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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?

Chemical Engineer
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2 Answers2

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To my knowledge there is not a general method for finding a Lyapunov function. In this case one can 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

$$ \dot{x}_1 = x_1 (C_1\,e^{-t} -1) $$

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.

Kwin van der Veen
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  • 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
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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.

Matthew C
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