Question

Consider the system encountered in the study of pendulum motion, $$ \begin{aligned} &x^{\prime}=y \\ &y^{\prime}=-\sin x \end{aligned} $$ at its equilibrium points \((0,0)\) and \((\pi, 0)\). (a) Let \(z_{1}=x, z_{2}=y\). Show that the system becomes $$ \mathbf{z}^{\prime}=\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right] \mathbf{z}+\left[\begin{array}{c} 0 \\ z_{1}-\sin z_{1} \end{array}\right] \text {. } $$ (b) Let \(z_{1}=x-\pi, z_{2}=y\). Show that the system becomes $$ \mathbf{z}^{\prime}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \mathbf{z}-\left[\begin{array}{c} 0 \\ z_{1}-\sin z_{1} \end{array}\right] $$ (c) Show that the system is almost linear at both equilibrium points.

Short answer

And how does this show that the system is almost linear at both equilibrium points? Answer: By rewriting the given system in terms of \(z_1\) and \(z_2\), we find that the system becomes equivalent to the given matrices and non-linear terms when \(z_1 = x\) and \(z_2 = y\) in part (a), and when \(z_1 = x - \pi\) and \(z_2 = y\) in part (b). This transformation allows us to express the system in matrix form, showing how the variables in the system interact. At equilibrium points \((0, 0)\) and \((\pi, 0)\), the non-linear terms \(z_1 - \sin z_1\) (or \(z_1 - \sin (z_1 + \pi)\)) become very close to 0 for small values of \(z_1\) around the equilibrium points. This closeness to 0 indicates that the system behaves almost linearly near the equilibrium points. The system being almost linear at equilibrium points simplifies the analysis of the system's behavior in the vicinity of these points.

Step-by-step solution

Rewrite the given system in terms of z1 and z2 for (a)

We are given the system: $$ \begin{aligned} &x^{\prime}=y \\\ &y^{\prime}=-\sin x \end{aligned} $$ Let \(z_1 = x\) and \(z_2=y\). Now we rewrite the system in terms of \(z_1\) and \(z_2\): $$ \begin{aligned} &z_{1}^{\prime}=z_{2} \\\ &z_{2}^{\prime}=-\sin z_{1} \end{aligned} $$

Rewrite the system in matrix form for (a)

Now we rewrite this system as a matrix equation, that is \(\mathbf{z}'=A\mathbf{z}+B\) where \(\mathbf{z}=\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\): $$ \mathbf{z}^{\prime}=\left[\begin{array}{rr} 0 & 1 \\\ -1 & 0 \end{array}\right] \mathbf{z}+\left[\begin{array}{c} 0 \\\ z_{1}-\sin z_{1} \end{array}\right] $$ This is the desired form for part (a).

Rewrite the given system in terms of z1 and z2 for (b)

Now, let \(z_1 = x - \pi\) and \(z_2 = y\). We need to rewrite the original system in terms of \(z_1\) and \(z_2\). Recall that \(z_1 = x-\pi\), so \(x = z_1+\pi\). Now we rewrite the system in terms of \(z_1\) and \(z_2\): $$ \begin{aligned} &z_{1}^{\prime}=z_{2} \\\ &z_{2}^{\prime}=-\sin (z_{1}+\pi) \end{aligned} $$

Rewrite the system in matrix form for (b)

Now we rewrite this system as a matrix equation, that is \(\mathbf{z}'=A\mathbf{z}+B\) where \(\mathbf{z}=\begin{bmatrix} z_1 \\ z_2 \end{bmatrix}\): $$ \mathbf{z}^{\prime}=\left[\begin{array}{ll} 0 & 1 \\\ 1 & 0 \end{array}\right] \mathbf{z}-\left[\begin{array}{c} 0 \\\ z_{1}-\sin (z_{1}+\pi) \end{array}\right] $$ This is the desired form for part (b).

Show that the system is almost linear at equilibrium points

The equilibrium points are \((0, 0)\) and \((\pi, 0)\). In both cases, we can see that the non-linear term \(z_1 - \sin z_1\) (or \(z_1 - \sin (z_1 + \pi)\)) becomes very close to 0 for small values of \(z_1\) around the equilibrium points. This indicates that the system behaves almost linearly near the equilibrium points, satisfying the condition for almost linearity.

Key concepts

Differential Equations

In the study of pendulum motion, differential equations play a central role. These equations describe how the system's variables, such as position and velocity, change over time. Given the system of equations: \[ \begin{aligned} &x^{\prime}=y \ &y^{\prime}=-\sin x \end{aligned} \] we see the essence of pendulum motion captured in a mathematical format. Here, the variables involved represent the angular position \(x\) and the angular velocity \(y\) of the pendulum. - The equation \(x^{\prime}=y \) suggests that the rate of change of \(x\) (i.e., position) is the velocity \(y\).- The equation \(y^{\prime}=-\sin x\) indicates that the rate of change of \(y\) (i.e., velocity) depends on the negative sine of \(x\), which relates to the restoring force of the pendulum.Understanding differential equations is key to analyzing dynamic systems like pendulums. Each differential equation can represent a physical law or principle that dictates how the system evolves.

Equilibrium Points

Equilibrium points occur where the system's variables do not change over time, effectively resulting in a state of rest or constant motion. In our pendulum system, the equilibrium points are \((0,0)\) and \((\pi, 0)\). At these points, the pendulum is either perfectly vertical or inverted, and the velocity is zero.To find these equilibrium points, set the derivatives \(x^{\prime} = 0\) and \(y^{\prime} = 0\). This reveals that:- At \((0, 0)\), the pendulum hangs straight down with no motion.- At \((\pi, 0)\), the pendulum is straight up and also at rest, though this point is less stable.Recognizing equilibrium points allows us to analyze the stability of the system, as they form the basis for understanding how nearby movement behaves and whether it moves back to equilibrium or away in oscillation.

Linearization

Linearization involves approximating a nonlinear system with a linear one around its equilibrium points. This is crucial because linear systems are easier to solve and analyze. In the context of the pendulum system, we take small deviations from the equilibrium points and simplify the equations:- For the equilibrium point \((0, 0)\), the system becomes: \[ \mathbf{z}^{\prime}=\begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \mathbf{z} + \begin{bmatrix} 0 \ z_{1}-\sin z_{1} \end{bmatrix} \] - For the equilibrium point \((\pi, 0)\), after substituting \(z_1 = x - \pi\), it turns into:\[ \mathbf{z}^{\prime}=\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \mathbf{z} - \begin{bmatrix} 0 \ z_{1}-\sin (z_{1}+\pi) \end{bmatrix} \]These linear approximations simplify the study of the pendulum's behavior close to equilibrium, allowing us to apply linear stability analysis and gain insight into the nature of the motion around these points.

Matrix Representation

The matrix representation of a system of differential equations provides a clear, concise way of expressing the relationships between variables and their rates of change. For our pendulum example, matrix representation elegantly encapsulates the system's dynamics.- For \((0, 0)\), the matrix form is: \[ \mathbf{z}^{\prime} = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \mathbf{z} + \begin{bmatrix} 0 \ z_{1} - \sin z_{1} \end{bmatrix} \]- For \((\pi, 0)\), it is given by:\[ \mathbf{z}^{\prime} = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \mathbf{z} - \begin{bmatrix} 0 \ z_{1} - \sin (z_{1} + \pi) \end{bmatrix} \]Matrices allow us to use powerful mathematical tools like eigenvalues and eigenvectors to study system behavior, such as predicting stability and oscillation patterns. Understanding matrix representation is thus essential for deeply analyzing pendulum motion or any similar systems.