Content deleted Content added
m fmt |
No edit summary |
||
Line 9:
:<math>\begin{cases} g : \R ^n \to \R ^n \\ \dot{y} = g(y) \end{cases}</math>
with an equilibrium point at <math>y=0</math> is a [[scalar function]] <math>V:\R^n\to\R</math> that is continuous, has continuous first derivatives, is strictly positive, and for which <math>-\nabla{V}\cdot g</math> is also strictly positive. The condition that <math>-\nabla{V}\cdot g</math> is strictly positive is sometimes stated as <math>-\nabla{V}\cdot g</math> is
===Further discussion of the terms arising in the definition===
Line 18:
for some smooth <math>g:\R^n \to \R^n.</math>
An equilibrium point is a point <math>y^*</math> such that <math>g\left(y^*\right) = 0.</math> Given an equilibrium point, <math>y^*,</math> there always exists a coordinate transformation <math>x = y - y^*,</math> such that:
:<math>\begin{cases} \dot{x} = \dot{y} = g(y) = g\left(x + y^*\right) = f(x) \\ f(0) = 0 \end{cases}</math>
Thus, in studying equilibrium points, it is sufficient to assume the equilibrium point occurs at <math>0</math>.
Line 46:
===Stable equilibrium===
If <math>V</math> is a Lyapunov function, then the equilibrium is [[Stability theory|Lyapunov stable]]. The converse is also true, and was proved by [[José Luis Massera|J. L. Massera]].
===Globally asymptotically stable equilibrium===
Line 60 ⟶ 58:
==Example==
Consider the following differential equation with solution <math>x</math> on <math>\R
:<math>\dot x = -x.</math>
Considering that <math>x^2</math> is always positive around the origin it is a natural candidate to be a Lyapunov function to help us study <math>x</math>. So let <math>V(x)=x^2</math> on <math>\R </math>. Then,
:<math>\dot V(x) = V'(x) \dot x = 2x\cdot (-x) = -2x^2< 0.</math>
| |||