Inf-sup estimates for the Stokes problem in a periodic channel
aa r X i v : . [ m a t h . A P ] J un INF-SUP ESTIMATES FOR THE STOKES PROBLEM IN APERIODIC CHANNEL
JON WILKENING ∗ Abstract.
We derive estimates of the Babu˘ska-Brezzi inf-sup constant β for two-dimensionalincompressible flow in a periodic channel with one flat boundary and the other given by a periodic,Lipschitz continuous function h . If h is a constant function (so the domain is rectangular), we showthat periodicity in one direction but not the other leads to an interesting connection between β and the unitary operator mapping the Fourier sine coefficients of a function to its Fourier cosinecoefficients. We exploit this connection to determine the dependence of β on the aspect ratio of therectangle. We then show how to transfer this result to the case that h is C , or even C , by achange of variables. We avoid non-constructive theorems of functional analysis in order to explicitlyexhibit the dependence of β on features of the geometry such as the aspect ratio, the maximum slope,and the minimum gap thickness (if h passes near the substrate). We give an example to show thatour estimates are optimal in their dependence on the minimum gap thickness in the C , case, andnearly optimal in the Lipschitz case. Key words.
Incompressible flow, Stokes equations, Babu˘ska-Brezzi inf-sup condition, gradient,divergence, Sobolev space, dual space
AMS subject classifications.
1. Introduction.
Many problems of industrial and biological importance in-volve fluid flow in narrow channels with moving boundaries [7, 11]. Examples includethe flow of oil in journal bearings or between moving machine parts, the flow of air be-tween disk drive platters and read-write heads, or the flow of mucus under a crawlinggastropod [13]. A primary objective in all these problems is to solve for the pressurerequired to maintain incompressibility. Indeed, it is the pressure that determines theload sustainable by a journal bearing, and that provides propulsion against viscousdrag forces in peristaltic locomotion. However, only the gradient of pressure entersdirectly into the Stokes or Navier-Stokes equations; thus, regardless of the methodused to solve the equations, the pressure must be determined via its gradient.The fundamental fact that makes it possible to extract p from ∇ p is that thegradient is an isomorphism from L (Ω), the space of mean-zero square integrablefunctions, onto the subspace of linear functionals in H − (Ω) that annihilate the di-vergence free vector fields u ∈ H (Ω) ; see Section 2 below. The inf-sup constant β (or rather, its inverse) gives a bound on the norm of the inverse of this operator.Thus the magnitude of p (and our ability to estimate errors in p ) depends to a largeextent on the size of β − . However, to the author’s knowledge, every existing proof(e.g. [4, 8]) that β − is finite relies on Rellich’s compactness theorem to extract a sub-sequence whose lower order derivatives converge, making it impossible to determinehow large β − might be or how it depends on Ω. The proof in [4] also uses the closedgraph theorem, which, like Rellich’s theorem, leads to constants that depend on Ω inan uncontrollable way. These proofs are appropriate for pathological domains withbulbous regions connected by thin, circuitous pathways; however, for “nice domains”,it should be possible to obtain better estimates of the constants — existing theoremsare of limited practical use. ∗ Department of Mathematics and Lawrence Berkeley National Laboratory, University of Cali-fornia, Berkeley, CA 94720 ( [email protected] ). This work was supported in part by theDirector, Office of Science, Advanced Scientific Computing Research, U.S. Department of Energyunder Contract No. DE-AC02-05CH11231.
JON WILKENING Γ x = 0 Γ u = u = x = L y h ( x ) − ∆ u + ∇ p = f ∇ · u = 0 h = max ≤ x ≤ L h ( x ) h = min ≤ x ≤ L h ( x ) > β for two-dimensional incompressible flow in a periodic channel with one flat boundary and theother given by a periodic, Lipschitz continuous function h ( x ). Our goal is to determinehow β − depends on features of the geometry such as the aspect ratio, the maximumslope, and the minimum gap thickness (if h passes near the substrate). Althoughthese requirements on Ω are fairly restrictive, such geometries do cover a wide rangeof interesting applications.Our interest in this problem arose in the course of deriving a-priori error estimatesfor Reynolds’ lubrication approximation (and its higher order corrections) with con-stants that depend on Ω in an explicit, intuitive way; see [12] and also [7, 10, 11] forbackground on lubrication theory. These a-priori estimates were used by the authorand A. E. Hosoi to monitor errors in the lubrication approximation while studyingshape optimization of swimming sheets over thin liquid films; see [13].
2. Preliminaries.
In this section we briefly review the weak formulation of theStokes equations, emphasizing the role played by the Babu˘ska-Brezzi inf-sup condi-tion; see e.g. [2, 3, 6] for a more detailed account.Consider the two-dimensional, x -periodic Lipschitz domain Ω shown in Figure 2.1:Ω = { ( x, y ) : x ∈ T, < y < h ( x ) } , h ∈ C , ( T ) , T = [0 , L ] p . (2.1)The case of non-zero Dirichlet boundary conditions may be reduced to the homoge-neous case by subtracting off an appropriate function to transfer the inhomogeneityfrom the boundary conditions to the body force f ; see e.g. [2]. We treat Ω and T as C ∞ manifolds by identifying the pointsΩ : (0 , y ) ∼ ( L, y ) 0 < y < h (0) ,T : 0 ∼ L (2.2)and adding a coordinate chart to each that “wraps around”. In particular: a functionin C k (Ω) or C k ( T ) is understood to have k continuous periodic derivatives; ∂ Ω =Γ ∪ Γ ; ∂T = ∅ ; the support of a function φ ∈ C kc (Ω) vanishes near Γ and Γ butnot necessarily at x = 0 and x = L ; and the Sobolev spaces H k (Ω) and H k (Ω) are thecompletions of C k (Ω) and C kc (Ω) in the k · k k norm, and thus contain only x -periodicfunctions with appropriate smoothness at x = 0 , L .In the weak formulation of the Stokes equations, we seek the velocity u and NF-SUP ESTIMATES FOR THE STOKES PROBLEM IN A PERIODIC CHANNEL p in the