L q -solutions to the Cosserat spectrum in bounded and exterior domains

Transcription

1 L q -solutions to the Cosserat spectrum in bounded exterior domains by Stephan Weyers, November 2005

2 Contents Introduction 3 Part I: Preliminaries 12 1 Notations 12 2 The space Ĥ1,q () 15 3 The space Ĥ2,q () 17 4 The spaces A q () B q () 21 5 Theorems about differentiable functions 23 6 Existence of ζ C0 k (Rn ), ζ = 0, ζ = N 31 7 Regularity theorems 33 8 The asymptotic behavior of harmonic functions in exterior domains 39 9 The density of H k,q () B q () in B q () The compactness of the imbedding H 1,q () B q () B q () in exterior domains 61 Part II: The Cosserat spectrum Definition of the operator Z q its fundamental properties The eigenvalues in B q () for exterior domains The eigenvalues in B q () for bounded domains The Cosserat spectrum Regularity of the solutions Explicit solutions for B 1 R n \B 1 89 Part III: reen s function reproducing kernels Existence of reen s function Existence of reproducing kernels in B q () Relationship of reen s function to reproducing kernels 107 1

3 20 Explicit calculation for B Appendix 116 A Proofs 116 B Spectral theory of compact operators in real Banach spaces 129 2

4 Abstract In the present paper we consider for λ R the existence of non-trivial solutions of the following problems: 1. classical Cosserat spectrum 2. weak Cosserat spectrum u C 2 () n C 0 () n u = λ div u u = 0 u Ĥ1,q () n u, φ = λ div u, div φ φ Ĥ1,q () n where R n is a bounded or an exterior domain Ĥ1,q () (cf. Definition 2.1) is the suitable space for weak solutions. This problem was investigated firstly by Eugene Francois Cosserat. It is a special case of the Lame equation describes the displacement of a homogeneous isotropic linear static elastic body without exterior forces. In this paper we characterize the weak Cosserat spectrum for bounded or exterior domains R n (n 2) 1 < q < (for the definition of Ĥ2,q () cf. Definition 3.1) Theorem Let n 2, 1 < q <, k N, k 2, k > n q let Rn be a bounded or an exterior domain with C k The set W := { λ R : there is 0 u Ĥ1,q () n, such that for all is finite or countably infinite. 2. For λ R\{1, 2} the space φ Ĥ1,q () n holds u, φ = λ div u, div φ V λ := { u Ĥ1,q () n : u, φ = λ div u, div φ φ Ĥ1,q () n} is finite dimensional. 3. For every sequence (λ m ) W with λ m λ l for m l holds λ m 2 (m ) } 4. { s : s Ĥ2,q ()} V 1 3

5 Therefore λ = 1 is an eigenvalue of infinite multiplicity λ = 2 is an accumulation point of eigenvalues of finite multiplicity. In this generality the result is new. E. F. Cosserat [Co1-Co9] studied the classical Cosserat spectrum for certain types of domains like a ball, a spherical shell or an ellipsoid. In chapter 16 we use their approach for explicit solutions. eneral results are due to Mikhlin [Mi, 1973], who investigated the Cosserat spectrum for n = 3 q = 2, Kozhevnikov [Ko2, 1993], who treated bounded domains in the case n = 3 q = 2. Kozhevnikov s proof is based on the theory of pseudodifferential operators. Faierman, Fries, Mennicken Möller [FFMM, 2000] gave a direct proof for bounded domains, n 2 q = 2. Michel Crouzeix gave 1997 a simple proof for bounded domains, in case n = 2, 3 q = 2. In this paper we use the idea of Crouzeix to proof the results for bounded exterior domains, n 2 1 < q <. The following regularity theorem is new: Theorem Let n 2, 1 < q <, k N, k 2, k > n q let Rn be either a bounded or an exterior domain with C k+3. Assume that u Ĥ1,q (), λ R\{1, 2} Then u, φ = λ div u, div φ 1. u Ĥ1, q () n u H k, q () n2 for all 1 < q <, 2. u C k (), 3. u = λ div u for all φ Ĥ1,q () n It is amazing, that the eigenspaces of eigenvalues λ / {1, 2} don t depend on q! Further we get important results for the classical Cosserat spectrum: λ = 2 is an accumulation point of eigenvalues, too. λ = 1 is also a classical eigenvalue, because for s C0 () with u := s holds u = div u. Now we like to describe, how we proved these results. Starting point was the paper [Si] of Christian. Simader. He proved, that in the case of the upper half space H = {(x, x n ) R n : x n > 0} there exists exactly two eigenvalues, namely λ = 1 λ = 2. Simader used the paper [MueR] the decomposition (cf. Theorem 4.2) L q (H) = A q (H) B q (H) He was able to solve explicitly in Ĥ1,q (H) n the equation T q (p), φ H = p, div φ H φ Ĥ1,q (H) n 4

6 for a given p L q (H), proved by direct calculations, that for p 0 A q (H) holds for p h B q (H) div T q (p 0 ) = p 0 div T q (p h ) = 1 2 p h We tried to carry over these results to the slightly perturbated half space H w = {(x, x n ) R n : x n > w(x )} (for w C0 2(Rn 1 )) to domains with compact boundary. Therefore we considered for suitable µ > 0 { ρ µ C0 0, if t 4µ (R), 0 ρ µ 1, ρ µ (t) = 1, if t 2µ the isomorphism f : H w H, f(x) = ( x, x n w(x )ρ µ (x n ) ) the Piola transform (cf. [Cia, p.37pp]) P : Ĥ 1,q (H w ) n Ĥ1,q (H) n, (P v)(y) = [ detf (f 1 (y)) ] 1 f (f 1 (y)) v(f 1 (y)) converted the inner products P 1 v, P 1 φ H w respectively div (P 1 v), div (P 1 φ) H w by means of the transformation rule in inner products v, φ H + B 1(v, φ) respectively div v, div φ H + B 2(v, φ) By the properties of Piola s transform B 2 defines a compact operator, if div v B q (H), but we are not able to prove that for B 1 too. As another approach we searched for a relationship of reen s function of the Laplace operator (cf. Definition 17.4) to the reproducing kernel R in B q () (cf. Definition 18.2), because with Definition Let n 2 let R n be either a bounded or an exterior domain with C Let T q : L q () Ĥ1,q () n be defined by (cf. Theorem 2.9) T q (p), φ = p, div φ for all φ Ĥ1,q () n 2. Let Z q : L q () L q (), Z q (p) := div (T q p) holds 5

7 Theorem Let n 2 let R n be either a bounded or an exterior domain with C 1. Assume λ R. Then there is u Ĥ1,q () n with u, φ = λ div u, div φ for all φ Ĥ1,q () n if only if there is p L q () with In this case one can choose p = div u. λ Z q (p) = p If R n is a bounded domain, for p L q () formally holds with u := T q (p) u(x) = (x, y) ( u)(y) dy = (x, y) ( p)(y) dy Z q (p)(x) = div u(x) = = = p(y) ( xi )(x, y) ( yi p)(y) dy ( xi )(x, y) } {{ } =0 yi xi (x, y) dy p(y) N i (y) dω y + p(y) yi xi (x, y) dy If therefore yi xi (x, y) 1 R(x, y) 2 was a compact operator, the assertion about the Cosserat spectrum would follow by the spectral theorem for compact Hermitian operators. We couldn t prove that directly. There are results about the relationship of reen s function of the bilaplace operator 2 to the reproducing kernel in B 2 () (see [ELPP, Theorem 4.3, p.113]), but we couldn t find the relationship above. After solving the Cosserat spectrum in another way we can prove the relationship indirectly: 6

8 Theorem 19.1 Let n 2, 1 < q <, k N, k > 1 + n q let Rn be a bounded domain with C 2+k. Let (x, y) = S(x y) + h(x, y) be reen s function of the Laplace operator in let R be the reproducing kernel in B q (). Then Z q (p)(x) = p(x) + p(y) yi xi h(x, y) dy a.e. for p B q () Therefore is a compact operator. yi xi h(x, y) + 1 R(x, y) 2 For the unit sphere B 1 (chapter 20) for the half space one can prove this result directly. In this cases reproducing kernel reen s function are known explictly. It is an interesting question, whether it is possible to prove this directly in general, too. Finally we found the paper [Cr] of Michel Crouzeix. His sketch of a proof for bounded domains q = 2 is very short. He proved, that for p B 2 () holds 1 Z 2 (p) 1 2 p 1,2; C p 2; ( ) 1. It suffices to prove ( ) for p H k,2 () B 2 (), because H k,2 () B 2 () is dense in B 2 () with respect to 2;. 2. Obviously Z 2 (p) 1 2 p 2; C p 2; 3. Choose a sufficiently smooth ζ : R n R with ζ = 0, ζ = N (outer unitary normal) 1 Crouzeix considered only p L 2 () with p dx = 0. That means no loss of generality, because for arbitrary p L 2 () for φ H 1,2 0 () holds: p p dx, div φ = p, div φ Furthermore only for p L q () with p dx = 0 there is a constant C > 0, which doesn t depend on p, such that p, div φ p q; C sup 0 φ H 1,q φ 0 () q (see [St, Satz 8.2.1, p.256]). For our purpose the restriction to L q -functions with mean value zero is not necessary. 7

