Traces and Duality Lemma

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1 Traces and Duality Lemma Recall the duality lemma with H / ( ) := γ 0 (H ()) defined as the trace space of H () endowed with minimal extension norm; i.e., for w H / ( ) L ( ), w H / ( ) = min{ ŵ H () ŵ H (), γ 0 ŵ = w}, H ( ) := dual to H / ( ) =: H / ( )! = γ ν (H(div, )). Any q H(div, ) (i.e. q L (, R ), div q L ()) defines γ ν q H ( ) by (γ ν q)(w) =: q ν, w = (q ŵ + ŵ div q)dx for w H / ( ) and ŵ H () with γ 0 ŵ = w. (Side note: implies γ ν q H ( ) q H(div,).) q ν, w q ŵ + div q ŵ q H(div,) ŵ H () Duality Lemma. (a) There exists exactly one such that for all q H (; R n ) γ ν L(H(div, ); H ( )) γ ν q = (γ 0 q) ν a.e. on. (b) Let, denote the duality brackets of H ( ) H / ( ). All q H(div, ) and v H () satisfy the formula ( ) γ ν q, γ 0 v = v div q + q v dx.

2 (c) The operator γ ν is surjective and ker γ ν = H 0 (div, ) := D(; R n ) H(div). (d) (Duality lemma) For all t H ( ) with t(w) =: t, w for w H / ( ), it holds t H ( ) = t, w w H / ( ), w H / = = v H (), γ 0 v 0 inf ϕ H () = inf q H(div,). q H(div,), γν q=t t, γ 0 v v ϕ H () Proof. Proof of (a). Let q H(div, ). denotes the unique weak solution of For all v H / ( ) ˆv H () ˆv + ˆv = 0 in, γ 0ˆv = v on. Then v H / ( ) = ˆv H (). Define X q (v) := (ˆv div q + q ˆv)dx The repeated application of the Cauchy Schwarz inequality shows X q (v) ˆv L () div q L () + q L () ˆv L () q H(div) ˆv H () = q H(div) v H ( ). Hence X q : H / ( ) R is linear and bounded. Thus for any q H(div, ) there exists g(q) H ( ) with X q = g(q). Define γ ν : q g(q). This operator is linear. The last inequality shows that the operator is also bounded, more precisely γ ν L(H(div,);H ( )). Moreover, for all functions v H / ( ) and q H (, R n ), an integration by parts leads to (γ 0 q) ν, v = (γ 0 q) ν, γ 0ˆv = (ˆv div q + q ˆv)dx,

3 Thus (γ 0 q) ν, v = γ ν q, v. Hence, for all v H / ( ) it holds (γ 0 q) ν γ ν q, v = 0, i.e., (γ 0 q) ν = γ ν q H ( ). This implies (a). Moreover,, extends the scalar product (, ) in L ( ) for smooth functions. Proof of (b). For all v H () and q H(div, ) it holds ( ) γ ν q, γ 0 v = q ˆv + ˆv div q dx, where ˆv H () is such that ˆv + ˆv = 0 and γ 0 v = γ 0ˆv in the weak sense. Since ker γ 0 = H0() and v ˆv H0(), this equals ( ) γ ν q γ 0 vds = q v + v div q dx and implies (b). Proof of (c). For all q D(; R n ), γ ν q = (γ 0 q) ν = 0 a.e. by (a). Hence, H 0 (div, ) = D(; R n ) H(div) ker γν. The proof of ker γ ν H 0 (div, ) is more technical and can be found in the literature, i.e., in [Girault, V. and Raviart, P. A., Finite Element Methods for Navier-Stokes Equations, Springer-Verlag, Berlin, Heidelberg, New York (986)]. It remains to show the surjectivity of γ ν. Given any t H ( ), the functional T : H () R, v t, γ 0 v is linear and bounded, written T H (). The Riesz representation z H () of T in the Hilbert space H () satisfies z, H () = T ( ). For ϕ D(), it follows for γ 0 ϕ = 0 that z, ϕ H () = T (ϕ) = t, γ 0 ϕ = 0. This proves z + z = 0 in the weak sense. In particular, q := z L (; R n ) and div z = z leads to div q = z L (). Hence, q H(div, ) and q H(div) = ( div q + q ) / = ( z + z ) / = z H () = T (H ()). 3

4 For any v H (), it follows ( ) γ ν q, γ 0 v = q v + v div q dx = = z, v H () = T (v) = t, γ 0 v. ( ) z v + vz dx This implies γ ν q t, γ 0 v = 0 for all v H (), which is γ ν q t = 0 in H ( ). Consequently, t = γ ν q R(γ ν ). Proof of (d). For any t H ( ) let z and q be as above in the proof of (c). Then t H ( ) = t, γ 0ˆv ˆv H () γ 0ˆv H / = ( ) with t, γ 0ˆv = γ ν z, γ 0ˆv = z, ˆv H () z H () ˆv H (). Since ˆv H () =, this implies t H ( ) z H (). Conversely, t, γ 0 z = z, z H () = z H () proves t H ( ) z H (). This concludes the proof and characterizes H ( ) completely. Primal PMP with test functions in H () without (BC) leads to b(u, t; v) = a(u, v) t, v! = F (v) for all v H (). (P) Theorem. u solves (PMP) (u, γ ν u) solves (P). Proof. v H0() implies t, v = 0. Hence u solves (PMP). Let u H0() solve (PMP), then p := u H(div, ) leads to t := γ ν p H ( ) so that, for all v H (), it follows it follows t, v = p ν, γ 0 (v) = ( p v + v div p)dx = a(u, v) F (v). }{{}}{{} = u = f Define H 0 (div, ) := {q H(div, ) γ ν q = 0}. 4

5 Interface trace spaces For a shape-regular triangulation T of R n into simplices define H ( T ) := {(t K ) K T K T H ( K) q H(div, ) K T, γ ν (q K ) = t K H ( K)} endowed with the norm (t K ) K T H ( T ) := min{ q H(div,) K T, γ ν (q K ) = t K } and H (T ) := {v L () K T, v K H (K)} = H (K) K T with (v K ) K T H (T ) := K T v K H (K). Given t (t K ) K T H ( T ) and v H (T ) define t, v T := K T t K, v K K. There exists q H(div, ) such that t K = γ ν (q K ) H ( K) for all K T and t, v T = K T (q v + v div q)dx = K (q NC v + v div q)dtx q H(div,) v H (T )! = t H ( T ) v H () Define { b: (H 0 () H ( T )) H (T ) R b(u, t; v) := ((u, t), v) a NC (u, v) t, v T 5

6 Theorem. u solves (PMP) and t = (t K ) K T = (γ ν ( u K )) K T if and only if (u, t) H0() H ( T ) solves for all v H (T ). b(u, t; v) = F (v) Proof. The Proof is left as an exercise. Remark. t, v T = 0 for v H 0(). inf- Condition This section is devoted to some immediate estimation for β > 0. Recall X := X X := H0() H ( T ), Y := H (T ) and the bounded bilinear form b : X Y R with b(u, t; v) = b((u, t), v) = a NC (u, v) t, v T (u, t) X, v Y. For any (u, t) S(X) and v S(Y ) the Cauchy-Schwarz inequality leads to b(u, t; v) u v NC + t H ( T ) v H (T ) u + t v H ( T ) NC + v H (T ). With v NC v H (T ) the choice of (u, t) and v finally shows, that b(u, t; v). Given (u, t) S(X) set M := b(u, t; ) H (T ). For u 0 choose v := u/ u H (T ) to obtain t, u T = ( ) t, u K = q u + u div q dx = u q νds = 0. K T Hence, b(u, t; v) = a NC(u, u) u H (T ) = u u + u. The Friedrichs inequality implies u C F () u with C F width()/π. This leads to b(u, t; v) u + C F () M. 6

7 Hence u M + CF (). () Given t let q H(div, ) have minimal extension norm in H(div, ) with q ν = t on K for all K T. The duality lemma leads to some v H (T ) with v H (T ) = and t H ( T ) = t, v T (i.e. v is the normed Riesz representation of t, T in H (T )). This implies whence u v NC + t H = a NC (u, v) + t H = b(u, t; v) M, The inequalities () and () show that = u + t H ( T ) M ( + This leads to t H ( T ) u M. () + C F ()) + M ( + C F ()). ( + C F ()) + + C F () M = b(u, t; ) H (T ). Since this holds for all (u, t) S(X), it implies 0 < 3 + C F () + + C F () β = inf x S(X) v S(H (T )) b(u, t; v). Splitting Lemmas Splitting Lemma I. Given real Hilbert spaces X, X, X := X X, {0} Y Y and bounded bilinear forms b j : X j Y for j =,, let b : X Y R, (x, x ; y) b (x, y) + b (x, y). Suppose (A) 0 < β := (A) 0 < β := (A3) b X Y = 0. inf x S(X ) y S(Y ) inf x S(X ) y S(Y ) b (x, y ), b (x, y), 7

