Find a Lyapunov function and prove that an almost linear closed-loop system is stable

Question

I would like to find a Lyapunov function and prove the following closed-loop system is stable:

$\dot{x}=(A-BK)x+(z(u)+\epsilon)$,

where a function of control input $z(u)$ satisfies $\|z\| < \rho \|x\|$ with $\rho$ being a positive small constant, $K$ is an optimal feedback gain from LQR, and $\epsilon$ is small perturbation. You can see this system as an almost linear system, which has small nonlinear disturbance.

Here's my solution (but it fails to achieve to show $x$ converges). I would be very happy if you give me any advice or exhibit another solution:

Overall, I'm going to prove the stability by defining a Lyapunov function $V=x^\top P x$ such that the time-derivative is upper bounded by $V$ itself. Then I can apply the Comparison lemma to the inequality to demonstrate $\lim_{t \to \infty}\|x\|$ converges to an error ball. Note that $P$ is positive semi-definite from the Riccati equation of the LQR. The reason why I use the Comparison lemma is just because I learn this lemma in this semester. Of course, you do not have to use this lemma.

Proof:

$\dot{V} =\dot{x}^\top P x + x^\top P \dot{x}$

$=x^\top(A-BK)^\top P x + x^\top P(A-BK)x+2x^\top P(z+\epsilon) $

Since the first and second terms come from the Riccati equation, we can rewrite the sum using weights $R, Q$ as follows (watch this video if you do not understand this formulation),

$=x^\top(-Q-K^\top R K) x +2x^\top P(z+\epsilon) \cdots$ (i).

To apply the Comparison lemma, I make this time-derivative upper bounded by $V$.

Let $w=\Lambda x$ where $\Lambda$ satisfies $P=\Lambda^\top\Lambda$. Then $V=w^\top w= \|w\|^2$, and using the inequality $\|z\| < \rho \|x\|$,$\|\epsilon\|<\epsilon_m$ the time-derivative (i) becomes $\dot{V} =x^\top(-Q-K^\top R K) x +2x^\top P(z +\epsilon) $

$\leq \lambda_{max}(-Q-K^\top R K)\|x\|^2+2\|x^\top P\|(\rho\|x\|+\epsilon_{m})$

$\leq \lambda_{max}(-Q-K^\top R K)\|\Lambda^{-1}w \|^2+2\|w^\top \Lambda\|(\rho\|\Lambda^{-1}w\|+\epsilon_{m})$

$\leq \lambda_{max}(-Q-K^\top R K)\sigma(\Lambda^{-1})^2\|w \|^2+2\sigma(\Lambda)(\rho\sigma(\Lambda^{-1})\|w \|^2+\epsilon_m\|w \|) \cdots$ (ii),

where $\sigma(\cdot)$ denotes the maximum singular value, and $\lambda_{max}(-Q-K^\top R K)$ is the maximum eigenvalue of the negative semi-definite (later it is denoted only as $\lambda$).

Now we have the following inequality

$\dot{V}\leq \sigma(\Lambda^{-1})^2(\lambda +2\sigma(\Lambda)\frac{\rho}{ \sigma(\Lambda^{-1}) })\|w \|^2+2\sigma(\Lambda)\epsilon_{m}\|w \|$

$\rightarrow$ $\dot{V} \leq \alpha V+2\beta\sqrt{V} \cdots$ (iii)

where $\alpha=\sigma(\Lambda^{-1})^2(\lambda +2\sigma(\Lambda)\frac{\rho}{ \sigma(\Lambda^{-1}) })$, and $\beta=\sigma(\Lambda)\epsilon_{m} $. Finally, we can apply the Comparison lemma to (iii) by defining $W=\sqrt{V}, \dot{W}=\dot{V}/2\sqrt{V}$ to get

$2\sqrt{V}\dot{W} \leq \alpha WW+ 2\beta \sqrt{V}$

$\rightarrow$$\dot{W} \leq \frac{\alpha}{2}W+\beta \cdots$ (iv)

By comparison lemma, the inequality (iv) boils down to,

$W(t) \leq W_{0}\exp(\frac{\alpha}{2}(t-t_{0}))-\frac{2\beta}{\alpha}$

Thus, when $\lambda +\sigma(\Lambda)\frac{\rho}{ \sigma(\Lambda^{-1}) }<0$ such that $\frac{\alpha}{2}<0 $, $W(t)=\sqrt{V(t)}=\|w\|$ converges to the error ball $\epsilon_{error}=-\frac{2\beta}{\alpha}=-\frac{2\sigma(\Lambda)\epsilon_{m}}{\sigma(\Lambda^{-1})^2(\lambda +\sigma(\Lambda)\frac{\rho}{ \sigma(\Lambda^{-1}) })}$

Q.E.D

However, as you can see, this whole proof only proves $\|w\|=\|\Lambda x\|$ converge, not $\|x\|$.

How could I fix this process? May the Lyapunov function or using the comparison lemma not be appropriate? Please give me any advice! Thank you.

Answer 1

You have the system

$$ \dot{x}=(A-BK)x+\rho(\Vert x\Vert+\epsilon)e(x) $$

with $e(x)=\frac{x}{\Vert x\Vert}$. First we can recognize that $\dot{x}$ is undefined for $x=0$. For $x\neq 0$ you get

$$ \begin{align} \dot{x}&=(A-BK)x+\rho(\Vert x\Vert+\epsilon)\frac{x}{\Vert x\Vert}\\ &=(A-BK)x+\rho I x+\epsilon\frac{x}{\Vert x\Vert}\\ &=(A-BK + \rho I)x+\epsilon d \end{align} $$ with identity matrix $I$ and disturbance $$ d=\frac{x}{\Vert x\Vert} $$ Note that $\Vert d\Vert = 1$ is bounded so your system can be viewed as a linear time invariant (LTI) system with bounded input. As such, it is stable if and only if $(A-BK+\rho I)$ is Hurwitz.

Since stable LTI systems are always input to state stable, $x$ will then converge to some error ball. You can use standard techniques for LTI systems to compute an estimate of that error ball.