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

MATH215 December 2013

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  

   Other MATH215 Exams

  •  April 2013  •  December 2011  •  December 2013  •  

Question 06 (b)
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: $\left\{{\begin{array}{c}{\frac {dx}{dt}}=x\left(3-x-y\right)\\{\frac {dy}{dt}}=y\left(1+x-y\right)\end{array}}\right.$ (b) Sketch a phase portrait for the nonlinear system by first sketching a few trajectories near each critical point.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
To find some characteristic trajectories near each critical point, you'll need to compute the eigenvectors of the linearized systems - this will tell you what direction the trajectories move.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

We will need to analyze the linearized system at each critical point. At (x,y) = (0,0) $J={\begin{pmatrix}3&0\\0&1\end{pmatrix}}$ which has eigenvectors $(1,0)^{T}$ and $(0,1)^{T}$ with corresponding eigenvalues 3 and 1 respectively. The local solution structure centred at (0,0) is ${\begin{pmatrix}x\\y\end{pmatrix}}=C_{1}{\begin{pmatrix}1\\0\end{pmatrix}}e^{3t}+C_{2}{\begin{pmatrix}0\\1\end{pmatrix}}e^{t}$ Along both eigenvectors, the trajectories move away, with the $(1,0)^{T}$ direction being dominant, since it is multiplied by the larger term, e3t. At (x,y) = (0,1) $J={\begin{pmatrix}2&0\\1&-1\end{pmatrix}}.$ To find the eigenvector corresponding to $\lambda =2$, we seek the solution to $(J-2I){\vec {v}}={\vec {0}}$ so that ${\begin{pmatrix}0&0\\1&-3\end{pmatrix}}{\vec {v}}={\vec {0}}$ Or $v_{1}-3v_{2}=0$, so we take ${\vec {v}}=(3,1)^{T}$. To find the eigenvector corresponding to $\lambda =-1$, we seek the solution to $(J-(-1)I){\vec {v}}={\vec {0}}$ so that ${\begin{pmatrix}1&0\\1&0\end{pmatrix}}{\vec {v}}={\vec {0}}$ Or v1 = 0, v2 = anything, so we take ${\vec {v}}=(0,1)^{T}$. The local solution structure centred at (0,1) is ${\begin{pmatrix}x\\y\end{pmatrix}}=C_{1}{\begin{pmatrix}3\\1\end{pmatrix}}e^{2t}+{\begin{pmatrix}0\\1\end{pmatrix}}e^{-t}$ The solution grows exponentially along the $(3,1)^{T}$ direction and decays exponentially along the $(0,1)^{T}$ direction. Starting anywhere other than along the $(0,1)^{T}$ vector, the solution will approach the $(3,1)^{T}$ vector. At (x,y) = (3,0) $J={\begin{pmatrix}-3&-3\\0&4\end{pmatrix}}.$ To find the eigenvector corresponding to $\lambda =-3$, we seek the solution to $(J-(-3)I){\vec {v}}={\vec {0}}$ so that ${\begin{pmatrix}0&-3\\0&7\end{pmatrix}}{\vec {v}}={\vec {0}}$ Or v2 = 0, v1 = anything, so we take ${\vec {v}}=(1,0)^{T}$. To find the eigenvector corresponding to $\lambda =4$, we seek the solution to $(J-4I){\vec {v}}={\vec {0}}$ so that ${\begin{pmatrix}-7&-3\\0&0\end{pmatrix}}{\vec {v}}={\vec {0}}$ Or $-7v_{1}-3v_{2}=0$, so we take ${\vec {v}}=(-3,7)^{T}$. The local solution structure centred at (3,0) is ${\begin{pmatrix}x\\y\end{pmatrix}}=C_{1}{\begin{pmatrix}1\\0\end{pmatrix}}e^{-3t}+{\begin{pmatrix}-3\\7\end{pmatrix}}e^{4t}$ The solution grows exponentially along the $(-3,7)^{T}$ direction and decays exponentially along the $(1,0)^{T}$ direction. Starting anywhere other than along the $(1,0)^{T}$ vector, the solution will approach the $(-3,7)^{T}$ vector. At (x,y) = (1,2) $J={\begin{pmatrix}-1&-1\\2&-2\end{pmatrix}}$, which as we saw in part (a) has complex eigenvalues with negative real part. By plugging in a point, say $(1,0)^{T}$ into the local linear system we can determine the direction of the spiral. ${\begin{pmatrix}-1&-1\\2&-2\end{pmatrix}}{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}-1\\2\end{pmatrix}}$ which points up and to the left at (1,0). Therefore the spiral is counterclockwise. This is all put together in the very rough global phase portrait.