A posteriori error estimation for elliptic problems

Transcription

1 A posteriori error estimation for elliptic problems Praveen. C praveen@math.tifrbng.res.in Tata Institute of Fundamental Research Center for Applicable Mathematics Bangalore March 4, / 0

2 Partial differential equation u = f in Ω u = 0 on Ω Weak formulation Find u H0 1 (Ω) such that a(u, v) = l(v) v H0 1 (Ω) (1) where a(u, v) = u v, l(v) = fv Ω Ω Galerkin method Find u h V h H 1 0 (Ω) such that a(u h, v h ) = l(v h ) v h V h () / 0

3 Example: u = 1 π tan 1 (y/x) ndofs=11 ndofs=441 3 / 0

4 A priori error estimate If we take V h = X 1 h = {v C 0 ( Ω) : v P 1 } and if u H (Ω) then we have the error estimate u u h 1,Ω C ( T h h u, ) 1 Ch u,ω This is not precisely computable since it depends on the exact solution u. We can estimate the element seminorms u, by using the numerical solution u h and some finite difference approximation, etc. Then h u h, indicates the error from element. Identify elements with largest error indicator and refine them. Or equi-distribute the error: given some error tolerance ɛ, let N = number of elements. Then refine all elements for which h u h, > ɛ CN 4 / 0

5 L a posteriori error estimate Let Ω be a convex polygonal domain and let u and u h be the solutions to (1) and (), respectively. Then u u h 0,Ω C ( T h h η ) 1 where η = h f + u h 0, + 1 h 1 n u h 0, \Γ (3) where n u h denotes the jump across in the normal derivative n u h. Proof: We use the duality argument which was also used to prove the L a-priori error estimate. Let e h = u h u be the error and let ϕ be the solution of the adjoint problem, ϕ = e h in Ω and ϕ = 0 on Γ: find ϕ H0 1 (Ω) such that a(v, ϕ) = e h v v H0 1 (Ω) Note that ϕ H (Ω) and ϕ,ω C e h 0,Ω. Ω 5 / 0

6 Taking v = e h in adjoint problem, the error can be written as e h 0,Ω = a(e h, ϕ) = a(u h u, ϕ) = a(u h, ϕ) (f, ϕ) = = = [( u h, ϕ) (f, ϕ) ] [( u h f, ϕ) + (n u h, ϕ) ] [( u h f, ϕ) + 1 ( n u h, ϕ) \Γ ] where the factor of 1 appears in the second term since each interior edge belongs to two elements. Since a(e h, v h ) = a(u h u, v h ) = 0 v h V h we can replace ϕ with ϕ I h ϕ to obtain e h 0,Ω = a(e h, ϕ I h ϕ) [ u h + f ϕ I h ϕ + 1 ] n u h \Γ ϕ I h ϕ \Γ 6 / 0

7 We have the interpolation error estimates ϕ I h ϕ Ch ϕ, ϕ I h ϕ Ch 3 ϕ, (4) which leads to the desired result since e h 0,Ω C [h u h + f ϕ, + 1 ] h 3 n u h \Γ ϕ, = C h η ϕ, ( C h η ) 1 ( ϕ, ) 1 ( ) 1 ( ) 1 = C h η ϕ,ω C h η ϕ,ω ( ) 1 C h η e h 0,Ω 7 / 0

8 To show the second inequality in (4) we will show that for any w H 1 () ( ) w C h 1 w + h 1 w 1, which then leads to ( ) ϕ I h ϕ C h 1 ϕ I hϕ + h 1 ϕ I h ϕ 1, ( ) C h 1 h ϕ, + h 1 h ϕ, Ch 3 ϕ, To prove (5) we map any element to the reference element ˆ by the affine transformaton x = F (ˆx) = B ˆx + b ; then (5) w h d 1 ŵ ˆ, d =, 3 (6) From the trace theorem, we have continuity of trace operator which means ŵ ˆ C ŵ 1, ˆ C( ŵ ˆ + ŵ 1, ˆ) (7) 8 / 0

9 Now we convert the norms to the original element using a previous result ŵ m, ˆ C B m det B 1 w m, w H m () Note that det B = > Cρ d Cκ d h d where we use the fact that we have a regular triangulation. Moreover recall that B h ˆρ Hence for m = 0, 1 we get ŵ ˆ Ch d w and ŵ 1, ˆ Ch1 d w 1, (8) Combining (6), (7), (8) we obtain ( ) w Ch d 1 ( ŵ ˆ + ŵ 1, ˆ) C h 1 w + h 1 w 1, 9 / 0

