Using the Maple software package:

(a) Consider the square plate in Example 6.2 . 1 for a mesh spacing of \(h=4 \mathrm{~cm}\) (that is, \(h=a / n\), where \(a=12 \mathrm{~cm}\) and \(n=3\) ). However, the bottom and side boundaries each have a temperature of \(100^{\circ} \mathrm{C}\) and the top boundary is at \(0^{\circ} \mathrm{C}\). Therefore show that \(u_{11}=u_{21}=87.5^{\circ} \mathrm{C}\) and \(u_{12}=u_{22}=62.5^{\circ} \mathrm{C}\).

(b) Repeat the calculation for a mesh spacing of \(h=3 \mathrm{~cm}\) (that is, \(n=4\) ) and find the steady-state temperature at the internal mesh points. Hint: The number of mesh points can be reduced if symmetry is considered.

(c) The analytical solution is given by

\[
\begin{aligned}
u(x, y)= & \frac{400}{\pi} \sum_{n=0,1 \ldots}^{\infty} \frac{1}{(2 n+1) \sinh (2 n+1) \pi} \\
& \times\left\{\frac{\sin (2 n+1) \pi x}{a} \sinh \frac{(2 n+1) \pi(a-y)}{a}ight. \\
& \left.+\frac{\sin (2 n+1) \pi y}{a}\left[\sinh \frac{(2 n+1) \pi(a-x)}{a}+\sinh \frac{(2 n+1) \pi x}{a}ight]ight\} .
\end{aligned}
\]

Show that this analytical solution yields the values \(u_{11}=u_{21}=88.1^{\circ} \mathrm{C}\) and \(u_{12}=\) \(u_{22}=61.9^{\circ} \mathrm{C}\) for the given mesh in part (a). Compare the numerical results of part (b) with the values obtained from the analytical solution.