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Chapter 11: Problem 61

Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium. $$ A=\left[\begin{array}{ll} 1 & 3 \\ 2 & 3 \end{array}\right] $$

Short Answer

Expert verified
The equilibrium at \((0,0)\) is an unstable node, with eigenvalues \(2 + \sqrt{7}\) and \(2 - \sqrt{7}\), both positive.

Step by step solution

01

Eigenvalue Calculation Preparation

To analyze the stability of the equilibrium point at \((0,0)\), we need to find the eigenvalues of matrix \(A\). The matrix is given as \(A = \begin{bmatrix} 1 & 3 \ 2 & 3 \end{bmatrix}\). Eigenvalues \(\lambda\) are found by solving the characteristic equation \(\text{det}(A - \lambda I) = 0\).
02

Form the Characteristic Equation

Calculate \(A - \lambda I\): \[\begin{bmatrix} 1-\lambda & 3 \ 2 & 3-\lambda \end{bmatrix}\].The determinant is found by: \[\text{det}(A - \lambda I) = (1-\lambda)(3-\lambda) - 2 \cdot 3\].
03

Solve the Determinant

Simplifying the determinant, we get: \[(1-\lambda)(3-\lambda) - 6 = 3 - \lambda - 3\lambda + \lambda^2 - 6 = \lambda^2 - 4\lambda - 3\]. So, the characteristic equation is \(\lambda^2 - 4\lambda - 3 = 0\).
04

Find the Eigenvalues

Solve \(\lambda^2 - 4\lambda - 3 = 0\) using the quadratic formula: \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = -3\). Calculating, we find:\[\lambda = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2}\]. Thus, the eigenvalues are \(\lambda_1 = 2 + \sqrt{7}\) and \(\lambda_2 = 2 - \sqrt{7}\).
05

Analyze Stability

Both eigenvalues are real and positive, \(\lambda_1 = 2 + \sqrt{7}\) and \(\lambda_2 = 2 - \sqrt{7}\). Since they are positive, the equilibrium point \((0,0)\) is an unstable node.

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Key Concepts

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

Equilibrium Point
In mathematics and specifically in differential equations and dynamic systems, the term **equilibrium point** refers to a state where the system is in balance. At this point, there is no change over time, meaning that if the system starts at the equilibrium point, it will stay there unless perturbed.

In the context of our exercise, the equilibrium point given is \(0,0\). This point becomes the focus of stability analysis because understanding whether it will remain stable or become unstable under perturbations can give us insights into the behavior of the entire system.

Examining the equilibrium point is crucial because:
  • It represents where the forces in the system counteract each other perfectly.
  • Learning about its stability helps us predict the system's behavior if it begins near this point.
  • If an equilibrium point is stable, small changes do not lead to a significant deviation from this point.
Eigenvalues
Eigenvalues are critical in determining the behavior of a system near an equilibrium point. They are associated with a square matrix and provide insights into how the system behaves when slightly disturbed.

In our matrix \(\begin{bmatrix}1 & 3 \2 & 3 \\end{bmatrix}\), the eigenvalues help us to decide how solutions to the system evolve over time. We derive eigenvalues by solving the characteristic equation, which we'll explore next.

Some basic thoughts about eigenvalues include:
  • They indicate the direction and rate of change in system dynamics.
  • A real positive eigenvalue signifies that the system will move away from the equilibrium, indicating instability.
  • Complex eigenvalues with a positive real part similarly suggest instability.
Characteristic Equation
The **characteristic equation** is an essential polynomial equation derived from a matrix that is crucial for finding eigenvalues. It is obtained by setting the determinant of \(A - \lambda I\), where \(A\) is a matrix and \(\lambda\) represents eigenvalues, equal to zero.

In our example, we calculated the determinant of \(\begin{bmatrix}1-\lambda & 3 \2 & 3-\lambda \\end{bmatrix}\), guiding us to the characteristic equation \(\lambda^2 - 4\lambda - 3 = 0\).

Key points about the characteristic equation are:
  • It provides all possible eigenvalues for a system's matrix.
  • Solving it usually involves factoring or using the quadratic formula.
  • The nature of the roots (the eigenvalues) informs system stability.
Unstable Node
An **unstable node** in a dynamical system indicates an equilibrium point where nearby solutions diverge away as time progresses. This occurs when all eigenvalues are real and positive, which means that minor deviations from the equilibrium will tend to grow.

In our case, the computed eigenvalues \(2 + \sqrt{7}\) and \(2 - \sqrt{7}\) are both real, and since their values are positive, \(\lambda_1 > 0\) and \(\lambda_2 > 0\), the equilibrium \(0,0\) is classified as an unstable node.

Understanding why a node is unstable includes:
  • Real positive eigenvalues cause the system's trajectories to move away from the equilibrium.
  • This behavior is akin to a ball perched at the top of a hill, where any slight push rolls the ball downhill.
  • Knowing its instability helps in modeling and predicting the long-term behavior of systems.

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Most popular questions from this chapter

Write each system of differential equations in matrix form. $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{3}-2 x_{1} \\ \frac{d x_{2}}{d t}=-x_{1} \end{array} $$

Assume that the diagonal elements \(a_{i i}\) of the community matrix of a species assemblage in equilibrium are negative. Explain why this assumption implies that species \(i\) exhibits selfregulation.

A very simple two-compartment model for gap dynamics in a forest assumes that gaps are created by disturbances (wind, fire, etc.) and that gaps revert to forest as trees grow in the gaps. We denote by \(x_{1}(t)\) the area occupied by gaps and by \(x_{2}(t)\) the area occupied by adult trees. We assume that the dynamics are given by $$ \begin{array}{l} \frac{d x_{1}}{d t}=-0.2 x_{1}+0.1 x_{2} \\ \frac{d x_{2}}{d t}=0.2 x_{1}-0.1 x_{2} \end{array} $$ (a) Find the corresponding compartment diagram. (b) Show that \(x_{1}(t)+x_{2}(t)\) is a constant. Denote the constant by \(A\) and give its meaning. [Hint: Show that \(\frac{d}{d t}\left(x_{1}+x_{2}\right)=0 .\) ] (c) Let \(x_{1}(0)+x_{2}(0)=20\). Use your answer in (b) to explain why this equation implies that \(x_{1}(t)+x_{2}(t)=20\) for all \(t>0\). (d) Use your result in (c) to replace \(x_{2}\) in (11.44) by \(20-x_{1}\), and show that doing so reduces the system \((11.44)\) and \((11.45)\) to $$ \frac{d x_{1}}{d t}=2-0.3 x_{1} $$ with \(x_{1}(t)+x_{2}(t)=20\) for all \(t \geq 0\). (e) Solve the system (11.44) and (11.45), and determine what fraction of the forest is occupied by adult trees at time \(t\) when \(x_{1}(0)=2\) and \(x_{2}(0)=18\). What happens as \(t \rightarrow \infty\) ?

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. $$ A=\left[\begin{array}{rr} 2 & -1 \\ 0 & 3 \end{array}\right] $$

Let $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(2-x_{1}\right)-x_{1} x_{2} \\ \frac{d x_{2}}{d t}=x_{1} x_{2}-x_{2} \end{array} $$ (a) Graph the zero isoclines. (b) Show that \((1,1)\) is an equilibrium. Use the graphical approach to determine its stability. 24\. Let $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(2-x_{1}^{2}\right)-x_{1} x_{2} \\ \frac{d x_{2}}{d t}=x_{1} x_{2}-x_{2} \end{array} $$ (a) Graph the zero isoclines. (b) Show that \((1,1)\) is an equilibrium. Use the graphical approach to determine its stability.
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