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Chapter 5: Problem 8

Write a system of linear first-order ODEs whose solutions have the following qualitative behaviors: (a) \((0,0)\) is a stable node with eigenvalues \(\lambda_{1}=-1\) and \(\lambda_{2}=-2\). (b) \((0,0)\) is a saddle point with eigenvalues \(\lambda_{1}=-1\) and \(\lambda_{2}=3\). (c) \((0,0)\) is a center with eigenvalues \(\lambda=\pm 2 i\). (d) \((0,0)\) is an unstable node with eigenvalues \(\lambda_{1}=2\) and \(\lambda_{2}=3\). Hint: Use the fact that \(\lambda_{1}\) and \(\lambda_{2}\) are eigenvalues of a matrix \(A\), then $$ \begin{aligned} &\lambda_{1}+\lambda_{2}=\operatorname{Tr} \mathrm{A}=a_{11}+a_{22} \\ &\lambda_{1} \lambda_{2}=\operatorname{det} \mathrm{A}=a_{11} a_{22}-a_{12} a_{21} . \end{aligned} $$ Note that there will be many possible choices for each of the above.

Short Answer

Expert verified
Matrices with required behaviors are: \[A_{a} = \begin{pmatrix} -1 & 0 \ 0 & -2 \end{pmatrix}, A_{b} = \begin{pmatrix} 2 & 1 \ 3 & 0 \end{pmatrix}, A_{c} = \begin{pmatrix} 0 & -2 \ 2 & 0 \end{pmatrix}, A_{d} = \begin{pmatrix} 4 & 1 \ 2 & 1 \end{pmatrix}\]}

Step by step solution

01

Understand Eigenvalues and Stability

Identify the type of critical point given by the eigenvalues. Each case represents a type of behavior at \(0,0\). \(\lambda_1\) and \(\lambda_2\) will guide the stability and type of solutions.
02

Case (a): Stable Node

Given: \(\lambda_{1}=-1\) and \(\lambda_{2}=-2\). The trace \(\operatorname{Tr} A = -1 - 2 = -3\) and determinant \(\operatorname{det} A = (-1)(-2) = 2\). One possible choice of matrix \(A\) is: \[A = \begin{pmatrix} -1 & 0 \ 0 & -2 \end{pmatrix}\]
03

Case (b): Saddle Point

Given: \(\lambda_{1}=-1\) and \(\lambda_{2}=3\). The trace \(\operatorname{Tr} A = -1 + 3 = 2\) and determinant \(\operatorname{det} A = (-1)(3) = -3\). One possible choice of matrix \(A\) is: \[A = \begin{pmatrix} 2 & 1 \ 3 & 0 \end{pmatrix}\]
04

Case (c): Center

Given: \(\lambda = \pm 2i\). The trace \(\operatorname{Tr} A = 0\) since the imaginary parts cancel each other, and determinant \(\operatorname{det} A = 4\) since \(\pm 2i\) squared is \(-4\) (considering complex roots). One possible choice of matrix \(A\) is: \[A = \begin{pmatrix} 0 & -2 \ 2 & 0 \end{pmatrix}\]
05

Case (d): Unstable Node

Given: \(\lambda_{1}=2\) and \(\lambda_{2}=3\). The trace \(\operatorname{Tr} A = 2 + 3 = 5\) and determinant \(\operatorname{det} A = (2)(3) = 6\). One possible choice of matrix \(A\) is: \[A = \begin{pmatrix} 4 & 1 \ 2 & 1 \end{pmatrix}\]

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Key Concepts

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

Eigenvalues and Stability
Understanding the behavior of solutions to systems of linear first-order ordinary differential equations (ODEs) primarily involves analyzing eigenvalues and their impact on stability.
Eigenvalues, denoted as \(\lambda_1\) and \(\lambda_2\), are critical to determining the nature of the equilibrium points (critical points) within the system.
The stability and type of solutions near these points are directly linked to the values of the eigenvalues.
Here are the key points regarding eigenvalues and stability:
  • If both eigenvalues are real and negative, the equilibrium is a stable node.
  • If one eigenvalue is positive and the other is negative, it results in a saddle point, indicating instability.
  • When eigenvalues are purely imaginary, we get a center, depicting neutral stability with circular motion.
  • If both eigenvalues are real and positive, the system shows an unstable node, indicating divergence from the equilibrium.
Stable Nodes
A stable node occurs when both eigenvalues are real and negative. This situation implies that as time progresses, the solutions to the system of ODEs tend to approach the equilibrium point, reflecting stability.
In mathematical terms, given \(\lambda_1 = -1\) and \(\lambda_2 = -2\):
The trace \(\text{Tr}\bs{A} = -3\) and the determinant \(\text{det}\bs{A} = 2\) help us to form the matrix \(\bs{A}\):

