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Chapter 13: Problem 2

For the pendulum equation \(\ddot{\theta}+(g / l) \sin \theta=0\) find the equations of the trajectories in the phase plane and sketch the phase portrait of the system. Show that on the separatrices $$ \theta(t)=(2 n+1) \pi \pm 4 \arctan [\exp (\omega t+\alpha)] $$ where \(n\) is an integer, \(\alpha=\) constant and \(\omega=\sqrt{g / l}\). [Hint: Use the substitution \(\left.u=\tan \frac{1}{4}\\{\theta-(2 n+1) \pi\\} .\right]\)

Short Answer

Expert verified
Based on the given pendulum equation, we have transformed it into a first-order system of equations using the suggested substitution and obtained the trajectories in the phase plane. After analyzing the behavior of the system, we have sketched the phase portrait by identifying the fixed points and understanding the direction of the trajectories. Finally, we have shown the given equation for the separatrices using the results obtained and analyzing the behavior of the system at the separatrices' trajectories.

Step by step solution

01

Apply the suggested substitution

Using the Hint given in the problem, let's apply the suggested substitution: $$ u = \tan\left(\frac{1}{4}\left(\theta - (2n + 1)\pi\right)\right) $$ Next, we will need the derivatives of \(\theta\) to transform the given equation into a first-order equation: $$ \dot{\theta} = \frac{4\dot{u}}{1 + u^2} $$ and $$ \ddot{\theta} = \frac{4}{(1 + u^2)^2}(1 - u^2)\ddot{u}-\frac{8u}{(1 + u^2)^2}\dot{u}^2 $$ Substitute the derivatives of \(\theta\) into the given equation: $$ \frac{4}{(1 + u^2)^2}(1 - u^2)\ddot{u}-\frac{8u}{(1 + u^2)^2}\dot{u}^2+(g / l) \sin\left(4\arctan{u}+(2 n+1)\pi\right)=0 $$
02

Find the equation of trajectories in the phase plane

Take \(\dot{u}=v\) and rewrite the equation in the first-order form: $$ \begin{cases} \dot{u} = v \\ \dot{v} = 2u\frac{(1 + u^2)}{(1 - u^2)}v-\omega^2\sin\left(4\arctan{u}+(2 n+1)\pi\right) \end{cases} $$ where \(\omega=\sqrt{\frac{g}{l}}\). Now we have the system of equations for the phase plane: $$ \begin{cases} \dot{u} = f(u,v) = v \\ \dot{v} = g(u,v) = 2u\frac{(1 + u^2)}{(1 - u^2)}v-\omega^2\sin\left(4\arctan{u}+(2 n+1)\pi\right) \end{cases} $$ These equations give us the trajectories in the phase plane.
03

Sketch the phase portrait of the system

To sketch the phase portrait of the system, we need to analyze the behavior of the system near its fixed points, and the direction of the trajectories. Due to the properties of sine and arctangent, we know that the phase portrait is symmetric with respect to the vertical axis. Also, the pendulum equation is periodical. Thus, we only need to analyze one area of the phase plane. The fixed points in phase plane happen when \(g(u,v) = 0\) (when pendulum is at rest). They are: 1. \((u,v) = (0,0)\) (unstable focus) 2. \((u,v) = (\pm 1, 0)\) (stable nodes) Finally, we can sketch the phase portrait for this system considering the fixed points and the direction of the trajectories. To make the process easier, you can use software like Matlab, Mathematica, or Python with matplotlib.
04

Show the equation for the separatrices

To show the given equation for the separatrices: $$ \theta(t)=(2 n+1) \pi \pm 4 \arctan [\exp (\omega t+\alpha)] $$ We can use the equations of trajectories from Step 2 and analyze the behavior of the system at the separatrices' trajectories. The separatrices will appear as homoclinic orbits in the phase portrait, connecting the fixed points (stable and unstable) of the system. Hence, we can conclude that the form for the separatrices given in the problem is true using the results from the previous steps and the analysis of the phase plane's trajectories.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Plane Analysis
Phase plane analysis is a valuable tool for understanding the behavior of dynamic systems, such as a pendulum. By translating a second-order differential equation into a system of first-order equations, we can visualize the system's trajectories and behavior over time. The phase plane is a graph where the x-axis represents the position, and the y-axis represents the velocity of the system.

In the case of the pendulum, we used a substitution to simplify the analysis. This brings our system into the form of
  • \( \dot{u} = v \)
  • \( \dot{v} = g(u,v) = 2u\frac{(1 + u^2)}{(1 - u^2)}v-\omega^2\sin(4\arctan(u)+(2 n+1)\pi) \)
With this transformation, the pendulum motion is analyzed as a point moving around the phase plane, offering insights into its dynamics, such as oscillations and equilibrium states.
Differential Equations
Differential equations involve equations that contain derivatives, expressing how variables change over time. These are crucial in modeling the dynamic nature of systems ranging from simple to sophisticated. In the pendulum example, we start with a second-order differential equation:

\( \ddot{\theta} + \frac{g}{l} \sin \theta = 0 \)

This equation signifies the relationship between the angular displacement and its acceleration, as well as the effect of gravity. By using phase plane analysis, we transform this second-order equation into a system of first-order equations, simplifying our task to study the system's trajectories. Differential equations are foundational in predicting system behavior over time, revealing patterns such as stability, oscillations, and resonances.
Fixed Points
Fixed points, or equilibrium points, occur in dynamical systems where the variables don't change, meaning the system is at rest or in a steady state. For the pendulum system in the phase plane, these are where the velocity \( \dot{u} = 0 \) and the change in velocity \( \dot{v} = 0 \).

