Question

In Exercises \(1-4\) find the equations of the trajectories of the given undamped equation. Identify the equilibrium solutions, determine whether they are stable or unstable, and plot some trajectories. $$ \begin{array}{|l|l|} \hline \mathrm{C} / \mathrm{G} & y^{\prime \prime}+y|y|=0 \end{array} $$

Short answer

#Answer# 1. The equilibrium solution for the undamped equation $y^{\prime \prime}+y|y|=0$ is $y=0$. 2. The equilibrium solution is unstable for all values of $y$. 3. The trajectories of the equation show that: a) For positive values of $y$, the trajectory is concave down. b) For negative values of $y$, the trajectory is concave up.

Step-by-step solution

Rewrite the given equation and find the equilibrium solutions

First, we rewrite the given equation as: $$y^{\prime \prime}=-y|y|$$ Now, the equilibrium solutions are the values of \(y\) that make the second derivative, \(y^{\prime \prime}\), equal to 0. Thus, we have: $$-y|y|=0$$ This equation holds when either \(y=0\) or \(|y|=0\). In either case, the equilibrium solution is \(y=0\).

Determine the stability of the equilibrium solutions

To determine the stability of the equilibrium solutions, we will examine the behavior of the second derivative, \(y^{\prime \prime}\), around the equilibrium points. Consider the case when \(y>0\). In this case, \(|y|=y\), and the equation becomes: $$y^{\prime \prime}=-y^2$$ This equation shows that when \(y>0\), the second derivative is always negative, implying the solution will move away from the equilibrium point. Thus, the equilibrium solution \(y=0\) is unstable for \(y>0\). Now, consider the case when \(y<0\). In this case, \(|y|=-y\), and the equation becomes: $$y^{\prime \prime}=y^2$$ This equation shows that when \(y<0\), the second derivative is always positive, implying the solution will move away from the equilibrium point. Thus, the equilibrium solution \(y=0\) is also unstable for \(y<0\). Therefore, the equilibrium solution \(y=0\) is unstable.

Plot some trajectories of the given equation

Now, let's plot some trajectories of the given equation: $$y^{\prime \prime}=-y|y|$$ To do this, we want to find the expressions for \(y^{\prime \prime}\) for various values of \(y\): For positive values of \(y\), \(y^{\prime \prime}=-y^2\): - For \(y>0\), \(y^{\prime \prime}<0\), implying the trajectory will be concave down. For negative values of \(y\), \(y^{\prime \prime}=y^2\): - For \(y<0\), \(y^{\prime \prime}>0\), implying the trajectory will be concave up. To plot the trajectories, we will consider four points around the equilibrium solution \(y=0\): 1. \(y = 0.5\) 2. \(y = -0.5\) 3. \(y = 1\) 4. \(y = -1\) These four points will allow us to see the behavior of the trajectories around the equilibrium solution, and verify that the solution is unstable for all values of \(y\).

Key concepts

Equilibrium Solutions

When studying differential equations, particularly those representing physical systems, identifying points where the system is at rest or in a steady state is critical. These points are known as equilibrium solutions. For equation

$$y^{\'\prime}=-y|y|$$

we find that equilibrium solutions occur at values of \(y\) which cause the second derivative \(y^{\'\prime}\) to vanish. In this case, the equilibrium solution is simply \(y=0\), where the system experiences no further change unless perturbed. Understanding equilibrium helps predict system behavior without having to solve the entire differential equation. It acts as a snapshot where the forces in the system are in balance.

Stability Analysis

Following the discovery of equilibrium solutions, the next step is often stability analysis. The stability of an equilibrium solution determines whether the system, once disturbed slightly, will return to its equilibrium state or move far away from it. In the example,

$$y^{\'\prime}=-y|y|$$

we assess stability by examining the second derivative on either side of the equilibrium point, i.e., for \(y>0\) and \(y<0\). We discovered that for \(y>0\) the second derivative is negative, while for \(y<0\), it is positive. This implies that the solution will diverge from \(y=0\) if perturbed in either direction, classifying \(y=0\) as an unstable equilibrium point.

Second Derivative Test

The second derivative test is a widely used tool to determine the concavity of graphs of functions and can be employed for stability analysis in differential equations. By inspecting the sign of the second derivative, we can ascertain if a local maximum or minimum — or neither — is present at a certain point. Here's a closer look:

• For \(y>0\): \(y^{\'\prime}=-y^2\), which indicates concavity down, hinting at instability as the trajectories lead away from equilibrium.
• For \(y<0\): \(y^{\'\prime}=y^2\), suggesting concavity up and similarly demonstrating instability as trajectories also lead away from equilibrium.

Conclusively, this test reveals that not only is \(y=0\) an equilibrium point, but it is also an unstable one, confirming prior assessments realized in stability analysis.

Undamped Motion

Undamped motion refers to systems that oscillate without any loss of energy, meaning the amplitude of the oscillation does not decrease over time. In the context of the given differential equation,

$$y^{\'\prime}=-y|y|$$

this system appears to depict undamped motion as there are no terms present that would suggest a diminishment of the system's energy. However, despite this label, the specific form of nonlinearity in our equation creates an atypical undamped behavior where the solution does not oscillate about the equilibrium point. Rather, the force driven by \(|y|\) leads to increased separation from equilibrium, differing from traditional undamped systems where variables typically oscillate symmetrically about the equilibrium.