spaces X = H (Ω) , M = L (Ω) = n p ∈ L (Ω) : Z Ω p dA = 0 o , (2.3)respectively, such that a ( u , v ) + b ( v , p ) = h f , v i (2.4a) b ( u , q ) = 0 (2.4b)for all v ∈ X and q ∈ M , where the body force f may be any linear functional in thedual space X ′ = H − (Ω) and a ( u , v ) = Z Ω ∇ u : ∇ v dA, b ( u , p ) = − Z Ω p ∇ · u dA. (2.5)We endow M with the L norm k · k and X with the energy norm (i.e. the H semi-norm) k u k a = p a ( u , u ), which is equivalent to the H norm k u k = p k u k + k u k a due to the Poincar´e-Friedrichs inequality (see Lemma A.2): k u k ≤ h √ k u k a , ( u ∈ X ) , h = max ≤ x ≤ L h ( x ) . (2.6)Next we define the operators B : X → M ′ and B ′ : M → X ′ via h B u , p i = b ( u , p ) = h B ′ p, u i , ( B = div , B ′ = grad) . (2.7) B and B ′ are clearly bounded and satisfy k B k = k B ′ k = sup p ∈ ˙ M sup u ∈ ˙ X | b ( u , p ) |k p k k u k a ≤ √ , (2.8)where ˙ M = M \ { } and ˙ X = X \ { } . We note that if u ∈ X then ∇ · u ∈ M , i.e. thedivergence of u has zero mean; hence, V := ker B = { u ∈ X : ∇ · u = 0 } . (2.9)The Babu˘ska-Brezzi inf-sup condition ∃ β > p ∈ ˙ M u ∈ ˙ X | b ( u , p ) |k p k k u k a ≥ β (2.10)is precisely the condition required for B ′ to be an isomorphism onto its range withinverse bounded by k ( B ′ ) − k ≤ β − . Once we know the range of B ′ is closed, we maytake the polar of the equation ran( B ′ ) = ker( B ) = V to concluderan( B ′ ) = V = { f ∈ X ′ : h f , u i = 0 whenever u ∈ V } . (2.11)As V is naturally isomorphic to ( X/V ) ′ , we see that e B : X/V → M ′ : ( u + V ) B u is the adjoint of the composite map M B ′ −→ V ∼ = −→ ( X/V ) ′ , and is therefore itself anisomorphism with the same bound on the inverse. Identifying X/V with V ⊥ = { u ∈ X : a ( u , v ) = 0 whenever v ∈ V } , (2.12) JON WILKENING we learn that the restriction of B to V ⊥ is an isomorphism onto M ′ , which wouldbe essential to the analysis of the Stokes equations if the right hand side of (2.4b)were inhomogeneous. Other interesting solutions of B u = ϕ with ϕ ∈ M ′ (requiringe.g. u ∈ L ∞ (Ω) ∩ X or ∇ × u = 0 rather than u ∈ V ⊥ ) are studied in [1]. Finally,we define A : X → X ′ and ˜ A : V → V ′ via h A u , v i = a ( u , v ) , ( u , v ∈ X ) , h ˜ A u , v i = a ( u , v ) , ( u , v ∈ V ) . (2.13)Both are isometric isomorphisms in the k · k a norm.The weak solution ( u , p ) of (2.4) must satisfy u ∈ V so that B u = 0. But then A u + B ′ p = f requires ( ∗ ) ˜ A u = ˜ f , ( † ) B ′ p = f − A u , (2.14)where ˜ f = f | V ∈ V ′ and we note that ( f − A u ) ∈ range( B ′ ) = V iff u satisfies ( ∗ ).Since ˜ A and B ′ are isomorphisms onto their ranges, a unique solution of (2.4) existsand we have the estimates k u k a = k ˜ f k V ′ ≤ k f k X ′ = sup u ∈ ˙ X |h f , u i|k u k a , k p k ≤ β − k f k X ′ . (2.15)In summary, the inf-sup condition (2.10) is the key to analyzing the weak formulationof the Stokes equations — it is equivalent to the assertion that the gradient B ′ is anisomorphism from M = L (Ω) onto the polar set V of linear functionals in X ′ thatannihilate the divergence free vector fields u ∈ V .It is instructive to compare the inf-sup condition written in the form β k p k ≤ k B ′ p k X ′ = k∇ p k − ≤ √ k p k ( p ∈ L (Ω)) , (2.16)to the Poincar´e-Friedrichs inequality for mean-zero functions: k p k ≤ C k∇ p k ⇒ (1 + C ) − / k p k ≤ k∇ p k ≤ k p k ( p ∈ H (Ω)) . (2.17)Whereas (2.17) is easy to prove for p ∈ H (Ω) (with C = √ h in our case), it ismore challenging to prove for mean zero functions p ∈ H (Ω). The usual proof [2, 5]relies on Rellich’s theorem that H (Ω) is compactly embedded in L (Ω). As a result,the proof does not tell us how large the constant C might be or how it depends on Ω.Similarly, the usual proof [4] of (2.16) makes use of Rellich’s theorem that L (Ω) iscompactly embedded in H − (Ω) = H (Ω) ′ ; however, there is an added complicationnot present in proving (2.17): it must first be established that k p k ≤ C ( k p k − + k∇ p k − ) , ( p ∈ L (Ω)) . (2.18)This can be done in our case (if h ∈ C , ( T )) by flattening out the boundary andconstructing appropriate extension operators from H − (Ω) to H − ( T × R ) to reducethe problem to a case that can be solved using the Fourier transform; see Duvaut andLions [4] and also Nitsche [9], who used a similar technique to prove Korn’s inequality.In this paper, we show how to bypass (2.18) and prove (2.16) directly without invokingRellich’s theorem , which allows us to determine how the constant β depends on Ω. Wepresent two versions of the proof: one assuming h ∈ C , ( T ), and the other assumingonly that h ∈ C , ( T ), i.e. that h is a periodic, Lipschitz continuous function. Ourproof does rely on the boundary of Ω being the graph of a function h ( x ); however, wefeel this is a sufficiently important case to warrant a separate analysis. We sketch aproof of (2.17) that avoids Rellich’s theorem in Appendix B for comparison. NF-SUP ESTIMATES FOR THE STOKES PROBLEM IN A PERIODIC CHANNEL
3. A rectangular channel.
In the following theorem, we prove that B ′ in (2.7)is an isomorphism onto its range (cid:0) with β = min(1 , HL ) (cid:1) when Ω is the x -periodicrectangle R = T × (0 , H ) of height H . In Sections 4 and 5, we will transfer thisresult to a general x -periodic domain Ω by a change of variables. It is useful in thischange of variables to know that the constant C in Theorem 3.1 (and especially inCorollary 3.4) does not diverge as H approaches zero.The periodicity of the domain in one direction but not the other leads to aninteresting relationship between the inf-sup condition and the unitary operator map-ping the Fourier sine coefficients of a function of one variable to its Fourier cosinecoefficients. By studying this operator, we can obtain explicit estimates of β and itsdependence on L/H .Recall that every u ∈ H ( R ) must be zero (in the trace sense) on the top andbottom walls but not necessarily on the side walls, where it is only required to beperiodic. Such a function can be expanded in a sine or cosine series in the y -directionand differentiated term by term. (If u ∈ H ( R ) is not zero on the top and bottomwalls, only the cosine series can be differentiated term by term). Theorem 3.1.
