The effect of predation on a 2-year cicada
population in equilibrium-state
ir. S. Schaap
May 2, 2007
Contents
1 Introduction
2
2 Model for the 2-year cicada’s and their predators
2.1 Two Year classes, without predation . . . . . . . .
2.2 Two Year classes, with predation . . . . . . . . . .
2.2.1 Derivation of the prey equations . . . . . .
2.2.2 Derivation of the predator equation . . . .
2.3 rescaling P (t) . . . . . . . . . . . . . . . . . . . . .
2.4 the scaled dynamical system . . . . . . . . . . . . .
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3 Equilibrium state for the system without predator influence
4 Stability analysis for the system with predator influence
4.1 the Jacobian M (a1 ) . . . . . . . . . . . . . . . . . . . . . .
4.2 M (0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 v(a1 ) and v ∗ (a1 ) . . . . . . . . . . . . . . . . . . . . . . . .
4.4 M 0 (0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Conditions on gi and ci for the equilibrium state . . . . . .
4.6 the sign of λ0 (0) . . . . . . . . . . . . . . . . . . . . . . . . .
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1
Introduction
Periodical cicada’s are insects of 25-50mm who can be found in eastern North
America. They have life-cycles of 13 or 17 years and belong to the genus magicicada’s. The cicada juveniles, called nymphs, live underground of root fluids.
The reproduction is strangely enough synchronized: the cicada’s are nearly absent for 12 or 16 years, then they emerge and reproduct. After six to ten weeks
the eggs hatch, the nymphs drop from the trees, burrow underground and look
for a rootlet to feed themselves. See also [2].
The question is why the cicada’s do not reproduct every year. Why is it that of
the thirteen year-classes (or the seventeen year-classes) only one survives?
This abstract investigates the influence of the natural predator of the periodical
cicada’s, the birds, on this fenomena. The problem is simplified by considering
2-year cicada’s instead of 13-year or 17-year cicada’s.
2
Model for the 2-year cicada’s and their predators
We will look at the reproduction of cicada’s. We assume that there are two states
in which they can occur: The state in which the insect is a nymph (dutch: larve)
and the state in which the insect is an adult cicada.
2.1
Two Year classes, without predation
The number of nymphs at time t will be denoted by N0 (t) and the number of
adults at time t will be denoted by N1 (t).
The chance a nymph will survive is given by the function h0 . The number of
offsprings per adult is given by the function h1 . The functions h0 and h1 will
in the future be referred to as sensitivity functions.
In the model hi are updated every year. The sensitivity functions depend on
the environmental variable I. The dependence is given by the following formula
(known as the Ricker type density dependency):
hi (I) = σi e−gi I , where σi and gi are some given constants.
or the so called Beverton-Holt type density dependency:
hi (I) = σi
1
1 + gi I
We assume that the environmental variable depends linearly on N0 and N1 , i.e.
I(t) = c0 N0 (t) + c1 N1 (t).
This gives us the following model.
2
The system of equations which describes the cicada reproduction without predation is given by:
N0 (t + 1)
=
h1 (I(t))N1 (t)
(1)
N1 (t + 1)
=
h0 (I(t))N0 (t)
(2)
I(t)
= c0 N0 (t) + c1 N1 (t)
σi e−gi I (Ricker type) or
hi (I) =
1
σi 1+g
(Beverton-Holt type)
iI
2.2
2.2.1
(3)
(4)
Two Year classes, with predation
Derivation of the prey equations
The nymphs are preyed upon by birds. This means that of the N0 (t) nymphs
only a certain amount of them will survive, let’s say Ñ0 (t). So Ñ0 (t) = F ·N0 (t),
for some 0 < F < 1. We will now derive this survival probability F .
The birds will only have six to ten weeks to prey upon the nymphs, before the
nymphs are too deep in the ground to eat them. Because this time-period of
six to ten weeks is rather small compared to a whole year, it makes sense to
fix the variable t (i.e. we will consider t as a constant!) and to introduce a
