Numerical Methods on Partial Differential Equation Md. Mashiur Rahman Department of Physics University of Chittagong Laplace Equation

FDM for Laplace Equation Finite Difference Method: Transforms Derivatives into Finite Difference. Region of Problem must be divided into some small grids. This process is known as Discretization. Solution of Laplace Equation : [ SFPF ] [ DFPF ] Solution at Grid-point (i, j) Md. Mashiur Rahman Thursday, April 25, 2019 Department of Physics, CU. Slide: 02/10

FDM for Laplace Equation SFPF DFPF x y x y Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 03/10

FDM for Laplace Equation Well-posed PDE: Laplace equation with Dirichlet Boundary condition y SFPF C2 C1 C3 C5 C4 C7 C6 C9 C8 C11 C10 C13 C12 C15 C14 C16 DFPF u2 u1 u3 u5 u4 u9 u7 u6 u8 x Known: Boundary conditions Ck Unknown: uk  ui,j Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 04/10

FDM for Laplace Equation Well-posed PDE: Laplace equation with Dirichlet Boundary condition y C2 C1 C3 C5 C4 C7 C6 C9 C8 C11 C10 C13 C12 C15 C14 C16 u2 u1 u3 u5 u4 u9 u7 u6 u8 SFPF DFPF SFPF x Iteration Known: Boundary conditions Ck Unknown: uk  ui,j Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 05/10

Iteration Procedures Jacobi (/dʒə’koʊbi/) Method: German Mathematician (1804-1851) Jacobi (/dʒə’koʊbi/) Method: A system of n linear equations: Initial Guess: k-th Iterated values: k-th iterated value is evaluated by taking the values obtained in the (k-1)-th iteration. Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 06/10

Iteration Procedures Jacobi Method for Laplace Equation: German Mathematician (1804-1851) Jacobi Method for Laplace Equation: [ SFPF ] k-th Iterated values: German Mathematician Gauss: 1777-1855 Seidel: 1821-1896 Gauss-Seidel Method for Laplace Equation: Jacobi method calculates k-th iteration taking values obtained from the (k-1)-th iteration. (i-1, j)-th grid point is calculated ahead of (i, j)-th grid point. So, (i-1, j)-th grid point has a better approximated value. GS method looks for the most recent values during iteration. GS method is faster than Jacobi method. Md. Mashiur Rahman Thursday, April 25, 2019 Department of Physics, CU. Slide: 07/10

Iteration Procedures Jacobi Method for Laplace Equation: German Mathematician (1804-1851) Jacobi Method for Laplace Equation: [ SFPF ] k-th Iterated values: German Mathematician Gauss: 1777-1855 Seidel: 1821-1896 Gauss-Seidel Method for Laplace Equation: Successive Over-Relaxation (SOR) Method for Laplace Equation: Change in ui,j in k-th iteration. Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 08/10

Iteration Procedures Jacobi Method for Laplace Equation: German Mathematician (1804-1851) Jacobi Method for Laplace Equation: k-th Iterated values: Gauss-Seidel Method for Laplace Equation: German Mathematician Gauss: 1777-1855 Seidel: 1821-1896 Successive Over-Relaxation (SOR) Method for Laplace Equation: Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 09/10

Iteration Procedures Jacobi Method for Laplace Equation: German Mathematician (1804-1851) Jacobi Method for Laplace Equation: k-th Iterated values: Gauss-Seidel Method for Laplace Equation: German Mathematician Gauss: 1777-1855 Seidel: 1821-1896 Successive Over-Relaxation (SOR) Method for Laplace Equation:  is called Accelerating Factor and converges fast for 1<  <2 It’s difficult to estimate the best value of . Under Relaxation 0<  <1 Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 010/10

Problem Solve 2D Laplace equation for u(x, y) in a region bounded by the following boundary conditions: u(0, y) = 0 for 0  y  4 u(4, y) = 12 + y for 0  y  4 u(x, 0) = 3x for 0  x  4 u(x, 4) = x2 for 0  x  4 u(0, y) = 12 + y means x = 4 line C9 =12, C8 =13, C7 =14, C6 =15, C5 =16 y C2 C1 C3 C5 C4 C7 C6 C9 C8 C11 C10 C13 C12 C15 C14 C16 u2 u1 u3 u5 u4 u9 u7 u6 u8 u(x, 4) = x2 means y = 4 line C1 =0, C2 =1, C3 =4, C4 =9, C5 =16 u(0, y) = 0 means x = 0 line C13 =0, C14 =0, C15 =0, C16 =0, C1 =0 u(x, 0) = 3x means y = 0 line C13 =0, C12 =3, C11 =6, C10 =9, C9 =12 Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 011/10

Problem Initial values of u(x, y): [SFPF] y 1 4 16 9 14 15 12 13 6 3 4 16 9 14 15 12 13 6 3 u2 u1 u3 u5 u4 u9 u7 u6 u8 Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 012/10

Problem Initial values of u(x, y): [SFPF] [DFPF] y 1 4 16 9 14 15 12 4 16 9 14 15 12 13 6 3 u2 u1 u3 u4 u9 u7 u6 u8 [DFPF] Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 013/10

Problem Initial values of u(x, y): [SFPF] [DFPF] [SFPF] y 1 4 16 9 14 4 16 9 14 15 12 13 6 3 u2 2.5 10 u4 9.5 u6 u8 [DFPF] [SFPF] Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 014/10

Problem Initial values of u(x, y): [SFPF] [DFPF] [SFPF] y 1 4 16 9 14 4 16 9 14 15 12 13 6 3 5.63 2.5 10 2.88 9.5 9.88 6.13 [DFPF] [SFPF] Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 015/10

