Let us consider the following dynamical system: $$ \dot{X} = A\cdot X + B\cdot u$$ where $X,B \in \mathbb{R}^{n\times 1}$. The considered system is linear, but I think that $A\cdot X$ can be eventually replaced with $f(X) \in \mathbb{R}^{n\times 1}$. I try to define some Lyapunov candidate functions for the general case. However some mistake must be made because I obtain a discontinuous control law able to stabilize some systems which I know are NOT stabilizable. However, below is the reasoning:

Problem description

Let $$P_+(X) = (X+B)^T (X+B)\cdot 1_{n\times n} + (X+B) (X+B)^T$$ and $$ P_-(X) = (X-B)^T (X-B)\cdot 1_{n\times n} + (X-B) (X-B)^T $$ Then $P_+(X)$, for $X\neq -B$, is a real symmetric matrix with eigenvalues $\{2\cdot \|X+B\|^2, \|x+B\|^2,..., \|X+B\|^2\}$ hence $P_+$ is positively defined. Also $P_-(X)$ for $X\neq B$ is a positively defined matrix.

Define the following Lyapunov candidate function: $$V(X) = \begin{cases}V_1(X) = X^T\cdot P_+(X)\cdot X, \hspace{0.5cm} X^T \cdot B \geq 0\\ V_2(X) = X^T\cdot P_-(X)\cdot X, \hspace{0.5cm} X^T \cdot B < 0 \end{cases}$$ Then $V(X) > 0$ for all $X \neq 0$.

Furthermore $V_1(X)\biggr\rvert_{X^T\cdot B = 0} = \|X\|^2(\|X\|^2+\|B\|^2) + \|X\|^4 = V_2(X)\biggr\rvert_{X^T\cdot B = 0}$ hence $V(X)$ is continuous.

Evaluating the derivative of $V$ one obtains: $$\dot{V}(X) = \begin{cases} 2\cdot X^T\cdot P_+(X)\cdot \left( AX+Bu\right) + X^T\cdot \dot{P}_+(X) \cdot X \hspace{0.5cm} X^T\cdot B \geq 0\\ 2\cdot X^T\cdot P_-(X)\cdot \left( AX+Bu\right) + X^T\cdot \dot{P}_-(X) \cdot X \hspace{0.5cm} X^T\cdot B< 0\end{cases}$$ where let $$ F_1 = X^T\cdot P_+(X)\cdot B = X^T B \cdot (\|X+B\|^2 + \|X\|^2 + \|B\|^2) + (X^T B)^2 + \|X\|^2\|B\|^2$$ then

$$ F_2 = X^T\cdot P_-(X)\cdot B = X^T B \cdot (\|X-B\|^2 + \|X\|^2 + \|B\|^2) - (X^T B)^2 - \|X\|^2\|B\|^2$$ Also let

$$ E_1 = X^T\cdot P_+(X)\cdot AX \hspace{0.5cm} E_2 = X^T\cdot P_-(X)\cdot AX$$

Finally $$X^T\cdot \dot{P}_+(X) \cdot X = X^T\cdot \left( 2(X+B)^T\cdot (AX+Bu) 1_{n\times n} + (AX+Bu)(X+B)^T + (X+B)(AX+Bu)^T\right) \cdot X $$ hence let $$ G_1 = ... \hspace{0.5cm} H_1 = 2\cdot X^TB \cdot (\|B\|^2+\|X\|^2 ) + 2\cdot \|B\|^2\|X\|^2 + 2\cdot (B^TX)^2$$ and $G_2$ and $H_2$ are obtained in a similar fashion for $\dot{P}_-(X)$. $$ G_2 = ... \hspace{0.5cm} H_2 = 2\cdot X^TB \cdot (\|B\|^2+\|X\|^2 ) - 2\cdot \|B\|^2\|X\|^2 - 2\cdot (B^TX)^2$$

Follows: $$\dot{V}(X) = \begin{cases}2\cdot E_1(X) + G_1(X) + (2F_1 + H_1)\cdot u \hspace{0.5cm} X^TB \geq 0\\ 2\cdot E_2(X) + G_2(X) + (2F_2 + H_2)\cdot u \hspace{0.5cm} X^TB < 0 \end{cases}$$

Control law

Let $$u(x) = \begin{cases} \frac{-V_1(X) - 2\cdot E_1(X) - G_1(X)}{2\cdot F_1(X) + H_1(X)} \hspace{0.5cm} &x^T\cdot B \geq 0\\ \frac{-V_2(X) - 2\cdot E_2(X) - G_2(X)}{2\cdot F_2(X) + H_2(x)} \hspace{0.5cm} & x^T\cdot B < 0 \\ \end{cases}$$

In this way one obtains $$ \begin{cases} \dot{V}_1(X) + V_1(X) = 0 \hspace{0.5cm} &X^T\cdot B \geq 0\\ \dot{V}_2(X) + V_2(X) = 0 \hspace{0.5cm} &X^T\cdot B < 0 \end{cases} $$ which means $V_1 \to 0$ and $V_2 \to 0$ right? What am I doing wrong? I obtain a stabilization law for ... every system with $B \neq 0$? This is obviously wrong ... but I do not know where is the mistake.

  • $\begingroup$ Thank you for your attention sir! As you can see $2F_2 + H_2 < 0$ for all $X^T\cdot B < 0$ and $2F_1 + H_1 > 0$ for all $X^T\cdot B \geq 0$ except for $X = 0$ $\endgroup$ – C Marius Aug 7 '18 at 19:50
  • $\begingroup$ Your expressions for $G_1(X)$ and $G_2(X)$ are missing. I also got a different results, namely $H_1 = 2(x^\top A^\top (x + B) \|x\|^2 + x^\top A\,x\,(x + B)^\top x)$ and $H_2 = 2(x^\top A^\top (x - B) \|x\|^2 + x^\top A\,x\,(x - B)^\top x)$. But even if these results are the correct ones I still think that the rest of the question seems to hold. $\endgroup$ – Kwin van der Veen Aug 7 '18 at 22:41
  • $\begingroup$ @Kwin $H_1$ and $H_2$ are some functions of $X$ that multiply the input $u$ and $G_1$ and $G_2$ are some functions such that $X^T\dot{P}_+(X)\cdot X = G_1(X) + H_1(X)\cdot u$ and similarly for $G_2, H_2$. Is this what you have considered? In my count, no coefficient of $u$ has $A$ in it ... $\endgroup$ – C Marius Aug 7 '18 at 23:00
  • $\begingroup$ You are right. I switched my results of $H$ and $G$. So I have $H_1 = 2(B^\top (x + B) \|x\|^2 + x^\top B\,(x + B)^\top x)$ and $H_2 = 2(B^\top (x - B) \|x\|^2 + x^\top B\,(x - B)^\top x)$. $\endgroup$ – Kwin van der Veen Aug 7 '18 at 23:07
  • $\begingroup$ Probably if you expand them they get the same shape as mine. I expressed them in this form to show that $H_1 > \|X\|^2 \cdot \|B\|^2$ for $X^TB > 0$ and similar for $H_2 < -\|X\|^2\cdot \|B\|^2$ for $X^T\cdot B < 0$. That is to show that we are not dividing by $0$ $\endgroup$ – C Marius Aug 7 '18 at 23:11

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