How do I analyze the stability of a system with Lyapunov method?

Question

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

$\dot{x_1}=x_2+x_3$

$\dot{x_2}=-asinx_1-bx_2$

$\dot{x_3}=-asinx_1-x_3$

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:

$V(x)=x^TPx$

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:

$V(x)=\frac{1}{2}x^Tx+a(1-cosx_1)$

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?

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.

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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.

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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).

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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.