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I'm trying to prove stability of the following dynamic system but I think my Mathematics knowledge is not deep enough.

My dynamic system consists of a state vector $x \in \mathbb{R}^n$. The system changes the value of $x_i$ as in the following equation,

$x'_i = (x_i + KT(\sum_{j \ne i}^n{\frac{1}{T - |\Delta_{i,j}|}}- \sum_{j \ne i}^n{\frac{1}{|\Delta_{i,j}|}})) \mod{T}$,

where $\Delta_{i,j} = x_i - x_j$, $K$ and $T$ are constants, and I assume |$\Delta_{i,j}| \in (0, T)$ .

The system changes the vector $x$ in circular order. For example, $x_1$ is changed first. Then, $x_2$ is changed and so on. Until $x_n$ is changed, the system re-starts with $x_1$ again.

I've ever proved only the stability of a linear dynamic system by using eigenvalues. However, with this equation form, I think eigenvalues do not work because it's non-linear dynamic system.

Any help?

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As far as I know, to prove the stability you should at first find the equilibrium if it does exist. Note that the origin here is not an equilibrium here, moreover if you write your equation in the form $\dot{\mathbf{x}} = f(\mathbf x)$ then $f$ is undefined in quite many points: origin and hyperspaces $|x_i-x_j| = T$. –  Ilya Aug 1 '11 at 10:54
    
Thank you @Gortaur. I forgot to specify that $|x_i - x_j| \in (0, T)$ to prevent such undefined value. –  Supasate Aug 1 '11 at 11:20
    
anyway, what kind of stability would you like to verify? If we talk about the stability of an equilibrium they you should first find it. Btw, why you wrote about the order of change of values of $\mathbf x$? If you have a system of ODEs then all components of $\mathbf x$ change "simultaneously". –  Ilya Aug 1 '11 at 11:34
    
Because, in my system, only one component ($x_i$) is changed at a time. The system reaches the equilibrium when the intervals of each consecutive components ($x_i$ and $x_{i+1}$) are equal and each $x_i$ stays still. –  Supasate Aug 1 '11 at 11:54
    
Actually, I simulated this system and found that it always converges. However, I cannot prove it mathematically. –  Supasate Aug 1 '11 at 11:56

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