Discontinuous Petrov-Galerkin Methods

Discontinuous Petrov-Galerkin Methods Friederike Hellwig 1st CENTRAL School on Analysis and Numerics for Partial Differential Equations, November 12, 2015

Motivation discontinuous Petrov-Galerkin (dpg) methods invented by L. Demkowicz, J. Gopalakrishnan (2010) motivated by search for optimal test functions dpg can be viewed as mixed method with nonstandard test space (typically broken test functions) or minimal residual method main idea: too many test functions ensure an inf-sup condition, let computer handle redundant dofs applications: linear elasticity, Stokes equations, Maxwell,... F. Hellwig 1/27

Outline 1 General Framework of dpg Methods Inf-Sup Conditions dpg as Mixed Method and Minimal Residual Method Built-in A Posteriori Estimate 2 The Primal dpg Method for the Poisson Model Problem Problem Formulation Proof of (H1) and (H2) F. Hellwig 2/27

Notation X Hilbert space with norm X F. Hellwig 4/27

Notation X Hilbert space with norm X Y Hilbert space with norm Y and scalar product (, ) Y F. Hellwig 4/27

Notation X Hilbert space with norm X Y Hilbert space with norm Y and scalar product (, ) Y for any normed space W, let S(W ) := {w W w W = 1} F. Hellwig 4/27

Notation X Hilbert space with norm X Y Hilbert space with norm Y and scalar product (, ) Y for any normed space W, let S(W ) := {w W w W = 1} b : X Y R a bounded bilinear form with b = sup x S(X ) sup y S(Y ) b(x,y ) F. Hellwig 4/27

The Continuous Problem Suppose that the problem x X : b(x, ) = F in Y is well-posed with unique solution. F. Hellwig 5/27

The Continuous Problem Suppose that the problem x X : b(x, ) = F in Y is well-posed with unique solution. The standard theory on mixed finite element methods [Boffi, Brezzi, Fortin] shows that this is equivalent to the inf-sup condition 0 < β := inf sup x S(X ) y S(Y ) b(x,y ) (H1) and non-degeneracy {0} = N := {y Y b(,y ) = 0 in X }. F. Hellwig 5/27

The Discrete Problem? Question: How to discretize this problem? F. Hellwig 6/27

The Discrete Problem? Question: How to discretize this problem? For any chosen finite-dimensional subspaces X h X, Y h Y, the well-posedness of the discrete problem x h X h : b(x h, ) = F in Y h is equivalent to the discrete inf-sup condition 0 < β h := inf sup x h S(X h ) y h S(Y h ) b(x h,y h ) (H2) and non-degeneracy {0} = N h := {y h Y h b(,y h ) = 0 in X h }. F. Hellwig 6/27

The Discrete Problem? (2) 0 < β h := inf sup x h S(X h ) y h S(Y h ) b(x h,y h ) {0} = N h := {y h Y h b(,y h ) = 0 in X h } Spaces X h and Y h need to be well-balanced to satisfy these two conditions. For fixed X h, a big Y h makes it easier to satisfy discrete inf-sup condition but harder to guarantee discrete non-degeneracy. F. Hellwig 7/27

Trial-to-Test-Operator The idea of discontinuous Petrov-Galerkin methods (dpg methods): choose only the discrete trial space X h X, compute a discrete test space Y with (H2) and non-degeneracy. Define the trial-to-test-operator T : X Y by (Tx,y ) Y = b(x,y ) for any y Y. F. Hellwig 8/27

dpg methods Variational formulation continuous problem x X : b(x, ) = F in Y F. Hellwig 9/27

dpg methods Variational formulation continuous problem x X : b(x, ) = F in Y idealized dpg x h X h : b(x h, ) = F in T (X h ) F. Hellwig 9/27

dpg methods Variational formulation continuous problem x X : b(x, ) = F in Y idealized dpg x h X h : b(x h, ) = F in T (X h ) X h and T (X h ) of idealized dpg method automatically satisfy the discrete inf-sup condition and non-degeneracy Operator T requires solution to infinite-dimensional problem not computable Remedy: discrete trial-to-test-operator T h : X Y h for some discrete space Y h Y. Define T h : X Y h similar to the continuous operator T by (T h x,y h ) Y = b(x,y h ) for any y h Y h. F. Hellwig 9/27

