Interpolatory curl-free wavelets on bounded domains and characterization of Besov spaces

Transcription

1 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 RESEARCH Oen Access Interolatory curl-free wavelets on bounded domains and characterization of Besov saces Yingchun Jiang Corresondence: guiliniang@6 com School of Mathematics and Comutational Science Guilin University of Electronic Technology Guilin P R China Abstract Based on interolatory Hermite slines on rectangular domains the interolatory curl-free wavelets and its duals are first constructed Then we use it to characterize a class of vector-valued Besov saces Finally the stability of wavelets that we constructed are studied MR000 Subect Classification: 4C5; 4C40 Keywords: interolatory curl-free wavelets bounded domains Besov saces Introduction Due to its otential use in many hysical roblems like the simulation of incomressible fluids or in electromagnetism curl-free wavelet bases have been advocated in several articles and most of the study focus on the cases of R and R 3 [-4] However it is reasonable to study the corresonding wavelet bases on bounded domains because of some ractical use At the same time the stability and the characterization of function saces are also necessary in some alications such as the adative wavelet methods In recent years divergence-free and curl-free wavelets on bounded domains begin to be studied [5-8] In articular [8] use the truncation method to obtain interolatory sline wavelets on rectangular domains from [3] Insired by this we mainly study the interolatory 3D curl-free wavelet bases on the cube and its alications for characterizing the vector-valued Besov saces In Section we first give the construction of interolatory curl-free wavelets and its duals on the cube The characterization of a class of vector-valued Besov saces are given in art 3 Finally we also study the stability of the corresonding curl-free wavelets Now we begin with some notations and formulae which will be used later on Let ξ and ξ stand for two cubic Hermite slines: ξ x =: 3x x 3 X [ 0 x 3x x 3 X [0 x ξ x =:x x x 3 X [ 0 xx x x 3 X [0 x Similarly the quadratic Hermite slines are defined as ξ x =: 6x 6x X [ 0] ξ x =: 4x3x X [ 0] x 4x3x X [0 x 0 Jiang; licensee Sringer This is an Oen Access article distributed under the terms of the Creative Commons Attribution License htt://creativecommonsorg/licenses/by/0 which ermits unrestricted use distribution and reroduction in any medium rovided the original work is roerly cited

2 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page of 3 Let Z 0 =: {0 } Z =: { } Z =: {0 } and Z 3 =: { } For each 0 define the scaling functions on [0]: Let V {V ξ m;k =: ξ m x kx [0] k Z 0 ; ξ ;k =: ξ x kx [0] = ξ x k k Z ; ξ ;k =: ξ x kx [0] k Z 0 { } { } =: san ξ ;k ξ ;k : k Z0 V =: san ξ ;k ξ ;k : k Z k Z 0 then } and {V } are two MRAs on L [0] [8] The corresonding duals given in the sense of distributions: ξ =: δ 0 f ξ ;k ξ =: X [ 0] f ξ ;k = f k k Z 0 ; ξ =: δ 0 = k k The inerolating multi-wavelets η ± m;k f ξ ;k f xdx k Z ; ξ =: δ 0 f ξ ;k = f k k Z 0 ; = f k k Z 0 ξ ± m;k are on [0] as well as the wavelet saces are defined by { } η ± m;k x =:η± m;k xx [0] = η ± m;k x k Z ; W± =: san η ± m;k : m =k Z with η m =: ξ m m =;η =: ξ ξ ; η =: ξ Here and after h k = h -k The corresonding duals are given by η =: δ δ 0 δ 8 δ 0 8 δ η =: 3 4 δ δ δ 8 δ 0 8 δ and f η ± m;k = d dx ξ η =: X[ 0 ] X [ ] 4 δ 0 4 δ η =: δ 4 δ 0 4 δ 3 X [0] f k η ± m Moreover there is the following differential relations ;k x = ξ ;k d dx ξ = ξ ; d ; ; dx ξ ;k = ξ x ξ;k x k Z 3 ; d dx ξ ;0 x = ξ ; x ;k k Z0 ; d dx η d dx ξ ;k = d dx ξ ;k d dx ξ ;k m;k = η m;k k Z dx ξ d ;k = d ξ ;k dx : Curl-free wavelets on the cube For u x y z = u u u 3 T the 3D curl-oerator is defined as curl u = u 3 3 u 3 u u 3 u u T Let I { 3} =: I 0 define scaling functions ϕ I m x x x 3 =: with ξ I μv = { ξ μ v I ξ μ v / I 3 ξm I v v x v m =m m m 3 T { } 3 v=

