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

2. 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

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

4. 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

5. 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

6. Iteration Procedures Jacobi (/dʒə’koʊbi/) Method:
German Mathematician
( )
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

7. Iteration Procedures Jacobi Method for Laplace Equation:
German Mathematician
( )
Jacobi Method for Laplace Equation:
[ SFPF ]
k-th Iterated values:
German Mathematician
Gauss:
Seidel:
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

8. Iteration Procedures Jacobi Method for Laplace Equation:
German Mathematician
( )
Jacobi Method for Laplace Equation:
[ SFPF ]
k-th Iterated values:
German Mathematician
Gauss:
Seidel:
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

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

10. Iteration Procedures Jacobi Method for Laplace Equation:
German Mathematician
( )
Jacobi Method for Laplace Equation:
k-th Iterated values:
Gauss-Seidel Method for Laplace Equation:
German Mathematician
Gauss:
Seidel:
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

11. 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

12. Problem Initial values of u(x, y): [SFPF] y 1 4 16 9 14 15 12 13 6 34 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

13. Problem Initial values of u(x, y): [SFPF] [DFPF] y 1 4 16 9 14 15 124 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

14. Problem Initial values of u(x, y): [SFPF] [DFPF] [SFPF] y 1 4 16 9 144 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

15. Problem Initial values of u(x, y): [SFPF] [DFPF] [SFPF] y 1 4 16 9 144 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

16. 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

17. Problem 1st Iterated values of u(x, y): Gauss-Seidel Method y 1 4 16 94 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

18. 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

19. Problem 1st Iterated values of u(x, y): Gauss-Seidel Method y 1 4 16 94 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

20. Problem y 1 4 16 9 14 15 12 13 6 3 u2 u1 u3 Gauss-Seidel Method4 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

21. Problem y 1 4 16 9 14 15 12 13 6 3 u2 u1 u3 Gauss-Seidel Method4 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

22. Problem y 1 4 16 9 14 15 12 13 6 3 u2 u1 u3 Gauss-Seidel Method4 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

23. 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

24. 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  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

25. 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

26. 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

27. 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) = x
u(4, y) = y
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