UPPSALA UNIVERSITY Departent of Inforation Technology Division of Scientific Coputing Solutions to exa in Finite eleent ethods II 14-6-5 Stefan Engblo, Daniel Elfverson Question 1 Note: a inus sign in front of the A is issing on the exa. (a) Multiplying the PDE by a test-function v which vanishes at the boundary and integrating by parts, we get (A u, v) + (b u, v) + (cu, v) = (f, v). Define V = H 1 (Ω), the variational forulation reads: find u V such that a(u, v) =: (A u, v) + (b u, v) + (cu, v) = (f, v) := l(v), v V. For the Galerkin approxiation we consider a subspace V h V, find u h V h such that a(u h, v) = l(v), v V h. (b) The assuptions are (A u, u) a ( u, u), c 1 b > c >, and f L (Ω). Note also that We get = (n bu, 1) Ω = ( (bu ), 1) Ω = (( b)u, u) Ω + (b u, u) Ω. (1) a(u, u) = (A u, u) + (b u, u) + (cu, u) a ( u, u) + (cu, u) 1 (( b)u, u) a ( u, u) + c (u, u) = a u + c u in(a, c ) u V = u V, () for coercivity, a(u, v) = (A u, v) + (b u, v) + (cu, v) A L (Ω) u v + b L (Ω) u v + c L (Ω) u v C ab u v + C c u v C u V v V (3) for continuity, and l(v) = (f, v) = f v f v V, (4) which together show that the Lax-Milgra lea can be applied. (c) We have the basic Galerkin orthogonality a(u u h, v) = v V h. (5) We start by proving Cea s lea and then use the interpolation estiate u u h V 1 a(u u h, u u h ) = C a a(u u h, u v) We obtain for a Poincaré constant C p, C a u u h V u v V. (6) u u h V C a u v V C ac p (u πu) C ac p C π hin(,p) u H 3 (Ω). (7) 1
(d) First is to show that the solution the the variational proble iniize the functional F (v) = 1 a(v, v) l(v). We obtain F (u + v) = 1 a(u + v, u + v) F (u + v) = 1 (a(u, u) + a(u, v) + a(v, v)) l(u) l(v) = F (u) + a(u, v) l(u) + 1 a(v, v) F (u). (8) Then we ust show that a iniizer to F (v) solves the variational proble. Let g(ɛ) = F (u + ɛv), we obtain Differentiating g with respect the ɛ, we have g(ɛ) = 1 a(u + ɛv, u + ɛv) l(u + ɛv). (9) g (ɛ) = a(u, v) ɛa(v, v) l(v) (1) and setting ɛ = we have the variational forulation. Note that V h V, we can concludes the proof. Question (a) We have the syste 1 L 1 (1) L 1 (r) L 1 (s) L 1 (t) L 1 (rs) L 1 (rt) L 1 (st) L 1 (rst) c 1 L (1) L (r) L (s) L (t) L (rs) L (rt) L (st) L (rst) c L 3 (1) L 3 (r) L 3 (s) L 3 (t) L 3 (rs) L 3 (rt) L 3 (st) L 3 (rst) c 3 e 1 = = L 4 (1) L 4 (r) L 4 (s) L 4 (t) L 4 (rs) L 4 (rt) L 4 (st) L 4 (rst) c 4 L 5 (1) L 5 (r) L 5 (s) L 5 (t) L 5 (rs) L 5 (rt) L 5 (st) L 5 (rst) c 5 L 6 (1) L 6 (r) L 6 (s) L 6 (t) L 6 (rs) L 6 (rt) L 6 (st) L 6 (rst) c 6 L 7 (1) L 7 (r) L 7 (s) L 7 (t) L 7 (rs) L 7 (rt) L 7 (st) L 7 (rst) c 7 L 8 (1) L 8 (r) L 8 (s) L 8 (t) L 8 (rs) L 8 (rt) L 8 (st) L 8 (rst) c 8 (11) Which is 1 1 c 1 1 1 c 1 1 1 1 c 3 = 1 1 c 4 1 1 c 5 (1) 1 1 1 1 c 6 1 1 1 1 1 1 1 1 c 7 1 1 1 1 (b) We have u u h h u v h + v u h h (13) for v V h. For the first ter choose v as sup v Vh u v h. For the second ter we have v u h h 1 a h(v u h, v u h ) = 1 a h(v u + u u h, v u h ) c 8 = 1 (a h(v u, v u h ) + a h (u u h, v u h )) (14) = 1 (C α v u h v u h h + a h (u, v u h ) l(v u h ))
