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Chapter 2: Problem 6

Involve equations of the form \(d y / d t=f(y)\). In each problem sketch the graph of \(f(y)\) versus \(y,\) determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. $$ d y / d t=-2(\arctan y) /\left(1+y^{2}\right), \quad-\infty

Short Answer

Expert verified
Answer: The equilibrium point at \(y = 0\) is asymptotically stable.

Step by step solution

01

Understand the given differential equation

The given differential equation is: $$ \frac{dy}{dt} = -2\frac{\arctan{y}}{1+y^2} $$ This is an autonomous first-order differential equation because there is no explicit time dependence in the equation.
02

Sketch the graph of \(f(y)\) versus \(y\)

To sketch the graph of \(f(y) = -2\frac{\arctan{y}}{1+y^2}\) versus \(y\), we can consider the behavior of \(f(y)\) at some key points. 1. When \(y = 0\), we have \(f(0) = -2\frac{\arctan{0}}{1+0^2} = 0\). 2. When \(y \rightarrow \infty\), since \(\arctan{y}\) approaches \(\frac{\pi}{2}\), we have \(f(y) \rightarrow -2\frac{\frac{\pi}{2}}{1+y^2} = 0\). 3. When \(y \rightarrow -\infty\), since \(\arctan{y}\) approaches \(-\frac{\pi}{2}\), we have \(f(y) \rightarrow -2\frac{-\frac{\pi}{2}}{1+y^2} = 0\). From these results, we can observe that the function \(f(y)\) is a continuous and bounded function in the range \(-\infty < y < \infty\). It starts at zero for \(y \rightarrow -\infty\), goes to zero at \(y=0\), and finally approaches zero as \(y \rightarrow \infty\). Now we sketch the graph of \(f(y)\) versus y, using the above insights.
03

Determine the critical (equilibrium) points

Critical points (equilibrium points) occur when \(\frac{dy}{dt} = 0\). This implies the following equation: $$ -2\frac{\arctan{y}}{1+y^2} = 0 $$ Since the only case when the equation can be true is when the numerator is zero (\(\arctan{y} = 0\)), the only equilibrium point is \(y = 0\).
04

Classify each critical point as asymptotically stable or unstable

Now we will determine if the equilibrium point is asymptotically stable or unstable by examining the behavior of \(f(y)\) in the vicinity of \(y = 0\). For \(y > 0\), we have \(\arctan{y} > 0\), and since \(1+y^2 > 1\), then \(\frac{\arctan{y}}{1+y^2} > 0\). Therefore, \(f(y) = -2\frac{\arctan{y}}{1+y^2} < 0\). This means that if \(y > 0\), then \(\frac{dy}{dt} < 0\), indicating that \(y\) will decrease towards the equilibrium point. For \(y < 0\), we have \(\arctan{y} < 0\), and since \(1+y^2 > 1\), then \(\frac{\arctan{y}}{1+y^2} < 0\). Therefore, \(f(y) = -2\frac{\arctan{y}}{1+y^2} > 0\). This means that if \(y < 0\), then \(\frac{dy}{dt} > 0\), indicating that \(y\) will increase towards the equilibrium point. Since both conditions \(y > 0\) and \(y < 0\) show that the system will converge to the equilibrium point at \(y = 0\), we can conclude that this point is asymptotically stable.

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Key Concepts

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

Equilibrium Points
In the context of autonomous differential equations, equilibrium points are the values of the variable where the rate of change is zero. For the exercise given, the differential equation is \(\frac{dy}{dt} = -2\frac{\arctan{y}}{1+y^2}\). To find the equilibrium points, we set \(\frac{dy}{dt} = 0\). This means solving:
  • \(-2\frac{\arctan{y}}{1+y^2} = 0\)
  • Here, this can only be true if \(\arctan{y} = 0\).
This leads us to the conclusion that the only equilibrium point is \(y = 0\). Equilibrium points can be thought of as places where the function "settles" or "balances out." They are critical in understanding how the behavior of solutions to the differential equation evolves over time.
Asymptotic Stability
Asymptotic stability at an equilibrium point is determined by whether small deviations from the equilibrium point will eventually return to it. In this exercise, we have found one equilibrium point at \(y = 0\). To analyze stability, examine the behavior of \(f(y)\) nearby:
  • For \(y > 0\), \(\arctan{y} > 0\), thus \(\frac{dy}{dt} < 0\). This implies that \(y\) decreases towards zero.
  • For \(y < 0\), \(\arctan{y} < 0\), resulting in \(\frac{dy}{dt} > 0\). Hence, \(y\) increases towards zero.
Because \(y\) approaches zero from both positive and negative values, this equilibrium point at \(y = 0\) is asymptotically stable. It ensures that small perturbations in \(y\) will lead back to the equilibrium, indicating a naturally stabilizing behavior.
Phase Line Analysis
Phase line analysis provides a visual way of analyzing the stability and direction of trajectories in a differential equation with respect to equilibrium points. For our equation \(\frac{dy}{dt} = -2\frac{\arctan{y}}{1+y^2}\), the phase line helps illustrate these:
  • Identify equilibrium points on a y-axis line, here simply \(y = 0\).
  • Add arrows indicating the direction of \(y\) movement based on \(\frac{dy}{dt}\). For \(y > 0\), the arrows point left (since \(\frac{dy}{dt} < 0\)), and for \(y < 0\), arrows point right (\(\frac{dy}{dt} > 0\)).
Through this phase line, you can observe that all trajectories, whether starting from positive or negative \(y\), move towards the equilibrium point at \(y = 0\). This tool simplifies the understanding of how the system inevitably stabilizes around its equilibrium, reflecting in real-world terms how a system may naturally return to a steady state.

