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Chapter 11: Problem 17

Find the general solution of each given system of differential equations and sketch the lines in the direction of the eigenvectors. Indicate on each line the direction in which the solution would move if it starts on that line. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} -2 & 0 \\ -3 & 1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$

Short Answer

Expert verified
The general solution is \( \mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} + c_2 e^{t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \).

Step by step solution

01

Identify the Matrix

The system of differential equations can be expressed in matrix form as \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \), where \( A = \begin{bmatrix} -2 & 0 \ -3 & 1 \end{bmatrix} \). The vector \( \mathbf{x}(t) = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \) represents the variables of the system.
02

Find the Eigenvalues of the Matrix

The eigenvalues \( \lambda \) of the matrix \( A \) satisfy the characteristic equation \( \det(A - \lambda I) = 0 \). For the given matrix, this becomes: \[ \det \begin{bmatrix} -2 - \lambda & 0 \ -3 & 1 - \lambda \end{bmatrix} = (\lambda + 2)(\lambda - 1) = 0 \] Solving this gives the eigenvalues \( \lambda_1 = -2 \) and \( \lambda_2 = 1 \).
03

Find the Eigenvectors for Each Eigenvalue

For each eigenvalue \( \lambda_i \), find the corresponding eigenvector \( \mathbf{v}_i \) by solving \( (A - \lambda_i I)\mathbf{v}_i = \mathbf{0} \). - For \( \lambda_1 = -2 \): Solve \( \begin{bmatrix} 0 & 0 \ -3 & 3 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This gives \( v_1 = 0 \) and \( v_2 \) is free, so an eigenvector is \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \).- For \( \lambda_2 = 1 \): Solve \( \begin{bmatrix} -3 & 0 \ -3 & 0 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This gives \( v_2 = 0 \) and \( v_1 \) is free, so an eigenvector is \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \).
04

Form the General Solution

The general solution of the system can be written as a linear combination of the eigenvectors, each multiplied by an exponential function of their respective eigenvalue:\( \mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 0 \ 1 \end{bmatrix} + c_2 e^{t} \begin{bmatrix} 1 \ 0 \end{bmatrix} \), where \( c_1 \) and \( c_2 \) are arbitrary constants.
05

Sketch Direction Lines and Their Solutions

Eigenvector \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \) corresponds to solutions moving along the \( x_2 \) axis, decreasing since \( e^{-2t} \) decreases over time. Eigenvector \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) corresponds to solutions moving along the \( x_1 \) axis, increasing since \( e^{t} \) increases over time. Draw these lines in the direction of their respective eigenvectors and indicate the direction of increasing or decreasing motion along these lines.

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Key Concepts

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

Eigenvalues
Eigenvalues are fundamental in understanding systems of differential equations. They provide insight into how a system behaves over time. For a matrix, eigenvalues are the special numbers that reveal information about the transformation it represents. When solving differential equations in the form \( \frac{d\mathbf{x}}{dt} = A\mathbf{x} \), discovering the eigenvalues of matrix \( A \) becomes crucial.

The characteristic equation \( \det(A - \lambda I) = 0 \) helps us find eigenvalues. Here, \( I \) is the identity matrix, and \( \lambda \) represents an eigenvalue. For example, for a matrix \( A = \begin{bmatrix} -2 & 0 \ -3 & 1 \end{bmatrix} \), solving the characteristic equation \( (\lambda + 2)(\lambda - 1) = 0 \) gives the eigenvalues \( \lambda_1 = -2 \) and \( \lambda_2 = 1 \).

Understanding eigenvalues helps describe how each component of a solution grows or decays. In this case, the eigenvalue \( -2 \) suggests a decaying behavior, while \( 1 \) leads to growth. This foundational step sets the stage for further analysis and interpretation of the system's dynamics.
Eigenvectors
Eigenvectors accompany eigenvalues and further aid in understanding the behavior of differential equations. Once we have the eigenvalues, we need to find the eigenvectors. These special vectors do not change direction when the matrix transformation is applied, only potentially their magnitude.

