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In a recent topic I've studied on complex analysis I had to study the differential system on the torus $\mathbb T^2:$ $$\begin{cases}\frac{\partial}{\partial y}u-\frac{\partial}{\partial x}v=\sin(y)-\cos(x),\\\\ \frac{\partial}{\partial x} u+\frac{\partial}{\partial y}v=0,\end{cases}$$ with the conditions $$\int_0^{2\pi}\int_0^{2\pi}u(x,y)\mathrm d x\mathrm dy=\int_0^{2\pi}\int_0^{2\pi}v(x,y)\mathrm d x\mathrm dy=0.$$

In particular it seemed to me that this system was explicitly solvable and to do so i relied on the inhomogeneous Cauchy Riemann equations (swapping the coordinates $x\leftrightarrow y) $ and i basically followed this link. Unfortunately my calculations didn't lead nowhere..

I am asking two things..

Is that way followable to finish the problem, and if so can you help me in finishing the proof?

More importantly: Are there smarter ways to do the problem?

Thanks in advance..

-Guido-

EDIT

I've got the following question related to the previous post so I'm writing it as an edit to this question.

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

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

satisfies

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

Thanks for your patience.

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The second question has an interest on its own. I suggest to have it as a separate thread so that we can still upvote and our answers could be accepted. –  uforoboa Aug 23 '12 at 17:39
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1 Answer 1

up vote 3 down vote accepted

A hint: Introduce new unknown functions $\bar u$, $\,\bar v$ by means of $$u(x,y)=-\cos y+\bar u(x,y)\ ,\qquad v(x,y)=\sin x+\bar v(x,y)$$ and check the resulting conditions for $\bar u$, $\,\bar v$.

Following the hint one finds that $\bar u$ and $\bar v$ are harmonic functions on the torus ${\mathbb T}^2$. A nonconstant harmonic function cannot take a maximum in an interior point of its domain; but as ${\mathbb T}^2$ is a compact manifold $\bar u$ and $\bar v$ would have to. Therefore both $\bar u$ and $\bar v$ have to be constant. Since $\int_0^{2\pi}\cos y\ dy=\int_0^{2\pi}\sin x\ dx=0$ it follows from the last condition that $\bar u$ and $\bar v$ have to be identically zero.

This brings us to the conclusion that necessarily $u(x,y)=-\cos y$ and $v(x,y)=\sin x$.

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So i presume you would like to conclude that $\tilde u$ and $\tilde v$ are identically zero, being harmonic and with $0$ integral. Am I right? but unfortunately i cannot conclude by myself. Can you help me? –  guido giuliani Aug 22 '12 at 14:13
    
@guido giuliani: See my edit. –  Christian Blatter Aug 22 '12 at 15:25
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