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Chapter 16: Problem 7

a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identify the signs of \(y^{\prime}\) and \(y^{\prime \prime}\) c. Sketch several solution curves. $$y^{\prime}=(y-1)(y-2)(y-3)$$

Short Answer

Expert verified
Equilibrium points are \( y=1, 2, 3 \). \( y=2 \) is stable, \( y=1, 3 \) are unstable.

Step by step solution

01

Finding Equilibrium Points

To find the equilibrium points, we set the derivative equal to zero: \( y' = (y-1)(y-2)(y-3) = 0 \). This gives us three equilibrium points: \( y = 1, 2, \) and \( 3 \).
02

Determine Stability of Equilibrium Points

To determine the stability, examine the sign of \( y' \) around each equilibrium point. For \( y=1 \): to the left \( (y-1) < 0 \) and \( y'- < 0 \); to the right \( (y-1) > 0 \) and \( y'- > 0 \). Thus, \( y=1 \) is unstable. For \( y=2 \): to the left \( (y-2) < 0 \) and \( y'+ > 0 \); to the right \( (y-2) > 0 \) and \( y'- < 0 \). Thus, \( y=2 \) is stable. For \( y=3 \): to the left \( (y-3) < 0 \) and \( y'+ > 0 \); to the right \( (y-3) > 0 \) and \( y'- > 0 \). Thus, \( y=3 \) is unstable.
03

Constructing the Phase Line

A phase line summarizes the flow of solutions. For \( y=1 \), solutions move away; for \( y=2 \), solutions move towards it; and for \( y=3 \), solutions move away. This means the phase line will have arrows pointing away from \( y=1 \) and \( y=3 \), and towards \( y=2 \). Such an analysis confirms that \( y=2 \) is the only stable equilibrium.
04

Identify Signs of Derivatives

Determine the sign of \( y' \) in the intervals: \( y < 1 \), \( 1 < y < 2 \), \( 2 < y < 3 \), and \( y > 3 \). Also, for concavity \( y'' = d(y')/dy \), calculate \( y'' = 3y^2 - 12y + 11 \) and determine its sign within these intervals.
05

Sketch Solution Curves

Use information from the phase line and derivative signs to sketch solution curves. Start near equilibrium points. For \( y < 1 \) and \( y>3 \), solutions move away. Between \( 1 \) and \( 3 \), see solutions move towards \( y=2 \) for stability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Points
Equilibrium points in a differential equation are where the derivative equals zero. They represent constant solutions where the state is balanced. For the equation given as: \[ y' = (y-1)(y-2)(y-3) \] By setting \( y' = 0 \), we solve for the values of \( y \). This yields the equilibrium points: \( y = 1 \), \( y = 2 \), and \( y = 3 \). Each of these values results in a zero product in the derivative equation, indicating no change over time at those points. Understanding these points is crucial as they give a snapshot where the system is in stasis, providing a basis for further stability analysis.
Stability Analysis
Stability analysis examines whether solutions near equilibrium points will return to equilibrium or diverge away. - **Stable equilibrium**: If solutions approach the equilibrium point as time progresses, the point is stable. - **Unstable equilibrium**: If solutions diverge away, it's called unstable.For each equilibrium point, we check the sign of \( y' \) on the intervals just around those points: - **At \( y = 1 \):** The sign analysis reveals that solutions move away both to the left and right, marking it as unstable. - **At \( y = 2 \):** The solutions approach from both sides, making it stable. - **At \( y = 3 \):** Just like \( y = 1 \), solutions diverge, qualifying it as unstable. Recognizing the stability of equilibrium points helps predict long-term behavior of a system.
Phase Line
A phase line is a straightforward visual tool to summarize the behavior of solutions along the differential equation's axis. By placing equilibrium points on a line and utilizing arrows to show the movement of solutions, it illuminates stability characteristics clearly. To construct the phase line:- Draw a horizontal line and mark our equilibrium points: \( y = 1 \), \( y = 2 \), \( y = 3 \). - Between these points, illustrate the direction of \( y' \): - To the left and right of \( y = 1 \) and \( y = 3 \), use arrows pointing outward, indicating an unstable equilibrium. - Towards \( y = 2 \), draw arrows pointing inward, capturing the stable nature.The phase line succinctly shows how the nature of these points affects the flow and is valuable for visualizing the overall system's behavior.
Solution Curves
Solution curves graphically depict how solutions evolve over time from different starting points. They help to see how a system progresses and affects by its equilibrium points and their stability. These curves are sketched using insights from:- **Equilibrium Points**: Start curves close to these points to observe how they influence solutions. - **Phase Line**: The direction indicators assist in shaping the flow of the curve. Solutions illustrate moving towards a stable equilibrium, or away if unstable. For this exercise:- **For \( y < 1 \) and \( y > 3 \):** Solutions should curve away from both \( y = 1 \) and \( y = 3 \), as these points are unstable.- **Between \( 1 < y < 2 \) and \( 2 < y < 3 \):** Curves bend towards \( y = 2 \), displaying its stability. Sketching helps connect theoretical stability analysis to practical system evolution, making abstract ideas tangible.

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Most popular questions from this chapter

Use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. $$y^{\prime}=y e^{x}, \quad y(0)=2, \quad d x=0.5$$

In 1925 Lotka and Volterra introduced the predator-prey equations, a system of equations that models the populations of two species, one of which preys on the other. Let \(x(t)\) represent the number of rabbits living in a region at time \(t,\) and \(y(t)\) the number of foxes in the same region. As time passes, the number of rabbits increases at a rate proportional to their population, and decreases at a rate proportional to the number of encounters between rabbits and foxes. The foxes, which compete for food, increase in number at a rate proportional to the number of encounters with rabbits but decrease at a rate proportional to the number of foxes. The number of encounters between rabbits and foxes is assumed to be proportional to the product of the two populations. These assumptions lead to the autonomous system $$\begin{aligned}\frac{d x}{d t} &=(a-b y) x \\\\\frac{d y}{d t} &=(-c+d x) y \end{aligned}$$ where \(a, b, c, d\) are positive constants. The values of these constants vary according to the specific situation being modeled. We can study the nature of the population changes without setting these constants to specific values. Show, by differentiating, that the function $$C(t)=a \ln y(t)-b y(t)-d x(t)+c \ln x(t)$$ is constant when \(x(t)\) and \(y(t)\) are positive and satisfy the predatorprey equations.

Find the family of solutions of the given differential equation and the family of orthogonal trajectories. Sketch both families. a. \(x d x+y d y=0\) b. \(x d y-2 y d x=0\)

Use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. $$y^{\prime}=2 x e^{x^{2}}, \quad y(0)=2, \quad d x=0.1$$

An executive conference room of a corporation contains \(4500 \mathrm{ft}^{3}\) of air initially free of carbon monoxide. Starting at time \(t=0,\) cigarette smoke containing \(4 \%\) carbon monoxide is blown into the room at the rate of \(0.3 \mathrm{ft}^{3} / \mathrm{min} .\) A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of \(0.3 \mathrm{ft}^{3} / \mathrm{min}\). Find the time when the concentration of carbon monoxide in the room reaches \(0.01 \%\)
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