Problem 6 - Advanced Engineering Mathematics

Question

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. $$m \frac{d v}{d t}=m g-k v$$

Step-by-Step Solution

Answer

The critical point \( v_c = \frac{mg}{k} \) is asymptotically stable.

Step 1: Identify the Critical Points

To find the critical points, we set the derivative equal to zero. That means setting \( \frac{dv}{dt} = 0 \). Solve \( mg - kv = 0 \) to obtain the critical point \( v_c \). We rearrange to get \( kv = mg \), leading to \( v_c = \frac{mg}{k} \).

Step 2: Linearize the Equation Around the Critical Points

We linearize around the critical point by examining the sign of \( \frac{df}{dv} \) where \( f(v) = mg - kv \). First, differentiate \( f(v) \) to get \( \frac{df}{dv} = -k \).

Step 3: Determine the Stability

The sign of \( \frac{df}{dv} \) at the critical point \( v_c \) indicates stability. Since \( \frac{df}{dv} = -k \) and \( k > 0 \), \( \frac{df}{dv} \) is negative. This implies that the critical point is asymptotically stable.

Key Concepts

Critical Points in Differential Equations

Critical points are essential when studying differential equations, especially autonomous ones. These points signify where the system's derivatives, or rate of change, become zero. In the context of our ongoing exercise, the first-order differential equation:

• Form: \( m \frac{d v}{d t}=m g-k v \)
• Critical points found by setting \( \frac{dv}{dt} = 0 \)
• Solve for \( v \) resulting in \( v_c = \frac{mg}{k} \)

This means at a critical point, the system has a specific behavior that might be of particular long-term interest, as the rate of change of\( v \) is zero. It's from these positions that one studies the stability of the solution.

Asymptotic Stability in Systems

The stability of critical points is critical since it tells us how the system behaves over time when perturbed slightly. If a critical point is asymptotically stable:

• The system returns to this point as time progresses
• Small deviations won't change long-term outcomes, they "decay" back to stability

In our exercise, because the derivative \( \frac{df}{dv} = -k \) is negative, it implies asymptotic stability at the critical point \( v_c \). This occurs because negative derivatives near a critical point pull the function back towards equilibrium.

Linearization Near Critical Points

Linearization is a technique that simplifies the study of differential equations around critical points. It involves approximating the system with a linear equation by calculating the first derivative of the function at the point:

• For our problem, \( f(v) = mg - kv \), leading to \( \frac{df}{dv} = -k \)
• The negative value of \( \frac{df}{dv} \) tells us about stability

By linearizing, complex nonlinear equations give us a clearer picture of the behavior around a critical point. This simplification is particularly powerful when determining if a system will return to its stable state over time.

Understanding First-Order Differential Equations

First-order differential equations involve functions and their first derivatives. They are fundamental in areas requiring modeling of change, such as physics or biology. For our autonomous equation:

• The function involves one derivative, \( \frac{dv}{dt} \)
• Behaviour deduced without explicit solution-solving

The beauty and utility of first-order differential equations lie in their simplicity and versatility as seen in our autonomous equation example, where variables like \( v \) change over time under the influence of terms like \(-kv\) resulting in detailed system behavior analysis.