Sliding mode stability

Question

Given the scalar system

$$ \dot{x} = -\text{sgn}(x) \tag{1} $$

with

$$ \text{sgn}(x) = \begin{cases} -1 & x < 0 \\ 0 & x = 0 \\ 1 & x > 0 \,. \end{cases} $$

What is an easy method to check for stability? And: Say I use quasi sliding mode with

$$ \dot{x} = -\tanh(a x) \,. \tag{2} $$

If it is shown that $(2)$ is stable for all $a > 0$, does that also show stability of $(1)$ because

$$ \lim_{a \rightarrow \infty} \tanh(a x) = \text{sgn}(x) \,? $$

Edit: For the general question, it is required that both original and aproximation function have exactly the same equilibria, i.e. that the approximation doesn't add any new equilibira.

Answer 1

I doubt that the method with the limit is always possible, but I am also looking forward to an answer that explains if it is applicable at all or if it is applicable under certain conditions.

This is the reason why this is not an answer for the applicability of the limit for the stability analysis of this system. I will rather show a direct method to investigate the stability of the nonlinear system.

First, note that $x=0$ is an equilibrium point of the nonlinear system. Then integrate the equation exploiting that $\text{sgn}(x)$ can be treated as a constant.

$x(t)=-\text{sgn}(x)t+c$

Edit: In order to motivate this solution you can look at three cases $x>0$, $x=0$ and $x<0$. For the first case, we obtain $$\dot{x}=-1 \implies x(t) = -t+ c_1.$$ The case $x=0$ leads to $$\dot{x}=0 \implies x(t) = c_2.$$

The last case results in

$$\dot{x} = +1 \implies x(t) = t + c_3$$

Combining all cases gives the previous expression.

If $x(t=0)=x_0$ then we can look at the two cases $x_0>0$ and $x_0<0$. Can you continue?

Answer 2

The answer to the second question is generally NO. Consider a family $f_a(x) = x (x - \frac{1}{a}) (x + \frac{1}{a})$, $a > 0$. One has $\lim\limits_{a \to \infty} f_a(x) = x^3$, uniformly for $x$ in compact subsets of $\mathbb{R}$. The equilibrium $0$ is (even asymptotically) stable for any $\dot{x} = f_a(x)$, whereas it is unstable for the limiting equation $\dot{x} = x^3$.

Answer 3

Since this is a first order system of the form

$$ \dot{x} = f(x) $$

then any function, that goes through the origin and satisfies the constraint $x\,f(x) < 0\ \forall\ x\neq0$, would make the system asymptotically stable.

To show this one can use the following Lyapunov function

$$ V(x) = x^2 $$

such that

$$ \dot{V}(x) = 2\,x\,f(x) $$

so applying the constraint yields $\dot{V}(x)<0\ \forall\ x\neq0$.