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Chapter 2: Problem 32

A critical point \(c\) of an autonomous first-order DE is said to be isolated if there exists some open interval that contains \(c\) but no other critical point. Discuss: Can there exist an autonomous DE of the form given in (1) for which every critical point is nonisolated? Do not think profound thoughts.

Short Answer

Expert verified
Yes, every critical point can be nonisolated if the function \( f(y) = 0 \) for all \( y \).

Step by step solution

01

Understanding Critical Points and Definition

A critical point of an autonomous differential equation is a point where the derivative is zero: \( f(y) = 0 \). The problem asks if every critical point can be nonisolated, meaning each critical point has other critical points arbitrarily close to it.
02

Exploring the Concept of Continuous Functions

Consider the function \( f(y) \) that defines the DE. For a critical point to be nonisolated, \( f(y) = 0 \) must hold over intervals, not just discrete points. This would mean that \( f(y) \) is zero over an interval, rather than at isolated points.
03

Construct a Function with Non-Isolated Critical Points

An example of such a function is \( f(y) = 0 \) for all \( y \), which is trivially zero everywhere. This means that the derivative \( \frac{dy}{dt} = 0 \), implying the system has infinite nonisolated critical points everywhere on the real line.
04

Conclusion on Existence

Yes, it is possible for an autonomous DE to have every critical point nonisolated by defining \( f(y) \) as zero over all \( y \). This function results in the critical points forming a continuum, since they are present at every point in any interval over the entire real line.

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Key Concepts

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

Critical Points
In autonomous differential equations, critical points are of great importance. A critical point, often denoted by a variable like \( c \), is where the derivative of the function \( f(y) \) equals zero: \( f(y) = 0 \). This indicates that at these points, there is no change in \( y \) with respect to time, meaning that the solution is stationary at these points.
Critical points are key to understanding the behavior of solutions to differential equations, as they often represent equilibrium states of a system. These are places where the dynamical system can potentially rest or change its stability. Recognizing these points allows us to analyze the potential behavior at particular states and determine the stability of these solutions.
Isolated Points
Isolated points are a specific type of critical point in the context of differential equations. An isolated critical point \( c \) in an autonomous differential equation is a point where there exists some open interval around \( c \) that contains no other critical points. This means within the neighborhood surrounding \( c \), \( f(y) = 0 \) holds only at \( c \).
Isolated points are distinct in that they stand alone, without other critical points clustering around them. They can be very useful in analyzing systems because they indicate points where the behavior of the system doesn't get influenced by other critical states nearby. This isolation often simplifies the study of the stability of a system around such critical points, as they are easier to explore individually.
First-Order Differential Equations
A first-order differential equation involves the derivatives of a function in terms of the first derivative only, without higher derivatives. For autonomous differential equations of the form \( \frac{dy}{dt} = f(y) \), the function \( f(y) \) does not explicitly depend on time \( t \). Instead, the change in \( y \) is driven by \( y \) itself.
This simplicity is a core reason for studying first-order equations, as they often form the basis for understanding more complex systems. In autonomous systems, these first-order equations help us understand the fundamental behavior of the system over time, often leading us to focus on critical and isolated points. By examining \( f(y) \) and its zeros (the critical points), we gain valuable insights into where the system may stabilize or change state, showing the flow and stability of the system's solutions.

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Most popular questions from this chapter

Consider the competition model defined by $$ \begin{aligned} &\frac{d x}{d t}=x(2-0.4 x-0.3 y) \\ &\frac{d y}{d t}=y(1-0.1 y-0.3 x) \end{aligned} $$ where the populations \(x(t)\) and \(y(t)\) are measured in the thousands and \(t\) in years. Use a numerical solver to analyze the populations over a long period of time for each of the cases: (a) \(x(0)=1.5, \quad y(0)=3.5\) (b) \(x(0)=1, \quad y(0)=1\) (c) \(x(0)=2\) \(y(0)=7\) (d) \(x(0)=4.5\), \(y(0)=0.5\)

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ L \frac{d i}{d t}+R i=E ; \quad i(0)=i_{0}, L, R, E, \text { and } i_{0} \text { constants } $$

A differential equation governing the velocity \(v\) of a falling mass \(m\) subjected to air resistance proportional to the square of the instantaneous velocity is $$ m \frac{d v}{d t}=m g-k v^{2}, $$ where \(k>0\) is the drag coefficient. The positive direction is downward. (a) Solve this equation subject to the initial condition \(v(0)=v_{0}\) (b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the \(\mathrm{DE}\) in Problem 39 in Exercises \(2.1\). (c) If distance \(s\), measured from the point where the mass was released above ground, is related to velocity \(v\) by \(d s / d t=v(t)\), find an explicit expression for \(s(t)\) if \(s(0)=0\).

A tank in the form of a right circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3, when friction and contraction of water at the hole are ignored, the height \(h\) of water in the tank is described by $$ \frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h}, $$ where \(A_{w}\) and \(A_{h}\) are the cross-sectional areas of the water and the hole, respectively. (a) Solve for \(h(t)\) if the initial height of the water is \(H\). By hand, sketch the graph of \(h(t)\) and give its interval \(I\) of definition in terms of the symbols \(A_{w}, A_{h}\), and \(H\). Use \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) (b) Suppose the tank is \(10 \mathrm{ft}\) high and has radius \(2 \mathrm{ft}\) and the circular hole has radius \(\frac{1}{2}\) in. If the tank is initially full, how long will it take to empty?

Rocket Motion Suppose a small single-stage rocket of total mass \(m(t)\) is launched vertically and that the rocket consumes its fuel at a constant rate. If the positive direction is upward and if we take air resistance to be linear, then a differential equation for its velocity \(v(t)\) is given by $$ \frac{d v}{d t}+\frac{k-\lambda}{m_{0}-\lambda t} v=-g+\frac{R}{m_{0}-\lambda t^{\prime}} $$ where \(k\) is the drag coefficient, \(\lambda\) is the rate at which fuel is consumed, \(R\) is the thrust of the rocket, \(m_{0}\) is the total mass of the rocket at \(t=0\), and \(g\) is the acceleration due to gravity. See Problem 21 in Exercises \(1.3\). (a) Find the velocity \(v(t)\) of the rocket if \(m_{0}=200 \mathrm{~kg}\), \(R=2000 \mathrm{~N}, \lambda=1 \mathrm{~kg} / \mathrm{s}, g=9.8 \mathrm{~m} / \mathrm{s}^{2}, k=3 \mathrm{~kg} / \mathrm{s}\), and \(v(0)=0\) (b) Use \(d s / d t=v\) and the result in part (a) to find the height \(s(t)\) of the rocket at time \(t\).
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