Problem 7.8#
We seek the inverse of the mapping defined in 7.55,
\[\begin{equation*}
x(\xi) = \sum_{a=1}^2 x_{A+a-1} N_a ( \xi)
\end{equation*}\]
where
\[\begin{equation*}
N_a(\xi) = \frac{1}{2} \left[ 1 + (-1)^a \xi \right] .
\end{equation*}\]
To do this, let us write out the mapping explicitly:
\[\begin{align*}
x(\xi) &= x_{A} N_1 ( \xi) + x_{A+1} N_2 ( \xi) \quad, \\
x(\xi) &= \frac{1}{2} \Big( x_{A} \left[ 1 - \xi \right] + x_{A+1} \left[ 1 + \xi \right] \Big).
\end{align*}\]
Multiplying by 2 and collecting terms this rearranges to
\[\begin{align*}
2 x = x_{A} + x_{A+1} + \xi \left( x_{A+1} - x_{A} \right)
\end{align*}\]
and therefore
\[\begin{align*}
\xi = \frac{ 2 x - x_{A} - x_{A+1} } { x_{A+1} - x_{A} } = \frac{ 2 x - x_{A} - x_{A+1} } { \Delta x_{A} }
\end{align*}\]
as required. We can then see that at \(x = x_A\) and \(x = x_{A+1}\) this becomes
\[\begin{align*}
\xi(x_A) &= \frac{ x_{A} - x_{A+1} } { x_{A+1} - x_{A} } = - 1 \\
\xi(x_{A+1}) &= \frac{ x_{A+1} - x_{A} } { x_{A+1} - x_{A} } = +1 \quad .
\end{align*}\]