\[ \bs{A} = \begin{pmatrix}-1 & 0\0 & -2 \ \end{pmatrix}\]
The negative eigenvalues indicate that perturbations in the system decay over time, thus making the equilibrium point stable.
Saddle Points
A saddle point is characterized by having one positive and one negative real eigenvalue. This implies that the equilibrium point is inherently unstable.
For instance, with eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = 3\):
The trace is \(\text{Tr}\bs{A} = 2\) and the determinant is \(\text{det}\bs{A} = -3\), leading to a possible matrix \(\bs{A}\):

\[ \bs{A} = \begin{pmatrix} 2 & 1\3 & 0 \ \end{pmatrix} \]
The mixed signs of the eigenvalues indicate that while one trajectory moves towards the equilibrium, another moves away, forming a saddle-like structure.
Center Points
Center points occur when the eigenvalues are purely imaginary, indicating neutral stability. Unlike nodes, the solution trajectories form closed orbits around the equilibrium, leading to periodic solutions.
Consider the eigenvalues \(\lambda = \pm 2i\):
The trace is \(\text{Tr}\bs{A} = 0\) and the determinant is \(\text{det}\bs{A} = 4\), creating the matrix \(\bs{A}\):

\[ \bs{A} = \begin{pmatrix} 0 & -2\2 & 0 \ \end{pmatrix} \]
This situation reflects that any deviations from the equilibrium result in circular trajectories, highlighting the neutral stability typical of center points.
Unstable Nodes
An unstable node is defined by having both real eigenvalues positive, leading to a situation where all solutions diverge away from the equilibrium point.
Given the eigenvalues \(\lambda_1 = 2\) and \(\lambda_2 = 3\):
The trace and determinant are \(\text{Tr}\bs{A} = 5\) and \(\text{det}\bs{A} = 6\) respectively, yielding the matrix \(\bs{A}\):

\[ \bs{A} = \begin{pmatrix} 4 & 1\2 & 1 \ \end{pmatrix} \]
The positive eigenvalues indicate that any small perturbations grow exponentially over time, clearly marking the equilibrium point as unstable.

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Most popular questions from this chapter

The following equations were given by J. S. Griffith \((1971, \mathrm{pp} .118-122)\), as a model for the interactions of messenger RNA \(M\) and protein \(E\) : $$ \dot{M}=\frac{a K E^{m}}{1+K E^{m}}-b M, \quad \dot{E}=c M-d E . $$ (See problem 25 in Chapter 7 for an interpretation.) (a) Show that by changing units one can rewrite these in terms of dimensionless variables, as follows $$ \dot{M}=\frac{E^{m}}{1+E^{m}}-\alpha M, \quad \dot{E}=M-\beta E . $$ Find \(\alpha\) and \(\beta\) in terms of the original parameters. (b) Show that one steady state is \(E=M=0\) and that others satisfy \(E^{m-1}=\alpha \beta\left(1+E^{m}\right)\). For \(m=1\) show that this steady state exists only if \(\alpha \beta \leq 1\). (c) Case 1. Show that for \(m=1\) and \(\alpha \beta>1\), the only steady state \(E=M=0\) is stable. Draw a phase-plane diagram of the system. (d) Case 2. Show that for \(m=2\), at steady state $$ E=\frac{1 \pm\left(1+4 \alpha^{2} \beta^{2}\right)^{1 / 2}}{2 \alpha \beta} $$ Conclude that there are two solutions if \(2 \alpha \beta<1\), one if \(2 \alpha \beta=1\), and none if \(2 \alpha \beta>1\). (e) Case 2 continued. For \(m=2\) and \(2 \alpha \beta<1\), show that there are two stable steady states (one of which is at \(E=M=0\) ) and one saddle point. Draw a phase-plane diagram of this system.