For all q ∈ L ( R ) , k q k ≤ C k ∂ x q k − + C k ∂ y q k − , (3.1) where C = max (cid:16) , L H (cid:17) , C = 9 , and k f k − = sup u ∈ H ( R ) |h f,u i|k u k a .Proof . We may expand any q ∈ L ( R ) and u ∈ H ( R ) in a Fourier series q ( x, y ) = X n ∈ Z (cid:18) a n + ∞ X j =1 a nj √ πjyH (cid:19) e πinxL = X n ∈ Z (cid:18) ∞ X j =1 b nj √ πjyH (cid:19) e πinxL ,u ( x, y ) = X n ∈ Z (cid:18) c n + ∞ X j =1 c nj √ πjyH (cid:19) e πinxL = X n ∈ Z (cid:18) ∞ X j =1 d nj √ πjyH (cid:19) e πinxL so that k q k = X Z × N LH | a nj | = X Z × N LH | b nj | , (3.2) k u k a = X Z × N LH h(cid:16) πnL (cid:17) + (cid:16) πjH (cid:17) i | c nj | = X Z × N LH h(cid:16) πnL (cid:17) + (cid:16) πjH (cid:17) i | d nj | . (3.3)Here N = { }∪ N and the sums are over ordered pairs ( n, j ). Let us denote ( Z × N ) ′ = Z × N \ { (0 , } . We claim that A A k ∂ y q k − ↓ ց = k q k = X ( Z × N ) ′ LH (2 πn/L ) | a nj | (2 πn/L ) + ( πj/H ) + X Z × N LH ( πj/H ) | a nj | (2 πn/L ) + ( πj/H ) ≤ ≥ k q k = X Z × N LH (2 πn/L ) | b nj | (2 πn/L ) + ( πj/H ) + X Z × N LH ( πj/H ) | b nj | (2 πn/L ) + ( πj/H ) = տ ↑k ∂ x q k − B B (3.4) JON WILKENING
Here A , A , B and B are labels to represent the indicated sums. The horizontalassertions clearly hold (since q ∈ L ( R ) ⇒ a = 0) while the vertical assertions followfrom the Cauchy-Schwarz inequality and a particular choice of u to show that two ofthe upper bounds are least upper bounds: h ∂ x q, u i = Z R ( q )( − ∂ x u ) dA = X Z × N LH b − n,j (cid:16) − πinL d nj (cid:17) ≤ B / k u k a , h ∂ x q, u i = X ( Z × N ) ′ LH a − n,j (cid:16) − πinL c nj (cid:17) ≤ A / k u k a , h ∂ y q, u i = Z R ( q )( − ∂ y u ) dA = X Z × N LH a − n,j (cid:16) − πjH d nj (cid:17) ≤ A / k u k a , h ∂ y q, u i = X Z × N LH b − n,j (cid:16) πjH c nj (cid:17) ≤ B / k u k a ,d nj = (2 πin/L ) ¯ b − n,j (2 πn/L ) + ( πj/H ) ⇒ k u k a = B / , h ∂ x q, u i = B , (3.5) d nj = − ( πj/H ) ¯ a − n,j (2 πn/L ) + ( πj/H ) ⇒ k u k a = A / , h ∂ y q, u i = A . (3.6)The choices of c nj analogous to (3.5) and (3.6) do not generally lead to functions u that satisfy the boundary conditions on the top and bottom walls; hence, we cannotreplace the inequalities in (3.4) by equalities.The theorem will be proved if we can show that θ ( A + A ) + (1 − θ )( B + B ) ≤ C B + C A (3.7)for some θ ∈ [0 , θ , so weset θ = 1 here for simplicity. We will prove (3.7) by slicing the lattices ( Z × N ) ′ and Z × N into vertical strips and showing that A ,n ≤ C B ,n + ( C − A ,n , ( n ∈ Z ) , (3.8)where the subscript n indicates that only the terms in strip n should be included inthe sum, e.g. A , = P ∞ j =0 LH (6 π/L ) | a j | / [(6 π/L ) + ( πj/H ) ]. Since A , = 0,the n = 0 case holds trivially. If we freeze n ∈ Z \ { } , we find that1 L Z L q ( x, y ) e − πinxL dx = a n + ∞ X k =1 a nk √ πkyH = ∞ X j =1 b nj √ πjyH . (3.9)Thus, the coefficients a nk and b nj are related to each other by a unitary transformation a nk = ∞ X j =1 E kj b nj , ( n ∈ Z , k ≥ . (3.10)The entries of E can be computed explicitly: for j ≥ E kj = ( R √ πjη ) dη, k = 0 R πjη ) cos( πkη ) dη, k ≥ ) = √ / ( jπ ) , k = 0 , j odd4 j ( j − k ) π , k > , j − k odd0 , otherwise (3.11) NF-SUP ESTIMATES FOR THE STOKES PROBLEM IN A PERIODIC CHANNEL n ∈ Z \ { } frozen and dividing (3.8) by LH , we must show that P ∞ k =0 (2 πn/L ) | a nk | (2 πn/L ) +( πk/H ) ≤ C P ∞ j =1 (2 πn/L ) | b nj | (2 πn/L ) +( πj/H ) +( C − P ∞ k =1 ( πk/H ) | a nk | (2 πn/L ) +( πk/H ) . This is accomplished via the following lemma using ν = 2 | n | H/L and ν = 2 H/L . Lemma 3.2.
Suppose b ∈ ℓ ( N ) and let a = Eb ∈ ℓ ( N ) , where E maps theFourier sine coefficients of a function to its Fourier cosine coefficients; see (3.11)above. Then for ν > there holds ∞ X k =0 ν ν + k | a k | ≤ C ∞ X j =1 ν ν + j | b j | + ( C − ∞ X k =1 k ν + k | a k | , (3.12) with C = max (cid:0) , ν − (cid:1) and C = 9 . If ν ≥ ν > , C = max (cid:0) , ν − (cid:1) also works. Proof . It suffices to show that (3.12) holds whenever b is a unit vector in ℓ ( N ).The general case follows by re-scaling this result. We will split each sum into terms oflow and high index and use different arguments to handle the two cases. Let k ≥ j ≥ k = k + 1 and j = j + 1. If we discard terms on the right hand side with j ≥ j and k ≤ k , we obtain a sufficient condition for (3.12) to hold. Also, on theleft hand side, P ∞ k = k ν ν + k | a k | ≤ ν k P ∞ k = k k ν + k | a k | , so it suffices to show that k X k =0 ν ν + k | a k | ≤ C j X j =1 ν ν + j | b j | + (cid:18) C − − ν k (cid:19) ∞ X k = k k ν + k | a k | . (3.13)Next, we see that (3.13) will hold if we can show that α ≤ C ν ν + j β + ( C − − ν /k ) k ν + k (1 − α ) , (3.14)where α = P k k =0 | a k | , β = P j j =1 | b j | , and 1 − α = P ∞ k = k | a k | . Note that β here is not the inf-sup constant β , but rather a measure of the relative weight of lowfrequency modes in comparison to high frequency modes in a sine series expansion.Solving (3.14) for α , we require α ≤ C C ν /k j /ν β + 1 − ν /k C . (3.15)Our goal is to show that for each ν > j ≥ k ≥ C ≤ max(9 , (9 / ν − ) and C ≤ b ∈ ℓ ( N ); ( b determines a , α and β ). C and C can then be increasedif necessary to the values stated in the lemma without violating (3.12).We now use the fact that a and b are unit vectors related by a known unitarytransformation to obtain a bound on α in terms of β . Let S , T , x , y , z be thesub-matrices and sub-vectors S = E (0 : k , j ) , T = E (0 : k , j : ∞ ) , E (0 : k , :) = [ S, T ] .z = a (0 : k ) , x = b (1 : j ) , y = b ( j : ∞ ) , z = Sx + T y. (3.16)We have α = k z k and β = k x k = p − k y k . Since k S k ≤ k y k ≤
1, theestimate k z k ≤ k Sx k + k T y k gives α ≤ β + t, t = k T k ≤ . (3.17) JON WILKENING If t <
