Question

In Exercises \(1-4\), without the aid of technology, using only your algebra skills, sketch the nullclines and find the equilibrium point(s) of the assigned system. Indicate the flow of the the vector field along each nullcline, similar to that shown in Figure 13.1. Check your result with pplane6. If the Symbolic Toolbox is available, use the solve command to find the equilibrium point(s). $$ \begin{aligned} &x^{\prime}=2 x+y \\ &y^{\prime}=4 x+2 y \end{aligned} $$

Short answer

The nullcline is \(y = -2x\). The line \(y = -2x\) is the equilibrium.

Step-by-step solution

Identify Nullclines

Nullclines are found by setting each equation's derivative to zero; find when \(x' = 0\) and \(y' = 0\). For \(x' = 2x + y = 0\), solving for \(y\) gives \(y = -2x\). For \(y' = 4x + 2y = 0\), solving for \(y\) gives \(y = -2x\). Both nullclines are identical: \(y = -2x\).

Find Equilibrium Point

The equilibrium points occur where both derivatives are zero simultaneously. Substitute the nullcline into each derivative: from \(x' = 2x + (-2x) = 0\) and \(y' = 4x + 2(-2x) = 0\). These confirm points along the line \(y = -2x\). Substitute in any value \(x\), \(y = -2x\) does not give new information, implying solutions along the line. Since the nullcline and the equilibrium condition coincide, any point \((x, -2x)\) is part of the equilibrium line.

Analyze Flow Along Nullclines

Evaluate the vector field direction by picking a test point close to the nullcline. Consider \(x = 1\), substitute in \(y = -2\). Compute the derivatives: \(x' = 2(1) - 2 = 0\), \(y' = 4(1) + 2(-2) = 0\). At \((1, -2)\), both derivatives are zero confirming equilibrium. Repeat for \(x = 0\) and \(y = 0\) to see direction doesn't change along the line.

Conclusion on System's Behavior

The line \(y = -2x\) is a line of equilibrium as derivatives remain zero. Therefore, no movement occurs along this line, indicating equilibrium for any \(x\). The vector field flow aligns with this understanding and lacks movement perpendicular to the nullcline, suggesting linear stability along the line.

Key concepts

Nullclines

Nullclines are essential in understanding the behavior of dynamic systems. They are specific curves in the phase plane where either one or both of the derivatives in a system of differential equations equal zero. Essentially, nullclines help us identify changes in the system's dynamics.
In our system, we have two equations:

• \( x' = 2x + y \)
• \( y' = 4x + 2y \)

To find the nullclines, we set each derivative to zero:

• For \( x' = 0 \), solve \( 2x + y = 0 \), resulting in \( y = -2x \).
• For \( y' = 0 \), solve \( 4x + 2y = 0 \), also giving \( y = -2x \).

In this case, both nullclines coincide, producing a single line in the phase plane where the behavior of the system changes.

Algebra Skills

Your algebra skills are crucial when analyzing differential equations. They allow you to manipulate and solve the equations to find important features like the nullclines and equilibrium points.
For the given system, applying algebraic manipulation allows us to identify the common nullclines and equilibrium. By setting \( 2x + y = 0 \) to find the \( x' \) nullcline, directly solving for \( y \) gives \( y = -2x \). Similarly, setting \( 4x + 2y = 0 \) for the \( y' \) nullcline results in the same equation.
Using substitution or elimination methods are part of these skills, enabling seamlessly finding relationships like \( y = -2x \) across both equations. This practice not only finds equilibrium points but helps verify consistency across solutions.

Vector Field

The vector field in a system of differential equations visually represents the direction and strength of the system's changes at any point in the state space. It consists of arrows indicating the trajectory that the system follows from any given point. Understanding the vector field helps forecast system behavior over time.
In our exercise, the vector field can be observed by analyzing test points near the nullcline \( y = -2x \). For example, choosing the point \( (1, -2) \) where the vector components are calculated as \( x' = 0 \) and \( y' = 0 \), shows no movement. This zero vector indicates that points along this line are equilibrium points with no flow in the vector field direction, aligning with the nullcline results.

System Stability

System stability relates to how a system behaves when slightly perturbed from its equilibrium state. Finding whether a system will return to equilibrium (stable), move away (unstable), or have other behavior is key in dynamic analysis.
Here, since the nullclines coincide along a straight line \( y = -2x \) and all vector field evaluations result in zero, indicating no directional change, the system is linearly stable along this nullcline.
This stability means points slightly perturbed from this line will return to it, maintaining equilibrium along \( y = -2x \). Such analysis provides insight into the long-term behavior and reliability of the system in maintaining equilibrium despite minor disturbances.