Problem 38 - Calculus for Biology and Medicine

Question

Denote the size of a population by \(N(t)\), and assume that \(N(t)\) satisfies $$ \frac{d N}{d t}=N\left(1-\frac{N}{K}\right)-N \ln N $$ where \(K\) is a positive constant. (a) Show that if \(K>1\), then there exists a nontrivial equilibrium \(N^{*}>0\) that satisfies $$ 1-\frac{N^{*}}{K}=\ln N^{*} $$ (b) Assume now that the nontrivial equilibrium \(N^{*}\) is a function of the parameter \(K\). Use implicit differentiation to show that \(N^{*}\) is an increasing function of \(K\).

Step-by-Step Solution

Answer

If \(K>1\), a nontrivial equilibrium \(N^*>0\) exists. \(N^*\) increases with \(K\).

Step 1: Understanding Equilibrium Points

An equilibrium point is where the population size does not change over time. Mathematically, this means solving \(\frac{dN}{dt} = 0\). Given the equation \(\frac{dN}{dt}=N\left(1-\frac{N}{K}\right)-N \ln N\), we set it to zero: \(N\left(1-\frac{N}{K}\right)-N \ln N = 0\). Simplifying, we divide both sides by \(N\) (assuming \(N eq 0\)): \(1-\frac{N}{K}= \ln N\). This defines the condition for a nontrivial equilibrium \(N^*>0\).

Step 2: Showing Existence of Nontrivial Equilibrium

Consider the function \(f(N) = 1 - \frac{N}{K} - \ln N\). We seek \(N^* > 0\) such that \(f(N^*) = 0\). At \(N = 1\), \(f(1) = 1 - \frac{1}{K}\). Since \(K > 1\), \(f(1) > 0\). As \(N \rightarrow 0^+\) or \(N \rightarrow \infty\), \(f(N)\) becomes negative. By the Intermediate Value Theorem, there exists some \(N^* > 0\) such that \(f(N^*) = 0\).

Step 3: Implicit Differentiation

Given \(1 - \frac{N^*}{K} = \ln N^*\), we want to find \(\frac{dN^*}{dK}\). Differentiating both sides with respect to \(K\): \(-\frac{dN^*}{dK} \cdot \frac{1}{K} + \frac{N^*}{K^2} = \frac{1}{N^*} \cdot \frac{dN^*}{dK}\). Simplifying, we get \(-\frac{1}{K} \cdot \frac{dN^*}{dK} = \frac{1}{N^*} \cdot \frac{dN^*}{dK} - \frac{N^*}{K^2}\).

Step 4: Solving for Derivative

Rearrange the differentiated equation to find \(\frac{dN^*}{dK}\): \(-\frac{1}{K} \cdot \frac{dN^*}{dK} - \frac{1}{N^*} \cdot \frac{dN^*}{dK} = -\frac{N^*}{K^2}\) which simplifies to \((\frac{1}{K} + \frac{1}{N^*}) \cdot \frac{dN^*}{dK} = \frac{N^*}{K^2}\). Thus, \(\frac{dN^*}{dK} = \frac{N^*}{K^2 (\frac{1}{K} + \frac{1}{N^*})}\).

Key Concepts

Equilibrium Points

In population dynamics, equilibrium points represent scenarios where the population size remains constant over time. For a given population size function, specifically here noted as \(N(t)\), an equilibrium point is obtained when the rate of change of the population, \(\frac{dN}{dt}\), is zero. This effectively signifies a balance where births equal deaths – the population neither grows nor shrinks.

In the equation \(\frac{dN}{dt}=N\left(1-\frac{N}{K}\right)-N \ln N\), setting \(\frac{dN}{dt} = 0\) leads us to the condition \(1-\frac{N}{K}= \ln N\). To ensure this situation describes a population, we focus on a positive solution, meaning \(N^* > 0\), which defines a nontrivial equilibrium point, particularly meaningful when \(K > 1\). Such a point is the size at which the population stays stable without tending toward zero or infinite growth.

The trick here is realizing that this balance requires solving a transcendental equation, often best approached using numerical methods or further mathematical insights like the Intermediate Value Theorem (IVT), ensuring solutions exist under specific conditions.

Implicit Differentiation

Implicit differentiation is a powerful tool in calculus used when a function is not given explicitly in terms of one independent variable. In our context, we tackle the equation for equilibrium \(1 - \frac{N^*}{K} = \ln N^*\) with \(N^*\) depending implicitly on \(K\) (the carrying capacity).

By differentiating both sides with respect to \(K\), we can find out how the equilibrium points \(N^*\) change as the carrying capacity \(K\) varies. This involves taking the derivative while respecting each variable's relationship to the other, essentially treating \(N^*\) as a function of \(K\).

• Start by differentiating both sides. On the left, the chain rule applies, resulting in \(-\frac{1}{K} \frac{dN^*}{dK} + \frac{N^*}{K^2}\).
• On the right, differentiate with respect to \(K\) using properties of natural logarithms, yielding \(\frac{1}{N^*} \frac{dN^*}{dK}\).

By re-arranging the resulting expression, you can isolate \(\frac{dN^*}{dK}\), solving the relationship between \(N^*\) and \(K\). This process helps reveal whether increases in \(K\) correspond to increases in the stable population size \(N^*\), a typical demand when managing sustainable populations.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is an essential concept in calculus that supports finding solutions to equations within specific intervals. This theorem states that if a continuous function, say \(f(x)\), changes signs over an interval \([a, b]\), then there exists at least one point \(c\) within that interval where \(f(c) = 0\).

Applying IVT to the problem involves examining the function \(f(N) = 1 - \frac{N}{K} - \ln N\). With \(K > 1\), we observed that at \(N = 1\), \(f(1) = 1 - \frac{1}{K} > 0\), indicating the function is positive. Meanwhile, as \(N\) approaches zero or infinity, \(f(N)\) becomes negative. The change from positive to negative values as \(N\) varies confirms there is at least one \(N^* > 0\) where \(f(N^*) = 0\).

Thus, IVT guarantees the existence of nontrivial equilibrium points for the population within the domain, particularly highlighting the presence of a solution without explicitly solving the complex equation. This theorem acts as a bridge in proving existence, critical for validating models in theoretical ecology and real-world applications.