Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r|r-2|(r-3), \quad d \theta / d t=-1 $$

Short Answer

Expert verified
Based on the given autonomous system of differential equations in polar coordinates, we analyzed the radial and angular differential equations to find all periodic solutions, limit cycles, and their stability characteristics. We found that there are two equilibrium points (\(r=2\) and \(r=3\)) that correspond to stable and unstable limit cycles, respectively. We determined that the periodic solutions have a period of \(-2\pi\). The limit cycles are stable for \(r=2\) and unstable for \(r=3\).

Step by step solution

01

Analyze the radial differential equation

The radial differential equation is given as: $$ \frac{dr}{dt} = r|r-2|(r-3) $$ To find the equilibrium points, we set \(\frac{dr}{dt} = 0\) and solve for \(r\) in terms of the equilibrium points: $$ r|r-2|(r-3) = 0 $$ This equation has three solutions for \(r\): \(r = 0\), \(r = 2\), and \(r = 3\). These correspond to the equilibrium points of the system.
02

Analyze the angular differential equation

The angular differential equation is given as: $$ \frac{d\theta}{dt} = -1 $$ As there are no equilibrium points for this equation, the angular dynamics are a simple rotation with a constant angular velocity \(-1\). Now combine this information with the radial dynamics for periodic solutions.
03

Discuss the limit cycles and their stability characteristics

For \(r=0\), the system collapses to the origin and does not show any oscillatory behavior or limit cycles. Hence, it is asymptotically stable. For \(r=2\) and \(r=3\), these equilibria can be considered as periodic solutions since they correspond to circles moving in the angular direction with constant angular velocity \(-1\). Their period can be computed as follows: $$ T(\theta) = \frac{d\theta}{dt} = \frac{\Delta\theta}{-1} = \frac{2\pi}{-1} = -2\pi $$ Thus, we have two periodic solutions with period \(-2\pi\). For limit cycles, we need to look at the behavior of the radial dynamics in the vicinity of the equilibrium points. - Near \(r=2\), the dynamics is given as \(\frac{dr}{dt}=-r(r-1)(2-r)\), indicating a stable limit cycle with a radius of \(r = 2\). - Near \(r=3\), the dynamics is given as \(\frac{dr}{dt}=r(3-r)(r-1)\), indicating an unstable limit cycle with the radius of \(r = 3\). In conclusion, we have found the periodic solutions, limit cycles (r=2, stable; r=3, unstable), and characterized their stability.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Periodic Solutions
Periodic solutions in the context of autonomous systems in polar coordinates revolve around finding closed and repeating paths of the system over time. Here, we examine the system defined by the given equations. A solution is periodic if it repeats after a certain interval called the period. In our exercise, we analyze the radial equation: \[ \frac{dr}{dt} = r|r-2|(r-3) \]To identify periodic solutions, we first find where the derivative is zero, indicating no change in the radius over time, hence these are equilibrium points. By solving \( r|r-2|(r-3) = 0 \), we find the equilibrium points at \( r = 0 \), \( r = 2 \), and \( r = 3 \). This implies potential periodic paths when combined with angular dynamics. For the angular motion described by:\[ \frac{d\theta}{dt} = -1 \]The angular component spins continuously and does not affect the periodicity due to its constant rate. Combining this with the equilibrium points of the radial component, \( r = 2 \) and \( r = 3 \) result in periodic circular motions around the origin. Since each complete revolution takes \(-2\pi\) due to the negative angular velocity, these are indeed periodic solutions.
Limit Cycles
Limit cycles are closed trajectories representing isolated periodic solutions indicative of the system's long-term dynamic behavior. In polar coordinates, these cycles are often visualized as concentric circles. For the equation given \( \frac{dr}{dt} = r|r-2|(r-3) \), limit cycles are found by examining the behavior of \( r \) as it approaches these equilibrium points.- **Near \( r=2 \):**The radial growth is given by \( \frac{dr}{dt} = -r(r-1)(2-r) \). This suggests that trajectories starting nearby will spiral towards \( r = 2 \). Consequently, this is a stable limit cycle as it's attracting nearby paths.- **Near \( r=3 \):**Here, \( \frac{dr}{dt} = r(3-r)(r-1) \) describes the radial expansion, illustrating that trajectories move away from \( r = 3 \). Therefore, this is an unstable limit cycle, pushing nearby trajectories away.In summary, our system reveals two noteworthy limit cycles: a stable limit cycle at \( r = 2 \) and an unstable limit cycle at \( r = 3 \). Both revolve around central paths, defining distinct behaviors.
Stability Characteristics
Understanding the stability characteristics in polar autonomous systems is crucial for predicting system behavior over time. Stability tells us whether a system will return to an equilibrium state or diverge away when disturbed. Looking at our radial equation \( \frac{dr}{dt} = r|r-2|(r-3) \): - **Equilibrium Point \( r = 0 \):**This corresponds to the origin in polar coordinates. Since all trajectories collapse here without oscillation, it exhibits asymptotic stability, meaning any perturbation will return it to \( r=0 \).- **Equilibrium Point \( r = 2 \):**As discussed under limit cycles, \( r = 2 \) represents a stable limit cycle. Here, small disturbances result in the system returning to this path, showcasing it as an attractive periodic orbit.- **Equilibrium Point \( r = 3 \):**Since \( r = 3 \) is an unstable limit cycle, disturbances will lead trajectories away from this radius. Hence, this point functions as a repeller, marking it as an unstable point.By examining the sign changes in \( \frac{dr}{dt} \) around each equilibrium, we deduce bumpy paths as well as predictable behaviors. Stability in each context predicates how systems respond to tiny nudges, crucial for forecasting real-world phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Verify that \((0,0)\) is a critical point, show that the system is almost linear, and discuss the type and stability of the critical point \((0,0)\) by examining the corresponding linear system. $$ d x / d t=x+y^{2}, \quad d y / d t=x+y $$

The motion of a certain undamped pendulum is described by the equations $$ d x / d t=y, \quad d y / d t=-4 \sin x $$ If the pendulum is set in motion with an angular displacement \(A\) and no initial velocity, then the initial conditions are \(x(0)=A, y(0)=0\) (a) Let \(A=0.25\) and plot \(x\) versus \(t\). From the graph estimate the amplitude \(R\) and period \(T\) of the resulting motion of the pendulum. (b) Repeat part (a) for \(A=0.5,1.0,1.5,\) and \(2.0 .\) (c) How do the amplitude and period of the pendulum's motion depend on the initial position \(A^{7}\) Draw a graph to show each of these relationships. Can you say anything about the limiting value of the period as \(A \rightarrow 0 ?\) (d) Let \(A=4\) and plot \(x\) versus \(t\) Explain why this graph differs from those in parts (a) and (b). For what value of \(A\) does the transition take place?

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-x-0.5 y)} \\ {d y / d t=y(2-0.5 y-1.5 x)}\end{array} $$

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=x-y^{2}, \quad d y / d t=y-x^{2} $$

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r(1-r)(r-2), \quad d \theta / d t=-1 $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free