question: Determine whether the origin (0,0) is a critical point and whether it is stable, unstable, or asymptotically stable for the given system of nonlinear equations for a damped pendulum. Answer: The origin (0,0) is a critical point and it is asymptotically stable.

Step-by-step solution

Determine the origin as a critical point

To determine the critical points, we set the derivatives in the given equation to 0: $$ \frac{dx}{dt} = y = 0 \text{, } \frac{dy}{dt} = -y - \sin x = 0. $$ Thus, \(x = 0\) and \(y = 0\) imply that the origin (0, 0) is a critical point.

Analyze the Liapunov function

The given Liapunov function is \(V(x,y) = x^2 + y^2\). To find \(f(x,y)\), we calculate the derivative of \(V\) w.r.t time: $$ \frac{dV}{dt} = \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial y}\frac{dy}{dt} = 2x \cdot y + 2y(-y - \sin x) = 2xy - 2y^2 - 2y\sin x. $$ Since \(V(x,y)\) is positive definite, we need to check if \(f(x,y)\) takes on positive and negative values in any domain containing the origin. If \(x>0, x-\sin x > 0\), with \(y\) positive and small enough such that \(y^2\) can be ignored, \(f(x,y)>0\). If \(x<0, x-\sin x < 0\), with \(y\) positive and small enough such that \(y^2\) can be ignored, \(f(x,y)<0\). Therefore, the given \(V(x,y)\) is not a Liapunov function.

Show the origin as a stable critical point using the energy function

We are now given the energy function \(V(x,y) = \frac{1}{2}y^2 + (1 - \cos x)\). We find \(f(x,y)\): $$ \frac{dV}{dt} = \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial y}\frac{dy}{dt} = y \cdot (\sin x) + y(-y - \sin x) = -y^2 \leq 0. $$ Since the function \(f(x,y)\) is always non-positive, and its derivative is zero at the origin, it shows that the origin (0,0) is a stable critical point.

Construct a better Liapunov function for asymptotic stability

We are given the Liapunov function: \(V(x,y) = \frac{1}{2}(x+y)^2 + x^2 + \frac{1}{2}y^2\). We find \(f(x,y)\): $$ \frac{dV}{dt} = \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial y}\frac{dy}{dt} = (x+y)y + 2x\cdot y - y[(x+y)+\sin x]. $$ Now, we use Taylor's formula with the remainder: $$ \sin x = x - \alpha x^3/3! $$ where \(0 < \alpha < 1\) for \(-\pi/2 < x < \pi/2\). Let \(x = r\cos\theta\) and \(y = r\sin\theta\), then the given function transforms into: $$ \hat{V}(r\cos\theta, r\sin\theta) = -r^2[1+h(r,\theta)], $$ where \(|h(r,\theta)|<1\) if \(r\) is sufficiently small. Since the Liapunov function has the necessary properties, we can conclude that the origin (0,0) is an asymptotically stable critical point.

Key concepts

Damped Pendulum

A damped pendulum is a classic physical system in which a pendulum's motion experiences a resistance force that acts in opposition to its oscillation. This resistance, or damping, typically arises from friction or air resistance, leading to a gradual reduction in the amplitude of swings over time.

When we analyze a damped pendulum mathematically, we use differential equations to describe its motion. For instance, the system of equations given in the exercise characterizes the behavior of the damped pendulum in terms of dimensionless variables, which simplifies the analysis while preserving the core dynamics of the system.

Understanding the motion of a damped pendulum involves recognizing the impact of both the gravitational force, which tends to bring the pendulum to its lowest energy state (straight down), and the damping force, which continuously extracts energy from the system, eventually bringing the pendulum to a stop.

Critical Point Stability

The term critical point stability refers to the behavior of a dynamical system in the vicinity of a critical point or an equilibrium point. A critical point occurs when all the derivatives of the system equate to zero, indicating no change or a steady-state condition.

In our exercise, demonstrating the origin as a critical point was step one. This point corresponds to the pendulum being upright with no motion (i.e., zero angular displacement and zero angular velocity). The stability of this point is essential to understand because it tells us whether the pendulum will remain at rest (stable) or diverge away (unstable) after a slight disturbance.

Liapunov Function and Stability

Liapunov's method helps in assessing the stability of critical points. A Liapunov function is essentially an energy-like scalar function that, if found to always decrease over time, can indicate a stable critical point. By evaluating the Liapunov function and its derivatives, we can infer whether the origin is stable, which was precisely the focus of steps 2 and 3 in the solution.

Asymptotic Stability

The concept of asymptotic stability pertains to a system's behavior over an extended period. When a critical point is asymptotically stable, any small deviation from this point will not only cease to increase but will also eventually decay back to the equilibrium over time. It's a stronger condition than mere stability, as it guarantees that the system will return to the critical point and not merely stay close without oscillating indefinitely.

For our damped pendulum, asymptotic stability would mean that after nudging the pendulum, no matter how slightly, it would come to rest in the upright position, without continually oscillating around it. In the exercise, the last step required constructing a new Liapunov function showing that the energy of the system continuously decreases, thus confirming asymptotic stability at the origin. This was achieved by utilizing a transformation into polar coordinates and examining the behavior of the Liapunov function for sufficiently small radii around the critical point.