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B
as a one to one mapping from the set of
equation indices
\{ 1 , \ldots , m \}
into the set of variable
indices
\{ 1 , \ldots , n \}
.
If
m = 2
and
n = 2
,
there are two possible basis functions
B^1
and
B^2
defined by
\[
\begin{array}{cc}
B^1 (1) = 1 & B^1 (2) = 2 \\
B^2 (1) = 2 & B^2 (2) = 1
\end{array}
\]
In the case
m = 2
and
n = 3
,
there are six possible basis functions.
Of these, the functions that map the argument 1 to the value 1 are
defined by
\[
\begin{array}{cc}
B^1 (1) = 1 & B^1 (2) = 2 \\
B^2 (1) = 1 & B^2 (2) = 3
\end{array}
\]
What are the other four basis functions ?
B^3
,
B^4
,
B^5
and
B^6
can be defined by
\[
\begin{array}{cc}
B^3 (1) = 2 & B^3 (2) = 1 \\
B^4 (1) = 2 & B^4 (2) = 3 \\
B^5 (1) = 3 & B^5 (2) = 1 \\
B^6 (1) = 3 & B^6 (2) = 2
\end{array}
\]
b1 and a12
are known numbers and we have the equation
x1 = b1 - a12 * x2
B1 and A12
are known numbers and we have the equation
x1 = B1 - A12 * x2
(x1, x2)
of the first equation is also a solution of the
second equation, then
B1 = b1 and A12 = a12.
(Hint, see equation 3.2 and 3.3 in the text.)
x2,
b1 - a12 * x2 = B1 - A12 * x2
b1 - B1 = (a12 - A12) * x2
x2, hence
(a12 - A12) = 0
x2 = 2
and x2 = 1 to conclude that the following must hold
b1 - B1 = 2 * (a12 - A12)
b1 - B1 = 1 * (a12 - A12)
(a12 - A12) = 0
a12 = A12. In addition, we have
b1 - B1 = (a12 - A12) * x2 = 0 * x2 = 0
b1 = B1.