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Derivative Matrices 141
Let us consider now second-order problems. Let q be a polynomial in Pn−2, n ≥ 2.
For σ1, σ2 ∈ R, one is concerned with finding p ∈ Pn such that
(7.4.9)
⎧
⎪⎨
⎪⎩
−p ′′ (η (n)
i ) = q(η (n)
i ) 1 ≤ i ≤ n − 1,
p(η (n)
0 ) = σ1,
p(η (n)
n ) = σ2.
This corresponds to find the solution of a linear system. We show the case n = 3:
(7.4.10)
⎡
⎢
⎣
1 0 0 0
− ˜ d (2)
10 − ˜ d (2)
11 − ˜ d (2)
12 − ˜ d (2)
13
− ˜ d (2)
20 − ˜ d (2)
21 − ˜ d (2)
22 − ˜ d (2)
23
0 0 0 1
⎤⎡
p(η
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎦⎢
⎣
(n)
0 )
p(η (n)
1 )
p(η (n)
⎤
⎥
2 ) ⎥
⎦
p(η (n)
3 )
=
⎡
σ1
⎢
⎢q(η
⎢
⎣
(n)
1 )
q(η (n)
⎥
⎥.
⎥
2 ) ⎥
⎦
Let r be the solution of (7.4.10) with σ1 = σ2 = 0. Thus, we can eliminate two
unknowns and reduce the system to
(7.4.11)
⎡
⎣ − ˜ d (2)
11 − ˜ d (2) ⎤⎡
12
⎦
− ˜ d (2)
21 − ˜ d (2)
22
⎣ r(η(n)
⎤
1 )
⎦ =
r(η (n)
2 )
⎡
⎣ q(η(n)
⎤
1 )
⎦.
q(η (n)
2 )
Finally, we recover p from the relation p(x) = r(x) + 1
2 (1 − x)σ1 + 1
2 (1 + x)σ2, ∀x ∈ Ī.
Of course, this procedure applies for any n ≥ 2. We note that the matrix in (7.4.11) is
not the square of some first derivative matrix.
Problem (7.4.9) is generalized as follows:
(7.4.12)
⎧
⎪⎨
⎪⎩
−p ′′ (η (n)
i ) + (Bp ′ + Ap)(η (n)
i ) = q(η (n)
i ) 1 ≤ i ≤ n − 1,
p(η (n)
0 ) = σ1,
p(η (n)
n ) = σ2,
where A and B are continuous functions in Ī.
σ2
⎤