Predator-Prey Dynamics and Population Models Explained

Kliah Soto Jorge Munoz Francisco Hernandez

x = 0dx dt= 0 x = 0 dy dt= cy and y = 0 dx dt= ax y = 0 dy dt= 0

dx dt= ax bxy = 0 dy dt= cy + dxy = 0 results in (0,0) and (c d,a b)

Solve the system of equations: dt= cy + dxy ax bxy =y( c + dx) x(a by) dy dx=dy dt/dx a by y dy = c + dx dx x a by y c + dx x = dy dx alny by +clnx dx = k

Solution curve with all parameters = 1 Pink: prey x Blue: predator y

dx dt= ax bxy dy dt= cy + dxy eyz dz dt= fz + gyz

Case 1: if z=0 then we have the 2 dimensional case Case 2: y=0 dx dt= ax dy dt= 0 dz dt= fz

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 = = / species z approaches zero as t goes to infinity, and species x exponentially grows as t approaches infinity. dx dt dt ax 1 1 = dz dx fz ax f = z Kx a

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 dt= 0 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: dt species y and z will approach zero as t approaches infinity. ( + ( ) dz dz dy z f gy = = / ) dy dt y c ez + c ez f gy = dz dy z y + c ez f gy = d z dy z y + = ln ln c z ez f y gy K

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 dt= ax bxy = 0 dy dt= cy + dxy eyz = 0 dz dt= fz+ gyz = 0

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) dx dt= ax bxy dy dt= cy + dxy eyz When y=0 System reduces to: dz dt= fz + gyz 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 dt= as bsy = s(a by) y =a dy dt= cy + dsy eyz = y( c + ds ez) z =ds c dz dt= fz + gyz = z( f + gy) y =f b e g Equilibrium point at (s,a/b=f/g,(ds-c)/e)

dx f f f dx = + + x y z = = ( , , ) ax bxy f x y z dt x y z dt dy g g g = + + x y z dt x y z dy = + = ( , , ) cy dxy eyz g x y z dy h h h dt = + + x y z dt x y z dz = + = ( , , ) fz gyz h x y z Where the partial derivatives are evaluated at the equilibrium point dt yd 0 0 a by xb = + f ( , , ) J x y z c dx ez ye + zg gy

Number of eigenvalues: Dimension of the manifold Manifold is tangent to the eigenspace spanned by the eigenvectors of their corresponding eigenvalues Real part of the eigenvalues Positive: Unstable Negative: Stable Zero: Center

0 0 a 0 Eigenvalues: a, -c, -f Eigenvectors: {1,0,0}, {0,1,0}, {0,0,1} = ) 0 , 0 , 0 ( 0 J c 0 0 f One-dimensional unstable manifold: curve x-axis Two-dimensional stable manifold: surface yz- Plane

10 50000 5 40000 30000 1 2 3 4 5 20000 5 10000 10 5 10 15 20 Unstable x-axis Stable yz-Plane

ac Eigenvalues ( 0 / 0 bc d / ) ga fb b = f ( / , / ) 0 , / 0 / J c d a b ad b ae b i ac + 0 0 / ga b Eigenvectors: 1 2 d + + 2 2 2 2 2 cd ( , 1 { ) ( , 2 ) } fb ag ab b df abdfg a dg 2 b c ab ce id ac , 1 { } 0 , bc

( / ) ga fb b One-Dimensional invariant curve: Stable if ga<fb Unstable ga>fb i ac 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 dt= ax bxy dy dt= cy + dxy eyz 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. dz dt= fz + gyz Eigenvalues for (c/d, a/b,0) ( / ) ga fb b i ac