Math 5110
Midterm 2
Problems:
1. In class we discussed several 2-species models including predator-prey and competitive systems.
However, species may derive mutual benefit from their association; this type of interaction is known
as mutualism. In 1976, May suggested the following set of equations to describe a pair of mutualist
populations:
N1
dN1
= rN1 1 −
,
dt
κ1 + αN2
dN2
N2
= rN2 1 −
,
dt
κ2 + βN1
(1)
where N1 and N2 are the population densities of the two species. We will assume that the parameters
r, κ1 , κ2 , α, and β are all non-negative (≥ 0).
(a) Explain (in words) why these equations describe a mutualistic interaction.
(b) Assume that αβ < 1. Determine the behavior of the two species population. Does it appear to
approach an equilibrium? Is this equilibrium stable? How do you know?
(c) Why is it necessary that αβ < 1? What happens to the populations when αβ > 1? Justify
your answer.
Solution:
(a) If we relabel the quantity κ1 + αN2 and call it γ(N2 ), then we can rewrite the equation for
population one in the following manner:
dN1
N2
= rN1 1 −
.
(2)
dt
γ(N2 )
Thinking back to the logistic model, recall that γ is therefore the (population dependent) carrying capacity, or the density of species one that the environment can sustain. It is easy to see
that γ is an increasing function of N2 , that is, as N2 increases, γ increases. This means that
an increase in the population of species two increases the population of species one that may
be sustained. The consequence of this is that dN1 /dt is an increasing function of N2 (as the
denominator of the negative term gets bigger). A completely analogous argument says that
an increase in population one leads to an increase in the carrying capacity of population two.
Therefore dN2 /dt is an increasing function of N1 . Therefore, both populations have positive
impacts on one another, and maybe considered mutualistic.
(b) Setting the dN1 /dt = 0, a quick bit of algebra gives the following relations which define the
N1 null clines:
N1 = 0, or N1 = κ1 + αN2 .
(3)
Doing the same for the dN2 /dt equation yields the N2 null clines:
N2 = 0,
or
N2 = κ2 + βN1 .
(4)
An equilibrium occurs when an N1 null cline intersects an N2 null cline. Three such intersections are immediately apparent:
N1 = 0, and N2 = 0,
N1 = 0, and N2 = κ2 ,
N1 = κ2 , and N2 = 0.
(5)
(6)
(7)
A bit more algebra is required to find the final equilibrium. First, we use the equation
N1 = κ1 + αN2 ,
(8)
and substitute into the equation for the N2 null cline to get
N2 = κ2 + β (κ1 + αN2 ) .
(9)
This may be solved to give N2 = (κ2 + βκ1 ) / (1 − αβ). Finally we plug back into the equation for the N1 null cine to give N1 = (κ1 + ακ2 ) / (1 − αβ). This is the final equilibrium and
we can see that it is positive (and physical) because by assumption, αβ < 1. We will denote
this equilibrium by (N1∗ , N2∗ )
We will now calculate the Jacobian matrix. It is given by
 
rN12 α
1
r 1 − κ12N
2
+αN2
(κ1 +αN2 ) 
J =
(10)
rN22 β
2N2
r
1
−
.
2
κ
+βN
(κ +βN )
2
1
2
1
Plugging in our four equilibria, we see
r 0
J(0, 0) =
,
0 r
−r rα
J(κ1 , 0) =
,
0
r
r
0
J(0, κ2 ) =
,
rβ −r
−r rα
∗
∗
J(N1 , N2 ) =
.
rβ −r
(11)
(12)
From this we can readily see that at (0, 0), the jacobian has a positive trace, and a positive
determinant. This is an unstable node. At (0, κ2 ) and , the jacobian has a zero trace and a
negative determinant. These are saddles. Finally, at (N1∗ , N2∗ ), the jacobian has a negative
trace and a determinant of ∆ = r2 (1 − αβ). By the assumptions of the problem, this is positive
and therefore the fixed point is a stable node. Because all other equilibrium are unstable,
we suspect that all population trajectories will approach this stable node, which represents
an equilibrium of coexistence between the two species. The phase plane in Figure 1a was
generated with PPlane when α = 1 and β = 1/2. It confirms that all initial conditions appear
to approach this stable equilibrium.
(c) All of the algebra from the previous section is still valid. However, we see that when αβ > 1,
then the non-trivial equilibrium (N1∗ , N2∗ ) becomes negative. Furthermore, the determinant of
the jacobian at this point becomes negative, which implies that this fixed point is now a saddle.
Because there are no stable fixed points in the problem, population trajectories must either
grow without bound, or approach a limit cycle. We have no reason to suspect there is any limit
cycles in this problem, and therefore expect that the two populations will grow without bound.
