Linearization
June 01, 2023
Given an system,
$$ \begin{align}\label{eq:sys} \dot{x}(t) &= f(x(t),u(t),t)\\ y(t) &= g(x(t),u(t),t) \end{align} $$If $x^*$ is an equilibrium point of an uncontrolled system, this implies that $\dot{x}(t) = 0$. Let $x^*$ be an equilibrium point of $\eqref{eq:sys}$. The deviation from the equilibrium point $x^*$ will be
$$ \begin{equation} \label{eq:deltax} \Delta x(t) = x(t) - x^* \end{equation} $$By differentiating $\eqref{eq:deltax}$,
$$ \frac{d}{dt}\Delta x(t) = f(x(t),u(t) $$The right hand side could be expanded using Taylor's series as follows.
$$
\begin{equation*}
\frac{d}{dt}\Delta x(t) = f(x^*,0)+\frac{\partial}{\partial x} f(x,u)\Big|_{x=x^*,u=0} \Delta x(t) + \frac{\partial}{\partial u}f(x,u)\Big|_{x = x^*,u=0} u(t)+ O(\cdot)
\end{equation*}
$$
Then, the Linear Time-Invariant (LTI) system is expressed as follows.
$$ \begin{equation}\label{eq:deltaxd} \Delta\dot{x}(t) = A \Delta x(t) + B u(t) \end{equation} $$where,
$$ \begin{align} A &= \frac{\partial}{\partial x} f(x,u)\Big|_{x=x^*,u=0}\\ B &= \frac{\partial}{\partial u}f(x,u)\Big|_{x = x^*,u=0} \end{align} $$Hartman-Grobman Theorem
The set of solutions of $\eqref{eq:sys}$ in the neighborhood of a hyperbolic equilibrium point $x^*$ is homeomorphic to that of the linearized system $\eqref{eq:deltaxd}$ in the neighborhood of the origin.