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.

 

[blue_raver22]

I would like to clarify and expand on the content mentioned in the question. The content is discussing a proof by mathematician Vladimir Arnol'd, which states that for a given vector field v and a stationary point x0, the solution of the equation (1) with the initial condition (2) is unique. This means that if we have a solution \varphi of (1) such that \varphi(t0) = x0, then \varphi(t) will always be equal to x0.

To understand this proof, we need to understand the equations (1) and (2). Equation (1) is a differential equation that describes the rate of change of a variable x with respect to time. The variable x belongs to a set U, and the function v(x) describes the velocity of x at any given point. Equation (2) defines the initial condition, where \varphi is the solution of (1) at time t0 and x0 is the initial value of x.

The proof begins by assuming that x0 is a stationary point of the vector field v, meaning that v(x0) = 0. This means that at x0, the velocity of x is equal to 0, and thus x0 is a point where x does not change with respect to time. The proof then goes on to show that if we have any other solution \varphi of (1) with the initial condition (2), then \varphi(t) will always be equal to x0. This means that x0 is the only solution for the given initial condition, and thus the solution is unique.

To prove this, Arnol'd uses the fact that the vector field v is differentiable and that v(0) = 0. This means that the function v is smooth and has a derivative at x = 0. Using this, Arnol'd shows that for sufficiently small values of x, the magnitude of v(x) is always less than a positive constant k multiplied by the magnitude of x. This means that as x gets closer to 0, the rate of change of x decreases. This is a crucial step in the proof as it allows Arnol'd to show that \varphi(t) will always be equal to x0.

In conclusion, the content mentioned in the question is a proof by Vladimir Arnol'd that shows the uniqueness of solutions for a given differential equation (1) with a specific