Lecture 5
The Principle of Competitive
Exclusion
Reading: The treatment here was inspired by that in Section 4.11 of Martin Braun’s
(terrific) book
Martin Braun (1993), Di↵erential Equations and Their Applications, Vol
11 in Springer Texts in Applied Mathematics, ISBN 0-387-97894-1,
but this is a much-studied problem and similar material appears in Section 3.5 of
Murray’s book
James D. Murray (2002), Mathematical Biology I: An introduction, 3rd
edition, Springer Interdisciplinary Applied Mathematics Series, Vol. 17,
ISBN 0-387-95223-2.
Available online at http://bit.ly/OnlineMurrayVol1.
and Section 2.5 of Britton’s
Nicholas F. Britton (2003), Essential Mathematical Biology, Springer
Undergraduate Mathematics Series, ISBN 978-85233-536-6.
5.1
Two competing species
This part of the course involves an ecological model for two species that compete for
a common resource, for example, two species of rats infesting the same building. The
Principle of Competitive Exclusion says that typically only one of the two species can
survive indefinitely, while the other will be driven to extinction. Working out when
and why this principle holds will also provide an example of rigorous phase-plane
analysis.
We’ll study a model that involves two kinds of beings that interact by competing
for something—food, say, or nesting sites—whose availability is limited. We’ll refer
5.1
to the two populations as N1 (t) and N2 (t) and imagine them to evolve according to1
✓
◆
dN1
N 1 + a1 N 2
= r 1 N1 1
dt
K1
✓
◆
dN2
N 2 + a2 N 1
= r 2 N2 1
.
(5.1)
dt
K2
Both ODEs are modified versions of the logistic growth law
✓
◆
dN
N
= rN 1
,
dt
K
so the growth rates rj (which have units of 1/time), the carrying capacities Kj (units
of “beasts”) and the dimensionless coefficients aj are all positive real numbers. The
novelty here is that where the logistic law for N1 (t) would include a populationlimiting term of the form N1 /K1 , the system (5.1) has (N1 + a1 N2 )/K1 instead.
That is, members of the second species appear in dN1 /dt like extra members of
the first species and, up to a constant of proportionality a1 , limit the growth of N1
accordingly.
The motivation behind the model in Eqn. (5.1) is especially clear when the
limiting resource is food. Suppose that N1 represents the population of the larger
animals, and N2 that of the smaller. If one of the smaller animals needs half as
much food as one of the larger ones, then we’d set:
• a1 = 1/2 as each animal of species two appear like half a animal from species
one;
• a2 = 2 as each animal of species one eats as much food as two animals from
species two;
• K2 = 2K1 as the environment can support twice as many of the smaller beast.
5.1.1
Dimensionless form
As usually, the first thing we’ll do is reformulate the problem in terms of dimensionless variables. Although there are many ways to do this, the following choices
N1
N2
u=
,
v=
and
⌧ = r1 t
K1
K2
are certainly very natural: we scale the populations by their carrying capacities and
use the intrinsic growth rate r1 to define a dimensionless time. Familiar calculations
(see, e.g., Section 1.4) then lead to the system
du
dv
= f (u, v) = u (1 u ↵v) and
= g(u, v) = ⇢v (1 v
u) ,
(5.2)
d⌧
d⌧
where
✓ ◆
✓ ◆
K2
K1
r2
↵ = a1
,
= a2
and ⇢ = .
K1
K2
r1
This reduces the original set of six parameters down to three and, as we’ll see, the
long-term behaviour of the system depends on only two of them, ↵ and .
1
The notation in Eqn. (5.1) is close, but not identical, to that used in Murray’s Section 3.5.
5.2
v
v
v
(0,1/α )
(0,1/α )
α≤β<1
α<1<β
1<α≤β
(0,1)
(0,1)
(0,1)
(u* , v*)
(0,0)
(0,1/α )
(1,0)
(1/β,0)
u
(0,0)
(1/β,0)
(1,0)
u
(u* , v* )
(0,0)
(1/β,0)
(1,0)
u
Figure 5.1: The null clines for Eqn. (5.2): du/d⌧ = 0 on the blue curves, while
dv/d⌧ = 0 on the red ones. The equilibria at (0, 0), (1, 0) and (0, 1) are always
present, but the one with u? > 0 and v? > 0 is present only in cases of weak
(↵  < 1) or strong (1 < ↵  ) competition.