1, this can be used to derive a bound on α of the form (3.15). However,we can obtain a sharper estimate as follows. First, we compute the singular valuedecomposition S = U Σ V ∗ and rotate the rows of T by a unitary operator Q suchthat U ∗ [ S, T ] (cid:20) V Q (cid:21) = σ · · · t · · · . . . ... ... . . . ...0 σ k · · · t k · · · , (3.18)where σ k + t k = 1 for 0 ≤ k ≤ k . We assume here that j ≥ k + 1; otherwise we willnot be able to derive a sufficient condition for (3.15) to hold, for if S has more rowsthan columns, we can produce a unit vector a = [ z ; 0] with S ∗ z = 0 so that b = E ∗ a yields α = 1 and β = 0. Next we define ˜ z = U ∗ z , ˜ x = V ∗ x , ˜ y = Q ∗ y so that α = k X k =0 | ˜ z k | , β = j X j =1 | ˜ x j | , − β = ∞ X j =1 | ˜ y j | , ˜ z k = σ k ˜ x k +1 + t k ˜ y k +1 (3.19)and, by Lemma A.1 below, | ˜ z k | ≤ − t k | σ k ˜ x k +1 | + 1 t k | t k ˜ y k +1 | = (1 + t k ) | ˜ x k +1 | + t k | ˜ y k +1 | . (3.20)Hence, majorizing t k by k T k = t max = p − σ and summing over k , we obtain α ≤ (1 + t ) β + t (1 − β ) = β + t, t = k T k ≤ . (3.21)Thus, (3.15) holds if we define C and C via (cid:16) − ν /k C (cid:17) = t and (cid:16) C C ν /k j /ν (cid:17) = 1: C ( ν ) = 1 + j /ν − t , C ( ν ) = 1 + ν /k − t . (3.22)Next we look for choices of k and j that lead to a window of values of ν over which C and C remain small. We need enough such windows to cover the positive realline ν >
0. The trade-off is that choosing j ≫ k makes t small but also makes oneof the numerators in (3.22) large. We consider 3 cases: • Case 1: (0 < ν ≤ / k = j = 1 so that t = p − E = . C ( ν ) = (1 + ν ) / (1 − t ) ≤ (5 / / (1 − t ) = 2 . ≤ / ,C ( ν ) = (1 + ν − ) / (1 − t ) = C ν − ≤ (9 / ν − , (0 < ν ≤ / . (3.23) • Case 2: (1 / ≤ ν ≤ S for all pairs of small integers k and j satisfying k ≤ j ≤ k ≤
400 to determine t = p − σ for each pair. We then choose a threshold C thresh and find the values ν min and ν max such that C ( ν min ) = C thresh and C ( ν max ) = C thresh . We then discard all cases with ν max < ν min and sort the remaining intervals[ ν min , ν max ] by their first entry. Finally, we discard all intervals for which ν min of thenext interval is smaller than ν max of the previous interval (to avoid redundancy). Theresults with C thresh = 4 . C thresh = 8 . C thresh < . NF-SUP ESTIMATES FOR THE STOKES PROBLEM IN A PERIODIC CHANNEL C ( ν ) and C ( ν ) with C thresh = 8 .
9. The correspond-ing table with C thresh = 5 . k j t ν min ν max ( ν ) and C ( ν ) ν C ( ν )C ( ν ) max(9,(9/4) ν −2 ) C thresh = 5.9C thresh = 8.9 Fig. 3.1: Plot of C ( ν ) and C ( ν ) over the range0 . ≤ ν ≤ ν min ≤ ν ≤ ν max in Table 3.1. • Case 3: ( ν ≥ k = ⌊ ν/ √ ⌋ , j = 3 k and bound t by the Frobeniusnorm: t ≤ k T k F = k X k =0 ∞ X j = j | E kj | = 8 π ∞ X j = j δ j, odd j + 16 π k X k =1 ∞ X j = j j δ j − k, odd ( j − k ) (3.24) ≤ π Z ∞ j − x dx + 8 π k Z ∞ j − x ( x − k ) dx, ( j − j − π ( j −
1) + 4 π (cid:20) κκ − κ + 1 κ − (cid:21) , (cid:18) κ = j − k > j k = 3 (cid:19) ≤ π + 4 π (cid:20)
38 + 12 log 2 (cid:21) = ( . , (cid:18) k ≥ (cid:22) √ (cid:23) = 57 , j ≥ (cid:19) . Here we represent sums of decreasing functions sampled at even or odd integers bystaircases of width two and half the height of the function at the right endpoint. Eachchoice of k and j will cover the range √ k ≤ ν < √ k + 1); over this range, wehave νk ≤ (cid:18) k + 1 k (cid:19) (cid:18) νk + 1 (cid:19) ≤ √ , j ν ≤ (cid:18) j k (cid:19) (cid:18) k ν (cid:19) ≤ (3) 1 √ √ C and C are bounded by / − . = 8 . ν > C ≤ max(9 , (9 / ν − ) and C ≤
9, as claimed.
Corollary 3.3.
For all q ∈ L ( R ) , k q k − ≤ L π (cid:13)(cid:13) ∂ x q (cid:13)(cid:13) − + H π (cid:13)(cid:13) ∂ y q (cid:13)(cid:13) − .Proof . Arguing as in (3.4)–(3.6), it is readily shown that k q k − = X Z × N LH | b nj | (2 πn/L ) + ( πj/H ) ≤ L π X Z × N LH (2 πn/L ) | b nj | (2 πn/L ) + ( πj/H ) + ∞ X j =1 LH | b j | ( πj/H ) . The first term on the right hand side is simply L π (cid:13)(cid:13) ∂ x q (cid:13)(cid:13) − while the second satisfies ∞ X j =1 LH | b j | ( πj/H ) ≤ H π ∞ X j =1 LH | b j | = H π ∞ X j =1 LH | a j | ≤ H π (cid:13)(cid:13) ∂ y q (cid:13)(cid:13) − , (3.26)0 JON WILKENING where the middle equality follows from the fact that a = 0. Corollary 3.4.
Suppose q ∈ L ( R ) and ζ ∈ L ∞ ( T ) . Then for any aspect ratio H/L , we have k ∂ y ( ζq ) k − ≤ C M (cid:0) k ∂ x q k − + k ∂ y q k − (cid:1) , (3.27) where C = 9 and M = k ζ k ∞ .Proof . Since ζ does not depend on y , the Fourier coefficients of ˜ q = ζq are relatedto the those of q via column-by-column convolution with the Fourier coefficients of ζ :˜ a nk = X m ∈ Z ˆ ζ n − m a mk , ˜ b nj = X m ∈ Z ˆ ζ n − m b mj , ( n ∈ Z , k ≥ , j > . (3.28)Since multiplication by ζ is bounded in L ( T ) by M , convolution with ˆ ζ is boundedin ℓ ( Z ) by M . Thus, by (3.4), we have k ∂ y ( ζq ) k − ≤ LH X k> X n ∈ Z | ˜ a nk | ≤ M LH X k> X n ∈ Z | a nk | ! . (3.29)The key point is that entries a nk with k = 0 are absent from the right hand side. Thequantity in parentheses may be written as A + A just as in (3.4), but omitting the k = 0 terms from A . Thus, it suffices to show that (3.12) holds with C replaced by C if the k = 0 term is omitted from the sum on the left. For ν ≥
1, the result hasalready been proved without omitting this term. But for ν <
1, we see that C = 0and C = 2 suffice (since ν ≤ k for k ≥ C = C = 9 works for all ν >
4. Curved boundaries.
We now perform a change of variables to transfer theresult of Theorem 3.1 from the rectangle R to a domain Ω bounded on one side by aperiodic, Lipschitz continuous function h ∈ C , ( T ): H LR, ξ, η F Lh Ω , x, y x = ξ, y = h ( ξ ) H η, dx dy = h ( ξ ) H dξ dη,ξ = x, η = Hh ( x ) y, dξ dη = Hh ( x ) dx dy,∂∂x = ∂∂ξ − ηh h x ∂∂η , ∂∂y = Hh ∂∂η , ∂∂ξ = ∂∂x + yh h x ∂∂y , ∂∂η = hH ∂∂y . (4.1)The main challenges involve avoiding lower order terms that have to be dealt withusing Rellich’s compactness theorem, balancing the sources of error to avoid excessiveoverestimation of the constants in the error bounds, and dealing with various subtletiesof the dual space H − (Ω) such as the fact that if p ∈ L (Ω) and ζ ∈ L ∞ (Ω) then k ζp k − need not be smaller than k ζ k ∞ k p k − . For clarity, we postpone the case that h is only Lipschitz continuous to Section 5 and begin with the simplifying assumption h ∈ C , ( T ). The aspect ratio of the rectangle R plays an essential role in the Lipschitzcase but only a minor role (improving our estimate of β ) here. Theorem 4.1.
