Numerical Solutions to Partial Differential Equations Zhiping Li - PowerPoint PPT Presentation

Numerical Solutions to Partial Differential Equations Zhiping Li LMAM and School of Mathematical Sciences Peking University

Error Estimates of Finite Element Solutions Aubin-Nische technique and error estimates in L 2 -norm Dual Variational Problem The Relations Between the Errors in L 2 and H 1 norms 1 It follows from the interpolation error estimates on regular affine family of finite element function spaces (see Theorem 7.7) that � v − Π h v � m , Ω ≤ C h k +1 − m | v | k +1 , Ω , m = 0 , 1 . 2 In other words, under the same conditions, the error of the finite element interpolation in the L 2 (Ω)-norm is one order higher than that in the H 1 (Ω)-norm. 2 / 21

Error Estimates of Finite Element Solutions Aubin-Nische technique and error estimates in L 2 -norm Dual Variational Problem The Relations Between the Errors in L 2 and H 1 norms 3 By the C´ ea lemma, the error of the finite element solution u h in H 1 (Ω)-norm is optimal. However, the error in L 2 (Ω)-norm thus obtained � u − u h � 0 , Ω ≤ � u − u h � 1 , Ω ≤ C � u − Π h u � 1 , Ω , is obviously not optimal. 4 Under certain additional conditions, optimal L 2 (Ω)-norm error estimate for FE solutions can be obtained by applying the Aubin-Nische technique based on the dual variational problem. 3 / 21

Error Estimates of Finite Element Solutions Aubin-Nische technique and error estimates in L 2 -norm Dual Variational Problem Dual Variational Problem and Relations of Errors in L 2 and H 1 norms 1 Consider the variational problem � Find u ∈ V such that ∀ v ∈ V , a ( u , v ) = f ( v ) , where V ⊂ H 1 (Ω), the bilinear form a ( · , · ) and the linear form f ( · ) satisfy the conditions of the Lax-Milgram lemma. 2 Let V h be a closed linear subspace of V , and u h ∈ V h satisfy the equation ∀ v h ∈ V h . a ( u h , v h ) = f ( v h ) , 3 Define the dual variational problem: � Find ϕ ∈ V such that a ( v , ϕ ) = ( u − u h , v ) , ∀ v ∈ V , where ( · , · ) is the L 2 (Ω) inner product. 4 / 21

Error Estimates of Finite Element Solutions Aubin-Nische technique and error estimates in L 2 -norm Dual Variational Problem Dual Variational Problem and Relations of Errors in L 2 and H 1 norms Lemma Let ϕ ∈ V be the solution of the dual variational problem, and let ϕ h ∈ V h satisfy the equation a ( v h , ϕ h ) = ( u − u h , v h ) , ∀ v h ∈ V h . Then, we have � u − u h � 2 0 , Ω ≤ M � u − u h � 1 , Ω � ϕ − ϕ h � 1 , Ω . 5 / 21

Error Estimates of Finite Element Solutions Aubin-Nische technique and error estimates in L 2 -norm Dual Variational Problem Dual Variational Problem and Relations of Errors in L 2 and H 1 norms proof : Take v = u − u h in the dual variational equation, and by the facts that a ( u − u h , v h ) = 0, ∀ v h ∈ V h and a ( · , · ) is bounded, we are lead to � u − u h � 2 0 , Ω = a ( u − u h , ϕ ) = a ( u − u h , ϕ − ϕ h ) ≤ M � u − u h � 1 , Ω � ϕ − ϕ h � 1 , Ω . � 6 / 21

An Optimal Error Estimate in L 2 -Norm Theorem Let the space dimension n ≤ 3 . Assume that the solution ϕ of the dual variational problem (7.3.10) is in H 2 (Ω) ∩ V , and satisfies � ϕ � 2 , Ω ≤ C � u − u h � 0 , Ω . h > 0 T h (Ω) be a family of regular class C 0 type Let { ( K , P K , Σ K ) } K ∈ � (1) Lagrange affine equivalent finite elements. Then, the L 2 (Ω) -norm error of the finite element solutions of the variational problem (7.1.1) satisfy � u − u h � 0 , Ω ≤ C h � u − u h � 1 , Ω . Furthermore, if the solution u of the variational problem (7.1.1) is in H 2 (Ω) ∩ V , then � u − u h � 0 , Ω ≤ C h 2 | u | 2 , Ω , Here C in the three inequalities represent generally different constants which are independent of h.

Error Estimates of Finite Element Solutions Aubin-Nische technique and error estimates in L 2 -norm Optimal Error Estimates in L 2 -Norm Proof of the Optimal Error Estimate in L 2 -Norm c 1 By the Sobolve embedding theorem, W m + s , p (Ω) → C s (Ω), ֒ → C (¯ ∀ s ≥ 0, if m > n / p . In particular, H 2 (Ω) ֒ Ω), if n ≤ 3. 2 Thus, by applying the error estimates for finite element solutions in H 1 (Ω) norm (see Theorem 7.10 with k = 1 and s = 0) to the dual variational problem (7.3.10), we obtain � ϕ − ϕ h � 1 , Ω ≤ Ch | ϕ | 2 , Ω . 8 / 21

