Post on 25-Feb-2016
description
A Three Dimensional Lotka-Volterra
SystemKliah Soto
Jorge MunozFrancisco Hernandez
Two-Dimensional Case
Extreme Cases
x 0 dxdt
0
x 0 dydt
cy
and
y 0 dxdt
ax
y 0 dydt
0
Equilibria
dxdt
ax bxy 0
dydt
cy dxy 0
results in(0,0)and
(cd
,ab
)
Solution Curves Solve the system of equations:
dydx
dydt
/dxdt
cy dxyax bxy
y( c dx)x(a by)
a byy
dy c dxx
dx
a byy
dy c dxx
dx
a ln y by c ln x dx k
Solution Curve Solution curve with all parameters = 1
Pink: prey xBlue: predator y
Three Dimensional Case
dxdt
ax bxy
dydt
cy dxy eyz
dzdt
fz gyz
Extremities Case 1: if z=0 then we have the 2
dimensional case Case 2: y=0
dxdt
ax
dydt
0
dzdt
fz
In the absence of the middle predator y, we are left with:
We combine it to one fraction and use separation of variables:
af
Kxz
dxax
dzfz
axfz
dtdx
dtdz
dxdz
11
/
fzdtdz
axdtdx
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 1 0
1
2
3
4
Case 3: x=0
dxdt
0
dydt
cy eyz
dzdt
fz gyz
In the absence of the prey x, we are left with:
dydt
cy eyz
dzdt
fz gyz
We combine it to one fraction and use separation of variables:
Kgyyfezzc
dyygyfzd
zezc
dyygyfdz
zezc
ezcygyfz
dtdy
dtdz
dydz
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:
dxdt
ax bxy 0
dydt
cy dxy eyz 0
dzdt
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:
fzdtdz
axdtdx
x=0 and y=0 since a and f are positive. Again equilibrium (0,0,0).
dxdt
ax bxy
dydt
cy dxy eyz
dzdt
fz gyz
When we consider:
)( gyfzgyzfzdtdz
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:
dxdt
as bsy s(a by) y ab
dydt
cy dsy eyz y( c ds ez ) z ds ce
dzdt
fz gyz z( f gy) y fg
Equilibrium point at (s,a/b=f/g,(ds-c)/e)
Linearize the System by finding the Jacobian
gyfzgyeezdxcyd
xbbyazyxJ
0
0),,(
),,(
),,(
),,(
zyxhgyzfzdtdz
zyxgeyzdxycydtdy
zyxfbxyaxdtdx
zzhy
yhx
xh
dtdy
zzgy
ygx
xg
dtdy
zzfy
yfx
xf
dtdx
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
fc
aJ
000000
)0,0,0( Eigenvalues:
◦ a, -c, -f Eigenvectors: {1,0,0}, {0,1,0}, {0,0,1}
Solution:
5 10 15 20
10 00 0
20 00 0
30 00 0
40 00 0
50 00 0
1 2 3 4 5
10
5
5
10
Unstable x-axis Stable yz-Plane
Equilibrium at (c/d, a/b, 0) Eigenvalues
Eigenvectors:
bgafbaebad
dbcbadcJ
/00/0/
0/0)0,/,/(
ac
aci
bfbga
/)(
}0,,1{
}1)2(,)(,1{ 222222
2
bcacid
ceabdgaabdfgdfbab
cbdagfb 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 predatorYellow 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.
dxdt
ax bxy
dydt
cy dxy eyz
dzdt
fz gyz
aci
bfbga
/)(
Eigenvalues for (c/d, a/b,0)