9 4. define for p H k,2 () B 2 () u := T q (p) H 1,2 0 ()n w := u ζ 1 2 p ζ 5. One can show that w H 1,2 0 () Therefore w H 2,2 () w = 2 u ζ + u ζ 1 2 p ζ L2 () w 2,2; C p 2; 6. Furthermore w ζ (div u 1 2 p) H1,2 0 () 7. Then (div u 1 2 p) 2; ( w ζ) 2; + [ w ζ (div u 1 2 p)] 2; finally ( ). C p 2; 8. Then for bounded domains by Rellich s imbedding theorem Z 2 1 2I is a compact operator, by the spectral theorem for compact self-adjoint operators the assertion follows. Now we describe for each part of the proof, which additional work was necessary to carry over the results to exterior domains the case 1 < q <. 1. is proved in chapter 9. We need elliptic regularity theorems (Theorem ) for exterior domains the asymptotic behavior of harmonic functions (Lemma 8.9). 2. follows immediately by Definition 11.1 Theorem For ζ C k 0 (Rn ) to hold we need in Theorem 6.1 C k+1. It is possible, that it suffices to assume that C k. But our proof is very elementary. 4. The definition of w was the ingenious idea of Crouzeix. 5. is proved in Lemma 12.2 respectively We need p C 0 () u C 1 () n. This is shown in Lemma 12.1 respectively 13.1 by means of Sobolev s imbedding theorems. Therefore we must assume C k+2 with k > n q in Theorem In Lemma 12.3 respectively 13.3 we need again the regularity of p u a few theorems about differentiable functions (chapter 5). 8

10 7. Then the estimate follows by Theorem Analogously to Rellich s imbedding theorem for bounded domains the imbedding H 1,q () B q () in B q () is compact in exterior domains (Theorem 10.1). The proof is based on the asymptotic behavior of harmonic functions (Theorem 8.7). For real Banach spaces the spectral theorem B.9 is applicable. Finally we derive Theorem The regularity of the solutions (Theorem 15.5) can be proved as follows: If u Ĥ 1,q () n, λ R\{1, 2} u, φ = λ div u, div φ for all φ Ĥ1,q () n then with p := div u by Theorem holds p B q (), λ Z q (p) = p With µ := λ 1 λ 2 R we derive by ( ) respectively Theorem p = µ (Z q (p) 12 ) p H 1,q () For 1 < q < n with q = nq n q holds p Lq () then p = µ (Z q (p) 12 ) p H 1,q () By induction we derive p H 1,s () for a certain n < s <, whence p C 0 (). Because of the asymptotic behavior of B q -functions in exterior domains (Theorem 8.12) we derive further p H 1, q () C 0 () 1 < q < u H 1, q () n2 C 1 () 1 < q < For the regularity of higher derivatives we use use the density of H k,q () B q () in B q () with respect to q; (Theorem ) the inequality (Lemma 15.3) Z q (π) 1 2 π k,q; C k π k 1,q; Then we can prove p = µ k (Z s 1 2 I ) k p H k,s () for all n < s <. Therefore Theorem 15.5 holds. 9

11 Part I: Preliminaries 10

12 1 Notations For x 0 R n, 0 < r < R we denote B r (x 0 ) := {x R n : x x 0 < r} B r := {x R n : x < r} A r,r (x 0 ) := {x R n : r < x x 0 < R} A r,r := {x R n : r < x < R} Further we define for an open R n k N C k () := {f C k () : for α N n 0, α k there is f (α) C 0 () with f (α) = D α f} C k 0 () := {f C k () : supp(f) } If 1 < q < we always use the notation For f L q () denote q := ( f q; := q q 1 ) 1 f q q dx Usually we don t strictly distinguish between a function the corresponding equivalence class in L q (). For example the notation f L q () C 0 () means, that there is a (unique) continuous representative. Let H k,q () := {u : R u measurable, D α u L q () for all α k} With u k,q; := D α u q q; α k H k,q () is a Banach space. Let 1 q for u H k,q () Underlined terms always denote vectors Often we use the notation H k,q 0 () := C 0 () k,q; u := (u 1,..., u n ) u H 1,q () instead of u H 1,q () n or u L q () instead of u L q () n 11

13 if no confusion could arise. Further we use the notations f, g := fg dx f, g := f, g := f, g := i,j=1 2 f, 2 g := i,j=1 ( i f) ( i g) dx f i g i dx ( i f j ) ( i g j ) dx ( i j f) ( i j g) dx f q; := f q; := 2 f q; := f q; := ( f i q q; ) 1 q ( i f q q; ) 1 q i j f q q; i,j=1 j f i q q; i,j=1 1 q 1 q if the expressions are well defined. The inner product in R n we denote most of the time by x y := x i y i if x, y R n but sometimes we write x, y := x i y i if x, y R n An exterior domain is a domain R n with R n \ compact 0 R n \. A B means: A, B R n open, A bounded A B Let f ; := sup f(x) x 12

14 S n 1 := {x R n : x = 1} = B 1 ω n := S n 1 n 1 If X is a real normed vector space, we denote by { X := F F (x) : X R sup x 0 x < } its dual space. The property (A) denotes for R n : (A) There is an open K R n with = R n \K 13

15 2 The space Ĥ1,q () Definition 2.1. Let n 2, 1 q < let R n satisfy (A). Then Ĥ 1,q () := {u : R u measurable, u L q ( B R ) R > 0, u L q () for each η C0 (R n ) holds ηu H 1,q 0 ()} Definition 2.2. Let n 2, 1 q < let R n satisfy (A). Then Ĥ 1,q 0 () := {u : R measurable u Lq ( B R ) R > 0, u L q () there exists a sequence (u i ) C0 () so that u u i q, BR 0 R > 0 u u i q, 0} Theorem 2.3. (a) H 1,q 0 Let n 2, 1 q < let R n satisfy (A). Then () Ĥ1,q 0 () Ĥ1,q () (b) For u Ĥ1,q () by u q, a norm is defined on Ĥ1,q (). (c) Equipped with q, -norm Ĥ1,q () is a Banach space being reflexive for 1 < q <. If q = 2 then Ĥ1,2 () is a Hilbert space with inner product u, v for u, v Ĥ1,2 (). (d) Ĥ1,q 0 () is a closed subspace of Ĥ1,q () Ĥ1,q 0 () = C () q, 0 Proof. see [Si/So, Theorem I.2.2, p.27] Theorem 2.4. Let n 2, 1 q < let R n satisfy (A). Then Ĥ 1,q () = {u : R measurable u L q ( B R ) R > 0, u L q () there exists a sequence (u i ) C0 () so that u u i q, BR + u u i q, BR 0 R > 0} Proof. see [Si/So, Theorem I.2.4, p.29] Theorem 2.5. Then Let n 2, 1 q < let R n be open bounded. Ĥ 1,q () = H 1,q 0 () Proof. Easy consequence of Definition

16 Theorem 2.6. Let 2 n q < let R n satisfy (A). Then Ĥ 1,q () = Ĥ1,q 0 () Proof. see [Si/So, Theorem I.2.7, p.31] Theorem 2.7. Let n 2 let R n be an exterior domain. Suppose 1 q < n. Choose r > 0 with R n \ B r let { ϕ r C (R n 0, if x r ), 0 ϕ r 1, ϕ r (x) = 1, if x 2r Then 1. Ĥ 1,q 0 () Ĥ1,q () Ĥ1,q 0 () Ĥ1,q () 2. Ĥ 1,q () = Ĥ1,q 0 () {αϕ r : α R} in the sense of a direct decomposition. Proof. see [Si/So, Theorem I.2.16, p.36] Theorem 2.8 (Variational inequality in Ĥ1,q ()). Let R n (n 2) be either a bounded or an exterior domain let C 1. Let 1 < q <. Then there exists a constant C q = C(q, ) > 0 so that u q, C q sup 0 φ Ĥ1,q () u, φ φ q, u Ĥ1,q () Proof. see [Si/So, Theorem II.1.1, p.45] Theorem 2.9 (Functional representation in Ĥ1,q ()). Let R n (n 2) be either a bounded or an exterior domain let C 1. Let 1 < q <. Then for every F Ĥ1,q () there exists a unique u Ĥ1,q () so that F (φ) = u, φ φ Ĥ1,q () Furthermore with C q by Theorem 2.8 holds { } Cq 1 u q, sup F (φ) : φ Ĥ1,q () φ q, 1 u q, Proof. see [Si/So, Theorem II.1.2, p.45] 15

