### Examensarbete

### Stable Coexistence of Three Species in Competition

### Linn´

### ea Carlsson

LiTH - MAT - EX - - 2009 / 02 - - SE

### Stable Coexistence of Three Species in Competition

Applied Mathematics, Link¨opings Universitet Linn´ea Carlsson

LiTH - MAT - EX - - 2009 / 02 - - SE

Examensarbete: 15 hp Level: C

Supervisor: Dr. Kurt Hansson,

Applied Mathematics, Link¨opings Universitet Examiner: Dr. Kurt Hansson,

Applied Mathematics, Link¨opings Universitet Link¨oping: june 2009

Matematiska Institutionen 581 83 LINK ¨OPING SWEDEN june 2009 x x http://urn.kb.se/resolve?urn=urn:nbn:se:liu:diva-18807 LiTH - MAT - EX - - 2009 / 02 - - SE

Stable Coexistence of Three Species in Competition

Linn´ea Carlsson

This report consider a system describing three competing species with pop-ulations x, y and z. Sufficient conditions for every positive equilibrium to be asymptotically stable have been found. First it is shown that conditions on the pairwise competitive interaction between the populations are needed. Actually, these conditions are equivalent to asymptotic stability for any two-dimensional competing system of the three species. It is also shown that these alone are not enough, and that a condition on the competitive interaction between all three populations is also needed. If all conditions are fulfilled, each population will survive on a long-term basis and there will be a stable coexistence.

ordinary differential equations, competing species, coexistence, asymptotic stability, Routh’s criterion

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ISSN 0348-2960 ISRN ISBN Spr˚ak Language Svenska/Swedish Engelska/English Rapporttyp Report category Licentiatavhandling Examensarbete C-uppsats D-uppsats ¨ Ovrig rapport Avdelning, Institution Division, Department Datum Date

Abstract

This report consider a system describing three competing species with populations x, y and z. Sufficient conditions for every positive equilibrium to be asymptotically stable have been found. First it is shown that conditions on the pairwise competitive interaction between the populations are needed. Actually, these conditions are equivalent to asymptotic stability for any two-dimensional competing system of the three species. It is also shown that these alone are not enough, and that a condition on the competitive interaction between all three populations is also needed. If all conditions are fulfilled, each population will survive on a long-term basis and there will be a stable coexistence.

Keywords: ordinary differential equations, competing species, coexistence, asymptotic stability, Routh’s criterion

Acknowledgements

I would like to thank my supervisor during this work, Dr. Kurt Hansson. I am very thankful for his support, guidance and help during the entire work, and for giving me the opportunity to do this in the first place. I also would like to thank my parents for their encouragement. Contents Abstract 6 Acknowledgements 6 1. Introduction 7 1.1. Background 7 2. Competing species 9 2.1. The model 9

2.2. Two competing species 10

3. Stable coexistence of three species 11

3.1. Positive solutions 11 3.2. Asymptotic stability 12 3.3. Routh’s criterion 12 3.4. Conditions 13 4. Conclusions 19 References 19

7 1. Introduction

This report consider three competing species of animals, plants, or bacteria, for in-stance, that live in the same environment. We will determine when all three populations are able to survive and live together, to coexist on a long-term basis. When is that possible? What are the conditions for stable coexistence?

1.1. Background. Describing ecological systems in nature is not simple. Often they involve several species and their interactions among each other and their growth rates are often complicated. To create an approximation to an actual population, we need mathematical models that are as accurate and realistic as possible, but still manageable from a mathematical point of view.

First, let x(t) be the number of individuals in a population at time t, t ≥ 0, and suppose that the population only changes by the occurrence of births and deaths. In other words, there is no immigration or emigration from outside the environment where the popula-tion lives. Then let β(t), the birth rate funcpopula-tion, and δ(t), the death rate funcpopula-tion, be defined as follows

• β(t) is the number of births per unit of population per unit of time at time t. • δ(t) is the number of deaths per unit of population per unit of time at time t.