10 Clement interpolation The standard interpolation requires function to be in H. For functions which are only in H 1, we have to use other types of interpolation operations. Let T h be a shape-regular triangulation of Ω. Given a node x j, let ω j = { T h : x j }, ω = {ω j : x j } The number of triangles that belong to ω j and ω is bounded. Since T h is shape-regular, the area of ω can be bounded as ω c(κ)h For v H 1 we cannot evaluate the functions pointwise since they may not be continuous. We will use a local averaging procedure to associate a function value to each node of the mesh. Let X 1 h denote the space of piecewise P 1 functions, and let {ϕ j } j be the standard hat basis functions. 10 / 0

11 Clement interpolation Clement interpolation Let T h be a shape-regular triangulation of Ω. Then there exists a linear mapping C h : H 1 (Ω) X 1 h such that for all T h v C h v m, Ch 1 m v 1, ω m = 0, 1 v C h v 0, Ch 1 v 1, ω Proof: Given a nodal point x j, let Q j : L (ω j ) P 0 be the L projection onto the constant functions, i.e., Q j v = v = Q j v = 1 ω j ω j ω j It follows by applying Bramble-Hilbert lemma that ω j v v Q j v 0,ωj Ch j v 1,ωj (9) 11 / 0

12 Clement interpolation where h j is the diameter of ω j. In order to cope with homogeneous Dirichlet boundary conditions on Γ D Ω we can modify the operator and set { 0 if x j Γ D Q j v = Q j v otherwise Using a proof similar to Poincare-Friedrichs inequality we obtain v Q j v = v 0,ωj Ch j v 1,ωj if x j Γ D 0,ωj Next we define the Clement interpolation as C h v = j ( Q j v)ϕ j X 1 h The shape functions {ϕ j } j form a partition of unity. Hence v C h v = j vϕ j j ( Q j v)ϕ j = j (v ( Q j v))ϕ j 1 / 0

13 Clement interpolation and hence v C h v 0, j v ( Q j v))ϕ j 0, j v Q j v 0,ωj C j h j v 1,ωj Ch v 1, ω The case of m = 1 is left for further studies. Error norm on : We make use of inequality (5). ( ) v C h v 0, C h 1 v C hv 0, + h 1 v C h v 1, ( ) C h 1 h v 1, ω + h 1 v 1, ω = Ch 1 v 1, ω 13 / 0

14 We next derive an error estimate for the Galerkin solution assuming that the true solution u H 1 0 (Ω) only. H 1 semi-norm error estimate Let T h be a shape-regular triangulation. Then the Galerkin solution satisfies the a-posteriori error estimate u u h 1,Ω C where η is given by equation (3). ( T h η Proof: We start by using a duality argument to compute the semi-norm of the error ( (u u h ), w) Ω L(w) u u h 1,Ω = sup = sup w H0 1(Ω) w 1,Ω w H0 1(Ω) w 1,Ω ) 1 14 / 0

15 Then L(w) = ( (u u h ), w) Ω = (f, w) Ω ( u h, w) Ω = [(f, w) ( u h, w) ] = = [( u h + f, w) 1 ( n u h, w) \Γ ] [ (f, w) ( u h, w) (n u h, w) \Γ ] The Galerkin solution satisfies a(u u h, w h ) = ( (u u h ), w h ) Ω = 0 w h V h 15 / 0

16 Let us take w h = C h w so that L(w) = L(w C h w) [( u h + f, w C h w) 1 ] ( n u h, w C h w) \Γ = C = C [ u h + f w C h w + 1 n u h \Γ w C h w ] [h u h + f w 1, ω + 1 h 1 n u h \Γ w 1, ω ] [h u h + f + 1 h 1 n u h \Γ ] w 1, ω = C η w 1, ω C ( η ) 1 ( ) 1 ( w 1, ω C η ) 1 w 1,Ω which yields the desired result. 16 / 0

17 Galerkin method with P 1 functions In this case u h 0 so that η = h f + 1 h 1 n u h Compute η for each T h Sort the values {η } in decreasing order. Select some fraction of elements with highest value of η and flag them for division Divide each element (into two or four); divide neighbouring elements to avoid hanging nodes. If dividing into two elements, select largest edge of and divide it. 17 / 0

18 Example: u = 1 π tan 1 (y/x) 18 / 0

19 Example: u = 1 π tan 1 (y/x) ndofs=131 ndofs=31 19 / 0

20 Example: u = 1 π tan 1 (y/x) 10 1 Uniform Adaptive 10 L error ndof Uniform refinement shows convergence rate / 0