Question 10.5: IRREGULAR BOUNDARY Solve uxx + uyy = 0 in the region shown i...

Based on the grid shown in Figure 10.9, Equation 10.3

u_{i-1, j}+u_{i+1, j}+u_{i, j-1}+u_{i, j+1}-4 u_{i j}=0         (10.3)

can be applied at mesh points (1,1),(2,1), and (1,2) because all four neighboring points at those mesh points are on the grid. Using the boundary conditions provided, the resulting difference equations are

\begin{aligned} & 1+4+u_{12}+u_{21}-4 u_{11}=0 \\ & u_{11}+7+9.5+u_{22}-4 u_{21}=0 \\ & 1+u_{11}+\frac{1}{3}+u_{22}-4 u_{12}=0 \end{aligned}          (10.23)

However, at (2,2), we need to use Equation 10.22.

\frac{1}{\alpha(\alpha+1)} u_{i+1, j}+\frac{1}{\beta(\beta+1)} u_{i, j+1}+\frac{1}{\alpha+1} u_{i-1, j}+\frac{1}{\beta+1} u_{i, j-1}-\frac{\alpha+\beta}{\alpha \beta} u_{i, j}=0        (10.22)

Using the equation of the slanting segment, we find that the point at the (2,3) location is a vertical distance of \frac{1}{3} from the (2,2) mesh point. But since h=\frac{1}{2}, we have \beta=\frac{2}{3}. On the other hand, \alpha=1 since the neighboring point to the right of (2,2) is itself a mesh point. With these, and knowing u_{32}=9 by the given boundary condition, Equation 10.22 yields

\frac{9}{2}+\frac{1}{\frac{2}{3} \cdot \frac{5}{3}}(2)+u_{12}+\frac{1}{\frac{5}{3}} u_{21}-\frac{1+\frac{2}{3}}{\frac{2}{3}} u_{22}=0

Combining this with Equation 10.23, and simplifying, we find

\left[\begin{array}{cccc} -4 & 1 & 1 & 0 \\ 1 & -4 & 0 & 1 \\ 1 & 0 & -4 & 1 \\ 0 & 1.2 & 1 & -5 \end{array}\right]\left\{\begin{array}{l} u_{11} \\ u_{21} \\ u_{12} \\ u_{22} \end{array}\right\}=\left\{\begin{array}{c} -5 \\ -16.5 \\ -1.3333 \\ -12.6 \end{array}\right\} \begin{array}{cc} & u_{11}=3.3354 \\ \text { Solve }& u_{21}=6.0666 \\ \Rightarrow & u_{12}=2.2749 \\ & u_{22}=4.4310 \end{array}