The Domain of Analyticity of Solutions to the Three-Dimensional Euler Equations in a Half Space
aa r X i v : . [ m a t h . A P ] J u l THE DOMAIN OF ANALYTICITY OF SOLUTIONS TO THE THREE-DIMENSIONALEULER EQUATIONS IN A HALF SPACE
IGOR KUKAVICA AND VLAD VICOL
Abstract.
We address the problem of analyticity up to the boundary of solutions to the Euler equationsin the half space. We characterize the rate of decay of the real-analyticity radius of the solution u ( t ) interms of exp R t k∇ u ( s ) k L ∞ ds , improving the previously known results. We also prove the persistence of thesub-analytic Gevrey-class regularity for the Euler equations in a half space, and obtain an explicit rate ofdecay of the radius of Gevrey-class regularity. Introduction
The Euler equations on a half space for the velocity vector field u ( x, t ) and the scalar pressure field p ( x, t ),where x ∈ Ω = { x ∈ R : x > } and t ≥
0, are given by ∂ t u + ( u · ∇ ) u + ∇ p = 0 , in Ω × (0 , ∞ ) , (E.1) ∇ · u = 0 , in Ω × (0 , ∞ ) , (E.2) u · n = 0 , on ∂ Ω × (0 , ∞ ) , (E.3)where n = (0 , , −
1) is the outward unit normal to ∂ Ω = { x ∈ R : x = 0 } . We consider the initial valueproblem associated to (E.1)–(E.3) with a divergence free initial datum u (0) = u , in Ω . (E.4)The local existence and uniqueness of H r -solutions, with r > / , T ∗ )holds (cf. [12, 15, 23, 32, 36]), and lim T ր T ∗ R T k curl u ( t ) k L ∞ dt = ∞ , if T ∗ < ∞ (cf. [6]); additionally thepersistence of C ∞ smoothness was proven by Foias, Frisch, and Temam [17]. In this paper we address thesolutions of the Euler initial value problem evolving from real-analytic and Gevrey-class initial datum (up tothe boundary), and characterize the domain of analyticity. We emphasize that the radius of real-analyticitygives an estimate on the minimal scale in the flow [22, 25], and it also gives the explicit rate of exponentialdecay of its Fourier coefficients [18].In a three dimensional bounded domain, the persistence of analyticity was proven by Bardos and Be-nachour [3] by an implicit argument (see also Alinhac and M´etivier [1]). In [2, 7] the authors give anexplicit estimate on the radius of analyticity, but which vanishes in finite time (independent of T ∗ ). How-ever, the proof of persistency [3] can be modified to show that the radius of analyticity decays at a rateproportional to the exponential of a high Sobolev norm of the solution (see also [1]). On the three dimen-sional periodic domain (or equivalently on R ) this is the same rate obtained by Levermore and Oliver in[30], using the method of Gevrey-class regularity. This Fourier based method was introduced by Foias andTemam [18] to study the analyticity of the Navier-Stokes equations. For further results on analyticity see[1, 13, 16, 20, 21, 24, 25, 28, 29, 27, 33, 35]. Explicit and even algebraic lower bounds for the radius ofanalyticity for dispersive equations were obtained by Bona, Gruji´c, and Kalisch in [9, 10] (see also [8, 11]).In [26] we have proven that in the periodic setting, or on R , the analyticity radius decays algebraicallyin the Sobolev norm k curl u ( t ) k H r , with r > /
2, and exponentially in R t k∇ u ( s ) k L ∞ , for all t < T ∗ . Inthe present paper we show that the algebraic dependence on the Sobolev norm holds in the case when thedomain has boundaries (cf. Theorem 2.1), thereby improving the previously known results. The interioranalyticity in the case of the half-space, for short time (independent of T ∗ ), was treated in [35]. We note Mathematics Subject Classification.
Primary: 76B03; Secondary: 35L60.
Key words and phrases.
Euler equations, analyticity radius, Gevrey class. that the shear flow example of Bardos and Titi [5] (cf. [14, 38]) may be used to construct explicit solutionsto the three-dimensional Euler equations whose radius of analyticity is decaying for all time.Additionally we prove the persistence of sub-analytic Gevrey-class regularity up to the boundary (cf. [18,31]) for the Euler equations on the half space. To the best of our knowledge this was only known for theperiodic domain cf. [26, 30], but not for a domain with boundary. The methods of [1, 3, 27, 35] rely essentiallyon the special structure of the complex holomorphic functions, and do not apply to the non-analytic Gevrey-class setting.The presence of the boundary creates several difficulties that do not arise in the periodic setting. Inparticular we cannot use Fourier-based methods, nor can we use the vorticity formulation of the equations.Instead we need to estimate the pressure, which satisfies (cf. [36]) the elliptic Neumann problem − ∆ p = ∂ j u i ∂ i u j , in Ω × (0 , ∞ ) , (P.1) ∂p∂n = ( u · ∇ ) u · n = 0 , on ∂ Ω × (0 , ∞ ) , (P.2)since n = (0 , , − Main Theorems
The following statement is our main theorem addressing the analyticity of the solution. Theorem 2.2below concerns the Gevrey class persistence.
Theorem 2.1.
