Science:Math Exam Resources/Courses/MATH215/December 2011/Question 07 (c)

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MATH215 December 2011

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

Other MATH215 Exams

• April 2013 • December 2011 • December 2013 •

Question 07 (c)
Consider the system of equations $\begin{aligned} x^{'} & = x (2 - x - y) \\ y^{'} & = y (1 - x) \end{aligned}$ (c) Classify each critical (fixed) point, and sketch the phase portrait of the linearized system near each critical point.

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Hint 1
For each of the three equilibrium points, calculate the eigenvalues of the corresponding Jacobian matrix.

Hint 2
For your plots, find the eigenvectors corresponding to these eigenvalues. Remember that solutions which start on an eigenvector, stay on this eigenvector.

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Solution
Using the Jacobian from 7 (b) we find the linearization of the system in the critical points. Critical point $y_{0} = (0, 0)$ $J (0, 0) = (\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}) .$ This matrix is diagonal and so the diagonal elements are the eigenvalues This matrix is positive definite since its two eigenvalues $1, 2$ are positive. All positive eigenvalues indicate that the critical point $(0, 0)$ is an unstable node. The eigenvector for the eigenvalue $1$ is $(\begin{matrix} 0 \\ 1 \end{matrix})$ . The eigenvector for the eigenvalue $2$ is $(\begin{matrix} 1 \\ 0 \end{matrix})$ . The linearized critical point in the phase space is on the following figure: Critical point $y_{0} = (2, 0)$ $J (2, 0) = (\begin{matrix} - 2 & - 2 \\ 0 & - 1 \end{matrix}) .$ The eigenvalues are $- 2, - 1$ , because a triangular matrix has its eigenvalues on the diagonal. The eigenvalues are both negative and hence, the matrix is negative definite and this critical point is a stable node. The eigenvector for the eigenvalue $- 2$ is $(\begin{matrix} 1 \\ 0 \end{matrix})$ because of the triangular shape of the matrix. For the other eigenvector we need to work a little more and calculate the null space of $J (2, 0) - (- 1 I)$ : $(J (2, 0) + 1 I) (v_{1}, v_{2})^{T} = (\begin{matrix} - 1 & - 2 \\ 0 & 0 \end{matrix}) (\begin{matrix} v_{1} \\ v_{2} \end{matrix}) = (\begin{matrix} - v_{1} - 2 v_{2} \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ Hence, $v_{1} = - 2 v_{2}$ We find the the eigenvector for $- 1$ is $(\begin{matrix} - 2 \\ 1 \end{matrix})$ . The linearized critical point in the phase space is on the following figure: Critical point $y_{0} = (1, 1)$ $J (1, 1) = (\begin{matrix} - 1 & - 1 \\ - 1 & 0 \end{matrix}) .$ For the eigenvalues we calculate $d e t (\begin{matrix} - 1 - λ & - 1 \\ - 1 & - λ \end{matrix}) = λ^{2} + λ - 1 = 0$ $λ = \frac{- 1 \pm \sqrt{5}}{2}$ The eigenvalue for $\frac{- 1 - \sqrt{5}}{2}$ is the null space of $(\begin{matrix} - 1 + \frac{1 + \sqrt{5}}{2} & - 1 \\ - 1 & \frac{1 + \sqrt{5}}{2} \end{matrix}) (\begin{matrix} v_{1} \\ v_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ $v_{1} = \frac{1 + \sqrt{5}}{2} v_{2}$ so that the eigenvector is $(\begin{matrix} 1 + \sqrt{5} \\ 2 \end{matrix})$ . The eigenvalue for $\frac{- 1 + \sqrt{5}}{2}$ is the null space of $(\begin{matrix} - 1 + \frac{1 - \sqrt{5}}{2} & - 1 \\ - 1 & \frac{1 - \sqrt{5}}{2} \end{matrix}) (\begin{matrix} v_{1} \\ v_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ $v_{1} = \frac{1 - \sqrt{5}}{2} v_{2}$ and the eigenvector is $(\begin{matrix} 1 - \sqrt{5} \\ 2 \end{matrix})$ . Because one eigenvalue is positive, the other negative, the critical point is not asymptotically stable. Along the eigenvector for $\frac{- 1 + \sqrt{5}}{2}$ , the solution moves away from the critical point, along the eigenvector for $\frac{- 1 - \sqrt{5}}{2}$ , the solution moves towards the critical point. Since the eigenvalues do not have the same sign, the matrix is indefinite and the equilibrium is a saddle point. See the following figure: