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 $u(r) =$ some constant. Munson's fluids textbook states this (5th ed., p. 243).
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 $u(r)$ is a continuous function?
I can't think of any counterexamples offhand. Intuitively, it seems to me that it might be possible for the velocity to dip below the average in some region and exceed the average in another region such that $\alpha = 1$.
I'm afraid integral equations are well outside my expertise, and the literature I've read on Fredholm integral equations does not seem to be helpful. Any assistance proving or disproving the uniqueness of this solution would be appreciated!
If $u(r)$ behaves like $A\delta_0+B\delta_R$, where $\delta_u$ is the Dirac delta distribution centered in $u$, then: $$\alpha = \frac{2}{R}\left(\frac{2B}{A+B}\right)^3, $$ that can be made arbitrarily close to $1$ by carefully choosing $A$ and $B$, so the converse implication does not hold.
By density, continuity is not a real issue. Almost the same counterexample holds by taking $u(r)$ as: $$ u(r) = A e^{-Nr^2} + B e^{-N(R-r)^2}.$$