Higher-Order Compact Finite Element Method

Transcription

1 Higher-Order Compact Finite Element Method Major Qualifying Project Advisor: Professor Marcus Sarkis Amorn Chokchaisiripakdee Worcester Polytechnic Institute

2 Abstract The Finite Element Method (FEM) is a scheme that can approximate the solution of boundary value problems. We study the fundamentals of FEM and construct a MATLAB code to approximate the error of the solution for each refinement and compute the rate of convergence of this discretization. Then, we study the article of Bramble and Schatz [2] to construct MATLAB code to approximate better solution by averaging the FEM s solution. 1

3 Acknowledgement I want to thank Professor Marcus Sarkis for his guidance, his continual support and his teaching. I would like to thank my friend, Thanacha Pi Choopojcharoen, for his guidance in MATLAB. And I want to thank my mother for her support all four terms. 2

4 Contents 1 Introduction The Model Problem Domain and Boundary Boundary Value Problem Weak form of FEM Triangulation Linear and Quadratic triangular The Reference Triangle Basis function Mesh Grid Local vertices Elements Dirichlet boundary Q Loval vertices Element Dirichlet boundary Finite Element Method Left-hand side Hat functions of P Hat functions of P Hat function of Q Right-hand side FEM of P FEM of P FEM of Q More accuracy by averaging Bramble and Schatz Averaging P 2 in 2D Case Case Case Case

5 3.3 Averaging Q Numerical Experiments Numerical experiments of P Numerical experiments of P Numerical experiments of Q Conclusion 41 4

6 Chapter 1 Introduction The Finite Element Method is a scheme that can approximate the solution of boundary value problem described by a partial differential equation (PDE). In this paper, we focus on polygonal R 2 domain. To compute the solution, we divide our domain into many subdomains called finite element. The simplest elements to compute are triangle and square, and they are the ones we consider here. Then we construct stiffness matrices associated to the Poisson problem. 1.1 The Model Problem In our model problem, we start with definition of our domain and boundaries. Then, we introduce basis function to solve the Poisson problem Domain and Boundary Denote the polygon domain in R 2 by Ω. Two boundary conditions, Dirichlet Γ D and Neumann Γ N. Γ D Ω Γ N Figure 1.1: Domain Ω Γ D is the Dirichlet boundary where the solution is given. 5

7 Γ N is the Neumann boundary where the normal derivative is given Boundary Value Problem Consider the Laplace operator, or Laplacian The boundary value problem is u = 2 u x + 2 u 2 y 2 u + cu = f in Ω u = g 0 n u = g 1 on Γ D on Γ N u is an unknown function defined on the domain Ω c is a non-negative constant value. In this paper, we set c = 0 f is a given function on Ω g 0, g 1 are given on two different parts of boundary Weak form of FEM We now use the weak form of FEM. To get start, we introduce the Green s Theorem. The theorem states that ( u)v + u v = ( n u)v Ω Ω Γ Our boundary contains with Dirichlet and Neumann ( u)v + u v = Ω Ω ( n u)v + Γ N ( n u)v Γ D Since c = 0 so that u = f u v = fv + Ω Ω g 1 v + Γ N ( n u)v Γ D Since the value on Γ D is given, we set v = 0 on Γ D Therefore, we have a new formula of our problem u v = fv + Ω Ω g 1 v Γ N In this project we assume that Γ N =, that is, the boundary of Ω is all Dirichlet boundary. u v = fv Ω Ω 6

8 1.2 Triangulation As we mentioned in domain and boundary, we have to divide our domain into subdomains. We focus on linear and quadratic function of two variables P 1 and P 2. K Figure 1.2: Subdomains of Ω Linear and Quadratic triangular Define P 1 : Linear Triangular and P 2 : Quadratic Triangular A linear function of two variables is p(x, y) = a i + b i x + c i y A quadratic function of two variables is p P 1 = {a i + b i x + c i y a i, b i, c i R} p(x, y) = a i + b i x + c i y + d i xy + e i x 2 + f i y 2 p P 2 = {a i + b i x + c i y + d i xy + e i x 2 + f i y 2 a i, b i, c i, d i, e i, f i R} The Reference Triangle Given a subdomain K in Ω, define F k is a function mapping ˆK K K is a triangle on Ω ˆK is a reference triangle with ˆv 1 =(0,0), ˆv 2 =(1,0), and ˆv 3 =(0,1) 7

