Existence of solutions to Maurer-Cartan equation I am considering the following equation:
$$\label{1}\tag{1}
\frac{\partial f}{\partial\theta_{2}}-f=f\frac{\partial g}{\partial\theta_{1}},
$$
where $f,g\in C^{\infty}(S^{1}\times S^{1})$ are smooth periodic functions in $2$ variables $\theta_{1},\theta_{2}$. Clearly, any pair of functions of the form $(0,g)$ with $g\in C^{\infty}(S^{1})$ is a solution to the equation \eqref{1}. Does there exist a solution $(f,g)$ where $f$ is not identically zero?
Context: The equation \eqref{1} is the Maurer-Cartan equation of an $L_{\infty}$-algebra that governs a certain deformation problem. Namely, I am considering the deformations of a given coisotropic submanifold, and I wonder if they are necessarily constrained to a certain locus. This amounts to the question I asked above.
Note: The linearized Maurer-Cartan equation is 
$$\tag{2}\label{2}
\frac{\partial f}{\partial\theta_{2}}-f=0,
$$
and it only admits the zero solution $f\equiv 0$. Indeed, since $f$ is continuous with compact domain, it attains a maximum $M=f(p)$ and a minimum $m=f(q)$. But then \eqref{2} implies
\begin{align}
&M=f(p)=\frac{\partial f}{\partial\theta_{2}}(p)=0,\\
&m=f(q)=\frac{\partial f}{\partial\theta_{2}}(q)=0.
\end{align}
 A: If $g \in C^\infty(S^1\times S^1)$, then so is $\partial_1 g$. Where $f \ne 0$, the equation can be re-written as 
$$\frac {\partial_2 f}f =\partial_1 g + 1$$
which integrates to $$f(\theta_1, \theta_2) = Ae^{\partial_1 \int g(\theta_1, \theta_2)\,d\theta_2+\theta_2}$$
If $f$ is never $0$, this holds everywhere and is obviously not periodic because of the extra $\theta_2$ exponent. But even more, note that because $S^1 \times S^1$ is compact and $h(\theta_1,\theta_2) := Ae^{\partial_1 \int g(\theta_1, \theta_2)\,d\theta_2}$ is continuous, $h(S^1 \times S^1)$ is a compact subset of $\Bbb R\setminus \{0\}$. And no matter what interval $[\alpha, \alpha + 2\pi)$ you choose to measure $\theta_2$ over, it also is bounded, making $e^{\theta_2}$ bounded away from $0$. Thus if $f$ is non-zero anywhere, it is the product of two functions bounded away from $0$, and so can never approach $0$ from that point, so either $f$ would have to be non-zero everywhere, or else $f$ would be discontinuous.
So you are correct. There are no $f,g \in C^\infty(S^1\times S^1)$ satisfying the equation with $f \not \equiv 0$. 
