I am a student and I am studying Lyapunov stability. I am considering the following system:




and I want to study its stability at the origin. I am having some troubles understanding how to operate.

I have understood that the first thing to do is to choose a candidate Lyapunov function, but in this case I dont' understand how to choose it.

I know it has to be positive definite, so I should choose something like:


con $P$ positive definite and symmetric.

But in my case I don't find really easy finding a lyapunov function.

I have a sketch of solution for this, and as Lyapunov function has been chosen:


but to be honest I have not clear why.

I have studied that it can be used the Graient Method to find a Lyapunov function. I have seen that this method starts directly from the gradient of the Lyapunov function, imposing that it has to be less or equal to zero, so negative semi definite, but I have not clear how to use it.

I have also seen that it is possible to use an indirect method, which consists in considering a linear approximation of the original nonlinear model, and sudying the stability of the linearized model with the eigenvalues of the Jacobian matrix.

This method seems much simpler than choosing a Lyapunov function and working in the direct way, but can I always use it? I have seen that the equilibrium point should be isolated, but how do I check this?

Can somebody please help me?

  • $\begingroup$ I updated the answer because it was only true for $b \neq 1$. $\endgroup$
    – SampleTime
    Jun 24, 2020 at 17:47
  • 2
    $\begingroup$ The Lyapunov function is $\frac12 x_2^2+\frac12 x_3^2+a(1-\cos x_1)$ $\endgroup$
    – AVK
    Jun 24, 2020 at 17:49

1 Answer 1


These are multiple question, I will try to answer some of them.

I assume that $a, b > 0$.

An equilibrium is isolated if there does not exist a small "ball" around it such that it is the only equilibrium in that ball. For example:

$$ \dot{x_1} = 0 $$

has no isolated equilibrium because no matter how close you look at an equilibrium, there are always infinitely many other equilibrium points close by. Similiar, the system

$$ \begin{align} \dot{x}_1 &= x_1 - x_2 \\ \dot{x}_2 &= x_1 - x_2 \\ \end{align} $$

has no isolated equilibrium: For example $x_1 = x_2 = 1$ is is an equilibrium, but so is $x_1 = x_2 = 1.1$ and $x_1 = x_2 = 1.01$ and $x_1 = x_2 = 1.00000001$ and so on. No matter how close you look, there are always "more" equilibrium points.

So to your example, you have:

$$ \begin{align} \dot{x}_1 &= x_2 + x_3 \\ \dot{x}_2 &= -a \sin(x_1) - b x_2 \\ \dot{x}_3 &= -a \sin(x_1) + x_3 \end{align} $$

To get the equilibrium points you set $\dot{x} = 0$ so you need $x_2 = -x_3$ from the first equation. Then you get

$$ \begin{align} \dot{x}_2 &= \dot{x}_3 = 0 \\ \implies -a \sin(x_1) - b x_2 &= -a \sin(x_1) + x_3 \\ \implies -b x_2 &= x_3 \\ \implies -b x_2 &= -x_2 \end{align} $$

We assumed that $b > 0$ and if $b \neq 1$ the last equation can only be true if $x_2 = 0$ and then $x_3 = 0$. If you insert that you get

$$ -a \sin(x_1) = 0 $$

so the equilbrium points are $x_1 = \sin(k \pi), k \in \mathbb{Z}, x_2 = 0, x_3 = 0$. Although that means you have infinitly many equilibrium points they are all isolated because the roots of $\sin(x_1) = 0$ are isolated. Conclusion: You can use Lyapunov method.

If $b = 1$ the system has no isolated equilibrium points.

You can use indirect method but it will only tell you local results and sometimes you can't use it (if you get zero eigenvalues).

You have the function

$$ V(x) = \frac{1}{2} x^T x + a(1 - \cos(x_1)) $$

but it is not a Lyapunov function for your system:

$$ \dot{V}(x) = - b x_2^2 + x_1 x_2 - x_3^2 + x_1 x_3 $$

If $x_3 = 0$ and $x_1 = 2 b x_2$ you get $\dot{V}(x) = b x_2^2$ and that is positive no matter how small you make $x_2$. So this function can't be used.


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