17 3 The space Ĥ2,q () Definition 3.1. Let n 2, 1 q < let R n satisfy (A). Then Ĥ 2,q () := {u : R measurable u, u L q ( B R ) R > 0, 2 u L q () for each η C0 (R n ) holds ηu H 2,q 0 ()} Definition 3.2. Let n 2, 1 q < let R n satisfy (A). Then Ĥ 2,q 0 () := {u : R u, u Lq ( B R ) R > 0, 2 u L q () there exists a sequence (u i ) C0 () so that u u i q, BR + u u i q, BR 0 R > 0 2 u 2 u i q, 0} Theorem 3.3. (a) H 2,q 0 Let n 2, 1 q < let R n satisfy (A). Then () Ĥ2,q 0 () Ĥ2,q () (b) For u Ĥ2,q () by 2 u q, a norm is defined on Ĥ2,q (). (c) Equipped with 2 q, -norm Ĥ2,q () is a Banach space being reflexive for 1 < q <. If q = 2 then Ĥ2,2 () is a Hilbert space with inner product 2 u, 2 v for u, v Ĥ2,2 (). (d) Ĥ2,q 0 () is a closed subspace of Ĥ2,q () Ĥ2,q 0 () = C () 2 q, 0 Proof. see [MueR, Satz II.1, p.126] Theorem 3.4. Let n 2, 1 q < let R n satisfy (A). Then Ĥ 2,q () = {u : R u, u L q ( B R ) R > 0, 2 u L q () there exists a sequence (u i ) C0 () so that u u i q, BR + u u i q, BR + 2 u 2 u i q, BR 0 for all R > 0} Proof. see [MueR, Lemma II.3, p.129] Theorem 3.5. Then Let n 2, 1 q < let R n be open bounded. Ĥ 2,q () = H 2,q 0 () 16

18 Proof. Easy consequence of Definition 3.1 Theorem 3.6. Let 2 n q < let R n satisfy (A). Then Ĥ 2,q () = Ĥ2,q 0 () Proof. see [MueR, Satz II.3, p.133] Theorem 3.7. Let n 2 let R n be an exterior domain. Suppose 1 q < n. Choose r > 0 with R n \ B r let { ϕ r C (R n 0, if x r ), 0 ϕ r 1, ϕ r (x) = 1, if x 2r Furthermore define ψ ri (x) := ϕ r (x)x i for all i = 1,.., n. Then 1. ϕ r, ψ ri Ĥ2,s () for all 1 < s <, 2. Ĥ 2,q 0 () Ĥ2,q () Ĥ2,q 0 () Ĥ2,q (), 3. Ĥ 2,q () = Ĥ2,q 0 () {αϕ r : α R} { n β iψ ri : β i R} for 1 < q < n 2, 4. Ĥ 2,q () = Ĥ2,q 0 () { n β iψ ri : β i R} for n 2 q < n. Proof. see [MueR, Lemma II.8, p.140] [MueR, Satz II.4, p.144] Theorem 3.8 (Variational inequality in Ĥ2,q ()). Let R n (n 2) be either a bounded or an exterior domain let C 2. Let 1 < q <. Then there exists a constant C q = C(q, n, ) > 0 so that u q, C q sup 0 φ Ĥ2,q () u, φ φ q, u Ĥ2,q () Proof. see [MueR, Hauptsatz, p.191] Theorem 3.9 (Functional representation in Ĥ2,q ()). Let R n (n 2) be either a bounded or an exterior domain let C 2. Let 1 < q <. Then for every F Ĥ2,q () there exists a unique u Ĥ2,q () so that F (φ) = 2 u, 2 φ φ Ĥ2,q () Furthermore there exists a constant D q = D(q, ) with { } Dq 1 2 u q, sup F (φ) : φ Ĥ2,q () 2 φ q, 1 2 u q, 17

19 Proof. see [MueR, Lemma III.15, p.164] Theorem 3.8 Lemma Let R n (n 2) be either a bounded or an exterior domain let 1 < q <. Then 2 u, 2 φ = u, φ for all u Ĥ2,q (), φ Ĥ2,q () Proof. (a) Consider ϕ r, ψ ri from Theorem 3.7. It holds j ϕ r (x) = k j ϕ r (x) = 0 for x r for x 2r Furthermore j ψ ri (x) = j ϕ r (x)x i + ϕ r (x)δ ij k j ψ ri (x) = k j ϕ r (x)x i + j ϕ r (x)δ ik + k ϕ r (x)δ ij Therefore j k ψ ri, j k ϕ r C 0 (A r 2,3r ) (b) Suppose R n is an exterior domain, R n \ B r. By Theorem for u Ĥ2,q () there are v Ĥ2,q 0 () f C (R n ) such that i j f C 0 (A r 2,3r ) C 0 () u = v + f There is by Definition 3.2 a sequence (v k ) C0 () with 2 (v k v) q; 0. Let φ Ĥ2,q (). Then 2 u, 2 φ = i,j=1 = lim k i,j=1 = lim i j v i j φ dx + k i,j=1 i,j=1 i j v k i j φ dx i i v k j j φ dx + i j f i j φ dx } {{ } () j ( i i f ) } {{ } j φ dx C0 () i i f j j φ dx C0 i,j=1 i,j=1 = v, φ + f, φ = u, φ (c) Suppose R n is a bounded domain. assertion follows as in (b) with f = 0. Then Ĥ2,q () = H 2,q () the 0 18

20 Theorem 3.11 (Functional representation in Ĥ2,q ()). Let R n (n 2) be either a bounded or an exterior domain let C 2. Let 1 < q <. Then for every F Ĥ2,q () there exists a unique u Ĥ2,q () so that F (φ) = u, φ φ Ĥ2,q () Furthermore there exists a constant K q = K(n, q, ) with { } Kq 1 u q, sup F (φ) : φ Ĥ2,q () φ q, 1 u q, Proof. Let F Ĥ2,q () be given. By Theorem 3.9 there exists a unique u Ĥ 2,q () a constant D q = D(q, ) > 0 with F (φ) = 2 u, 2 φ φ Ĥ2,q () { } Dq 1 2 u q, sup F (φ) : φ Ĥ2,q () 2 φ q, 1 By Lemma 3.10 it holds 2 u q, F (φ) = u, φ φ Ĥ2,q () D 1 q u q, K n Dq 1 2 u q, { } K n sup F (φ) : φ Ĥ2,q () 2 φ q, 1 = K n sup 0 φ Ĥ2,q () K 2 n sup = sup 0 φ Ĥ2,q () F (φ) 2 φ q, F (φ) φ q, { F (φ) : φ Ĥ2,q () φ q, 1 } Remark By Theorem 3.9 Lemma 3.10 we derive, that 2 q; q; are equivalent norms on Ĥ2,q (). 19

21 4 The spaces A q () B q () Definition 4.1. Let R n be either a bounded or an exterior domain with C 2. Let 1 < q <. Then A q () := { u : u Ĥ2,q ()} B q () := {h L q () : h, φ = 0 φ Ĥ2,q ()} Theorem 4.2. Let R n be either a bounded or an exterior domain with C 2. Let 1 < q <. Then L q () = A q () B q () in the sense of a direct decomposition Proof. see [MueR, Satz IV.2.1, p.201] Remark By Weyl s Lemma for h B q () holds (for a representative) h = 0. For bounded domains even holds B q () = {h L q () : h = 0} For exterior domains on the other h there are harmonic L q -functions, which are not in B q () (see [MueR] or Lemma 4.4). Lemma 4.3. Let R n be either a bounded or an exterior domain with C 2. Let 1 < q < h B q () H 1,q (). Then for each φ Ĥ1,q () h, φ = 0 Proof. (a) For φ Ĥ1,q 0 () there is a sequence (φ k ) C 0 φ k φ q ; 0 () such that Then h, φ = lim k h, φ k = lim k h, φ k = 0 (b) For φ = ϕ r by Theorem 2.7 holds ϕ r C0 ()n ϕ r Ĥ2,q (). Then h, ϕ r = h, ϕ r = 0 (c) By Theorem the assertion follows. 20

22 Lemma 4.4. Let R n be an exterior domain with C 2. Let ϕ r, ψ ri as in Theorem 3.7. Let 1 (n 2)ω n z 2 n, z 0, n 3 S(z) := ln z, z 0, n = 2 Then for all i, j = 1,..., n holds 1 2π 0, z = 0, n 2 j S, ψ ri = δ ij Proof. For z 0 holds Then j S(z) = 1 ω n z j z n j S, ψ ri = j S, (x i ϕ r ) = S, [ δ ij ϕ r + x i ( j ϕ r ) ] == mϕ r C 0 () δ ij S, ϕ r S, x i ( j ϕ r ) == S=0 δ ij S, ϕ r B2r = δ ij S, ϕ r B2r +δ ij = δ ij 1 ω n B 2r l=1 B 2r ( l S)(z) ϕ r (z) } {{ } =1 l=1 z l z n z l z dω z z l z dω z = δ ij 21

23 5 Theorems about differentiable functions Lemma 5.1. Let n 2, R > 0, h > 0 Z + R,h := {x = (x, x n ) R n : x < R, 0 < x n < h} Z R,h := {x = (x, x n ) R n : x < R, h < x n < 0} Z R,h := {x = (x, x n ) R n : x < R, x n < h} E R := {x = (x, x n ) R n : x < R, x n = 0} Suppose Let Then { F (x) := f C 0( ) Z + R,h C 1( ) Z + R,h f(x, x n ), if 0 x n < h 3f(x, x n ) + 4f(x, xn 2 ), if h < x n < 0 F C 1 (Z R,h ), F Z + = f R,h Proof. (a) There are f i C 0 (Z + R,h ), i = 1... n with f i Z + = i f R,h (b) For i = 1... n 1 x < R let (h k ) R with 0 < h k < R x, h k 0. Let 0 < x n < h. Then (x + h k e i, x n ) Z + R,h For x n 0 we get f(x + h k e i, x n ) f(x, x n ) = hk 0 ( i f)(x + te i, x n ) dt f(x + h k e i, 0) f(x, 0) = = mean value theorem hk 0 f i (x + te i, 0) dt f i(x + ζ k e i, 0)h k with ζ k between 0 h k (this implies ζ k 0 ) Therefore f(x + he i, 0) f(x, 0) lim = f i (x, 0) h 0 h (c) For i = n x < R, 0 < x n < h 2 let (h k) R with 0 < h k < h 2, h k 0. Then f(x, x n + h k ) f(x, x n ) = For x n 0 we get f(x, h k ) f(x, 0) = hk 0 22 hk 0 ( n f)(x, x n + t) dt f n (x, t) dt = f n (x, ζ k )h k