Fix r > / . Let u ∈ H r (Ω) be divergence-free and uniformly real-analytic in Ω . Thenthe unique solution u ( t ) ∈ C (0 , T ∗ ; H r (Ω)) of the initial value problem associated to the Euler equations (E.1) – (E.4) is real-analytic for all time t < T ∗ , where T ∗ ∈ (0 , ∞ ] denotes the maximal time of existence ofthe H r -solution. Moreover, the uniform radius of space analyticity τ ( t ) of u ( t ) satisfies τ ( t ) ≥ C (1 + t ) exp (cid:18) − C Z t k∇ u ( s ) k L ∞ ds (cid:19) , (1) where C > is a constant that depends only on r , while C has additional dependence on u as described in (11) below. Remark 1.
The lower bound (1) improves the rate of decay from Bardos and Benachour [3] on a boundeddomain (which can be inferred to be proportional to exp R t k u ( s ) k H r ds ), and it matches the rate of decaywe obtained in [26] on the periodic domain. Remark 2.
The proof of Theorem 2.1 also works in the case of the half-plane (recall that in two dimensions T ∗ may be taken arbitrarily large, cf. [32, 37]) with the same lower bound on the radius of analyticity of thesolution. Since in two dimensions k∇ u ( t ) k L ∞ grows at a rate of C exp( Ct ), for some positive constant C , theestimate (1) shows that the rate of decay of the analyticity radius is at least C exp( − C exp( Ct )), for some C >
0. This recovers the two-dimensional rate of decay obtained by Bardos, Benachour and Zerner [4] on abounded domain and by the authors of this paper on the torus [26]. It would be interesting if one could provea similar lower bound to (1) but where the quantity R t k∇ u ( s ) k L ∞ ds is replaced by R t k curl u ( s ) k L ∞ ds . Inparticular, such an estimate would imply in two dimensions that the radius of analyticity decays as a singleexponential in time. NALYTICITY OF SOLUTIONS TO THE EULER EQUATIONS IN A HALF SPACE 3
Recall (cf. [31]) that a smooth function v is uniformly of Gevrey-class s , with s ≥
1, if there exist
M, τ > | ∂ α v ( x ) | ≤ M | α | ! s τ | α | , (2)for all x ∈ Ω and all multi-indices α ∈ N . When s = 1 we recover the class of real-analytic functions, andfor s ∈ (1 , ∞ ) these functions are C ∞ smooth but might not be analytic. We call the constant τ in (2) theradius of Gevrey-class regularity. The following theorem shows the persistence of the Gevrey-class regularityfor the Euler equations in a half-space. Theorem 2.2.
Fix r > / . Let u be uniformly of Gevrey-class s on Ω , with s > , and divergence-free.Then the unique H r -solution u ( t ) of the initial value problem (E.1) – (E.4) on [0 , T ∗ ) is of Gevrey-class s , forall t < T ∗ , and the radius τ ( t ) of Gevrey-class regularity of the solution satisfies the lower bound (1) . Proofs of Theorem 2.1 and Theorem 2.2
For a multi-index α = ( α , α , α ) in N , we denote α ′ = ( α , α ). Define the Sobolev and Lipshitzsemi-norms | · | m and | · | m, ∞ by | v | m = X | α | = m M α k ∂ α v k L , (3)and | v | m, ∞ = X | α | = m M α k ∂ α v k L ∞ , where M α = | α ′ | ! α ′ ! = (cid:18) α + α α (cid:19) . (4)The need for the binomial weights M α in (3) shall be evident in Section 5 where we study the higher regularityestimates associated with the Neumann problem (P.1)–(P.2). For s ≥ τ >
0, define the space X τ = { v ∈ C ∞ (Ω) : k v k X τ < ∞} , where k v k X τ = ∞ X m =3 | v | m τ m − ( m − s . Similarly let Y τ = { v ∈ C ∞ (Ω) : k v k Y τ < ∞} , where k v k Y τ = ∞ X m =4 | v | m ( m − τ m − ( m − s . Remark 3.
The above defined spaces X τ and Y τ can be identified with the classical Gevrey- s classesas defined in [31]. On the full space or on the torus, the Gevrey- s classes can also be identified with D (( − ∆) r/ exp ( τ ( − ∆) / s )) (cf. [18, 26, 30]).We shall prove Theorems 2.1 and 2.2 simultaneously by looking at the evolution equation in Gevrey- s classes with s ≥
1. If u is of Gevrey-class s in Ω, with s ≥
1, then there exists τ (0) > u ∈ X τ (0) ,and moreover τ (0) can be chosen arbitrarily close to the uniform real-analyticity radius of u , respectivelyto the radius of Gevrey-class regularity. Let u ( t ) be the classical H r -solution of the initial value problem(E.1)–(E.4).With the notations of Section 2 we have an a priori estimate ddt k u ( t ) k X τ ( t ) = ˙ τ ( t ) k u ( t ) k Y τ ( t ) + ∞ X m =3 (cid:18) ddt | u ( t ) | m (cid:19) τ ( t ) m − ( m − s . (5) IGOR KUKAVICA AND VLAD VICOL
Fix m ≥
3. In order to estimate ( d/dt ) | u ( t ) | m , for each | α | = m we apply ∂ α on (E.1) and take the L -innerproduct with ∂ α u . We obtain12 ddt k ∂ α u k L + < ∂ α ( u · ∇ u ) , ∂ α u > + < ∇ ∂ α p, ∂ α u > = 0 . (6)On the second term on the left, we apply the Leibniz rule and recall that < u · ∇ ∂ α u, ∂ α u > = 0. For thethird term on the left of (6) we note that since n = (0 , , −
1) and u · n = 0 on ∂ Ω, we have that ∂ α u · n = 0for all α such that α = 0. Together with ∇ · u = 0 in Ω this implies that < ∇ ∂ α p, ∂ α u > = 0 whenever α = 0. Using the Cauchy-Schwarz inequality and summing over | α | = m we then obtain ddt | u | m ≤ X | α | = m X β ≤ α,β =0 M α (cid:18) αβ (cid:19) k ∂ β u · ∇ ∂ α − β u k L + X | α | = m,α =0 M α k∇ ∂ α p k L . Combined with (5), the above estimate shows that ddt k u ( t ) k X τ ( t ) ≤ ˙ τ ( t ) k u ( t ) k Y τ ( t ) + C + P , (7)where the upper bound on the commutator term is given by C = ∞ X m =3 X | α | = m X β ≤ α,β =0 M α (cid:18) αβ (cid:19) k ∂ β u · ∇ ∂ α − β u k L τ m − ( m − s , and the upper bound on the pressure term is P = ∞ X m =3 X | α | = m,α =0 M α k∇ ∂ α p k L τ m − ( m − s . In order to estimate C we use the following lemma, the proof of which is given in Section 4 below. Lemma 3.1.