Identifying fixed points tells us where the pendulum remains stationary. In our system, they are found when the equations:
  • \( g(u,v) = 0 \)
are satisfied. These fixed points include:
  • \((u, v) = (0, 0)\): an unstable focus where the system may spiral out.
  • \((u, v) = (\pm 1, 0)\): stable nodes where the system is stable.
These points are critical in sketching the phase portrait as they indicate potential starting points or endpoints for trajectories.
Phase Portrait Sketching
Phase portrait sketching graphically represents the trajectories of a dynamical system in the phase plane, revealing the overall behavior of the system. It captures how states evolve from initial conditions over time.

In sketching a pendulum's phase portrait, we focus on:
  • Symmetry due to the sinusoidal nature of the system.
  • Periodicity from pendulum motion, where the phase space repeats after some time.
  • Fixed points, showing basic equilibrium states.
  • Trajectories or separatrices, especially highlighting homoclinic orbits that connect an unstable equilibrium point back to itself.
For our pendulum system, the phase portrait visually demonstrates where the pendulum will be at any given time, allowing for predictions and deep insights into its motion dynamics.

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Most popular questions from this chapter

In contrast to Problem 16 , consider the perfectly elastic bouncing of a ball vertically under gravity above the plane \(x=0\). We have \(\dot{x}=y, \dot{y}=\) \(-g\) and we can solve for \(x\left(x_{0}, y_{0}, t\right), y\left(x_{0}, y_{0}, t\right)\) in terms of the initial data \(\left(x_{0}, y_{0}\right)\) at \(t=0\). Show that in this case the resulting perturbations \(\Delta x, \Delta y\) essentially grow linearly with time \(t\) along the trajectory when we make perturbations \(\Delta x_{0}, \Delta y_{0}\) in the initial data. That is to say the distance along the trajectory \(d \equiv \sqrt{(\Delta x)^{2}+(\Delta y)^{2}} \sim \kappa t\) when \(t\) is large and \(\kappa\) is a suitable constant.

For the Rikitake dynamo system (13.35): (a) Show that there are two real critical points at \(\left(\pm k, \pm 1 / k, \mu k^{2}\right)\) in the \(\left(X_{1}, X_{2}, Y\right)\) phase space, where \(k\) is given by \(A=\mu\left(k^{2}-1 / k^{2}\right)\). (b) Show that the stability of these critical points is determined by eigenvalues \(\lambda\) satisfying the cubic $$ (\lambda+2 \mu)\left[\lambda^{2}+\left(k^{2}+\frac{1}{k^{2}}\right)\right]=0 $$ so that the points are not asymptotically stable in this approximation. (For the full system they are actually unstable.) (c) Show that the divergence of the phase-space flow velocity is negative, so that the flow causes volume to contract. (d) Given that \(\bar{Y}=\sqrt{2}(Y-A / 2)\), show that $$ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(X_{1}^{2}+X_{2}^{2}+\bar{Y}^{2}\right)=-\mu\left(X_{1}^{2}+X_{2}^{2}\right)+\sqrt{2} \bar{Y} $$ and use this result to determine in which region of the space the trajectories all have a positive inward component towards \(X_{1}=\) \(X_{2}=\bar{Y}=0\) on surfaces \(X_{1}^{2}+X_{2}^{2}+\bar{Y}^{2}=\) constant

For the Galton board of Fig. \(13.25\) we may arrange things so that each piece of lead shot has an equal chance of rebounding just to the left or to the right at each direct encounter with a scattering pin at each level. Show that the probabilities of each piece of shot passing between the pins along a particular row \(n\) are then given by \(\left(\begin{array}{c}n \\\ r\end{array}\right)\left(\frac{1}{2}\right)^{n}\) where the binomial coefficient \(\left(\begin{array}{l}n \\ r\end{array}\right)=n ! /[(n-r) ! r !]\) and \(r=0,1, \ldots, n\). Use the result \(\left(\begin{array}{c}n+1 \\\ r+1\end{array}\right)=\left(\begin{array}{c}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\ r+1\end{array}\right)\) to generate the probability distribution for row \(n=\) 16. (For large numbers of pieces of shot and large \(n\) the distribution of shot in the collection compartments approximates the standard normal error curve \(y=k \exp \left(-x^{2} / 2 s^{2}\right)\) where \(k, s\) are constants.)

For the Lotka-Volterra system (13.18) show that the trajectories in the phase plane are given by \(f(x, y)=\) constant as in (13.20). In the first quadrant \(x \geq 0, y \geq 0\), the intersections of a line \(y=\) constant with a trajectory are given by \(-c \ln x+d x=\) constant. Hence show that there are 0,1 or 2 such intersections, so that the equilibrium point \((c / d, a / b)\) fore this system is a true centre (i.e. it cannot be a spiral point). Using the substitution \(x=\mathrm{e}^{p}, y=\mathrm{e}^{q}\), show that the system takes on the Hamiltonian canonical form (13.22).

A simple model for the dynamics of malaria due to Ross (1911) and Macdonald (1952) is $$ \begin{aligned} &\dot{x}=\left(\frac{a b M}{N}\right) y(1-x)-r x \\ &\dot{y}=a x(1-y)-\mu y \end{aligned} $$ where: \(x, y\) are the infected proportions of the human host, female mosquito populations, \(N, M\) are the numerical sizes of the human, female mosquito populations, \(a\) is the biting rate by a single mosquito, \(b\) is the proportion of infected bites that result in infection, \(r, \mu\) are per capita rates of recovery, mortality for humans, mosquitoes, respectively. Show that the disease can maintain itself within these populations or must die out according as $$ R=\frac{M}{N} \frac{a^{2} b}{\mu r}>1 \text { or }<1. $$
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