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Suppose we have:

$\frac{dx}{dt} = 14x-\frac{1}{2}x^2-xy$

$\frac{dy}{dt} = 16y-\frac{1}{2}y^2-xy$

My textbook outlines the following steps:

For the critical point (0,32) of the above system, compute the Jacobian as

$J(0,32) = \left( \begin{array}{ccc} -18 & 0\\ -32 & -16\\ \end{array} \right)$

which corresponds to the linearization

$\frac{du}{dt} = -18u$

$\frac{dv}{dt} = -32u - 16v$

The matrix $J(0,32)$ has characteristic equation $(-18-\lambda)(-16-\lambda)=0$ with eigenvalues $\lambda_1 = -18$ and $\lambda_2 = -16$. Because both eigenvalues are negative, it follows that (0,0) is a nodal sink for the linearized system, and hence (0,32) is also a stable nodal sink for the original system.

Now my question is the part in bold. Why does the Jacobian of the critical point (0,32) tell us anything about (0,0)? In particular, why does knowing anything about the stability of (0,0) help in analyzing the critical point (0,32)?

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Suppose you have the system $$ \dot y_1=f(y_1,y_2),\\ \dot y_2=g(y_1,y_2) $$ and $\hat{y}=(\hat{y}_1,\hat{y}_2)$ is a critical point. Then you use Taylor's series to write $$ \dot y_1=f(\hat{y}_1,\hat{y}_2)+f_{y_1}(\hat{y})(y_1-\hat{y}_1)+f_{y_2}(\hat{y})(y_2-\hat y_2)+\ldots\\ \dot y_2=g(\hat{y}_1,\hat{y}_2)+g_{y_1}(\hat{y})(y_1-\hat{y}_1)+g_{y_2}(\hat{y})(y_2-\hat y_2)+\ldots\\ $$ We use the fact that $\hat{y}$ is a critical point and drop all the terms of Taylor's series that shown as dots above: $$ \dot y_1=f_{y_1}(\hat{y})(y_1-\hat{y}_1)+f_{y_2}(\hat{y})(y_2-\hat y_2),\\ \dot y_2=g_{y_1}(\hat{y})(y_1-\hat{y}_1)+g_{y_2}(\hat{y})(y_2-\hat y_2), $$ which is a linearization of the original system around $\hat{y}$. Note that $\hat y$ is still a critical point here. Now you make a change of variables $$ x_1=y_1-\hat y_1\\ x_2=y_2-\hat y_2 $$ and get $$ \dot x_1=f_{y_1}(\hat{y})x_1+f_{y_2}(\hat{y})x_2\\ \dot x_2=g_{y_1}(\hat{y})x_1+g_{y_2}(\hat{y})x_2, $$ whose critical point is now $(0,0)$.

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