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

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)  •  

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  •  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.

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
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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Using the Jacobian from 7 (b) we find the linearization of the system in the critical points. Critical point $\ y_{0}=(0,0)$ $J(0,0)={\begin{pmatrix}2&0\\0&1\end{pmatrix}}.$ 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 ${\begin{pmatrix}0\\1\end{pmatrix}}$. The eigenvector for the eigenvalue $\ 2$ is ${\begin{pmatrix}1\\0\end{pmatrix}}$. The linearized critical point in the phase space is on the following figure: Critical point $\ y_{0}=(2,0)$ $J(2,0)={\begin{pmatrix}-2&-2\\0&-1\end{pmatrix}}.$ 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 ${\begin{pmatrix}1\\0\end{pmatrix}}$ 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 $\displaystyle J(2,0)-(-1I)$: $(J(2,0)+1I)(v_{1},v_{2})^{T}={\begin{pmatrix}-1&-2\\0&0\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}={\begin{pmatrix}-v_{1}-2v_{2}\\0\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}$ Hence, $\ v_{1}=-2v_{2}$ We find the the eigenvector for $\ -1$ is ${\begin{pmatrix}-2\\1\end{pmatrix}}$. The linearized critical point in the phase space is on the following figure: Critical point $\ y_{0}=(1,1)$ $J(1,1)={\begin{pmatrix}-1&-1\\-1&0\end{pmatrix}}.$ For the eigenvalues we calculate $det{\begin{pmatrix}-1-\lambda &-1\\-1&-\lambda \end{pmatrix}}=\lambda ^{2}+\lambda -1=0$ $\lambda ={\frac {-1\pm {\sqrt {5}}}{2}}$ The eigenvalue for ${\frac {-1-{\sqrt {5}}}{2}}$ is the null space of ${\begin{pmatrix}-1+{\frac {1+{\sqrt {5}}}{2}}&-1\\-1&{\frac {1+{\sqrt {5}}}{2}}\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}$ $v_{1}={\frac {1+{\sqrt {5}}}{2}}v_{2}$ so that the eigenvector is ${\begin{pmatrix}1+{\sqrt {5}}\\2\end{pmatrix}}$. The eigenvalue for ${\frac {-1+{\sqrt {5}}}{2}}$ is the null space of ${\begin{pmatrix}-1+{\frac {1-{\sqrt {5}}}{2}}&-1\\-1&{\frac {1-{\sqrt {5}}}{2}}\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}$ $v_{1}={\frac {1-{\sqrt {5}}}{2}}v_{2}$ and the eigenvector is ${\begin{pmatrix}1-{\sqrt {5}}\\2\end{pmatrix}}$. 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: