Question on proof from Arnol'd

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The relevant equations:

(1) [tex]\dot{x}[/tex] = v(x), x [tex]\in[/tex] U

(2) [tex]\varphi[/tex] = x0, t0 [tex]\in[/tex] R x0[tex]\in[/tex] 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 [tex]\varphi[/tex] is any solution of (1) such that [tex]\varphi[/tex](t0) = x0, then [tex]\varphi[/tex](t)[tex]\equiv[/tex] 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.

Answers and Replies

  • #2
Science Advisor
Homework Helper
Apply the mean value theorem between x and 0.
  • #3
*smacks self in head*
  • #4
Science Advisor
So the claim is that v is Lipschitz continuous on a neighbourhood of zero. Are you sure that v is merely assumed to be differentiable? I suspect that you need it to be C^1, i.e. its derivative must also be continuous. This is so because a differentiable function is Lipschitz iff its derivative is bounded, in which case the Lipschitz constant is the supremum of the absolute values of the derivatives. Any C^1 function is locally Lipschitz, but this need not hold for merely differentiable functions.