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1. System Formulation and Equilibrium

1.1 System Formulation

The general continues time and time invariant systems have the following formulations:

\[ \dot{x}(t) = f(x(t), u(t)), \left\{\begin{aligned} &x(0) = x_0\\ &t_0 = 0 \end{aligned}\right. \]

If the system \(u(t) = 0\) for all the time, we called the system as autonomous system. The autonomous system have the following formulation:

\[ \dot{x}(t) = f(x(t)) \]

And if the system is a linear system, we can use a linear form to express this system:

\[ \dot{x}(t) = Ax(t) + Bu(t) \]

1.2 Equilibrium

Give the definition of equilibrium of a system:

Note

Giving the equilibrium state \(\bar x\), \(\exists u = \bar u\), at \(t \geq 0\), that makes \(f(\bar x, \bar u) = 0\), such a point \((\bar x, \bar u)\) is called equilibrium pair, and \(\bar x\) is an isolated equilibrium.

For a general system, the system is equilibria if: \(\exists \delta > 0\) that makes \(\bar x\) is the only equilibrium contained in the region \(B_\delta(\bar{x}) = \{x\in \mathbb{R}^n:||x-\bar{x}||\leq \delta\}\).

To find the equilibrium pairs, we can consider the following 2 conditions:

  1. The system is a linear system,

    \[ \begin{aligned} &u(t) = \bar u, t \geq 0\\ &A\bar x + B\bar u = 0 \\ \Rightarrow &A\bar x = -B \bar u \end{aligned} \]
    • if \(det(A) \neq 0\), \(\bar x = -A^{-1}B\bar u\)
    • if \(det(A) = 0\), depend on \(\bar u\), \(\bar x = 0\) or \(\bar x = \infty\)

    Example

    \[ \begin{aligned} &\dot{x}(t) = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}x(t) + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(t) \\ &\Rightarrow \left\{\begin{aligned} \dot{x}_1(t) &= x_2(t)\\ \dot{x}_2(t) &= u(t)\\ \end{aligned}\right. \end{aligned} \]

    To find the equilibrium, we let \(u(t) = \bar u\), for \(t \geq 0\), thus, we have:

    \[ \begin{aligned} &\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} \bar{x}_1\\ \bar{x}_2 \end{bmatrix} =- \begin{bmatrix} 0\\ 1 \end{bmatrix} \bar{u} \\ &\Rightarrow \left\{ \begin{aligned} &\text{if } \bar{u} \neq 0 \Rightarrow \bar{x} \text{ not exists }\\ &\text{if } \bar{u} = 0 \Rightarrow \bar{x} = \begin{bmatrix} \alpha\\ 0 \end{bmatrix}, \alpha \in \mathbb{R} \end{aligned}\right. \end{aligned} \]

  2. For a nonlinear system with the form \(\dot{x}(t) = f(x(t), u(t))\), we can give the perturbance of the system near the equilibrium pair:

    \[ \begin{aligned} \delta x(t) &= x(t)- \bar{x}\\ \delta u(t) &= u(t)- \bar{u} \end{aligned} \]

    Thus, we can get:

    \[ \begin{aligned} \dot{x}(t) &= f(x(t), u(t)) \\ \delta\dot{x}(t) &= f(\delta x(t) + \bar{x}, \delta u(t) + \bar{u}) \\ &= \varphi(\delta x(t), \delta u(t)) \end{aligned} \]

    And \((0,0)\) is an equilibrium pair of \(\varphi(\delta x(t), \delta u(t))\). \(\varphi(0,0) = f(0+\bar{x}, 0+\bar{u}) = 0\).

    To find the equilibrium pair of the system, we still make \(u(t) = \bar u\), \(\forall t\). Thus,

    \[ \dot{x}(t) = f(x(t), \bar{u}) = \varphi_{\bar{u}}(x(t)) \]

    Solve the \(\varphi_{\bar{u}}(x(t)) = 0\), we can find the equilibrium points or ranges.

    Example

    \[ \dot{x}(t) = x^3(t) + u(t)x(t) \]

    Let \(u(t) = \bar{u}, \forall t\),

    \[ \dot{x}(t) = \varphi_{\bar{u}}(x(t)) = -x^3(t) +\bar{u}x(t) \]

    And let \(\varphi_{\bar{u}}(x(t)) = 0\),

    \[ \begin{aligned} &-\bar{x}^3 + \bar{u}\cdot\bar{x} = 0\\ &-\bar{x}(\bar{x}^2 - \bar{u}) = 0 \\ \Rightarrow & \left\{ \begin{aligned} &\text{if } \bar{u} \leq 0 \Rightarrow \bar{x} = 0 \\ &\text{if } \bar{u} > 0 \Rightarrow \bar{x} = 0 \text{ or } \bar{x} = \pm \sqrt{\bar u} \end{aligned}\right. \end{aligned} \]

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