Uniqueness of Filippov sliding vector field on the intersection of two surfaces in \(\mathbb R^3\) and implications for stability of periodic orbits
The following piecewise smooth system in \(\mathbb{R}^{3}\) is considered \[ \frac{dx}{dt} = f(x),\quad f(x) = f_{i}(x) \text{ for } x\in \mathbb{R}_{i}, \tag{1} \] where for \(i=1,2,3,4, \mathbb{R}_{i}\) are open, disjoint and connected sets, and (locally) \(\mathbb{R}^{3}=\mathbb{R}_{1}\cup\cdots \cup \mathbb{R}_{4}\). Moreover, each \(f_{i}\) is smooth (at least \(C^{1}\)) in an open neighborhood of each \(\mathbb{R}_{i}\) and \(\mathbb{R}^{3}\setminus \mathbb{R}_{1}\cup\cdots \cup \mathbb{R}_{4}\) has zero (Lebesgue) measure. The surface \(\Sigma =\Sigma_{1}\cup\Sigma_{2}\), \(\Sigma_{j}=\{x\in \mathbb{R}^{3}: h_j(x) =0\}, j=1,2\), with \(\nabla h_{j}\neq 0, h_{j}\in C^k, k\geq 2\) and \(\nabla h_1,\nabla h_2\) are linearly independent for \(x\in\Sigma\). Although, in general, there is no unique Filippov sliding vector field on \(\Sigma\) under natural conditions it is proved that all Filippov sliding vector fields are orbitally equivalent to \(\Sigma\). The implication of this result in the important case of a periodic orbit is examined.
scientific article; zbMATH DE number 6521646
scientific article