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

Short Answer

Expert verified
Based on the analysis of the given autonomous system in polar coordinates, there is one stable limit cycle with r=2, which represents a clockwise rotation around the origin with constant negative angular velocity. The system's trajectories will converge to this limit cycle over time. The equilibrium points r=0 and r=1 are both unstable, and there are no other periodic solutions or limit cycles since there are no other stable equilibrium points.

Step by step solution

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01

Analyze radial part of the equation

To analyze the radial part, focus on the dr/dt equation: $$\frac{dr}{dt}=r(1-r)(r-2)$$. Set dr/dt equal to zero and solve for r to find its equilibrium points: $$ r(1-r)(r-2)=0 $$ This equation has three solutions (equilibrium points): r = 0, r = 1, and r = 2.
02

Determine stability of equilibrium points

To determine the stability of each equilibrium point, find the first derivative of the radial function with respect to r and substitute each equilibrium point into the derived equation. Differentiate the equation $$r(1-r)(r-2)$$ with respect to r: $$ \frac{d}{dr}(r(1-r)(r-2)) = (1-r)(r-2) + r(1-2r) - r^2(1-r) $$ Now, substitute each equilibrium value into the derived equation to determine the stability: For r = 0: $$ (1-0)(0-2) + 0(1-0) - 0^2(1-0) = 2 > 0 $$ For r = 1: $$ (1-1)(1-2) + 1(1-2) - 1^2(1-1) = 1 > 0 $$ For r = 2: $$ (1-2)(2-2) + 2(1-4) - 2^2(1-2) = -4 < 0 $$ With these results, we can conclude that: - The equilibrium point r=0 is unstable (source) since its derivative is greater than 0. - The equilibrium point r=1 is unstable (source) since its derivative is greater than 0. - The equilibrium point r=2 is stable (sink) since its derivative is less than 0.
03

Analyze angular part of the equation

Now let's analyze the angular part of the equation: $$\frac{d\theta}{dt}=-1$$. This equation implies that the angular velocity is constant, and the trajectory will be a simple rotation around the origin with negative (clockwise) angular velocity.
04

Determine periodic solutions, limit cycles, and stability

Based on the previous steps, we can conclude that: - There are three equilibrium points r=0, r=1, and r=2, with the latter being stable and the first two being unstable. - The stable equilibrium point r=2 represents a limit cycle of rotation around the origin with constant negative angular velocity. - The system's trajectories will converge to this limit cycle (r=2) as time goes on, regardless of their initial conditions. - The limit cycle is stable since its derivative is less than 0. - There are no other periodic solutions or limit cycles since there are no other stable equilibrium points.

Key Concepts

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

Equilibrium Points
Equilibrium points are essential in understanding the behavior of dynamical systems, such as autonomous systems expressed in polar coordinates. In our scenario, the radial equation \( \frac{dr}{dt} = r(1-r)(r-2) \) helps identify the equilibrium points. Equilibrium points occur where the derivative with respect to time is zero, meaning there is no change in \( r \). By setting \( r(1-r)(r-2) = 0 \), we determine the equilibrium points: \( r = 0 \), \( r = 1 \), and \( r = 2 \). These points indicate where the system's state could remain constant over time, under certain conditions.
Stability Analysis
Stability analysis examines what happens to trajectories when they are near an equilibrium point. To determine stability, we look at the behavior of the system when it experiences small perturbations around these points. We calculate the first derivative of the radial equation with respect to \( r \) and substitute each equilibrium point to assess its stability.
- For \( r = 0 \) and \( r = 1 \), the derivatives are positive. Therefore, these are unstable equilibrium points or sources. Trajectories move away from these points.
- For \( r = 2 \), the derivative is negative, indicating stability. It acts as a sink, with trajectories converging to this point. Stability analysis thus helps predict how trajectories evolve over time.
Limit Cycle
A limit cycle is a closed trajectory in phase space that represents a periodic solution. In our system, the stable equilibrium point \( r = 2 \) acts as a limit cycle. This means the system's trajectories will spiral towards \( r = 2 \), creating a stable pattern of periodic behavior. Importantly, once on the limit cycle, trajectories will remain on it as time progresses, making it a robust and recurring behavior in the system. This single limit cycle determines how the system behaves for a wide range of initial conditions.
Polar Coordinates
Polar coordinates provide a way to analyze dynamical systems by expressing them in terms of rotational movement. Here, the radial and angular components of motion are separated, making it easier to evaluate rotational dynamics.
- The term \( \frac{d\theta}{dt} = -1 \) indicates constant angular velocity. This results in a uniform, clockwise rotation around the origin, simplifying the analysis of rotational behavior. - The radial component, \( \frac{dr}{dt} \), could be analyzed for radii \( r \), affecting how far from the origin the trajectory remains. In this system, over time, regardless of initial conditions, the trajectory reaches the stable limit cycle at radius \( r = 2 \). The use of polar coordinates helps break down complex motion into more understandable rotational components.

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Most popular questions from this chapter

(a) Find the eigenvalues and eigenvectors. (b) Classify the critical point \((0,0)\) as to type and determine whether it is stable, asymptotically stable, or unstable. (c) Sketch several trajectories in the phase plane and also sketch some typical graphs of \(x_{1}\) versus \(t .\) (d) Use a computer to plot accurately the curves requested in part (c). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{ll}{3} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x}\)

(a) Find all the critical points (equilibrium solutions). (b) Use a computer to draw a direction field and portrait for the system. (c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable, stable, or unstable, and classify it as to type. $$ d x / d t=(2+x)(y-x), \quad d y / d t=y\left(2+x-x^{2}\right) $$

(a) Find the eigenvalues and eigenvectors. (b) Classify the critical point \((0,0)\) as to type and determine whether it is stable, asymptotically stable, or unstable. (c) Sketch several trajectories in the phase plane and also sketch some typical graphs of \(x_{1}\) versus \(t .\) (d) Use a computer to plot accurately the curves requested in part (c). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{cc}{2} & {-\frac{5}{2}} \\\ {\frac{9}{5}} & {-1}\end{array}\right) \mathbf{x}\)

Consider the eigenvalues given by equation ( 39 ). Show that $$\left(\sigma_{1} X+\sigma_{2} Y\right)^{2}-4\left(\sigma_{1} \sigma_{2}-\alpha_{1} \alpha_{2}\right) X Y=\left(\sigma_{1} X-\sigma_{2} Y\right)^{2}+4 \alpha_{1} \alpha_{2} X Y$$ Hence conclude that the eigenvalues can never be complex-valued.

Consider again the pendulum equations (sce Problem \(21)\) $$ d x / d t=y, \quad d y / d t=-4 \sin x $$ If the pendulum is set in motion from its downward equilibrium position with angular velocity \(v,\) then the initial conditions are \(x(0)=0, y(0)=v\) (a) Plot \(x\) versus \(t\) for \(v=2\) and also for \(v=5 .\) Explain the differing motions of the pendulum that these two graphs represent. (b) There is a critical value of \(v,\) which we denote by \(v_{c}\), such that one type of motion occurs for \(vv_{c}\). Estimate the value of \(v_{c} .\)

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