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42 1. Finite Difference Method for the Poisson Equation

ˆ
Then a solution of A h u h = −A h ˆ u h + f satisfies
(1) (2) (1) (2)
(1) − |f| ∞ w + |ˆ u h | ∞ w ≤ u h ≤|f| ∞ w + |ˆ u h | ∞ w ,
h h h h

(1)

(2)
(2) |u h | ∞ ≤ w
|f| ∞ + w
|ˆ u h | ∞ .

h ∞ h ∞
Under the assumptions (1.32) (1)–(3), (4) ,and (1.40) the matrix norm
∗
· ∞ induced by |· | ∞ satisfies

(1)
A −1 ≤ w
.

h ∞ h ∞
Proof: Since −|f| ∞ 1 ≤ f ≤|f| ∞ 1 and the analogous statement for ˆu h
(1) (2)
is valid, the vector v h := |f| ∞ w + |ˆ u h | ∞ w − u h satisfies
h h
ˆ
A h v h ≥|f| ∞ 1 − f − A h (|ˆ u h | ∞ 1 − ˆ u h ) ≥ 0 ,
ˆ
wherewehave also used −A h ≥ 0 in the last estimate. Therefore, the right
inequality of (1) implies from Theorem 1.12 that the left inequality can be
proven analogously. The further assertions follow immediately from (1).
Because of the inverse monotonicity and from (1.32) (5) the vectors pos-
(i)
tulated in Theorem 1.14 have to satisfy w ≥ 0 necessarily for i =1, 2.
h
Thus stability with respect to · ∞ of the method defined by (1.31) as-
suming (1.32) (1)–(3), (4)* is guaranteed if a vector 0 ≤ w h ∈ R M 1 and a
constant C> 0 independent of h can be found such that
A h w h ≥ 1 and |w h | ∞ ≤ C. (1.41)

Finally, this will be proven for the five-point stencil discretization (1.1),
2
2
(1.2) on the rectangle Ω = (0,a) × (0,b)for C = 1 (a + b ).
16
For this reason we define polynomials of second degree w 1 ,w 2 by
1 1
w 1 (x):= x(a − x) and w 2 (y):= y(b − y) . (1.42)
4 4
It is clear that w 1 (x) ≥ 0 for all x ∈ [0,a]and w 2 (y) ≥ 0 for all y ∈ [0,b].
Furthermore, we have w 1 (0) = 0 = w 1 (a)and w 2 (0) = 0 = w 2 (b), and
1 1

w (x)= − and w (y)= − .
1
2
2 2
Therefore w 1 and w 2 are strictly concave and attain their maximum in a
2
b
and , respectively. Thus the function w(x, y):= w 1 (x)+ w 2 (x)satisfies
2
−∆w =1 in Ω ,
(1.43)
w ≥ 0 on ∂Ω .
Now let w h ∈ R M 1 be, for a fixed ordering, the representation of the grid
function w h defined by
(w h )(ih, jh):= w(ih, jh) for i =1,... ,l − 1 ,j =1,... ,m − 1 .