We want to find $u \in C^2 $ that minimises the integral:
$$\iint_{1 \leq x^2 + y^2 \leq 4} \left(\left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2\right) \mathrm d x \, \mathrm d y $$
with constant boundary conditions, i.e., $u(\{x^2 + y^2 = 1 \}) = \text{const}$ and $u(\{x^2 + y^2 = 4\}) = \text{const}$.
In the task there was the following hint:
... using the symmetry of the problem, reduce the problem to a $1$-dimensional one.
So, my guess is that at least one of the solutions should be a rotational-invariant function. But I have no ideas how to prove that. Any thoughts?
In polar coordinates
$$ E(u)=\iint_{1\leq x^2+y^2\leq 4}\left\|\nabla u\right\|^2\,d\mu =\int_{0}^{2\pi}\int_{1}^{2}\rho\left[\left(\frac{\partial u}{\partial\rho}\right)^2+\frac{1}{\rho^2}\left(\frac{\partial u}{\partial\theta}\right)^2\right]\,d\rho\,d\theta$$ hence given some function $u$ meeting the wanted constraints, its radial rearrangement $$\tilde{u}(x_0,y_0)=\frac{1}{2\pi\sqrt{x_0^2+y_0^2}}\oint_{x^2+y^2=x_0^2+y_0^2}u(x,y)\,dx\,dy$$ has a smaller energy, due to $\frac{\partial\widetilde{u}}{\partial\theta}=0$ and Wirtinger's inequality. In particular it is enough to find the minimum of $\int_{1}^{2} r f'(r)^2\,dr$ for a function fulfilling $f(1)=A$ and $f(2)=B$. Assuming that $f'$ has a constant sign, by the Cauchy-Schwarz inequality $$ \int_{1}^{2}\frac{dr}{r}\int_{1}^{2}r f'(r)^2\,dr \geq \left(\int_{1}^{2}f'(r)\,dr\right)^2=(B-A)^2$$ hence $\int_{1}^{2}rf'(r)^2\,dr\geq \frac{(B-A)^2}{\log 2}$ and the optimal solution is of the $K_1+K_2\log\rho$ kind.