Page 38 - Numerical Methods for Chemical Engineering
P. 38
Existence and uniqueness of solutions 27
[2]
N
[1]
Given v ∈ and a linearly independent basis {b , b ,..., b [N] },what is the set of scalar coeffi-
cients {c 1 , c 2 ,..., c N } that represent vas the basis set expansion v = c 1 b [1] + c 2 b [2] + ··· + c N b [N] ?
To answer this question, we note that we obtain a set of N linear equations by forming the
[k]
dot products of v with each b ,
[1] [1] [1] [2] [1] [N] [1]
c 1 b · b + c 2 b · b +· · · + c N b · b = b · v
[2] [1] [2] [2] [2] [N] [2]
c 1 b · b + c 2 b · b +· · · + c N b · b = b · v
.
.
. (1.136)
[N] [1] [N] [2] [N] [N] [N]
c 1 b · b + c 2 b · b +· · · + c N b · b = b · v
In matrix-vector form, this system is written as
[1] [1] [1] [2] [1] [N] [1]
b · b b · b ... b · b b · v
c 1
[2] [1] [2] [2] [2] [2]
b · b b · b ... b · b c 2 b · v
[N]
. . . .
. = (1.137)
. . . . .
.
. . . .
[N] [1] [N] [2] [N] [N] [N]
b · b b · b ... b · b c N b · v
To compute the scalar coefficients in this manner, we must solve a system of N linear equa-
[1]
3
[2]
tions, requiring ∼N FLOPs. However, if we were to use a basis set {w , w ,..., w [N] }
that is orthogonal, i.e., the dot product between any two unlike members is zero,
2
[i] [ j] 2 w [i] , i = j
w · w = w [i] δ ij = (1.138)
0, i = j
the coefficients are obtained far more easily,
2 [1]
w w · v
[1] c 1 [ j]
2 w · v
[2]
w [2] c 2 w
· v
2
c j =
. . [ j]
. = w
.
. . . .
. j = 1, 2,..., N
2 [N]
w [3] c N w · v
(1.139)
For this reason, the use of orthogonal basis sets, and of orthonormal basis sets
[1]
[k]
[2]
{u , u ,..., u [N] } that in addition have all |u |= 1, is common. For an orthonormal
basis, we have simply
N
c j = u [ j] · v v = u [ j] · v u [ j] (1.140)
j=1