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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..



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 $\mathbb T^2$

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


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

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
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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