The limit α→0 of the α -Euler equations in the half plane with no-slip boundary conditions and vortex sheet initial data
A. V. Busuioc, D. Iftimie, M. C. Lopes Filho, H. J. Nussenzveig Lopes
aa r X i v : . [ m a t h . A P ] F e b THE LIMIT α → OF THE α -EULER EQUATIONS IN THE HALF PLANEWITH NO-SLIP BOUNDARY CONDITIONS AND VORTEX SHEET INITIALDATA A. V. BUSUIOC, D. IFTIMIE, M. C. LOPES FILHO AND H. J. NUSSENZVEIG LOPES
Abstract.
In this article we study the limit when α → α -Euler system inthe half-plane, with no-slip boundary conditions, to weak solutions of the 2D incompressible Eulerequations with non-negative initial vorticity in the space of bounded Radon measures in H − . Thisresult extends the analysis done in [4, 13]. It requires a substantially distinct approach, analogousto that used for Delort’s Theorem, and a new detailed investigation of the relation between (no-slip)filtered velocity and potential vorticity in the half-plane.
1. Introduction
This article concerns the limit α → α -Euler equations in the half-plane, with no-slipboundary conditions, with initial velocity in L and initial vorticity whose singular part is a non-negative bounded Radon measure. The present work is a natural continuation of research containedin [4, 13], where the respective authors proved convergence, first for initial velocity in H , see [13]and then for initial vorticity in L p , p >
1, see [4], both for flows in bounded, smooth domains.The extension to initial vorticities in the space of Radon measures requires a substantial change intechnique. The previous results are based on energy estimates and boundary correctors [13] or onthe compactness of the velocity sequences obtained from boundedness of the corresponding vorticityin a suitable space [4]. For the present work, a compensated compactness argument is required,involving a subtle cancellation property of the nonlinearity, in the spirit of Delort’s celebratedexistence result, see [8]. Let us mention that the limit α → α -Euler equations,the key novelty of our result.More precisely, much of our analysis focuses on the fine properties of the operator B , introducedin Definition 5, which maps the potential vorticity q to the filtered velocity u . This is a classicalpseudo-differential operator of order −
3, given by B = ( I + α A ) − K H , where A is the half-planeStokes operator with no-slip boundary conditions and K H is the Biot-Savart operator for the half-plane. It decomposes naturally into an interior part, which is easy to understand, and a boundarypart, similar to a Poisson integral, which is more delicate. The analysis of the boundary part makesuse of Fourier methods, one of the main reasons why we restrict ourselves to half-plane flows.From a broader point-of-view, the α -Euler equations are a regularization of the Euler equa-tions, obtained by averaging the transporting velocity at scale √ α . It is the inviscid limit of thesecond-grade fluid model, see [7], the equation for geodesics in the group of volume-preserving dif-feomorphisms with a natural metric, see [15] and a variant of the vortex blob method, a standardnumerical method for discretizing 2D inviscid flows. The desingularized velocity is obtained fromthe physical one by inverting the elliptic operator ( I − α P ∆), which, in a domain with boundary,requires boundary conditions. The no-slip boundary conditions are the most natural, but Navier-type conditions have also been used (see [5, 6]). Choosing no-slip makes the vanishing α problem esemble the vanishing viscosity limit, an important open problem. In this setting, the vanishing α limit could present some of the complications of the vanishing viscosity limit, such as boundarylayers and spontaneous small-scale generation, see [2]. This similarity between the present problemand vanishing viscosity is the chief motivation for the present work. The results obtained to date,including those we present here, suggest that these two limits behave in sharply distinct ways, butit is not entirely clear why that might be the case.The remainder of this article is organized as follows. Still in the Introduction, we briefly stateour main results. In Section 2 we fix notation, we introduce elementary facts of Potential Theoryin the half-plane and compute some Fourier transforms. In Section 3 we introduce the operator B ,which maps potential vorticity to filtered velocity. In Section 4 we sketch the proof of Theorem 1,the existence result for α fixed. This is an adaptation to the case of the half-plane of a similar resultin the full-plane case, see [16]. In Section 5 we introduce the decomposition of the operator B ininterior and boundary parts. In Section 6 we derive precise estimates for the boundary potentialsassociated with the operator B . In Section 7 we apply the results obtained to prove Theorem 2,adapting Schochet’s argument, see [17], and the argument used in [14]. Finally, in the last section,we present some concluding remarks and a few open problems.Let us continue with some notation. We will denote by H the half-plane H = { x ∈ R ; x > } . The initial-value problem for the α -Euler equations with no-slip boundary conditions on H aregiven by: ∂ t ( u − α ∆ u ) + u · ∇ ( u − α ∆ u ) + P j ( u − α ∆ u ) j ∇ u j = −∇ p, in R + × H div u = 0 , in R + × H ,u = 0 , on R + × ∂ H ,u (0 , · ) = u , on { t = 0 } × H . Above, u − α ∆ u is called the unfiltered velocity, u is the filtered velocity and p is the scalar pressure.Taking the curl of the α -Euler equations, in two dimensions, gives rise to an active scalar transportequation given by(1) (cid:26) ∂ t q + u · ∇ q = 0 ,q (0 , · ) = q , where u is related to the potential vorticity q through the following system:(2) div u = 0 , curl( u − α ∆ u ) = qu (cid:12)(cid:12) ∂ H = 0 . The scalar quantity q ≡ curl( I − α ∆) u is called the potential vorticity associated to the velocity u .The equations in (1)-(2) are the potential vorticity equations, i.e. the vorticity formulation of the α -Euler equations.Let BM ( H ) be the set of bounded Radon measures on H and recall that the norm of a measure µ in BM ( H ) is given by the total variation | µ | ( H ). Set ˙ H − ( H ) = { curl w | w ∈ L ( H ) } , which wenote in passing is a proper subset of H − ( H ). We have that ˙ H − ( H ) is a Banach space with the norm k q k ˙ H − = inf {k w k L | q = curl w } . Let P denote the Leray projector in L ( H ) onto divergence freevector fields which are tangent to the boundary of H . Note that k q k ˙ H − ≡ k P w k L ( H ) , independentlyof w ∈ L such that curl w = q . e will now state our main results. Theorem 1 (Existence) . Assume that q ∈ BM ( H ) ∩ ˙ H − ( H ) . Then there exists a global solution u ∈ C b ( R + ; H w ( H )) ∩ C ( R + × H ) , q ∈ L ∞ ( R + ; BM ( H )) of the α -Euler equations with initial data q . In addition we have the energy inequality (3) k u ( t ) k L + α k∇ u ( t ) k L ≤ k u k L + α k∇ u k L ∀ t ≥ . and the bound (4) k q k L ∞ ( R + ; BM ( H )) ≤ k q k BM ( H ) . Above C b is the space of bounded continuous functions and H w denotes the space H endowedwith the weak topology. Theorem 2 (Convergence) . Assume that q ∈ (cid:0) BM + ( H ) + L ( H ) (cid:1) ∩ ˙ H − ( H ) is independent of α .Let u α , q α be a global solution of the α –Euler equations with initial data q , as obtained in Theorem 1.Then there exists a vortex sheet solution v ∈ L ∞ ( R + ; L ( H )) , with ω = curl v ∈ L ∞ ( R + ; BM ( H )) ,of the incompressible Euler equations with initial vorticity ω = q , and a subsequence u α k , q α k suchthat u α k ⇀ v weak- ∗ L ∞ ( R + ; L ) and q α k ⇀ ω weak- ∗ L ∞ ( R + ; BM ( H )) .
2. Notations and some preliminary results
We begin by fixing notation. The constant C denotes a generic constant whose value may changefrom one line to another. If ( a, b ) ∈ R then we denote ( a, b ) ⊥ ≡ ( − b, a ).We will use standard notation for function spaces: L p (Lebesgue space), W m,p (Sobolev space), L , ∞ (Lorentz space), BM (bounded Radon measures), etc. All function spaces are defined on H unless otherwise specified. The notation L pσ denotes the space of L p divergence free vector fieldstangent to the boundary endowed with the L p norm. We define in a similar manner L , ∞ σ .Recall the Leray projector P , i.e. the L orthogonal projector from L to L σ . It is well-knownthat P can be extended to a bounded operator from L p to L pσ for all 1 < p < ∞ . The Stokesoperator A is defined as A = − P ∆. Various regularity properties for the Stokes operator in L p spaces on a half-plane were proved in [3]. We note, in particular, that, for any α > I + α A isinvertible on L pσ ( H ) with values in W ,p ( H ) ∩ W ,p ( H ), see Section 3 of [3]. We denote this inverseby ( I + α A ) − .Let us start with a very simple H estimate. Lemma 3.
