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I've got the following question related to my previous post and I was suggested to write a new question, so there you are.

The question is the following: prove that if $f$ is smooth, periodic and with zero average, then the solution to the system on $\mathbb T^2$

$$\begin{cases}\frac{\partial}{\partial y}u'-\frac{\partial}{\partial x}v'=0,\\\\ \frac{\partial}{\partial x} u'+\frac{\partial}{\partial y}v'=f(x,y),\end{cases}$$


$$\int_{0}^{2\pi}\int_0^{2\pi}\left(-\cos (y)u'(x,y)+\sin(x)v'(x,y)\right)\mathrm dx\mathrm dy=0.$$

I tried to apply similar techniques to those explained to me in the other question, however it seems that in this case I am not supposed to solve the system explicitly. In particular I cannot see where the condition on the zero mean of $f$ comes into play. Can anybody help me?

Many thanks


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up vote 1 down vote accepted

It seems to follow if you multiply the first equation by $\cos x + \sin y$ and integrate by parts.

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