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2. Stability of Linear Systems

2.1 Stability Definitions

Since we have find the trajectory of the equilibrium in the last part, it gives \(\dot{x}(t) = \varphi(x(t))\), take the equilibrium \(\bar x\) into the equation, we can get \(\varphi(\bar x) = 0\), this is a function of \(\bar x\).

  1. When \(\bar u\) is not considered in calculating the equilibrium, \(\bar x\) is a stable equilibrium if \(\forall \varepsilon > 0\), \(\exists \delta > 0\), \(\forall x_0 \in B_\delta(\bar{x})\), and \(x_0\) satisfy \(||x_{x_0}(t)-\bar{x}||\leq \varepsilon, \forall t \geq 0\). For a second order system, this should be like the figure:

    And the equilibrium is asymptotically stable (A.S.) if:

    • \(\bar x\) is stable
    • \(\exists \delta > 0\) that when \(\lim\limits_{t\to \infty}||x_{x_0}(t)-\bar{x}|| = 0\), \(\forall x_0 \in B_\delta(\bar x)\) Thus, \(\bar x\) is convergent and A.S.

    Example

    \[ \begin{aligned} \dot x_1(t) &= x_1^2(t) - x_2(t)\\ \dot x_2(t) &= 2x_1(t)x_2(t) \end{aligned} \]

    \(\bar x = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) is an equilibrium but not a stable equilibrium.

    Quote

    A.S is a local property

    Example

    \[ \begin{aligned} \dot x (t) &= x(t)(x(t) - 1)u(t)\\ u(t) &= 1, t\geq 0 \end{aligned} \]

    Calculate the equilibrium:

    \[ \begin{aligned} \dot x(t) &= x(t)(x^2(t) -1)\\ \bar x (\bar x^2 -1) &= 0\\ \Rightarrow &\left\{ \begin{aligned} \bar x &= 0, \text{ A.S.} \\ \bar x &= \pm 1 \text{ unstable} \end{aligned}\right. \end{aligned} \]

    \(D=(-1,1)\) is the region of attraction for \(\bar x = 0\).

    To make the system A.S. globally, \(\bar x\) should meet the condition: \(\lim_{t\to \infty}||x_{x_0}(t)-\bar{x}|| = 0\), \(\forall x_0 \in \mathbb{R}^n\).

  2. Considering the stability properties depend on \(\bar u\), we can give a similar example below:

    Example

    \[ \begin{aligned} \dot x(t) &= x^3(t) +u(t)x(t) \\ u(t) &= \bar u, t\geq 0 \end{aligned} \]

    We can plot \(\dot x\) as a function of \(x\), \(f(x, u) = -x^3 + \bar u x\).

    \(u \leq 0\) \(u > 0\)
    \(x = 0\) is globally A.S. (G.A.S.) \(x = 0\) is unstable

  3. For the A.S. system, there have the property of convergence rate, give the definition of exponential convergence rate (aka. exponential stability) for the system \(\dot x(t) = \varphi (x(t))\),

    • Giving \(\bar x\) that is an equilibrium of \(\varphi(\bar x ) = 0\), if \(\exists \delta >0\), \(\forall x_0 \in B_\delta (\bar x)\), \(\exists a >0\) and \(\lambda > 0\), that makes \(||x_{x_0} - \bar x || \leq a||x_0 - \bar x||e^{-\lambda t}, t\geq 0\), then \(\bar x\) is an exponentially stable equilibrium
    • exponential stability (E.S.) \(\in\) A.S.

2.2 Stability of Linear Systems

For a linear system, there have following features:

  1. The system is G.A.S. if it satisfied A.S. conditions
  2. All equilibrium shares the same stability properties, stability is a property of the system
  3. The system is E.S. when it is A.S.

Here we prove the above points, giving a linear system:

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

suppose \(\bar x\) is an equation associated with \(u(t) = \bar u\), when \(t \geq 0\), there have:

\[ \begin{aligned} &A\bar x +B\bar u = 0 \\ \Rightarrow &\left\{ \begin{aligned} \delta x(t) = x(t) - \bar x \\ \delta u(t) = u(t) - \bar u \end{aligned} \right.\\ \end{aligned} \]
\[ \begin{aligned} \delta \dot x(t) &= A (\delta x(t)+\bar x) + B(\delta u(t) +\bar u) \\ &= A\delta x(t) + B \delta u(t) \end{aligned} \]

And since we does not consider the perturbance from the inputs,

\[ \begin{aligned} &\delta \dot x(t) = A\delta x(t)\\ \Rightarrow &\delta x(t) = e^{At}\delta x(t) \end{aligned} \]

in this equation, \(e^{At} = \sum_{i=0}^n K_i e^{\lambda_i t}\), where \(\lambda_i\) is the eigen value of \(A\), the system is G.A.S. when \(\lambda_i < 0\).

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