# Chapt_16 - 16 16 Numerical Solutions of Partial...

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16 16 Numerical Solutions of Partial Differential Equations EXERCISES 16.1 Laplace’s Equation 1. The fgure shows the values oF u ( x, y ) along the boundary. We need to determine u 11 and u 21 . The system is u 21 +2+0+0 4 u 11 =0 1+2+ u 11 +0 4 u 21 or 4 u 11 + u 21 = 2 u 11 4 u 21 = 3 . Solving we obtain u 11 =11 / 15 and u 21 =14 / 15. 2. The fgure shows the values oF u ( x, y ) along the boundary. We need to determine u 11 , u 21 , and u 31 . By symmetry u 11 = u 31 and the system is u 21 +0+0+100 4 u 11 u 31 +0+ u 11 + 100 4 u 21 0+0+ u 21 + 100 4 u 31 or 4 u 11 + u 21 = 100 2 u 11 4 u 21 = 100 . Solving we obtain u 11 = u 31 = 250 / 7 and u 21 = 300 / 7. 3. The fgure shows the values oF u ( x, y ) along the boundary. We need to determine u 11 , u 21 , u 12 , and u 22 . By symmetry u 11 = u 21 and u 12 = u 22 . The system is u 21 + u 12 +0+0 4 u 11 0+ u 22 + u 11 4 u 21 u 22 + 3 / 2+0+ u 11 4 u 12 3 / 2+ u 12 + u 21 4 u 22 or 3 u 11 + u 12 u 11 3 u 12 = 3 2 . Solving we obtain u 11 = u 21 = 3 / 16 and u 12 = u 22 =3 3 / 16. 4. The fgure shows the values oF u ( x, y ) along the boundary. We need to determine u 11 , u 21 , u 12 , and u 22 . The system is u 21 + u 12 +8+0 4 u 11 u 22 + u 11 4 u 21 u 22 +0+16+ u 11 4 u 12 u 12 + u 21 4 u 22 or 4 u 11 + u 21 + u 12 = 8 u 11 4 u 21 + u 22 u 11 4 u 12 + u 22 = 16 u 21 + u 12 4 u 22 . Solving we obtain u 11 / 3, u 21 =4 / 3, u 12 =16 / 3, and u 22 =5 / 3. 832

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16.1 Laplace’s Equation 5. The fgure shows the values oF u ( x, y ) along the boundary. ±or Gauss-Seidel the coeﬃcients oF the unknowns u 11 , u 21 , u 31 , u 12 , u 22 , u 32 , u 13 , u 23 , u 33 are shown in the matrix 0 . 25 0 . 2 5 00000 . 25 0 . 25 0 . 2 5 0000 0 . 2 5 000 . 2 5 . 2 5 . 25 0 . 25 0 0 0 . 25 0 . 25 0 . 25 0 . 25 0 00 . 25 0 . 2 5 . 25 . 25 0 0 0 . 25 0 . 25 0 . 25 0 . 25 . 25 0 . 25 0 The constant terms in the equations are 0, 0, 6 . 25, 0, 0, 12 . 5, 6 . 25, 12 . 5, 37 . 5. We use 25 as the initial guess For each variable. Then u 11 =6 . 25, u 21 = u 12 =12 . 5, u 31 = u 13 =18 . 75, u 22 = 25, u 32 = u 23 =37 . 5, and u 33 =56 . 25 6. The coeﬃcients oF the unknowns are the same as shown above in Problem 5. The constant terms are 7 . 5, 5, 20, 10, 0, 15, 17 . 5, 5, 27 . 5. We use 32 . 5 as the initial guess For each variable. Then u 11 =21 . 92, u 21 =28 . 30, u 31 =38 . 17, u 12 =29 . 38, u 22 =33 . 13, u 32 =44 . 38, u 13 =22 . 46, u 23 =30 . 45, and u 33 =46 . 21. 7. (a) Using the di²erence approximations For u xx and u yy we obtain u xx + u = 1 h 2 ( u i +1 ,j + u i,j +1 + u i 1 ,j + u i,j 1 4 u ij )= f ( x, y ) so that u i +1 ,j + u i,j +1 + u i 1 ,j + u i,j 1 4 u ij = h 2 f ( x, y ) . (b) By symmetry, as shown in the fgure, we need only solve For u 1 , u 2 , u 3 , u 4 , and u 5 . The di²erence equations are u 2 +0+0+1 4 u 1 = 1 4 ( 2) u 3 +0+ u 1 +1 4 u 2 = 1 4 ( 2) u 4 u 2 + u 5 4 u 3 = 1 4 ( 2) 0+0+ u 3 + u 3 4 u 4 = 1 4 ( 2) u 3 + u 3 +1+1 4 u 5 = 1 4 ( 2) or u 1 =0 . 25 u 2 +0 . 375 u 2 . 25 u 1 . 25 u 3 . 375 u 3 . 25 u 2 . 25 u 4 . 25 u 5 . 125 u 4 . 5 u 3 . 125 u 5 . 5 u 3 . 625 . 833
16.1 Laplace’s Equation Using Gauss-Seidel iteration we fnd u 1 =0 . 5427, u 2 . 6707, u 3 . 6402, u 4 . 4451, and u 5 . 9451.

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Chapt_16 - 16 16 Numerical Solutions of Partial...

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