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Chapter 11: Problem 44

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 complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. $$ A=\left[\begin{array}{rr} -1 & -5 \\ 4 & -3 \end{array}\right] $$

Short Answer

Expert verified
The equilibrium \((0,0)\) is a stable spiral.

Step by step solution

01

Determine Eigenvalues

First, find the eigenvalues of the matrix \( A \). The eigenvalues are found from the characteristic equation \( \det(A - \lambda I) = 0 \), where \( \lambda \) is the eigenvalue, and \( I \) is the identity matrix.
02

Set Up the Characteristic Equation

For the matrix \( A = \begin{bmatrix} -1 & -5 \ 4 & -3 \end{bmatrix} \), the characteristic equation is \( \det \begin{bmatrix} -1 - \lambda & -5 \ 4 & -3 - \lambda \end{bmatrix} = 0 \).
03

Calculate the Determinant

The determinant is calculated as: \[ (-1-\lambda)(-3-\lambda) - (-5)(4) = \lambda^2 + 4\lambda + 11 \]This is the characteristic polynomial.
04

Solve the Characteristic Equation

Solve the quadratic equation \( \lambda^2 + 4\lambda + 11 = 0 \) using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 4, c = 11 \).
05

Apply the Quadratic Formula

Substitute into the quadratic formula: \[ \lambda = \frac{-4 \pm \sqrt{16 - 44}}{2} = \frac{-4 \pm \sqrt{-28}}{2} \]\[ \lambda = -2 \pm i \sqrt{7} \].
06

Analyze the Eigenvalues

The eigenvalues are complex \( \lambda = -2 \pm i \sqrt{7} \). The real part of the eigenvalues is \( -2 \).
07

Determine Stability

Since the real part of the eigenvalues is negative, the equilibrium at \((0,0)\) is stable. The presence of complex eigenvalues indicates spiral behavior.
08

Classify the Equilibrium

The equilibrium is a stable spiral since the real parts of the eigenvalues are negative.

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Key Concepts

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

Stability Analysis
Understanding stability in differential equations helps us determine how systems behave over time. In simple terms, stability analysis involves examining whether solutions of a system converge to a point (equilibrium) as time progresses. For a matrix system like \( \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) \), stability is often assessed by checking the real parts of its eigenvalues.
  • If all real parts are negative, solutions tend to approach the equilibrium, indicating stability.
  • If any real part is positive, the equilibrium is unstable since solutions diverge.
Analyzing eigenvalues, particularly their real parts, is crucial in determining overall system behavior.
Eigenvalues
Eigenvalues are critical in analyzing linear transformations represented by matrices. Essentially, an eigenvalue \( \lambda \) relates to a special set of vectors known as eigenvectors, which, when the matrix is applied, only get scaled but not a directionally altered. To find the eigenvalues of a matrix \( A \), one uses the characteristic equation formed as: \[ \det(A - \lambda I) = 0 \]Eigenvalues tell us how the dynamical system evolves, particularly around an equilibrium point.
Equilibrium Classification
Once we calculate eigenvalues, their nature helps in classifying equilibrium points of differential equations. Here's how classification generally works:
  • Stable focus or spiral: Occurs if complex eigenvalues have negative real parts. The system returns to equilibrium in a spiraling fashion.
  • Unstable focus or spiral: If complex eigenvalues have positive real parts, the system spirals out of equilibrium.
  • Center: Occurs when eigenvalues are purely imaginary with zero real parts, indicating neutral stability with circular solution paths around the equilibrium.
Understanding these classifications provides insights into how solutions behave around equilibrium points.
Complex Eigenvalues
Complex eigenvalues arise when solving the characteristic equation leads to square roots of negative numbers. They are typically expressed as \( \alpha \pm \beta i \), where \( \alpha \) is the real part and \( \beta \) is the imaginary part.
  • The real part \( \alpha \) determines whether solutions grow or decay over time.
  • The imaginary part \( \beta \) relates to oscillatory behavior, indicating how solutions spiral around equilibrium.
In stability analysis, complex eigenvalues often signal rotational behavior, manifesting as spirals in the solution graph.
Characteristic Equation
The characteristic equation comes into play when determining eigenvalues. Each matrix has an associated characteristic polynomial derived from the determinant equation:\[ \det(A - \lambda I) = 0 \]This polynomial is typically quadratic for 2x2 matrices, as in the given example problem. Solving this equation reveals the eigenvalues, guiding us on the system's dynamic nature. The quadratic formula, \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is used to solve it, which can yield real or complex solutions. Determining these eigenvalues helps in analyzing system behavior and classification of equilibrium.

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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. $$ A=\left[\begin{array}{rr} -1 & 0 \\ 0 & -4 \end{array}\right] $$

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): $$ \begin{array}{l} \frac{d N}{d t}=3 N\left(1-\frac{N}{10}\right)-2 P N \\ \frac{d P}{d t}=P N-4 P \end{array} $$ (a) Explain why the prey evolves according to $$ \frac{d N}{d t}=3 N\left(1-\frac{N}{10}\right) $$ in the absence of the predator. Investigate the long-term behavior of solutions to (11.87). (b) Find all equilibria of (11.86), and use the eigenvalue approach to determine their stability. (c) Use a graphing calculator to sketch the solution curve of \((11.86)\) in the \(N-P\) plane when \(N(0)=2\) and \(P(0)=2 .\) Also, sketch \(N(t)\) and \(P(t)\) as functions of time, starting with \(N(0)=2\) and \(P(0)=2\).

Suppose that \(x(t)+y(t)\) is a conserved quantity. If $$ \frac{d x}{d t}=-3 x+2 x y $$ find the differential equation for \(y(t)\).

Consider $$ \begin{array}{l} \frac{d x_{1}}{d t}=2 x_{1}-x_{2} \\ \frac{d x_{2}}{d t}=-x_{2} \end{array} $$ Determine the direction vectors associated with the following points in the \(x_{1}-x_{2}\) plane, and graph the direction vectors in the \(x_{1^{-}}\) \(x_{2}\) plane: \((2,0),(1.5,1),(1,0),(0,-1),(1,1),(0,0)\), and \((-2,-2)\).

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.
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