3 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page 3 of 3 The corresonding wavelets are ψ I em x x x 3 =: 3 ϑe I v m v v x v e E 3 m =m m m 3 T { } 3 v= Here and after E 3 denotes the non-zero aexes of the unite cube and { ξ ϑlμv I I = μv l =0 ημv I l = Let ϕ I m;k =: ϕi m;k X [0] 3 which is the tensor roduct of corresonding interolatory scaling functions on the interval The corresonding duals are given similarly Furthermore define ϕ mi;k =: ϕ I 0\{i} m;k δ i { } V =: san ϕ mi;k : m { } 3 k i 3 and the roection oerators: =: {3} δ {3} δ {} δ 3 =: {} δ {} δ {3} δ 3 Lemma [8] For smooth functions f : [0 ] 3 R x i I f = I\{i} f i I x i Proosition For f C curl; [0 ] 3 =: { υ C[0 ] 3 3 :curl υ C[0 ] 3 3 } there has curl f = curl f Proof Note that Lemma then curl f = {} f3 = δ {} f f x x 3 f 3 {3} f δ x 3 {3} f {3} f δ 3 x x x {} which is curl f by definition f 3 x 3 x δ {3} f f δ 3 x x x 3 {3} f x {} f 3 δ Proosition is imortant because it tells us that kees curl-free roerty In general vector-valued wavelets and wavelet saces are given resectively by emi;k =: ψ I0\{i} em;k δ i { ψ } W =: san emi;k : e E 3 m { }3 i 3 k 0 For e E 3 and m Î { } 3 we define em;k =: gradψ I0 em;k = Clearly curl em;k = 0 To give a decomosition for W take non emi;k =: ψi 0\{i} em;k δ i k ψ I0 em;k x δ ψ I0 em;k x δ ψ I0 em;k x δ 3 k 3

4 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page 4 of 3 for i Î I 0 \{i e } Here we choose i e such that e ie = Proosition The vector-valued function system { em;k non emi;k e E 3 m { }3 i = i e k k } is comlete in W Proof It is sufficient to show the statement for = 0 Let f f ψ em;0k = f ψ non emi;0k = 0 W 0 satisfy for all e E 3 m { }3 k } i 3and i i e Here the inner roduct is in L [0] 3 Without loss of generality one assumes i e = Then f ψ non emi;0k = 0 leads to f ψ I 0\{} em;0k = f 3 ψ I 0\{3} em;0k =0 By the definition of ψ I and differential relations one knows f ψ I 0 em;0k = f 3 ψ I 0 em;0k =0 x x 3 Moreover f ψ em;0k =0reduces to f ψ I 0 em;0k = 0 Nowitfollowsthat x f ψ I 0\{} em;0k = 0 from i e =Finally f ψ emi;0k = 0 and f = 0 follows from the definition of W 0 To give the bi-orthogonal decomosition we define ψ em;k = ψ I 0\{i e } em;k δ ie : Assume I ={i i e i } then curl non emi;k =curlψi 0\{i} em;k δ i = ε ψ I 0\{ii e } em;k δ i ε x i with ε = ε = and ε ε = - Now define ψ I 0\{i} em;k δ i e ψ non emi;k =: curl ε ψ I 0\{ii e } em;k δ i : Here the derivatives are meant in the sense of distributions Now we state the main result: Proosition 3 The set { em;k non emi;k e E 3 m { }3 i = i e 0 k k } is a bi-orthogonal wavelet basis of L [0] 3 3 with duals defined in and Proof According to Proosition one only need show ψ non i e m i; k ψ em;k = 0; ψ non ii em;k ψ e m i; k = 0; ψ iii em;k ψ e m ; k = δ ee δ mm δ δ kk ;

