Problem 20

Question

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d x}{d t}=x^{5}-x $$

Step-by-Step Solution

Verified

Answer

The equilibria are at $ x = -1, 0, 1 $. $ x = -1, 1 $ are unstable; $ x = 0 $ is stable.

1Step 1: Reformat the Differential Equation

Write the given differential equation in a format suitable for analysis: $ \frac{dx}{dt} = x^5 - x $. Here, we recognize that the differential equation is autonomous, meaning it does not explicitly depend on the independent variable $ t $.

2Step 2: Find Equilibria Points

Equilibria occur where $ \frac{dx}{dt} = 0 $. Set $ x^5 - x = 0 $. Factor the equation: $ x(x^4 - 1) = 0 $. This gives us solutions $ x = 0 $ and $ x^4 = 1 $, leading to $ x = \pm 1 $. Therefore, the equilibria are at $ x = 0, 1, $ and $ -1 $.

3Step 3: Analyze Stability Using Phase Line or Test Points

To classify the stability of each equilibrium, analyze the sign of $ \frac{dx}{dt} $ in the intervals between and outside of the equilibria:$ (-\infty, -1), (-1, 0), (0, 1), (1, \infty) $. Use test points to determine the sign of $ x^5 - x $ in these intervals.

4Step 4: Determine Stability

- In $ (-\infty, -1) $, choose $ x = -2 $ then $ (-2)^5 - (-2) = -32 + 2 = -30 $ (negative).- In $ (-1, 0) $, choose $ x = -0.5 $ then $ (-0.5)^5 + 0.5 = -0.03125 + 0.5 \approx 0.46875 $ (positive). - In $ (0, 1) $, choose $ x = 0.5 $ then $ (0.5)^5 - 0.5 = 0.03125 - 0.5 = -0.46875 $ (negative).- In $ (1, \infty) $, choose $ x = 2 $ then $ 2^5 - 2 = 32 - 2 = 30 $ (positive).Hence, equilibria: $ x = -1 $, and $ x = 1 $ are unstable; $ x = 0 $ is stable.

5Step 5: Create Vector Field Plot

Graph the vector field of $ \frac{dx}{dt} = x^5 - x $ using a software tool. Use the stability information:- At $ x = -1, 1 $, the vectors should point away from the equilibrium (unstable).- At $ x = 0 $, the vectors should point toward the equilibrium (stable).This visualizes the solutions' behavior near equilibrium points.

Key Concepts

Equilibria in Differential EquationsStability Analysis of EquilibriaUnderstanding Vector Field Plots

Equilibria in Differential Equations

In the context of differential equations, equilibria refer to specific values of a variable where the rate of change, or the derivative, equals zero. For our equation, $ \frac{dx}{dt} = x^5 - x $, equilibria occur where this derivative behaves like a still point.
To find these points, you set $ x^5 - x = 0 $. This can be rearranged to $ x(x^4 - 1) = 0 $. So, the solutions to this equation lead us to equilibria at points $ x = 0, 1, $ and $ -1 $. These values represent moments where the system isn't changing, and can be thought of as potential resting places for the system's behavior.
Being able to calculate these points is crucial, and sets the stage for deeper analysis, such as stability assessment.

Stability Analysis of Equilibria

Determining whether an equilibrium is stable or unstable in differential equations is an essential part of understanding the system's dynamics. Stable equilibria are like valleys where things settle, while unstable ones are like peaks that repel.
To assess the stability, we look at the sign of the derivative, $ \frac{dx}{dt} $, in intervals around the equilibria: $(-\infty, -1), (-1, 0), (0, 1), (1, \infty)$. For each range, we choose test points to evaluate $ x^5 - x $. This helps us determine whether the solution is moving towards or away from each equilibrium.

In $(-\infty, -1)$, the derivative is negative, indicating motion away from $ x = -1 $.
In $(-1, 0)$, the derivative is positive, showing motion towards $ x = 0 $.
In $(0, 1)$, the derivative is negative, indicating motion towards $ x = 0 $.
In $(1, \infty)$, the derivative is positive, showing motion away from $ x = 1 $.

This analysis concludes that $ x = 0 $ is a stable equilibrium, while $ x = -1 $ and $ x = 1 $ are unstable equilibria.

Understanding Vector Field Plots

A vector field plot for a differential equation graphically represents the direction and magnitude of the rate of change across different points. For the equation $ \frac{dx}{dt} = x^5 - x $, such a plot would help us visualize how solutions behave concerning our equilibria.
In a vector field plot, small arrows indicate the direction the system is likely to move in as time changes. At stable equilibria, arrows point toward the equilibrium, while at unstable points, they point away. This intuitive visual representation makes it easier to understand the dynamics at play without crunching numbers.
With our identified equilibria:

At $ x = 0 $ (stable), arrows on the plot point toward this equilibrium, suggesting solutions settle here.
At $ x = -1 $ and $ x = 1 $ (unstable), arrows point away, indicating solutions are driven away.

Vector field plots are extremely useful tools, offering a comprehensive overview of how a system behaves around its critical points.

Problem 19

Problem 21

Other exercises in this chapter

Problem 19

whose size at time $t$ is denoted by $N(t)$, grows according to $$ \frac{d N}{d t}=0.3 N \quad \text { with } N(0)=20 $$ Solve this differential equation, a

View solution

Problem 19

For each of the Problems 13-24 you should determine whether the problem needs to be solved using separation of variables or integrating factors (some of the pro

View solution

Problem 21

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify

View solution

Problem 21

A reversible chemical reaction between chemicals $A$ and $B$ produces a product C: $A+B \rightleftharpoons$. We modeled this reaction in Section 8.3.3 usi

View solution