Problem 56 - Calculus for Biology and Medicine

Question

Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=x^{3} e^{-x} $$

Step-by-Step Solution

Answer

The equilibrium at \( x = 0 \) is stable.

Step 1: Understand the System

We are given the differential equation \( \frac{d x}{d t} = x^3 e^{-x} \). The goal is to determine the stability of the equilibrium point at \( x = 0 \).

Step 2: Identify the Equilibrium

An equilibrium point is found when \( \frac{d x}{d t} = 0 \). In this case, setting \( x^3 e^{-x} = 0 \), we find that \( x = 0 \) is the only equilibrium since \( e^{-x} eq 0 \) for any real \( x \).

Step 3: Determine the Derivative

To determine stability, we examine the sign of the derivative of \( \frac{d x}{d t} \) with respect to \( x \) around the equilibrium point. We calculate \( f(x) = x^3 e^{-x} \) and find its derivative \( f'(x) = 3x^2 e^{-x} - x^3 e^{-x} \).

Step 4: Evaluate the Derivative at Equilibrium

Evaluate \( f'(x) \) at \( x = 0 \):\[ f'(0) = 3(0)^2 e^{0} - (0)^3 e^{0} = 0. \]The derivative at \( x = 0 \) is zero, which doesn't help determine stability directly. Thus, we need to use higher-order terms.

Step 5: Use a Higher-Order Derivative Test

Since the first derivative test is inconclusive, we use the first non-zero higher-order derivative. We compute:\( f''(x) = (-6x + 3x^2) e^{-x} - (3x^2 - x^3) e^{-x} \). At \( x = 0 \),\[ f''(0) = (-6 \cdot 0 + 3 \cdot 0^2) e^{0} = 0, \]we move to the third derivative,\( f'''(x) = (-6 + 6x - 6x^2 + x^3) e^{-x} \).\At \( x = 0 \),\[ f'''(0) = -6 e^{0} = -6. \]

Step 6: Analyze the Third Derivative Result

Since \( f'''(0) = -6 < 0 \), the equilibrium point \( x = 0 \) is considered stable because the higher-order derivative test shows that \( x = 0 \) acts like a local minimum.

Key Concepts

Differential Equations

Differential equations, like the one presented here, play a crucial role in modeling dynamic systems. They express how the rate of change of a variable, such as position or concentration, depends on the variable itself. This particular equation, \( \frac{d x}{d t} = x^3 e^{-x} \), suggests how the value of \( x \) changes over time. Understanding how these changes occur is essential for predicting the system's behavior, such as population growth or chemical reactions. To solve a differential equation, one often seeks solutions termed "equilibrium points," where the rate of change equals zero. Here, this condition leads us to analyze stability at different points. Without direct computation, one can rely on qualitative methods, like phase portraits or numerical simulations, although analytical analysis as shown here is often possible. In summary, differential equations offer a way to describe the dynamics of systems. They are a foundational concept in fields ranging from physics to finance, being essential for analyzing system stability and evolution.

Equilibrium Points

Equilibrium points in a differential equation are locations where the derivative equals zero, indicating no change over time at that precise point. For our example, setting \( \frac{d x}{d t} = 0 \) leads to finding equilibrium at \( x = 0 \). These points are crucial for determining the system's long-term behavior as they can represent stable, unstable, or semi-stable conditions. In analyzing equilibrium points, we need to determine their nature:

• Stable equilibrium, where small disturbances reduce over time, returning the system to equilibrium, similar to a ball in a bowl.
• Unstable equilibrium, where perturbations grow, taking the system away from the point, akin to a ball on a hill.
• Semi-stable equilibrium, which depends on the direction of disturbances.

The character of these points can be inferred through techniques like linearization or examining higher-order derivatives if the linear test is inconclusive. Understanding these points helps predict how a system reacts to internal and external changes, hence providing insight into its inherent stability.

Higher-Order Derivatives

When the first derivative at an equilibrium point is zero, as in our example solution, resorting to higher-order derivatives can provide insight into the stability. This method involves evaluating the second or third derivative at the equilibrium to check their signs, which can indicate stability. In this problem, the first derivative test proved unstable, making it necessary to analyze higher derivatives:

• The second derivative, \( f''(x) \), was calculated as \((-6x + 3x^2) e^{-x} - (3x^2 - x^3) e^{-x}\). Evaluating this at \( x = 0 \) also equaled zero, leading us to further derivatives.
• The third-order derivative, \( f'''(x) = (-6 + 6x - 6x^2 + x^3) e^{-x} \), provided the necessary insight.
• Finally, \( f'''(0) = -6 \), a negative value, indicated stability as it resembles a local minimum.

Higher-order derivatives thus play a fundamental role when initial simple derivative tests yield inconclusive results. By revealing subtle curvature features of a system at equilibrium, these derivatives help ascertain the behavior around critical points, enhancing our understanding of the system's stability characteristics.