Let f ∈ L . Then u = ( I + α A ) − P f ∈ H ∩ H and we have the following estimate: k u k L + α k∇ u k L ≤ k f k L . In particular, the operator ( I + α A ) − P is continuous from L to H .Proof. We know from the results of [3] that u ∈ H ∩ H . We have that u − α ∆ u + ∇ π = f for some π . We multiply the above relation by u and integrate by parts using that u is divergencefree and vanishes at the boundary. We get that k u k L + α k∇ u k L = Z f · u ≤ k f k L k u k L ≤ k f k L ( k u k L + α k∇ u k L ) so k u k L + α k∇ u k L ≤ k f k L . (cid:3) he Fourier transform in R is denoted by F : F ( f )( ξ ) = Z R e − ix · ξ f ( x ) dx The Fourier transform in R is denoted by F R or b g = F R g where g is defined on R : F R g ( s ) = b g ( s ) = Z R e − ist g ( t ) dt. For functions of two variables we will use the partial Fourier transform in the first variable andwe will denote it by F , or also F f = e f . That is, for functions f defined on R or on H we define(5) F ( f )( ξ , x ) = e f ( ξ , x ) ≡ Z R e − ix · ξ f ( x , x ) dx We define in the same manner F the partial Fourier transform in the second variable.We denote by G α the Green’s function of the operator I − α ∆ in R , i.e. (6) G α ( x ) = F − (cid:0)
11 + α | ξ | (cid:1) . We have that(7) G α ( x ) = 1 α G (cid:0) x √ α (cid:1) where G ( x ) = F − (cid:0)
11 + | ξ | (cid:1) . is a function who is exponentially decaying at infinity and has a logarithmic singularity at the origin.The Green’s function of the Laplacian in R is denoted by G ( x ) = 12 π ln | x | . We shall also use the following function(8) H α ( x ) = αG α ( x ) + G ( x ) = αG α ( x ) + 12 π ln | x | . A scalar function ω ∈ L p ( R ) gives rise to a divergence-free vector field u on R whose curl is ω through the Biot-Savart law: u = K ∗ ω , with the (Biot-Savart) kernel K , given by:(9) K ( x ) = ∇ ⊥ G ( x ) = x ⊥ π | x | . We also need to introduce the following smoothed-out kernel(10) K α = K ∗ G α . We recall now several well-known (inverse) Fourier transforms. For all a > F R (cid:0) t + a (cid:1) = πa e − a | s | , (12) F − R (cid:0) s + a (cid:1) = 12 a e − a | t | and(13) F − R ( e − a | s | ) = aπ ( t + a ) . ifferentiating with respect to a the relation above also yields the following inverse Fourier trans-form:(14) F − R ( | s | e − a | s | ) = a − t π ( t + a ) . Applying F to (6) and using (12) we get that F G α ( ξ , x ) = F − (cid:0)
11 + α | ξ | (cid:1) ( ξ , x ) = 12 αξ α e − ξ α | x | . where we used the notation ξ α = r ξ + 1 α . Differentiating with respect to x yields(15) F ∂ G α ( ξ , x ) = − α e − ξ α x ∀ x > . Using (11) we compute(16) F ∂ G ( ξ , x ) = x π F (cid:0) x + x (cid:1) = 12 e − x | ξ | ∀ x > . From (8), (15) and (16) we conclude that for all x > F ∂ H α ( ξ , x ) = α F ∂ G α ( ξ , x ) + F ∂ G ( ξ , x ) = 12 ( e − x | ξ | − e − x ξ α ) . In the remainder of this work we will frequently need to consider the odd extension to R ofa scalar defined on H , as well as the Biot-Savart law induced by the extension. To this end weintroduce the following notation: if q ∈ C ∞ ( H ) then its odd extension will be denoted q = q ( x ) andis given by q ( x ) = ( q ( x ) if x > − q ( x , − x ) if x < . If q ∈ BM ( H ) then we will also need to consider its odd extension, still denoted q . Let ϕ ∈ C ( R ) be a test function and split ϕ into its odd and even parts: ϕ = ϕ o + ϕ e . Then h q, ϕ i ≡ h q, ϕ o i . Of course q ∈ BM ( R ). Furthermore, if q ∈ ˙ H − ( H ) then q ∈ ˙ H − ( R ) = { curl w | w ∈ L ( R ) } .Let us consider the Biot-Savart velocity field in R induced by the odd extension of a measure q ∈ BM ( H ), namely K ∗ q , where K was introduced in (9). The odd symmetry of q induces acovariant symmetry under which the first component ( K ∗ q ) is even while the second component( K ∗ q ) is odd, with respect to x . Hence K ∗ q (cid:12)(cid:12) H is divergence free, tangent to the boundary of H , and its curl in H is q . This vector field can be written as K ∗ q (cid:12)(cid:12) H = Z R K ( x − y ) dq ( y ) = Z H K H ( x, y ) dq ( y ) , where(18) K H ( x, y ) ≡ ( x − y ) ⊥ π | x − y | − ( x − y ) ⊥ π | x − y | where y = ( y , − y ) is the image of y .The Biot-Savart law in the half-plane H is the integral operator acting on q given by(19) K H [ q ] = Z H K H ( x, y ) dq ( y ) , nd K H is the Biot-Savart kernel in the half-plane. Note that K H [ q ] = K ∗ q (cid:12)(cid:12) H .
3. Finding velocity from potential vorticity: the solutionoperator
Let u be the velocity in the half-plane induced by a potential vorticity q , the solution of thesystem of equations (2). The aim of this section is to produce and understand the solution operatorfor u in terms of q .We begin with an estimate for the Biot-Savart law induced by a measure q ∈ BM ( H ). Lemma 4.
Let q ∈ BM ( H ) and consider K H [ q ] as introduced in (19) . Then K H [ q ] ∈ L , ∞ σ and k K H [ q ] k L , ∞ ≤ C k q k BM ( H ) for some universal constant C > .If, in addition, q ∈ ˙ H − ( H ) then K H [ q ] ∈ L σ and k K H [ q ] k L = k q k ˙ H − ( H ) .Proof. Recall that, since q ∈ BM ( H ), its odd extension q belongs to BM ( R ). Since K ∈ L , ∞ ( R )it follows from the Young inequality in Lorentz spaces that K ∗ q ∈ L , ∞ ( R ). Therefore K H [ q ] ∈ L , ∞ σ ( H ) and the estimate follows from said Young’s inequality.Assume, additionally, that q ∈ ˙ H − ( H ). Then q = curl w with w ∈ L . From the properties of theLeray projector P , we know that P w ∈ L σ , i.e. P w is divergence free and tangent to the boundaryof H . Furthermore w − P w is a gradient, so that curl P w = q . We infer that K H [ q ] = P w ∈ L and k K H [ q ] k L = k P w k L = k q k ˙ H − ( H ) . This completes the proof of the lemma. (cid:3) We claim that the solution of (2), u , satisfies( I + α A ) u = K H [ q ] , where A = − P ∆ is the Stokes operator.Indeed, we have that curl[( I − α ∆) u ] = q = curl K H [ q ]so ( I − α ∆) u and K H [ q ] differ by a gradient. Since the Leray projection vanishes for gradient fieldsand reduces to the identity on divergence free vector fields which are tangent to the boundary, itfollows that K H [ q ] = P K H [ q ] = P [( I − α ∆) u ] = P u − α P ∆ u = u + α A u, as desired.Since u vanishes on the boundary of H we find that(20) u = ( I + α A ) − K H [ q ] . From Lemma 4 we have that, if q ∈ ˙ H − ( H ) then K H [ q ] ∈ L ( H ). Hence, from Lemma 3, itfollows that u ∈ L ( H ). Definition 5.
The solution operator for the system of equations (2) , denoted B = B ( q ) , is givenby B : ˙ H − ( H ) → L ( H ) q ( I + α A ) − K H [ q ] . In view of Lemma 3 we actually have B ( q ) ∈ H ∩ H , if q ∈ ˙ H − ( H ), and that B is continuousfrom ˙ H − to H . With this notation the solution of (2) is u = B ( q ). . The existence result In this section we will establish Theorem 1. The strategy is standard, so we will only give asketch of the proof. We emphasize that, in this section, α >
Proof of Theorem 1.
