4. Lyapunov Theory

4.1 Lyapunov Function

To introduce the Lyapunov Theory, we can start from an example:

Example

The dynamic equation of a spring-damping system is:

\[ \begin{aligned} m\ddot x(t) &= -K(x(t))h(x(t))+u(t)\\ K(x) &= K_0x +K_1x^3\\ h(x) &= b\dot x|\dot x| \end{aligned} \]

Give the state variable,

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

There have,

\[ \left\{\begin{aligned} &\dot x_1(t) = x_2(t)\\ &\dot x_2(t) = \frac{1}{m}(-bx_2(t)|x_2(t)| - K_0x_1(t) - K_1x_1^3(t)+u(t)) \end{aligned}\right. \]

When \(u(t) = 0\), at \(t\geq 0\),

\[ \left\{\begin{aligned} &\bar x_2 = 0\\ &K_0\bar x_1 + K_1 \bar x_1^3 = 0 \end{aligned}\right. \Rightarrow \left\{\begin{aligned} &\bar x_1 = 0\\ &\bar x_2 = 0 \end{aligned}\right. \]

The system is not differentiable, to find the stability of this system, Lyapunov Theory should be applied. Introducing the Lyapunov function which represents the energy of the system:

\[ \begin{aligned} V(x) &= \frac{1}{2}mx_2^2 \\ V(x) &= \int_0^{x_1}(K_0n+K_1n^3)dn\\ &=\frac12 K_0x_1^2 + \frac14 K_1x_1^4 \end{aligned} \]

It turns out \(V(\begin{bmatrix} 0\\0 \end{bmatrix}) = 0\), and \(V(\begin{bmatrix} x_1\\x_2 \end{bmatrix}) > 0, \forall \begin{bmatrix} x_1\\x_2 \end{bmatrix} \neq \begin{bmatrix} 0\\0 \end{bmatrix}\)

Derivate the Lyapunov function,

\[ \begin{aligned} \dot V(x) &= \frac{d}{dt}V(\begin{bmatrix} x_1(t)\\x_2(t) \end{bmatrix}) \\ &= \frac{\partial V}{\partial x_1}\dot x_1 + \frac{\partial V}{\partial x_2}\dot x_2 \\ &= \nabla V \cdot \begin{bmatrix} \dot x_1\\ \dot x_2 \end{bmatrix} \\ &= -bx_2^2|x_2| \leq 0\\ \bar x &= \begin{bmatrix} 0\\0 \end{bmatrix}, \text{Stable Equilibrium} \end{aligned} \]

The Lyapunov function can be drawn like below

Now, we can give the definition of Lyapunov function: \(V: \mathbb R^n \to \mathbb R\), \(V \in C^1\)

• \(V\) is positive definite (PD) in \(\bar x\) if \(V(\bar x) = 0\) and \(V(x) > 0\), \(\forall x \in D\{\bar x\}\) where \(D\) is an open neighbor of \(\bar x\), if \(D = \mathbb R^n \Rightarrow V\) is globally positive definite
• \(V\) is negative definite (ND) in \(\bar x\) if \(-V\) is positive definite in \(\bar x\)
• \(V\) is positive semi-definite (PSD) in \(\bar x\) if \(V(\bar x) = 0\) and \(V(x) \geq 0, \forall x \in D\) where in an open neighbor of \(\bar x\)
• \(V\) is negative semi-definite (NSD) if \(-V\) is positive semi definite

Radially unbounded globally positive definite function

\[ \lim_{||x|| \to \infty} V(x) =+\infty \]

Example

\[ V(x) = \frac{x_1^2}{1+x_1^2} +x_2^2 \]

\(\bar x=0\), \(V(x)\) is globally positive definite

\[ \begin{aligned} x &= \begin{bmatrix} x_1 \\ 0 \end{bmatrix}\\ \lim_{|x_1|\to +\infty} \frac{x_1^2}{1+x_1^2} &= 1 \end{aligned} \]

Lever surface of \(V\) positive definite is: \(\{x\in \mathbb R^n:V(x) = \bar V\}\)

4.2 Lipchitz Continuous

\[ \dot x(t) = \varphi(x(t)) \]

\(\bar x\) is a equilibrium of \(\varphi(\bar x = 0)\), where \(\varphi: \mathbb R^n \to \mathbb R^n\) is lipchitz in a neighbor \(D\) of \(\bar x\).

