Problem 7.6#
The local element solution has the spatial mapping defined as
\[\begin{equation*}
x(\xi) = \sum_{a=1}^2 x_{A+a-1} N_a ( \xi)
\end{equation*}\]
for which the linear shape functions are
\[\begin{equation*}
N_a(\xi) = \frac{1}{2} \left[ 1 + (-1)^a \xi \right]
\end{equation*}\]
Let us expand the mapping with these shape functions by writing the summation explicitly
\[\begin{equation*}
x(\xi) = x_{A} N_1 ( \xi) + x_{A+1} N_2 ( \xi)
\end{equation*}\]
and now substituting in the shape functions:
\[\begin{align*}
N_1(\xi) &= \frac{1}{2} \left[ 1 - \xi \right] \\
N_2(\xi) &= \frac{1}{2} \left[ 1 + \xi \right]
\end{align*}\]
the spatial mapping is then
\[\begin{equation*}
x(\xi) = \frac{x_{A}}{2} \left[ 1 - \xi \right] + \frac{x_{A+1}}{2} \left[ 1 + \xi \right]
\end{equation*}\]
Verifying the values at \(\xi = \pm 1\):#
At \(\xi = -1 \) the mapping becomes
\[\begin{equation*}
x(\xi) = \frac{2 x_{A}}{2} = x_{A}
\end{equation*}\]
as required. Similarly at \(\xi = 1 \) the mapping becomes
\[\begin{equation*}
x(\xi) = \frac{2 x_{A+1}}{2} = x_{A+1}
\end{equation*}\]