There exists a sufficiently large constant
C > such that C ≤ C ( C + C k u k Y τ ) , where C = | u | , ∞ | u | + | u | , ∞ | u | + τ | u | , ∞ | u | , and C = τ | u | , ∞ + τ | u | , ∞ + τ | u | , ∞ + τ / k u k X τ . The following lemma shall be used to estimate P . The proof is given in Section 5 below. Lemma 3.2.
There exists a sufficiently large constant
C > such that P ≤ C ( P + P k u k Y τ ) , where P = | u | , ∞ | u | + | u | , ∞ | u | + τ | u | , ∞ | u | + τ | u | , ∞ | u | , and P = τ | u | , ∞ + τ | u | , ∞ + τ | u | , ∞ + τ / k u k X τ . Let r > / ddt k u ( t ) k X τ ( t ) ≤ ˙ τ ( t ) k u ( t ) k Y τ ( t ) + C k u ( t ) k H r (1 + τ ( t ) )+ C k u ( t ) k Y τ ( t ) (cid:16) τ ( t ) k∇ u ( t ) k L ∞ + ( τ ( t ) + τ ( t ) ) k u ( t ) k H r + τ ( t ) / k u ( t ) k X τ ( t ) (cid:17) . (8)If τ ( t ) decreases fast enough so that for all 0 ≤ t < T ∗ we have˙ τ ( t ) + Cτ ( t ) k∇ u ( t ) k L ∞ + C ( τ ( t ) + τ ( t ) ) k u ( t ) k H r + Cτ ( t ) / k u ( t ) k X τ ( t ) ≤ , (9) NALYTICITY OF SOLUTIONS TO THE EULER EQUATIONS IN A HALF SPACE 5 then (8) implies that ddt k u ( t ) k X τ ( t ) ≤ C k u ( t ) k H r (1 + τ (0) ) , and therefore k u ( t ) k X τ ( t ) ≤ k u k X τ (0) + C τ (0) Z t k u ( s ) k H r ds = M ( t ) , for all 0 ≤ t < T ∗ , where C τ (0) = 1 + τ (0) . Since τ must be chosen to be a decreasing function, a sufficientcondition for (9) to hold is that˙ τ ( t ) + Cτ ( t ) k∇ u ( t ) k L ∞ + Cτ ( t ) / (cid:16) C ′ τ (0) k u ( t ) k H r + M ( t ) (cid:17) ≤ , (10)where C ′ τ (0) = τ (0) / + τ (0) / . For simplicity of the exposition we denote G ( t ) = exp (cid:18) C Z t k∇ u ( s ) k L ∞ ds (cid:19) , where the constant C > k u ( t ) k H r ≤ k u k H r G ( t ). It then follows that(10) is satisfied if we let τ ( t ) = G ( t ) − / (cid:18) τ (0) − / + C Z t (cid:16) C ′ τ (0) k u ( s ) k H r + M ( s ) (cid:17) G ( s ) − ds (cid:19) − / . The lower bound (1) on the radius of analyticity stated in Theorem 2.1 is then obtained by noting that τ (0) − / + C Z t (cid:16) C ′ τ (0) k u ( s ) k H r + M ( s ) (cid:17) G ( s ) − ds ≤ τ (0) − / + C Z t (cid:16) C ′ τ (0) k u k H r + k u k X τ (0) + sC τ (0) k u k H r (cid:17) ds ≤ C (1 + t ) , (11)and therefore τ ( t ) ≥ G ( t ) − / C t . The last inequality in (11) above gives the explicit dependence of C on u . This concludes the a priori estimates that are used to prove Theorem 2.1. The proof can be made formal by considering an approximatingsolution u ( n ) , n ∈ N , proving the above estimates for u ( n ) , and then taking the limit as n → ∞ . We omitthese details. 4. The commutator estimate
Before we prove Lemma 3.1 we state and prove two useful lemmas about multi-indexes, that will be usedthroughout in Sections 4 and 5 below.
Lemma 4.1.
We have (cid:18) αβ (cid:19) M α M − β M − α − β ≤ (cid:18) | α || β | (cid:19) (12) for all α, β ∈ N with β ≤ α .Proof. Using (4) we have that (cid:18) α ′ β ′ (cid:19) M α M − β M − α − β = (cid:18) | α ′ || β ′ | (cid:19) , and hence the left side of (12) is bounded by (cid:18) | α ′ || β ′ | (cid:19)(cid:18) α β (cid:19) . IGOR KUKAVICA AND VLAD VICOL
The lemma then follows from (cid:18) ni (cid:19)(cid:18) mj (cid:19) ≤ (cid:18) n + mi + j (cid:19) , for any n, m ≥ n ≥ i and m ≥ j , which in turn we obtain by computing the coefficient in frontof x i + j in the binomial expansions of (1 + x ) n (1 + x ) m and (1 + x ) m + n . (cid:3) The second lemma allows us to re-write certain double sums involving multi-indices.
Lemma 4.2.
Let { x λ } λ ∈ N and { y λ } λ ∈ N be real numbers. Then we have X | α | = m X | β | = j,β ≤ α x β y α − β = X | β | = j x β X | γ | = m − j y γ . (13)The proof of the above lemma is omitted: it consists of re-labeling of the terms on the left side of (13).Now we proceed by proving the commutator estimate. Proof of Lemma 3.1.
We have C = ∞ X m =3 m X j =1 C m,j , where we denoted C m,j = τ m − ( m − s X | α | = m X | β | = j,β ≤ α M α (cid:18) αβ (cid:19) k ∂ β u · ∇ ∂ α − β u k L . (14)We now split the right side of the above equality into seven terms according to the values of m and j , andprove the following estimates. For low j , we claim ∞ X m =3 C m, ≤ C | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ , (15) ∞ X m =3 C m, ≤ C | u | , ∞ | u | + Cτ | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ , (16)for intermediate j , we have ∞ X m =6 [ m/ X j =3 C m,j ≤ Cτ / k u k X τ k u k Y τ , (17) ∞ X m =7 m − X j =[ m/ C m,j ≤ Cτ / k u k X τ k u k Y τ , (18)and for high j , ∞ X m =5 C m,m − ≤ Cτ | u | , ∞ k u k Y τ , (19) ∞ X m =4 C m,m − ≤ Cτ | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ , (20) ∞ X m =3 C m,m ≤ C | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ . (21)Due to symmetry we shall only prove (15)–(17) and indicate the necessary modifications for (18)–(21). NALYTICITY OF SOLUTIONS TO THE EULER EQUATIONS IN A HALF SPACE 7
Proof of (15): The H¨older inequality, (14), and Lemma 4.1 imply that ∞ X m =3 C m, = X | α | =3 X | β | =1 ,β ≤ α (cid:0) M β k ∂ β u k L ∞ (cid:1) (cid:0) M α − β k ∂ α − β ∇ u k L (cid:1) M α M − β M − α − β (cid:18) αβ (cid:19) + ∞ X m =4 X | α | = m X | β | =1 ,β ≤ α (cid:0) M β k ∂ β u k L ∞ (cid:1) (cid:18) M α − β k ∂ α − β ∇ u k L ( m − τ m − ( m − s (cid:19) × M α M − β M − α − β (cid:18) αβ (cid:19) m − τ ≤ C X | α | =3 X | β | =1 ,β ≤ α (cid:0) M β k ∂ β u k L ∞ (cid:1) (cid:0) M α − β k ∂ α − β ∇ u k L (cid:1) + Cτ ∞ X m =4 X | α | = m X | β | =1 ,β ≤ α (cid:0) M β k ∂ β u k L ∞ (cid:1) × (cid:18) M α − β k ∂ α − β ∇ u k L ( m − τ m − ( m − s (cid:19) mm − . (22)The first sum on the far right side of (22) can be estimated by C | u | , ∞ |∇ u | ≤ C | u | , ∞ | u | . Since m ≥
4, Lemma 4.2 implies that the second term on the far right side of (22) is bounded by Cτ | u | , ∞ ∞ X m =4 |∇ u | m − ( m − τ m − ( m − s ≤ Cτ | u | , ∞ k u k Y τ , concluding the proof of (15).Proof of (16): As in the proof of (15) above, we have ∞ X m =3 C m, ≤ C X | α | =3 , X | β | =2 ,β ≤ α τ m − (cid:0) M β k ∂ β u k L ∞ (cid:1) (cid:0) M α − β k ∂ α − β ∇ u k L (cid:1) × M α M − β M − α − β (cid:18) αβ (cid:19) + C ∞ X m =5 X | α | = m X | β | =2 ,β ≤ α (cid:0) M β k ∂ β u k L ∞ (cid:1) (cid:18) M α − β k ∂ α − β ∇ u k L ( m − τ m − ( m − s (cid:19) × M α M − β M − α − β (cid:18) αβ (cid:19) m − m − s τ . (23)Using Lemma 4.2, the first sum on the right of (23) can be estimated from above by C | u | , ∞ |∇ u | + Cτ | u | , ∞ |∇ u | ≤ C | u | , ∞ | u | + Cτ | u | , ∞ | u | . On the other hand, since s ≥ | β | = 2, and | α | = m ≥
5, we have by Lemma 4.1 that M α M − β M − α − β (cid:18) αβ (cid:19) m − m − s ≤ (cid:18) m (cid:19) m − m − ≤ C. By Lemma 4.2, the second sum on the right of (23) is thus bounded by Cτ ∞ X m =5 | u | , ∞ |∇ u | m − ( m − τ m − ( m − s ≤ Cτ | u | , ∞ k u k Y τ . This proves the desired estimate.Proof of (17): We first observe that the H¨older inequality and the Sobolev inequality give k ∂ β u · ∇ ∂ α − β u k L ≤ C k ∂ β u k / L k ∆ ∂ β u k / L k∇ ∂ α − β u k L . IGOR KUKAVICA AND VLAD VICOL
Therefore we can bound the right hand side of (17) as follows ∞ X m =6 [ m/ X j =3 C m,j ≤ ∞ X m =6 [ m/ X j =3 X | α | = m X | β | = j,β ≤ α (cid:18) M β k ∂ β u k L τ j − ( j − s (cid:19) / τ / A α,β,s × (cid:18) M β k ∂ β ∆ u k L τ j − ( j − s (cid:19) / (cid:18) M α − β k ∂ α − β ∇ u k L ( m − j − τ m − j − ( m − j − s (cid:19) , where A α,β,s = M α M − β M − α − β (cid:18) αβ (cid:19) ( j − s/ ( j − s/ ( m − j − s ( m − s ( m − j − . By Lemma 4.1, we have that for m ≥ ≤ j ≤ [ m/ A α,β,s ≤ C (cid:18) mj (cid:19)(cid:18) m − j − (cid:19) − s m − j − j − s/ ( j − s/ ≤ C (cid:18) m − j − (cid:19) − s +1 j s/ . Since s ≥ A α,β,s ≤ C . Together with Lemma 4.2 and the discreteH¨older inequality this shows that ∞ X m =6 [ m/ X j =3 C m,j ≤ Cτ / ∞ X m =6 [ m/ X j =3 (cid:18) | u | j τ j − ( j − s (cid:19) / (cid:18) | ∆ u | j τ j − ( j − s (cid:19) / × (cid:18) |∇ u | m − j ( m − j − τ m − j − ( m − j − s (cid:19) . The discrete Young and H¨older inequalities then give ∞ X m =6 [ m/ X j =3 C m,j ≤ Cτ / k u k X τ k u k Y τ , concluding the proof of (17).To prove (18)–(21) we proceed as in the proofs of (15)–(17) above, with the roles of j and m − j reversed.Instead of estimating k ∂ β u · ∇ ∂ α − β u k L with k ∂ β u k L ∞ k ∂ α − β ∇ u k L we instead bound k ∂ β u · ∇ ∂ α − β u k L ≤ k ∂ β u k L k ∂ α − β ∇ u k L ∞ . We omit further details. This concludes the proof of Lemma 3.1. (cid:3) The pressure estimate
In the proof of the Lemma 3.2 we need to use the following higher regularity estimate on the solution ofthe Neumann problem associated to the Poisson equation for the half-space.
Lemma 5.1.
Assume that p is a smooth solution of the Neumann problem − ∆ p = v in Ω ,∂p∂n = 0 on ∂ Ω , with v ∈ C ∞ (Ω) . Then there is a universal constant C > such that k ∂ ∂ α p k L ≤ C X s,t ∈ N , | β | = m − β ′ − α ′ =(2 s, t ) (cid:18) s + ts (cid:19) k ∂ β v k L , (24) NALYTICITY OF SOLUTIONS TO THE EULER EQUATIONS IN A HALF SPACE 9 for any m ≥ and any multiindex α ∈ N with | α | = m and α = 0 . Additionally, if α ≥ then k ∂ ∂ α p k L ≤ C X s,t ∈ N , | β | = m − β ′ − α ′ =(2 s +1 , t ) (cid:18) s + ts (cid:19) k ∂ β v k L , (25) k ∂ ∂ α p k L ≤ C X s,t ∈ N , | β | = m − β ′ − α ′ =(2 s, t +1) (cid:18) s + ts (cid:19) k ∂ β v k L , (26) where C > is a universal constant. We emphasize that the constant C in the above lemma is independent of α and m . In (24) we have aresumming over the set { β ∈ N : | β | = m − , ∃ s, t ∈ N such that β ′ − α ′ = (2 s, t ) } and similar conventions are used in (25), (26), and throughout this section. Proof.
In order to avoid repetition, we only prove (24) and indicate the necessary changes for (25) and (26).Let ∆ ′ = ∂ + ∂ be the tangential Laplacian. Using induction on k ∈ N we obtain the identity ∂ k +23 p = ( − ∆ ′ ) k +1 p − k X j =0 ∂ j ( − ∆ ′ ) k − j v, and upon applying ∂ to the above equation ∂ k +33 p = ∂ ( − ∆ ′ ) k +1 p − k X j =0 ∂ j +13 ( − ∆ ′ ) k − j v. Therefore given | α | = m , with α = 2 k + 1 ≥
1, we have ∂ ∂ α p = ∂ k +23 ∂ α ′ p = ( − ∆ ′ ) k +1 ∂ α ′ p + k X j =0 ( − k − j +1 ∂ j ( ∂ + ∂ ) k − j ∂ α ′ v, (27)and if α = 2 k + 2 ≥
2, we have ∂ ∂ α p = ∂ k +33 ∂ α ′ p = ∂ ( − ∆ ′ ) k +1 ∂ α ′ p + k X j =0 ( − k − j +1 ∂ j +13 ( ∂ + ∂ ) k − j ∂ α ′ v. (28)Since n = (0 , , − g = ( − ∆ ′ ) k ∂ α ′ p satisfies the Neumann problem − ∆ g = ( − ∆ ′ ) k ∂ α ′ v in Ω ,∂g∂n = 0 on ∂ Ω . Using the classical H -regularity argument for the Neumann problem we then have k ∆ ′ g k L ≤ C k ( − ∆ ′ ) k ∂ α ′ v k L , and k ∂ ∆ ′ g k L ≤ C k ∂ ( − ∆ ′ ) k ∂ α ′ v k L , for a positive universal constant C . Combining the above estimates with (27), (28), and the identity( ∂ + ∂ ) m w = m X s =0 (cid:18) ms (cid:19) ∂ s ∂ m − s w, we obtain k ∂ ∂ α p k L ≤ C k X j =0 k ∂ j ( ∂ + ∂ ) k − j ∂ α ′ v k L ≤ C k X j =0 k − j X s =0 (cid:18) k − js (cid:19) k ∂ s + α ∂ k − j − s + α ∂ j v k L (29)if α = 2 k + 1 ≥
