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144 Polynomial Approximation of Differential Equations
(7.4.20) s(x) := 1
2σ1 + 2σ2 + σ3 − σ4 + (3σ2 − 3σ1 − σ3 − σ4)x
4
+ (σ4 − σ3)x 2 + (σ1 − σ2 + σ3 + σ4)x 3
, ∀x ∈ Ī.
Then, r := p − s ∈ Pn+2, is expressed as follows (see also funaro and heinrichs
(1990)):
(7.4.21) r(x) =
n−1
j=1
[p(η (n)
j ) − s(η (n)
j )]
1 − x 2
1 − (η (n)
j ) 2
˜ l (n)
j (x), ∀x ∈ Ī,
where we noted that r(±1) = r ′ (±1) = 0. Deriving this relation four times and
evaluating at the nodes, finally leads to the (n − 1) × (n − 1) system
(7.4.22)
n−1
j=1
1
1 − (η (n)
j ) 2
[1 − (η (n)
i ) 2 ] ˜ d (4)
ij − 8η(n) ˜d i
(3)
ij − 12 ˜ d (2)
ij r(η (n)
j ) = q(η (n)
i ),
Once r is computed, we determine p from the expression (7.4.20).
Let us analyze another example. We wish to find p ∈ Pn+2, n ≥ 2, such that
(7.4.23)
⎧
⎪⎨
⎪⎩
p IV (η (n)
i ) = q(η (n)
i ) 1 ≤ i ≤ n − 1,
p(η (n)
0 ) = σ1, p(η (n)
n ) = σ2,
p ′′ (η (n)
0 ) = σ3, p ′′ (η (n)
n ) = σ4,
1 ≤ i ≤ n − 1.
where q ∈ Pn−2 and σk ∈ R, 1 ≤ k ≤ 4. Again, it is sufficient to solve a (n−1)×(n−1)
linear system in the unknowns r(η (n)
j ), 1 ≤ j ≤ n − 1, where r := p − s and s ∈ P3
is now given by
(7.4.24) s(x) := 1
6σ1 + 6σ2 − 3σ3 − 3σ4 + (6σ2 − 6σ1 + σ3 − σ4)x
12
+ (3σ3 + 3σ4)x 2 + (σ4 − σ3)x 3
, ∀x ∈ Ī.