dpg methods Variational formulation continuous problem x X : b(x, ) = F in Y idealized dpg x h X h : b(x h, ) = F in T (X h ) practical dpg x h X h : b(x h, ) = F in T h (X h ) X h and T (X h ) of idealized dpg method automatically satisfy the discrete inf-sup condition and non-degeneracy Operator T requires solution to infinite-dimensional problem not computable Remedy: discrete trial-to-test-operator T h : X Y h for some discrete space Y h Y. Define T h : X Y h similar to the continuous operator T by (T h x,y h ) Y = b(x,y h ) for any y h Y h. F. Hellwig 9/27

dpg methods Variational formulation continuous problem x X : b(x, ) = F in Y idealized dpg x h X h : b(x h, ) = F in T (X h ) practical dpg x h X h : b(x h, ) = F in T h (X h ) X h and T h (X h ) Y h satisfy non-degeneracy but not necessarily the discrete inf-sup condition continuous inf-sup condition holds Y h big enough leads to the discrete inf-sup condition 0 < inf sup x h S(X h ) y h S(Y h ) b(x h,y h ) = inf sup x h S(X h ) y h S(T h (X h )) b(x h,y h ). F. Hellwig 9/27

dpg methods...seen as minimization Variational formulation Minimal Residual Method continuous problem x X : b(x, ) = F in Y x arg min ξ X b(ξ, ) F Y idealized dpg x h X h : b(x h, ) = F in T (X h ) x h arg min ξh X h b(ξ h, ) F Y practical dpg x h X h : b(x h, ) = F in T h (X h ) equivalence by calculation of Gâteaux derivative F. Hellwig 9/27

dpg methods...seen as minimization Variational formulation Minimal Residual Method continuous problem x X : b(x, ) = F in Y x arg min ξ X b(ξ, ) F Y idealized dpg x h X h : b(x h, ) = F in T (X h ) x h arg min ξh X h b(ξ h, ) F Y practical dpg x h X h : b(x h, ) = F in T h (X h ) x h arg min ξh X h b(ξ h, ) F Y h equivalence by calculation of Gâteaux derivative F. Hellwig 9/27

A priori estimate Theorem Suppose (H1) and (H2) hold. Then the solution x h X h to the practical dpg method satisfies x x h X b β h min x ξ h X. ξ h X h F. Hellwig 10/27

A posteriori estimate for idealized dpg Idealized dpg method has built-in a posteriori estimate β x x h X b(x h, ) F Y b x x h X, i.e. x x h X b(x h, ) F Y. Is such an estimate possible for the practical dpg method as well? F. Hellwig 11/27

Fortin Operator For an a posteriori estimate of the practical dpg method, we need some Fortin operator Π : Y Y h. Existence of Π : Y Y h linear and bounded projection with b(x h, (1 Π)Y ) = 0 Lemma Suppose (H1) holds. Then (H2) and (H3) are equivalent. (H3) Proof. (H3) = (H2). Continuity of Π implies 1/ y Y Π / Πy Y and (H1) shows 0 < β inf Π sup x h S(X h ) y Y,y 0 inf sup b(x h,y ) y Y Π inf b(x h,y h ) y h Y. sup x h S(X h ) y Y,y 0 b(x h,πy ) Πy Y x h S(X h ) y h Y h,y h 0 F. Hellwig 12/27

Fortin Operator (2) Lemma Suppose (H1) holds. Then (H2) and (H3) are equivalent. Proof. (H2) = (H3). F. Hellwig 13/27

Fortin Operator (2) Lemma Suppose (H1) holds. Then (H2) and (H3) are equivalent. Proof. (H2) = (H3). Condition (H2) and non-degeneracy imply that B : T h (X h ) X h,btx h := b(,tx h ) is an isomorphism. Let G : Y X h,gy := b(,y ). Define Π := B 1 G : Y T h (X h ) linear and bounded. Π(Tx h ) = B 1 b(,tx h ) = B 1 BTx h = Tx h y Y satisfies b(,y ) = G (y ) = B (Πy ) = b(,πy ) in X h F. Hellwig 13/27

A posteriori estimate Theorem Suppose (H1) and (H3) hold. Then any ξ h X h satisfies β x x h X Π b(x h, ) F Y + F (1 Π) Y h 2 b Π x x h X. F. Hellwig 14/27

A posteriori estimate Theorem Suppose (H1) and (H3) hold. Then any ξ h X h satisfies β x x h X Π b(x h, ) F Y + F (1 Π) Y h 2 b Π x x h X. Proof. (H1) and (H3) imply β x x h X F b(ξ h, ) Y F b(ξ h, ) Y sup y S(Y ) F ((1 Π)y ) + sup y S(Y ) and F (Πy ) b(ξ h,πy ). F. Hellwig 14/27

