Problem 11 - Advanced Engineering Mathematics

Question

Solving $$\begin{array}{l} x(20-0.4 x-0.3 y)=0 \\ y(10-0.1 y-0.3 x)=0 \end{array}$$ we see that critical points are \((0,0),(0,100),(50,0),\) and \((20,40) .\) The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0.08(20-0.8 x-0.3 y) & -0.024 x \\ -0.018 y & 0.06(10-0.2 y-0.3 x) \end{array}\right)$$ and so $$\begin{array}{ll} \mathbf{A}_{1}=\mathbf{g}^{\prime}((0,0))=\left(\begin{array}{cc} 1.6 & 0 \\ 0 & 0.6 \end{array}\right) & \mathbf{A}_{2}=\mathbf{g}^{\prime}((0,100))=\left(\begin{array}{cc} -0.8 & 0 \\ -1.8 & -0.6 \end{array}\right) \\ \mathbf{A}_{3}=\mathbf{g}^{\prime}((50,0))=\left(\begin{array}{cc} -1.6 & -1.2 \\ 0 & -0.3 \end{array}\right) & \mathbf{A}_{4}=\mathbf{g}^{\prime}((20,40))=\left(\begin{array}{cc} -0.64 & -0.48 \\ -0.72 & -0.24 \end{array}\right) \end{array}$$ since \(\operatorname{det}\left(\mathbf{A}_{1}\right)=\Delta_{1}=0.96>0, \tau=2.2>0,\) and \(\tau_{1}^{2}-4 \Delta_{1}=1 >0,\) we see that (0,0) is an unstable node. since \(\operatorname{det}\left(\mathbf{A}_{2}\right)=\Delta_{2}=0.48>0, \tau=-1.4<0,\) and \(\tau_{2}^{2}-4 \Delta_{2}=0.04 >0,\) we see that (0,100) is a stable node. since \(\operatorname{det}\left(\mathbf{A}_{3}\right)=\Delta_{3}=0.48>0, \tau=-1.9<0,\) and \(\tau_{3}^{2}-4 \Delta_{3}=1.69 >0,\) we see that (50,0) is a stable node. since \(\operatorname{det}\left(\mathbf{A}_{4}\right)=-0.192< 0\) we see that (20,40) is a saddle point.

Step-by-Step Solution

Answer

(0,0) is an unstable node; (0,100) and (50,0) are stable nodes; (20,40) is a saddle point.

Step 1: Identify the system of equations

The system of equations given is: 1. \( x(20 - 0.4x - 0.3y) = 0 \) 2. \( y(10 - 0.1y - 0.3x) = 0 \) The goal is to find the critical points by solving these equations simultaneously.

Step 2: Find the critical points

Each equation is set to zero, implying potential critical points:1. For \( x(20 - 0.4x - 0.3y) = 0 \), we have \( x = 0 \) or \( 20 - 0.4x - 0.3y = 0 \).2. For \( y(10 - 0.1y - 0.3x) = 0 \), we have \( y = 0 \) or \( 10 - 0.1y - 0.3x = 0 \).Evaluating these results in critical points: \((0,0), (0,100), (50,0), (20,40)\).

Step 3: Set up the Jacobian Matrix

The given Jacobian matrix for the system is: \[ \mathbf{g}^{\prime}(\mathbf{X}) = \begin{pmatrix} 0.08(20-0.8x-0.3y) & -0.024x \ -0.018y & 0.06(10-0.2y-0.3x) \end{pmatrix} \]This matrix is used to analyze the local stability of the system at each critical point.