24 with 0 < ζ k < h k. Therefore f(x, h) f(x, 0) lim = f n (x, 0) h 0 h h>0 (d) Obviously F Z + C 1 (Z + R,h R,h ) F Z R,h C 1 (Z R,h ) (e) Let i = 1... n 1 x < R. Then F (x + he i, 0) F (x, 0) lim h 0 h = lim h 0 f(x + he i, 0) f(x, 0) h = f i(x, 0) (b) (f) Let x < R. Then Furthermore F (x, h) F (x, 0) lim h 0 h h>0 F (x, h) F (x, 0) lim h 0 h h<0 f(x, h) f(x, 0) = lim h 0 h h>0 = f n(x, 0) (c) 3f(x, h) + 4f(x, h 2 = lim ) f(x, 0) h 0 h h<0 3f(x, h) + 3f(x, 0) = lim + 4f(x, h 2 ) 4f(x, 0) h 0 h h h<0 3f(x, h ) 3f(x, 0) = lim h 0 h 2f(x, h ) 2f(x, 0) h h >0 = (c) f n (x, 0) (g) So for i = 1,..., n 1 holds f i (x, x n ), x n 0 i F (x) = 3f i (x, x n ) + 4f i (x, xn 2 ), x n < 0 That is f n (x, x n ), x n 0 n F (x) = 3f n (x, x n ) 2f n (x, xn 2 ), x n < 0 F C 0 (Z R,h ) n Theorem 5.2. Let R n be open bounded with C 1. Then f C 1 () if only if there exists f C 1 0 (Rn ) such that f = f 23

25 Proof. Let f C 1 () (the inversion is trivial). We denote again by f C 0 () the continuation of f. There are f i C 0 () with f i = i f (a) For x 0 there exists an open V x0 R n with x 0 V x0. Further there is R > 0 a diffeomorphism φ x0 : V x0 Z R,R such that holds: φ x0 (V x0 ) = E R φ x0 (V x0 ) = Z + R,R. Let Ṽx 0 := φ 1 x 0 (Z R 2, R ) 4 (b) With Ṽx by (a) (x ) holds: x Ṽ x Because of the compactness of there are x 1,..., x N such that with Ṽi := Ṽx i holds N Let Ṽ0 :=. Then N i=0 Choose ϕ i C 0 (Ṽi) for i = 0, 1,..., N such that Ṽ i Ṽ i N ϕ i (x) = 1 i=0 for all x Define Then for i, j = 1,..., N holds j g i g i := ϕ i f g 0 C 1 0(R n ) = ( j ϕ i )f + ϕ i ( j f) = g (i) j with g (i) j := ( j ϕ i )f + ϕ i f j C 0 () Define for i = 1,..., N h i (x) := g i ( φ 1 i (x) ) for all x Z + R i,r i E Ri = φ i (V i ) Then Z j h i + R i 2, R i 4 = h i C (Z 0 + R i 2, R i 4 [ k=1 g (i) k ( φ 1 i ) ) ][ j ( φ 1 i ) ] Z k + R i 2, R i 4 24

26 By Lemma 5.1 there is a h i C 1 (Z R i 2, R i 4 Z ) such that hi + R i 2, R i 4 By the Definition of the continuation in Lemma 5.1 we see h i C 1 0(Z Ri,R i ) = h i Let Then f i (y) := h i (φ i (y)) for all y V i = φ 1 i (Z Ri,R i ) f i C 1 0(V i ) fi Vi = g i At least f := g 0 + N f i C0(R 1 n ), f = f Lemma 5.3. Let f C 1 (R) with f(0) = 0 f (t) L for all t R. Further let R n be open 1 < q <. Then for u H 1,q () holds f(u) H 1,q () f(u) = f (u) u Proof. see [SiDL, Satz 6.14] Lemma 5.4. Let R n be open bounded. Let u H 1,q () (strictly: let u be a representative) Z(u) := {x : u(x) = 0} Then i u(x) = 0 for almost all x Z(u) for all i = 1,..., n Proof. see [SiDL, Satz 6.15] Theorem 5.5. Let 1 q < let R n be open bounded. Suppose u C 0 () H 1,q () u = 0. Then u H 1,q 0 () 25

27 Proof. (a) Choose ϕ C (R), 0 ϕ 1, ϕ(t) = ϕ( t), ϕ(t) = { 0,if t 1 1,if t 2 For k N let Then f k (t) := f k (t) = 0 t 0 ϕ(ks) ds t 1 k Furthermore t f k (t) t 0 f k (t) = ϕ(kt) ( (1 ϕ(ks)) ds min t, 2 ) k (k ) { 0,if t = 0 1,if t 0 (b) Let By Lemma 5.3 holds u k (x) := f k (u(x)) u k H 1,q () Because u = 0, u C 0 () compact, there are k with u(x) 1 k for all x \ k Therefore u k (x) = 0 x \ k u k H 1,q 0 () (c) Now u k (x) u(x) = f k (u(x)) u(x) (a) ( min u(x), 2 ) 0 k for all x therefore by the dominated convergence theorem u k u q; 0 (k ) (d) By Lemma 5.4 for almost every x Z(u) holds [ ] i u(x) i u k (x) = i u(x) 1 ϕ(k u(x)) = 0 For x \Z(u) we get by (a): i u(x) i u k (x) 0 26

28 Again by the dominated convergence theorem [ i u i u k q q dx = i u(x) 1 ϕ(k u(x))] dx [ q = i u(x) 1 ϕ(k u(x))] dx 0 \Z(u) (e) By the closedness of H 1,q 0 () in H1,q () it follows u H 1,q 0 () Theorem 5.6. Let R n be open with C 1. Suppose u C 1 (R n ) u = 0. Then there is λ : R with ( u)(x) = λ(x)n(x) for all x where N(x) is the outer normal in x Proof. Let x. Then there are open U R n (x U) O R n 1, there is a one to one map φ : O U with φ C 1 (O; R n ) φ(s) = x with a suitable s O. That is u(φ(t)) = 0 t O Differentiating this equation we derive by the chain rule 0 = ( u)(φ(t)), φ (t) t O t i Especially for t = s 0 = ( u)(x), φ (s) t i It is known that for the tangent space holds ( ) φ φ T x ( ) = span (s),..., (s) t 1 t n 1 Therefore there is a λ(x) R such that N(x) T x ( ) ( u)(x) = λ(x)n(x) Lemma 5.7. Let 1 < q < let H := {(x, x n ) R n : x n > 0} be the upper half space. Suppose u H 1,q 0 (H) C0 (H). Then u = 0 H 27

29 Proof. Let u k C 0 (H) with u u k 1,q;H 0. Then u k (x, x n ) = u k (x, x n ) u k (x, 0) = } {{ } =0 As a convergent sequence (u k ) is bounded xn 0 ( n u k )(x, t)dt u k 1,q; C k N For x n a we get by Hölder s inequality u k (x, x n ) u k (x, x n ) q dx x r [ a 0 ] 1 ( n u k )(x, t) q q 1 dt a q a x r 0 R n 1 0 ( n u k )(x, t) q dt dx a q q ( n u k )(x, t) q dt dx a q q a 0 Therefore x r C q a q 1 u k (x, x n ) q dx dx n C q a q r > 0 [ a 0 x r For (k ) we derive u(x, x n ) q dx dx n ] 1 q [ a 0 x r [ a + 0 u k (x, x n ) q dx dx n ] 1 q + x r Ca + u k u q;h (u u k )(x, x n ) q dx dx n ] 1 q [ a 0 x r u(x, x n ) q dx dx n ] 1 q Ca a > 0 r > 0 Define for r > 0 0 x n f r (x n ) := x r Then f r is continuous in [0, [ for r > 0 a 0 u(x, x n ) q dx f r (x n ) dx n C q a q a > 0 28

30 Therefore x r For a 0 we get u(x, 0) q dx = f r (0) = 1 a = 1 a x r a 0 a 0 f r (0) dx n (f r (0) f r (x n )) dx n + 1 a a max f r(0) f r (x n ) + C q a q 1 0 x n a u(x, 0) q dx 0 r > 0 Because u is continuous by a stard argument we finally derive u(x, 0) = 0 x R n 1 0 f r (x n ) dx n Theorem 5.8. Let 1 < q < let R n be open with C 1. Suppose f C 1 () H 1,q 0 (). Then f = 0 Proof. Let x 0 be arbitrary but fixed. There is an open x 0 V R n a C 1 -diffeomorphism φ : V Z R,R, (R > 0) with φ(v ) = E R, φ(v ) = Z + R,R (for the definition of Z R,R, E R, Z + R,R see Lemma 5.1). Without loss of generality we can assume φ C 1 (V ; R n ) φ 1 C 1 (Z R,R ; R n ). Choose ϕ C 0 (V ) such that ϕ(x 0) = 1. By Lemma A.13 we get obviously by Theorem 5.2 By Lemma A.14 holds Define Then By Lemma 5.7 follows Therefore g := ϕf H 1,q 0 ( V ) g C 1 0(V ) h := g φ 1 H 1,q 0 (Z+ R,R ) C1 0(Z R,R ) h(x) := { h(x), x ZR,R 0, x R n \Z R,R h H 1,q 0 (H) C0 (H) h = 0 H 0 = h(φ(x 0 )) = h(φ(x 0 )) = g(x 0 ) = ϕ(x 0 )f(x 0 ) = f(x 0 ) 29