8 Then b satisfies an inf- condition with β > 0 and 0 < β β (β + b ) + β β β. Example (Application to primal dpg for PMP). Let X := H 0(), X := H ( T ) and Y := H 0() H (T ) =: Y. Ad (A). Show that β = inf u H0 (), u = v H0 (), v + v = a(u, v) = + C F (). Proof of is as above. For utilize the first Dirichlet eigenpair (λ, Φ ) with Φ = and Φ = λ / so that C F () = λ and Φ H (T ) = + λ. Consequently β a(φ, v) v H0 (), v H () = Φ. Since the eigenvectors (Φ j ) j N form an L -orthonormal and a-orthogonal basis of H0() the remum is attained by v := Φ / Φ H (). This leads to β Φ Φ Φ H () = Φ Φ H () Ad (A). By duality lemma it holds β := inf t S(H ( T )) v S(H (T )) = t, v T = λ + λ = Ad (A3). The Cauchy-Schwarz inequality implies b = u S(H0 ()) v S(H (T )) a NC (u, v) = + λ + C F (). inf t H ( T ) =. t S(H ( T )) u S(H0 ()) v S(H (T )) u v. The proof of b is left as an exercise. This leads to the inf- estimate + C F () + ( + C F () + ) = 3 + C F () + + C F () β. 8

9 Proof of the first splitting lemma. Given (x, x ) S(X X ) let s := x X and x X = s for 0 s. Then (A) and (A3) imply β s b (x, ) Y = b(x, x ; ) Y =: M. Moreover, (A), the definition of b and triangle inequality show that β s b (x, ) Y b(x, x ; ) Y + b (x, ) Y M + b s. Consequently f(s) := max{β s, β s b s} M. It remains to compute min f := min 0 s f(s) M. Since (x, x ) S(X) is arbitrary, this lead to β 0 β. The monotony of β s and β s b s shows that the minimizer s exists in (0, ) with ( b + β )s = β s. Set κ := β /(β + b ), so s = κ ( s ), whence s = κ/ + κ. Consequently, concludes the proof. β 0 = β κ + κ Splitting Lemma II. In addition to the notation of the first splitting lemma with (A)-(A), pose Y := {y Y b (, y) = 0 in X } (then (A3) follows and characterizes maximal Y in (A3)) and Then and N := {y Y b (, y ) = 0 in X } = {0}. N := {y Y b(, y) = 0 in X} = 0 β := β β β + β + b + (β + β + b ) 4β β β. 9

10 Example (Application to primal dpg for PMP). Given v Y, then for any q H(div, ) follows 0 = (v div q + q NC v) dx. Hence, any α =, and ϕ H () satisfy 0 = (v ϕ/ α + ϕ e α NC v) dx. Hence, NC v is the weak gradient of v L (), i.e. v H (). Consequently, v q ν ds = 0 for all q H(div, ). This implies v = 0 on, whence v H 0(). Consequently, Moreover, Y = {v H (T ) t H ( T ), t, v T = 0} = H 0(). N := {w H 0() a NC (, w) = 0 in H 0()} = {0}. Recall = β = b and / + CF () and compute β 0 β = ( + CF ()) + + (3 + CF ()) 4( + CF ()) = 3 + C F () C F () + 8CF () Proof of the second splitting lemma. Since Y is a closed subspace of the Hilbert space Y, there is an orthogonal decomposition Y = Y Y with Y = Y. Then 0 < inf x S(X ) y S(Y ) b (x, y) = inf x S(X ) y S(Y ) b (x, y ) = β. Any y Y with b (, y ) = 0 in X belongs to Y, whence y Y Y = {0}. Consequently, b X Y satisfies inf- condition with β and is nondegenerate. General theory of bilinear forms shows β = inf y S(Y ) x S(X ) b (x, y ) > 0. 0