Step 4: Evaluate the Jacobian at each critical point

We calculate the Jacobian at each critical point:1. \( \mathbf{A}_{1} = \mathbf{g}^{\prime}((0,0)) = \begin{pmatrix} 1.6 & 0 \ 0 & 0.6 \end{pmatrix} \)2. \( \mathbf{A}_{2} = \mathbf{g}^{\prime}((0,100)) = \begin{pmatrix} -0.8 & 0 \ -1.8 & -0.6 \end{pmatrix} \)3. \( \mathbf{A}_{3} = \mathbf{g}^{\prime}((50,0)) = \begin{pmatrix} -1.6 & -1.2 \ 0 & -0.3 \end{pmatrix} \)4. \( \mathbf{A}_{4} = \mathbf{g}^{\prime}((20,40)) = \begin{pmatrix} -0.64 & -0.48 \ -0.72 & -0.24 \end{pmatrix} \)

Step 5: Determine the stability of each critical point

Using the Jacobian matrices, determine stability by the eigenvalue method:1. \((0,0)\): \( \Delta_{1} = 0.96 > 0, \tau_{1} = 2.2 > 0, \tau_{1}^{2} - 4\Delta_{1} = 1 > 0\). Unstable node.2. \((0,100)\): \( \Delta_{2} = 0.48 > 0, \tau_{2} = -1.4 < 0, \tau_{2}^{2} - 4\Delta_{2} = 0.04 > 0\). Stable node.3. \((50,0)\): \( \Delta_{3} = 0.48 > 0, \tau_{3} = -1.9 < 0, \tau_{3}^{2} - 4\Delta_{3} = 1.69 > 0\). Stable node.4. \((20,40)\): \( \Delta_{4} = -0.192 < 0\). Saddle point.

Key Concepts

Jacobian Matrix

In the context of stability analysis for dynamical systems, the Jacobian matrix plays a significant role. It is essentially a matrix of partial derivatives and provides a linear approximation of a system near a critical point. For any system described by functions of multiple variables, the Jacobian matrix helps us understand the system's local behavior.

In our case, the Jacobian matrix is denoted as \(\mathbf{g}^{\prime}(\mathbf{X})\). The matrix contains the derivatives of the given system equations with respect to the variables, calculated at various critical points. By evaluating the Jacobian at critical points, we can determine how small perturbations grow or shrink, which is crucial to the stability of dynamical systems.

The Jacobian helps predict whether a given critical point acts as a node or a saddle and if it is stable or unstable. For example, at point \((0,0)\), the Jacobian is \(\begin{pmatrix} 1.6 & 0 \ 0 & 0.6 \end{pmatrix}\), providing the information needed to assess stability tendencies there.

Critical Points

Critical points in a dynamical system are specific values of the variables where the system doesn't change. Think of them as equilibrium points where the rate of change of the variables is zero. Identifying these points is the first step in analyzing the system's stability.

For the equations \(x(20 - 0.4x - 0.3y) = 0\) and \(y(10 - 0.1y - 0.3x) = 0\), setting these equal to zero allows us to find potential values of \(x\) and \(y\) where the system is in equilibrium. This results in the critical points: \((0,0)\), \((0,100)\), \((50,0)\), and \((20,40)\).

Understanding and identifying these points helps us assess how the system behaves near these values and determine overall system stability.

Eigenvalue Method

The eigenvalue method is a critical mathematical tool used to study the stability of critical points in a dynamical system. After obtaining the Jacobian matrix, the next step is to find its eigenvalues. These eigenvalues help in comprehending how perturbations evolve near critical points.

By examining the eigenvalues, we can determine the nature of each critical point. The determinant of the Jacobian \(\Delta\) and the trace \(\tau\) are used in the classification. For instance, when \(\Delta > 0\) and \(\tau^2 - 4\Delta > 0\), the system is stable at that point. Conversely, if \(\tau > 0\), it indicates instability.

This method is invaluable because it translates the problem into a linear algebra one, providing insight into system stability through mere matrix calculations.

Saddle Point

A saddle point represents a critical point where a dynamical system behaves unstably in some directions and stably in others. It is analogous to a mountain pass, directing away in certain paths while converging in others.

Mathematically, a saddle point is identified by a Jacobian matrix with a determinant \(\Delta\) less than zero. This indicates that the system neither purely stabilizes nor destabilizes. In our exercise, the critical point \((20,40)\) is a saddle point because its Jacobian matrix determinant \(\Delta_4\) is less than zero.

Recognizing saddle points is crucial in stability analysis, as they highlight regions where small changes can amplify, possibly leading to significant deviations in system behavior.