11
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 ($r=1$). $\endgroup$ – Carlo Beenakker Jun 8 '13 at 16:42
  • $\begingroup$ 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.) $\endgroup$ – Thibaut Demaerel Jun 8 '13 at 17:04
  • $\begingroup$ I believe it is a very interesting question if the answer is still positive in the case of a more realistic turbulent flow. $\endgroup$ – Alex Gavrilov Feb 28 '18 at 16:26
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.