For a given eigenvalue \( \lambda_i \), the eigenvector is found using the equation \( (A - \lambda_i I)\mathbf{v}_i = \mathbf{0} \). In simple terms, we solve this equation to find vectors that show how solutions to the differential equations behave. For our matrix \( A \), we have two eigenvectors:
  • For \( \lambda_1 = -2 \), the eigenvector is \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \), which suggests a transformation along the \( x_2 \) axis.
  • For \( \lambda_2 = 1 \), the eigenvector is \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \), indicating changes along the \( x_1 \) axis.
Using eigenvectors, we can predict the trajectory of points in the system's phase space. This informs us on how the system evolves, providing a comprehensive view of its orientation.
General Solution
The general solution of a differential equation system describes all possible behaviors of the system over time. It is constructed by combining the eigenvectors with their respective exponential growth or decay functions derived from the eigenvalues. This combination is represented as:

\[ \mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 0 \ 1 \end{bmatrix} + c_2 e^{t} \begin{bmatrix} 1 \ 0 \end{bmatrix} \]
Here, \( c_1 \) and \( c_2 \) are arbitrary constants representing initial conditions. This is important because any specific initial state can be expressed with a particular set of constants.

The exponential terms \( e^{-2t} \) and \( e^{t} \) indicate how each component of the solution changes over time. The term \( e^{-2t} \) implies decay, while \( e^{t} \) suggests growth. By graphing these solutions, you can visualize how they move and change, offering a clear picture of the system's dynamic behavior. Understanding the general solution is vital in predicting how the system's state evolves with time, especially when visualizing stability and direction.

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Most popular questions from this chapter

We consider differential equations of the form $$ \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) $$ where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of A will be real, distinct, and nonzero. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a sink, a source, or a saddle point. $$ A=\left[\begin{array}{rr} 0 & -2 \\ -1 & 3 \end{array}\right] $$

An unrealistic feature of the Lotka-Volterra model is that the prey exhibits unlimited growth in the absence of the predator. The model described by the following system remedies this shortcoming (in the model, we assume that the prey evolves according to logistic growth in the absence of the predator; the other features of the model are retained): $$ \begin{array}{l} \frac{d N}{d t}=N\left(1-\frac{N}{K}\right)-4 P N \\ \frac{d P}{d t}=P N-5 P \end{array} $$ Here, \(K>0\) denotes the carrying capacity of the prey in the absence of the predator. In what follows, we will investigate how the carrying capacity affects the outcome of this predator-prey interaction. (a) Draw the zero isoclines of \((11.88)\) for (i) \(K=10\) and (ii) \(K=\) \(3 .\) (b) When \(K=10\), the zero isoclines intersect, indicating the existence of a nontrivial equilibrium. Analyze the stability of this nontrivial equilibrium. (c) Is there a minimum carrying capacity required in order to have a nontrivial equilibrium? If yes, find it and explain what happens when the carrying capacity is below this minimum and what happens when the carrying capacity is above this minimum.

Find all equilibria of each system of differential equations and use the analytical approach to determine the stability of each equilibrium. $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1} x_{2}-x_{2} \\ \frac{d x_{2}}{d t}=x_{1}+x_{2} \end{array} $$

The classical Lotka-Volterra model of predation is given by $$ \begin{array}{l} \frac{d N}{d t}=a N-b N P \\ \frac{d P}{d t}=c N P-d P \end{array} $$ where \(N=N(t)\) is the prey density at time \(t\) and \(P=P(t)\) is the predator density at time \(t .\) The constants \(a, b, c\), and \(d\) are all positive. (a) Find the nontrivial equilibrium \((\hat{N}, \hat{P})\) with \(\hat{N}>0\) and \(\hat{P}>0\). (b) Find the community matrix corresponding to the nontrivial equilibrium. (c) Explain each entry of the community matrix found in (b) in terms of how individuals in this community affect each other.

We consider differential equations of the form $$ \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) $$ where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of A will be real, distinct, and nonzero. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a sink, a source, or a saddle point. $$ A=\left[\begin{array}{rr} -3 & 1 \\ 1 & -2 \end{array}\right] $$
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