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40 1. Finite Difference Method for the Poisson Equation

Motivated by the example above we call a point r ∈{1,... ,M 1 } far from
the boundary if (1.36) holds, and close to the boundary if (1.37) holds, and
the points r ∈{M 1 +1,... ,M} are called boundary points.

∗
Theorem 1.10 We consider (1.31) under the assumption (1.32) .
If f ≤ 0,then
max (˜ u h ) r ≤ max (ˆ u h ) r . (1.38)
r∈{1,...,M} r∈{M 1 +1,...,M}

Proof: We use the same notation and the same arguments as in the
proof of Theorem 1.9. In (1.35) in the last estimate equality holds, so that
∗
no sign conditions for ¯u are necessary. Because of (4) the maximum will
also be attained at a point close to the boundary and therefore also at
∗
its neighbours. Because of (6) a boundary point also belongs to these
neighbours, which proves the assertion.
From the maximum principles we immediately conclude a comparison
principle:
∗
Lemma 1.11 We assume (1.32) or (1.32) .
Let u h1 , u h2 ∈ R M 1 be solutions of
ˆ
A h u hi = −A h ˆ u hi + f for i =1, 2
i
for given f , f ∈ R M 1 , ˆ u h1 , ˆ u h2 ∈ R M 2 , which satisfy f ≤ f , ˆ u h1 ≤
1 2 1 2
ˆ u h2 .Then
u h1 ≤ u h2 .

ˆ
Proof: From A h (u h1 −u h2 )= −A h (ˆ u h1 − ˆ u h2 )+f −f we can conclude
2
1
with Theorem 1.9 or 1.10 that
max (u h1 − u h2 ) r ≤ 0 .
r∈{1,...,M 1 }

This implies in particular the uniqueness of a solution of (1.31) for
arbitrary ˆ u h and f and also the regularity of A h .
In the following we denote by 0 and 0 the zero vector and the zero
matrix, respectively, where all components are equal to 0. An immediate
consequence of Lemma 1.11 is the following
Theorem 1.12 Let A h ∈ R M 1 ,M 1 be a matrix with the properties (1.32)
(1)–(3) (i), (4) ,and u h ∈ R M 1 .Then
∗
A h u h ≥ 0 implies u h ≥ 0 . (1.39)

Proof: To be able to apply Lemma 1.11, one has to construct a matrix
ˆ
A h ∈ R M 1 ,M 2 such that (1.32)* holds. Obviously, this is possible. Then one