Kliah Soto
Jorge Munoz
Francisco Hernandez
dx
x  0
0
dt
dy
x  0
 cy
dt
and

dx
y  0
 ax
dt
dy
y  0
0
dt
dx
 ax  bxy  0
dt
dy
 cy  dxy  0
dt
results in
(0,0)
and
c a
( , )
d b

Solve the system of equations:
dy dy dx cy  dxy y(c  dx)

/


dx dt dt
ax  bxy
x(a  by)
a  by
c  dx
dy 
dx
y
x

a  by
dy 
y

c  dx
dx
x
aln y  by  c ln x  dx  k

Solution curve with all parameters = 1
Pink: prey x
Blue: predator y
dx
 ax  bxy
dt
dy
 cy  dxy  eyz
dt
dz
  fz  gyz
dt


Case 1: if z=0 then we have the 2
dimensional case
 Case 2: y=0
dx
 ax
dt
dy
0
dt
dz
  fz
dt
In the absence of the middle predator y, we are left with:
dx
 ax
dt
dz
  fz
dt
We combine it to one fraction
and use separation of
variables:
dz dz dx  fz
 /

dx dt dt
ax
1
1
  fz dz   ax dx
z  Kx

f
a
species z approaches zero as t
goes to infinity, and species x
exponentially grows as t
approaches infinity.

The blue curve
represents the prey,
while the red curve
represents the predator.
4
3
2
1
0
2
4
6
8
10

Case 3: x=0
dx
0
dt
dy
 cy  eyz
dt
dz
  fz  gyz
dt

In the absence of the prey x, we are left with:
dy
 cy  eyz
dt
dz
  fz  gyz
dt
We combine it to one fraction and
use separation of variables:
dz dz dy
z ( f  gy)
 / 
dy dt dt y (c  ez)
 c  ez
 f  gy
dz 
dy
z
y
 c  ez
 f  gy
d
z

 z
 y dy
c ln z  ez  f ln y  gy  K
species y and z will approach zero as
t approaches infinity.

The blue curve
represents the top
predator, while
the red curve
represents the
middle predator.
yz
1.0
0.8
0.6
0.4
0.2
t
1
2
3
4
5

Set all three equations equal to zero to
determine the equilibria of the system:
dx
 ax  bxy  0
dt
dy
 cy  dxy  eyz  0
dt
dz
  fz  gyz  0
dt

dx
 ax  bxy
dt



dy
 cy  dxy  eyz
dt
dz
  fz  gyz
dt



When x=0:
Either y=0 or z=-c/e
z has to be positive so we
conclude that y=0 making the last
equation z=0.
Equilibrium at (0,0,0)
When y=0
System reduces to:
dx
 ax
dt
dz
  fz
dt
x=0 and y=0 since a and f are positive. Again equilibrium (0,0,0).
When we consider:
dz
  fz  gyz  z ( f  gy )
dt
Either z= 0 or –f+gy =0. Taking the first case will result in the trivial
solution again as well as the equilibrium from the two dimensional case.
(c/d,a/b,0)
Using parameterization we set x=s and the last equilibrium is:
dx
a
 as  bsy  s(a  by)  y 
dt
b
dy
ds  c
 cy  dsy  eyz  y(c  ds  ez)  z 
dt
e
dz
f
  fz  gyz  z( f  gy)  y 
dt
g

Equilibrium point at
(s,a/b=f/g,(ds-c)/e)
dx
 ax  bxy  f ( x, y, z )
dt
dy
 cy  dxy  eyz  g ( x, y, z )
dt
dz
  fz  gyz  h( x, y, z )
dt
 dx
 dt 

 dy
 
 dt
 dy
 
 dt
f
f
f
x
y
z
x
y
z
g
g
g
x
y
z
x
y
z
h
h
h
x
y
z
x
y
z
Where the partial derivatives
are evaluated at the equilibrium
point
 xb
0 
a  by


J ( x, y, z )   yd
 c  dx  ez
 ye 
 0
zg
 f  gy

Real part of the
eigenvalues
◦ Positive: Unstable
◦ Negative: Stable
◦ Zero: Center

Number of eigenvalues:
◦ Dimension of the
manifold

Manifold is tangent to
the eigenspace spanned
by the eigenvectors of
their corresponding
eigenvalues

Eigenvalues:
◦ a, -c, -f


Eigenvectors:
{1,0,0}, {0,1,0}, {0,0,1}


0 
a 0


J (0,0,0)  0  c 0 
0 0  f 
One-dimensional unstable manifold:
curve  x-axis
Two-dimensional stable manifold:
surface yz- Plane
10
50 000
5
40 000
30 000
1
2
20 000
5
10 000
5

10
Unstable x-axis
15
20
10

Stable yz-Plane
3
4
5
ac
0
 0  bc / d

J (c / d , a / b,0)  ad / b
0
 ae / b 
 0
0
 f  ga / b

Eigenvectors:
{1, ( fb  ag)
{1,
d
1
2 cd
2
2
2
2
,
(
ab

b
df

2
abdfg

a
dg
)
}
b 2c
ab2ce
id ac
,0}
bc

Eigenvalues
( ga  fb) / b
 i ac
( ga  fb) / b
 i ac



One-Dimensional invariant curve:
◦ Stable if ga<fb
◦ Unstable ga>fb
Two-Dimensional center manifold
Three-dimensional center
manifold
◦ If ga=fb
Blue represents the prey.
Pink is the middle predator
Yellow is the top predator
(2,2,2)
All parameters equal 1 a = 0.8
Blue represents the prey.
Yellow is the middle predator
Pink is the top predator
(2,2,2)
a=1.2 , b=c=d=e=f=g=1
Blue represents the prey.
Pink is the middle predator
Yellow is the top predator

All parameters 1
initial condition
(1,2,4)
dx
 ax  bxy
dt

The only parameters that have an
effect on the top predator are a, g,
f and b.
◦ Large values of a and g are beneficial
while large values of f and b
represent extinction.



The parameters that affect the
middle predator are c, d and e.

They do not affect the survival of z.
The survival of the middle predator
is guaranteed as long as the prey is
present.
The top predator is the only one
tha faces extinction when all
species are present.
dy
 cy  dxy  eyz
dt
dz
  fz  gyz
dt
Eigenvalues for (c/d, a/b,0)
( ga  fb) / b
 i ac