1 Discretizing BVP with Finite Element Methods.

Transcription

1 1 Discretizing BVP with Finite Element Methods In this section, we will discuss a process for solving boundary value problems numerically, the Finite Element Method (FEM) We note that such method is a main tool in solving PDE 11 Finite Elements in 1D We consider the 1D poisson equation with both Dirichlet and Nuemann boundary conditions, u (x) = f(x) in (, 1) u() = u (1) = (1) Weak Form Let v(x) be any smooth function with v() = Integration by parts yields < f, v > := = [ = u v = Here, a(u, v), is called a bilinear functional f(x)v(x) dx u (x)v(x) dx ] 1 + u v dx u v dx =: a(u, v) Define the subspace V = { v L 2 (, 1) : a(v, v) < and v() = } (2) The solution of (1) satisfies This is the the weak form of (1) u V such that a(u, v) =< f, v > for all v V (3) 1

2 Questions 1 Does any solution of (3) solve (1)? 2 Is u unique? 3 Does (3) always have a solution? Theorem 11 (1) Suppose f C ([, 1]) and u C 2 ([, 1]) satisfy (3), then u is solves Proof By hypothesis < f, v >= a(u, v) = ( u )v dx + u (1)v(1), (4) for every v V C 1 ([, 1]) Thus < u + f, v >= for all v V C 1 ([, 1]) such that v(1) = Suppose w = u + f Then there exists a point ˆx such that w(ˆx) By the continuity of w, there exists and interval x ˆx < δ, where w is all the same sign Without loss of generality, we assume w > on this interval We chose a specific function in V C 1 ([, 1]) such that v(1) =, v(x) = Then we arrive at the contradiction { (x (ˆx + δ)) 2 (x (ˆx δ)) 2 x ˆx δ otherwise } (5) (u + f)v dx > (6) Thus -u = f in (, 1), and the PDE is satisfied The left boundary condition is satisfied because u V To show that u satisfies the right boundary condition, we plug a specific function in V, namely v(x) = x, into the weak formulation: But, because u = f in (, 1), Therefore, u (1) = < f, x >= < f, x >= ( u )x dx + u (1) (7) ( u )x dx (8) 2

3 Remark 11 For f and u sufficiently smooth, we have shown that the solution to (3) is equivalent to that of (1) Remark 12 The boundary condition u() = is called essential and must appear explicitly in the definition of our space V The boundary condition u (1) = is callednatural and it is incorporated implicitly u(x) = essential Dirichlet u (x) = natural Nuemann (9) Ritz-Galerkin Approximation Define a finite-dimensional subspace, S V Consider the weak problem, (3), restricted to S, u s S such that a(u s, v) =< f, v > for all v S (1) Questions 1 Does such a u s always exist? 2 Is u s unique? 3 If u s does exist, how does it relate to the solution of (1)? Theorem 12 (Existence and Uniqueness) solution For f L 2 (, 1), (1) has a unique Proof Since S is finite dimensional it has a finite dimensional basis S = Span{φ j } n (11) If u s S, then u s = n α jφ j We can restate (1) as, find α such that or ( n ) a α j φ j, v =< f, v > v S (12) ( n ) a α j φ j, φ j =< f, φ j > for i = 1,, n (13) 3

4 Define n α j a(φ j, φ j ) =< f, φ j > for i = 1,, n (14) K = (K ij ) = a(φ j, φ i ) and f = (f i ) =< f, φ i > (15) Then (1) is rewritten as a linear system Kα = f (16) And the question of existence and uniqueness is reduced to that of a square linear system Now, suppose K is singular, then there exists a β such that Kβ = Furthermore, Now, define ˆv = n β jφ j, then < Kβ, β > R n = < Kβ, β > R n= β T Kβ = (17) n ( n ) β i a φ j, φ i i=1 ( n = a β j φ j, = a(ˆv, ˆv) = i=1 n ) β i φ i (ˆv ) 2 dx = Thus, ˆv = in (, 1), meaning ˆv is constant But, ˆv S V, so ˆv() = The constant must be zero: ˆv in (, 1) Because {φ i } n i=1 is a linearly independent set, β = as well So K must be a non-singular matrix and Kα = f has a unique solution Thus, u s = is a unique solution to (1) n α j φ j (18) Remark 13 We have not proved that (3) has a unique solution, yet We have only proved that if (3) has solution u C 2 ([, 1]), then u solves (1), and that (1) always has a unique solution u s S 4

