--- title: "Laplace: five-point formula, 2x2" subject: "Numerical Method" chapter: 7 key: "semester_4/numerical_method:laplace-five-point-iterative" source: https://www.bctengineer.com/q/semester_4/numerical_method/laplace-five-point-iterative site: BCT Engineering --- # Laplace: five-point formula, 2x2 Numerical Method · Chapter 7: Numerical solution of Partial differential Equation - Asked in: 2068 Magh, 10 marks - 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-five-point-iterative - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-7 ## Question Given the values of u(x, y) on the boundary of the square as shown in the figure below, evaluate the function u(x, y) satisfying the Laplace ∇²u = 0 at the pivotal points, using standard five point formula iterative method. [Figure: interior points u1 (top-left), u2 (top-right), u3 (bottom-left), u4 (bottom-right). Top boundary: 1000, 1000, 1000 (left to right). Left boundary top to bottom: 1000 (corner), 2000, 2000, 1000 (corner). Right boundary: 500 (next to u2 row), 0 (next to u4 row), 0 (bottom). Bottom boundary: 500, 0.] ## Short answer Using the standard five-point formula for Laplace's equation and iterative method (Gauss-Seidel), the interior nodal values are u1 = 1208.3333, u2 = 791.6667, u3 = 1041.6667, u4 = 458.3333. ## Answer Interior points: u₁ (top-left), u₂ (top-right), u₃ (bottom-left), u₄ (bottom-right). The standard five-point formula for ∇²u = 0 is u(i,j) = ¼[u(i−1,j) + u(i+1,j) + u(i,j−1) + u(i,j+1)] Reading the figure: top 1000, 1000 (above u₁, u₂); left 2000, 2000 (beside u₁, u₃); right 500, 0 (beside u₂, u₄); bottom 500, 0 (below u₃, u₄). Corner values are not used. #### Equations ```text u₁ = ¼(3000 + u₂ + u₃) u₂ = ¼(1500 + u₁ + u₄) u₃ = ¼(2500 + u₁ + u₄) u₄ = ¼(0 + u₂ + u₃) ``` #### Iteration (Gauss–Seidel / Liebmann) Initial value 875 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 | u₃ | u₄ | |---|---|---|---|---| | 0 | 875.0000 | 875.0000 | 875.0000 | 875.0000 | | 1 | 1187.5000 | 890.6250 | 1140.6250 | 507.8125 | | 2 | 1257.8125 | 816.4062 | 1066.4062 | 470.7031 | | 3 | 1220.7031 | 797.8516 | 1047.8516 | 461.4258 | | 4 | 1211.4258 | 793.2129 | 1043.2129 | 459.1064 | | 5 | 1209.1064 | 792.0532 | 1042.0532 | 458.5266 | | 6 | 1208.5266 | 791.7633 | 1041.7633 | 458.3817 | | 7 | 1208.3817 | 791.6908 | 1041.6908 | 458.3454 | | 8 | 1208.3454 | 791.6727 | 1041.6727 | 458.3364 | | 9 | 1208.3364 | 791.6682 | 1041.6682 | 458.3341 | | 10 | 1208.3341 | 791.6670 | 1041.6670 | 458.3335 | | 11 | 1208.3335 | 791.6668 | 1041.6668 | 458.3334 | | 12 | 1208.3334 | 791.6667 | 1041.6667 | 458.3333 | | 13 | 1208.3333 | 791.6667 | 1041.6667 | 458.3333 | Converged to 4 decimals after 13 iterations. #### Direct solution of the equations (check) ```text u₁ = 1208.3333 u₂ = 791.6667 u₃ = 1041.6667 u₄ = 458.3333 ``` **Answer: u₁ = 1208.3333, u₂ = 791.6667, u₃ = 1041.6667, u₄ = 458.3333** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. u1 = ? A. (2000 + u2 + u3)/4 B. (1000 + u2 + u3)/4 C. (3000 + u2 + u3)/4 D. (4000 + u2 + u3)/4 Answer: C. Top 1000 + left 2000. (Key point: Equation u1) 2. u3 = ? A. (2000 + u1 + u4)/4 B. (1500 + u1 + u4)/4 C. (500 + u1 + u4)/4 D. (2500 + u1 + u4)/4 Answer: D. Left 2000 + bottom 500. (Key point: Equation u3) 3. When is iteration stopped? A. When successive values agree to the required decimals B. After exactly two iterations C. When all values are equal D. When u1 is largest Answer: A. Iteration stops once changes are below tolerance. (Key point: Convergence) 4. What is u4? A. 791.6667 B. 458.3333 C. 1041.6667 D. 1208.3333 Answer: B. Solution: 1208.3333, 791.6667, 1041.6667, 458.3333. 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