Good day,

This is my first question, I hope all information is given. If not, feel free to ask. Currently I am reading the paper "Stability of relative equilibria in the problem of N+1 vortices" by Cabral and Schmidt. See here http://epubs.siam.org/doi/abs/10.1137/S0036141098302124 (you need University Access to see it for free)

There is a passage where I'm stuck. It's at page 243 for those who want to look at it. But it can bee seen separately from the rest of the paper and I'm now giving you everything you need.

Let $0 \in \mathbb{R}^{2(N+1)}$ be an equilibrium of a given dynamical system. The claim is that 0 is Lyapunov stable. They give a short proof of it and suggest to use a Hamiltonian as a Lyapunov function. (For Definitions see at the end) Remember that a Hamiltonian fulfills $\dot H = \frac{d}{dt} H = 0 $, so we fulfilled already the 3rd property of a Lyapunov-function.

Okay, let $x,y \in \mathbb{R}^{N+1}$ now they use $$H(x,y)= \frac{1}{2} (x_0,x_2,x_N) A (x_0,x_2,x_N)^T + a_1 x_1^2 + a_3 x_3^2 + ... + a_{N-1} x_{N-1}^2 + \frac{1}{2} (y_0,y_2,y_N) B (y_0,y_2,y_N)^T + b_3 y_3^2 + ... + b_{N-1} y_{N-1}^2 $$ as a Lyapunov-function, where $A,B \in \mathbb{R}^3$, obviously $H(0,0)=0$.

Now they argument in the following way:

$\rightarrow$ "All coefficients $a_i,b_i$ ($i=1,...,N-1$) of the quadratic terms are positive." (This is okay)

$\rightarrow$ "It remains to show that the two quadratic forms are also positive definite." (Now they compute the eigenvalues of these matrices and show that they are strictly greater than 0, this is also okay for me)

$\rightarrow$"Equilibrium is Liapunov stable"

Now my problem: There isn't a term of $y_1^2$ in $H(x,y)$ so I don't get how they can follow Lyapunov stability. The definition (below) says that a Lypaunov function has to fulfill $H(x,y) > 0$ for $x \in U \backslash \{0\}$ with U an appropriate neighborhood. I imagine that I can easily vary $y_1$ and I stay at zero. So like that: Let $y' := (y_0,y_2,...,y_N)$ then write $H(x,y',y_1)$ and so we have $H(0,0,y_1)=0$ for all $y_1 \in \mathbb{R}$, so also for $y_1 \in U$.

It could only work, I think, if U is neighborhood of $x_e$ without $y_1$ so I kind of delete this axis. But I don't think this is allowed, U is suggested to be full-dimensional in the definition.

An equivalent way to show $H(x,y) > 0$ in this nbh is by showing that $D^2 H (0)$ is positive definite ($D^2$ denotes the Hessian). But obviously the $y_i$ line and row denotes a zero and so we have an eigenvalue zero that destroys the needed positive definite.

Okay. These were my approaches. I'm thankful for every help, let it be another approach/proof or a corrections to mine.

Thanks a lot, Marvin

$\bigg [$ Now the results I learned by S. Wiggins, Introduction to Applied Nonlinear Dynamical Systems and Chaos.

$\rightarrow$ Definition: Let $(X,\phi)$ be a dynamical system and $x_e$ an equilibrium. Then $x_e$ is called stable in the sense of Lyapunov, if $$\forall \epsilon >0\hspace{0.1cm} \exists\delta > 0 \hspace{0.1cm}: \hspace{0.1cm} \|x-x_e\|_X < \delta \Rightarrow \forall t\geq 0 : \| \phi(t,x)-x_e \|_X<\epsilon. $$

$\rightarrow$ Theorem: Let $\dot x=f(x), x\in \mathbb{R}^n, f(x_e)=0$. It exists a so called Lyapunov-function $V \in C^1(U \subset \mathbb{R}^n, \mathbb{R})$, where U is a neighborhood of $x_e$ and V fulfills the following proporties $$ \begin{cases} V(x_e)=0 \\ V(x)>0 \hspace{0.1cm} \textrm{for} \hspace{0.1cm} x \in U\backslash\{x_e\} \\ \langle \nabla V(x),f(x)\rangle \leq 0 \hspace{0.1cm} \textrm{for} \hspace{0.1cm} x \in U\backslash\{x_e\} \end{cases}$$ Then $x_e$ is stable in the sense of Lyapunov. $\bigg ]$

  • $\begingroup$ look for papers on "positive semi-definite lyapunov functions". $\endgroup$ – Piyush Grover Aug 17 '15 at 12:02
  • $\begingroup$ Thank you for your answer. I looked at "Semi-Definte Lyapunov functions stability and stabilizability" by Chabour and Kalitine, see at math.univ-metz.fr/~chabour/Articles/… It states at Theorem 1: If in a nbh U of the origin there existis a $C^1$ function $V:U \to \mathbb{R}^+$ such that: i. $V(x) \geq 0$ for all $x \in U$, $V(0)=0$. ii. $\dot V(x) \leq 0$ for all $x \in U$. iii. The origin is asymptotically stable wrt $Y_0=\{x \in U : V(x)=0 \}$ then the origin is a Lyapunov stable equilibrium point for the system. $\endgroup$ – Fritz Aug 17 '15 at 14:36
  • $\begingroup$ I will try this and see if I can get something. In my case I have to prove that every point on the $y_1$-Axis "goes" to the origin. $\endgroup$ – Fritz Aug 17 '15 at 14:39

With your setting, it seems you are right and this cannot be a Lyapunov function, but I cannot check the original paper, which could have some relevant context. Note that different authors provide slightly different definitions of Lyapunov function.

Aside from formally checking the conditions of the definition, there is something fundamentally awkward to me in using a Hamiltonian as Lyapunov function. The physical intuition of energy conservation seems completely contradictory with the notion of dissipation.

  • $\begingroup$ Thank you for your answer. Yeah you are right, I think there might be missing something. I have written the author a mail and asked him about my problem and he answered me the following (translated): "It is right that Lypaunov stability can't be assumed if a coefficient is missing. It would only work when there is a rotation or another symmetry because then there is no global stability in this problem." And the original system was really transformed with a rotation to get the new one. $\endgroup$ – Fritz Aug 17 '15 at 14:45
  • $\begingroup$ He answered even more: "The coordinates are chosen such that the center of the mass is in the origin and the rotation prevents that there is vortex on the $x_1$-axis and so $y_1$ must be constant and can be excluded. $\endgroup$ – Fritz Aug 17 '15 at 14:46
  • $\begingroup$ Yeah I didn't think the situation would get important, but it is in the following way: There are N+1 vortices on a ring with a center mass. Now this whole situation is transformed via a rotation. We look at a rotating coordinate system. But I don't get why this $y_1$ is constant and so can get excluded. $\endgroup$ – Fritz Aug 17 '15 at 14:49
  • $\begingroup$ @marvin I am no expert in differential geometry but I guess that, if the system has some symmetry, it can be formulated on a low-dimensional manifold so that, in practice, there are "less" variables and some of them can be neglected. $\endgroup$ – Miguel Aug 18 '15 at 7:40
  • $\begingroup$ @marvin Please note that the question remains "unanswered" if the answer is not accepted and/or upvoted $\endgroup$ – Miguel Aug 20 '15 at 11:17

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