5 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page 5 of 3 ψ non non iv e m i ; k ψ e m i ; k = δ e e δ m m δ i i δ δ k k The identity i holdsobviouslyfori i e Fori = i e sincei i e theni e i e which means e e Finally the result i follows from the bi-orthogonality of ψ I 0\{i} e m ; k and ψ I 0\{i} em;k f Note that curl g = curl f g Thenii follows from curl grad = 0 Furthermore ψ em;k ψ e m ; k = ψ I0 em;k x ψ I0\{i e } e m ; k = ψ ie I0 em;k x ie ψ I0\{i e } e m ; k Then by the fact d dx η m = η m and the bi-orthogonality of ψ I 0 em ψ I 0 em one obtains ψ em;k ψ e m ; k = ψ I 0 em;k ψ I 0 e m ; k = δ ee δ mm δ δ kk Now it remains to rove iv which is equivalent to ε ψ I 0\{i i e } e m ; k δ i ε ψ I 0\{i } e x i m ; k δ ie ε ψ I 0\{i i e } e m ; k δ i = δ e e δ m m δ i i δ δ k k :3 It is easily roved when e = e :Infactsincei e = i e one can assume i = i and i = i because i = i leads to 3 obviously In that case the left-hand side of 3 reduces to ε ε ψ I\{i i e } e m ; k ψ I\{i i e } e m ; k = δ m m δ δ k k which is the desired To the end it is sufficient to rove that for e e that is ε ψ I 0\{i i e } e m ; k δ i ε ψ I 0\{i } e x i m ; k δ ie ε ψ I 0\{i i e } e m ; k δ i =0 :4 Note that i {i i i e } Then the conclusion is obvious when i = i Wheni = i then {i i e } = {i i e } and the left-hand side of 4 reduces to ε ε ψ I 0\{i i e } e m ; k ψ I 0\{i i e } e m ; k = 0 Hence one only need to show 4 when i = i e However 4 becomes ψ I 0\{i } e x i m ; k ψ I 0\{i i e } e m ; k =0 :5 in that case Since {i i i e } = {i i i e } = I two cases should be considered: i = i i = i e i e = i and i = i i = i e i e = i Using d dx η m = η m the left-hand side of 5 is ψ I 0\{i } e m k ψ I 0\{i i e } e x i m k = ψ I 0\{i } e m ; k ψ I 0\{i } e m ; k =0 in the first case; In the second one the left-hand side of 5 becomes ψ I 0\{i } e x m ; k ψ I 0\{i i } e m ; k According to the differential relation ψ I 0\{i } e i x m ; k is i

6 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page 6 of 3 a linear combination of ψ I 0\{i i } e m ; k By the bi-orthogonality of ψ I 0\{i i } e m ; k and ψ I 0\{i i } e m ; k one receives the desired conclusion 3 Characterization for Besov saces We shall characterize a class of vector-valued Besov saces in this section For 0 < q and s > 0 the Besov sace B s q L is the set of all f Î L Ω such that f B s q L =: { s ω m f L } l q < with m =[s] andω m f - L Ω the classical m-order modulus of smoothness The corresonding norm is defined by f B s q L =: f L f B s q L Our Besov sace is defined as B s q L [0 ] 3 =: { f B s q L [0 ] 3 3 : with the norm f B s ql [0] 3 =: 3 3 fib s q L [0] 3 f i x i= x f i B s q L [0 ] 3 i =3; = i} i= 3 =i Clearly curl f B s q L [0 ] 3 3 when f B s q L [0 ] 3 B s q L [0] 3 The following lemma is easily roved by the definition of modulus of smoothness: Lemma 3 Iff x gx B s q L R then f x gx B s q L R For α =α 0 and a =a k k define n α l s =: s α l l q Lemma 3 [8] If φ B σ L R n is comactly suorted 0 < q and 0 <s <s then β k φ k k α k φ k k 0 B s q L [0] n B s q L [0] n where =: {k : su -k [0] n } s n β l α l s Theorem 3 Let ϕ mi;k em;k and non emi;k < s < and 0 < q then one has be defined in Section If 0

