A brief introduction to finite element methods

Transcription

1 CHAPTER A brief introduction to finite element methods 1. Two-point boundary value problem and the variational formulation 1.1. The model problem. Consider the two-point boundary value problem: Given a constant a and a function f(x, find u(x such that { u +au = f(x, < x < 1, u( =, u (.1 (1 = The variational formulation. If u is the solution to (.1 and v(x is any (sufficiently regular function such that v( =, then integration by parts yields u vdx+ auvdx = u (1v(1+u (v(+ = fvdx. Let us introduce the bilinear form and define A(u,v = u (xv (xdx+ (u v +auvdx, V = { v L 2 ([,1] : A(v,v < and v( = }. Then we can see that the solution u to (.1 is characterized by u V such that A(u,v = which is called the variational or weak formulation of (.1. auvdx f(xv(xdx v V. (.2 Theorem 1.1. Suppose f C ([,1] and u C 2 ([,1] satisfies (.2. Then u solves (.1. Proof. Let v V C 1 ([,1]. Then integration by parts gives fvdx = A(u,v = u vdx+ auvdx+u (1v(1. (.3 Thus, (f +u auvdx = for all v V C 1 ([,1] such that v(1 =. Let w = f +u au C ([,1]. If w, then w(x is of one sign in some interval [b,c] [,1], with b < c. Choose v(x = (x b 2 (x c 2 in [b,c] and v outside 1

2 2. A BRIEF INTRODUCTION TO FINITE ELEMENT METHODS [b,c]. But then wvdx which is a contradiction. Thus u +au = f. Now apply (.3 with v(x = x to find u (1 =. So u solves (.1. Weremarkthattheboundaryconditionu( = iscalledessential asitappears in the variational formulation explicitly, i.e., in the definition of V. This type of boundary condition is also called Dirichlet boundary condition. The boundary condition u ( = is called natural as it is incorporated implicitly. This type of boundary condition is often referred to by the name Neumann. 2. The finite element method 2.1. Meshes. Let M h be a partition of [,1]: = x < x 1 < x 2 < < x n 1 < x n = 1. The points {x i } are called nodes. Let h i = x i be the length of the i-th subinterval [,x i ]. Define h = max 1 i n h i Finite element spaces. We shall approximate the solution u(x by using the continuous piecewise linear functions over M h. Introduce the linear space of functions V h = { v C ([,1] : v( =, It is clear that V h V. v [xi 1,x i] is a linear polynomial, i = 1,,n }. ( The finite element method. The finite element discretization of (.2 reads as: Find u h V h such that A(u h,v h = f(xv h (xdx v h V h. ( A nodal basis. For i = 1,,n, define φ i V h by the requirement that φ i (x j = δ ij = the Kronecker delta, as shown in Fig. 1: φ i (x φ n (x x 1 x i x i+1 1 x 1 x i x n 1 1 Figure 1. piecewise linear basis function φ i.

3 φ i = φ n = x 2. THE FINITE ELEMENT METHOD 3 h i, x x i, x i+1 x h i+1, x i < x x i+1,, x < or x > x i+1, { x xn 1 h n, x n 1 x 1,, x < x n 1. For any v h V h, let v i be the value of v h at the node x i, i.e., then v i = v h (x i, i = 1,2,,n, v h = v 1 φ 1 (x+v 2 φ 2 (x+ +v n φ n (x The finite element equations. Let 1 i n 1. u h = u 1 φ 1 +u 2 φ 2 + +u n φ n, u 1,,u n R, where u i = u h (x i. Let v h = φ i, i = 1,,n in (.5, then we obtain an algebraic linear system in unknowns u 1,u 2,,u n : Denote by A(φ 1,φ i u 1 +A(φ 2,φ i u 2 + +A(φ n,φ i u n = k ij = A(φ j,φ i = φ j φ i +aφ jφ i dx, f i = i = 1,,n. f(xφ i dx, f(xφ i dx, (.6 and K = ( k ij n n, F = ( f i n 1, U = ( u i n 1, then (.6 can be rewritten as: KU = F (.7 Here K is called the stiffness matrix. It is clear that k ij = if x i and x j are not adjacent to each other. Therefore K is sparse. Next we compute k ij = A(φ j,φ i. Note that xl k ij = A(φ j,φ i = φ jφ i +aφ j φ i dx. On the interval [,x l ], the following integrals are involved. k l 11 := xl k l 12 := xl l=1 φ l 1 φ l 1 +aφ l 1φ l 1 dx, k l 21 := xl φ l φ l 1 +aφ lφ l 1 dx, k l 22 := φ l 1 φ l +aφ l 1φ l dx, xl x l φ l φ l +aφ lφ l dx, Some simple calculations yield k l 11 = kl 22 = 1 h l + a 3 h l, k l 12 = kl 21 = 1 h l + a 6 h l.

4 4. A BRIEF INTRODUCTION TO FINITE ELEMENT METHODS The matrix K l := (k ij 2 2 is called the element stiffness matrix on [,x l ]. The stiffness matrix K can be assembled by using the element stiffness matrices as follows. k22 i +ki+1 11 = a k ii = h i h i+1 3 (h i +h i+1, i = 1,,n 1, k22 n = 1 + a h n 3 h n, i = n, k i 1,i = k i,i 1 = k i 12 = 1 h i + a 6 h i, i = 2,,n. Combining the above equations and (.6 yields [ a(h1+h 2 ] h h u1 2 + ( ah h u2 2 = f 1, ( ahi 6 1 h ui 1 i + [ a(h i+h i+1 ] h i + 1 h ui i+1 + ( ah i h ui+1 i+1 = f i i = 2,,n 1, ( ahn 6 1 h un 1 n + [ ah n ] h un n = f n The interpolant. Given u C ([,1], the interpolant u I V h of u is determined by u I = u(x i φ i. It is clear that u I (x i = u(x i, i =,1,,n, and u I (x = x i x u( + x u(x i for x [,x i ]. h i h i Denote by τ i = [,x i ] and by g L2 (τ i = ( x i g 2 dx 1/2. Theorem 2.1. u u I L2 (τ i 1 π h i u L2 (τ i, (.8 u u I L2 (τ i 1 π 2h2 i u L2 (τ i, (.9 u u I L 2 (τ i 1 π h i u L 2 (τ i. (.1 Proof. We only prove (.8 and leave the others as an exercise. We first change (.8 to the reference interval [,1]. Let ˆx = (x /h i and let ê(ˆx = u(x u I (x. Note that ê( = ê(1 = and k = u I is a constant. The inequality (.8 is equivalent to that is ê 2 L 2 ([,1] = 1 h i u u I 2 L 2 (τ i h i π 2 u 2 L 2 (τ i = 1 π 2 ê +kh i 2 L 2 ([,1], ê 2 L 2 ([,1] 1 π 2 ê 2 L 2 ([,1] + 1 π 2 kh i 2 L 2 ([,1]. (.11 Sinceê( = ê(1 =, it followsfromthe Fourierexpansionê(ˆx = n=1 a nsinnπˆx and the Parseval s identity that ê 2 L 2 ([,1] 1 π ê 2 2 L 2 ([,1], which implies (.11. This completes the proof of (.8.

5 2. THE FINITE ELEMENT METHOD A priori error estimate. Introduce the energy norm From the Cauchy inequality, v = A(v,v 1/2. A(u,v u v. By taking v = v h V h in (.2 and subtracting it from (.5, we have the following fundamental orthogonality Therefore A(u u h,v h = v h V h. (.12 u u h 2 = A(u u h,u u h = A(u u h,u u I u u h u u I, It follows from Theorem 2.1 that [ n ] 1/2 u u h u u I = ( u u I 2 L 2 (τ +a u u i I 2 L 2 (τ i [ n (( h 2 u ( 2 h 4 u ] 1/2 L π 2 (τ +a i 2 L π 2 (τ i = h [ ( ( h 2 1+a π π 1/2 (u dx] 2. We have proved the following error estimate. Theorem 2.2. u u h h ( ( h 2 1/2 u 1+a π π L2([,1]. Since the above estimate depends on the unknown solution u, it is called the a priori error estimate A posteriori error estimates. We will derive error estimates independent of the unknown solution u. Let e = u u h. Then A(e,e = A(u u h,e e I = = f (e e I dx (f au h (e e I dx u h (e e I dx u h (e e I dx Since u h is constant on each interval (,x i, A(e,e = (f au h (e e I dx f au h L 2 (τ e e i I L 2 (τ i h i π f au h L2 (τ i e L2 (τ i. au h (e e I dx

6 6. A BRIEF INTRODUCTION TO FINITE ELEMENT METHODS Here we have used Theorem 2.1 to derive the last inequality. Define the local error estimator on the element τ i = [,x i ] as follows Then η i = 1 π h i f au h L2 (τ i. (.13 ( n 1/2 ( n 1/2 e 2 ηi 2 e ηi 2 e. That is, we have the following a posteriori error estimate. Theorem 2.3 (Upper bound. ( n 1/2 u u h ηi 2. (.14 Now a question is if the above upper bound overestimates the true error. To answer this question we introduce the following theorem that gives a lower bound of the true error. Theorem 2.4 (Lower bound. Define φ τi = ( x i ((φ 2 +aφ 2 dx 1/2. Let (f au h i = 1 xi h i (f au h dx and osc i = 1 π h i f au h (f au h i L2 (τ. i Then As a consequence ( n 5π 2 u u h 6+6ah 2 i η i 5+ 3 ( 6+6ah 2 1 osc i i 2 5 5π 2 u u h τi. (.15 ( ( max η i /2 osc i,. (.16 Proof. For simplicity, denote by R = f au h and R i = (f au h i. For any ψ V, it is clear that A(e,ψ = Rψdx u h ψ dx (.17 Define ψ i (x = φ i 1 (xφ i (x if x τ i and ψ i (x = otherwise. Choose ψ = ψ i R i. By simple calculations, we obtain R i ψdx = 1 6 R i 2 L 2 (τ i, 3 ψ L 2 (τ i = 3h i ψ L 2 (τ i = R i L 2 (τ i. From (.17, We have, which implies A(e,ψ = Rψdx = (R R i ψdx+ 1 6 R i 2 L 2 (τ i. R i 2 L 2 (τ 6 e i τi ψ τi +6πosc i ψ L 2 (τ i (( = 12+ 6ah2 1 2 i e τi + π 3 osc i h 1 i R i 5 5 L2 (τ, i ( h i R i L 2 (τ i 12+ 6ah2 1 i 2 5 e τi + π 3 osc i 5

7 3. EXERCISES 7 Now the proof is completed by using η i 1 π h i R i L 2 (τ i +osc i. We remark that the term osc i is of high order compared to η i if f and a are smooth enough on τ i. Example 2.5. We solve the following problem by the linear finite element method. The true solution (see Fig. 2 is u +1u = 1, < x < 1, u( = u(1 =. u = 1 1 (1 e1x +e 1(1 x 1+e 1 If we use the uniform mesh obtained by dividing the interval [,1] into 18 subintervals of equal length, then the error u u h On the other hand, if we use a non-uniform mesh as shown in Fig. 2 which also contains 18 subintervals, then the error u u h is smaller than that obtained by using the uniform mesh.. x Figure 2. Example 2.5. The finite element solution and the mesh. 3. Exercises Exercise.1. Prove (.9 and (.1. Exercise.2. Use Example 2.5 to verify numerically the a posteriori error estimates in Theorem 2.3 and (.16 in Theorem 2.4.