Problem 7.15#
\[\begin{equation*}
\end{equation*}\]
Starting with the definition for the local capacity matrix in Eqn 7.104
\[\begin{equation*}
m_{ab}^A = \int_{x_A}^{x_{A+1}} N_a(x) \, N_b(x) \: \mathrm{d}x \quad,
\end{equation*}\]
for which the shape functions in the reference coordinates, \(\xi\), are defined in 7.56 as
\[\begin{equation*}
N_a = \frac{1}{2}\left[ 1 + (-1)^a\xi \right] \quad.
\end{equation*}\]
We may pull back the integration to a reference element using the mapping in 7.55 such that the bounds of integration go from \([x_A,\, x_{A+1}] \rightarrow [-1, \, 1] \):
\[\begin{align*}
m_{ab}^A &= \int_{x_A}^{x_{A+1}} N_a(x) \, N_b(x) \: \mathrm{d}x \quad, \\
&= \int_{-1}^{1} N_a(\xi) \, N_b(\xi) \, \frac{\mathrm{d}x}{\mathrm{d}\xi} \mathrm{d}\xi \quad, \\
&= \frac{1}{4} \int_{-1}^{1} \left[ 1 + (-1)^a\xi \right] \,\left[ 1 + (-1)^b\xi \right] \, \frac{\Delta x_A}{2} \mathrm{d}\xi \quad,
\end{align*} \]
where we have used \(\frac{\mathrm{d}x}{\mathrm{d}\xi} = \frac{\Delta x_A}{2} \) defined in Eqn 7.57. Evaluating this integral yields
\[\begin{align*}
m_{ab}^A &= \frac{\Delta x_A}{8} \left[ \xi + \frac{(-1)^a + (-1)^b}{2}\xi^2 + \frac{(-1)^{a+b}}{3} \xi^3 \right]^{1}_{-1} \\
&= \frac{\Delta x_A}{4}\left[ 1 + \frac{(-1)^{a+b}}{3} \right]
\end{align*}\]
as required.