5.1.2
Null clines and equilibria
When analysing systems such as (5.2) it’s often instructive to sketch the null clines,
which are curves along which one or the other of the time derivatives vanish. Thus
for the model of Eqn. (5.2) there are two sets of null clines, one of which is defined
by
du
= 0 ) u(1 u ↵v) = 0 ) u = 0 or (1 u ↵v) = 0.
d⌧
The latter of the two conditions defines a line on the (u, v) plane and one can sketch
a line by finding two points on it. Here (0, 1/↵) and (1, 0) are a suitable pair, so
the du/d⌧ = 0 null cline is the union of the v axis (the set u = 0) and the line that
passes through (0, 1/↵) and (1, 0).
Very similar calculations show that the other null cline is defined by
dv
= 0 ) ⇢v(1
d⌧
v
u) = 0 ) v = 0 or (1
v
u) = 0,
so this null cline is the union of the u axis and the line passing through (0, 1) and
(1/ , 0). Both sets of null clines are illustrated in Figure 5.1. Equilibria occur at
places where both derivatives are zero and so appear in the phase portrait as places
where the dv/d⌧ and du/d⌧ null clines intersect.
5.2
Analysis of the model
The main question one wants to ask about the model in Eqn. (5.2) concerns longterm dynamics: Can the two species coexist indefinitely or will one of them eventually go extinct? This turns out to be closely related to the number of equilibria the
system has and the stability types of those equilibria. I’ll begin, then, by working
out the community matrix, which is


@u f @v f
1 2u ↵v
↵u
A =
=
.
(5.3)
@ u g @v g
⇢ v
⇢(1 2v
u)
5.3
(u? , v? )
Linearisation

(0, 0)
(1, 0)
(0, 1)

1 0
0 ⇢
1
↵
both positive
unstable node
< 1) or
> 1)
saddle or stable node
one (when ↵ < 1) or
two (when ↵ > 1)
negative
0
⇢
⇢
Classification
one (when
two (when
negative
1
↵
0 ⇢(1
)

Eigenvalues
saddle or stable node
Table 5.1: The three equilibria of (5.2) that are always present along with the
linearisation (5.3) evaluated at (u? , v? ) and the resulting classifications.
Whatever values the parameters of the system (5.2) have, there are always three
equilibria—at (u? , v? ) = (0, 0), (0, 1) and (1, 0)—that correspond to extinction of
one or both of the species. The community matrices at these points and their
corresponding stability types are summarised in Table 5.1.
5.2.1
Coexistence: equilibria with u? , v? > 0
There may also be one more equilibrium that represents coexistence of the two
species and so has positive populations for both. It occurs when the null clines
(1
u
↵v) = 0
and
(1
v
u) = 0
intersect in the region where u? > 0 and v? > 0. These lines can be rewritten in the
following way
1 u
v=
and
v=1
u
↵
and as they must intersect at the equilibrium (u? , v? ), we have
1
u?
↵
= v? = 1
u?
or
1
u? = ↵
↵ u?
so
u? =
which means that the equilibrium will be
✓
◆
1 ↵ 1
(u? , v? ) =
,
.
1 ↵ 1 ↵
1 ↵
.
1 ↵
(5.4)
One could check the stability of this equilibrium by simply plugging the expressions from (5.4) into the community matrix (5.3) and hacking away, but this involves
needless struggle. In this case we know that
1
u?
↵v? = 0
so
5.4
1
2u?
↵v? =
u?
and similarly
1
v?
u? = 0
so
⇢(1
2v?
u? ) =
⇢v? .
Thus at the equilibrium given by (5.4) we have the simpler community matrix


1 2u? ↵v?
↵u?
u?
↵u?
A =
=
.
(5.5)
⇢ v?
⇢(1 2v?
u? )
⇢ v?
⇢v?
This form of the linearisation makes it much easier to work out the stability type
of the equilibrium. We have
Tr(A) =
u?
⇢v? < 0,
and
det(A) = ⇢u? v?
