On a disk $\Omega \subseteq \mathbb{R}^2$ is the with middle point $(0,0)$ and radius 1. Is there a harmonic function such that: $u(0,0) = −1, u(1,0) = 3, u(0,1) = 6, u(−1,0) = 2, u(0,−1) = 7$ ?
I think the answer is yes. But how can I argue it riguously but easily? Using Poisson kernel? The I have to find such function $f(e^{2\pi it})$ taking values 3, 6, 2, 7 at some points and whose integral is -1.
Kelenner's suggest is probably the quickest way to handle the problem: let $u$ be the real part of the interpolating polynomial $P$ with given data: $$P(0)=-1, \ P(1)=3, \ P(i) = 6, \ P(-1) = 2, \ P(-i) = 7$$ Wolfram Alpha gives $$ P(z) = \frac14(22z^4 +(1-i)z^3 -8z^2 + (1+i)z -4) $$ which is easily seen to have the required values, so $u=\operatorname{Re} P$ is a harmonic function with the given conditions.