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Magazine Chapter 8: Problem 54
Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=x^{2}-x^{3} $$
Short Answer
Expert verified
The equilibrium at \( x=0 \) is semi-stable as it changes behavior depending on direction.
Step by step solution
01
Identify Equilibrium Points
To find the equilibrium points, set the given differential equation \( \frac{d x}{d t} = x^2 - x^3 \) equal to zero. Thus, solve \( x^2 - x^3 = 0 \). Factoring the equation, we get \( x^2(1-x) = 0 \). The solutions to this equation are \( x = 0 \) and \( x = 1 \). These are the equilibrium points.
02
Find the Derivative of the Differential Equation
To determine the stability of the equilibrium at \( x=0 \), we need to analyze the derivative of \( \frac{d x}{d t} \) with respect to \( x \). Differentiate \( f(x) = x^2 - x^3 \) to get \( f'(x) = 2x - 3x^2 \).
03
Evaluate the Derivative at the Equilibrium Point
Evaluate \( f'(x) = 2x - 3x^2 \) at the equilibrium point \( x = 0 \). Substitute \( x = 0 \) into the derivative to get \( f'(0) = 2(0) - 3(0)^2 = 0 \).
04
Analyze Stability Based on Derivative
Since \( f'(0) = 0 \), the stability cannot be determined using linear stability analysis alone, as the first derivative test is inconclusive. We must instead look at the behavior of \( \frac{d x}{d t} = x^2 - x^3 \) around \( x = 0 \). For small positive \( x \), \( x^2 > x^3 \) leading to \( \frac{d x}{d t} > 0 \) indicating increase, while for small negative \( x \), the system does not apply since \( x^2 \) and \( x^3 \) are both positive, meaning the solution is not valid for negative \( x \). Therefore, \( x = 0 \) behaves as a semi-stable point, as the derivative changes sign but is primarily positive around \( x = 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical expressions that describe the relationship between a function and its derivatives. They are used extensively in science and engineering to model how quantities change over time or space. A key component of differential equations is finding how solutions behave over time, which often involves analyzing stability and equilibria.
In the exercise, the differential equation is given as \( \frac{d x}{d t} = x^2 - x^3 \). This is a first-order differential equation, which means the highest derivative is of the first order. The goal is to find the values for \( x \) that do not change over time. These points are known as equilibrium points. By solving \( x^2 - x^3 = 0 \), we identify that the equilibrium points are \( x = 0 \) and \( x = 1 \).
Understanding the behavior of such equations helps predict the stability of systems in fields like physics, biology, and economics.
In the exercise, the differential equation is given as \( \frac{d x}{d t} = x^2 - x^3 \). This is a first-order differential equation, which means the highest derivative is of the first order. The goal is to find the values for \( x \) that do not change over time. These points are known as equilibrium points. By solving \( x^2 - x^3 = 0 \), we identify that the equilibrium points are \( x = 0 \) and \( x = 1 \).
Understanding the behavior of such equations helps predict the stability of systems in fields like physics, biology, and economics.
Linear Stability Analysis
Linear stability analysis is a method used to determine the stability of equilibrium points in differential equations. When given an equilibrium point, this technique involves analyzing the derivative (or linearization) of the differential equation to determine how perturbations from the equilibrium behave over time.
In the context of the original problem, step two involved finding the derivative of \( \frac{d x}{d t} = x^2 - x^3 \), which is \( f'(x) = 2x - 3x^2 \). By evaluating this derivative at an equilibrium, we ascertain whether small deviations will cause the system to return to the equilibrium or move away from it. However, near the equilibrium point \( x = 0 \), the derivative becomes \( f'(0) = 0 \), which makes the analysis indecisive.
Because the derivative was zero, linear stability analysis alone couldn't definitively determine stability in this scenario, prompting a deeper look into the behavior of the function around the equilibrium.
In the context of the original problem, step two involved finding the derivative of \( \frac{d x}{d t} = x^2 - x^3 \), which is \( f'(x) = 2x - 3x^2 \). By evaluating this derivative at an equilibrium, we ascertain whether small deviations will cause the system to return to the equilibrium or move away from it. However, near the equilibrium point \( x = 0 \), the derivative becomes \( f'(0) = 0 \), which makes the analysis indecisive.
Because the derivative was zero, linear stability analysis alone couldn't definitively determine stability in this scenario, prompting a deeper look into the behavior of the function around the equilibrium.
Equilibrium Points
Equilibrium points of a differential equation occur where the rate of change is zero, meaning the system is in a state of balance. In simpler terms, for a function \( f(x) \), equilibrium points satisfy the condition \( f(x) = 0 \). Identifying these points is essential to understanding the behavior of the system over time.
In the problem considered, we identify equilibrium points by solving \( x^2 - x^3 = 0 \), giving us \( x = 0 \) and \( x = 1 \) as the equilibrium points.
Determining the stability of these points requires further analysis. The stability at \( x = 0 \) was evaluated by substituting into the derivative \( f'(x) \). Since this evaluated to zero, linear stability tests were inconclusive. Therefore, the behavior of \( x^2 - x^3 \) around \( x = 0 \) showed that the equilibrium was semi-stable. At this point, small positive deviations from zero tend to grow, indicating instability, while no behavior is defined for negative \( x \). This complex behavior often requires manually examining signs and values of the system to fully comprehend the equilibrium dynamics.
In the problem considered, we identify equilibrium points by solving \( x^2 - x^3 = 0 \), giving us \( x = 0 \) and \( x = 1 \) as the equilibrium points.
Determining the stability of these points requires further analysis. The stability at \( x = 0 \) was evaluated by substituting into the derivative \( f'(x) \). Since this evaluated to zero, linear stability tests were inconclusive. Therefore, the behavior of \( x^2 - x^3 \) around \( x = 0 \) showed that the equilibrium was semi-stable. At this point, small positive deviations from zero tend to grow, indicating instability, while no behavior is defined for negative \( x \). This complex behavior often requires manually examining signs and values of the system to fully comprehend the equilibrium dynamics.
