Lyapunov stabilty, elementary question Let’s say I have a system 1/(T1s+1) or any other n-th order polynomial and a PI controller (KP and TI). I already know that the system is stable but for, let’s say, educational purposes (not homework), I want to test the stability using different stability criterions. 
My question is:
Can I check if the system is stable or not by using Lyapunov stability criterion; similar to Nyquist or Routh-Hurwitz stability criterion? And how :) ? I was unable to find any examples on that or similar system. 
I must admit that I am a beginner in the topic of stability, but as I see it Nyquist, Bode, Routh-Hurwitz they are more engineering orientated (i.e. in a few steps I can easily test the stability), while Lyapunov is not :( 
Yet in almost every book Lyapunov’s direct method is considered one of the best for determining the stability of dynamical systems!
 A: Short answer: Yes, you can.
Long answer: Here is how.
I remember we often used Lyapunov's direct method when studying nonlinear system analysis and control. It hinges on the following idea:
If the total energy of a system continuously decreases, then the system, whether linear or nonlinear, must eventually settle down to an equilibrium point.
Consider the linear time-invariant system from your example, i.e.
$$G(s) = \displaystyle\frac{1}{T_1s+1}, \quad T_1 \in \mathbb{R}_{>0}.$$
Then the time-domain description of the system is given by the following linear differential equation:
$$\dot{x}(t) + \frac{1}{T_1}x(t) = \frac{1}{T_1}u(t)$$
We would like to check the zero-input stability of the system with an arbitrary initial state, i.e. $u(t) = 0 \ \forall t \in \mathbb{R}_{\geq 0}$ and $x(0) = x_0$. Then we have
$$\dot{x}(t) = -\frac{1}{T_1}x(t)$$
with the equilibrium point at $x = 0$. Select a Lyapunov candidate function $V(x)$ as the square of the distance from the origin times a non-zero constant, e.g.
$$V(x) = \frac{1}{2}x^2.$$
Check that the above $V(x)$ satisfies the criterion for being a positive definite function:
$$V(x) \geq 0 \quad \forall x \in \mathbb{R}, \quad \text{and} \quad V(x) = 0 \iff x = 0.$$
Observe that $V(x)$ is a scalar function, which can be interpreted as the "energy function" of the system. Now, to check if the system is stable, we compute the "rate of change of energy", i.e. the derivative of $V(x)$:
$$\begin{aligned}\dot{V}(x) &= x\dot{x}\\
&= x\left(-\frac{1}{T_1}x\right)\\
&= -\frac{1}{T_1}x^2,\end{aligned}$$
which is negative definite. That is,
$$ \dot{V}(x) \leq 0 \quad \forall x \in \mathbb{R}, \quad \text{and} \quad \dot{V}(x) = 0 \iff x = 0.$$
So we may say that the system passively dissipates energy. Since


*

*$V(x)$ is positive definite,

*$\dot{V}(x)$ is negative definite, and

*$V(x) \to \infty$ as $\|x\| \to \infty$,


then we conclude that the system is (globally) asymptotically stable at the origin in the Lyapunov sense.
