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). $$ \begin{aligned} &x^{\prime}=x+2 y-4 \\ &y^{\prime}=2 x-y-2 \end{aligned} $$

Short answer

The equilibrium point is \((\frac{8}{5}, \frac{6}{5})\). Sketch the nullclines and flow based on calculations.

Step-by-step solution

Identify the Nullclines

Nullclines are the curves where the derivative of the variables is zero. For the given system, we set each equation to zero.For \(x' = x + 2y - 4 = 0\), solve for \(y\):\[ y = 2 - \frac{x}{2} \]For \(y' = 2x - y - 2 = 0\), solve for \(y\):\[ y = 2x - 2 \]

Find the Equilibrium Points

Equilibrium points occur where both derivatives are zero simultaneously. Set the two expressions for \(y\) equal to each other:\[ 2 - \frac{x}{2} = 2x - 2 \]Solve for \(x\):\[ 2 + 2 = 2x + \frac{x}{2} \]\[ 4 = \frac{5x}{2} \]\[ x = \frac{8}{5} \]Substitute \(x = \frac{8}{5}\) back into one of the expressions for \(y\) (for example, \(y = 2 - \frac{x}{2}\)):\[ y = 2 - \frac{8}{10} = \frac{12}{10} = \frac{6}{5} \]Thus, the equilibrium point is \(\left(\frac{8}{5}, \frac{6}{5}\right)\).

Determine Vector Field Flow Along Nullclines

Check the direction of the vector field on each nullcline.For the nullcline \(y = 2 - \frac{x}{2}\) (\(x' = 0\)), determine the sign of \(y'\):Substitute \(y = 2 - \frac{x}{2}\) into \(y' = 2x - y - 2 = 0\):\[ y' = 2x - \left(2 - \frac{x}{2}\right) - 2 \]Simplify and check the sign about the nullcline to understand the flow direction.For the nullcline \(y = 2x - 2\) (\(y' = 0\)), determine the sign of \(x'\):Substitute \(y = 2x - 2\) into \(x' = x + 2y - 4\):\[ x' = x + 2(2x - 2) - 4 \]Simplify and check the sign about the nullcline to understand the flow direction.

Sketch the Nullclines and Equilibrium Point

Using the expressions derived for the nullclines, plot them on a coordinate plane. The nullclines \(y = 2 - \frac{x}{2}\) and \(y = 2x - 2\) intersect at the equilibrium point \(\left(\frac{8}{5}, \frac{6}{5}\right)\). Draw arrows indicating the flow direction determined in Step 3. This provides a qualitative understanding of the vector field.

Key concepts

Understanding Nullclines

When working with differential equations, nullclines are vital components that help us analyze the system's behavior. A nullcline is essentially where one of the derivatives of a system becomes zero.

• For example, if we have a system of equations, setting one equation to zero would give us its nullcline.
• This means the nullcline for a particular equation shows where its rate of change is zero.

In our exercise, for the equation \(x' = x + 2y - 4\), setting it to zero gives us the nullcline \(y = 2 - \frac{x}{2}\). Similarly, the equation \(y' = 2x - y - 2\) becomes zero at its nullcline \(y = 2x - 2\). Understanding these lines helps in identifying how the system behaves since they define boundaries where the dynamics change.
Nullclines are like paths in a field of arrows, guiding us where the system slows down or halts, which is crucial for finding equilibrium points.

Identifying Equilibrium Points

Equilibrium points in a system of differential equations are where all derivatives simultaneously equal zero, representing a state where the system doesn't change over time. In simpler terms, it's where everything comes to a standstill. To locate these points, we look for intersections of nullclines, as they mark where both components have zero change.
For instance, by equating the nullclines \(y = 2 - \frac{x}{2}\) and \(y = 2x - 2\), we determine the equilibrium point. Solving these equations, we find the intersection point at \(\left(\frac{8}{5}, \frac{6}{5}\right)\).
Why is this important? Knowing where the system can rest helps us predict long-term behavior and stability. Understanding equilibrium can be analogous to finding balance – knowing where these points are helps provide insight into this stability.

Harnessing Algebra Skills

Algebra skills are incredibly useful when analyzing systems of differential equations, especially for sketching nullclines and finding equilibrium points without computational tools. Fairly often, algebra requires only simple manipulation and solving, which can provide insightful analysis of even seemingly complex systems.

• Nullclines: By rearranging terms and solving linear equations, you can identify nullclines. This acts as a foundation for visual understanding.
• Equilibrium Points: Set multiple equations equal and use basic algebraic techniques to find where they intersect. This step forms the crux of stability analysis.

Understanding these algebraic underpinnings enhances a deeper grasp of the whole system. Practicing these skills makes the process of analysis more intuitive, building confidence in tackling similar problems independently.

Visualizing Vector Fields

Vector fields give us the best visual grasp of how a differential equation behaves over space, displaying the direction and magnitude of velocities across points. When studying these fields, nullclines are crucial, as they act as guides revealing where the system might change behavior.

• Flow Direction: By examining how vectors align along nullclines, one can determine which way the system will tend to move in different zones.
• Intersection Influence: Where nullclines intersect at equilibrium points, the vector field's behavior reflects how such points attract, repel, or create complex flows.

This visual strategy aids both understanding and prediction of the system's future dynamics. As a sketch or in a software simulation, viewing vector fields can bridge theoretical understanding with tangible insights, making these mathematical abstractions more relatable.