Population Model Homework: Determine Equilibrium Solutions

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SUMMARY

The discussion centers on determining equilibrium solutions for a logistic population model with harvesting. The model is defined by the differential equation dP/dt = kP(1 - (P/a)) - EP, where k = 1 and a = 4. The user attempted to factor the equation incorrectly, leading to erroneous equilibrium solutions. The correct equilibrium solutions occur when dP/dt = 0, yielding P(0) = -E and 1/4, but the negative solution is physically impossible.

PREREQUISITES
  • Understanding of logistic population models
  • Familiarity with differential equations
  • Knowledge of equilibrium analysis in population dynamics
  • Basic grasp of harvesting models in ecology
NEXT STEPS
  • Study the implications of harvesting on logistic growth models
  • Learn about stability analysis of equilibrium points in differential equations
  • Explore the effects of varying the effort constant E on population dynamics
  • Investigate alternative models for population growth beyond the logistic model
USEFUL FOR

Students studying population dynamics, ecologists modeling fish populations, and anyone interested in mathematical biology and differential equations.

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Homework Statement



The simplest useful model for fisher comes from the logistic model for population growth, together with a harvest h which is proportional to the current population P, that is,

h=EP,

where the constant E is called the effort. E measures the fraction of the population harvested, so that 0 <= E <= 1. This gives the model

dP/dt = kP(1-(P/a) - EP,

where P(t) is the number of these fish at time t year and k (the natural growth rate) and a( the carrying capacity) are constants for a particular fish polulation. In what follows take k= 1 and a= 4, for simplicity.

a) determine the equilibrium solutions for a given effort E.

my attempt.

dP/dt = P - P^2/4 - EP

dP/dt = ( P + E) (P -1/4)

equilibrium solutions are when dP/dt=0

so P(0)= -E, 1/4.
 
Last edited:
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tylersmith7690 said:

Homework Statement



The simplest useful model for fisher comes from the logistic model for population growth, together with a harvest h which is proportional to the current population P, that is,

h=EP,

where the constant E is called the effort. E measures the fraction of the population harvested, so that 0 <= E <= 1. This gives the model

dP/dt = kP(1-(P/a) - EP,

You miss a parentheses: dP/dt = kP(1-(P/a)) - EP

tylersmith7690 said:
where P(t) is the number of these fish at time t year and k (the natural growth rate) and a( the carrying capacity) are constants for a particular fish polulation. In what follows take k= 1 and a= 4, for simplicity.

a) determine the equilibrium solutions for a given effort E.

my attempt.

dP/dt = P - P^2/4 - EP

dP/dt = ( P + E) (P -1/4)

The last equation is wrong.

tylersmith7690 said:
equilibrium solutions are when dP/dt=0

so P(0)= -E, 1/4.

Because of the faulty factorisation, the result is not correct (and physically impossible).

ehild
 

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