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An Example in Two Dimensions 285
(13.2.5) Kn :=
⎡
1 0 0 0 0 0 0 0
⎤
0
⎢0
⎢
⎢0
⎢
⎢0
⎢
⎢0
⎢
⎢0
⎢
⎢0
⎣
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0 ⎥
0 ⎥
0 ⎥
0 ⎥.
⎥
0 ⎥
0 ⎥
⎦
0
0 0 0 0 0 0 0 0 1
Actually, for any 1 ≤ i ≤ n −1 and 1 ≤ j ≤ n −1, the transformation associated with
Kn maps the point (η (n)
i ,η (n)
j ) into the point (η (n)
j ,η (n)
i ). Therefore, the matrix-vector
multiplication ¯pn → An¯pn is equivalent to performing n − 1 second-order partial
derivatives with respect to the variable x, and the multiplication ¯pn → KnAnKn¯pn
corresponds to n − 1 second-order partial derivatives with respect to the variable y.
The sum of the two contributions gives a discretization of the Laplace operator.
We observe that, in the construction of the matrix D ⋆ n, we implicitly assumed that the
unknown polynomial pn vanishes on ∂Ω. This condition is obtained by eliminating
all the entries corresponding to the boundary nodes. The same trick was applied in
(7.4.11) for n = 3 in the one-dimensional case.
A theoretical analysis of the convergence of pn to U as n → +∞ can be developed.
We outline the proof in the Legendre case (α = β = 0). We start by recasting (13.1.1) in
a weak form (see section 9.3), which is obtained by multiplying the differential equation
by a test function φ such that φ ≡ 0 on ∂Ω, and integrating on Ω. With the help of
Green’s formula one gets
(13.2.6)
Ω
∂U ∂φ
∂x ∂x
+ ∂U
∂y
∂φ
dxdy =
∂y
Ω
fφ dxdy =: F(φ).
Similarly, problem (13.2.2) can be formulated in a variational setting. In fact, by formula
(3.5.1) with w ≡ 1, integration by parts yields