Problem 81 - Calculus for Biology and Medicine

Question

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. $$ \frac{d N}{d t}=N \ln \left(\frac{2}{N}\right) \quad N>0 $$

Step-by-Step Solution

Answer

The equilibria are \( N=0 \) (unstable) and \( N=2 \) (stable).

Step 1: Identifying Equilibrium Points

Equilibrium points occur when the right-hand side of the differential equation is zero. Set \( \frac{dN}{dt} = N \ln\left(\frac{2}{N}\right) = 0 \). The solutions to this are when either \( N = 0 \) or \( \ln\left(\frac{2}{N}\right) = 0 \). Solving \( \ln\left(\frac{2}{N}\right) = 0 \) gives \( \frac{2}{N} = 1 \), so \( N = 2 \). Thus, the equilibrium points are \( N = 0 \) and \( N = 2 \).

Step 2: Calculating the Derivative

To determine the stability of the equilibria, calculate the derivative of the right-hand side with respect to \( N \). The expression is \( f(N) = N \ln\left(\frac{2}{N}\right) \). Using the product rule and the derivative of a logarithm, \( f'(N) = \ln\left(\frac{2}{N}\right) - 1 \).

Step 3: Evaluating Stability at Each Equilibrium

Evaluate \( f'(N) \) at the equilibrium points. For \( N = 2 \), \( f'(2) = \ln(1) - 1 = -1 \), indicating a stable equilibrium since the derivative is negative. For \( N = 0 \), consider that as \( N \to 0^+ \), \( \ln\left(\frac{2}{N}\right) \to \infty \), hence \( f'(N) \to \infty \), which means \( N = 0 \) is an unstable equilibrium.

Key Concepts

Stability Analysis

Stability analysis is a critical step in understanding how solutions to differential equations behave over time. It helps in determining whether small disturbances will cause the solution to return to its equilibrium state or diverge away from it. In other words, it tells us if an equilibrium point is stable or unstable.

When performing a stability analysis, we focus on the sign of the derivative of the function near the equilibrium point. - If the derivative is negative, small deviations will diminish, implying the equilibrium is stable. - Conversely, if the derivative is positive, any small disturbance will grow, indicating instability.In this exercise, the derivative of the differential equation at various equilibrium points was calculated. For example, at the equilibrium point \( N = 2 \), the derivative was found to be negative, \( f'(2) = -1 \), confirming that this point is stable. On the other hand, as \( N \) approaches zero, the derivative tends towards infinity, which flags the point \( N = 0 \) as unstable.

Equilibrium Points

Equilibrium points are values of \( N \) where the rate of change \( \frac{dN}{dt} \) is zero. This means the system is at rest or in a steady state at these points. Finding equilibrium points involves setting the differential equation to zero and solving for \( N \).

In our exercise, the equation \( \frac{dN}{dt} = N \ln\left(\frac{2}{N}\right) \) was solved to find equilibrium points. By setting \( N \ln\left(\frac{2}{N}\right) = 0 \), we identified two possibilities: - \( N = 0 \)- \( \ln\left(\frac{2}{N}\right) = 0 \) which gives \( N = 2 \)These are the values where the system exhibits no net change, meaning they are equilibrium points. Identifying these points is the first essential step before proceeding to analyze the system's stability.

Linearization

Linearization is the process of approximating a nonlinear system by a linear one near the equilibrium points. This is useful because linear systems are much simpler to analyze, especially when it comes to studying stability.

To perform linearization, we calculate the derivative of the function, which represents the slope of the tangent line at the equilibrium point. This slope helps in making predictions about the system's behavior.In our differential equation \( N \ln\left(\frac{2}{N}\right) \), the derivative \( f'(N) = \ln\left(\frac{2}{N}\right) - 1 \) was used. Evaluating this at the equilibrium points gives us estimates of how the function behaves when slightly nudged from these points. - For \( N = 2 \), a decreasing slope confirms a stable point.- Meanwhile, near \( N = 0 \), the slope is overwhelmingly large, indicative of an unstable point.By replacing the nonlinear curves with linear approximations, linearization simplifies the task of stability analysis near each equilibrium, providing clear insights into the system's dynamic responses.