--- title: "Laplace: 3x3 grid 500/1000/500" subject: "Numerical Method" chapter: 7 key: "semester_4/numerical_method:laplace-3x3-0-500-1000" source: https://www.bctengineer.com/q/semester_4/numerical_method/laplace-3x3-0-500-1000 site: BCT Engineering --- # Laplace: 3x3 grid 500/1000/500 Numerical Method · Chapter 7: Numerical solution of Partial differential Equation - Asked in: 2068 Bhadra, 10 marks; 2068 Magh, 10 marks (asked 2 times) - Video: [MA25C09| Liebmann's Iteration Process | Elliptic Equation | Solution of Laplace equation in Tamil](https://www.youtube.com/watch?v=2iR7JIQVV2s) — Mathematics Kala - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/numerical_method/laplace-3x3-0-500-1000 - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-7 ## Question Solve the elliptic equation uxx + uyy = 0 for the square mesh with the boundary values as shown. [Figure: 3x3 interior points u1 u2 u3 / u4 u5 u6 / u7 u8 u9. Top boundary left to right: 0 (corner), 500, 1000, 500, 0 (corner). Bottom boundary: 0 (corner), 500, 1000, 500, 0 (corner). Left boundary top to bottom: 1000, 2000, 1000. Right boundary top to bottom: 1000, 2000, 1000.] ## Short answer Use the five-point Laplace formula with Gauss-Seidel iteration; symmetry reduces the unknowns. The values are u1=u3=u7=u9=937.5, u2=u8=1000.0, u4=u6=1250.0, and u5=1125.0. ## Answer Number the interior points row-wise: u₁ u₂ u₃ / u₄ u₅ u₆ / u₇ u₈ u₉. For Laplace's equation ∇²u = 0 on a square mesh, each interior value equals the mean of its four neighbours (standard five-point formula): u(i,j) = ¼[u(i−1,j) + u(i+1,j) + u(i,j−1) + u(i,j+1)] Boundary values next to the interior points: top 500, 1000, 500 (above u₁, u₂, u₃) and the same along the bottom (below u₇, u₈, u₉); left 1000, 2000, 1000 (beside u₁, u₄, u₇) and the same along the right (beside u₃, u₆, u₉). Corner values are not used. #### Equations ```text u₁ = ¼(1500 + u₂ + u₄) u₂ = ¼(1000 + 2u₁ + u₅) u₄ = ¼(2000 + 2u₁ + u₅) u₅ = ¼(0 + 2u₂ + 2u₄) ``` The figure is symmetric about the vertical and horizontal centre lines, so u₁ = u₃ = u₇ = u₉, u₂ = u₈ and u₄ = u₆. Only u₁, u₂, u₄ and u₅ are unknown. #### Iteration (Gauss–Seidel / Liebmann) Initial value 1000 for every unknown (close to the mean of the boundary values); each new value is used at once in the next equation. | iteration | u₁ | u₂ | u₄ | u₅ | |---|---|---|---|---| | 0 | 1000.0000 | 1000.0000 | 1000.0000 | 1000.0000 | | 1 | 875.0000 | 937.5000 | 1187.5000 | 1062.5000 | | 2 | 906.2500 | 968.7500 | 1218.7500 | 1093.7500 | | 3 | 921.8750 | 984.3750 | 1234.3750 | 1109.3750 | | 4 | 929.6875 | 992.1875 | 1242.1875 | 1117.1875 | | 5 | 933.5938 | 996.0938 | 1246.0938 | 1121.0938 | | 6 | 935.5469 | 998.0469 | 1248.0469 | 1123.0469 | | 7 | 936.5234 | 999.0234 | 1249.0234 | 1124.0234 | | 8 | 937.0117 | 999.5117 | 1249.5117 | 1124.5117 | | 9 | 937.2559 | 999.7559 | 1249.7559 | 1124.7559 | | 10 | 937.3779 | 999.8779 | 1249.8779 | 1124.8779 | | … | … | … | … | … | | 21 | 937.4999 | 999.9999 | 1249.9999 | 1124.9999 | | 22 | 937.5000 | 1000.0000 | 1250.0000 | 1125.0000 | Converged to 4 decimals after 22 iterations. #### Direct solution of the equations (check) ```text u₁ = 937.5000 u₂ = 1000.0000 u₄ = 1250.0000 u₅ = 1125.0000 ``` **Answer: u₁ = 937.5000, u₂ = 1000.0000, u₃ = 937.5000, u₄ = 1250.0000, u₅ = 1125.0000, u₆ = 1250.0000, u₇ = 937.5000, u₈ = 1000.0000, u₉ = 937.5000** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What symmetry simplifies the 3x3 problem? A. Symmetry about both centre lines gives u1=u3=u7=u9, u2=u8, u4=u6, so only four unknowns B. All nine values are equal C. Only u5 is unknown D. The corners equal the centre Answer: A. The symmetric boundary values imply symmetric solution values. (Key point: Symmetry) 2. What is the equation for u2? A. u2 = (500 + u1 + u5)/4 B. u2 = (1000 + 2u1 + u5)/4 C. u2 = (1000 + u1 + u3 + u5)/4 with u3 = 0 D. u2 = (2000 + 2u1 + u5)/4 Answer: B. Top boundary above u2 is 1000; neighbours u1 and u3 (=u1) and u5. (Key point: Equation for u2) 3. Why does u5 = (2u2 + 2u4)/4? A. Its boundary neighbours are all zero by assumption B. It uses the corner values C. It has no boundary neighbour; its neighbours are u2, u8=u2, u4 and u6=u4 D. It is always the average of 2 and 4 Answer: C. The centre point is surrounded by interior points only. (Key point: Equation for u5) 4. What are u1, u2, u4, u5? A. 1000, 1000, 1250, 1125 B. 937.5, 1000, 1125, 1250 C. 1250, 1125, 937.5, 1000 D. 937.5, 1000, 1250, 1125 Answer: D. Solving gives the values and the iteration converges to them. (Key point: Result) ## Close variants set in other exams - [Laplace: 3x3 grid 400/800/600/1200](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alaplace-3x3-0-400-800) — Numerical Method, chapter 7, asked 6× - [Laplace: 3x3 grid 200/300/400/500](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alaplace-3x3-200-300-400-500) — Numerical Method, chapter 7, asked 5× - [Elliptic: 2x2 grid 10/20/40/50](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aelliptic-2x2-10-20-40-50) — Numerical Method, chapter 7, asked 5× - [Laplace: 2x2 grid 10/15/20](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alaplace-2x2-10-15-20-25) — Numerical Method, chapter 7, asked 6× ## Related in Numerical Method - [Laplace: 2x2 grid 3000/2000/500](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alaplace-2x2-3000-grid) — Numerical Method, chapter 7 - [Laplace: 2x2 grid 100/200/50](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alaplace-100-200-50-grid) — Numerical Method, chapter 7, asked 2× - [Elliptic: 0≤x,y≤3 boundary](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alaplace-uxx-0-3-boundary) — Numerical Method, chapter 7 - [Laplace: five-point formula, 2x2](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Alaplace-five-point-iterative) — Numerical Method, chapter 7