Question

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

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

Step-by-step solution

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.

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.

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 \(y = 0\).

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.

Key concepts

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.