Are there Lyapunov stability conditions that focus on a subset of the available state equations? Essentially, I have a system of nonlinear ODEs and I only care that some of the variables converge to zero and not others. For example, consider the contrived system:

\begin{align} \dot{y_1} =& -y_1 + e^{-t}y_2 \\\\ \dot{y_2} =& \frac{5}{2}\pi\cos(10\pi t) + e^{-t}y_1\\\\ y_1(0) =& 1\\\\ y_2(0) =& 1 \end{align}

This has a solution seen here:

Plot of the solution of the above equations

Clearly, $y_1$ converges to $0$ and $y_2$ does not. Now, normally Lyapunov stability conditions come in the form of something like:

If $V (y, t)$ is locally positive definite and decrescent, and $\dot{V} (y, t) \leq 0$ locally in y and for all t, then the origin of the system is uniformly locally stable (in the sense of Lyapunov).


$$ \dot{V}\bigg\rvert_{\dot{y}=f(y,t)} = \frac{\partial V}{\partial t} + \frac{\partial V}{\partial y} f. $$

However, this doesn't work in the above case because $y_2$ continues to oscillate around some nonzero value. In truth, I don't care about $y_2$, I only care about $y_1$, so is there a transformation or alternative Lyapunov stability theorem that only focuses on this quantity?


This is known as partial stability and was pioneered by Rumyantsev. You may want to read:

  • Rumyantsev, V. V. "On the stability of motion relative to some of the variables." Vestnik Moskov. Gos. Univ., Ser. Mat., Mekh., Fiz., Astron., Khim. 5 (1957).
  • Rumyantsev, V. V. "On the stability with respect to a part of the variables." Nonlinear mechanics and stability (1970): 243-265.
  • Vorotnikov, V. I. "On the theory of partial stability." Journal of Applied Mathematics and Mechanics 59.4 (1995): 525-531.
  • Vorotnikov, V. I.. Partial stability and control. Springer Science & Business Media (2012).

In short, if you have

$$ \begin{align} \dot{x}_1 &= f_1(x_1, x_2) \\ \dot{x}_2 &= f_2(x_1, x_2) \end{align} $$

with $f_1(0, x_2) = 0$ for all $x_2$ and you only care whether $x_1\rightarrow 0$ then you can show that with $V(x_1)$ with $V(0) = 0$ and $V(x_1) > 0$ for all $x_1 \neq 0$ if you also can show that

$$ \dot{V}(x_1, x_2) < 0 $$

for all $x_1$ and $x_2$. The problem is (as always) that you first need to find $V$.

Edit: I just noticed that for your system the equilibrium condition is not met so unfortunatly this won't work for your system. But there are many other ways to check partial stability which can be found in the book by Vorotnikov. Maybe one of these can be used here.

  • $\begingroup$ The reason this doesn't work for this system is I think because $|y_2(0)|$ can be chosen arbitrarily large which can cause $y_1(t)$ to take an arbitrarily long time before it is within some bound of $0$. $\endgroup$ May 12 '21 at 11:38

For the given contrived system one can do a coordinate transformation. When checking the stability of $y_1$ a natural set of coordinates would be $x_1 = y_1$ and $x_2 = \dot{y}_1$, yielding

\begin{align} \dot{x}_1 &= x_2, \\ \dot{x}_2 &= -2\,x_2 - \left(1 - e^{-2\,t}\right) x_1 + \frac{5}{2}\pi\cos(10\,\pi\,t)\,e^{-t}, \end{align}

which is a linear time varying inhomogeneous system. This isn't super trivial to analyse, but I think it is possible to show that this is exponentially stable.

The transformation back to the original coordinates yields $y_1 = x_1$ and $y_2 = e^t \left(x_1 + x_2\right)$. This transformation for $y_2$ at $t\to\infty$ is not well defined, which explains why even though $x_1,x_2\to0$ as $t\to\infty$ does not imply that $y_2\to0$.

  • $\begingroup$ Nice, from this we can simplify more: we can drop the "input" because it is bounded and converges to zero as $t \rightarrow 0$. So it is sufficient to show the system $\dot{x}_1 = x_2, \dot{x}_2=-2x_2-(1-e^{-2t})x_1 + u$ is input to state stable then also $(x_1,x_2)\rightarrow (0,0)$ . However LTV systems with bounded $B(t)$ matrix are input to state stable if the system without inputs ($u=0$) is exponentially stable. So the last missing piece would be to show $\dot{x}_1 = x_2, \dot{x}_2=-2x_2-(1-e^{-2t})x_1$ is exponentially stable. $\endgroup$
    – SampleTime
    May 14 '21 at 13:39
  • 1
    $\begingroup$ @SampleTime I initially tried to find the state transition matrix, which could then also be used by plugging that into the convolution integral to find the forced response. Plugging the homogeneous differential equation into Wolfram Alpha did yield an analytical solution containing Bessel functions of the first and second kind. $\endgroup$ May 14 '21 at 16:36
  • $\begingroup$ That should do it because Wolfram Alpha says that $\lim_{t\rightarrow \infty} \sqrt{e^{-2 t}} K_0(\sqrt{e^{-2 t}}) = \lim_{t\rightarrow \infty} \sqrt{e^{-2 t}} I_0(\sqrt{e^{-2 t}}) = \lim_{t\rightarrow \infty} e^{-2 t} K_1(\sqrt{e^{-2 t}}) = \lim_{t\rightarrow \infty} e^{-2 t} I_1(\sqrt{e^{-2 t}}) = 0$. $\endgroup$
    – SampleTime
    May 14 '21 at 20:42
  • $\begingroup$ The $I_n(z)$ is the modified Bessel function of the first kind and $K_n(z)$ of the second kind. So that does show $x_1,x_2 \rightarrow 0$. $\endgroup$
    – SampleTime
    May 14 '21 at 20:44
  • $\begingroup$ @SampleTime I believe linear time varying systems with bounded $|A(t)|$ won't have finite time blow up. Therefore, given that the in-homogeneous part goes to zero and $A(t)$ becomes constant as $t\to\infty$, it can be sufficient to look at the eigenvalues of $A(\infty)$ in order to determine stability in this case (since the two eigenvalues are both -1). Though, I am not 100% if this approach would be sufficiently rigorous in general. $\endgroup$ May 14 '21 at 23:05

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