I am working a problem to determine the stability of the given system:

$\ddot{q} + \alpha\dot{q} + \beta q^{3}$

$\alpha \neq 0$

$\beta \neq 0$

With the given candidate lyapunov function:

$V(q,\dot{q}) = \frac{\dot{q^{2}}}{2} + \frac{\beta q^{4}}{4} $

I have worked out the problem but I am having difficulty understanding my results.

From the state space representation

$ f(x) = \begin{bmatrix} x_{2} \\ -\alpha x_2 - \beta x_1^{3} \\ \end{bmatrix} = 0 $

There is an equilibrium point at $x_e =[0\ 0]^T$

I checked the suitability of the candidate lyapunov function by checking

$[\partial xV(x)]\cdot f(x) \leq 0 $ for all $ x \in D $

Which reduced down to:

$-\alpha x_{2}^{2} \leq 0 $

Which holds for all x, provided $\alpha > 0$, so the function is fine as a candidate. I then checked if $x_{e}$ was locally stable using the method below:

$V(x) > V(x_{e}) $ for all $x \in D \setminus \{x_e\}$

$\frac{\dot{q^{2}}}{2} + \frac{\beta q^{4}}{4} > 0$

Which the lyapunov function satisfies this condition.

Then for local asymptotic stability I looked at the following condition

$\partial{x}V(x)f(x) < 0$ for all $x \in D \setminus \{x_e\}$

$-\alpha x_{2}^{2} < 0$

Which led me to believe that the system was locally asymptotically stable, but when I used the indirect method of lyapunov the system linearized about the equilibrium was not Hurwitz.

Clearly I am missing some understanding here.

  • 2
    $\begingroup$ What about $x_2=0$ and $x_1\neq0$, this satisfies $x\neq0$? $\endgroup$ Oct 31 '17 at 16:54
  • 1
    $\begingroup$ Also, since you wrote $\beta \neq 0$, notice that if $\beta < 0$, your $V(x)$ is not a valid candidate Lyapunov function anymore. $\endgroup$
    – SampleTime
    Oct 31 '17 at 17:28

Given your system as

$$ \begin{split} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= -\alpha x_2 - \beta x_1^3 \,. \end{split} $$

Using the indirect method of linearization around the equilibrium $\textbf{x}_e = \textbf{0}$, we get

$$ \dot{\textbf{x}} = A(\textbf{x}) |_{\textbf{x} = \textbf{0}} \textbf{x} = \begin{bmatrix} 0 & 1 \\ -3\beta x_1^2 & -\alpha \end{bmatrix}\Bigg\rvert_{\textbf{x} = \textbf{0}} \textbf{x} = \begin{bmatrix} 0 & 1 \\ 0 & -\alpha \\ \end{bmatrix} \textbf{x}\,. $$

Notice that the eigenvalues of $A(\textbf{0})$ are located at $\lambda_1 = 0$, $\lambda_2 = -\alpha$.

This means we have the special case here that one eigenvalue lies on the imaginary axis. This means that no conclusion about stability can be drawn using the indirect method. The system could be either stable or unstable.

This is the classical case where investigating stability by looking at the linearization of the system fails.

To check for stability, you can use the direct method as you already did. Notice however that the function you used only shows stability, not asymptotic stability since $\dot{V}(\textbf{x})$ is only negative semi definite.

However, you can use LaSalle's theorem to show the system is actually asymptotically stable (see the comment by @Kwin van der Veen). Let

$$ S=\{\mathbf{x}\,| \, \dot{V}(\mathbf{x}) = 0 \} \,. $$

Since for any $\mathbf{x} \in S$ with $\mathbf{x} \neq \mathbf{0}$ we get $\dot{x}_2 \neq 0$, no trajectory other than the trivial one can stay in $S$. Thus, by LaSalle's theorem, the origin is asymptotically stable.

  • 3
    $\begingroup$ It can also be noted that if from linearization it follows that the equilibrium point is stable, that the system is (locally) exponentially stable. While if $V(x)>0\,\forall\,x\neq x_e$ and $\dot{V}(x)<0\,\forall\,x\neq x_e$ then the system is (globally) asymptotically stable (only if $\dot{V}(x) \leq -k\,V(x)$ with $k>0$ then you also have exponentially stability). $\endgroup$ Oct 31 '17 at 16:51
  • $\begingroup$ How is my function $-\alpha x_{2}^{2} < 0$ only positive semi-definite? That's the part I don't understand. Here my current knowledge leads me to think that the equation is always less that 0 because the equilibria $x_{e} = [0 \ 0]^T$ is excluded from the evaluation (via $\setminus \{x_{e}\}$). $\endgroup$ Oct 31 '17 at 17:38
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    $\begingroup$ It is negative semi-definite (not positive) because it is zero everywhere along the line $x_2 = 0$. For example, $\dot{V}(\smash{\begin{bmatrix}1 & 0\end{bmatrix}^T}) = 0$ but to show global asymptotic stability, it has to be negative everywhere exept the origin. $\endgroup$
    – SampleTime
    Oct 31 '17 at 17:43
  • 1
    $\begingroup$ Correct. Since $\dot{V}$ is zero for $x_2 = 0$ and $x_1$ arbitrary, $\dot{V}$ is negative semi-definite. This is enough to show stability in the sense of Lyapunov, because $\dot{V} \leq 0$ for $x \neq 0$. However, to show asymptotic stability, you need $\dot{V} < 0$ for $x \neq 0$ (negative definite). $\endgroup$
    – SampleTime
    Oct 31 '17 at 18:11
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    $\begingroup$ You can also apply LaSall's invariance principle in order to still potentially show (global) asymptotic stability. $\endgroup$ Oct 31 '17 at 18:24

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