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Derivative Matrices 139
(7.4.1)
⎧
⎨
⎩
p ′ (η (n)
i ) = q(η (n)
i ) 1 ≤ i ≤ n,
p(η (n)
0 ) = σ.
This is equivalent to a linear system in the unknowns p(η (n)
i ), 0 ≤ i ≤ n. For example,
when n = 3, with the notations of section 7.2, we get
(7.4.2)
⎡
⎢
⎣
1 0 0 0
˜d (1)
10
˜d (1)
20
˜d (1)
30
˜d (1)
11
˜d (1)
21
˜d (1)
31
˜d (1)
12
˜d (1)
22
˜d (1)
32
˜d (1)
13
˜d (1)
23
˜d (1)
33
⎤⎡
p(η
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎦⎢
⎣
(n)
0 )
p(η (n)
1 )
p(η (n)
⎤
⎥
2 ) ⎥
⎦
p(η (n)
3 )
=
⎡
σ
⎢
⎢q(η
⎢
⎣
(n)
1 )
q(η (n)
⎤
⎥
⎥.
⎥
2 ) ⎥
⎦
q(η (n)
3 )
If r ∈ Pn is the solution of (7.4.2) with σ = 0, we note that r satisfies
(7.4.3)
⎡
⎢
⎣
˜d (1)
11
˜d (1)
21
˜d (1)
31
˜d (1)
12
˜d (1)
22
˜d (1)
32
˜d (1)
13
˜d (1)
23
˜d (1)
33
⎤⎡
r(η
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎦⎣
(n)
1 )
r(η (n)
⎤
⎥
2 ) ⎥
⎦ =
r(η (n)
3 )
⎡
q(η
⎢
⎣
(n)
1 )
q(η (n)
⎤
⎥
2 ) ⎥
⎦ .
q(η (n)
3 )
Therefore, we reduced the initial (n + 1) × (n + 1) system to a n × n system. The
final solution is then obtained by setting p = r + σ.
Another interesting problem is stated as follows. For q ∈ Pn and γ ∈ R, γ = 0,
find p ∈ Pn such that
(7.4.4)
⎧
⎨
⎩
p ′ (η (n)
i ) = q(η (n)
i ) 1 ≤ i ≤ n,
p ′ (η (n)
0 ) + γp(η (n)
0 ) = q(η (n)
0 ) + γσ.
We observe that now the boundary constraint is not exactly imposed. Actually, we
are trying to enforce both the equation and the boundary condition at the point η (n)
0 .
Using the Lagrange polynomials introduced in theorem 3.2.1, an equivalent formulation
is: