Some New Friedrichs-Type Inequalities in Domains with Microinhomogeneous Structure

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1 LICENTIATE T H E SIS Some New Friedrichs-Type Inequalities in Domains with Microinhomogeneous Structure Yulia Koroleva Luleå University of Technology

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3 Some New Friedrichs-Type Inequalities in Domains with Microinhomogeneous Structure. Yulia Koroleva Department of Mathematics Lulea University of Technology SE Lulea,Sweden & Department of Differential Equations Faculty of Mechanics and Mathematics Moscow Lomonosov State University Moscow ,Russia

4 Key words and phrases. Mathematics, Partial Differential Equations, Functional Analysis, Spectral Theory, Homogenization Theory, Friedrichs-type Inequalities Tryck: Universitetstryckeriet, Luleå ISSN: ISBN Luleå

5 Abstract This Licentiate thesis is aimed to derive and discuss some new Friedrichs-type inequalities for functions which belong to the Sobolev space H 1 in domains with microinhomogeneous structure and which vanish on a part of the boundary. The H 1 with a constant classical Friedrics inequality holds for functions from the space depending only on the measure of the domain. It is well known that if the function has not zero trace on the whole boundary,but only on a subset of the boundary of a positive measure (more precisely,of a positive harmonic capacity),then the Friedrichs inequality is still valid. Moreover,in such a case the constant in the inequality increases when the measure (the capacity) of the set where the function vanishes,tends to zero. In particular,in this thesis we study the corresponding behavior of the constant in our new Friedrichs-type inequalities. In paper A we prove a Friedrichs-type inequality for functions,having zero trace on the small pieces of the boundary of a two-dimensional domain,which are periodically alternating. The total measure of the set where the function vanishes, tends to zero. It turn out that for this case the capacity is positive and hence the constant in the Friedrichs inequality is bounded. Moreover,we describe the precise asymptotics of the constant in the derived Friedrichs inequality as the small parameter characterizing the microinhomogeneous structure of the boundary,tends to zero. Paper B is devoted to the asymptotic analysis of functions depending on a small parameter,which characterizes the microinhomogeneous structure of the domain where the functions are dened. We consider a boundary-value problem in a twodimensional domain perforated nonperiodically along the boundary in the case when the diameter of circles and the distance between them have the same order. In particular,we prove that a Dirichlet problem is the limit for the original problem. Moreover,we use numerical simulations to illustrate the results. We also derive the Friedrichs inequality for functions vanishing on the boundary of the cavities and prove that the constant in the obtained inequality is close to the constant in the inequality for functions from H 1. iii

6 iv ABSTRACT In paper C we consider a boundary-value problem in a three-dimensional domain, which is periodically perforated along the boundary in the case when the diameter of the holes and the distance between them have the same order. We suppose that the Dirichlet boundary condition holds on the boundary of the cavities. In particular, we derive the limit (homogenized) problem for the original problem. Moreover, we establish strong convergence in H 1 for the solutions of the considered problems to the corresponding solutions of the limit problem. Moreover, we prove that the eigenelements of the original spectral problem converge to the corresponding eigenelements of the limit spectral problem. We apply these results to obtain that the constant in the derived Friedrichs inequality tends to the constant of the classical inequality for functions from H 1, when the small parameter describing the size of perforation tends to zero.

7 Preface This Licentiate thesis is composed of three papers (A-C) together with an introduction which put these papers into more general frame. [A] G. A. Chechkin, Yu. O. Koroleva and L. -E. Persson. On the precise asymptotics of the constant in the Friedrich`s inequality for functions, vanishing on the part of the boundary with microinhomogeneous structure. J. Inequal. Appl., 2007, Article ID 34138, 13 pages, [B] G. A. Chechkin, Yu. O. Koroleva, A. Meidell and L. -E. Persson. On the Friedrichs inequality in a domain perforated nonperiodically along the boundary. Homogenization procedure. Asymptotics in parabolic problems. Russ. J. Math. Phys., 16 (2009), No.1, [C] Yu. O. Koroleva. On the Friedrichs inequality in a cube perforated periodically along the part of the boundary. Homogenization procedure. Research Report, No.2, ISSN: , Department of Mathematics, Lulea University of Technology, 25 pages, Moreover, the following publications have inuenced the overview in the introduction and paper [C]: [D] Yu. O. Koroleva, On the constant in the Friedrichs inequality, In: Book of Abstracts of the International Conference Dierential Equations and Related Topics dedicated to the Centenary Anniversary of Ivan G. Petrovskii (XXII Joint Session of Petrovskii Seminar and Moscow Mathematical Society) (May 21-26, 2007, Moscow, Russia), Moscow University Press, Moscow, 2007, [in Russian] [E] R. R. Gadyl'shin, Yu. O. Koroleva and G. A. Chechkin, On the Convergance of the Solutions and Eigenelements of the BoundaryValue Problem in the Domain Perforated Along the Boundary, to appear in Dier. Equ., [in Russian] v

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9 Acknowledgement First of all I want to thank my main supervisors Professor Lars-Erik Persson and Professor Gregory A. Chechkin for their attention to my work,their valuable remarks and suggestions and for their constant support and help. Moreover,I thank my co-supervisors Dr. Annette Meidell and Peter Wall for helping and supporting me in various ways e.g. concerning numerical and application aspects on my work. I am also deeply grateful to PhD Student John Fabricius for helping me with several practical things. I also want to pronounce that the agreement about scientic collaboration and PhD education between Moscow Lomonosov State University and Lulea University of Technology has been very important. In particular,i thank both of these Universities for nancial support. Finally,I thank my family for all support and just being there. vii

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11 " $ & ' * $ &, ' 2 $ $ 4 & 4 2 : 4 ' &.? * A & & A &? $ $. 4 2 & * H I K, $, 2 N K I P I Q 2 N R I H I T 5 & A 5 V 2 N W I Y I [ & \ & ] N W I Q * ` I H I ` $ & 2 2 V c. & & & I : I e f h i N e f k i N e f l i N e m n i N e m f i N e m o i N e q n i N e q f i N e q q i N e q k i N e h m i e h q i t I u A 5 ' * $ &,. A * & &. A 2 * &.? $ $ 0. x 4 5 & *? 4 5 &, " 2 & ' 4. N., & 4 ' &? * A & 4 & ' 4? 4 5 & * $ & 4 5 & 4 5 & ' * & A 4 & & * A I W & " & N , & & ' 4? 4 5 & * & ' &.., $ $ ', & 4 & A ? : & 4, * &? 2 4 &. & A 4 2 : A, ' & 4. N A 5 A " & : & 4 ] & c &. &. &?, &. & t 0 5 & 4 5 &., $ $ ', & 4 & 4 &. 4 ] & I 5 & 5, : & 2 ] 4 2 ' A & &?. A 5 ' * $ &, & $ $. 4 2 & c. & & & I : I e i N e f h i N e m n i N e m i N e m i N e q k i N e q l i N e h m i e h q i t I 5 &, 2 &. $ 4. * 4 2 & &. & 0 V. A * 2 & * & &. A 2 * &.? $ $ 0. x 4 5 &. $ 4 2.? * " $ & ' * $ &, ' 2 $ $ 4 & 4 2 : * A A " & : & & A &. ' 2 :. $ 4 2.? * " $ & ' * $ &, & ˆ & A 4 2 " & * A I 5 & 4 ' &? $ 2, 2 4 * " $ & ' * $ &,. & ' & 4 5 & 4 2 * & 4 0 & & 4 5 &, &. &.? ' 4.? 4 5 & * ˆ & & 4 * A & 2 : 2 $ ' * $ &,. I 5 & ' ' &. e m n i N e m i N e m h i e m o i & $ & $ 4 & 4 2? 4 5 & K 2 2 A 5 $ & 4 * A & &, 4 5 & Š 2 & c * 2 t 4 ' &? * A I u, & &. 4 2, 4 &.? 4 5 & 4 &? 4 5 & A " & : & A & 0 & & * 4 2 & & 4 5 &.., ' & A 5., $ $ A, ' & 4? 4 5 & *. 5 2 V. 2 4 ' 2 4 I Œ e m n i 4 5 & 4 5 A. 2 &. * " $ & ' * $ &, ˆ & & 4 4 ' &.? * A N. 4 4 &., $ $ $ 4 & 4 2 : ' 2 & A &.? 4 5 & * I 5 & * & 5 " 2? 4 5 &. $ 4 2.?., & ' * $ &,. 0 5 & 4 5 &., $ $ ', & 4 & 4 &. 4 ] & & &. A 2 * & I Œ ' ' & 4 5 &., $ $ ', & 4 & &. A 2 * &. 4 5 & ' & 2? A 5 : &.? 4 ' &.? 4 5 & * A I Œ ' 4 2 A $ N 4 5 & 4 5 * 4 2. &. 4 2, 4 &.? 4 5 & 2 ˆ & & A & * & 4 0 & & 4 5 &. $ 4 2.? 4 5 & 2 : 2 $ ' * $ &,. 4 5 &. $ 4 2? 4 5 & A &. ' 2 : $ 2, 2 4 ' * $ &, I W & " & N 4 5 & &. ' & A 4 $ ' ' & 4 2 &.? 4 5 &. & ' * $ &,. I Œ 4. 5 $ * & 4 & ' ' &? 4 5 & Ž , & ' &. & 4 & A, ' $ & 4 & A $.. 2 Ž A 4 2? 4 5 & 5, : & 2 ] & ' * $ &,. N & ' & 2 : 4 5 & 4 2? 4 5 &., $ $ ', & 4 &. A 5 A 4 & 2 ] 2 : 4 5 &? & & A? 4 5 & ' & 2 2 A $ A 5 : &? 4 5 & * A I Œ e m i., & * " $ & ' * $ &,. 2, $ 4 2 2, &. 2 $, ' 2 $ $ 4 & 4 2 : * A & &. 4 2 & I Œ ' 4 2 A $ N ' " & & 5, : & 2 ] & ' * $ &,. & ' & 4 5 &., ' A.? 4 5 & Ž. 4 & 2 : & " $ &? 4 5 & A &. ' 2 :. ' & A 4 $ ' * $ &, A. 2 & & & A & $ $? ' & 2 2 A 2 4 I 5 2.., ' A. 0. A. 4 A 4 & * 4 5 & 4 5. ' ' $ 2 & 4 &. 4 2, 4 & 4 5 & 4 &? 4 5 & A " & : & A &? 4 5 &. $ 4 2? 4 5 & 2 : 2 $ ' * $ &, & A &. ' 2 :. $ 4 2? 4 5 & 5, : & 2 ] & ' * $ &, 0 5 & 4 5 &., $ $ ', & 4 & 4 &. 4 ] & I Œ e m i., & * " $ & ' * $ &, ' 2 $ $ 4 & 4 2 : 4 ' &.? * A & & A. 2 & &? 4 5 & Q ' $ A & & 2, &. 2 $, 2 I Œ , & & *? 4 5 &, 2 A ? 4 0 ' 4. I ` &? 4 5 &, 5. ' & $ ' & 2 2 A, 2 A. 4 A 4 & I Œ 4 A $ * & ' 2 $ $ 4 & 4 2 :. ' 4. ' & 2 2 A $ $ * 4 & 5 $ &. c? 4 5 & A. &? * &, ' &? 4 & 0 $ $ 2. 2 & t I Œ 4 5 & Ž. 4 A. & 0 & 5 " & * &

12 , , 2 A 2 5, : & &.. 4 A 4 &? 4 5 & * N $ & &. & A & 0 & 5 " & 4 0, 2. A & A 4 & 4 5 : & 5 $ &. I Œ 4 5 &. & A A. & 4 5 & 4 5. A. 2 & ' * $ &,. 2 * 4 5 * &. *, 2. 2 & * & & * &. *, 2. I Œ ' 4 2 A $ N 4 5 & 4 5. ' &. & 4 & A, ' $ & 4 & A $.. 2 Ž A 4 2? 4 5 & 5, : & 2 ] & ' * $ &,. A A & 2 : 4 5 & 2 & ' & & A & 4 5 & 4 2? 4 5 &., $ $ ', & 4 &. A 5 A 4 & 2 ] 2 : 4 5 &? & & A? 4 5 & ' & 2 2 A $ A 5 : &? 4 5 & * A I W & " & N 4 5 & A &. ' 2 :. ' & A 4 $ ' * $ &,. 0 & & A. 2 & & * 4 5 5, : & 2 ] 4 2 &. ' & A 4 2 " & A " & : & A & 4 5 & &,. 0 & & ' " & I & 4 0 &, & &., ' A.? 4 5 &. $ 4 2.?., & ' * $ &, ' 2 $. A 2 $ $ 4 2 : * A & & A. 4 A 4 & 2 e k i N e l i N e o i N e f n i N e f f i N e f i N e m f i N e m i N e q i e i I Š 4 5 &, & N., & 4 0 2, &. 2 $ * " $ & ' * $ &,. 0 & & A. 2 & & 2 e l i N e o i N e f i N e m i N e q i e i I Œ e m f i 4 5 &., ' A & ' A. 4 A 4 &? 4 5 &. $ 4 2? 4 5 & [ 2.. & 4 2 2, $ 4 2 2, &. 2 $ $ & & A. & 0 5 & 4 5 & * A $ 4 & 4 & ' & 2 2 A $ $. & V 2 : 2 4 ' 2 4 I W & " & N A, ' $ & 4 & & '. 2.? 4 5 & & 2 : & & $ &, & 4.? 4 5 & Q ' $ A & ' & 4 2 A $ 2 & ? & & 4 $ $ 4 & 4 2 : K 2 2 A 5 $ & 4 &, A &.? A &? 4 5 & A $ 2 & N 0 & & A. 4 A 4 & 2 e k i e f n i I Œ e f n i. ' & A 2 $ ' * $ &, K 2 2 A 5 $ & 4 A & $ 4 & $.? A & 0. A. 2 & &. 5, : & 2 ] & I Œ , & & ' 2 & A &.? 4 5 & * K 2 2 A 5 $ & 4 A " & 4 5 &., & &?., $ $. 2 ] &. ' 2 & A & &, A I Œ e k i 4 5 & A. & A &. ' 2 : 4 $ 2, 2 4 ' * $ &, &, Š 2 & c * 2 t * A & I 5 & 4 5 ' " & * e k i e f n i & 2 : 2 $ ' * $ &, 5. $ & 2 : & " $ &.?, $ 4 2 ' $ 2 A 2 4 & 4 0 I W & " & N 2 e k i 4 5 & 4 5 A. 4 A 4 & & ' $ 2 A 2 4 $ 4 5 & $ & 2 : 4 &,.? 4 5 &., ' A & '. 2.? 4 5 & & 2 : & " $ &. 4 5 & & 2 : &? A & A. & 0 5 & 4 5 & * " $ & ' * $ &, 5. &, Š 2 & c * 2 t * A & $ 4 & $.? A &? A $ 2 & 5 " 2 : * 2 4. & A 4 2 I Œ N &. & & 2 : & " $ &. A " & : & & A &. ' 2 : $ 2, 2 4 & 2 : & " $ &.?, $ 4 2 ' $ 2 A 2 4 & I Œ e f q i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e f q i 0 & & $. 2. A.. & 2 e f f i I Œ ' ' & H 0 &. & 4 5 & A " & : & A &? 4 5 & & 2 : & " $ &.?., & * " $ & ' * $ &, ' 2 $ $ 4 & 4 2 : 4 ' &.? * A * & ' & & 4 $ 2 4 & &. 4 2 :? A 4 A A & 2 : 4 5 & A & Š 2 & 2 A 5. 2 & $ 2 4 I 5 &? $ $ 0 2 : &. 4 2, 4 & 2. V 0 c. & & & I : I e h o N T 5 I Œ Œ Œ N i t. 4 5 & Š 2 & 2 A 5. 2 & $ 2 4?? A 4 2.?, 4 5 & u * $ & ". ' A & H 1 () :

13 & ª «c I t u 2 dx K 0 u 2 dx, 0 5 & & 4 5 & A. 4 4 & ' &. $ 4 5 &, 2 K 0 I Œ 4. 5 $ * & 4 & & Š 2 & 2 A 5. 2 & $ 2 4 Ž * 4 2 & * I Š 2 & 2 A 5.? * &, &? $ $ 0 2 :?, x u 2 dx K u 2 dx + u 2 ds,. & & e f i I Q 4 & N &, : ' 5 e q h i 4 5 &. A 2 & 4 A & & : 2 " &? 4 5 & " $ 2 2 4? 4 5 &? $ $ 0 2 : [ 2 A \ 2 & $ 2 4 x 2 u 2 dx K u 2 dx + 1 udx, m 0 5 & & 4 5 & A. 4 4 & ' &. 4 5 &, 2 m. W, 4 5 &, 4 2 A 2. 5 " & * & & 2 4 & &. 4 & & $ 2 4? " 2. &.. I 5 & &? & $ 4? 2 ˆ & & 4 &. $ & Š 2 & 2 A 5. 4 ' & [ 2 A & \ 4 ' & 2 & $ &. 5 " & * & & * 4 2 & I Œ A & 0 2 $ $ ' &. & 4., &? 4 5 &,. 4 2, ' 4 4? 4 5 &. & &. $ 4. I Œ e l o i 2 & A 4 '?? 4 5 & Š 2 & 2 A 5. 4 ' & 2 & $ &. 4 5 & [ 2 A & \ 4 ' & 2 & $ &. 2., &. *. ' A &.? Wp 1() & : 2 " & N 0 5 & & R n 2. * & A & A 4 & ' &, 2 N n =2, 3. 5 & [ 2 A & \ Š 2 & 2 A 5. 2 & $ &.? v Wω 1,p () = {v Wp 1() : v ω =0}, 1 <p<, 0 & & & 2 " & & 4 5 &.., ' &, &. &? ω 2. ' " & ' &, 2 c? & Ž & & & I : I e h k i t &. ' & A 4 4 ω. W & " & N 4 5 & & ' $ 2 A 2 4 & ' & & A &? 4 5 & A & 2, & 4 &? ω 0. : 2 " & I 5 &? $ $ 0 2 : 2 4 & &. 4 2 :? A 4 0. & 2 " & x 4 5 & A & 2 : &? 4 5 & [ 2 A & \ 2 & $ A &. &. 0 5 & ω. 5 2 V. I Œ ' 4 2 A $ N 0 5 & ω. 5 2 V. 4 ' 2 4 N 4 5 & 2. A.. & 2 & $ " $ 2 I & & A A & 4 & A 4 & &, ' $ & 0. : 2 " & 4. ' ' ? A 4 I ± & 4 & & A. & p>n 4 5 & A & ' & & 4? 4 5 & 2, & 4 &? ω,. 2 A & Wp 1 () C 0 ()? p>n. ± & $., & & ' ' & e k n i N 0 5 & &., & ' ' & 4 2 &.? 4 5 & * &. 4 A & Š 2 & 2 A 5. 2 & $ & &. 4 2 & I & & 4 5 &? $ $ 0 2 : " $ Š 2 & 2 A 5. 2 & $ A. 2 & & x c I ² t min u L p() = λ pq () > 0 2, u Lq() 0 5 & & 4 5 &, 2 2,, 2. 4 V & " & $ $? A 4 2. u Wp 1() " : 4 5 & 0 5 $ & *., & ' 4? 2 4 I & & , & <p< 4 5 & " $ &? q 2. & 4 &, 2 & * 4 5 & &, * & 2 : 4 5 & &,? 4 5 & u * $ & ". ' A &. x 1 q np n p, 2? p<n, 2? 1 q<+, p = n, 2? 1 q +, p>n.

