E.2 Exercise 2

Ex.1 Given the continues system

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

Show that,

1. the \((\dot x_1, \dot x_2) = (0,0)\) is an equilibriym point

   Answer

   Let \((\dot x_1, \dot x_2) = (0,0)\), we can get:

   \[ \left\{\begin{aligned} \dot 0 &= -x_1 + x_2^2 \\ \dot 0 &= -x_2 \end{aligned}\right. \]

   \(\Rightarrow\) \(x_2 = 0\), \(x_1 = 0\), (0,0) is an equilibriym point.

2. and study its stability through the linearized system

   Answer

   Linearize the system: \(\delta \dot x = A \delta x\), where, \(A = \frac{\partial f}{\partial x}|_{\bar x}\)

   \[ \begin{bmatrix} \delta \dot x_1\\ \delta \dot x_2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \delta x_1\\ \delta x_2 \end{bmatrix} \]

   \(\lambda_{1,2} = -1\), The system is A.S. at equilibrium point \((0,0)\).

   Warning

   1. \(s < 0 \Rightarrow\), locally A.S. at equilibrium point
   2. \(s \leq 0, \exists s = 0\), no conclusion
   3. \(\exists s > 0\), unstable

Given the function \(V(x) = \frac12(x_1^2 +x_2^2)\),

1. Check that it is in fact a suitable Lyapunov function

   Answer

   1. The function is PD
   2. The function is continous
   3. Radially unbounded

   \(V(x) \succ 0\), \(V(x) \in C^1\), \(||V|| \in \mathbb R^n \to \mathbb R^n\), so it is a suitable Lyapunov function.

2. and study the stability of the system with the given function

   Answer

   \[ \begin{aligned} \dot V(x) &= x_1 \dot x_1 + x_2 \dot x_2 \\ &= x_1 (-x_1 + x_2^2) + x_2(-x_2) \\ &= -x_1^2 -x_2^2 + x_1x_2^2 \end{aligned} \]

   From above equation, we cannot say the stability of the system, to find the stability, we can make an approximation near the equilibrium point,

   \[ \begin{aligned} \dot V(x) &\approx -x_1^2 - x_2^2 + \vartheta(\bar x^3) \\ &\approx -x_1^2 - x_2^2 \end{aligned} \]

   The system is A.S. near the equilibrium point. This conclusion has no difference with the linearization method, to get the region of attraction, we can let:

   \[ \dot V(x) = -x_1^2 + x_2^2(1 - x_1) \]

   if \(x_1 \leq 1\), \(\dot V(x) < 0\)

   \[ D = \{(x_1, x_2)\in \mathbb R | x_1 < 1 \} \]

   Extension

   We can add the initial state into the system,

   \[ \begin{aligned} x_1(t) &= x_1(0)e^{-\tau} + (x_2(0)e^{-\tau})^2 \\ x_2(t) &= x_2(0)e^{-\tau} \end{aligned} \]

   \[ \begin{aligned} x_1(t) &= x_1(0)e^{-\tau} + \int_0^tV(\tau)e^{t-\tau}dt \\ x_2(t) &= x_2(0)e^{-\tau} \end{aligned} \]

   \[ \begin{aligned} x_1(t) &= x_1(0)e^{-t} + (e^{-t}-e^{-2t})(x_2(0))^2 \\ x_2(t) &= x_2(0)e^{-t} \end{aligned} \]

   if \(t \to +\infty\), \(e^{-t} \to 0\), \(e^{-2t} \to 0\), \(x_1(+\infty) = 0, x_2(+\infty) = 0\), the system is G.A.S.