⇢↵ u? v? = ⇢u? v? (1
↵ )
whose sign is the same as that of (1 ↵ ). This means that stable coexistence
(a stable equilibrium with both u? and v? positive) is possible only when ↵ < 1,
which makes good ecological sense, as in this case the the e↵ect of each species on
the other is small or, equivalently, they compete only weakly. In the case of strong
competition—manifested here by ↵ > 1, or equivalently, (1 ↵ ) < 0—there is an
equilibrium where both populations are positive, but it’s a saddle and so is unstable.
5.2.2
Three parameter regimes
Our problem thus has three interesting parameter regimes: they’re illustrated in
Figure 5.1. Note that we can assume, without loss of generality, that ↵  as we
can always swap the roles of u and v to make this true.
(i) ↵ 
< 1. This is sometimes called the weak competition regime in that
the influence of the two species on each other is modest: each experiences the
impact of an individual of the other as something less than a whole (dimensionless) animal. In this case coexistence is possible (there is an equilibrium
with both u? and v? positive) and it is stable.
Further, if ↵,
⌧ 1 one can see that the equilibrium in Eqn. (5.4) has
u? ⇡ 1 ⇡ v ?
which means that the populations are essentially the same as they would be
if the other species were absent. This makes good ecological sense: very weak
competition has only a very small e↵ect.
(ii) ↵ < 1 < . In this case coexistence is not possible: the equilibrium given by
the formula in Eqn. (5.4) involves at least one negative population. Further,
the equilibrium (1, 0) is a stable node, while the equilibrium at (0, 1) is a
saddle. As we will see below, this implies that almost every initial condition
leads to the extinction of species two. This makes ecological sense in that
species two has only a modest detrimental e↵ect (↵ < 1) on species one, while
species one has a large detrimental e↵ect ( > 1) on species two.
5.5
(iii) 1 < ↵  , the strong competition regime. Coexistence is once again possible, but as (1 ↵ ) < 0 the analysis in the previous section shows that the
corresponding equilibrium is unstable, while those at (0, 1) and (1, 0) are both
stable. In this case one species will always drive the other to extinction, but
the identity of the winner depends in detail on the initial data.
Finally, careful readers will have noticed that that I have skipped all the parameter
regimes where one, or both, of ↵ and equal one. That’s because these are rather
special cases:
• If exactly one of ↵ and is one, the equilibrium (5.4) coincides with either
the one at (0, 1) or the one at (1, 0). Linear stability analysis then is then
uninformative and one needs more elaborate approaches.
• If ↵ = = 1 then the two species are indistinguishable as far as the model
is concerned: animals feel the impact of members of the opposite species in
exactly the same way as they feel that from members of their own. In this
case the whole line segment connecting (0, 1) and (1, 0) consists of equilibria
and linear stability analysis is again uninformative.
5.2.3
A brief mathematical interlude
We’ll conclude our study of the Principle of Competitive Exclusion by working
through a rigorous argument that establishes the claims made about the regime
↵ < 1 < , but before we can do this we need two lemmas from Braun. They
sometimes enable one to make strong statements about the long-time behaviour of
the solutions of ODEs, even when one cannot solve the equations explicitly. As in
Section 4.2.2, these results apply to systems of ODEs of the form ẋ = f (x), where
x 2 Rn and f : Rn ! Rn , where
2
3
f1 (x1 , . . . , xn )
6 f2 (x1 , . . . , xn ) 7
6
7
f (x) = 6
(5.6)
7
..
4
5
.
fn (x1 , . . . , xn )
with fj : Rn ! R.
Lemma 5.1 (Bounded functions and limits). Let g(t) be a monotonic increasing
(decreasing) function of time g(t) : [t0 , 1) ! R with g(t)  c (respectively, g(t) c)
for some constant c 2 R. Then the limit
g? = lim g(t)
t!1
exists and satisfies g?  c (respectively, g?
c).
Lemma 5.2 (Asymptotic limits are equilibria). If x(t) is a solution to a system of
ODEs such as that in Eqn. (5.6) and
lim x(t) = x?
t!1
then x? must be an equilibrium. That is, f (x? ) = 0 which is equivalent to fj (x? ) = 0
for all 1  j  n.
