Problem 19 - Advanced Engineering Mathematics

Question

In Problems \(11-20\), classify (if possible) each critical point of the given plane autonomous system as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point. \(x^{\prime}=x\left(10-x-\frac{1}{2} y\right)\) \(y^{\prime}=y(16-y-x)\)

Step-by-Step Solution

Answer

Critical points are: (0,0) unstable node, (0,16) and (10,0) saddle points, (4,12) unstable spiral point.

Step 1: Identify Critical Points

To find the critical points, set \(x'\) and \(y'\) both equal to zero.\[x'=x(10-x-\frac{1}{2}y)=0\y'=y(16-y-x)=0\]From these equations, the critical points are found when either \(x = 0\) or \(y = 0\), or when the factors are zero:1. \(10-x-\frac{1}{2}y = 0\)2. \(16-y-x = 0\)Let's solve these equations for critical points.

Step 2: Solve for Critical Points

First, let's solve \(10-x-\frac{1}{2}y=0\) and \(16-y-x=0\).From the first equation: \[y = 2(10-x)\]Substitute into the second equation:\[16 - y - x = 0 \Rightarrow 16 - (2(10-x)) - x = 0\]\[16 - 20 + 2x - x = 0\]\[x = 4\]Substitute \(x = 4\) into \(y = 2(10-x)\):\[y = 2(10-4) = 12\]So the critical point is \((4, 12)\). Consider also the cases when \(x = 0\) and \(y = 0\) leading to additional critical points \((0, 0)\), \((0, 16)\), and \((10, 0)\).

Step 3: Linearize the System

To classify the critical points, we need to linearize the system around each critical point by finding the Jacobian matrix. The Jacobian of the system is:\[J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix}\]Calculate the partial derivatives:\[\frac{\partial x'}{\partial x} = 10 - 2x - \frac{1}{2}y, \quad \frac{\partial x'}{\partial y} = -\frac{1}{2}x\]\[\frac{\partial y'}{\partial x} = -y, \quad \frac{\partial y'}{\partial y} = 16 - y - x\]

Step 4: Evaluate the Jacobian at Each Critical Point

Plug each critical point into the Jacobian to find the behavior:1. At \((0, 0)\):\[ J = \begin{bmatrix} 10 & 0 \ 0 & 16 \end{bmatrix} \]Eigenvalues are \(\lambda_1 = 10\), \(\lambda_2 = 16\); this is an unstable node.2. At \((0, 16)\):\[ J = \begin{bmatrix} 2 & 0 \ -16 & 0 \end{bmatrix} \]Eigenvalues are \(\lambda_1 = 2\), \(\lambda_2 = -16\); this is a saddle point.3. At \((10, 0)\):\[ J = \begin{bmatrix} -10 & -5 \ 0 & 6 \end{bmatrix} \]Eigenvalues are \(\lambda_1 = -10\), \(\lambda_2 = 6\); this is a saddle point.4. At \((4, 12)\):\[ J = \begin{bmatrix} 2 & -2 \ -12 & 0 \end{bmatrix} \]Determine eigenvalues: \(\lambda^2 - 2\lambda + 24 = 0\); has complex roots implying the system forms a spiral.

Step 5: Classify Point (4, 12)

Solving \(\lambda^2 - 2\lambda + 24 = 0\) yields complex roots, implying a spiral point.Calculate roots:\[\lambda = \frac{2 \pm \sqrt{4 - 96}}{2} = 1 \pm i\sqrt{23}\]Real part is positive, indicating an unstable spiral point.

Key Concepts

Autonomous System

An autonomous system refers to a type of differential equation where the variables are independent of time. In simpler terms, the equations do not change as time progresses, which makes them "autonomous". For example, in the given problem, we have a system described by two equations: \(x' = x(10-x-\frac{1}{2}y)\) and \(y' = y(16-y-x)\). These equations depend only on \(x\) and \(y\), not on time \(t\).

Autonomous systems are commonly used to model real-world phenomena where the rules of the system remain constant over time, such as predator-prey models in ecology or the supply-demand curves in economics. The main advantage of working with autonomous systems is the simplicity in finding critical points, thereby helping in understanding the system's behavior around those points.

Jacobian Matrix

The Jacobian matrix is a crucial tool in analyzing a system of equations, especially when you need to investigate the behavior near critical points. In the context of a two-variable system, it is a 2×2 matrix of first-order partial derivatives. For our system, the Jacobian matrix \(J\) is given by:
\[J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} \]
Calculating these derivatives for our problem:

• \(\frac{\partial x'}{\partial x} = 10 - 2x - \frac{1}{2}y\)
• \(\frac{\partial x'}{\partial y} = -\frac{1}{2}x\)
• \(\frac{\partial y'}{\partial x} = -y\)
• \(\frac{\partial y'}{\partial y} = 16 - y - x\)

The Jacobian matrix helps classify the critical points by determining the stability and type of equilibrium points, using the eigenvalues derived from it.

Eigenvalues

Eigenvalues are fundamental when analyzing the stability of critical points in a system of differential equations. They are values obtained from the characteristic equation formed from the Jacobian matrix. In layman's terms, they give insight into how perturbations around a critical point will behave. For a 2×2 matrix \(A\), the eigenvalues \(\lambda\) are solutions of the equation \(\text{det}(A - \lambda I) = 0\), where \(I\) is the identity matrix.

In our problem, evaluating the Jacobian at each critical point yields different sets of eigenvalues which help us in classifying these points. For instance:

• At \((0, 0)\), eigenvalues \(\lambda_1 = 10\) and \(\lambda_2 = 16\) indicate an unstable node.
• At \((0, 16)\) and \((10, 0)\), different eigenvalue combinations signify a saddle point.
• At \((4, 12)\), complex eigenvalues imply a spiral point.

The sign and nature (real or complex) of the eigenvalues tell us about the type and stability of the critical point.

Spiral Point

A spiral point is a type of critical point in a dynamical system characterized by complex eigenvalues with a non-zero imaginary component. These complex roots generally indicate a rotational type of motion around the point, resembling a spiral. Depending on the real part of the eigenvalues, the spiral can either be stable (spiraling inward) or unstable (spiraling outward).

For the critical point \((4, 12)\) in this system, the characteristic equation yields complex roots \(1 \pm i\sqrt{23}\). Since the real part of these roots is positive, this point is classified as an unstable spiral point. The system exhibits a spiraling motion away from the critical point, implying a tendency to diverge rather than converge.

Unstable Node

An unstable node is another type of critical point where both eigenvalues are real and positive. This configuration leads to trajectories moving away from the critical point in both directions. In our exercise, the point \((0, 0)\) is identified as an unstable node because the eigenvalues \(\lambda_1 = 10\) and \(\lambda_2 = 16\) are both positive.

In practical terms, an unstable node could model situations where initial conditions rapidly diverge, leading to exponential growth or other forms of instability. Systems with unstable nodes require careful control to prevent undesirable exponential deviation from equilibrium.

Saddle Point

A saddle point is a type of equilibrium where the system experiences both stability and instability, depending on the direction of approach. In mathematical terms, a saddle point is classified by one positive and one negative real eigenvalue. This means trajectories will diverge in one direction and converge in another.

For example, in our system, the critical points \((0, 16)\) and \((10, 0)\) are identified as saddle points due to their mixed-sign eigenvalues. Such points often signify scenarios where specific perturbations move towards the equilibrium, while others move away, creating a saddle-like shape in the phase portrait.
Understanding saddle points is crucial in systems like mechanical structures where certain forces stabilize portions while others destabilize them, making visualization and management of forces necessary for system integrity.