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Chapter 9: Problem 6

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. $$y^{\prime}=y-\sqrt{y}, \quad y > 0$$

Short Answer

Expert verified
Equilibrium values are 0 (unstable) and 1 (stable). The phase line indicates stability, and solution curves confirm this dynamics.

Step by step solution

01

Identify Critical Points

To find critical points, set the derivative equal to zero: \[ y' = y - \sqrt{y} = 0 \]Solving this gives \( y = \sqrt{y} \).Rearrange: \( \sqrt{y} = y \) implies \( y = 0 \) or \( y = 1 \) are critical points.
02

Determine Stability of Equilibrium Points

To determine stability, analyze the sign of \( y' \) in different intervals. Consider intervals: - For \( y < 0 \), \( y' < 0 \) because both \( y \) and \( \sqrt{y} \) are real and valid only if \( y > 0 \). - For \( 0 < y < 1 \), \( y' = y - \sqrt{y} > 0 \). - For \( y > 1 \), \( y' = y - \sqrt{y} > 0 \).Conclusion: \( y = 0 \) is unstable, and \( y = 1 \) is stable.
03

Construct the Phase Line

Draw a vertical line (phase line) and mark critical points \( y = 0 \) and \( y = 1 \). Show arrows:- Pointing away from \( y = 0 \), indicating instability.- Pointing towards \( y = 1 \), indicating stability.
04

Identify Signs of First and Second Derivatives

For \( y' = y - \sqrt{y} \):- Compute the second derivative \( y'' \) by differentiating \( y' \):\[ y'' = 1 - \frac{1}{2\sqrt{y}} \]At equilibrium:\( 0 < 1 - \frac{1}{2 \sqrt{y}} < 1 \) for \( y = 1 \), imply \( y'' > 0 \) indicating concave up at \( y = 1 \).
05

Sketch Solution Curves

Sketch several solution curves with the following characteristics:- Start some curves in the interval \( 0 < y < 1 \), which increase and approach \( y = 1 \).- Start some curves above 1, they decrease towards equilibrium at \( y = 1 \).- Use the phase line as a guide for direction of curves.- The curve for initial condition \( y(0) = 0 \) will move away from \( y = 0 \) due to instability, trending towards \( y = 1 \).

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Key Concepts

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

Equilibrium Points
Differential equations often help us understand how certain values, termed as equilibrium points, play a pivotal role in analyzing the behavior of solutions. Let's dive into what equilibrium points are. For our differential equation
  • Start by setting the derivative to zero, in this case, it's setting the expression \( y' = y - \sqrt{y} \) to zero.
  • This simplifies to \( y = \sqrt{y} \), indicating our critical points are where the behavior of solutions transitions or changes nature.
  • Simplifying further provides us the values: \( y = 0 \) and \( y = 1 \).
These points are where the derivative vanishes, and the system is momentarily in equilibrium; no change occurs at these points. Whether the system remains there or evolves further depends on their stability, which takes us to our next section.
Stability Analysis
Understanding whether an equilibrium point is stable or unstable is crucial for predicting the system's behavior. Let's break down the process:
  • For stability analysis, examine the sign of \( y' \) at different intervals around the equilibrium points.
  • For our equation, if \( y < 0 \), \( y' < 0 \) isn't physically valid since \( y > 0 \) by definition; hence we focus on \( 0 < y < 1 \) and \( y > 1 \).
  • In the range \( 0 < y < 1 \), \( y' > 0 \) meaning solutions move away from \( y = 0 \), indicating it's unstable.
  • In the range \( y > 1 \), still \( y' > 0 \) but behaviors around \( y = 1 \), with \( y' \) moving towards it, indicate stability at \( y = 1 \).
By understanding the sign changes in \( y' \), we deduce the stability behavior of solutions around these points. Stability analysis helps us visually on constructs like a phase line where this stability and instability is clearly shown.
Phase Plane
The phase plane is a visual tool used in stability and equilibrium analysis, offering insights on how solutions behave over time for different initial values.
  • In one-dimensional systems like ours, the phase line鈥攅ssentially a vertical line鈥攊s used instead of a plane.
  • Critical points are plotted: \( y = 0 \) and \( y = 1 \).
  • For \( y = 0 \), arrows point away, demonstrating that solutions do not stay here due to instability.
  • For \( y = 1 \), arrows point towards it, indicating stability since solutions tend to settle here over time.
The phase line provides a clear picture of system dynamics. Solution curves can be sketched based on this line, predicting behavior鈥攕uch as whether they tend to increase or decrease鈥攊n specific regions simulating the system. This adds a layer of intuition on top of analytical methods, making it easier to anticipate behavior in varied differential scenarios.

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Most popular questions from this chapter

The autonomous differential equations represent models for population growth. For each exercise, use a phase line analysis to sketch solution curves for \(P(t),\) selecting different starting values \(P(0) .\) Which equilibria are stable, and which are unstable? $$\frac{d P}{d t}=P(1-2 P)$$

What integral equation is equivalent to the initial value problem \(y^{\prime}=f(x), y\left(x_{0}\right)=y_{0} ?\)

Consider another competitive-hunter model defined by $$\begin{aligned}\frac{d x}{d t} &=a\left(1-\frac{x}{k_{1}}\right) x-b x y \\\\\frac{d y}{d t} &=m\left(1-\frac{y}{k_{2}}\right) y-n x y\end{aligned}$$ where \(x\) and \(y\) represent trout and bass populations, respectively. a. What assumptions are implicitly being made about the growth of trout and bass in the absence of competition? b. Interpret the constants \(a, b, m, n, k_{1},\) and \(k_{2}\) in terms of the physical problem. c. Perform a graphical analysis: i) Find the possible equilibrium levels. ii) Determine whether coexistence is possible. iii) Pick several typical starting points and sketch typical trajectories in the phase plane. iv) Interpret the outcomes predicted by your graphical analysis in terms of the constants \(a, b, m, n, k_{1},\) and \(k_{2}\) Note: When you get to part (iii), you should realize that five cases exist. You will need to analyze all five cases.

Solve the differential equations. \(e^{2 x} y^{\prime}+2 e^{2 x} y=2 x\)

Solve the initial value problems. \((x+1) \frac{d y}{d x}-2\left(x^{2}+x\right) y=\frac{e^{x^{2}}}{x+1}, \quad x>-1, \quad y(0)=5\)
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