Frobenius theorem and the size of integral manifold

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 $$-\epsilon for some positive real number $$\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 $$\mathbb{R}^2$$?

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

1 Answer

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, $$|g|\le \pi$$.

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