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Most popular questions from this chapter

Consider the sequence \(\phi_{n}(x)=2 n x e^{-n x^{2}}, 0 \leq x \leq 1\) (a) Show that \(\lim _{n \rightarrow \infty} \phi_{n}(x)=0\) for \(0 \leq x \leq 1\); hence $$ \int_{0}^{1} \lim _{n \rightarrow \infty} \phi_{n}(x) d x=0 $$ (b) Show that \(\int_{0}^{1} 2 n x e^{-x x^{2}} d x=1-e^{-x}\); hence $$ \lim _{n \rightarrow \infty} \int_{0}^{1} \phi_{n}(x) d x=1 $$ Thus, in this example, $$ \lim _{n \rightarrow \infty} \int_{a}^{b} \phi_{n}(x) d x \neq \int_{a}^{b} \lim _{n \rightarrow \infty} \phi_{n}(x) d x $$ even though \(\lim _{n \rightarrow \infty} \phi_{n}(x)\) exists and is continuous.

Solve the differential equation $$ \left(3 x y+y^{2}\right)+\left(x^{2}+x y\right) y^{\prime}=0 $$ using the integrating factor \(\mu(x, y)=[x y(2 x+y)]^{-1}\). Verify that the solution is the same as that obtained in Example 4 with a different integrating factor.

Harvesting a Renewable Resource. Suppose that the population \(y\) of a certain species of fish (for example, tuna or halibut) in a given area of the ocean is described by the logistic equation $$ d y / d t=r(1-y / K) y . $$ While it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level, and possibly even driven to extinction. Problems 20 and 21 explore some of the questions involved in formulating a rational strategy for managing the fishery. In this problem we assume that fish are caught at a constant rate \(h\) independent of the size of the fish population. Then \(y\) satisfies $$ d y / d t=r(1-y / K) y-h $$ The assumption of a constant catch rate \(h\) may be reasonable when \(y\) is large, but becomes less so when \(y\) is small. (a) If \(hy_{0}>y_{1},\) then \(y \rightarrow y_{2}\) as \(t \rightarrow \infty,\) but that if \(y_{0}r K / 4,\) show that \(y\) decreases to zero as \(l\) increases regardless of the value of \(y_{0}\). (c) If \(h=r K / 4\), show that there is a single cquilibrium point \(y=K / 2\) and that this point is semistable (see Problem 7 ). Thus the maximum sustainable yield is \(h_{m}=r K / 4\) corresponding to the equilibrium value \(y=K / 2 .\) Observe that \(h_{m}\) has the same value as \(Y_{m}\) in Problem \(20(\mathrm{d})\). The fishery is considered to be overexploited if \(y\) is reduced to a level below \(K / 2\).

Epidemics. The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli in 1760 on smallpox. In more recent years many mathematical models have been proposed and studied for many different diseases. Deal with a few of the simpler models and the conclusions that can be drawn from them. Similar models have also been used to describe the spread of rumors and of consumer products. Daniel Bemoulli's work in 1760 had the goal of appraising the effectiveness of a controversial inoculation program against smallpox, which at that time was a major threat to public health. His model applies equally well to any other disease that, once contracted and survived, confers a lifetime immunity. Consider the cohort of individuals born in a given year \((t=0),\) and let \(n(t)\) be the number of these individuals surviving \(l\) years later. Let \(x(t)\) be the number of members of this cohort who have not had smallpox by year \(t,\) and who are therefore still susceptible. Let \(\beta\) be the rate at which susceptibles contract smallpox, and let \(v\) be the rate at which people who contract smallpox die from the disease. Finally, let \(\mu(t)\) be the death rate from all causes other than smallpox. Then \(d x / d t,\) the rate at which the number of susceptibles declines, is given by $$ d x / d t=-[\beta+\mu(t)] x $$ the first term on the right side of Eq. (i) is the rate at which susceptibles contract smallpox, while the second term is the rate at which they die from all other causes. Also $$ d n / d t=-v \beta x-\mu(t) n $$ where \(d n / d t\) is the death rate of the entire cohort, and the two terms on the right side are the death rates duc to smallpox and to all other causes, respectively. (a) Let \(z=x / n\) and show that \(z\) satisfics the initial value problem $$ d z / d t=-\beta z(1-v z), \quad z(0)=1 $$ Observe that the initial value problem (iii) does not depend on \(\mu(t) .\) (b) Find \(z(t)\) by solving Eq. (iii). (c) Bernoulli estimated that \(v=\beta=\frac{1}{8} .\) Using these values, determine the proportion of 20 -year-olds who have not had smallpox.

Involve equations of the form \(d y / d t=f(y) .\) In each problem sketch the graph of \(f(y)\) versus \(y\), determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable (see Problem 7 ). $$ d y / d t=y^{2}\left(y^{2}-1\right), \quad-\infty
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