# How to pick a Lyapunov function and prove stability?

I am currently trying to learn how to determine the stability of a solution using Lyapunov's Method for autonomous systems.

Say we are given the nonlinear system: $$\dot{x_1}(t)=-x_1(t) + x_1(t)x_2(t)$$ $$\dot{x_2}(t)=-x_2(t)$$ And we want to show that the solution $$x(t)=0$$ is asymptotically stable (I know it is).

We need to pick a Lyapunov function $$V(x)$$ such that $$V(x)$$ is positive definite.

And we need $$\dot{V}(x)$$ to be negative definite to prove asymptotic stability.

I tried $$V(x)=\frac{1}{2}({x_1}^2 +{x_2}^2)$$

Where

$$\dot{V}(x)={x_1}\dot{x_1}+{x_2}\dot{x_2}=-{x_1}^2 +{x_1}^2{x_2} -{x_2}^2$$

As far as I can tell, in this case $$\dot{V}(x)$$ is not negative definite. So what am I missing? If $$V(x)$$ is positive definite & $$\dot{V}(x)$$ is indefinite, do I need to choose a new Lyapunov function? Or do I have to look at different ranges in $$x$$ to determine stability (global vs local stability).

When it comes to selecting Lyapunov functions, how do you know you have a correct function?

To my knowledge there is not a general method for finding a Lyapunov function. In this case one solve the differential equations and use that to find a Lyapunov function. Namely $$x_2$$ is decoupled from $$x_1$$ and can be shown to have the following solution

$$x_2(t) = C_1\,e^{-t},$$

where $$C_1$$ is a constant and depends on the initial condition of $$x_2$$. Substituting the above equation into the expression for $$\dot{x}_1$$ gives

$$\dot{x} = x_1 (C_1\,e^{-t} -1)$$

which is a separable differential equation, namely

$$\frac{dx_1}{x_1} = (C_1\,e^{-t} -1) dt.$$

Integrating on both sides gives

$$\log(x_1) = -C_1\,e^{-t} -t+C_2.$$

Solving for $$x_1$$ gives

\begin{align} x_1(t) &= e^{-C_1\,e^{-t} -t+C_2}, \\ &= C_3\,e^{-C_1\,e^{-t} -t}, \\ &= C_3\,e^{-t}\,e^{-C_1\,e^{-t}}, \end{align}

or when using the definition for $$x_2$$ then it can also be expressed as $$x_1(t)=C_3\,e^{-t}\,e^{-x_2}$$. So the quantities $$x_2$$ and $$x_1\,e^{x_2}$$ will both decay exponentially fast, so the following Lyapunov function can be used

$$V(x) = x_2^2 + x_1^2\,e^{2\,x_2},$$

for which it can be shown that its derivative is

$$\dot{V}(x) = -2\,x_2^2 - 2\,x_1^2\,e^{2\,x_2}.$$

I will leave proving that $$V(x)$$ is radially unbounded to you.

• Great answer. I just want clarify one thing. How did you know how my original Lyapunov function was not a good candidate? Is it because it was indefinite? – Chemical Engineer Oct 26 '18 at 16:54
• @ChemicalEngineer Yes, namely to show assymptotic stability you need that $V(x)$ is positive definite and radially unbounded and $\dot{V}(x)$ is negative definite. Some times you can use LaSalle which states that if $\dot{V}(x)\leq0\ \forall\,x\neq0$ then it can still be assymptotically stable under certain conditions. It can also be noted that if $\dot{V}(x)\leq-\alpha\,V(x)$ with $\alpha>0$ you also have exponential stability, which is the case for this Lyapunov function. – Kwin van der Veen Oct 26 '18 at 17:04

If you compute the discriminant of $$-x_1^2+x_1^2x_2-x_2^2$$ you get $$D= 4-4x_2-4x_1^2$$ which at $$(0,0)$$ is positive. Since both partials are negative, this implies $$(0,0)$$ is a local maximum. So $$\dot{V}$$ is indeed negative near zero.

• But this system is globally assymptotically stable, so there should be a Lyapunov function which proofs that. – Kwin van der Veen Oct 26 '18 at 15:13