construct different ode systems but with the same lyapunov function

Question

I am thinking of whether there are some ode systems that are different with each other, suppose all of them have zero as an equilibrium point. Moreover, they have a common lyapunov function that can be used to show the zero solution is stable.

Could you give such example?

Thanks!

Answer 1

There are many such systems. Take for example for the first system

$$ \dot{x}=-x $$

and for the second system

$$ \dot{y}=-2y $$

You can take as common Lyapunov function $V(x)=V(y)=\frac{1}{2}x^2$ which works for both systems. More generally you could take

$$ \dot{x}=-ax^k $$

where $k$ is an odd, strictly positive integer and $a>0$. Use again $V(x)=\frac{1}{2}x^2$ you get $\dot{V}(x)=-ax^{k+1}$ which is negative definite because $k+1$ is even.

So for example $\dot{x}=-x$ and $\dot{y}=-3y^5$ have a common Lyapunov function.

These are just some examples, there are many more of course.

Answer 2

Let us consider a family of Hurwitz stable matrices $\{A_1,A_2,\ldots\}$ which may be finite or even infinite.

A result by Liberzon, Hespanha, and Morse proved that it there is basis change $Q$ such that $Q^{-1}A_iQ$ are upper-triangular for all $i$, then there exists a symmetric positive definite matrix $P$ such that $V(x)=x^TPx$ is a Lyapunov function for the system $\dot{x}(t)=A_{\sigma(t)}x(t)$ where $\sigma$ is any function from $\mathbb{R}_{\ge0}$ to $\{1,...\}$.

This can be straightforwardly generalized to continuously parametrized matrices of the form $A(\rho)$ provided that this matrix is uniformly bounded for all possible values of the parameter $\rho$.