Error Estimates of Finite Element Solutions Aubin-Nische technique and error estimates in L 2 -norm Optimal Error Estimates in L 2 -Norm Proof of the Optimal Error Estimate in L 2 -Norm 3 Therefore, by the lemma on the dual problem and � ϕ � 2 , Ω ≤ C � u − u h � 0 , Ω , we have � u − u h � 0 , Ω ≤ C h � u − u h � 1 , Ω . 4 Applying again Theorem 7.10 with k = 1 and s = 0 to � u − u h � 1 , Ω , we are lead to � u − u h � 0 , Ω ≤ C h 2 | u | 2 , Ω . � 9 / 21

Error Estimates of Finite Element Solutions Aubin-Nische technique and error estimates in L 2 -norm Optimal Error Estimates in L 2 -Norm Remarks on the Optimal Error Estimate in L 2 -Norm 1 The key to increase the L 2 -norm error estimate by an order is � ϕ � 2 , Ω ≤ C � u − u h � 0 , Ω , which does hold, if the coefficients of the second order elliptic operator are sufficiently smooth, and Ω is a convex polygonal region or a region with sufficiently smooth boundary. 2 In the general case, if we have � ϕ − ϕ h � 1 , Ω ∝ h α � u − u h � 0 , Ω and � u − u h � 1 , Ω ∝ h α , then, � u − u h � 0 , Ω ∝ h 2 α . 3 Generally, we expect the convergence rate of finite element solutions in the L 2 -norm is twice of that in the H 1 -norm. 10 / 21

Error Estimates of Finite Element Solutions Break of conformity and the Consistency Error Consistency Error and the First Strang Lemma Nonconformity and Consistency Error The conformity of finite element methods is often broken, so it is necessary to extend abstract error estimates accordingly. Numerical quadratures break the conformity and introduce consistency error. 11 / 21

Error Estimates of Finite Element Solutions Break of conformity and the Consistency Error Consistency Error and the First Strang Lemma First Strang Lemma — Abstract Error Estimate Including Consistency Error Theorem Let V h ⊂ V , and let the bilinear form a h ( · , · ) defined on V h × V h be uniform V h -elliptic, i.e. there exists a constant ˆ α > 0 independent of h such that α � v h � 2 , ∀ v h ∈ V h . a h ( v h , v h ) ≥ ˆ Then, there exists a constant C independent h such that | a ( v h , w h ) − a h ( v h , w h ) | � � � � u − u h � ≤ C inf � u − v h � + sup � w h � v h ∈ V h w h ∈ V h | f ( w h ) − f h ( w h ) | � + sup . � w h � w h ∈ V h 12 / 21

Error Estimates of Finite Element Solutions Break of conformity and the Consistency Error Consistency Error and the First Strang Lemma Proof of the First Strang Lemma 1 Since V h ⊂ V and a ( u , v ) = f ( v ), ∀ v ∈ V , we have a ( u − v h , u h − v h )+ a ( v h , u h − v h ) − f ( u h − v h ) = 0 , ∀ v h ∈ V h . 2 Since a h ( u h , v h ) = f h ( v h ), ∀ v h ∈ V h , we have ∀ v h ∈ V h . a h ( u h − v h , u h − v h ) = f h ( u h − v h ) − a h ( v h , u h − v h ) , 3 Therefore, by the uniform V h -ellipticity of a h ( · , · ) on V h , we have α � v h − u h � 2 ≤ a h ( u h − v h , u h − v h ) ˆ = a ( u − v h , u h − v h ) + { a ( v h , u h − v h ) − a h ( v h , u h − v h ) } + { f h ( u h − v h ) − f ( u h − v h ) } . 13 / 21

Error Estimates of Finite Element Solutions Break of conformity and the Consistency Error Consistency Error and the First Strang Lemma Proof of the First Strang Lemma 4 Hence, by the boundedness of the bilinear form a ( · , · ), and | f ( w h ) − f h ( w h ) | | f h ( u h − v h ) − f ( u h − v h ) | ≤ sup w h ∈ V h � u h − v h � � w h � | a ( v h , w h ) − a h ( v h , w h ) | | a ( v h , u h − v h ) − a h ( v h , u h − v h ) | ≤ sup � u h − v h � , � w h � w h ∈ V h we are lead to α � u h − v h � ≤ M � u − v h � ˆ | a ( v h , w h ) − a h ( v h , w h ) | | f ( w h ) − f h ( w h ) | + sup + sup . � w h � � w h � w h ∈ V h w h ∈ V h 5 Since � u − u h � ≤ � u − v h � + � u h − v h � , the conclusion of the α − 1 , 1 + ˆ α − 1 M } . theorem follows for C = max { ˆ � 14 / 21

Error Estimates of Finite Element Solutions Break of conformity and the Consistency Error Non-Conformity and the Second Strang Lemma Use of Non-Conforming Finite Element Function Spaces 1 The conformity will be broken, if a non-conforming finite element is used to construct the finite element function spaces. 2 In such a case, V h � V , therefore, � · � , f ( · ) and a ( · , · ) must be extended to � · � h , f h ( · ) and a h ( · , · ) defined on V + V h . 3 For example, if V = H 1 � 0 (Ω) and a ( u , v ) = Ω ∇ u · ∇ v dx , we may define � 1 / 2 � � | v h | 2 v h �→ � v h � h := , 1 , K K ∈ T h (Ω) � � ( u h , v h ) �→ a h ( u h , v h ) := ∇ u h · ∇ v h dx . K K ∈ T h (Ω) 15 / 21