# Question on proof from Arnol'd

osnarf
The relevant equations:

(1) $$\dot{x}$$ = v(x), x $$\in$$ U

(2) $$\varphi$$ = x0, t0 $$\in$$ R x0$$\in$$ U

Let x0 be a stationary point of a vector field v, so that v(x0) = 0. Then, as we now show, the solution of equation (1) satistying the initial condition (2) is unique, ie., if $$\varphi$$ is any solution of (1) such that $$\varphi$$(t0) = x0, then $$\varphi$$(t)$$\equiv$$ x0. There is no loss of generality in assuming that x0 = 0. Since the field v is differentiable and v(0) = 0, we have:

|v(x)| < k|x| for sufficiently small |x| =/= 0, where k > 0 is a positive constant.

Huh? i was good up until the inequality.

If somebody has the book, it is section 2.8 of the first chapter.