During a time interval, [t, t + ∆t], the numbers of births and deaths is approximately given by β(t)x(t)∆t respectively δ(t)x(t)∆t. Hence, the change ∆x in the population during this time interval is

∆x ≈ β(t)x(t)∆t − δ(t)x(t)∆t =⇒ ∆x_{∆t} ≈ (β(t) − δ(t)) x(t)
Now let ∆t → 0. Then we get the following differential equation

(1) dx

dt = (β − δ)x

which is the general population equation and in which we write β = β(t), δ = δ(t) and x = x(t).

In many situations it is observed that the birth rate decreases when the population itself increases. Depending on what kind of population is considered, there are different reasons. Limited food supply or overcrowding are two of them. Now suppose that the birth rate β is a linear function of the population size x, β = β0− β1x, where β0 and

β1 are positive constants. If the death rate δ = δ0 remains constant, then equation (1)

becomes

dx

dt = (β0− β1x − δ)x Let r = β0− δ0 and a = β1. Then we get

(2) dx

dt = rx − ax

2

which is called the logistic equation if both r and a are positive.

Now rewrite the logistic equation as in the following logistic inital value problem

(3) dx

where k = a and M = r/a. Since r, a > 0, also k, M > 0. The solution to (3) is

(4) x(t) = M x0

x0+ (M − x0) e−kMt

Since we are considering populations of species, assume that x0 is positive. We see that

if x0 = M , then we have x(t) ≡ M, a constant-valued population. If 0 < x0 < M then

we see from (3) that x′_{(t) > 0 and}

x(t) = M x0

x0+ (M − x0) e−kMt

< M x0 x0

= M If x0 > M , however, then we see from (3) that x′(t) < 0 and

x(t) = M x0
x0+ (M − x0) e−kMt
> M x0
x0
= M
We also have
lim
t_{→+∞}x(t) =
M x0
x0+ 0
= M

Thus, a logistic population does not increase or decrease without boundaries. Instead, it approaches the limiting population M as t → +∞. M is also called the carrying capacity of the environment, because it is the maximum population that can be supported by the environment on a long-term basis.

If we now consider two species with logistic populations, x(t) respectively y(t), that live in the same environment, we get the following equations

dx/dt = r1x − a1x2− b1xy

dy/dt = r2y − a2y2− b2xy

where the coefficients r1, r2, a1and a2are positive. The xy-terms describe the interaction

between the two populations. If b1 = b2 = 0, then the equations describe two separate

logistic populations with no interaction between each other.

If both b1 and b2 are negative, then the rates of growth of both populations increase

as the effect of their interaction. This interaction is described as cooperation, both populations are gained by the interaction.

If the interaction coefficients, however, have different signs, then one population is hurt and the other population is gained by their interaction. This interaction is described as predation. If, for example, b1 > 0 and b2< 0, then x(t) is a prey population and y(t) is

a predator population.

Finally, if both b1 and b2 are positive, then the rates of growth of both populations

decrease as the effect of their interaction. Hence, both populations are hurt. This interaction is described as competition and that is the interaction considered in this report.

9 2. Competing species

2.1. The model. From section 1.1 we have a model for two competing species. Now we need a mathematical model for three competing species that is as realistic as possible. First let the three species have populations x(t), y(t) and z(t) at time t, t ≥ 0, and assume that in absence of other species, each population will be logistic and bounded,

dx/dt = r1x − a1x2

dy/dt = r2y − a2y2

dz/dt = r3z − a3z2

Then assume that the competition between the species has rates of decline that are proportional to their products with the other two species, and we get the following competition system, dx/dt = r1x − a1x2− b1xy − c1xz = x(r1− a1x − b1y − c1z) = f1(x, y, z) dy/dt = r2y − b2xy − a2y2− d2yz = y(r2− b2x − a2y − d2z) = f2(x, y, z) dz/dt = r3z − c3xz − d3yz − a3z2 = z(r3− c3x − d3y − a3z) = f3(x, y, z) (5)

where the coefficients r1, r2, r3, a1, a2, a3, b1, b2, c1, c3, d2 and d3 are all positive and the

functions f1(x, y, z), f2(x, y, z) and f3(x, y, z) are in our case given by the expressions

above.

