Science:Math Exam Resources/Courses/MATH215/December 2013/Question 06 (a)

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: $\left\{{\begin{array}{c}{\frac {dx}{dt}}=x\left(3-x-y\right)\\{\frac {dy}{dt}}=y\left(1+x-y\right)\end{array}}\right.$ (a) Determine all the critical points and classify their type and stability.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If $\displaystyle x'=0$ then either $x=0$ or $x=3-y$. Using $\displaystyle x=0$ in $y'=0$ we get $y=0,1$. This gives two critical points: (0,0) and (0,1). Using $\displaystyle x=3-y$ 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 ${\begin{pmatrix}x'\\y'\end{pmatrix}}={\begin{pmatrix}3x-x^{2}-xy\\y+xy-y^{2}\end{pmatrix}}$ Labelling $\displaystyle f(x,y)=3x-x^{2}-xy$ and $\displaystyle g(x,y)=y+xy-y^{2}$, the Jacobian of the system at $(x,y)$ is $J={\begin{pmatrix}f_{x}&f_{y}\\g_{x}&g_{y}\end{pmatrix}}={\begin{pmatrix}3-2x-y&-x\\y&1+x-2y\end{pmatrix}}$ 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 $\lambda ^{\pm }={\frac {-3\pm {\sqrt {9-16}}}{2}}=-3/2\pm {\sqrt {7}}i/2.$ 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).