31 6 Existence of ζ C0 k (R n ), ζ = 0, ζ = N Theorem 6.1. Let n 2, k N let R n either be a bounded or an exterior domain with C k+1. Then there is ζ C0 k (R n ), ζ = 0, ζ = N where N is the unitary outer normal of. Proof. (a) Denote M :=. For x 0 M there is an open x 0 Ṽ Rn h : Ṽ R, h C k+1 (Ṽ ) such that ( h)(x 0 ) 0 M Ṽ = {x Ṽ : h(x) = 0} Like in Theorem 5.6 one can prove ( h)(x) = ± ( h)(x) N(x) x M Ṽ Because ( h)(x 0 ) 0 there is an open x 0 V Ṽ such that ( h)(x) 0 x V h h N both are continuous in V M So we can assume without loss of generality ( h)(x) = + ( h)(x) N(x) x M V Define Then for x M V h := h h h(x) = 0 ( h)(x) = h(x) }{{} =0 ( C k (V ) x M V 1 h(x) ) + h h } {{ } =N(x) = N(x) (b) Because M is compact by (a), there are open V 1,..., V m R n h i C k (V i ) such that ( h i )(x) 0 x V i i = 1,..., m Furthermore M V i = {x V i : h i (x) = 0} m M V i 30

32 h i (x) = N(x) Choose φ k C0 (Rn ) such that 0 φ k 1, supp(φ k ) V k, x M V i m φ k (x) = 1 k=1 x M Define For x M holds ( f)(x) = = {j:x V j } {j:x V j } f j := φ j h j C k 0 (V j ) C k 0 (R n ) f := f(x) = m f j C0 k (R n ) j=1 m j=1 x V j [φ j (x)h j (x)] = φ j (x)n(x) = φ j (x) h j (x) = 0 } {{ } =0 {j:x V j } [ ] ( φ j )(x) h j (x) +φ j (x) ( h j )(x) } {{ } } {{ } =0 =N(x) m φ j (x)n(x) = N(x) j=1 31

33 7 Regularity theorems Theorem 7.1. Let k Z, k 0 let R n (n 2) satisfy (A) C 2+k. Let 1 < q < let p Ĥ1,q (). Let x 0 R 0 > 0 assume that there is f H k,q ( B R0 (x 0 )) so that p, φ = f, φ for all φ Ĥ1,q ( B R0 (x 0 )) Then there exists 0 < R k < R 0 so that p BRk (x 0 ) H2+k,q ( B Rk (x 0 )) with a constant C k = C(k, R k, R 1,, q) > 0 ) p 2+k,q; BRk (x 0 ) C k ( f k,q; BR0 (x 0 ) + p 1,q; BR0 (x 0 ) Proof. see [Si/So, Theorem II.8.8, p.91] Theorem 7.2. Let 1 < q <, x 0 R n, R > 0 p H 1,q (B R (x 0 )). Let k Z, k 0 let f H k,q (B R (x 0 )). Further assume Then for 0 < R 1 < R holds p, φ = f, φ for all φ C 0 (B R (x 0 )) p BR1 (x 0 ) H2+k,q (B R1 (x 0 )) there is a constant C k = C(k, R, R 1, q) > 0 so that ( ) p 2+k,q;BR1 (x 0 ) C k f k,q;br (x 0 ) + p 1,q;BR (x 0 ) Proof. see [Si/So, Theorem II.8.9, p.91] Theorem 7.3. Let R > 0, k Z, k 0 1 < q <. Let p L q (B s \B R ) for all s > R p L q (R n \B R ). Further assume that there is f H k,q (R n \B R ) so that p, φ = f, φ for all φ C 0 (R n \B R ) Then there exists R < R k < so that j i p H k,q (R n \B Rk ) for i, j = 1,..., n with C k = C(n, k, R, R k, q) > 0 holds i,j=1 i j p k,q;r n \B Rk C k ( f k,q;r n \B R + p 1,q;B2Rk \B R ) Proof. see [Si/So, Corollary II.8.12, p.94] 32

34 Theorem 7.4. Let k N 0, k 0 let R n be a bounded domain with C k+2. Suppose 1 < q <. Assume u H 1,q 0 () that there is f H k,q () such that u, φ = f, φ φ C 0 () Then u H 2+k,q () there exists a constant C k = C(, q, k, n) > 0 such that u 2+k,q; C k ( f k,q; + u 1,q; ) Proof. (a) Let x. Then u, φ = f, φ φ C 0 ( B 1 (x)) C 0 () Because Ĥ1,q ( B 1 (x)) = H 1,q 0 ( B 1 (x)) = C 0 ( B 1(x)) 1,q ; B 1 (x) u, φ = f, φ φ Ĥ1,q ( B 1 (x)) therefore by Theorem 7.1 there is 0 < R x < 1 a constant C 1 = C 1 (k, q,, x) such that u H2+k,q ( B Rx (x)) BRx (x) ( ) u 2+k,q; BRx (x) C 1 f k,q; B1 (x) + u 1,q; B1 (x) C 1 ( f k,q; + u 1,q; ) (b) Let x. Then there is R x > 0 such that B 2Rx (x). Obviously u H1,q (B 2Rx (x)) B2Rx (x) u, φ = f, φ φ C0 (B 2Rx (x)) By Theorem 7.2 we derive u H2+k,q (B Rx (x)) BRx (x) there is a constant C 2 = C 2 (k, q, x, ) > 0 such that u 2+k,q;BRx (x) C 2 ( f k,q; + u 1,q; ) (c) Because is compact we find M, N N open balls V i R n as in (a) U j as in (b) such that N V i N V i M j=1 U j 33

35 Choose ϕ j, C 0 (U j), φ i C 0 (V i) with 0 ϕ j, φ i 1 M N ϕ j (x) + φ i (x) = 1 j=1 x By (a) (b) holds u H 2+k,q (U j ), u 2+k,q;Uj C 2,j ( f k,q; + u 1,q; ) Uj u H 2+k,q ( V i ), u 2+k,q; Vi C 1,i ( f k,q; + u 1,q; ) Vi (d) Let now z C0 (Rn ) α N n 0, α 2 + k. Then M N u (D α z) dx = u D α ϕ j z + φ i z dx j=1 = M j=1 M = ( 1) α Define Then with holds Therefore that is j=1 U j U j u D α (ϕ j z) } {{ } C 0 (U j) dx + N ( ) D α u ϕ j z dx + ( 1) α Uj V i u D α (φ i z) } {{ } C0 ( V i) N V i ) D (u g j (x) := α, x U j Uj 0 else ) D (u h i (x) := α, x V i Vi 0 else h := g j, h i L q () N M φ i h i + ϕ j g j j=1 u (D α z) dx = ( 1) α D α u = h L q () u H 2+k,q () L q () hz dx α 2 + k dx ( ) D α u Vi φ i z dx 34

36 by (c) u 2+k,q; M N C q 2,j + j=1 C q 1,i 1 q ( f k,q; + u 1,q; ) Theorem 7.5. Let k N 0, k 0 let R n be an exterior domain with C k+2 R n \ B R0. Suppose 1 < q <. Assume u Ĥ1,q () that there is f H k,q () such that u, φ = f, φ φ C 0 () Then u H 1+k,q () there exists a constant C k = C(, q, k, n) > 0 R k > R 0 such that ( ) u 1+k,q; C k f k,q; + u q; BRk + u q; Proof. (a) As in Theorem 7.4 one can show u H 1+k,q ( B R ) R > R 0 BR u 1+k,q; BR C 1 ( f k,q; B2R + u 1,q; B2R ) (b) By Theorem 7.3 there is R k > R 0 C 2 > 0 with u H 1+k,q (R n \B Rk ) R n \B Rk u 1+k,q;R n \B Rk C 2 ( f k,q;r n \B R0 + u 1,q;B2Rk \B R0 ) (c) Finally by (a),(b) Lemma A.5 the assertion follows. Theorem 7.6. Let k N 0, k 0 let R n be either a bounded or an exterior domain with C 2+k. Let u Ĥ1,q () u H k,q (). Then u H k+1,q () there is a constant C k = C(k,, q, n) > 0 a R k > 0 such that ( ) u 1+k,q; C k u k,q; + u q; BRk + u q; 35

