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I have a problem solving the following subject:

Determine the stability properties and convergence on the origin using Lyapunov Direct Method:

$\dot{x_1} = x_2 - \frac{x_1 \cos{t}}{k_0 + \sin{t}} \\ \dot{x_2} = - x_2 - x_1(k_0 + \sin{t}) , k_0 > 1$

My problem is that I haven't been able to find the correct Lyapunov candidate function so that:

  1. $V(0) = 0 $

  2. $V(x) > 0 $ when $x \neq 0$

  3. $\dot{V}(x) \leq 0$

So far I've used: $V = \frac{x_2^2 x_1^2}{2}$ and $V = \frac{x_1^2}{2} + \frac{x_2^2}{2}$, I don't even know where to go next... I am a total novice at this...

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    $\begingroup$ Then your first $V$ is not positive definite, so can already be excluded. $\endgroup$ Feb 1, 2018 at 15:11
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    $\begingroup$ Since its a linear time varying system, you could consider the transition matrix. This you can also consider to construct a Lyapunov function $\endgroup$
    – Carlos
    Feb 3, 2018 at 7:58
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    $\begingroup$ @MrYouMath: The topic is treated e.g. in Khalil H.K., Nonlinear Systems, Prentice Hall and Rugh W.J., Linear System Theory, Prentice Hall. $\endgroup$
    – Carlos
    Feb 3, 2018 at 12:08
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    $\begingroup$ There are several procedures. Specificly for periodic coefficients, a procedure based of Floquet theory is useful, as you already mentioned: Let $\Phi(t,t_0)$ the state transition matrix, i.e., the solution is given by $x(t) = x(t_0)\Phi(t,t_0)$. (It can be obtained from Peano-Baker series). When $A(t+T) = A(t)$ it can be defined a matrix $B$ by $\exp(BT) = \Phi(T,0)$, and another matrix $P(t) = \exp(Bt)\Phi(0,t)$. It can be shown that $\Phi(t,\tau) = P^{-1}(t)\exp((t-\tau)B)P(\tau)$. The origin of the LTV system is exponentially stable if $B$ is Hurwitz. $\endgroup$
    – Carlos
    Feb 3, 2018 at 12:08
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    $\begingroup$ Another, more general approach, might be $P(t) = \int_t^{\infty} \Phi^T(\tau,t)Q(\tau)\Phi(\tau,t)\mathrm d\tau$. Then a Lyapunov function candidate is $V=x^TP(t)x$ and the origin is exponentially stable, if $Q = Q^T>0$. $\endgroup$
    – Carlos
    Feb 3, 2018 at 12:08

2 Answers 2

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This is a solution using the direct method of Lyapunov. (using other procedures might be more straightforward here)

The system is nonautonomous (time variant), therefore, a time variant Lyapunov function might be suited for the problem.

Consider the Lyapunov function:

$V = a(t)\frac{1}{2}x_1^2 + b(t)\frac{1}{2}x_2^2$,

where $0<a(t)<a_1$, $\forall t$ and $0<b(t)<b_1$, $\forall t$, with constants $a_1,\,b_1$. The parameters $a(t)$ and $b(t)$ must be determined. ($a$ and $b$ must be positive and bounded that $V$ is positive definite.)

The derivative along the trajectories of the system

$\dot x_1 = x_2 - \frac{x_1 \cos{t}}{k_0 + \sin{t}} \\ \dot x_2 = - x_2 - x_1(k_0 + \sin{t}) , k_0 > 1$

is:

\begin{align} \dot V & = \dot a \frac{1}{2}x_1^2 + a x_1 \dot x_1 + \dot b \frac{1}{2}x_2^2 + b x_2 \dot x_2 \\ & = \dot a \frac{1}{2}x_1^2+ \dot b \frac{1}{2}x_2^2\\ & \quad+ a x_1 ( x_2 - \frac{x_1 \cos{t}}{k_0 + \sin{t}}) \\ & \quad+ bx_2(- x_2 - x_1(k_0 + \sin{t}) )\\ & = \dot a \frac{1}{2}x_1^2+ \dot b \frac{1}{2}x_2^2\\ & \quad + a x_1 x_2 - a x_1^2\frac{ \cos{t}}{k_0 + \sin{t}} \\ & \quad - b x_2^2 - x_1x_2 b (k_0 + \sin{t}) ) \\ &= -( a\frac{ \cos{t}}{k_0 + \sin{t}} -\dot a \frac{1}{2})x_1^2 - (b-\dot b \frac{1}{2})x_2^2 \\ & \quad+ (a-b(k_0 + \sin{t}) ))x_1x_2 \end{align}

