# Proving that a second order system is not exponentially stable

This is a homework exercise that I've been struggling to solve. Any help is appreciated.

Consider the system:

$$\dot{x}_1 = x_2 \\ \dot{x}_2 = -x_1 - g(t)x_2$$

Where g(t) is continuously differentiable and $$0 < k_1 \leq g(t) \leq k_2$$ for all $$t \geq 0$$.

The first part of this exercise asks to prove that the origin is exponentially stable. I solved this part by considering a candidate Lyapunov function $$V(x) = \frac{1}{2}(x_1^2 + x_2^2)$$, and proving that for $$a = 2$$, the inequality

$$k_1||x||^a \leq V(x) \leq k_2 ||x||^a$$

holds. Although I'm not really sure if this is the right way to prove it, my attempt was to consider that $$\dot{V}(x)$$ can be written as $$\dot{V}(x) = -x^TPx$$,

where P = $$\begin{bmatrix}0 & 1 \\ -1 & g(t)\end{bmatrix}$$,

by using the fact that $$0 < k_1 \leq g(t) \leq k_2$$, and by calculating the eigenvalues of P and showing that it satisfies:

$$\lambda_{min}(P)x^Tx \leq x^TPx \leq \lambda_{max}(P)x^Tx$$, for $$k_2 \geq 2$$.

The second part of the exercise asks if the system is exponentially stable if $$g(t)$$ were not bounded. As an example, the exercise uses $$g(t) = 2 + e^t$$. From this part on, I couldn't imagine how to proceed and show that the origin (I guess) is not exponentially stable. In this specific case, how do I prove if the system is or isn't exponentially stable for $$g(t) = 2 + e^t$$?

• $P=\begin{bmatrix}0&-1\\1&g(t)\end{bmatrix}$. More to the point: if $x_2=0$ then $x^TPx=0$, so the first inequality in the last display does not hold for $g(t)\equiv2$, for example (then $\lambda_{min}(P)=\lambda_{max}(P)=1$). – user539887 Apr 30 at 5:34

I managed to solve this problem. By using Barbashin-Krasovskii Theorem, it is possible to determine if the origin is exponentially stable in case the following properties hold:

1. V is $$C^1$$
2. $$k_1||x||^a \leq V(x) \leq k_2||x||^a$$
3. $$\dot{V}(x) \leq -k_3 ||x||^a$$
4. $$k_1, k_2, k_3, a > 0$$.

Looking at the Lyapunov candidate function $$V(x)$$, it is possible to rewrite it as $$V(x) = \frac{1}{2}x^TPx$$, or $$V(x) = \frac{1}{2}||x||^2$$. Considering the second part of Barbashin-Krasovskii Theorem, it is already possible to write it, for this problem, as:

$$k_1||x||^a \leq \frac{1}{2}||x||^2 \leq k_2||x||^a$$

which is true using: $$k_1 = k_2 = \frac{1}{2}$$ and $$a = 2$$. It is also valid for $$0 < k_1 \leq g(t) \leq k_2$$.

For the third part, we just need $$\dot{V}(x) \leq -k_3||x||^2$$, since we already defined $$a = 2$$. By calculating $$\dot{V}(x)$$, we concluded that $$\dot{V}(x) = -g(t)x^2_2$$. Since $$g(t) \leq k_2$$, it is possible to rewrite $$\dot{V}(x)$$ as $$\dot{V}(x) \leq -k_2x^2_2$$. Therefore, if we choose $$k_3 = k_2$$, for example, the third condition of the theorem is also satisfied and the origin is exponentially stable.

For the second item of the exercise, if $$g(t) = 2 + e^t$$, then the system can be written as:

$$\dot{x}_1 = x_2 \\ \dot{x}_2 = -x_1 - x_2(2 + e^t)$$

This system has the following solution: $$x_1(t) = -(1+e^{-t}), x_2 = e^{-t}$$. Now, checking the provided solution for $$t \rightarrow \infty$$, $$x_1(t)$$ = -1 and $$x_2(t) = 0$$. This shows that the origin is not asymptotically stable, thus not exponentially stable.

• Are you sure that $\dot{V}(x)\leq-k_2x_2^2$ implies exponential stability? – Ilbant May 3 at 8:04
• As @Ilbant wrote, this is wrong because your $\dot{V}$ is only negative semi-definite. Note that because of this you can't even say that the system is asymptotically stable (and because it is a time-varying system, LaSalle won't work here). – SampleTime May 14 at 18:38
• It's easy to see that for $V(x)=x^TPx$, $P=I_{2\times2}$, and for $A=\begin{bmatrix}0 & 1 \\ -1 & -g(t)\end{bmatrix}$, then $A^TP+PA\leq0$ for $g(t)\ge0$, $\forall t\ge0$ which implies uniform stability. What if you use $P=\begin{bmatrix}1 & \tfrac{1}{4} \\ \tfrac{1}{4} & 1\end{bmatrix}$? I think you can find bounds on $g(t)$ to have exponential stability. Since $g(t)$ is continuously differentiable, you may also find a time-varying Lyapunov based on $g(t)$ and get different bounds. Probably you will need bounds on the derivative of $g(t)$. – Ilbant May 15 at 8:54