7 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page 7 of 3 β mik ϕ mi;0k mi k 0 0 emi =i e k β l mi em mi i =ie αem;k c α emi non l s em;k 0 α em c em k l s αemi;k non non emi;k B s ql [0] 3 Proof It is enough to rove the following inequality: i β mik ϕ mi;0 k k 0 ii α c em;k ψ em;k 0 k iii α non emi;k ψ non emi;k 0 k B s ql [0] 3 B s ql [0] 3 B s ql [0] 3 β mi l ; α em c l s ; α non emi l s i = i e Let h =: β mik ϕ mi;0 k and h ν be the νth comonent of h Then for μ i k 0 h i = β mi;k ϕm I0\{i} 0 x k h i = β mi;k 0 x μ k 0 k 0 Since ξ m B L R B L R and ξm B ϕ I\{i} m are in the Besov sace x μ B L R 3 ϕ I 0\{i} m 0 x k x μ L R thenboth ϕm I\{i} due to Lemma 3 Moreover Lemma 3 imlies h i B s q L [0] 3 β l mi and h i x β l mi Note that h ν = μ B s q L [0] 3 0 for ν i Finally the first inequality follows from the definition Let g =: α c em;k ψ em;k and g ν be the νth comonent of g v 3 Then 0 k g v = αem;k c 0 k g v = α c em;k x μ 0 k Similar to the above x v ψ I 0 em ψ I 0 em x k x v x μ x v ψ I 0 em x v x μ ψ I 0 g v B s q L [0] 3 α em l s α em l s x k em B and and L R 3 According to Lemma 3 g v x α l em s μ B s q L [0] 3 Finally one receives the second inequality and the last one follows analogously

8 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page 8 of 3 Let W μ τ D denotes the Sobolev sace with regularity exonent μ and domain D Moreover E d f W μ τ D =: inf P d f P W μ τ D Furthermore let s k = - [0] 3 k fork Z 3 the boundary cases are: when thereisonlyonek i = i 3 s k is defined as relacing the ith osition of - [k k ] [k k ] [k 3 k 3 ] by [ - ]; when k i = k i = i for i i Î { 3} both the ositions i and i are relaced by [ - ]; finally σ =: [ ] [ ] [ ] Lemma 33 [8] Let n n τ μ<s < d s > 0 μ N 00< q τ 0 N 0 Then s n n τ μ Ed f W μ τ σk k Z n l l 0 q f B s q L [0] n The following lemma can be easily roved but it is imortant for roving Theorem 3: Lemma 34 The following relations hold: = f k f k k 8 f k 8 f k ; f η ;k f η ;k = f k f η ;k = k k f η ;k = f k 3 4 f k 4 [f kf k ]; k f tdt f tdt 4 f k 4 f k ; k k k f tdt 4 f k 4 f k f k Theorem 3 Let 3 < s < 3 and 0 < q Then f β mi = f f ϕ mi;0k k 0 αc em = ψ em;k α non 0k emi = ψ non emi;k 0 i = i e satisfy 0k βmil α non emi mi em i =ie l s α c l em s Proof One only need to show the following inequality: i β mi l ii α non emi l s f iii α em c l s B s ql [0] 3 ; f f B s ql [0] 3 ; B s ql [0] 3 f B s ql [0] 3

9 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page 9 of 3 f Note that β mi;k = ϕ mi;0k = f i ϕ I 0\{i} m; 0 Then one assumes i = without loss of k generality and roves first β l m = f ϕ {3} m; 0 k k 0 l f B s ql [0] 3 By the embedding roerty B s q L D Wτ μ D for s > μ and τ f W μ τ D f B s q L D 3: When m = m 3 = f ϕ {3} m; 0 k f L σ0k δ for m =or f L σ0k for m = Using 3 one obtains β l m k 0 f L σ 0 k δ k 0 f L σ 0 k m = m = f B s q L [0] 3 f When m 3 = similarly for m = we obtain f ϕ {3} m; 0 k = f 0 ϕ {} m; x 0 k 0 f x m = 3 W σ 0 k δ 3 0 f x m = 3 W σ 0 k B s ql [0] Note that < s < 3 then the same arguments as above lead to i For h B s q L [0 ] 3 we first claim that there are only the following two cases: a h ψ {i } em;k E 3 h W σ k or E 3 h W σ k δ i ; b h ψ {i } em;k E3 h L σ k or E 3 h L σ k δi In fact by the vanishing moment roerty of the dual wavelets that is η m;k P =0P 3 ; η m;k P =0P Then h ψ {i } em;k = h P ψ {i } em;k for each P Î Π Henceife i =ore i =0butm i = the differential relation imlies that { h ψ {i } em;k h P ψ 0 x em;k h P W σ k δ i e i =0m i =; i h P W σ k ow = Moreover the a art follows from the definition of E 3 h W D; Ife i = 0 and m i

