Problem 22 - 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. $$\frac{d^{2} y}{d t^{2}}+6 \frac{d y}{d t}+9 y=0$$

Step-by-Step Solution

Answer

Using the substitution \(u=y\) and \(v=\frac{dy}{dt}\), we can rewrite the given differential equation as a first-order system: \(\begin{cases} \frac{du}{dt} = v \\ \frac{dv}{dt} = -6v - 9u \end{cases}\) Analyzing the phase portrait, we find that the origin (0,0) is a stable node, with all trajectories tending towards it in the long run. The u-nullcline is the horizontal line at \(v=0\), and the v-nullcline is the line \(v=-\frac{3}{2}u\).

Step 1: Use the substitution

Let's use the substitution \(u=y\) and \(v=\frac{dy}{dt}\) for the given equation. Our aim is to rewrite the differential equation in terms of \(u\) and \(v\).

Step 2: Rewrite the equation in terms of u and v

Since we know \(v=\frac{dy}{dt}\), this also means that \(\frac{dv}{dt}=\frac{d^2y}{dt^2}\). Now, we can rewrite the given differential equation: \(\frac{d^2y}{dt^2} + 6\frac{dy}{dt} + 9y = 0 \) as \(\frac{dv}{dt} + 6v + 9u = 0\)

Step 3: Find the first-order differential equations

Now we can write the first-order system by finding the corresponding differential equations for \(u\) and \(v\). We have: 1. \(\frac{du}{dt} = v\) (since by definition \(u = y\), from the substitution rule) 2. Rearrange \( \frac{dv}{dt} + 6v + 9u = 0\) to get \(\frac{dv}{dt} = -6v - 9u\) Thus, we have the first-order system: \(\begin{cases} \frac{du}{dt} = v \\ \frac{dv}{dt} = -6v - 9u \end{cases}\)

Step 4: Analyzing the phase portrait for the first-order system

Now that we have the first-order system, let’s analyze the phase portrait. To do this, we will look at the nullclines, which are the curves in the \(u-v\) plane where the derivatives are zero. 1. When \(\frac{du}{dt} = 0\), we have \(v = 0\). This means our u-nullcline is a horizontal line through the v-axis at v=0. 2. When \(\frac{dv}{dt} = 0\), we have \(-6v - 9u = 0\), which simplifies to \(v = -\frac{3}{2}u\). This is our v-nullcline, which is a line passing through the origin with a slope of -3/2. By analyzing the slopes of the first-order differential equations in each of the four regions separated by the nullclines, we can determine the direction field and the behavior of the trajectories. Combining this information, we can sketch the phase portrait of the system. For this particular system, it turns out the origin (0,0) is a stable node, and all trajectories tend towards it in the long run.

Key Concepts

Phase Portrait

A phase portrait is an insightful tool that graphically represents the trajectory solutions of a system of differential equations in the phase plane. In our problem, we deal with a converted first-order system derived from a second-order differential equation. This graphical representation helps us understand the system's long-term behavior and stability.

The phase plane consists of axes corresponding to variables of the system, in this case, the variables are \(u\) and \(v\). Each point in the phase plane represents a unique state of the system at a particular time. The trajectories, or paths, that these points follow demonstrate how the state of the system evolves.

• Nullclines help divide the phase plane into regions indicating different directional dynamics.
• Stable nodes, as discussed later, show where the system's trajectories converge indicating stable solutions.

By plotting the different trajectories, we obtain a visual representation that captures all possible behaviors of the dynamic system.

System of Differential Equations

Systems of differential equations involve more than one differential equation and often describe how a set of quantities change over time. In our exercise, we transformed a single second-order differential equation into a system of first-order differential equations using substitutions. This is a common method used to simplify and solve complex differential equations.

The transformation was achieved by:

• Substituting \(u = y\) and \(v = \frac{dy}{dt}\).
• Rewriting the original differential equation to express derivatives as \(\frac{du}{dt}\) and \(\frac{dv}{dt}\).

The resulting first-order system:\[\begin{align*}\frac{du}{dt} &= v \\frac{dv}{dt} &= -6v - 9u\end{align*}\]provides us with two linked differential equations that describe how \(u\) and \(v\) change with time. By analyzing solutions to these equations, we can predict the behaviors and future states of the system.

Nullclines

Nullclines are critical for understanding the dynamics of a system of differential equations. They denote where the rate of change of one of the variables is zero.

In this exercise, nullclines help us delineate regions in the phase plane:

• The \(u\)-nullcline corresponds to \(\frac{du}{dt} = 0\), which happens when \(v = 0\). This is a horizontal line along the \(u\)-axis.
• The \(v\)-nullcline corresponds to \(\frac{dv}{dt} = 0\). Solving \(-6v - 9u = 0\) gives us \(v = -\frac{3}{2}u\), a line through the origin with a slope of \(-3/2\).

These lines divide the phase plane into sections where the trajectories behave differently. Observing how trajectories interact with nullclines can give vital insights into the system's stability and long-term behavior.

Stable Node

A stable node in the context of a phase portrait signifies a point where trajectories tend to converge over time, indicating stability. Such points typically represent equilibrium solutions where the system doesn't change anymore as time progresses.

In our specific system, the origin (0,0) serves as a stable node. This means:

• All trajectory paths in the phase plane eventually aim towards the origin.
• The closer the system is to this point, the slower the trajectories evolve towards it.

What makes the node 'stable' is the tendency for any nearby starting point to move towards it.
Understanding stable nodes is crucial because they offer insight into the system's natural state or equilibrium and show how disturbances are handled — essentially showing the system's robustness in returning to equilibrium. This is incredibly useful for predicting the system's behavior over a long period.