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I'm trying to make linearization for the following system:

$y_1'=(1+y_1)\sin y_2, y_2'=1-y_1-\cos y_2$ in the critical point $y_1=y_2=0$.

I'm not sure that I know what I need to do.

I believe that I need to solve the homogeneous system: $(1+y_1)\sin y_2=0, 1-y_1-\cos y_2$=0 so I got that $y=-1$ and $\sin y_2=0$, but the first option is not possible, so I got that $y_2= k\pi $ for all $k \in \mathbb{z}$ and $y_1=2$ or $y_1=0$.

For $y_1=0$, I get that the linear approximation equations are:

$y_1= \epsilon z_1 +o( \epsilon)$, $y_2=\pi k +\epsilon z_2 +o( \epsilon)$, but I don't manage to reach an equation for $z_1$ and $z_2$, Is this the way to do that?


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up vote 1 down vote accepted

You need to calculate the Jacobi matrix. If you have the system $$ \dot y_1=f(y_1,y_2),\quad \dot y_2=g(y_1,y_2), $$ then the Jacobi matrix is given by $$ DF(y_1,y_2)=\begin{pmatrix} f'_{y_1} && f'_{y_2}\\ g'_{y_1} && g'_{y_2} \end{pmatrix} $$ Then the linearization is a neighborhood of a stationary point $(\hat{y}_1,\hat{y}_2)$ is given by $$ \dot z=DF(\hat{y}_1,\hat{y}_2)z,\quad z=(z_1,z_2), $$ which is a linear system with a constant matrix.

In general, you obtain this linearized system if you make the change of the variables $$ z_1=y_1-\hat{y}_1,\quad z_2=y_2-\hat{y}_2 $$ and use the Taylor expansion around $(\hat{y}_1,\hat{y}_2)$, dropping all the terms of $o(|z|)$. This can be justified, however, only if $DF(\hat{y}_1,\hat{y}_2)$ has no eigenvalues with zero real part.

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Your solutions can be

$$y_1=0 \qquad y_2=k\pi \qquad k=0,\pm 1,\pm 2,\ldots$$

For $y_1=y_2=0$ the system takes the simple form

$$y_1'=y_2 \qquad y_2'=-y_1$$

and you can follow a similar approach with a standard Taylor expansion for the other points.

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