MA 3431 Plankton and Whales Model
Jovanna Baptista and Megan Lally

Introduction

The Plankton and Whales model is a modified predator-prey model, based on the interactions between whales and their food supply, plankton.
To begin, the fixed points of the model were determined and the stability of the hyperbolic fixed point was evaluated.
Next, the stability of non-hyperbolic points was investigated using Bifurcation Theory.

Background

p' = p(k-p-h/(1+p))
h' = h(p/(1+p) - ah)

In this model, the population of plankton is denoted by p, and h represents the population of whales.  The constants k and a are positive.  In h', a is related to the death rate of whales.  In p', k is the growth rate of plankton.  Notice the term p/(1+p) appears in h'; this is logical, because the population of whales will increase as their food supply p becomes more abundant.  A similar term in p', h/(1+p), indicates that the population of plankton depletes as their predator h increases.

Analysis

Fixed Points

The fixed points of the model occur when p' and h' equal zero.  The two simple fixed points are:

Steady State 1:  p=0, h=0          Corresponds to two extinct populations
Steady State 2:  p=k, h=0          Corresponds to whales extinct, and plankton growing at a logistic rate

To find the stability of the steady states, the Jacobian must be evaluated at each fixed point.  The general form of the Jacobian for the Plankton and Whales model is as follows:

k-2p-h/(1+p2)         -p/1+p
JAC=
h/(1+p)2                      p/1+p -2ah

The Jacobian evaluated at Steady State 1 yields the eigenvalues k, 0.  This is a non-hyperbolic fixed point, so its stability can not be determined using linearization.  This steady state will be examined using phaseportraits along with the third steady state.

The Jacobian evaluated at Steady State 2 results in the eigenvalues -k and k/(1+k).  Because k must be positive, this steady state is an unstable saddle point.

The complex third steady state, which corresponds to competetive coexistence, occurs when

k-p-h/(1+p)=p/(1+p)-ah=0.

Because it is difficult to directly solve for values of p and h under these conditions, this steady state can be analyzed without determining these values.

This steady state should correspond to a periodic solution due to the nature of predator-prey models:

As the population of the prey increases, the predators have more food and can grow at a faster rate.  However, theincrease in the predator population causes the population of prey to decrease.  The depletion of prey leads to a shortage of food for the predators, thus decreasing the predator population.  The cycle begins again when the prey, in absence of a large predator population, grow at a positive rate.
A periodic solution is indicative of a Hopf bifurcation, which can only occur if the fixed point is non-hyperbolic.

Investigation of non-hyperbolic fixed points

If Steady State 3 is a Hopf bifurcation, the eigenvalues must be purely imaginary.  This requires that the trace of the Jacobian be zero and the determinant be greater than zero at this fixed point.

Since there is no explicit value or equation for p at this fixed point, it is convenient to transform our model into the following:

p' = p k1(p,h)
h' = h k2(p,h)

where k1(p,h)=(k-p-h/(1+p)) and k2(p,h)=(p/(1+p) -ah)

The evaluation of the Jacobian is as follows:

k1(p,h) + p(dk1/dp)         p(dk1/dh)
JAC=
h(dk2/dp)                             k2(p,h) + h(dk2/dh)

In order to solve for this fixed point,  k1(p,h) and k2(p,h) must equal zero.  The Jacobian, with the partial derivatives included:

-p + hp/(1+p)2          -p/(1+p)
JAC=
h/(1+p)2                       -ah

Before the trace and determinant can be evaluated, some substitutions can be performed that will eliminate a and h.  The substitutions are derived by solving k1(p,h)=0 and k2(p,h)=0 for h:

h1 = (k-p)(1+p)
h2 = p/(a+ap)

h1 is substituted into column one of the Jacobian, and h2 is substituted into the second row of the second column.  The Jacobian simplifies to:

-p(2p+1-k)/(1+p)             -p/(1+p)
JAC=
-(p-k)/(1+p)                         -p/(1+p)

Using this Jacobian, the trace set equal to zero is:

p(-2p - 2 + k)/(1+p) = 0

This can only be true when either p = 0 or k = 2p+2.  Since Steady State 1 already includes the case where p = 0, this case will be neglected.  However, k = 2p+2 can be substituted into the Jacobian to determine the value of the determinant:

p/(1+p)                            -p/(1+p)
JAC=
(p+2)/(1+p)                         -p/(1+p)

The determinant equals 2p/(1+p) which must be greater than zero, as p > 0 in this biological model.
Therefore, the conditions are met for a Hopf bifurcation.

Bifurcation Point

As previously stated, it is difficult to solve directly for p'=0 and h'=0 when neither p nor h are allowed to equal zero, as in the following equality:

eq 1: k-p-h/(1+p)=0
eq 2:  p/(1+p)-ah=0

If eq 1 is solved for h, then substituted into eq 2, the following cubic equation is the result:

eq 3: ap3 + (2a-ak)p2 + (1-2ak+a)p - ak = 0

To determine the values of a, p, and k, use eq 3=0, eq 3'=0, and eq 3''=0.  This is logical, because at a steady state the rate of change must equal zero.  Using these three equations, the the results of the three unknowns are a=1/27, k=8, and p=2.

When the parameters a and k equal 1/27 and 8, respectively, Hopf bifurcation occurs.  The change in stability can be illustrated by varying these parameters and examining solution curves.

The following four phaseportraits illustrate solution curves for various values of a and k, the parameters of the Plankton and Whales model.  As promised, this is also where the stability of Steady State 1 is analyzed.  Note that regardless of the values of a and k, the solution curves never end at the origin.

CASE 1 (SOURCE):  This diagram corresponds to a<1/27, k<8.  Notice that solution curves spiral outward in a periodic manner.  This indicates Hopf bifurcation.

CASE 2 (SINK):  In this phaseportrait, a>1/27, k<8.  The solution curves tend to a stable steady state.

CASE 3 (SINK):  The parameters in this phaseportrait are a<1/27, k>8.  The solution curves tend to a stable steady state, in a manner similar to case 2.

CASE 4 (SOURCE):  This diagram has the parameters a>1/27, k<8.  Here, the solution curves are like those in case1, where solutions spiral outward.

Conclusion

The Plankton Whale model encompasses three states of equlibria between the two populations, corresponding to the three steady states:

• Both species are extinct -  In this case, the whales and plankton are both extinct.  However, since this steady state is unstable, it can only occur if both initial populations are zero.
• One species is extinct - This model only allows for the predator population, whales, to be extinct.  When this occurs, the plankton grow logistically, their growth limited only by space and resources.  It is logical that only the whales could be extinct because the model assumes that the prey, plankton, is the predators only food source.
• Both species survive - Here, the two species coexist.  It was shown that depending upon the parameter values of a and k, the two populations could either approach a constant value, or enter into a cyclic state of existence.
Although more realistic than the Lotka-Volterra model, the Plankton and Whales model is still quite simple.  It does not allow for the extinction of both species over time.  This is unrealistic because for very small plankton populations, it is quite possible that the whale population would starve.  In addition, the model makes the assumption that the predator-prey relationship is exclusive.  Even if plankton is a whale's main food source, it is likely that a whale would find an alternate food source before starvation.  Also, there may exist other predators that deplete the plankton population.  The model neglects these possibilities.  Incorporating these factors would make the model more realistic, but it is likely that the model would drastically increase in complexity.

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