Estimate for vector fields tangential to the boundary

Question

Let $\xi:D\to\mathbb{R}^2$ be a smooth vector field defined on the unit ball $D=\lbrace x\in\mathbb{R}^2:|x|\leq 1 \rbrace\subset\mathbb{R}^2$ always tangential to its boundary, i.e. $x\cdot\xi(x) = 0$ for all $x\in\partial D.$

Can we then find a bound of the type: $\exists C>0$ such that $ |x\cdot\xi(x)| \leq C \text{dist}(x,\partial D)|\xi(x)| $?

Answer 1

No. Here is a counterexample. Let us denote $r = \|x\|$ and define $\xi(x) = (1-r^2)x$.

$$x.\xi(x) = (1-r^2)r^2, \quad d(x, \partial D) = 1-r, \quad \|\xi(x)\| = (1-r^2)r. $$ We would have the bound $(1-r^2)r^2 \leq C (1-r)(1-r^2)r$ and thus $$ r \leq C (1-r). $$ There is no such constant as we can see by letting $r$ go to 1.