5.6
v
(0,1/α )
(0,1)
C
B
A
(0,0)
(1/β,0)
(1,0)
u
Figure 5.2: The phase portrait for the case ↵ < 1 < : the positive quadrant (where
all ecologically-sensible solutions lie) is divided up into three regions, as shown. The
arrows on the border of region B are meant to indicate the direction that trajectories
starting on the border will flow, but don’t indicate the speed of the flow.
5.2.4
Rigorous treatment of the case where ↵ < 1 <
Our aim here will be to establish
Proposition 5.3. Suppose ↵ < 1 < in Eqn. (5.2). Then if u(0) > 0 and v(0)
we can conclude
lim (u(⌧ ), v(⌧ )) = (1, 0).
0
⌧ !1
That is, the second species is driven to extinction whenever some animals of
the first species are present. Our proof will involve the three regions A, B and C
illustrated in Figure 5.2 and we will establish each of the following:
(a) Any solution (u(⌧ ), v(⌧ )) whose initial condition (u(0), v(0)) lies in region A
must eventually leave the region.
(b) If a solution (u(⌧ ), v(⌧ )) lies within region B at some time ⌧ = ⌧0 , then
(u(⌧ ), v(⌧ )) will remain in the B for all ⌧
⌧0 (B is a so-called trapping
region). Further, any solution that remains within B for all times ⌧ ⌧0 must
have lim⌧ !1 (u(⌧ ), v(⌧ )) = (1, 0).
(c) Finally, any solution (u(⌧ ), v(⌧ )) whose initial condition lies in region C must
eventually either
• leave C and enter B or
• remain in C for all time and have lim⌧ !1 (u(⌧ ), v(⌧ )) = (1, 0).
The regions A, B, and C are bounded by segments of null clines, but don’t have
any null clines in their interiors. This means that the functions f (u, v) = du/d⌧ and
g(u, v) = dv/d⌧ are non-zero inside the regions and so a simple argument based on
the Intermediate Value Theorem says that these functions can’t change sign inside
5.7
Region
du
d⌧
dv
d⌧
A
B
C
+
+
+
Table 5.2: The signs of the derivatives in the regions illustrated in Figure 5.2
A, B or C and so the derivatives have a definite sign (either positive or negative)
everywhere in their interiors. This observation is the cornerstone of our proof and
so I’ve tabulated the signs of the derivatives in Table 5.2.
In the arguments below I’ll make frequent use of the term trajectory: it means
a curve traced out by a solution (u(⌧ ), v(⌧ )) and a key tool in phase plane analysis
is the observation that trajectories can’t intersect: this follows from the uniqueness
of solutions to ODEs.
Part (a): trajectories leave A
We’ll prove by contradiction that solutions starting in A must leave the region.
Assume there’s some initial condition (u0 , v0 ) 2 A whose corresponding solution
(u(⌧ ), v(⌧ )) remains in A forever. Since we know that du/d⌧ > 0 throughout A
we can conclude that u(⌧ ) is a monotone increasing function and, as the region
is bounded (if (u, v) 2 A then 0 < u < 1/ and 0 < v < 1) we have further
that u(⌧ ) is a bounded monotone function. Lemma 5.1 thus applies and says that
lim⌧ !1 u(⌧ ) exists and lies within the limits imposed by the boundaries of A. A
very similar argument then tells us that lim⌧ !1 v(⌧ ) exists too. Further, we know
lim⌧ !1 (u(⌧ ), v(⌧ )) lies within A or on its boundary.
But then, as lim⌧ !1 (u(⌧ ), v(⌧ )) exists, we know from Lemma 5.2 that the limit
must be an equilibrium of (5.2). There is only one equilibrium point lying in A or
on its boundary—the one at (1, 0)—and that’s a saddle. The Hartman-Grobman
Theorem thus assures us that almost all nearby trajectories flow away from the
saddle. The exceptions are the trajectories that flow along the v-axis, but our
assumption that u(0) > 0 rules these possibilities out. Thus we have reached a
contradiction: a trajectory that stays in A forever must converge to an equilibrium
point, but the only one accessible from A is unstable and the solutions that interest
us can’t converge to it.