Fix q ∈ BM ( H ) ∩ ˙ H − . We wish to obtain a solution of the α − Eulerequations with initial potential vorticity q . The strategy will be to choose a sequence of smoothapproximations q n to q and solve the α − Euler equations with q n as initial potential vorticities.This results in a sequence of smooth, time-dependent, potential vorticities q n and vector fields u n which we subsequently pass to a weak limit. We will provide sufficient estimates for q n and u n toshow that such a weak limit is a weak solution of the α − Euler equations and has, as initial potentialvorticity, q .The initial velocity u is determined through the solution operator: u = B ( q ), where B wasintroduced in Definition 5. Furthermore, since q ∈ ˙ H − we have, using Lemma 3, that u ∈ H ∩ H .We begin by constructing the smooth approximations. Let q be the odd extension of q to thefull plane as described in Section 2; of course q ∈ BM ( R ) and, also, q ∈ ˙ H − ( R ). Consider theBiot-Savart law in R for q : K ∗ q , where K is the Biot-Savart kernel given in (9). Since q ∈ ˙ H − we have that K ∗ q ∈ L ( R ). We mollify q with an even, smooth, compactly supported mollifier φ n , and we denote by q n = q ∗ φ n the resulting mollified potential vorticity. Let q n = q n (cid:12)(cid:12) H . Clearlywe have that k q n k L ≤ k q k BM ( H ) , k q n k ˙ H − ≤ k q k ˙ H − . In addition, since clearly we have q n ⇀ q weak − ∗ BM ( R ), it follows that q n ⇀ q weak − ∗ BM ( H ) . By construction we have K ∗ q n → K ∗ q strongly in L ( R ). Therefore(21) K H [ q n ] → K H [ q ] strongly in L ( H ) . Let u n = B ( q n ). Using (21) and Lemma 3 we find that(22) k u n k L + α k∇ u n k L → k u k L + α k∇ u k L as n → ∞ .Because q ∈ ˙ H − ( R ) we deduce that, for each fixed n , q n ∈ L ( R ). Hence q n ∈ L and thereforewe have that K H [ q n ] ∈ H ∩ L σ , so that u n = ( I + α A ) − K H [ q n ] belongs to H ∩ H . For such initialdata there exists a global solution u n in H of the α –Euler equations having u n as initial velocity.To see this one can use, for instance the method employed in [13]. Since q n = curl( u n − α ∆ u n )is transported by u n and div u n = 0, it follows that k q n ( · , t ) k L is a conserved quantity. Because q n (0) = q n is bounded in L we infer that q n is bounded in L ∞ ( R + ; L ) and(23) k q n k L ∞ ( R + ; L ) = k q n k L ≤ k q k BM ( H ) . In addition, if we multiply the equation of u n by u n and integrate by parts we get the classicalenergy equality:(24) k u n ( t ) k L + α k∇ u n ( t ) k L = k u n (0) k L + α k∇ u n (0) k L ∀ t ≥ . In particular, u n is bounded in L ∞ ( R + ; H ).Given the boundedness of u n in L ∞ ( R + ; H ) and of q n in L ∞ ( R + ; L ), we can extract subsequencesrelabeled u n and q n which converge u n ⇀ u in L ∞ ( R + ; H ) weak ∗ and(25) q n ⇀ q in L ∞ ( R + ; BM ( H )) weak ∗ . s the equations are nonlinear, this convergence is not sufficient to pass to the limit in the weakform of the α -Euler equations. We now proceed as follows. Let p ∈ (1 ,
2) and q ∈ (2 , ∞ ) be dualindexes: p + q = 1. We will prove that ∂ t u n is bounded in L p independently of n . Let ϕ ∈ L qσ .Recall that A = − P ∆ is the Stokes operator. Set ψ = ( I + α A ) − ϕ . Then ψ ∈ W ,q ∩ W ,q ∩ L qσ and k ψ k W ,q ≤ C k ϕ k L q (see [3]).The velocity formulation of the α –Euler equations can be written as follows (see [11]): ∂ t ( u n − α ∆ u n ) + u n · ∇ u n − α X j,k ∂ j ∂ k ( u nj ∂ k u n ) + α X j,k ∂ j ( ∂ k u nj ∂ k u n ) − α X j,k ∂ k ( ∂ k u nj ∇ u nj ) + ∇ p n = 0 . We apply the Leray projector P above and multiply by ψ . We get(26) h ∂ t ( u n + α A u n ) , ψ i = −h P (cid:2) u n · ∇ u n − α X j,k ∂ j ∂ k ( u nj ∂ k u n ) + α X j,k ∂ j ( ∂ k u nj ∂ k u n ) − α X j,k ∂ k ( ∂ k u nj ∇ u nj ) (cid:3) , ψ i = −h u n · ∇ u n − α X j,k ∂ j ∂ k ( u nj ∂ k u n ) + α X j,k ∂ j ( ∂ k u nj ∂ k u n ) − α X j,k ∂ k ( ∂ k u nj ∇ u nj ) , ψ i . Observe that h ∂ t ( u n + α A u n ) , ψ i = h ∂ t u n , ( I + α A ) ψ i = h ∂ t u n , ϕ i . Next we estimate the first of the four terms in (26): |h u n · ∇ u n , ψ i| = (cid:12)(cid:12)Z H u n · ∇ u n · ψ (cid:12)(cid:12) ≤ k u n k H k ψ k L ∞ ≤ C k u n k H k ψ k W ,q ≤ C k u n k H k ϕ k L q . To estimate the second and third terms in (26) we integrate by parts, using that u n and ψ vanishat the boundary. We deduce that: |h α∂ j ∂ k ( u nj ∂ k u n ) , ψ i| = (cid:12)(cid:12) α Z ∂ j ∂ k ( u nj ∂ k u n ) · ψ (cid:12)(cid:12) = (cid:12)(cid:12) α Z u nj ∂ k u n · ∂ j ∂ k ψ (cid:12)(cid:12) ≤ C k u n k L qq − k ∂ k u n k L k ∂ j ∂ k ψ k L q ≤ C k u n k H k ϕ k L q and |h α∂ j ( ∂ k u nj ∂ k u n ) , ψ i| = (cid:12)(cid:12) α Z ∂ k u nj ∂ k u n · ∂ j ψ (cid:12)(cid:12) ≤ α k u n k H k∇ ψ k L ∞ ≤ C k u n k H k ψ k W ,q ≤ C k u n k H k ϕ k L q . The same estimate holds true for the last term in (26). We have thus obtained the following bound |h ∂ t u n , ϕ i| ≤ C k u n k H k ϕ k L q ∀ ϕ ∈ L qσ . Since L qσ is dual to L pσ (see [3]), we deduce that k ∂ t u n k L p ≤ C k u n k H . Therefore ∂ t u n is bounded in L ∞ ( R + ; L p ). By the Ascoli theorem and the compact embedding L p ֒ → W − ,ploc , we infer, passing to subsequences as necessary, that:(27) u n → u strongly in C ( R + ; W − ,ploc ) . ext we have that: Claim 6.
The operator B is bounded from BM ( H ) to W ,ploc ( H ) for all p < Proof of Claim:
From the results of [3] we know that, for all 1 < r < ∞ , the operator ϕ ( I + α A ) − ϕ is bounded from L rσ to L r . By interpolation, we infer that it is also bounded from L , ∞ σ to L , ∞ . The embedding L , ∞ ֒ → L ploc implies that ϕ
7→ ∇ ( I + α A ) − ϕ is bounded from L , ∞ σ to L ploc . A similar argument shows that the operators ϕ
7→ ∇ ( I + α A ) − ϕ and ϕ ( I + α A ) − ϕ arebounded from L , ∞ σ to L ploc . We conclude that the operator ( I + α A ) − is bounded from L , ∞ σ ( H ) to W ,ploc ( H ) for all p < K H [ q ] ∈ L , ∞ σ for all q ∈ BM ( H )and B ( q ) = ( I + α A ) − K H [ q ]. (cid:3) Since q n is bounded in BM ( H ) we find, in view of Claim 6, that u n is bounded in L ∞ ( R + ; W ,ploc ( H ))for all p <
2. Then, interpolation together with the uniform convergence (27) yield u n → u strongly in C ( R + ; W s,ploc ) ∀ s < , p < . Using Sobolev embeddings we further deduce that u n → u strongly in C ( R + × H ) . Recalling the weak convergence of q n expressed in (25) we finally deduce that u n q n → uq in D ′ ( R ∗ + × H )so div( u n q n ) → div( uq ) in D ′ ( R ∗ + × H ) . We infer that q is a solution of the α –Euler equations. Moreover, the bound (4) follows from (23)and (25).It remains to prove the bound (3). We proceed in the following manner. From (27) we deducethat u n ( t ) → u ( t ) in D ′ for all t . Since u n ( t ) is bounded in H we infer that u n ( t ) ⇀ u ( t ) weaklyin H for all t . Therefore k u ( t ) k L ≤ lim inf n →∞ k u n ( t ) k L and k∇ u ( t ) k L ≤ lim inf n →∞ k∇ u n ( t ) k L for all t ≥
0. We finally deduce from (22) and (24) that k u ( t ) k L + α k∇ u ( t ) k L ≤ lim inf n →∞ ( k u n ( t ) k L + α k∇ u n ( t ) k L )= lim inf n →∞ ( k u n (0) k L + α k∇ u n (0) k L )= k u k L + α k∇ u k L for all t ≥
0. This proves (3) and completes the proof of Theorem 1. (cid:3)
5. Finding velocity from potential vorticity: interior andboundary parts
In Section 3 we found the solution operator which gives the velocity in terms of potential vorticity.The resulting expression, however, is not explicit enough for us to pass to the limit α →
0. Here wewill produce a representation formula for u in terms of q in which we write u as a sum of a vectorfield u int , constructed using the method of images but with possibly non-vanishing boundary value,and a vector field u bdry which corrects the boundary condition.We will use the notation introduced in Section 2, particularly (6), (19) and (10). et us start by introducing(28) u int ≡ G α ∗ ( K ∗ q ) ≡ K α ∗ q. Then ( I − α ∆) u int = K ∗ q and div u int = 0 in R . Moreover, since G α is radial it follows that u int inherits the symmetry properties of K ∗ q . Inparticular, the second component of u int is odd with respect to x , so that u int vanishes at x = 0.Let(29) u bdry = u − u int . Then u bdry is divergence free in H and tangent to the boundary of H . Moreover,curl( I − α ∆) u bdry = curl( I − α ∆) u − curl( I − α ∆) u int = q − curl( K ∗ q ) = 0 in H . Thus there exists some scalar function p such that u bdry − α ∆ u bdry + ∇ p = 0 in H . We denote by g the trace of the first component of u bdry , u bdry , on the boundary of H :(30) g ( x ) ≡ u bdry ( x ,
0) = − u int ( x ,
0) = − G α ∗ ( K ∗ q ) ( x , . We conclude that w = u bdry satisfies the following system of equations w − α ∆ w + ∇ p = 0 in H (31) div w = 0 in H (32) w (cid:12)(cid:12) x =0 = g ( x )(33) w (cid:12)(cid:12) x =0 = 0 . (34)We will find a formula for the solution of this problem through a method which was employed bySolonnikov [18] to find the Green’s function of the evolutionary Stokes operator in the half-space.Let ∗ denote the convolution in the first variable and ∗ H denote the convolution on H defined asfollows. If ϕ, ψ are functions defined on H , the convolution ϕ ∗ H ψ is a function defined on H whosevalue is ϕ ∗ H ψ ( x , x ) = Z R Z x ϕ ( y ) ψ ( x − y ) dy Next we introduce a pair of functions which will appear frequently in what follows:(35) η = η ( x , x ) = x − x | x | , and(36) η = η ( x , x ) = x x | x | . With this notation we have the following result.
