Problem 23 - Differential Equations and Linear Algebra

Question

Convert the given differential equation 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. $$\frac{d^{2} y}{d t^{2}}+16 y=0$$

Step-by-Step Solution

Answer

We have converted the given second-order differential equation \(\frac{d^{2}y}{dt^{2}}+16y=0\) to a first-order system using the substitution \(u=y, v=\frac{dy}{dt}\), resulting in the system \(\frac{du}{dt} = v, \frac{dv}{dt} = -16u\). The phase portrait has concentric circles centered at the origin, representing different energy levels, with a clockwise flow around the origin.

Step 1: Perform Substitution

Given \(u = y\) and \(v = \frac{dy}{dt}\), we can rewrite the second-order differential equation as: \(u = y\) \(\frac{du}{dt} = \frac{dy}{dt} = v\) Now, differentiate \(v\) with respect to \(t\): \(\frac{dv}{dt} = \frac{d^{2}y}{dt^{2}}\) So, we have: \(\frac{d^{2}y}{dt^{2}}+16y = \frac{dv}{dt} + 16u = 0\)

Step 2: Separate into Two First-Order Equations

Now that we've performed the substitution, we can separate the equation into two first-order equations: 1) \(\frac{du}{dt} = v\) 2) \(\frac{dv}{dt} = -16u\)

Step 3: Analyze the First-Order System

The given first-order system consists of the following two equations: \(\frac{du}{dt} = v\) \(\frac{dv}{dt} = -16u\) This is a linear first-order system. We should notice that both of these equations are decoupled since each derivative is expressed only in terms of the other variable. This makes the system easier to analyze and the phase portrait can be determined independently.

Step 4: Determine the Phase Portrait for the First-Order System

The phase portrait will show the relationship between \(u\) (y-axis) and \(v\) (x-axis) through their derivatives. For the given system, it's useful to rewrite the second equation in terms of the variables: \(v = -\frac{1}{16} \cdot \frac{dv}{du}\) The phase portrait can be visualized by sketching the directions of the flow given by the equation. Since the system is linear, it is symmetric about the origin along the \(u\)-axis. Consequently, the phase portrait will consist of a set of concentric circles centered at the origin, with radius proportional to the energy in the system. The flow will be clockwise for this equation. In summary, we have successfully converted the given second-order differential equation to a first-order system and determined its phase portrait. The phase portrait consists of a set of concentric circles centered at the origin representing different energy levels, with clockwise flow around the origin.

Key Concepts

Understanding Phase Portraits

Phase portraits are a visual representation of a dynamical system in the phase plane. They help us understand the behavior of solutions to differential equations. Imagine each point in the phase plane as representing a possible state of the system. By observing how points move around the phase plane, we can see how the system evolves over time.

When examining a system of differential equations, we plot trajectories that follow the direction of the derivatives. For our system:

• The horizontal axis represents the variable \(v\), which is the derivative of \(y\).
• The vertical axis is the variable \(u\), corresponding to \(y\) itself.

In the analyzed first-order system, the phase portrait showed concentric circles centered at the origin.

• These circles indicate harmonic motion, typical of systems like simple pendulums.
• Each circle corresponds to a different initial condition or energy level.

The directionality of movement on the phase portrait helps identify whether the system is stable, with solutions remaining bounded over time, or unstable if they diverge. In our case, solutions follow the symmetric circles, indicating stable behavior with periodic oscillation.

Introduction to Linear Systems

Linear systems in the context of differential equations involve writing the system as a set of equations where each equation is linear. This means that all terms involving the dependent variable and its derivatives are first-degree polynomials.

In our differential equation conversion:

• We started with a second-order differential equation \(\frac{d^2 y}{dt^2} + 16y = 0\).
• After substitution, we created a linear system with: \(\frac{du}{dt} = v\) and \(\frac{dv}{dt} = -16u\).

This transformation simplifies the original problem:

• Each first-order equation carries clear meaning and significance, focusing on the rate of change for each variable.
• By working with first-order systems, we can more easily apply powerful tools for analysis, such as phase portraits and eigenvalue techniques.

Linear systems like this one are particularly valuable because they can be solved analytically. This allows us to predict the system's behavior based on initial conditions, often leading to a deeper understanding of the physical phenomena described by the differential equations.

Differential Equation Conversion Process

Converting a higher-order differential equation into a first-order system is a valuable technique in solving and analyzing complex problems.

To achieve this conversion:

• Begin by introducing new variables to represent the original function and its derivatives. In our example, we used \(u = y\) and \(v = \frac{dy}{dt}\).
• Rewrite the second-order differential equation using these new variables. This step often involves differentiating one of the substitutions, as with \(\frac{du}{dt} = v\) and \(\frac{dv}{dt} = -16u\) derived from \(\frac{d^2 y}{dt^2} + 16y = 0\).

This transformation has several advantages:

• It reduces the complexity of the problem by breaking it into simpler, more manageable equations.
• It allows the use of graphical methods, like phase portraits, to visualize solutions and predict long-term behavior.

Carrying out these steps provides a structured approach to tackling systems of differential equations, offering insights and easing the path to finding solutions, especially when dealing with linear, time-invariant systems. By following such a systematic approach, one can effectively transform and solve a wide array of differential equations.