14 µ 5 &. ' 5 & 2 A $ $ & B R2 \B R1,R 2 >R 1 > 0 0. A 5. &.. Œ , & N N u W p 1() ε = R2 R 1 1. Œ 4 0. V 0 & $ 2 & ? 2. * $ $ N 4 5 & 4 5 &, 2 2,, 2 c I ² t & 2 $ $.,, & 4 2 A? A 4 2 I H $ :. ' * $ &, 2. A 5. ' 5 & 2 A $ $ & & & $ 2 &? p =2. Œ e k n i ' " & 4 5 4? q>p 4 5 & & & ε = ε(q,p,n). A ? ε< ε 2 $? A & 4 &,, ' 2 4? c I ² t $ &? q p 4 5 & & 4 &,, ' 2 4? c I ² t &. ' 5 & 2 A $ $ & 2. 2 $? A 4 2 I H $ :.? A 4 0. &. 4 * $ 2. 5 & $ & A. & 0 5 & 2. ' $ & $ & I Œ e h k i., & : & & $ 4 5 & &,. A A & 2 : 4 5 & Š 2 & 2 A 5. 4 ' & 2 & $ &.?? A 4 2.?, 2 ˆ & & 4. ' A &. n 2, &. 2 $ A * & 0 & & &. 4 * $ 2. 5 & I Œ 4 5 & '?. 4 & A 5 2 &?, 4 5 & 4 5 &? 2. ' & 2, & 4 2 A $ 2 & $ &. 0.. & I Œ ' 4 2 A $ N 4 5 & ' ' & 4 5 & $ 0 & &. 4 2, 4 &.? 4 5 & * &. 4 A * 4 2 & I Œ & * &. 4 A A 5 4 ' &.? 2 & $ &. & 2 " &. & $ 4 5, 2 A A ' A 2 4 c? & Ž & & N & I : I e h k N T 5 I Œ N ² i t? 4 5 &. & & & 4 5 &? A 4 2 " 2. 5 &. I 5 &, 2 &. $ & Š 2 & 2 A 5. 2 & $ $. 2? $ 2? 4 5 & 5, 2 A A ' A ' " & I 5 2., & & Š 2 & 2 A 5. 2 & $ $. " $ 2 c &., & $.., ' & : &, & 4? 4 5 &, 2 t?? A 4 2.?, 4 5 &. ' A & H 1 (), " : & 0 5 $ & *? 4 5 &, 2 * 4. 4., $ $ ' 2 & A &.? 2 4 I Œ ' ' & 0 & : 2 " & 2 & A 4 '?? 4 5 & Š 2 & 2 A 5. 2 & $ 2 4?? A 4 2.?, H 1, " : 4 5 & *? 4 5 & A " & & ' & 2 2 A A. & & A & $ A $. & & & & A $.. 2 A $ Š 2 & 2 A 5. 2 & $ 2 4 I Œ e l i N e l i e l h i., & 2. A & 4 & $ : &.? Š 2 & 2 A 5. 4 ' & [ 2 A & \ 4 ' & 2 & $ &. 0 & & ' " & 2. A.. & I Œ e l h i 4 5 & 2 & A 4 '?.?., & 2. A & 4 & [ 2 A & \ Š 2 & 2 A 5. 2 & $ &.? A $..? A?, 2 : ' ' 2, 4 2.? 4 5 & u * $ & ". ' A & H 1 () 0 & & ' " & x c I» t g 2 (x) dx C F g(x) 2 dx, g W 0 (J k ), c I ¼ t g 2 (x) dx C P K J k K K J k K 2 g(x) 2 dx + C p g(x) dx, g W (J k ). & & & 4 &.. 5 ' & & : $. 2, ' $ 2 {J k } A 2 $ 4 2 : $ 4 2.? k. Œ 4 A ? 4 2 : $ & , &. 2 $. ' A &? 4 & 4 5 & & & 2, &. 2 $. ' A & I 5 &. ' A & W (J k ) A. 2. 4? 4 5 &? A 4 2. N A 5 $ A $ $ * & $ : 4 H 1 (K) & A 5 K J k, $ & 4 5 &. ' A & W 0 (J k ) A. 2. 4? $ $? A 4 2.?, W (J k ), A 5 5 " & ] & 4 A &. 4 5 & 4 &. 2 &.? K. 5 &. &. ' A &. & A?, ' ' 2, 4 2.? 4 5 & A &. ' 2 : u * $ & ". ' A &. N W 0 (J k ) H 1 () W 0 (J k ) H 1 (). W & " & N 2 e l h i $. 4 5 & * &. 4 A &. & 2 & $ &. 0 & & ' 2 4 & & 2. A & 4 & Š 2 & 2 A 5. 2 & $ & 4 & & , A 5 2. * & $ 2 &? 4 5 & 2 & A 4 2. I Œ 4 0 2, &. 2 $ 4 5 & & 2, &. 2 $. ' A & ' " & & A. 4 4 & ' &. $ 4 5 &. &? 4 5 & 2 Ž,, C " & 4 5 & F A V &..? 2 & 2 & A 4 2 I Œ N & 2 " & & ' &. $ C P * 4 5? A " & A " &, 2.. ` & V &

15 & ª ½ 2 & & ' ' & e l h i 0. 4 & 4 & 4 5 & '?? 4 5 & 2. A & 4 & [ 2 A \ Š 2 & 2 A 5. 2 & $ &.? 4 5 & A. & ' 2 & A & 0 2. & A. 4 4? A & & Ž 2 4 & " $, &, & , & : & & $ A. & I Œ 4. 5 $ * & 4 & & &. $ 4. * 4 2 & 2 e l h i A * & ' ' $ 2 & & $. 2.? A?, 2 :, & 2 A $, & 4 5. c & I : I? 4 5 & 2. A 4 2. R $ & V 2, & 4 5 t I Y $ 2 & N 4 5 & " $ 2 2 4? c I» t 0. &. 4 * $ 2. 5 & 2 e l i? W 0 (J k ) A :? ' 2 & A & 0 2. & $ 2 &? A 4 2. I Œ e l i &. $ 4 0. : & & $ 2 ] & 4. ' A &. A :? 5 2 : 5 & & ' $, 2 $? A 4 2. I Q 4 & N 2 e f i 4 5 & 2 & $ &. c I» t c I ¼ t 0 & & & 2 " &? ' 2 & A & 0 2. & H 2? A , & 4 0 2, &. 2 $, 2. I 5 & Š 2 & 2 A 5. 4 ' & 2 & $ ' & A 2 $ A. &? 4 5 &? $ $ 0 2 :, $ 4 2 2, &. 2 $ 4 ' & 2 & $ 2 4 x R n x n+ p(n 2) 2 u p dx 2 p C R n u(x) 2 dx, 0 5 & & u(x) C0 (Rn ),n>2, 2 p 5 & 2n n 2 A $.. 2 A $ 2 & $ Ž. 4 ' * $ & c '? t * R I I 2 ¾ ² 2 e o i ' " & 2 ¾ ² 2 e n i I H $. 4 5 & Ž. 4 0 & 2 : 5 4 & " &. 2?. 2 & $ & 4 5 2,. & $? &..? $ $ 0. x 0 5 & & 0 x r F (x) p dx f(x) 0, p>1, r 1 F (x) = x 0 ( ) p p x r (xf(x)) p dx, r 1 f(t) dt. r>1, F(x) = 0 x f(t) dt. r<1.. 2 : & $ r = p a>1 u(x) =F (x) 0 & * &? $ $ 0 2 :. A $ $ & 2 & $ ˆ & & 4 2 $?, x c I t 0 u p dx xp a ( ) p p u p x a dx, 1 <p<,a<p 1, p 1 a A 5 2. " &. &? $?, ' ' $ 2 A 4 2. I & $ 2 4 * " 2. $ 5 $? u(x) AC[0, ), u(0) = 0, 0 5 & & AC[0, ) & A $..? *. $ 4 & $ A 4 2. [0, )? A 4 2. I Y $ c I t 5 $. 2? $ 2? u(x) =0. Œ 4 0. À I & AÁ. 0 5 Ž. 4 : & & $ 2 ] & 4 5 & 2 & $ 2 4 c I t 4, 2. 2 R n c. & & e k f i t I & ' " & ? 2. * &, Q 2 '. A ] * N 1 <p<,a<p 1, 4 5 &? u C0 () 4 5 & 2 & $ 2 4 u(x) p ϱ(x) p+a dx C u(x) p ϱ(x) a dx 5 $. N 0 5 & & ϱ(x) =dist(x, ) & $ : & & $ 2 ] & $ 4 & * H I? & 2 e k i 4, &  $ & *? 4 & * H I ± & * c. & & e l k i t 4, : & & $ 2 ] &  $ & A I H? 4 & $ : &, * &? A, ' $ &, & 4 &. $ 4. : & & $ 2 ] " & * & & ' " & I Œ & & N 4 ' & 2 & $ &. & 4 $,

16 Ã. & & & I : I 4 5 & * V. e l i N e o i e k m i 4 5 & &? & & A &. : 2 " & 4 5 & & I Œ " 2 & 0? 4 5 & &. $ 4.? & & $., & , &. &? $ &. 4 2, 4 &.? 4 5 & * &. 4 A , &, $ 4 2 2, &. 2 $ 4 ' & 2 & $ &. 0 & & ' " & 2 e f i N e n i e l i I 5 & A $.. 2 A $ 4 ' & 2 & $ 2 4 A $. * & : & & $ 2 ] & &? $ $ 0 2 :, $ 4 2 2, &. 2 $?, x c I Ä t V (x) u(x) p dx 1 p C 1 q W (x) u(x) q dx, R n 0 5 & & u(x) C0 (Rn ),V(x) 0,W(x) 0, p,q>0, 4 5 & A. 4 4 C & ' &. $ V (x) W (x) I 5 & & &. & " & $ &. $ 4. $.? , A 5, & : & & $ A. & c. & & & I : I 4 5 & * V., & 4 2 & * " & 4 5 & &? & & A &. : 2 " & 4 5 & & t I & & 0 &. 4, & ? 4 5 & A. & q = p = ' " & * Å I W ] c. & & e h k i t & A. & W (x) =1 4 5 &? $ $ 0 2 : A * 4 5 & A &... A 2 & 4? 4 5 & " $ 2 2 4? c I Ä t x F R n V (x) dx βcap(f, ), 0 5 & & cap (F, ) = inf{ u(x) 2 dx}, F 2. A, ' A 4. & & 2 Ž,, 2. 4 V & " & 4 5 &. & 4 {u(x) C0 (), u 1 2 & 2 : 5 * 5? F, 0 u 1}. Œ ' 4 2 A $ N? n =1 4 5 & ' & A 2. & &. $ 4 &..? $ $ 0. x 5 & 2 & $ q(x)u 2 (x) dx c 0 u 2 (x) dx, 5 $. 2? $ 2? 4 5 & & & A. 4 4 sup x (0, ) x x 0 5 & & C 1. A q(x) dx C 1. u(0) = 0, ± & ' A & & $ &. * 4 5? 4 ' &? Š 2 & 2 A 5. 4 ' & & " & 2, ' 4 4?, ' ' $ 2 A 4 2. I Œ ' ' & H 0 & A. 4 A &., ' A.? 4 5 & A & Š 2 & 2 A 5. 4 ' & 2 & $ 2 4?? A 4 2.?, 4 5 & u * $ & " u ' A & H 1 N A 5 " 2. 5., $ $ ' 2 & A &.? 4 5 & * I 5 2. &. $ 4 0. &. A 2 * & * 2 & 2 e h i I 5 & ' ' &. T & & " 4 & 4 2 " &. 4 2 : 4 2.?., & & 0 * " $ & ' * $ &, ' &? 4 2 $ : 4 5 & * I " $ & ' * $ &,. 2 ' &? 4 &, 2. 0 & &. 4 2 & *, 4 5. c. & & & I : I e m i e i N e m q i e q o i N e q i N e i N e h n i e h l i e k i e l f i t I Œ 4 5 &. & ' ' &., : ' & & 2 ˆ & & 4 V 2.? ' &? 4 2? $ 2 &. 0 & $ $. $ 2 & 2 ˆ & & 4 2 $ ' & 4. I Œ 4 5 & ' &. & A & * 2 & &. A 2 * & 4 5 & * 4 2 & &. $ 4. I " $ & ' * $ &,. 2 ' &? 4 &, & A. & 0 5 & 4 5 & A " &. & * 4 & ' & 2 2 A $ $ , $ $ ' & 2 ' " 2 & &. 2 ] &? 4 5 & A " & &., & &. 4 5 & ' & 2? ' &? 4 2 N & 2 0 V.? ` I H I ` $ & 2 2 V 5 &. 4 & 4. c. & & & I : I e m i N e k q i

17 ª Æ e k i t I 5 & A. & 0 5 & 4 5 &. 2 ] &? 4 5 & A " &. 5., A 5 $ &.. & 4 5 & 4 5 & ' & 2? ' &? A. 2 & & 2 e q i N e q i e i I 5 & 0 V e q i c. & & $. 4 5 & & e i t 0. & 2 A 4 & 4 $. 2.? * " $ & ' * $ &,. 2 ' &? 4 &, & A. & 0 5 & 4 5 & 2, & 4 &? 5 $ &. 2., A 5 $ & & 4 5 & 2. 4 A & * & 4 0 & & 4 5 &, I H 5, : & 2 ] & &, 0. ' " &? * " $ & ' * $ &, &, * A & & 4 & $ * K 2 2 A 5 $ & 4 * A & *? A " &. I W & " & N 4 5 & 4 5 A. 4 A 4 & 4 5 & Ž. 4 4 &,? 4 5 &., ' A & '. 2? 4 5 & 2 : 2 $ ' * $ &, * 4 2 & &. 4 2, 4 &.? 4 5 & 2 ˆ & & A & * & 4 0 & & 4 5 & 2 : 2 $. $ & Ž. 4 4 &,? 4 5 &., ' A & '. 2 I 5 & & $.. ' & A 4 $ ' ' & 4 2 &.?. A 5 * " $ & ' * $ &,. * 4 2 &., & &. 4 2, 4 &.? 4 5 & 2 ˆ & & A & * & 4 0 & & 4 5 & & 2 : & & $ &, & 4.? 4 5 & 2 : 2 $ ' * $ &,. 4 5 & & 2 : & & $ &, & 4.? 4 5 & A &. ' 2 : 5, : & 2 ] & ' * $ &,. I Œ e h l i 4 5 & K 2 2 A 5 $ & 4 ' * $ &, 0. A. 2 & & 2, ' & 2 2 A $ $ * 4 & A " &. I Œ ' 4 2 A $ N 4 5 & 0 & V A " & : & A & 2? 4 5 & L. $ &,.? 2 A " & : & A &? 5, 2 A A ' A 2 4? 4 5 & A " &. 0. ' " & I " $ & ' * $ &,. 2, ' &? 4 2 $ :.? A & 0 & &. 4 2 & 2 e q o i N e h i N e h N T 5 I Œ N» i e k k i I Œ e h N T 5 I Œ N» i 4 5 & & * " $ & ' * $ &, 2, 2 ' &? 4 & $ : A $. & A " & I Œ ' 4 2 A $ N 4 5 & 2?, A " & : & A &? 4 5 &. $ 4 2.? 4 5 & 2 : 2 $ ' * $ &,. 2 A, ' A 4. *, 2. N 2 " $ " 2 : 4 5 & A " & N &. $ 4 2.? $ 2, 2 4 ' * $ &, 0. ' " & I Œ e k k i * " $ & ' * $ &, & 2, 2 2 " 2 & * ' &? 4 & 0 $ $ I W & " & N 4 5 & 0 & V A " & : & A &? 4 5 &. $ 4 2.? 4 5 & 5, : & 2 ] & ' * $ &, &. $ 4 2? $ 2, 2 4 ' * $ &, 0. ' " & I W & " & N 4 5 & &. 4 2, 4 &.? 4 5 & 4 &? 4 5 & A " & : & A &?., &. ' & A 2 $ A. &. 0 & & * 4 2 & I Œ e q o i 4 5 & 4 5 A. 2 & & 4 5 & A V &,. 2 & " & ' * $ &, N A 5 2. & 4 &. 2? ' * $ &, 2. & * Å I ` I W A 5 & V Y I È I 5. $ " c. & & e h i t N, &. ' & A 2 Ž A $ $ N * Y I u A 5 &. [ $ & A 2 c. & & e k k i t x 4 5 & ' $ & x n =0, ' &? 4 & * ε ' & 2 2 A $ $ * 4 &., $ $ 5 $ & , & 4 &. r(ε) < ε 2,. & ' 4 &. n 2, &. 2 $ * &, 2 G *, 2. G + G. 5 & * & 5 " 2? 4 5 &. $ 4 2.? * " $ & ' * $ &,. 2 u ε G & 0 5 & ε 0+. Œ e q o i $ :. ' * $ &, 0. A. 2 & &? É A V Ê. 2 & " & & $ : & A V &.. h(ε) $ &. * & 2 : A $ 2 &.? 5 & 2 : 5 4 h(ε) c. & & Š 2 : &. ² t I Ë Ì Í Î Ï Ð Ñ Ò 5 &. 2 & " & & A. & n =2 I 5 &? A &. $ 4 2? 4 5 & & 4 2 u ε Δu + u = f. * & A & &, * A &.? A &? 4 5 &. 2 & " & N & I : I N 4 5 & K 2 2 A 5 $ & 4 A & *? G & " & I 5 &. $ 4 2 u A " & : &. N 2., &. &. & N 4 ε u + 2 G + 4 u 2 G 0 5 & ε, h(ε) r(ε) 4 & 4 I Œ 4 0. ' " & 2 e q o i &. &? A 4 2. & 4 5 &. $ 4 2.? 4 5 & & 4 2 * " &. 4 2.? 2 : 4 5 & K 2 2 A 5 $ & 4 A & *

18 Ó Ë Ì Í Î Ï Ð Ô Ò 5 &. 2 & " & & A. & n =3 I? G & " & N * 4 $ : 4 5 & ' $ & x n =0,, '. 2. & A A 2 : &? $ $ 0 2 : x u + n μ(u+ u ) =0, 0 5 & & 0 μ + 2.., & A. 4 4 n & 4 &, $ &. ' & A 4 4 G + G. W & " & N 4 5 & & ' & & A &? 4 5 & A. 4 4 μ 4 5 &? A 4 2. r(ε) h(ε) 0 5 & ε 0? 2, &. 2 n 3 0. & 2 " & 2. A.. & I Œ ' 4 2 A $ N , & μ =0 2? 4 5 & 5 $ &. 5 " & É 4., $ $ 2, & 4 &. Ê & É 4 $ : Ê Õ μ>0 2? 4 5 & 5 $ &. & É * 2 : & : 5 Ê * 4 É 4 4 $ : Ê c & $ 4 & A. & μ = t I 5 &. & A. &. & & A 4 $. & ' 4 & *., & A A $? A 4 2. r (ε),h (ε) 4 5 & " $ &.? μ & : 2 " & *? A 4 2 $ $ V 2 : " &. 2, 2 $. 4 5 & & Ž 2 4 2? A ' A 2 4 I 0 & " & N? 2, &. 2 n =2., & 2 A $ 4 2 &. 2. & & $ :. ' * $ &, , & & &, 2 ' & I 5 & ' ' & e h i 2. & " 4 & &., ' A * & 5 " 2? 4 5 &. $ 4 2.? * " $ & ' * $ &, 2, 2 ' &? 4 & $ :, 2? $ ˆ & & 4 4 ' &.? A & *? A " &. I & & 4 5 & 4 5. A. 2 & * &, 2 2 R n A 2 & 4 $., 4 5 *,. & 4 γ, A 5 2. A $ $ & A 4 2?., &. & 4.. Q & & ' 2 4. P j (j =1,...,N(ε), N(ε) d 0 ε 1 n,d 0 = const) * & $ : 4 γ, 0 5 & & ε.. $ & 4 &.., $ $ ' " & ', & 4 & I K & 4 & * G j (a j ε). & 4 * & $ : 2 : 4, , 4 5 * G j (a j ε), P j G j (a j ε), 4 5 & 2, & 4 &? G j (a j ε) & $. 4 a j ε,a j ε C 0 ε, C 0 = const > 0,G j (a j ε) G i (a i ε)=? i j. 5 & 4 5. A. 2 & $ $ '.. 2 * $ & A. &.? * & 5 " 2? a j 0 5 & ε ε 0. Q & 4. &. A 2 * & 4 5 & &. $ 4.? e h i 2, & & 4 2 $ I 5 & & 4 5 &? $ $ 0 2 : x G ε = N(ε) j=1 G j (a j ε), ε =\ G ε,s ε = N(ε) j=1 G j (a j ε), S ε = N(ε) j=1 G j (a j ε), Γ ε = ε \ S ε.

