Consider the ODE system $$ x'=-x+xy.\qquad y'=y+x^2. $$ It has equilibrium $(0,0)$.

What's the linearization in $(0,0)$? The linearization of the second equation is just $y'=y$. But what's the linearization of the first one? Or is this already linear?

Is the linearization matrix just $$ A=\begin{pmatrix}y-1 & 0\\1 & 0\end{pmatrix}? $$


Not quite, the system is not linear since it contains non-linear terms such as $x^2$. Also,

$$A =\begin{pmatrix}\frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y}\\\frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y}\end{pmatrix} = \begin{pmatrix}y-1 & x\\2x & 1\end{pmatrix}$$

Evaluating at $(0,0)$ gives $$A = \begin{pmatrix}-1 & 0\\0 & 1\end{pmatrix}$$

  • $\begingroup$ Thats the Jacobimatrix. $\endgroup$ – Rhjg Jan 15 '16 at 21:22
  • $\begingroup$ Which is how you linearize the system. $\endgroup$ – fosho Jan 15 '16 at 21:25

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