time-variable τ for the local time-change. We choose τ such that τ = 0 when
the nymphs have just appeared, and that τ = 1 at the moment they are for
the first time too deep in the ground to be preyed upon (so approximately τ
denotes the time per six to ten weeks).
For notational simplicity we will write N0 (τ ) for N0 at time t increased with τ .
Note that N0 (τ ) 6= N0 (t + τ ).
When τ is given, for example τ = 1, we need an additional notation, because
N0 (1) would suggest that we have N0 at time t = 1. Therefore we write
N0 (τ )|τ =1 .
Let P (t) be the number of prey at time t. Locally around t we assume the
following relation:
dN0 (τ )
= −a1 N0 (τ )P (t)
dτ
The constant a1 is known as the attack rate. In this relation P (t) is considered
constant, whereas N0 is not considered constant. Integration of both sides
therefore gives:
N0 (τ ) = N0 (τ )|τ =0 e−a1 P (t)τ
The nymphs that survive predation were named Ñ0 . By definition Ñ0 =
N0 (τ )|τ =1 , so after six to ten weeks the nymphs that have survived are given by
Ñ0 (t)
=
N0 (τ )|τ =1
=
N0 (τ )|τ =0 e−a1 P (t)
=
N0 (t)e−a1 P (t)
3
The number of adults at time t + 1 was represented by formula 2 on page 3, but
due to predation it now looks like
N1 (t + 1)
2.2.2
=
h0 (I(t))Ñ0 (t)
=
h0 (I(t))N0 (t)e−a1 P (t)
Derivation of the predator equation
But how does P (t) evolve? We assume that the population of the predators is
composed of three groups:
1. the predators of last year which have survived,
2. the number of offspring due to standard consumption,
3. the number of offspring due to consumption of cicada nymphs.
the surviving predators
So we assume that food has direct influence on the number of offspring.
When we are not considering a maximum age, and when we assume that the
chance of survival, F say, stays constant irrespective of the age of the individual
predator we have F P (t) survivors of year t in year t + 1.
standard consumption
We will now look at the second group, consisting of the number of offspring, J1
say, due to standard consumption. The amount of food will be denoted by V
and for simplicity we assume there will not be any food produced during time,
V only decreases due to consumption by the predators. There are furthermore
no other animals which consume from the same food. The amount of food V
at the beginning of a year will be the same as the amount of food in the beginning of the previous year. With these assumptions it makes sense to write the
following differential equation:
dV (τ )
= −a0 P (t)V (τ )
dτ
where t is considered fixed and τ this time denotes the time passed in years
after time t. After integration of both sides we get:
V (τ )
= V (τ )|τ =0 e−a0 P (t)τ , where a0 is a certain constant
and V (τ )|τ =0 the initial condition.
The effect of the standard food on the offspring is modelled by:
dJ1 (τ )
dτ
= βV (τ )P (t), with β a certain constant.
4
(5)
We integrate this equation over the interval [0, 1] and then substitute the solution for V
Z 1
J1 (τ )|τ =1 = βP (t)
V (τ )dτ
τ =0
1
Z
=
V (τ )|τ =0 e−a0 P (t)τ dτ
βP (t)
τ =0
Z
=
1
βP (t)V (τ )|τ =0
e−a0 P (t)τ dτ
τ =0
=
=
=
−βP (t)V (τ )|τ =0 −a0 P (t)τ 1
[e
]τ =0
a0 P (t)
βV (τ )|τ =0
(1 − e−a0 P (t) )
a0
K0 (1 − e−a0 P (t) ).
Here we have used the definition of K0
K0
βV (τ )|τ =0
a0
=
(6)
additional consumption
We have already modelled the change of the number of nymphs N0 during the
six to ten weeks they are attacked by birds:
N0 (τ ) = N0 (τ )|τ =0 e−a1 P (t)τ , with τ per six to ten weeks and τ = 0 at time t
If we assume that the amount of extra offspring, J2 say, produced by the birds
is proportional to the amount of nymphs eaten we can model the extra offspring
J2 due to this additional consumption by:
dJ
= αN0 (τ )P (t) , with α some constant.
dτ
We integrate this differential equation over the interval [0, 1] and substitute the
above formula for N0
Z 1
J(1) = αP (t)
N0 (τ )dτ
0
Z
= αP (t)
1
N0 (τ )|τ =0 e−a1 P (t)τ dτ
0
=
=
−αP (t)N0 (τ )|τ =0 −a1 P (t)τ 1
[e
]0
a1 P (t)
αN0 (τ )|τ =0
(1 − e−a1 P (t) )
a1
5
Substituting N0 (τ )|τ =0 = h1 (I(t))N1 (t) we get
J(1)
=
α
