I'm having trouble solving the following PDE problem. We're in the open unit ball in the plane, centered at the origin,
$$B=\{(x,y)\in \mathbb{R}^{2},\ \ x^2+y^2<1\}$$
The following boundary problem is given
$$\Delta u(x,y)=0,\ \ \text{in}\ B,\ \ \ u(x,y)=\sin(x)\ \ \text{on}\ \partial B$$
The problem is the following: Use the mean value theorem to find the value $u(0,0)$.
I'm not so sure how to deal with this one. Help would be much appreciated!
Since, $u(x) = \sin x$ on the boundary, of $B,$ By the mean value property together with the change of variables $y =-z$ that is $d\sigma(z)= d\sigma(y)$ $$u(0)=\frac{1}{2 \pi} \int_{\partial B}u(z) d \sigma(z) =\frac{1}{2 \pi} \int_{\partial B}\sin(z) d \sigma(z) = -\frac{1}{2 \pi} \int_{\partial B}\sin(y) d \sigma(y) = -u(0)$$
That is $$u(0) =0$$