19 ù ª 5 &? $ $ 0 2 : * " $ &. ' * $ &,. 0 & & A. 2 & & x { c I Ø t Δu ε = f 2 ε, u ε ν =0 S ε, u ε =0 Γ ε, { c I Ù t Δu ε = f 2 ε, u ε ν + βu ε =0 S ε, u ε =0 Γ ε, β(x) β 0 = const > 0, { c I ¾ t Δu ε = f 2 ε, u ε =0 ε, 0 5 & & ν & 4 &. 4 5 & &, $ 4 Q & 4 * &. $ 4 2? 4 5 &? $ $ 0 2 : ' S ε *. $ &, x v 0 { c I t Δv 0 = f 2, v 0 =0, 0 5 & & f 2.., 4 5? A 4 2?,. 5 &? $ $ 0 2 : &. 4 2, 4 &. 0 & & * 4 2 & & A. & 0 5 & & 0 & V u. $ 4 2? 4 5 & ' * $ &, c I Ø t N &. $ 4 2? c I t x ε v 0 Ú Û Ð Ü Ï Ð Ý I Ò Þ ß à á â ã ã ä å æ ç è é ç S ε è æ ë =. ê é ë ì u ε v 0 2 H 1 ( K ε) 1(max a j ε j )n ε 1 n, λ k ε λ k 2 C 1 (max a j ε) n ε 1 n. j Þ í à á â ã ã ä å æ ç è é ç S ε é ë ì ç è é ç ç è æ ë ä å î æ ï ð ñ ã æ ç ã G j (a j ε), ë ð ë ò ó ë ç æ ï ã æ ô ç ó ë õ ö ó ç è, ó ã æ ã ã ç è æ ë ð ï æ ø ä é ç ð d 1 ε 2 n, é ë ì ç è é ç G j (a j ε) d 2 (a j ε) n 1 è æ ë. ê é ë ì u ε v 0 2 H 1 ( K ε) 2 max(a j ε j )n 1 ε 2 n, λ k ε λ k 2 C 1 max(a j ε) n 1 ε 2 n. j æ ï æ {λ k ε }, {λk },k =1, 2,... é ï æ ç è æ ã æ ø ä æ ë ô æ ã ð ñ æ ó õ æ ë ú é ä æ ã ð ñ ç è æ ô ð ï ï æ ò ã û ð ë ì ó ë õ ã û æ ô ç ï é û ï ð î æ å ã á 5 & A " & : & A & &. A 2 * & 2 & 4 5 & &, 0. ' " & & A. & 0 5 & & 0 & V. $ 4 2? 4 5 & ' * $ &, c I Ù t &. $ 4 2? u ε v 4 5 & $ 2, 2 4 ' * $ &, c I t x 0 Ú Û Ð Ü Ï Ð Ý I ² Ò â ã ã ä å æ ç è é ç G j (a j ε) K(a j ε) n 1 é ë ì ç è é ç, η ε (max a j ε)ε 1 0 ö è æ ë è æ ë ε 0. ê u ε v 0 2 j H 1 ( 0 ö è æ ë ε) ε 0, é ë ì u ε v 0 2 H 1 ( ε) K 3η n 1 ε,

20 ù ù ý é ë ì ) n 1 λ k ε λ k 2 C 2 (max(a j ε)ε 1. j æ ï æ {λ k ε }, {λk },k =1, 2,... é ï æ ç è æ ã æ ø ä æ ë ô æ ã ð ñ ç è æ æ ó õ æ ë ú é ä æ ã ð ñ ç è æ ô ð ï ò ï æ ã û ð ë ì ó ë õ ã û æ ô ç ï é û ï ð î æ å ã á Š 4 5 &, & N 4 5 & 4 5. A. 2 & & 4 5 & A. & 0 5 & γ 2., & 5 ' & ' $ & {x : x 1 =0}, γ= {x : x 1 =0}, Q= {x : 1 2 <x 1 < 1 2,j = 1,...,n},G ε = (a ε G 0 +εz). & & 2., , 4 5 * N G 0 z Z a ε Cε, C = const, G 0 Q, Z &. & 4? " & A & : & A, ' & 4. I Œ N , & a ε ε 1 C 0 = const > & ε 0 C 0 G 0 Q. Q & &, 2 * & & ' &. & 4 & &? $ $ 0 2 : 2 x ε c I t ε = + ε γ+ ε Π ε γ ε ε, 0 5 & & { Π ε = x : x 1 < ε } {, Π ε =Π ε \ G 2 ε,γ ε ± = x : x 1 = ± ε }, 2 { + ε = x : x 1 > ε } {, ε 2 = x : x 1 < ε }, G ε = G ε 2. &. 2 & N S ε = G ε, Γ ε = ε \ S ε,l ε = G ε,l ε = G ε. 5 & & 4 5 & ' * $ &, { 2 c I ² t v = f \ γ, v =0, [v] γ =0, [ v γ ν] = μv γ, 0 5 & & μ = β 0 C0 n 1 G 0. Œ? 2. 0 & V. $ 4 2? 4 5 & ' * $ &, c I Ù t 2 u ε ε, : 2 " & *?, $ c I t N v 2.. $ & ' * $ &, c I ² t N 4 5 & 4 5 &? $ $ 0 2 : 5 & &, 5 $. x Ú Û Ð Ü Ï Ð Ý I» Ò ÿ ä û û ð ã æ ç è é ç a ε ε 1 C 0 = const > 0 ö è æ ë ç è é ç è æ ë l ε d 3 ε. ê é ë ì u ε v H 1 ( ε) K 4 ( ε + aε ε 1 C 0 ), ε 0, é ë ì λ k ε λ k 2 ( C 3 ε + aε ε 1 C 0 2). æ ï æ {λ k ε}, {λ k },k =1, 2,... é ï æ ç è æ ã æ ø ä æ ë ô æ ã ð ñ æ ó õ æ ë ú é ä æ ã ð ñ ç è æ ô ð ï ï æ ò ã û ð ë ì ó ë õ ã û æ ô ç ï é û ï ð î æ å ã á 5 & K 2 2 A 5 $ & 4 ' * $ &,? 4 5 & [ 2.. & $. A. 2 & & I Q & 4 * & 4 5 & 0 & V. $ 4 2? c I ¾ t $ & 4 u ε v * & 4 5 &. $ 4 2? 4 5 & ' * $ &, c I t I 5 & 4 5 &? $ $ 0 2 : 5 & &, 2. " $ 2 x Ú Û Ð Ü Ï Ð Ý I ¼ Ò ÿ ä û û ð ã æ ç è é ç lim max ε 0 j max ε 0 j lim (a j ε) n 2 ε 1 n =0, ö è æ ë ε 1 ln a j ε 1 =0, ö è æ ë n 3, n =2.

21 ù ñ ª n 3, ç è æ ë u ε v 0 2 H 1 ( K ε) 5(max a j ε j )n 2 ε 1 n, λ k ε λ k 2 C 4 (max a j ε) n 2 ε 1 n, j é ë ì ó ñ n =2, ç è æ ë uε v0 2 H 1 ( K ε) 6 max ε 1 ln a j ε 1, j λ k ε λ k 2 C 5 max ε 1 ln a j ε 1. j æ ï æ {λ k ε}, {λ k },k =1, 2,... é ï æ ç è æ ã æ ø ä æ ë ô æ ã ð ñ æ ó õ æ ë ú é ä æ ã ð ñ ç è æ ô ð ï ï æ ò ã û ð ë ì ó ë õ ã û æ ô ç ï é û ï ð î æ å ã á W & " & N 4 5 & ' * $ &, c I ¾ t 0. A. 2 & & &, 2 ε, & 4 &, 2 & *?, $ c I t & 4 5 &.., ' 4 2 G 0 = {x : x <a}, a j ε = a ε. 5 & $ $. ' '. & lim ε 0 an 2 ε ε 1 n = C 1 = const > 0, 0 5 & n 3, lim ε 0 ε 1 ln a ε 1 = C 2 = const > 0, 0 5 & n =2. Œ A. & 4 5 & $ 2, 2 4 ' * $ &, &? $ $ 0 2 : & x c I» t v = f 2 [ ] +, [v] γ =0, v γ x 1 = μ 1 v γ,v=0, 0 5 & & = {x : x 1 < 0}, + = {x : x 1 > 0}, μ 1 = (n 2)a n 2 ω(n)c 1, 2? n 3, μ 1 =2πC 2, 2? n =2, 0 5 & & ω(n) &. &? 4 5 & 2 4. ' 5 & & 2 R n. 5 &? $ $ 0 2 : &. 4 2, 4 &. & * 4 2 &? A. & x Ú Û Ð Ü Ï Ð Ý I Ò ÿ ä û û ð ã æ ç è é ç a ε ε 1 C 0 = const > 0 ö è æ ë ç è é ç è æ ë l ε d 3 ε. ê é ë ì ö è æ ë n 3 é ë ì u ε v L2( ε) K 7 ( ε + C1 a n 2 ε ε 1 n ), λ k ε λ k 2 C 6 ( ε + C1 a n 2 ε ε 1 n 2) ε 0, é ë ì ( u ε v L2(ε) K 8 ε + C2 (ε ln a ε ) 1 ), ) λ k ε λ k 2 C 7 (ε + C 2 (ε ln a ε 2 ) 1 ö è æ ë n =2. ù æ ï æ {λ k ε }, {λk },k =1, 2,... é ï æ ç è æ ã æ ø ä æ ë ô æ ã ð ñ æ ó õ æ ë ú é ä æ ã ð ñ ç è æ ô ð ï ï æ ã û ð ë ì ó ë õ ã û æ ô ç ï é û ï ð î æ å ã á Š 4 5 &, & N 4 5 & K 2 2 A 5 $ & 4 ' * $ &, c I ¾ t 0. A. 2 & & 2 e h i &, 2 ε, & Ž & * c I t &., & $.., ' 4 2. x { a 2 n ε ε n & ε 0, 2? n 3, ε ln a ε 0, 0 5 & ε 0, 2? n =2.

22 & & 4 5 & , & a j ε = a ε. Œ A. & 4 5 & $ 2, 2 4 ' * $ &, 2..? $ $ 0. x { v = f 2,v=0, v + = f 2 +,v=0 +. Œ A. & 4 5 &? $ $ 0 2 : &. 4 2, 4 &.? 4 5 & 4 &? 4 5 & A " & : & A & 5 $ x Ú Û Ð Ü Ï Ð Ý I Ä Ò ç ó æ ì ã ç è é ç é ë ì ö è æ ë é ë ì ũ ε v ± 2 L 2( ε) K 9a 2 n ε ε n 1, λ k ε λk 2 C 8 a 2 n ε ε n 1 n 3 é ë ì ũ ε v ± 2 L 2( ε) K 10ε ln a ε, λ k ε λ k 2 C 9 ε ln a ε ö è æ ë n =2. ù æ ï æ ì æ ë ð ç æ ã é ë æ ç æ ë ã ó ð ë ð ñ ó ë ũ ε u ε, é ë ì {λ k ε}, {λ k } é ï æ ç è æ ã æ ø ä æ ë ô æ ã ð ñ æ ó õ æ ë ú é ä æ ã ð ñ ç è æ ô ð ï ï æ ã û ð ë ì ó ë õ ã û æ ô ç ï é û ï ð î æ å ã á " $ & ' * $ &,. 2 ' &? 4 &, 2.? $ 2 & 2 ˆ & & 4 2 $ ' & 4. 0 & & A. 2 & & & I : I 2 e h n i N e h f i N e k l i N e k o i N e l n i e l f i I 5 &, : ' 5 e k l i & $ * " $ & ' * $ &,.? $ 2 & & $ $ 2 ' 4 2 A & 4 2.? * 2 4 & I 5 & Ž. 4 5 $?? 4 5 & * V & $ &. $ " * 2 $ 2 4 ' * $ &, $ & 4 5 &. & A & 2. & " 4 & & ' ' & 4 2 &.? 4 5 & 0 & V. $ 4 2. I W & " & N 4 5 & 4 5 A. 2 &. 5, : & 2 ] 4 2??, 2 $?. 2 $ 2 & * " $ & ' * $ &,. * 4 5 2, Ž & : 2 & * 2, A 5 & $. I Œ ' 4 2 A $ N ' " & &. $ 4 2.? 4 5 &. & * " $ & ' * $ &,. & A $. & 4. $ 4 2? A & 4 2 $ 2, $ 2 & ' * $ &, 2 ' &? 4 &, 2 I 5 & K 2 2 A 5 $ & 4 ' * $ &, 2 ' &? 4 &, 2.? $ 2 & & A. 2 & & & I : I 2 e k l i N e k o i e l n i I W & " & N? 4 5 & 2 " &. 4 2 : 4 2.?. 2 $ 2 & ' * $ &,. 2. & & A &?, Ž & : 2 & *. 4 2 :? 4 5 & A " & : & A &? 4 5 & & 2 : & " $ &.? $ 2 & K 2 2 A 5 $ & 4 ' * $ &,. 2. A 5, 2. A * &? & ' ' & e l f i I & &, 2.? $ $ A. 2 & & x s =\ I(s) i=1 F (s) i, 0 5 & & R n 2. * 2 4, 2 F (s) i,i =1,...I(s) <, s N 2. Ž 2 4 &, * &? A $. &, 2. N A 5 & A 4 2 & 2. & & $..., & & 2, & 4 &.? 4 5 &. & 4. F (s) i, 4 5 & 2. 4 A & * & 4 0 & & 4 0 & 2 : 5 * 2 :. & & 2. 4 A & * & 4 0 & & 4 5 &. & & *? 4 & 4 ] & 0 5 & s. Œ , &? $ $ 0 2 : $ 2 & ' * $ &, & x c I ¼ t n j=1 d dx j f j (x, u s (x), u s (x)) f 0 (x, u s (x), u s (x)) = = λ s g 0 (x, u s (x)), x s, u s (x) = 0, x s.

23 ª «Œ 4 0. ' " & & ' * $ &, * & $ & $ 2, 2 4? c I ¼ t &., & A &? A 4 2. g 0,f j,j=0,...,n: c I t n d dx j f j (x, u(x), u(x)) f 0 (x, u(x), u(x)) + c 0 (x, u(x)) = j=1 = λg 0 (x, u(x)), x, u(x) = 0, x. W & " & N 4 5 & A " & : & A &? 4 5 & & 2 : & & $ &, & 4.? 4 5 & 5, : & 2 ] & ' * $ &, & & 2 : & & $ &, & 4.? 4 5 & A &. ' 2 : $ 2, 2 4 ' * $ &, 0. &. 4 * $ 2. 5 & x $ & 4 * & 4 5 & & 2 : & " $ &? c I ¼ t λ s λ * & 4 5 & & 2 : & " $ &? c I t $ & 4 u s (x) ũ(x) * & 4 5 & A &. ' 2 : & 2 : &? A 4 2. I 5 & lim λ s = λ 4 5 &. & & s A & {u s (x)} A " & : &. 0 5 & s=1 s 4 ũ(x). 4 : $ 2 Wr 1 ()? r<m 0 & V $ 2 W Œ e h f i & : & & 4 & m() $ 2 & K 2 2 A 5 $ & 4 ' * $ &, & 2, ' & 2 2 A $ ' &? 4 2 & 4 5 & * I Œ ' 4 2 A $ N A ? 4 5 & & 2. 4 & A &? 4 5 & $ 2, 2 4 ' * $ &, 0. & 2 " & N, & " & N 4 5 & 0 & V A " & : & A &? 4 5 &. $ &. 4 * $ 2. 5 & I u, & & L 2 A A & 2 : 4 5 &., ' A * & 5 " 2?. 2 $ 2 & ' * $ 2 A & 4 2 2, Ž & : 2 & * 0. A. 2 & & 2 e h n i I Œ ' 4 2 A $ N 4 5 & 0 & V A " & : & A & 2? 4 5 &. $ &,.? 4 5 & L A " & : & A &? 4 5 & 5, 2 2 A A ' A 2 4?., $ $. & ' " & I Œ e m i N e m q i N e q m i e k i., & ' * $ &, , ' &? & & A. 2 & & 2. A.. & I Œ ' 4 2 A $ N 2 e m i. 4 A A ' 4 2 $ 2 ˆ & & 4 2 $ & & 2, , ' &? 4 2 I 5 & 4 5 A. 2 & &, 2 G & 4 5 & & 2, &. 2 $ Y A $ 2 2. ' A & 2 & ' & & 4 2 & 4 2 A $ $ * 4 &, " & A 4. U 1,U 2,... ' * * 2 $ 2 4. ' A & (,A,P ) " $ &. 2 G. 5 &? $ $ 0 2 :. & 4 0. & Ž & x G n = G \ n S(U k,ε), k=1 0 5 & & S(x, ε) & 4 &. 4 5 &. ' 5 & &? 2. ε A & 4 & 4 x. 5 &? $ $ 0 2 : $ 2 & $ A $ ' & 4 L : C0 (T ) L 2(T ), 0. A. 2 & & I & & , & S T R d, 0 5 & & S 2.. & 4?, 2. Γ= S. H $. 4 5 &? $ $ 0 2 : & &. & 2 e m i x X(S) ={x X D (T ):supp x S},F(S) ={f L 2 (T ),suppf S}, X + (Γ) = {x X(Γ),x= L g, supp g T \ S}, H = L 2 (,A,P) 2. 2 $ * & 4. ' A &?, " 2 * $ &. I 5 & 4 5 A. 2 & & 4 5 &? $ $ 0 2 :. 4 A A * " $ & ' * $ &, &, * x { 2 c I Ä t Lξ = η S (η, x) =(ξ Γ,x) x X + (Γ), 0 5 & & 4 5 &, ' ' 2 :. η : x(s) H, ξ Γ : X + (Γ) H & $ 2 & A 4 2.?, &. 2 & N 4 5 &? $ $ 0 2 : & 4 &, A ' * $ &, &,

24 ê µ *?, 4 5 & 4 5 &. 2 & x { 2 c I Ø t Lξ n =0 G n, (ξ, x)=(ξ,x) x X + ( G n ), 0 5 & & G n S, ξ &. $ 4 2? c I Ä t ξ n = ξ 2 S \ G n. 5 &, 2 &. $ &? $ $ 0 2 : 5 & &, x Ú Û Ð Ü Ï Ð Ý I Ø Ò è æ ö æ é ô ð ë ú æ ï õ æ ë ô æ ξ n u ó ë P, é á ã á ó ñ lim n nεd n = α, 0 α<. ð ï æ ð ú æ ï ó ñ lim n nεd n = é ë ì H =(,A,P) è ð ì ã lim n εd n = β, 0 β, è æ ó å ó ç ñ ä ë ô ç ó ð ë ç è æ ë ξ n u ó ë ç è æ ë ð ï å ð ñ L 2 (,A A,P P ). ê u ó ã ç è æ ã ð ä ç ó ð ë ð ñ ç è æ ñ ð ð ö ó ë õ û ï ð î æ å { ó ë c I Ù t Lu = Cη S, (u, x) =(ξ Γ,x) x X + (Γ), ö è æ ï æ C(t) =1 e αw dϕ(t) ó ñ α< é ë ì å æ é ë ã ç è æ ú ð ä å æ ð ñ ç è æ ä ë ó ç w ã û è æ ï æ ó ë d R d,ϕ(t) ó ã ç è æ ô ð ë ç ó ë ä ð ä ã ì æ ë ã ó ç ñ ä ë ô ç ó ð ë ð ñ ç è æ ú æ ô ç ð ï ã U 1,U 2,... ð ë (,A,P ) C(t) =χ Aβ (t) ó ã é ë ó ë ì ó ô é ç ð ï ñ ä ë ô ç ó ð ë ö ó ç è A β (t) ={t : s, ϕ(s) > 0,t S(s, β)} ö è æ ë α =. Œ e k i 4 5 & & & 2 : & " $ &.? 4 5 & K 2 2 A 5 $ & 4 * " $ & ' * $ &,? 4 5 & Q ' $ A 2 2, , $ * 4 &. ' 5 & 2 A $ *. 4 A $ &. I & & 0 & A. 2 &. & & A &? 2 & ' & & 4 $, $ * 4 & A & 4 &. {w 1,...,w N } 2, 2 2 R 3 &, " &?, M. & & A &? * $ $. {B k = B(w k,r)}, A & 4 &. 4 {w k } Ž & 2 2 r = α m. Œ 4 2. $..., & N m β,β<3. W & " & N 4 5 & 4 4 $, &. &? {B k } 4 & & m. 5 & 4 5 & 4 &. * μ i ({w k }) 4 5 & j 4 5 & 2 : & " $ &? 4 5 & K 2 2 A 5 $ & 4 ' * $ &,? 4 5 & Q ' $ A 2 2 M \ B k ; & A 5 μ A * & A. 2 & &., " 2 * $ & j μ j (w 1,...,w N ) 4 5 & ' A 4. ' A & M M c 4 2, &. t I Š β = e h i μ j ({w}) A " & : &. I & I & j 4 5 & 2 : & " $ & μ V? 4 5 & u A 5  2 : & ' & 4 j H = Δ +4παV 2 W N 0 5 & & Å & 4 &. 4 5 & * 4 2 &. 2 4? & A 5 " 2 * $ & w k. 5 & e k i & Ž & &. $ 4 &. 4 * $ 2. 5 & A & 4 $ $ 2, & &,? 4 5 &, " 2 * $ &. μ j ({w k }) 0 5 & m. Œ 4.. & μ j ({w k })=μ V j + m β 2 1 S j, R.. 2 " 2 * $ &.? ] & S, & 4 5 & " 2 j A & & ' & 2 : 4 5 & j 4 5 & 2 : &? A 4 2? Δ M. 5 & 4 5. $. ' &. & 4 & ' ' 2, 4 & A. 4 A 4 2? 4 5 & R & &? A 4 2 G(x; y)? Δ M\ Bk, ' & A 2. & &. 4 2, 4 &.? 4 5 & 2 ˆ & & A & * & 4 0 & & R 2 4. ' ' 2, 4 2 I 5 &., ' A * & 5 " 2? 4 5 &. $ 4 2 2, 2, $ ' &? 4 & $ : 4 5 & * & 2 e m q i c. & & $. 4 5 & & e q m i t I Œ 4 0. ' " & &. $ 4 2.? 4 5 & 2 : 2 $ ' * $ &, ,. 4 A 4 & A " & : & 0 & V $ & u * $ & ". ' A & H 1 () &. $ 4 2?, c & 4 &, A t ' * $ &, , : & 2 ] & * A & *? 4 5 &, 2 I

