Problem 8 - Advanced Engineering Mathematics

Question

Find all critical points of the given plane autonomous system. $$ \begin{aligned} &x^{\prime}=y^{2}-x \\ &y^{\prime}=x^{2}-y \end{aligned} $$

Step-by-Step Solution

Answer

The critical points of the system are (0, 0) and (1, 1).

Step 1: Identify the system of equations

We are given the system: \( x' = y^2 - x \) and \( y' = x^2 - y \). Our task is to find the critical points of this system.

Step 2: Set derivatives to zero for equilibrium

Critical points occur where both derivatives are zero. Set \( x' = y^2 - x = 0 \) and \( y' = x^2 - y = 0 \).

Step 3: Solve the first equation

From \( y^2 - x = 0 \), we express \( x \) in terms of \( y \): \( x = y^2 \).

Step 4: Solve the second equation

Using \( x = y^2 \) in \( x^2 - y = 0 \) gives \( (y^2)^2 - y = 0 \). Simplifying, we have \( y^4 - y = 0 \).

Step 5: Factor and solve for y

Factor the equation \( y^4 - y = 0 \) as \( y(y^3 - 1) = 0 \). This gives solutions \( y = 0 \) or \( y^3 = 1 \).

Step 6: Find y solutions

Solving \( y^3 = 1 \), we find \( y = 1 \). Thus, the possible values for \( y \) are \( y = 0 \) or \( y = 1 \).

Step 7: Determine corresponding x values

Using \( x = y^2 \) with \( y = 0 \) and \( y = 1 \):- If \( y = 0 \), \( x = 0^2 = 0 \).- If \( y = 1 \), \( x = 1^2 = 1 \).

Step 8: Identify the critical points

The critical points of the system are \((x, y) = (0, 0)\) and \((x, y) = (1, 1)\).

Key Concepts

Critical Points

In a system of differential equations, critical points (often referred to as equilibrium points) are crucial for understanding the system's behavior. These points are where the derivative of each function in the system equals zero. This means that at these points, the system is at rest, with no change occurring in either variable over time.

To find critical points, set each equation in a system of differential equations equal to zero. For instance, for the given system:

• The first equation is: \( x' = y^2 - x = 0 \)
• The second equation is: \( y' = x^2 - y = 0 \)

Start by expressing one variable in terms of the other from one equation and substitute it into the second equation. This approach allows you to isolate potential values for your variables and identify where both derivatives simultaneously reach zero, marking these as the critical points.

Equilibrium Points

Equilibrium points in a differential equation system are locations where the system doesn’t change. They can be thought of as resting points where the system is stable under certain conditions.

In the equation system given, the equilibrium points are found using the conditions:

• \( x' = y^2 - x = 0 \)
• \( y' = x^2 - y = 0 \)

By solving these, we look for all values of \( x \) and \( y \) that satisfy both conditions simultaneously. The solution to this specific system yields two such equilibrium points, at \((x, y) = (0, 0)\) and \((x, y) = (1, 1)\). These points can represent stable states of the system where any perturbation will be balanced out by the system dynamics.

System of Differential Equations

A system of differential equations consists of multiple equations involving derivatives of several variables. It describes how each variable changes with respect to one or more independent variables. These systems are common in modeling dynamic systems in physics, biology, and economics.

Consider the given system:

• \( x' = y^2 - x \)
• \( y' = x^2 - y \)

This system is autonomous because the derivatives do not explicitly depend on any independent variable like time, but only on the variables\( x \) and \( y \) themselves.

Solving a system of differential equations often involves finding the equilibrium points, which provide insights into the long-term behavior of the system. These equilibrium points help in determining the stability and oscillatory patterns within the system and are foundational for predicting system behavior under different conditions.