Ex. 2: Given the following parametric system

\[ \left\{\begin{aligned} \dot x_1 &= x_1(k^2 - x_1^2 - x_2^2) + x_2(k^2 + x_1^2 + x_2^2)\\ \dot x_2 &= -x_1(k^2 + x_1^2 + x_2^2) + x_2(k^2 - x_1^2 - x_2^2)\\ \end{aligned}\right. \]

and the function \(V(x) = \frac12(x_1^2 + x_2^2)\),

1. Check that V(x) is a Lyapunov function,

2. study the stability of the origin of the system for \(k = 0\)

3. and for \(k \neq 0\)

   Answer

   \[ \begin{aligned} \dot V(x) &= x_1 \dot x_1 + x_2 \dot x_2 \\ &= x_1^2(k^2 - x_1^2 - x_2^2) + x_1x_2(k^2 + x_1^2 + x_2^2) - x_1x_2(k^2 + x_1^2 + x_2^2) + x_2^2(k^2 - x_1^2 - x_2^2) \\ &= (x_1^2+x_2^2)(k^2 - x_1^2 - x_2^2) \end{aligned} \]

   1. if \(k=0\), \(\dot V(x) \leq 0\), the system is G.A.S.

   2. if \(k\neq 0\), \(k^2 - x_1^2 - x_2^2 < 0\), \(k^2 < x_1^2 + x_2^2\), the system is A.S. and the region of attraction is the area outside of the circle \(K^2 = x_1^2 + x_2^2\). Inside the circle, the system is unstable.

Ex. 3: Given the continuous time system

\[ \left\{\begin{aligned} \dot x_1 &= x_2\\ \dot x_2 &= -x_2 - \alpha x_1 - (x_1^2 + x_2^2)x_2 \end{aligned}\right. \]

with \(\alpha > 0\) and the function \(V(x) = \alpha x_1^2 + x_2^2\),

1. Check that \(V(x)\) is a Lyapunov function,
2. study the stability of the origin of the system

3. and characterize the trajectories of the state around the origin (linearized system’s eigenvalues)

   Answer

   \[ \begin{aligned} \dot V(x) &= 2\alpha x_1 \dot x_1 + 2x_2 \dot x_2 \\ &= 2\alpha x_1 x_2 + 2 x_2 \left[ -x_2 - \alpha x_1 - (x_1 + x_2)^2 x_2 \right] \\ &= -2x_2^2 (1 + (x_1 +x_2)^2) \end{aligned} \]

   \(\dot V(x) = 0, \forall x_1\), when \(x_2 = 0\)

   KLS: \(x_2 = 0\),

   \[ \left\{\begin{aligned} \dot x_1 &= 0\\ \dot x_2 &= -0 - \alpha x_1 - 0 \end{aligned}\right. \]

   \(\dot x_2 = 0\) if and only if when \(x_1 = 0\), the system is G.A.S.

   Use the linearization method, the linearized system is:

   \[ \begin{aligned} \delta \dot x_1 = \delta x_2 \\ \delta \dot x_2 = \alpha \delta x_1 - \delta x_2 \end{aligned} \]

   \(A = \begin{bmatrix} 0&1\\ -\alpha & -1 \end{bmatrix}\), \(det(A - s\mathbf I) = 0\),

   \[ s_{1,2} = \frac{-1 \pm \sqrt{1-4\alpha}}{\alpha} \]

   1. \(1-4\alpha < 0\), \(\Rightarrow\) A.S. with complex conjugate eigenvalues
   2. \(1-4\alpha > 0\),
      • \(\alpha = 0\), \(s_1 = -1\), \(s_2 = 0\)
      • \(\alpha \neq 0\), \(s_1 < s_2 < 0\), \(\Rightarrow\) A.S. with real eigenvalues.

Ex. 4: Given the following discrete time system

\[ \left\{\begin{aligned} x_1(k+1) &= x_2(k) \cos(x_1(k))\\ x_2(k+1) &= x_1(k) \cos(x_2(k)) \end{aligned}\right. \]

1. Study the stability of the origin of the linearized system,

   Answer

   \[ \begin{aligned} &\left\{\begin{aligned} \delta x_1(k+1) &= - \bar x_2 \sin(\bar x_1) \delta x_1(k) + \cos(\bar x_1) \delta x_2(k)\\ \delta x_2(k+1) &= \cos(\bar x_2) \delta x_2(k) + (-\bar x_1 \sin(\bar x_2)) \delta x_2(k) \end{aligned}\right. \\ \Rightarrow &\left\{\begin{aligned} \delta x_1(k+1) &= \delta x_2(k)\\ \delta x_2(k+1) &= \delta x_1(k) \end{aligned}\right. \end{aligned} \]

   \(A = \begin{bmatrix} 0&1\\1&0 \end{bmatrix}\), \(det(-A+z\mathbf I) = 0\), \(z^2 -1 = 0\), \(z_{1,2} = \pm 1\), we cannot get any conclusion from the linearization.

