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$\dot{x}=x\left ( 3-x-2y \right )$ $\dot{y}=y\left ( 2-x-y \right )$

In matrix form: $\begin{bmatrix} \dot{x}\\ \dot{y} \end{bmatrix}$ $=\begin{bmatrix} \left ( 3-x-2y \right ) &0 \\ 0& \left ( 2-x-y \right ) \end{bmatrix}$

$\begin{bmatrix} x\\ y \end{bmatrix}$

To solve for the eigenvalues, we compute the Jacobian of the matrix $A=\begin{bmatrix} \left ( 3-x-2y \right ) &0 \\ 0& \left ( 2-x-y \right ) \end{bmatrix}$

The eigenvalues are $\lambda_{1}=3,\lambda_{2}=2$

The eigenvectors associated with $\lambda_{1}$:

$A-\lambda_{1}I=\begin{bmatrix} \left ( 3-x-2y \right )-3 &0 \\ 0&\left ( 2-x-y \right )-3 \end{bmatrix}$

Then, $\left [ A-\lambda_{1} I|0 \right ]$, we get $\vec{V}=\left \langle 0,0 \right \rangle$

which is incorrect. Could anyone tell me where am I going wrong here?

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  • $\begingroup$ Try solving $Av=\lamba_1v$, where $A$ has been evaluated at $x=0=y$. $\endgroup$ – Mar5bar Apr 3 '16 at 11:46
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What you're doing isn't making any sense, since it's a nonlinear system. It doesn't help that have written it "in matrix form", since the matrix depends on $x$ and $y$.

What you have to do first is to find the equilibrium points, then linearize around each of them. Then you can look at the eigenvalues of the Jacobian matrix at each equilibrium to find out whether it's stable or not.

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  • $\begingroup$ What do you mean by linearise about each of the equilibrium points? I haven't solved for any eigensolution of any non-linear matrix before. Could I request for an example? $\endgroup$ – Mathematicing Apr 3 '16 at 12:04
  • $\begingroup$ Here, for example: math.stackexchange.com/questions/729655/how-do-i-linearize $\endgroup$ – Hans Lundmark Apr 3 '16 at 14:41
  • $\begingroup$ But if this problem is part of some course, you should have seen some explanation of the procedure in your textbook and/or in class. $\endgroup$ – Hans Lundmark Apr 3 '16 at 14:42
  • $\begingroup$ How silly!( of me of course). My notes refer to this as 'computing the small disturbance from fixed point'. $\endgroup$ – Mathematicing Apr 3 '16 at 23:36
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    $\begingroup$ For each fixed point you get a linear system which approximates the behaviour of the nonlinear system near that point. And for these linear systems, you just compute eigenvalues and eigenvectors as usual. $\endgroup$ – Hans Lundmark Apr 4 '16 at 12:03
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You will need to express $\dot x, \dot y$ as linear functions of $x$ and $y$, for example $$\left[\begin{array}{c}\dot x\\\dot y\end{array}\right] = A\left[\begin{array}{c}x\\y\\1\end{array}\right]$$ for a matrix $A$. You can for example make a first order Taylor expansion of the original expressions (it is probably the easiest way to do). So assume you are close to the point $\left[\begin{array}{c} x_0\\ y_0\end{array}\right]$ and make a linear Taylor approximation there.

Note : The $A$ matrix should be a function of this point $\left[\begin{array}{c} x_0\\ y_0\end{array}\right]$.

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