Proposition 7.
The solution w = ( w , w ) of the problem (31) – (34) is given by the followingformula: w = − αg ∗ ∂ G α + 2 απ g ∗ ∂ G α ∗ H η w = − απ g ∗ ∂ G α ∗ H η . emark 8. We note that, if ϕ , ψ and ρ are functions on H , then ϕ ∗ ( ψ ∗ H ρ ) = ( ϕ ∗ ψ ) ∗ H ρ as long as ϕ is independent of x . Proof.
We will use the notations F and ˜ introduced in (5). Taking the divergence of (31) yields∆ p = 0. We apply F and deduce that ( ∂ x − ξ ) e p = 0. So e p is a linear combination of e ± x | ξ | withcoefficients functions of ξ . Because e p can’t exhibit exponential growth at infinity, we infer that e p ( ξ , x ) = C ( ξ ) e − x | ξ | . Let now A = w + ∇ p. Because p is harmonic, relation (31) implies that A − α ∆ A = 0. We apply as above F and deducethat ( ∂ x − ξ − α ) e A = 0. Since e A can’t exhibit exponential growth at infinity either, we infer thatthere exists C ( ξ ) and C ( ξ ) such that e A ( ξ , x ) = e − x ξ α (cid:18) C ( ξ ) C ( ξ ) (cid:19) . We conclude that(37) e w ( ξ , x ) = F ( A − ∇ p )= e − x ξ α (cid:18) C ( ξ ) C ( ξ ) (cid:19) − (cid:18) iξ ∂ x (cid:19) (cid:2) C ( ξ ) e − x | ξ | (cid:3) = e − x ξ α (cid:18) C ( ξ ) C ( ξ ) (cid:19) − e − x | ξ | C ( ξ ) (cid:18) iξ −| ξ | (cid:19) . Applying the Fourier transform F to (32)–(34) we get iξ e w + ∂ x e w = 0(38) e w ( ξ ,
0) = b g ( ξ )(39) e w ( ξ ,
0) = 0 . (40)Using (37) in the three equations above yields a linear system of three equations in the unknowns C , C and C . This system can be easily solved allowing in return to compute e w . Plugging (37)in (39) gives C ( ξ ) = b g ( ξ ) + iξ C ( ξ )and in (40) gives C ( ξ ) = −| ξ | C ( ξ )so that e w ( ξ , x ) = e − x ξ α (cid:18)b g ( ξ )0 (cid:19) + C ( ξ ) (cid:18) iξ −| ξ | (cid:19) ( e − x ξ α − e − x | ξ | ) . Using this in (38) yields the following value for C : C ( ξ ) = iξ ξ − | ξ | ξ α b g ( ξ ) = − iα ξ | ξ | ( | ξ | + ξ α ) b g ( ξ ) . So e w = e − x ξ α (cid:18)b g ( ξ )0 (cid:19) + α (cid:18) | ξ | iξ (cid:19) ( | ξ | + ξ α )( e − x ξ α − e − x | ξ | ) b g ( ξ ) . bserving that e − x ξ α − e − x | ξ | | ξ | − ξ α = Z x e − y ξ α e −| ξ | ( x − y ) dy we finally find the following formula for e w :(41) e w = e − x ξ α (cid:18)b g ( ξ )0 (cid:19) + (cid:18) i ξ | ξ | − (cid:19) iξ b g ( ξ ) Z x e − y ξ α e −| ξ | ( x − y ) dy . We would now like to take the inverse Fourier transform in the first variable to find the formulafor w . We deal first with the simplest term above, that is the term e − x ξ α b g ( ξ ). Using (15) we havethat(42) F − [ e − x ξ α b g ( ξ )] = g ∗ F − ( e − x ξ α ) = − αg ∗ ∂ G α where we recall that ∗ denotes the convolution in the first variable.Now let us express the other terms appearing in (41). Taking the second component of (41)yields e w = − iξ b g ( ξ ) Z x e − y ξ α e −| ξ | ( x − y ) dy . We apply F − above. Recalling (13) and (15) we can write w = −F − ( b g ( ξ )) ∗ Z x F − (cid:0) e − y ξ α (cid:1) ∗ F − (cid:0) iξ e −| ξ | ( x − y ) (cid:1) dy = − g ∗ Z x ( − α∂ G α ( · , y )) ∗ ∂ F − (cid:0) e −| ξ | ( x − y ) (cid:1) dy = − g ∗ Z x ( − α∂ G α ( · , y )) ∗ ∂ ρ ( · , x − y ) dy , with ρ ( x , x ) ≡ x / | x | .Now, ∂ ρ = − η , with η as in (36), so that(43) w = 2 απ g ∗ ∂ G α ∗ H ∂ ρ = − απ g ∗ ∂ G α ∗ H η . We can deduce in a similar fashion using (14) and (15) that F − h | ξ | b g ( ξ ) Z x e − y ξ α e −| ξ | ( x − y ) dy i = g ∗ Z x F − (cid:0) e − y ξ α (cid:1) ∗ F − (cid:0) | ξ | e −| ξ | ( x − y ) (cid:1) dy = g ∗ Z x ( − α∂ G α ( · , y )) ∗ π η ( · , x − y ) dy = − απ g ∗ ∂ G α ∗ H η , with η as in (35).Putting together relations (41), (42), (43) and the above equalities completes the proof of theproposition. (cid:3) To be able to give a complete, explicit, formula for u in terms of q it remains to express theboundary data g defined in (30) in terms of q . We proceed in the following manner.Recall that g ( x ) = − ( G α ∗ ( K ∗ q ) )( x , . Recall that K = ∇ ⊥ G , where G = π ln | x | is the Green function of the Laplacian in R . Therefore g ( x ) = − ( G α ∗ ( ∇ ⊥ G ) ∗ q )( x , . ow, since G α is the Green function of I − α ∆ in R we have that ( I − α ∆) G α = δ so G α = α ∆ G α + δ .Recall the vector field H α introduced in (8). We have, then: G α ∗ ∇ ⊥ G = ( α ∆ G α + δ ) ∗ ∇ ⊥ G = α ∆ G α ∗ ∇ ⊥ G + ∇ ⊥ G = αG α ∗ ∇ ⊥ ∆ G + ∇ ⊥ G = αG α ∗ ∇ ⊥ δ + ∇ ⊥ G = ∇ ⊥ ( αG α + G )= ∇ ⊥ H α . We infer that g ( x ) = ( ∂ H α ∗ q )( x , ∂ H α and q are both odd with respect to x we finally get that g ( x ) = − Z H ∂ H α ( x − y , y ) q ( y ) dy. From (28) we have that(44) u int = G α ∗ x ⊥ π | x | ∗ q = K α ∗ q where K α = G α ∗ x ⊥ π | x | = G α ∗ ∇ ⊥ G = ∇ ⊥ H α . We deduce from relations (29), (44) and Proposition 7 the following formula for the solution of(2).
Proposition 9.
The solution of (2) is given by u = u int + u bdry where u int is the interior part u int ≡ K α ∗ q and u bdry is the contribution of the boundary which takes the form u bdry ≡ − αg ∗ ∂ G α + 2 απ g ∗ ∂ G α ∗ H η (45) u bdry ≡ − απ g ∗ ∂ G α ∗ H η and (46) g ( x ) = − Z H ∂ H α ( x − y , y ) q ( y ) dy. Above, η and η were introduced in (35), (36). . Boundary part estimates The purpose of this section is to obtain estimates for the boundary correction term u bdry in theinterior of H . We will show that, for any ε >
0, if x > ε then u bdry is bounded, uniformly withrespect to x and α ≪
1, in terms of the total variation of the potential vorticity.We begin with a monotonicity result.
Lemma 10.
We have that ∂ H α ≥ in the half-plane H .Proof. The function G , which is a Bessel potential, satisfies: G ( x ) = 14 π Z ∞ t e − π | x | t e − t π dt, see [19, page 132].Assume that x >
0. We use (7) and (8) to deduce that ∂ H α = α∂ G α + x π | x | = 1 √ α ∂ G (cid:0) x √ α (cid:1) + x π | x | = − x α Z ∞ t e − π | x | tα e − t π dt + x π | x | ≥ − x α Z ∞ t e − π | x | tα dt + x π | x | = 0 . (cid:3) Next, we will show that the boundary value of u bdry , g , see Proposition 9, can be estimated in L by the total variation of the potential vorticity. Proposition 11.
Assume that q ∈ BM ( H ) and let g be the boundary value of u bdry , as definedthrough (46) . Then g ∈ L ( R ) and we have k g k L ( R ) ≤ k q k BM ( H ) . Proof.
According to Lemma 10 we know that ∂ H α ( x − y , y ), the kernel in (46), is non-negative.We take the absolute value in (46), we integrate and we use the Fubini theorem to deduce that Z R | g ( x ) | dx ≤ Z H (cid:16)Z R ∂ H α ( x , y ) dx (cid:17) d | q | ( y ) . We find, for y >
0, that: Z R ∂ H α ( x , y ) dx = F ∂ H α (0 , y ) = 12 − e − y √ α ≤ , where we used (17). Therefore Z R | g ( x ) | dx ≤ Z H d | q | ( y ) = k q k BM ( H ) , as we wished. (cid:3) We proceed with two technical lemmas. emma 12. Let η be a function homogeneous of degree γ with γ < − and smooth on R \ { } .There exists a constant C = C ( η ) such that for all x = 0 we have that k η ( · , x ) k L ( R ) ≤ Cx γ + . Proof.
We have that | η ( x ) | ≤ C | x | γ so Z R | η ( x ) | dx ≤ C Z R | x | γ dx = C Z R ( x + x ) γ dx = Cx γ +12 Z R ( t + 1) γ dt where we changed variables x = tx . (cid:3) Lemma 13.
There exists a universal constant C such that, for all x > , we have: k ∂ G α ( · , x ) k L ∞ ( R ) ≤ Cαx e − x √ α , (47) k ∂ G α ( · , x ) k L ( R ) ≤ Cα √ x e − x √ α , (48) and k ∂ ∂ G α ( · , x ) k L ( R ) ≤ Cαx √ x e − x √ α . Proof.
Using (15) we obtain k ∂ G α ( · , x ) k L ∞ ( R ) ≤ π kF ∂ G α ( · , x ) k L ( R ) = 14 πα Z R e − x √ ξ + α dξ = 14 πα Z | ξ | < √ α e − x √ ξ + α dξ + 14 πα Z | ξ | > √ α e − x √ ξ + α dξ ≤ πα Z | ξ | < √ α e − x √ α dξ + 14 πα Z | ξ | > √ α e − x | ξ | dξ = 12 πα √ α e − x √ α + 12 παx e − x √ α ≤ Cαx e − x √ α , where we used that the function se − s is bounded on R . This establishes (47).To prove (48) we use the Plancherel theorem and relation (15) to write k ∂ G α ( · , x ) k L ( R ) = 1 √ π kF ∂ G α ( · , x ) k L ( R ) = 12 α √ π k e − x √ ξ + α k L ( dξ ) . ext we have: k e − x √ ξ + α k L ( dξ ) = Z R e − x √ ξ + α dξ = Z | ξ | < √ α e − x √ ξ + α dξ + Z | ξ | > √ α e − x √ ξ + α dξ ≤ Z | ξ | < √ α e − x √ α dξ + Z | ξ | > √ α e − x | ξ | dξ = 2 √ α e − x √ α + 1 x e − x √ α ≤ Cx e − x √ α . This proves (48).Similarly, the Plancherel theorem and relation (15) imply that k ∂ ∂ G α ( · , x ) k L ( R ) = 1 √ π kF ∂ ∂ G α ( · , x ) k L ( R ) = 12 α √ π k ξ e − x √ ξ + α k L ( dξ ) . Furthermore we have: k ξ e − x √ ξ + α k L ( dξ ) = Z R ξ e − x √ ξ + α dξ = Z | ξ | < √ α ξ e − x √ ξ + α dξ + Z | ξ | > √ α ξ e − x √ ξ + α dξ ≤ Z | ξ | < √ α ξ e − x √ α dξ + Z | ξ | > √ α ξ e − x | ξ | dξ = 23 α √ α e − x √ α + e − x √ α (cid:0) αx + 1 √ αx + 12 x (cid:1) ≤ Cx e − x √ α . This completes the proof. (cid:3)
With these estimates in hand we can now establish the main result in this section.
Proposition 14.
Let u bdry be given as in Proposition 9. There exists a universal constant C > such that | u bdry ( x ) | ≤ C k q k BM ( H ) α x − ∀ x ∈ H . Proof.
We first estimate the first component of u bdry , given by u bdry = − αg ∗ ∂ G α + 2 απ g ∗ ∂ G α ∗ H η , see (45), where η was defined in (35).We use Young’s inequality and (47) to bound the first term:(49) k − αg ∗ ∂ G α ( · , x ) k L ∞ ( R ) ≤ α k g k L k ∂ G α ( · , x ) k L ∞ ( R ) ≤ C k g k L e − x √ α x o bound the second term in the expression for u bdry we note that η ( x ) = x − x | x | = ∂ ( x / | x | ) ≡ ∂ ζ . We write this term as2 απ g ∗ ∂ G α ∗ H η ( x ) = 2 απ Z x g ∗ ∂ G α ( · , y ) ∗ η ( · , x − y ) dy = 2 απ Z x g ∗ ∂ G α ( · , y ) ∗ η ( · , x − y ) dy + 2 απ Z x x g ∗ ∂ G α ( · , y ) ∗ η ( · , x − y ) dy ≡ I + I . We use Lemmas 12 and 13 and the Young inequality to bound I as follows: k I ( · , x ) k L ∞ ( R ) ≤ Cα Z x k g k L k ∂ G α ( · , y ) k L ( R ) k η ( · , x − y ) k L ( R ) dy ≤ C k g k L Z x e − y √ α √ y ( x − y ) − dy ≤ C k g k L x − Z x e − y √ α √ y dy = C k g k L x − α Z x √ α e − t √ t dt ≤ C k g k L x − α . To estimate I we write first I = 2 απ Z x x g ∗ ∂ G α ( · , y ) ∗ η ( · , x − y ) dy = 2 απ Z x x g ∗ ∂ G α ( · , y ) ∗ ∂ ζ ( · , x − y ) dy = 2 απ Z x x g ∗ ∂ ∂ G α ( · , y ) ∗ ζ ( · , x − y ) dy . We can now bound as above with the help of Lemmas 12 and 13 and the Young inequality: k I ( · , x ) k L ∞ ( R ) ≤ Cα Z x x k g k L k ∂ ∂ G α ( · , y ) k L ( R ) k ζ ( · , x − y ) k L ( R ) dy ≤ C k g k L Z x x e − y √ α y √ y √ x − y dy ≤ C k g k L x − e − x √ α Z x x √ x − y dy ≤ C k g k L e − x √ α x . he estimates for I and I obtained above yield the following bound:(50) | απ g ∗ ∂ G α ∗ H x − x | x | | ≤ C k g k L α x + C k g k L e − x √ α x . Now, by Proposition 11 we have that k g k L is bounded by k q k BM ( H ) . In addition, the estimate e − z ≤ √ z applied for z = x √ α shows that the second term on the rhs above is bounded by the firstterm. Therefore, (49), (50) yield the desired estimate for u bdry .Next recall that the second component of u bdry is given by u bdry = − απ g ∗ ∂ G α ∗ H η , where η is as in (36).Noticing that η = x x | x | = − ∂ (cid:0) x | x | (cid:1) we can use the same analysis leading up to (50) to estimate u bdry .This concludes the proof. (cid:3)