9 v 3 ˆv 3 K v 1 v 2 F k ˆK ˆv 1 ˆv 2 Figure 1.3: The function map to reference triangle For any point (x,y) in K, (x, y) = F k (ˆx, ŷ) ( ) ) x (ˆx = F y k ŷ Define B k = ( ) x = y ( ) x2 x 1 x 3 x 1 y 2 y 1 y 3 y 1 ( ) x = y ( x2 x 1 x 3 x 1 ( x1 The area of the triangle K = detb k /2 y 1 y 2 y 1 y 3 y 1 ) (1 ˆx ŷ) + ) ) (ˆx ŷ ( x2 y 2 + ) ˆx + ( x1 y 1 ( x3 y 3 ) ) ŷ Basis function The polynomials are defined on each triangle in the mesh, which form a set of basis functions for the triangular element. Each basis function has a value of 1 at its node and a value of 0 at the other nodes giving it a pyramid shape over the triangle. Define v h be a piecewise linear and continuous function on Ω so we can also define a basis function { 0 i j φ i (x i ) = δ ij = 1 i = j 8

10 Figure 1.4: The hat function (image credit: It is easy to see that w h (x) = i N w h (v i )φ i (x) where N is the index set of interior nodes, since we consider w h (x) = 0 for x Ω. 1.3 Mesh Grid Our domain is a square 1-by-1 to be easier to compute. We refine our domain into mesh grid that we choose L = 1,2,3,... Here is the mesh grid for L = Figure 1.5: The global vertices for P 1 for L = 1 9

11 1.3.1 Local vertices Figure 1.6: The global vertices for P 2 for L = 1 To order the number of vertices, we have to choose the direction clockwise or counterclockwise. In this report, we use counter-clockwise for P 1. In P 2, three vertices in the middle of each edges are ordered in opposite of vertices. ˆ3 ˆ2 ˆ1 ˆ1 ˆ2 ˆ3 Figure 1.7: The ordering of vertices for P 1 ˆ3 ˆ2 ˆ6 ˆ1 ˆ5 ˆ4 ˆ4 ˆ5 ˆ1 ˆ6 ˆ2 ˆ3 Figure 1.8: The ordering of vertices for P 2 For example, we map vertices of triangle in global to local by ˆ3 ˆ6 ˆ ˆ4 ˆ5 5 ˆ2 Figure 1.9: Example of mapping vertices 10

12 1.3.2 Elements To compute each element, we order our element at the figure below Figure 1.10: The ordering of element for L = 1 By using the ordering of vertices above, we get the ordering of vertices in each element. Element v 1 v 2 v Table 1.1: The order of vertices for P 1 Element v 1 v 2 v 3 v 4 v 5 v Table 1.2: The order of vertices for P 2 11

13 1.3.3 Dirichlet boundary As we mentioned above, we are focusing only zero Dirichlet boundary condition, so all the vertices are zero at the boundary. However, we can adapt to have Neumann boundary if we want to. Here is the ordering of Dirichlet global boundary nodes and the Tables 1.3 and 1.4 of the ordering of vertices Figure 1.11: The ordering of Dirichlet boundaries for L = 1 Dirichlet v 1 v Table 1.3: The ordering of vertices for P 1 Dirichlet v 1 v 2 v Table 1.4: The ordering of vertices for P 2 12

14 1.4 Q 3 We introduce new element which is easy to compute for square domain. It is a square element. As we define our mesh grid, coordinate, element, and Dirichlet boundary above. We do the same thing in our Q 3. Here is the global ordering of vertices for L = Loval vertices Figure 1.12: The ordering of vertices for Q 3 We also define the order of each element in the same way. Here is the example of first element for L = 1 13 ˆ 14 ˆ 15 ˆ 16 ˆ ˆ9 10 ˆ 11 ˆ 12 ˆ ˆ5 ˆ6 ˆ7 ˆ8 ˆ1 ˆ2 ˆ3 ˆ Element Figure 1.13: The ordering of vertices for Q 3 Now our element is a square with four points on boundary of each element. 13

15 Figure 1.14: The ordering of element for L = 1 Element v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v 9 v 10 v 11 v 12 v 13 v 14 v 15 v Dirichlet boundary Table 1.5: The ordering of vertices for Q 3 For the Dirichlet boundary, we also order in counter-clockwise Figure 1.15: The ordering of Dirichlet boundaries for L = 1 for Q 3 14

