Is it necessary for a Lyapunov Candidate to be Differentiable at an Equilibrium Point?

Question

For example, given a general nonlinear system where we want to show that the error system is stable $e=x-x_d$ is it necessary for the Lyapunov candidate to be continuously differentiable at the equilibrium point since we are only concerned about what happens around the equilibrium point?

At the end of the day I would like to show that $\vert\vert e\vert\vert$ is a valid Lyapunov candidate. However, this is not differentiable at $e=0$ since the derivative is $\frac{e^{T}\dot{e}}{\vert\vert e\vert\vert}$.

My current digging into this topic has found the following paper: Vector Norms as Lyapunov Functions for Linear Systems by Kiendl. However, this paper is restricted to linear systems (plus other restrictions).

Note: this is a follow up to a question I asked yesterday but I believe this is better said and is more direct.

Thanks for your input

Answer 1

Well, nondifferentiable Lyapunov functions are very common in the analysis of discontinuous control systems and the situation will be very simple if the right-hand side of your equation is continuous.

Suppose the right-hand side of your equation is continuous (otherwise you would have mentioned it, lol). Then by the set-valued Lie derivative, the Lyapunov function $\|e\|$ is monotonically decreasing if $\max(\partial_{x}\|e\|\dot{e})\leq0$, where $\partial_{x}\|e\|$ is the generalized gradient of $\|e\|$. Hence, if $\dot{e}=0$ at $e=0$, then your Lyapunov function works.

Answer 2

Definition 1（Lie Derivative)Lie derivative of a function $V:\mathbb{R}^n\to\mathbb{R}$ along vector field f：$\mathbb{R}^n:→\mathbb{R}^n$is:

$$ {L}_f V(x)\triangleq\left(\dfrac{\partial V}{\partial x}(x)\right)^T f(x) $$ Therefore, the Lie derivative characterizes the time-causing evolution of the value of V along the solution trajectory of $\dot{x}=f(x)$.

Thus the lyapunov theorem using Lie derivative could be stated as follows.

Let $D \subset \mathbb{R}^n$” be a set containing an open neighborhood of the origin. If there exists a PD function $V:D\to\mathbb{R}$ Such that

$${L}_f V(x)\leq 0$$ is NSD, then the origin is stable. If in addition

$${L}_f V(x)<0$$ is ND, then the origin is asymptotically stable.

In your case, I think you have a Lyapunov function given by $$V = \|{e}\|$$ which is continuous, positively defined, and non-differentiable at the origin. Then you can check the Lie derivative of $V$ follows the gradient of the function $f(e)=\|e\|$ which means $$\dfrac{\partial V}{\partial e}(e)=\frac{e^{T}\dot{e}}{\Vert e\Vert}$$ and the Lie derivative is $$L_fV=\dfrac{\partial V}{\partial e}(e)\|e\|=\frac{e^{T}\dot{e}}{\Vert e\Vert}\|e\|$$

Then we only need to make sure or design the controller to let the r.h.s of the Lie derivative of V is PD, then the system $\|e\|\rightarrow 0$.

And this is what wjxjmj said $\max(\partial_{x}\|e\|\dot{e})\leq0$.