# Solving Laplace's equation using mean value property

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.

$$u(x)=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B} u(y)\,d\sigma(y)=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B} y\,d\sigma(y)$$

• That integral would give you the value of $u$ at $0$. Commented Apr 14, 2023 at 14:57
• Because $0$ is the center of the sphere. The mean value property tells you that the value at the center equals the mean on the sphere. Commented Apr 14, 2023 at 15:02
• Any by the way, the integral is $0$. Here I am always assuming that the sphere is centered at the origin. Commented Apr 14, 2023 at 15:12
• Exactly. There are other methods to solve that, though. Try looking for "Poisson kernel". Commented Apr 14, 2023 at 15:34
• Yes. Right. It is actually even simpler. It is clear that $-\Delta x=(-\Delta x_1, -\Delta x_2)=(0,0).$ So $x$ is a solution to the PDE. It is obvious that $x$ satisfies the boundary condition. Conclusion $x$ is a solution to the boundary value problem. Now, that boundary value problem has only one solution. Therefore $x$ is the only solution. Commented Apr 19, 2023 at 13:00

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 $$y\mapsto -y$$).