16 Element v 1 v 2 v 3 v Table 1.6: The ordering of vertices for Q 3 15

17 Chapter 2 Finite Element Method In this section, we decide to set all the boundary to be zero Dirichlet. Therefore, we get: Find u V h (Ω) such that u v = fv v V h (Ω), Ω Ω where V h (Ω) is the space of piecewise linear and continuous functions which are vanish on Ω. We study this equation in two parts, left-hand side and right-hand side. 2.1 Left-hand side We denote the i, j position of the stiffness matrix by Recall our solution u is of the form Ω φ i φ j u = j N u j φ j then substitute u Ω u φ i = j N We compute element K in Ω φ K i φ K j = detb K K Then, we use a change of variable to obtain φ K i φ K j = detb K K u j Ω φ j φ i ( φ K i F K ) ( φ K j F K ) ˆK (B T ˆ K ˆφî) (B T ˆ K ˆφĵ) ˆK 16

18 Therefore, the entries of the stiffness matrix is given by φ K i φ K j = detb K C K ˆ ˆφî ˆ ˆφĵ K ˆK where ( ) c C K = B 1 K B T K = K 11 c K 12 c K 21 c K Hat functions of P 1 The basis function on ˆK is of the form: ˆφî(ˆx, ŷ) = a 1 + b 1ˆx + c 1 ŷ Each ˆφî has three unknown coefficients a i, b i, c i. For instance, for ˆφˆ1 (1, 0) = 0, and ˆφˆ1 (0, 1) = 0. So, this implies that ˆφˆ1 we want ˆφˆ1 (0, 0) = 1, a 1 + b 1ˆxˆ1 + c 1ŷˆ1 = 1 a 1 + b 1ˆxˆ2 + c 1ŷˆ2 = 0 a 1 + b 1ˆxˆ3 + c 1ŷˆ3 = 0 so we can write it in matrix form as 1 ˆxˆ1 ŷˆ1 a ˆxˆ2 ŷˆ2 b 1 = 0 1 ˆxˆ3 ŷˆ3 c 1 0 For all the three basis function in triangle, we get the equation of the form 1 ˆxˆ1 ŷˆ1 a 1 a 2 a ˆxˆ2 ŷˆ2 b 1 b 2 b 3 = ˆxˆ3 ŷˆ3 c 1 c 2 c Solve for coefficient matrix, we get each ˆφ i in the form ˆφˆ1 = 1 ˆx ŷ ˆφˆ2 = ˆx ˆφˆ3 = ŷ 17

19 2.1.2 Hat functions of P 2 The basis function on ˆK is of the form: ˆφî(ˆx, ŷ) = a 1 + b 1ˆx + c 1 ŷ + d 1ˆxŷ + e 1ˆx 2 + f 1 ŷ 2 Each ˆφî has three unknown coefficients a i, b i, c i. For instance, for ˆφˆ1 we want ˆφˆ1 (0, 0) = 1, ˆφˆ1 (1, 0) = 0, ˆφˆ1 (0, 1) = 0, ˆφˆ1 (1/2, 1/2) = 0, ˆφˆ1 (0, 1/2) = 0, and ˆφˆ1 (1/2, 0) = 0. So, this implies that a 1 + b 1ˆxˆ1 + c 1ŷˆ1 + d 1ˆxˆ1ŷˆ1 + e 1ˆx 2ˆ1 + f 1ŷ 2ˆ1 = 1 a 1 + b 1ˆxˆ2 + c 1ŷˆ2 + d 1ˆxˆ2ŷˆ2 + e 1ˆx 2ˆ2 + f 1ŷ 2ˆ2 = 0 a 1 + b 1ˆxˆ3 + c 1ŷˆ3 + d 1ˆxˆ3ŷˆ3 + e 1ˆx 2ˆ3 + f 1ŷ 2ˆ3 = 0 a 1 + b 1ˆxˆ4 + c 1ŷˆ4 + d 1ˆxˆ4ŷˆ4 + e 1ˆx 2ˆ4 + f 1ŷ 2ˆ4 = 0 a 1 + b 1ˆxˆ5 + c 1ŷˆ5 + d 1ˆxˆ5ŷˆ5 + e 1ˆx 2ˆ5 + f 1ŷ 2ˆ5 = 0 a 1 + b 1ˆxˆ6 + c 1ŷˆ6 + d 1ˆxˆ6ŷˆ6 + e 1ˆx 2ˆ6 + f 1ŷ 2ˆ6 = 0 so we can write it in matrix form as 1 ˆxˆ1 ŷˆ1 ˆxˆ1ŷˆ1 ˆx 2ˆ1 1 ˆx 2 ŷ 2 ˆx 2 ŷ 2 ˆx 2ˆ2 1 ˆx 3 ŷ 3 ˆx 3 ŷ 3 ˆx 2ˆ3 1 ˆx 4 ŷ 4 ˆx 4 ŷ 4 ˆx 2ˆ4 1 ˆx 5 ŷ 5 ˆx 5 ŷ 5 ˆx 2ˆ5 1 ˆx 6 ŷ 6 ˆx 6 ŷ 6 ˆx 2ˆ6 ŷ 2ˆ1 ŷ 2ˆ2 ŷ 2ˆ3 ŷ 2ˆ4 ŷ 2ˆ5 ŷ 2ˆ6 a 1 b 1 c 1 d 1 e 1 f = For all the six basis function in triangle, we get the equation of the form 1 ˆxˆ1 ŷˆ1 ˆxˆ1ŷˆ1 ˆx 2ˆ1 ŷ 2ˆ1 a 1 a 2 a 3 a 4 a 5 a ˆx 2 ŷ 2 ˆx 2 ŷ 2 ˆx 2ˆ2 ŷ 2ˆ2 b 1 b 2 b 3 b 4 b 5 b ˆx 3 ŷ 3 ˆx 3 ŷ 3 ˆx 2ˆ3 ŷ 2ˆ3 c 1 c 2 c 3 c 4 c 5 c 6 1 ˆx 4 ŷ 4 ˆx 4 ŷ 4 ˆx 2ˆ4 ŷ 2ˆ4 d 1 d 2 d 3 d 4 d 5 d 6 = ˆx 5 ŷ 5 ˆx 5 ŷ 5 ˆx 2ˆ5 ŷ 2ˆ5 e 1 e 2 e 3 e 4 e 5 e ˆx 6 ŷ 6 ˆx 6 ŷ 6 ˆx 2ˆ6 ŷ 2ˆ6 f 1 f 2 f 3 f 4 f 5 f Solve for coefficient matrix, we get each ˆφ i in the form ˆφˆ1 = (1 ˆx ŷ)(1 2ˆx 2ŷ) ˆφˆ5 ˆφˆ6 ˆφˆ2 ˆφˆ3 = ˆx(2ˆx 1) = ŷ(2ŷ 1) ˆφˆ4 = 4ˆxŷ = 4ˆx(1 ˆx ŷ) = 4ŷ(1 ˆx ŷ) 18

