Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 6B (Mathematical Biology)

Table of contents
- i 6B
  - Solution i
- ii 6B
  - Solution ii
- iii 6B
  - Solution iii

i

Solution

At a nonzero fixed point, both bracketed per-capita growth rates vanish:

- ϵ_{1} + α N_{2}^{*} = 0, ϵ_{2} - α N_{1}^{*} = 0.

Hence the coexistence equilibrium of this Lotka-Volterra predator-prey model is

(N_{1}^{*}, N_{2}^{*}) = (\frac{ϵ _{2}}{α}, \frac{ϵ _{1}}{α}) .

ii

Solution

Since

N_{i} = N_{i}^{*} e^{x_{i}}

, division of each population equation by

N_{i}

gives

\overset{x}{˙}_{1} = ϵ_{1} (e^{x_{2}} - 1), \overset{x}{˙}_{2} = - ϵ_{2} (e^{x_{1}} - 1) .

Define the Hamiltonian function

H (x_{1}, x_{2}) = ϵ_{2} (e^{x_{1}} - x_{1} - 1) + ϵ_{1} (e^{x_{2}} - x_{2} - 1)

and the antisymmetric matrix

K = (0 - 1 10) .

Then

\nabla H = (ϵ_{2} (e^{x_{1}} - 1) ϵ_{1} (e^{x_{2}} - 1)), \dot{x} = K \nabla H .

iii

Solution

Along a solution, the chain rule and antisymmetry of

K

give

\frac{d H}{d t} = (\nabla H)^{T} \dot{x} = (\nabla H)^{T} K \nabla H = 0,

because

v^{T} K v = 0

for every real vector

v

. Therefore

H (x_{1} (t), x_{2} (t)) is constant .

The closed level curves around the strict minimum at the coexistence equilibrium are the periodic population cycles of the Lotka-Volterra predator-prey model.

 Ancestors (9)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version