5 Predator prey model

Question 1

Predator prey model
assumptions and model

Answer 1

N(t)= number of preys (for example rabbits) at time t,
P(t) = number of predators (for example foxes) at time t.

The prey population grows if there is no predator.
ˆ The predators die out if there is no prey.
ˆ The rate of contact between predator and prey is proportional to N(t)P(t)
Contact is good for the predator but bad for the prey.

dN(t)/dt= aN(t)-bN(t)P(t)

dP(t)/dt=-dP(t)+cN(t)P(t)

Question 2

Predator prey model change of vars
u(t) = (c/d)N(t),
v(t) = (b/a)P(t),
τ = at, α =d/a

dN(t)/dt= aN(t)-bN(t)P(t)

dP(t)/dt=-dP(t)+cN(t)P(t)

model becomes

Answer 2

du(τ )/dτ
= u(τ )(1 − v(τ ))

dv(τ )/dτ
= −αv(τ )(1 − u(τ ))

SS same for new system
change of variables rescales system and rescales SS
if not possible to get an explicit sol we study the dynamics and implicit sols
we can solve these above

Question 3

Can we SOLVE?
change of vars predator prey model

du(τ )/dτ
= u(τ )(1 − v(τ ))

dv(τ )/dτ
= −αv(τ )(1 − u(τ ))

Answer 3

Implicit sols
How about

du(τ)/dv(v)
= [u(τ )(1 − v(τ ))]/[−αv(τ )(1 − u(τ ))]
dv/du=-α[(v-uv)/(u-uv)]
seperate vars

v − log v = −α(u − log u) + c.

Question 4

what trick can we use with dv/dt and du/dt in predator prey?

Answer 4

Even if one cannot find an explicit solution for the ODEs of interest, First, we
can imagine ourselves sitting in the two-dimensional (u, v) space, with the differential equations
telling us which direction to move. By analysing the sign of the derivatives of the ODEs above,
one can see how the system evolves over time, and draw the dynamics.
We can, in fact, go beyond this and find an implicit solution using the following trick:

find dv/du and solve

Question 5

solving pred prey dv/du
v − log v = −α(u − log u) + c.
what does the contour map given initial values of u and v to calculate c?

Answer 5

rather than approaching any fixed point, the trajectories of the model cycle around a path.
change of vars model
nullclines:
u nullclines: u=0 and v=1
v-nullclines: v=0 u=1
drawing these and noting the signs of derivatives for u-v plane diagram
cycles around anticlockwise

Trajectories are spirals or cycles depending on the stability

Thus, it is interesting to calculate the average value of u and of v around one
cycle.

Question 6

did go through? Thus, it is interesting to calculate the average value of u and of v around one
cycle. predator prey

Answer 6

That is, we conclude that ¯v = 1 (which is not obvious if you look at trajectories. By a similar
calculation, ¯u = 1.

average value ¯v= (1/T) integral v(t).dt where T is the time it takes to go around,

we dont know u,v or T but use du/dt to integrate so we get it in terms of integral that makes ¯v

Question 7

Example two ODES: Predator-Prey model
How do we analyse the steady states of the rescaled model?

du(τ )/dτ
= u(τ )(1 − v(τ ))

dv(τ )/dτ
= −αv(τ )(1 − u(τ ))

Answer 7

du(τ )/dτ
= f(u(τ ),v(τ ))

dv(τ )/dτ
= g(u(τ ),v(τ ))

2 SS
(0,0) and (1,1)
to analyse using partial derivatives
∂f/∂u = 1-v ∂f/∂v = -u
∂g/∂u =αv ∂g/∂v = -α(1-u)

J_(0,0) =
(1 0)
(0 -α )
eigenvalues λ=1 and λ=-α so UNSTABLE SADDLE

J_(1,1) =
(0 -1)
(α 0 )
trace=0
det =α>0
λ^2+α=0
λ=±√-α so λ=±i√α
neither unstable or stable CENTRE

Question 7

predator prey model ss for
orignal model
and change of vars model

Answer 7

dN/dt=0 and dP/dt=0
give:
(N,P)=(0,0)
(N,P)=(d/c,a/b)

rescaled model
dN/dt=0 and dP/dt=0
give:
(u,v)=(0,0)
(u,v)=(1,1)