Keri Molloy

2022-02-18

Let $\left(\stackrel{―}{x},\stackrel{―}{y},\stackrel{―}{z}\right)$ be an equilibrium of autonomous first-order differential equation:

Is it possible to say that $\left(\stackrel{―}{x},\stackrel{―}{y},\stackrel{―}{z}\right)$ is unique equilibrium point of the following system:

vazen2bl

A point ${x}^{\cdot }$ is called an equilibrium (stationary point/critical point) of a system of differential equations:
${\stackrel{\to }{x}}^{\prime }=\stackrel{\to }{f\left({x}_{1},\dots ,{x}_{n}\right)}$
if $x\cdot \in {\mathbb{R}}^{n}$ is a vector, such:
$\stackrel{\to }{f\left({x}_{1}^{\cdot },\dots ,{x}_{n}^{\cdot }\right)}=0$
For your specific example, letting the expressions:
${\stackrel{\to }{x}}^{\prime }=\stackrel{\to }{F}\left({x}_{1},{x}_{2},{x}_{3}\right)-\left(\stackrel{\to }{x}-{\stackrel{\to }{x}}^{\cdot }\right)=0$
which means that ${\stackrel{\to }{x}}^{\cdot }$ is indeed an equilibrium of the alternative system of differential equations that you mentioned.

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