Lyapunov stability of nonlinear system

Mar 2020
6
0
Norway
Problem 3.5, system 3 in "Nonlinear Control" by Khalil:
Determine whether the origin is stable, asymptotically stable or unstable
\(\displaystyle \dot{x}=\begin{bmatrix}x_2+x_3\\-\sin(x_1)-x_3\\-\sin(x_1)+x_2 \end{bmatrix} \)

When linearizing we get a system matrix whose eigenvalues have zero real part, so linearization fails to conclude about the nonlinear system.

Trying the Lyapunov candidate, \(\displaystyle V(x)=\frac{1}{2}\left(x_1^2 + x_2^2 + x_3^2\right)\) leads to
\(\displaystyle \begin{align}\dot{V}&=x_1(x_2+x_3)+x_2(-\sin(x_1)-x_3)+x_3(-\sin(x_1)+x_2)\\&=x_2(x_1-\sin(x_1)) + x_3(x_1 - \sin(x_1))\end{align}\)
which is not negative (semi) definite.

From what I understand, you cannot conclude instability from the failure of a certain Lyapunov candidate. So, how to proceed ?
 

SDK

Sep 2016
793
540
USA
Problem 3.5, system 3 in "Nonlinear Control" by Khalil:
Determine whether the origin is stable, asymptotically stable or unstable
\(\displaystyle \dot{x}=\begin{bmatrix}x_2+x_3\\-\sin(x_1)-x_3\\-\sin(x_1)+x_2 \end{bmatrix} \)

When linearizing we get a system matrix whose eigenvalues have zero real part, so linearization fails to conclude about the nonlinear system.
Correct
Trying the Lyapunov candidate, \(\displaystyle V(x)=\frac{1}{2}\left(x_1^2 + x_2^2 + x_3^2\right)\) leads to
\(\displaystyle \begin{align}\dot{V}&=x_1(x_2+x_3)+x_2(-\sin(x_1)-x_3)+x_3(-\sin(x_1)+x_2)\\&=x_2(x_1-\sin(x_1)) + x_3(x_1 - \sin(x_1))\end{align}\)
which is not negative (semi) definite.

From what I understand, you cannot conclude instability from the failure of a certain Lyapunov candidate. So, how to proceed ?
This last statement is also correct. Construction of a Lyapunov function is generally a difficult problem. It requires looking at the higher order term in the expansion of the flow. To start, can you explain why you thought $V(x) = \frac{1}{2} \left|x\right|^2$ would be a Lyapunov function for this system?
 
Mar 2020
6
0
Norway
To start, can you explain why you thought $V(x) = \frac{1}{2} \left|x\right|^2$ would be a Lyapunov function for this system?
I have no specific reason other than that I know that the candidate is a positive definite function whose derivative is a function of all the states.
 

SDK

Sep 2016
793
540
USA
I have no specific reason other than that I know that the candidate is a positive definite function whose derivative is a function of all the states.
I see. Here is a hint. Consider the (autonomous) planar ODE:
\[
\dot x =
\begin{pmatrix}
x_2 \\
h(x_1)
\end{pmatrix}
\]
where $h$ is continuous. This always produces an integrable system with the Hamiltonian given by
\[
H(x) = \frac{1}{2} x_2^2 + \int_0^{x_1} h(s) \ ds
\]
which can just be checked directly via the fundamental theorem of calculus. Now, if we add some friction as a linear function, the ODE becomes
\[
\dot x =
\begin{pmatrix}
x_2 \\
-\gamma x_2 + h(x_1)
\end{pmatrix}
\qquad \gamma > 0
\]
and the system is now dissipative. The proof of this is very easy since the Hamiltonian in the integrable case is precisely the Lyapunov function required. Also note that all equilibria are preserved but in the old case they were all center type and in the new system they are all stable.

Now, a well known example of this setting is the damped pendulum which is the case where $h(x) = -\sin(x_1)$ (after normalizing the mass). If we also take $\gamma = 1$ we obtain
\[
\dot x =
\begin{pmatrix}
x_2 \\
-x_2 - \sin(x_1)
\end{pmatrix}
\]
and the Lyapunov function is given by $V(x) = -\frac{1}{2} x_2^2 + (1 - \cos(x_1))$.

Notice that this vector field is extremely similar to your situation. In particular, the nonlinear part of your system occurs only in the 2nd and 3rd coordinates, each of which is a function of only the 1st coordinate.
 
Last edited:
Mar 2020
6
0
Norway
I tried the Lyapunov function
\(\displaystyle V(x) = \frac{1}{2}\left(x_2^2 + x_3^2\right) + \int_0^{x_1} \sin(s)ds\)
which is positive definite and has a derivative
\(\displaystyle \begin{align}\dot{V}(x) &= x_2\dot{x}_2 + x_3 \dot{x}_3 + \sin(x_1)\dot{x}_1\\ &=x_2(-\sin(x_1)-x_3)+x_3(-\sin(x_1)+x_2)+\sin(x_1)(x_2+x_3)\\&=0\end{align}\)
So, we have a \(\displaystyle \dot{V}\) that is identically zero. So, we can conclude that the origin is stable ?
 
Last edited:
Mar 2020
6
0
Norway
This is how the system behaves for an initial condition \(\displaystyle x_0=(0, 0.1, 0.1)\)
Figure_1.png
I think the origin is a center.
 

SDK

Sep 2016
793
540
USA
I tried the Lyapunov function
\(\displaystyle V(x) = \frac{1}{2}\left(x_2^2 + x_3^2\right) + \int_0^{x_1} \sin(s)ds\)
which is positive definite and has a derivative
\(\displaystyle \begin{align}\dot{V}(x) &= x_2\dot{x}_2 + x_3 \dot{x}_3 + \sin(x_1)\dot{x}_1\\ &=x_2(-\sin(x_1)-x_3)+x_3(-\sin(x_1)+x_2)+\sin(x_1)(x_2+x_3)\\&=0\end{align}\)
So, we have a \(\displaystyle \dot{V}\) that is identically zero. So, we can conclude that the origin is stable ?
Yes the origin is stable. Keep in mind $\dot V = 0$ is not quite enough to conclude this though. You also need the level sets of $V$ to be bounded. But its easy to see that they are in this case and combined with the fact the $\dot V = 0$ you can conclude that the origin is stable. This is also enough to conclude that it can't be asymptotically stable. Do you see why?
 
Mar 2020
6
0
Norway
The origin can't be asymptotically stable. Do you see why?
I guess that you are hinting to the fact that \(\displaystyle V \) cannot decrease towards zero when \(\displaystyle \dot{V} \) is zero. And since \(\displaystyle V \) is positive definite it is zero only at the origin.