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Chapter 7: Problem 7

Find all equilibria and use the local stability criterion to determine if each is locally stable or unstable. Then construct the phase plot. (a) $$y^{\prime}=5-3 y$$ (b) $$y^{\prime}=y(a-y)^{2}(b-y), \quad b>a \geqslant 0$$ (c) $$y^{\prime}=\frac{a}{b+y}-1, \quad a>b>0, y \geqslant 0$$

Short Answer

Expert verified
(a): Stable at y=5/3. (b): Stable at y=a. (c): Stable at y=a-b.

Step by step solution

01

Find Equilibria for (a)

To find the equilibria, set \( y' = 0 \): \[ 0 = 5 - 3y \]Solve for \( y \): \[ y = \frac{5}{3} \].The equilibrium point for equation (a) is \( y = \frac{5}{3} \).
02

Analyze Local Stability for (a)

Calculate the derivative of \( y' = 5 - 3y \) with respect to \( y \): \[ \frac{d}{dy}(5 - 3y) = -3 \].Since \( -3 < 0 \), the equilibrium \( y = \frac{5}{3} \) is stable because the derivative is negative.
03

Find Equilibria for (b)

Set \( y' = 0 \): \[ 0 = y(a-y)^{2}(b-y) \]The equilibrium points are determined from \[ y = 0, \quad y = a, \quad y = b \].So, the equilibria for (b) are \( y = 0, a, b \).
04

Analyze Local Stability for (b)

Find the derivative \[ \frac{d}{dy}(y(a-y)^{2}(b-y)) = (a-y)^2(b-y) - 2y(a-y)(b-y) - y(a-y)^2 \]. Evaluate at each equilibrium:1. At \( y = 0 \): The derivative, \( ab \), is positive, so unstable.2. At \( y = a \): The derivative, \( -a(a-b) \), is negative (since \( b > a \)), so stable.3. At \( y = b \): The derivative, \( b(b-a) \), is positive, so unstable.
05

Find Equilibria for (c)

Set \( y' = 0 \): \[ 0 = \frac{a}{b+y} - 1 \]Solve for \( y \): \[ a = b + y \quad \Rightarrow \quad y = a - b \].The equilibrium point for (c) is \( y = a - b \).
06

Analyze Local Stability for (c)

Take the derivative of \( y'(y) = \frac{a}{b+y} - 1 \):\[ \frac{d}{dy}\left(\frac{a}{b+y} - 1\right) = -\frac{a}{(b+y)^2} \].Evaluate at \( y = a-b \): \[ \frac{d}{dy}|_{y=a-b} = -\frac{a}{a^2} = -\frac{1}{a} \], which is negative. Hence, \( y=a-b \) is stable.

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Key Concepts

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

Equilibria
Equilibria are key points in a system where the state doesn't change over time. For each differential equation, these points are found by setting the derivative equal to zero. This means finding where the rate of change is zero.
This often results in solving an equation for specific variables, giving the equilibrium points.
  • For equation (a), the equilibrium is found by setting \( y' = 5 - 3y = 0\) and solving for \( y \), resulting in \( y = rac{5}{3} \).
  • In equation (b), setting \( y' = y(a-y)^2(b-y) = 0\) results in possible values \( y = 0, a, ext{ and } b \) as equilibria.
  • For equation (c), the equilibrium is when \( y' = rac{a}{b+y} - 1 = 0\), leading to \( y = a - b \).
Finding equilibria is essential for examining how a system behaves over time.
Phase Plot
Phase plots are visual tools that illustrate how a system behaves over time near its equilibrium points. These plots graphically display the direction of movement (or flow) in phase space.
By examining these plots, you can visually identify stability characteristics at different points.
  • In equation (a), the phase plot will show arrows pointing towards the equilibrium point \( y = rac{5}{3} \), indicating stability.
  • For equation (b), the phase plot will illustrate flow directions at three points: flowing away from \( y = 0 \) and \( y = b \), as both are unstable. Flow towards \( y = a \), showing stability there.
  • Finally, for equation (c), the phase plot reveals flow lines converging towards \( y = a - b \), indicating a stable equilibrium.
These visualizations are excellent for understanding the dynamical system and predicting future states.
Local Stability Criterion
The local stability criterion helps determine whether an equilibrium point is stable or unstable by analyzing the sign of the derivative at that point.
If the derivative is negative, the equilibrium is usually deemed stable; if positive, unstable.
  • For equation (a), the derivative \( -3 \) suggests that the equilibrium \( y = rac{5}{3} \) is stable.
  • In equation (b), each equilibrium has a localized derivative check: \( y=0 \) (unstable, positive derivative), \( y=a \) (stable, negative derivative), and \( y=b \) (unstable, positive derivative).
  • Equation (c)'s equilibrium at \( y = a-b \) shows a negative derivative value, implying stability.
This criterion gives us clear insights into how solutions will evolve around each equilibrium point.

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Most popular questions from this chapter

Describe the motion of a particle with position \((x, y)\) as \(t\) varies in the given interval. \(x=\sin t, \quad y=\cos ^{2} t, \quad-2 \pi \leqslant t \leqslant 2 \pi\)

(a) Use Euler's method with step size 0.2 to estimate \(y(0.4),\) where \(y(x)\) is the solution of the initial-value problem \(y^{\prime}=x+y^{2}, y(0)=0\) (b) Repeat part (a) with step size 0.1 .

mRNA transcription The intermediate molecule mRNA arises in the decoding of DNA: it is produced by a process called transcription and it eventually decays. Suppose that the rate of transcription is changing exponentially according to the expression \(e ^ { b t } ,\) where \(b\) is a positive constant and mRNA has a constant per capita decay rate of \(k\) . The number of mRNA transcript molecules \(T\) thus changes as \(\frac { d T } { d t } = e ^ { b t } - k T\) Although the form of this equation is similar to that from Exercise \(43 ,\) the first term on the right side is now time varying. As a result, the differential equation is no longer separable; however, the equation can be solved using the change of variables \(y ( t ) = e ^ { k t } T ( t ) .\) Solve the differential equation using this technique.

A system of differential equations is given. (a) Use a phase plane analysis to determine the values of the constant \(a\) for which the sole equilibrium of the differential equations is locally stable. (b) Obtain an expression for each equilibrium (it may be a function of the constant \(a\) ). \(x^{\prime}=a y^{2}-x+1, \quad y^{\prime}=2(1-y)\)

Find the solution of the differential equation that satisfies the given initial condition. $$\frac { d y } { d x } = \frac { \ln x } { x y } , \quad y ( 1 ) = 2$$
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