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Chapter 9: Problem 2

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

Short Answer

Expert verified
Based on our analysis of the given system of autonomous differential equations in polar coordinates, we can conclude that there are periodic solutions with period \(2\pi\) for any radius value, and there is one stable limit cycle at \(r=1\). The stability characteristics of the periodic solutions remain constant, while the stability of the limit cycle is determined by the stability of the fixed point \(r=1\).

Step by step solution

01

Identify the given system

We have the following system of autonomous differential equations in polar coordinates: $$ \frac{dr}{dt} = r(1-r)^2, \quad \frac{d\theta}{dt} = -1 $$ Step 2: Analyze the behavior of the angle
02

Analyze the behavior of the angle

The equation for the angle is independent of the radius and only depends on time. The equation is given by: $$ \frac{d\theta}{dt} = -1 $$ This equation represents a simple angular motion with constant angular speed equal to \(-1\). Step 3: Analyze the behavior of the radius
03

Analyze the behavior of the radius

Now, let's analyze the equation for the radius: $$ \frac{dr}{dt} = r(1-r)^2 $$ We can observe that the equation is nonlinear, yet tractable. First, let's find its fixed points by setting \(\frac{dr}{dt} = 0\). The fixed points are: $$ r = 0 \quad \text{and} \quad r = 1 $$ Step 4: Determine the stability of the fixed points
04

Determine the stability of the fixed points

We can study the stability of the fixed points by analyzing the sign of the derivative \(\frac{dr}{dt}\) near each fixed point: 1. If \(0 < r < 1\), \(\frac{dr}{dt} > 0\), indicating that the radius will increase. Hence, \(r=0\) is an unstable fixed point. 2. If \(r > 1\), \(\frac{dr}{dt} < 0\), indicating that the radius will decrease. Hence, \(r=1\) is a stable fixed point. Step 5: Identify periodic solutions and limit cycles
05

Identify periodic solutions and limit cycles

Since \(\frac{d\theta}{dt} = -1\), there will always be periodic solutions with period \(2\pi\), regardless of the radius value. As for the limit cycles, we have one limit cycle at \(r=1\). Step 6: Determine the stability characteristics of the periodic solutions and limit cycles
06

Determine the stability characteristics of the periodic solutions and limit cycles

Based on our analysis above, we conclude that: 1. There are periodic solutions with period \(2\pi\) for any radius value. 2. There is one limit cycle at \(r=1\), which is stable. The stability characteristics of the periodic solutions do not change, while the stability of the limit cycle is determined by the stability of the fixed point \(r=1\).

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Key Concepts

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

Polar Coordinates
Polar coordinates are a two-dimensional coordinate system where each point in the plane is described by a distance from a reference point and an angle from a reference direction.
In this coordinate system, we often use the letter \(r\) for the radius (the distance from the origin) and \(\theta\) for the angle.
They are particularly useful in autonomous systems when dealing with problems that exhibit symmetry or rotational qualities.
  • The system given in the exercise uses polar coordinates, with equations for both radius \(r\) and angle \(\theta\).
  • Such systems can simplify differential equations by separating the motion into radial and angular components.
This makes it easier to study the behavior of systems in circular paths or radial motion.
Understanding polar coordinates is crucial for analyzing autonomous systems composed of circular motion or those that naturally fit into rotational frameworks.
Limit Cycles
Limit cycles represent closed trajectories in the phase space of a dynamical system.
These are essentially loops or cycles that solutions will approach, regardless of their initial state, over time.
In our exercise, the limit cycle occurs at \(r = 1\).
  • The equation \(\frac{dr}{dt} = r(1-r)^2\) suggests that when \(r = 1\), the growth rate becomes zero, indicating a steady behavior.
  • This limit cycle is common in models with oscillatory or cyclic behavior, such as predator-prey systems or chemical reactions with feedback loops.
By finding limit cycles, we can predict points of equilibrium in repetitive processes, which is a foundation for further stability analysis.
Recognizing limit cycles helps in understanding the long-term behavior of autonomous systems.
Periodic Solutions
Periodic solutions refer to solutions of differential equations that repeat at regular intervals.
They are characterized by a specific period, the time it takes for the process to complete one full cycle and restart.
In the given scenario, the periodic solutions are determined by the angular motion \(\frac{d\theta}{dt} = -1\).
  • This equation results in solutions with a period of \(2\pi\), as the motion represents a complete circular rotation.
  • The periodic nature here is crucial for predicting the dynamic behavior of systems that repeat over time, like the rotation of planets or oscillations in mechanical systems.
Periodic solutions allow us to simplify complex systems by focusing on their repetitive cycles, providing insights into their continuous evolution.
Understanding these solutions is fundamental for predicting system behavior in systems with inherent repetitive motions.
Stability Analysis
Stability analysis involves examining the behavior of solutions to differential equations as time progresses.
It tells us whether small deviations will cause solutions to return to a steady state (stable) or deviate further away (unstable).
In the problem, stability analysis is used to evaluate the fixed points and the limit cycle.
  • At \(r = 0\), we found instability as the radius increased for values of \(0 < r < 1\).
  • Conversely, \(r = 1\) is stable since deviations shrink and the system returns to this state.
By using stability analysis, predictions can be made regarding how systems react to small disturbances.
This is crucial in many fields, such as engineering and biology, where system control and behavior must be managed.