20 2.1.3 Hat function of Q 3 For Q 3, it is different from P 1 and P 2. We now have an element that is a square which in each direction x and y is a polynomial degree 3 in x and y respectively. We first focus in 1D. ˆφî(ˆx) = a 1 + b 1ˆx + c 1ˆx 2 + d 1ˆx 3 Each ˆφî has four unknown coefficients a i, b i, c i, d i For instance, for ˆφî we want ˆφˆ1 (0) = 1, (1/3) = 0, ˆφˆ1 (2/3) = 0, and ˆφˆ1 (1) = 0. So, this implies that ˆφˆ1 so we can write it in matrix form as 1 ˆxˆ1 ˆx 2ˆ1 1 ˆxˆ2 ˆx 2ˆ2 1 ˆxˆ3 ˆx 2ˆ3 1 ˆxˆ4 ˆx 2ˆ4 a 1 + b 1ˆxˆ1 + c 1ˆx 2ˆ1 + d 1ˆx 3ˆ1 = 1 a 1 + b 1ˆxˆ2 + c 1ˆx 2ˆ2 + d 1ˆx 3ˆ2 = 0 a 1 + b 1ˆxˆ3 + c 1ˆx 2ˆ3 + d 1ˆx 3ˆ3 = 0 a 1 + b 1ˆxˆ4 + c 1ˆx 2ˆ4 + d 1ˆx 3ˆ4 = 0 ˆx 3ˆ1 ˆx 3ˆ2 ˆx 3ˆ3 ˆx 3ˆ4 a 1 b 1 c 1 d 1 1 = For all the four basis function in x-axis, we get the equation of the form 1 ˆxˆ1 ˆx 2ˆ1 ˆx 3ˆ1 a 1 a 2 a 3 a ˆxˆ2 ˆx 2ˆ2 ˆx 3ˆ2 b 1 b 2 b 3 b 4 1 ˆxˆ3 ˆx 2ˆ3 ˆx 3ˆ3 c 1 c 2 c 3 c 4 = ˆxˆ4 ˆx 2ˆ4 ˆx 3ˆ4 d 1 d 2 d 3 d Solve for coefficient matrix, we get each ˆφ i in the form Also for y-axis, ˆφˆ1 ˆφˆ1 = 9/2(1 ˆx)(1/3 ˆx)(2/3 ˆx) ˆφˆ2 ˆφˆ3 ˆφˆ4 = 27/2(ˆx)(1 ˆx)(2/3 ˆx) = 27/2(ˆx)(1 ˆx)(ˆx 1/3) = 9/2(ˆx)(ˆx 1/3)(ˆx 2/3) = 9/2(1 ŷ)(1/3 ŷ)(2/3 ŷ) ˆφˆ2 ˆφˆ3 ˆφˆ4 = 27/2(ŷ)(1 ŷ)(2/3 ŷ) = 27/2(ŷ)(1 ŷ)(ŷ 1/3) = 9/2(ŷ)(ŷ 1/3)(ŷ 2/3) For 2D problem, there are 16 nodes in each elements which are ˆφîĵ (ˆx, ŷ) = ˆφ î (ˆx) ˆφ ĵ (ŷ) for î, ĵ = ˆ1, ˆ2, ˆ3, ˆ4 19