5 12 Error Esitmates What can we say about the error, (u u s )? First we define some useful tools Energy Norm Under certain conditions, the bilinear function is used to define the energy norm on V, v e = a(v, v) (19) Why is this a norm? What are the certain conditions? Energy Inner Product Again, under similar conditions, the bilinear functional is used to define the energy inner product on V, < u, v > e = a(u, v) (2) Why is this an inner product? We say that u is energy orthogonal to v, if < u, v > e = Lemma 13 (Orthogonality Relation) The error is energy orthogonal to S, or a(u u s, w) = w S (21) Proof Algebra gives a(u, v) = < f, v > v V S a(u s, v) = < f, v > v S A(u u s, v) = v S (22) Theorem 14 Proof For any v S, u u s e = min v V u v e (23) u u s 2 e = a(u u s, u u s ) = a(u u s, u v) + a(u u s, v u s ) (by the bilinearity of a(, )) = a(u u s, u v) (by the orthogonality relation) u u s e u v e (by Cauchy-Schwartz) (24) Cea s Lemma Thus, u u s e u v e for all v S 5

6 Remark 14 Function u s is the member of S that minimizes the error in the energynorm, u u s e The key ingredient is that a(u, u) is a norm and a(u, v) is an inner product Approximation Assumption How small is u u s e? This depends on S Assume that f L 2 (, 1) and u C 2 ([, 1]) Furthermore, assume that for any w C 2 ([, 1]), we have ɛ >, such that min w v e ɛ w (25) v S Remark 15 We will demonstrate the validity of this assumption for a specific S later Theorem 15 If (25) holds, then u u s ɛ u u s e ɛ 2 u = ɛ 2 f (26) Proof Let w be the solution of (3) with f replaced by (u u s ) That is, w satisfies a(w, v) =< u u s, v > v V (27) In particular, replace v by (u u s ) V, a(w, u u s ) =< u u s, u u s > v V (28) Now, a(u, v) = a(v, u), by the symmetry of a(, ), so a(w, u u s ) = a(u u s, w) (by symmetry) = a(u u s, w v), v S (by orthogonality relation) u u s e w v e (by Cauchy Schwartz) (29) Now we apply the approximation assumption, (25) u u s 2 u u s e w v e ɛ u u s e w e, v S (by app assumption) = ɛ u u s e u u s e (because w = u u s ) (3) or u u s ɛ u u s e Apply the approximation assumption once more to get Aubin-Nitsche Trick u u s ɛ u u s e ɛ 2 u = ɛ 2 f (31) Remark 16 We have shown 6

7 a) Strong form (1) Weak form (3) b) Discrete weak form always has a solution, u s c) Discrete weak solution, u s, is the best possible in the energy norm d) Under a suitable hypothesis, u u s e ɛ f, we have u u s ɛ 2 f Now we choose an actual S and explore 2 Linear Finite Element Subspaces Again we consider the PDE u (x) = f(x) in (, 1) u() = u (1) = (1) We define a mesh {x i } n i=, a set of nodes in Ω such that = x < x 1 < < x n = 1 Also, we define the mesh width, h i = (x i x i 1 ), i = 1,, n, over each interval in Ω Then we define a finite dimensional space of functions V h := { u(x) : u() =, u(x) is piecewise linear on C i := (x i 1, x i ), i = 1,, n }, (2) 7

8 and which has a nodal basis, φ i (x) = x x i 1 x i x i 1 x (x i 1, x i ) x i+1 x x i+1 x i, x (x i, x i+1 ) elsewhere φ n (x) = { x xn 1 x n x n 1 x (x n 1, x n ) elsewhere for i = 1,, n 1 (3) } (4) We also define the infinite dimensional space, Note that V h V V := { v L 2 (, 1) : a(v, v) <, v() = } (5) Definition 21 Let u C ([, 1]) such that u() = Then the function u I := n u(x i ) φ i (x) (6) i=1 is called the interpolant of u Clearly, u I V h Moreover, if v V h, then v = v I The mapping Π h : u u I (7) is a projection onto V h Theorem 21 (Interpolation Error) Let h = max i (x i x i 1 ) i = 1,, n Then where c is a constant independent of h and u u u I e ch u, u V, (8) 8

9 Proof We prove this piece-wise That is, we prove that (a) xj x j 1 [(u u I ) ] 2 dx c(x j x j 1 ) 2 xj x j 1 [u ] 2 dx (9) Let ɛ = u u I denote the error Since u I = on (x j 1, x j ), (a) is equivalent to xj We make the following substitutions Now, (a) is equivalent to x j 1 [ɛ ] 2 dx c(x j x j 1 ) 2 xj x = x j 1 + s(x j x j 1 ) s (, 1) dx = (x j x j 1 )ds ˆɛ(s) = ɛ(x j 1 + s(x j x j 1 )) ˆɛ s = (x j x j 1 )ɛ 2ˆɛ s 2 = (x j x j 1 ) 2 ɛ [ˆɛ ] 2 ds c x j 1 [ɛ ] 2 dx (1) [ˆɛ ] 2 ds (11) The new estimate does not involve h j = (x j x j 1 ) Note that ˆɛ() = ˆɛ(1) = By Rolle s Theorem, there is some ξ (, 1) such that ˆɛ (ξ) = Thus, ˆɛ (s) = ˆɛ (s) = s ˆɛ (s) ds ξ s (1)ˆɛ (s) ds ξ ( s ) 1/2 ( s 1/2 ds) [1] 2 ds [ˆɛ (s)] 2 (by Cauchy-Schwartz) ξ ξ ( s = s ξ 1/2 [ˆɛ (s)] 2 ds s ξ 1/2 ( Squaring both sides and integrating gives ξ ) 1/2 ) 1/2 [ˆɛ (s)] 2 ds [ˆɛ (s)] 2 ds ˆɛ 2 s ξ ds ˆɛ ˆɛ 2 9

