Let $u: \mathbb{R^2} \to \mathbb{R}$, $u \in C^2$ be a solution to $\Delta u = 0$ in $B_1(0)$. Moreover $u(x) = \alpha x_1^2$ on the boundary $\partial B_1(0)$.
(i) Calculate $\alpha$ under the condition that $u(0)=1$.
(ii) Calculate $\nabla u(0)$.
I know that I have to use the Poisson integral formula, which is given by
Let $u \in C^2\overline{(B_R(0))}$ be a harmonic function. Than
$u(x) = \frac{R^2-|x|^2}{R S_n} \int_{\partial B_R(0)} \frac{u(y)}{|x-y|^n} ds$, $~~~$ $x \in B_R(0)$.
But if I try to set it up for my problem, I end up with a division by zero since $|x-y|^n$=$|0-0|^2$. I'm sure I mistake y here for something it isn't.
Using the Poisson integral formula, we get that \begin{align} u(0) =&\ \frac{R}{2\pi} \int_{\partial B_R(0)} \frac{u(y)}{|y|^2}\ dS = \frac{1}{2R} \int_{\partial B_R(0)} u(y)\ dS\\ =&\ \frac{1}{2} \int^{2\pi}_0 u(R\cos\theta, R\sin\theta)\ d\theta = \frac{\alpha }{2} \int^{2\pi}_0 R^2\cos^2\theta\ d\theta = \frac{\alpha \pi R^2}{2}=1 \end{align} which means $\alpha = 2/\pi R^2$.
Next, observe \begin{align} \nabla u(x) =&\ \frac{-x}{\pi R}\int_{\partial B_R(0)} \frac{u(y)}{|x-y|^2}\ dS + \frac{R^2-|x|^2}{2\pi R}\int_{\partial B_R(0)} u(y) \nabla_x\left(\frac{1}{|x-y|^2}\right)\ dS\\ =&\ \frac{-x}{\pi R}\int_{\partial B_R(0)} \frac{u(y)}{|x-y|^2}-\frac{R^2-|x|^2}{2\pi R}\int_{\partial B_R(0)} u(y)\frac{(x-y)}{|x-y|^4}\ dS \end{align} which means \begin{align} \partial_1 u(0) =&\ \frac{R}{2\pi}\int_{\partial B_R(0)} \frac{u(y)}{|y|^4} y_1\ dS = \frac{1}{2\pi R^3} \int_{\partial B_R(0)} u(y) y_1\ dS\\ =&\ \frac{\alpha}{2\pi R} \int^{2\pi}_0 u(R\cos\theta, R\sin\theta)\cos\theta\ d\theta = \frac{\alpha R}{2\pi} \int^{2\pi}_0 \cos^3\theta\ d\theta =0 \end{align} and \begin{align} \partial_2 u(0) = \frac{\alpha R}{2\pi} \int^{2\pi}_0 \cos^2\theta \sin\theta\ d\theta=0.\ \end{align}