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In class, we are given the Lyapunov Stability Theorem, which reads: For an LTI autonomous system $\dot x = Ax$, the following statements are equivalent:

  1. The system is asymptotically stable.
  2. All eigenvalues of matrix A have a negative real part.
  3. For every symmetric and positive definite matrix $Q$, there exists a unique solution $P$ to the Lyapunov equation $PA + A^TP = -Q$, where $P$ is positive definite.

Then the note gives one example where there exists a $Q$ such that a unique positive definite $P$ is found. It directly states that the system is asymptotically stable. And another example where a positive definite $P$ is given and found a positive definite $Q$, also states that the system is asymptotically stable.

What I am confused is: The 3) condition reads "'FOR ALL' p.d. $Q$, there exists a unique p.d. P that satisfies the Lyapunov condition, then the system is asymptotically stable". How can, in the example given, only finding one $Q$ with unique $P$ can conclude that the system is asymptotically stable? Also, how can finding one $P$ with unique $Q$ can conclude the system is asymptotically stable?

In my opinion, it is impossible to use condition 3) to conclude if the system is asymptotically stable or not because it is impossible to enumerate all $Q$. Did I misunderstand anything with the Lyapunov Stability Theorem here?

Thank you so much for reading my questions and the answers!

For reference, I have taken two classes that stated this theorem with an example. Here are the screenshots. The first two images are from the first lecture with an example of "given a $P$, find a $Q$ and the system is asymptotically stable". The next three images are from the second lecture, saying the same theorem, with an example of "given a $Q$, find a $P$ and the the system is asymptotically stable".

Appreciatively, William

Lecture 1 theorem: Lecture 1-1 Lecture 1 example: Lecture 1-2

Lecture 2 theorem: Lecture 2-1 Lecture 2-2 Lecture 2 example: Lecture 2-3

I found a proof from the same prof's higher level class proving that these three statements are equivalent, which further confuses me why it's sufficient to have one satisfying example and conclude the asymptotical stability.

proof-1 proof-2

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  • $\begingroup$ Do you have a reference for this? I can't find a version of the theorem stated this way via googling, and Wikipedia has a different and simpler statement (en.wikipedia.org/wiki/…). $\endgroup$ Oct 11, 2022 at 1:07
  • $\begingroup$ Hello Qiaochu, thank you for your reply and the reference. I have attached the reference from my two lectures. $\endgroup$ Oct 11, 2022 at 10:58
  • $\begingroup$ This theorem states that the three statements are equivalent. Hence, in case the system is stable, pick any $Q$ that satisfies the conditions, then you can find a $P$ such that the system is stable. So you don't have to check for every $Q$, just pick one and see whether the matrix inequality holds for some $P$ satisfying the condition. $\endgroup$
    – seaver
    Oct 11, 2022 at 11:03
  • $\begingroup$ It is easy to show by the way. Suppose the system is stable, then we have that for some $Q$, $P$ the following holds: $PA + A^\top P = -Q$, we can without an issue multiply both sides with a positive scalar $\alpha$, i.e., $\alpha PA + \alpha A^\top P = -\alpha Q$. Now define $\tilde{Q}=\alpha Q$ and $\tilde{P}=\alpha P$, then we have $\tilde{P}A + A^\top \tilde{P} = -\tilde{Q}$. as $\alpha$ is arbitrary, it must hold for all these $Q$'s, therefore the phrasing in the statement. :) $\endgroup$
    – seaver
    Oct 11, 2022 at 11:08
  • $\begingroup$ Hello seaver, thank you for the insight! It helps me from 1) to 3). How about the other way around? If I want to show the system is asymptotically stable, how can I use condition 3) to do that? Since condition 3) states for "every", I don't know how to enumerate all the $Q$ and $P$ so that condition 3) is satisfied. $\endgroup$ Oct 11, 2022 at 12:14

2 Answers 2

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I suspect that your professor might have combined the single case (one positive definite $Q$) together with the theorem of the Lyapunov equation.

Namely, the equation $P\,A+A^\top P=-Q$ arises from the Lyapunov function $V(x)=x^\top P\,x$ combined with the dynamics $\dot{x}=A\,x$. Using the dynamics in the Lyapunov function when evaluating it time derivative yields

\begin{align} \dot{V}(x) =& x^\top P\,\dot{x} + \dot{x}^\top P\,x, \\ =& x^\top P\,A\,x + x^\top A^\top P\,x, \\ =& x^\top (P\,A + A^\top P)\,x. \end{align}

So if $P\,A+A^\top P=-Q$ then $\dot{V}(x) = -x^\top Q\,x$. So if you can find a single pair of positive definite $Q$ and $P$ that satisfy this, then you have a radially unbound positive definite Lyapunov function for equilibrium $x=0$ of $\dot{x}=A\,x$ which has a negative definite time derivative. This is sufficient to show global asymptotic stability.

The theorem of the Lyapunov equation states that if $\dot{x}=A\,x$ is globally asymptotically stable, then for every positive definite $Q$ there is a positive definite $P$ such that $P\,A+A^\top P=-Q$. However, the previous paragraph showed that if a single pair of positive definite $Q$ and $P$ that satisfies $P\,A+A^\top P=-Q$ the system can be shown to be globally asymptotically stable. Thus from the theorem of the Lyapunov equation it follows that if one can find a single pair it implies that for every positive definite $Q$ there exists a positive definite $P$ that also solves $P\,A+A^\top P=-Q$.

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I think there is just a typo in your Lecture 1. There the theorem says "for every $Q$..." but the statement in Lecture 2 says "for any $R$..." (I have no idea why the notation was switched here), which seems more consistent with the statement on Wikipedia.

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  • $\begingroup$ I thought both "for every" and "for any" are "$\forall$", aren't they? Did I misunderstand? $\endgroup$ Oct 11, 2022 at 17:37
  • $\begingroup$ @WilliamLin: hmm. I think it's ambiguous. "$\exists$" seems to be what your examples imply (finding only one $Q$), also seems to be what seaver is implying in the comments, and seems more consistent with the statement on Wikipedia. You can find a discussion of the ambiguity of "for any" here: math.stackexchange.com/questions/430646/… $\endgroup$ Oct 11, 2022 at 17:45
  • $\begingroup$ At this point I think you should just ask your professor what the correct statement is. $\endgroup$ Oct 11, 2022 at 17:45
  • $\begingroup$ I indeed found proof of the same prof's higher level class that it's "$\forall$". I attached it to my question. It confuses me then why finding one example is sufficient to conclude its asymptotic stability... $\endgroup$ Oct 11, 2022 at 17:59

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