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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 $g(t) = 2 + e^t$?

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  • $\begingroup$ $P=\begin{bmatrix}0&-1\\1&g(t)\end{bmatrix}$. 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 $\lambda_{min}(P)=\lambda_{max}(P)=1$). $\endgroup$ – user539887 Apr 30 at 5:34
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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.

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    $\begingroup$ Are you sure that $\dot{V}(x)\leq-k_2x_2^2$ implies exponential stability? $\endgroup$ – Ilbant May 3 at 8:04
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    $\begingroup$ 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). $\endgroup$ – SampleTime May 14 at 18:38
  • $\begingroup$ 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 $g(t)$. $\endgroup$ – Ilbant May 15 at 8:54

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