7. Passing to the limit α → We are now ready to prove the main result of this work, the convergence, passing to subsequencesas needed, of solutions of the α -Euler equations in the half-plane, with no-slip boundary conditions,to solutions of the Euler equations, assuming a sign condition on the singular part of the initial(potential) vorticity. In other words, we show that, if v is any weak limit, in a sense to be madeprecise, of solutions u α of the α -Euler equations in the half-plane, with an initial potential vorticity q ∈ (cid:0) BM + ( H ) + L ( H ) (cid:1) ∩ ˙ H − ( H ), such that u α vanishes on ∂ H , then v is a weak solution of theincompressible Euler equations in H , with initial vorticity q . The proof is based on the proof ofthe well-known Delort Theorem, see [8], on the existence of vortex sheet evolution for the 2D Eulerequations when vorticity has a distinguished sign. From a technical point-of-view we make use ofthe idea of symmetrization of the kernels involved in recovering velocity from (potential) vorticity,as was done in [17]. Below, the gradient is taken in the spatial variable and not in time.Let us begin by describing a weak formulation of the α -Euler equations which will be useful forour purposes. Fix T >
0. Let q ∈ BM ( H ). Let u α ∈ C b ( R + ; H w ( H )) ∩ C ( R + × H ) be the solutionwith initial potential vorticity q obtained in Theorem 1. Recall that q α ∈ L ∞ ( R + ; BM ( H )).Let ϕ ∈ C ∞ c ([0 , T ) × H ) be a test function. Then ϕ ∈ C ∞ c ([0 , T ) × H ε ) for some ε >
0, where H ε denotes the half-plane { x > ε } . We multiply the vorticity formulation of the α -Euler equation by ϕ and integrate by parts to obtain Z T Z H ∂ t ϕ dq α dt + Z H ϕ (0 , x ) dq ( x ) = − Z T Z H u α · ∇ ϕ dq α dt = − Z T Z H u α,int · ∇ ϕ dq α dt − Z T Z H u α,bdry · ∇ ϕ dq α dt where we used the decomposition u α = u α,int + u α,bdry introduced in Proposition 9. From Proposition14 we get that k u α,bdry k L ∞ ( H ε ) ≤ C ε k q α k BM ( H ) , for some C ε > α , for 0 < α <
1. From the bound (4) on the BM ( H )-norm of q α we infer that(51) (cid:12)(cid:12)(cid:12)Z T Z H u α,bdry · ∇ ϕ dq α dt (cid:12)(cid:12) ≤ C ε T k q k BM ( H ) k∇ ϕ k L ∞ . e write next Z T Z H u α,int · ∇ ϕ dq α dt = Z T Z H K α ∗ q α ( x ) · ∇ ϕ ( t, x ) dq α ( x ) dt = Z T Z Z H × R K α ( x − y ) · ∇ ϕ ( t, x ) dq α ( x ) dq α ( y ) dt = Z T Z Z H × H [ K α ( x − y ) − K α ( x − y )] · ∇ ϕ ( t, x ) dq α ( x ) dq α ( y ) dt ≡ Z T Z Z H × H N α ( t, x, y ) dq α ( x ) dq α ( y ) dt, where y = ( y , − y ) is the image of y ; above we have used that q α is the odd extension of q α to R .Next we symmetrize and write Z T Z H u α,int · ∇ ϕ dq α dt = Z T Z Z H × H N α ( t, x, y ) + N α ( t, y, x )2 dq α ( x ) dq α ( y ) dt ≡ Z T Z Z H × H H αϕ ( t, x, y ) dq α ( x ) dq α ( y ) dt. Since K α is odd we have that(52) H αϕ ( x, y ) = 12 K α ( x − y ) · [ ∇ ϕ ( t, x ) − ∇ ϕ ( t, y )] − K α ( x − y ) · ∇ ϕ ( t, x ) − K α ( y − x ) · ∇ ϕ ( t, y ) . We have established that u α satisfies the following weak formulation:(53) Z T Z H ∂ t ϕdq α dt + Z T Z Z H × H H αϕ ( t, x, y ) dq α ( x ) dq α ( y ) dt + Z T Z H u α,bdry · ∇ ϕ dq α dt + Z H ϕ (0 , x ) dq ( x ) = 0 . We now recall the weak vorticity formulation of the incompressible Euler equations in the half-plane for vortex sheet regularity. For a further discussion of weak vorticity formulations in otherdomains see [14] and [12].
Definition 15.
Let ω ∈ BM ( H ) ∩ ˙ H − ( H ) . We say ω ∈ L ∞ (0 , T ; BM ( H ) ∩ ˙ H − ( H )) is a solutionof the weak vorticity formulation of the incompressible Euler equations, with initial data ω , if, forany ϕ ∈ C ∞ c ([0 , T ) × H ) , the following identity holds true: Z T Z H ∂ t ϕdωdt + Z T Z Z H × H H ϕ ( t, x, y ) dω ( x ) dω ( y ) dt + Z H ϕ (0 , x ) dω ( x ) dt = 0 . The auxiliary test function H ϕ is given by H ϕ = H ϕ ( t, x, y ) ≡ K H ( x, y ) · ∇ ϕ ( t, x ) + K H ( y, x ) · ∇ ϕ ( t, y )2 . In view of the expression for K H , see relation (18), it follows that(54) H ϕ ( t, x, y ) = 12 K ( x − y ) · [ ∇ ϕ ( t, x ) − ∇ ϕ ( t, y )] − K ( x − y ) · ∇ ϕ ( t, x ) − K ( y − x ) · ∇ ϕ ( t, y ) . e will need some properties of the kernels K and K α , which we collect in the following propo-sition. Proposition 16.
There exists a universal constant
C > such that, for all x ∈ R \ { } , we havethat | K α ( x ) | ≤ C | x | and | K α ( x ) − K ( x ) | ≤ C √ α | x | . In particular, for any θ > we have that K α α → −−→ K uniformly in the set | x | > θ .Proof. The first bound was proved in [1, pages 703 and 715]. To prove the second bound we recallthat F ( K ) = F (cid:0) x ⊥ π | x | (cid:1) = − i ξ ⊥ | ξ | . Then F ( K α ) = F ( K ∗ G α ) = F ( K ) F ( G α ) = − i ξ ⊥ | ξ | (1 + α | ξ | ) . We can now estimate k| x | ( K α − K ) k L ∞ ( R ) ≤ π kF (cid:2) | x | ( K α − K ) (cid:3) k L ( R ) = 12 π k ∆ F ( K α − K ) k L ( R ) = 12 π (cid:13)(cid:13)(cid:13)(cid:13) ∆ (cid:0) αξ ⊥ α | ξ | (cid:1)(cid:13)(cid:13)(cid:13)(cid:13) L ( R ) = √ α π (cid:13)(cid:13)(cid:13)(cid:13) ∆ (cid:0) ξ ⊥ | ξ | (cid:1)(cid:13)(cid:13)(cid:13)(cid:13) L ( R ) . This completes the proof of the proposition. (cid:3)
We will also need some special properties of the functions H αϕ and H ϕ . We will see that, in theanalysis of convergence of the nonlinear terms in the proof of Theorem 2, time will be treated as aparameter. Thus we omit the dependence on t in both H αϕ and H ϕ . Lemma 17.
We have that a) The function H ϕ is smooth on H × H \ supp( ϕ ) × supp( ϕ ) , supported in H ε × H ∪ H × H ε and vanishes on the boundary of H × H and at infinity. In addition, H αϕ is also supported in H ε × H ∪ H × H ε . b) The functions H αϕ and H ϕ are uniformly bounded. More precisely, there exists a constant C ε such that (55) | H αϕ ( x, y ) | ≤ C ε k ϕ k W , ∞ and | H ϕ ( x, y ) | ≤ C ε k ϕ k W , ∞ . c) For all θ > we have that H αϕ α → −−→ H ϕ uniformly in the set | x − y | > θ . d) There exists a non-negative function F continuous on H × H , vanishing at infinity and suchthat | H αϕ | ≤ F for all α and | H ϕ | ≤ F .Proof. We prove first a). The function N ( x, y ) = [ K ( x − y ) − K ( x − y )] · ∇ ϕ ( x )is clearly smooth on H × H \ supp( ϕ ) × supp( ϕ ) and supported in H ε × H (recall that ϕ is supportedin H ε ). In particular it vanishes if x ∈ ∂ H . It also vanishes if y ∈ ∂ H because for such a y we havethat y = y . Obviously(56) | N ( x, y ) | ≤ C (cid:0) | x − y | + 1 | x − y | (cid:1) |∇ ϕ ( x ) | ≤ C |∇ ϕ ( x ) || x − y | . e infer that N vanishes at infinity. Since(57) H ϕ ( x, y ) = N ( x, y ) + N ( y, x )2we deduce that H ϕ have all the properties listed in a). We observe in a similar manner that H αϕ issupported in H ε × H ∪ H × H ε . This completes the proof of part a).To prove part b), we recall the definition of H αϕ given in (52) and use Proposition 16 to bound | H αϕ ( x, y ) | ≤ C |∇ ϕ ( x ) − ∇ ϕ ( y ) || x − y | + C |∇ ϕ ( x ) || x − y | + C |∇ ϕ ( y ) || y − x |≤ C k∇ ϕ k L ∞ + C |∇ ϕ ( x ) | + |∇ ϕ ( y ) | x + y ≤ C k∇ ϕ k L ∞ + Cε k∇ ϕ k L ∞ ≤ C ε k ϕ k W , ∞ where we also used that supp ϕ ⊂ H ε . The same argument works for H ϕ so this proves b).To prove c), we subtract (54) from (52) and use the last estimate from Proposition 16 to bound | H αϕ ( x, y ) − H ϕ ( x, y ) | ≤ C √ α |∇ ϕ ( x ) − ∇ ϕ ( y ) || x − y | + C √ α |∇ ϕ ( x ) || x − y | + C √ α |∇ ϕ ( y ) || y − x | ≤ C √ α |∇ ϕ ( x ) − ∇ ϕ ( y ) || x − y | + C √ α |∇ ϕ ( x ) | ε + C √ α |∇ ϕ ( y ) | ε ≤ C √ α k∇ ϕ k L ∞ | x − y | + C √ α k∇ ϕ k L ∞ ε . If we assume that | x − y | > θ > | H αϕ ( x, y ) − H ϕ ( x, y ) | ≤ C √ α (cid:16) k∇ ϕ k L ∞ θ + k∇ ϕ k L ∞ ε (cid:17) α → −−→ | H αϕ ( x, y ) | ≤ C |∇ ϕ ( x ) | + |∇ ϕ ( y ) || x − y | ≤ C |∇ ϕ ( x ) | + |∇ ϕ ( y ) | + 1 | x − y | . Recalling the uniform bound (55) one can easily check that the function F ( x, y ) = min (cid:16) C ε k ϕ k W , ∞ , C |∇ ϕ ( x ) | + |∇ ϕ ( y ) | + 1 | x − y | (cid:17) has all the required properties. This completes the proof of the lemma. (cid:3) Remark 18.
The properties of H ϕ above have been discussed and used in [14] and, in fact, theyhold for more general domains. See [12] for a thorough account.We are now ready to establish our main result. Proof of Theorem 2.
Let us fix q ∈ (cid:0) BM + ( H ) + L ( H ) (cid:1) ∩ ˙ H − ( H ), independent of α . Let u α , q α solve the α -Euler equations as given by Theorem 1. Since q ∈ ˙ H − ( H ) there exists f ∈ L suchthat curl f = q . We know that u α, = ( I + α A ) − P f so, by Lemma 3, we have that k u α, k L + α k∇ u α, k L ≤ k f k L . The energy inequality (3) now implies that k u α ( t ) k L + α k∇ u α ( t ) k L ≤ k f k L ∀ t ≥ o u α is bounded in L ∞ ( R + ; L σ ) uniformly in α . Then there exists some v ∈ L ∞ ( R + ; L σ ) and somesubsequence of u α , which we do not relabel, such that(58) u α ⇀ v in L ∞ ( R + ; L σ ) weak ∗ as α → q α is bounded in L ∞ ( R + ; BM ( H )) uniformly in α . So there existssome ω ∈ L ∞ ( R + ; BM ( H )) and some subsequence of q α , which again we do not relabel, such that(59) q α ⇀ ω in L ∞ ( R + ; BM ( H )) weak ∗ as α → u α → curl v in D ′ ( R ∗ + × H )as α →
0. Moreover curl ∆ u α → curl ∆ v in D ′ ( R ∗ + × H )and, therefore, α curl ∆ u α → D ′ ( R ∗ + × H )as α →
0. We infer that q α = curl u α − α curl ∆ u α → curl v in D ′ ( R ∗ + × H )as α →
0. Comparing this with (59) and using the uniqueness of limits in the sense of the distribu-tions we infer that ω = curl v . Recalling that v ∈ L ∞ ( R + ; L σ ) we further deduce that(60) ω ∈ L ∞ ( R + ; H − ) . It follows that(61) Z T Z H ∂ t ϕdq α dt → Z T Z H ∂ t ϕdωdt as α →
0. This is the only linear term in (53) which we need to analyze, given that ω ≡ q .Putting together relations (51), (53) and (55) we deduce |h ∂ t q α , ϕ i D ′ , D | ≤ C ε k ϕ k W , ∞ ≤ C ε k ϕ k H . This implies that k ∂ t q α k L ∞ (0 ,T ; H − ( H ε )) ≤ C ε , so ∂ t q α is bounded in L ∞ (0 , T ; H − loc ). By the Ascoli theorem, we find, passing to subsequences asneeded, that q α → ω in C ( R + ; H − loc ) . In particular, for all t ≥
0, we have that q α ( t ) → ω ( t ) in H − loc . Given that q α ( t ) is bounded in BM ( H ), we deduce that q α ( t ) ⇀ ω ( t ) weak- ∗ BM ( H ).Let us now address the nonlinear terms in (53).We apply first Proposition 14 and use that supp ϕ ⊂ [0 , T ) × H ε to bound(62) (cid:12)(cid:12)Z T Z H u α,bdry · ∇ ϕ dq α dt (cid:12)(cid:12) ≤ CT k∇ ϕ k L ∞ k u α,bdry k L ∞ ((0 ,T ) × H ε ) k q α k L ∞ (0 ,T ; BM ( H )) ≤ CT k∇ ϕ k L ∞ k q k BM ( H ) α ε − α → −−→ . Therefore the nonlinear term in the second line of (53) vanishes as α → e now pass to the limit in the nonlinear term in the first line of (53), namely Z T Z Z H × H H αϕ ( t, x, y ) dq α ( x ) dq α ( y ) dt. From Proposition 9 we have that H αϕ is bounded uniformly in t and α . Since q α is bounded withrespect to α in L ∞ (0 , T ; BM ( H )) it follows that (cid:12)(cid:12)(cid:12)(cid:12)Z Z H × H H αϕ ( t, x, y ) dq α ( x ) dq α ( y ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ k H αϕ k L ∞ k q k BM ( H ) . Hence, by the Lebesgue Dominated Convergence theorem, it suffices to pass to the limit for a fixedtime t , hereafter omitted. From (60) we know that that ω ( t ) ∈ H − ( H ) a.e. in time, so we canassume in the sequel that ω ∈ H − ( H ).We will prove that(63) Z Z H × H H αϕ ( t, x, y ) dq α ( x ) dq α ( y ) α → −−→ Z Z H × H H ϕ ( t, x, y ) dω ( x ) dω ( y )We know that q α ( t ) ⇀ ω ( t ) weak- ∗ BM ( H ). Since | q α ( t ) | is bounded in BM ( H ), it admits asub-sequence, relabeled | q α ( t ) | , which converges weak- ∗ BM ( H ) to some measure µ ∈ BM + ( H ).This sub-sequence depends on the time t and we cannot assume that we can choose the same sub-sequence for all times t . But since the limit we seek to find in (63) does not depend on µ , proving theconvergence on this time-dependent sub-sequence implies the convergence for the whole sequencewithout extracting the time-dependent sub-sequence.A crucial property of the measures ω and µ is that they are continuous: ω ( { x } ) = µ ( { x } ) = 0for all x ∈ H . Claim 19.
The measures ω , | ω | and µ are continuous on H . Proof of Claim.
