Discrete Dirichlet problem. What’s wrong with the solution

linear algebralinear-transformationsmatrices

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

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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?

Best Answer

The maximum principle doesn't state that all boundary values will be larger than non-boundary ones. The only thing it says is that the maximum will be achieved on the boundary. For example consider the identity function on $[0,1]$.

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