10 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page 0 of 3 { h ψ {i } em;k = h P ψ {i } h P L em;k σ k δi e i =0m i =; h P L σ k ow and the b art follows Now one is ready to estimate α non emi;k and α c em;k By the definition of knows α non emi;k =: f ψ non emi;k = curl f ψ I 0\{ii e } em;k δ i = fi f i e ψ {i } em;k fi ψ {i } x ie x i x em;k ie fie x i ψ {i } em;k non ψ one emi Define βemi;k non =: fi ψ {i } em;k and γemi;k non x =: fie ψ {i } ie x em;k Then it is i sufficient to show βemi non l s f B s ql [0] 3 and γ non emi l s f B s ql [0] 3 f i Let h =: in our claim Then βemi;k non x E 3 h W σ k E 3 h W σ k δ i ie or βemi;k non E 3 h L σ k E 3 h L σ k δi By α l s =: n s α l l q and Lemma 33 one receives that βemi non l s f i B s q L [0] 3 Similarly γ non emi l s x ie f f B s ql [0] 3 B holds and α non s ql [0] 3 emi l s f B s ql [0] 3 f Finallytoestimateαem;k c =: ψ em;k = f ie ψ I 0\{i e } em;k one assumes without loss of generality that i e =andαem;k c = f ψ {3} Notethat d em;k dx η m = η m and d dx ξ α = ξ Then c em;k f ψ {l} x em;k with i l Î { 3} and i l when e = i or e 3 =orm =orm 3 = Similar to the last case one obtains αem;k c f E 3 W x σ k or f E 3 L σ k i x i and iii is roved in these cases Now it remains to show iii when e = e 3 = 0 and m = m 3 =: For each P Î Π let gx x 3 be a rimitive of Px x 3 ie gx x x 3 =: Px x x 3 dx moreover g x = k =: f x = k if k = and g x =: f x = = if k = 0 Since e = and η m has vanishing moments of order then we obtain

11 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page of 3 f g ψ {3} f em;k = g ψ {3} em;k f g ψ {3} em;k δ x k = ; f g ψ {3} em;kδ x f g ψ {3} em;kδ k =0 f g ψ = x em;k {3} k = ; f g ψ {3} em;kδ x k =0 f g x x k = L σ k δ ; f g x x k =0 L σ k f P x k = = L σ k δ ; f P x k =0 L σ k Therefore we have f ψ {3} em;k = f g ψ {3} em;k f E 3 L σ k δ x E 3 f x L σ k k = ; k =0 The desired result follows from Lemma 33 It should be ointed out that there is no common range for s in Theorems 3 and 3 Indeed this is a big shortcoming However we need only one estimate in many cases 4 The stability of curl-free wavelet bases In this art we shall rove that the single-scale wavelet bases that we have constructed in Section are stable The following lemma is the classical result of functional analysis: Lemma 4 LetX be a Banach sace and x x x n X be linearly indeendent Then there exists a constant C > 0 such that for any scalars a a a n one has α x α x α n x n C α α α n Lemma 4 [8] Let X be a Banach sace and f i f i in i X be linearly indeendent for each i = m then the tensor roducts {f x f x f mm x m } i { n i } i = m are also linearly indeendent Theorem 4 The function system { 3 ψ } em;k 3 ψ non emi;k e E 3 m { }3 k k i = i e generates a Riesz basis for W with Riesz bounds indeendent of Proof By Proosition one need only show the stability of the function system Let ω =: d c em;k ψ em;k d non emi;k ψ non emi;k W emk emk i =i e