Computation of dpg Solution {Φ 1,...,Φ J } basis of X h {Ψ 1,...,Ψ K } basis of Y h A R K J, A kj := b(φ j,ψ k ), k = 1,...,K, j = 1,...,J, matrix for bilinear form M R K K, M kl := (Ψ k,ψ l ) Y, k,l = 1,...,K, matrix for scalar product on Y b R K, b k := F (Ψ k ), k = 1,...,K, vector for right-hand side For x R J coefficient vector for solution x h = J j=1 x j Φ j, the linear system of equation reads A M 1 Ax = A M 1 b. Bigger Y h guarantees inf-sup condition, but computation will be more expensive! Concept of broken test functions leads to block-diagonal M. F. Hellwig 15/27

1 General Framework of dpg Methods Inf-Sup Conditions dpg as Mixed Method and Minimal Residual Method Built-in A Posteriori Estimate 2 The Primal dpg Method for the Poisson Model Problem Problem Formulation Proof of (H1) and (H2) F. Hellwig 16/27

Traces Theorem (Traces of H 1 -functions) For U R n open and bounded Lipschitz domain, there exists continuous, linear γ 0 : H 1 (U) L 2 ( U) with γ 0 w = w U for all w H 1 (U) C 0 (U). F. Hellwig 17/27

Traces Theorem (Traces of H 1 -functions) For U R n open and bounded Lipschitz domain, there exists continuous, linear γ 0 : H 1 (U) L 2 ( U) with γ 0 w = w U for all w H 1 (U) C 0 (U). Set H 1/2 ( U) := γ 0 (H 1 (U)) and H 1/2 ( U) := (H 1/2 ( U)). F. Hellwig 17/27

Traces (2) Theorem (Normal traces of H(div )-functions) For U R n open and bounded Lipschitz domain, there exists continuous, linear γ ν : H(div,U) H 1/2 ( U), with γ ν q = q U ν for all q C (U; R n ). Any q H(div,U) and w H 1 (U) satisfies γ ν q,γ 0 w U = (q, w ) U + (div q,w ) U. F. Hellwig 18/27

Traces (3) T regular triangulation, T := T T T the skeleton F. Hellwig 19/27

Traces (3) T regular triangulation, T := T T T the skeleton For q H(div, T ) define γ T ν q T T H 1/2 ( T ) by γ T ν q := (t T ) T T, with t T := γ ν (q T ) for all T T. F. Hellwig 19/27

Traces (3) T regular triangulation, T := T T T the skeleton For q H(div, T ) define γ T ν q T T H 1/2 ( T ) by γ T ν q := (t T ) T T, with t T := γ ν (q T ) for all T T. H 1/2 ( T ) := γ T ν H(div, Ω) is a Hilbert space with minimal extension norm t H 1/2 ( T ) = min{ q H(div) q H(div, Ω),γ T ν q = t}. F. Hellwig 19/27

Traces (3) T regular triangulation, T := T T T the skeleton For q H(div, T ) define γ T ν q T T H 1/2 ( T ) by γ T ν q := (t T ) T T, with t T := γ ν (q T ) for all T T. H 1/2 ( T ) := γν T H(div, Ω) is a Hilbert space with minimal extension norm t H 1/2 ( T ) = min{ q H(div) q H(div, Ω),γν T q = t}. For t T T H 1/2 ( T ) and v H 1 (T ) define t,v T := t T,γ 0 v T. T T F. Hellwig 19/27

Poisson Model Problem Ω R 2 open Lipschitz domain with polygonal boundary Seek u : Ω R with u = f in Ω, u = 0 on Ω. F. Hellwig 20/27

Weak formulation multiply equation with test function v, integrate by parts on Ω test function v H 1 0 (Ω) fv dx = Ω Ω u v dx Ω v u ν ds F. Hellwig 21/27

Weak formulation multiply equation with test function v, integrate by parts on Ω test function v H0 1(Ω) boundary integral vanishes fv dx = Ω Ω u v dx Ω v u ν ds } {{ } =0 F. Hellwig 21/27

Weak formulation multiply equation with test function v, integrate by parts on T T test function v H 1 (T ) boundary integral? fv dx = T T u v dx T v u ν ds F. Hellwig 21/27

Weak formulation multiply equation with test function v, integrate by parts on T T test function v H 1 (T ) boundary integral introduces new variable t fv dx = T T u v dx T v u }{{} ν ds t T F. Hellwig 21/27

Weak formulation multiply equation with test function v, integrate by parts on T T test function v H 1 (T ) boundary integral introduces new variable t sum over all elements fv dx = u v dx T T T v u }{{} ν ds t T Primal dpg formulation seeks u H 1 0 (Ω),t H 1/2 ( T ) with (f,v ) Ω = ( u, NC v ) Ω t,v T for all v H 1 (T ). F. Hellwig 21/27

