Is the Proof of Uniqueness in Arnol'd's Vector Field Correct?

[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.

 

[HallsofIvy]

Apply the mean value theorem between x and 0.

 

[osnarf]

*smacks self in head*
thanks

 

[Landau]

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.