Give the definition of Lipchitz Continuous,

\[ \exists L: ||\varphi(x)-\varphi>(y)||\leq L||x-y||, \forall x,y \in D \]

Where, \(\varphi\) is continuous in \(D\)

Example

• \(\varphi (x) = x^{\frac13}\) is not lipchitz
• \(\varphi(x) = |x|\) is lipchitz
• \(\varphi \in C^1 \Rightarrow \varphi\) is lipchitz

4.3 Lyapunov Theory

Giving Lyapunov function \(V\), \(\exists V(x) \in C^1\), that are PD in \(\bar x\), and \(\dot V\) is NSD in \(\bar x\), then \(\bar x\) is stable.

Info

Proof

\(\bar x = 0\)

\(\delta\) is the minimum distance of \(\bar x\) from the boundary of \(V\)

For the Lyapunov Theory, there have following conditions:

• A.S. \(\to\) if \(\dot V(x)\) is negative definite in \(\bar x\)
• G.A.S \(\to\) globally positive definite \(V(x)\) in \(\bar x\), and globally negative definite \(\dot V(x)\) in \(\bar x\), Radially unbounded \(V(x)\)
• Instability \(\to\) positive definite \(\dot V (x)\) in \(\bar x\)

Example

\[ \begin{aligned} \dot x_1(t) &= -x_1(t)\\ \dot x_2(t) &= x_2(t)(x_1(t) -1)\\ \bar x &= \begin{bmatrix} 0\\0 \end{bmatrix} \end{aligned} \]

\[ \begin{aligned} V(x) &= \frac12(a_1x_1^2 +a_2x_2^2), a_1,a_2 \geq0 \\ \dot V(x) &= a_1 x_1(-x_1) + a_2 x_2^2(x_1 - 1) \\ &= -a_1x_1^2 - a_2x_2^2(1-x_1)\\ D&=\{x:x_1 < 1\} \end{aligned} \]

\(\bar x\) is A.S.

4.4 Conclusion on Lyapunov Theory

For the system,

\[ \begin{aligned} \dot x(t) = \varphi (x(t)) \\ \end{aligned} \]

\(\bar x\) is equilibrium \(\Rightarrow\) \(\varphi (\bar x) = 0\),

• \(\varphi: \mathbb R^n \to \mathbb R^n\)
• Lipchitz on \(D\) in the neighbor of \(\bar x\)

If \(\exists V(x) \in C^1\), \(\bar x\) is PD,

1. \(\bar x\) is stable if \(\dot V(x)\) is NSD
2. \(\bar x\) is A.S. if \(\dot V(x)\) is ND
3. \(\bar x\) is unstable if \(\dot V(x)\) is PD

If \(V(x)\) is globally PD in \(\bar x\), \(\bar x\) is G.A.S. when:

• \(V(x)\) is radially unbounded
• \(\dot V(x)\) is globally ND in \(\bar x\)

4.5 Candidate Lyapunov Functions

To choose a suitable Lyapunov function, we should meet the condition below:

• Lyapunov function should be an energy function

One possible option is: \(V(x) = (x-\bar x)^TP(x-\bar x)\)

\[ \begin{aligned} P = P^T \succ \mathbf 0 &\Rightarrow V^TPV > 0, \forall V \neq 0\\ &\Rightarrow V^TPV = 0, V = 0 \end{aligned} \]

Example

\[ \left\{\begin{aligned} \dot x_1 &= x_1 x_2 - x_1\\ \dot x_2 &= -x_1^2 - x_2^3 \end{aligned}\right. \]

\(\bar x = \begin{bmatrix} 0\\0 \end{bmatrix}\) is an equilibrium point, find the stability.

1. Linearization

   \[ \left\{\begin{aligned} \delta \dot x_1 &= -\delta x_1\\ \delta \dot x_2 &= 0 \end{aligned}\right. \Rightarrow A = \begin{bmatrix} -1 & 0\\ 0 & 0 \end{bmatrix} \]

2. Lyapunov Theory

   \[ V(x) = \frac12 (x_1^2 + x_2^2) \]

   Which is globally PD in \(\bar x = 0\) and radially unbounded

   \[ \begin{aligned} \dot V(x) &= x_1 \dot x_1 + x_2 \dot x_2 \\ &= x_1^2 x_2 - x_1^2 -x_2 x_1^2 - x_2^4 \\ &= -(x_1^2 + x_2^4) \leq 0 \end{aligned} \]