Suppose h ∈ C , ( T ) and < h ≤ h ( x ) ≤ h for ≤ x ≤ L .Then for every p ∈ L (Ω) we have (cid:13)(cid:13) p (cid:13)(cid:13) , Ω ≤ β − (cid:13)(cid:13) ∇ p (cid:13)(cid:13) − , Ω , β − = 94 (cid:0) M (cid:1) (cid:18) h h (cid:19) / max (cid:18) , Lh , h h (cid:19) , (4.2) NF-SUP ESTIMATES FOR THE STOKES PROBLEM IN A PERIODIC CHANNEL where M = max (cid:16)(cid:13)(cid:13) h x (cid:13)(cid:13) ∞ , (cid:13)(cid:13) hh xx (cid:13)(cid:13) ∞ (cid:17) . Remark 4.2.
The quantity hh xx arises naturally in the study of Reynolds’lubrication approximation and its higher order corrections on a periodic domain [12]. Remark 4.3.
In many practical applications, the aspect ratio
L/h is large while M ≪ h /h ≈
1; in this regime, (4.2) shows that β − scales linearly with L/h .If the geometry has a narrow gap so that h /h ≫
1, we learn that β − dependson the gap size as h − / . This dependence is shown to be optimal in Example 4.5below. We do not know if the quadratic dependence on M is optimal; it seems to bean unavoidable artifact of changing variables to a rectangular geometry. Proof of Theorem 4.1 . The coordinate transformation ( x, y ) = F ( ξ, η ) defined in(4.1) provides a one-to-one correspondence between functions p ∈ L (Ω), u ∈ H (Ω)and their counterparts ˜ p = p ◦ F ∈ L ( R ), ˜ u = u ◦ F ∈ H ( R ):˜ p ( ξ, η ) = p (cid:18) ξ, h ( ξ ) H η (cid:19) , ˜ u ( ξ, η ) = u (cid:18) ξ, h ( ξ ) H η (cid:19) . (4.3) F does not map L (Ω) to L ( R ); however, the norm of p ∈ L (Ω) does not decreaseif we add a constant to p to enforce R Ω h − p dA = 0 instead of R Ω p dA = 0. ByTheorem 3.1, this new p satisfies (cid:13)(cid:13) p (cid:13)(cid:13) , Ω ≤ (cid:13)(cid:13)(cid:13)(cid:16) h h (cid:17) / p (cid:13)(cid:13)(cid:13) , Ω = h H (cid:13)(cid:13) ˜ p (cid:13)(cid:13) ,R ≤ C h H (cid:13)(cid:13) ∂ ξ ˜ p (cid:13)(cid:13) − ,R + C h H (cid:13)(cid:13) ∂ η ˜ p (cid:13)(cid:13) − ,R , (4.4)where C = max(16 , L H ) and C = 9. But since the right hand side does not changewhen a constant is added to ˜ p , the original p also satisfies this equation (dropping theintermediate inequalities). We can relate the action of ∇ ξ ˜ p on ˜ u to that of ∇ x p on u : h ∂ ξ ˜ p, ˜ u i R = (cid:28) Hh p, (cid:16) − ∂ x − yh h x ∂ y (cid:17) u (cid:29) Ω = H (cid:10) ∂ x p, h − u (cid:11) Ω + H D ∂ y p, h x yh u E Ω , (4.5) h ∂ η ˜ p, ˜ v i R = (cid:28) Hh p, − hH v y (cid:29) Ω = h ∂ y p, v i Ω , (4.6)where we used ∂ x ( h − ) + ∂ y ( yh − h x ) = 0 in (4.5). If we had not introduced thefactor of h − / in (4.4), this cancellation would not have occurred and the proofwould become much more complicated; see Remark 4.4 below. It will be shown inLemmas A.3, A.4 and A.5 that H (cid:13)(cid:13) h − u (cid:13)(cid:13) a, Ω ≤ C (cid:13)(cid:13) ˜ u (cid:13)(cid:13) a,R , C = max (cid:18) Hh , (cid:0) M (cid:1) H h (cid:19) , (4.7) H (cid:13)(cid:13)(cid:13) h x yh u (cid:13)(cid:13)(cid:13) a, Ω ≤ C (cid:13)(cid:13) ˜ u (cid:13)(cid:13) a,R , C = max (cid:18) M Hh , (cid:0) M + 6 M (cid:1) H h (cid:19) , (4.8) (cid:13)(cid:13) v (cid:13)(cid:13) a, Ω ≤ C (cid:13)(cid:13) ˜ v (cid:13)(cid:13) a,R , C = max (cid:18) h H , (cid:0) M (cid:1) Hh (cid:19) . (4.9)If h only belongs to C , ( T ), then (4.8) does not hold and we have to replace the lastterm in (4.5) by (cid:10) ∂ y ( h x p ) , Hyh − u (cid:11) Ω , which requires a more difficult analysis; seeSection 5 below. Combining (4.5)–(4.9), we obtain (cid:12)(cid:12) h ∂ ξ ˜ p, ˜ u i R (cid:12)(cid:12) ≤ (cid:16) C (cid:13)(cid:13) ∂ x p (cid:13)(cid:13) − , Ω + C (cid:13)(cid:13) ∂ y p (cid:13)(cid:13) − , Ω (cid:17)(cid:13)(cid:13) ˜ u (cid:13)(cid:13) a,R , (cid:12)(cid:12) h ∂ η ˜ p, ˜ v i R (cid:12)(cid:12) ≤ C (cid:13)(cid:13) ∂ y p (cid:13)(cid:13) − , Ω (cid:13)(cid:13) ˜ v (cid:13)(cid:13) a,R . (4.10)2 JON WILKENING
It follows that (cid:13)(cid:13) ∂ ξ ˜ p (cid:13)(cid:13) − ,R ≤ C (cid:13)(cid:13) ∂ x p (cid:13)(cid:13) − , Ω + 32 C (cid:13)(cid:13) ∂ y p (cid:13)(cid:13) − , Ω , (cid:13)(cid:13) ∂ η ˜ p (cid:13)(cid:13) − ,R ≤ C (cid:13)(cid:13) ∂ y p (cid:13)(cid:13) − , Ω , (4.11)which, together with (4.4), gives (cid:13)(cid:13) p (cid:13)(cid:13) , Ω ≤ β − (cid:16)(cid:13)(cid:13) ∂ x p (cid:13)(cid:13) − , Ω + (cid:13)(cid:13) ∂ y p (cid:13)(cid:13) − , Ω (cid:17) , β − = h H max (cid:18) C C , C C + C C (cid:19) . Next, we choose H = h so that3 C C ≤ (cid:0) M (cid:1) C , C C ≤ (cid:0) M + 9 M (cid:1) C , C C ≤ h h + 18 M . Finally, we observe that h h ≤ max (cid:16) , h h (cid:17) regardless of whether h h ≥
4. As aresult, C C ≤ (8 + 2 M ) max (cid:16) , h h (cid:17) and β − ≤ h h max (cid:8) M ) , M + 9 M (cid:9)
916 max (cid:18) , L h , h h (cid:19) , (4.12)which yields (4.2) when we majorize the terms in braces by 9(1 + M ) . Remark 4.4.
