This is a followup to another question I asked recently. This is a slight modification to that question.
In the fluid mechanics of pipe flow, it is sometimes stated that the velocity profile $u(r)$ which corresponds to a kinetic energy coefficient of 1 is always uniform, so $u(r) = \overline{u}$ (a constant).
The kinetic energy coefficient is defined as $$\alpha \equiv \frac{2}{R^2} \int_0^R \left(\frac{u(r)}{\overline{u}}\right)^3 r dr$$
where $$\overline{u} \equiv \frac{\int_A u(r) dA}{A} = \frac{2 \int_0^R u(r) r dr}{R^2}$$ is the average velocity.
It is obvious that $u(r) = \overline{u}$ returns $\alpha = 1$. How do you prove or disprove the converse, i.e., that $\alpha = 1$ implies $u(r) = \overline{u}$ if $u(r) \geq 0$ is a continuous function with $\frac{d u}{d r}|_{r = 0} = 0$? (The greater than or equal part and the symmetry condition are the only parts that differs with the previous question.)
I have tried to construct polynomials with these properties to disprove the converse. In all cases I've tried, $u(r)$ went negative if $u(r) \neq \overline{u}$ and $\alpha = 1$. This is despite trying to force the function to be positive, e.g., by setting $\frac{d u}{d r}|_{r = R}$ equal to a large negative number and $u(0)$ equal to a positive number. I am not sure if there is a way to prove that there are no strictly positive solutions aside from $u(r) = \overline{u}$.
