Recent content by dmytro

Found the answer. According to the theorem 1.1 from this paper, there indeed exists sufficiently small \beta such that \dot{x}=\left[A(t)+B(t)\right]x has negative maximum lyapunov exponent.

I have a linear time-varying linearly perturbed ODE of the form: \dot{x} = [A(t)+B(t)]x where A(t) is a bounded lower-triangular matrix with negative functions on the main diagonal, i.e. 0>a^0\ge a_{ii}(t). The matrix B(t) is bounded, so that ||B(t)|| \le \beta. The question is...

This seems to be a very relevant discussion: https://www.physicsforums.com/showthread.php?t=588101

Look at it this way: suppose you calculate the amounts of all the ingredients in the bread recipe. If you mix all the ingredients in a blender and put the thing into the oven, it's now easy to calculate the probability that you'll get a nice crispy bread, just use this formula: P(\text{nice...

Thanks for the clues. I can't seem to figure out how to make use of that theorem though... However, I found a solution for a special case of W. If one picks W^T with only first non-zero column, then the product \Gamma = \frac{\partial f}{\partial y}W^T also has only first column not equal to...

edit: f:\mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^n should be f:\mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}^n \frac{\partial f}{\partial x}: \mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^{n\times n} should be \frac{\partial f}{\partial x}: \mathbb{R}^n\times\mathbb{R}^m \to...

Quick version: I have a vector field f:\mathbb{R}^n\oplus\mathbb{R}^m \to \mathbb{R}^n of two arguments x \in \mathbb{R}^n, y \in \mathbb{R}^m, which has the following properties: The jacobian matrix of f wrt to the first argument \frac{\partial f}{\partial x}: \mathbb{R}^n\oplus\mathbb{R}^m...