37 Proof. For φ C0 () holds u, φ = u, φ the assertion follows by Theorem 7.4 Theorem 7.5. Theorem 7.7. Let k N 0, k 0 let R n be a bounded domain with C 4+k. Suppose 1 < q <. Assume u H 2,q 0 () that there is f H k,q () such that u, φ = f, φ for all φ C 0 () Then u H 4+k,q () Proof. see [SiLec, Theorem 9.12, p.157] with m = 2. Theorem 7.8. Let k N 0, k 0 let R n be an exterior domain with C 4+k. Suppose 1 < q <. Assume u Ĥ2,q () that there is f H k,q () such that u, φ = f, φ φ C 0 () Then for every r > 0 holds u H 4+k,q ( B r ) Br Proof. The Proof is exactly the same as in [SiLec, Theorem 9.12 respectively 9.11, p.156]. Theorem 7.9. Let R n be a bounded domain with C 1. Let 1 < s < assume that p Ĥ1,s () = H 1,s (). Let 1 < q < assume that S q (p) := 0 sup 0 φ C 0 p, φ () φ q ; < Then p H 1,q 0 () with C q > 0 by Theorem 2.8 holds p q; C q S q (p) Proof. see [Si/So, Theorem II.5.1, p.66] 36

38 Theorem Let R n be an exterior domain with C 1. Let 1 < s < suppose p Ĥ1,s (). Let 1 < q < assume that S q (p) := sup 0 φ C r () p, φ φ q ; < (where C r () := {φ 0 + cϕ r : φ 0 C 0 (), c R} with ϕ r as in Theorem 2.7) Then p Ĥ1,q () with C q > 0 by Theorem 2.8 holds p q; C q S q (p) Proof. see [Si/So, Theorem II.5.3, p.67] 37

39 8 The asymptotic behavior of harmonic functions in exterior domains Lemma 8.1. Let U R n be open. Assume that ρ > 0, x R n such that B ρ (x) U. Let p : U R be a harmonic function. Then ( j p)(x) n(n + 1) ω n 1 ρ n+1 B ρ p(x + y) dy Proof. With p = 0 holds j p = 0 too. Let 0 < ε < ρ. Because of the mean value property we derive 1 ( j p)(x) = ( j p)(x + y) dy auß 1 = p(x + y) y j B ε B ε B ε B ε y dω y n = ε n ε n 1 p(x + εξ) ξ j dω ξ = n p(x + εξ) ξ j dω ξ ω n S n 1 ε ω n Therefore Next finally ρ 0 ε ω n n ( jp)(x) = S n 1 ε n ω ρ n n ( jp)(x) dε = 0 1 n + 1 ρn+1 ω n n ( jp)(x) = ( j p)(x) p(x + εξ) ξ j dω ξ S n 1 ε n 1 S n 1 p(x + εξ) ξ j dω ξ dε n(n + 1) ω n 1 ρ n+1 B ρ B ρ p(x + y) y j y dy p(x + y) dy Lemma 8.2. Let R n be open let p L q () (1 q < ) be harmonic in. Assume that 1 such that d := dist( 1, ) > 0. Then i p q;1 n + 1 R p q; for all 0 < R < d 2 Proof. By Lemma 8.1 we have for x 1 0 < R < d 2 ( j p)(x) n + 1 p(x + y) dy R B R n + 1 R = n + 1 R B R ( B R 1 ( B R 1 q 38 B R B R ) 1 p(x + y) q q dy BR q 1 q ) 1 p(x + y) q q dy

40 This implies ( 1 ( j p)(x) q dx ) 1 q n + 1 R = n + 1 R n + 1 R = n + 1 R ( B R 1 q ( B R 1 q ( B R 1 q p q; 1 B R B R B R 1 ) 1 p(x + y) q q dy dx ) 1 p(x + y) q q dx dy ) 1 p(z) q q dz dy Lemma 8.3. Let U R n be open. Assume that ρ > 0, x R n such that B ρ (x) U. Let p : U R be a harmonic function. Then for all i, j, k = 1,..., n holds n(n + 1)2 2 n+2 ( i j p)(x) ω n ρ n+2 p(x + y) dy ( i j k p)(x) n(n + 1)3 ω n 2 n 3n+3 ρ n+3 B ρ B ρ p(x + y) dy Proof. (a) By Lemma 8.1 we derive ( i j p)(x) n(n + 1) ω n 2 n+1 ρ n+1 B ρ 2 (x) ( j p)(y) dy for y B ρ 2 (x) ( j p)(y) n(n + 1) ω n 2 n+1 ρ n+1 B ρ 2 (y) p(z) dz Then ( i j p)(x) n2 (n + 1) 2 ω n 2 n2 (n + 1) 2 ω n 2 2 2(n+1) ρ 2(n+1) 2 2(n+1) ρ 2(n+1) B ρ 2 (x) B ρ 2 (x) B ρ 2 (y) B ρ(x) p(z) dz dy p(z) dz dy = n2 (n + 1) 2 ω n 2 = n(n + 1)2 ω n 2 n+2 ρ n+2 2 2(n+1) B ρ 2(n+1) ρ (x) 2 B ρ(x) B ρ(x) p(z) dz p(z) dz 39

41 (b) By Lemma 8.1 (a) we get ( i j k p)(x) n(n + 1)2 ω n 3 n+2 ρ n+2 B 2ρ3 (x) ( k p)(y) dy for y B 2ρ (x) 3 ( k p)(y) n(n + 1) ω n 3 n+1 ρ n+1 B ρ 3 (y) p(z) dz Therefore ( i j k p)(x) n2 (n + 1) 3 ω n 2 n2 (n + 1) 3 ω n 2 = n2 (n + 1) 3 ω n 2 = 3 2n+3 ρ 2n+3 3 2n+3 ρ 2n+3 3 2n+3 ω n ρ 2n+3 n n(n + 1)3 ω n 2 n 3n+3 ρ n+3 B 2ρ3 (x) B 2ρ3 (x) B ρ 3 (y) B ρ(x) ( ) 2ρ n 3 B ρ(x) B ρ(x) p(z) dz p(z) dz dy p(z) dz dy p(z) dz Lemma 8.4. in R n \B R. Let Let n 2, R > 0, 1 q < let u L q (R n \B R ) be harmonic ( ) x v(x) := x 2 n u x 2 for all x B 1 \{0} R 1R be the Kelvin transform. Then v ist harmonic in B 1 \{0} R B v(y) q y q(n 2) 2n dy = u(y) q dy R n \B R Proof. (a) For v = 0 see e.g. [ABR, p.62]. (b) Let r > R. Then by Lemma A.6 holds u(x) AR,r q dx x= y ( ) y === 2 y q u 1 y 2 dy y 2n = = A 1r, 1 R A 1r, 1 R A 1r, 1 R ( y u y 2 ) y 2 n y n 2 q v(y) q y q(n 2) 2n dy 1 dy y 2n 40

42 For r the assertion follows. Lemma 8.5. in R n \B R. Let Let n 2, R > 0, 1 q < let u L q (R n \B R ) be harmonic ( ) x v(x) := x 2 n u x 2 for all x B 1 \{0} R be the Kelvin transform. Then there is a constant C(n, q) > 0 such that for all 0 < r < 1 R B r v(y) dy C(n, q) r n q +2 (B r v(y) q y q(n 2) 2n dy = C(n, q) r n q +2 R n \B 1r u(x) q dx 1 q ) 1 q r Proof. (a) Assume 1 < q <. Then B v(y) dy = v(y) y q(n 2) 2n q 8.4 = = = B r ( B r R n \B 1r R n \B 1r R n \B 1r y 2n q(n 2) q dy ) 1 ( v(y) q y q(n 2) 2n q dy Br u(x) q dx u(x) q dx u(x) q dx 1 q 1 q 1 q ( C(n, q) r y 2n q(n 2) q 1 2n q(n 2) n+ q 1 C(n, q) r n(q 1)+2n q(n 2) q C(n, q) r n+2q q ) q 1 q ) q 1 q dy (b) Assume q = 1. Then B r v(y) dy = B r v(y) y (n 2) 2n y 2+n dy r 2+n B r v(y) y (n 2) 2n dy 8.4 = r R 2+n u(x) dx n \B 1r 41

43 Lemma 8.6. in R n \B R. Let Let n 2, R > 0, 1 q < let u L q (R n \B R ) be harmonic ( ) x v(x) := x 2 n u x 2 for all x B 1 \{0} R be the Kelvin transform. Then there is a ṽ C (B 1 ) such that R ṽ = 0 in B 1 R ṽ B 1R \{0} = v Proof. (a) Let Let Then ρ C (R), 0 ρ 1, ρ(t) = ρ k (x) := ρ (k x ) ρ k C (R n ), 0 ρ k 1, ρ k (x) = there is a constant C > 0 such that { 0,if t 1 1,if t 2 k N { 0,if x 1 k 1,if x 2 k ρ k k C i j ρ k k 2 C (b) Let φ C0 (B 1 ). Then ρ k φ C0 (B 1 \{0}) R R 0 = v, (ρ k φ) B 1R = v, φ ρ k B 1R + 2 v, ρ k φ B 1R + v, ρ k φ B 1R We estimate v, φ ρ k B 1R φ Ck B 2 2k v(x) dx Lemma 8.5 φ C k 2 C(n, q) ( ) n 2 q +2 u k q;r n \B R = K(n, q, φ, u) ( ) n 1 q k 0 (k ) Similarly we derive v, ρ k φ B 1R 0 (k ) (c) Obviously ρ k v φ v φ a.e. in B 1 R ρ k v φ φ v L 1 (B 1 ) R } {{ } by Lemma