One might hope to improve (4.2) by working directly with k p k in (4.4) instead of via (cid:13)(cid:13) h − / p (cid:13)(cid:13) . The main difference is that (4.5) acquires a lowerorder term (cid:10) ∂ ξ ( h / ˜ p ) , ˜ u (cid:11) R = H (cid:10) ∂ x p, h − / u (cid:11) Ω + H (cid:10) ∂ y p, yh x h − / u (cid:11) Ω + H (cid:10) p, h x h − / u (cid:11) Ω that would normally be dealt with by invoking a compactness argument to bound k p k − , Ω by a constant times k∇ p k − , Ω . This is not acceptable in the current calcula-tion as this constant depends on Ω, and hence h . It is possible to bound k p k − , Ω interms of k ˜ p k − ,R and then use Corollary 3.3. But the final step of bounding k∇ ξ ˜ p k − ,R by k∇ x p k − , Ω brings us back to the proof given above. The following example showsthat the power of h − / in the formula (4.2) for β − is the best possible. Example 4.5.
Suppose 0 < h < h ( x ) thattransitions smoothly and symmetrically between h for x ∈ [3 / , / ∪ [7 / ,
1] and 1for x ∈ [1 / , / ∪ [5 / , / , Ω , Ω , and Ω be the regions under the curve h with x ∈ [0 , / x ∈ [3 / , / x ∈ [1 / , /
8] and x ∈ [7 / , p ( x, y ) be the continuous, piecewise linear function that equals − , 1 on Ω ,and satisfies p x = ± p y = 0 on Ω and Ω . Ω h
38 7812 p x = 16 p x = − Ω Ω p = − p = 1 Then for any u ∈ H (Ω), we have (cid:12)(cid:12) h ∂ y p, u i (cid:12)(cid:12) = 0 and (cid:12)(cid:12) h ∂ x p, u i (cid:12)(cid:12) ≤ Z Ω ∪ Ω | u ( x, y ) | dA ≤ p area(Ω ∪ Ω ) k u k , Ω ∪ Ω ≤ h / (cid:0) h / √ (cid:1) k u y k , Ω ∪ Ω ≤ √ h / k u k a, Ω , (4.13) NF-SUP ESTIMATES FOR THE STOKES PROBLEM IN A PERIODIC CHANNEL (cid:13)(cid:13) ∇ p (cid:13)(cid:13) − , Ω ≤ √ h / while k p k , Ω ≥ /
2, showing that β − in (4.2) mustbe at least (cid:0) √ (cid:1) − h − / , i.e. the power h − / is optimal.
5. Lipschitz boundaries.
In this section we show how to modify the proof ofTheorem 4.1 to handle the case that h only belongs to C , ( T ). The main differenceis that yh − h x u no longer belongs to H (Ω) in (4.5), so a different strategy is requiredto deal with the term (cid:10) ∂ y p, Hyh − h x u (cid:11) Ω . The idea is to show that when h − h x isgrouped with p , this term can be made small in comparison to the other two termsin (4.5) by choosing the aspect ratio of the rectangle R small enough. The loss of apower of h / in the estimate of β − when M is not small is discussed in Remark 5.3below. Theorem 5.1.
Suppose h ∈ C , ( T ) and < h ≤ h ( x ) ≤ h for ≤ x ≤ L .Then for every p ∈ L (Ω) we have (cid:13)(cid:13) p (cid:13)(cid:13) , Ω ≤ β − (cid:13)(cid:13) ∇ p (cid:13)(cid:13) − , Ω , β − = 2 max (cid:18) , L √ h h , Lh M (cid:19) max(1 , M ) h h , (5.1) where M = k h x k ∞ .Proof . As before, (4.4) holds for all p ∈ L (Ω): (cid:13)(cid:13) p (cid:13)(cid:13) , Ω ≤ C h H (cid:13)(cid:13) ∂ ξ ˜ p (cid:13)(cid:13) − ,R + C h H (cid:13)(cid:13) ∂ η ˜ p (cid:13)(cid:13) − ,R , C = 916 max (cid:18) , L H (cid:19) , C = 9 . (5.2)We now transform the problematic term in (4.5) back to the ξ, η coordinate system: h f, ˜ u i R − h g , ˜ u i R := h ∂ ξ ˜ p, ˜ u i R − (cid:10) ∂ η ( h − h x ˜ p ) , η ˜ u (cid:11) R = (cid:10) ∂ x p, Hh − u (cid:11) Ω , (5.3) h g, ˜ v i R := h ∂ η ˜ p, ˜ v i R = h ∂ y p, v i Ω . (5.4)So we can bound k p k , Ω in terms of f and g and we can bound ( f − g ) and g in termsof k∇ x p k − , Ω ; thus, we need a bridge from f to ( f − g ) and g . By Corollary 3.4 andLemma A.6, | h g , ˜ u i R | ≤ k ∂ η ( h − h x ˜ p ) k − ,R k η ˜ u k a,R ≤ (cid:16) M h − k∇ ξ ˜ p k − ,R (cid:17) (cid:18) H k ˜ u k a,R (cid:19) , ⇒ k g k − ,R ≤ θ (cid:16) k f k − ,R + k g k − ,R (cid:17) , θ = 4 Hh M. (5.5)As a result, k f k ≤ k f − g k + 2 θ ( k f k + k g k ), which implies k f k ≤ k f − g k + 4 θ k g k , ( θ ≤ / . (5.6)Equation (5.2) now becomes k p k , Ω ≤ C h H k f − g k − ,R + (4 θ C + C ) h H k g k − ,R , ( HM ≤ h / . (5.7)From (5.3) and (5.4) we see that | h f − g , ˜ u i | ≤ k ∂ x p k − , Ω k Hh − u k a, Ω , | h g, ˜ v i | ≤ k ∂ y p k − , Ω k v k a, Ω . (5.8)4 JON WILKENING
By Lemmas A.3 and A.5 below, we then have k f − g k − ,R ≤ C k ∂ x p k − , Ω , C = max (cid:18) Hh , (1 + 16 M ) H h (cid:19) , (5.9) k g k − ,R ≤ C k ∂ y p k − , Ω , C = max (cid:18) h H , (1 + 9 M ) Hh (cid:19) . (5.10)It follows from (5.7) that (cid:13)(cid:13) p (cid:13)(cid:13) , Ω ≤ β − (cid:16)(cid:13)(cid:13) ∂ x p (cid:13)(cid:13) − , Ω + (cid:13)(cid:13) ∂ y p (cid:13)(cid:13) − , Ω (cid:17) , β − = h H max (cid:0) C C , (4 θ C + C ) C (cid:1) . Finally, we choose H = min (cid:0) h , M h (cid:1) so that if M ≥ / h H C ≤ max „ , M + 1 « h h ≤ h h , h H C ≤ max „ h h M , (1 + 9 M ) h h « ≤ M h h and if M ≤ / h H C ≤ max „ , M « h h ≤ h h , h H C ≤ max „ h h , (1 + 9 M ) h h « ≤ h h . Moreover, C = max (cid:16) , L h , L h M (cid:17) and 4 θ C + C ≤ (cid:16) , L h M (cid:17) re-gardless of whether M ≤ /
8. Combining these results, we obtain4 C h H C ≤ (cid:18) , L h , L h M (cid:19) h h , (4 θ C + C ) h H C ≤ (cid:18) , L h M (cid:19) max(1 , M ) h h . (5.11)Formula (5.1) for β − follows by taking the square root of the maximum of theseexpressions after increasing the constants and consolidating terms. Remark 5.2.