12 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page of 3 Then ω L [0] 3 3 ω H s [0] 3 3 for s > 0 Since H s [0 ] 3 =B s L [0 ] 3 then ω L [0] 3 3 ω B s L [0] 3 3 Moreover one receives ω L [0] 3 3 emk i =i e d non emi;k emk d c em;k due to Theorem 3 Now it remains to rove the lower bound Let σ k =: [0 ] 3 k k Z σ 3 then k = [0 ] 3 Wetakeanexamlefore = k Z and m = ψ em;k = ξ has ;k ξ ;k ξ ;k η ;k 3 ξ ψ non em;k = ξ ;k ξ ;k η ;k 3 00 η ;k 3 ξ ;k ξ ;k η ;k 3 ;k ξ ;k ψ non em;k = 0 ξ ;k ξ ;k η ;k 3 0 For each fixed k Z 3 by the characteristics of suorts Lemma 4 and 4 one σ k ω dx = σ k d c em;k ψ em;k i= d non em;k ψ non em;k dnon em;δ ψ non d non emμ;kδ δ non emμ;kδ δ d c em;kδ i em;k dc em;kδ δ em;kδ δ em;kδ μ= e =00m =i =i ek C d c em;k d c d em;kδ i c d em;kδδ non μ= i= d non emμ;kδ d non emμ;kδ δ k Z 3 σ k d non emμ;kδ ψ non emμ;kδ d c em;k em;k d non emi;k non emi;k dx em;k e =00m =i =i ek Finally the lower estimation follows from ω L [0] 3 3 = ω dx C emk d non emi;k i =i e d non em;kδ d c d em;k non emi;k emk d c em;k { Corollary 4 The system 3 } ψ em;k e E 3 m { }3 k is a Riesz basis for W { =: san 3 ψ em;k e E 3 m { }3 k Proof Notethat ω = emk d c em;k ψ em;k emk i =i e d non emi;k } with bounds indeendent of non emi;k W and curl grad = 0 Then the desired result follows from the fact that ω is curl-free if and only if for all d non emi;k = 0

13 Jiang Journal of Inequalities and Alications 0 0:68 htt://wwwournalofinequalitiesandalicationscom/content/0//68 Page 3 of 3 Acknowledgements This is suorted by the 863 Proect of ChinaNo 0AA0005 the roect of Guangxi Innovative TeamNo 0GAG000 the National Natural Science Foundation of China No and the fund of Education Deartment of Guangxi No00M ZD05 006LX7 Cometing interests The authors declare that they have no cometing interests Received: 7 December 0 Acceted: 3 March 0 Published: 3 March 0 References Deriaz E Perrier V: Towards a divergence-free wavelet method for the simulation of D/3D turbulent flows J Turbul 73: Deriaz E Perrier V: Orthogonal Helmholtz decomosition in arbitrary dimension using divergence-free and curl-free wavelets Al Comut Harmon Anal 6: doi:006/acha Bittner K Urban K: On interolatory divergence-free wavelets Math Comut doi:0090/s Urban K: Wavelet bases in Hdiv and Hcurl Math Comut 7034: Stevenson R: Divergence-free wavelet bases on the hyercube Al Comut Harmon Anal doi:006/acha Stevenson R: Divergence-free wavelet bases on the hyercube: Free-sli boundary conditions and alications for solving the instationary Stokes equations Math Comut doi:0090/s Harouna SK Perrier V: Divergence-free and curl-free wavelets on the square for numerical simulations Math Models Methods Al Sci Prerinthtt://halinriafr/hal /PDF/errier-kadridf 8 Zhao J: Interolatory Hermite slines on rectangular domains Al Math Comut doi:006/ amc doi:086/09-4x-0-68 Cite this article as: Jiang: Interolatory curl-free wavelets on bounded domains and characterization of Besov saces Journal of Inequalities and Alications 0 0:68 Submit your manuscrit to a ournal and benefit from: 7 Convenient online submission 7 Rigorous eer review 7 Immediate ublication on accetance 7 Oen access: articles freely available online 7 High visibility within the field 7 Retaining the coyright to your article Submit your next manuscrit at 7 sringeroencom