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Let $X =(X_0,X_1)\in \mathbb{R}^2$ and $Y=(Y_0,Y_1)\in \mathbb{R}^2$ be two vector fields on $\mathbb{R}^2$ such that $X,Y$ are independent on each tangent plane and $[X,Y]:=XY-YX=0$.
Then by Frobenius theorem, the partial differential equation on $\mathbb{R}^2$ given by $\frac{d}{ds}f=X_0(f(s,t),g(s,t)),\frac{d}{ds}g=X_1(f(s,t),g(s,t))$,
$\frac{d}{dt}f=Y_0(f(s,t),g(s,t)),\frac{d}{dt}g=Y_1(f(s,t),g(s,t))$,
$ (f(0,0),g(0,0))=(0,0)$
has a solution in $-\epsilon<s<\epsilon,-\epsilon<t<\epsilon$ for some positive real number $\epsilon$.

My question is if $f,g$ are maximal solutions (i.e $f,g$ cannot be extended) then is the image of $(f(s,t),g(s,t))$ equal to the whole $\mathbb{R}^2$?

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I'm especially curious about the case when $X_0,X_1,Y_0,Y_1$ are polynomials.

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1 Answer 1

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Your equations are equivalent to the $1$-form equations $$ \mathrm{d}f = X_0(f,g)\,\mathrm{d}s + Y_0(f,g)\,\mathrm{d}t \quad \text{and}\quad \mathrm{d}g = X_1(f,g)\,\mathrm{d}s + Y_1(f,g)\,\mathrm{d}t $$ The Frobenius compatibility condition for these two $1$-form equations is indeed the condition that $[X,Y]=0$, where, in the $uv$-plane $$ X = X_0(u,v)\,\frac{\partial}{\partial u} + X_1(u,v)\,\frac{\partial}{\partial v}\quad \text{and}\quad Y = Y_0(u,v)\,\frac{\partial}{\partial u} + Y_1(u,v)\,\frac{\partial}{\partial v}\,. $$

Note that, setting $$ \begin{aligned} X_0(u,v)&=\phantom{-}e^{-u}\cos(v),&& Y_0(u,v)=e^{-u}\sin(v),\\ X_1(u,v)&=-e^{-u}\sin(v),&& Y_1(u,v)=e^{-u}\cos(v), \end{aligned} $$ we get an example (not polynomial, though), for which it turns out that there are maximal solutions (f,g) satisfying your initial conditions for which the range of $(f,g)$ is not all of $\mathbb{R}^2$. (Maximal solutions are not unique.). The point is that a solution satisfying your initial condition also satisfies $$ s = e^f\cos g - 1\quad\text{and}\quad t = e^f\sin g, $$ and these equations are equivalent to the complex equation $$ 1+ s + i t = e^{f+ig}, $$ so $f+i g$ must be a branch of $\log (1 + s+it)$, which cannot be global. There is a unique solution on the complement of the ray $t=0$ and $s\le-1$, and if you look at that solution, you will see that $f+ig$ does not map this open set onto the complex plane, in fact, $|g|\le \pi$.

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  • $\begingroup$ I think $X$ and $Y$ aren't linearly independent where $sin(v)=cos(v)$. $\endgroup$
    – George
    Mar 17 at 20:37
  • $\begingroup$ Should we choose $Y_1=e^{-u}\cos(v)$? $\endgroup$
    – George
    Mar 17 at 21:07
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    $\begingroup$ @George: You are correct. That minus sign was an error on my part (caused by my miscopying the formula). I have fixed it now. $\endgroup$ Mar 18 at 18:38

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