Sketch the nullelines in the xy phase plane, identify steady states, and draw directions of arrows on the nullclines for the following systems of first- order equations: (a) \(\frac{d x}{d t}=y^{2}-x^{2}\), $$ \frac{d y}{d t}=x-1 \text {. } $$ (b) \(\frac{d x}{d t}=x\left(y^{2}-y\right)\), $$ \frac{d y}{d r}=x-y \text {. } $$ (c) \(\frac{d x}{d t}=x^{2}+y\), $$ \frac{d y}{d t}=-y \text {. } $$ (d) \(\frac{d x}{d t}=-x y\), \(\frac{d y}{d t}=(1+x)(1-y)\). (e) \(\frac{d x}{d t}=x^{2}-y\), $$ \frac{d y}{d t}=y^{2}-x \text {. } $$ (f) \(\frac{d x}{d t}=\frac{-x y}{1+x}+x\), \(\frac{d y}{d t}=\frac{x y}{1+x}-y .\) (g) \(\frac{d x}{d t}=x y(1-x)+C\), \(\frac{d y}{d t}=y\left(1-\frac{y}{x}\right) .\)

For the following first-order ordinary differential equations, sketch solution curves \(y(t)\) by first plotting the tangent vectors specified by the differential equations: (a) \(\frac{d y}{d t}=y^{2}\). (d) \(\frac{d y}{d t}=y e^{(r-1)}\) (b) \(\frac{d y}{d t}=1-\frac{y}{1+y}\). (e) \(\frac{d y}{d t}=\sin y \cos y\). (c) \(\frac{d y}{d t}=y(y-2)\).

Sketch the phase-plane behavior of the following systems of linear equations and classify the stability characteristic of the steady state at \((0,0)\) : (a) \(\frac{d x}{d t}=-2 y\), $$ \frac{d y}{d t}=x \text {. } $$ (b) \(\frac{d x}{d t}=3 x+2 y\), $$ \frac{d y}{d t}=4 x+y . $$ (c) \(\begin{aligned} \frac{d x}{d t} &=2 x+y, \\ \frac{d y}{d t} &=x+2 y . \end{aligned}\) (d) $$ \begin{aligned} &\frac{d x}{d t}=5 x+8 y \\ &\frac{d y}{d t}=-3 x-5 y \end{aligned} $$ (e) $$ \begin{aligned} &\frac{d x}{d t}=-4-2 y \\ &\frac{d y}{d t}=3 x-y \end{aligned} $$ (f) \(\begin{aligned} \frac{d x}{d t} &=x-4 y \\ \frac{d y}{d t} &=x+y \end{aligned}\)

In this problem we examine a continuous plant-herbivore model. We shall define \(q\) as the chemical state of the plant. Low values of \(q\) mean that the plant is toxic; higher values mean that the herbivores derive some nutritious value from it. Consider a situation in which plant quality is enhanced when the vegetation is subjected to a low to moderate level of herbivory, and declines when herbivory is extensive. Assume that herbivores whose density is \(I\) are small insects (such as scale bugs) that attach themselves to one plant for long periods of time. Further assume that their growth rate depends on the quality of the vegetation they consume. Typical equations that have been suggested for such a system are $$ \begin{aligned} \frac{d q}{d t} &=K_{1}-K_{2} q I\left(t-I_{0}\right) \\ \frac{d l}{d t} &=K_{3} I\left(1-\frac{K_{4} I}{q}\right) \end{aligned} $$ (a) Explain the equations, and suggest possible meanings for \(K_{1}, K_{2}, I_{0}, K_{3}\), and \(K_{4}\). (b) Show that the equations can be written in the following dimensionless form: $$ \begin{aligned} &\frac{d q}{d t}=1-K q I(I-1), \\ &\frac{d I}{d t}=\alpha I\left(1-\frac{I}{q}\right) . \end{aligned} $$ (c) Find qualitative solutions using phase-plane methods. Is there a steady state? What are its stability properties? (d) Interpret your solutions in part (c).
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