here is a way to solve a three-diagonal finite difference scheme with an
linear
number of algebraic operations. Suppose some finite difference scheme takes
the
form
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|
(Factorization1)
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with the
properties
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(Factorization3)
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(Factorization4)
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We look for a recursive
relationship
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(Factorization2)
|
with some coefficients
that we determine through the substitution into the scheme
(
Factorization1
):
We express all the
through
the
using the relationship
(
Factorization2
):
Hence, to satisfy the (
Factorization1
) we must
have
and
Such conditions lead to the recursive
relationships
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(Factorization5)
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(Factorization6)
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Under conditions
(
Factorization3
),(
Factorization4
)
we
have
for all
.
Therefore, we start
with
and produce
according to the
(
Factorization5
),(
Factorization6
).
At the final step of the iteration we use the boundary
condition
Afterward, with
already known we compute
with recursion in opposite direction
according to the (
Factorization2
).
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