The only way out of this impasse is to conclude that solutions don’t stay in A
forever, but instead leave in finite time. This means that they must cross one of
A’s boundaries. They can’t go through the ones with u = 0 or v = 0 because those
two are segments of trajectories and we know that trajectories can’t intersect. Thus
solutions that begin in A must exit by hitting the boundary between A and B. And
once they hit this boundary, they cross into B as we know that du/d⌧ > 0 all along
it.
5.8
Part (b): trajectories get trapped in B and converge to (1, 0)
Begin by imagining that at some time ⌧0 0 we know that (u(⌧0 ), v(⌧0 )) is in the
interior of region B. Now suppose, for contradiction, that at some later time T > ⌧0
the solution (u(T ), v(T )) is outside B. Then there would need to have been an earlier
moment—call it ⌧1 , with ⌧0 < ⌧1 < T —when the trajectory (u(⌧ ), v(⌧ )) crosses the
boundary2 of B. But this is impossible.
As we saw in the previous part of the argument: no trajectory can leave B
through those parts of its boundary that coincide with the u and v axes, as these
are segments of other trajectories. Thus a trajectory exiting B would need to flow
out through either the boundary between regions A and B or the boundary between
B and C. But in both cases, the ODEs are such that trajectories that start on
B’s boundary flow inwards, as is illustrated in Figure 5.2. The boundary between
regions A and B is a piece of the dv/d⌧ = 0 null cline and so trajectories starting
there have, initially, du/d⌧ > 0 and dv/d⌧ = 0 and flow horizontally to the right.
Similarly, the boundary between regions B and C is part of the du/d⌧ = 0 null cline
and so trajectories starting there have, initially, du/d⌧ = 0 and dv/d⌧ < 0, so flow
vertically downward. These observations provide the contradiction we sought: we
assumed that a trajectory that began in B could eventually escape, but then we
established that it is impossible for any trajectory to cross B’s boundary from the
inside to the outside.
A trajectory that starts or arrives in B is thus trapped there forever after. Table 5.2 tells us that such a trapped trajectory has du/d⌧ > 0 and dv/d⌧ < 0, so
u(⌧ ) is a bounded, monotone increasing function while v(⌧ ) is bounded and monotone decreasing. This means that both have limits and, as in the previous section,
lim⌧ !1 (u(⌧ ), v(⌧ )) must be an equilibrium. As the Hartman-Grobman Theorem
rules out the one at (0, 1), the only remaining possibility is (u? , v? ) = (1, 0).
Part (c): trajectories that start in C converge to (1, 0)
In region C we know both du/d⌧ < 0 and dv/d⌧ < 0 and, as u(⌧ ) 0 and v(⌧ ) 0,
both are bounded monotone decreasing functions. If a trajectory remains in C
forever (and some certainly do) then now-familiar arguments imply that
lim (u(⌧ ), v(⌧ )) = (1, 0).
⌧ !1
If, on the other hand, the trajectory eventually leaves C then it must pass through
the boundary with region B and become trapped and then, as we proved in Part (b),
converge to (1, 0).
5.3
Afterword
It’s possible to make too much of the Principle of Competitive Exclusion. The
arguments we’ve just made depend very strongly on properties of curves in the plane
2
The sophisticated reader will recognize that we are, implicitly, invoking the Jordan Curve
Theorem here.
5.9
and won’t apply when three or more competing species are present, though similar
results are available. Also, experimental evidence suggests that while Eqn. (5.1)
provides a good model for some forms of interspecific conflict, it doesn’t work for all,
not even when the two species are very similar: Francisco Ayala and his collaborators
report3 on extensive experiments involving pairs of species of Drosophila (fruit flies)
and find that in most cases alternative models involving equations of the form
✓
◆
(Nj + ↵jk Nk + j Nj2 )
dNj
= r j Nj 1
dt
Kj
or
"
dNj
= r j Nj 1
dt
✓
Nj
Kj
◆✓ j
↵jk
✓
Nk
Kj
◆#
with ✓j > 0 provided better fits to the data. The use of Eqn. (5.1) to model economic
or ideological competition requires a very strenuous leap of the imagination.
3
Francisco J. Ayala, Michael E. Gilpin and Joan G. Ehrenfeld (1973), Competition between
species: Theoretical models and experimental tests, Theoretical Population Biology, 4(3):331–356.
DOI 10.1016/0040-5809(73)90014-2
5.10