2.1.1. Equilibrium and stability.

Definition 1. Consider a differential equation

(6) dx/dt = f (x); x = (x1, . . . , xn); f : W → Rn; W ⊂ Rn is open

Suppose f is C1. Then a point xe∈ W is called an equilibrium point of (6) if f (xe) = 0.

xe is also called critical point or stationary point.

Definition 2. An equilibrium point xe is called stable if there for every > 0 exists a

δ > 0 such that if kx(0) − xek < δ, then kx(t) − xek < for all t ≥ 0

Definition 3. An equilibrium point xe is called asymptotically stable if there exists a

δ > 0 such that if kx(0) − xek < δ, then limt→+∞kx(t) − xek = 0

Definition 4. A critical point is called unstable if it is not stable or asymptotically stable.

Definition 5. An equilibrium xe to equation (6) is called a sink if all eigenvalues of the

derivative Df (xe) have negative real parts.

Theorem 1. If xe is a sink and an equilibrium point to equation (6), then xe is

asymp-totically stable. [3, p. 181]

Theorem 2. If the real part of any of the eigenvalues of derivative Df (xe) is positive,

2.2. Two competing species. For two species the competing system will be (7)

dx/dt = r1x − a1x2− b1xy = x(r1− a1x − b1y) = g1(x, y)

dy/dt = r2y − b2xy − a2y2= y(r2− b2x − a2y) = g2(x, y)

This system has four critical points, (0, 0), (0, r2/a2), (r1/a1, 0) and (xe, ye). The fourth

point represent the possibility of coexistence and it is the solution to the following linear system

a1x + b1y = r1

b2x + a2y = r2

Theorem 3. A positive equilibrium (xe, ye) to (7) is asymptotically stable if a1a2> b1b2

and unstable if a1a2 < b1b2.

Proof. Linearization of the system (7) gives Dg = J =

r1− 2a1x − b1y −b1x

−b2y r2− b2x − 2a2y

and for the equilibrium point (xe, ye) we get

Dg(xe, ye) = Je=

−a1xe −b1xe

−b2ye −a2ye

According to Theorem 1 the critical point (xe, ye) is asymptotically stable if the real

parts of all eigenvalues of Je are negative. The eigenvalues of Je are given by the roots

of the characteristic equation, p(λ) = 0, where p(λ) = |Je− λI| = −a1xe− λ −b1xe −b2ye −a2ye− λ = λ 2 − trace(Je)λ + |Je|

The roots are λ1 = − (a1xe+ a2ye) 2 + s (a1xe+ a2ye)2 4 − (a1a2− b1b2)xeye λ2= − (a1xe+ a2ye) 2 − s (a1xe+ a2ye)2 4 − (a1a2− b1b2)xeye The real part of λ2 is always negative and the real part of λ1 is negative if

(a1xe+ a2ye) 2 > s (a1xe+ a2ye)2 4 − (a1a2− b1b2)xeye Hence, we have (a1xe+ a2ye)2 4 > (a1xe+ a2ye)2 4 − (a1a2− b1b2)xeye which is equivalent to 0 > −(a1a2− b1b2)xeye

and since xe, ye> 0 we have

11 We conclude that (xe, ye) is asymptotically stable if a1a2> b1b2. We also conclude that

if a1a2< b1b2, the real part of λ1 is positive, and then according to Theorem 2, (xe, ye)

is unstable.