Duality lemma Theorem Any t H 1/2 ( T ) satisfies t H 1/2 ( T ) t,v T sup. v H 1 (T ),v 0 v H 1 (T ) F. Hellwig 22/27

Duality lemma Theorem Any t H 1/2 ( T ) satisfies t H 1/2 ( T ) t,v T sup. v H 1 (T ),v 0 v H 1 (T ) Proof. Let v H 1 (T ) on each T T weak solution to v + v = 0 in T and v ν = t T on T. With q := NC v H(div, Ω), it holds t H 1/2 ( T ) q H(div), div q = v, and q H(div) = v H 1 (T ). Integration by parts shows t,v T = (q, NC v ) Ω + (div q,v ) Ω = q 2 H(div). F. Hellwig 22/27

Inf-Sup Condition Theorem The spaces X := H 1 0 (Ω) H 1/2 ( T ), Y := H 1 (T ) and the bilinear form b : X Y, b(u,t; v ) := ( u, NC v ) Ω t,v T satisfy (H1). F. Hellwig 23/27

Inf-Sup Condition Theorem The spaces X := H 1 0 (Ω) H 1/2 ( T ), Y := H 1 (T ) and the bilinear form b : X Y, b(u,t; v ) := ( u, NC v ) Ω t,v T satisfy (H1). Proof. The Friedrichs inequality implies u L 2 (Ω) ( u, v ) Ω sup v H0 1(Ω),v 0 v H 1 (T ) b(u,t; v ) = sup. v H0 1(Ω),v 0 v H 1 (T ) The duality lemma and the triangle inequality show t H 1/2 ( T ) b(u,t; v ) sup + sup v Y,v 0 v H 1 (T ) v Y,v 0 ( u, NC v ) Ω. v H 1 (T ) F. Hellwig 23/27

Discretization Recall X = H0 1 (Ω) H 1/2 ( T ), Y = H 1 (T ). The discrete spaces read X h := S0 1 (T ) P 0(E) X, Y h := P 1 (T ) Y. F. Hellwig 24/27

Discrete Inf-Sup Condition Theorem The discrete spaces X h and Y h satisfy (H2). F. Hellwig 25/27

Discrete Inf-Sup Condition Theorem The discrete spaces X h and Y h satisfy (H2). Proof. given x h = (u C,t 0 ) X h, let q RT RT 0 (T ), γ T ν (q RT ) = t 0 choose v 1 = div q RT + ( u C Π 0 q RT ) ( mid(t )) Y h integration by parts shows b(u C,t 0 ; v 1 ) = ( u C q RT, NC v 1 ) Ω (div q RT,v 1 ) Ω = u C Π 0 q RT 2 L 2 (Ω) + div q RT 2 L 2 (Ω) Since P 0 (T ) is orthogonal to ( mid(t )) in L 2 (Ω), v 1 2 H 1 (T ) (1 + h2 max)b(u C,t 0 ; v 1 ). F. Hellwig 25/27

Discrete Inf-Sup Condition (2) Proof. recall b(u C,t 0 ; v 1 ) = u C Π 0 q RT 2 L 2 (Ω) + div q RT 2 L 2 (Ω) Helmholtz decomposition leads to α C S0 1 (T ), β CR CR 1 (T ) with u C Π 0 q RT = α C + Curl NC β CR. Orthogonality in Helmholtz and integration by parts shows (u C α C ) 2 L 2 (Ω) = ( (u C α C ),q RT ) Ω = (u C α C,div q RT ) Ω (u C α C ) L 2 (Ω) div q RT L 2 (Ω). triangle inequality implies u C L 2 (Ω) (u C α C ) L 2 (Ω) + α C L 2 (Ω) div q RT L 2 (Ω) + u C Π 0 q RT L 2 (Ω) b(u C,t 0 ; v 1 ) 1/2 and t 0 H 1/2 ( T ) q RT H(div) Π 0 q RT L 2 (Ω) + div q RT L 2 (Ω) b(u C,t 0 ; v 1 ) 1/2. F. Hellwig 26/27

Summary idea of dpg: choose discrete trial space, compute discrete test space idealized dpg: inf-sup stable, but not practical practical dpg inf-sup stable for Y h big enough practical dpg has built-in a priori and a posteriori error control application to Poisson as primal dpg with broken test functions continuous inf-sup follows from stability of non-broken functions discrete inf-sup utilizes discrete Helmholtz decomposition F. Hellwig 27/27