1, and k ∂ ∂ α p k L ≤ C k X j =0 k ∂ j +13 ( ∂ + ∂ ) k − j ∂ α ′ v k L ≤ C k X j =0 k − j X s =0 (cid:18) k − js (cid:19) k ∂ s + α ∂ k − j − s + α ∂ j +13 v k L (30)if α = 2 k + 2 ≥
2. To simplify (29) above, let t = k − j − s ≥ β = (2 s + α , k − j − s + α , j ) =(2 s + α , t + α , α − − s − t ) ∈ N . Since | α | = m and α = 2 k + 1, we have | β | = m −
1, and byre-indexing the sums, (29) can be re-written as k ∂ ∂ α p k L ≤ C X s,t ∈ N , | β | = m − β ′ − α ′ =(2 s, t ) (cid:18) s + ts (cid:19) k ∂ β v k L , The above estimate also holds for α = 2 k + 2 with the substitution β = (2 s + α , k − j − s + α , j + 1),thereby simplifying the upper bound (30), and concluding the proof of (24).To prove (25) we proceed as above and obtain k ∂ ∂ α p k L = k ∂ α +11 ∂ α ∂ k +23 p k L ≤ C k X j =0 k − j X s =0 (cid:18) k − js (cid:19) k ∂ α +2 s +11 ∂ α +2 k − j − s ∂ j v k L (31)if α = 2 k + 2 ≥
2, and k ∂ ∂ α p k L = k ∂ α +11 ∂ α ∂ k +33 p k L ≤ C k X j =0 k − j X s =0 (cid:18) k − js (cid:19) k ∂ α +2 s +11 ∂ α +2 k − j − s ∂ j +13 v k L (32)if α = 2 k + 3 ≥
3. In (31) we let t = k − j − s ≥ β = ( α + 2 s + 1 , α + 2 t, j ) = ( α + 2 s + 1 , α +2 t, α − − s − t ), since α = 2 k + 2 and | α | = m . Similarly in (32) we let β = ( α + 2 s + 1 , α + 2 t, j + 1) =( α + 2 s + 1 , α + 2 t, α − − s − t ), since α = 2 k + 3 and | α | = m . The above substitutions and re-indexingprove (25). Upon permuting the first and second coordinates, this also proves (26). (cid:3) Remark 4.
We note that Lemma 5.1 does not give an estimate for k ∂ ∂ α p k L and k ∂ ∂ α p k L if α = 1. Inthis case we note that the function g = ∂ α ′ p satisfies the Neumann problem − ∆ g = ∂ α ′ v in Ω ,∂g∂n = 0 on ∂ Ω . The classical H -regularity argument then gives k ∂ ∂ α p k L = k ∂ ∂ ∂ α ′ p k L ≤ C k ∂ α ′ v k L , and k ∂ ∂ α p k L = k ∂ ∂ ∂ α ′ p k L ≤ C k ∂ α ′ v k L , for a positive universal constant C > NALYTICITY OF SOLUTIONS TO THE EULER EQUATIONS IN A HALF SPACE 11
We note that Lemma 5.1 is different from the classical higher regularity estimates (cf. [19, 31, 36]) for theNeumann problem in the fact that the constant C in (24)–(26) does not increase with m . The dependenceon m is encoded in the sums with binomial weights on the right side of (24)–(26).The following lemma shows that only a factor of m is lost in the above higher regularity estimates if each k ∂ β v k L term is paired with a proper binomial weight. This explains the definition of the homogeneousSobolev norms | · | m in (3). Lemma 5.2.
There exists a positive universal constant C such that [ β / X s =0 [ β / X t =0 (cid:18) β + β − s − tβ − s (cid:19)(cid:18) s + ts (cid:19) ≤ Cm (cid:18) β + β β (cid:19) (33) for any m ≥ and any multi-index β = ( β , β , m − − β − β ) ∈ N . Additionally, if β ≥ we have [( β − / X s =0 [ β / X t =0 (cid:18) β + β − s − − tβ − s − (cid:19)(cid:18) s + ts (cid:19) ≤ Cm (cid:18) β + β β (cid:19) , (34) while if β ≥ we have [ β / X s =0 [( β − / X t =0 (cid:18) β + β − s − t − β − s (cid:19)(cid:18) s + ts (cid:19) ≤ Cm (cid:18) β + β β (cid:19) , (35) where C is a universal constant. We note that in particular the constant C is independent of m and β . Proof.