The inequalities in Theorem 3 has a natural interpretation. The coefficients a1 and

a2 represent the inhibiting effect on the natural growth of the populations, and the

coefficients b1 and b2 represent the effect of competition between the populations.

If the measure of competition, b1b2, is less than the measure of inhibition, a1a2, then

there exist an asymptotically stable coexistence that is approached by the solution as t → ∞. In this case both species survive. If the measure of competition is greater than the measure of inhibition, either x(t) or y(t) approaches zero as t → ∞. In this case the two species cannot coexist, one survives and the other becomes exstinct.

3. Stable coexistence of three species

In this section we determine the neccessary and sufficient conditions for asymptotic stability at all positive critical points for a system describing three competing species. 3.1. Positive solutions. To the competition system (5) there are eight critical points, but only one of these points can be a stable coexistence of three competing species. That is the critical point where all the populations x, y and z are non-zero. Let this critical point be (xe, ye, ze). It is the solution to the following linear system

a1x + b1y + c1z = r1 b2x + a2y + d2z = r2 c3x + d3y + a3z = r3

Theorem 4. If the initial values x(0) = x0, y(0) = y0, z(0) = z0 of the competition

system (5) are positive, then a solution (x(t), y(t), z(t)) will remain positive for all t ≥ 0. Proof. Rewrite the competition system (5) as follows and conclude that sincef1(x,y,z)

x ,

f2(x,y,z)

y

and f3(x,y,z)

z are continuous functions they will attain a maximum and minimum on every

compact set K ⊆ R3 _{according to the Extreme Value Theorem [2, p. 41, Theorem 4].}

Hence, if |x| , |y| , |z| ≤ N,

x′/x = r1− a1x − b1y − c1z ≥ −k

y′/y = r2− b2x − a2y − d2z ≥ −m

z′/z = r3− c3x − d3y − a3z ≥ −n

where k, m and n are positive and real, and depend on N . Integration of both sides gives

Rt
0x′/x dt = ln x(t) − ln x(0) ≥
Rt
0 −kdt = −kt
Rt
0y′/y dt = ln y(t) − ln y(0) ≥
Rt
0−mdt = −mt
Rt
0z′/z dt = ln z(t) − ln z(0) ≥
Rt
0 −ndt = −nt
thus,
eln x(t)−ln x0 _{= x(t)/x}
0≥ e−kt
eln y(t)−ln y0 _{= y(t)/y}
0≥ e−mt
eln z(t)−ln z0 _{= z(t)/z}
0 ≥ e−nt

and since the initial values x0, y0 and z0 are positive, we have as long as

|x(t)| , |y(t)| , |z(t)| ≤ N that

x(t) ≥ x0e−kt> 0

y(t) ≥ y0e−mt> 0

z(t) ≥ z0e−nt > 0

Since N > 0 is arbitrary, we conclude that the solutions x(t), y(t) and z(t) will remain

positive for t ≥ 0.

3.2. Asymptotic stability. At first we examine what conditions are neccessary for (xe, ye, ze) to be an asymptotically stable sink. Linearization of the competition

sys-tem (5) gives Df = A = r1− 2a1x − b1y − c1z −b1x −c1x −b2y r2− b2x − 2a2y − d2z −d2y −c3z −d3z r3− c3x − d3y − 2a3z

and for the equilibrium point (xe, ye, ze) we get

Df (xe, ye, ze) = Ae= −a1xe −b1xe −c1xe −b2ye −a2ye −d2ye −c3ze −d3ze −a3ze

According to Theorem 1 the critical point (xe, ye, ze) is asymptotically stable if the real

parts of all eigenvalues of Ae are negative. The eigenvalues of Ae are given by the roots

of the characteristic equation, p(λ) = 0, where (8) p(λ) = |Ae− λI| = −a1xe− λ −b1xe −c1xe −b2ye −a2ye− λ −d2ye −c3ze −d3ze −a3ze− λ = −λ3+trace(Ae)λ2−Mλ+|Ae| where M = M11+ M22+ M33= a2yea3ze− d2yed3ze+ a1xea3ze− c1xec3ze+ a1xea2ye−

b1xeb2ye, is the sum of the principal minors of Ae, and trace(Ae) = −a1xe− a2ye− a3ze.