21 2.2 Right-hand side Our computation of the right-hand side is We approximate this integration by where Then, Therefore, our Finite Element is u j or where K Ω fv K Ω fv c i K c i = f(x i ) φ i φ j φ K i φ K j = detb K (φ K i F K ) (φ K j F K ) ˆK j Ω φ j φ i = K Au = b c i A = φ j φ i and b = Ω K K φ i φ j c i K φ i φ j 2.3 FEM of P 1 Recall our stiffness matrix where ˆK xx = ˆK K φ K i φ K j = detb K C K ˆ ˆφi ˆ ˆφ j ˆK ( ) c C K = B 1 K B T K = K 11 c K 12 c K 21 c K 22 x ˆφ i x ˆφ j, ˆKyy = y ˆφ i y ˆφ j, ˆKxy = x ˆφ i y ˆφ j, ˆKyx = ˆK Then, our stiffness matrix with respect to the reference triangle is φ K i φ K j = detb K (c K ˆK 11 xx + c K ˆK 22 yy + c K ˆK 12 xy + c K ˆK 21 yx ) K ˆK ˆK y ˆφ i x ˆφ j 20

22 Recall our P 1 basis functions has the following ordering ˆφˆ1 = 1 ˆx ŷ, ˆφˆ2 = ˆx, ˆφˆ3 = ŷ We start with construct Stiffness matrix. ( ) ( ) ( ) ˆ ˆφˆ1 = ˆ ˆφˆ2 1 = ˆ ˆφˆ3 0 = 1 ˆK xx = ˆKyy = ˆK xy = ˆKyx = The part detb K and c K ij depend on element K For the right-hand side of FEM, we can compute the Mass matrix Mass ˆK = ˆφ i ˆφ j Then, for P 1, it is Mass ˆK = And we compute for each element and sum in our matrix b ˆK 2.4 FEM of P 2 Recall our basis functions of P ˆφˆ = (1 ˆx ŷ)(1 2ˆx 2ŷ), ˆφˆ2 = ˆx(2ˆx 1), ˆφˆ3 = ŷ(2ŷ 1) 21

23 ˆφˆ4 = 4ˆxŷ, ˆφˆ5 = 4ˆx(1 ˆx ŷ), ˆφˆ6 = 4ŷ(1 ˆx ŷ) We take the gradient of our basis functions and construct stiffness matrix as P 1 ( ) ( ) ( ) 4ˆx + 4ŷ 3 4ˆx 1 0 ˆ ˆφˆ1 = ˆ ˆφˆ2 4ˆx + 4ŷ 3 = ˆ ˆφˆ3 0 = 4ŷ 1 ( ) ( ) ( ) 4ŷ ˆ ˆφˆ4 = ˆ ˆφˆ5 4ˆx = 4ŷ 4 8ˆx 4ŷ ˆ ˆφˆ6 4 4ˆx 8ŷ = 4ˆx ˆK xx = ˆK xy = ˆK yy = ˆK yx = For the Right-hand side of FEM, we can compute by integrate Mass matrix Mass ˆK = ˆφ i ˆφ j Then, for P 2, it is Mass ˆK = And we compute for each element and sum in our matrix b ˆK 2.5 FEM of Q 3 Recall our basis functions of Q 3 ˆφîĵ (ˆx, ŷ) = ˆφ î (ˆx) ˆφ ĵ (ŷ) 22

24 where The gradient of Q 3 is 9/2(1 ˆx)(1/3 ˆx)(2/3 ˆx) î=1 27/2(ˆx)(1 ˆx)(2/3 ˆx) î=2 ˆφî(ˆx) = 27/2(ˆx)(1 ˆx)(ˆx 1/3) î=3 9/2(ˆx)(ˆx 1/3)(ˆx 2/3) î=4 9/2(1 ŷ)(1/3 ŷ)(2/3 ŷ) ĵ=1 27/2(ŷ)(1 ŷ)(2/3 ŷ) ĵ=2 ˆφĵ(ŷ) = 27/2(ŷ)(1 ŷ)(ŷ 1/3) ĵ=3 9/2(ŷ)(ŷ 1/3)(ŷ 2/3) ĵ=4 ( ) ˆ ˆφîĵ (ˆx, ŷ) = ˆx ˆφî(ˆx) ˆφĵ(ŷ) ˆφî(ˆx) ŷ ˆφĵ(ŷ) î,ĵ =1,2,3,4 To compute stiffness matrix ( ) ( ) ˆ ˆφîĵ (ˆx, ŷ) ˆ ˆx ˆφî(ˆx) ˆφĵ(ŷ) ˆφ ˆmˆn (ˆx, ŷ) = ˆx ˆφ ˆm (ˆx) ˆφˆn (ŷ) ˆφî(ˆx) ŷ ˆφĵ(ŷ) ˆφ ˆm (ˆx) ŷ ˆφˆn (ŷ) = ( ˆx ˆφî(ˆx) ˆφĵ(ŷ))( ˆx ˆφ ˆm (ˆx) ˆφˆn (ŷ)) + ( ˆφî(ˆx) ŷ ˆφĵ(y))( ˆφ ˆm (ˆx) ŷ ˆφˆn (ŷ)) = ˆx ˆφî(ˆx) ˆx ˆφ ˆm (ˆx) + ˆφĵ(ŷ) ˆφˆn (ŷ) + ˆφî(ˆx) ˆφ ˆm (ˆx) + ŷ ˆφĵ(ŷ) ŷ ˆφˆn (ŷ) Therefore our stiffness matrix of Q 3 is 16-by16. For the right-hand side, we can directly integrate our equation with the basis functions fv c ij f ˆφ ij Ω K K where c ij = f(x i, y j ) There will be 16-by-16 as well as our stiffness matrix for Q 3. 23

25 Chapter 3 More accuracy by averaging By the article of J.H. Bramble and A.H. Schatz [2], Higher Order Local Accuracy by Averaging in the Finite Element Method, we can define a better approximation of any given point x in Ω by averaging u h (x) of the neighborhood of that point. We denote this approximation by ũ h. 3.1 Bramble and Schatz In the article of Bramble and Schatz [2], they give an example of subspaces generated by the B-splines of Schoenberg. For t is real number, define { 1 t 1/2, χ = 0 t > 1/2 and for l an integer, set convolution l 1 times. Example, for l = 2 see Figure 3.1. For l = 3 see Figure 3.2. ψ (l) 1 (t) = χ χ χ. Let ψ (l) 1 be the one-dimensional smooth spline of order l defined by above equation. For l = r 2, r 2 given, find k 0,k 1,..., k r 2 by r 2 k j j=0 The constants k j are defined as k j = k j, j = 0,..., r 2. ψ (r 2) 1 (y)(y + j) 2m dy = R 1 k 0 = k 0 and k j = k j/2, j = 0,..., r 2, where the k j, j = 0,..., r 2. We can compute new solution by averaging as (K h u h )(hγ) = 24 { 1 if m = 0, 0 if m = 1,..., r 2. γ,δ Z N a j γ δ dj δ

26 where d j δ = k β ψ r 2 (δ β η)φ j (η)dη β Z R In the article of Bramble and Schatz [2], they already gave us the Table of k j and l j\r /12 37/ / /24-23/ / / / / Table 3.1: k j, l = r 2, t = r Averaging P 2 in 2D By the Table k j of Bramble and Schatz, we can construct a coefficient area k i,j where k i,j = k ik j. For P 2, we set r = 3, then our k 0 = 13/12 and k 1 = k 1 = 1/24 for defining ψ (1) Figure 3.1: The graph of coefficients in 1 j= 1 k jψ (1) 1 (x j) In 2D, we define the area of coefficients k i,j as 1 1 i= 1 j= 1 k i,jψ (1) 1 (x i)ψ (1) 1 (x j) 25

27 where k i,j = k ik j and ψ (1) 1 is 1 for 1 2 x 1 2 and zero otherwise. k 1,1 k 0,1 k 1,1 k 1,0 k 0,0 k 1,0 k 1, 1 k 0, 1 k 1, 1 To compute the weights for averaging the solution, we have consider 4 cases, case3 case4 case1 case Case 1 As we can see in Figure 3.3, we show that to average the point (0, 0), we must find all the weight d(m, n). d(4,-4) d(4,4) d(0,0) d(-4,-4) d(-4,4) 26

28 Figure 3.3: The first case of the weights d(m, n) We define d(m, n) as d(m, n) = k ik jψ(ˆx i)ψ(ŷ j) ˆφ m,n (ˆx, ŷ)dˆxdŷ, where ˆφ m,n is a basis function of ˆP 2. Example: to find d(0, 0), we use triangles that are connected to the point d(0, 0) Figure 3.4: The area to find d(0, 0) Each part of triangle can be computed by mapping ˆφ m,n to basis functions ˆφ i. d( 2, 0) d( 1, 0) d(0, 0) ˆ3 ˆ6 ˆ1 d( 1, 1) d(0, 1) ˆ4 ˆ5 ˆ2 d(0, 2) Figure 3.5: The mapping from ˆφ m,n to basis functions ˆφ i By Figure 3.4, we can see the support of ˆφ 0,0. We note that the sum of d(m, n) = 1. 27

29 3.2.2 Case 2 As we can see in Figure 3.6, we can compute d(m, n) in the same method as case 1. d(4,-4) d(4,4) d(0,0) d(-4,-4) d(-4,4) Figure 3.6: The second case of the weights d(m, n) and show the support of ˆφ 0, Case 3 As we can see in Figure 3.7, we can compute d(m, n) in the same method as case 1. d(4,-4) d(4,4) d(0,0) d(-4,-4) d(-4,4) Figure 3.7: The third case of the weights d(m, n) and show the support of ˆφ 0,0 28

30 3.2.4 Case 4 As we can see in Figure 3.8, we can compute d(m, n) in the same method as case 1. d(4,-4) d(4,4) d(0,0) d(-4,-4) d(-4,4) Figure 3.8: The fourth case of the weights d(m, n) and show the support of ˆφ 0,0 3.3 Averaging Q 3 By the Table 3.1 k j of Bramble and Schatz [2], we can construct a coefficient graph of k j. In 2D, we have r = 4, our k 0 = 37/30, k 1 = k 1 = 23/180 and k 2 = k 2 = 1/90 for defining ψ (2) Figure 3.9: The graph of coefficients in 2 j= 2 k jψ (2) 1 (x j) 29

31 To compute the weight, we recall the basis function of Q Figure 3.10: The basis function of Q 3 9/2(1 x)(1/3 x)(2/3 x) i = 1 27/2(x)(1 x)(2/3 x) i = 2 ˆφ i (x) = 27/2(x)(1 x)(x 1/3) i = 3 9/2(x)(x 1/3)(x 2/3) i = 4 We consider the computation for 1D. Because of the square element of Q 3, we can find the weight in 2D by cross-product. We define the weight as d i = k iψ (2) 1 (ˆx i)dˆx R For example, in each element by the Figure 3.11, we show that d(m, n) can be computed by cross-product. We map the node to the basis function of Q 3. d(1,4) d(4,4) d(1,1) d(4,1) Figure 3.11: The weight of Q 3 30

32 Chapter 4 Numerical Experiments In this section, we consider an example where the exact solution is given by u = sin(πx)sin(πy)(x + πy) We note that we can average the solution u h to obtain ũ h only at nodes which are not too close to the boundary. 4.1 Numerical experiments of P 1 The domain Ω = (0, 1) 2 and h = 1 2 L. The Figures are the finite element solutions with L = 1, 2,..., 5 respectively and f = u. By the Table 4.1, we denote N 1 to be a set of all nodes of P 1 and define the maximum error as MaxErr = max i N 1 u h (v i ) u(v i ) where u h is the FEM solution and u is the exact solution. 31

33 Figure 4.1: The solution of P 1 for L = 1. MaxErr = Figure 4.2: The solution of P 1 for L = 2. MaxErr = Figure 4.3: The solution of P 1 for L = 3. MaxErr =

34 Figure 4.4: The solution of P 1 for L = 4. MaxErr = Figure 4.5: The solution of P 1 for L = 5. MaxErr = For the rate of convergence, we define that Rate = MaxErr L 1 MaxErr L 33

35 L\Error M axerr Rate Table 4.1: The error value of each L for P Numerical experiments of P 2 For P 2, the domain Ω = (0, 1) 2 and h = 1 2 L. The Figures are the finite element solutions with L = 1, 2,..., 5 respectively and f = u. We denote N 2 to be a set of all nodes of P 2. The maximum error of P 2 is given by MaxErr = max i N 2 u h (v i ) u(v i ) Since the averaging can be computed only the nodes that are not too close to the boundary, we denote Ñ2 to be a set of the nodes of P 2 that can be averaged. By the Table 4.2, we define MaxAvgErr = max ũ h (v i ) u(v i ) i Ñ2 where ũ h is the solution by averaging MaxErrNode = max u h (v i ) u(v i ) i Ñ2 34