44 By the dominated convergence theorem we get v, ρ k φ B 1R v, φ B 1R (k ) By (b) we derive for (k ) v, φ B 1R = 0 finally by Weyl s Lemma the assertion. Theorem 8.7. Let n 2, R > 0, 1 q <. Then there is a constant C = C(n, q, R) > 0 such that for each x R n \B 2R for every harmonic u L q (R n \B R ) holds u(x) C u 1;R n \B R x 3,if q = 1 C u q;r n \B R x 2,if 1 < q 2 C u q;r n \B R x 1,if 2 < q < if n = 2 C u 1;R n \B R x n 1,if q = 1 u(x) C u q;r n \B R x n,if 1 < q n n 1 C u q;r n \B R x 1 n,if n n 1 < q n n 2 C u q;r n \B R x 2 n,if n n 2 < q < if n 3 Proof. (a) Let ( ) x v(x) := x 2 n u x 2 for all x B 1 \{0} R be the Kelvin transform. By Lemma 8.6 there is a harmonic continuation of v in, which we again denote by v. Because of the mean value property Lemma B 1 R 8.5 holds for 0 < r < 1 R v(0) 1 v(y) dy B r B r 1 B r C(n, q) r n q +2 R n \B 1r = n r n C(n, q) r n q +2 ω n 43 R n \B 1r u(x) q dx 1 q u(x) q dx } {{ } 0 (r 0) 1 q

45 which tends to 0 for (r 0), if So we derive n q + 2 n 0 that is n q(n 2) v(0) = 0 if n = 2 v(0) = 0 if n 3 q n n 2 (b) Similarly by Lemma 8.1 Lemma 8.5 we get n(n + 1) 1 ( j v)(0) ω n r n+1 v(y) dy B r n(n + 1) 1 ω n r n+1 C(n, q) r n q +2 which tends to 0 for (r 0), if n q R n \B 1r u(x) q dx } {{ } 0 (r 0) + 2 (n + 1) 0 that is n q(n 1) 1 q So we derive ( j v)(0) = 0 if n 2 q n n 1 (c) Similarly by Lemma 8.3 Lemma 8.5 we get ( i j v)(0) which tends to 0 for (r 0), if Therefore n q n(n + 1)2 1 ω n r n+2 B 2r v(y) dy n(n + 1)2 1 ω n r n+2 C(n, q) r n q +2 R n \B 1 2r + 2 (n + 2) 0 that is 1 q ( i j v)(0) = 0 if n 2 q = 1 u(x) q dx } {{ } 0 (r 0) 1 q 44

46 (d) For x B 1 R there exists by Taylor s theorem a x, b x, c x B 1 R v(x) = v(0) + v(a x ), x such that v(x) = v(0) + v(0), x (Hessv)(b x) x, x v(x) = α 2 (D α v)(0) α! x α + α =3 (D α v)(c x ) α! x α (e) Let n 3 Lemma 8.5 n n 2 < q <. Let x B 1. Then by the mean value property 2R v(x) 1 B 1 2R K 1 (n, R) B 1 (x) 2R B 1R v(y) dy v(y) dy K 2 (n, R, q) u q;r n \B R For y R n \B 2R therefore holds u(y) = y 2 n v ( ) y y 2 y 2 n K 2 (n, R, q) u q;r n \B R n n (f) Let n 3 n 1 < q n 2 or let n = 2 2 < q <. Then by (a) v(0) = 0 for x B 1 holds by Lemma 8.1 Lemma 8.5 2R n(n + 1) ( j v)(x) (2R) n+1 v(y) dy ω n K 3 (n, R) B 1R B 1 2R (x) v(y) dy K 4 (n, R, q) u q;r n \B R therefore by (d) v(x) = v(0) + v(a }{{} x ), x =0 v(a x ) x K 5 (n, R, q) u q;r n \B R x For y R n \B 2R therefore holds u(y) = y 2 n v ( ) y y 2 y 1 n K 4 (n, R, q) u q;r n \B R 45

47 n (g) Let n 2 1 < q n 1. Then by (a) (b) v(0) = 0, v(0) = 0 for x B 1 holds by Lemma 8.3 Lemma 8.5 2R n(n + 1)2 ( i j v)(x) (4R) n+2 v(y) dy ω n K 6 (n, R) B 1R v(y) dy B 1 2R (x) K 7 (n, R, q) u q;r n \B R therefore by (d) v(x) = v(0) + v(0), x + 1 }{{} } {{ } 2 (Hessv)(b x) x, x =0 = x 2 ( i j v)(b x ) 2 K 8 (n, R, q) u q;r n \B R x 2 i,j=1 For y R n \B 2R therefore holds u(y) = y 2 n v ( ) y y 2 y n K 8 (n, R, q) u q;r n \B R (h) Let n 2 q = 1. Then by (a), (b) (c) v(0) = 0, i v(0) = 0 i j v(0) = 0, for x B 1 holds by Lemma 8.3 Lemma 8.5 2R ( i j k v)(x) n(n + 1)3 2 n 3 n+3 (2R) n+3 ω n v(y) dy K 9 (n, R) B 1R v(y) dy B 1 2R (x) K 10 (n, R) u 1;R n \B R therefore by (d) analogously to (f) (g) For y R n \B 2R therefore holds v(x) K 11 (n, R) u 1;R n \B R x 3 u(y) = y 2 n v ( ) y y 2 y n 1 K 11 (n, R) u 1;R n \B R 46

48 Theorem 8.8. Let n 2, R > 0, 1 q <. Then there is a constant C = C(n, q, R) > 0 a function f q L q (R n \B 2R ) such that for each x R n \B 2R for every harmonic u L q (R n \B R ) holds u(x) C u q;r n \B R f q (x) Proof. For k Z holds k n L q (R n \B 2R ) 2R r n 1+q(k n) dr < n 1 + q(k n) < 1 n < q(n k) Therefore n = 2 : 3 L q (R n \B 2R ) q > 2 3 n = 2 : 2 L q (R n \B 2R ) q > 1 n = 2 : 1 L q (R n \B 2R ) q > 2 n 3 : n 1 L q (R n \B 2R ) q > n n+1 n 3 : n L q (R n \B 2R ) q > 1 n 3 : 1 n L q (R n \B 2R ) q > n n 1 n 3 : 2 n L q (R n \B 2R ) q > n n 2 the assertion follows by Theorem 8.7 Lemma 8.9. Let R n be open let p L q () (1 q < ) be harmonic in. Assume that 1 such that d := dist( 1, ) > 0. Then p H,q ( 1 ) 1 Proof. (a) It is well known that p C (). We prove by induction p Hk,q ( ) for every k N every with dist(, ) > 0. 47

49 (b) k = 1 : By Lemma 8.2 we derive for with dist(, ) > 0 i p q; 4(n + 1) dist(, ) p q; < (c) k k + 1 : For k N may hold D α p L q ( ) for every α = k every with dist(, ) > 0. Let 0 with dist( 0, ) > 0. Choose := {x : dist(x, ) > 1 2 dist( 0, )} Then 0, dist(, ) > 0 dist( 0, ) > 0. Because D α p = 0 in by assumption D α p L q ( ) we derive by Lemma 8.2 i D α p q;0 4(n + 1) dist( 0, ) Dα p q; < Lemma If n 3 Let n 2, C > 0, R > 1 let p be harmonic in R n \B R. 2 p(x) C x 2 n for all x R, then there is a constant C > 0 such that for all x 2R all i = 1,..., n. 2. If n 2 i p(x) C x 1 n p(x) C x 1 n for all x R, then there is a constant C > 0 such that for all x 2R all i = 1,..., n. 3. If n 2 i p(x) C x n p(x) C x n for all x R, then there is a constant C > 0 such that for all x 2R all i = 1,..., n. i p(x) C x n 1 48

50 Proof. (a) Let n 3 for all x R. Let There is s < q < with Then p(x) C x 2 n s := for all r > R. By Lemma 8.2 we derive n n 2 n (n 3)q p L q (R n \B 3R ) 4 p L s (A r,4r ) i p s;a2r,3r 4(n + 1) r Hölder 4(n + 1) with q s >1 r p s;ar,4r = ( p q dx A r,4r 4(n + 1) r ) s q [ A r,4r q s q A r,4r p s dx 1 s ] 1 s Because n(q s) sq = C 1 (n) r r n(q s) qs p q;ar,4r 1 n(q s) sq nq ns sq nq s(q + n) nq n + q n n 2 q(n 2) n + q q(n 3) n for r > R > 1 holds i p s;a2r,3r C 1 (n) p q;ar,4r p q q;a r,4r C q A r,4r x q(2 n) dx = C q ω n = C q ω n n + q(2 n) 4r r t n 1+q(2 n) dt [ (4r) n+q(2 n) r n+q(2 n)] = C 2 (n, q) r n+q(2 n) Therefore i p s s;a 2r,3r C 3 (n, q) r s(n+q(2 n)) q 49

