Problem 10 - Advanced Engineering Mathematics

Question

In Problems, find all critical points of the given plane autonomous system. $$ \begin{aligned} &x^{\prime}=x^{3}-y \\ &y^{\prime}=x-y^{3} \end{aligned} $$

Step-by-Step Solution

Answer

Critical points are (0, 0), (1, 1), and (-1, -1).

Step 1: Set the Equations to Zero

To find the critical points of the system, set both equations equal to zero:\[ x' = x^3 - y = 0 \]\[ y' = x - y^3 = 0 \]

Step 2: Solve the First Equation

From the first equation, \( x^3 - y = 0 \), you can solve for \( y \):\[ y = x^3 \]

Step 3: Substitute into the Second Equation

Substitute \( y = x^3 \) from Step 2 into the second equation:\[ x - (x^3)^3 = 0 \]This simplifies to:\[ x - x^9 = 0 \]

Step 4: Factor the Equation

Factor the equation \( x - x^9 = 0 \) to find values of \( x \):\[ x(1 - x^8) = 0 \]Thus, \( x = 0 \) or \( 1 - x^8 = 0 \).

Step 5: Solve for x-values

From \( 1 - x^8 = 0 \), we get:\[ x^8 = 1 \]Since \( x^8 = 1 \), possible solutions for \( x \) are \( x = 1 \) and \( x = -1 \) considering real solutions within \([-1, 1]\).

Step 6: Find Corresponding y-values

For each \( x \) value found, find the corresponding \( y \) using the relation \( y = x^3 \):- If \( x = 0 \), then \( y = 0^3 = 0 \).- If \( x = 1 \), then \( y = 1^3 = 1 \).- If \( x = -1 \), then \( y = (-1)^3 = -1 \).

Step 7: List the Critical Points

The critical points are the pairs \((x, y)\) we found:- \((0, 0)\)- \((1, 1)\)- \((-1, -1)\)

Key Concepts

Plane Autonomous System

In mathematics, a plane autonomous system is a system of differential equations that does not explicitly depend on the independent variable, often time. The equations are in two dimensions, which is why it's called a plane. These systems are crucial in understanding dynamics as they model a variety of phenomena in natural sciences and engineering.
For example, the system given by:

• \( x' = x^3 - y \)
• \( y' = x - y^3 \)

These equations describe the rate of change of two variables \( x \) and \( y \), each independent of time. The behavior of such systems can lead to equilibrium points, oscillations, and chaotic behavior, which is why finding critical points is essential.

System of Equations

A system of equations is a set of equations with multiple variables that are solved simultaneously. In our problem, we have two equations:

• \( x' = x^3 - y \)
• \( y' = x - y^3 \)

Solving a system of equations means finding the values of the variables that satisfy all equations simultaneously.
The goal is to find all possible combinations of \( x \) and \( y \) that make both equations equal zero at the same time. This forms the basis for understanding and predicting the behavior of the system modeled by these equations.

Factorization

Factorization is the process of breaking down an equation into simpler components called factors that can be multiplied together to achieve the original equation. It's a vital step in finding solutions to polynomial equations.
In the context of our problem, we encounter the equation:

• \( x - x^9 = 0 \)

By factoring this equation, we express it as:

• \( x(1 - x^8) = 0 \)

This factorization reveals potential solutions quickly, setting each factor to zero gives us a simple way to find the values of \( x \) that satisfy the equation.

Real Solutions

Real solutions are solutions to an equation that are real numbers, as opposed to complex numbers. When solving polynomial equations within a real number context, like our exercise, we look for solutions that fit within the real number line.
The factorized equation \( x(1 - x^8) = 0 \) yields real solutions because the powers and coefficients involved point to values that can be naturally expressed as real numbers. In our example:

• \( x = 0 \)
• \( x = 1 \)
• \( x = -1 \)

These solutions are real and they fall within the understandable range of values, making them meaningful in the context of most physical and real-world systems.