Let we have a problem:
$$ \min \big\{\int_B |\nabla u|^2 dx: \ u\in\mathcal{C}^2(\bar B,\mathbb{R}^2), u|_{\partial B}=x \big\}
$$
which is equivalent to solve the Laplace's equation
$$
\begin{cases}
-\Delta u (x)=0 \\
u|_{\partial B}=x
\end{cases}
$$
My question here is – can we apply the Mean Value property of the harmonic functions to obtain the solution using the boundary condition? I am not sure how to work it out.
$$
u(x)=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B} u(y)\,d\sigma(y)=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B} y\,d\sigma(y)
$$
Best Answer
We made some progress on this question in the comments. We reached the conclusion that the mean value property can only be used to obtain the value of $u$ at the center of the ball $B$. If such center is at $0$, then $$ u(0)=\frac{1}{\lvert\partial B\rvert}\int_{\partial B}y\, d\sigma(y)=0, $$ and the integral vanishes by symmetry (it turns into $-$ itself under the transformation $y\mapsto -y$).