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Suppose that $U \subset \mathbb{R}^2$ is nonempty, open, connected and bounded. Consider a Poisseuille flow in the pipe $U \times \mathbb{R}$. That is: a time-independent incompressible flow of the form: $$v:U \times \mathbb{R} \rightarrow \mathbb{R}^3: (x,y,z) \mapsto (0,0,w(x,y))$$ which satisfies: $$\frac{\partial^2w}{\partial x^2} + \frac{\partial^2w}{\partial y^2} = k < 0$$ $$w \left.\right|_{\partial U} = 0$$ with $k$ some constant involving the pressure-gradient along the pipe's axis and the viscosity.
If we consider pipes with fixed cross-section, $$\int_U \mathbb{d}x \mathbb{d}y = C , $$ is it true that the flow rate $\int_U w(x,y) \mathbb{d}x \mathbb{d}y$ is maximized only if $U$ is a disc?

If the answer is negative, consider instead pipes with a fixed boundary length, $$\int_{\partial U} \mathbb{d}l = L,$$ is it now true that the flow rate is maximized only if $U$ is a disc?

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I don't know about a mathematically rigorous proof, but the answer is certainly "yes"; an example that can be solved exactly is that of an elliptical cross-section: the ratio of the flow through an elliptical and circular crosssection, at constant area, is $$\gamma=2(r+1/r)^{-1}$$ with $r$ ratio of the two axis of the ellipse; this is maximal for a circle ($r=1$). –  Carlo Beenakker Jun 8 '13 at 16:42
    
I know, and we can solve the flow problem in many other geometries too (since $h(x,y) := w(x,y) - \frac{k}{4}(x^2+y^2)$ is harmonic, we can use complex-analytic tools like the Riemann-mapping theorem. That's why I think that an actual proof should not be too difficult for experts.) –  Thibaut Demaerel Jun 8 '13 at 17:04

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up vote 6 down vote accepted

The answer is positive, this is a consequence of a result by Talenti ["Elliptic Equations and Rearrangements", Annali SNS 3 (1976)]. Let $k>0$, $-\Delta u= k$ in $U$, $u=0$ on $\partial U$, and consider the ball $B$ having the same volume as $U$. If $v$ is the solution of $-\Delta v=k$ in $B$, $v=0$ on $\partial B$, then $u^* \leq v$ in $B$, where $u^*$ is the spherically symmetric rearrangement of $u$. Since $u$ and $v$ are positive functions, and rearrangements preserve norms, one has $\|u\|_1 = \|u^*\|_1 \leq \|v\|_1$, which is the claim.

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