We know that ω ∈ H − ( H ) so it is a continuous measure. It follows, therefore, that | ω | is also a continuous measure.Next let us prove that µ is a continuous measure. Since q α verifies the transport equation (1), weknow from the DiPerna-Lions theory (see [9, Theorem III.2]) that q α ( t ) is the image measure of q by a measure preserving flow map X ( t ): q α ( t ) = q ◦ X ( t ).By hypothesis q ∈ BM + ( H ) + L ( H ) so there exists q ∈ BM + ( H ) and q ∈ L ( H ) such that q = q + q . Let q α = q ◦ X ( t ) and q α = q ◦ X ( t ) so that q α = q α + q α . Since the flow map X ( t )is volume preserving, the measures q α and q α are bounded in BM ( H ). Extracting sub-sequencesif necessary, we can assume without loss of generality that q α α → −− ⇀ ω weak ∗ in BM ( H ) q α α → −− ⇀ ω weak ∗ in BM ( H )and | q α | α → −− ⇀ ω weak ∗ in BM ( H ) . Since q ∈ L ( H ) and X ( t ) is volume preserving, we observe that the sequence q α = q ◦ X ( t )is equi-integrable. The Dunford-Pettis theorem implies that ω ∈ L ( H ). Similarly ω ∈ L ( H ).Recall also that q ∈ BM + ( H ) so q α ∈ BM + ( H ) which in turn implies that ω ∈ BM + ( H ).Since q α converges to ω weak ∗ in BM ( H ) we have that ω = ω + ω . We know that ω is acontinuous measure. But ω is also continuous as a measure because it is an L function. Weconclude that ω = ω − ω is a continuous measure too.Next we bound | q α | = | q α + q α | ≤ | q α | + | q α | = q α + | q α | . his inequality is preserved by the weak ∗ limit in BM ( H ). At the limit we get that µ ≤ ω + ω . Since ω is an L function, it defines a continuous measure. Therefore the rhs above is a continuousmeasure. Recalling that µ ≥ µ must be a continuous measure. Thiscompletes the proof of the claim. (cid:3) We continue now with the proof of the convergence stated in (63).Let 0 < θ < ε/ χ ∈ C ∞ ( R × R ; [0 , χ θ ( x, y ) = 1 if | x − y | < θ/ χ θ ( x, y ) = 0 if | x − y | > θ . We decompose(64) Z Z H × H H αϕ dq α ( x ) dq α ( y ) − Z Z H × H H ϕ dω ( x ) dω ( y ) = I + I + I − I where I = Z Z H × H H αϕ χ θ dq α ( x ) dq α ( y ) I = Z Z H × H ( H αϕ − H ϕ )(1 − χ θ ) dq α ( x ) dq α ( y ) I = Z Z H × H H ϕ (1 − χ θ ) dq α ( x ) dq α ( y ) − Z Z H × H H ϕ (1 − χ θ ) dω ( x ) dω ( y )and I = Z Z H × H H ϕ χ θ dω ( x ) dω ( y )From Lemma 17 c) we know that ( H αϕ − H ϕ )(1 − χ θ ) → α → I = Z Z H × H ( H αϕ − H ϕ )(1 − χ θ ) dq α ( x ) dq α ( y ) α → −−→ . From Lemma 17 a) we know that H ϕ (1 − χ θ ) is continuous on H × H and vanishes at the boundaryand at the infinity. Since q α ⊗ q α → ω ⊗ ω in BM ( H × H ) we deduce that(66) I = Z Z H × H H ϕ (1 − χ θ ) dq α ( x ) dq α ( y ) − Z Z H × H H ϕ (1 − χ θ ) dω ( x ) dω ( y ) α → −−→ . Next, let η ε ∈ C ( H ; [0 , η ε ( x ) = 1 if x > ε and η ε ( x ) = 0 if x < ε/
2. Thefunction η ε ( x ) + η ε ( y ) is greater than 1 on H ε × H ∪ H × H ε . We know from Lemma 17 a) that H αϕ is supported in H ε × H ∪ H × H ε , so | H αϕ | ≤ ( η ε ( x ) + η ε ( y )) | H αϕ | . Using Lemma 17 d) we can furtherbound | H αϕ | ≤ ( η ε ( x ) + η ε ( y )) F. This allows to estimate the term I as follows: | I | = (cid:12)(cid:12)Z Z H × H H αϕ χ θ dq α ( x ) dq α ( y ) (cid:12)(cid:12) ≤ Z Z H × H ( η ε ( x ) + η ε ( y )) F χ θ d | q α | ( x ) d | q α | ( y ) . From the localization properties of the supports of η ε and χ θ we observe that the function ( η ε ( x )+ η ε ( y )) F χ θ vanishes at the boundary of H × H (even in a neighborhood of this boundary). It alsovanishes at infinity because F vanishes at infinity. Using that | q α | ⊗ | q α | → µ ⊗ µ in BM ( H × H )we deduce that Z Z H × H ( η ε ( x ) + η ε ( y )) F χ θ d | q α | ( x ) d | q α | ( y ) α → −−→ Z Z H × H ( η ε ( x ) + η ε ( y )) F χ θ dµ ( x ) dµ ( y ) o(67) lim sup α → | I | ≤ Z Z H × H ( η ε ( x ) + η ε ( y )) F χ θ dµ ( x ) dµ ( y ) . Similarly(68) | I | = (cid:12)(cid:12)Z Z H × H H ϕ χ θ dω ( x ) dω ( y ) (cid:12)(cid:12) ≤ Z Z H × H ( η ε ( x ) + η ε ( y )) F χ θ d | ω | ( x ) d | ω | ( y ) . We deduce from (64), (65), (66), (67) and (68) that(69) lim sup α → (cid:12)(cid:12)Z Z H × H H αϕ dq α ( x ) dq α ( y ) − Z Z H × H H ϕ dω ( x ) dω ( y ) (cid:12)(cid:12) ≤ Z Z H × H ( η ε ( x ) + η ε ( y )) F χ θ dµ ( x ) dµ ( y ) + Z Z H × H ( η ε ( x ) + η ε ( y )) F χ θ d | ω | ( x ) d | ω | ( y ) . We let now θ →
0. We clearly have that χ θ converges pointwise to the characteristic function ofthe diagonal, denoted by x = y , and is uniformly bounded. By the Lebesgue dominated convergencetheorem we infer thatlim θ → Z Z H × H ( η ε ( x ) + η ε ( y )) F χ θ dµ ( x ) dµ ( y ) = Z Z H × H ( η ε ( x ) + η ε ( y )) F ( x, y ) x = y dµ ( x ) dµ ( y )= 2 Z H η ε ( x ) F ( x, x ) µ ( { x } ) dµ ( x )= 0 , where we used the Fubini theorem and Claim 19. We can show in the same manner thatlim θ → Z Z H × H ( η ε ( x ) + η ε ( y )) F χ θ d | ω | ( x ) d | ω | ( y ) = 0so taking the limit θ → α → (cid:12)(cid:12)Z Z H × H H αϕ dq α ( x ) dq α ( y ) − Z Z H × H H ϕ dω ( x ) dω ( y ) (cid:12)(cid:12) = 0 . This shows the convergence stated in (63). Putting together (61), (62) and (63) we obtain thatthe limit α → Z T Z H ∂ t ϕ dωdt + Z T Z Z H × H H ϕ dω ( x ) dω ( y ) dt + Z H ϕ (0 , · ) dq = 0 . Therefore ω satisfies the weak vorticity formulation in Definition 15 with initial vorticity q . Thisconcludes the proof of Theorem 2. (cid:3)
8. Comments and conclusions
We have shown that solutions of the α -Euler equations on the half-plane, under no-slip boundaryconditions, converge in the vanishing α limit to a weak solution of the incompressible 2D Eulerequations when the initial potential vorticity is independent of α and a bounded Radon measure ofdistinguished sign in ˙ H − . Several comments are in order.First, we emphasize that, for the weak solution of the 2D Euler equations which we are producingthrough the vanishing α limit, the test functions are supported in the interior of H . This is incontrast to the weak solutions obtained by two of the authors in [14], for which the test functionsmerely vanished at the boundary of the half-plane, but their normal derivatives were not necessarilyzero. Those weak solutions were called boundary coupled weak solutions and it was shown in [14],see Theorem 2, that the method of images is valid for vortex sheet weak solutions if and only if hey are boundary coupled. Thus the weak solutions discussed here may not give rise to a weaksolution in the full plane through the method of images. This issue is under further investigationby the authors.Second, we comment on a significant technical difference in the proof of convergence, with respectto the proof of the Delort theorem, namely that the potential vorticities q α are not a priori boundedin L ∞ ( R + ; H − ( H )). In the case of the 2D Euler equations the approximate vorticities obeyed thisbound and this led to an a priori estimate on the mass of small balls: Z B ( x ; r ) ω n ( t, · ) dy ≤ C | log r | − / which, in turn, implied no Dirac masses in the limit. In our vanishing α limit we use, instead, that q α ⇀ ω , u α ⇀ u , so that, by linearity, ω = curl u . Since u ∈ L we find ω ∈ H − and, thus, noDirac masses.In this work we have discussed only the case of flow in the half-plane. It would be interestingto study the α → Acknowledgments.
A.V. Busuioc and D. Iftimie thank the Franco-Brazilian Network in Mathematics (RFBM) and H.J.Nussenzveig Lopes thanks the PICS
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Adriana Valentina Busuioc:
Universit´e de Lyon, Universit´e de Saint-Etienne – CNRS UMR 5208 Institut CamilleJordan – Facult´e des Sciences – 23 rue Docteur Paul Michelon – 42023 Saint-Etienne Cedex 2, France.Email: [email protected]
Drago¸s Iftimie:
Universit´e de Lyon, CNRS, Universit´e Lyon 1, Institut Camille Jordan, 43 bd. du 11 novembre,Villeurbanne Cedex F-69622, France.Email: [email protected]
Web page: http://math.univ-lyon1.fr/~iftimie
Milton C. Lopes Filho:
Instituto de Matem´atica, Universidade Federal do Rio de Janeiro, Cidade Universit´aria – Ilhado Fund˜ao, Caixa Postal 68530, 21941-909 Rio de Janeiro, RJ – Brasil.Email: [email protected]
Web page:
Helena J. Nussenzveig Lopes:
Instituto de Matem´atica, Universidade Federal do Rio de Janeiro, Cidade Universit´aria– Ilha do Fund˜ao, Caixa Postal 68530, 21941-909 Rio de Janeiro, RJ – Brasil.Email: [email protected]
Web page: