Math 660-Lecture 15: Finite element spaces (I)

Math 660-Lecture 15: Finite element spaces (I) (Chapter 3, 4.2, 4.3) Before we introduce the concrete spaces, let s first of all introduce the following important lemma. Theorem 1. Let V h consists of piecewise polynomials. Then, V h H 1 (Ω) V h C 0 ( Ω) and V h H 2 (Ω) V h C 1 ( Ω) Here C 0 simply means continuous functions. Here, one only needs to show that H 1 C 0. The piecewise polynomial and continuous condition obviously implies in H 1 (if not continuous, the weak derivative will be infinity at the interface. Such distributions are not in L 2.) For H 1 C 0, similarly, if not continuous, the weak derivative doesn t exist. Remark: Consider an open set Ω with smooth boundary. For general Sobolev spaces, if p > n, then Wp k C k [n/p] 1,γ where 0 γ < 1. For n = 1, the continuity can be ensured if p = n. Hence, H 2 = W2 2 for n = 2 is in C 0,γ for any γ < 1 but not C 1 in general. H 1 = W2 1 doesn t imply C0 and there are counterexamples. However, if we talk about piecewise polynomials, the above theorem is true. 1 1D spaces Consider the interval Ω = [a, b] and a subdivision (triangulation) T h = {[x j, x j+1 ]} and K = [x j, x j+1 ]. Define P r (K) = {v : v is a polynomial with degree r on K} Consider the linear function P 1 (K). P 1 (K) can be determined uniquely by the values at the node. Hence, if we make up a finite element space V h such that each function in V h when restricted on K belongs to P 1 (K), then V h is C 0 ([a, b]) and therefore in H 1. Consider P 2 (K). It has three coefficients. Then, it s impossible to construct C 1 functions on the whole Ω since there will be four requirements. The function on each K can be determined by the values at the endpoints and the value at the midpoint. Functions generated in such a way is still in H 1 (Ω). Using P 3 (K) functions, we can construct a finite element space V h C 1 (Ω) and thus in H 2 (Ω). 1

See the pictures on P81. Example: solve the 1d problem u (x) = f with u(0) = u (0) = u(1) = u (1) = 0. Solution. We first of all get the weak formulation. Multiply a test function v and integrate: 1 0 f(x)v(x)dx = u vdx = u v 1 0 u v dx = (u v u v ) 1 0+ u v dx First of all, for the integral u v dx to make sense, we need both u and v to be in H 2. Secondly, if we require v H0 2 H2 (which means v, v both vanish at the endpoints), then we don t have the boundary terms. Hence, a suitable weak formulation is: Find u H0 2 ([0, 1]) such that 1 a(u, v) = u v dx = f(x)v(x)dx, v H0 1. 0 Now considering using Galerkin s FEM. We need a finite element space to be in H 2. If we use piecewise polynomials, then the functions should be in C 1. From what we just studied, clearly, we should use P 3 (K). Then, for node x j, we need two functions: φ j and ϕ j such that (1). φ j (x i ) = δ ij, φ j (x i) = 0; (2). ϕ j (x i ) = 0 and ϕ j (x i) = δ ij. Each of them are cubic on the interval. The finite element space is then V h = { j a j φ j + j b j ϕ j } These basis functions can be constructed using Hermite interpolation. You ll finish this in HW3. 2 2D spaces: Triangular elements Ω R 2 and T h = {K} is a collection of triangles for the triangulation. P r (K) = {v v is a polynomial of deg r on K}. Lemma 1. dim(p r ) = C 2 r+2 = (r+2)(r+1) 2 2

Proof. the basis is x i 1 xj 2. 0 i + j r. Consider we divide r units into 3 distinct groups. Let s say the number of the units in the first group is i and number of units in the second group is j. Clearly, one such division corresponds to uniquely an (i, j) pair. The number of such divisions are equivalent to choosing two from r +2 positions and the chosen two positions will be the partition boundary. 2.1 Examples of finite element spaces Consider the following finite element space: V h = {v is defined on Ω : v K P 1 (K)} Since dim(p 1 (K)) = 3, then there are 3 degrees of freedom. a i, 1 i 3 be the vertices of K. Then, we claim: Let Theorem 2. Given α i, 1 i 3, there is uniquely a function v P 1 (K) such that v(a i ) = α i. In other words, the function on K is uniquely determined by its values at the vertices. Proof. The number of unknowns (dimension, which is 3) equals the number of conditions given. Hence, we have a 3 by 3 linear system. For such a system, the solution exists uniquely is equivalent to saying the homogeneous solution is trivial. In other words, it suffices to show that if v = C 1 + C 2 x 1 + C 3 x 2 is zero on a i, then v = 0. The line a 1 a 2 has an equation of the form d 1 x 1 + d 2 x 2 + d 3 = 0. Since v vanishes on this edge, we must have v = C(x)(d 1 x 1 + d 2 x 2 + d 3 ). The degree of v is at most 1, and therefore C(x) = C. Further, a 3 is not on the line, we must have d 1 a 3 1 + d 2a 3 2 + d 3 0. Hence, C = 0. Let s consider constructing the basis functions of V h. By the theorem just proved, we need to specify the values at the nodes only. Let λ i P 1 (K) and λ i (a j ) = δ ij. λ 1 = µ(d 1 x 1 + d 2 x 2 + d 3 ) where d 1 x 1 +... is the equation of a 2 a 3. Then, µ is determined uniquely by the condition λ 1 (a 1 ) = 1. Now, suppose we are given v(a i ). We define the finite element space to be 3 V h = {v : v K = v(a i )λ i }. i=1 3

Theorem 3. V h C 0 (Ω). By this definition, w V h is continuous at the nodes. Consider one edge e. Suppose e is the intersection between K 1 and K 2. On e, we define g = w K1 w K2. g = c 1 x 1 + c 2 x 2 + c 3 is linear on e. It is zero at the endpoints. Then, it must be zero on the whole line containing e since it is linear. (g itself may not be a zero function.) Hence, w is continuous at the edge. Consider V h = {v : v K P 2 (K), K T h } The dimension of P 2 (K) is 6. Let a ij = 1 2 (ai + a j ) be the midpoint of a i and a j. Then, we have Theorem 4. v P 2 (K) is uniquely determined by the values at a i, 1 i 3 and a ij, i < j. Proof. Again, the unknowns (dimension) equals the conditions. suffices to show that if v vanishes at these points, then it s zero. First of all, it vanishes on a 1 a 2 because a 1D quadratic function is zero if it s zero on three points. Then, v a 1 a 2 = h(x)λ 3 where λ 3 is the basis function in the previous example (linear function and is only 1 at a 3 ). Since v is quadratic, h is linear. Since v also vanishes on a 1 a 3, then h vanishes on a 1 a 3. Hence, h = Cλ 2. v = Cλ 2 λ 3. Finally, v also vanishes on a 23, then C has to be zero. Clearly, λ i (2λ i 1) is a quadratic function that is only 1 at a i and vanishes at the other five points. 4λ i λ j is the quadratic function that is 1 only at a ij. Hence, in general, It v K = 3 v(a i )λ i (2λ i 1) + v(a ij )4λ i λ j i<j i=1 Then, we actually only have V h C 0 (Ω). V h is not in C 1. P 2 (K) is not enough for us to make C 1 functions. To have C 1 functions, we actually need P 5 (K). dim(p 5 (K)) = 21. Theorem 5. v P 5 (K) is uniquely determined by D α v(a i ), α 2 and v(a ij )/ n. 4

Proof. Let s show that if on K, for v these values are zero, then v = 0. On a 2 a 3, noting first that v = v/ s = 2 v/ s 2 = 0 at a 2 and a 3 (where s is the arc length of a 2 a 3 ), we conclude that v = 0 on this edge since there are 6 conditions for a five degree 1D polynomial. Secondly, g = v/ n as a function on a 2 a 3 is a polynomial of degree at most 4. g(a 23 ) = 0 by the given condition. g = g/ s = 0 at a 2 and a 3 by the fact that D α v = 0. Hence g = v/ n should be zero. (we have used 10 conditions). Then, we must have that v = λ 2 1 h. Similarly, λ2 2, λ2 3 can divide v as well. These polynomials don t have common factors. Then, λ 2 1 λ2 2 λ2 3 can divide v. Since v is of only 5 degree, we must have v = 0. Using what we have proved, we can construct V h = {v : v K P 5 (K)} such that if w V h, then w has continuous D α w at the vertices and continuous w/ n at the midpoints. Then, we can show that w is C 1 on one edge. Then, V h C 1 (Ω). 5