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Most popular questions from this chapter

In this problem we show how small changes in the coefficients of a system of linear equations can affect the nature of a critical point when the eigenvalues are equal. Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{-1} & {1} \\ {0} & {-1}\end{array}\right) \mathbf{x} $$ Show that the eigenvalues are \(r_{1}=-1, r_{2}=-1\) so that the critical point \((0,0)\) is an asymptotically stable node. Now consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{-1} & {1} \\ {-\epsilon} & {-1}\end{array}\right) \mathbf{x} $$ where \(|\epsilon|\) is arbitrararily small. Show that if \(\epsilon>0,\) then the eigenvalues are \(-1 \pm i \sqrt{\epsilon}\), so that the asymptotically stable node becomes an asymptotically stable spiral point. If \(\epsilon<0,\) then the roots are \(-1 \pm \sqrt{|\epsilon|},\) and the critical point remains an asymptotically stable node.

(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. $$ \text { The van der Pol equation: } \quad d x / d t=y, \quad d y / d t=\left(1-x^{2}\right) y-x $$

(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. $$ d x / d t=(2+x)(y-x), \quad d y / d t=(4-x)(y+x) $$

(a) A special case of the Lienard equation of Problem 8 is $$ \frac{d^{2} u}{d t^{2}}+\frac{d u}{d t}+g(u)=0 $$ where \(g\) satisfies the conditions of Problem 6 . Letting \(x=u, y=d u / d t,\) show that the origin is a critical point of the resulting system. This equation can be interpreted as describing the motion of a spring-mass system with damping proportional to the velocity and a nonlinear restoring force. Using the Liapunov function of Problem \(6,\) show that the origin is a stable critical point, but note that even with damping we cannot conclude asymptotic stability using this Liapunov function. (b) Asymptotic stability of the critical point \((0,0)\) can be shown by constructing a better Liapunov function as was done in part (d) of Problem 7 . However, the analysis for a general function \(g\) is somewhat sophisticated and we only mention that appropriate form for \(V\) is $$ V(x, y)=\frac{1}{2} y^{2}+A y g(x)+\int_{0}^{x} g(s) d s $$ where \(A\) is a positive constant to be chosen so that \(V\) is positive definite and \(\hat{V}\) is negative definite. For the pendulum problem \([g(x)=\sin x]\) use \(V\) as given by the preceding equation with \(A=\frac{1}{2}\) to show that the origin is asymptotically stable. Hint: Use \(\sin x=x-\alpha x^{3} / 3 !\) and \(\cos x=1-\beta x^{2} / 2 !\) where \(\alpha\) and \(\beta\) depend on \(x,\) but \(0<\alpha<1\) and \(0<\beta<1\) for \(-\pi / 2

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