25 ª ½ 5 & ' ' & 2. & " 4 & 4., ' A $. 2.?? A 4 2. & ' & 2 :., $ $ ', & 4 & &. A 2 * 2 : 4 5 &, 2 A 2 5, : & &.. 4 A 4 &? 4 5 &, 2 N 0 5 & & 4 5 &? A 4 2. & & Ž & I ± & A. 2 & * " $ & ' * $ &, 2 & A 4 : $ & ' &? 4 & ' & 2 2 A $ $ $ : 4 5 & * & A. & 0 5 & 4 5 & 2, & 4 &.? A 2 A $ &. 4 5 & 2. 4 A & * & 4 0 & & 4 5 &, &? 4 5 &., & & I 5 &, 2 &. $ 4 * 4 2 & 2 ' ' & & A " & : & A &? 4 5 &. $ 4 2.? 4 5 & 2 : 2 $ ' * $ &, &. $ 4 2? 4 5 & A &. ' 2 : $ 2, 2 4 ' * $ &,. 0 & $ $. 4 5 & A " & : & A &? 4 5 & & 2 : & & $ &, & 4.? 4 5 & &. ' & A 4 2 " &. ' & A 4 $ ' * $ &,. I W & " & N 0 & ' ' $ &. $ 4 4 &. 4 2, 4 & 4 5 & A & 0 Š 2 & 2 A 5. 4 ' & 2 & $ 2 4 I Œ 4 5 & ' ' & T 0 & A. 2 & ' * $ &, & & 2, &. 2 $ A * & N ' & 2 2 A $ $ ' &? 4 & $ : ' 4? 4 5 & * & A. & 0 5 & 4 5 & 2, & 4 &? 5 $ &. 4 5 & 2. 4 A & * & 4 0 & & 4 5 &, 5 " & 4 5 &., & & I H $ :. ' * $ &,? 4 5 & 4 0 2, &. 2 $, 2 0. A. 2 & & 2 e k i c ' & 2 2 A A. & t 2 e m k i c ' & 2 2 A A. & t I & & 0 &. ' '. & & K 2 2 A 5 $ & 4 * A & & *? A " &. I 5 &., ' A * & 5 " 2? 4 5 &. $ 4 2.? 4 5 & 2 : 2 $ ' * $ &, & 2 & 4 2 $ I Œ ' 4 2 A $ N 0 & & 2 " & 4 5 & $ 2, 2 4 c 5, : & 2 ] & t ' * $ &,? 4 5 & 2 : 2 $ ' * $ &, I W & " & N 0 & &. 4 * $ : A " & : & A & 2 H 1? 4 5 &. $ 4 2.? 4 5 & A. 2 & & ' * $ &, & A &. ' 2 :. $ 4 2? 4 5 & $ 2, 2 4 ' * $ &, I W & " & N 0 & ' " & & & 2 : & & $ &, & 4.? 4 5 &. ' & A 4 $ ' * $ &,. A " & : & & & 2 : & & $ &, & 4.? 4 5 &. ' & A 4 $ $ 2, 2 4 ' * $ &, I &. 2 & N 0 & : 2 " & 4 5 & 2 & A 4 '?? 4 5 & Š 2 & 2 A 5. 2 & $ 2 4?? A 4 2. * & $ : 2 : & u * $ & ". ' A & H 1, " :., $ $ A " &.? 4 5 & ' &? 4 2 I W & " & N 0 & ' " & & 2 ˆ & & A & * & 4 0 & & 4 5 & A & & A $.. 2 A $ Š 2 & 2 A 5. 2 & $ &. 4 ] & 0 5 & 4 5 &., $ $ ', & 4 & 4 &. 4 ] & I ± & & 2 " & ? A 4 *. 2 : 4 5 & A " & : & A &? 4 5 & & 2 : & " $ &.? 4 5 & 2 : 2 $ 4 5 & 5, : & 2 ] & ' * $ &, N A. 2 & & ' ' & I Š 2 $ $ N 0 &, & , & & 0 ' & ' * $ &,. & ' &. & 4 & Q 2 A & & 4 5 &. 2. N A 5 A * &? 2 4 & &. 4 2? 4 5 & &. & A ?. A : Ž & $?, 4 5 &, 4 2 A. I

26

27 ! " # % ' ) ' ), % ' 9. ;. < < = > ) ' B C 3 4. E = > C K C > 4 M > 4 4 N R S = 3, U W % N K N ; C X C Z 1 = ;. 4. C > 6 [ ] ^ _ ` ^ a b c d ^ b _ e f ` f g h j a ] c e k l a o j ] p f b ^ p d q q _ t k u w y y z { u > 1 K C, { '! w # ' } ' ) 3 = 4 N N = 3 U = ; 1 C ƒ 3 N K ; C 0 C. >. 4 5 N 0 C ˆ 3 C, N = 3. C > 6 d q q _ t l b o j t a o o t k u " z z Œ { 6 N ' 6 " W Ž '! # ) ' ˆ ' C ; U = C X 6 1 ; = 3 ƒ C K = 4. > N R N = 3 U = ; 1 C ƒ 3 N K ; C 0 > 6 j h j a f ` f t l ] f ] o b o a ^ ` š a c f ` o e t } N > M N 6 " z z y '! 1 > >. #! # ) ' ˆ ' C ; U = C X 6 S N 0 N = 4. N R = K N = 3 U X = ; 1 C < 3 N K ; C = , K N = 3 U M 4. R N 3 ƒ N. > > C Z 1 = =, N 0 < C 3 R N 3 = 4 C, = ; 4 5 C K N = 3 U 6 f q t l b o t [ b k u " z z y { 6 N ' 6 " w 1 > >. #! Œ # ) ' 1 ' C ; U = C X ˆ ' ) ' 5 C M 5 6 ) X C 3 N < C 3 = 4 N 3 >. 4 5 K N = 3 U M 4. > N C > M = ; C, > M C 6 l b o t b a o ` k u " z z z { 6 N ' 6 z Ž W Œ " y ' u ;. > > ; = «l b o j t [ ] o a f k u " z z z { 6 N ' 6 " W w z ' {! Ž # ' '. E 0 C 4 N X 6 ) > U 0 < 4 N 4. M > N R 4 5 C C. C ; C 0 4 > N R K N = 3 U X = ; 1 C < 3 N K ; C 0 > R N C % M 5 3 C 3 N < C 3 = 4 N = ; = 3 C < N 4 4. = ; ; N M = ;. C, = > 0 = ; ; > C 4 6 j t e j ` f _ t l b o t l b o t ` ± t k u w y y Ž { 6 N ' 6 Ž Ž ³ W Ž w ' u ;. > > ; = µ ] q o t l b o j t l b o j t j e f t k u w y y Ž { 6 N ' 6 Ž Ž W Ž Œ y ' {! ³ # ' Y ' N 3. > N X 6 N = 3 U X = ; 1 C < 3 N K ; C = M U C R 3 C Z 1 4 ; U M 4 U < C N R K N = 3 U M 4. > 6 l b o j t ¹ t k º» ¼ u w y y w { 6 N ' ³ 6 ³ W Ž ' u ;. > > ; = ¹ t l b o j t º» ¼ u w y y w { 6 N ' ³ 6 z ³ ³ W " y y ' {! # ' Y ' N 3. > N X 6 B N < = 3 = 0 C 4 C 3 = > U 0 < 4 N 4. M = K N = 3 U X = ; 1 C < 3 N K ; C 0 R N C = < ; = M. 6 l b o t b a o ` k ¾ u w y y " { 6 N ' 6 Œ w y W Œ ' u ;. > > ; = «l b o j t [ ] o a f k ¾ u w y y " { 6 N ' 6 ³ " W Œ ' {! z # ' Y ' N 3. > N X 6 = = < ; = M. 4 5 R 3 C Z 1 4 ; < C 3. N,. M = ; ; U = ; 4 C = K N = 3 U M 4. > 6 ] _ t d b p t [ b k ¼ ¼ u w y y w { 6 N ' 6 W Œ ' u ;. > > ; = «] _ t l b o j t k º» ¼ u w y y w { 6 N ' ³ 6 z ³ ³ W " y y ' {! " y # ' Y ' N 3. > N X 6 = 1 ; = 3 ; U < C K C, K N = 3 U X = ; 1 C < 3 N K ; C 0 R N C = < ; = = M U C 3 6 ` à a c t c b š ^ k ¼ u w y y w { 6 N ' 6 " y ³ " W " y ³ ' u ;. > > ; = «` à a c t Ä Å t k ¼ u w y y w { 6 N ' 6 " " y W " " ' {! " " # ' Y ' N 3. > N X 6 B 5 C = > U 0 < 4 N 4. M > N R 4 5 C C. C ; C 0 4 > N R 4 5 C = < ; = = M U C R 3 C Z 1 4 ; U N > M. ; ; = K N = 3 U M 4. > 6 µ t Æ t d b p t ` t b c ` f k a c t Ç Ç ¹ k ¼ È» u w y y " { 6 N ' " y 6 ³ " ³ W ³ w " '! " w # ' Y ' N 3. > N X 6 = 0 N, C ; K N = 3 U X = ; 1 C < 3 N K ; C 0 R N 3 = < ; = M. 4 5 R 3 C Z 1 4 ; U = ; 4 C = 4 U < C N R K N = 3 U M 4. > 6 d f e q o t d ^ b _ t k ¼ u w y y { 6 N ' " 6 " W w Ž '! " # ' Y ' N 3. > N X 6 ) > U 0 < 4 N 4. M > C > 4. 0 = 4 C > N R 4 5 C M X C 3 M C 3 = 4 = C C,. 0 >. = ; K N = 3 U X = ; 1 C < 3 N K ; C = <., ; U = ; 4 C = K N = 3 U M 4. > 6 ` ¹ t l b o t j t k u w y y { 6 N ' w 6 w ³ W w z ' u ;. > > ; = «` ¹ t l b o j t Î t k u w y y { 6 N ' w 6 w w w W w w ' {! " # ' Y ' N 3. > N X ' ' ˆ =, U ; > C > < C M N R 4 5 C = < ; = M. 4 5 R 3 C Z 1 4 ; U = ; 4 C = K N = 3 U M 4. > 6 h j a ] c t l b o j t j e f t k º º u " z z z { 6 N ' 6 ³ W Œ ' u ;. > > ; = «h j a ] c t l b o j t j e f t k º º u " z z z { 6 N ' 6 w ³ w W w ³ ³ ' {! " Œ # % ' ' 6 Ð ' Ñ Ò ' Ó 5 = N 6 ƒ M = 3 C W 9 3. C, 3. M 5 > C Z 1 = ;. 4. C > R N 3 ƒ. C M C. > C H 9 M 4. > 6 [ a c t ^ o t d ^ b _ t Ô q o ` ` ± b o ` ] ^ k È u w y y { 6 N ' Œ W Ž 6 Ž W ³ '! " Ž # ˆ ' 2 ) ' 5 C M 5 6 N = 3 U W = ; 1 C ƒ 3 N K ; C 0 > R N 3 = > C M W N 3, C 3 ª ; ;. < 4. M ª Z 1 = H > M. ; ; = N = 3 U 4. > 6 «[ ] ^ _ b f f ` b _ b c o ` b _ ` à a c a ^ o ` b _ Ä Å b o ` ] ^ f u ª, ' ' ' 3 = N X { 6 N X N >. K. 3 > E > C N R } = 4 5 C 0 = 4. M > 6 %. K C 3. X. >. N R 4 5 C ) M =, C 0 U N R % M. M C > N R 4 5 C Õ % % u Y } % H ) % % % { 6 u " z { 6 z Œ W " y '! 1 > >. # Æ

28 Ó Ø Ù Ú Û Ü Ý! " ³ # ˆ ' ) ' 5 C M 5 6 < = 3 4. = ; ; U Þ C, 0 C 0 K 3 C 6 ß a o a ^ à ` ¹ t l b o t Ô ¹ f j t k [ ] š ] f ` ¹ ` c f k u " z z { 6 y W w 1 > >. #! " # ˆ ' ) ' 5 C M 5 6 % < C M 4 3 = ; < 3 N < C 3 4. C > N R C ; ;. < 4. M < 3 N K ; C = <., ; U N > M. ; ; = K N = 3 U M 4. > 6 «ß ] ^ p b c e b _ a c ] ¹ _ a f ã ] c [ ] ^ _ b f f ` b _ b c o ` b _ ` à a c ä a ^ o ` b _ Ä Å b o ` ] ^ f k u ª, ' ' ' 3 = N X { 6 N X N >. K. 3 > E > C R N 3 } = 4 5 C 0 = 4. M > N R 4 5 C %. K C 3. X. >. N R 4 5 C ) M =, C 0 U N R %. M C > N R 4 5 C % N X. C 4 ƒ 3 C > > 6 u " z z { 6 " z ³ W w y y 1 > >. #! " z # ˆ ' ) ' 5 C M 5 6 ª ; ;. < 4. M N = 3 U = ; 1 C ƒ 3 N K ; C 0 >. 4 5 = <., ; U ) ; 4 C = N = 3 U 4. > 6 ƒ 5 ' B 5 C >. > ' } N > M N % 4 = 4 C X C 3 >. 4 U 6 " z z w 1 > >. #! w y # ˆ ' ) ' 5 C M 5 6 ) X C 3 N R K N = 3 U X = ; 1 C < 3 N K ; C 0 >. 4 5 = 1 ; = 3 < C K = 4. N R 4 5 C K N = 3 U M 4. > 6 l b o j t ¹ t k º u " z z { 6 N ' Ž 6 z z W " Œ y ' u ;. > > ; = «Æ f f t d b p t ` t k ¹ t k l b o j t k ¾» u " z z { 6 N ' " 6 " z " W w w w ' {! w " # ˆ ' ) ' 5 C M 5 6 B 5 C = > U 0 < 4 N 4. M C Þ < >. N R 4 5 C > N ; 1 4. N R = K N = 3 U X = ; 1 C < 3 N K ; C = <., ; U = ; 4 C = 4 U < C N R K N = 3 U M 4. > 6 h c t a ` ^ t Ç t Ç t å t a o c ] ä š f ] æ ] k º» u w y y z { 6 w W ³ ' u ;. > > ; = «Î t l b o j t ` t k [ a ç ] c k u " z z ³ { 6 N ' Ž 6 w y W w z ' {! w w # ˆ ' ) ' 5 C M 5 6 ) > U 0 < 4 N 4. M C Þ < >. > N R 4 5 C C. X = ; 1 C > C. R M 4. > N R C ; ;. < 4. M N < C 3 = 4 N =, N U è ;. 5 4 é M M 4 3 = 4 C, 0 = > > C C = C K N = 3 U ' B 5 C 4 N,. 0 >. = ; M = > C 6 Ç ± š t Æ ] f f t d b p t [ b a c t l b o t k» u w y y Œ { 6 N ' 6 " Ž " W w y ' u ;. > > ; = Ç ± š t l b o j t k» u w y y Œ { 6 N ' 6 y Œ W Ž ' {! w # ˆ ' ) ' 5 C M 5 6 B ' ƒ ' 5 C M 5 = 6 ' ê ) <. M C Õ ' C } =. N 6 S N 0 N = N 0 > N 0 ; U ƒ C 3 R N 3 = 4 C, ) ; 4 5 C N = 3 U 6 4 N = < < C = a f c a o a b ^ p µ ] ^ o ` ^ ä ] f e ^ b ` b _ e f o a f k a c t ß ë µ ä ß ì k u w y y z { 6 N ' " " '! w # ˆ ' ) ' 5 C M 5 ' ' ˆ =, U ; > 6 ) K N = 3 U X = ; 1 C < 3 N K ; C 0 R N C = < ; = M = <., ; U M 4 U < C N R K N = 3 U M 4. = 0 1 ; 4.,. 0 >. = ;, N 0 6 ` ¹ t l b o t t k u " z z z { 6 N ' w 6 w ³ " W w ³ ' u ;. > > ; = ` ¹ t l b o j t Î t k u " z z z { 6 N ' w 6 w ³ " W w ³ ' {! w Œ # ˆ ' ) ' 5 C M 5 ' ' ˆ =, U ; > 6 K N = 3 U X = ; 1 C < 3 N K ; C 0 > R N C = < ; = K N C,, N 0 > M 3 5 N 0 N C N 1 > > M C N R 4 5 C K N = 3. C > 6 d o b l b o j t ` ^ t k Ä ^ æ _ t a c t k È ¼ u w y y ³ { 6 N ' w 6 w ³ W w '! w Ž # ˆ ' ) ' 5 C M 5 ª ' Y ' N = C ) > U 0 < 4 N 4. M > N R 4 5 C % < C M N R = N = 3 U = ; 1 C ƒ 3 N K ; C < C 3. N,. M = <., ; U ) ; 4 C = N = 3 U 4. > 6 ^ o t ` à a c t Ä Å t k u w y y " { 6 N ' " W w 6 " " " W " w w '! w ³ # ˆ ' ) ' 5 C M ' H ' Ð N 3 N ; C X = 6 ) ' } C., C ; ; ' ª ' ƒ C 3 > > C 9 3. C, 3. M 5 C Z 1 = ;. 4 =, N 0 < C 3 R N 3 = 4 < C 3. N,. M = ; ; U = ; 4 5 C K N = 3 U ' S N 0 N = 4. < 3 N M C, 1 3 C ' ) > U 0 < 4 N 4. M < = 3 = K N ;. M < 3 N K ; C 0 > 6 Æ f f t Î t l b o j t j e f t 6 º u w y y z { 6 N ' " 6 " W " Ž '! w # ˆ ' ) ' 5 C M ' H ' Ð N 3 N ; C X = ' ª ' ƒ C 3 > > C < 3 C M. > C = > U 0 < 4 N 4. M > N R 4 5 C M > C 9 3. C, 3. M 5 ï C Z 1 = ;. 4 U R N 3 R M 4. > 6 X > 4 5 C < = 3 4 N R 4 5 C K N = 3 U M 3 5 N 0 N C N 1 > > M C 6 Î t Ç ^ a Å b _ t d q q _ t k w y y ³ 6 ) 3 4. M ; C Y " 6 " < = C > 6 w y y ³ '! w z # ˆ ' ) ' 5 C M 5 H ' ) ' H ; E 6 K N = 3 U X = ; 1 C < 3 N K ; C 0 > R N 3 C ; ;. < 4. M C Z 1 = 4. > = <., ; U M 4 U < C N R K N = 3 U M 4. > 6 f q t l b o t [ b k u " z z { 6 N ' Ž 6 " Ž W " Ž Œ ' u ;. > > ; = «Æ f f t l b o j t c š t k u " z z { 6 N ' Ž 6 " ³ W " ³ Œ ' {! y # ˆ ' ) ' 5 C M 5 H ' ) ' H ; E 6 = ƒ 3 N K ; C 0 N R N = 3 U ) X C 3 R N 3 = % U > 4 C 0 N R ª ; = > 4. 4 U B 5 C N 3 U 6 f q t l b o t [ b k» u " z z { 6 N ' 6 " " ' u ;. > > ; = «Æ f f t l b o j t c š t k» u " z z { 6 N ' 6 " " W " " ' {! " # ˆ ' ) ' 5 C M 5 H ' ) ' H ; E 6 = > U 0 < 4 N 4. M > N R > N ; 1 4. > C. X = ; 1 C > N R 4 5 C K N = 3 U X = ; 1 C < 3 N K ; C 0 > = <., ; U = ; 4 C = K N = 3 U M 4. > R N C > U > 4 C 0 N R C ; = > 4. M. 4 U 6 d o o ` d b p t [ b ± t ` ^ a ` k µ _ t ` t ` f t l b o t [ b o t k Ç ð t a c t k Æ a ^ p t ` ^ a ` k l b o t d q q _ t k ¾ u " z z Ž { 6 N ' " 6 Œ W " Œ '! w # ˆ ' ) ' 5 C M 5 ª ' ' H > 4 3 N X > E = U = 6 C 5 = X. N 1 3 N R = N, U N 0 ; U ƒ C 3 R N 3 = 4 C, ) ; 4 5 C N = 3 U 6 «ß ] ] ] ã d ¹ f o c b o f ] ã o j a Ç ^ o a c ^ b o ` ] ^ b _ µ ] ^ ã a c a ^ a ñ ` à a c ä a ^ o ` b _ Ä Å b o ` ] ^ f b ^ p Æ a _ b o a p h ] q ` f ò p a p ` b o a p o ] o j a µ a ^ o a ^ b c e d ^ ^ ` š a c f b c e ] ã Ç š b ^ å t a o c ] š f ` ` ë ð ð Î ] ` ^ o a f f ` ] ^ ] ã a o c ] š f ` ` a ` ^ b c b ^ p l ] f ] l b o j a b o ` b _ ] ` a o e ì u } = U w w w ³ 6 w y y " 6 } N > M N 6 1 > >. = { 6 6 } N > M N X C 3 >. 4 U ƒ 3 C > > 6 } N > M N 6 w y y " ' Y % Œ w " " y y ³

