Question 12.1.1: Determine the steady-state heat distribution in a thin squar...

Place the sides with the zero boundary conditions along the x- and y-axes. Then the problem is expressed as

\frac{\partial^{2} u}{\partial x^{2}}(x, y)+\frac{\partial^{2} u}{\partial y^{2}}(x, y)=0

for (x, y) in the set R = {(x, y) | 0 < x < 0.5, 0 < y < 0.5 }. The boundary conditions are

u(0, y) = 0, u(x, 0) = 0, u(x, 0.5) = 200x, and u(0.5, y) = 200y.

If n = m = 4, the problem has the grid given in Figure 12.7, and the difference equation (12.4) is

2\left[\left(\frac{h}{k}\right)^{2}+1\right] w_{i j}-\left(w_{i+1, j}+w_{i-1, j}\right)-\left(\frac{h}{k}\right)^{2}\left(w_{i, j+1}+w_{i, j-1}\right)=-h^{2} f\left(x_{i}, y_{j}\right)                             (12.4)

4 w_{i, j}-w_{i+1, j}-w_{i-1, j}-w_{i, j-1}-w_{i, j+1}=0

for each i = 1, 2, 3 and j = 1, 2, 3.

Expressing this in terms of the relabeled interior grid points w_{i}=u\left(P_{i}\right) implies that the equations at the points P_{i} are:

\begin{array}{rr}P_{1}: & 4 w_{1}-w_{2}-w_{4}=w_{0,3}+w_{1,4} \\P_{2}: & 4 w_{2}-w_{3}-w_{1}-w_{5}=w_{2,4} \\P_{3}: & 4 w_{3}-w_{2}-w_{6}=w_{4,3}+w_{3,4}, \\P_{4}: & 4 w_{4}-w_{5}-w_{1}-w_{7}=w_{0,2} \\P_{5}: & 4 w_{5}-w_{6}-w_{4}-w_{2}-w_{8}=0 \\P_{6}: & 4 w_{6}-w_{5}-w_{3}-w_{9}=w_{4,2} \\P_{7}: & 4 w_{7}-w_{8}-w_{4}=w_{0,1}+w_{1,0} \\P_{8}: & 4 w_{8}-w_{9}-w_{7}-w_{5}=w_{2,0} \\P_{9}: & 4 w_{9}-w_{8}-w_{6}=w_{3,0}+w_{4,1},\end{array}

where the right sides of the equations are obtained from the boundary conditions.
In fact, the boundary conditions imply that

\begin{aligned}&w_{1,0}=w_{2,0}=w_{3,0}=w_{0,1}=w_{0,2}=w_{0,3}=0 \\&w_{1,4}=w_{4,1}=25, \quad w_{2,4}=w_{4,2}=50, \quad \text { and } \quad w_{3,4}=w_{4,3}=75\end{aligned}

So the linear system associated with this problem has the form

\left[\begin{array}{rrrrrrrrr}4 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\-1 & 4 & -1 & 0 & -1 & 0 & 0 & 0 & 0 \\0 & -1 & 4 & 0 & 0 & -1 & 0 & 0 & 0 \\-1 & 0 & 0 & 4 & -1 & 0 & -1 & 0 & 0 \\0 & -1 & 0 & -1 & 4 & -1 & 0 & -1 & 0 \\0 & 0 & -1 & 0 & -1 & 4 & 0 & 0 & -1 \\0 & 0 & 0 & -1 & 0 & 0 & 4 & -1 & 0 \\0 & 0 & 0 & 0 & -1 & 0 & -1 & 4 & -1 \\0 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & 4\end{array}\right]\left[\begin{array}{c}w_{1} \\w_{2} \\w_{3} \\w_{4} \\w_{5} \\w_{6} \\w_{7} \\w_{8} \\w_{9}\end{array}\right]=\left[\begin{array}{r}25 \\50 \\150 \\0 \\0 \\50 \\0 \\0 \\25\end{array}\right] .

The values of w_{1}, w_{2}, \ldots, w_{9} , found by applying the Gauss-Seidel method to this matrix,
are given in Table 12.1.

These answers are exact, because the true solution, u(x, y) = 400xy, has

\frac{\partial^{4} u}{\partial x^{4}}=\frac{\partial^{4} u}{\partial y^{4}} \equiv 0

and the truncation error is zero at each step.

Table 12.1

\begin{array}{c|c}\hline i & w_{i} \\\hline 1 & 18.75 \\2 & 37.50 \\3 & 56.25 \\4 & 12.50 \\5 & 25.00 \\6 & 37.50 \\7 & 6.25 \\8 & 12.50 \\9 & 18.75 \\\hline\end{array}