MATH215 December 2013
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Question 06 (a)

Suppose the interaction between two species ($x=x(t)$ is the “prey” while $y=y(t)$ is the predator) can be modelled by the autonomous system:
 $\left\{{\begin{array}{c}{\frac {dx}{dt}}=x\left(3xy\right)\\{\frac {dy}{dt}}=y\left(1+xy\right)\end{array}}\right.$
(a) Determine all the critical points and classify their type and stability.

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

At a critical point, both dx/dt and dy/dt must be 0. Be careful to consider all possible cases.

Hint 2

To determine the stability, consider the eigenvalues of the Jacobian matrix
 $J={\begin{pmatrix}f_{x}&f_{y}\\g_{x}&g_{y}\end{pmatrix}}$

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Solution

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If $\displaystyle x'=0$ then either $x=0$ or $x=3y$.
 Using $\displaystyle x=0$ in $y'=0$ we get $y=0,1$. This gives two critical points: (0,0) and (0,1).
 Using $\displaystyle x=3y$ in $y'=0$ we get $y=0,2$. In this case, x=3 when y=0, and x=1 when y=2. This gives two more critical points (1,2) and (3,0).
Considering the full system, we have
 ${\begin{pmatrix}x'\\y'\end{pmatrix}}={\begin{pmatrix}3xx^{2}xy\\y+xyy^{2}\end{pmatrix}}$
Labelling $\displaystyle f(x,y)=3xx^{2}xy$ and $\displaystyle g(x,y)=y+xyy^{2}$, the Jacobian of the system at $(x,y)$ is
 $J={\begin{pmatrix}f_{x}&f_{y}\\g_{x}&g_{y}\end{pmatrix}}={\begin{pmatrix}32xy&x\\y&1+x2y\end{pmatrix}}$
The Jacobian evaluated at the critical points gives us information about the local stability of the critical points.
At (x,y) = (0,0)
 $J(0,0)={\begin{pmatrix}3&0\\0&1\end{pmatrix}}$ which has eigenvalues of +1 and +3. Thus (0,0) is an unstable node (both eigenvalues are real and positive).
At (x,y) = (0,1)
 $J(0,1)={\begin{pmatrix}2&0\\1&1\end{pmatrix}}$ which has eigenvalues of +2 and 1 (being lower triangular, the eigenvalues are the diagonal entries). As the eigenvalues are real and of opposite sign, (0,1) is a saddle point.
At (x,y) = (3,0)
 $J(3,0)={\begin{pmatrix}3&3\\0&4\end{pmatrix}}$ having eigenvalues of 3 and +4 (the matrix is upper triangular and its eigenvalues are the diagonal entries). We thus have (3,0) is a saddle point.
At (x,y) = (1,2)
 $J(1,2)={\begin{pmatrix}1&1\\2&2\end{pmatrix}}$. The eigenvalues satisfy the characteristic polynomial $\lambda ^{2}+3\lambda +4=0$ so $\lambda ^{\pm }={\frac {3\pm {\sqrt {916}}}{2}}=3/2\pm {\sqrt {7}}i/2.$ These eigenvalues have a nonzero imaginary part giving rise to a spiral/ellipse. As there real components are negative, we have a stable spiral at (1,2).


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