Question

Consider the following system of differential equations representing a prey and predator
population model
x y
dt
dx
= −
2

x y
dt
dy
= +
i) Identify all the real critical points of the system
ii) Obtain the type and stability of these critical points.

Answer 1

$\frac{d x}{ d t} = x − x y \ ,\frac{ d y}{ d t} = y + x y$

If we look at the first equation we note that the population x only grows owing to itself (+ x term) and diminishes in interacting with population y (- xy term)

On the other hand, looking at the second equation we note that the population y grows owing to itself (+ y term) and also grows while interacting with population x (+ xy term).

Owing to the two paragraphs above, population x is the population of a prey while population y is the population of a predator. Hence what we have here is a predator-prey model.

To find equilibrium points of the above model we solve the system of equations obtained by setting the right hand sides of the two equations in the system simultaneously equal to zero as follows:

$x − x y = f ( x , y ) = 0 \ \ ...…….( 1 )$

$y + x y = g ( x , y ) = 0\ \ ...……. ( 2 )$

Then from (1) get $x − x y = 0 ⟹ x ( 1 − y ) = 0 ⟹ x = 0 \ a n d\ y = 1$

Start with $x = 0$ and substitute in (2) to get $y + 0 ( y ) = 0 ⟹ y = 0.$

Hence our first equilibrium point is $E_ 1 = ( 0 , 0 ) .$

Next set $y = 1$ in (2) to arrive at $1 + x = 0 ⟹ x = − 1.$ Since populations cannot be negative we disregard this value of x and any equilibrium point that may follow.

Therefore our only equilibrium point is$E _1 = ( 0 , 0 ) .$

To study the stability of E₁ we first construct the Jacobian matrix J of the system as follows:

$J=\begin{bmatrix} f_x& f_y \\ g_x & g_y \end{bmatrix}=\begin{bmatrix} 1-y & -x \\ y & 1+x \end{bmatrix}$

This is following by evaluating the Jacobian matrix at the values of the equilibrium point(s) to get

$J(E_1)=\begin{bmatrix} 1-0 & -0 \\ 0 & 1+0 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

One may observe that the matrix above has all zeros as its non-diagonal elements. Hence its eigenvalues will be its diagonal elements. Hence the eigenvalues of $J ( E_ 1 )$are 1 and 1.

Since both eigenvalues are positive, the equilibrium point$E_ 1 = ( 0 , 0 )$ is unstable