Problem 42 - Differential Equations and Linear Algebra

Question

Convert the given DE to a first-order system using the substitution \(u=y, v=\frac{d y}{d t},\) and determine the phase portrait for the resulting system. $$\frac{d^{2} y}{d t^{2}}+\frac{d y}{d t}-12 y=0$$

Step-by-Step Solution

Answer

The given second-order differential equation is converted into a first-order system using the substitution \(u=y\) and \(v=\frac{dy}{dt}\). The resulting system of equations is: \[\begin{cases} \frac{du}{dt} = v\\ \frac{dv}{dt} = 12u - v \end{cases}\] The equilibrium point is (0, 0) and the phase portrait exhibits counterclockwise rotations around the equilibrium point in the first and third quadrants.

Step 1: Convert the given DE to a first-order system

Use the substitutions: \(u = y\) and \(v = \frac{dy}{dt}\). We need to relate \(u\) and \(v\) to the original equation, so we rewrite it using the substitutions: \(\frac{d(u)}{dt} = v\) \(\frac{d(v)}{dt} = \frac{d^2y}{dt^2} = 12u - v\) Now, we have the following system of equations: \[\begin{cases} \frac{du}{dt} = v\\ \frac{dv}{dt} = 12u - v \end{cases}\]

Step 2: Find the equilibrium points

An equilibrium point, \((u^*,v^*)\), occurs when the derivatives of both variables are zero: \[\begin{cases} \frac{du}{dt} = 0\\ \frac{dv}{dt} = 0 \end{cases}\] Substitute the given equations into this system: \[\begin{cases} v = 0\\ 12u - v = 0 \end{cases}\] Solve for the equilibrium points (u, v): \[\begin{cases} u = 0 \\ 12(0) - v = 0 \\ v = 0 \end{cases}\] Thus, there is only one equilibrium point (\(u^*\), \(v^*\)) = (0, 0).

Step 3: Determine the phase portrait

To determine the phase portrait, we will analyze the behavior of the system around the equilibrium point: - When u > 0, the system will have positive values for \(\frac{du}{dt}\) and negative values for \(\frac{dv}{dt}\). This means that the solutions will travel counterclockwise in the first quadrant. - When u < 0, the system will have negative values for \(\frac{du}{dt}\) and positive values for \(\frac{dv}{dt}\). This indicates that the solutions will travel counterclockwise in the third quadrant. The phase portrait will show a counterclockwise rotation of the trajectories around the equilibrium point (0, 0) in both the first and third quadrants. In conclusion, upon converting the given second-order differential equation into a system of first-order differential equations, we found that their phase portrait showcases counterclockwise rotations around the equilibrium point (0, 0).

Key Concepts

First-Order System

A differential equation of first-order involves only the first derivative of an unknown function. These first-order systems are essential tools in analyzing more complex, higher-order scenarios found in models. To convert a second-order differential equation to a first-order system, substitutions are often used.

• In our case, we used u = y and v = \(\frac{dy}{dt}\).

This reduces our original second-order equation to two first-order equations. By rewriting the original second-order equation with these substitutions, we get:

• \( \frac{du}{dt} = v \)
• \( \frac{dv}{dt} = 12u - v \)

This system of equations now allows us to study the system behavior using simpler mathematics. A vital aspect of first-order systems is paving the way to analyze the dynamics and trajectories of solutions.

Phase Portrait

A phase portrait visually represents the trajectories that solutions take in a dynamic system of differential equations. It's like a roadmap showing how solutions change over time. Understanding the phase portrait is crucial for observing the overall behavior of the system near the equilibrium points.
The given system produces a phase portrait that depicts counterclockwise motion around the equilibrium point (0, 0). This visualization gives insight into how the system's variables interact:

• For regions where u > 0, the phase arrows indicate a combination of increasing v and decreasing u.
• For regions where u < 0, the direction is such that the solutions rotate back toward the origin.

These rotations hint at self-correcting behavior, a property prevalent in systems that try, through feedback mechanisms, to return to equilibrium.

Equilibrium Point

An equilibrium point in a system of differential equations is where the system remains unchanged over time, meaning all derivatives are zero.
In our system's case, to find this equilibrium point, we solve the equations such that:

• \( \frac{du}{dt} = 0 \)
• \( \frac{dv}{dt} = 0 \)

By substituting from the first-order system, we found:

• u = 0
• v = 0

The single equilibrium point is (0, 0), indicating that when both u and v are zero, no changes occur. This information is vital since the behavior of other solutions is often discussed concerning these equilibrium points. Whether they move toward, away from, or circle these points indicates stability characteristics of the system.

Second-Order Differential Equation

Second-order differential equations involve the second derivative of a function and are frequently used to model real-world phenomena such as mechanical systems, electrical circuits, and more. They provide a more profound insight into systems that change not just with their current states, but also with how those states evolve.
The given differential equation was:

• \( \frac{d^{2} y}{dt^{2}} + \frac{d y}{dt} - 12y = 0 \)

This equation can pose challenges when solved directly. Converting it to a first-order system simplifies understanding its dynamic behavior. By translating the second-order terms into first-order terms through substitution, we bridge the gap between complex processes and their simpler mathematical representation. This approach illuminates new perspectives when analyzing the nature of solutions, equilibrium, and overall system dynamics.