Do circular pipes maximize flow rate?

Question

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?

Answer 1

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.

Answer 2

Let me sketch a simple ad hoc proof: Let $w:U \subset \mathbb{R}^2 \to \mathbb{R}$ be a function such as described above. Assume that its boundary condition $w|_{\partial U}=0$ concretely means that the formal partial integration $$\int_U |\nabla w|^2 = -\int_U w \Delta w\left(=-k\int_U w\right)$$ holds (without boundary term). Let $R>0$ be so that $U$ and $D(0,R)$ have the same area and let $w':D(0,R) \to \mathbb{R}$ be the spherical rearrangement of $w$. By virtue of the Pólya-Szegó inequality, we have $$\int_{D(0,R)}\left(\frac{1}{2}|\nabla w'|^2+kw'\right)\leq \int_U\left(\frac{1}{2}|\nabla w|^2+kw\right)=\frac{k}{2}\int_U w\quad(1)$$ Setting $w_0:D(0,R) \to \mathbb{R}^2:(x,y) \mapsto -\frac{k}{4}(R^2-x^2-y^2)$, one can verify that $\Delta w_0 =k$ and $w_0$ is the minimizer of the energy functional $f \mapsto \int_{D(0,R)}\left(\frac{1}{2}|\nabla f|^2+kf\right)$.* Then $$\int_{D(0,R)}\left(\frac{1}{2}|\nabla w'|^2+kw'\right)\geq \int_{D(0,R)}\left(\frac{1}{2}|\nabla w_0|^2+kw_0\right)\overset{P.I.}{=} \frac{k}{2}\int_{D(0,R)}w_0\quad (2)$$ Combining (1) and (2) (and not forgetting that $k<0$) we get $$\int_U w \leq \int_{D(0,R)} w_0.$$

(*) To prove that $w_0$ is a minimizer of that energy-functional $E$, it suffices to establish that $E(w_0+tg) = E(w_0) + t^2\int_U |\nabla g|^2$.