3.3. Routh’s criterion.

Theorem 5. All the roots of the real polynomial f(z) have negative real parts if and only if in the carrying out of Routh’s algorithm all the elements of the first column of Routh’s scheme are different from zero and of like sign. [4, p. 180]

To carry out of Routh’s algorithm, first introduce the following notation for a real polynomial f (z)

f (z) = a0zn+ b0zn−1+ a1zn−2+ b1zn−3+ . . . , (a0 6= 0)

and then let f1(z) = a0zn− a1zn−2+ . . . and f2(z) = b0zn−1− b1zn−3+ . . .

Use Euclidean algorithm to construct a generalized Sturm chain f1(z), f2(z), f3(z), . . . , fm(z) :

f3(z) = f2(z)q1(z) − f1(z) = c0zn−2− c1zn−4+ c2zn−6− . . .

f4(z) = f3(z)q2(z) − f2(z) = d0zn−3− d1zn−5+ d2zn−7− . . .

.. .

13 Then form the Routh scheme:

a0 a1 a2 . . .

b0 b1 b2 . . .

c0 c1 c2 . . .

d0 d1 d2 . . .

. . . .

To determine when all the roots of the characteristic polynomial p(λ) from (8), i.e. the eigenvalues of Ae, have negative real parts we apply Routh’s algorithm to

p(λ) = −λ3+ trace(Ae)λ2− Mλ + |Ae|

from which we construct

p1(λ) = −λ3+ M λ, p2(λ) = trace(Ae)λ2− |Ae|

Then by applying Euclidean algorithm to p1(λ) and p2(λ) we get

p3(λ) = (|Ae| − M trace(Ae

))λ trace(Ae)

, p4(λ) = |Ae|

The generalized Sturm chain p1(λ), p2(λ), p3(λ), p4(λ) forms the following Routh scheme

−1 M 0

trace(Ae) |Ae| 0 |Ae|−M trace(Ae)

trace(Ae) 0 0

|Ae| 0 0

From Routh´s criterion, Theorem 5, we know that all eigenvalues of Ae have negative

real parts if and only if the elements in the first column is different from zero and of like sign. Since trace(Ae) < 0, it follows that (xe, ye, ze) is an asymptotically stable

equilibrium if

(9) _{|A}e| < 0, M − |Ae|

trace(Ae)

> 0 Since |Ae|

trace(Ae) > 0, it follows from the second inequality in (9) that M = M11+M22+M33

has to be positive.

3.4. Conditions. What conditions on the coefficients a1, a2, a3, b1, b2, c1, c3, d2 and d3

are neccessary and sufficient for the inequalities in (9) to be fulfilled for all positive critical points (xe, ye, ze)? Do they have any natural interpretation similiar to the

two-dimensional case in section 2.2? From the inequalities (9) we have

(10) _{|A}e| = (a1d2d3+ c1a2c3+ b1b2a3− a1a2a3− c1b2d3− b1d2c3)xeyeze< 0

and since trace(Ae) = −a1xe− a2ye− a3ze< 0, the second inequaility (9) can be written

as M trace(Ae) − |Ae| < 0, which is equivalent to

(11)

