I am trying to solve for $y(x)$ in terms of $f(x)$ in a convenient space (eg. $\dot{H}^2(\mathbb{T})$-zero mean). Here is the ode: $y(x)+y(x)y'(x)=f(x)$. I think a contraction mapping argument will work, but this is not clear to me. In trying to compute the $\dot{H}^2(\mathbb{T})$ norm, I get $\int_{\mathbb{T}} (1+ \frac{5}{2}y')(y'')2\, dx = \int_{\mathbb{T}}y'' f''\, dx$. Here I am just trying to show that if $\vert h'' \vert _{\dot{H}^2(\mathbb{T})}\leq C$, where $C$ is a constant then the operator $T(y):=y(x)+y(x)y'(x)$ maps the ball of radius $C$ into a ball of radius $C$. But $T$ as defined above is not a contraction.... May I have some suggestions, please?

Thank you! Rosa