11 Given any (x, x ) S(X X ) there exists a unique solution y Y to b (, y ) = x, X. From Riesz isomorphism follows b(, y ) X = x X. Then for any y Y β y Y b (, y ) X = x X. Since β > 0 and N = {0}, b X Y satisfies inf- conditions and is nondegenerate, whence there exists a unique solution y Y to Consequently, Altogether b (, y ) =, x X b (, y ) in X. β Y Y b (, y ) X x X + b y Y. b(x, y + y ) = b (x, y + y ) + b (x, y + y ) On the other hand, = x X + b (x, y ) = x X + x Y =. y + y Y = y Y + y Y β ( x X + b x X /β ) + x X /β ( β = ( x X, x X ) b β ) ( ) β x X b β β β ( + b /β) x X is bounded from above by the maximal eigenvalue Λ of the matrix ( ) b /β β. b /β This implies β + b β Λ b(x, y + y ) y + y Y b(x, ) Y. Since x S(X) is arbitrary, this proves β Λ. The formula follows from explicit calculations of the above matrix.

12 Discretization Define for k N 0 S k+ 0 (T ) X = H 0() P k (E) X = H ( T ) X h := S k+ 0 P k (E) Y h := P k+d (T ) Y = H (T ). Suggest d = dimension of domain and all k N 0. This lecture studies d = for n = space dimensions and k = 0. Remark ( on P 0 (E) H ( T )). Given any t 0 P 0 (E). Let τ RT RT 0 (T ) H(div, ) satisfy E E : t 0 = τ RT ν E on E. Then for all v H (T ). t 0, v T = K T K (τ RT K ν K )v ds Discrete duality lemma. For any t 0 P 0 (E) there exists exactly one p RT RT 0 (T ) H(div, ) such that for all K T and E E(K) (ν E ν K E )t 0 = (p RT ν K ) E. Then t 0 H ( T ) p RT H(div,) + h max π t 0 H ( T ). Proof. Recall that t 0 H ( T ) is the minimum of all q H(div,) for any q H(div, ) with (ν E ν K E )t 0 = (q ν K ) E for all K T, E E(K). (3) This proves the first inequality. Given any q H(div, ), (p RT q) ν K = 0 on K defined by the integration-by-parts formula. In particular 0 = (p RT q) ν K ds = div(p RT q) dx for all K T. K K

13 Consequently, div p RT = Π 0 div q a.e. in. An integration by parts shows for any v H (T ) that (p RT q) NC v dx = (v Π 0 v) div(q p RT ) dx h max /π v NC ( Π 0 ) div q L (). Set v(x) := (Π 0 p RT ) (x mid(k)) + /4(div p RT ) x mid(k) with NC v = Π 0 p RT + /(div p RT ) + / div p RT ( mid(t )) = p RT in the previous estimate to deduce p RT L () = q p RT dx + (p RT q) NC v dx whence q L () p RT L () + h max π p RT L () ( Π 0 ) div q L (), (4) p RT L () q L () + h max π ( Π 0) div q L (). This and (4) imply with λ = h max /π p RT H(div,) ( q L () + λ ( Π 0 ) div q L ()) + Π 0 div q L () ( + λ ) q L () + ( + /λ )λ ( Π 0 ) div q L () ( + λ )( q L () + ( Π 0) div q L () + Π 0 div q ). In other words, p RT H(div,) / + h max/π is a lower bound of q H(div,) for all q with (3). By definition of t 0 H / ( T ) as the minimum, this shows p RT H(div,) + h max /π t 0 H ( T ). Annulation property for P := INC loc : H (T ) H (T ) projection onto P (T ) defined by I loc NCv K := E E(K) E (v K ) dsψ E K P (K) for any v H (T ), K T. 3

14 Given any v H (T ), ( P )v Y = v INC locv + v INC locv + κ h max v NC. Consequently, the Kato lemma implies P = P + κ h max. Mean value property of the gradients Π 0 NC INC locv for all v H (T ) leads to the annulation property K T K t 0 (v K I loc NCv K ) ds = t 0, v P v T. Hence, for all x h = (u c, t 0 ) X h and v H (T ), it follows b(x h, v P v) = a NC (u c, v P v) t 0, v P v T = 0. The abstract theory asserts discrete inf- condition with β P β h b. This shows β + κ h max β h. The a posteriori analysis involves F ( P ) Y computable but not of higher order. κ h T f L (T ), which is 4

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