10 Corollary 22 Let u s be the solution of the discrete weak problem with S = V h u u S (ch) u u S e (ch) 2 u (12) Proof Theorem from last lecture with ɛ = (ch) u u I 2 e = τ i u u I 2 e,τ i τ i 1 2 h2 j u 2 τ i 1 2 h2 τ i u 2 τ i Taking a square-root gives u u I e 1 2 h u (13) Relationship to Finite Difference Method Recall the weak form: find u h V h such that a(u h, v h ) = f, v h v h V h, (14) with basis {φ j } n This becomes: find uh = n α jφ j such that n a α j φ j, φ i = f, φ i i = 1,, n n a(φ j, φ i )α j = f, φ i Define matrix (K) ij = a(φ j, φ i ) For i = j, we have xj+1 xj [ ] [φ j] dx = dx + x j 1 x j 1 x j x j = + x j x j 1 x j+1 x j = h j h j+1 1 xj+1 x j [ 1 x j+1 x j ] 2 dx

11 For i j, i = j + 1, we have xj+1 x j ( ) x ( ) xj xj+1 x dx = h j+1 h j+1 xj+1 x j = 1 h j+1 1 h 2 j+1 dx We have a similar result for i = j 1 The stencil becomes ( 1, h j 1 + 1, 1 h j h j+1 h j+1 ) (15) Which, for a equispaced grid(h = h j = h j+1 ), gives our finite-difference stencil, but scaled by h The right-hand side is xj ( ) x xj+1 ( ) xj 1 xj+1 x f, φ j = f(x) dx + f(x) dx (16) x j 1 h j x j h j+1 By the Generalized Mean-Value Theorem, if f is continuous, we have [ xj ( ) x xj+1 ( ) ] xj 1 xj+1 x f, φ j = f(ξ) dx + dx x j 1 h j x j h j+1 ( ) hj + h j+1 = f(ξ), 2 for some ξ (x j 1, x j+1 ) Note that, when h = h j = h j+1, this is not just the rghit-hand side of the finite-difference method, scaled by h Its different To summarize, we look at the j-th equations of both methods: 2 h j + h j+1 Finite Element: u j u j 1 + u ( ) j u j+1 hj + h j+1 = f(ξ) h j h j+1 2 [ Finite Difference: uj u j 1 + u ] j u j+1 = f j h j h j+1 Accumulation via Elements Recall the system Ku = f, (17) where K ij = a(φ j, φ i ) and f i = f, φ i To compute the finite element system we make use of a reference element, τ, to do all necessary integration For 1D linears, we define 11

12 the reference element to be the unit interval, τ = [, 1] Associated with each element, e j, is a linear map from e j onto τ We map the basis with support on e j onto the reference element In the 1D linear case, {φ l } j l=j 1 gets mapped onto { ˆφ l } 1 l=, where ˆφ l (x) = { φj l+l (x j 1 + (x j x j 1 )x) x (, 1) else l =, 1 (18) The local-to-global map, i(j, l), is a table that specifies what nodes each element has Index j specifies the element number, and index l specifies which node we want from that element Element j Node l 1 e 1 1 e e (19) e e e j (j 1) j An interpolant can be evaluated element by element v I (x) = e j ( 1 l= v(x j 1+l )φ l (x) ) (2) 12

13 The bilinear form can be evaluated element by element, in small local stiffness matrices a(u, v) = e j a ej (u, v) u, v V h, u I = u, v I = v, (21) where a ej (u, v) = (u )(v x ) dx, Let s = e j x i(j,1) x i(j,) ( 1 1 ) ( 1 ) = (x i(j,1) x i(j,) ) 1 u i(j,l) ˆφ l (s) v i(j,l) ˆφ l (s) ds = ( ui(j,) u i(i,1) ) T K ej ( vi(j,) v i(i,1) l= ), l= with local stiffness matrix, [K ej ] lm = ˆφ l ˆφ m ds l, m =, 1 (22) The global stiffness matrix is the sum of the local stiffness matrices, injected into the global structure Notice that the adjacent local matrices overlap Example: Quadratic Elements Consider discretizing the same PDE with V h being piecewise quadratic polynomials 13

14 we could resuse the basis from the linear example, and add second order functions to each interval Reference element Nodal 14

15 Reference element 15