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Consider the system $$ \dot{x}=y,\qquad\dot{y}=x+x^2-y. $$ It has two equilibria, namely $(0,0)$ and $(-1,0)$.

I would like to linearize the system in both equilibria.

My start is to set $\Delta x=x-x_0,\qquad \Delta y=y-y_0$. Then $$ \dot{\Delta x}=\dot{x}-\dot{x_0}=y-y_0=\Delta y $$ and $$ \dot{\Delta y}=\dot{y}-\dot{y_0}=\Delta x-\Delta y+x^2-x_0^2. $$

How can I get rid of the summand $x^2-x_0^2$?

Can I approximate the function $f(x)=x^2$ by Taylor, getting $x^2-x_0^2\approx 2x_0\Delta x$?

I then would get the linearization matrices $$ \begin{pmatrix}0 & 1\\ 1 & -1\end{pmatrix}\text{ for }(0,0) $$ and, similarly, $$ \begin{pmatrix}0 & 1\\-1 & -1\end{pmatrix}\text{ for }(-1,0). $$

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  • $\begingroup$ You also can write $x^2 - x_0^2 = (x-x_0)(x+x_0) = \Delta x ( (x-x_0) + 2x_0) = \Delta x (\Delta x + 2x_0)$ and after that throw away all terms of higher order in $\Delta x$. $\endgroup$ – Evgeny Jan 3 '16 at 13:42
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I will do the general case.

Let $F:\mathbb{R}^n\to\mathbb{R}^n$ be a smooth map. Consider the differential equation on $\mathbb{R}^n$ given by $$\dot{z} = F(z).$$ An equilibrium point is a point $z_0\in\mathbb{R}^n$ where $F(z_0)=0$. This implies that $z(t)=z_0$ is a solution of the differential equation. The linearization of the differential equation at any point is given by taking the Taylor series of $F$ in the right hand side and cutting it off after the linear term. If you are at an equilibrium point, then the constant term is zero and you get $$\dot{z}=D_{z_0}F(z).$$ This gives a good approximation for the behavior of solutions of the original differential equation near an equilibrium point.

In your special case, we have $$F\pmatrix{x\\y} = \pmatrix{y\\x^2+x-y}.$$ Therefore, the matrices you obtain are $$D_{(0,0)}F = \pmatrix{0&1\\1&-1},\qquad D_{(-1,0)}F=\pmatrix{0&1\\-1&-1},$$ which agrees with your solution.

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  • $\begingroup$ I won't say that OP has error in the procedure. His derivation is perfectly right in case when $(x_0(t), y_0(t))$ is non-equilibrium trajectory and is still valid when it is an equilibrium. It's just another way to derive these equations. $\endgroup$ – Evgeny Jan 3 '16 at 13:33
  • $\begingroup$ @Evgeny Right, edited thanks. $\endgroup$ – Daniel Robert-Nicoud Jan 3 '16 at 13:45
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No, that is wrong. $x_0$ and $y_0$ are constants, only $Δx$ and $Δy$ are functions of $t$. Thus \begin{align} (Δx)˙=\frac d{dt}(x_0+Δx(t))&=y_0\quad+\quad Δy \\ \\ (Δy)˙=\frac d{dt}(y_0+Δy(t)) &=(x_0+Δx)+(x_0+Δx)^2−(y_0+Δy) \\ &=x_0+x_0^2-y_0\quad+\quad (1+2x_0)Δx-Δy\quad+\quad Δx^2 \end{align}

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    $\begingroup$ I do not see why you get $y_0+\Delta y$ for the first equation. $\endgroup$ – Rhjg Jan 3 '16 at 13:37

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