29 Ø Ù Ú Û Ü Ý! # ˆ ' ) ' 5 C M 5 6 ) ' ' ƒ. = 4 > E. ) ' % ' % 5 = 0 = C X 6 ó ] ] æ a ^ ` ± b o ` ] ^ t l a o j ] p f b ^ p d q q _ ` b o ` ] ^ f k B 3 > ; = 4. N R } = 4 5 C 0 = 4. M = ; } N 3 = < 5 > w ' ƒ 3 N X., M C 6 Y «) 0 C 3. M } = 4 5 C 0 = 4. M = ; % N M. C 4 U 6 w y y ³ '! # } ' Y ' 5 C 3, 4 > C X 6 ) > U 0 < 4 N 4. M > N R 4 5 C C. X = ; 1 C N R 4 5 C = < ; = M C N < C 3 = 4 N =, N = 1 ; = 3 ; U < C K C, K N = 3 U ' l b o t b a o ` k ¾ u w y y Œ { 6 N ' w 6 w z z W y ³ ' u ;. > > ; = l b o j t [ ] o a f ¾ u w y y Œ { 6 N ' " w 6 w ³ y W w ³ ' {! Œ # '. N 3 C > M 1 9 ' } 1 3 = C 3 0 C C 4 3 C X 1, ê =. ; ; C 1 3 > Y 6 Y Y 6 «[ ] ^ _ ` ^ a b c q b c o ` b _ p ` à a c a ^ o ` b _ a Å b o ` ] ^ f b ^ p o j a ` c b q q _ ` b o ` ] ^ f t µ ] a æ a p a c b ^ a a ` ^ b c k Æ a f t [ ] o a f l b o j t k ô ¾ u " z w { 6 N ' w W 6 z W " 6 " Œ W " ³ '! Ž # ' N ' S. ; K C l a o j ] p f ] ã l b o j a b o ` b _ j e f ` f ' Ñ. ; C U 6 C N 3 E 6 " z z '! ³ # ) ' = 0 ; = 0. ' õ '. 6 N = 3 U 5 N 0 N = 4. R N 3 C ; ;. < 4. M < 3 N K ; C 0 > 6 Î t l b o j t c a d q q _ t k u " z ³ { 6 N ' 6 Œ " W Ž " '! # Ò ' = ö X. ; = 6 C = 3 C ; ;. < 4. M C Z 1 = = <., ; U N > M. ; ; = = ; 4 C = K N = 3 U M 4. > 6 d f e q o ] o ` t d ^ b _ t k È u w y y " { 6 N ' W 6 w ³ z W y ³ '! z # B ' C ; C M M 5. N 6 B 5 C 4 5. M E C 1 ê > %. C X C 6 d ^ ^ t l b o t c b d q q _ t k º ¾ u " z ³ { 6 N ' 6 Ž W y w '! y # % ' 9. ;. < < = > 6 ' ˆ ' } = ê U = ) ' B C 3 4. E = > 6 % 5 = 3 < S = 3, U W % N K N ; C C Z 1 = ;. 4. C > 6 µ t Æ t k l b o j t k d b p t ` t b c ` f k ¼ ¼» u w y y { 6 N ' ³ 6 W Ž '! " # Ð ' 9 3. C, 3. M 5 > 6 % < C M 4 3 = ; 4 5 C N 3. C 5 = ; K K C > M 5 3 E 4 C 3 H < C 3 = 4 N 3 = 1 R,. C % < C M 4 3 = ; ; C X C 3 4. = ; N < C 3 = 4 N 3 ' Y 6 Y Y 6 l b o j t d ^ b _ t k º» u " z { 6 Ž W ³ 6 Ž Œ W ³ " '! w # ' ' ˆ =, U ; > C C. X = ; 1 C = > U 0 < 4 N 4. M > R N 3 < C 3. N,. M = ; ; U M ; = 0 < C, 0 C 0 K 3 C > 6 d _ æ a ¹ c b d ^ b _ t k º u " z z { 6 N ' " 6 W " z ' u ;. > > ; = «o t a o a c ¹ c æ l b o j t Î t k º u " z z z { 6 N ' " 6 " W " ' {! # ' ' ˆ =, U ; > 6 ) > U 0 < 4 N 4. M > N R 4 5 C C. X = ; 1 C > N R = K N = 3 U X = ; 1 C < 3 N K ; C = <., ; U N > M. ; ; = K N = 3 U M 4. > 6 ` à a c t c b š ^ t k ¼ u " z z z { 6 N ' 6 Œ y W Œ Œ " 6 " z z z ' u ;. > > ; = «` à a c t Ä Å t k ¼ u " z z z { 6 N ' 6 Œ y W Œ Œ " ' {! # ' ' ˆ =, U ; > 6 S N 0 N = 4. = > U 0 < 4 N 4. M > R N 3 = 0 C 0 K 3 C. 4 5 M ; N > C ; U > < = M C, M ; = 0 < 4 > 6 j t e j ` f _ t l b o t l b o t ` ± t k º u w y y " { 6 N ' " w 6 " Œ ³ W " Ž z ' u ;. > > ; = «µ ] q o t l b o j t l b o j t j e f t k º u w y y " { 6 N ' " w 6 " ³ Ž Œ W " ³ ³ Ž ' {! Œ # ' ' ˆ =, U ; ê > C ) > U 0 < 4 N 4. M > N R 4 5 C ª. X = ; 1 C > R N 3 ƒ C 3. N,. M = ; ; U ; = 0 < C, } C 0 K 3 C > 6 o t a o a c f ¹ c æ l b o j t Î t k º u " z z z { 6 N ' " y 6 " W " '! Ž # ' ' ˆ =, U ; ê > 6 B 5 C } C 4 5 N, N R } = 4 M N R ) > U 0 < 4 N 4. M ª Þ < >. = 1 ; = 3 ; U W ƒ C K C, N = 3 U W = ; 1 C ƒ 3 N K ; C 0 R N C = < ; = M C H < C 3 = 4 N 3 6 ] š c a t l b o t c ` _ ] ± j t k º È u w y y { 6 N ' Œ 6 W w ' u ;. > > ; = Ò ' } = 4 5 ' % M. u ' ' { 6 º È u w y y Œ { 6 N ' Œ 6 Œ ³ z W Ž y z ' {! ³ # ' ' ˆ =, U ; ê > 6 1 ' H ' Ð N 3 N ; C X = ˆ ' ) ' 5 C M C X C 3 M C N R 4 5 C % N ; 1 4. > ª. C ; C 0 4 > N R 4 5 C N = 3 U W = ; 1 C ƒ 3 N K ; C 4 5 C N 0 ƒ C 3 R N 3 = 4 C, ) ; 4 5 C N = 3 U 6 4 N = < < C = ` à a c t Ä Å t k u w y y z { 6 u " Ž < = C > { '! # Ò ' ƒ ' ˆ = 3 M. = ) N 3 C 3 N Y ' ƒ C 3 = ; ) ; > N 6 S = 3, U C Z 1 ;. 4. C > % N 0 C ª ; ;. < 4. M ƒ = 3 = K N ;. M ƒ 3 N K ; C 0 > 6 Î t ` à a c t Ä Å t k º u " z z { 6 N ' w 6 " W ³ Ž '! z # ˆ ' S ' S = 3, U 6 [ ] o a ] ^ b h j a ] c a ] ã ó ` _ ¹ a c o k l b o j t t k u " z w y { 6 N ' Ž 6 " W " ³ '! Œ y # ˆ ' S ' S = 3, U 6 N 4 C > > N 0 C < C 3 = ; M = ; M 1 ; 1 > 6 ú C Z 1 = ;. 4 U K C 4 4 C 3 = ; > 6 l a f f a ^ æ a c ] ã l b o j t u " z w Œ { 6 " Œ y W " Œ Ž '! Œ " # ˆ ' S ' S = 3, U 6 Ò ' ª '. 4 4 ; C N N, ˆ ' ƒ N ö ; U = 6 Ç ^ a Å b _ ` o ` a f k C, ' 6 = 0 K 3., C X C 3 >. 4 U ƒ 3 C > > 6 " z Œ w '! Œ w # ˆ ' ) ' Y N >. R ê U 6 H ' ) ' H ; E ) ' % ' % 5 = 0 = C X 6 ) > U 0 < 4 N 4. M C Þ < >. > N R > N ; 1 4. > N R 4 5 C. 3. M 5 ; C 4 < 3 N K ; C 0 R N 3 C ; ;. < 4. M C Z 1 = 4. > > U > 4 C 0 > R N C 4 5 C N 3 U N R C ; = > 4. 4 = < C 3 R N 3 = 4 C,, N 0 6 ] _ t d b p t [ b Æ k È u " z Œ { 6 N ' Œ 6 " y Ž w W " y Ž Ž 1 > >. #! Œ # S ' Ð = M S N 0 N > = 4. C > < 3 N K ; C 0 >, C. 3. M 5 ; C 4 = X C M, C > < C 4. 4 > 4 3 N 1 > 6 h j a f a û ä a a e _ a k ^ ` š a c f ` o e b c ` f Y 6 " z Ž '! Œ # S ' Ð = M ' } 1 3 = 4 6 ª > 4. 0 = 4. C ; ê C 3 3 C 1 3, >, C > < 3 N K ; C 0 >, C. 3. M 5 ; C 4 N 1 = < < = 3 =. 4 4 C 3 0 C C 4 3 C 6 Ç ^ g c ] æ c a f f ` ^ [ ] ^ ä _ ` ^ a b c ` à a c a ^ o ` b _ Ä Å b o ` ] ^ f b ^ p h j a ` c d q q _ ` b o ` ] ^ f t b c o ` b _ ` à a c a ^ o ` b _ Ä Å b o ` ] ^ f b ^ p o j a µ b _ _ f ] ã b c ` b o ` ] ^ f t ] _ a Ç Ç t Ä f f b e f ` ^ ó ] ^ ] c ] ã Ä ^ ^ ` ] a å ` ] c æ ` t ß ` c j b f a c t ß ] f o ] ^ t ß b f a _ t ß a c _ ` ^ t Ž Ž " W Ž z Ž 6 " z z '

30 ý Ø Ù Ú Û Ü Ý! Œ Œ # ' Ð N E. ; = > 5 X. ;. 6 ) ' } C > E 5. ' ª ' ƒ C 3 > > ü a ` æ j o a p [ ] c Ç ^ a Å b _ ` o ` a f ã ] c Ç ^ o a æ c b _ h c b ^ f ã ] c f ` o j c ] p o ý a c ^ a _ f 6 N X = % M. M C ƒ 1 K ;. > 5 C 3 > 6 M ' 6 w y y z '! Œ Ž # 1 ' H ' Ð N 3 N ; C X = C M > C 9 3. C, 3. M 5 C Z 1 = ;. 4 U ' Ç ^ g ß ] ] ] ã d ¹ f o c b o f ] ã o j a Ç ^ o a c ^ b o ` ] ^ b _ µ ] ^ ã a c a ^ a ñ ` à a c a ^ o ` b _ Ä Å b o ` ] ^ f b ^ p Æ a _ b o a p h ] q ` f ò p a p ` b o a p o ] o j a µ a ^ o a ^ b c e d ^ ^ ` š a c f b c e ] ã Ç š b ^ å t a o c ] š f ` ` ë ð ð Ç Ç Î ] ` ^ o a f f ` ] ^ ] ã a o c ] š f ` ` a ` ^ b c b ^ p l ] f ] l b o j a b o ` b _ ] ` a o e ì ë l b e þ ÿ ä þ k þ k l ] f ] k Æ f f ` b ì k } N > M N X C 3 >. 4 U ƒ 3 C > > 6 } N > M N 6 w y y ³ 6 " Œ W " Œ 1 > >. #! Œ ³ # ) ' Ð 1 C 3 6 ü a ` æ j a p ] ¹ ] _ a š q b a f 6 Ò N Ñ. ; C U % > 6 " z Œ '! Œ # ) ' Ð 1 C 3 6 ' } = ;. 3 = ' ª ' ƒ C 3 > > 6 h j a ó b c p e Ç ^ a Å b _ ` o e t d ¹ ] o ` o f ó ` f o ] c e b ^ p f ] a Æ a _ b o a p Æ a f _ o f 6 U, = X C 4 C ; > E U % C 3 X. > ƒ 1 K ;. > S N 1 > C 6 ƒ. ; > 6 w y y ³ '! Œ z # ) ' Ð 1 C 3 ' ª ' ƒ C 3 > > 6 ü a ` æ j a p Ç ^ a Å b _ ` o ` a f ] ã ó b c p e h e q a 6 Ñ N 3 ;, % M. 4. M 6 C Ò C 3 > C U = < N 3 C S Ð 6 w y y '! Ž y # % ' ) ' = 0 N X 6 X C 3 M C N R > N ; 1 4. > N R 4 5 C 3 > 4 K N = 3 U X = ; 1 C < 3 N K ; C 0 R N 3 Z 1 = >. C = 3 < = 3 = K N ;. M C Z 1 = 4. N 0 >. 4 5 C 3 C, K N = 3 U 6 l b o t ` ± t [ a _ ` ^ a ` ^ t l a j t ¼ u " z { 6 N ' w 6 Ž y W Ž 1 > >. #! Ž " # ' ' = 6 ), C C 3 = 4 C Z 1 = >. C = M 5 ; C 4 < 3 N K ; C 0 R N 3, N 0 >. 4 5 C 3 C, K N = 3 U ' B 5 C M = > C N R > 1 3 R = M C,. > 4 3. K 1 4. N R è 3 > é 6 h c t Ç ^ f o t c ` _ t l b o t l a j t k È u " z z { 6 " y W " " Œ 1 > >. #! Ž w # } ' N K N } ' ª ' ƒ C 3 C 6 ) > U 0 < 4 N 4. M C 5 = X. N 3 N R ª ; = > 4. M N, U Ñ. 4 5 = % 1 3 R = M C S = % 0 = ; ; % 4 1 M E C. > 6 l b o j t l ] p a ` ^ æ [ a c ` b _ d ^ b _ t È È u " z { 6 N ' 6 Ž y z W Ž w '! Ž # } ' N K N } ' ª ' ƒ C 3 C 6 N = 3 U S N 0 N = 4. N R M C 3 4 C ; ;. < 4. M = ; < 3 N K ; C 0 > R N 3 M U 3. M = ; K N,. C > 6 ß t ] t l b o j t a c t þ k º º u " z z w { 6 N ' 6 z z W w Ž 6 " z z w '! Ž # } ' N K N 6 H ' ) ' H ; E 6 } ' ª ' ƒ C 3 C B ' ) ' % 5 = < N > E N X = 6 S N 0 N = 4. N R % N ; 1 4. > N R N = 3 U = ; 1 C ƒ 3 N K ; C 0 N 0 > 6 ƒ C 3 R N 3 = 4 C, ) ; } R N ;, > 6 d ^ ^ t t [ ] c t q a c t ` f b k µ _ t ` t k Ç a c t k È u " z z ³ { 6 N ' W 6 Ž " " Ž w z '! Ž Œ # ' ) ' } = 3 M 5 E N ª ' = ' Ð > ; N X 6 ß ] ^ p b c e b _ a c ] ¹ _ a f ` ^ ] b ` ^ f ` o j ` ^ a å c b ` ^ a p ß ] ^ p b c ` a f k = 1 E N X = 1 0 E = 6 Ð. C X 6 " z ³ 1 > >. #! Ž Ž # ' ) ' } = 3 M 5 E N ª ' = ' Ð > ; N X 6 ó ] ] æ a ^ ` ± a p l ] p a _ f ] ã l ` c ] ` ^ j ] ] æ a ^ a ] f l a p ` b 6 = 1 E N X = 1 0 E = 6 Ð. C X 6 w y y Œ ' u ;. > > ; = «ó ] ] æ a ^ ` ± b o ` ] ^ ] ã q b c ä o ` b _ p ` à a c a ^ o ` b _ a Å b o ` ] ^ f ' ƒ 3 N 3 C > } = 4 5 C 0 = 4. M = ; ƒ 5 U >. M > 6 Ž ' } ) «. 3 E 5 = 1 > C 3 6 N > 4 6 w y y Ž ' {! Ž ³ # ' ˆ ' } = ï U = 6 h j a t t ] ¹ ] _ a š q b a f 3 =, X ' ƒ 3 C > > 3 =, 6 " z Œ 1 > >. #! Ž # ' ˆ ' }. E 5 =. ; E N 6 N = 3 U X = ; 1 C < 3 N K ; C 0 >. 4 C 3 C, K N = 3 U R N 3 > C M N 3, C 3 C ; ;. < 4. M,. C 3 4. = ; N < C 3 = 4 N 3 > 6 h j a ] c t ^ t ^ t d ^ b _ t d q q _ t k u " z Ž { 6 N ' " z W " " y» u " z Ž z { 6 N ' w 6 ³ Œ W '! Ž z # ' ƒ ' }. E 5 =. ; N X 6 b c o ` b _ ` à a c a ^ o ` b _ Ä Å b o ` ] ^ f 6 = 1 E = 6 } N > M N 6 " z 1 > >. #! ³ y # ) ' Y ' = = 3 N X 6 B 5 C C, = ; M 5 = 3 = M 4 C 3 N R C Þ 4 3 C < 4 N R 4 5 C 9 3. C, 3. M 5 C Z 1 = ;. 4 > < 5 C 3. M = ; < ; C ; = U C 3 > 6 c ] ¹ _ t l b o t d ^ b _ t k È u w y y y { 6 " ³ " W " z y ' u ;. > > ; = «Î t l b o j t ` t k º È u w y y y { 6 N ' Œ 6 ³ W Ž ' {! ³ " # Ò ' C M = > 6 % 1 3 C 0 C 4 5 N, C < N C > N 1 3, C ; C > C Z 1 = 4. > = 1 Þ, C 3. X C C > < = 3 4. C ; ; C, 1 4 U < C C ; ;. < 4. Z 1 C 6 X N. C, C ; = X = 3. = 4. C ; ; C 6 d ^ ^ t ] _ b [ ] c t q t ` f b k º u " z Ž w { 6 y Œ W Ž w '! ³ w # ' H <. M ) ' Ð 1 C 3 6 ó b c p e ä h e q a Ç ^ a Å b _ ` o ` a f S = 3 ; N 6 " z z y '! ³ # H ' ) ' H ; E 6 ) ' % ' % 5 = 0 = C X ˆ ' ) ' N >. 6 = > U 0 < 4 N 4. M C Þ < >. > N R > N ; 1 4. > N R. 3. M 5 ; C 4 < 3 N K ; C 0 R N 3 C ; ;. < 4. M C Z 1 = 4. = < C 3 R N 3 = 4 C,, N 0 6 Ç ^ g [ ] ^ _ ` ^ b c b c o ` b _ ` à a a ^ o ` b _ Ä Å b o ` ] ^ f b ^ p o j a ` c d q q _ ` b o ` ] ^ f t N ; ; C C, C 9 3 M C % C = 3 6 Y Y Y 6 " z ³ '! ³ # H ' ) ' H ; E 6 ) ' % ' % 5 = 0 = C X ˆ ' ) ' N >. 6 l b o j a b o ` b _ q c ] ¹ _ a f ` ^ a _ b f o ` o e b ^ p j ] ] æ a ^ ` ± b o ` ] ^ k % 4 1,. C } = 4 5 C 0 = 4. M > 4 > ) < < ;. M = 4. > 6 w Ž ' N S N ; ; ƒ 1 K ;. > N ' 6 ) 0 > 4 C 3, = 0 6 " z z w '! ³ Œ # % ' H = = 6 ) < < 3 N Þ. 0 = 4. N R ˆ 3 C ê > 9 M = C. 4 5 } U H K > 4 = M ; C > 6 Ç ^ g å a ] a o c e b ^ p d ^ b _ e f ` f ] ã l b ^ ` ã ] _ p f k ß a c _ ` ^ k q c ` ^ æ a c a c _ b æ k ÿ k º ¼ ¼» u " z { 6 w " w W w w Œ '

31 Ø Ù Ú Û Ü Ý! ³ Ž # } ' 1 ' ƒ ; = C M X C 3 M C N R 4 5 C > N ; 1 4. > N R 1 ; = 3 ; U < C K C, K N = 3 U X = ; 1 C < 3 N K ; C 0 > R N C = < ; = M. 6 l b o t b a o ` k ¾ º u w y y w { 6 N ' Ž 6 Ž ³ W ³ ³ ' u ;. > > ; = «l b o j [ ] o a f k ¾ º u w y y w { 6 N ' Ž 6 ³ z W y ' {! ³ ³ # ª ' % M 5 C W ƒ = ; M. = 6 N = 3 U X = ; 1 C < 3 N K ; C 0 N 0 > M < C 3 R N 3 = 4 C, = ; ; > 6 [ ] ^ _ ` ^ a b c q b c o ` b _ p ` à a c a ^ o ` b _ a Å b o ` ] ^ f b ^ p o j a ` c b q q _ ` b o ` ] ^ f t µ ] t p a c b ^ a a ` ^ t k Æ a f t [ ] o a f l b o j t k ¾ 6 u " z w { 6 N ' 6 y z W w Œ '! ³ # Y ' ' % E 3 U E 6 l a o j ] p f ] ã b ^ b _ e f ` f ] ã ^ ] ^ _ ` ^ a b c a ` q o ` ¹ ] ^ p b c e ä š b _ a q c ] ¹ _ a f k 9. 0 = 4 ;. 4 ' 6 } N > M N 6 " z z y ' u ;. > > ; = 4. «º ¼» 6 ) 0 C 3. M } = 4 5 C 0 = 4. M = ; % N M. C 4 U 6 ƒ 3 N X., M C 6 Y 6 " z z ' {! ³ z # Y ' ' % E 3 U E 6 ) > U 0 < 4 N 4. M K C 5 = X. N 1 3 N R > N ; 1 4. > N C = 3 C ; ;. < 4. M < 3 N K ; C 0 < M C,, N 0 > 6 l b o t ¹ t k º u " z z { 6 N ' " y 6 Ž ³ W z y ' u ;. > > ; = «Æ f f ` b ^ d b p t ` t ¹ t l b o j t k u " z z { 6 N ' " 6 " z " W w y z ' {! y # Y ' ' % E 3 U E 6 C M 4. > R N C = X C 3 N C = M 5 ; C 4 < 3 N K ; C 0 < M C,, N 0 > 6 c t l b o t j t k u " z z Ž { 6 N ' Œ 6 Ž ³ Œ W Ž z ' u ;. > > ; = «c b ^ ` b ^ l b o j t Î t k u " z z Ž { 6 N ' Œ 6 ³ Œ W ³ ³ ' {! " # Y ' ' % E 3 U E 1 ' ' = 0 ; C C X = 6 X C 3 M C N R ª. X = ; 1 C > ª. R M 4. > N R C = M 5 ; C 4 ƒ 3 N K ; C 0 N 0 >. 4 5 C W ˆ 3 N = 3 U 6 c b ` ^ ` b ^ l b o j t Î t k u w y y { 6 N ' Ž 6 z z W " y " " 6 w y y '! w # % ' ' % N K N ; C X 6 ] a b q q _ ` b o ` ] ^ f ] ã ã ^ o ` ] ^ b _ b ^ b _ e f ` f o ] b o j a b o ` b _ q j e f ` f k B 3 > ; = 4. > N R } = 4 5 C 0 = 4. M = ; } N 3 = < 5 > ' ) 0 C 3. M } = 4 5 C 0 = 4. M = ; % N M. C 4 U 6 ƒ 3 N X., M C 6 Y 6 z y 6 " z z " '! # % ' ' % N K N ; C X 6 a _ a o a p q c ] ¹ _ a f ` ^ o j a o j a ] c e ] ã ã ^ o ` ] ^ f q b a f b ^ p æ a ^ a c b _ ` ± a p ã ^ ä o ` ] ^ f k = 1 E = 6 } N > M N 6 " z z 1 > >. #! # ' B C 0 = C B 5 C N 3 U 1 0 C 3. M = ; = ; U >. > N R 4 5 C = X. C 3 % 4 N E C > ª Z 1 = 4. > 6 µ ] a æ a b c k l g ^ ` š a c f ` o e ] ã l b c e _ b ^ p k " z ³ '! Œ # Ò ' } ' B 5 N 0 = > 6 % 1 3 ; ê = ; U > 1 0 C 3. Z 1 C, C > 0 C 4 5 N, C >, ê C ; C 0 4 > 5 U K 4., C > C 4 0. Þ 4 C > 6 j t t ` f f a c o b o ` ] ^ k ^ ` š a c f ` o a ` a c c a a o l b c ` a µ c ` a ë b c ` f ì k " z ³ ³ '! Ž # } ' N 5 3 = ;. E C. > M 3 C 4 C ƒ M = 3 C W 9 3. C, 3. M 5 > C Z 1 = ;. 4. C > R N 3 M R N 3 ) < < 3 N Þ. 0 = 4. > N R 4 5 C % N K N ; C X % < = M C H 1, [ a c t ^ o t d ^ b _ t Ô q o ` ` ± b o ` ] ^ k È u w y y Œ { 6 N ' ³ 6 z w Œ W z Œ w '! ³ # ) ' C K N 6 S = 3, U C Z 1 = ;. 4. C > Y 0 N 0 > ˆ C 3 = C N 0 > 6 c ] t d a c t l b o j t ] t k º È È u " z z { 6 " " " " " z y ' 0,λ! # Ð ' N >., = 6 9 M 4. = ; = ; U >. > ' Æ a q c ` ^ o ] ã o j a f ` o j ë ÿ ì a p ` o ` ] ^ t µ _ b f f ` f ` ^ l b o j ä a b o ` f t % < C 3 C 3 ; = 6 C 3 6 " z z Œ '! z # Ñ ' Ó 5 S ' õ C, 3. M 5 > ƒ M = 3 C 4 U < C C Z 1 = ;. 4. C > 6 Î t l b o j t d ^ b _ t d q q _ t k ¼ u w y y Œ { 6 Œ w W Œ Œ " '

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35 Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2007, Article ID 34138, 13 pages doi: /2007/34138 Research Article On the Precise Asymptotics of the Constant in Friedrich s Inequality for Functions Vanishing on the Part of the Boundary with Microinhomogeneous Structure G. A. Chechkin, Yu. O. Koroleva, and L.-E. Persson Received 3 April 2007; Revised 28 June 2007; Accepted 23 October 2007 Recommended by Michel Chipot We construct the asymptotics of the sharp constant in the Friedrich-type inequality for functions, which vanish on the small part of the boundary Γ ε 1. It is assumed that Γ ε 1 consists of (1/δ) n 1 pieces with diameter of order O(εδ). In addition, δ = δ(ε) andδ 0as ε 0. Copyright 2007 G. A. Chechkin et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction The domain is an open bounded set from the space R n. The Sobolev space H 1 () is defined as the completion of the set of functions from the space C () bythenorm (u 2 + u 2 )dx. The space H 1 () is the set of functions from the space H 1 (), with zero trace on. Let ε = 1/N, N N, be a small positive parameter. Consider the set Γ ε which depends on the parameter ε. The space H 1 (,Γ ε ) is the set of functions from H 1 (), vanishing on Γ ε. The following estimate is known as Friedrich s inequality for functions u H 1 (): u 2 dx K 0 u 2 dx, (1.1) where the constant K 0 depends on the domain only and does not depend on the function u. Inequality (1.1) is very important for several applications and it may be regarded as a special case of multidimensional Hardy-type inequalities. Such inequalities has attracted a lot of interest in particular during the last years; see, for example, the books [1 3] and

36 2 Journal of Inequalities and Applications the references given therein. We pronounce that not so much is known concerning the best constants in multidimensional Hardy-type inequalities and the aim of this paper is to study the asymptotic behavior of the constant in [4] for functions vanishing on a part of the boundary with microinhomogeneous structure. In particular, such result are useful in homogenization theory and in fact this was our original interest in the subject. The paper is organized as follows. In Section 2, we present and discuss our main results. In Section 3, these results are proved via some auxiliary results, which are of independent interest. In Section 4, we consider partial cases, where it is possible to give the asymptotic expansion for the constant with respect to ε. 2. The main results It is well known (see, e.g., [5]) that the Friedrich s inequality (1.1) is valid for functions u H 1 (,Γ ε )andk 0 = O(1/capΓ ε ), where we denote by cap F the capacity of F R n in R n : { cap F = inf ϕ 2 ( dx : ϕ C 0 R n ) }, ϕ 1onF. (2.1) R n Remark 2.1. Friedrich s inequality, when the functions vanishes on a part of the boundary is sometimes called Poincaré s inequality, but we prefer to say Friedrich s or Friedrich s type inequality keeping the name Poincaré s inequality for the following (see, e.g., [6]): ( 2 u 2 dx udx) + u 2 dx, u H 1 (). (2.2) Further, it will be shown later on that K 0 is uniformly bounded under special assumptions on Γ ε inthecasewhenmesγ ε 0asε 0. Consider now the domain R 2 with smooth boundary of the length 1 such that = Γ ε 1 Γ ε 2, mesγ ε 1i = εδ(ε), Γ ε 1 = ( Γ ε ) 1i, Γ ε 2 = ( Γ ε ) 2i, Γ ε 1i Γ ε 2j =, i i mes ( ( ) Γ ε 1i Γ ε ) 1 2i = δ(ε), δ(ε) = o as ε 0, lnε (2.3) where Γ ε 1i and Γ ε 2i are alternating (see Figure 2.1). Here Γ ε = Γ ε 1. Our first main result reads as follows. Theorem 2.2. Suppose n = 2.Foru H 1 (,Γ ε 1), the following Friedrich s inequality holds true: u 2 dx K ε u 2 dx, Kε = K 0 + ϕ(ε), (2.4) where K 0 is a constant in Friedrich s inequality (1.1) for functions u H 1 (), and ϕ(ε) ( lnε ) 1/2 +(δ(ε) lnε ) 1/2 as ε 0.

37 G. A. Chechkin et al. 3 δ εδ Figure 2.1. Plane domain. δ εδ Figure 2.2. Spatial domain. For the case n 3, the geometrical constructions are similar. We assume that = S Γ, S Γ =, Γ belongs to the hyperplane x n = 0, and Γ = Γ ε 1 Γ ε 2, Γ ε 1 Γ ε 2 =. Denote by ω a bounded domain in the hyperplane x n = 0, which contains the origin. Without loss of generality ω, where ={ x : 1/2 <x i < 1/2, i = 1,...,n 1}, x = ( x,x n ). Let ω ε be the domain { x : x/ε ω}. Denote by Γ the integer translations of ω ε onthehyperplaneinx i direction, i = 1,...,n 1. Finally, Γ ε 1 ={x : x/δ Γ} Γ, Γ ε 2 = Γ \ Γ ε 1 (see Figure 2.2). In other words, Γ ε 1 is a translations of vectors mδ(ε)e i (m Z, i = 1,...,n 1) of a set diameter εδ(ε) contained in a ball of radius δ(ε). Here we assume that δ(ε) = o(ε n 2 )asε 0. Also we suppose that Γ ε = Γ ε 1 S. In this case our main result reads as follows. Theorem 2.3. Suppose n 3.Foru H 1 (,Γ ε 1 S), the following Friedrich s inequality is valid: u 2 dx K ε u 2 dx, K ε = K 0 + ϕ(ε), (2.5) where K 0 is a constant in Friedrich s inequality (1.1) for functions u H 1 (), and ϕ(ε) ε n/2 1 +(δ(ε)ε 2 n ) 1/2 as ε 0.

38 4 Journal of Inequalities and Applications Thus the precise dependence of the constant in Friedrich s inequality of the small parameter ε will be established. Hence, it is possible to construct the lower and the upper bounds for K ε. 3.Proofsofthemainresultsandsomeauxiliaryresults In Sections 3.1 and 3.2 we discuss, present, and prove some auxiliary results, which are of independent interest but also crucial for the proof of the main results in Section The relation between the constant in Friedrich s inequality and the first eigenvalue of a boundary value problem. Let be some bounded domain with smooth boundary, Γ ε. Suppose that the function u belongs to H 1 (). Consider the following problem: Δu = λ ε u in, u = 0 onγ ε, u ν = 0 on \ Γ ε. (3.1) Definition 3.1. The function u H 1 (,Γ ε ) is a solution of problem (3.1), if the following integral identity is valid: u vdx= λ ε uv dx (3.2) for all functions v H 1 (,Γ ε ). The operator of problem (3.1) is positive and selfadjoint (it follows directly from the integral identity). According to the general theory (see, e.g., [7]), all eigenvalues of the problem are real, positive, and satisfy 0 λ 1 ε λ2 ε, λk ε as k. (3.3) Here we assume that the eigenvalues λ k ε are repeated according to their multiplicities. Denote by μ ε the following value: μ ε = inf v 2 dx v 2 dx. (3.4) v H 1 (,Γ ε)\{0} We need the following lemma (see the analogous lemma in [4]). Lemma 3.2. The number μ ε is the first eigenvalue λ 1 ε of the problem (3.1). For the convenience of the reader we present the details of the proof. Proof. It is sufficient to show that there exists such eigenfunction u 1 of problem (3.1), corresponding to the first eigenvalue λ 1 ε, that it satisfies μ ε = u 1 2 dx ( u 1 ) 2. (3.5) dx

39 G. A. Chechkin et al. 5 Let {v (k) } be a minimization sequence for (3.4), that is, v (k) H 1( ),Γ ε, v (k) 2 L = 1, 2() v (k) 2 dx μ ε, ask. (3.6) It is obvious that the sequence {v (k) } is bounded in H 1 (,Γ ε ). Hence, according to the Rellich theorem, there exists a subsequence of {v (k) }, converging weekly in H 1 (,Γ ε )and strongly in L 2 (). For this subsequence, we keep the same notation {v (k) }.Wehavethat Using the following formula: v(k) + v (l) 2 2 v (k) v (l) 2 L 2() <η as k,l>k 0(η). (3.7) = 1 L 2() 2 v (k) 2 L + 1 2() 2 v (l) 2 L v (k) v (l) 2() 2 2, (3.8) L 2() we obtain that v(k) + v (l) 2 2 L 2() > 1 η 4. (3.9) From the definition of μ ε we conclude that v 2 dx μ ε v 2 L 2() (3.10) for all function v H 1 (,Γ ε ). Inequalities (3.9)and(3.10) give the following estimate: ( v (k) + v (l) ) 2 ( dx > μ ε 1 η ). (3.11) 2 4 If k,l>k 0 (η), it follows that v (k) 2 dx < μ ε + η, v (l) 2 dx < μ ε + η. (3.12) Hence, ( v (k) v (l) ) 2 dx 2 = 1 2 μ ε + η 2 v (k) 2 dx μ ε + η 2 v (l) 2 dx ( μ ε 1 η ) ( = η 4 1+ μ ε 4 ( v (k) + v (l) ) 2 dx 2 ) 0, η 0. (3.13)

40 6 Journal of Inequalities and Applications Finally, according to the Cauchy condition the sequence {v (k) } convergestosomefunction v H 1 (,Γ ε ) in the space H 1 (,Γ ε ), and v 2 dx = μ ε, v 2 L 2() = 1. (3.14) Assume that v H 1 (,Γ ε ) is an arbitrary function. Denote ( v + tv ) 2 dx g(t) = v + tv 2. (3.15) L 2() The function g(t)iscontinuouslydifferentiable in some neighborhood of t = 0. This ratio has the minimum, which is equal to μ ε. Using the Fermat theorem, we obtain that 0 = g t=0 = 2 v 2 ( L 2() v, v ) dx 2 v vdx v 2 dx v 4 L 2() ( = 2 v, v ) (3.16) dx 2μ ε v vdx. Thus, we have proved that ( v, v ) dx = μ ε v vdx (3.17) for v H 1 (,Γ ε ), that is, v satisfies the integral identity (3.1). This means that we have found a function v such that v 2 dx ( ) v 2 dx v 2 = inf dx v H 1 (,Γ ε)\{0} v 2 = μ ε. (3.18) dx Keeping in mind that u 1 = v we conclude that μ ε = λ 1 ε. The proof is complete. Lemma 3.3. The following Friedrich inequality holds true: U 2 dx K ε U 2 dx, U H 1( ),Γ ε, (3.19) where K ε = 1/λ 1 ε. Proof. From Lemma 3.2 we get that λ 1 U 2 dx ε U 2 dx for any U H 1( ),Γ ε. (3.20) Using this estimate, we deduce that U 2 dx 1 λ 1 U 2 dx. ε (3.21) Denoting by K ε the value 1/λ 1 ε, we conclude the statement of our lemma. In the following section we will estimate λ 1 ε.

41 G. A. Chechkin et al Auxiliary boundary value problems. Assume that f L 2 () and consider the following boundary value problems: Δu ε = f in, u ε = 0 onγ ε, u ε ν = 0 onγε 2, Δu 0 = f in, u 0 = 0 on. (3.22) (3.23) Note that for n = 2 we assume that Γ ε = Γ ε 1 and for n 3 we assume that Γ ε = Γ ε 1 S. Problem (3.23) is the homogenized (limit as ε 0) problem for problem (3.22) (a proof of this fact can be found in [4, 8], see also [6]). Consider now the respective spectral problems: Δu k ε = λk ε u k ε in, u k ε = 0 onγ ε, u k ε ν = 0 onγε 2, Δu k 0 = λk 0u k 0 in, (3.24) u k 0 = 0 on. Next, let us estimate the difference 1/λ k ε 1/λk 0. We will use the method introduced by Oleĭnik et al. (see [9, 10]). Let H ε, H 0 be separable Hilbert spaces with the inner products (u ε,v ε ) Hε and (u,v) H0, and the norms u ε Hε and u H0, respectively; assume that ε is a small parameter, A ε (H ε ), A 0 (H 0 ) are linear continuous operators and ImA 0 V H 0,whereV is a linear subspace of H 0. (C1) There exist linear continuous operators R ε : H ε H 0 such that for all f V we have (R ε f,r ε f ) Hε c( f, f ) H0 as ε 0, where c = const > 0 does not depend on f. (C2) The operators A ε, A 0 are positive, compact and selfadjoint in H ε, H 0, respectively, and sup ε A ε (Hε) < +. (C3) For all f V we have A ε R ε f R ε A 0 f Hε 0asε 0. (C4) The sequence of operators A ε is uniformly compact in the following sense: if a sequence f ε H ε is such that sup ε f ε (Hε) < +, then there exist a subsequence f ε and a vector w 0 V such that A ε f ε R ε w 0 Hε 0asε 0. Assume that the spectral problems for the operators A ε, A 0 are A ε u k ε = μk ε u k ε, k = 1,2,..., μ 1 ε μ2 ε > 0, ( u l ε,u m ) ε = δlm, A 0 u k 0 = μ k 0u k 0, k = 1,2,..., μ 1 0 μ 2 0 > 0, ( u l 0,u m ) 0 = δlm, (3.25)

42 8 Journal of Inequalities and Applications where δ lm is the Kronecker symbol and the eigenvalues μ k ε, μ k 0 are repeated according to their multiplicities. The following theorem holds true (see [9]). Theorem 3.4 (Oleĭnik et al. [9, 10]). Suppose that the conditions (C1) (C4) arevalid. Then μ k ε converges to μ k 0 as ε 0, and the following estimate takes place: μ k ε μk 0 c 1/2 sup f N(μ k 0,A 0), f H0 =1 A ε R ε f R ε A 0 f Hε, (3.26) where N(μ k 0, A 0 ) ={u H 0, A 0 u = μ k 0u}. Assume also that k 1, s 1, are the integer numbers, μ k 0 = = μ0 k+s 1 and the multiplicity of μ k 0 is equal to s. Then there exist linear combinations U ε of the eigenfunctions u k ε,...,uε k+s 1 to problem (3.22) such that for all w N(μ k 0, A 0 ) we get U ε R ε w Hε 0 as ε 0. To use the method of Oleĭnik et al. [9, 10], we define the spaces H ε and H 0 and the operators A ε, A 0,andR ε in an appropriate way. Assume that H ε = H 0 = V = L 2 (), and R ε is the identity operator. The operators A ε, A 0 aredefinedinthefollowingway:a ε f = u ε, A 0 f = u 0,whereu ε, u 0 are the solutions to problems (3.22)and(3.23), respectively. Let us verifythe conditions (C1) (C4). The condition (C1) is fulfilled automatically because R ε is the identity operator, c = 1. Let us verify the selfadjointness of the operator A ε. Define A ε f = u ε, A ε g = v ε, f,g L 2 (). Because of the integral identity of problem (3.22) the following identities are valid: Hence, fv ε dx = v ε u ε dx = gu ε dx. (3.27) ( Aε f,g ) L = ( 2() u ε,g ) L = 2() u ε gdx= v ε u ε dx = fv ε dx = ( ) f,v ε L = ( 2() f,a ε g ) (3.28) L. 2() The selfadjointness of the operator A 0 canbeprovedinananalogousway. It is easy to prove the positiveness of the operator A ε : ( Aε f, f ) L 2() = ( u ε, f ) L 2() = u ε fdx= uε 2 dx 0, (3.29) and u ε 2 dx > 0if f 0. The positiveness of A 0 may be proved in the same way. Next, we prove that A ε, A 0 are compact operators: let the sequence { f θ } be bounded in L 2 (). It is evident that the sequence {A ε f θ }={u ε,θ } is bounded in H 1 (,Γ ε ) and the sequence {A 0 f θ }={u 0,θ } is bounded in H 1 (). Note that {u ε,θ } is bounded uniformly on ε (for a proof see [4]). Because of compact embedding of the space H 1 () tol 2 (), we conclude that A ε and A 0 are compact operators. Moreover, A (Hε) ε u H1 ε (,Γ ε) < < + (3.30)

43 G. A. Chechkin et al. 9 and, consequently, sup A (Hε) ε < < +. (3.31) ε Let us verify the condition (C3). The operator R ε is the identity operator and, thus, it is sufficient to prove that for all f L 2 ()wehavethat A ε f A 0 f L2() 0asε 0, that is, u ε u 0 L2() 0asε 0. It is enough to prove that u ε u 0 in H 1 (). (The week convergence in H 1 () gives the strong convergence in L 2 ().) The sequence u ε is uniformly bounded in H 1 (). Consequently, there exists a subsequence u ε,suchthatu ε u in H 1 (). Further we will set that u ε is the same subsequence. Let us show that u u 0, that is, for all v H 1 (), fvdx= u vdx. (3.32) The integral identity for problem (3.22) gives that fvdx= u ε vdx. (3.33) Because u is a week limit of u ε in H 1 (), the following is valid: u ε vdx u vdx when ε 0 v H 1 (), (3.34) and this gives us the desired result, because the integrals u vdx and fvdx do not depend on ε. Let us verify the condition (C4). Consider the sequence { f ε }, which is bounded in L 2 (). Then A ε f ε H1 (,Γ ε) = u ε H1 (,Γ ε) const, that is, the sequence {A ε f ε } is compact in L 2 () and, consequently, there exists a subsequence ε such that A ε f ε w 0 as ε 0, where w 0 L 2 (). (3.35) Hence, we have that A ε f ε w 0 L2() 0asε 0. Thus, the conditions (C1) (C4) are valid. It is evident that λ k ε = 1/μ k ε, λ k 0 = 1/μ k 0. Using the estimate (3.26)wehave 1 λ k 1 ε λ k 0 sup f N(λ k 0,A 0), f L2 ()=1 A ε f A 0 f H1 (,Γ ε) = sup u ε u H1 0 (,Γ, (3.36) ε) f N(λ k 0,A 0), f L2 ()=1 where u ε, u 0 are the solutions of problems (3.22)and(3.23), respectively. The following inequality was established in [4]: ( ( ) u ε u H1 0 (,Γ K f (με ) 1/2 δ 1/2 ) ε) L 2() +, (3.37) με

44 10 Journal of Inequalities and Applications where μ ε = inf u H1 (,Γ ε 1)\{0}( u 2 dx/ u 2 dx) and the constant K depends only on the domain. Moreover, the following asymptotics was proved in [4] (thecasen = 2) and in [11](thecasen 3) (see also [8, 12]): ( ) π 1 lnε + O lnε μ ε = 2, ifn = 2, ε n 2 σ n 2 c ω + O ( (3.38) ε n 1), ifn>2, as ε 0. Here σ n is the area of the unit sphere in R n,andc ω > 0 is the capacity of the (n 1)- dimensional disk ω (see [13, 14]). Define the following value: ( ( ) (με ) 1/2 δ 1/2 ) ϕ(ε) K +. (3.39) με Note that if n = 2, it implies that μ ε 1/ lnε and δ = O(1/ lnε ). Consequently, we have δ = o ( ) 1/ lnε lnε = o(1) μ ε π (3.40) as ε 0. Inthesameway,ifn>2 it yields that δ o(1) μ ε (3.41) as ε 0. Using this asymptotics, we deduce (lnε) 1/2 + ( δ(ε) lnε ) 1/2 ( + o (lnε) 1/2 + ( δ(ε) lnε ) 1/2), ifn = 2, ϕ(ε) = K ε n/2 1 + ( δ(ε)ε n 2) 1/2 ( + o ε n/2 1 + ( δ(ε)ε n 2) 1/2), ifn>2, (3.42) as ε 0. Finally, due to (3.36), (3.37), and (3.38)wegetthat 1 λ 1 1 ε λ 1 ϕ(ε), (3.43) 0 where ϕ(ε)hastheasymptotics(3.42) Proofs of the main results. Proof of Theorem 2.2. Actually, because of estimate (3.19), Friedrich s inequality u 2 dx 1 λ 1 u 2 dx, u H 1( ),Γ ε ε (3.44)

45 G. A. Chechkin et al. 11 is valid. The estimate (3.43) implies that 1 λ 1 1 ε λ 1 + ϕ(ε). (3.45) 0 By rewriting inequality (3.44), using the established relations between λ 1 ε and λ 1 0 we find that u 2 dx 1 ( ) 1 λ 1 u 2 dx ε λ 1 + ϕ(ε) u 2 dx. (3.46) 0 According to our notations, 1/λ 1 0 = K 0. Thus, for u H 1 (,Γ ε ), the following Friedrich inequality holds true: u 2 dx K ε u 2 dx, K ε = K 0 + ϕ(ε), (3.47) where ϕ(ε) ( lnε ) 1/2 +(δ(ε) lnε ) 1/2 as ε 0, if n = 2. Hence, the proof is complete. Proof of Theorem 2.3. The proof is completely analogous to that of Theorem 2.2: using inequalities (3.19), (3.42), and (3.43), we obtain the asymptotics of the constant K ε,hence weleaveoutthedetails.noteonlythatinthiscaseϕ(ε) ε n/2 1 +(δ(ε)ε n 2 ) 1/2 as ε Special cases In this section, we consider domains with special geometry. Let G be the boundary of the unit disk G centered at the origin. Assume that ω ε = {(r,θ): r = 1, δ(ε)(π/2) < θ < δ(ε)(π/2)} is the arc, where (r,θ) are the polar coordinates. Suppose also that η ε = ( εδ,εδ). Denote γ ε = G \ Γ ε,whereγ ε is the union of the sets obtained from η ε by rotation about the origin through the angle επ and its multiples. For simplicity we assume here that ε = 2/N, N N.Let be an arbitrary conformal mapping of a disk with radius exceeding 1, let be the image of the unit disk G and let Γ ε 1 = (Γ ε ), Γ ε 2 = (γ ε ). For this domain we have the following theorem (see [15 17]). Theorem 4.1. Suppose that n = 2, the domain istheimageoftheunitdiskgas defined at the beginning of the section and δ lnε 0 as ε 0.Foru H 1 (,Γ ε 1), the following Friedrich inequality holds true: u 2 dx K ε u 2 dx, K ε = K 0 + ϕ(ε), (4.1) where K 0 is a constant in Friedrich s inequality (1.1) for functions u H 1 (), andϕ(ε) = δ lnsinε ( ψ 0 / ν) 2 ds+ o(δ) as ε 0, where ψ 0 is the first normalized in L 2 () eigenfunction of the problem and = 1. Δψ 0 = K 0 ψ 0, x ; ψ 0 = 0, x, (4.2)

46 12 Journal of Inequalities and Applications If the mapping is identical, then the following statement takes place (see [15 18]). Theorem 4.2. Suppose that n = 2, the domain is the unit disk and δ lnε 0 as ε 0.For u H 1 (,Γ ε 1), the following Friedrich inequality holds true: u 2 dx K ε u 2 dx, ( K ε = K 0 1+2δlnsinε +2δ 2 (lnsinε) 2 + o ( (4.3) δ 2)) as ε 0,whereK 0 is a constant in Friedrich s inequality (1.1) for functions u H 1 (). Note that nonperiodic geometry are considered in [19]. In the paper the author constructed and verified the asymptotic expansions of eigenvalues. Keeping in mind these results it is possible to obtain sharp bounds for the constant in Friedrich s inequality. Acknowledgments The research presented in this paper was initiated at a research stay of G. A. Chechkin at Luleå University of Technology, Luleå, Sweden, in June The final version was completed when G. A. Chechkin was visiting Laboratoire J.-L. Lions de l Université Pierreet Marie Curie, Paris, France, in March-April The authors thank the referees for several valuable comments and suggestions, which have improved the final version of the paper. The work of the first author was partially supported by RFBR ( ). The work of the first and the second authors was partially supported by the program Leading Scientific Schools (HIII ). The work of the second author was partially supported by RFBR. References [1] A. Kufner and L.-E. Persson, Weighted Inequalities of Hardy Type, World Scientific, River Edge, NJ, USA, [2] A. Kufner, L. Maligranda, and L.-E. Persson, The Hardy Inequality. About Its History and Related Results, Vydavetelsky Servis, Pilsen, Germany, [3] B. Opic and A. Kufner, Hardy-Type Inequalities, vol. 219 of Pitman Research Notes in Mathematics Series, Longman Scientific & Technical, Harlow, UK, [4] G. A. Chechkin, Averaging of boundary value problems with singular perturbation of the boundary conditions, Russian Academy of Sciences. Sbornik. Mathematics, vol. 79, no. 1, pp , 1994, translation in Matematicheskiĭ Sbornik, vol. 184, no. 6, pp , [5] V. G. Maz ya, Prostranstva S. L. Soboleva, Leningrad University, Leningrad, Russia, [6] G. A. Chechkin, A. L. Piatnitski, and A. S. Shamaev, Homogenization: Methods and Applications, vol. 234 of Translations of Mathematical Monographs, American Mathematical Society, Providence, RI, USA, [7] V. S. Vladimirov, The Equations of Mathematical Physics, Nauka, Moscow, Russia, 3rd edition, [8] G. A. Chechkin, On the estimation of solutions of boundary-value problems in domains with concentrated masses periodically distributed along the boundary. The case of light masses, Mathematical Notes, vol. 76, no. 6, pp , 2004, translation in Matematicheskie Zametki, vol. 76, no. 6, pp , 2004.

47 G. A. Chechkin et al. 13 [9] O. A. Oleĭnik, A. S. Shamaev, and G. A. Yosifian, Mathematical Problems in Elasticity and Homogenization, vol. 26 of Studies in Mathematics and Its Applications, North-Holland, Amsterdam, The Netherlands, [10] G. A. Iosif yan, O. A. Oleĭnik, and A. S. Shamaev, On the limit behavior of the spectrum of a sequence of operators defined in different Hilbert spaces, Russian Mathematical Surveys, vol. 44, no. 3, pp , 1989, translation in Uspekhi Matematicheskikh Nauk, vol. 44, no. 3(267), pp , [11] R. R. Gadyl shin and G. A. Chechkin, A boundary value problem for the Laplacian with rapidly changing type of boundary conditions in a multi-dimensional domain, Siberian Mathematical Journal, vol. 40, no. 2, pp , 1999, translation in Sibirskiĭ Matematicheskiĭ Zhurnal, vol. 40, no. 2, pp , [12] A. G. Belyaev, G. A. Chechkin, and R. R. Gadyl shin, Effective membrane permeability: estimates and low concentration asymptotics, SIAM Journal on Applied Mathematics,vol.60,no.1, pp , [13] N. S. Landkof, Foundations of Modern Potential Theory, Springer, New York, NY, USA, [14] G. Pólya and G. Szegö, Isoperimetric Inequalities in Mathematical Physics, Annals of Mathematics Studies, no. 27, Princeton University Press, Princeton, NJ, USA, [15] R. R. Gadyl shin, On the asymptotics of eigenvalues for a periodically fixed membrane, St. Petersburg Mathematical Journal, vol. 10, no. 1, pp. 1 14, 1999, translation in Algebra I Analiz, vol. 10, no. 1, pp. 3 19, [16] R. R. Gadyl shin, A boundary value problem for the Laplacian with rapidly oscillating boundary conditions, Doklady Mathematics, vol. 58, no. 2, pp , 1998, translation in Doklady Akademii Nauk, vol. 362, no. 4, pp , [17] D. I. Borisov, Two-parametrical asymptotics for the eigenevalues of the laplacian with frequent alternation of boundary conditions, Journal of Young Scientists. Applied Mathematics and Mechanics, vol. 1, pp , 2002 (Russian). [18] D. I. Borisov, Two-parameter asymptotics in a boundary value problem for the Laplacian, Mathematical Notes, vol. 70, no. 3-4, pp , 2001, translation in Matematicheskie Zametki, vol. 70, no. 4, pp , [19] D. I. Borisov, Asymptotics and estimates for the eigenelements of the Laplacian with frequent nonperiod change of boundary conditions, Izvestiya. Seriya Matematicheskaya, vol. 67, no. 6, pp , 2003, translation in Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, vol. 67, no. 6, pp , G. A. Chechkin: Department of Differential Equations, Faculty of Mechanics and Mathematics, Moscow Lomonosov State University, Moscow , Russia address: chechkin@mech.math.msu.su Yu. O. Koroleva: Department of Differential Equations, Faculty of Mechanics and Mathematics, Moscow Lomonosov State University, Moscow , Russia address: korolevajula@mail.ru L.-E. Persson: Department of Mathematics, Luleå Universityof Technology, Luleå, Sweden address: larserik@sm.luth.se

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51 ISSN , Russian Journal of Mathematical Physics, Vol. 16, No. 1, 2009, pp c Pleiades Publishing, Ltd., On the Friedrichs Inequality in a Domain Perforated Aperiodically along the Boundary. Homogenization Procedure. Asymptotics for Parabolic Problems G. A. Chechkin*, **, Yu. O. Koroleva*, ***, A. Meidell**, and L.-E. Persson***, ** *Department of Mechanics and Mathematics, Moscow State University, Vorob evy gory, Moscow, Russia **Narvik University College, Postboks 385, 8505 Narvik, Norway ***Department of Mathematics, Luleå University of Technology, SE Luleå, Sweden chechkin@mech.math.msu.su, korolevajula@mail.ru, Annette.Meidell@hin.no, larserik@sm.luth.se Received December 11, 2008 Abstract. This paper is devoted to the asymptotic analysis of functions depending on a small parameter characterizing the microinhomogeneous structure of the domain on which the functions are defined. We derive the Friedrichs inequality for these functions and prove the convergence of solutions to corresponding problems posed in a domain perforated aperiodically along the boundary. Moreover, we use numerical simulation to illustrate the results. DOI: /S INTRODUCTION The boundary value problems in a domain perforated along the boundary were studied by several authors (see, for instance, [1, 5, 7, 16, 17]). The authors considered periodic perforation and random perforation along the boundary. In these papers and monographs, one can find proofs of homogenization theorems, convergence results, and estimates for the rate of convergence for boundary value problems and for the related spectral problems. In [16, Chap. I, Sec. 3], the method of potentials was used to prove homogenization theorems for the problem posed in a domain with perforation along a closed curve. The uniform convergence in absolute value of the solution of the original problem to asolution of the limit problem incompact subdomains disjoint from the chosen curve was proved there. The papers [15, 22] are also ofinterest. Here the problem inadomain divided by aperiodically perforated interior wall is investigated and the weak convergence of the solution of the original problem to asolution of the homogenized problem isproved and, in some cases, estimates for the rate of convergence are obtained. In [5, 7, 19], problems in domains with random structure of perforation were treated (for details concerning the random structure, see [14, 8]). In [5, 7], the perforation almost surely has the diameter equal to the distance between holes, whereas, in [19], the perforation disappears inthe limit and gives nocontribution to the homogenized problem. In the present paper, we consider a boundary value problem inadomain perforated aperiodically along the boundary for the case in which the diameter of circles and the distance between them are of the same order. Moreover, we derive the Friedrichs inequality for the functions vanishing on the boundary of the cavitiesinthe aperiodic case (see the similar paper [6], in which the function vanishes on small alternating parts of afixed boundary). Itshould benoted that weuse a general scheme ofproving the convergence, which was developed in [2, 4, 9, 20]. The paper is organized as follows. In Section 2,wegive geometrical settings anddiscuss the main results. In Section 3, weprove the Friedrichs inequality. Section 4isdevoted to investigations The work of the first author was partially supported by RFBR. The work of the first and the second authors was partially supported by the program Leading Scientific Schools (NSh ). 1

52 2 G. A. CHECHKIN et al. Fig. 1. Domain perforated along the boundary. Fig. 2. The division of the domain ε. of asymptotics of solutions to boundary value problems in a domain of the above type, and we also prove the convergence of solutions, as well as the convergence of corresponding eigenelements, to those for the problem in a domain perforated aperiodically along the boundary. In Section 5, we apply numerical simulations for the parabolic problem in a domain of the above type with inhomogeneous structure along the boundary. In the appendix, we prove an auxiliary result that we need concerning properties of solutions to boundary value problems in a domain perforated aperiodically along the boundary. 2. GEOMETRIC SETTINGS Denote by the rectangle [0,a] [0,b]. Assume that ε 1 is a small parameter, and introduce another small parameter by the rule ε i 1 = α i ε, α i = const, 0 <α i < 1. Let Bε i be asequence of disjoint open circles with radii ε i 1. The centers of the circles belong to the segment 0 x a, y = ε, and the distance between the centers of two neighboring circles Bε i and Bε i+1 is equal to ε i 2 = β i ε, β i = const, α 1 β 0 < B, α i + α i+1 β i < B, where B = const, i N. Denote by B ε the union i Bε. i Define the domain ε =\ B ε (see Fig. 1). We use the following notation: Γ 1 = {(x, y) :y = b, 0 x a}, Γ 3 = {(x, y) :x = a, 0 y b}, Γ 2 = {(x, y) :x =0, 0 y b}, Γ 4 = {(x, y) :y =0, 0 x a}. Thus, =Γ 1 Γ 2 Γ 3 Γ 4 and Γ ε = B ε. Denote by H 1 ( ε, Γ ε ) the set of functions in H 1 ( ε ) with zero trace onγ ε. Similarly, by H 1 (, Γ 4 ) we denote the set of functions in H 1 () with zero trace on Γ 4. We also use the space C0 (, Γ 4 ), which consists of the functions in C () vanishing onaneighborhood of Γ 4.Inthe same way, the symbol C0 ( ε, Γ ε )stands for the set of functions in C ( ε ) vanishing in a neighborhood of Γ ε. Remark 2.1. Further, we identify functions belonging toh 1 ( ε, Γ ε )and functions in H 1 () vanishing on B ε. Similarly, we identify functions belonging tol 2 ( ε )and functions in L 2 () vanishing on B ε. We use the same notation for functions in L 2 () restricted to ε. Our aim is to derive the Friedrichs inequality for functions u ε H 1 ( ε, Γ ε )and use it tostudy the behavior of solutions to an elliptic problem andtoarelated parabolic problem. 3. FRIEDRICHS INEQUALITY In this section, we derive the Friedrichs inequality for functions belonging toh 1 ( ε, Γ ε ). Theorem 3.1. The inequality u 2 ε(x, y) dx dy M ε u ε (x, y) 2 dx dy ε (3.1) holds for any functions u ε H 1 ( ε, Γ ε ), where the constant M does not depend on ε. RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

53 ON THE FRIEDRICHS INEQUALITY 3 Fig. 3. The division of the layer. Fig. 4. Tangent lines. Proof. Let us represent the domain ε in the form ε = i Π i+ ε 1 Π i ε 1 Π i ε 2, where Π i+ ε 1 is the domain of width 2ε i 1 bounded from below by the ith upper semicircle and bounded from above by the line y = b, Π i ε 1 is the domain of width 2ε i 1 bounded from above by the ith lower semicircle and bounded from below by the line y =0, and, finally, Π i ε 2 is the rectangle of width ε i 2 ε i 1 ε i+1 1 if ε i 2 >ε i 1 + ε i+1 1 (and of width 0ifε i 2 = ε i 1 + ε i+1 1 )andof height b (see Fig. 2). Denote by Π ω,i+ ε 1 the domain {(x, y) Π i+ ε 1,y<ω}, where ω<b.we also write Γ ω,i ε 1 = {(x, y) Π i+ ε 1,y= ω} if ω ε and Γ ω,i ε 1 = {(x, y) Π i ε 1,y= ω} if ω<ε (see Fig. 3). The sets Π ω,i ε 2 and Γ ω,i ε 2 are defined in a similar way. Assume first that u ε C0 ( ε, Γ ε ).Consider the point (x, ω) Π i+ ε 1, ω>ε, and the corresponding point (x, y 0 ) Γ ε belonging to the ith upper semicircle. The Newton Leibniz formula yields ω u ε u ε (x, ω) =u ε (x, ω) u ε (x, y 0 )= y dy, since u ε (x, y 0 )=0. By the Cauchy Schwarz Bunyakovskii inequality, ( ω u 2 u ε ε(x, ω) = y 0 y dy ) 2 ω ω In this case, integrating inequality (3.2) with respect to x over Γ ω,i ε 1 It follows from (3.3) that Γ ω,i ε 1 Γ ω,i ε 1 y 0 y 0 ( uε ) 2 ω dy ω u ε 2 dy. (3.2) y y 0, we obtain u 2 ε(x, ω) dx ω u ε 2 dx dy. (3.3) Π ω,i+ ε 1 u 2 ε(x, ω) dx b u ε 2 dx dy. (3.4) Π i+ ε 1 Finally, integrating (3.4) with respect to ω over [ε, b], we see that u 2 ε dx dy b 2 u ε 2 dx dy. (3.5) Π i+ ε 1 Π i+ ε 1 RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

54 4 G. A. CHECHKIN et al. The case in which (x, ω) Π i ε 1,ω < ε can be considered similarly. Thus, weomit the details. In this case, an analogous inequality can be obtained, namely, Π i ε 1 u 2 ε dx dy b 2 Π i ε 1 u ε 2 dx dy. (3.6) Consider now the rectangle Π i ε 2 and assume that the circle Bε i touches it from the left. For Π 0 ε 2, we consider the circle Bε. 1 Assume that ω>ε(the case in which ω<εis analogous). Draw the tangents to the circle Bε i from the ends of the segment Γ ω,i ε 2 (see Fig. 4). Connect all points of the segment Γ ω,i ε 2 to the boundary of Bε i in such a way that all the lines meet at the intersection point of the tangents. We obtain a beam of lines with the directions l(x). Let (x, ω) Γ ω,i ε 2 and ( x, ỹ) Bε i l(x, ω). Itfollows from the geometrical constructions that the distance between the points (x, ω) and ( x, ỹ) isless than or equal to (ω ε) 2 +(α i + β i ) 2 ε 2. Since u ε ( x, ỹ) =0, the Newton Leibniz formula yields (x,ω) u ε u ε (x, ω) = dl. ( x,ỹ) l Squaring the formula and using the Cauchy Schwartz Bunyakovsky inequality,we obtain the inequality u 2 ε(x, ω) (ω ε) 2 +(α i + β i ) 2 ε 2 (x,ω) ( x,ỹ) ( uε l ) 2 (x,ω) dl ci ω u ε 2 dl, where the constant c i does not depend on ε. Keeping in mind the estimates for α i and β i, we conclude that there is a constant C 1 such that c i <C 1 for all i. Integrating the last inequality with respect to x over Γ ω,i ε 2 and replacing the right-hand side by a greater integral, weobtain u 2 ε(x, ω) dx C 1 ω u ε 2 dx dy. (3.7) Π ω,i ε 2 Π ω,i+ Π ω,i ε 1 Γ ω,i ε 2 ε 1 ( x,ỹ) It follows from (3.7) that Γ ω,i ε 2 u 2 ε dx C 1 b u ε 2 dx dy. (3.8) Π i ε Π i+ 2 ε 1 Π i ε 1 Integrating inequality (3.8) with respect to ω over [0,b], we see that u 2 ε dx dy C 1 b 2 u ε 2 dx dy. (3.9) Π i ε Π i 2 ε Π i+ 2 ε 1 Π i ε 1 Adding inequalities (3.5), (3.6), and (3.9), weobtain u 2 ε dx dy C 1 b 2 ε ε u ε 2 dx dy. (3.10) Approximating the functions in H 1 ( ε, Γ ε ) by smooth functions, weconclude that inequality (3.10) is valid for u ε H 1 ( ε, Γ ε ). Inequality (3.1) follows directly from (3.10). The theorem is proved. RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

55 ON THE FRIEDRICHS INEQUALITY 5 4. CONVERGENCE AND ESTIMATES In this section, weprove the convergence of eigenelements for the following spectral problem: Δu ε = λ ε u ε in ε, u ε =0 on Γ ε, (4.1) u ε ν =0 on, to the corresponding eigenelements for the limit problem Δu 0 = λ 0 u 0 in, u 0 =0 on Γ 4, (4.2) u 0 ν =0 on Γ 1 Γ 2 Γ 3, as ε 0, as well as the convergence of solutions to the boundary value problem ΔU ε = λu ε + F in ε, U ε =0 on Γ ε, (4.3) U ε ν =0 on, to the solution of the following homogenized problem: ΔU 0 = λu 0 + F in, U 0 =0 on Γ 4, U 0 ν =0 on Γ 1 Γ 2 Γ 3, as ε 0 This means that problem (4.2) is the limit (homogenized) problem for (4.1) and (4.4) is the limit (homogenized) problem for (4.3). Denote by (, ) 0 and (, ) 1 the inner products in spaces L 2 () and H 1 () corresponding to the norms 0 and 1, respectively. Further, denote by (X, Y) the inner product of the vectors X and Y. Consider weak solutions to problems (4.1), (4.2), (4.3), and (4.4) (see, for instance, [23]). The function U ε H 1 ( ε, Γ ε )is a weak solution to the boundary value problem (4.3) if the integral identity ( U ε, v) dx dy λ U ε vdxdy= F vdxdy (4.5) ε ε ε holds for any v H 1 ( ε, Γ ε ); a function U 0 H 1 (, Γ 4 )isreferred to as a weak solution to the boundary value problem (4.4) if the integral identity ( U 0, v) dx dy λ U 0 vdxdy= F vdxdy (4.6) is valid for any v H 1 (, Γ 4 ).Similarly, a function u ε H 1 ( ε, Γ ε )isaneigenfunction for the spectral problem (4.1) corresponding toan eigenvalue λ ε if the integral identity ( u ε, v) dx dy = λ ε u ε vdxdy ε ε holds for any v H 1 ( ε, Γ ε ); finally, a function u 0 H 1 (, Γ 4 )isaneigenfunction for the spectral problem (4.2) corresponding toan eigenvalue λ 0 if the integral identity ( u 0, v) dx dy = λ 0 u 0 vdxdy is valid for any v H 1 (, Γ 4 ). Note that, due toremark 2.1, the integral identity (4.5) can be represented as follows: ( U ε, v) dx dy λ U ε vdxdy= F vdxdy. (4.7) The main goal of this section is to prove the following theorems. RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No (4.4)

