Question 16.14: Solve the Poisson equation ∇²u = uxx + uyy = x + y for the s...
Solve the Poisson equation \nabla^2 u=u_{x x}+u_{y y}=x+y for the square mesh, whose edges, x = 0, x = 0.8, y = 0, y = 0.6 are kept at the temperature shown in the following figure. Find the values of u(x, y) at the nodal points of the rectangular region with mess length 0.2. Use Gauss–Seidel iterative method to compute values at nodal points until the difference between successive values at each point is less than 0.005.
The edges of the rectangular region are x = 0, x = 0.8, y = 0, y = 0.6 with the mess length 0.2. So, values of x and y are as follows
\begin{array}{ll} x & y \\ x_0=0 & y_0=0 \\ x_1=0.2 & y_1=0.2 \\ x_2=0.4 & y_2=0.4 \\ x_3=0.6 & y_3=0.6 \\ x_4=0.8 \end{array} (16.79)
Let the value of u\left(x_i, y_j\right)=u_{i j} , we have
u_{10}=8.4, u_{20}=9.6, u_{30}=10.1, u_{01}=7.3, u_{02}=5
u_{13}=5.5, u_{23}=6.7, u_{33}=8.5, u_{41}=10.4, u_{42}=9.7 (16.80)
On replacing the derivative terms with central differences in the Poisson equation u_{x x}+u_{y y}=x+y \text { at the point, }\left(x_i, y_j\right) , we have
\frac{u_{i+1, j}-2 u_{i, j}+u_{i-1, j}}{h^2}+\frac{u_{i, j+1}-2 u_{i, j}+u_{i, j-1}}{h^2}=x_i+y_j
(or) u_{i+1, j}+u_{i-1, j}+u_{i, j+1}+u_{i, j-1}-4 u_{i, j}=h^2\left(x_i+y_j\right)
For i = 1, 2, 3 and j = 1, 2, we obtain
\begin{array}{ll} \text { At }(1,1) & u_{21}+u_{01}+u_{12}+u_{10}-4 u_{11}=(0.2)^2\left(x_1+y_1\right) \\ \text { At }(2,1) & u_{31}+u_{11}+u_{22}+u_{20}-4 u_{21}=(0.2)^2\left(x_2+y_1\right) \\ \text { At }(3,1) & u_{41}+u_{21}+u_{32}+u_{30}-4 u_{31}=(0.2)^2\left(x_3+y_1\right) \\ \text { At }(1,2) & u_{22}+u_{02}+u_{13}+u_{11}-4 u_{12}=(0.2)^2\left(x_1+y_2\right) \\ \text { At }(2,2) & u_{32}+u_{12}+u_{23}+u_{21}-4 u_{22}=(0.2)^2\left(x_2+y_2\right) \\ \text { At }(3,2) & u_{42}+u_{22}+u_{33}+u_{31}-4 u_{32}=(0.2)^2\left(x_3+y_2\right) \end{array}
Using the values of x_i, y_j, u_{i j} from Eqs. (16.79) and (16.80), we get
\begin{aligned} &u_{21}+u_{12}-4 u_{11}=-15.684 \\ &u_{31}+u_{11}+u_{22}-4 u_{21}=-9.576 \\ &u_{21}+u_{32}-4 u_{31}=-20.468 \\ &u_{22}+u_{11}-4 u_{12}=-10.476 \\ &u_{32}+u_{12}+u_{21}-4 u_{22}=-6.668 \\ &u_{22}+u_{31}-4 u_{32}=-18.16 \end{aligned}
On solving the above system of simultaneous linear equations by Gauss–Seidel with initial approximation [0,0,0,0,0,0]^{ T } , the following iterations \left[u_{11}, u_{21}, u_{31}, u_{12}, u_{22}, u_{13}\right] are obtained
| Iteration 1 | |||||
| 6.882734 | 3.410375 | 3.599250 | 5.960563 | 3.374250 | 3.921000 |
| Iteration 2 | |||||
| 8.170926 | 6.147813 | 4.887688 | 8.375891 | 6.152828 | 5.664375 |
| Iteration 3 | |||||
| 8.583406 | 7.090093 | 5.826235 | 9.083534 | 7.695209 | 6.681129 |
| Iteration 4 | |||||
| 8.730324 | 7.432755 | 6.216864 | 9.328539 | 8.262747 | 7.301361 |
| Iteration 5 | |||||
| 8.783635 | 7.557571 | 6.362414 | 9.416968 | 8.469549 | 7.540903 |
| Iteration 6 | |||||
| 8.803042 | 7.603040 | 6.415641 | 9.449129 | 8.544883 | 7.628991 |
| Iteration 7 | |||||
| 8.810111 | 7.619602 | 6.435042 | 9.460842 | 8.572325 | 7.661131 |
| Iteration 8 | |||||
| 8.812686 | 7.625636 | 6.442111 | 9.465109 | 8.582321 | 7.672842 |
| Iteration 9 | |||||
| 8.813623 | 7.627833 | 6.444686 | 9.466662 | 8.585963 | 7.677108 |
| Iteration 10 | |||||
| 8.813966 | 7.628634 | 6.445624 | 9.467228 | 8.587290 | 7.678662 |
The final solution is as follows
\begin{array}{lll} u_{11}=7.678662 & u_{21}=8.587290 & u_{31}=9.467228 \\ u_{12}=6.445624 & u_{22}=7.628634 & u_{32}=8.813966 \end{array}