Problem 52
Question
Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=x^{3}-x^{5} $$
Step-by-Step Solution
Verified Answer
The equilibrium at \( x = 0 \) is semi-stable.
1Step 1: Identify the Equilibrium Points
To find the equilibrium points, set \( \frac{d x}{d t} = 0 \). This gives us the equation \( x^3 - x^5 = 0 \). Factor this equation to get \( x^3(1 - x^2) = 0 \). So, the equilibrium points are where \( x^3 = 0 \) and \( 1 - x^2 = 0 \). This leads to equilibrium points at \( x = 0 \) and \( x = \pm 1 \).
2Step 2: Determine Stability at x=0
To determine the stability of the equilibrium point at \( x = 0 \), analyze the sign of the derivative of \( \frac{d x}{d t} \). Calculate the derivative: \( \frac{d}{d x}(x^3 - x^5) = 3x^2 - 5x^4 \). Substitute \( x = 0 \) into the derivative: \( 3(0)^2 - 5(0)^4 = 0 \). Since the derivative is neither positive nor negative at \( x = 0 \), we cannot conclude stability from this alone. Thus, continue with the analysis.
3Step 3: Conduct the Sign Analysis
Next, perform a sign analysis of \( \frac{d x}{d t} = x^3 - x^5 \) around \( x = 0 \). Consider intervals around \( x = 0 \):- For \( x \) just greater than 0, \( x^3 - x^5 \) is positive because \( x^3 > x^5 \). - For \( x \) just less than 0, \( x^3 - x^5 \) is negative because \( x^3 < x^5 \) (both terms are negative).The sign change analysis shows that \( \frac{d x}{d t} \) flips sign as \( x \) moves through 0, indicating that the slope changes direction: negative before and positive after. This suggests a semi-stable equilibrium at \( x = 0 \).
Key Concepts
Equilibrium PointsStability AnalysisSign Analysis
Equilibrium Points
In the world of differential equations, equilibrium points are essential to understanding the behavior of systems. An equilibrium point occurs where the rate of change becomes zero, essentially indicating a state of rest or balance.
To identify these points in a given differential equation, one needs to set the derivative equal to zero. In the exercise, the differential equation is given as \( \frac{d x}{d t} = x^{3} - x^{5} \).
By setting \( x^{3} - x^{5} = 0 \), we find the potential equilibrium points.
Factorizing gives \( x^{3}(1 - x^{2}) = 0 \), leading to solutions:
To identify these points in a given differential equation, one needs to set the derivative equal to zero. In the exercise, the differential equation is given as \( \frac{d x}{d t} = x^{3} - x^{5} \).
By setting \( x^{3} - x^{5} = 0 \), we find the potential equilibrium points.
Factorizing gives \( x^{3}(1 - x^{2}) = 0 \), leading to solutions:
- \( x^{3} = 0 \) which implies \( x = 0 \)
- \( 1 - x^{2} = 0 \) which implies \( x = \pm 1 \)
Stability Analysis
Once the equilibrium points are found, the next challenge is determining their stability. Stability indicates whether a system will return to equilibrium after a slight disturbance or diverge away. Analyzing the stability around the found equilibrium points requires examining the derivative of the original function.
In this scenario, we first determine the derivative of \( x^{3} - x^{5} \), which is \( 3x^{2} - 5x^{4} \).
Next, substitute the equilibrium point \( x = 0 \) into this derivative. The result is 0, neither positive nor negative, which means the derivative test alone cannot conclusively determine the stability at this point.
This leads us to conduct further analysis, as stability of \( x = 0 \) remains indeterminate through this method alone.
In this scenario, we first determine the derivative of \( x^{3} - x^{5} \), which is \( 3x^{2} - 5x^{4} \).
Next, substitute the equilibrium point \( x = 0 \) into this derivative. The result is 0, neither positive nor negative, which means the derivative test alone cannot conclusively determine the stability at this point.
This leads us to conduct further analysis, as stability of \( x = 0 \) remains indeterminate through this method alone.
Sign Analysis
Sign analysis is a technique used to investigate the behavior of a function around a particular point by examining its sign in small intervals. This approach helps understand the directionality of the slope, thereby providing clues about stability.
To apply this in our exercise, we analyze the equation \( x^3 - x^5 \) around \( x=0 \).
Consider two scenarios:
To apply this in our exercise, we analyze the equation \( x^3 - x^5 \) around \( x=0 \).
Consider two scenarios:
- For \( x \) values just slightly greater than 0, where \( x^3 > x^5 \), the function becomes positive, indicating that the system is increasing.
- Conversely, for \( x \) values slightly less than 0, \( x^3 < x^5 \) results in a negative value, showing a decrease in the system.
Other exercises in this chapter
Problem 51
Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=x^{4} $$
View solution Problem 51
In Problems, solve each differential equation with the given initial condition. $$ \frac{d y}{d x}=\frac{y+1}{x-1}, \text { with } y(2)=5 $$
View solution Problem 52
In Problems, solve each differential equation with the given initial condition.$$ \frac{d u}{d t}=\frac{\sin t}{u+1}, \text { with } u(0)=3 $$
View solution Problem 53
Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=x^{3}+x^{4} $$
View solution