# Proving a nonlinear system of derivatives is not asymptotically stable to the origin

I'm trying to prove that a system on nonlinear differential equations is not asymptotically stable to the origin.

The overall problem is an inverted pendulum on a cart that can be controlled with an input u, and uses the variables $\theta, \dot{\theta}, \ddot{\theta}, y, \dot{y}, \ddot{y}$.

I converted the problem into State Space via

$X = \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix} = \begin{bmatrix}\theta\\ \dot{\theta} \\ y \\ \dot{y} \end{bmatrix}$

The State Space becomes

$\dot{X} = f(X) + g(X)*u$

The output is $y = x_3$

I am trying to prove that the system either is or is not minimum phase.

I am stuck trying to show that

$\begin{bmatrix}\dot{x_1}\\\dot{x_2}\end{bmatrix} = \begin{bmatrix}\ \dot{\theta}\\ \ddot{\theta} \end{bmatrix} = \begin{bmatrix}\ x_2 \\ \frac{mgL\sin(x_1)}{(J+mL^2)} \end{bmatrix}$

is not asymptotically stable to the origin. Generally, if the system is linear, I can look at the real part of the eigenvalues of the A matrix to determine stability. For proving a nonlinear system asymptotically stable to the origin, I can use a Lyapunov function to show asymptotic stability.

I am unsure how to prove this reduced system is not asymptotically stable. By inspection, the system is not asymptotically stable to the origin (the origin being $\theta = 0 =>$ pendulum inverted straight up. The system contains no damping, and would oscillate forever if the initial conditions are not the origin.

Would it be enough to state that because $(\theta,\dot{\theta}) = (\pi,0)$ is a solution $$\begin{bmatrix}\dot{x_1}\\\dot{x_2}\end{bmatrix} = \begin{bmatrix}\ 0 \\ 0 \end{bmatrix}$$

the system is not asymptotically stable to the origin?

• If you linearise at the origin, one of the eigenvalues is strictly positive. – copper.hat Mar 19 '18 at 4:08
• Yes, but would that apply to the nonlinear system? I'm trying to disprove the asymptotic stability of the nonlinear system, not a linearized version. If you take a nonlinear system and linearize it about the origin and the linearized system has a strictly positive eigenvalue, does that prove the nonlinear system is also not asymptotically stable? – Cled1990 Mar 19 '18 at 5:25
• Look up Lyapunov's first criterion. – copper.hat Mar 19 '18 at 5:52
• A linearisation is a first order approximation of the nonlinear dynamics. So very close to the linearized equilibrium the dynamics of the nonlinear system will be the same as the linearized system. Since you are only interested in local stability of the origin, then analyzing the linearized system should give the same answer (if no eigenvalue lie of the imaginary axis) as the nonlinear system. – Kwin van der Veen Mar 19 '18 at 7:43
• Your solution at $(\pi,0)$ only implies that the origin cannot be globally asymptotically stable (GAS) but it does not rule out that the origin is locally asymptotical stable. Nonlinear systems can have an arbitrary amount of equilibrium points (discrete and continuous) which might be asymptotically stable or unstable. – MrYouMath Mar 19 '18 at 10:11

## 1 Answer

As other user already proposed you can linearize the system at the origin to obtain:

$$\Delta \dot{\boldsymbol{x}}=\begin{bmatrix}0 & 1\\ \dfrac{mgL}{J+mL^2} & 0\end{bmatrix}\Delta \boldsymbol{x}.$$

The characteristic polynomial of this equation is given by

$$\chi(s) =s^2-\dfrac{mgL}{J+mL^2}.$$

The roots of this characteristic polynomial are $$s_{1,2}=\pm\sqrt{\dfrac{mgL}{J+mL^2}}.$$

Hence, by Lyapunov's indirect method via linearization, we see that the origin of the nonlinear system is unstable because one of the eigenvalues has a strictly positive real part.

An alternative way is to use a Lyapunov function for instability.

A possible choice could be the indefinite function

$$V(\boldsymbol{x}) = x_1x_2$$ $$\implies \dot{V}=\dot{x}_1x_2+x_1\dot{x}_2=x_2^2+x_1\dfrac{mgL}{J+mL^2}\sin x_1$$

For $D=\{\boldsymbol{x}\in \mathbb{R}^2\, | -\pi<x_1<\pi\, \}$ it is possible to find an arbitrarly small $\boldsymbol {x}_0 \in D$ such that $V(\boldsymbol{x}_0)>0$. Additionally we have $\dot{V}>0$ for $\boldsymbol{x}\in D$. Hence, by Lyapunov's Theorem for instability the origin is unstable (See: Hassan Khalil's book Nonlinear Systems Theorem 4.3).