The phase plane in Figure 1b was generated with PPlane when α = 1 and β = 2. It confirms
that all initial conditions appear to diverge to infinity as both populations grow without bound.
r = 1 k1 = 1 k2 = 1
a = 1 b = 0.5
r=1
x ’ = r x (1 − x/(k1 + a y))
y ’ = r y (1 − y/(k2 + b x))
5
5
4.5
4.5
4
4
3.5
3.5
3
3
2.5
2.5
y
y
x ’ = r x (1 − x/(k1 + a y))
y ’ = r y (1 − y/(k2 + b x))
2
2
1.5
1.5
1
1
0.5
0.5
0
k1 = 1 k2 = 1
a=1 b=2
0
0
0.5
1
1.5
2
2.5
x
3
3.5
4
4.5
5
(a)
0
0.5
1
1.5
2
2.5
x
3
3.5
4
4.5
5
(b)
Figure 1: Phase plane of the mutualistic species model. Both simulations were run with κ1 = κ2 = r = 1.
Panel (a) was generated with α = 1 and β = 1/2. Panel (b) was generated with α = 1 and β = 2.
2. Consider the following predator-prey model that I have (just now) proposed to study the interactions
between baleen whales and their main food source, krill (tiny shrimp). The equations (in nondimensional form) are
ds
= αs (1 − s) − βsw ,
dt
dw
1−w
= γw
.
dt
s
(13)
Here, w is the population density of whales, and s is the population density of krill (tiny shrimp).
All three parameters (α, β, and γ) are non-negative.
(a) To avoid complications, assume that s 6= 0 for this part. Find both equilibria of the model.
What conditions must the parameters meet in order to ensure the existence of an equilibrium
where both species can coexist. Is the coexistence equilibrium stable? How do you know?
(b) Generate a phase plane and some population trajectories illustrating the behavior from part
(a). Now, what happens if the parameters fail to meet the condition you found in part (a)?
What appears to be the fate of whales and krill in this case? Generate some more population
trajectories and interpret your findings biologically.
(c) We can rewrite the equation for the change in whale population density as
dw
= λ(s, w)w,
dt
where λ is the population dependent growth rate of whales. The model that I have proposed has
assumed something very unreasonable (biologically) about λ(s, w), what is the unreasonable
assumption?
Solution:
(a) Setting ds/dt = 0, we have the equation for the shrimp null clines:
αs (1 − s) = βsw.
(14)
Doing the same for dw/dt, we arrive at the equation for the whale null clines:
γw
(1 − w) = 0.
(15)
s
This equation is satisfied whenever w = 1 or w = 0. These are the two whale null clines. We
now consider the case w = 0. Plugging this back into the equation for the shrimp null clines,
we have
αs (1 − s) = 0.
(16)
This is clearly true whenever s = 0 or s = 1. We ignore the case that s = 0, and we have our
first equilibrium: (s, w) = (1, 0). We now return to the case that w = 1. Substituting as before
we, we arrive at
s (α − αs − β) = 0.
(17)
This is true if s = 0, which we will ignore, or when s = 1 − β/α. This gives us our second
equlibrium (s, w) = (1 − β/α, 1). This is the only equalibrium where both species have a
non-zero population. In order for there to be a positive population of shrimp, it must be the
case that 1 − β/α > 0, which is equivalent to α > β.
We now calculate the jacobian of the system:
α (1 − s) − αs − βw
−βs
.
(18)
J=
γ(1−w)−γw
−γw(1−w)
s2
s
Plugging in just the positive coexistence equilibrium, this simplifies to
β − α αβ (β − α)
J(1 − β/α, 1) =
.
γα
0
β−α
(19)
Because we know that α > β, we know that the diagonal elements of this matrix are negative,
and therefore the trace is negative while the determinant is positive. This equilibrium is stable.
(b) In Figure 2a, I show a phase plan generated by PPlane when α = 2 and β = 1 . You can
see that the coexistence equilibrium appears to be stable regardless of the initial condition.
Figure 2b shows a phase plane when α = 1 and β = 2. You can see that the coexistence
equilibrium is no longer positive, and is thus unphysical. It is also unstable. All trajectories
appear to converge to the equilibrium where w = 1 and s = 0. Notice that β is the parameter
that quantifies the negative impact that the whale population has on the krill population. If β
is too big, the whales will “overfish” the krill population and push them to extinction. In this
case, “too big” means bigger than the parameter which describes the krill growth rate, α.
(c) Rearranging the equation for dw/dt, we see that we can define the population dependent
growth rate by
1−w
.
(20)
λ=γ
s
Notice that this means the growth rate of whales is a decreasing function of their food supply,
s. As s goes to infinity, the growth rate of whales goes to zero. As s goes to zero, the growth
rate of whales goes to infinity. Clearly this is a very biologically suspect way for a prey species
to affect its predators.
a=2 b=1
c=1
a=1 b=2
c=1
x ’ = a x (1 − x) − b x y
y ’ = c y (1 − y)/x
5
5
4.5
4.5
4
4
3.5
3.5
3
3
2.5
2.5
y
y
x ’ = a x (1 − x) − b x y
y ’ = c y (1 − y)/x
2
2
1.5
1.5
1
1
0.5
0.5
0
0
0
0.5
1
1.5
2
2.5
x
(a)
3
3.5
4
4.5
5
0
0.5
1
1.5
2
2.5
x
3
3.5
4
4.5
5
(b)
Figure 2: Phase plane of the predator-prey species model. Both simulations were run with γ = 1. Panel
(a) was generated with α = 2 and β = 1. Panel (b) was generated with α = 2 and β = 1.