Due to symmetry we only give the proof of (33). Estimates (34) and (35) are proven mutatis-mutandi .First we recall that given α, γ ∈ N , with γ ≤ α , we have (cid:18) αγ (cid:19) ≤ (cid:18) | α || γ | (cid:19) . Using the above inequality we get [ β / X s =0 [ β / X t =0 (cid:18) β + β − s − tβ − s (cid:19)(cid:18) s + ts (cid:19)(cid:18) β + β β (cid:19) − ≤ [ β / X s =0 [ β / X t =0 (cid:18) β + β − s − tβ − s (cid:19)(cid:18) β + β β (cid:19) − ≤ [ β / X s =0 [ β / X t =0 (cid:18) s + ts (cid:19) − . The lemma is then proven if we find a constant C such that [ β / X s =0 [ β / X t =0 (cid:18) s + ts (cid:19) − ≤ C ( β + β ) . Without loss of generality we may assume that β , β ≥
4. We split the above sum into [ β / X s =0 [ β / X t =0 (cid:18) s + ts (cid:19) − ≤ [ β / X t =0 (cid:18) t (cid:19) − + (cid:18) t + 11 (cid:19) − + [ β / X s =0 (cid:18) ss (cid:19) − + (cid:18) s + 1 s (cid:19) − + [ β / X s =2 [ β / X t =2 (cid:18) s + ts (cid:19) − = T + T + T . (36)It is clear that T + T ≤ C ( β + β ) . (37) We estimate T by appealing to the Stirling estimate (cf. [34, p. 200]) e / √ n (cid:16) ne (cid:17) n < n ! < e √ n (cid:16) ne (cid:17) n . This implies s ! t !( s + t )! ≤ e / r sts + t s/t ) t t/s ) s . Thus we obtain T ≤ C [ β / X s =2 [ β / X t =2 √ t t/s ) s . (38)Since s ≥
2, the Binomial Theorem implies (cid:18) ts (cid:19) s ≥ (cid:18) s (cid:19) (cid:18) ts (cid:19) , and by (38) we have T ≤ C [ β / X s =2 [ β / X t =2 √ t t ≤ C [ β / X s =2 ∞ X t =2 t / ! ≤ Cβ . Since β + β ≤ m −
1, the above inequality, (36), and (37) complete the proof of the lemma. (cid:3)
Proof of Lemma 3.2.
First, note that since p satisfies the elliptic Neumann problem (P.1)–(P.2) we may useLemma 5.1 to estimate higher derivatives of ∂ p as X | α | = m,α =0 M α k ∂ ∂ α p k L ≤ C X | α | = m,α =0 X s,t ∈ N , | β | = m − β ′ − α ′ =(2 s, t ) M α (cid:18) s + ts (cid:19) k ∂ β ( ∂ i u k ∂ k u i ) k L . By re-indexing the terms in the parenthesis, the right side of the above inequality may be re-written as X | β | = m − β / X s =0 [ β / X t =0 (cid:18) β + β − s − tβ − s (cid:19)(cid:18) s + ts (cid:19) k ∂ β ( ∂ i u k ∂ k u i ) k L . Using the estimate (33) of Lemma 5.2 we bound the above expression by Cm X | β | = m − M β k ∂ β ( ∂ i u k ∂ k u i ) k L and therefore ∞ X m =3 X | α | = m,α =0 M α k ∂ ∂ α p k L τ m − ( m − s ≤ C ∞ X m =3 X | β | = m − M β k ∂ β ( ∂ i u k ∂ k u i ) k L mτ m − ( m − s . (39)On the other hand, higher derivatives of ∂ p are estimated using the decomposition X | α | = m,α =0 M α k ∂ ∂ α p k L = X | α | = m,α =1 M α k ∂ ∂ α p k L + X | α | = m,α ≥ M α k ∂ ∂ α p k L . (40)By Remark 4, the first term on the right of (40) is bounded by C X | α | = m,α =1 M α k ∂ α ′ ( ∂ i u k ∂ k u i ) k L = C X | β | = m − ,β =0 M β k ∂ β ( ∂ i u k ∂ k u i ) k L . (41) NALYTICITY OF SOLUTIONS TO THE EULER EQUATIONS IN A HALF SPACE 13
Using estimate (25), the second term on the right side of (40) is estimated by C X | α | = m,α ≥ X s,t ∈ N , | β | = m − β ′ − α ′ =(2 s +1 , t ) M α (cid:18) s + ts (cid:19) k ∂ β ( ∂ i u k ∂ k u i ) k L . By re-indexing the above expression equals C X | β | = m − ,β ≥ β − / X s =0 [ β / X t =0 (cid:18) β − β − s − tβ − s − (cid:19)(cid:18) s + ts (cid:19) k ∂ β ( ∂ i u k ∂ k u i ) k L , and using (34) it is bounded from above by Cm X | β | = m − ,β ≥ M β k ∂ β ( ∂ i u k ∂ k u i ) k L . (42)Therefore, by (40), (41), and (42), we have ∞ X m =3 X | α | = m,α =0 M α k ∂ ∂ α p k L τ m − ( m − s ≤ C ∞ X m =3 X | β | = m − M β k ∂ β ( ∂ i u k ∂ k u i ) k L mτ m − ( m − s . (43)By symmetry, we also get ∞ X m =3 X | α | = m,α =0 M α k ∂ ∂ α p k L τ m − ( m − s ≤ C ∞ X m =3 X | β | = m − M β k ∂ β ( ∂ i u k ∂ k u i ) k L mτ m − ( m − s . (44)Combining (39), (43), (44), and the Leibniz rule we obtain P ≤ C ∞ X m =3 X | β | = m − M β k ∂ β ( ∂ i u k ∂ k u i ) k L mτ m − ( m − s ≤ C ∞ X m =3 m − X j =0 P m,j , (45)where P m,j = mτ m − ( m − s X | β | = m − X | γ | = j,γ ≤ β M β (cid:18) βγ (cid:19) k ∂ γ ∂ i