Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Exercise 17.4.10 Suppose a marine fish population when not subject to harvest is reasonably modeled by $$u^{\prime}(t)=r \times u(t) \times(1-u(t))$$ with time measured in years. Suppose a harvest procedure is initiated, and that a fraction, \(h,\) of the existing population is harvested every year. The harvest is not a fixed amount each year, but depends on the number of fish available. The growth rate will be the difference between the natural birth-death process and the harvest and may be modeled by $$ u^{\prime}(t)=r \times u(t) \times(1-u(t))-h \times u(t) $$ a. Assume \(h=r\) (the harvest rate equals the low density growth rate) Substitute \(h=r\) in Equation 17.17, and simplify. Show that $$u(t)=\frac{1}{r t+1 / u_{0}} \quad \text { where } \quad u(0)=u_{0}$$ is a solution for this model. What will be the eventual annual fish harvest under this harvest strategy? b. Assume \(h=\frac{3}{4} r\) in Equation 17.17 and simplify. Draw a direction field or phase plane for this model. What will be the eventual annual fish harvest under this harvest strategy? c. Assume \(h=\frac{1}{2} r\) in Equation \(17.17,\) and simplify. Draw a direction field for this model. What will be the eventual annual fish harvest under this harvest strategy? d. Which of the three strategies will provide the largest long term harvest?

Step-by-Step Solution

Answer

The largest long-term fish harvest is achieved when \( h = \frac{3}{4}r \).

Step 1: Substitute and Simplify for h = r

Given the equation \( u'(t) = r \cdot u(t) \cdot (1-u(t)) - h \cdot u(t) \) with \( h = r \), substitute to get: \[ u'(t) = r \cdot u(t) \cdot (1-u(t)) - r \cdot u(t) \]. Simplifying gives \[ u'(t) = -r \cdot u(t)^2 \]. This is a separable differential equation.

Step 2: Solve the Differential Equation for Part a

Rearrange terms for separation of variables: \( \frac{du}{u^2} = -r \cdot dt \). Integrate both sides to get \( -\frac{1}{u} = -rt + C \), where \( C \) is the constant of integration. Simplifying, \( \frac{1}{u} = rt + C \). Solve for \( u(t) \), \( u(t) = \frac{1}{rt + C} \). Given \( u(0) = u_0 \), \( \frac{1}{u_0} = C \). Hence, \( u(t) = \frac{1}{rt + 1/u_0} \).

Step 3: Evaluate the Annual Fish Harvest for h = r

The annual fish harvest equals the rate \( h \) times the remaining population \( u(t) \), i.e., \( h \cdot u(t) = r \cdot u(t) = \frac{r}{rt + 1/u_0} \). As \( t \to \infty \), \( u(t) \to 0 \), so the annual harvest approaches zero in the long term.

Step 4: Substitute and Simplify for h = 3/4 r

Substitute \( h = \frac{3}{4}r \) in the equation \( u'(t) = r \cdot u(t) \cdot (1-u(t)) - \frac{3}{4} r \cdot u(t) \). Simplifying gives \( u'(t) = \frac{1}{4} r \cdot u(t) \cdot (1-u(t)) \).

Step 5: Analyze the Differential Equation for h = 3/4 r

For this simpler growth rate, draw a direction field to visualize behavior. The equilibrium occurs when \( u'(t) = 0 \), yielding equilibrium points at \( u(t) = 0 \) or \( u(t) = 1 \). The direction field converges to \( u(t) = 1 \). Thus, the eventual annual harvest is: \( h \cdot u(t) = \frac{3}{4}r \).

Step 6: Substitute and Simplify for h = 1/2 r

Substitute \( h = \frac{1}{2}r \) in the equation \( u'(t) = r \cdot u(t) \cdot (1-u(t)) - \frac{1}{2} r \cdot u(t)\). Simplifying gives \( u'(t) = \frac{1}{2} r \cdot u(t) \cdot (1-u(t)) \).

Step 7: Analyze the Differential Equation for h = 1/2 r

This again calls for a direction field. Equilibrium occurs at \( u(t) = 0 \) or \( u(t) = 1 \) as \( t \to \infty \), the population grows towards \( u(t) = 1 \). Thus, the annual harvest becomes \( \frac{1}{2}r \) as approached in the direction field.

Step 8: Determine Strategy with Largest Long Term Harvest

Compare eventual harvests for each scenario: For \( h = r \), harvest approaches zero. For \( h = \frac{3}{4}r \), harvest stabilizes at \( \frac{3}{4}r \), and for \( h = \frac{1}{2}r \), harvest stabilizes at \( \frac{1}{2}r \). Therefore, \( h = \frac{3}{4}r \) provides the largest long-term harvest.

Key Concepts

Separable Differential Equations

A separable differential equation is a type of differential equation where variables can be separated on opposite sides of the equation. This makes it easier to integrate and solve. In the context of harvest models, the fish population equation becomes separable when simplified. For example, after substituting the harvest rate equal to the growth rate, we simplify the equation to \[ u'(t) = -r \cdot u(t)^2 \]. This equation is separable because the right-hand side depends solely on the function of the dependent variable \(u(t)\). To solve, rearrange terms to separate the variables \(u\) and \(t\): \[ \frac{du}{u^2} = -r \cdot dt \]. Then, integrate both sides. This results in \[ -\frac{1}{u} = -rt + C \], where \(C\) is a constant of integration. Solving for \(u(t)\) gives the solution \[ u(t) = \frac{1}{rt + 1/u_0} \]. This technique transforms complex systems into manageable forms, leading to solutions that describe the system's dynamics.

Harvest Models

In mathematical biology, harvest models are used to describe populations that are subject to harvesting. This is often important in fishery management or wildlife conservation, where certain portions of the population are removed regularly. In the provided problem, the equation \[ u'(t) = r \cdot u(t) \cdot (1-u(t)) - h \cdot u(t) \] describes the modified fish population dynamics, where a fraction \(h\) of the population is harvested each year. The aim is to balance between the natural growth \( r \cdot u(t) \cdot (1-u(t)) \) and harvesting \( h \cdot u(t) \) so that the population remains sustainable. The exercise explores various harvesting rates \(h\) such as when \( h = r, \frac{3}{4}r, \frac{1}{2}r\). Each rate changes how the population approaches equilibrium and the long-term fish harvest availability.Understanding these models helps in planning sustainable harvesting strategies, preventing resource depletion, and maintaining ecological balance.

Equilibrium Points

Equilibrium points in a differential equation system represent states where the system doesn't change over time. Mathematically, these points occur when the derivative is zero, i.e., \( u'(t) = 0 \).For the harvest model equation discussed, the equilibrium points occur at \( u(t) = 0 \) or \( u(t) = 1 \).- **\( u(t) = 0 \)**: This is an extinction point, where no fish are left in the population.- **\( u(t) = 1 \)**: Represents full carrying capacity of the environment, acting as a stable population level.In the context of different scenarios, these points indicate how the population will behave over time with varying harvest rates. For example, when \( h = \frac{3}{4}r \), the direction field indicates that the population stabilizes around the full capacity, leading to sustainable harvesting.Identifying equilibrium helps predict long-term outcomes, inform management decisions, and ensure ecological conservation by maintaining an optimal population size.