which iplies that v u h h 1 ( C α v u h + a ) h(u, v u h ) l(v u h ) v u h h 1 ( ) (15) a h (u, w) l(w) C α v u h + sup w V h Added the bound for the first and second ter we have for C = 1 + Cα. Question 3 u u h h (1 + C α ) inf u v h + 1 v V h sup a h (u, w) l(w) w V h ( ) (16) a h (u, w) l(w) C inf u v h + sup v V h w V h (a) Lets try to bound the gradient of the solution in ters of data. We have the variational for: find u V = H 1 (Ω) We obtain ɛ( u, v) + (b u, v) = (f, v) v V. (17) ɛ u = ɛ( u, u) + (b u, u) = (f, v) f v C P F f v. (18) where we used that = (n bu, 1) Ω = ( (bu ), 1) Ω = (( b)u, u) Ω + (b u, u) Ω. (19) together with b =. (b) First fro the continuous proble we have that u solves: find u V = H 1 (Ω) such that ɛ( u, v) + (b u, v) = (f, v) v V. () For the reaining part we have, ( ɛ u + b u, ɛ v + b v)) = (f, ɛ v + b v). (1) Also since, u H (Ω) V, we have that ɛ : H (Ω) L (Ω), but ɛ T : H 1 (T ) L (T ) is not true in general, so we can not test for all v V. The eleentwice operator ɛ T : V h L (T ) is true since all v V h are polynoials of finite diension. Multiplying the original equation with a test function in w L (Ω) defined as w T = ɛ v + b v L (T ) on eleent T, we have that (ɛ u, w) + (b u, w) = (f, w) ( ɛ u + b u, w) = (f, w), () for all v V h. Using the first and second part together we obtain that that u H V satisfies a h (u, v) = l(v) v V h. (3) 3
Question 4 (a) A typical Picard iteration for the proble is With u k as given we find by linearity that (c(u k ) u k+1 ) = f. (c(u k ) (u k u k+1 )) = r k, with hoogeneous Dirichlet boundary conditions. (b) The suggested iteration is clearly consistent since it is satisfied identically by any solution u to the PDE itself. Using the boundary conditions we readily find the variational forulation find u k+1 H 1 (Ω) such that for all v H 1 (Ω). This iplies the FEM ( v, a(u k ) u k+1 ) = (v, f) + ( v, b(u k ) u k ) A k ξ k+1 = b + B k ξ k, A k (i, j) = ( ϕ i, a(u k ) ϕ j ), B k (i, j) = ( ϕ i, b(u k ) ϕ j ), b i = (ϕ i, f), U k = j ξ k j ϕ j. (c) By taking v = u k+1 in the variational forulation we produce the energy estiate α u k+1 ( u k+1, a(u k ) u k+1 ) = (u k+1, f) + ( u k+1, b(u k ) u k ) u k+1 f + β u k+1 u k C f u k+1 + β u k+1 u k. Hence a suitable a priori bound is u k+1 α 1 C f + α 1 β u k, which if κ := α 1 β < 1 can be iterated to give α 1 C f + κα 1 C f + κ u k 1... (1 + κ +... + κ k )α 1 C f + κ k+1 u α 1 C f 1 κ + u. (d) Here we show that the LM lea is applicable, but the proble is syetric so the RR theore is also applicable. We have l(v) = (v, f) + ( v, b(u k ) u k ) ax( f, β u k ) v H1 (Ω). a(u, v) = ( v, a(u k ) u) A k v u A k v H 1 (Ω) u H 1 (Ω), where a(u k ) A k (recall that a is bounded). Finally, a(u, u) α u α C u, = a(u, u) α in(1, 1/C ) u H 1 (Ω). 4
(e) We arrive at the Newton ethod by linearising around the previous iterate u k. Put u k+1 = u k +δ. Then this is achieved by a(u k+1 ) u k+1 = a(u k +δ) (u k +δ) = a(u k ) (u k + δ) + δa (u k ) u k + O(δ ). The variational forulation becoes find u k+1 H 1 (Ω) such that ( v, a(u k ) u k+1 ) + ( v, u k+1 a (u k ) u k ) = (v, f) + ( v, b(u k ) u k ) + ( v, u k a (u k ) u k ) for all v H 1 (Ω). The FEM becoes with atrices A and B as before and with (A k + C k )ξ k+1 = b + (B k + C k )ξ k, C k (i, j) = ( ϕ i, ϕ j a (U k ) U k ). 5