36 Figure 4.6: The solution of P 2 for L = 1. MaxErr = Figure 4.7: The solution of P 2 for L = 2. MaxErr = MaxAvgErr = MaxErrNode = Figure 4.8: The solution of P 2 for L = 3. MaxErr = e-4. MaxAvgErr = MaxErrNode = e-4. 35

37 Figure 4.9: The solution of P 2 for L = 4. MaxErr = e-5. MaxAvgErr = e-4. MaxErrNode = e-5. Figure 4.10: The solution of P 2 for L = 5. MaxErr = e-6. MaxAvgErr = e-6. MaxErrNode = e-6. For the rate of convergence, we define that RateAvg = MaxAvgErr L 1 MaxAvgErr L 36

38 RateNode = MaxErrNode L 1 MaxErrNode L L\Error M axerr Rate M axavgerr RateAvg M axerrn ode RateN ode e e e e e e e e Table 4.2: The error value of each L for P Numerical experiments of Q 3 For Q 3, the domain Ω = (0, 1) 2 and h = 1 2 L. The Figure are the finite element solutions with L = 1, 2,..., 5 respectively and f = u. We denote N 3 to be a set of all nodes of Q 3. The maximum error of Q 3 is given by MaxErr = max i N 3 u h (v i ) u(v i ) we denote Ñ3 to be a set of the nodes of Q 3 that can be averaged. By the Table 4.3, we define the maximum error of Ñ3 nodes MaxAvgErr = max ũ h (v i ) u(v i ) i Ñ3 MaxErrNode = max u h (v i ) u(v i ) i Ñ3 37

39 Figure 4.11: The solution of Q 3 for L = 1. MaxErr = Figure 4.12: The solution of Q 3 for L = 2. MaxErr = e-4. Figure 4.13: The solution of Q 3 for L = 3. MaxErr = e-5. MaxAvgErr = e-5. MaxErrNode = e-5. 38

40 Figure 4.14: The solution of Q 3 for L = 4. MaxErr = e-6. MaxAvgErr = e-7. MaxErrNode = e-6. Figure 4.15: The solution of Q 3 for L = 5. MaxErr = e-7. MaxAvgErr = e-8. MaxErrNode = e-7. The rate of convergence is given by RateAvg = MaxAvgErr L 1 MaxAvgErr L 39

41 RateNode = MaxErrNode L 1 MaxErrNode L L\Error M axerr Rate M axavgerr RateAvg M axerrn ode RateN ode e e e e e e e e e e Table 4.3: The error value of each L for Q 3 40

42 Chapter 5 Conclusion In this Chapter, we conclude our result of the errors in each refinement. In P 1, by the Table 4.1, we have the rate of convergence approximate to 4. Next, we compare the rate of convergence of P 2. The rate of convergence is 16 by the Table 4.2. That is, in P 2, the FEM generates a better solution at some nodes due to super-convergence for structured mesh. In the corollary of Bramble and Schatz states that let Ω 0 Ω 1 Ω and N 0 = N/2 + 1, at points hγ Ω 0 and γ Z 2 sup hγ Ω 0 γ Z u(hγ) a j γ αd j α C(h2r 2 u 2r 2+N0,Ω 1 ) 2 α In P 2, we have r = 3 so that the rate is h 4 which is the same result from our FEM. And it does not give us a better result. Therefore, we introduce the Q 3 refinement of FEM. By the Table 4.3, the rate of convergence is approximated to 16 and the rate of convergence by averaging is approximated to 64. As Bramble and Schatz s theorem, we have r = 4 which give us h 6 as same as the rate of convergence that we have from the averaged Q 3 so that the averaging method shows that our test has more accuracy. Furthermore, we can expect that for Q 4 or Q 5 can give us a better result by using FEM solution. 41

43 Bibliography [1] Sayas Francisco-Javier. A gentle introduction to the Finite Element Method [2] J.H. Bramble, A.H. Schatz. Higher order local accuracy by averaging in the finite element method. Math. Comp., 31:94-111,1977. [3] Braess, Dietrich. Finite Elements Theory, Fast Solvers, and Applications in Elasticity Theory. 3rd ed. Cambridge: Cambridge UP, Print 42