Science:Math Exam Resources/Courses/MATH215/December 2011/Question 07 (c)/Solution 1

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: