Cubic Population Model with steady states

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SUMMARY

The discussion centers on the Cubic Population Model represented by the differential equation dN/dt = cN(N-k)(1-N), where c>0 and 0 PREREQUISITES

  • Understanding of differential equations and stability analysis
  • Familiarity with population dynamics concepts
  • Knowledge of cubic functions and their properties
  • Basic calculus, particularly regarding derivatives and equilibrium points
NEXT STEPS
  • Study stability analysis in nonlinear differential equations
  • Learn about phase plane analysis for population models
  • Explore the implications of equilibrium points in ecological models
  • Investigate numerical methods for solving differential equations
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Ecologists, mathematicians, and researchers in population dynamics who are analyzing population models and their long-term behaviors.

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Cubic Population Model with steady states !

I am unsure as what this question means:

Consider the cubic population model: dN/dt = cN(N-k)(1-N) where c>0 and 0<k<1

If the the initial populations is N_0 describe without proof the future of the population, distinguish the various cases on the size of N_0 relative to the steady states N_1, N_2 and N_3.


Now I have found and classified the 3 steady states. But am not sure how to proceed. Solving the equation does not seem to be feasible so what do I do ?
 
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I presume you have determined that the steady states are N=0, N= k, and N= 1.

Now, look at what happens between those values.

We can write the equation as dN/dt= (-1)(N- 0)(N- k)(N- 1).

If N< 0, all three of those factors are negative. dN/dt is the product of 4 negative numbers so dN/dt is positive. N moves toward 0.

If 0< N< k, N-0 is positive while the other two factors are still negative. dN/dt is the product of one positive and three negative numbers so dN/dt is negative. N moves down toward 0 (N= 0 is a "stable" equilibrium).

If k< N< 1, both N- 0 and N- k are positive while N-1 is still positive. dN/dt is the product of two positive and two negative numbers so dN/dt is positive. N moves away from k toward 1. (k is an "unstable equilibrium".)

Finally, if N> 1, all terms, except that original (-1), are positive so dN/dt is the product of three positive and one negative term. N moves down toward 1. (1 is a "stable equilibrium".)
 


HallsofIvy said:
If N< 0, all three of those factors are negative. dN/dt is the product of 4 negative numbers so dN/dt is positive. N moves towards 0.

You mean towards 1 right ? positive = unstable ?
 

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