   \(\dot V(x)\) is globally ND in \(\bar x = 0\)

Example

For the pendulum system, give the state space function:

\[ \left\{\begin{aligned} \dot x_1 &= x_2 \\ \dot x_2 &= -\frac{g}{L} \sin(x_1) - \frac{K}{mL^2}x_2 + \frac{1}{mL^2}\bar u \end{aligned}\right. \]

\(\bar u = 0\), \(\bar x = \begin{bmatrix} 0\\0 \end{bmatrix}\) is an equilibrium point, find the stability.

\[ \begin{aligned} V(x) &= T(x) + U(x)\\ &= \frac12 m (x_2L)^2 + mg(L-L\cos(x_1)) \\ \end{aligned} \]

\(V(\bar x) = 0\), \(D = \{x \in \mathbb R^2, |x_1| < 2\pi\}\)

\[ \begin{aligned} \dot V(x) &= mL^2 x_2 \dot x_2 + mgL \sin(x_1) \dot x_1 \\ &= -Kx_2^2 \end{aligned} \]

is ND in \(\bar x = 0\), \(\bar x\) is stable, but we cannot find the region of attraction.

4.6 Krasowski - La Salle Theorem

To solve the example above, we introduce the Krasowski - La Salle theorem, it gives:

\(\dot x = \varphi(x)\), \(\varphi(x)\) is Lipchitz on \(D\), \(\bar x\) is a equilibrium, \(\varphi(\bar x) = 0\), let:

\(V: D \to \mathbb R \in C^1\) is PD in \(\bar x\) on \(D\) and \(\dot V(x)\) is NSD in \(\bar x\) on \(D\)

If set \(S = \{x\in D: \dot V(x) = 0\}\) does not contain any perturbed trajectory of the system, \(\bar x\) is an A.S. equilibrium.

\[ \dot V (x) = -Kx_2^2 \]

\(S = \{x\in D: x_2 = 0\}\)

• \(\dot x_1 = 0 \to x_1(t) = \bar x_1, \forall t\)

  \(\begin{aligned}0 = \dot x_2 &= -\frac{g}{L} \sin(x_1) \\ &\Rightarrow \sin(\bar x_1) = 0 \\ &\Rightarrow x_1 = h\pi \\ &\Rightarrow x = 0, \pm1, \pm2 \dots\end{aligned}\)

\(\Rightarrow\) in \(D = \{x: |x_1|<\pi\}\), \(\bar x\) is A.S.

If \(V\) is globally PD in \(\bar x\) and radially unbounded, \(\dot V(x)\) is NSD on \(D = \mathbb R^n\) and \(S = \{x \in \mathbb R^n: \dot V(x) = 0\}\),

• If \(S\) does not contain any perturbed trajectory \(\bar x\) is G.A.S.

Example

\[ \left\{\begin{aligned} &\dot x_1(t) = x_2(t)\\ &\dot x_2(t) = \frac{1}{m}(-bx_2(t)|x_2(t)| - K_0x_1(t) - K_1x_1^3(t)+u(t)) \end{aligned}\right. \]

\(\bar x = \begin{bmatrix} 0\\0 \end{bmatrix}\) is an equilibrium point, find the stability.

\[ \begin{aligned} V(x) = \frac12 m x_2^2 + \frac12 K_0 x_1^2 + \frac14 K_1 x_2^4 \\ \dot V(x) = -bx_2^2|x_2| \end{aligned} \]

is ND in \(\bar x = 0\), \(\bar x\) is stable.

\[ \begin{aligned} S &= \{x \in \mathbb R^2: \dot V(x) = 0\} \\ &= \{\begin{bmatrix} \alpha \\ 0 \end{bmatrix}, \alpha \in \mathbb R \} \end{aligned} \]

\(x_2(t) = 0 \Rightarrow \dot x_1 = x_2 = 0 \Rightarrow x_1(t) = \bar x_1\)

\(\begin{aligned}0 = \dot x_2 &= \frac{1}{m} (-K_0 \bar x_1 - K_1 \bar x_1^3) \\ &\Rightarrow \bar x_1 (K_0 + K_1 \bar x_1^2 = 0) \\ &\Rightarrow \bar x_1 = 0\end{aligned}\)

\(\bar x = \begin{bmatrix} 0\\0 \end{bmatrix}\) is G.A.S.

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