2. Use the following Lyapunov function to study the stability of the system, \(V(x) = x_1^2(k) + x_2^2(k)\),

   Answer

   \[ \begin{aligned} \Delta V(x) &= V(x(k+1)) - V(x(k))\\ &= (x_2(k)\cos(x_1(k)))^2 + (x_1(k)\cos(x_2(k)))^2 - x_1^2(k) - x_2^2(k) \\ &= -x_1^2\sin^2(x_2) - x_2^2 \sin^2(x_1) \leq 0 \end{aligned} \]

   if \(x_1 = 0\), \(\Delta V = 0, \forall x_2\),

   if \(x_2 = 0\), \(\Delta V = 0, \forall x_1\),

   • Case 1:

     \[ \left\{\begin{aligned} x_1(0) &= 0 \\ x_2(0) &= \bar x_2 \end{aligned}\right. \]

     We take one step forward,

     \[ \left\{\begin{aligned} x_1(1) &= \bar x_2 \\ x_2(1) &= 0 \end{aligned}\right. \to \left\{\begin{aligned} x_1(2) &= 0 \\ x_2(2) &= \bar x_2 \end{aligned}\right. \]

     The system is stable but not A.S.,

   • Case 2:

     \[ \left\{\begin{aligned} x_1(0) &= \bar x_1 \\ x_2(0) &= 0 \end{aligned}\right. \to \left\{\begin{aligned} x_1(1) &= 0 \\ x_2(1) &= \bar x_1 \end{aligned}\right. \]

     The system is also only stable but not A.S.

Ex. 8: Given the following system

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

and the Back-stepping formula (given at the exam)

\[ u = - \frac{dV_1(x_1)}{dx_1} g_1 (x_1) - k(x_2 - \phi(x_1)) + \frac{d\phi(x_1)}{dx_1}(f_1(x_1) + g_1(x_1)x_2) \]

Determine a control law stabilizing the origin using the Back-stepping method.

Answer

Backstepping can be applied to the system with the form:

\[ \left\{\begin{aligned} \dot x_1 &= f_1(x_1) + g_1(x_1)x_2 \\ \dot x_2 &= u \end{aligned}\right. \]

We cannnot apply the backstepping to the system directly, but we can apply the extended backstepping to the system, the extended backstepping have the form:

\[ \left\{\begin{aligned} \dot x_1 &= f_1(x_1) + g_1(x_1)x_2 \\ \dot x_2 &= f_2(x_1, x_2) + g(x_1, x_2) u \end{aligned}\right. \]

In this system, we have:

• \(f_1 = x_1^3\), \(g_1 = 1\),
• \(f_2 = x_2^2\), \(g_2 = 1\)

With the change of varibale,

\[ u_a = f_2 + g_2 u \]

The system becomes:

\[ \left\{\begin{aligned} \dot x_1 &= f_1(x_1) + g_1(x_1)x_2 \\ \dot x_2 &= u_a \end{aligned}\right. \]

For the system above, we have:

\[ u_a = x_2^2 + u \]

To design the parameter,

1. \[ \begin{aligned} &\phi(x_1): x_2 = \phi(x_1) | \phi(0) = 0 \\ &\phi(x_1) = -\gamma x_1, \gamma > 0 \end{aligned} \]
2. \[ \begin{aligned} V_1(x_1) &= \frac12 x_1^2 \\ \dot V_1(x_1) &= -x_1^4 - \gamma x_1^2 \end{aligned} \]

   and \(u_a = -x_1 - k\gamma x_1 + \gamma x_1^3 -kx_2 - \gamma x_2\), \(u(t) = u_a - x_2^2\), is G.A.S in \((0,0)\)

Pros	Cons
G.A.S	No guarantees for the performance
	\(\phi\), \(V\) is hard to design
	Requests a good model
	Need to care about saturation problem

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