56 6 G. A. CHECHKIN et al. Theorem 4.1. Assume that F L 2 () and K is an arbitrary compact set in the complex plane C such that K contains no eigenvalues of the limit problem (4.2). In this case, the following statements hold. 1. There is a number ε 0 > 0 such that, for any ε<ε 0 and any λ K, a solution to problem (4.3) does exist, and this solution is unique. Moreover, the uniform (with respect to ε and λ) estimate U ε 1 C F 0 (4.8) is valid, where C does not depend on U ε and F. 2. The convergence U ε U as ε 0 (4.9) holds for the solutions to problems (4.3) and (4.4), respectively. Theorem 4.2. Let λ 0 be an eigenvalue of multiplicity N for the limit spectral problem (4.2). Then the following statements hold. 1. There exist exactly N eigenvalues (counted according to their multiplicities) of the original problem (4.1) that converge to the eigenvalue λ If λ 1 ε,...,λ N ε are the eigenvalues of problem (4.1) which converge to λ 0 and if u 1 ε,...,u N ε are the corresponding eigenfunctions (orthogonal and normalized) in L 2 (), then, for any sequence ε k 0, there exists a subsequence ε k 0 such that k u j ε u j as ε = ε k 0, where u 1 0,...,u N 0 are eigenfunctions for the spectral problem (4.2) corresponding to λ 0 which are orthogonal and normalized in L 2 (). The proof of these theorems uses the approach developed in [2, 4, 11 13, 20] for diverse types of singular perturbations. However, we must modify this approach to fit it into the geometric structure of our domain. Let us now prove some auxiliary statements which are needed in our analysis. The proof of the following lemma is similar tothat of the convergence theorem in[3] and [4]. Lemma 4.1. Let v ε be a sequence of functions in H 1 ( ε, Γ ε ). Assume that v ε v weakly in H 1 () as ε 0. Then v H 1 (, Γ 4 ). Proof. Introduce the following sets: Γ ω = ε {(x, y) :y = ω}, i.e., Γ ω = i Π ω = ε {(x, y) :y<ω}, i.e., Π ω = i ( Γ ω,i ε 1 Γ ω,i ) ε 2, ( Π ω,i ε 2 Π ω,i+ ε 1 Π ω,i ) ε 1. Assume first that v ε C0 ( ε, Γ ε ). Adding inequalities (3.3) and (3.7) with respect to i, we find that vε(x, 2 ω) dx C 2 ω v ε 2 dx dy C 2 ω v ε 2 dx dy. (4.10) Γ ω Π ω ε Approximating the functions from H 1 ( ε, Γ ε ) by smooth functions, weconclude that inequality (4.10) is valid for v ε H 1 ( ε, Γ ε ). Using (4.10) and keeping in mind the uniform boundedness of the sequence v ε, we obtain vε(x, 2 ω) dx ωc 3. (4.11) Γ ω Since the embedding of H 1 ( ε, Γ ε )inl 2 (Γ 4 )iscompact (see, for instance, [23] and [24]), we can pass to the limit in (4.11) as ε 0and see that (v ) 2 (x, ω) dx ωc 3. (4.12) Γ ω RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

57 ON THE FRIEDRICHS INEQUALITY 7 Since ω is an arbitrarily small positive number and v H 1 (), it follows from (4.12) that v =0on Γ 4. The lemma is proved. Denote by K ε the set of disjoint open circles K j ε of radius r j,α j <r j 1, r j + r j+1 β j, and centered at the center of B j ε. Write 1 ε =\ K ε. Introduce the following cutoff functions: ψ j (s) C (R), 0 ψ j 1, which vanishes for s α j ε and is equal to 1for s r j ε, and Ψ ε (x, y) = i ( ) i ) 2 ψ i (x ε β j +(y ε) 2. (4.13) j=0 We can readily see that Ψ ε =0inB ε and Ψ ε = 1 in 1 ε. The following lemma is used in our further analysis. Lemma 4.2. Suppose that v H 1 (, Γ 4 ). Then the sequence vψ ε converges to the function v strongly in H 1 ( ε ). Proof. It follows from the definition of the function Ψ ε that vψ ε vin H 1 ( ε ). We must prove the convergence vψ ε 2 1 v 2 1 as ε 0 (4.14) to prove the strong convergence of vψ ε to the function v in H 1 ( ε ). Assume first that v C0 (, Γ 4 ). Then vψ ε 2 1 = v 2 Ψ 2 ε dx dy + v 2 Ψ 2 ε dx dy + v 2 Ψ ε 2 dx dy +2 Ψ ε v( v, Ψ ε ) dx dy. ε ε ε ε Due toproperties of the function Ψ ε, it is clear that v 2 Ψ 2 ε dx dy + ε v 2 Ψ 2 ε dx dy ε v 2 dx dy + ε v 2 dx dy = v 2 1 ε (4.15) as ε 0. We claim that ε v 2 Ψ ε 2 dx dy tends to zero as ε 0. Taking into account the properties of the function Ψ ε, we obtain v 2 Ψ ε 2 dx dy = v 2 ψ i 2 dx dy. ε i Kε i\bi ε (4.16) Let us estimate ψ i 2. Obviously, ψ i x = i (x ε j=0 β j) (x ε i j=0 β j) 2 +(y ε) 2 ψ i, ψ i y = (y ε) (x ε ψ i. i j=0 β j) 2 +(y ε) 2 Therefore, ψ i 2 C 4 as (x, y) K i ε \ B i ε. (4.17) Using inequalities (4.16) and (4.17) and the condition v O(ε) on K i ε\b i ε (since v C 0 (, Γ 4 ) H 1 (, Γ 4 )), wesee that ε v 2 Ψ ε 2 dx dy C 5 1 ε ε2 ε 2 0 as ε 0. (4.18) RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

58 8 G. A. CHECHKIN et al. In a similar way, we can prove that ε vψ ε ( v, Ψ ε ) dx dy C 6 1 ε ε2 ε 0 as ε 0. (4.19) Estimates (4.15), (4.18), and (4.19) imply (4.14) if v C0 (, Γ 4 ).Approximating any function in H 1 ( ε, Γ ε ) by smooth functions in C0 (, Γ 4 ),weconclude that inequality (4.14) holds for v H 1 ( ε, Γ ε ). The lemma is proved. Proof of Theorem 4.1. Let us first prove the estimate U ε 1 C 7 ( U ε 0 + F 0 ), (4.20) where C 7 does not depend on ε and λ. Substituting v = U ε into the integral identity (4.7), we obtain U ε 2 1 C 8 ( U ε U ε 0 F 0 ) C 7 ( F 0 + U ε 0 ) U ε 1, where C 7 does not depend on ε and λ. It is clear that formula (4.20) follows from this inequality. 1. The existence and uniqueness of asolution to Problem (4.3) follows from estimate (4.8). Indeed, the integral identity (4.7) can be represented as follows: ( U ε, v) dx dy + This means that identity (4.21) is equivalent to the relation U ε vdxdy (λ +1) U ε vdxdy= F vdxdy. (4.21) (U ε,v) 1 (λ + 1)(U ε,v) 0 =(F, v) 0. (4.22) The Riesz theorem (see, for example, [26]) says that there is an operator A for which (U ε,v) 0 = (AU ε,v) 1. Hence, identity (4.22) is equivalent to (U ε,v) 1 (λ + 1)(AU ε,v) 1 =(F, v) 0. This equation is solvable because the kernel of the operator [I (λ + 1)A] is trivial (since the solution to problem (4.3) is unique; the uniqueness of solutions to (4.3) follows from (4.8)). Now let us derive estimate (4.8). Assume that estimate (4.8) fails to hold, i.e., there exist sequences ε k, F k, and λ k such that ε k 0ask, F k L 2 (), and the inequality U εk 1 >k F k 0 (4.23) holds for the solutions to problem (4.3) with ε = ε k,λ= λ k,f= F k. Without loss of generality, we may assume that the sequence U ε is normalized in L 2 (), i.e., U εk 0 =1. (4.24) (If U εk 0 1, then, dividing (4.20) by U εk 0 and writing V εk = U εk / U εk 0, we obtain the similar inequality V εk 1 C 9 ( V εk 0 + ) 1 ( ) F U εk 0 C 10 Vεk 0 + F 0, 0 where V εk 0 =1. The value max (C 9,C 9 / U εk 0 ) can be taken for C 10. Hence, we can assume that U εk 0 =1from the very beginning.) RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

59 Now using (4.20) and (4.23), we obtain ON THE FRIEDRICHS INEQUALITY 9 U εk 1 C 11, F k 0 < C 11 k as k. (4.25) As isknown, a bounded subset of H 1 is weakly compact in H 1, and hence there exists asubsequence {k } of indices {k} and U and λ such that λ k λ K, U εk U in H 1 () as k +. (4.26) The inequalities (4.24) and (4.26) give U 0. Suppose that v is an arbitrary fixed function inc0 (, Γ 4 ). Then v H 1 ( ε, Γ ε )for any small value of ε. Itis clear that vψ ε H 1 ( ε, Γ ε ), where the function Ψ ε is defined in (4.13). Substitute vψ εk as a test function, U = U εk, λ = λ k, and F = F k in the integral identity (4.7). We obtain ( Uεk, (vψ εk ) ) dx dy λ k U εk vψ εk dx dy = F k vψ εk dx dy. (4.27) Passing to the limit in the integral identity (4.27) as ε k 0 and using (4.25), (4.26), and Lemma 4.2, we obtain the integral identity ( U, v) dx dy = λ U vdxdy. It followsfrom the density of the embeddingofc0 (, Γ 4 )inh 1 (, Γ 4 ) thatthis identity remains valid for any v H 1 (, Γ 4 ). It follows from Lemma 4.1 that U H 1 (, Γ 4 ). Since U 0 and v is an arbitrary function inh 1 (, Γ 4 ), it follows that λ K is an eigenvalue of the limit problem (4.2). However, we have assumed that K contains no eigenvalues of the limit problem (4.2). This contradictionproves estimate (4.8). 2. Let λ K be an arbitrary fixed number. Assume that {ε k } 0ask +. We can choose asubsequence weakly convergent in H 1 from asequence bounded in H 1. Hence, there is a subsequence {k } of indices {k} and a function U such that U εk U in H 1 () (and strongly in L 2 ()) as k +. Recalling Lemma 4.1, wesee that the limit function U is in H 1 (, Γ 4 ). Passing to the limit in the integral identity (4.27) (using the same reasoning as in the proof of part 1 of the theorem), we have ( U, v) dx dy = λ U vdxdy+ F vdxdy, which coincides with the integral identity of problem (4.4). Since the solution to problem (4.4) is unique, we conclude that U = U 0. Moreover, it follows from (4.26) that U ε U 0 strongly in L 2 () and weakly in H 1 () as ε 0. Substituting v = U εk in the integral identity (4.21) and using the convergence of the function in L 2 (), we obtain U ε 2 0 U = FU ε dx dy FU 0 dx dy = F (U ε U 0 ) dx dy 0 as ε 0. (4.28) Formula (4.28) and the weak convergence of U εk to U 0 in H 1 () imply the strong convergence in this space. Hence, the convergence (4.9) holds, because ε k is an arbitrary sequence. The theorem is proved. The following lemma can be proved by using the lines of the proof of assertion 2of Theorem 4.1. RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

60 10 G. A. CHECHKIN et al. Lemma 4.3. Assume that λ ε is an eigenvalue of problem (4.1) which converges to an eigenvalue λ 0 of the limit problem (4.2) as ε 0. Letu ε be an eigenfunction corresponding to the eigenvalue λ ε and normalized in L 2 (). Then the sequence {ε k } k=1 0 admits a subsequence {ε k m } m=1 0 such that u εkm u 0 1 0, where u 0 is an eigenfunction corresponding to the eigenvalue λ 0 and normalized in L 2 (). The mutual orthogonality of the eigenfunctions of problem (4.2), a similar fact for problem (4.1), and Lemma 4.3 give the following statement. Lemma 4.4. Let the multiplicity of an eigenvalue λ 0 for problem (4.2) be equal to N. Suppose that the eigenvalues λ ε of problem (4.1) converge to λ 0. Then the multiplicity of λ ε is less than or equal to N. Below we use the following lemma, which is proved in the appendix. Lemma 4.5. Let U 0,U ε be solutions to problems (4.3) and (4.4), respectively. Assume that K is an arbitrary compact set on the complex plane C and K contains no eigenvalues of the spectral problem (4.2). Then the following representations hold: U 0 = n=1 (F, u n 0 ) 0 λ n 0 λ un 0, U ε = n=1 (F, u n ε ) 0 λ n ε λ un ε, (4.29) where u n 0,u n ε are eigenfunctions and λ n 0,λ n ε are the eigenvalues of problems (4.2) and (4.1), respectively. Proof of Theorem 4.2. Let 0 <λ 1 0 λ 2 0 be eigenvalues of problem (4.2), let u 1 0,u 2 0,... be the corresponding eigenfunctions that are orthogonal and normalized in L 2 (), let be the eigenvalues of problem (4.1), and let 0 <λ 1 ε λ 2 ε u 1 ε,u 2 ε,... be the corresponding eigenfunctions which are orthogonal and normalized in L 2 ( ε )(hence, in L 2 () as well) and λ p 0 <λ 0 = λ p+1 0 = = λ p+n 0 <λ p+n+1 0, where λ 0 0 := 0. Due tolemmas 4.3 and 4.4, it is sufficient toshow that N ε N, where N ε stands for the multiplicity of the eigenvalues λ i ε convergent toλ 0 as ε 0and N for the multiplicity of the eigenvalue λ 0. Suppose the contrary. This means that there is a sequence {ε k } k=1 0for which N ε k <N. Without loss of generality,wecan assume that N εk = M<Nand that λ p+1 ε,...,λ p+m ε (4.30) are eigenvalues ofproblem(4.1) convergenttoλ 0 as ε 0.DuetoLemma 4.3, there is a subsequence {ε km } m=1 0such that u p+i ε km u p+i 0 0, i =1,...,M, RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

61 ON THE FRIEDRICHS INEQUALITY 11 where u p+i stand for the eigenfunctions of problem (4.2) which are orthogonal and normalized in L 2 (). Without loss of generality, we assume that u p+i = u p+i 0, i.e., that the following convergence relation holds: Define u p+i ε km u p+i 0 0 0, i =1,...,M. (4.31) F := u p+m+1 0, G 0 (λ) :=(U 0 (,λ),u p+m+1 0 ) 0, G ε (λ) :=(U ε (,λ),u p+m+1 0 ) 0, where U 0 and U ε are the solutions to problems (4.3) and (4.4), respectively. Let S(t, z) be the circle of radius t centered at the point z on the complex plane. There is a T>0 such that the circle S(t, λ 0 ) contains no eigenvalues of the limit problem except for λ 0 whenever t<t (because the set of eigenvalues of the limit problem (4.2) has no finite accumulation points). By Theorem 4.1 andlemmas 4.3 and 4.4, the convergence relation G εkm (λ) G 0 (λ), λ S(t, λ 0 ), (4.32) is valid for any t T as m. As isknown (see also the appendix), the eigenfunctions {u j 0 } and {uj ε} form the orthogonal basis in L 2 () and in L 2 ( ε ),respectively. It follows from Lemma 4.5 that U δ = n=1 The representation (4.33) implies the relations (F, u n δ ) 0 λ n δ λ un δ, δ 0. (4.33) where G 0 (λ) = g ε (λ) = 1 λ 0 λ, G ε(λ) = p s=1 (u p+m+1 0,u s ε) 0 2 λ s ε λ M i=1 + (u p+m+1 0,u p+i ε ) 0 2 λ p+i ε s=p+m+1 λ (u p+m+1 + g ε (λ), (4.34) 0,u s ε) 0 2 λ s. (4.35) ε λ We claim that the function g εkm (λ) isholomorphic with respect to λ S(t, λ 0 )for any t<t. Note that the circle S(T,λ 0 ) contains no eigenvalues except for those in (4.30) whenever ε km is too small (because otherwise there is another subsequence for which these eigenvalues converge to some λ S(T,λ 0 ), λ λ 0,but this leads to a contradiction, since the circle S(T,λ 0 ) contains no eigenvalues of problem (4.2) except for λ 0 ).Consequently, g ε (λ) (T t) 1 (u p+m+1 0,u s ε km ) 0 2. (4.36) s=1 Since the system {u s ε km } s=1 forms an orthogonal basis in L 2 ( ε ), the Parseval Steklov formula (see, for instance, [25]) F 2 L = 2( ε) F i 2, F i =(F, u i ε km ) 0, i=1 holds for any F L 2 (). It follows from (4.36) and from the Parseval Steklov formula that g ε (λ) (T t) 1 u p+m =(T t) 1, i.e., the series (4.35) converges uniformly with respect to λ S(t, λ 0 ).Hence, by Weierstrass theorem, g ε is a holomorphic function of λ S(t, λ 0 ). RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

62 12 G. A. CHECHKIN et al. Consequently, using (4.32), (4.34) and the Lebesgue dominated convergence theorem, we obtain G εkm (λ) dλ G 0 (λ) dλ, m, (4.37) S(t,λ 0) S(t,λ 0) for t<t.by (4.34) and by the Cauchy residue theorem, wehave G 0 (λ) dλ =2πi, (4.38) S(t,λ 0) S(t,λ 0) S(t,λ 0) M G εkm (λ) dλ =2πi (u p+m+1 0,u i+p ε km ) 0 2. (4.39) It follows from (4.39) and (4.31) that: M G εkm (λ) dλ 2πi (u p+m+1 0,u i+p 0 ) 0 2 =0 (4.40) as m. Identities (4.38) and (4.40) contradict the convergence relation (4.37). Hence, N = M. The theorem isproved. Remark 4.1. Using Theorems 4.1and4.2,weconclude that the followingfriedrichsinequalities with sharp constants hold true for u ε H 1 ( ε, Γ ε )and u 0 H 1 (, Γ 4 ): u 2 ε dx dy M ε u ε 2 dx dy, u 2 0 dx dy M 0 u 0 2 dx dy, ε ε where the constant M 0 does not depend on ε and i=1 i=1 M ε M 0 0 as ε 0. This follows from the relations M ε =1/λ 1 ε and M 0 =1/λ 1 0,which can be proved by using the variational principle for problems (4.1) and (4.2), respectively. Remark 4.2. Results similar to those obtained in this section and in the previous ones can also be obtained if ε is an arbitrary domain perforated aperiodically along the boundary (see Fig. 5). Fig. 5. Domain ε. 5. NUMERICAL SIMULATIONS FOR THE CORRESPONDING PARABOLIC PROBLEM In the previous section, weconsidered an elliptic boundary value problem inadomain perforated aperiodically along the boundary. This section isdevoted to numerical studying the asymptotic behavior of solutions to the corresponding parabolic problem. Consider the following heat problem v ε t = σδv ε in ε, v ε =0 on Γ ε, v ε (5.1) ν =0 on, v ε (, 0) = g( ) in ε, RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No

63 ON THE FRIEDRICHS INEQUALITY 13 Fig. 6. The computed temperature distribution obtained by using the finite-element program ANSYS 11 for t =2.5. and the corresponding limit problem v 0 t = σδv 0 in, v 0 =0 on Γ 4, v 0 ν =0 on Γ 1 Γ 2 Γ 3, v 0 (, 0) = g( ) in, (5.2) where g stands for asmooth functionand σ for the thermal diffusivity. The fact that problem (5.2) is the homogenized (limit) for problem (5.1) can be verified in just the same way as in Section 4.In fact, it can be shown that the solutions v ε and v 0 are the limits (or at least sub-limits) of sequences of the form M M v ε,m (x, t) = u i ε(x)d i ε(t) and v 0,M (x, t) = u i 0(x)d i 0(t), i=1 respectively, as M +. Here u i ε and v0 i are eigenfunctions of problems (2.1) and (2.2), respectively, and v ε,m (,t)and v 0,M (,t)are the projections of the solutions v ε (,t)and v 0 (,t) to the finite-dimensional subspace spanned by the eigenfunctions { uε} i M i=1 and { u0} i M, respectively (see, i=1 e.g., [10]). Using convergence results for the eigenvalues, one can prove that v ε (x, t) converges (in an appropriate norm) to v 0 (x, t) at every chosen point x.alternatively, one can use Rothe s method of discretization intime (see, for instance, [21]) to prove the convergence of solutions of problem (5.1) to asolution of problem (5.2). As an illustration of our theoretical results, wecan use numerical simulations. The asymptotic behavior of the solutions to aparabolic problem mustbesimilar to the behavior of the solutions to the corresponding elliptic problem. Asanexample, we consider the case with σ =1, g(x) =1 for all x, a= b =10, ε =1/2, ε i 1 =(2/3) ε =1/3, and ε i 2 =2ε =1. In this case, the solution of the limit problem becomes ( ( v 0 (x, t) = e (2k+1) 2t 20 π) 4 ( ( (2k +1) )) ) sin πx 2. (2k +1)π 20 k=0 RUSSIAN JOURNAL OF MATHEMATICAL PHYSICS Vol. 16 No i=1