((b1b2− a1a2)ye+ (c1c3− a1a3)ze)a1x2e+

((b1b2− a1a2)xe+ (d2d3− a2a3)ze)a2y2e+

((c1c3− a1a3)xe+ (d2d3− a2a3)ye)a3z2e+

Let xe, ye and ze, one at a time, → +∞. Then for the leading terms to be negative we

must have

(b1b2− a1a2)ye+ (c1c3− a1a3)ze < 0

(b1b2− a1a2)xe+ (d2d3− a2a3)ze< 0

(c1c3− a1a3)xe+ (d2d3− a2a3)ye< 0

Now let xe, ye and ze → +∞ respectively and it follows that

b1b2− a1a2 < 0

c1c3− a1a3 < 0

d2d3− a2a3 < 0

Let

m11= a2a3− d2d3, m22= a1a3− c1c3, m33= a1a2− b1b2

To be sure that M trace(Ae) − |Ae| < 0 for every positive (xe, ye, ze) we see that m11,

m22 and m33 have to be positive. Hence, a neccessary condition for all positive critical

points to be asymptotically stable is that

(12) m11> 0

(13) m22> 0

(14) m33> 0

As we saw in section 2.2, these are the conditions for stable coexistence for competition systems of two of the populations at a time; of population y and z in (12), of population x and z in (13) and of population x and y in (14).

As we also saw in section 2.2, each of these inequalities has a natural interpretation. In inequality (12) the coefficients a2 and a3 represent the inhibiting effect on the natural

growth of population y and population z, and the coefficients d2 and d3 represent the

effect of competition between population y and population z.

If the measure of competition, d2d3, is less than the measure of inhibition, a2a3, then

there exist an asymptotically stable coexistence. The other two inequalities have the same natural interpretation, but with population x and z in (13) respectively population x and y in (14).

Now let

m0 = c1b2d3+ b1d2c3− 2a1a2a3

then, from (11) we have

m0xeyeze < m33a1x2eye+ m22a1x2eze+ m33a2xey2e+ m11a2ye2ze+ m22a3xeze2+ m11a3yeze2

Since xe, ye, ze> 0, the inequality can be divided on both sides by xeyeze

m0< m33a1 xe ze + m22a1 xe ye + +m33a2 ye ze + m11a2 ye xe + m22a3 ze ye + m11a3 ze xe

Let n1 = xe/ye, n2 = ye/ze, n3 = ze/xe. Notice that n1n2n3 = 1 and that since

xe, ye, ze > 0, also n1, n2, n3 > 0. Hence, m0 < m22a1n1+ m11a2 n1 | {z } g1(n1) + m33a2n2+ m22a3 n2 | {z } g2(n2) + m11a3n3+ m33a1 n3 | {z } g3(n3) = g(n1, n2, n3)

15 By minimizing each of the functions g1(n1), g2(n2) and g3(n3), the sum g(n1, n2, n3) is

also minimized. We have
m22a1n1+ m11a2
n1
| {z }
g1(n1)
−2√m11m22a1a2 =√m22a1n1−r m11a2
n1
2
≥ 0
where√m22a1n1−
q_{m}
11a2
n1
2
= 0 ⇐⇒√m22a1n1 =
q_{m}
11a2
n1 ⇐⇒ n1 =
q_{m}
11a2
m22a1
Hence, g1
q_{m}
11a2
m22a1

= 2√m11m22a1a2 is the minimum of g1(n1). In a similar way it is

shown that g2
q_{m}
22a3
m33a2
= 2√m22m33a2a3 and g3
q_{m}
33a1
m11a3
= 2√m11m33a1a3 are the

minima of g2(n2) respectively g3(n3). Since

q
m11a2
m22a1
q
m22a3
m33a2
q
m33a1
m11a3 = 1 the constraint
n1n2n3 = 1 is fulfilled. Hence,
g1(n1) = m22a1n1+
m11a2
n1 ≥ 2
√
m11m22a1a2
g2(n2) = m33a2n2+
m22a3
n2 ≥ 2
√_{m}
22m33a2a3
g3(n3) = m11a3n3+
m33a1
n3 ≥ 2
√_{m}
11m33a1a3
Now let
A1 = m11a1, A2= m22a2, A3 = m33a3

then it follows that

m0 < 2 p A1A2+ p A2A3+ p A1A3 We have (n1, n2, n3) = q m11a2 m22a1, q m22a3 m33a2, q m33a1 m11a3

, which correspond to the critical point (xe, ye, ze) = r m11 a1 ,r m22 a2 ,r m33 a3 From (11) we have M trace(Ae)−|Ae| = −(m33ye+m22ze)a1x2e−(m33xe+m11ze)a2ye2−(m22xe+m11ye)a3ze2+m0xeyeze

from which it follows that

|Ae| = M trace(Ae)+(m33ye+m22ze)a1x2e+(m33xe+m11ze)a2ye2+(m22xe+m11ye)a3z2e−m0xeyeze

From (9) we have |Ae| < 0. Hence,

M trace(Ae)+(m33ye+m22ze)a1x2e+(m33xe+m11ze)a2y2e+(m22xe+m11ye)a3ze2< m0xeyeze

which is equivalent to

(m11yeze+ m22xeze+ m33xeye)(−a1xe− a2ye− a3ze) + (m33ye+ m22ze)a1x2e+

(m33xe+ m11ze)a2y2e+ (m22xe+ m11ye)a3z2e < m0xeyeze

and from which it follows that

Finally, divide both sides of the inequality by xeyeze and it follows that

−A1− A2− A3 < m0

The reasoning above proves the following theorem.

Theorem 6. Every positive equilibrium (xe, ye, ze) to a competing system (5) is

asymp-totically stable if
(15) m11> 0, m22> 0, m33> 0
and
(16) _{−A}1− A2− A3 < m0 < 2(
p
A1A2+
p
A2A3+
p
A1A3)

Since −A1− A2− A3 < m0 in (16) is equivalent to |Ae| < 0, we know that if

−A1 − A2 − A3 = m0, then also |Ae| = 0 and Ae = Df (xe, y,ze) is not invertible.

If Df (xe, y,ze) is not invertible, we can not use Theorem 1. Instead, notice that |Ae|

is the constant term in the characteristic polynomial p(λ) in (8). From Intermediate Value Theorem [1, p. 479, Theorem 1] we know that if |Ae| > 0, which is equivalent to

−A1− A2− A3 > m0, there is at least one root of p(λ), i.e. eigenvalue, with a positive

real part. Then, from Theorem 2, it follows that there exist unstable critical points if −A1− A2− A3 > m0.

Now consider m0 < 2 √A1A2+√A2A3+√A1A3

in (16) and the matrices Ae= −a1xe −b1xe −c1xe −b2ye −a2ye −d2ye −c3ze −d3ze −a3ze , C = a1 b1 c1 b2 a2 d2 c3 d3 a3

where C is a matrix with the coefficients in Ae. Notice that if we let

C = a1 1/S S S a2 1/S 1/S S a3

the conditions in (15) are fulfilled for all S > 0 if a1, a2, a3> 1. We also have

m0 = S3+1/S3−2a1a2a3 and 2 √A1A2+√A2A3+√A1A3= 2 p (a2a3− 1)(a1a3− 1)a1a2+ 2p(a1a2− 1)(a1a3− 1)a2a3+ 2 p (a1a2− 1)(a2a3− 1)a1a3.

It follows that m0→ +∞ if S → +∞. Hence, the conditions in (15) are not enough for

the inequality in (16) to be fulfilled. The following example show that if (15) is fulfilled but m0 > 2 √A1A2+√A2A3+√A1A3 there exist unstable critical points.

Example 1. Let C = 2 1 S 1 2 1 1/S 1 2 We have m11= 2 · 2 − 1 · 1 = 3 > 0 m22= 2 · 2 − S · 1/S = 3 > 0 m33= 2 · 2 − 1 · 1 = 3 > 0

17
and
m0 = S · 1 · 1 + 1 · 1 ·_{S}1 − 2 · 2 · 2 · 2 = S +_{S}1 − 16
−A1− A2− A3 = −3 · 2 − 3 · 2 − 3 · 2 = −18
2 √A1A2+√A2A3+√A1A3= 2 √6 · 6 +√6 · 6 +√6 · 6= 36
We get
(xe, ye, ze) =
r
3
2,
r
3
2,
r
3
2
!
If m0 = 2 √A1A2+√A2A3+√A1A3 ⇐⇒ S +_{S}1 − 16 = 36 ⇐⇒ S = 26 ± 15√3, we
get
Ae=
−√6 _{−}√_{2}6 _{−}(26±15
√
3)√6
2
−√26 −
√
6 _{−}√_{2}6
−_{2}_{(}_{26±15}√6√
3) −
√
6
2 −
√
6

with the following eigenvalues for both S = 26 + 15√_{3 and S = 26 − 15}√3
λ1 = −3

√

6, λ2,3 = ±i

3√6 2

One eigenvalue has negative real part and the other two have real part zero. This indicates that S = 26 − 15√3 and S = 26 + 15√3 are the values where transitions between stable and unstable equilibrium occurs. This situation is confirmed by the following plots.

If 0 < S < 26 − 15√3, then m0 > 2 √A1A2+√A2A3+√A1A3. If 26 − 15√3 <

S < 26 + 15√3 then m0 < 2 √A1A2+√A2A3+√A1A3 and the condition in (16) is

fulfilled, and if S > 26 + 15√3, then m0 > 2 √A1A2+√A2A3+√A1A3. Figure 1 is

a plot over the complex eigenvalues of Ae when S runs over the interval [26 − 15

√ 3, 26 + 15√3] with stepsize √3. Figure 2 is a plot over the complex eigenvalues of Ae when S

runs over the interval [26 + 15√3, 26 + 40√3] with stepsize √3. Figure 3 is a plot over the complex eigenvalues of Ae when S runs over the interval ]0, 26 − 15

√ 3]. x K10 K5 0 5 10 K3 K2 K1 1 2 3

x K10 K5 0 5 10 K4 K3 K2 K1 1 2 3 4

Figure 2. _{Eigenvalues of A}_{e} _{when S ≥ 26 + 15}√_{3}

x K10 K5 0 5 10 K4 K3 K2 K1 1 2 3 4

Figure 3. _{Eigenvalues of A}_{e} _{when 0 < S ≤ 26 − 15}√_{3}

From Figure 2, Figure 3 and Theorem 2 we conclude that there exist unstable critical points if m0 > 2 √A1A2+√A2A3+√A1A3

.

19 4. Conclusions

For a competing system (5) describing three species with populations x, y and z, suf-ficient conditions for all positive critical points to be asymptotically stable have been found, (15) and (16). The inequalities in (15) are equivalent to the conditions for as-ymptotic stability for any two-dimensional competing system of the populations. We have seen that the conditions in (15) alone are not sufficient, and that asymptotic stability also depends on the competitive interaction between all populations. (16) de-termine a lower bound and an upper bound for m0, the difference between the measure

of competition and the measure of inhibition for all three populations.

If both (15) and (16) are fulfilled, each population will survive on a long-term basis and there will be a stable coexistence.

References

[1] Arne Persson, Lars-Christer B¨oiers, (2005), ANALYS I EN VARIABLER, Studentlitteratur, Lund. [2] Arne Persson, Lars-Christer B¨oiers, (2005), ANALYS I FLERA VARIABLER, Studentlitteratur,

Lund.

[3] Morris W. Hirsch, Stephen Smale, (1974), DIFFERENTIAL EQUATIONS, DYNAMICAL SYS-TEMS, AND LINEAR ALGEBRA, ACADEMIC PRESS, INC, New York.

[4] F.R. Gantmacher, (1959), Applications of the Theory of Matrices, Vol. II, Wiley, New York. [5] C. Henry Edwards, David E. Penney, (2008), Differential Equations COMPUTING AND

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