Inequality (5.5) is the key to this proof. For fixed u , both h f, ˜ u i and h g , ˜ u i in (5.3) scale like H while h g, ˜ u i in (5.4) is independent of H . Because ofthe way k ˜ u k a,R depends on H , it follows that if R = T × H , R = T × H , and H < H , then k g k − ,R ≤ H H k g k − ,R , k f k − ,R ≤ H H k f k − ,R , k g k − ,R ≤ H H k g k − ,R . Thus, k g k and θ k g k are both O ( H ) quantities and the surprising aspect of (5.5)is that the O ( H ) term θ k f k is sufficient to help θ k g k bound k g k . Remark 5.3.
We believe the optimal bound in the Lipschitz case should scale like β − ∼ h − / , just as in the C , case; however, proving this would require eliminating(or at least finding a better bound for) the cross term 4 θ C h H k g k − ,R in (5.7). As itstands, θ C and H − k g k each contribute a factor of h − to this cross term due tothe requirement HM ≤ h /
8, which yields β − ∼ h − . We suspect that the functions p that require C to diverge as H → p for which k f − g k ≪ k f k in (5.6), but we have not found a way to make this idea rigorous. Appendix A. Useful Lemmas.
In this section we gather several results thatare either elementary but used frequently in our proofs or are tedious and distractfrom the main argument.
Lemma A.1.
Suppose γ , . . . , γ n are positive real numbers such that P n γ − j ≤ .Then | w + · · · + w n | ≤ P nj =1 γ j | w j | , for all w ∈ C n . NF-SUP ESTIMATES FOR THE STOKES PROBLEM IN A PERIODIC CHANNEL Proof . This is a consequence of the Cauchy-Schwarz inequality: (cid:12)(cid:12)(cid:12)P j w j (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)P j (cid:16) γ − / j (cid:17) (cid:16) γ / j w j (cid:17)(cid:12)(cid:12)(cid:12) ≤ (cid:16)P j γ − j (cid:17) (cid:16)P j γ j | w j | (cid:17) . (A.1) Lemma A.2. (Poincar´e-Friedrichs inequality). If Ω has the geometry of Fig-ure 2.1 with h ∈ C , ( T ) and if R is the x -periodic rectangle of width L and height H , then k u k , Ω ≤ h k u y k , Ω , k ˜ u k , Ω ≤ H π k ˜ u y k ,R , (cid:16) u ∈ H (Ω) , ˜ u ∈ H ( R ) (cid:17) . (A.2) The former inequality also works over the subregion Ω ∪ Ω in Example 4.5 with h replaced by the maximum height of that subregion, namely h .Proof . The latter inequality follows by expanding ˜ u = P d nj √ πinxL ) sin πjyH and comparing the formulas for k u k ,R and k u y k ,R . If u ∈ C c (Ω), the former in-equality follows by integrating | u ( x, y ) | ≤ (cid:12)(cid:12)R y u y ( x, y ′ ) dy ′ (cid:12)(cid:12) ≤ y R h/ | u y ( x, y ′ ) | dy ′ , (cid:0) ≤ y ≤ h ( x ) (cid:1) | u ( x, y ) | ≤ (cid:12)(cid:12)(cid:12)R hy u y ( x, y ′ ) dy ′ (cid:12)(cid:12)(cid:12) ≤ ( h − y ) R hh/ | u y ( x, y ′ ) | dy ′ , (cid:0) h ( x ) ≤ y ≤ h ( x ) (cid:1) over the lower and upper halves of Ω, respectively, and combining the results. Theresult for u ∈ H (Ω) then follows by a standard density argument. Lemma A.3.
Suppose h ∈ C , ( T ) , R = T × H and u ∈ H (Ω) . Then (cid:13)(cid:13) Hh − u (cid:13)(cid:13) a, Ω ≤ C k ˜ u k a,R , (A.3) where ˜ u expresses u in the ξ , η coordinate system of R and C is given by (4.7) or(5.9).Proof . Using the change of variables formulas (4.1) and (4.3), we obtain (cid:13)(cid:13) ∂ x ( Hh − u ) (cid:13)(cid:13) , Ω = Z R H (cid:16) − h − h x ˜ u + h − ˜ u ξ − h − ηh h x ˜ u η (cid:17) hH dξ dη ≤ γ Hh − M k ˜ u k ,R + γ Hh − k ˜ u ξ k ,R + γ H h − M k ˜ u η k ,R , (A.4) (cid:13)(cid:13) ∂ y ( Hh − u ) (cid:13)(cid:13) , Ω = Z R H h (cid:16) Hh ˜ u η (cid:17) hH dξ dη ≤ H h − k ˜ u y k ,R , (A.5)where M = k h x k ∞ , h = min ≤ x ≤ L h ( x ), and γ − + γ − + γ − ≤
1. Combining theseand using the Poincar´e-Friedrichs inequality (with 9 instead of π ), we find that (cid:13)(cid:13) Hh − u (cid:13)(cid:13) a, Ω ≤ max (cid:18) γ Hh , (cid:16) (cid:16) γ + γ (cid:17) M (cid:17) H h (cid:19) k ˜ u k a,R , (A.6)which yields (4.7) with ~γ = (9 , ,
2) and (5.9) with ~γ = (36 , / , Lemma A.4.
Suppose h ∈ C , ( T ) , R = T × H and u ∈ H (Ω) . Then (cid:13)(cid:13) Hh − yh x u (cid:13)(cid:13) a, Ω ≤ C k ˜ u k a,R , C = max (cid:18) M Hh , (cid:0) M + 6 M (cid:1) H h (cid:19) , (A.7)6 JON WILKENING where M = max (cid:16)(cid:13)(cid:13) h x (cid:13)(cid:13) ∞ , (cid:13)(cid:13) hh xx (cid:13)(cid:13) ∞ (cid:17) .Proof . Using the change of variables formulas (4.1) and (4.3) as well as thePoincar´e-Friedrichs inequality, we obtain (cid:13)(cid:13)(cid:13) ∂ x (cid:16) Hyh x uh (cid:17)(cid:13)(cid:13)(cid:13) , Ω = Z R h − ηh x h ˜ u + 2 ηh (cid:16) hh xx (cid:17) ˜ u + ηh x h (cid:16) ˜ u ξ − ηh x h ˜ u η (cid:17)i hH dξ dη ≤ γ + γ ) HM h k ˜ u k ,R + γ HM h k ˜ u ξ k ,R + γ H M h k ˜ u η k ,R , ≤ γ HM h k ˜ u ξ k ,R + (cid:16) γ + 4 π ( γ + γ ) (cid:17) H M h k ˜ u η k ,R (A.8) (cid:13)(cid:13)(cid:13) ∂ y (cid:16) Hyh x uh (cid:17)(cid:13)(cid:13)(cid:13) , Ω = Z R (cid:16) Hh x h ˜ u + Hηh x h ˜ u η (cid:17) hH dξ dη (A.9) ≤ δ HM h k ˜ u k ,R + δ H M h k ˜ u η k ,R ≤ (cid:16) δ δ (cid:17) H M h k ˜ u η k ,R , where P γ − j ≤ δ − + δ − ≤
