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Using the Jacobian from 7 (b) we find the linearization of the system in the critical points.
Critical point
![{\displaystyle J(0,0)={\begin{pmatrix}2&0\\0&1\end{pmatrix}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8fffed41df80290675ac24ee5d10ed048b06ca0e)
- This matrix is diagonal and so the diagonal elements are the eigenvalues
- This matrix is positive definite since its two eigenvalues
are positive. All positive eigenvalues indicate that the critical point is an unstable node.
- The eigenvector for the eigenvalue
is .
- The eigenvector for the eigenvalue
is .
- The linearized critical point in the phase space is on the following figure:
![Phase space for linearization of critical point (0,0)](//wiki.ubc.ca/images/thumb/9/9a/Fig_215_2011_7c1.jpg/400px-Fig_215_2011_7c1.jpg)
Critical point
![{\displaystyle J(2,0)={\begin{pmatrix}-2&-2\\0&-1\end{pmatrix}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/a6364ec381a29eba2072166f12a4375cfbe98cbe)
- The eigenvalues are
, 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
is 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
:
Hence,
- We find the the eigenvector for
is .
- The linearized critical point in the phase space is on the following figure:
Critical point
![{\displaystyle J(1,1)={\begin{pmatrix}-1&-1\\-1&0\end{pmatrix}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/1e471c065afab86f9aa71de888448783479ef741)
- For the eigenvalues we calculate
- The eigenvalue for
is the null space of
so that the eigenvector is .
- The eigenvalue for
is the null space of
and the eigenvector is .
- Because one eigenvalue is positive, the other negative, the critical point is not asymptotically stable. Along the eigenvector for
, the solution moves away from the critical point, along the eigenvector for , 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:
![Phase space for linearization of critical point (1,1)](//wiki.ubc.ca/images/thumb/6/62/Fig_215_2011_7c3.jpg/400px-Fig_215_2011_7c3.jpg)
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