h1 (I(t))N1 (t)(1 − e−a1 P (t) )
a1
=
K1 h1 (I(t))N1 (t)(1 − e−a1 P (t) ) with K1 =
α
.
a1
the predator equation
We combine the above results to formulate the predator equation
P (t + 1) = F P (t) + K0 (1 − e−a0 P (t) ) + K1 h1 (I(t))N1 (t)(1 − e−a1 P (t) )
We state the set of equations which now fully describes the cicada-prey model
below.
The system of equations which describes the cicada reproduction effected by
predation, is given by:
N0 (t + 1)
=
h1 (I(t))N1 (t)
(7)
−a1 P (t)
N1 (t + 1)
=
h0 (I(t))N0 (t)e
P (t + 1)
=
F P (t) + K0 (1 − e−a0 P (t) )
+K1 h1 (I(t))N1 (t)(1 − e−a1 P (t) )
I(t)
= c0 N0 (t) + c1 N1 (t)
−gi I
hi (I)
2.3
= σi e
(8)
(9)
(10)
(11)
rescaling P (t)
Before we will investigate the influence of the birds on the cicada population we
will rescale P (t). Given is the equation
P (t + 1) = F P (t) + K0 1 − e−a0 P (t) +
+K1 h0 (I(t))N0 (t) 1 − e−a1 P (t)
(12)
Let us call P̄ the equilibrium solution of equation (12) for a1 = 0. So P̄ is by
definition the unique solution of
P̄
=
K0 1 − e−a0 P̄
1−F
We define
p(t) =
(13)
P (t)
P̄
and therefore
P (t)
=
6
p(t)P̄
(14)
We substitute the above transformation (equation 14) and the implicit relation
for P̄ (equation 13) in equation 12 to get
p(t + 1)P̄ = F p(t)P̄ + K0 1 − e−a0 p(t)P̄ +
+K1 h0 (I(t))N0 (t) 1 − e−a1 p(t)P̄
K0 p(t + 1) = F p(t) +
1 − e−a0 p(t)P̄ +
P̄
K1
+
h0 (I(t))N0 (t) 1 − e−a1 p(t)P̄
P̄
1 − e−a˜0 p(t)
p(t + 1) = F p(t) + (1 − F )
+
1 − e−a˜0
+K̃1 h0 (I(t))N0 (t) 1 − e−a˜1 p(t) ,
(15)
1 (1−F )
where a˜0 = a0 P̄ , a˜1 = a1 P̄ , and K̃1 = K K(1−e
−a0 P̄ ) . Equation 15 is the scaled
0
version of equation 12. Note that when p(t) is in equilibrium for a1 = 0 we have
that p(t) = 1 (see equation 14).
In the future we will use the scaled equation in stead of the unscaled equation.
For simplicity reasons we will omit the tildes and p(t) will be again written as
P (t + 1) (that is, with a capital P ).
2.4
the scaled dynamical system
According to [1] the formula’s of the system do not change when we scale the
constants gi and ci such that g0 + g1 = 1 and c0 + c1 = 1. For the sensitivity
functions we will from now on use the Beverton-Holt type sensitivity functions.
The (scaled) system of equations which describes the cicada reproduction affected by predation, is given by:
N0 (t + 1)
N1 (t + 1)
P (t + 1)
=
=
=
h1 (I(t))N1 (t)
h0 (I(t))N0 (t)e
(16)
−a1 P (t)
1 − e−a0 P (t)
F P (t) + (1 − F )
+
1 − e−a0
+K1 h0 (I(t))N0 (t)(1 − e−a1 P (t) )
I(t)
(17)
(18)
= c0 N0 (t) + c1 N1 (t)
1 + gi I¯
hi (I) =
(Beverton-Holt type)
1 + gi I
g0 + g1 = 1
(19)
c0 + c1
(22)
=
1
7
(20)
(21)
3
Equilibrium state for the system without predator influence
When the birds do not influence the evolution of the cicada population (i.e.
a1 = 0) the steady state of the dynamical system is known. We will give the
steady state in the following lemma.
Lemma 3.1 When a1 = 0 the system of equations (16)-(22) has the solution
N̄0
1
¯
= I
N̄1
1
P̄ = 1
Proof
The Ni are in equilibrium state, i.e.
¯
N̄0
N̄0
0
h1 (I)
=
1
¯
h0 (I)
0
N̄1
N̄1
Or equivalently
−1
1
1
−1
N̄0
0
=
N̄1
0
This system has eigenvector
N̄0
N̄1
1
=s
,
1
¯ because I = c0 N0 + c1 N1 , and c0 + c1 = 1, and therefore
Note that s = I,
I¯ = c0 N̄0 + c1 N̄1
= c0 s + c1 s
= s(c0 + c1 )
= s
So we have
(23)
N̄0
1
¯
=I
.
N̄1
1
4
Stability analysis for the system with predator
influence
When a1 = 0 we understand the behaviour of the system (see [1]). In this section we will see what happens with the stability of the system when we make
8
a1 a little positive, i.e. we have a small influence of the birds on the population
of the cicada’s.
We will consider the case when the eigenvalue of the jacobian of the system
equals −1. We will then derive an expression for λ0 (0). If λ0 (0) happens to be
negative, then |λ| > 1, and hence the system becomes unstable. In this scenario
the birds will have a destabilizing effect on the cicada’s for λ = −1. If λ0 (0) > 0
the system becomes stable, because then |λ| < 1. In this scenario the birds will
have a stabilizing effect on the cicada’s for λ = −1.
Let us start. We define F by




h1 (I)y
x
−a
z

h0 (I)xe 1
F : a, y  7−→ 
−a0 z
1−e
−a
z
1
z
))
F z + (1 − F ) 1−e−a0 + K1 h0 (I)x(1 − e
where
I = c0 x + c1 y and hi given by (20)
The system can now be written as
xa1 (k + 1) = F(a, xa1 (k))
The steady state will be given by
x̄a1 = F(a1 , x̄a1 ).
We will consider a point close to the steady state, i.e.
xa1 (k) = x̄a1 + y(k), where y(k) is small.
We substitute this in the system
xa1 (k + 1)
=
x̄a1 + y(k + 1)
=
x̄a1 + y(k + 1)
y(k + 1)
y(k + 1)
F(a1 , xa1 (k))
F(a1 , x̄a1 + y(k))
∂F
(a1 , x̄a1 )y(k) (Taylor in x̄a1 )
≈ F(a1 , x̄a1 ) +
∂x
∂F
(a1 , x̄a1 )y(k)
≈
∂x
∂F
≈ M (a1 )y(k), where M (a1 ) =
(a1 , x̄a1 )
∂x
(24)
To investigate the influence of a1 on the stability of the system it suffices to
look at the stability of the linear system y(k + 1) = M (a1 )y(k). Thus we need
to investigate what happens with the eigenvalue λ(a1 ) of the jacobian M (a1 )
when a1 becomes sligtly unzero. We use the fact that an eigenvalue of M (0)
9
has value −1 for the steady state. We will now derive a formula to calculate the
sign of λ0 (a1 ).
The following notation will be used: eigenvector v(a1 ), eigenvalue λ(a1 ), and adjoint eigenvector v ∗ (a1 ), i.e. M (0)v(0) = λ(0)v(0) and v ∗ (0)M (0) = λ(0)v ∗ (0).
We will give a formula for λ0 (0) in the following lemma.
Lemma 4.1 Let M (a1 ) be a continuous differentiable square matrix, let v and
v ∗ be the corresponding eigenvector and adjoint eigenvector, and let λ be a simple
eigenvalue of M (a1 ) (i.e. v ∗ v 6= 0). Then
λ0 (0)
=
v ∗ (0)M 0 (0)v(0)
v ∗ (0)v(0)
(25)
Proof
M (a1 )v(a1 ) =
dM (a1 )v(a1 )
=
da1
M 0 (a1 )v(a1 ) + M (a1 )v 0 (a1 ) =
M 0 (0)v(0) + M (0)v 0 (0)
v ∗ (0) (M 0 (0)v(0) + M (0)v 0 (0))
v ∗ (0)M 0 (0)v(0) + v ∗ (0)M (0)v 0 (0)
∗
0
∗
0
v (0)M (0)v(0) + λ(0)v (0)v (0)
λ(a1 )v(a1 )
dλ(a1 )v(a1 )
da1
0
λ (a1 )v(a1 ) + λ(a1 )v 0 (a1 )
= λ0 (0)v(0) + λ(0)v 0 (0)
=
v ∗ (0) (λ0 (0)v(0) + λ(0)v 0 (0))
= v ∗ (0)λ0 (0)v(0) + v ∗ (0)λ(0)v 0 (0)
= λ0 (0)v ∗ (0)v(0) + λ(0)v ∗ (0)v 0 (0)
v ∗ (0)M 0 (0)v(0) = λ0 (0)v ∗ (0)v(0)
v ∗ (0)M 0 (0)v(0)
= λ0 (0)
v ∗ (0)v(0)
So we will need M 0 (0), M (0), v(0), and v ∗ (0) to compute λ0 (0).
4.1
the Jacobian M (a1 )
The Beverton-Holt sensitivity functions can be rewritten as follows:
hi (I) =
1 + gi I¯
1 + gi I¯
=
1 + gi I
1 + gi c0 x + gi c1 y
The partial derivatives of the sensitivity functions are
∂hi (I)
∂x
=
¯ i c0
−(1 + gi I)g
(1 + gi c0 x + gi c1 y)2
10
=
=
∂hi (I)
∂y
=
=
=
2
−gi c0
1 + gi I¯
1 + gi I¯ 1 + gi c0 x + gi c1 y
−gi c0
(hi (I))2
1 + gi I¯
¯ i c1
−(1 + gi I)g
(1 + gi c0 x + gi c1 y)2
2
−gi c1
1 + gi I¯
1 + gi I¯ 1 + gi c0 x + gi c1 y
−gi c1
(hi (I))2
1 + gi I¯
We use these derivatives for the partials of F(a1 , xa1 )
∂F(a1 , xa1 )1
∂x
=
=
=
∂F(a1 , xa1 )1
∂y
∂F(a1 , xa1 )1
∂z
∂(h1 (I)y)
∂y
∂h1 (I)
= h1 (I) +
y
∂y
g1 c1
=
1−
h1 (I)y h1 (I)
1 + g1 I¯
=
=
=
∂F(a1 , xa1 )2
∂x
∂F(a1 , xa1 )2
∂y
∂(h1 (I)y)
∂x
∂h1 (I)
y
∂x
−g1 c0
(h1 (I))2 y
1 + g1 I¯
∂(h1 (I)y)
∂z
0
∂(h0 (I)xe−a1 z )
∂x
∂h0 (I) −a1 z
−a1 z
= h0 (I)e
+
xe
∂x g0 c0
=
1−
h0 (I)x h0 (I)e−a1 z
1 + g0 I¯
=
=
=
=
∂(h0 (I)xe−a1 z )
∂y
∂h0 (I) −a1 z
xe
∂y
−g0 c1
(h0 (I))2 xe−a1 z
1 + g0 I¯
11
∂F(a1 , xa1 )2
∂z
∂F(a1 , xa1 )3
∂x
∂F(a1 , xa1 )3
∂y
∂F(a1 , xa1 )3
∂z
∂(h0 (I)xe−a1 z )
∂z
= −a1 h0 (I)xe−a1 z
=
=
−a0 z
−a1 z
∂ F z + (1 − F ) 1−e
+
K
h
(I)x(1
−
e
)
−a
1
0
1−e 0
∂x
∂ (K1 h0 (I)x(1 − e−a1 z ))
=
∂x
∂h0 (I)
= K1 h0 (I) +
x (1 − e−a1 z )
∂x
g0 c0
= K1 h0 (I) 1 −
h
(I)x
(1 − e−a1 z )
0
1 + g0 I¯
−a0 z
−a1 z
∂ F z + (1 − F ) 1−e
+
K
h
(I)x(1
−
e
)
1 0
1−e−a0
=
∂y
∂ (K1 h0 (I)x(1 − e−a1 z ))
=
∂y
∂h0 (I)
= K1
x(1 − e−a1 z )
∂y
g0 c1
= −K1
(h0 (I))2 x(1 − e−a1 z )
1 + g0 I¯
−a0 z
−a1 z
∂ F z + (1 − F ) 1−e
+
K
h
(I)x(1
−
e
)
−a
1
0
1−e 0
=
∂z
(1 − F )a0 −a0 z
= F+
e
+ a1 K1 h0 (I)xe−a1 z
1 − e−a0
We will denote the jacobian of F(a1 , xa1 ) as M (a1 ):

g1 c1
−g1 c0
2
(h
(I))
y
1
−
h
(I)y
h1 (I)
0
1
1
¯
¯
1+g1 I
1+g1 I




g0 c0
g0 c 1
2
−a1 z
1 − 1+g
h
(I)x
h0 (I)e−a1 z
− 1+g
−a1 h0 (I)xe−a1 z

0
¯
¯ (h0 (I)) xe
0I
0I




)a0 −a0 z
g0 c0
g0 c1
2
−a1 z
+
h
(I)x
h0 (I)(1 − e−a1 z ) −K1 1+g
) F + (1−F
e
 K1 1 − 1+g
0
¯
¯ (h0 (I)) x(1 − e
1−e−a0
0I
0I
+a1 K1 h0 (I)xe−a1 z
12











4.2
M (0)
For a1 = 0 M (a1 ) takes the following form

−g1 c0 ¯
g1 c 1 ¯
I
1 − 1+g
¯I
1+g1 I¯
1I



g0 c 0 ¯
−g0 c1 ¯
I
M (0) =  1 − 1+g
¯I
1+g0 I¯
0I


0
0
F+
4.3
0

0






(1−F )a0 −a0
e
1−e−a0
v(a1 ) and v ∗ (a1 )
We use the first row of M (0) − λI, i.e. of M (0) + I to find
 1+g1 c0 I¯ 
¯ 
¯
1 I−g1 c1 I
− 1+g
− 1+g1 c1 I¯
¯ 1 c0 I¯
1+g1 I−g



= s
=
s
1
1
0
0

v(0)
(26)
with s 6= 0 a constant. We use the first column of M (0) + I to find




¯
¯
¯
¯
¯
(1+g1 I)(1+g
(1+g1 I)(1+g
0 c1 I)
0 I−g0 c0 I)
− (1+g
−
¯
¯
¯
¯
¯
I)
I−g
c
I)
I)(1+g
I)(1+g
c
(1+g
1 0
1
1 1
0
0




v ∗ (0) = t 
 = t
 (27)
1
1
0
0
with t 6= 0 a constant. We will choose s = t = 1.
4.4
M 0 (0)
The derivative of M 0 (0) will be given in the following lemma.
Lemma 4.2 Given the function M (a1 ) =
spect to a1 is given by
M 0 (a1 )
=
∂F
∂x (a1 , x̄a1 )
the derivative with re-
∂2F
∂2F
∂xa1
(a1 , x̄a1 ) +
(a1 , x̄a1 )
2
∂a1 ∂x
∂x
∂a1
(28)
Proof
0
M (a1 )
=
=
∂
∂F
(a1 , x̄a1 )
∂a1 ∂x
∂2F
∂2F
∂xa1
(a1 , x̄a1 ) +
(a1 , x̄a1 )
∂a1 ∂x
∂x2
∂a1
(29)
13
So, in order to derive the terms of
∂2F
∂xa1
∂2F
(0, x̄0 )
(0, x̄0 ) +
2
∂a1 ∂x
∂x
∂a1
M 0 (0) =
we apparently need to be a little patient. The first term of M 0 (0) is attained in
a pretty straightforward way.
Lemma 4.3
∂2F
∂a1 ∂x (a1 , x̄a1 )

=
0
0



g0 c0
h
(I)x
h0 (I)ze−a1 z
 − 1 − 1+g
0
¯
0I



g0 c0
K1 1 − 1+g
h
(I)x
h0 (I)ze−a1 z
¯ 0
0I
0
g0 c1
(h (I))2 xze−a1 z
1+g0 I¯ 0
−(1 − a1 z)h0 (I)xe−a1 z
g0 c 1
2
−a1 z
−K1 1+g
¯ (h0 (I)) xze
0I
K1 (1 − a1 z)h0 (I)xe−a1 z
Thus when a1 = 0 we have
Corrollary 4.4

∂2F
(0, x̄0 )
∂a1 ∂x
0
0
0



g0 c0 ¯
g0 c 1 ¯
− 1 − 1+g
I
−I¯
¯I
= 
I
1+g0 I¯
0



g0 c0 ¯
g0 c1 ¯
K1 1 − 1+g
I
−K1 1+g
K1 I¯
¯
¯I
0I
0I








In order to find the second term of M 0 (0) we need to do some more work.
2
We will start by deriving the terms ∂∂xF2 (0, x̄0 ). Secondly we will find an expression for ∂∂ax̄10 . Multiplication of these two will give us the second term of M 0 (0).
Lemma 4.5

∂2F
(0, x̄0 )
∂x∂x
g1 c 0 2 ¯
2( 1+g
¯) I
1I

=  − 2g0 c0¯ (1 −
1+g0 I
∗
g0 c0 I¯
)
1+g0 I¯
14
−g1 c0
(1
1+g1 I¯
−g0 c1
(1
1+g0 I¯
¯
g1 c1 I
− 2 1+g
¯) ∗
1I


g0 c0 I¯
(30)
− 2 1+g
¯) ∗ 
0I
∗
∗









2
∂ F
(0, x̄0 )
∂y∂x
∂2F
(0, x̄0 )
∂z∂x

= 

−g1 c0
(1
1+g1 I¯
−g0 c1
(1
1+g0 I¯
0

0
=
∗
0
0
∗
¯
g1 c1 I
− 2 1+g
¯)
1I
¯
g0 c0 I
− 2 1+g
¯)
0I
∗

∗
∗ 
∗
−2g1 c1
g1 c1 I¯
(1 − 1+g
¯)
1+g1 I¯
1I
g0 c1 2 ¯
2( 1+g0 I¯ ) I
∗
∗


∗ 
∗
(31)
(32)
Note: Several elements are omitted by replacing them by an asterisk. This is
done because we will not need them later on in the computations.
Proof
We will use the Jacobian of F(a1 , xa1 ) which we have calculated earlier on.
∂
−g1 c0
∂
2
DF(a1 , xa1 )1,1 =
(h1 (I)) y
∂x
∂x 1 + g1 I¯
−g1 c0 ∂
(h1 (I))2 y
=
1 + g1 I¯ ∂x
−g1 c0
∂
=
2h1 (I) (h1 (I))y
∂x
1 + g1 I¯
2
−g1 c0
= 2
(h1 (I))3 y
1 + g1 I¯
∂
∂
−g1 c0
2
(h1 (I)) y
DF(a1 , xa1 )1,1 =
∂y
∂y 1 + g1 I¯
−g1 c0 ∂
=
((h1 (I))2 y)
1 + g1 I¯ ∂y
−g1 c0
∂
2
(h
(I))y
+
(h
(I))
=
2h
(I)
1
1
1
∂y
1 + g1 I¯
−g1 c0
−g1 c1
3
2
=
2
(h
(I))
y
+
(h
(I))
1
1
1 + g1 I¯
1 + g1 I¯
−g1 c0
g1 c1
h1 (I)y (h1 (I))2
=
1−2
¯
1 + g1 I
1 + g1 I¯
∂
∂
−g1 c0
2
DF(a1 , xa1 )1,1 =
(h
(I))
y
1
∂z
∂z 1 + g1 I¯
= 0
∂
∂
DF(a1 , xa1 )1,2 =
DF(a1 , xa1 )1,1
∂x
∂y
∂
∂
g1 c1
DF(a1 , xa1 )1,2 =
(1 −
h
(I)y)h
(I)
1
1
∂y
∂y
1 + g1 I¯
∂
g1 c1 ∂
=
(h1 (I)) −
((h1 (I))2 y)
∂y
1 + g1 I¯ ∂y
−g1 c1
g1 c1
−g1 c1
2
2
2
=
(h1 (I)) −
2h1 (I)
(h1 (I)) y + (h1 (I))
1 + g1 I¯
1 + g1 I¯
1 + g1 I¯
15
=
∂
DF(a1 , xa1 )1,2
∂z
=
=
∂
DF(a1 , xa1 )2,1
∂x
=
=
=
=
∂
DF(a1 , xa1 )2,1
∂y
=
=
=
=
∂
DF(a1 , xa1 )2,1
∂z
=
=
=
∂
DF(a1 , xa1 )2,2
∂x
∂
DF(a1 , xa1 )2,2
∂y
=
=
=
=
∂
DF(a1 , xa1 )2,2
∂z
=
=
−g1 c1
g1 c1
1
−
h
(I)y
(h1 (I))2
1
1 + g1 I¯
1 + g1 I¯
∂
g1 c1
h
(I)y)h
(I)
(1 −
1
1
∂z
1 + g1 I¯
0 ∂
g0 c0
−a1 z
(1 −
h0 (I)x)h0 (I)e
∂x
1 + g0 I¯
∂
g0 c0 ∂
2
(h0 (I)) −
((h0 (I)) x) e−a1 z
∂x
1 + g0 I¯ ∂x
g0 c0
−g0 c0
−g0 c0
2
2
2
(h0 (I)) −
2h0 (I)
(h0 (I)) x + (h0 (I))
e−a1 z
1 + g0 I¯
1 + g0 I¯
1 + g0 I¯
−g0 c0
g0 c0
2
h0 (I)x (h0 (I))2 e−a1 z
1−
1 + g0 I¯
1 + g0 I¯
∂
g0 c0
−a1 z
h0 (I)x)h0 (I)e
(1 −
∂y
1 + g0 I¯
∂
g0 c0 ∂
2
(h0 (I)) −
((h0 (I)) x) e−a1 z
∂y
1 + g0 I¯ ∂y
−g0 c1
g0 c0
−g0 c1
2
2
(h0 (I)) −
2h0 (I)
(h0 (I)) x e−a1 z
1 + g0 I¯
1 + g0 I¯
1 + g0 I¯
−g0 c1
g0 c0
h0 (I)x (h0 (I))2 e−a1 z
1−2
1 + g0 I¯
1 + g0 I¯
∂
g0 c0
−a1 z
1−
h0 (I)x )h0 (I)e
∂z
1 + g0 I¯
g0 c0
∂
1−
h0 (I)x h0 (I) (e−a1 z )
∂z
1 + g0 I¯
g0 c0
−a1 1 −
h0 (I)x h0 (I)e−a1 z
1 + g0 I¯
∂
DF(a1 , xa1 )2,1
∂y
∂
−g0 c1
2
−a1 z
(h
(I))
xe
0
∂y 1 + g0 I¯
−g0 c1 ∂
(h0 (I))2 xe−a1 z
¯
1 + g0 I ∂y
2
g0 c1
2
(h0 (I))3 xe−a1 z
1 + g0 I¯
∂
−g0 c1
2
−a1 z
(h
(I))
xe
0
∂z 1 + g0 I¯
−g0 c1
∂
(h0 (I))2 x (e−a1 z )
¯
∂z
1 + g0 I
2
16
= a1
g0 c1
(h0 (I))2 x(e−a1 z )
1 + g0 I¯
Lemma 4.6
¯

∂ x̄0
∂a1
¯
1 c0 I)(1+g0 I)
− (1+g1+2g
¯
0 g1 I

 (1+g1 (1+c0 )I)(1+g
¯
¯ 
0 I) 
 −


1+2g0 g1 I¯
1
1−e−a0
¯
K
I
1−F 1−e−a0 (1+a0 ) 1
=
(33)
Proof
−1 ∂F
∂x
We will firstly show that ∂aa11 = 1 − ∂F
∂x (a1 , x̄a1 )
∂a1 (a1 , x̄a1 ).
We start with the expression for the steady state and then take the derivative
with respect to a1
x̄a1
∂ x̄a1
∂a1
∂ x̄a1
∂a1
= F(a1 , x̄a1 )
∂
=
(F(a1 , x̄a1 ))
∂a1
∂F
∂xa1
∂F
(a1 , x̄a1 )
=
(a1 , x̄a1 ) +
∂a1
∂x
∂a1
(34)
∂x
Combining the two terms containing ∂aa11 and after having taken the inverse of
1 − ∂F
∂x (a1 , x̄a1 ) we get the above equation.
Now

∂F
(0, x̄0 )
∂a1

0
=  −h0 (I)xze−a1 z 
K1 h0 (I)xze−a1 z a =0,x=x̄(0)
1


0
=  −I¯ 
K1 I¯
Furthermore
1−
∂F
(0, x̄0 )
∂x
=
1 − M (0)

−g1 c0
1+g1 I¯
=



1− 1−


I¯
g0 c 0 ¯
I
1+g0 I¯
0
17
g1 c 1 ¯
I
1+g1 I¯
0

−g0 c1 ¯
I
1+g0 I¯
0
0
a0 e
F + (1 − F ) 1−e
−a0






1−
−a0

1+
g1 c0 ¯
I
1+g1 I¯



=  −1 +


g0 c 0 ¯
I
1+g0 I¯
g0 c1 ¯
I
1+g0 I¯
1+
0
1+g1 (1+c0 ) ¯
I
1+g1 I¯



1+g c
 − 1+g00 I¯1 I¯



0






−a0
1 c0 ¯
− 1+g
I
1+g1 I¯
1+g0 (1+c1 ) ¯
I
1+g0 I¯
0
0
a0 e
(1 − F ) − (1 − F ) 1−e
−a0

0



0



0

=
g1 c1 ¯
I
1+g1 I¯
−1 +
−a0
(1+a0 )
(1 − F ) 1−e1−e−a
0
0
And because if

A
=
a11
a12


 a21


0
0



0 


a33
a22
0
then
A
−1

a22
a11 a22 −a12 a21
−a12
a11 a22 −a12 a21
0


= 


−a21
a11 a22 −a12 a21
a11
a11 a22 −a12 a21


0 


0
0
¯
1+g0 (1+c1 )IC
1+g0 I¯
¯
1+g1 c0 IC
1+g1 I¯

1
a33
we have

1−
∂F
(0, x̄0
∂x
−1
=








0
¯
1+g0 c1 IC
1+g0 I¯
¯
1+g1 (1+c0 )IC
1+g1 I¯
0
0
0
1
1−e−a0
(1−F ) 1−e−a0 (1+a0 )
with
C
=
=
¯ + g1 I)
¯
(1 + g0 I)(1
¯ + g0 (1 + c1 )I)
¯ − (1 + g1 c0 I)(1
¯ + g0 c1 I)
¯
(1 + g1 (1 + c0 )I)(1
¯
¯
(1 + g0 I)(1 + g1 I)
¯ I¯
(1 + 2g0 g1 I)
Therefore we have

−1
∂F
∂F
1−
(0, x̄0 )
(0, x̄0 )
∂x
∂a1
=
18
¯
¯
1 c0 I)(1+g0 I)
− (1+g1+2g
¯
0 g1 I

 (1+g1 (1+c0 )I)(1+g
¯
¯ 
0 I) 
 −
¯


1+2g0 g1 I
1
1−e−a0
¯
1−F 1−e−a0 (1+a0 ) K1 I







We are now able to give M 0 (0).
Lemma 4.7
m011

m021
=
∗
m012
m022
∗

M 0 (0)

∗
∗ ,
∗
with
m011
=
m012
=
m021
=
m022
=
2
g1 c0
1 + g0 I¯
¯
2(1
+
g
c
I)
I¯
1 0
1 + 2g0 g1 I¯
1 + g1 I¯
g1 c0
2g1 c1 I¯
¯
−1
+(1 + g1 (1 + c0 )I)
1 + g1 I¯
1 + g1 I¯
1 + g0 I¯
g1 c1 I¯
¯ g1 c0
−
1
−
(1
+
g
c
I)
2
1
0
1 + 2g0 g1 I¯
1 + g1 I¯
1 + g1 I¯
g1 c1 I¯
¯ 2g1 c1
+(1 + g1 (1 + c0 )I)
−
1
1 + g1 I¯ 1 + g1 I¯
g0 c0
g0 c0
1 + g0 I¯
¯
−
1
2(1
+
g
c
I)
−
1 0
1 + 2g0 g1 I¯
1 + g0 I¯ 1 + g0 I¯
¯
g0 c1
g0 c0 I¯
¯ 2g0 c0 I − 1
+(1 + g1 (1 + c0 )I)
−1
+
¯
¯
1 + g0 I
1 + g0 I
1 + g0 I¯
1 + g0 I¯
2g0 c0
g0 c1
¯
−
(1
+
g
c
−
1
I)
1 0
1 + 2g0 g1 I¯
1 + g0 I¯ 1 + g0 I¯
2 !
g
c
g0 c1 I¯
0
1
¯
+(1 + g1 (1 + c0 )I)2
I¯ +
¯
1 + g0 I
1 + g0 I¯
−
Proof
Using
∂2F
∂xa1
∂xa1
∂ya1
∂za1
∂2F
∂2F
∂2F
(0,
x̄
)
=
(0,
x̄
)
+
(0,
x̄
)
+
(0, x̄0 )
0
0
0
2
∂x
∂a1
∂x∂x
∂a1
∂y∂x
∂a1
∂z∂x
∂a1
together with corrollary 4.4, lemma 4.5, and lemma 4.6 we get
M 0 (0)
=

