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

  • $\begingroup$ $P=\begin{bmatrix}0&-1\\1&g(t)\end{bmatrix}$. 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

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.

  • 1
    $\begingroup$ Are you sure that $\dot{V}(x)\leq-k_2x_2^2$ implies exponential stability? $\endgroup$ – Ilbant May 3 at 8:04
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.