Equilibrium of autonomous first-order differential equation: ‎

159 Views Asked by At

Let ‎$ ‎‎(\bar{x},‎\bar{y},‎\bar{z})‎‎‎‎ $‎ be an equilibrium of autonomous first-order differential equation: ‎\begin{equation}‎ ‎\begin{cases}‎ ‎\dot{x}=‎ f_1(x,y,z)‎‎‎,\\‎ ‎\dot{y}=‎ ‎f_2(x,y,z)‎‎, ‎\\‎ ‎\dot{z}=‎ f_3(x,y,z)‎. ‎\end{cases}‎ ‎\end{equation}‎ Is it possible to say that $‎(\bar{x},‎\bar{y},‎\bar{z})‎‎‎$ is unique equilibrium point of the following system: ‎\begin{equation}‎ ‎\begin{cases}‎ ‎\dot{x}=‎ f_1(x,y,z)-(x-‎\bar{x}‎),\\‎ ‎\dot{y}=‎ ‎f_2(x,y,z)-(y-‎\bar{y}‎)‎‎, ‎\\‎ ‎\dot{z}=‎ f_3(x,y,z)-(z-‎\bar{z}‎)‎. ‎\end{cases}‎ ‎\end{equation}‎

1

There are 1 best solutions below

3
On

A point $x^*$ is called an equilibrium (stationary point/critical point) of a system of differential equations :

$$\vec{x}' = \vec{f}(x_1,\dots,x_n)$$

if $x^* \in \mathbb R^n$ is a vector, such :

$$\vec{f}(x^*_1,\dots,x^*_n)=0$$

For your specific example, letting the expressions :

$$\vec{x}' =\vec{F}(x_1,x_2,x_3)=\vec{f}(x_1,x_2,x_3) - (\vec{x}-{\vec{x^*}})$$

we have that :

$$\vec{F}(x_1,x_2,x_3)=\vec{f}(x^*_1,x^*_2,x^*_3)-(\vec{x^*}-\vec{x^*})=0$$

which means that $\vec{x^*}$ is indeed an equilibrium of the alternative system of differential equations that you mentioned.