Question

Harvesting a Renewable Resource. Suppose that the population \(y\) of a certain species of fish (for example, tuna or halibut) in a given area of the ocean is described by the logistic equation $$ d y / d t=r(1-y / K) y . $$ While it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level, and possibly even driven to extinction. Problems 20 and 21 explore some of the questions involved in formulating a rational strategy for managing the fishery. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population \(y:\) The more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by \(E y,\) where \(E\) is a positive constant, with units of \(1 /\) time, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic equation is replaced by $$ d y / d t=r(1-y / K) y-E y $$ This equation is known as the Schaefer model after the biologist, M. B. Schaefer, who applied it to fish populations. (a) Show that if \(E0 .\) (b) Show that \(y=y_{1}\) is unstable and \(y=y_{2}\) is asymptotically stable. (c) A sustainable yield \(Y\) of the fishery is a rate at which fish can be caught indefinitely. It is the product of the effort \(E\) and the asymptotically stable population \(y_{2} .\) Find \(Y\) as a function of the effort \(E ;\) the graph of this function is known as the yield-effort curve. (d) Determine \(E\) so as to maximize \(Y\) and thereby find the maximum sustainable yield \(Y_{m}\).

Short answer

The maximum sustainable yield (Y_m) of the Schaefer model is given by: $$ Y_m = \frac{2 rK^2}{9} $$

Step-by-step solution

Find the equilibrium points

To find the equilibrium points, we need to set the rate of change of the fish population \((d y / d t)\) equal to zero and solve for \(y\). $$ 0 = r(1-y / K) y - E y $$ Now, let's solve for \(y\). First, we can factor out \(y\) from the equation: $$ 0 = y\left(r(1 - y/K) - E\right) $$ This gives us two possible solutions for \(y\): $$ y_1 = 0 $$ and $$ r(1-y_2 / K) - E = 0 $$ To solve for \(y_2\), we can rearrange the equation: $$ \frac{y_2}{K} = 1 - \frac{E}{r} $$ and then multiply by \(K\): $$ y_2 = K(1 - \frac{E}{r}) $$ Under the condition \(E < r\), we have two equilibrium points: \(y_1 = 0\) and \(y_2 = K(1 - E / r) > 0\).

Analyze the stability of equilibrium points

To analyze the stability, we will examine the sign of \(d y / d t\) at each equilibrium point. To do so, let's rewrite the equation of \(d y / d t\) in the form: $$ \frac{d y}{d t} = r(1 - \frac{y}{K})y - Ey $$ At \(y = y_1 = 0\): $$ \frac{d y}{d t} = r(1 - 0)y_1 - E y_1 = 0 $$ Since the population remains constant at \(y_1\), \(y = y_1\) will be unstable. At \(y = y_2 = K(1 - E / r)\), since \(y_2 > 0\), the population will grow. Thus, \(y = y_2\) is asymptotically stable.

Determine the sustainable yield as a function of effort

The sustainable yield \(Y\) is defined as the product of the effort \(E\) and the asymptotically stable population \(y_2\). Therefore, $$ Y = E y_2 = E K(1 - \frac{E}{r}) $$ This function represents the yield-effort curve.

Maximize the sustainable yield and find the maximum sustainable yield

To maximize the sustainable yield \(Y\), we need to find the value of \(E\) that maximizes \(Y\). To do so, we'll take the derivative of \(Y\) with respect to \(E\) and set it to zero: $$ \frac{d Y}{d E} = K(1 - \frac{E}{r}) - 2 \frac{E K}{r} $$ By setting \(\frac{d Y}{d E} = 0\), we get: $$ K(1 - \frac{E}{r}) = 2 \frac{E K}{r} $$ Solve for \(E\): $$ E\left(\frac{3}{r}\right) = K $$ and $$ E = \frac{r}{3} K $$ Now, we can find the maximum sustainable yield, \(Y_m\), by plugging this value of \(E\) into the yield-effort curve: $$ Y_m = \left(\frac{r}{3} K\right) K\left(1 - \frac{\frac{r}{3} K}{r}\right) $$ Simplify the expression: $$ Y_m = \frac{rK^2}{3} \cdot \frac{2}{3} $$ Therefore, the maximum sustainable yield is given by: $$ Y_m = \frac{2 rK^2}{9} $$

Key concepts

Equilibrium Points

In the context of the Schaefer Model, equilibrium points are vital as they indicate stable states where the population of fish doesn't change over time. To find these equilibrium points, we set the rate of change of the fish population, \(\frac{dy}{dt}\), to zero. This is the first step towards understanding the population dynamics in the model.

In the modified logistic equation \(\frac{dy}{dt} = r(1 - \frac{y}{K})y - Ey\), solving this for \(y\) by setting \(\frac{dy}{dt} = 0\) results in two potential solutions:

• \(y_1 = 0\): This point represents the extinction equilibrium where there are no fish left. It's important to recognize this as an unstable state because if the population approaches zero, any disturbance can halt recovery.
• \(y_2 = K(1 - \frac{E}{r})\): This is the positive equilibrium point, which exists only if \(E < r\). It implies a balance between natural growth and harvesting, making it asymptotically stable. This means that if the population is disturbed slightly around this point, it will return to this equilibrium over time.

Analyzing the stability of these points helps in managing the fish population sustainably. For minimal disturbance, scientists aim to operate around the stable equilibrium.

Logistic Equation

The logistic equation describes how populations grow in an environment with limited resources. At its heart, it illustrates the reality that no population can expand indefinitely due to environmental constraints. In the context of fishery management, this equation is modified to incorporate harvesting effects in the Schaefer Model:

• \(\frac{dy}{dt} = r(1 - \frac{y}{K})y - Ey\): The first term, \(r(1 - \frac{y}{K})y\), represents logistic growth where \(r\) is the intrinsic growth rate, \(K\) is the carrying capacity, and \(y\) is the population size. This term shows that the growth rate decreases as the population approaches the carrying capacity \(K\).
• The second term, \(Ey\), introduces the harvesting effect, where \(E\) is the effort applied to harvest the fish. This term shows a reduction in population directly proportional to fishing effort \(E\) and size of the population \(y\).

This adjusted equation provides a more realistic picture for fisheries, as it suggests a balance between growth and harvesting can be achieved. This balance forms the basis for determining sustainable management practices to ensure long-term fish population viability.

Sustainable Yield

In fishery management, a key objective is to determine the sustainable yield, or the amount of fish that can be harvested without depleting the population over time. In the Schaefer Model, sustainable yield, denoted by \(Y\), is defined as the product of the fishing effort \(E\) and the stable equilibrium population \(y_2\):

\[Y = E \cdot K \left(1 - \frac{E}{r}\right)\]

This equation is known as the yield-effort curve and provides insights into how different levels of fishing effort affect the yield. Notably:

• At low levels of effort, yield increases almost linearly with effort because the fish population is large and can replenish quickly.
• As effort increases, the yield begins to decrease after reaching a maximum due to overfishing, reducing the population below its optimal level for most rapid growth.
• The task is to find the optimal \(E\) that maximizes \(Y\), leading to the maximum sustainable yield \(Y_m\). By taking the derivative of \(Y\) with respect to \(E\), setting it to zero, and solving, we find that \(Y_m\) not only maintains ecological balance but also maximizes the long-term harvest potential.

Thus, understanding this balance allows for effective and sustainable fishery management, ensuring fish populations remain stable for future generations.