Problem 1st Iterated values of u(x, y): Gauss-Seidel Method Since we know all values of u(x, y), we’ll use only SFPF. We start from u1, then u2, u3, ………… u9. y 1 4 16 9 14 15 12 13 6 3 5.63 2.5 10 2.88 9.5 9.88 6.13 Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 016/10

Problem 1st Iterated values of u(x, y): Gauss-Seidel Method y 1 4 16 9 4 16 9 14 15 12 13 6 3 u2 u1 u3 u5 u4 u9 u7 u6 u8 k u1 u2 u3 u4 u5 u6 u7 u8 u9 2.50 5.63 10.0 2.88 6.00 9.88 3.00 6.13 9.50 1 2.38 5.60 9.87 Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 017/10

Problem y 1 4 16 9 14 15 12 13 6 3 u2 5.63 5.60 u1 2.5 2.38 u3 10 9.87 u5 u4 u9 u7 u6 u8 Gauss-Seidel Method x Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 018/10

Problem 1st Iterated values of u(x, y): Gauss-Seidel Method y 1 4 16 9 4 16 9 14 15 12 13 6 3 u2 u1 u3 u5 u4 u9 u7 u6 u8 k u1 u2 u3 u4 u5 u6 u7 u8 u9 2.50 5.63 10.0 2.88 6.00 9.88 3.00 6.13 9.50 1 2.38 5.60 9.87 2.85 6.12 9.87 3.00 6.16 9.51 Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 019/10

Problem y 1 4 16 9 14 15 12 13 6 3 u2 u1 u3 Gauss-Seidel Method 4 16 9 14 15 12 13 6 3 u2 5.63 5.60 u1 2.5 2.38 u3 10 9.87 Gauss-Seidel Method 2nd Iteration u4 2.88 2.65 u5 6.0 6.12 u6 9.88 9.87 u7 3.0 3.00 u8 6.13 6.16 u9 9.5 9.51 x Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 020/10

Problem y 1 4 16 9 14 15 12 13 6 3 u2 u1 u3 Gauss-Seidel Method 4 16 9 14 15 12 13 6 3 u2 5.63 5.60 5.59 u1 2.5 2.38 2.36 u3 10 9.87 Gauss-Seidel Method 3rd Iteration u4 2.88 2.65 2.87 u5 6.0 6.12 u6 9.88 9.87 u7 3.0 3.00 3.01 u8 6.13 6.16 u9 9.5 9.51 x Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 021/10

Problem y 1 4 16 9 14 15 12 13 6 3 u2 u1 u3 Gauss-Seidel Method 4 16 9 14 15 12 13 6 3 u2 5.63 5.60 5.59 u1 2.5 2.38 2.36 2.37 u3 10 9.87 Gauss-Seidel Method 4th Iteration u4 2.88 2.65 2.87 u5 6.0 6.12 6.13 u6 9.88 9.87 u7 3.0 3.00 3.01 u8 6.13 6.16 u9 9.5 9.51 x Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 022/10

Problem 4th Iterated values are same as the 3rd Iterated values. Gauss-Seidel Method k u1 u2 u3 u4 u5 u6 u7 u8 u9 2.50 5.63 10.0 2.88 6.00 9.88 3.00 6.13 9.50 1 2.38 5.60 9.87 2.85 6.12 6.16 9.51 2 2.36 5.59 2.87 3.01 3 2.37 4 4th Iterated values are same as the 3rd Iterated values. Solutions are: Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 023/10

Solve by Yourselves Solve 2D Laplace equation for u(x, y) in a region bounded by 0  x  4 and 0  y  4 . Boundary conditions are: u(0, y) = 0 u(4, y) = 8 + 2y u(x, 0) = x2/ 2 u(x, 4) = x2 Take h = k = 1 Find the values correct to three decimal places. Solve 2D Laplace equation for u(x, y) in a region bounded by 0  x  0.4 and 0  y  0.4 . Boundary conditions are: u(0, y) = 0 u(x, 0) = 0 u(x, 0.4) = 200x u(0.4, y) = 200y Take h = k = 0.1 Find the values correct to three decimal places. Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 024/10

Solve by Yourselves Solve 2D Laplace equation for u(x, y) in a region bounded by 1  x  2 and 0  y  1 . Boundary conditions are: u(0, y) = 0 u(x, 0) = 0 u(x, 1) = x u(2, y) = 2y Take h = k = 0.25 Find the values correct to three decimal places. Solve 2D Laplace equation for u(x, y) in a region bounded by 0  x  1 and 0  y  1 . Boundary conditions are: u(0, y) = 0 u(x, 0) = 0 u(x, 1) = x u(1, y) = y Take h = k = 0.25 Find the values correct to three decimal places. Compare with the exact solution: u(x, y) = xy Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 025/10

Solve by Yourselves Solve 2D Laplace equation for u(x, y) in a region illustrated in the following figure: y 10 5 u2 u1 u3 u5 u4 u9 u7 u6 u8 Find the values correct to three decimal places. Iterate using Jacobi and then, GS methods; make comments on the iteration. Md. Mashiur Rahman Tuesday, June 09, 2015 Department of Physics, CU. Slide: 026/10

Solve by Yourselves Solve 2D Laplace equation for u(x, y) in a region bounded by 0  x  4 and 0  y  4 . Boundary conditions are: u(0, y) = 10y u(x, 0) = 10x u(x, 4) = 40 - 10x u(4, y) = 40 - 10y Take h = k = 1 Find the values correct to three decimal places. y 8.7 12.1 9.0 12.8 21.0 17.0 18.6 21.9 19.7 11.1 u2 u1 u3 u5 u4 u9 u7 u6 u8 Solve 2D Laplace equation for u(x, y) : Find the values correct to three decimal places. Md. Mashiur Rahman Tuesday, July 09, 2015 Department of Physics, CU. Slide: 027/10