51 i p s s;r n \B 2R = So we get = i p s s;a ( 3 2 ) m 1 2R,( 3 2) m 2R m=1 ( (3 ) m 1 C 3 (n, q) R 2 m=1 [ ( 2 C 4 (n, q, R) 3 m=0 ) s q = i p s s;a 2 ( 3 2) m 1 R,3( 3 2) m 1 R m=1 ) s(n+q(2 n)) q >0 { }} { (q(n 2) n) ] m < } {{ } <1 i p L n n 2 (R n \B 2R ) i p is harmonic in R n \B R. Therefore 2 i p L n n 2 (R n \B R ) by Theorem 8.7 there is a constant C 5 = C 5 (n, q, R, p) > 0 such that (b) Let n 2 for all x R. Let There is s < q < with Similarly to (a) follows (c) Let n 2 for all x R. Let There is 1 = s < q < with Similarly to (a) follows ( i p)(x) C 5 x 1 n p(x) C x 1 n s := n n 1 n (n 2)q p L q (R n \B 3R ) 4 for all x 2R ( i p)(x) C x n for all x 2R p(x) C x n s := 1 n (n 1)q p L q (R n \B 3R ) 4 ( i p)(x) C x n 1 for all x 2R 50

52 Lemma Let n 2, R n be an exterior domain, let u be harmonic in. Let r > 0, ϕ r ψ ri as in Theorem For all R > 2r all k = 1,..., n holds B R u ϕ r dx = B R ( i u)(x) x i x dω x B R u ψ rk dx = B R u(x) x k x dω x B R ( i u)(x) x i x k x dω x 2. If there are R > r, C > 0 with u(x) C x 1 n for all x R, then u ϕ r dx = 0 3. If there are R > r, C > 0 with u(x) C x n for all x R, then for k = 1,..., n holds u ψ rk dx = 0 Proof. (a) Let R > 2r. Then by auß Theorem u ϕ r dx = ( u) ( ϕ r ) dx B R B R = ( u) ϕ } {{ } r dx B R =0 = ( i u)(x) B R x i x dω x ( i u)(x) ϕ r (x) } {{ } B R =1 x i x dω x (b) Let R > 2r k = 1,..., n. Again by auß Theorem we derive B R u ψ rk dx = B R ( i u)(x) i [x k ϕ r (x)] dx+ u(x) B R i [x k ϕ r (x)] x i x dω x 51

53 = ( u)(x) } {{ } B R =0 + u(x) B R = x k ϕ r (x) dx δ ik ϕ r (x) } {{ } =1 x i x dω x + B R ( i u)(x) (c) Assume that for all x R holds By Lemma 8.10 therefore B R x i x k x B R u(x) dω x + u(x) C x 1 n ( i u)(x) x k ϕ r (x) } {{ } =1 ( i u)(x) C x n for all x 2R. This implies u ϕ r dx = u ϕ r dx B 2R (a) = (2R) n 1 C (2R) n ω n x k ( i ϕ r )(x) } {{ } =0 B R u(x) x k x dω x x i x dω x+ x i x dω x = x i ( i u)(x) x dω x B 2R For R then u ϕ r dx = 0 (d) Assume that for all x R holds By Lemma 8.10 therefore u(x) C x n ( i u)(x) C x n 1 for all x 2R. This implies u ψ rk dx = u ψ rk dx B 2R (b) = ( i u)(x) B R x i x k x (2R) n 1 C (2R) n ω n + 52 x k dω x + u(x) x dω x B R (2R) n 1 C (2R) n 1 R ω n

54 For R therefore u ψ rk dx = 0 Theorem Let n 2, 1 < q <, R n be an exterior domain, let u L q () be harmonic in. Then u B q () if only if there are C > 0, R > 1 with u(x) C x n for all x R. Proof. (a) Assume that for all x R holds u(x) C x n By Lemma 8.11 then for k = 1,..., n holds u ψ rk dx = u ϕ r dx = 0 Further for φ C 0 () 0 = Therefore also for φ Ĥ2,q 0 () ( u) φ dx = 0 = By Theorem then 0 = u φ dx therefore (b) Assume that Let R > 0 with R n \ B R. Let 2 ( ) x v(x) := x 2 n u x 2 u φ dx u B q () u B q () 53 u φ dx φ Ĥ2,q () for each x B 1 \{0} R

55 be the Kelvin transform. By Lemma 8.6 there is a harmonic continuation of v in 0. This continuation we again denote by v. By Taylor s formula for every x B 1 2R there is a x B 1 such that 2R v(x) = v(0) + ( i v)(a x )x i Define Because i v is bounded in B 1 2R Let w(x) := v(x) v(0) = ( i v)(a x )x i there is a M > 0 with w(x) M x x 1 2R ( ) x (Kw)(x) := x 2 n w x 2 be the Kelvin transform of w. Then for x 2R holds Further by [ABR, S.62] So we derive by Lemma 8.11 But also holds (Kw)(x) x 2 n M 1 x therefore 0 = Kw ϕ r dx = u ϕ r dx = v(0) B 2r Kw = 0 Kw ϕ r dx = 0 for each x R n \B 2R = M x 1 n (Kw)(x) = u(x) x 2 n v(0) v(0) } {{ } =0 ( xi x 2 n) ( i ϕ r )(x) dx = v(0) ϕ r (x) x 2 n dx + v(0) } {{ } B 2r =0 = v(0) (2 n) x i x n x i x dω x B 2r x 2 n ( ϕ r )(x) dx B 2r = v(0) (2 n) (2r) n 1 (2r) 1 n ω n = v(0) (2 n) ω n ( xi x 2 n) x i ϕ r (x) } {{ } x dω x =1 54

56 For n 3 follows v(0) = 0 For n = 2 by part (a) of the proof of Theorem 8.7 holds By Taylor s formula for every x B 1 2R Define v(x) = v(0) + }{{} =0 h(x) := v(x) Because i j v is bounded in B 1 2R Let v(0) = 0 there is b x B 1 2R such that ( i v)(0)x i (Hessv)(b x) x, x ( i v)(0)x i = 1 2 (Hessv)(b x) x, x there is C > 0 with h(x) C x 2 x 1 2R ( ) x (Kh)(x) := x 2 n h x 2 be the Kelvin transform of h. Then for x 2R holds for each x R n \B 2R (Kh)(x) x 2 n C 1 x 2 = C x n Further by [ABR, S.62] So we derive by Lemma 8.11 Kh = 0 Kh ψ rk dx = 0 But also holds (Kh)(x) = u(x) x 2 n ( i v)(0) x i x 2 = u(x) ( i v)(0) x i x n therefore for all k = 1,..., n 0 = Kh ψ rk dx = u ψ rk dx } {{ } =0 ( i v)(0) 4.4 = ω n ( i v)(0) δ ik = ω n ( k v)(0) 55 x i x n ( ψ rk)(x) dx

57 Then That is for x 2R holds v(0) = 0 u(x) = (Kh)(x) x 2R u(x) = (Kh)(x) C x n 56

58 9 The density of H k,q () B q () in B q () Theorem 9.1. Let n 2, k N, k 2, 1 < q < let R n be a bounded domain with C 2+k. Suppose p B q (). Then there is a sequence (h m ) H k,q () B q () such that h m p q; 0 (m ) Proof. (a) Consider Friedrichs mollifier p ε for ε > 0. Because is bounded, we have p ε C 0 (R n ) H,q () ε > 0 By Theorem 3.11 there is s (ε) Ĥ2,q () such that s (ε), φ = p ε, φ φ Ĥ2,q () Because C 2+k, k 2, p ε H,q () s (ε), φ = p ε, φ φ C0 () we derive by Theorem 7.7 s (ε) H k,q () (b) For φ Ĥ2,q () we also get s (ε), φ = p ε, φ p, φ } {{ } =0 = p ε p, φ therefore by Theorem 3.8 s (ε) q; C q sup 0 φ Ĥ2,q () s (ε), φ φ q C q p ε p q; 0 (ε 0) (c) Let Then by (a) by (b) h (ε) := p ε s (ε) h (ε) H k,q () B q () h (ε) p q; p ε p q; + s (ε) q; (1 + C q ) p ε p q; 0 (ε 0) 57

59 Theorem 9.2. Let n 2, k N, k 2, 1 < q < let R n be an exterior domain with C 2+k. Suppose p B q (). Then there is a sequence (h m ) H k,q () B q () such that h m p q; 0 (m ) Proof. (a) Consider Friedrichs mollifier p ε for ε > 0. Let R n \ B R 2 Then for R > 0. dist(r n \B R, ) > dist(r n \B R, B R ) = R 2 2 Therefore by Lemma A.12 for each x R n \B R for every 0 < ε < R 2 holds p ε (x) = p(x) p ε L q (R n \B R ) is harmonic in R n \B R. By Lemma 8.9 we derive p ε R n \B 2R H,q (R n \B 2R ) Because p ε C (R n ) obviously for every r > 0 holds p ε Br H,q (B r ) Finally for every 0 < ε < R 2 holds p ε H,q (R n ) (b) By Theorem 3.11 there is s (ε) Ĥ2,q () such that s (ε), φ = p ε, φ φ Ĥ2,q () Because C 2+k, k 2, p ε H,q () s (ε), φ = p ε, φ φ C0 () we derive by Theorem 7.8 for each r > 0 for every 0 < ε < R 2 s (ε) Br H k,q ( B r ) Let φ C0 (Rn \B R ) Ĥ2,q (). Then for every 0 < ε < R 2 s (ε), φ = p ε, φ = 0 holds by (a) Therefore by Weyl s Lemma s (ε) L q () is harmonic in R n \B R. Again by Lemma 8.9 we derive s (ε) R n \B 2R H,q (R n \B 2R ) 58