Post on 16-Dec-2015
transcript
A Three Dimensional Lotka-Volterra
SystemKliah Soto
Jorge MunozFrancisco Hernandez
Two-Dimensional Case
Extreme Cases
x 0 dx
dt0
x 0 dy
dt cy
and
y 0 dx
dtax
y 0 dy
dt0
Equilibria
dx
dtax bxy 0
dy
dt cy dxy 0
results in
(0,0)
and
(c
d,a
b)
Solution Curves Solve the system of equations:
dy
dxdy
dt/dx
dt
cy dxyax bxy
y( c dx)x(a by)
a byy
dy c dxx
dx
a byy
dy c dxx
dx
aln y by c ln x dx k
Solution Curve Solution curve with all parameters = 1
Pink: prey x
Blue: predator y
Three Dimensional Case
dx
dtax bxy
dy
dt cy dxy eyz
dz
dt fz gyz
Extremities Case 1: if z=0 then we have the 2
dimensional case Case 2: y=0
dx
dtax
dy
dt0
dz
dt fz
In the absence of the middle predator y, we are left with:
We combine it to one fraction and use separation of variables:
a
f
Kxz
dxax
dzfz
ax
fz
dt
dx
dt
dz
dx
dz
11
/
fzdt
dz
axdt
dx
species z approaches zero as t goes to infinity, and species x exponentially grows as t approaches infinity.
Phase Portrait and Solution Curve when y=0
The blue curve represents the prey, while the red curve represents the predator.
0 2 4 6 8 10
1
2
3
4
Case 3: x=0
dx
dt0
dy
dt cy eyz
dz
dt fz gyz
In the absence of the prey x, we are left with:
dy
dt cy eyz
dz
dt fz gyz
We combine it to one fraction and use separation of variables:
Kgyyfezzc
dyy
gyfzd
z
ezc
dyy
gyfdz
z
ezc
ezcy
gyfz
dt
dy
dt
dz
dy
dz
lnln
)(
)(/
species y and z will approach zero as t approaches infinity.
Phase Portrait and Solution Curve when x=0
The blue curve represents the top predator, while the red curve represents the middle predator.
1 2 3 4 5t
0 .2
0 .4
0 .6
0 .8
1 .0
yz
Equilibria Set all three equations equal to zero to
determine the equilibria of the system:
dx
dtax bxy 0
dy
dt cy dxy eyz 0
dz
dt fz gyz 0
Cases of Equilibria 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:
fzdt
dz
axdt
dx
x=0 and y=0 since a and f are positive. Again equilibrium (0,0,0).
dx
dtax bxy
dy
dt cy dxy eyz
dz
dt fz gyz
When we consider:
)( gyfzgyzfzdt
dz
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
dtas bsy s(a by) y
a
bdy
dt cy dsy eyz y( c ds ez ) z
ds ce
dz
dt fz gyz z( f gy) y
f
g
Equilibrium point at (s,a/b=f/g,(ds-c)/e)
Linearize the System by finding the Jacobian
gyfzg
yeezdxcyd
xbbya
zyxJ
0
0
),,(
),,(
),,(
),,(
zyxhgyzfzdt
dz
zyxgeyzdxycydt
dy
zyxfbxyaxdt
dx
zz
hy
y
hxx
h
dt
dy
zz
gy
y
gx
x
g
dt
dy
zz
fy
y
fxx
f
dt
dx
Where the partial derivatives are evaluated at the equilibrium point
Center Manifold Theorem
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
Equilibrium at (0,0,0)
One-dimensional unstable manifold: curve x-axis
Two-dimensional stable manifold: surface yz- Plane
f
c
a
J
00
00
00
)0,0,0( Eigenvalues:
◦ a, -c, -f Eigenvectors: {1,0,0}, {0,1,0}, {0,0,1}
Solution:
5 1 0 1 5 2 0
1 0 0 00
2 0 0 00
3 0 0 00
4 0 0 00
5 0 0 00
1 2 3 4 5
10
5
5
10
Unstable x-axis Stable yz-Plane
Equilibrium at (c/d, a/b, 0) Eigenvalues
Eigenvectors:
bgaf
baebad
dbc
badcJ
/00
/0/
0/0
)0,/,/(
ac
aci
bfbga
/)(
}0,,1{
}1
)2(,)(,1{2
222222
bc
acid
ceabdgaabdfgdfbab
cb
dagfb cd
One-Dimensional invariant curve:◦ Stable if ga<fb◦ Unstable ga>fb
Two-Dimensional center manifold Three-dimensional center
manifold◦ If ga=fb
aci
bfbga
/)(
Stable Equilibrium ga<fb
All parameters equal 1 a = 0.8
Blue represents the prey.Pink is the middle predatorYellow is the top predator
(2,2,2)
a=1.2 , b=c=d=e=f=g=1
Unstable Equilibrium ga>fb
Blue represents the prey.Yellow is the middle predatorPink is the top predator
(2,2,2)
Blue represents the prey.
Pink is the middle predator
Yellow is the top predator
Three Dimensional Manifold ga=fb
All parameters 1 initial condition (1,2,4)
Conclusion 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.
dx
dtax bxy
dy
dt cy dxy eyz
dz
dt fz gyz
aci
bfbga
/)(
Eigenvalues for (c/d, a/b,0)