Question

Although 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. The following question 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 P: 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 Ep, 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: dP/dt = k (1- (P/N)P) - Ep(This equation is known as the Schaefer model after the biologist M.B. Schaefer.) (a) Show that there are two equilibrium points, P = 0 and P2 = N(1 - E/k). (b) Assume E 0. Also, classify the equilibrium points (source, sink, or node) only for P2. (c) Sketch several graphs of solutions. (d) 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 P2 obtained in part (a) above. Find Y as a function of the effort E. (FYI: The graph of this function is known as the yield-effort curve.) (e) Determine E so as to maximize Y and thereby find the maximum sustainable yield Ymax.

Answer 1

Answer:

See explanation

Step-by-step explanation:

Given:-

- The schaefer model is given as follows:

                                    $\frac{dp}{dt} = k*(1 - \frac{p}{N})*p - E*p$

Find:

Show that if E < k, then there are two equilibrium points p1 = 0 and p2 = N · (1 − e/k) > 0.

Solution:

- The equilibrium point is given by setting the ODE dp/dt to zero, we get:

                                 P*(k*( 1 - P / N) - E) = 0

                                 P1 = 0 or k*( 1 - P / N) - E = 0

                                 P1 = 0 or p2 = N · (1 − E/k)

                                 P2 > 0 when E > k & k > 0

Find:

- Assume E 0. Also, classify the equilibrium points (source, sink, or node) only for P2.

Solution:

- The stability of P2:

    - dp/dt has no further roots in the interval [ 0 ; N · (1 − E/k) ] , Hence, there is no particular sign change in this interval.

     - dp/dt > 0 for P = N · (1 − E/k) - E as dp/dt > 0 for p = E

     - dp/dt is a quadratic function of p with two roots p1 and p2, Hence, dp/dt changes signs at each root thus dp/dt < 0 for P = N · (1 − E/k) + E.

     - Thus, solutions close to the equilibrium P = (1 − E/k) moves closer to the equilibrium point reaching stability as a node.

Find:

- 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 P2 obtained in part (a) above. Find Y as a function of the effort E.

Solution:

- The sustainable yield (at least if po  > 0 ). is the yield we get in the asymptotically stable equilibrium as all solutions tend to this equilibrium.

                                      y = E . P2 = E.N(1 - E/k)

-  y is a quadratic function of E with roots in 0 and r and a negative second derivative $\frac{d^2y}{dE^2} = - 2*\frac{N}{k}$ . The graph is given qualitatively ( Attachment ).

Find:

Determine E so as to maximize Y and thereby find the maximum sustainable yield Ymax.

Solution:

- Maximize y with respect to E, take first derivative of y wrt E:

                       dy / dE = N* ( - 1 / k )*E + N*( 1 - E/k )

                                0 = N*( 1 - 2*E/k )

                                E = k / 2

- As y is a quadratic function with negative second derivative, hence The function y maximizes at E = k / 2. Hence, the maximum sustainable yield y_max is:

                               y_max = y ( k / 2 ) = k*N/ 2 * ( 1 - 1/2)

                               y_max =  k*N/ 2 * ( 1/2)

                               y_max = k*N/ 4

Note: The symbol k in attachment is N while r in attachment is k. Apologies for the inconvenience caused.