Ytukyg

Let $X(x_1, x_2) = (x_1 -x_2)partial_1 + (x_1 + x_2 + 1)partial_2$ be a vector field on $mathbb{R}^2$. I want to straighten it up around point $0$ . $X(0) = delta_2|_0 neq 0$. So it must be possible.

By straightening vector filed I mean finding a neighborhood $U$ of $0$ with equipped diffeomorphism $varphi : U to V subset mathbb{R}^2$ such that
$ varphi_* X = partial_1 $. Here $varphi_*$ is a push forward defined by

$$
varphi_* X(y) = Dvarphi_{|varphi^{-1}(y)} X(varphi^{-1}(y))
$$

Generally I just plug vector field into the prove of the straightening theorem.
First I compute flow $F(t;a)$ by solving associated ode $dot x = X(x) $ with an initial condition $x(0) = a$. Then for some $x in U$, where $U$ is a small neighborhood of $0$, I want to find point $(u,0)$ such that $ F(t_x;(u_x,0)) = x$. In the nice case this can be achieved by working with some equations, which will also restrict size of $U$. And then the straightening map can be defined by $varphi(x) = (t_x,u_x)$

However, the flow of $X$ turns out to have a rather complicated expression:

$$ F(t;x_1,x_2) =e^t binom{x_1 cos(t) - x_2 sin(t)}{x_2 cos(t) + x_1 sin(t)} +
frac{1}{2}e^t binom{cos(t) - sin(t) }{cos(t) + sin(t)} - binom{1/2}{1/2},
$$

And I son't вщтэе know how to solve the equations

$$
binom{x_1}{x_2} =F(t;u,0) = ue^t binom{cos(t)}{ sin(t)} +
frac{1}{2}e^t binom{cos(t) - sin(t) }{cos(t) + sin(t)} - binom{1/2}{1/2}
$$

So, I need to find better way to solve this problem.

Is there a trick to solve this equations? Are there a better (i.e. geometric) way which don't use flow equation explicitly?

Further thoughts:

I noticed that

$$
left(x_1 + frac{1}{2}right)^2 + left(x_2 + frac{1}{2}right)^2 =
u^2e^{2t} + frac{1}{2}ue^{2t} + frac{1}{2}e^{2t}
$$

Which can be rewritten as quadratic equation with
$0 = u^2 + frac{1}{2}u + a$ with

$$a = frac{1}{2} - frac{(x_1 + 1/2)^2 + (x_2 + 1/2)^2 }{e^{2t}}$$

On the other hand

$$x_1 - x_2 = u e^t (cos(t) - sin(t)) - e^tsin(t), $$
$$x_1 + x_2 + 1 = u e^t (cos(t) + sin(t)) + e^tcos(t); $$

Then the sum of squares is

$$(x_1 - x_2)^2 + (x_1 + x_2 + 1)^2 = 2u^2 e^{2t} + e^{2t} + ue^{2t},$$
which is essentially the same equation.

By dividing it into several smaller transformations and applying some straightforward observations, we can rectify the given vector field without any heavy lifting.

First, note that we can eliminate the constant term in the coefficient of $partial_{x^2}$ by making the affine coordinate change, e.g., $$y^i := 2 x^i + 1, quad i = 1, 2 ,$$
and in these coordinates, the vector field is
$$X = (y^1 - y^2) partial_{y^1} + (y^1 + y^2) partial_{y^2}.$$
We can recognize this vector field as the sum of the Euler vector field, $y^1 partial_{y^1} + y^2 partial_{y^2}$, and the vortex vector field, $-y^2 partial_{y^1} + y^1 partial_{y^2}$. Both of these are rotationally symmetric (the former is radial, and the latter is the standard infinitesimal generator of rotations of the plane), hence so is $X$. Thus, it's natural to absorb this symmetry into a new coordinate: The point $(x^1, x^2) = (0, 0)$ is $(y^1, y^2) = (1, 1)$, and since this is away from the origin (in $y^i$-coordinates) we can choose coordinates $(z^1, z^2)$ in some neighborhood of this point for which
$$y^1 = e^{z^1} cos z^2, qquad y^2 = e^{z^1} sin z^2 ;$$
in these coordinates, $$X = partial_{z^1} + partial_{z^2} .$$ But this vector field is constant, so a suitable linear transformation, e.g., $$z^1 = frac{1}{2}(w^1 + w^2), qquad z^2 = frac{1}{2}(w^1 - w^2),$$ gives coordinates in which $$boxed{X = partial_{w^1}}$$ as desired.

Remark If we make the usual identification $Bbb R^2 leftrightarrow Bbb C$, ${bf x} := (x^1, x^2) leftrightarrow x^1 + i x^2 ,$ we can write each of these transformations (and hence their composition) much more compactly: We have ${bf y} = 2 {bf x} + 1$, ${bf y} = exp {bf z}$, and ${bf z} = frac{1}{2} (1 - i) {bf w}$, and thus $${bf w} = (1 + i) log (2 {bf x} + 1) $$ for an appropriate choice of branch cut of $log$.

I think I found a solution:

Use line spanned by the vector $(1,1)$ to catch the integral curves. Then the expression turns into
$$
binom{x_1}{x_2} =F(t;u,u) = (u + 1/2)e^t binom{cos(t) - sin(t) }{cos(t) + sin(t)} - binom{1/2}{1/2}
$$

or equivalently

$$
y=binom{x_1 + 1/2}{x_2 + 1/2} = (u + 1/2)e^t binom{cos(t) - sin(t) }{cos(t) + sin(t)} = (u + 1/2)e^t left(begin{array}{cc}cos(t) & -sin(t) \ sin(t) & cos(t) end{array}right)binom{1}{1}.
$$

Note, that that the matrix on the right hand size represents rotation by the angle $t$. No other factors affect the sign of $y$. So it is safe to say that:
$$
t_x = mathrm{Arg}(y) = arctanleft(frac{x_1 + 1/2}{x_2 + 1/2}right) - frac{pi}{4}
$$

Now it is possible to express $u$ by

$$ sqrt{(x_1 + 1/2)^2 + (x_2 +1/2)^2} =| y | = sqrt{2}(u + 1/2)e^t $$

as

$$
u_x = frac{sqrt{(x_1 +1/2)^2 +(x_2 + 1/2)^2}}{sqrt{2} expBig(arctanbig(frac{2x_1 + 1}{2x_2 + 1}big) - pi/4Big)} -frac{1}{2}
$$
I think it is possible to take $U$ equal to an open ball of the radius $frac{1}{2}$ centered at $0$. And let $varphi(x) = (t_x,u_x)$ as was planed.

Im still will be thankful for any advice.