0

 
g0 c0 I¯
 − 1 − 1+g
¯
0I


∗
+
0
g0 c1 I¯
1+g0 I¯
∗

∗




∗ 


∗
g1 c0 2 ¯
2( 1+g
¯) I
1I
¯ + g0 I)
¯ 
−(1 + g1 c0 I)(1
2g0 c0
 − 1+g
¯ (1 −
¯
0I
1 + 2g0 g1 I
∗
19
g0 c0 I¯
)
1+g0 I¯
−g1 c0
(1
1+g1 I¯
−g0 c1
(1
1+g0 I¯
¯
g1 c1 I
− 2 1+g
¯) ∗
1I


g0 c0 I¯
− 2 1+g
¯) ∗ 
0I
∗
∗
+

¯
¯
(1 + g1 (1 + c0 )I)(1 + g0 I) 

1 + 2g0 g1 I¯
−g1 c0
(1
1+g1 I¯
−g0 c1
(1
1+g0 I¯
¯
g1 c1 I
− 2 1+g
¯)
1I
¯
g0 c0 I
− 2 1+g
¯)
0I
∗
−2g1 c1
g1 c1 I¯
(1 − 1+g
¯)
1+g1 I¯
1I
g0 c 1 2 ¯
2( 1+g0 I¯ ) I
∗
Thus the elements m011 , m012 , m021 , and m022 of M 0 (0) are stated as above. 4.5
Conditions on gi and ci for the equilibrium state
When the system is in its equilibrium state the eigenvalue of the linearized
system must equal −1. An eigenvalue 1 or a complex conjugate pair with norm
1 is not possible in our model. (see further [1]).
λ = −1 requires either c0 = c1 or g0 = g1 , indeed from the characteristic
polynomial of M (0) we derive:
g0 c1 I¯
g0 c0 I¯
g1 c1 I¯
g1 c0 I¯
1−
− 1−
1−
= 0
1−
1 + g1 I¯
1 + g0 I¯
1 + g0 I¯
1 + g1 I¯
¯ + g0 c0 I)
¯ − (1 + g0 c1 I)(1
¯ + g1 c0 I)
¯
(1 + g1 c1 I)(1
= 0
¯ + g0 I)
¯
(1 + g1 I)(1
¯ + g0 c0 I)
¯ − (1 + g0 c1 I)(1
¯ + g1 c0 I)
¯ = 0
(1 + g1 c1 I)(1
(g0 c0 − g0 c1 + g1 c1 − g1 c0 )I¯ = 0
(g0 (c0 − c1 ) + g1 (c1 − c0 ))I¯ = 0
(g0 − g1 )(c0 − c1 )I¯ = 0
4.6
the sign of λ0 (0)
When looking for the sign of λ0 (0) it suffices to consider v ∗ (0)M 0 (0)v(0), as
v ∗ (0)v(0) > 0. The adjoint eigenvector v ∗ (0), the matrix M 0 (0), and the eigenvector v(0) are given by formula’s 27, 35 and 26 respectively.
We will consider v ∗ (0)M 0 (0)v(0) in the two just mentioned cases, i.e. c0 = c1
and g0 = g1 .
c0 = c1
We substitute c0 = c1 = 0.5 and g1 = 1 − g0 in v ∗ (0)M 0 (0)v(0). We thereafter
collect terms with the same power of I. The coefficients are factorized when
possible. This results in:
v ∗ (0)M 0 (0)v(0)
=
K 4g02 (1 − g0 )3 I¯5
+g0 (1 − g0 )2 (9g02 + 11)I¯4
+(1 − g0 )(23g03 − 17g02 + 16g0 + 10)I¯3
+4(2 − g0 )(5g 2 − 6g0 + 5)I¯2
0
+(44g02
− 68g0 + 44)I¯
+16(1 − g0 ))
20
(35)
∗


∗ 
∗
with
K
=
−1
¯
¯ I¯
4(2 + (1 − g0 )I)(1 + 2g0 (1 − g0 ))(1 + (1 − g0 )I)
Note that, because 0 ≤ g0 ≤ 1, we have 23g03 −17g02 +16g0 +10 > 0, 5g02 −6g0 +5 >
0, and 44g02 − 68g0 + 44 > 0. Because I¯ > 0 we have v ∗ (0)M 0 (0)v(0) < 0.
g0 = g1
We substitute g0 = g1 = 0.5 and c1 = 1 − c0 in v ∗ (0)M 0 (0)v(0). We thereafter
collect terms with the same power of I. The coefficients are factorized when
possible. This results in:
v ∗ (0)M 0 (0)v(0)
=
L c20 I¯4
+c0 (c30 − 2c20 + 4c0 + 4)I¯3
+(−4c20 + 16c0 + 4)I¯2
+(−12c20 + 8c0 + 16)I¯
+16(1 − c0 ))
(36)
with
L =
−1
3
¯
(1 − c0 )I + (6 − 4c0 )I¯2 + (12 − 4c0 )I¯ + 8
Note that, because 0 ≤ c0 ≤ 1, we have c30 − 2c20 + 4c0 + 4, −4c20 + 16c0 + 4 > 0
and −12c20 + 8c0 + 16. Furthermore we have I¯ > 0, and so v ∗ (0)M 0 (0)v(0) < 0.
This means that λ0 (0) < 0, and therefore the system destabilizes when a1 becomes slightly unzero.
References
[1] Davydova, N.V., Diekmann, O., Gils, S.A. Year class coexistence or competitive exlusion for strict biennials? J.Math. Biol. 46, 95-131 (2003)
[2] Cooley, J., Marshall, D., O’Brien, M. UMMZ Periodical Cicada Page
(http://www.ummz.lsa.umich.edu/magicicada/Periodical/Index.html)
Ecology and Evolutionary Biology University of Connecticut
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