Arnold: ODEs, Ch. 1, section 2.2

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Main Question or Discussion Point

Attempting Problem 1 of Arnold: ODEs, Ch. 1, section 2.2, it seems I'm not understanding something pretty basic about what he means by "the solution of a differential equation". This is the kind of equation he's talking about:

[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?
 

Answers and Replies

  • #2
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Is the solution of the differential equation, in this context, a particular function gx which meets the condition that a particular expression substituted for v(x) in equation (1) is equal to

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

the value of x = gx0 here being determined by the particular gx?

So, if I've understood this, the question states that

[tex]\mathbf{v}(x_1)= \frac{\mathrm{d} }{\mathrm{d} t} \bigg |_{t=0} \text{ArcTan}(t) = 1[/tex]

(although not giving the specific expression for v(x1) that would be needed to derive this equation from scratch). And by this is meant that the above equation holds for one initial state, x1. However, when it gives the alternative solution x = ArcTan(t+1), here the x stands for a different initial state, x2, i.e. a different point of the phase space, in this case a different real number, which would have to replace x1 in v(x), thus:

[tex]\mathbf{v}(x_2)= \frac{\mathrm{d} }{\mathrm{d} t} \bigg |_{t=0} \text{ArcTan}(t+1) = \frac{1}{2}[/tex]

Specifically, [itex]x_1 = 0[/itex], and [itex]x_2 = \pi/4[/itex]. Is that right?

So, I think, the expressions x = ArcTan(t) and x = ArcTan(t+1) here mean, more explicity:

gx1 t = gt x1 = ArcTan(t),

gx2 t = gt x2 = ArcTan(t+1).

And the answer to the problem follows from the group properties of {gt}, as explained in Sec. 10.1; in particular, the closure of the real numbers under addition.
 
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  • #3
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(Hoping my interpretation in #2 of the question is correct...)

Proof (1). If [itex]g_x = \tan^{-1}[/itex] is a solution, then

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

We must show that there exists in the phase space of the equation a point, call it [itex]y[/itex], such that

[tex]\mathbf{v}(y) = \frac{\mathrm{d} }{\mathrm{d} t} \bigg|_{t=0} g_y (t) = \frac{\mathrm{d} }{\mathrm{d} t} \bigg|_{t=0} \tan^{-1}(t+1)[/tex]

[tex]= \frac{1}{1+(t+1)^2} \bigg|_{t=0} = \frac{1}{1+(t)^2} \bigg|_{t=1} = \frac{1}{2}.[/tex]

Then, by the theorem proved in Ch. 1, Problem 1.4.1, namely

[tex]\frac{\mathrm{d} }{\mathrm{d} t} \bigg|_{t=t_0} g_x t = \mathbf{v}(g^{t_0} x),[/tex]

the function [itex]g_y = \tan^{-1} \circ (+1)[/itex] is also a solution.

Let [itex]y = g_x 1 = g^{1} x[/itex]. Then

[tex]\mathbf{v}(y) = \frac{\mathrm{d} }{\mathrm{d} t} \bigg|_{t=0} g^t g^1 x = \frac{\mathrm{d} }{\mathrm{d} t} \bigg|_{t=1} g^t x = \tan^{-1}(t) \bigg|_{t=1} = = \frac{1}{1+(t)^2} \bigg|_{t=1} = \frac{1}{2},[/tex]

as was to be shown. [itex]\blacksquare[/itex]

Proof (2). Since the slope of the integral curves is the same everywhere along any horizontal line, a translation of the integral curve corresponding to any solution will be the integral curve corresponding to a solution, differing in x-intercept from the first unless the original solution was constant. [itex]\blacksquare[/itex]

EDIT: In the last line of--I fear now uneditable LaTeX--read "derivative at t=1 of arctan(t)" rather than simply "arctan(1)".
 
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