Bounding the action space

Recall that using normalized vector fields does not alter the existence of solutions. This is convenient because $ U(x)$ needs to be bounded to approximate it with a finite set of samples. It is useful to restrict the action set to obtain

$\displaystyle U(x) = \{ u \in {\mathbb{R}}^n \;\vert\;\; \Vert u\Vert \leq 1 \} .$ (8.60)

To improve performance, it is sometimes possible to use only those $ u$ for which $ \Vert u\Vert = 1$ or $ u = 0$; however, numerical instability problems may arise. A finite sample set for $ U(x)$ should have low dispersion and always include $ u = 0$.

Steven M LaValle 2012-04-20