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. $\bigg ]$