Now choose $b =\sin(t) + k_0 >0$ and $a = (\sin(t) + k_0)^2>0$, i.e. $\dot a = 2\cos(t)(\sin(t) + k_0) $, $\dot b = \cos(t)$:

\begin{align} \dot V & \leq - (\sin(t) + k_0 - \frac{\cos(t)}{2})x_2^2 \leq 0, \quad \text{for} \,\,k_0>\sqrt{5}/2\approx 1.1 \end{align} The derivative of $V$ is negative semi definite, i.e., the origin is stable.

Since the system is time varying, the standard invariance principle can not be considered to investigate asymptotic stability. Then, usually Barbalat's Lemma is considered. Roughly it says that: If a function $\phi(t)$, with $\lim_{t\to\infty} \phi(t)= c<\infty$, has a uniformly continuous derivative, then $\lim_{t\to \infty}\dot \phi(t) \to 0$. For the problem here, we can say that $V$ is lower bounded, $\dot V$ is negative semidefinite, (hence we know that V converges to a limit) and it is uniformly continuous since

$\ddot V = - (\sin(t) + k_0 - \frac{\cos(t)}{2})x_2(- x_2 - x_1(k_0 + \sin{t})) - (\cos(t) + k_0 - \frac{\sin(t)}{2})x_2^2 $

is bounded. (This can be concluded since $x_1$ and $x_2$ converge to finite limits.)

Hence, $\dot V \to 0$ for $t \to \infty$. For $x_2 \equiv 0$, $x_1$ must also be zero as it can be obtained from the system equations.

The origin is even exponentially stable.

In summary:

$V = \frac{(\sin(t) + k_0)^2}{2}x_1^2 + \frac{(\sin(t) + k_0)}{2}x_2^2$, positive definite and decrescent

$\dot V \leq - (\sin(t) + k_0 - \frac{\cos(t)}{2})x_2^2 \leq 0$, for $k_0>\sqrt{5}/2$, negative semi definit $\Rightarrow$ Lyapunov stable equilibrium

Barbalat's Lemma can be applied $\Rightarrow$ asymptotically stable equilibrium

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  • $\begingroup$ Thank you so much, you are my hero... $\endgroup$ Feb 5, 2018 at 21:37
  • $\begingroup$ @SergioPertuz: You‘re welcome $\endgroup$
    – Carlos
    Feb 6, 2018 at 7:44
  • $\begingroup$ +1: For the answer! How did you come up with the bound $k_0$ I would think that $k_0>\sqrt{5}/2$. Also the formulation of Barbalat’s Lemma is problematic as the limit $t\to\infty$ does not exist for your Laypunov function. I think you simply meant bounded. $\endgroup$
    – MrYouMath
    Feb 6, 2018 at 20:54
  • $\begingroup$ @MrYouMath : Thanks for your feedback! you‘re right with $\sqrt{5}/2$, i corrected it above. Also boundedness is the correct statement as it is necessary for the uniformly continuous derivative of $V$. $\endgroup$
    – Carlos
    Feb 7, 2018 at 19:00
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This is not an answer to the problem by using Lyapunov's direct method. I am looking forward to a solution using Lyapunov's direct method :D. Doing this with a Lyapunov function does seem to be tricky.

There is an alternative, but more computationally heavy, way which is applying Floquet theory (example). Because this system is a linear time-variant system with $2\pi$-periodic coefficients. You can also determine a non-elementary closed form solution (e.g. by using Maple) for this ODE which can be used to determine the eigenvalues of the monodromy matrix.

enter image description here

The $S,S',C,C'$ functions are called Mathieu functions. Numerical analysis of the eigenvalues of the monodromy matrix suggests that the system is asymptotically stable for $k_0 = 1.01,10,100, 1000$ because all eigenvalues lie inside the unit circle. I am pretty sure one can use the properties of the Mathieu functions to proof that the eigenvalues are always inside the unit circle.

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