1. Now we set ~γ = (cid:0) π , π , , (cid:1) and ~δ = (cid:0) , (cid:1) to obtain (A.7). Lemma A.5.
Suppose h ∈ C , ( T ) , R = T × H and v ∈ H (Ω) . Then k v k a, Ω ≤ C k ˜ v k a,R with C given by (4.9) or (5.10).Proof . Let M = k h x k ∞ . For any γ , γ satisfying γ − + γ − ≤
1, we have k ∂ x v k , Ω = Z R (cid:16) ˜ v ξ − ηh x h ˜ v η (cid:17) hH dξ dη ≤ γ h H k ˜ v ξ k ,R + γ HM h k ˜ v η k ,R , (A.10) k ∂ y v k , Ω = Z R (cid:16) Hh ˜ v η (cid:17) hH dξ dη ≤ Hh k ˜ v η k ,R . (A.11)It follows that k v k a, Ω ≤ C k ˜ v k a,R with C = max (cid:0) γ h H , (1 + γ M ) Hh (cid:1) . We obtain(4.9) using ~γ = (2 ,
2) and (5.10) using ~γ = (9 / , Lemma A.6.
On the ξ -periodic rectangle R , k η ˜ u k a ≤ H k ˜ u k a , (cid:0) ˜ u ∈ H ( R ) (cid:1) . (A.12) Proof . Using the Poincar´e-Friedrichs inequality, we find that k ∂ ξ ( η ˜ u ) k ,R ≤ H k ˜ u ξ k ,R , k ∂ η ( η ˜ u ) k ,R ≤ (cid:16) γ γ (cid:17) H k ˜ u η k ,R (A.13)provided γ − + γ − ≤
1. Choosing ~γ = (4 , / Appendix B. The Poincar´e-Friedrichs inequality on H (Ω) . In this sectionwe present a simple proof of the Poincar´e-Friedrichs inequality for H functions withzero mean. Our proof does not rely on Rellich’s compactness theorem, but doesrequire the boundary of Ω to be the graph of a Lipschitz continuous function h ; seeFigure 2.1 above. The main difference between the estimates k p k , Ω ≤ K k∇ p k , Ω , (cid:0) p ∈ H (Ω) (cid:1) , k p k , Ω ≤ β − k∇ p k − , Ω , (cid:0) p ∈ L (Ω) (cid:1) (B.1)proved below and in Theorems 4.1 and 5.1 above is that K ∼ h − / while β − ∼ h − / ;(we were only able to prove β − ∼ h − in the Lipschitz case). A narrow gap causes K NF-SUP ESTIMATES FOR THE STOKES PROBLEM IN A PERIODIC CHANNEL p in the gap region can lead to a large change in p across the gap with relatively little cost (in terms of k∇ p k , Ω ) due to the small areaof the gap region. The effect on β − is more severe than on K because, in additionto the small area of the gap region, the test functions ( u, v ) that ∇ p acts on belongto H (Ω) , i.e. they are zero on Γ and Γ . These boundary conditions cause u and v to be small in the gap region, which reduces their ability to penalize large gradientsof p there. This was illustrated in Example 4.5 above.To keep the equations dimensionally correct, we define the norm on H (Ω) to be k p k , Ω = L − k p k , Ω + k p k a, Ω = Z Ω | p | L + | p x | + | p y | dx dy, (B.2)i.e. we use L as a length scale to compare k p k to k p k a = k∇ p k . Theorem B.1.
Suppose h ∈ C , ( T ) and < h ≤ h ( x ) ≤ h for ≤ x ≤ L .Then for every p ∈ H (Ω) , we have L − k p k , Ω ≤ C k∇ p k , Ω , C = 1 + M π max (cid:18) , √ h h L (cid:19) r h h , (B.3) where M = k h x k ∞ . The constant K in (B.1) is given by K = (1 + C ) / .Proof . On the ξ -periodic rectangle R = T × (0 , H ), the expansion˜ p ( ξ, η ) = X n ∈ Z (cid:18) a n + ∞ X k =1 a nk √ πkηH (cid:19) e πinξL (cid:0) ˜ p ∈ H ( R ) (cid:1) (B.4)can be differentiated term by term and we have k ˜ p k ,R = X n,k LH | a nk | , k∇ ˜ p k ,R = X n,k LH (cid:20)(cid:16) πnL (cid:17) + (cid:16) πkH (cid:17) (cid:21) | a nk | . (B.5)Assuming ˜ p ∈ H ( R ), i.e. a = 0, we learn that k ˜ p k ,R ≤ L e C k∇ ˜ p k ,R , L e C = max n(cid:16) L π (cid:17) , (cid:16) Hπ (cid:17) o . (B.6)Now we transfer this result to Ω by the change of variables (4.1) and (4.3). To avoidRellich’s theorem, we estimate k p k , Ω ≤ (cid:13)(cid:13)(cid:13)(cid:16) h h (cid:17) / p (cid:13)(cid:13)(cid:13) , Ω = h H k ˜ p k ,R ≤ L e C h H k∇ ˜ p k ,R . (B.7)This inequality holds for all p such that ˜ p ∈ H ( R ). Arguing as in (4.4), we find thatif we drop the intermediate inequalities, (B.7) also holds for p ∈ H (Ω). Next, webound k∇ ˜ p k ,R in terms of k∇ p k ,R : k ˜ p ξ k ,R = Z Ω (cid:16) p x + yh h x p y (cid:17) Hh dx dy ≤ γ Hh k p x k , Ω + γ M Hh k p y k , Ω (B.8) k ˜ p η k ,R = Z Ω (cid:16) hH p y (cid:17) Hh dx dy ≤ h H k p y k , Ω , (B.9)where γ − + γ − = 1 and M = k h x k ∞ . It follows that L − k p k , Ω ≤ C k∇ p k , Ω , C = e C h H max (cid:16) γ Hh , h H + γ M Hh (cid:17) . (B.10)8 JON WILKENING
Next, we choose H = √ h h and minimize max (cid:0) γ , γ M ) over all choices of γ j such that γ − + γ − = 1. The result is γ = 1 + γ M = 14 (cid:0)p M + 4 + M (cid:1) ≤ (1 + M ) , (B.11)which yields C = π max (cid:16) , h h L (cid:17) h h (1 + M ) as claimed. Remark B.2.
Example 4.5 shows that the scaling C ∼ h − / is optimal: forthat function p , we have L − k p k , Ω ≥ L − h L h h (cid:18) h L (cid:16) L (cid:17) (cid:19) ≥ h h k p x k , Ω , (B.12)which shows that C in (B.3) is at least q h h . We do not know if the linear depen-dence of C on M is optimal — it seems to be an unavoidable artifact of changingvariables to a rectangular geometry. REFERENCES[1] J. Bourgain and H. Brezis. On the equation div Y = f and application to control of phases. J.Amer. Math. Soc. , 16(2):393–426, 2002.[2] D. Braess.
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