Let $D=\{(x_1,x_2)\in\mathbb{R}^n: x_1^2 + x_2^2 <1\}$. Suppose that exists $u$ harmonic in $D$, continuous in $\overline{D}$ and that coincides with $2x_1^2$ in $\partial D$. Find $u(0)$.
By the Representation Theorem for $u$ harmonic:
$$u(x) = -\int_{\partial \Omega} g(y)\frac{\partial G}{\partial v}(x,y) dS(y)$$ For our problem, $g(x_1,x_2) = 2x_1^2$ and $G$ would be the Green Function for the ball in $2$ dimensions: $$G(x_1,x_2) = -\frac{1}{2\pi}(\ln(|x_2-x_1|)-\ln(|x_1||x_2-x_1^*|))$$ where $x_1^* = \frac{x_1}{|x_1|^2}$
In $B(0,1)$, the normal $v$ is simply $(x_1,x_2)$, and we know $\frac{\partial G}{\partial v} = \nabla G\cdot v$.
$$\frac{\partial G}{\partial x_1} = -\frac{1}{2\pi}\left(\frac{1}{|x_2-x_1|}\frac{x_2-x_1}{|x_2-x_1|}-\frac{1}{|x_1||x_2-x_1^*|}\frac{\cdots}{\cdots}\right)$$
I didn't finish the derivative calculation because I think it's too big s this is not the way to solve this exercise. I'd have to multiply this entire thing by $2x_1^2$ and integrate for $x=(0,0)$. Doesn't this expression has a singulatiry on $(0,0)$ by the way?
Or should I simply see that $g(0,0) = 2(0)^2$ and thus the entire integrand is $0$ and thus $u(0) = 0$?
For a harmonic function, the value at a point is equal to the average around a circle centred at that point. In this case you want to take the average of $2 \cos^2(\theta)$ for $0 \le \theta \le 2\pi$, i.e. $$ \dfrac{1}{2\pi} \int_0^{2\pi} 2 \cos^2(\theta) \; d\theta $$