Problem 4 - Advanced Engineering Mathematics

Question

In Problems \(3-10\), without solving explicitly, classify the critical points of the given first-order autonomous differential equation as either asymptotically stable or unstable. All constants are assumed to be positive. $$ \frac{d x}{d t}=-k x \ln \frac{x}{k}, x>0 $$

Step-by-Step Solution

Answer

The critical point \( x = k \) is asymptotically stable.

Step 1: Identify Critical Points

To find the critical points, set \( \frac{d x}{d t} = 0 \). This gives the equation: \[-k x \ln \frac{x}{k} = 0\]. Since \( k \) and \( x \) are positive and not zero, the equation simplifies to \( \ln \frac{x}{k} = 0 \). The critical point is where \( \frac{x}{k} = 1 \), or equivalently, where \( x = k \).

Step 2: Determine Stability

To classify the stability, determine the sign and change of sign of \( \frac{d x}{d t} \) around the critical point \( x = k \). Consider \( \frac{d x}{d t} = -k x \ln \frac{x}{k} \). For \( x > k \), \( \frac{x}{k} > 1 \), so \( \ln \frac{x}{k} > 0 \) which makes \( \frac{d x}{d t} < 0 \). For \( x < k \), \( \frac{x}{k} < 1 \), thus \( \ln \frac{x}{k} < 0 \) which gives \( \frac{d x}{d t} > 0 \).

Step 3: Conclusion on Stability

The behavior around \( x = k \) indicates that if \( x \) is slightly greater than \( k \), the system moves back to \( k \) (since \( \frac{d x}{d t} < 0 \)). Likewise, if \( x \) is slightly less than \( k \), the system also moves towards \( k \) (because \( \frac{d x}{d t} > 0 \)). Hence, the critical point \( x = k \) is asymptotically stable.

Key Concepts

Critical Points

Critical points are fundamental in understanding the behavior of solutions to differential equations. They are the values where the derivative of the function is zero. In simpler terms, it's where the rate of change stops, and the solution becomes momentarily constant.
To find critical points in an autonomous differential equation like the one given, you set the derivative to zero. In the exercise, the equation is \[-k x \ln \frac{x}{k} = 0\].
This simplifies to where the natural logarithm becomes zero, that is, at \( x = k \). The critical point here is where the system settles into a state of no change temporarily.
Understanding the location and value of these critical points is crucial as they help us predict how the system evolves over time around these points.

Stability Analysis

Stability analysis dives into how the system behaves when slightly perturbed around the critical points. It helps us understand whether the system will return to the critical point or drift away, which relates directly to the concept of stability.
For the equation provided: \[\frac{d x}{d t} = -k x \ln \frac{x}{k}\],
we examine what happens when \( x \) is slightly more or less than \( k \), the critical point.

• When \( x > k \), the natural log term becomes positive, making \( \frac{d x}{d t} < 0 \). This means the system decreases and moves back towards \( k \).
• When \( x < k \), the natural log becomes negative, resulting in \( \frac{d x}{d t} > 0 \), causing \( x \) to increase towards \( k \).

These observations indicate that the system naturally moves back to the critical point, showing a tendency for stability around this point.

Asymptotic Stability

Asymptotic stability takes stability analysis a step further by examining if the system only returns to the critical point eventually. A system is asymptotically stable if any small deviation from a critical point results in the system adjusting itself back to that point over time.
Given the inferred behavior around our critical point \( x = k \):

• If the system starts with \( x > k \), it moves down back to \( k \).
• When starting from \( x < k \), it ascends towards \( k \).

This behavior shows that no matter where you start (in proximity to the critical point), the solutions tend to settle back to \( x = k \) as \( t \) approaches infinity.
Thus, the critical point at \( x = k \) is considered asymptotically stable, indicating that it's a steady resting place for the system's states over an extended period.