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

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

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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 \(r=1\).

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

(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. $$ d x / d t=-x+2 x y, \quad d y / d t=y-x^{2}-y^{2} $$

Consider the system $$ d x / d t=a x[1-(y / 2)], \quad d y / d t=b y[-1+(x / 3)] $$ where \(a\) and \(b\) are positive constants. Observe that this system is the same as in the example in the text if \(a=1\) and \(b=0.75 .\) Suppose the initial conditions are \(x(0)=5\) and \(y(0)=2\) (a) Let \(a=1\) and \(b=1 .\) Plot the trajectory in the phase plane and determine (or cstimate) the period of the oscillation. (b) Repeat part (a) for \(a=3\) and \(a=1 / 3,\) with \(b=1\) (c) Repeat part (a) for \(b=3\) and \(b=1 / 3,\) with \(a=1\) (d) Describe how the period and the shape of the trajectory depend on \(a\) and \(b\).

Construct a suitable Liapunov function of the form \(a x^{2}+c y^{2}\) where \(a\) and \(c\) are to be determined. Then show that the critical point at the origin is of the indicated type. $$ d x / d t=x^{3}-y^{3}, \quad d y / d t=2 x y^{2}+4 x^{2} y+2 y^{3} ; \quad \text { unstable } $$

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. $$ d x / d t=x-y^{2}, \quad d y / d t=x-2 y+x^{2} $$

Determine the critical point \(\mathbf{x}=\mathbf{x}^{0},\) and then classify its type and examine its stability by making the transformation \(\mathbf{x}=\mathbf{x}^{0}+\mathbf{u} .\) \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{0} & {-\beta} \\ {\delta} & {0}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right) ; \quad \alpha, \beta, \gamma, \delta>0\)

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