I'm reading Artin's Algebra and stumbled upon this miscellaneous question. Chapter 1 Excercise M11. I will post the problem here.

This is my progress up to now.

(a)

The lattice points $(0,0),(\pm 1,0),(0,\pm 1)$ are in the region $R$ and they map to $\{x_{0,0},x_{1,0},x_{0,1},x_{-1,0},x_{0,-1}\}$

Due to the given $\beta$ function all the neighbouring lattice points which are in the region $\bar{R}$ map to $0$ or $1$ depending whether they are above or below the x-axis. (See the definition of this function).

Therefore, we can get 5 linear equations which satisfy the Laplace equation. These linear equations have variables $x_{0,0},x_{1,0},x_{0,1},x_{-1,0},x_{0,-1}$

we get the following equations.

$$ x_{0,0} = \frac{1}{4} (x_{1,0}+x_{0,1}+x_{-1,0}+x_{0,-1})\\ x_{1,0} = \frac{1}{4} (x_{0,0} + 1)\\ x_{0,1} = \frac{1}{4} (x_{0,0} + 3)\\ x_{-1,0} = \frac{1}{4} (x_{0,0} + 1)\\ x_{0,-1} = \frac{1}{4} (x_{0,0}) $$

This is analogous to, $$ \begin{pmatrix} -4 & 1 & 1 & 1 & 1\\ -1 & 4 & 0 & 0 & 0\\ -1 & 0 & 4 & 0 & 0\\ -1 & 0 & 0 & 4 & 0\\ -1 & 0 & 0 & 0 & 4 \end{pmatrix} * \begin{pmatrix} x_{0,0} \\ x_{1,0} \\ x_{0,1} \\ x_{-1,0} \\ x_{0,-1} \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 3 \\1 \\ 0 \end{pmatrix} $$

By solving this I got the following results (I had to solve this on a computer since I couldn't find a way to row-reduce the coefficient matrix),

$$ \begin{pmatrix} x_{0,0} \\ x_{1,0} \\ x_{0,1} \\ x_{-1,0} \\ x_{0,-1} \end{pmatrix} = \begin{pmatrix} 0.41667 \\ 0.35417 \\ 0.85417 \\ 0.35417 \\ 0.10417 \end{pmatrix} $$

(b)

This is where I get stuck. It states that a harmonic function takes on its maximal value on the boundary. But by the results, I obtained in part (a), the boundary values are $x_{1,0},x_{0,1},x_{-1,0},x_{0,-1}$ and some of them are clearly less than $x_{0,0}$.

What's wrong with my solution?