- #1

Rasalhague

- 1,387

- 2

[tex]\dot{x} = \mathbf{v}(x) \enspace\enspace\enspace (1)[/tex]

where, if the definitions of section 1 are still in play, the left side is an abbreviation for

[tex]\frac{\mathrm{d} }{\mathrm{d} t} \bigg|_{t=0} g_xt,[/tex]

and [itex]g_x[/itex] is the motion of x (which one Wikipedia writer calls "the flow through x"), meaning g(_,x) :

**R**-->

**R**such that [itex]g_x t = g^t x = g(t,x)[/itex], and [itex]g[/itex] is the evolution function of the dynamical system. (

*NOTE: This was the equation he used to*)

*define*phase velocity. I'm guessing now he means that, in an actual differential equation, the left or right side would be replaced by a particular expression in terms of t (if left) or x (if right); in this problem, at least, it's the left that's replaced by an explicit expression.The problem is this: Let x = ArcTan(t) be a solution of equation (1). Prove that x = ArcTan(t+1) is also a solution.

Is the value of the solution supposed to be an evolution function [itex]g[/itex], so that [itex]g_xt = \text{ArcTan}(t)[/itex]? Is he saying that, in this particular example, it happens that [itex]g_x[/itex] = ArcTan(t) for every [itex]x[/itex] in the phase space? In that case,

[tex]\mathbf{v}(x) = \frac{\mathrm{d} }{\mathrm{d} t} \bigg|_{t=0} \text{ArcTan}(t) = \frac{1}{1+t^2} \bigg|_{t=0} = 1.[/tex]

But then he seems to be asking for a proof that

[tex]\mathbf{v}(x) = \frac{\mathrm{d} }{\mathrm{d} t} \bigg|_{t=0} \text{ArcTan}(t+1) = \frac{1}{1+(t+1)^2} \bigg|_{t=0} = \frac{1}{2}[/tex]

i.e. that 1 = 1/2. Also, by definition, [itex]g_x 0 = g^0 x = x[/itex]. But this doesn't hold if the initial state is anything other than x = 0, for the first solution, and x = Pi/4, for the second. So is "the solution" meant to select just one of the motions, rather than the whole family of them associated with the dynamical system? Do each of the suggested solutions decree a different initial condition?