u k · ∂ β − γ ∂ k u i k L . We split the right side of (45) into seven terms according to the values of m and j . For low j , we claim ∞ X m =3 P m, ≤ C | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ (46) ∞ X m =3 P m, ≤ C | u | , ∞ | u | + Cτ | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ (47) ∞ X m =5 P m, ≤ Cτ | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ (48) for intermediate j , we have ∞ X m =8 [ m/ − X j =3 P m,j ≤ Cτ / k u k X τ k u k Y τ (49) ∞ X m =6 m − X j =[ m/ P m,j ≤ Cτ / k u k X τ k u k Y τ (50)and for high j , we claim ∞ X m =4 P m,m − ≤ Cτ | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ (51) ∞ X m =3 P m,m − ≤ C | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ . (52)The above estimates are proven similarly to (15)–(21) in the proof of Lemma 3.1. Due to symmetry we havepresented there the proofs of the estimates where j ≤ m − j . For completeness of the exposition we providethe proofs of (50)–(52), where we have m − j < j .Proof of (50): We proceed as in the proof of (17) in Section 4. First, the H¨older and Sobolev inequalitiesimply that k ∂ γ ∂ i u k · ∂ β − γ ∂ k u i k L ≤ C k ∂ γ ∂ i u k k L k ∂ β − γ ∂ k u i k / L k ∆ ∂ β − γ ∂ k u i k / L . Therefore, ∞ X m =6 m − X j =[ m/ P m,j ≤ C ∞ X m =6 m − X j =[ m/ X | β | = m − X | γ | = j,γ ≤ β (cid:18) M γ k ∂ γ ∂ i u k k L ( j − τ j − ( j − s (cid:19) × (cid:18) M β − γ k ∂ β − γ ∂ k u i k L τ m − j − ( m − j − s (cid:19) / × (cid:18) M β − γ k ∆ ∂ β − γ ∂ k u i k L τ m − j − ( m − j − s (cid:19) / τ / B β,γ,s , where B β,γ,s = M β M − γ M − β − γ (cid:18) βγ (cid:19) m ( j − s ( m − j − s/ ( m − j − s/ ( j − m − s . By Lemma 4.1 we have that for m ≥ m/ ≤ j ≤ m − B β,γ,s ≤ C (cid:18) m − j (cid:19)(cid:18) m − j − (cid:19) − s m ( j − m − j − s/ ( m − j − s/ ≤ C (cid:18) m − j − (cid:19) − s ( m − j ) − s/ , since (cid:0) m − j (cid:1) ≤ C (cid:0) m − j − (cid:1) , when j ≥ m/
2. Therefore, B β,γ,s ≤ C ; hence, by Lemma 4.2 and the discrete H¨olderinequality, we have ∞ X m =6 m − X j =[ m/ P m,j ≤ Cτ / ∞ X m =6 m − X j =[ m/ (cid:18) | ∂ k u i | m − j − τ m − j − ( m − j − s (cid:19) / × (cid:18) | ∆ ∂ k u i | m − j − τ m − j − ( m − j − s (cid:19) / (cid:18) | ∂ i u k | j ( j − τ j − ( j − s (cid:19) . NALYTICITY OF SOLUTIONS TO THE EULER EQUATIONS IN A HALF SPACE 15
The discrete Young and H¨older inequalities then give ∞ X m =6 m − X j =[ m/ P m,j ≤ Cτ / k u k X τ k u k Y τ , concluding the proof of (50).Proof of (51): As above we use the H¨older inequality and obtain ∞ X m =4 P m,m − ≤ ∞ X m =4 X | β | = m − X | γ | = m − ,γ ≤ β M β (cid:18) βγ (cid:19) k ∂ γ ∂ i u k k L k ∂ β − γ ∂ k u i k L ∞ mτ m − ( m − s ≤ Cτ X | β | =3 X | γ | =2 ,γ ≤ β k ∂ γ ∂ i u k k L k ∂ β − γ ∂ k u i k L ∞ + Cτ ∞ X m =5 X | β | = m − X | γ | = m − ,γ ≤ β (cid:18) M γ k ∂ γ ∂ i u k k L mτ m − ( m − s (cid:19) × (cid:0) M β − γ k ∂ β − γ ∂ k u i k L ∞ (cid:1) M β M − γ M − β − γ (cid:18) βγ (cid:19) m − s . Using Lemma 4.1, Lemma 4.2, and s ≥
1, this shows that the far right side of the above chain of inequalitiesis bounded by Cτ | ∂ i u k | | ∂ k u i | , ∞ + Cτ | ∂ k u i | , ∞ ∞ X m =5 | ∂ i u k | m − mτ m − ( m − s ≤ Cτ | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ , thereby proving (51).Proof of (52): By the H¨older inequality we have ∞ X m =3 P m,m − ≤ ∞ X m =3 X | β | = m − M β k ∂ β ∂ i u k k L k ∂ k u i k L ∞ mτ m − ( m − s ≤ C | ∂ i u k | k ∂ k u i k L ∞ + Cτ k ∂ k u i k L ∞ ∞ X m =4 | ∂ i u k | m − mτ m − ( m − s ≤ C | u | , ∞ | u | + Cτ | u | , ∞ k u k Y τ , which gives the desired estimate. By symmetry, we may similarly prove (46)–(49), but in these cases weapply the H¨older inequality as k ∂ γ ∂ i u k · ∂ β − γ ∂ k u i k L ≤ k ∂ γ ∂ i u k k L ∞ k ∂ β − γ ∂ k u i k L , that is we reverse the roles of j and m − j . We omit further details. This concludes the proof of Lemma 3.2. (cid:3) Acknowledgments
Both authors were supported in part by the NSF grant DMS-0604886.
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