Consider a double integrator control system
$$\begin{cases}\dot{x}_1 = x_2,\\\dot{x}_2=u,\end{cases}$$
where $\dot{x} := \mathrm{d}x/\mathrm{d}t$. Then we apply the sliding mode control $u = -\beta(x)\mathrm{sgn}(s)$ with $s = a x_1+x_2$ ($a > 0$), in which $\beta(x) \geq a x_2 + \beta_0$ ($\beta_0 > 0$), and
$$\mathrm{sgn}(s)=\begin{cases}1, & s > 0,\\ 0, & s = 0,\\-1,& s<0.\end{cases}$$
By setting the Lyapunov function as $V = s^2/2$, any trajectory initially with $s \neq 0$ will reach $s = 0$ in finite time. This is because
$\dot{V} = s \dot{s} = s(a \dot{x}_1 + \dot{x}_2) = sa x_2 + su \leq |s|ax_2 + s [-\beta(x)\mathrm{sgn}(s)] = |s|(ax_2 - \beta(x)) \leq -\beta_0|s|.$
The above inequality also implies that: when reaching the manifold $s = 0$, the trajectory cannot leave it. On the manifold $s=0$, state $x_1$ follows $0 = s = ax_1 + x_2$, i.e., $\dot{x}_1 = -a x_1$, which means $(x_1, x_2)$ will finally converge to $(0,0)$ (along $s = 0$, since the trajectory cannot leave $s = 0$) as time goes to infinity.
However, here is the interesting thing. Do you really think $(x_1,x_2)$ will move along $s = 0$? Let's check the vector field at $(x_1, x_2)$, and it will gives
$$\begin{cases}\dot{x}_1 = x_2,\\\dot{x}_2=u=-\beta(x)\mathrm{sgn}(s)=0,\end{cases}$$
which never points along $s = ax_1 + x_2$ (recall that $a > 0$), and it will leave $s = 0$ instantly. This result contradicts "when reaching the manifold $s = 0$, the trajectory cannot leave it" implied by the Lyapunov function. What's happened on $s = 0$?
I read your post many times. I am not an expert in sliding mode control but I did learn a little bit about it. Sliding mode control brings your system(dynamics of the system expressed with a system of differential equations) to a sliding surface where some desirable properties are sought-after. Once the attraction to sliding surface is realized, the first system is turn into another system whose dynamics lie exclusively on the sliding surface.
In the case above, the system initially (outside the sliding surface) is expressed as :
dx1 = x2 dx2 = u
but once the system is brought to the sliding surface s = ax1+x2 = 0, it is confined there (assuming you have proved it , of course), Finally the dynamics of your system are described by the equation :
dx1 = -ax1
which means the original dynamics of the system do not describe the system motion no longer, but are turned to another dynamics specified by the sliding surface choice..